Problems in Operator Theory (Graduate Studies in Mathematics, V. 51)

Y. A. Abramovich C. D. Aliprantis

Graduate Studies in Mathematics Volume S 1

-=*

American Mathematical Society

Problems in

Operator Theory

Problems in

Operator Theory

Y. A. Abramovich Indiana University-Purdue University Indianapolis

C. D. Aliprantis Purdue University

Graduate Studies in Mathematics Volume 51

ja1it American Mathematical Society : ` Providence, Rhode Island

Editorial Board Walter Craig Nikolai Ivanov Steven G. Krantz David Saltman (Chair) 2000 Mathematics Subject Classiftcation. Primary 46Axx, 46Bxx, 46Gxx, 47Axx, 47Bxx, 47Cxx, 47Dxx, 47Lxx, 28Axx, 28Exx, 15A48, 15A18.

Library of Congress Cataloging-in-Publication Data Abramovich, Y. A. (Yuri A.) Problems in operator theory / Y. A. Abramovich, C. D. Aliprantis.

p. cm. - (Graduate studies in mathematics, ISSN 1065-7339; v. 51) Includes bibliographical references and index. ISBN 0-8218-2147-4 (alk. paper) 1. Operator theory-Problems, exercises, etc. I. Aliprantis, Charalambos D. Y. A. (Yuri A.). Invitation to operator theory. III. Title. IV. Series.

11. Abramovich,

QA329.A27 2002 515'.724'076-dc21

2002074421

Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society, 201 Charles Street, Providence, Rhode Island 02904-2294, USA. Requests can also be made by e-mail to reprint-psraissionlams.org. © 2002 by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America.

® The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability.

Visit the AMS home page at http://vvv.sms.org/

10987654321

070605040302

To the Memory of our Parents

Contents

Foreword

Chapter 1. Odds and Ends §1.1. Banach Spaces. Operators. and Linear Functionals §1.2. Banach Lattices and Positive Operators §1.3. Bases in Banach Spaces 61.4. Ultrapowers of Banach Spaces 61.5. Vector-valued Functions §1.6. Fundamentals of Measure Theory

xi 1 1

20 31

44

48 51

Chapter 2. Basic Operator Theory §2.1. Bounded Below Operators §2.2. The Ascent and Descent of an Operator §2.3. Banach Lattices with Order Continuous Norms §2.4. Compact and Weakly Compact Positive Operators

63

Chapter 3. Operators on AL- and AM-spaces 0.1. AL- and A lf-spaces 0.2. Complex Banach Lattices $3.3. The Center of a Banach Lattice 0.4. The Predual of a Principal Ideal

87

Chapter 4. Special Classes of Operators 44.1. Finite-rank Operators 4.2. Multiplication Operators

63 68 71

78

87

96 105 111

119 119

125

vii

viii

Contents

0.3. Lattice and Algebraic Homomorphisms

129

Fredholm Operators §4.5. Strictly Singular Operators

134

§4.4.

Chapter 5. Integral Operators §5.1. The Basics of Integral Operators §5.2. Abstract Integral Operators §5.3. Conditional Expectations and Positive Projections §5.4. Positive Projections and Lattice-subspaces

139 145

145 154 169 180

Chapter 6. Spectral Properties §6.1. The Spectrum of an Operator §6.2. Special Points of the Spectrum §6.3. The Resolvent of a Positive Operator §6.4. Functional Calculus

189

Chapter 7. Some Special Spectra §7.1. The Spectrum of a Compact Operator §7.2. Turning Approximate Eigenvalues into Eigenvalues §7.3. The Spectrum of a Lattice Homomorphism §7.4. The Order Spectrum of an Order Bounded Operator §7.5. The Essential Spectrum of a Bounded Operator

215

Chapter 8. Positive Matrices §8.1. The Banach Lattices and §8.2. Operators on Finite Dimensional Spaces §8.3. Matrices with Non-negative Entries 68.4. Irreducible Matrices §8.5. The Perron-Frobenius Theorem

243

Chapter 9. Irreducible Operators §9.1. Irreducible and Expanding Operators §9.2. Ideal Irreducibility and the Spectral Radius §9.3. Band Irreducibility and the Spectral Radius §9.4. Krein Operators and C(fl)-spaces

273

Chapter 10. Invariant Subspaces §10.1. A Smorgasbord of Invariant Subspaces §10.2. The Lomonosov Invariant Subspace Theorem

299

189 197 201

205

215 222

230

232 237

243 251

262 265

268

273 283

290 293

299

307

ix

Contents

§10.3. §10.4. §10.5.

§10.6. 110.7.

Invariant Ideals for Positive Operators Invariant Subspaces of Families of Positive Operators Compact-friendly Operators Positive Operators on Banach Spaces with Bases Non-transitive Algebras

Chapter 11. The Daugavet Equation §11.1. The Daugavet Equation and Uniform Convexity §11.2. The Daugavet Property in AL- and AM-spaces §11.3. The Daugavet Property in Banach Spaces §11.4. The Daugavet Property in C(c2)-spaces §11.5. Slices and the Daugavet Property §11.6. Narrow Operators &11.7. Some Applications of the Daugavet Equation Bibliography Index

310

317 320 329 331

335

335 352

356 359 365 369 372 375

Foreword

This book contains complete solutions to the more than six hundred exercises

in the authors' book: An Invitation to Operator Theory. American Mathematical Society. 2002. The problems have been spread over eleven chapters following the format of that book. Each problem is identified by a triplet of numbers x.y.z: x designates the chapter, y the section, and z the exercise. For instance. Problem 3.4.7 indicates Exercise 7 in Section 4 of Chapter 3. All solutions are based on the material covered in the text with frequent references to the results in the text. For example, a reference to Theorem 5.9 refers to Theorem 5.9 and a reference to Example 6.21 refers to Example 6.21

in the book An Invitation to Operator Theory. We have added an extra amount of material to many solutions in order to make this book as selfcontained as possible.

This problem book will be beneficial to students only if they use it "properly," that is to say. if students look at a solution of a problem only after

trying very hard to solve the problem. Students will do themselves great injustice by reading a solution without any prior attempt on the problem. It should be a real challenge to students to produce solutions which are different from the ones presented here. Due to the extra material incorporated into the problems, the book can be used as a companion supplement to any text used for the standard functional analysis graduate courses. In addition, this solution book can be used as a reference not only for mathematical subjects but also for other disciplines that rely on functional analytic or measure theoretic techniques.

xii

Foreword

We would like to express our most sincere thanks to all people who made constructive comments and corrections regarding the text and the problems. Special thanks are due to Professors Arkady Kitover and Vladimir Troitsky who read the solutions and made numerous suggestions and corrections.

A final thank you goes to Arlene O'Sean, the AMS Copy Editor, for her excellent job in editing the manuscript. Y. A. Abramovich and C. D. Aliprantis Indianapolis and West Lafayette, May 2002

Chapter 1

Odds and Ends

1.1. Banach Spaces, Operators, and Linear Functionals Problem 1.1.1. If Ilx+yli = IIxII+ IIyII for two vectors x and y in a normed space, then show that Ilax +;3yll = aIIxII + 43IIy1I for all scalars a.;I > 0.

Solution: If a > ;3, then aIIxII + (3Ilyll

Ilax + 3vll = IIa(x + y) + (;3 - a)yll allx+ yII - (a -.S)IIyiI = aIIxII +,3llyll

Hence, lax + i3yll = aIIxII + f3llyll holds for all a,,3 > 0.

I

Problem 1.1.2. Let T : V W be a surjective one-to-one operator between two vector spaces. Show that T-1 (the inverse of T) is a linear operator. Solution: Let S: W - V denote the inverse of T. That is, we have TS = IK and ST = Iv . Let v. uw E It' and let A be a scalar. For the additivity of S note that T[(Sv + Su,)] = TSty +TSw = v + w = TIS(v + u,)].

Since T is one-to-one, it follows that S(v+w) = S(v)+S(w). That is, 5 is additive. Similarly, for the homogeneity of S observe that T(S(Aw)) =

Au!

= ATSw = T(ASw).

and so S(Aw) = AS(w). Therefore. S is a linear operator.

0

Problem 1.1.3. If T : X -- Y is a bounded operator between normed spaces. then show that IITII = min{AI > 0: IITxII < MMIIxiI for all z E X } .

1. Odds and Ends

2

Solution: Let Alo = inf{,11 > 0: !1Tx11 < 1111rll for all r E X). and note that the infimum is. in fact. a minimum. Now if Al > 0 satisfies IITxII < '1MI1xfl for all IlTrll < M. and so IITII 5 Mo. On the other hand. r E X. then IHTI1 = IIT.:I < IITII ' Ilxii for each r E X implies AIo <- IITII. Thus. IITII = Mo.

Problem 1.1.4. Show that if X is a normed space and Y is a Banach space, then C(X, Y) equipped with the operator norm is a Banach space. Solution: We assume that X 0 {0}. Recall that the operator norm on C(X.Y) is the function II - II : C(X.Y) - R defined by IITII = S11pI1s11=1 IITzII Clearly, 1ITII > 0 for each T E C(X.Y) and 11TH = 0 if and only if T = 0. Also,

if T E C(X.Y) and a is a scalar. then

IlaTll = sup II(aT)xll = sup Ial ' IITrII = lal sup IITzii = IaI ' IITII 'Ixi=1

11=11=1

1x11=1

To establish the triangle inequality, let S,T E 4(X,Y). For each r E X with IITII = I we have 11(S + T)rll < IISxIi + IITrhl 5 IiSil + IITII, and so IIS + TII = sup II (S + T)xi1 5 IITII + IITII 1;.r11=1

Next. we shall show that C(X, Y) is a complete normed space. To this end, let {T } be a Cauchy sequence in C(X. Y). Fix e > 0 and then choose some no such that IIT - Tm II < e for all n, in > no. The inequality

IITnx - Tmxii 5 IIT - Tnll ' llxli 5 elixll (*) is true for all n, in > no. and this shows that (T,,r} is a Cauchy sequence in Y for each .r E X. Since Y is a Banach space. the sequence {Tnx} converges in Y for This defines a linear operator T: X -» Y. Letting each x E X. Let Tx = in -+ x in (*) yields IITnx - Trli S cllxll (**) for all n > no. From IITrhI < IIT ' - T,,xll + IITn,,ZII < (e + IITn0II)IixiI it follows

that T E C(X. Y). Now a glance at (**) guarantees that IITn - T11 < F for all it > no. That is. Tn -.T holds in C(X,Y). and so C(X.Y) is a Banach space. I

Problem 1.1.5. Prove the following converse of the preceding problem. If X is a non-trivial nonmed space, Y is another normed space and C(X, Y) is a Banach space. then Y is likewise a Banach space. Solution: Assume that C(X. Y) is a Banach space. Fix a non-zero continuous R and then choose some xo E X such that f (xo) = 1. linear functional f : X Now let {yn } be a Cauchy sequence of Y. For each it define the bounded operator T,,: X -+ Y by Tn (r) = f (r)y,, . From l Tnr - Trnrll = I f(r)I ' 11yn - i/mhi 5 II ya - Ymli ' 11f 11 ' Ilrii we see that IIT - T.115 11f 11 ' llyn - ym II for all it and in. This implies that {T,,} is a Cauchy sequence in £(X. Y). Since C(X, Y) is a Banach space, there exists some

T E C(X. Y) such that T,, -a T. In particular. we have !/n = f (xo)yn = Tn(xo) -+ T(ro) This shows that Y is a Banach space.

U

1.1. Banach Spaces, Operators, and Linear Plmctionals

3

Problem 1.1.6. Show that the complexifecation XX of a real vector space X under the operations (x1 + ty1) + (x2 + ty2) (a + z13)(x + 2y)

= x1 + X2 + z(yl + y2), and

= ax - /3y + t(Qx + ay)

is a complex vector space.

Solution:/ The addition axioms can be verified as follows: t

(x1 + zyl)+(x2 + ty2) = x1 +X2+ t(Yi + y2) =x2+x1 +t(y2 + yl) _ (x2+zy2)+(xi + tyl ),

[(x1 + zyl) + (x2 + zy2)] + (x3 +10)

_

[x1 + x2 + z(y1 + y2)] + (x3 + ty3)

(x1 + x2 + x3) + t(y1 + Y2 + Y3)

(x1 + zy1) + [(x2 + ty2) + (x3 + ty3)],

(xI +ty1)+(O+z0) = x1+ty1, (x1 +ty1)+[(-z1)+z(-y1)] = 0+10. The multiplication axioms can be verified as follows: [(a + 113)(y + tb)](x + 1y)

= [(ay - i3b) + t(Ol + ab))(x + ty) = [(ay -,36)x - ($'Y + ab)y] + 1[(a- -'36)y + (Th + ab)x] _ [a(yx - 6y) -,3(bx +'yy)) + t[/3(yx. - 6y) + a(6x + yy)] _ (a + ti3)[(tyx - 6y) + z(bx + yy)] _ (a + z$)[('y + z6)(x + ty)], ((a + 13) + (y + 16)](x + ty) = [(a + y) + 1(;3 + 6)1(x + ty) = [(a + y)x - (13 + 6)y] +:[(a + y)y + (13 + 6)x] = [(ax - )3y) + z(ay + Ox.)] + [(Ix - 6y) + t(yy + 6x.)] = (a + z3)(x + ty) + (y + z6)(x + zy), (a + z$) [(x1 + zyt) + (x2 + 2Y2)] = (a + 213)[(x1 + x2) + t(y1 + y2)]

= [a(x1 + x2) - 13(y1 + y2)] + z[a(yl + y2) + $(xl + x2)] = [(ax1 - 3y1) + z(ayl + $x1)] + [(ax2 -42) + t(ay2 + 13x2)] _ (a + 001 1 + zyl) + (a + 10)(X2 + ty2),

(1+20)(x+2y)=x+ty. Therefore, & is a complex vector space.

U

1. Odds and Ends

4

Problem 1.1.7. Let X be a real normed space with norm II

'

II

Show that

the function II 11c: Xc --, R defined via the formula -

llzllc= sup Ilxcos9+ysin9ll, z=x+zyEXc, OE [0,27r)

is a norm on X. that extends II

'

II

Also, show that

'(IIxII + Ilyfl) < IIzIIc S IIxIl + Ilyll

for each z = x + zy E X. Use the preceding inequality to verify that: (a) A sequence {zn}, where zn = xn + 2yn, satisfies Ilznllc -- 0 if and only if IIxnII -' 0 and IlynII

0.

(b) If X is a Banach space, then Xc is also a Banach space. Solution: Let z = x + zy E X. Since for each 0 E R there exists some 0 E [0, 27r] with cos 0 = cos 0 and sin 0 = sin 0, it should be clear that

IIzIIc = sup llxcoso+ysin9ll =supllxcoso+ysin9ll BER OE (0.2aj

From the inequality llx cos o + y sin 9ll 5 I cosof ' IIxII + I sin 0I - IIxII 5 IIxII + Ilyll, we

see that IIzIIc < IIxII + IIxII- In addition, from the inequalities

IIxII = Ilxcos0+xsin0ll <- IIzIIc and IIxII = llxcos z +ysin 2Il 5 IIzIIc we get IIxII + Ilyll 5 2IIzllc or

i(IIxII + IIxII) 5 IIzIIc. Therefore,

(IIxII + IIxII) < IIzIIc 5 IIxII + Ilyll

(*)

Clearly IIzIIc > 0 for each z E Xc, and from (*) we see that IIzIIc = 0 if and only if z = 0. For the triangle inequality note that if zi = x1 + zy1, z2 = X2 + zY2 E Xc, then z1 + Z2 = xi + X2 + z(yl + y2), and so from

II(xi +x2)cos9+(yi+y2)sinOll < llxicos9+ylsinOll+IIx2cos9+y2sinOlI 5 IIziIIc + IIz21lc, it follows that BERII(xi +x2)cos9+(yi + Y) smell < Ilzillc+llz211c-

Ilzi +z211c =

Next, we shall prove the homogeneity of II . IIc. To this end, fix z = x + zy E X. and a complex number a + zQ 0 0. Then (a + z$) z = ax - i3y + z(Qx + cry). Notice lies on the unit circle. So, there exists a unique that the point

7.i

angle 0 < w < 2a such that

and

cosw

sinw =

Q2+$2

Now observe that

acos0+# sin0

a2 +

=

(

v +7 cos 9 + as+ sin 9 1

a2 + 02 (sinw cos 0 + cos w sin 0)

a2+O2 sin(w+9) = Ia+t#lsin(w+9),

1.1. Banach Spaces, Operators. and Linear F inctionals

5

and similarly a sin O - i3 cos O =

a2 +1.32 (sin wsin B - cos w cos o)

-

a2 + 32 coS(w + 0) = - Ia + z(3I cos(w + 0)

.

Consequently, we have

II(ax- 3y)cos0+(;3x+ay)sinoll = I I (a cos 0 + ,I3 sin 0)x + (a sin o - ,3 cos 0)yII

= 11 [Ia + 01 sin(w + 0)] x - [Ia + z/3I cos(w + 0)]yII 0)11

1,31

,

from which it follows that II(a + z/3)zllc

supl1(ax - i3y) cos0 + (!3x + ay) sin01I

=

= sup la + z13I Ilx cos(w + 0) - y sin(w + 0)11 OER

= la + zfI sup IIx cos(w + 0) - y sin(w + 0) II OER

=

Ia+1/31' IIzllc

Properties (a) and (b) follow immediately from (*).

Problem 1.1.8. Let X be an arbitrary real vector space with complexification Xc = X ® zX. Establish the following properties. (a) If V is a subspace of X. then Vc = V ?, V = {x + zy : x, y E VI is a vector subspace of Xc.

(b) If X is also a normed space and V is a vector subspace of X, then the vector subspace Vc = V ® zV of Xc is closed in Xc if and only if V is a closed subspace of X.

(c) Let T : X - X be an operator. A subspace V of X is said to be invariant under T (or simply T -invariant) whenever T (V) C V.

A vector subspace V of X is T -invariant if and only if the vector subspace Vc = V ® zV of Xc is Tc-invariant. Solution: (a) Let z1 = r1 + zy1, z2 = .r2 + zy2 E VC and let a + z)3 be a complex number. Then we have

z1+z2 = (x1+zy1)+(x2+2y2)=x1+x2+z(y1+y2)EVc, and (a + zfl)zl

=

(ax1 - Qy1) + z(/3x1 + ay1) E Vc .

Therefore, Vc is a vector subspace of Xc.

(b) This follows from the fact that a sequence where z = x + zy,,, in Xc satisfies z x + zy in X,, if and only if x - x and y,, - y hold in X ; see the preceding problem.

1. Odds and Ends

(c) If T(V) C V, then T,(V,) VV is trivially true. Now assume T,,(VV) 9 VV and let x E V. Then x + zO E Vim, and so Tr (x + zO) = Tx + zTO = Tx + 10 E VV _ V ® W. This implies Tx E V, and thus T(V) 9 V.

U

Problem 1.1.9. Let X be a real vector space. It is very common to denote [x]. the vectors x + zy of Xc as column vectors Under this notation the vector space operations of Xc are given by

[yii +

[y2, = [yt + y2]

and

(a + zQ) [;I = I'3x + ay,

Now assume that X and Y are two real vector spaces. Show that a mapping T : Xc -+ Yc is a linear operator if and only if there exist two (uniquely determined) operators S, T : X - Y such that T has the matrix IT

representation T =

TJ ,

T (x + zy) =

where as usual

IT -Si T [Y] = [sx + Ty]

Also, show that if X and Y are normed spaces, then T is a bounded operator if and only if both S and T are bounded operators. Solution: A straightforward verification shows that every mapping represented by a matrix of the form IS

T J , where S, T : X -. Y are linear operators, defines

a linear operator from Xc to Yc. Now consider a linear operator T: Xc

Yc. For each x E X we can write

T(x+z0) =Tx+zSx, where Tx and Sx belong to Y. A direct computation shows that T and S define linear operators from X to Y. Moreover, the linearity of T implies T (x + zy)

= Tx + zTy = (Tx + zSx) + z(Ty + tSy)

= Tx-Sy+z(Sx+Ty) _ IS

T] [;}.

T -S Now assume that T = IS T J is a bounded operator. Then for each x E X we have

IITxII = II(Tx)cos0+(Sx)sinOll <_ IIT(x+z0)II

IITII' IIdII,

which shows that T is a bounded operator. Similarly, S is a bounded operator. Finally, suppose that S and T are bounded operators. Fix z = x + zy E X. and

1.1. Banach Spaces, Operators, and Linear Functionals

7

note that for each angle 9 we have II (Tx - Sy) cos 9 + (Sx + T y) sin Oil

= II T(x cos o + y sin 9) + S(x sin 9 - y cos 9) II IITII

llxcos9+ysinOll + IITII Ilxsin9 - ycosell

<- (IITII + IISII)IIzII,

and so IlTzll = sup ll (Tx - Sy) cos9 + (Sx + Ty) sin Oil 5 (IITII + IISII)llzll OER

This shows that T is norm bounded.

U

Problem 1.1.10. A bounded operator T : X

Y between normed spaces is said to be invertible if there exists (a uniquely determined) bounded operator

X satisfying ST = IX and TS = ly; the operator S is called the S: Y inverse of T and is denoted T-1. For an operator T : X -+ Y between real vector spaces establish the following properties.

(a) The operator T is surjective if and only if T,,: Xe -+ YY is surjective. (b) The operator T is one-to-one if and only if TT is one-to-one.

(c) The operator T is invertible if and only if TT is invertible. Also, = (T-1)c. show that if T is invertible, then (T,)-z

Solution: (a) Assume that T is surjective and let z = x + ty E Yr. Pick u, v E X such that Tu = x and Tv = y and note that TT(u + tv) = Tu + tTv = x + ty = z. That is. T,, is surjective. Conversely, if TT is surjective and y E Y. then there exists some z = u + zv E Xc such that Tax = Tu + tTv = y + W. So, Tu = y which proves that T is surjective.

(b) Assume that T is one-to-one. If z = x + zy satisfies Tx + sTy = TTz = 0, then Tx = Ty = 0, and sox = y = 0 or z = x + zy = 0. That is, T. is one-to-one. is one-to-one and Tx = 0, then T,,(x + z0) = 0, from which it Conversely, if follows that x + z0 = 0 or x = 0. Thus. T is one-to-one. (c) The first claim follows from (a) and (b) above. For the last part note that

ifx+zyEX, and u+zvE Yr, then (T-1).TT(x+zy) = (T-1),,(Tx+tTy)=T-1(Tx)+zT-1(Ty)=x+ty, and TT(T-1).(u+zv) = TT(T-lu+zT-lv) =T(T-1u) +zT(T-lv) = u+tv. = (T-1).. Therefore, T. is invertible and (T.)-1

Problem 1.1.11. Let X = Xi ® . . . ® X,, be the direct sum Banach space. Show that a mapping A: X - X is a bounded linear operator if and only if there exist (uniquely determined) bounded linear operators Ai3 : X3 -. Xi

(i, j = 1, 2, ... , n) such that A has the matrix representation A = [Aj],

1. Odds and Ends

8

where for each x = Si 0)

All

A12 A22

A21

Ax

ED xn we have

An2

An1 n

Aln

xl

A2n

X2

E Ej

Ann

xn

Ej1 Anjxj

...

n

1 Alixi 1 A2,7xi

n

(AI)zJ)e(A23xj)e...ED(AflJx). j=l j=l Solution: To see this, let A: X - X be a bounded operator. For each j define

j=l

the bounded operator T,: X. -+ X via the formula

0),

Tj(x_,)

where the vector xj occupies the j`h coordinate. We can write

T, J(x)) = A(0-0 041) ...®0®xj 90®...®O) =A, xj EA2jx, An easy argument shows that each A;j : Xj -- X; is a linear operator, and A is bounded if and only if every A,j is a bounded operator. Moreover, for each x, in X we have A(x)

=

j=l n

EA(0GO

0®r-j®0®

®0)

j=1 n

(`j-1 I

n

n

Anjxj) ® (u A2jxj) ®... ® (u Anjxj) jj=l jj=11

All A21

A12 A22

Ant

An2

...

-

Aln

xl

A2n

X2

Ann

xn

This establishes the desired representation of the operator A. The converse is straightforward and is left to the reader.

Problem 1.1.12. A (not necessarily surjective) operator T : X - Y between normed spaces is said to be an isometry (or more precisely a linear isometry) if IITxII = IIxil holds for each x E X. Show that a bounded operator T : X --+ Y between two normed spaces is Y** is likewise an an isometry if and only if its double adjoint T**: X** isometry.

Solution: Assume that T : X

Y is an isometry. Clearly, II TII = IIT' " II = 1 and <_ Ilx** II for all x" E X'*. Now fix x* E X * with IIx' II < 1. Then the so IIT''x'* II

formula y*(Tx) = (x,x*) defines a continuous linear functional on the range of T such that IIy* 11 <_ 1. Let y' also denote a continuous linear extension of y* to all

1.1. Banach Spaces, Operators. and Linear Functionals

9

of Y such that 11y11 5 1; clearly T' y' = x*. Now note that for each x" E X** we have

I (x',x")I = I (T'y',x")I = I (y T"x.')I <- Ily'll and so llx.,Il = supllx.11<1 I

IIT"r"II < IIT"x"ll

(x'.x")I c IIT".r"II. Thus. IIT"z"Il

=

IIx"II hold for

each x" E X". For the converse, assume that T' is an isometry. Then IlTxll = IIT"xII = llxll holds for each x E X so that T is an isometry.

Problem 1.1.13. Let T : X - X be a bounded operator on a normed space. The inequality IITx-TyII IITII-Ilx-yII implies that T is a uniformly continuous mapping. So. T has a unique (linear) continuous extension T : X -+ f C, where X denotes the norm completion of X. Establish the following properties:

(a) If T is an isometry on X, then so is t on k. (b) If T is an invertible isometry with inverse S, then S is also a (linear) isometry.

(c) If T is invertible. then so is t and (T)-1 = (T-1)'. (d) If ,\ - T is invertible on X for some scalar A, then A - T is likewise invertible on f C.

(e) If t is an invertible isometry on f C, is T necessarily an invertible isometry on X? Solution: (a) Assume that T is an isometry and let i E X. Pick a sequence {xn} in X such that Ilxn -.ill ~ 0. Then we have

IlTill = lim IITxnII = Rim IlxnIl = llill That is, t is an isometry.

(b) Assume that T : X - X is an isometry and that a bounded operator S: X - X satisfies ST = TS = I. i.e., S = T'1. Now notice that for each x E X we have Ilxll = IIT(Sx)II = IlSxll

This shows that S is an isometry.

(c) Assume that S: X

X is a bounded operator satisfying ST = TS = I

and let S denotes the unique continuous linear extension of S to all of X. Fix some I E X and choose a sequence {xn} in X satisfying Ilxn -ill 0. Then

STi = lim S(Txn) = lim STxn = lim xn n- oc and similarly TSi = i. These show that the operator S is the inverse of t and that (T)-1 = S = (T-1)'. (d) Note first that (A-T)' = A-T. Now by part (c), we see that if .\-T: X -. X is invertible, then (A - T)' = A - T : X - X is likewise invertible.

1. Odds and Ends

10

(e) Let X denote the vector space of all polynomials defined on [0, 1] equipped

with the sup norm. By the Stone-Weierstrass theorem, X = C[0.1]. Consider the operator T: X - X defined by Tp(t) = p(t2) for all p E X and all t E (0, 11. Clearly, T is a linear isometry. and T is not invertible since T is not surjective (its range consists of even polynomials). However, T: C[0,1] -. C[0,1] satisfies Tx(t) = x(t2) for all x E C[0, 1] and each t E [0, 1], which is an invertible isometry 1 on X.

Problem 1.1.14. Show that for bounded operators X T-+ Y S.+ Z between normed spaces (all over the same field) we have IISTII S IISII IITII. Use this inequality to show that:

(a) The map (S, T) * ST. from G(Y, Z) x C(X, Y) to C(X, Z), is jointly norm continuous. (b) The Banach space £(U, V) of all bounded operators from a normed space U to a Banach space V is a unital Banach algebra. Solution: Notice that if X -I- Y S., Z are bounded operators and x E X. then IISTxll <- IISII - IlTxll <- IISII

IITII - Ilxll

This implies IISTII <- IISII - IITII.

For the joint continuity of the map (S, T) --. ST use the inequality IIST - SoToll

= = <-

IIST -STo +STo - SoToll II S(T - To) + (S - So)TolI IIT - Toll - IISII+IIS - Sofi - IlToll

and the fact that convergent sequences are norm bounded.

1

Problem 1.1.15. Let X and Y be two nonmed spaces over the same field. Show that the mapping T T*, from £(X, Y) into £(Y', X'), is a linear isometry.

Solution: We have that IITII

=

sup IITxll = sup [ sup I(y',Tx)l] II=II=1

11x11=1 11Y *11=1

sup [ sup l (T'y*. x) I ] = sup II T'y' II

I1v11=1 IIxII=1

11v1I=1

IIT'll, as desired.

Problem 1.1.16. Show that every continuous linear functional defined on a subspace of a complex nonmed vector space X has a continuous linear extension to all of X with preservation of its norm. Solution: Let f : M -y C be a continuous linear functional defined on a vector subspace of a complex Banach space X. Then, by the discussion preceding Theorem 1.9, we know that there exists a unique continuous linear functional g: Alx - R

1.1. Banach Spaces. Operators, and Linear Functionals

11

satisfying IIxII = IIf II and f(x) = g(x) - 'g(11)

for each x E Al. Notice that MR is a vector subspace of the real normed space XR. So, by the classical Hahn-Banach theorem, there exists a continuous linear extension of g to all of XR, say g: XR - IR, such that 11911 = IIxII Now observe that the linear functional f : X -b C, defined by

f (x) = 9(x) - zy(zx), x E X. is a linear extension of f to all of X satisfying [Jill = II§II = IIgh = IIxII.

Problem 1.1.17. Let A be a non-empty subset of a normed space X and let xo E X. We say that there is a nearest (or a closest) point to x0 in A if there exists some ao E A such that Ilxo - aoll Establish the following.

Ilxo - all for all a E A.

IIxII is weakly lower semicontinuous. i.e., (a) The norm function x xa - x in X implies IIxII < liminf llx0II. (b) If X is a reflexive Banach space and A is a non-empty weakly closed subset of X, then for every point x0 E X there exists a nearest point to x0 in A.

(c) Assume that the norm of X is uniformly convex, that is, for each 0 < e < 2 there exists some 0 < b < 1 such that IIx - yli > E with x, y E Ux implies II"II < 1 - 6. A normed space is called uniformly convex if its norm is uniformly convex. If A is a non-empty, convex, and closed subset of a uniformly convex Banach space X, then for every xo E X there exists exactly one point in A that is nearest to xo. In other words, there exists a mapping it : X

A defined by 11x

- 7r(x)II = min IIx - yll YEA

The mapping 7r is called the nearest point mapping from X to A. (d) Let C be a non-empty closed convex subset of a uniformly convex Banach space X. Show that the nearest point mapping 7r is continuous.

Solution: (a) Assume xQ - x in X. and let r' E X' satisfy IIx' II S 1. Then for each a we have

so

Ix'(x)I = lim Ix'(x0)I <- liminf IIxaII. This implies IIxII = supjjr.11 <1 Ix'(x)I < liminf IIxQII.

(b) Assume that X is a reflexive Banach space. Also, we can assume that C A such that xo = 0. Put d = infaEA IIail, and then pick some sequence d. We can suppose that Ila,, II < d+ 1 holds for each n. Since X is reflexive. the closed balls of X are weakly compact sets, and so (using the Eberlein-Smulian theorem and passing to a subsequence if necessary). we can assume without loss of Ila,, II

1. Odds and Ends

12

generality that an -N'- a. Since A is weakly closed, a E A, and by part (a) we get d < flail <- lim Ilanll = d. Thus, Ilail = d, and so a E A is a nearest point to zero.

(c) Let X be a uniformly convex Banach space. The uniform convexity of the norm guarantees that X is a reflexive Banach space; see Problem 11.1.5. Now assume that A is a non-empty, closed, and convex subset of X, and that xo = 0 V A. Since A is automatically weakly closed, it follows from part (b) that there exists a nearest point u in A to zero. That is, Hull = min,EA Hall = d > 0. We claim that u is uniquely determined. To this end, assume that some v E A satisfies IIvII = d. We claim that u = v. To this end, assume by way of contradiction that u v, and consider the vectors x = u and y = d. Then IIaII = IIvII = 1, and clearly 0 < e = IIx - vII 5 2. So, by the uniform convexity of the norm, there exists some 0 < b < 1 such that II x y II < 1-6. This implies that the vector z = uzv E A satisfies IIzII <- (1 - 5)d < d, which is impossible. Consequently, u = v, and so there exists exactly one point in A that is closest to zero.

(d) Let x,, -p x in X. We must show that yn = zr(xn) a(x) = y. Translating and scaling appropriately, we can assume without loss of generality that x = 0 and 0. Then, by that IIvII = 1. Now suppose, contrary to our claim, that Ily,, - yll passing to a subsequence. we can also assume that there exists some e > 0 such that Ilyn - vii > 2e for all n. Since yn is the nearest point in C to xn we have llxn - vnll <- Ilxn - vii <- Ilxnll + Ilvli =Ilxnll + 1

1.

Similarly, since y is the nearest point in C to 0, we have

llxn-vnll>Ilvnll-Ilxnll>IIvII-Ilxnll=1-Ilxnll

1.

From these two estimates, it follows that Ilxn - vn ll -' 1 and hence ilvn ll -' 1. So, if we let Z. = vn/Ilvnll, then Iivn - znII = (1 - y' )Ilvnll - 0, and consequently fromIIzn - YII?Ilvn - vII - Ilvn - znII, we see that llzn-vII>efor an n>no. Since X is uniformly convex, there exists a 6 > 0 such that Iiz + vii < 2(1 - 5) provided IIzII S 1 and IIz - vII > e. In particular, Ilzn + yII < 2(1 - 6) for each n > no. Using again that Ilvn - znII - 0, the last inequality yields

Ilvn+vllsllyn-znII+Ilzn+vII
2(1-b)<2.

Therefore, for all n sufficiently large we have II 1- t II < 1. However, for any one of these sufficiently large n, the vector 1 belongs to C and this contradicts our assumption that the unit vector y E C is the nearest point to zero. Hence, yn y, and thus the nearest point mapping r is continuous.

Problem 1.1.18. For a non-empty closed convex subset C of a Hilbert space X establish the following properties of the nearest point map Tr: X --+ C.

(a) If x E C, then ir(x) = x. (b) If x 0 C, then 7r(x) E 8C.

(c) For each x E X and each y E C we have (x - zr(x), y - rr(x)) < 0. (d) For each x, y E X we have (zr(x) - pr(y), x - y) > 0.

1.1. Banadi Spaces, Operators, and Linear Punctionals

13

(e) For all x, y E X we have IIa(x) - r(V)II < IIx - yll.

(f) If C is a closed vector subspace of X, then for each x E X the vector x - ir(x) is orthogonal to C, i.e., (x - a(x),c) = 0 for all c E C.

Remark: Recall that a non-empty subset A of a topological space i is called

a retract of Il if there exists a continuous mapping f : SZ -> A, called a retraction of 0 onto A, such that f (a) = a for each a E A. The preceding properties show that 7r is is a retract of X onto C. Solution: (a) If x E C, then clearly ir(x) = x. (b) Let x f C and note that the arbitrary point xa = Ax + (1 - A)ir(x) on the line segment joining x and zr(x) satisfies JJx - xa JJ = (1 - A)Ilx - a(x) II < IIx - a(x)q for each 0 < A < I. So, if ir(x) is an interior point of C, then for small A the vector xx belongs to C and is closer to x than ir(x), a contradiction. Hence, 7r(x) E c3C.

(c)LetxEXandyEC. Foreach 0
- ir(x) II2 < = =

Ilx

- [ay + (1 - a)zr(x)] ll2

II Ix - , (x)] - a [y - r(x)] II2

IIx-7r(x)II2-2a(x-ir(x),y-x(x))+a2lly-x(x)II2.

This implies 0 < -2a(x - a(x), y - zr(x)) + a2lly - n(x)112 or

-2(x - ir(x).y - a(x)) +ally - ir(x)112 ? 0.

Letting a 10 yields -2(x - ir(x), y - a(x)) > 0 or (x - 7r(x), y - x(x)) < 0. (d) Let x, y E X. Replacing y in (b) with ir(y) E C, we get

(x - 'r(x), x(y) - zr(x)) < 0.

(*)

Also, replacing x by y and y by or(x) in (b), we get (y - 7r(y),7r(x) - r(y)) < 0 or (**)

(x(y) - y. x(y) - a(x)) < 0. Adding (*) and (+.*) yields 0

>

(x - 1r(x), x(y) - ir(x)) + (A(y) - y, x(y) - x(x))

(x - y, x(y) - 7r (x)) + IIx(y)

-

x(x)112.

Therefore, 0 5 Ilfi(y) - x(x)II2 5 (y - x, x(y) - x(x)) = (x(x) - x(y),x - y). (e) From the inequality in the preceding line and the classical Cauchy-Schwarz inequality, we get

IIA(y) - x(x)ll2 < (y - x, x(y) - r(x)) 5 IIy - xfi . IIx(y) - x(x)II This implies Ikkr(x) - 7r(y)II 5 fix - 'lI for all x,y E X.

1. Odds and Ends

14

(f) Replacing x by x - ir(x). we can assume from the outset that zr(x) = 0. Then for each c E C we have x = Ix - (x. c)c) + (x. c)c and x - (x. c)c 1 (x. c)c. and so

IIrI12 = I r - (x. c)cll2 + II (x. c)cll2 > Ilx - (x. C)c112 .

Therefore. (x. c)c. = 0 or (r. c) = 0. This implies x - rr(x) 1 C.

I

Problem 1.1.19. (Support Points). In this problem X will denote a real Banach space, H is a real Hilbert space. and C is a non-empty closed convex

subset of H. The function rr : H - C. is the nearest point mapping as defined in Problem 1.1.17. Recall that a point ao in a subset A of X is called a support point of A if there exists some non-zero x` E X* such that x`(ao) < x* (a) for each a E A. i.e.. if x' attains its minimum value on A at ao. Establish the following. (a) Only boundary points of a set can be support points. (b) If a closed convex subset of X has a non-empty interior, then every boundary point of the set is a support point. (c) A point co E 8C is a support point of C if and only if there exists some x V C such that ir(x) = co. (d) If H = 1Rk. then every boundary point of C is a support point. (e) (Bishop-Phelps [19]) The set of support points of C is dense in the boundary 8C of C.' (f) Give an example of a non-empty closed convex subset of f2 for which not every boundary point is a support point. Solution: (a) Assume that there exist a non-zero x' E X' and some ao E A such

that r'(ao) < x'(a) for all a E A. Also. suppose by way of contradiction that ao E A°. Then there exists some r > 0 such that Ilyll < r implies ao ± y E A. Therefore. x`(ao) < x'(ao t y) = x'(ao) ± x'(y) or ±x'(y) _> 0 for all y E X with Ilyll < r. This implies x' = 0. which is a contradiction. This establishes that ao E 0A.

(b) Let A be a closed convex subset of X such that A° 54 .0. We claim that A° is dense in A. To see this. let x E A° and fix some y E 8A. Choose an open neighborhood V of zero such that x + V C A and fix 0 < a < 1. Since y - I 'Q V is an open neighborhood of y. t here exists some z E A fl (y - *0 V). This implies (I - o)(y - z) E aV. Moreover, the convexity of A guarantees that the non-empty open set U = (1 - a )z + a(x + V) lies entirely in A. Consequently. from 1

ax +(1-a)y=ox +(1-a)(y-z)+(1-a)zEax+aV+(1-a)z=UCA. we see that ax + (l - o)y E A°. Letting a j 0. we infer that y E A°. That is. A° is dense in A. i This famous result of E. Bishop and R. R. Phelps is true for any real Banach space; for a complete discussion of the Bishop Phelps theorem see 14. Section 8.91. V. 1. l.omonoso (531 has proven recently that the Bishop-Phelps theorem fails for complex Banach spaces.

1.1. Banach Spaces. Operators. and Linear Functionals

15

Next, we claim that A° is a convex set. To see this. let it. r E A° and 0 < A < 1.

Put V = AA° + (1 - A)A° and note that V is a non-empty open set. Since A is convex, it follows that V C C. Clearly. Au + (1 - A)r E V. This implies that Au + (1 - A)i, is an interior point of A. i.e., Au + (1 - A)r E A°. Therefore. A° is a convex set.

Now assume that ao is a boundary point of A. Since ao 0 A°. it follows from the classical separation theorem that there exists some non-zero linear functional

x' E X` such that r'(ao) < x* (a) for all a E A°. Since A° is dense in A. we see that x'(ao) < x* (a) for all a E A. That is, ao is a support point of A. (c) Fix c41 E W. Assume first that there exists sonie r 0 C such that ir(r) = ce.

From part (c) of Problem 1.1.18. we know that (x - r(x). y - r(x)) < 0, i.e.. (r - cn. y - (a) < 0. for each y E C. Since x 0 co. the vector p = co - r 96 0. and the last inequality can be rewritten as (p. co) < (p, y) for all y E C'. This shows that co is a support point of C. For the converse., assume that r(j is a support point of C. So. there exists a non-zero vector p E H such that (p.co) < (p. c) holds for each r E C. We claim

that the vector x = co - p does not belong to C. Indeed, if r E C. then from (p. co) < (p, r) = (p. co - p) = (p. ro) - 11p112, we get 0 < 11p1I2 < 0. Which is a contradiction. Hence. r C. Now note that for each c E C we have (p. ro - c) < 0. and so 1Ir - CI12

= =

II(co - p) - c1I2 = II(co - c) - PII2 :Ico

-

2(p. ro - r) -r 11p1I2

PII2 = Px - ejl;2

holds for each r E C. This shows that -,(x) = c:o. (d) Let co be a boundary point of C. Choose it sequence

C Rk such that

x 4 C for each it and r, -- co. By the preceding part. for each it there exists some non-zero vector p E Rk such that holds for each it. Normalizing the sequence we can assume without loss of generality that {Ip II = I holds for each it. In this case, by passing to a subsequence.

we can assume without loss of generality that {pn} also converges. say pn - p. where p # 0. Now the continuity of r implies r(co) = co, and so from (* * *) we get p c > p co for all r E C. That is, ro is a support point of C. (e) Let co E 8C and pick a sequence

n and rn -y co. By part (c), each continuity of r to see that

of points such that r ¢ C' for each is a support point of C. Now use the

;r(co) = co.

(f) We let C' _ (I. the positive cone of 12. Clearly. C is a non-empty clued convex subset of f2. Moreover. CO = 0. So. every point of C is a boundary point.

W claim that a point c = (c1. C2....) E C is a support point of C if and only if c, = 0 for some i. To see this. let r = (cl. C2....) E C. Assume first that c, = 0 for some i. It is easy to see that the basic unit vector e, supports C at c. For the converse. assume

1. Odds and Ends

16

that c is a support point of C. So, there exists a non-zero vector p E £2 such that (p, x) > (p, c) for all x E C. Since x E C implies \x E C for all J1 > 0, it follows that (p, x) > 0 for each x E P2 . This implies that the vector p = (pl, P2...) satisfies pi > 0 for each i. Since p # 0, it follows that pi > 0 for some i. Therefore, from 0 E C, we get 0 < pic, < (p, c) < (p, 0) = 0, and so ci = 0. Hence, the support points of C are the non-negative vectors having at least one coordinate equal to zero. In particular, the latter shows that not every boundary point of C is a support point. I

Problem 1.1.20. (The Finite Dimensional Separation Theorem). Recall that a non-zero linear functional x' E X' separates two non-empty subsets A and B of a Banach space X if x* (a) < x* (b) for all a E A and all b E B. Show that every pair of disjoint non-empty convex subsets of Rk can always be separated.

Solution: Let A and B be two arbitrary non-empty convex subsets of Rk such that A fl B = 0. Let C = A - B and note that C is a non-empty convex subset of Rk such that 0 f C. We distinguish two cases. CASE I: 0 ¢ C. In this case, we apply Problem 1.1.18 with C in place of C. If we let x = 0 and

p = a(0) 0 0 in part (c) of that problem, then we have (0 - ir(O), y - a(0)) < 0 or -p (y - p) < 0 for each y E C. This implies 0 < 11p1I2 S p y for all y E C. In particular, letting y = a - b with a E A and b E B. we get p- b < p- a for all a E A and all b E B. That is, p separates A and B. CASE II: 0 E C.

In this case, we shall prove that 0 E 8C. If this is established, then from part (d) of Problem 1.1.19 we know that 0 is a support point of C. So, there exists a

non-zero vector p E Rk such that p y > p 0 = 0 for all y E C. This implies p a > p b for all a E A and all b E B. That is, p separates A and B. It remains to be shown that 0 E C implies 0 E 8C. In this case, we distinguish two subcases.

CASE IIa: CO = 0.

In this subcase, fix a point c E C and let B = c - C. Then B is a convex set, 0 E B and B° = c - CO = 0. We claim there do not exist k linearly independent vectors in B. To see this, assume by way of contradiction that {b1,. .. , bk } C B is a linearly independent set. Then each x E Rk can be written uniquely in the form x = albs + + akbk with ai E R. So, the function f : Rk - Rk, defined by f(x) = f(aibi + ... + akbk) = (a1,a2,...,ak),

is linear (and hence continuous). Since a; > 0 for each i in conjunction with the condition Ek 1 a, < I and the fact 0 E B implies

a1b1+...+akbk = a1b1+...+akbk+(1-E;` 1ai)OEB.

1.1. Banach Spaces, Operators, and Linear Functionals

17

we see that B contains the non-empty open set f-1(U), where rkjai < 11, U = {(al,a2,....ak) E (0,1)k: 0<

contradicting B° = 0. Thus, if we let {b1.... , bt} be a maximal linearly independent subset of B. then it follows that P < k and B C span{bl.... , be}. i.e., the set B is contained in a (closed) proper linear subspace. Hence, B C span{bl,.... bt}. This implies B = c - C = 0 and so 0. Therefore, 8C = C. and thus 0 E 8C. CASE IIb: COO 0. In this case, we need to invoke the following property of convex sets. (R) Let X be a finite dimensional vector space, let A be an open convex subset of X and let B be a convex subset of A. If B is dense in A, then B = A.

The proof is by induction on the dimension of X. If dim X = 1, then the result is trivial. So, for the induction step, assume that property (R) is true for any vector space of dimension n. and let A be an open convex subset of an (n + 1)-dimensional

vector space Y. Assume also that B is a convex subset of A that is dense in A. Translating A and B if necessary. we can assume that 0 E B. Fix a vector a E A and then choose a non-zero vector u E Y such that a - u = 0.

Let H = {yEY : ,y u = 0} and note that H is a vector subspace of dimension n. such that a E H n A. Also. H n A is an open convex subset of H. If the convex subset H n B of H n A is dense in H n A. then it will follow from our induction hypothesis that a E B for all a E A, i.e.. B = A. which is the desired conclusion. So, we shall finish the proof by showing that H rl B is dense in H fl A. To this end, let x E H n A and let e > 0. Also, consider the open half-spaces

H1={yEY:

and

H2={yEY:

Since x E A and x + Au E H, and .r - Au E H2 for all A > 0, there exists some Ao > 0 with Ao ilu II < 2 such that x + Aou E Hi n A and x - Aou E H2 n A. Since B is dense in A, there exist bl E HI n B and b2 E H2 fl B such that 11x +Aou - bx 11 < z

and Iix - Aou - b211 < 2. From u bl > 0 and b2 > u c2, it follows that there exists some 0 < a < 1 such that the vector b = abl + (1 - a)b2 E B satisfies u b = 0. i.e., b E H. Now note that lix - bil

= Ila(x + Aou - b1) + (1 - a)(x - Aou - b2) + [(1 - a)Aou - aAou] II

< allx+Aou-bill+(1-a)Ilx-Aou-b21I+(1-a)A,nIlull+aAollulI

< a2+(1-a)2+a2+(1-a)2 =c. This shows that H n B is dense in H fl A. as desired. An immediate conclusion of property (R) is the following. If a convex C in a finite dimensional vector space has a non-empty inte-

rior, then C = C° and 8C = 8C. Indeed. if CO 36 0, then CO is dense in C (see the proof of part (b) of Problem 1.1.19), and so C° is also dense in CO. This shows that C° = CO.

1. Odds and Ends

18

Finally, to finish the solution of the problem, notice that 0

C in conjunction

with C° 00 implies 0EC\C°=C\C =BC. Problem 1.1.21. Let Y be a closed subspace of a nonmed space X, and define the function III

I II

: X/Y - R by III[x]III = info IIxII YE[X

Establish the following.

(a) The function III -III is a norm on the quotient vector space X/Y. (b) The unit ball of X/Y under the quotient norm is

UX/y={[x]: xEUX}. (c) The quotient map x [x] is an open contraction. (d) If X is a Banach space, then (X/Y, III . II) is also a Banach space. Solution: (a) Clearly, III [x] 111 >- 0 for each x E X and [x] = 0 implies III [x] 111 = 0. Now assume that III [x] I II = 0. Then there exists a sequence { yn } C [x] such that

yn -+ 0. From x - yn E Y for each n. x - yn - x, and the closedness of Y, we get x E Y or [x] = 0. For the homogeneity of III .III, note that for each A 96 0 we have IIIA[x]III = III[Ax]III =

yEI =I

IIxII = IaI

m

l X11 a 11= al = f IIxII = IaI . III[x]III

For the triangle inequality, let u,v E X. Then for each x E [u] and y E [v], we have x + y E [u + v], and so

III[u+v]III S Ilx+xll <- IIxII+IIyII Taking infima for all x E [u] and y E [v], we obtain [[ 1 [u + v] ] II < III [u] I II + III [v] I II

(b) If x E UX, then III[x]lII <_ IIxII < 1, and so {[x]: x E Ux} C Ux1y. For the converse inclusion, let [x] E UX/y, i.e., III[x]lll = nfyE(=) IIxII < 1. This implies that there exists some y E [x] satisfying IIxfl < 1 or y E UX, and so [x] _ [y]. So, UX/ y C { [x] : x E Ux } is also true, and thus UX/ y = { [x] : x E Ux }.

(c) It should be clear that the quotient map x - ar(x) _ [x] is linear. Moreover,

from part (b), we have zr(UX) = UX/y. Rom this it easily follows that it is an open mapping. In addition, the inequality Illir(x)III < IIxI) for each x E X shows that the quotient map 7r is also a contraction. (d) Assume now that X is a Banach space and let { [xn] } be a Cauchy sequence in X/Y. By passing to a subsequence if necessary, we can assume without loss of generality that III[xn+1] - [xn]III = lll[xn+1 - xn]][[ < -1 holds for each n. Now for each n pick some yn E [xn+1 - xn] satisfying Ilyn [[ < z . Since X is a Banach space, the series E' 1 y,, is norm convergent in X; let y = 0= y, . Then the continuity of the quotient map implies

a n=1

([xk+1] - [xk]) = limo([xn] - [xl]) .

[xn+1 - xn] = lm

[tin] _

[y] _

n-1

oo

n=1

k=1

1.1. Banach Spaces, Operators, and Linear Functionals

Therefore, the sequence {[x]} is norm convergent in X/Y and This shows that X/Y is a Banach space.

19

[y] - [xl].

Problem 1.1.22. If V is a vector subspace of a normed vector space X, then its annihilator is the norm closed subspace of X* defined by

Vl={x'EX': x'(v)=0 for all vEV}. For a closed subspace Y of a normed space X establish that

(X/Y)' = Yl

and

Y* = X'/Y1

by proving the following statements.

(X/Y)'. defined via the formula

(1) The mapping m: Yl

O(x*)([x]) = x*(x) ,

x' E Yl, [x] E X/Y,

is a well-defined surjective linear isometry. (2) The mapping z¢: X'/Yl Y', defined via the formula

Vi([x']) = x* l,. (the restriction of x' to Y) for each [x'] E X'/Y1, is a well-defined surjective linear isometry. Solution: (1) We check first that 0 is well defined. To see this, let x' E Yl and [x] = [y], i.e., x - y E Y. Then x* (x) - x'(y) = x* (x - y) = 0 or x* (x) = x* (y), and so op is well defined. Clearly, ¢ is linear. To see that 0 is a surjective mapping, let ID E (X/Y)'. Since the quotient map

r: X -+ X/Y is continuous, it follows that the linear functional x' = T o r is also

continuous on X, i.e., x' E X'. From x'(x) = %Y(r(x)) and r(x) = 0 for each x E Y, we see that x' (x) = 0 for each x E Y, i.e., x' E Y-. Moreover,

O(x')([x]) = x'(x) = W(r(x)) =''([x]) holds for all x E X. That is. 4(x') =', and so 0 is also surjective. It remains to be shown that do is an isometry. To see this, let x' E Y1. Then for each [x] E U.,,,, and each y E [x] we have Io(x')([xl)I = Ix*(x)I = W WI <- IIx'II - IIyII II [x]I I for each [x] E U.,,,.. So. II0(x*) II < IIf II . On the other hand, for each x E Ux we have [x] E U,,,,., and thus

This implies I o(x') ([x]) I < 11x* 11

-

Ila(x')il 2: I¢(x')([xl)I = This implies that II0(x*) II > supzEfj I x' (x) I = 11x* 11. Therefore, 110(x*) 11 for each x' E Y1. Consequently, 0 is a surjective linear isometry.

= 11x' 11

(2) We shall verify first that w is well defined. To this end, let [z'] _ [x'] in X'/Y', that is, z' - x' E Y'. Consequently. for each vector y E Y we have z *(y) - x'(y) = (z' - x')(y) = 0, i.e., z'(y) = x'(y). This shows that the mapping tG is well defined. Clearly, 0 is linear.

1. Odds and Ends

20

To see that tG is surjective, let y' E Y. According to the Hahn-Banach theorem, there exists some z' E X' such that z'(y) = y'(y) for each y E Y. Any such z' E X' satisfies tii([z']) = y'. So, b is surjective. To finish the solution, it remains to be shown that tp is an isometry. To see this, fix [x'] E X'/Y1. Pick some z' E X' such that z'(y) = x'(y) for each y E Y (and so z' E [x*]) and Il z' Il = Ilx I Y II , where 11z11 is the norm of z' in X' and IIx'IYII is the norm of x'I,, in Y. Then for each vector y E Y with Ilyll = 1 we have II,([x'])II > It'([x`])(y)I = Ix'(y)I, and so

IIo([x*])II>sup {Ix'(y)I: y E Y and Ilyll=1}=llx IYII=Ilz'll>II[x']II On the other hand, for each if E [x'] and each y E Y we have II'0([x`])(y)II = Ix(y)I = Iu'(y)I 5 lull llvII This implies IIV([x'])(y)II < II[x']II Ilyll for all y E Y, from which it follows that I Hence IIW([x*])II = II[x']II, and the solution is finished. Ilt/'([x'])II < II[x']II.

1.2. Banach Lattices and Positive Operators Problem 1.2.1. Let C be a cone in a real vector space X. Show that the relation > on X defined by x > y whenever x - y E C is a vector space order

such that X+ = C. Solution: Recall that the cone C is characterized by the properties:

(a) C+CCC, (b) aC 9 C for all a > 0, and (c) C f1 (-C) = {0}. Now define the binary relation > on X by letting x > y whenever x - y E C. Next, we shall verify that 2 is a vector space order.

(1) (Reflexivity) We have x > x for each x E X. This follows immediately

fromx-x=OE C. (2) (Antisymmetry) Let x > y and y > z. That is, x - y E C and similarly

-(x-y)=y-xEC. By property(c),wegetx-y=Oorx=y.

(3) (Transitivity) Assume r > y and y > z. That is, x - y E C and y - z E C. From property (a), we see that x - z = (x - y) + (y - z) E C. So, x > z. (4) Let x > y, i.e., x-y E C. Then for each z E X we have (x+z)-(y+z) _

x-yEC,that is,x+z>y+z.

(5) Suppose x > y and let a > 0. Then x - y E C and using property (b), we get ax - ay = a(x - y) E C, so that ax 2 ay. Therefore, > is a vector space order.

Finally, notice that X+={xEX: x>0}={xEX: x-O=xEC}=C. I

1.2. Banach Lattices and Positive Operators

21

Problem 1.2.2. Let X and Y be two ordered vector spaces such that Y is Archimedean and the cone of X is generating. If T : X+ Y+ is an additive function (i.e., T(x + y) = Tx + Ty for all X. Y E X+), then show that T has a unique linear positive extension from X to Y. Moreover, if for each x E X we pick two vectors X1, X2 E X+ such that x = x1 - X2, then the unique extension T : X - Y of T satisfies

Tx = Tx1 -Tx2.

(*)

Solution: First notice that if r = X1 - x2 = yl - y2 with x,, y; E X+ for each i, then rl - r2 = Y1 - y2 or r1 + Y2 = Th + 12, and so by the additivity of T, we get

Tx1+Ty2=T(rl+y2)=T(yl+x2)=Ty1+Tx2. This implies Tx1 - Tx2 = Tyl - Ty2 and so t as given by (*) is well defined. From

(*) it should be obvious that Tx = Tx > 0 for each x E X+ and T(-x) = -Tx for all r E X. Next, we shall verify that T is linear. For the additivity of t notice that if x = x1 - 12 and y = y1 - y2 with the x; and y; all positive, then

T(r+y)

T(rl +Y1 - (r2+y2)) =T(xl+yl) -T(r2+y2) = Txl + Ty1 - Tx2 - Ty2 = (Txl - Tx2) + (Tyl - Ty2) =

= Tx+Ty. In particular, the additivity implies T(rx) = rT(x) for each rational number r and all x E X. For the homogeneity of T we need two properties. (i) T is monotone on X. That is, 0 < x < y implies Tx < Ty. Indeed. if 0 < x < y, then Ty = T(x + (y - x)) = Tx +T(y - x) > Tx. (ii) Let Z be an Archimedean ordered vector space, let x E Z+. and let y E Z. Assume that a sequence of real numbers converges to a, i.e., On -+

a. If anx < y (resp. Y!5

for each n. then ax < y (reap. y!5 ax).

To see this, fix k. Then for all sufficiently large n we have a - an < k, and so

(a - a )x < 1 x for all large n. Hence, ax - y < ax - anx = (a - an)x < kx or k(ax - y) < x for each k. Since X is Archimedean, we get ax - y < 0 or ax < y. Now let r E X + and a > 0. Pick two sequences of rational numbers {rn } and {tn } satisfying r T a and t j a. Using (i) and the homogeneity of T with respect to the rational numbers, we get T(rnx) < T(ax) < T(snx) = snTx.

From (ii), we get immediately that aTr < T(ax) < aTx, or aTx = T(ax). Finally. fix a > 0 and x = xl -12 E X with 11.x2 E X. Then we have

T(ax) = T(axl - axe) = T(ax1) - T(ax2) = a(Trl - Tx2) = aTx. and T(-ax) = T(ar2 - axl) = a(Tr2 - Tx1) = (-a)T(x). Thus. T is also homogeneous, and hence linear.

1. Odds and Ends

22

Problem 1.2.3. Show that order complete Riesz spaces and normed Riesz spaces are. Archimedean. Solution: Let E be an order complete Riesz space and assume that nx < y holds f o r all n and some x, y E E. Since E is order complete, u = sup{nx: n = 1, 2, ...}

exists in E. Now from nx = (n + 1)x - x < u - x. it follows that u < u - x or x < 0. Thus. E is Archimedean. Now assume that E is a normed Riesz space and let nx < y for all n and some

x, y E E. Notice that nx < y < y+ holds for all n, and from this we infer that 0 < nx+ < y+ for all n. Therefore, nHHx+II = llnx+I1 < 11y+II < oo also holds for each n. This implies 11x+11 = 0 or x+ = 0. Consequently x < x+ = 0, proving that E is Archimedean in this case too. I

Problem 1.2.4. Establish the following properties of order convergence.

(a) A net in a partially ordered set can have at most one order limit. (b) If xa T x or xa j. x in a partially ordered set, then xa -° x. (c) If a net {xa}aEA in a partially ordered set satisfies xa T (resp. xa 1)

and .ra ° x, then xa T x (resp. xa j x). (d) A positive operator T : X -> Y between two ordered vector spaces is order continuous if xa 10 in X implies Txa 10 in Y.

(e) A net {xa}aEA in a Riesz space satisfies xa ° 'x if and only if there exists another net {ua}AEA such that ua j 0 and for each A E A there exists some ao (depending on A) such that Ixa-x1 < ua

holds for each a > ao. (f) The lattice operations in any Riesz space are order continuous. Solution: (a) Let {xa}aEA be a net in a partially ordered set, and assume that .xa --9- z and xa - y. By the definition of order convergence there exist two nets

48}8EB and {z,},Er such that yq T x, z1 j x, and for each 0 E B and y E 1' there exists some a(3, y) E A such that y,3 < xa < zy holds for each a > a($, y). Similarly, there exist two nets {UA}aEA and {v,,}PEA, such that ua T y, v,, j y, and

for each A E A and p E AI there is some a'(A. p) E A such that ua < xa < v,, for each a > a'(A. p). Fix (3 E B, y E 1', A E A. and p E Al. Pick some index a E A with a > a((3, y) and a > a'(A, p). Then we have y8 < xa < v,, and ua < xa < z.y. This implies ua < z., and Y35 v,,

for all indices (3 E B. y E r, A E A, and p E Al. The latter inequalities yield x < y and y < x. That is, x = y, and the uniqueness of the order limits has been established.

(b) Assume that xa T x in a partially ordered set. Consider the two nets {ya }aEA and {za }aEA defined by ya = xa and za = x for each a E A. Since ya < xa < za for each a E A, it follows immediately that xa -a-- x. (c) Assume that a net {xa}aEA in a partially ordered set satisfies xa T and xa -2- x. Pick two nets {y8}$EB and {z., }7Er such that Y# T x, z, j x and for

1.2. Banach Lattices and Positive Operators

23

each 3 E B and -y E I' there exists some a(3. y) E A such that y3 < xa < z7 for each a > a(3,y). Now fix any index a E A and note that for each 13 E B and y E F there exists some index a' E A such that a' > a and a' > a(/3, -i). This implies x"' < X" < Z. and so xQ < z., for all a E A and all "p E F. From z., j x. we see that xQ < x for each a E A, i.e., the element x is an upper bound of the net {xQ}QEA.

If xQ < y holds for each a E A, then as above we see that yo < xQ < y, and so y3 < y for each 3 E B. Taking into account that y3 T .r, the latter shows that x < y. Therefore, x is the least upper bound of the net {xa}aEA, i.e., xQ T x. The decreasing case can be established in a similar manner. Y be a positive operator between two ordered vector spaces.

(d) Let T : X

Assume first that T is order continuous, and let xa j 0 in X. Then Txa I in Y. By part (b), we know that xa -4-a 0. and so. by the order continuity of T, we have

Txa -0- 0. By part (c) we get Txa 10 in Y. For the converse, assume that uQ j 0 in X implies Tu0 10 in Y. Let xQ -- x in X. Pick two nets {y313EB and {z,}. cr such that y3 T x, z7 j x and for each (3 E B and I E I' there exists some a(l3, y) E A such that y3 < xQ < z, holds for each a > a(8, y). Then the positivity of T implies that for all a > a(,3, y) we have Ty,3 < Txa < Tz.,. Moreover, our condition implies Ty3 T Tx and Tz., I Tx.

These show that Txa --' x in Y. (e) Let {xa }aEA be a net in a Riesz space. Assume first that xa 0 x. Pick two nets {yp}3EB and {z.,}.,Er such that Ya T x, z7 j x and for each 0 E B and 7 E I' there exists some a(3. y) E A such that y3 < xQ < z, holds for each a > a(a3. y).

Now note that for each a > a(0.1) we have xa - x < z- - x < z., - y3 and - (x0 - x) = x - xa < x - y3 < z., - y3, and so Ix,, - x1 < z., - yd. Therefore. if we let A = B x r (the product index set) and define the net {ua}AE,A by ua = z1 - y3 then Ix0 - x1 < uA for all a > a($, y) = a(A). for each A = For the converse, assume that there exists a net {ua}AEA such that ua 10 and for each A E A there exists some ao (depending on A) such that Ixa - xi < ua for each a > ao. The latter inequality is equivalent to x - u,\ < xa < x + uA for each

a > ao. So, if we put ya = x - uA and za = x + uA, then we easily infer that xQ 3y X.

(f) Assume that xa

in a R.iesz space. By (e) there exists some net

{UA}aEA such that u,, j 0 and for each A E A there exists some ao (depending

on A) such that Ix,, - xl < ua for each a > ao. Now use the inequalities IXQ - x+1 < Ix,, - xi, I x- - x- 15 Ix,, - xl, and IlxQl - jxlj < Ix,, - xj

and (e) to infer that

xQ x+. x- -x-. and IxQj -$- 1xl. Thus, the lattice operations x - x+. x - x-. and x- lxj are order continuous.

I

Problem 1.2.5. Show that a solid subset D of a Riesz space E is order closed if and only if {x019 D fl E{' and xQ j x in E imply x E D.

1. Odds and Ends

24

Solution: Let D be a solid subset in a Riesz space. If D is order closed and a net {xa} in D n E+ satisfies X. T x, then clearly x E D. Now assume that whenever a net {ya } C Dn E+ satisfies ya T y, then y belongs to D. Let {xa } be a net in D satisfying x. -Q-+ x. So, according to Problem 1.2.4(c), there exists another net {ua}aEA such that ua j 0 and for each A there exists some

ao such that Ixn - xl < ua for each a > ao. ):Yom x - ua < xa < Ixal, we see that (x - ua)+ < Ix0I for each a > ao. Since D is a solid set. it follows that {(x - ua)+} c D. Now from (x - uA)+ T x and our hypothesis, we infer that x E D. Hence, D is order closed. a

Problem 1.2.6. If J is an ideal in a Riesz space E, then show that the band Bi generated by J is given by

Bi ={xEE: 3 {x1,}CJ such that Moreover, show that for each x E E the principal band Bx generated by x is given by

Br= {y E E: lyl A nlxi T lyl } . Solution: Fix an ideal J in a Riesz space E and let

B={xEE: 3 Ix,,,}CJ with 0<xaT1x1}. Clearly, every band that includes J must include B. Thus, in order to establish that B,, = B, it suffices to show that B is a band. To this end, let x,y E B. Pick two nets Ix,,) and {ya} satisfying 0 < X,, T IW and 0:5 ya I lyI. From

l

Ix + yl A (xa + ya) T Ix + yl A (Ixl + Iyl) = Ix + yl

and lµlxa T IpUIxI = lµxl for each scalar p, it follows that B is a vector subspace. Also, if Izl _< Ixl, then from {Izl A xa) C J (for this use the fact that J is an ideal) and Izl A xa 1 (zi A Ixl = Izl. we see that z E B, and so B is an ideal. Now assume that a net {us} in B satisfies 0:5 uj T u in E. If we let

D = {v E J: v:5 us for some 3), then an easy argument shows that D T u. Therefore, considering D as a net in J, we infer that u E B. Now a glance at Problem 1.2.5 guarantees that B is a band, as desired.

To establish the formula for the principal band B, note first that Br = BE., and then let y E B. By the preceding case, there exists a net {xa } of Ex satisfying 0 < xo T Ivl For each index a, there exists some n such that xa < lyl A nlxl < Iyl I This implies lyl A nlxl T Iui, and from this the desired conclusion follows.

Problem 1.2.7. Show that in an Archimedean Riesz space the band generated by a non-empty set A coincides with Add = (Ad)d, i.e., BA = In particular, conclude from this that every band B in an Archimedean Riesz space satisfies B = Bda Add.

1.2. Banach Lattices and Positive Operators

25

Solution: It can be easily verified that S' is a band for every non-empty subset S of a Riesz space. To establish this, one must use just the order continuity of the lattice operations. Now assume that E is an Archimedean Riesz space, and let A he a non-empty

subset of E. Clearly, A C Add, and this implies that BA C Add. To finish the proof. we must verify that Add C_ BA. To this end, let 0 < x E Add and consider the set

D={zEBA:0
Clearly. D j. :and we claim that D , x.

To see this, suppose that a vector z E E satisfies a < z for all a E D. Then a < r A a also holds for all a E D. We must establish that x < z or, equivalently. that r = r A z. Assume by way of contradiction that u = x - x A z > 0. Clearly. u E Add. Since Ad f1 Add = {0}. we see that u Ad. This means that there exists

some y E A such that r _ lyl A u> 0. and clearly r E BA. Then v E D and 2v = t! + v < x A z + (r - x A z) = x. Now an easy inductive argument shows that nv < x for each n. Since E is Archimedean. the latter shows that tr = 0. which is a contradiction. Hence. u = 0 and so x = x A z < z. Thus, we have shown that D T x. Since D C BA and BA is a band, we get x E BA. Therefore. BA = Add

,

Problem 1.2.8. Show that a vector e > 0 in an Archimedean Riesz space is a weak unit if and only if IxI A e = 0 implies x = 0. Solution: Let e > U be a vector in an Archimedean R.iesz space E. Assume first that e is a weak unit (i.e.. Be = E), and let IxI A e = 0. This implies lxI A ne = 0 for all n. From Problem 1.2.6 we know that 0 = jxi A ne. T I rl. Since the order limits are uniquely determined, it follows that IxI = 0 or x = 0. (Note that this implication does require E to be Archimedean.) For the converse. assume that IxI A e = 0 implies x = 0. i.e., B = {0}. Since E is Archimedean, it follows from Problem 1.2.7 that Be = Bdd = {O}d = E, i.e., e is a weak unit.

Problem 1.2.9. Show that a positive operator T : X - Y between two partially ordered vector spaces is order continuous if and only if xa 10 in X implies Tx J. 0 in Y. Solution: Let T: X Y he a positive operator between ordered spaces. Assume first that T is order continuous and let xa j 0 in X. Since x -2- 0 is trivially true in X. we have Tx,, -9- 0 in Y. Also. since T is positive. we have 0 < Txa j in Y. To see that Tx,, 10. assume that Tx,, > y for all a and some y E Y. Pick two nets and {v,} in Y satisfying u < Tx, < va for each a, u,, T 0. and r 10. Now notice that y < Tx < v,, holds for all a. and from this we easily get that y < 0. Hence, Tx j 0. Now assume that x 10 in X implies Tx. j 0 in Y, and suppose za -Q+z in X. Pick two nets {aa} and {ba} in X satisfying as < za < b,, for each A, as T z and b 1 z. Now the positivity of T implies Tap, < Tz., < Tbj, Ta,, < 7'z < Tb,,. (*) and

1. Odds and Ends

26

as well as Taa T and Tba 1. Next, notice that b,\ - as 1 0 holds in X, and so (according to our hypothesis) Tba - Taa = T(ba - aa) 10 in Y. Now a glance at (*) guarantees that Taa T Tz and Tba 1 Tz in Y. proving that Tz.\ - - Tz in Y. That is, T is order continuous.

Problem 1.2.10. Show the following: (a) If 1? is a compact topological space, then in C(1t) each component of the constant function 1 is of the form Xv for some clopen set V. (b) If p is a finite measure, then in L1(µ) each component of the constant function 1 is of the form XA for some measurable set A. Solution: (a) If V is a clopen subset of St, then Xv is clearly a component of 1 (the constant function one on 1). Now let f E C(f2) be a component of 1, i.e., 1 A (1 - f) = 0. Therefore, 0 < f < 1 and min( f (w),1 - f (w)} = 0 for each w E ft. Hence, f = X1, for some subset V of Q. Since f is a continuous function, it is easy to see that V must be a clopen subset of Q.

(b) Let (X. E, p) be a finite measure space. As before, if f E L1(µ) is a component of 1, then f = XA for some subset A of X. Since f is measurable, it follows that A must be a measurable subset of X.

Problem 1.2.11. If E and F are Riesz spaces with F Dedekind complete, then show that the lattice operations of any two operators S,T E 4(E, F) are given by the Riesz-Kantorovich formulas

(S V T)(x) = sup{Sy + Tz: y, z E E+ and y + z = x }, and

(SAT)(x) = inf{Sy+Tz: y, z E E+ and y+z =x} for each XE E+. Solution: We prove the first formula only. From the well-known lattice identity S v T = T + (S - T) v 0, it suffices to establish the formula when T = 0. So, assume T = 0 and S E Cb(E, F). Define the mapping R: E+ F+ via Rx = sup Sy. o
Since F is Dedekind complete and S is order bounded, it follows that R is indeed a well-defined mapping from E+ to F+. We claim that R is additive.

Forthis, let u.v E E+. SinceO
Property, there exist 0 < yj < u and 0 < y2 < v satisfying yl + Y2 = y, and so Sy = Sy1 + Sy2 < Ru + Rv. This implies R(u + v) < Ru + Rv, and thus R(u + v) = Ru + Rv. That is, R is additive. Now by Problem 1.2.2, R has a unique linear positive extension (which we shall denote by R again) from E to F. To finish the proof, we shall show that R = S V 0 holds in Cb(E. F). To this end, note first that Sx < Rx for each x E E+, and hence S< R. Now assume that another operator Q E £,(E, F) satisfies Q > S and Q > 0. If x E E+,

1.2. Banach Lattices and Positive Operators

27

then for each 0 < y:5 r we have Sy < Qy < Qr, and so Rx = supo
Thus, Q _> R in Cb(E. F). That is, R is the least upper bound of S and 0 in

I

Lb(E. F) or R = S V 0. as desired.

Problem 1.2.12. For a nonmed Riesz space E establish the following properties.

(a) Its cone Et is norm (and hence weakly) closed. (b) The order inter-als of E are norm (and hence weakly) closed. (c) If a net {x(, } in E satisfies x T and Jl xn - x)[ -+ 0. then x0 T X. Solution: (a) To see that the cone E+ is closed, assume that a net {yx} in E+ satisfies yx - y in E. From Iyx-y+l < lyx-yl. it follows that Ilyx-y+ll < Ilyx-yIl and so yx --' y+. Thus. y = y+ > 0, proving that E+ is closed. Since. E+ is a convex set, it follows that E+ is also weakly closed.

(b) If r < y, then [.r. y) = (x + E+) n (y - E+). and so the order interval [r, y] is norm closed. Since [x. ,y) is also convex. it must also be weakly closed.

(c) Assume that a net note that the net

in E satisfies .r j and x,, - .r. If a is fixed, then

in E+ satisfies x3-r -j-+x-.r,,. So. z-r E El

and thus r > z for each a. This implies that x is an upper bound of the net {r,, }. Now assume that y 2 r for each cc. From y - r,, E E+ for each or and y - r,, - y - x. it follows that y -- r E F,'+ or y > Y. Thus, x = sup{Tq}.

Problem 1.2.13. Show that if xn - x holds in a Banach lattice, then there exist a subsequence {yn} of {.r0 and some u-> 0 satisfying Iyn - xj < iu for each it.

Solution: Assume that x -.r holds in a Banach lattice E. Pick a subsequence »z^ for each n. Then Iln ly,, - xl ll < 2 for each (y, ,j of Ix, ,j satisfying Ilyn it and since E is a Banach space. it follows that the series En n nlyn -.rI is norm

convergent in E. Let it = °_ n ly - xl E E. Now a glance at the preceding Problem 1.2.12 guarantees that nlyn - rI < it, that is, !y -TI < l u for each n. I

Problem 1.2.14. Show that every order bounded operator from a Banach lattice to a nonmed Riesz space is continuous. Solution: Let T: F - F be an order bounded operator from a lianac:h lattice to a normed Ri(-z space. Assume 1)v way of contradiction that T is not continuous. So. there exists a sequence {.rn } in E satisfying rn 0 in E and T.rn yc 0 in and some c > 0 such F. This means that there exist a subsequence of

that IITynII > c for each n (and. of course. y - 0 in F). Now. by the preceding and sonic it E E'' satisfying Problem 1.2.13. there exist a subsequence {:n} of Iznl < lu for each it. Since T is order bounded, there exists some w E F such that it. it follows that T([-u. it]) C [-w. u'). From IT(nzn)) < w or for each it. But then we have 0 < e < IIT--,.1+ <

11uc'l

--a 0 .

which is impossible. Hence. T is a continuous operator.

1. Odds and Ends

28

Problem 1.2.15. Show that for a given Riesz space E there is, up to an equivalence. at most one lattice norm which makes E a Banach lattice. Solution: Assume that a Riesz space E is a Banach lattice under two lattice 11, and II -112. Then the identity operator I: (E. 11 111) - (E, 11. 112) is positive and hence (by the preceding Problem 1.2.14) it must be a homeomorphism. norms 11

Problem 1.2.16. Show that a normed Riesz space E is a Banach lattice if and only if every increasing norm Cauchy sequence of E+ is norm convergent.

Solution: The `only if- part is trivial. To establish the "if" part let {zn } be a norm Cauchy sequence in E. It suffices to establish that {x.) has a convergent subsequence. Pick a subsequence {u} of { xn } satisfying 11 un+ l - un 11 < zi, for each n, and then consider the sequences {yn} and {zn} in E+ defined by n

n

yn =

Du1+1 - us)+

and

Z. =

E(ui+1 - uX

t=1

i=1

Clearly, OS yn T and from n+p

,

Ilyn+p - ynii

=

I

,

Lr

n+p

II(ui+1 - ui)+II

(u++l - ut)+1I

t=n+1

i=n+1

n+p

n+p

iiui+1 - aril <_

F, 2.

< 2^

we see that {yn) is a norm Cauchy sequence. Similarly, the increasing sequence { zn }

in E+ is norm Cauchy. By our hypothesis both sequences are norm convergent. Let yn -+ y and zn z. Next, notice that the lattice identity x = r+ - r- implies yn - Zn = un+1 - u1 for each n. In particular. it follows that un = yn-1 - zn-1 + u1 -+ y - z + ul. This guarantees that E is norm complete. I

Problem 1.2.17. Show that if E and F are two Banach lattices with F Dedekind complete, then the Dedekind complete Riesz space G,(E, F) of all regular operators from E to F equipped with the r-norm is a Banach lattice. Solution: The verification of the fact that 11.11, is a norm on Gr(E, F) is straightforward. Next. we shall verify that 11 11 is a lattice norm. Notice first that if T is a positive operator. then IIT1I, = IIITIII = 11T11. Moreover. if two operators S. T E L,(E, F) satisfy 0 < S < T and IIrhi = 111=111 < 1, then the inequalities Sri 5 Sixi <- Tint imply i1SXil = IIISniII < I1SIZi11 <_ IITixhII :5 11Ti1

So, IISII = sup,,=u<1 IlSrll 5 HT11. In particular, if S.T E C,(E, F) satisfy 1S1 S ITi.

then I1.11,= IIISIII <111T1I1=11T11,

1.2. Banach Lattices and Positive Operators

29

This proves that II - IIr is a lattice norm on Cr(E, F).

To establish that II II, is a complete norm, it suffices (according to Problem 1.2.16) to show that every increasing II . IIr-Cauchy sequence of positive operators is IIr-convergent in C,(E,F). To this end, let be an increasing 11,-Cauchy sequence of positive operators. From JJT,, - TmII : IITn T"'11'-, it II follows that {T,,} is a II - 11-norm Cauchy sequence. So, by Problem 1.1.4, there exists an operator T E C(E. F) satisfying IITn - TII 0. For each x E E+ we have 0 < Tnx T and IITnx - TxII - 0, and hence ProbII

-

lem 1.2.12(c) guarantees that Tnx T Tx for each x E E+. In particular, T is a positive operator (and so T E Cr(E, F)) and 0 < Tn < T holds for each n. This implies

IITn - Tlir = IIJTT - TIII = IIT - T,II - 0, proving that f-,, (E, F) under the r-norm is a Banach lattice.

Problem 1.2.18. Give an example of a regular operator T : E -' F between two Banach lattices with F Dedekind complete satisfying IITII < IiTIIr

Solution: Let E = F = R2 be equipped with its Euclidean norm ii(x,y)li = i defines an order bounded operator on E (x2 + y2) . The matrix A = 1 is

whose modulus is given by the matrix JAI = [1

1J

.

(For the computation of the

modulus of a matrix see also Problem 8.1.11.) Now let us compute the norms of these two matrices. To this end, fix some vector (x, y) E R2 with x2 + y2 = 1. Then we have 11A[xJ112=11[1

[]2

iJ

=

[1112 =

(x_y)2+(x+y)2 =

2(x2+y2)

= 2.

11

This implies JJAII = 52 For the modulus JAI of A note that: x II

IAI [Y]

it [11J [1112 -11 [x +

112

y] 112

= 2(x + y)2 = 2(1 + 2xy).

Now an easy computation shows that IIJAIII =

2(1 + 1) = 2. In particular, we

have

IIAII= So, A is a desired example.

<2=IIIAIII=IIAIir.

Problem 1.2.19. Show that a band projection P on a normed Riesz space is a contraction. Solution: Let P: E - E be a band projection on a normed Riesz space. That is, the range B = P(E) of P is a projection band. Hence, each x E E can be written in the form x = y + z with y E B and z E Bd. From y 1 z, it follows that I Pxi

= iyi s Iyi + IZI = Iy + Zi = Ixi,

1. Odds and Ends

30

and so IlPxll = IIIPxIII < Illxlll = IIxII for all x E E. Hence, IIPII < 1, that is, P is a contraction. (If P 0 0, then of course IIPII = 1.) 1

Problem 1.2.20. If E is a normed Riesz space and x' E E', then show that llx`II=IIx'llr=IIlx'III. Solution: From the inequalities Ixe(x)I <- Ix*I(IxI) <- II Ix'I II - IIxII,

we infer that Ilx'll < II Ix'I II. On the other hand, if IIxII <- 1 and e > 0 are fixed, then from Ix'I(Ixl) = sup{Ix'(y)I: lyl <- IxI }, we see that there exists some y E E with lyl < IxI such that Ix'l(Ixl) - e < lx'(y)l < IIx'II. Therefore, II Ix *I II = sup{Ix'1(IxI): IIxII <- 11:5 lix'll + e

for each e > 0. This implies II Ix' I II <- IIx' II and so II Ix' I II = Ilx' Il .

Problem 1.2.21. A Dedekind complete Riesz space t is said to be a Dedekind completion of a Riesz space E if there exists a lattice isomorphism

it : E - E such that i = sup{ir(v): v E E and rr(v) < i} = inf{7r(w): w E E and i < rr(w)}

for each i E E. We can identify E with 7r(E) so that E can be viewed as a Riesz subspace of E. It should be clear that only Archimedean Riesz spaces can have Dedekind completions. As a matter of fact, every Archimedean Riesz space E has a unique (up to a lattice isomorphism) Dedekind completion E6.

Now assume that E is a Banach lattice (so E is an Archimedean Riesz space) and let E6 be its Dedekind completion. On E6 define the norm

IIIiIII=inf{IIxII: xEE+ and111:5 xj. Show that Illxlll = IIxII for each x E E and that (E6, III .III) is a Banach lattice. Solution: The verification of the norm properties by III . III (including its lattice property) is straightforward and is omitted. Also, the equality Illxlll = IIxII for each x E E should be obvious. For simplicity, we shall denote I I I .III by II lI again. To check the norm completeness of E6, we use Problem 1.2.16. Assume that

a sequence {in } in E6 satisfies 0 < in T and IIin+1 -1.11 < sL for all n. For each n pick some yn E E+ such that 0 <- in+1 - in <- yn and Ilyn II < sL . If we let xn = E 1 yi, then {xn} is a norm Cauchy sequence in E and so {xn} is norm convergent in E, say to x = °_1 yn E E. From Problem 1.2.12(c), we know that xn T x in E. In particular, if we let zn = yi E E, then Ilznll -. 0. Letting io = 0 and choosing some yo E E such that yo > 1 i we see that n

n

in = F'(xi - it-0 < F, yi-1 = xn-1 + yo i=1

i=1

1.3. Bases in Banach Spaces

31

This implies in < x + yo for each n. Since E6 is Dedekind complete, there exists

some i E E6 such that i, j i. Next, note that for each n and p we have n+p-1

n+p-1

x

O < xn+p - in = E (i,+1 - xi) < E yi <: Yi = Z. . i=n

s=n

i=n

where the last inequality holds true by virtue of Problem 1.2.12(c). So, by letting p oo, we get 0 < i - in < zn for each n. This implies Il i - in ll < Il zn ll, and so P. -ill -' 0. Consequently, (E6, III III) is a Banach lattice.

Problem 1.2.22 (Nakano [60]). Let E be a Riesz space and let f E E. The null ideal of the functional f is the ideal Nf = {x E E: If I(IxI) = 0). The band C1 = N1 is called the carrier of f. Assume that E is Archimedean and f, g E En . Show that If I A I9I = 0 if and only if C1 A C9 = 10). Proof. We can assume 0 < f, g E En-. Suppose first that f n g = 0. Fix a vector 0 < x E C f = NN and let E > 0. From If ng] (x) = inf if (y)+g(x-y) : y E [O,x] } = 0, we see that there is a sequence {yn} C [0. x] such that f (yn) + g(x - yn) < 2-nE

for each n. If xn = At=, y, then note that xn j 0. Indeed, if 0 < z < xn for each n, then 0 < f(z) < f(xn) < f(yn) < 2-nE, and hence f(z) = 0. This shows that

zENfflCf={0} orz=0.

From the inequality x - xn = x - n I yi = V, 1(x - y,) < E1 1(x - yi), it follows that 0:5 g(x - xn) < En, g(x - y,) < E holds for each n. Since g is order continuous and x - xn j x, we get that g(x) = limn-o. 9(x - xn) < E for all e > 0. Thus, g(x) = 0. This implies Cf C N9, and consequently Cf f1 C. = {0}. For the converse, assume that Cf f1 C. = {0}. Since f is order continuous, it follows that Nf is a band. This implies C9 C CC = Njd = Nf. Consequently, if

05 x=y+zE N.99C.9, then 0:5 (f A g)(x) 5 (f A g)(y) + (f A g)(z) 5 g(y) + f(z) = 0.

That is, f A g = 0 on the order dense ideal N. ® C9. Since E is Archimedean, this guarantees that f A g = 0.

1.3. Bases in Banach Spaces Problem 1.3.1. Let X be a Banach space with a basis {xn}, and for each vector x E X let x = E 00 n=o anxn be its series representation with respect to the basis {xn}. Show that the function III . III: X - R, defined by n

IIIxIII =

sup i=1

Orix,ll

1. Odds and Ends

32

is a norm that is equivalent to the norm II II of X. Solution: We shall verify first that the function 111.111 is a norm. Clearly, II IxI ll > 0 for all z E X, and 1110111 = 0. Conversely, IIIxI1I = 0 is equivalent to 11E..1 a,xill = 0

for each n, which in turn (by the definition of the basis) is equivalent to a; = 0 for each i E N, that is, x = 0. For the homogeneity of 111 III note that if A is a scalar, then n

n

n

IIIAxIII = sup IIE Aa,xtll = sup IAI IIE a'x`II = Al Isup IIF a'x`lI = nEN nEN nEN i=1

i=1

IAI

Illxl II

i=1

For the triangle inequality note that if y = E I Onxn, then from n

n

n

IIE(a, + 3t)xill !5 IlEaixill + i=1

IlEoixill < lIIx111 + Illylll, i=1

i=1

it follows that

ll Ix + yI11 <- II Ixl1I + 111y111

The non-trivial part is to prove that 111-111 is a complete norm, i.e., that (X, II I -III) is a Banach space. To this end, let {zn} be a III - III-Cauchy sequence in X, where

zn = EL=I ai x,. From 11lzn - zmI11 = supkllEk I(ai -a;t)x,11, we see that n_ Q3

IQ,

I

IIE(Q; - Q; `)x, - E(Q; - a; `)sill

IIx,II

i=1

t=1

j

am)x`II + IIE(Qt t=1

t=1

<

- Qm)xill

2IIIZn-xm111

This implies that the sequence {Q }nEN is a Cauchy sequence of scalars for each j and so ar = limn-. a3 exists. To complete the proof of this part, we shall show that the series z = E;° 1 aixi is norm convergent in X and that Illzn - z111 -t 0. To this end, fix e > 0, and then pick a natural number no such that r

k

IIDa; i=I

- am)xill < sup IIE(Q: - a'n)x,II = Illzn - Zmlll < e rEN

(*)

i=1

for all n, m > no and all k E N. Letting n -' oo, we get 11E 1(a, - a; t)xi 115 e for all m > no and each k E N. To verify the norm convergence of the series I atxt, 1 a,x,. Then for m > n > no we have let or,, = m

Ilam - CnIl

m (a`-a'O)x'II+II

II

i=n+1

i=n+1 n

E aS0xtll

t=n+1 m

M i=1

t=t

M

< 2e+II 1 c °xill. i=n+ 1

t=n+1

1.3. Bases in Banach Spaces

33

Now use the fact that the series Es_1 a °xi is norm convergent in X to select some ko > no such that IIEmn+1 a" xill < e holds for all m > ko. So, for all m > n > ko we have Ilam - an II 5 3e. This shows that {an } is a II ' II-Cauchy sequence, and

hence z = limn-,, an aixi exists in X. Next, notice that by letting m - oo in (*), we get llF,k 1(ai - ai)xill < e for all k E N and all n > no. Therefore, for each n > no we have k

IIIzn-zlll=sui11D

-a')xiII<e,

i

i=1

so that IIIzn - zIII - 0. IIIxIII holds for each x E X, it follows from the Open Mapping Since IIxII Theorem that II ' II and III ' III are equivalent norms. So, there exists some constant c > 0 such that cilIxIII < IIxII S IIIxIII for each x E X. I

Problem 1.3.2. Show that a sequence {xn} of non-zero vectors in a Banach space X is a basis if and only if it satisfies the following properties:

(a) The linear span of the set {x1, x2, ... } is dense in X.

(b) There exists some K > 0 such that if m > n and Al i ... , An, are arbitrary scalars, then n

I1E,\tx=II i=1

m
Eatx,I1 i=1

Solution: Assume first that {xn} is a basis. Then (a) is obvious. For (b) note that if Pn : X - X denotes the natural projection defined by Pn (E'1 aixi) = E 1 aixi, then according to Theorem 1.37(b) there exists some constant K > 0 such that I I Pn l l < K holds for each n. Therefore, if m > n and \1 .... , A,n are arbitrary scalars, then n i=1

m

Aixi) II < IIPnII . Pn (E =1

m

m

KII

F,a,x, i=1

F aixi II i=1

holds.

For the converse, assume that conditions (a) and (b) are satisfied. Notice that if x = Ei=1 aixi = E°_° 1 J3ix then from (b) it follows that for each n we have n

m

II E(ai - aixi 11 < KII E(ai - Q,)xi II i=1

S=1

m

m

E six, i=1

3,-Ti 11

0

i=1

That is, E 1 aixi = E 1 Oixi for each n > 1. This easily implies ai = /3i for each i. Therefore, if x = Ei=1 aixi, then the sequence {an} is uniquely determined. Now let Y = {x E X : 3 a sequence {an} of scalars such that x = F, anxn } n=1

Clearly, Y is a vector subspace of X. To complete the solution, we need to show that Y is closed. If this is done, then by (a) we have Y = X and this will establish that {xn } is a basis of X.

1. Odds and Ends

34

To see that Y is closed, assume that a sequence {yn} C Y satisfies yn -+ y in

X. Let y,, =F_xla;xiand fix e>0. Fix some k E N such that Ilyn - yll < efor each n > k, and so IIyn - y,n II < 2e for all n, m > k. Rom IIy - yk II < E, we see that there exists some ro > k such that for all r > ro we have r (k) II v ax II i=1

Since for any t > r we have letting -4 oc we get

II

Er1(ai - am)xi II 5 KII r1(as` - a; `)x1 II, by

r

II Dc'; - am)xi II

KIIy,, - y.11

i=1

for all n, m, and r. Therefore, if j > 1 is fixed and m > n > k, then

I aj -

amI . IIx)II

= II u(ai - am)xi -

II

<

i=1

E(a= - am)Xi II

a - a;')x, 11 +

Il

I(a' - am)x, I!

i=1

2Klly, -y,nll <4Ke.

This guarantees that for each j (including j = 1) the sequence of real numbers {a!)

is a Cauchy sequence. Let ajI ,r - aj for each j. Letting n = k and m - co in (**), we get II Ei=1(ak - ai)xill < Ke for each r. Therefore, it follows from (*) that for r > ro we have r 1k' -

II

i=1

IIY

r

r

- F,

akxtII+I1E((Yk-oi)x,II

i=1

<e+WKe=(1+4K)e.

i=1

This shows that limr_" Er=1 a,x; = y or y = Ej_1 aix;, as desired. Problem 1.3.3. This problem describes a basis for the Banach spaces Lp[0,1] defined by with 1 < p < oe. Consider the sequence of functions hi (t) = 1 for each t E [0,1] and 1

h2k+c(t) _

-i 0

if


otherwise,

for k = 0, 1, 2.... and 1 < £ < 2k. The sequence {hn} is known as the Haar system. Establish the following. (a) hn # 0 for each n.

(b) The linear span of the set f{ = {h1, h2:...} is norm dense in LP[0,1].

35

1.3. Bases in Banach Spaces

(c) For any collection of scalars Al, ... , An, An+1 we have n+1

n

IlEathtllp.

llp

I

(d) The Haar system {hn} is a monotone basis in each Lp[O,1]. Solution: (a) This is obvious. The graphs of the three functions h2, h3, and h4 are shown in Figure 1.

(b) Let Y denote the linear span of the set of functions in the Haar system 7j = {h1, h2, ...}. We claim that the characteristic functions of the intervals of the form

[ t,

2

where 0 < e < 2k - 1 and k = 1, 2, ..., belong to Y. To see this, let [a,b] = [Z , ] for some 0 < e < 2k - 1. Also, let [c,d] _

[] if e is even, and let [c, d] _ [1, z ] if a is odd. We consider the first case; the second case can be established similarly. Define the function fo = X(a,b[ - X(c,dl A look at the definition of the Haar system reveals that fo E 7{. 1 14

1 The graph of h3

The graph of h2

The graph of h4

0-o

9

T 2

:2

-1i

:-0

-1}

:--o Figure 1

If cl is the midpoint of the interval [c,d], then 91 = X[c,c1) - Xlcl,dl E x. In turn, this implies that the function f1 = f0 + 91 = X(n.bl - X(e,.dl

belongs to Y. Now if c2 is the midpoint of the interval [cl,d], then the function 92 = XIc,.c,) - x(-,..j belongs to 11, and so the function f2 = fl + 92 = Xla,bl - 2Xlc2,dl

belongs to Y. Continuing in an inductive manner, we can construct a sequence {fn) in Y and a sequence {gn } n 71 such that

(t)

fn+1 = In - 9n+1 = Xla,bl - (n + 1)X(cit,id)

holds for each n. It is easy to see now that

f

Ilfn+1 - X[a,bJJI= [

0

1

1 (n+1)"X(c_+1,d1(t)dt]"

=2

z

H-00

0.

1. Odds and Ends

36

This implies fn - X(a,bj in Lp[0,1] and so X[a,bl E V. Therefore, the characteristic function of any interval of the form [2 0 < P < m < 2k, belongs to Y. Next, let [u, v] be any subinterval of [0,1] and let e > 0. Fix some natural number k such that < e, and then pick two integers 0 < e < m < 2k - 1 such that 2 < u < < b < .This yields fo lXl,, l(t)Ipdt < 2e, and so Xiu,,,l E V. The latter easily implies that Y = Lp[0,1]. (c) Assume 1 < p < oo. The function 9(t) = t", t > 0, is increasing and convex.

(This follows from 9'(t) = ptp-1 > 0 and 9"(t) = p(p - 1)tp-2 > 0 for all t > 0.) This implies that 0(t) = Itl , t E R, is also a convex function. For if x, y E R and a, 0 E [0,1] satisfy a + Q = 1, then

1/i(ax+fly)=Iax+Aylp5(alxl+QIyl)p
2IxIp
z

z

(*)

for all x,yER. Now let al, ... , An, Xn+1 be arbitrary scalars and consider the two functions f= 1 A1h; and g = E' a;h1 = f + n+1hn+1 Also, let [a, b] be the supportinterval of hn+1 Since the functions hl, . . . , hn are all constant on [a, b), it follows that there exists some scalar c such that f (t) = c for all t E [a, b]. This implies that the functions f and g agree outside [a, b], f = c on [a, b], and g takes the value

c + An+1 on the interval [a,2 °) and the value c - An+1 on [,2 b]. Now a glance at (*) yields b

I f (t) I p dt

=

Iclp(b - a) = 2lclp bb2a

<

IC + An+1lpb2a + IC - A n+1lpb fa = rb I9(t)Ip dt .

Ja

Consequently, f' I f (t)Ip dt < fo Ig(t)lP dt or o

n

n+1

IIEA*htIIp < IIEAihIlp. ,=1 ,=1 This shows that {hn} is a monotone basis of Lp[0,1].

(d) The properties (a), (b), and (c) guarantee (according to Problem 1.3.2) that {hn} is a basis in Lp[0,1] for each 1 < p < oo. Remark. It is considerably more difficult to show that the Haar system is also an unconditional basis in each Lp[0,1] with 1 < p < oo; for details see [72, p. 407]. In the Banach space L1[0,1] the Haar system fails to be an unconditional basis. See Corollary 11.61 in the text where we establish that L1 [0, 1] does not have any unconditional basis. Moreover, A. Pelczynski [63] has established the following general result. The Banach space L1 [0,1] is not embeddable in any Banach space with an unconditional basis.

37

1.3. Bases in Banac6 Spaces

For an alternative proof of the above result see [52, Proposition 1.d.1].

Problem 1.3.4. This problem describes a basis in the Banach space C[O, 1]. be the Haar system that was introduced in Problem 1.3.3. Consider Let in C[0,1] defined by the sequence of functions /'t

and s (t) = J hn_1(rr)dr for n = 2,3,4,...

sl = 1

.

0

is called the Schauder system. Show that The sequence monotone basis for C[O,1].

is a

Solution: Clearly, each sn is a continuous non-zero function. The graphs of the functions Si. s2, s3, 84, and s5 are shown in Figures 2 and 3. We need to verify that the sequence {sn} satisfies properties (a) and (b) of Problem 1.3.2. s2(t) = t 1

.

1

40

1

Figure 2

Figure 3 (a) Let Z denote the linear span of IS I, s2, ...}. We must show that the uniform closure Z of Z coincides with CIO, 11. According to the Hahn-Banach theorem, it suffices to show that every finite Borel signed measure on [0,1] that vanishes on Z also vanishes on CIO, 11. To this end, fix a finite Borel signed measure p on [0,1] that vanishes on Z. That is, assume that fo z(r) dp(r) = 0 for each z E Z. Before proving that p = 0, we need to establish a few facts. Consider a closed interval [a, b] of the form [ 2 , - ] , where 0 < t < 2k -1. From property (t) in the solution of part (b) of Problem 1.3.3, it follows that there exists a sequence { fn}

1. Odds and Ends

38

in the linear span of then Haar system satisfying

J0 tX(a.bl(r)dT-

jfi+i(r)drf <

(tt)

for each t E [0, 11. Since the function 9n(t) = fo f. (r) dr belongs to Z, it follows from (tt) that the function 9(t) = fo X(a,bl(T) dr belongs to Z. The latter implies that if [a, b] is any closed subinterval of [0, 1] with "dyadic" endpoints (i.e., a and b are of the form a = Jr and b = ), then the function fi(t) = fo X(a.bl (T) dr also belongs to Z. The graph of m is shown in Figure 4(a). In turn, the last conclusion implies that if 0 < a < b < c < d < 1 are dyadic points, then the function iP E C(0,1) defined by =

j

X(ab](T)dr- J X(.,.l (T)dr

also belongs to Z. Its graph is shown in Figure 4(b).

1

Figure 4 We are now ready to establish that µ = 0. Let (a, b) be an arbitrary open subinterval of [0, 1). Pick four sequences {an}, {bn}, {cn}, and (4) of dyadic

points such that a < an < b < c < do < b for each n, bn j a, and Cn j b, and consider the sequence of functions {t(, } C Z defined by

=

blaf

t X(a.b1(T) dClearly,

jXLc,d..l(r)dr.

iI',(t) X(a,b)(t) for each t E 10, 1]. So, by the Lebesgue Dominated Convergence Theorem, we see that i i µ((a b)) = X(a,b)(r) dt(r) = liM f 0. (r) dµ(r) = 0. Jo

If 9n(t) = 1 - fo X(o l(T)dr, then {9,} C Z and 9n(t) X{o}(t) for each t E [0, 11. As above, this implies p({O}) = limn fo 9n(r) d i(r) = 0. Similarly, µ({1}) = 0.

1.3. Bases in Banach Spaces

39

The above conclusions easily imply that p(J) = 0 for each open subset of [0111.

Since p (as a finite Borel signed measure on [0. 1]) is regular. we conclude that p(B) = 0 for each Borel subset B of [0.1]. That is. M = 0. as desired. (b) Assume that .\1, A2.... A,,. 1 .,is, and g functions f =

are arbitrary scalars, and consider the two

As, = f + An+lsn+1 It suffices to show

that n+1

n

IIEA'4'IIIlfIIx t t

IIgIIx.=I1AisilL.

Let [a, b] be the support-interval of s,,+1. That is. si+1(t) = 0 holds for all [a. b] and sn+1(t) > 0 for each a < t < b. This implies f (t) = g(t) for all [a, b], and/ so in order to establish that 11f IIx <- IIgII«., it is enough to show that

maxtE]a.b] If(t)1 < maxtEla.b] 19(01-

To this end. notice that for each 1 < i < n the function s; is linear over the interval [a. b]. This implies that there exist two constants m and 3 such that f (t) = net+13 for all t E [a, b].Now note that max:E(a.b] If (t)I = max{I f (a) I. If(b)I}. i.e., the maximum of the function If I over [a, b] is achieved at the endpoints of the

interval [a. b]. Since f (a) _ q(a) and f (b) = g(b), we infer that tmax]

If(t)I = coax{If(a)I.If(b) I} = max{Ig(a)I Ig(b)I} < tmax Ig(t)I

Thus, If IIx < IIyII,., and the solution is finished.

Problem 1.3.5 (Bessaga -Pelczynski (18]). Let be a sequence in an infinite dimensional Banach space satisfying x, t!'-. 0 and lim inf [jx ]j > 0. Show that

has a subsequence that is a basic sequence.

Solution: Assume that a sequence {xn } in an infinite dimensional Banach space X satisfies lim inf Ilxn II > 0 and x -w- 0. By passing to a subsequence, we can assume that IIxnII > c > 0 holds for each n and some constant c > 0. Obviously, we can suppose without loss of generality -X4 0. So. replacing xn by that IIrnII = 1 for each n. We claim that the following property is true: II

(P) If Y is a finite dimensional vector subspace of X and E > 0, then for each k there exists some n > k such that IIyII < (1 +e)Iiy+axnll for ally E Y and all scalars a.

To see this, assume that Y is finite dimensional, let f > 0, and fix k. ALSO. pick some 0 < 6 < a such that 26 < F. Since the closed unit ball of Y is norm compact. there exist unit vectors yl.... , ym in Y such that for each y E Y with IIyII = 1 there exists some i with Ily - yill < 2. For each 1 < i < rn choose some x; E X' satisfying 11 x! II = 1 and x; (y,) = 1. From .r,, -W- 0, it follows that there exists sonic n > k satisfying Ir; (xn) I < y for i = 1..... rn. Now let a be an arbitrary scalar and fix some y E Y with IIyII = 1. If IQI > 2. then we have

?I°I-IIull=ICI-i> 1.

1. Odds and Ends

40

On the other hand, if Ial < 2, then choose some yi E Y with Ily - yi II < that lax; (xn)I < 2 implies

>;(yi+axn) -

Ilyi+axnp -

Ily+axnll >

and note

l+ax;(xn)-i>t-Z-z=1-8.

This implies

1 < 1-'Illy+axnll <_ (1+26)fly+axn11 <_ (1+c)lly+axnli in this case too. Thus, we have established the validity of (P) for each unit vector y. The general case for an arbitrary y follows from the homogeneity. Fix some real number e > 0 and then choose a sequence {en} of positive real

numbers such that K = fln 1(1 + en) < 1 + e. Using property (P) and an easy inductive argument, we see that there exists a subsequence {zn} of {xn} satisfying IIzII <- (1+1En)flz+axn+11I

for all scalars a and all z in the vector subspace generated by {z1, . . . , zn}. Now note that if m > n, then for any choice of scalars a1, ... , an we have n

m

a`z'II. aiZi II

II

I

By Theorem 1.41, the subsequence {z} of {xn } is a basic sequence in X.

Problem 1.3.6. Assume that X is a reflexive Banach space with a basis {xn}. Show that its biorthogonal sequence {cn} is a basis in X'. Solution: We already know that {cn } is a basic sequence in X*; see the discussion at end of Section 1.3. What needs verification is the fact that the closed linear span

Yof{c,} is all of X°. If Y 96 X', then there exists (by virtue of the reflexivity of X) some non-zero x = En '=i anxn E X such that x(cn) = cn(x) = an = 0 holds for each n. But then x= anx,, = 0, a contradiction. This contradiction establishes that Y = X', and so the biorthogonal sequence {c*.} is a basis in X. it

Problem 1.3.7. Let E'_1 xn be a series in a Banach space X. (a) If X is finite dimensional, show that E', x,, converges unconditionally if and only if E', Ilxnll < oo. (b) Give an example of an unconditionally convergent series such that E n'=1 Ilxnll = cc.

=1 xn

r1O0

Solution: (a) We can assume that X = RA: for some k E N. Let xn = (xi , ... , xk ) and note that for each choice s = (31, 82, ...) of signs we have 0000

L. S nxn n=1

00

00

s n xj , n=1

00

F, s x2 , ... E s x n

n=1

n

,

n=1

k

This implies that the series En 1 snxn is norm convergent in X if and only if the series Ecc s,,x;' converges in R for each 1 < i:5 k. Now notice that the latter is equivalent to saying that El lx; I < oo holds for each 15 i < k. 1

1.3. Bases in Banach Spaces

41

If we consider X equipped with the L1-norm IIxIII = to see that

Ek1

Ixi1, then it is easy 00

II

n=1

xn II =

I

i=1 n=1

xi

I

- i=1 n=1 I xi I = n=1 i=1 I xi I = n=1

IIxn II 1

This inequality in connection with the preceding discussion shows that the series F,n 1 x; is unconditionally convergent if and only if F,°n°1= Ilxnll < 00-

(b) Let X = e2 and let en denote the standard unit coordinate vector. If xn = nen, then IIxn11 = 1,, and so E= 1 Ilxnll = E' 1 n = 00. On the other hand, it is easy to see that the series Fn 1 xn converges unconditionally in £2. 1

Problem 1.3.8. Show that for a series F,n l xn in a Banach space the following statements are equivalent. (1) The series E,°,01 xn converges unconditionally. (2) For any sequence {sn} of signs (i.e., Sn = ±1 for each n) the series M°O_1 Snxn is norm convergent.

(3) For every strictly increasing sequence {kn} of natural numbers the series En=1 xkn is norm convergent. (4) For each e > 0 there exists a natural number k such that for each finite subset A of N with min A > k we have IIEnEA xnII < C.

Solution: (1)

(2) Let {sn} be a sequence of signs, and assume by way of con-

tradiction that the series F,0 1 Snxn is not norm convergent. This means that there exist some e > 0 and a sequence {kn} of strictly increasing natural numbers such sixill > 2E holds for each n E N. We shall obtain a contradiction that by demonstrating the existence of a permutation 7r of N and two sequences {mn} and {rn} of natural numbers satisfying kn < Mn < r, and IIEi-m., xx(i)II > E for each n.

Let In = {k E N: kn < i < kn+1}. Now fix n E N. If si = l for each i E I., let 1rn(i) = i. Similarly, if sn = -I for each i E I,,, we let irn(i) = i. In these cases, we let ran = kn + 1 and rn = kn+1. We also let Io = { 1, 2, ... , kl } and iro(i) = i for each i E lo. Now assume that for some i E In we have si = 1 and for some j E In we have sj = -1. In this case, there exists some kn + 15 pn < kn+1 and some permutation 1 if i < Pn and s,,,,(i) = -1 if i > p, . Let sn = 1 for all an of In such that kn + 1 < i < pn and s,, = -1 for all pn < i < kn+1 In view of k,.+i

P.

r II E sixiII = IIu

X..(1) - E xn,,(i)II

2E,

it follows that either 104,+1 x*n(i)II > E or IIEk=P'+1 xxnli)II > . In the first case, we let ran = kn + 1 and rn = p,,, and in the second case ran = pn + 1 and rn = kn+1 . Now if it is the permutation of N that coincides on each In with 1rn,

1. Odds and Ends

42

r

then we have II

u

xn(t) II >

for each n. This shows that the series J:n 1

is not norm convergent, which

contradicts M.

(2) = (3) Letting s = 1 for each n, we get that E', x,a is norm convergent. Now assume that {k,,} is a strictly increasing sequence of natural numbers. Put tk.. = 1 for each n and ti = -1 if i # kn for each n, and note that

M y

[L Xn + n=1

n=1

This implies that

00

00

Xk

E tnZn n=1

En 1 xk., is norm convergent.

(4) If (4) is not true, then there exist some f > 0 and a sequence {An} (3) of non-empty subsets of N such that for each n we have max An < min An+l

and

it E x, II > C. IEAn

Let A = Jn An. Clearly. this is a subset of N, and so A = {kl < k2 <

}. It is easy to see that the sequence of partial sums of the series En 1 xk., is not a norm Cauchy sequence. This implies that the series Ene xk., is not norm convergent, a contradiction. 1

: (1) Assume by way of contradiction that the series En 1 xn is not (4) unconditionally convergent. This means that there exists a permutation 7r of N such that the series n 1 x,,(n) is not norm convergent. So, there exist some e > 0 and a strictly increasing sequence {kn} of natural numbers such that x*(i) 11 > E for each n. liEi=kn+1

Now fix k E N. Since it is a permutation of N. there exists some r such that {1,2.....k} C {7r(1).ir(2)...., rr(kr)}. So, if A = {rr(i): kr < i < kr+1}, then k

and

II E IEA

II = II E xx(i)

I)

> 'F.

i=kr+1

Thus, for each k E N there exists a non-empty finite subset A of N with min A > k e. This contradicts (4), and so (4) implies (1). 1 and IIFiEA

Problem 1.3.9. If a series E', xn in a Banach space converges unconditionally, then show that E' 1 xn = E° O_1 x,(,,) holds for each permutation

rrofN. Solution: Assume that a series En 1 x,, in a Banach space converges unconditionally, and let rr be a permutation of N. Put x = En 1 x,, and y = En l x,r(n), and fix > 0. By statement (4) of Problem 1.3.8 there exists some k such that 11 EIEA xi11 < E

for each finite subset A of N with min A > k. Next, choose some r > k such that lix - E xill < e, and then select some m > r such that: 1

1.3. Bases in Banach Spaces

43

(a) (1,2,...,r} C {7r(1),7r(2),....ir(m)}. (b) The finite subset A = {a(1), x(2), ... , ir(m)} \ {1, 2, ... , r} of N is nonempty.

(c) IIy - Ei

1 xn(i) II < E.

(The existence of such a number m follows from the fact that 7r is a permutation of N.) Clearly, min A > k. Now note that r

r

m

m

11X-Ex,II+IIEx,-Ex"m

IIx - Y11

i=1

< C+ 11y,

i=1

i=1

II+Il i=1

(')-yII

x'll+E <E+E+E=3E.

iEA

Since e > 0 is arbitrary, it follows that Jjx - vii = 0, that is, x = y, as desired.

I

Problem 1.3.10. Let FOO_1 xn be an unconditionally convergent series in a Banach space X. If (al, a2, ...) E e,,, then show that the series F1°O=1 anxn is norm convergent. Moreover, show that the mapping T : Q,,. -+ X, defined by

T (al, a2, ...)

anxn , n=1

is a bounded linear operator. Solution: We will consider the case of a real Banach space only. The complex case can be dealt with similarly. Let (al, a2, ...) E e . We can suppose that janj < 1 holds for each n. Fix e > 0. Since the series E°_1 xn is unconditionally convergent, statement 4 in Problem 1.3.8 implies immediately that there exists some m such that for any sequence {Sn} of signs and any non-empty finite subset F of N with min F > m we have II EnEF snxn II < E. Fix for a moment any such set F.

Pick x' E X' of norm one satisfying x' (EfEF anxn) = IIEfEF Next, consider the sequence of signs s = (81,827 ...) defined by sn = 1 if anx'(xn) > 0 and sn = -1 if anx'(xn) < 0. This implies anxnII.

IIE anxn nEF

11

=

x'(E nEF

snx'(xn)

anxn) 5 E Ianilx*(xn)I = nEF

nEF

x'(E snxn) < 11 F, snxnll < E. nEF

nEF

Therefore, the sequence of partial sums of the series F,' 1 anxn is a norm Cauchy sequence. Hence, the series En= 1 anxn is norm convergent. Now it is a routine matter to verify that the mapping T: t. X, defined by 00

T(a1, a2, ...) _ E anxn

,

n=1

is a linear operator. Finally, notice that IIE m+1 anxn II < c holds for all (al, a2, ...) E 1= satisfying Ian < 1 for each n. This implies that the linear operator carries the closed unit

44

1. Odds and Ends

ball of t , to a norm bounded subset of X, and so T : Qa, operator.

X is also a bounded

I

1.4. Ultrapowers of Banach Spaces Problem 1.4.1. For a filter.F of subsets of an infinite set A show that the following statements are equivalent.

(1) F is an ultrafilter.

(2) IfAUBE.F, then either AE.ForBE,F.

(3) IfAnB00 forallBE.F, then AE.F. Solution: (1)

(2) Assume that F is an ultrafilter and that A u B E.F. If A E F. then (2) is true. So, we can suppose that A g Y. Consider the collection Q = {C C A: A U C E .F }. A direct verification shows that Q is a filter, .F C Q, and that B E F. Since F is an ultrafilter, G = Y. Therefore, B E Q = F. (3) Assume that a subset A of A satisfies A n B # 0 for each B E .F (2) and suppose by way of contradiction that A g F. From A U A' = X E F and (2), it follows that A` E F. In particular, we have 0 = A n A` E F, a contradiction. Hence, A E.F. (3)

(1) Assume that a filter Q satisfies F C Q. Fix A E Q. If B E F. then

BE Q, and so AnB#Q for each BE F. Now (3) implies AEF,i.e.,QCT. Hence, Q = .F, and therefore .F is an ultrafilter.

I

Problem 1.4.2. If an ultrafilter U on a set A satisfies nAEU A 0 0, then show that there e x i s t s a unique b E A such that U= {ACA: b E A}. Solution: Let U be an ultrafilter on a set A such that nAEU A # 0. Fix some b E fAEU A and note that the set r = {A C A: d E A} is a filter of subsets of A satisfying U C -F. Since U is an ultrafilter, it follows that U =.F. If another point g E A satisfies U = {A C A: rt E A}, then {b}. {rl} E U. This implies 0 = {b} n {77} E U, which is impossible. Hence, there exists exactly one

oEAsuchthatU={A9A:6EA}.

a

Problem 1.4.3. Show that every filter coincides with the intersection of all ultrafilters that include it. Solution: Let ,F be a filter on an infinite set A and let A 1 F. It suffices to show that there exists an ultrafilter U such that F C U and A V U. To we this, start by observing that F n A° 0 0 holds for each F E F. Indeed, if for some F E F we have F n A° = 0. then from

F = F n X = (F n A) U (F n X) = F n A C A it follows that A E F, a contradiction. Hence, F n A` 0 0 for each F E.F.

1.4. Ultrapowers of Banach Spaces

45

Now consider the collection of sets

9={BCA: FnA`CB forsome FEF}. A direct verification shows that 9 is a filter satisfying F C 9 and A` E 9. If U is any ultrafilter containing 9, then F C_ U and A = (A`)` §E U, as desired.

Problem 1.4.4. If in a topological vector space xn I- x and yn u + y, then show that xn + yn -y+ x + y and Ax,, - -, Ax for each scalar A. Solution: Assume that xn IL, x and yn A+ y in a topological vector space. Let V be a neighborhood of zero, and then select another neighborhood W of zero with W + W C V. Now notice that

{nEN: xnEx+W}U{nEN: ynEx+W}C{nEN: xn+ynEx+y+V}. Since the sets {nEN : x, Ex + W } and {nEN : y nEx + W } belong to U, it follows that In E N : xn + yn E x + y + V } E U. This shows that xn + yn A' x + y. Next, suppose that .\ # 0. Clearly,

{nEN: AxnEAx+V}={nEN: xnEx+aV}EU. This shows that Ax, ld+ Ax.

Problem 1.4.5. Let Sl be a topological space, let A be a subset of St, and let {xn} be a sequence in 1 such that for some N E U we have xn E A for each n E N. Show that if xn . x, then x E A. Use this conclusion to show that the U-limits preserve inequalities in the following sense: If in a Banach lattice xn -y x, yn -U+ y, and for some N E U

we have xn>yn for each n E N, then x > y. Solution: Let A be a subset of a topological space [I and let a sequence (x,, I in Cl be such that for some N E U we have xn E A for each n E N and xn -4 x. If

V is a neighborhood of x, then the set J = {nEN: xn E V} belongs to U. In particular, we have N n J 96 0. Now if k E N n J, then xk E V n A. This shows that V n A 96 0 for each neighborhood V of x, i.e., x E A.

For the last part, notice that w > v in a Banach lattice E is equivalent to w - v E E+. Since E+ is norm closed, it follows from the preceding conclusion that

x - y = limn xn - limu yn = limu (xn - yn) E E+. This shows that x > Y.

1

Problem 1.4.6. For a non-zero bounded operator T : X --+ Y between two Banach spaces and a sequence {xn } in X establish the following.

0 in Y. (a) If xn -1- 0 in X, then Txn (b) If Txn - 0 in Y and limn xn 0 0 in X, then there exist some n > 0 and a subsequence {xk., } of {xn} satisfying Ilxkn II > q for each n and IITxk., II -+ 0.

Solution: (a) Assume that xn - U - 0 holds in X and let e > 0. Then the set A = In E N : Ilxn II < -} belongs to U. Clearly, A C In E N : IITxnII < E}, and therefore In E N: IITxnII < E} E U. This shows that Txn 'L+0 in Y.

1. Odds and Ends

46

(b) From limu xn # 0, it follows that there exists some r > 0 such that the set

A = {nEN: r?} does not belong to U. So, A` _ {nEN: IIxn1I > n} E U. Next, for each k the set Bk = (n E N : JjTxn I1 < 'I belongs to the free ultrafilter U. Put Ak = Bk ft A` E U; and so each Ak is an infinite set. Now if we select a sequence {mk} of strictly increasing natural numbers such that mk E Ak for each k, then IIx,nk 11 > n and IITxm,, II < k hold for each k, and the desired

I

conclusion follows.

Problem 1.4.7. Let Z be a normed space and let U be a free ultrafilter on N. I f, as usual, we let N u = { (zl, z2, ...) E e,o(Z) : U-4 then Nu is a closed vector subspace of Z, and so Zu = 4,(Z)/Nu is a normed space.

-

That is, in the definition of the ultrapowers we do not need Z to be complete.

Suppose that X is a normed space, Y is a subspace of X, and U is an ultrafilter on N. Also, let Nu = {(yl,y2.... ) E e,,,(Y): yn-LO} and xn-u 0}, and define J: Yu Xu by N u = {(xl,x2.... ) E 1, , .

J(y+Nu)=y+Nu for each y = (yi, y2, ...) E e,,,(Y). Show that J is a linear isometry-and so, under this isometry, YU can be considered as a vector subspace (which is norm closed if Y is also norm closed) of XU. Solution: We first check that J is a well-defined mapping. To this end, assume

that y, z E t (Y) satisfy y + Nu = z + Nu , i.e., y - z E Nu . This implies yn - zn -U- 0. Since e (Y) C e (X), it follows that y + .Nu = z + XU x. This shows that J is well defined. Clearly, J is a linear mapping, and it follows from Lemma 1.62 that for each y = (yi, y2, . . .) E

we have

IIJ(y+Nt)II =Ily+Nu II - hu 11ynu=Ily+Null.

I

Therefore, J is a linear isometry. as claimed.

Problem 1.4.8. For a bounded operator T : X

Y between two Banach

lattices the following statements are equivalent:

(1) T is a positive operator if and only if TU is a positive operator. (2) T is a lattice homomorphism if and only if Tu is a lattice homomorphism.

(3) If T is interval preserving, then so is Tu. Solution: (1) If Tu is positive, then T is clearly a positive operator. Now assume

that T is positive, and let [xJ > 0 in Xu. There exists some y E t(X) such that y E [x]. The positivity of T implies Tyn > 0 for each n. Therefore,

Tu[xJ =Tu[y] = [(Tyl,Ty2.... )] E Xu . This shows that Tu is a positive operator.

1.4. Ultrapowers of Banach Spaces

-Ir1

(2) If Tu is a lattice homomorphism. then T is clearly a lattice homomorphism. For the converse suppose that the operator T is a lattice homomorphism. Then for each x. U E f,,(X) we have Tu([r] V [y])

= 7u([x V yi) = =

T1

Y1

[(T(x V yl).T(x2 V y2)....)] = [(Txl V Tyl.Tx2 v Ty2....)] v (Ty1.Ty2... )] [(Tx1.Tx2.... )] V [(Ty1.Ty2....

= Tu([x]) V Tu([y]) This shows that 7j4 is a lattice

(3) Assume that the operator T is interval preserving and let x E !X (X) and y E t' (Y) satisfy 0 <_ [y] = [(y y2....)] < 7U[x] =

[(T.rt.Tx2... -)]

in 1 j4. Without loss of generality, we can suppose that 0 < y _< Tx,, holds for each n. Since the operator T is interval preserving. for each it there exists some

0 < zn < r such that T; = y,,. To complete the solution notice that the vector z = (z1. z2....) E f,. (X) satisfies () < [z] < [x] and 14[z] = [y]. This shows that TI, is interval preserving.

Problem 1.4.9. For a bounded operator T : X , Y between two Banach spaces establish the following properties.

(a) If T is surjecttve. then TI, is surjective. (b) If T has a closed range and Tu is surjective. then T is surjectite. (c) The operator T is invertible if and only ifTu is invertible -in which case we have (Tu)-' = (T-1)u Solution: (a) Assume first that T is surjective and let y = (yl.y2.... ) E By Corollary 2.15 there exists some constant c > 0 such that for each it there exists some vector x E X satisfying cjIyn11 and Txn = yn. Clearly, r = (rl.x2....) E 1, (X) and T4 [r] = [y]. Therefore. Tu is surjective. (b) Suppose that T has a closed range and that the operator Tu is surjective. Fix some y E Y and let y = (y. y. y....) E £ (Y). Since Trr is surjective, there exists some r E f,,;(X) such that T., [r] = [y]. That is. there exists some z = (zI. z2....) in VU such that Tx,, = y+a holds for each n. Frotn limo z,, = 0 (and passing to a subsequence if necessary), we can assume that zn 0 in Y. This implies T.c y.

Thus, y belongs to the closure of the range of T. Since T has a closed range. it follows that y E R(T). Therefore. R(T) = Y and so T is surjective. (c) Assume first that T is invertible. That is. there exists a bounded operator X satisfying T-17'= Ix and TT'"1 = l y. Consequently. (T-1)uTu = (Ix)u = Ix,, and Ti4(T^')u = Ili .

T-1: Y

This shows that T14: XU - }jr is invertible and that (Tu)-1 = (T-')u.

1. Odds and Ends

48

For the converse, assume that Tu is invertible. This implies immediately that T is one-to-one. Moreover, we claim that T has closed range. To see this, assume that Tx,, -. y in Y. Then we have

xn = (Tu)-'[(Tx.,Tx.,...)) -+ (Tu)-ly is a Cauchy sequence in X. If x --+ x holds in X, then in X. In particular, Tx - Tx in Y, and so y = Tx. Thus, T has a closed range. By part (c), T is surjective, and consequently invertible.

1.5. Vector-valued Functions Problem 1.5.1. Establish the following "product rule" for derivatives: Con-

sider two functions g: 0 -, C and f : 0 - X (where 0 is an open subset of C and X is a complex Banach space). If both functions f and g are differentiable at some Ao E 0, then show that the X -valued function A - [gf] (A) = g(A)f (A) is also differentiable at Ao and (gf)'(A0) = 9 (Ao)f(Ao) +g(Ao)f'(Ao) 0 f'(Ao) = 0, it follows Solution: From f (A) - f (Ao) _ ( X that limA.A° f (A) = f (Ao). That is, f is continuous at A0. Now note that A - ?f)(.ao) = 9O)IU)-gOo)J(Ao) = 9(A)-g(Ao)

A

Ao

This easily implies Alimao

A° o

= 9 (Ao)f (,\o) + g(Ao)f'(Ao)

So, g f is differentiable at Ao and (g f)'(Ao) = g'(A0) f (AO) + g(A0) f'(A0).

Problem 1.5.2. Morera'8 theorem is a converse of Cauchy's theorem and it states that: If a continuous function f : O - C (where 0 is an open subset of C) satisfies fc f (A) dA = 0 for each simple closed rectifiable curve C which lies with its interior in 0, then f is an analytic function. Show that Morera's theorem is valid for Banach-valued functions. Solution: Assume that a continuous function f : 0 -+ X (where 0 is an open subset of C and X is a complex Banach space) satisfies fc f (A) dA = 0 for each simple closed rectifiable curve C which lies with its interior in O. Fix an arbitrary continuous linear functional x' E X' and consider the continuous function x' o f : 0 C. Now if C is a simple closed rectifiable curve which lies with its interior in 0, then it follows from Theorem 1.69 that

j(fof)(A)dA=x'(jf(A)dA) = x'(0) = 0.

1.5. Vector-valued Functions

49

This, in connection with the standard Morera theorem, yields that x' o f is an analytic function for each x' E X'. That is. the function f : 0 X is weakly analytic. By Theorem 1.76. the function f is also analytic.

Problem 1.5.3. Let C be a rectifiable curve lying in C, and let X and Y be two complex Banach spaces. Also, let A: C - C(X, Y) be a continuous operator-valued function. Show that for each x E X we have

[JA(A)dA](x)

=

fAX)xdA.

Solution: Since the vectors on both sides belong to Y. it suffices to verify that for each x E X and each yr' E Y' we have

f

y'. [c J A(A) dA] (x)) = (y'. J A(A)x dA). In view of Theorem 1.69, the right-hand side equals fc y' (A(A)x) dA. Now consider

the functional y' 0 x on £(X. Y) defined by (y' 0 x. T) = (y', Tx) for each T in L(X. Y). So, on one hand. we have y`

x

f

c

A(A) dA)

_ (y [f A(A) dA] (x)}

and on the other hand. applying y` S x to the operator fc A(A) dA and again using Theorem 1.69, we get

\y'

X.

JC A(A)

dA) = j(i( . x. A(A)) dA =Jys(A(A)x)dA.

Therefore, X.

f

,

A(A) dA} = (y'. [

f A(A) dA] (x)) _ (y'. Jc `

for each x. E X and all y' E Y'. as desired.

A(A)x dA)

./

U

Problem 1.5.4. This is a generalization of Theorem 1.76. Let X be a Banach space and let G be a (not necessarily closed) norming subspace of X*, that is, there exists a constant 'y > 0 such that sup{lx*(x)l: x' E G and IIx*I1 < 1} > yjIx{I

for each x E X. Show that a continuous function f : 0 -+ X defined on an open subset 0 of the complex plane C is analytic if and only if the scalar-valued function x' o f is analytic for each x' E G. Solution: The proof mimics that of Theorem 1.76. The only difference is that instead of the inequality Ilxn - a%m 11 < 3 IAn - An, I we will obtain (using the analyticity of the functions x` o f for x' E G) that 'yllx,, - xn,I1 < IAn - Aml. This is enough to guarantee that {xn} is a Cauchy sequence. The other arguments in the proof remain the same.

1. Odds and Ends

50

Problem 1.5.5. For a continuous function A: 0 -, G(X, Y), where 0 is an open subset of C, show that the following statements are equivalent.

(a) The function A'- A(A) is analytic on O. (b) For each x E X the Y-valued function 0x(A) = A(A)x is analytic on O. (c) For every x E X and y` E Y` the standard complex-valued function O..b (A) = (A(A)x, y*) = y0 (A(A)x)

is analytic on O. Solution: To prove that the continuous function A(A) is analytic, it is sufficient to show (according to Problem 1.5.4) that there exists a norming subspace G of G(X,Y)' such that g' o A(A) is analytic for each g' E G. Let G = Y' ® X. Recall that each expression y' ® x can be considered as an element of £(X. Y)' by letting T ,-y y`(Tx). Obviously, G is norming and this proves the equivalence of (a) and 1 (c). The equivalence of (c) and (b) is a direct consequence of Theorem 1.76.

Problem 1.5.6. For each r > 0 let A,. = {A E C: IAI > r}. Assume that X is a complex Banach space and for some r > 0 a function f : A,. - X is analytic having the Laurent series expansion f (A) = F,'-. A"C,, on the open annulus A,.. Show that lima_,o f (A) = 0 if and only if c" = 0 for each n = 0, 1, 2,... . Solution: Assume first that limA, f (A) = 0. Fix any e > 0 and then choose some r1 > max{1,r} such that Ilf(A)II < e for each JAI > r1. If CR denotes the positively oriented circle with radius R and center zero, then according to Laurent's Theorem 1.78, for each R > r1 and each n > 0 we have

IIcnII=112;

JCR

aR+

dAl1<2*. ... 27rR=

R-

<e.

Since e > 0 is arbitrary, it follows that cn = 0 for all n > 0. Now assume that c = 0 for each n > 0. From Theorem 1.77, we know that limsup ^ IIc_nII <_ r. So, there exists some r2 > r such that ^ IIc_nlI < r2 for each n > 1. For each A E C with JAI > r2, let 6A = la l. Clearly, 0 < 6A < 1 and II < r - = 6n, it follows that for A E C with lima_ ba = 0. From II an II = I

g

JAI > r2 we have 00

Ilf(A)II = lIE An n=1

Since lima,,,

1

x `- III A. II << Eat = 00

11

n=1

1-da

n=1

= 0, the preceding inequality shows that lima.,o f (A) = 0.

I

Problem 1.5.7. Assume that two points a and b in a topological space X are joined by a curve C. If C is covered by a collection V of open sets, then show that there exist (not necessarily distinct) sets V1, V2, ... , V" in V such that:

1,6. Fundamentals of Measure Theory

51

(a) aEVi and bEV,,. (b) CC V,

n I;+1 # 0 for each i = 1, 2, ... , n -1.

(Such a collection V1, V2,- .. , V of open sets is called a chain of open sets covering the curve C.) Solution: Fix a continuous function x: [a, /3) -+ X such that x(a) = a, x((3) = b, and x ([a, , 3J) = C. Since x is continuous, it follows that {x-'(V): V E VI is an open cover of the closed interval [a, 03]. Now notice that for each V E V the open subset x-1(V) of [a, 3) is a union of open subintervals of [a,#). So, there exists a family of open subintervals 2 of [a, i3) such that I covers [a,,01 and for each I E I there exists some V E V with I C x-1(V). We claim that there exists a chain 11,12,... , Ik in I that covers [a, /3]. If this is the case, then for each 1 < r < k choose some V,. E V with 1, C r-1(V,) and note that Vi, V2,..., V, is a desired chain for the curve C. In view of the compactness of [a, 3), in order to establish our claim, it suffices to prove that whenever a closed interval is a finite union of open subintervals, then the open cover can be reduced to a chain. The proof is by induction on the number n of the subintervals of the open cover. For n = I the conclusion should be obvious. For the induction step, assume that every open cover of any closed interval consisting of at most n open subintervals can be reduced to a chain. Suppose that a closed interval [a, (3) is the union of a family I of n + 1 open subintervals of [a, 01. Pick some to E I such that 0 E lo, i.e., 10 = (-y, a] for some or < y < f3. Clearly, the family { 1 n [a, 7) : I E Z and I fl [a, yj 16 0} of open subintervals of [a, 7] covers [a, y1 and has at most n open subintervals. By our induction hypothesis, this open cover of [a, 7] can be reduced to a chain, say Il n [a, 7],12 n [a, 7J, ... , Ik n [a, 7]. To finish the induction argument, it remains to observe that 11, 12,... , Ik, Io is a chain for Z. 1

1.6. Fundamentals of Measure Theory Problem 1.6.1. Let E be a Riesz space with the countable sup property, and let D be a subset of E such that the supremum u = sup D exists. Does there exist a countable subset D1 of D such that for each x E D there is an element a E D1 such that a > x? (Any subset D, of D with this property is called cofinal in D.) Solution: The answer to this question is negative. That is, in general, cofinal countable subsets need not exist. Here is an example. Let E = C[0, 11. Since the Riemann integral is a strictly positive linear function

on C[0,11, it is easy to verify that E has the countable sup property. For each t E [0, 1] choose a function ft E E such that ft(t) = I and 0 < ft(s) < I for each s 0 t. Consider the set D = {ft : t E [0, 1]}. Obviously, sup D = 1 = X[0,11.

1. Odds and Ends

52

Now let D1 = {ft : n = 1.2....} be an arbitrary subset of D. Pick any t E [0. 11 that is distinct from each t,,. Since ft(t) = 1 and ft. (t) < 1 for each it. it should be clear that ft is not dominated by any function from D1. I

Problem 1.6.2. Let E be an Archimedean Riesz space satisfying the countable sup property and having a weak unit. Show that for each non-empty subset D of E there exists a countable subset Do of D such that Do has the same set of upper bounds as D. That is, whenever u E E satisfies it > x for each x E Do, then u> x for each x E D. Solution: Let e be a weak unit in E. Assume first that E is Dedekind complete. For each n consider the set D = D n ne = {r n ne: x E D}. This set is bounded from above and so u = sup D exists. Since E has the countable sup property, there is a countable subset A of D such that sup(A A ne) = u,,. We claim that the set Do = Un 1 A satisfies the desired property. To see this, let u E E satisfy it > a for all a E Do and fix some d E D. Fix for a moment some n. Then, of course, u > a n ne for each a E A,,, and so u > u holds for each n. This implies u> d n ne f o r each d E D and each n. From d n ne j d, we get u > d. Now let E be an arbitrary Riesz space. then consider E. the Dedekind completion of E. Clearly the weak unit e in E remains a weak unit in E. Also, it is easy to see that k has the countable sup property since E has it. Therefore, the general case is reduced to the previous one. The proof is finished. Also notice that the assumption that E has a weak unit is essential.

Problem 1.6.3. Let (St, E, ir) be an arbitrary measure space, and let E be the collection consisting of all subsets of fl that can be written in the form

AAM, where A E E and M C N for some N E E with rr(N) = 0. (As always, AAM = (A \ M) U (M \ A) is the symmetric difference of the sets A and M.) Verify that: (a)

(b)

E is a a-algebra. The function a: E - [0, oo], defined by ir(AAM) = 7r(A), is well defined and extends 7r from E to a measure on E.

(c) If some A E E,r satisfies rr*(A) < oo, then A E E. In particular, ? is a complete measure, that is, if some subset A of fZ satisfies

ir'(A)=0, then AEE. (d) If 7r is a-finite, then E = E, Solution: (a) Clearly, E C E. Now take an arbitrary set AAAI in t, where A E E and Al C N for some N E E with 7r(N) = 0. An easy verification shows that (AAM)` = A`AM = (A n M) U (A u AI)°

.

This implies that t is closed under complementation. To establish that E is closed under countable unions, we need to verify that

E={ D e 2n : 3 A. N E E such that ir(N) = 0 and A\ N C D C A U N}.

53

1.6. Fundamentals of Measure Theory

To see this, note that if A \ N C D C A u N holds with 7r(N) = 0, then letting M = D n N, we have D = (A \ N)DAI. On the other hand, if D = AiM, where

AEEandMCNforsome NEEwith ir(N)=0,then wehave

A\NCA\AlCD=(A\M)u(M\A)9AuN. We are now ready to show that E is closed under countable unions. To see this, let {Dn} C E be given, and put D = U', Dn. Pick two sequences {An} and {Nn} in E such that 7r(Nn) = 0 and An \ Nn C Dn C An U Nn holds for each n. If

A=Un°1AnEEandN=U'1N8EE.then ir(N)=0and 0C

x

"C

A\NC U(An\N8)C UDn=DC U(AnUNn)=AUN. n=1

n=1

n=1

This shows that D E E. Therefore, E is a o-algebra.

(b) Clearly, we have E C E C E, . Now let D E E. Pick A E E, N E E with 7r(N) = 0 and a subset M of N such that D = AOM. This implies that 7r(D) = 7r(A) = 7r'(A). Since 7r' is a measure on E, it follows that 7r' (and hence frr) is also a measure on E.

(c) Assume that a set A E E,, satisfies 7r'(A) < oo. Then for each n there exists some Bn E E such that 7r(Bn) < 7r* (A) + 1. Let B= nn= 1 Bn E E. From 7r(B) < 7r(Bn) < 7r'(A) + n < 7r(B) + n, it follows that 7r(B) = 7r'(A). Since 7r'(A) < oo, the set Al = B \ A satisfies 7r'(M) = 7r(B) - 7r'(A) = 0. Now choose some N E E with Al C N and 7r(N) = 0 and note that A = BOM. Therefore, A E E.

(d) We already know that t C E,,. Now assume that 7r is o-finite. Pick a sequence {SZn} C E with 7r(1,) < oe for each n and S2 =

Uoo

1 Stn.

If A E E,,, then by part (c) we have A n SZn E E for each n. This implies A = U' l A n Stn E E. Therefore, in this case, E = E,,, holds true.

Problem 1.6.4. Let J be an ideal in some Lo(7r)-space and let {xn} be an order bounded sequence in J. Show that xn - x holds in J if and only if xn(w) -+ x(w) for 7r-almost all w E Q.

Solution: Let {xn } be an order bounded sequence in some ideal J of Lo(7r). Fix some vector u E J such that IxnI < u holds for each n E N. Assume first that xn --+ x. This means that there exists some sequence (y.) C J satisfying yn j 0 and Izn -XI < yn for each n. If we pick a 7r-null set A such that Ixn(w)-x(w)I < yn(w) and yn(w) j 0 hold for each w 0 A, then xn(w) -' x(w) for each w V A. This shows x. that xn For the converse, suppose there is a 7r-null set A such that Ixn(w)I < u(w) < 00 and xn(w) x(w) for all w 0 A. Now define the two sequences {yn} and {zn} on i? \ A via the formulas yn(w) = Sup{x,(w): i > n} and zn(w) = inf{x{(w): i > n). Clearly, yn and zn are functions in Lo(7r) and in view of Iynl < u and Iznl < u, we see that {yn } and { zn } are sequences in J. Moreover, notice that zn (w) j x(w) and

1. Odds and Ends

54

yn(w) j x(w) f o r each w V A. Now if vn = yn - zn, then vn(w) j 0 for each w A (and so vn 10 in J) and Ixn(w) - x(w)j < vn(w) for all w V A. Thus, Ixn - xI vn in J for each n, and therefore xn -9- x in J.

Problem 1.6.5. If it is a -finite, then show that Lo(7r) is a Dedekind complete Riesz space with the countable sup property. Solution: Fix a sequence (On) C E satisfying 0 < a(S2n) < oc for each n and S2 = Un , Stn. Also, define p: Lo (a) -. [0, oo) by 00

P(X) =

1: z^ . n=1

xin.,

jn i+: dir.

It is easy to see that p has the following properties:

(1) A function x E Lo (a) satisfies p(x) = 0 if and only if x = 0.

(2) If x, y E Lo (a) satisfy 0 < x < y, then p(x) < p(y), i.e., p is strictly monotone.

(3) If 0 < xn j x holds in Lo (a), then P(Xn) j Ax).

Now let D be a non-empty subset of Lo(a) that is bounded from above in Lo(a). Since the collection of all finite suprema of D has the same upper bounds as D, we can assume that D is directed upward. Now fix some xo E D and consider the set Dl = {x E D : x > xo } - x0. Notice that sup D exists if and only sup Dl

exists and sup Dl = (sup D) - xo. So, replacing D with the set consisting of all finite suprema of Dl, we can assume without loss of generality that D is an upward directed subset of Lo (a).

Put m = sup:ED p(x) < oo, and then choose a sequence {xn} C D such that xn j and p(xn) j M. Since D is bounded from above, it follows that there exists some x E Lo (a) such that xn j x. This implies p(xn) T Ax). We claim that x = sup D. First, we shall show that x is an upper bound of D. To see this, let y E D. Then y V xn E D and y V xn j y V x. From xn < y V xn and p(y V xn) < m, it follows that p(y V x) = m. Now from 0 < x < y V x, p(y V x) = p(x) = m, and the strict monotonicity of p, we get x = y V x. This implies y < x, i.e., x is an upper bound of D. To see that x is the least upper bound of D, let z E Lo(a) be an upper bound of D, i.e., x < z for each x E D. Then xn < z also holds in Lo(a) for each n. Since x is the a-a.e. pointwise limit of the sequence {xn}, we see that x < z. Thus, x = sup D. The above arguments show that Lo(a) is a Dedekind complete Riesz space with the countable sup property.

Problem 1.6.6. Recall that a Riesz subspace G of a Riesz space L is said

to be order dense if for each 0 < x E L there exists some 0 < y E G satisfying 0 < y < x. Show that a Riesz subspace G of an Archimedean Riesz space L is order dense if and only if for each x E L+ we have {y E G+: y 5 x} j x in L.

1.6. Fundamentals of Measure Theory

55

Solution: Let G be an order dense Riesz subspace of an Archimedean Riesz space L, and let x E L. Assume by way of contradiction that sup{y E G: y!5 x} 54 x. This means that there exists some z < x such that y E G and y < x imply y < z.

Put u = x - z > 0, and then select some v E G with 0 < v < u < x. This implies

v
Problem 1.6.7. Consider the following property (P) for subsets D of Riesz spaces:

(P) If {xn} C D and An 10 in R. then Anxn -°-0.

Assume that a subset D of a Riesz space E satisfies property (P). Prove that the set consisting of all finite suprema of the set IDI = {Idl: d E D} also satisfies (P). Solution: Let D be a non-empty subset in a Riesz space E that satisfies property (P), and let b denote the set of all finite suprema of SDI. Assume that {An} is

a sequence of scalars such that An j 0, and let {yn} be a sequence in D. By definition of b for each n there is a finite subset {dn,, : 1 < i < kn} of D such that yn = Idn,l I V . . . V Idn,k I. Let {xn } denote the sequence in D defined by: d1.1 d1,2, .... dl.k, , d2,1 d2,2, ... , d2,k2.... .

For each n E N let rn = k1 +

+ kn, and then for each n satisfying rm -I < n < r,n

(we assume that ro = 0) let On = Am. Clearly On j 0 and therefore, by our hypothesis jinx,, -9-0. Let {z} be a sequence such that zn j 0 and IQnxn I < .;,n for each n. Put um = z,_ -I and note that um j 0 in E. Moreover, we have IAm1/ml = A,IyrI =

V

IAmZn! =

n=r,,,_,+1

V

1I3nxnl < zr.n_i = um .

n=r,,,-I+1

This shows that Anyn °-+ 0, as desired.

Problem 1.6.8. For a sequence {xn} in some Lo(zr)-space establish the following.

(a) If xn -s-' x and xn - y hold in Lo(ir). then x = y. (b) If xn -x in Lo(a) and {xn} is a positive sequence, then x > 0. Solution: (a) From Theorem 1.82 it follows that there exists a subsequence {xn} of {xn } such that zn -P+ x and zn y. Therefore, x(w) = y(w) for 7r-almost all

wEfl. (b) Again, there exists a subsequence {yn } of {xn } such that yn --» x. This implies that x(w) > 0 for 7r-almost all w E f2. I

Problem 1.6.9. We say that a sequence {xn} in some Lo(7r) is *-convergent to x (in symbols, xn - - x) if every subsequence of {xn} has a subsequence that converges pointwise 7r-almost everywhere to x.

1. Odds and Ends

56

Let (St, E, a) be a c-finite measure space and let A be a measurable set of finite measure. Also let 0: A - IR be a 7r-integrable function such that ¢(w) > 0 for ir-almost all w E A. Show that for each sequence {Bn} of measurable subsets of A the following statements are equivalent: +0.

(a) XBn

(b) XBn -'

0-

(c) fBn 0 d7r -P 0.

(d) ir(Bn) -+0. Solution: (a) b (b) This follows immediately from Theorem 1.82. (b)

(c) If fBn 0 da ,4 0, then there exists a subsequence {Cn } of {Bn }

and some a > 0 such that fc n 0 d7r > a holds f o r each n. Since XBn - - 1 - , there exists a subsequence { D } of {Cn } such that XDn -P- 0. But then, by the Lebesgue Dominated Convergence Theorem, we have a < fDn 0 d7r = fA OXDn d7r ---o 0, which is impossible. Hence, fen 0 d7r -' 0.

(d) If ir(Bn) f+ 0, then we can assume (by passing to a subsequence if (c) necessary) that there exists some n > 0 such that a(Bn) > tJ for each n. We claim that there is some constant a > 0 such that fBn 0 dir > a for each n. To see this, assume that there is no such constant a. This means that there exists a subsequence (C,,} of { Bn } satisfying fcn 0 d7r < s^ for every n E N. Put Dn = U =n C; and note that fDn 0 d7r F," O n fc, 0 d7r < 21-n for each n. If Dn I D, then from 0 < fD 0 d7r = limn-oC L. 0 da = 0, we get fD 0 d7r = 0. Since O(w) > 0 for zr-almost all w E A, we obtain ir(D) = 0. On the other hand, D1 C A implies w(DI) < oo, and from Cn C- Dn and the continuity of the measure 1r we get

0 = 7r(D) = lim zr(Dn) > limsupa(Cn) > Q > 0, n-oo n-oo which is a contradiction. This establishes that 1r(Bn) -+ 0.

(b) Let {Cn} be an arbitrary subsequence of {Bn}. Then we have (d) f Xcn d7r = ir(Cn) -' 0, and from this we infer that the sequence {Xcn } has a a-a.e. pointwise convergent subsequence. So, XBn -s-

1

Problem 1.6.10. Assume that (St, E, 7r) is a o-finite measure space and that a sequence {Stn} C E satisfies 9r(Stn) < oo for each n and Un 1 Stn = St.

Suppose also that 0: [0, oo) - [0, oo) is a continuous strictly increasing bounded function with 0(0) = 0; for instance, we can consider the functions

0(t) = I+e and 0(t) = 1 - a-t. Show that a sequence {xn} C Lo(rr) satisfies xn - 0 if and only if

l

fnk cb(IxnI)d7r = 0 for each k.

Solution: Let 0: [0, oo) -' [0, oo) be a continuous strictly increasing bounded function with 0(0) = 0. Fix some M > 0 such that 0 < 0(t) < M for each t > 0.

1.6. Fundamentals of Measure Theory

57

Suppose first that linen -a, f fk 9(Ix I) d1r = 0 for each k. Fix some A E E with

a(A) < oo, and let e > 0. Put 8, =

(Zk. and note that A n en j A. So, by Uk=1 the continuity of the measure, there exists some m such that rr(A\ e,n) < [Q(e)12. Now define An = {w E A: e} for each n, and observe that since 0 is strictly increasing, we have An = {w E A: p(Ixn(w)I) > Q(e)}. This implies Q(e)XA, <- XAQ(I xI ). Integrating yields Q(e)ir(An)

< f Q(IxnI)da < L\em A

<

Ef

1 f7r(A \ em) + k=1

m

Q(Ixnl) d7r

Slk

<

,

? f[Q(e)]2+E

k=1 Ilk

d(Ixnl)d7r.

So, limsupn_x Q(e)ar(An) < M[0(4F)12 or limsupn-,.r A(An) _< MO(e) for each

e > 0. Since e > 0 is arbitrary, 6(0) = 0 and b is continuous at zero, we get limsup,

7r(An) = 0. Consequently, limn_. rr(An) = 0, and thus xn -' 0.

^n

For the converse, assume that xn Jr- 0. Fix some k and let e > 0. Also, for each n let An = {w E Stk: Ixn(w)I > e}. Clearly. An = {w E Sak: Q(Ixn(w)I) > 40(e)}. According to our assumption we have limn-oo ir(An) = 0. Now note that

f

ilk

Q(I xn I )d7r =

f

A

Q(xn I)

dir

+f

k\A©(Ixn1) dir < M7r(An) +O(e)x(ftk)

-

This implies limsupn,. fslk 6(Ix. I) dir < O(e)ir(Ilk) Since e > 0 is arbitrary, it follows from the properties of the function O that limsllpn-,o fnk Q(Ixnl)dir = 0. Consequently, limn-, f nk Q(I xn I) dir = 0. and the solution is finished.

Problem 1.6.11. Let (11, E. Tr) be a o, finite measure. Fix a sequence {0k} in E such that 0 < 7r(Ilk) < oc for each k and 1 = Uk 1 Stk. Also, define the function p: Lo(ir) - (0, oo) by

x P( x)

=E k=1

1

1

a ilk)

L

-L1+jxjd7r.

Show that for a sequence {xn} C Lo(7r) and a function x E La(7r) the following statements are equivalent.

(a) xn --' x.

b For each k we have lim J(1k r 1+I:rn2n-X d7r = 0.

( )

(c) lim p(xn - x) = 0. n-Mc

Solution: (a)

(b) This follows immediately from Problem 1.6.10.

(c) Fix e > 0, and then pick some m such that Ek m+l 2-r < e. Also, pick some no such that n > 110 implies F,k 1 fllk -2T l+:ay d7r < e for all (b)

1. Odds and Ends

58

n>no. So. if n

n0. then

EJ

p(r - r)

k=1

1

1

'

ISn-4 ir + 1+.x,-.r dir

k

< f+e=2c. This shows that p(x,, - a) (c)

:

x.

F, flk A k=n1+1 1

r '-ri Ti1,) t+ rn "-Il dir 1

0.

(b) This implication is trivial.

0

Problem 1.6.12. Let (S. El. it) and (T. E2, v) be two measure spaces. If a function f : S x T -, R is jointly measurable, then f t) is E1-measurable for each t E T and f (s, ) is v-measurable for each s E S. Solution: For each subset D of S x T. each s E S. and each t E T we put:

D,={tET: (s.t)ED}

and

D'={sES: (s,t)ED}.

Consider the collection of subsets of S x T given by-

A={DCSxT: D,EE2 and D`EEI forall sES and tET}. A straightforward verification shows that A is a a-algebra. Moreover. from (A

x= B),

rB

1

0

if SEA if sOA

and

(AxB)t=

if tEB 0 if t0B. A

we see that A contains all measurable rectangles-. Therefore. E1 E2 C A. Now assulne that a fuI ctio11 f : S x T -- IR is jointly uieasttrahle. Let 13 be a Borel subset of R. and let s E S. Then f-I(B) E EI E2, and so [ f(s_.)] -1(B)

_ {t E T: f(,R.t) E B) _ [f-1(B),, E E2 This shows that the function f (s, ) is E2-measurable for each s E S. Similarly, f (', t) is E1-measurable for each t E T.

Problem 1.6.13. If (f2, E. r) is a a-finite measure space and E is an order dense ideal of Lo(7r), then show that for each f E Lo(r) there exists a sequence { f,l} C E n LI (7r) satisfying I ff,J < If for each n and f11(.o) --a f (,,o) for -,r-almost all w.

If. in addition. f E L' (7r). then show that the sequence J fn) can be chosen in E fl Li (7r) such that f,t j f. Solution: We claim that L1(7r) is an order dense ideal of LO (-,r). To see this, let 0 < f E Lo(r). Then for some e > 0 the r-pleasurable set A = {w E i2: f (w) > e} has positive measure. By the a-finitenes of n. there exists some B E E satisfying

B C A and 0 < r(B) < ,c. Now note that function y = 'i B E L1(r,) satisfies 0 < g < f. This shows that L1(7r) is order dense in Lo(ir). Since E is order dense. it follows that E n L1(r) is also an order dense ideal

(why?). Now let 0 < f E L0(r). Pick a net {f} C_ E n L, (-,r) with 0 < fQ I f in Lo(r). Since Lo(r) has the countable sup property (see Problem 1.6.5). there exists a sequence of indices {n,,} such that 0 < f,,,, 1 f.

1.6. Fundamentals of Measure Theory

59

Next, let f E Lo(ur) be an arbitrary function. Pick two sequences

and

in E fl L, (7r) such that 0 < g j f+ and 0 < h j f-. Now if On = gn - h,,, then On E E fl L, (7r) and If,, I _< If I hold for each n. To finish the solution, note that On (w) - f (w) for ir-almost all w E Q.

Problem 1.6.14. Let m : I' - [0, oc] be a measure on a semiring. Then for a function f E L1 (m) establish the following:

(1) f > 0 m-a.e. if and only if fA fdm > 0 holds for all A E IF with m(A) < oc. (2) f = 0 m-a.e. if and only if fA fdm = 0 holds for all A E t with m(A) < oc. Solution: (a) Let 1 be a semiring of subsets of a set Sl and assume that a function f E L1(m) satisfies f A fdm > 0 for all A E F with m(A) < oc. Consider the set C = {w E Q: f(w) < 0}. We must prove that m* (C) = 0. For this, it suffices to show that for each e > 0 the measurable set CE = {w E !1: f (w) < -e} has outer measure zero. Since If I > EXc., it follows that m' (CE) < oo. We shall verify that

m'(CE)=0 foreach e>0. To this end, fix e > 0 and let 6 > 0. Since f has an absolutely continuous integral, there exists some r) > 0 such that if A E

satisfies m'(A) < y, then

fA If I dm < b. Since m' (CE) < oo, there exists a sequence {A} of pairwise disjoint sets in 1' satisfying C, c A = U' 1 A and m'(A) = E°°_1 m(A) < m'(CE) + q. (Here we use the obvious fact that 1' is a semiring; see 18, Problem 15.2].) Clearly, m' (A \ CE) = m' (A) - m' (CE) < r/. This implies

O<_1 fdm = [J fdm- / fdm] ,

c

A

JA\C,JA J \C. IfIdm-J i_1 A <J

j

fdm

A

fdm

IfIdm<\C,

for each 6 > 0. Therefore, fc. fdm = 0, and from this (since f (w) < -e for each w E CE) we obtain m' (CE) = 0. Thus f > 0 m-a.e. The converse should be obvious.

(b) Assume that a function f E L1 (m) satisfies fA fdm = 0 for all A E F with

m(A) < oo. By part (a), f > 0 and f < 0 m-a.e. are both true. Hence, f = 0 m-a.e. The converse is trivial.

I

Problem 1.6.15. Let t be a semiring of subsets of set fi, and let two set functions 9, t : 1' -+ [0, oo) satisfy 0(A) _< (A) for each A E F. If 0 is finitely additive and 9 is also a measure.

is a measure (i.e.. countably additive), then show that

Solution: Assume that 0 and t; satisfy the stated properties. Since t; is a measure,

the outer measure ' oft is also a measure on the a-algebra of all c-measurable

1. Odds and Ends

60

sets that is an extension off . Now suppose that the sets A, A,, A2,. .. in I' satisfy

A = U°D_1 An and A, f1 Aj = 0 for i 0 j. Fix some e > 0. From the equality _°

{(A;) = la(A) < oo, it follows that there exists some k such that n

n

n

(A\ UA) = f'(A) -

e'(A;) _ (A) -

e(A;) < E

holds for all n > k. Now fix some integer n > k. Since r is a semiring, there exist pairwise disjoint sets C1,.. . , C,. in r such that A \ U 1 A; = U;_1 Cj or A = (U 1 A;) u (vi=1 C,). This, coupled with the finite additivity of 0, yields 8(A) = En=1 8(A;) +

Therefore,

0<8(A)-E8(A1)=E8(C,) 8=1

)=1

(A\UA1) < j=1

j=1

;=1

for all n > k. This shows that 8(A) = Ui_10(A;), and so 8 is a measure too.

Problem 1.6.16. Establish the following property of the product measure: If the measures p and v are a-finite, then for each set C E Eµx&, and for each e > 0 there exists a set of the form DV)=(CAD) Un=1 An x Bn, where An E El and Bn E E2 for each n, such that (p x < c. Solution: Without loss of generality we can assume that both measures are finite. By the definition of the outer measure we can find a sequence {An x Bn} of rectangles such that C C Un_ 1 A. x Bn and (µ x v) ((U' 1 An x B,,) \C) < Pick k large enough so that (p x v)(Un k+1 An x Bn) < 2. Then the set D = Un=1(An x Bn) satisfies the desired property. I 1.

Problem 1.6.17. If f E Lo(ir), then show that

max{esssup f,esssup(-f)} = esssupIfI = IIfII Solution: Let a = ess sup f, b = ess sup (-f), and c = esssup If 1. From f 5 If and -f < If I, it easily follows that a < c and b < c, and so a V b < c. If c = 0, then f = 0 a-a.e., and so a = b = c = 0 is trivially true. So, we can assume that

0
Fix 0 < 6 < c, and note that the measurable set D = {w E 11: if (w)I > 6} has positive measure. It follows that at least one of the measurable sets

D1={wEIl: f(w)>6} and D2={wESl: -f(w)>6} must have positive measure. This implies that either a > 6 or b > 6 must be true. Therefore, a V b > 6 for each 0 < 6 < c, and so a V b > c. Consequently, c = a V b, as desired.

Problem 1.6.18. Assume that (f', E, ir) is a non-atomic measure space and let {xn} C L' (7r). Also let x = sup. xn be the it-a.e. pointwise supnemum of the sequence {xn}. Show that x 0 L1(7r) if and only if there exists a pairwise disjoint sequence {An} 9 E such that F,n 1 fA xn d7r = oo.

1.6. Fundamentals of Measure Theory

61

Solution: Let {xn} be a sequence in L' (7r). We can suppose that each function xn is real-valued, i.e, xn : S2 -' R. Assume first that there exists a pairwise disjoint sequence {An} C E such that E.°° 1 fagzn da = oo. Since Fn_1 xnXA < x holds for each k, it easily follows that

E

n=1

!A xn da < fn x dzr for each k. This implies f1. x dir = oo, and therefore

x V L1(1r)

-

For the converse, assume that x V L1(a). Let E = {w E 11: x(w) = oo} and note that E E E. Suppose first that the function x is rr-a.e. finite, that is, 7r(E) = 0.

Put Al={tES2: x1(t)>1x(t)}and foreachn>1let An = It E 11 \ Uk=i Ak: xn(t) > ax(t)} .

J

AWe

Clearly, {An} is a disjoint sequence in E and Uk 1 Ak = S1. Consequently,

xdzr=zjxd7r=co.

now consider the case a(E) > 0. Using this and the non-atomicity of the measure, it is easy to construct a measurable function u : 12 - R such that 0 < u < x

and u 0 L1(7r). For instance, the non-atomicity of the measure guarantees the existence of a pairwise disjoint sequence {En} in E satisfying E C E and 9r(En) > 0 for each n. If we consider the "infinite step function" u = En 1 E7E XE,, then u is a real-valued measurable function such that 0 < u < x and u V L1(a). Finally, for each n let y, = u A x. and note that the sequence {yn} C L1 (7r) satisfies sup. yn = u. By the previous case, there exists a disjoint sequence {An} in E such that En'=I f Anyn da = oo. From xn > yn, we get F,n 1 fAnxn dir = 00, 0 and the solution is finished.

Chapter 2

Basic Operator Theory

2.1. Bounded Below Operators Problem 2.1.1. Let T : X - Y be a bounded below continuous operator between normed spaces. Then show the following:

(a) If X is not a Banach space. then T need not have a closed range. (b) The unique continuous linear extension of T to the norm completion of X is also bounded below. Solution: (a) Let X be a non-closed subspace of a Banach space Y and consider the operator T: X Y defined by Tx = .r for each x E X. Clearly. IITxii = Ilxii for each x E X and so T is bounded below. However. the range of T is X which is not clued in Y. (b) Assume IITIII > 7II.'II holds for all x E X and some ry > 0. Fixi in X (the norm completion of X), and then pick a sequence {xn } of X satisfying xn -+ i. Now note that

iITiII = lim

n-z

lim yllxnll = 7IIi1I n-x

So, t is bounded below.

Problem 2.1.2. Prove that an operatorT : X -a Y between two real nonmed spaces is bounded below if and only if Tc : Xc - Yc is likewise bounded below.

Solution: Let T : X - Y be an operator between two real normed spaces. Assume first that T is bounded below. So, there exists some -y > 0 such that 63

2. Basic Operator Theory

64

IITxII? vIrllforallzEX. Now ifz=x+tyEX,,.then wehave Z(IlTxll + IITVII) ? 1(71IxII + -rIlvll) IITczII = IITx+:Tyll ? z(IIxII + Ilvll) ? 21I=II

This shows that the operator TT: Xc -. Y,, is bounded below. For the converse, assume that TT : X. - YY is bounded below. Pick some constant -r > 0 satisfying IITczll ? 7llzll for all z = x + zy E X. In particular if

r E X, then IITxII = IITT(x+to)112711x+toll ='rllxll and so the operator T : X -. X is also bounded below.

Problem 2.1.3. Let T : X -+ Y be a bounded, one-to-one, and surjective operator between two Banach spaces. Show that T is an isometry if and only if T and T'1 are both contractions. (Recall that a bounded operator between two normed spaces is called a contraction if its norm is no more than one.) Solution: If T is an isometry, then clearly IITII = IIT-111 = I holds. For the converse, assume that T and T-1 are both contractions, i.e., that 11TI1 S I and <- 1 both hold. Now let x E X. Clearly, IITxII S 11TII IIxII : IIxII. On the other hand, note that 11x11= J1T-1(Tx),! < 11T" `11 IITxII <- IITxII. This implies IITxII = IIxII for each 11T-' 11

x E X. That is, T is an isometry.

Problem 2.1.4. Let T E C(X) be bounded below but not invertible. Show that there exists some e > 0 such that the operator A - T is not surjective whenever JAI < e.

Solution: Assume that the claim is false. Then there exists a sequence of scalars satisfying An -. 0 and such that A - T is surjective for each n. Now -T in C(X). Since -T is bounded below, it follows from notice that An - T Theorem 2.9 that -T (and hence T) is surjective. From the fact that T is bounded below and surjective. it follows that T is invertible, a contradiction. I

Problem 2.1.5. This is a generalization of the previous problem. Assume that X and Y are Banach spaces and let T E C(X, Y) be bounded below but not invertible. Then show that there exists some e > 0 such that the operator T - S is not surjective whenever S E C(X, Y) satisfies 1ISII < e. That is, show that the set of all bounded below and non-surjective operators from X to Y is an open subset of C(X, Y). Solution: If the claim is false, then there exists a sequence {S,,} of surjective operators satisfying S - 0 in C(X. Y) and such that T - Sn is surjective for each n. Since T - S - T in C(X. Y), it follows from Theorem 2.9 that T is also surjective. Since T is bounded below, T must be invertible, which is impossible.

Problem 2.1.6. Establish the following property that was used in the proof of Corollary 2.6. Assume that X is a vector space such that X = V 9W with

2.1. Bounded Below Operators

65

V finite dimensional. If Y is an arbitrary vector space, then any operator T : X -+ Y that is one-to-one on W satisfies dim N(T) < dim V. Solution: Assume that k = dim N(T) < oc. To show that N(T) has dimension at most k, it suffices to show that every collection of k + 1 vectors in N(T) is linearly dependent. So, let x1, x2, , xk+1 be k + 1 vectors in N(T). For each 1 < i < k + 1 choose vectors vi E V and wi E W such that xi = vi + wi. Since vl, v2, . , vk+l are linearly dependent, there exist scalars cl, c2, ... , ck+l not all Ek+1 c,wi, and so zero such that Ek+1 c,vi = 0. It follows that Ek+1 cixi = T(Ek+i ciw,) = T(Ek+1 c,xi) = 0. Since T is one-to-one on W, it follows that Fk+1i c,wi = 0. This implies Ek±1 cixi = 0. Thus, the vectors x1, x2, xk+l are linearly dependent, and consequently dim N(T) < dim V.

Problem 2.1.7. Show that the assumption dim N(T) < oo in the last conclusion of Corollary 2.6 cannot be replaced by the assumption that N(T) is complemented. Solution: We must show that there exists a Banach space X such that the collection B of all bounded operators T E C(X) having a closed range R(T) and a complemented null space N(T) is not open in C(X). Let X = e2, fix a number a E (0, 1). and consider the operators T and T on X defined as follows:

T(xl,x2,x3,x4,x5.... ) = and

Tn(x1, x2, x3, x4, x5, ...) = (x1, anx2, x3, an+1x4, x5,

an+2x6, x7 ...)-

It is obvious that T E B and that Tn -. T. It remains to notice that none of the operators Tn has a closed range. Indeed, each operator Tn is clearly one-to-one. Therefore if some Tn were to have a closed range, then this T. would be bounded from below. However, a direct inspection shows that this is not true.

Problem 2.1.8. Let T : X -' Y be a bounded operator between two Banach spaces. If the quotient vector space Y/R(T) is finite dimensional, then show that the range of the operator T is closed. Solution: Let T : X - Y be a bounded operator between two Banach spaces such that the quotient vector space Y/R(T) is finite dimensional, say of dimension n. Pick a basis { [yl], [y2], ... , [y,,] } in Y/R(T) and note that the vectors y1, y2, ... , yn are linearly independent in Y. Indeed, if Alyl + A2y2 + + Anyn = 0, then 0 holds in Y/R(T), and so Ai = 0 for each i = I,-, n. ,\1 [y1]+A2[y2]+ +A,, Let V be the n-dimensional vector subspace generated by {yl, y2,. .., y.) in Y. We claim first that R(T) fl V = {0}. To we this, let y E R(T) n V and write

y = j:° 1 a,yi. This implies [y] _ Fn j a,[yi] = 0 in Y/R(T), and so ai = 0 for each i = I_-, n. Therefore, y = 0, and thus R(T) fl V = {0}. Next, we assert that R(T) e V = Y. To this end, let y E Y. Pick scalars Ql, 02, ... , )3n such that [y] = E $.[yi] or, equivalently, [y - F_", Qiyi] = 0. This implies that the vector z = y - En 10,yi belongs to R(T). That is. y = z +' 1nQiyi E R(T) ® V. Hence 1

R(T) ® V = Y holds true.

2. Basic Operator Theory

66

Now Theorem 2.16 guarantees that R(T) is closed. That is, the operator T has a closed range.

Problem 2.1.9. Let T : X

Y be a bounded operator between Banach spaces such that its range R(T) is not closed. If X = V ® W, where V is closed and W is finite dimensional, then show that T : V - Y is not bounded below.

Solution: Assume that T : V -+ Y is bounded below, i.e., there exists some c > 0 such that cIIvii IITvII holds for each v E V. This implies that T(V) is a closed subspace of Y. Also T(X) = T(V)+T(W). and clearly T(W) is finite dimensional. Since the sum of a closed subspace and a finite dimensional subspace is closed (see Problem 3.4.1), we infer that T(X) is closed, which is impossible. This contradiction establishes that T cannot be bounded below on V. I

Problem 2.1.10. For a bounded operator T : X - Y between Banach spaces establish the following properties regarding the ranges of T and T*.

(a) The range of the operator T is dense if and only if the adjoint operator T*: Y* -> X' (defined as usual via the duality identity (Tx, y*) = (x.T*y*)) is one-to-one. (b) If the adjoint operator T* is one-to-one and has closed range, then show that T is surjective. (c) If T carries norm bounded closed subsets of X to closed subsets of Y, then T has closed range. Moreover, in this case, all powers of T also have closed ranges. Solution: (a) Assume that R(T) is dense and let T'y* = 0 for some y* E Y*. Then for each x E X we have (Tx, y*) = (x, T' y*) = (x, 0) = 0, i.e., y* vanishes on R(T). Since R(T) is dense, this implies y* = 0. Thus, T' is one-to-one.

For the converse, assume that T' is one-to-one. If R(T) is not dense in Y*, then by the Separation Theorem there exists some non-zero y* E Y* satisfying (y*.Tx) = (T*y*,x) = 0 for all x E X. This implies T'y* = 0, that is, T' is not one-to-one, a contradiction. Hence, R(T) is dense in Y. (b) According to Theorem 2.18 the range R(T) of the operator T is also closed and by (a) the range R(T) is dense. Hence, R(T) = Y and so T is surjective.

(c) Assume T is also one-to-one and let Tx -' y. If has a norm bounded subsequence, s a y (z,, 1. then y belongs to the closed set T Q zl, z2, ... }) 9 R(T).

The case IIx, II - oo cannot occur. For if Ilx, II - oo, then T( ) = =y 0, and so zero belongs to the closed set T({x E X : IIxII = 1}), contrary to the fact that T is one-to-one. To see that we can assume that T is one-to-one, we consider the closed subspace M = Ker T and the quotient Banach space X/M. Let 7r: X X/M be the quotient

map. Note that the function t: X/M --. Y defined by T(ir(x)) = Tx is a welldefined one-to-one bounded linear operator that satisfies R(T) = R(T). Moreover, we claim that t carries norm bounded closed subsets of X/M to closed subsets of

2.1. Bounded Below Operators

67

Y. To see this, let C be a bounded closed subset of X/1L1 and pick an open ball satisfying C C rUxl,tit = rir(U,v ). Since it is continuous, the set tr-'(C) is a closed subset of X. Now notice that if we consider the closed bounded subset K = (rUr) n -,r-'(C) of X, then t(C) = T(K), which (in view of our hypothesis) shows that t(C) is a closed subset of 1'.

Thus. we have shown that the one-to-one bounded operator t: X/11I - Y carries norm bounded closed subsets of X/:II to closed subsets of Y. By the oneto-one case considered above, t has a closed range. Since R(T) = R(T), the operator T likewise has a closed range, as desired. I

Problem 2.1.11. Let X and Y be two Banach spaces. and let T(X, Y) denote the open subset of £(X, Y) consisting of all isomorphisms from X to

Y. Show that the mapping T -- T-1. from T(X, Y) to T(Y, X). is (norm) continuous.

Solution: Fix some R E T(X.Y) and assume that an operator S E T(X.Y) satisfies IIS - RII < I RI I or 1 - IIR-' II IIS - RII > 0. From

S-'-R-' = R-' (R - S)S -' = R-' (R - S) [(S-' - R-') + R-' ]

= R-'(R-S)(S-'-R-')+R-'(R-S)R-', we get IIS-'-R'II < follows that

IIR

IIR-111

IIS-' -R- 'II <

- SII. It

JJR-

]I-RII IIS - RII, holds for all S E T(X.Y) sufficiently close to R. This implies that T'-. T-' is 11-IIR

norm continuous at R.

Problem 2.1.12. For a bounded operator T : X - Y between Banach spaces establish the following.

(a) The operator T has a closed range if and only if R(T') = N(T)-L. (b) If T has a closed range. then X'/R(T') = [N(T)]'.

Solution: (a) Assume first that R(T') = N(T)1. Then R(T') is closed. By Theorem 2.18. it follows immediately that R(T) is also closed. For the converse, assume that R(T) is closed. FYom Corollary 2.15 we know that there exists some constant c > 0 such that for each y E R(T) there exists some r E X with y = Ti and Ilxll cllyll

The inclusion R(T') C N(T)1 is obvious. Indeed, for each y' E Y' and each r E N(T) we have T'y'(x) = y'(Tx) = y'(0) = 0. and hence T'y' E N(T)1. For the reverse inclusion let r' E N(T)1. Define a scalar-valued mapping y' on R(T) via the formula

y'(Tx) = x'(x). Notice that if Tx = T:, then z - r E N(T) and r' E N(T)1 implies .r'(z) = r'(x). This shows that y' is well defined. Clearly, y' is also linear. Now for each y E R(T) pick some r E X with y = Tx and IIxll cllyll. and note that I1y*(y)II = II y'(Tx)II = I r'(x)I

I1x*11 ' IIxHI < cllr'll - Ilyll

2. Basic Operator Theory

68

This shows that y' is a continuous linear functional defined on R(T). If z' is any continuous linear extension of y' to all of Y', then for each x E X we have

T'z'(x) = z (Tx) = y'(Tx) = x'(x).

Thus. T'z' = x', and so x' E R(T'). This shows that N(T)1 C R(T') is also true, and therefore R(T') C N(T)l. (b) By Problem 1.1.22 we know that [N(T)1' = X'/N(T)1. Moreover, from part (a) it follows that R(T') = N(T)1. Therefore, [N(T))' = X'/R(T').

2.2. The Ascent and Descent of an Operator Problem 2.2.1. Let T : X - X be an operator on a vector space. If R(Tk) = R(Tk+i) holds for some k, then show that R(T) = R(Tk) for

allm>k.

Solution: The verification of the claim is by induction. Assume that R(Tk) _ R(Tk+n) for some n > 1. We must show that R(Tk) = R(Tk+n+1). Clearly, R(Tk+n+1) C R(Tk+n) For the reverse inclusion, let y E R(Tk+n) So, there exists some x E X such that y = Tk+nx. Since Tkx E R(Tk) = R(Tk+n), there exists some z E X such that Tkx = Tk+nz. This implies Tk+n+l(Tk-1z) E R(Tk+n+l), Y=Tk1nX = Tn(Tkx) = Tk(Tk+nz) =

and so R(Tk+n+1) = R(Tk+n) = R(Tk). Therefore, R(T-) = R(Tk) holds for all m > k.

Problem 2.2.2. Show that every operator on a finite dimensional vector space has finite ascent and descent. Solution: This follows easily from the fact that in a finite dimensional vector space there is neither a strictly increasing nor a strictly decreasing sequence of vector subspaces.

Problem 2.2.3. Consider the operator T : 1R3

00

1

1

0

T= 0 0

R3 defined by the matrix

0 0

Find the ascent and descent of T.

Solution: Note that 0

T2= 0 0

0 0 1

0

0 0

000

and T3=T2=010 0

0

0

2.2. The Ascent and Descent of an Operator

From N(T) _ {x = (x1, x2, x3) E R3 : Tx = (x3, x2, 0) = 0} and

69

(x1, 0, 0) : xi E R}

N(T2)={x=(xl,x2,x3) E R3: T2x=(0.x2,0)=0}={(x1,0,x3): xl,x2 E R}, it follows that N(T) is a proper subspace of N(T2). The identity T2 = T3 implies 1 N(T2) = N(T3). Therefore, a(T) = 2, and consequently b(T) = 2.

Problem 2.2.4. Give examples of bounded operators T : X -+ X on a Banach space such that:

(i) a(T) < oo and S(T) = oo. (ii) a(T) = oo and b(T) < oo. (iii) a(T) = oo and b(T) = oo. Solution: Let B, S: f2

e2 be the backward and forward shift, respectively.

That is, B(xi, x2, x3, ...) _ (x2, x3....)

and

S(x1, x2, x3, ...) _ (O, xl, x2, ...) .

(i) Clearly, S is one-to-one, and so N(S) = N(S2). This implies a(S) = 1. Since

R(Sk)={y=(yl,y2,y3.... )E12: yj=0 for each 1
N(Bk) _ {x = (xl, x2, x3, ...) E t2 : xi = 0 for each i > k} .

This implies that N(Bk) is a proper subspace of N(Bk+l) for each k. Hence, a(B) = oc. Since R(B) = R(B2) = 6, it follows that 6(B) = 1. (iii) Consider the operator T: t2 ® t2 -' e2 ® t2 defined by

= S(xl, x2, ...) ®B(zl, z2, ... = (O, xi, x2, ...) ®(x2, z3, ...) . That is. T = S ® B. Then it is easy to see that a(T) = b(T) = oo. T ((xl, x2, ...) ®(zi, z2, ...))

Problem 2.2.5. Let T : X -' X be a bounded operator on a Banach space with p = a(T) = b(T) < oo. Show that the adjoint operator T*: X' -+ X' also has finite ascent and descent and that a(T') = b(T') = p. Solution: Let V = N(TP) and W = R(TP). By Theorem 2.23 we know that T is nilpotent on V and invertible on W. Then X' = V'®W', T' is nilpotent on V' and invertible on W', and thus the pair (V', W') reduces T'. So, by Theorem 2.23, the operator T' has finite ascent and descent. Let q = a(T') = b(T' ). Since (T')P = (TP)' = 0 on V. it follows that q < p.

Similarly, T" has finite ascent and descent. Let r = a(T") = b(T"). By the above r < q. Now note that T' = 0 on V. This implies p < r < q < p. Consequently, p = q, and the solution is finished.

I

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70

Problem 2.2.6. Show that a bounded operator T : X

X on a Banach

space is nilpotent if and only if for each x E X there exists some n E 14 such

that 7x = 0. Solution: If T is nilpotent. then the condition is satisfied automatically. For the converse. assume for each .r. E X there exists some n E N (depending on .r) such that T"x = 0. This implies that

x

x=UN(T"). n=1

Since each N(T") is a closed vector subspace of X and X is complete. it follows from the Baire Category Theorem that for some k the closed vector subspace N(Tk)

has an interior point. This implies N(Tk) = X or Tk = 0 for some k. That is, T is a nilpotent operator.

Problem 2.2.7. If a nilpotent operator T : V index p, then show that a(T) = b(T) = p.

V on a vector space has

Solution: Let V be a vector space and let T : V V be a nilpotent operator of index p. Clearly, N(7P) = N(TPy1) = V and N(T1' ') 0 N(TP). This implies that T has finite ascent and that a(T) = p. From R(TP) = R(TP+1) = {0}. it follows that b(T) = p is also true.

Problem 2.2.8. For an operator T : V -' V on a vector space establish the following algebraic properties.

(a) For each n, m E N we have T" (N(T"+"`)) C R(T") n N(T-). (b) For each n. m E N the operator T" : N(T"+ n) - R(T") n N(T"') is surjective-and consequently the vector spaces N(T"+"')/N(T"') and R(T") n N(T"') are linearly isomorphic.

(c) If the ascent of T is finite, then R(TP) n N(r) = {0} for each nEN and all p > a(T). (d) If for some p, n E N we have R(T") n N(T") = {0}, then T has finite ascent and a(T) < p holds.

Solution: (a) Let v E N(7+'). Then Tm(T"v) = T"+mv = 0. This implies T'vv E N(T). and therefore T"u E R(T") n N(Tm). Consequently, we have T"(N(T"+"&)) E R(T") n N(7"1)

(b) To see that the operator 7": N(7+"') - R(T") n NJ") is surjective, let u E R(T")nN(T"'). This means Tu = 0 and u = Tv for some v E V. Hence, T"+'"v = T"(T"v) = Tu = 0, and so v E N(T"+m). The latter conclusion shows that the operatorT": N(T"+m) -' R(T") n N(TI) is surjective. Now if we define the operator S : N (T"++n) /N(Tm) - R(7'") n N(T"") via the formula S]v] = T"u. then it is easy to see that S is a surjective linear isomorphism.

Therefore. N(T"+m)/N(T) and R(7) n N(7) are linearly isomorphic vector spaces.

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71

(c) Assume that u E R(TP) fl N(T") for some n E N and some p > a(T). By part (b), there exists some vector v E N(TP+") such that u = T v. Since N(TP+") = N(TP) = N(T°(T)), it follows that v E N(TP). So. u = T v = 0 and hence R(TP) fl N(T") = {0}. (d) Now suppose that for some p, n E N we have R(TP) fl N(T") = {0}. From

part (b). we see that N(TP+")/N(TP) = {0} or N(TP) = N(TP+"). This implies N(TP) = N(TP+1) Consequently, T has finite ascent and a(T) < p holds.

Problem 2.2.9. Let T : V -+ V be an operator on a vector space. Show that for each n, m E N the mapping J: V/[N(T")+R(T"')] R(T")/R(T"+'"), defined by

J [t,] = T" t + R(T+'"). is a well-defined surjective linear isomorphism. Solution: We shall verify first that J is a well-defined mapping. To this end, let Jul = [v] E V/[N(T") + R(T'")], i.e, let u - v E N(T") + R(Tm). Pick X E N(T") and y E V such that u - v = x + Tm y, and note that Tnu - T"v = T"z + T"+my = 7M+my E R(T"+m).

This shows that J[u] = J[v], and thus J is well defined. Clearly, J is a surjective linear operator. To finish the solution, we must verify that J is one-to-one. To see this, assume

that J[v] = Tnv + R(T"+m) = 0, i.e. T"v E R(T"+'"). This implies the existence of some u E V such that Tnv = Tn+mu or T"(v - Tmu) = 0. So, the vector w = v-Tmu belongs to N(T"), and consequently v = w+Ttu E N(T")+R(Tm). That is, [v] = 0. and so .1 is one-to-one.

2.3. Banach Lattices with Order Continuous Norms Problem 2.3.1. Let S2 be a compact Hausdorff space. Show that the Banach lattice C(1l) has order continuous norm if and only if Q is a finite set. Solution: If 1 is a finite set, say S2 = { w 1 . . . . . . }, then C(f2) = R", and so C(Q) has order continuous norm. For the converse, assume that C(1) has order continuous norm. To finish the proof, it suffices to show that every point in St is an isolated point. To this end, fix some w E !1 and suppose by way of contradiction that w is an accumulation point. Consider the non-empty collection of functions

D. =

(f E C(1l) : 0 < f < I. f (w) = 1 and f vanishes outside of an open neighborhood of w)

Clearly. Dam, j and inf JE D. f (W') = 0 for each w' 0 w. Since w is an accumulation point, it follows that D4.. j 0. But then 11 f 11 x = 1 for each f E D,,, contradicts the

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72

order continuity of the norm II .11.. This contradiction establishes that each point w E fl is an isolated point, and so fl is a finite set. I

Problem 2.3.2. Show that in an Archimedean Riesz space E a positive vector is an atom if and only if it is a discrete vector. Also show that if u is an atom in E, then {Au: A E R} (the vector space generated by u) is a projection band. Solution: To begin with, we need two properties of Archimedean Riesz spaces. So, let E be a Riesz space. E is Archimedean if and only if en 10 in R implies enx 10 in E for each

xEE+. To see this, suppose first that E is Archimedean, en j 0 in R. and x E E+. Assume that y < enx holds for each n and some y E E. If k is a natural number, then en < k for all n sufficiently large, and therefore if we choose any such n, then

y<enx
If E is Archimedean and an -' a holds in R, then anx -2- ax in E for each xEE. To see this, for each n let en = Supk>n iak - Oil and note that En j 0. Hence, for each x E E we have Manx - axl = Ian- aUUxI < Enlxl 10We now return to our problem. We must keep in mind that the lattice operations of a Riesz space are order continuous. Assume that u > 0 is an atom in an

Archimedean Riesz space E and let 0 < x < u. Put a = sup{,Q > 0: Qx < u}. Clearly, a > 1, and (since E is Archimedean) we see that a < oo. Moreover,

ax 0. Since (v - nx)+ T v (here we use the above properties in connection with the order continuity of the lattice operations), it follows that (v - kx)+ > 0 for some k > 1. Clearly,

(v-kx)+=[u-(a+ )x]+
On the other hand, [u - (a + ')x]- > 0. (Otherwise, [u - (a + ')x]' = 0 implies (a + )x < u, contrary to the definition of a.) Next, observe that k [u - (a + ')x]- = [(a + 1)x - u]+ < 2ax < 2au.

So,ify=sv[u-(a+1)x]+andz=zQ[u-(a+k)x]-, then we have 0
assume 0 < Anu T x in E. Clearly, 0 < an T in R and, since E is Archimedean, An T \ in R. So, .1nu T Au, and hence x = Au, i.e., x E A. Thus, A is a band of E.

2.3. Banach Lattices with Order Continuous Norms

73

To see that A is a projection band, let 0 < x E E and put

a=sup{(i>0: ,Qu<x}. Since E is Archimedean, we have 0 _< a < oc and au < x. In view of the identity

x=au +(x-au), it suffices to show thatx - aulauorx - aulu. To this end, observe first that the definition of a implies

0< [x-(a+n)u] _ [(a+ 1)u-x]+<(1+a)u. Now from the fact that (1+a)u is an atom, 0 < [x- (a+,-,)u]+Au < (1+a)u, and

Qx-(a+1-)u]+nu)nQx-(a+,-,)u] ) =0,itfollowsthat [x-(a+I)u]+nu=0 for each n. Since a + n -i a and the lattice operations are order continuous, the latter implies (x - au) n u = (x - au)+ A u =O. Thus, x - au E Ad, and so A is a projection band.

Problem 2.3.3. Show that C[0,1] and L1 [0, 1] are atomless Banach lattices.

Solution: Let O < u E C[0,1]. Pick 0 < to < 1 and e > 0 with 0 < to - e < to + e < 1 such that u(t) > 0 holds for each to - e < t < to + e. Now choose two non-zero functions f,g E CIO, 11 such that 0 < f,g < 1, f is supported by the interval [to - e, to] and g is supported by the interval [to, to + e]. Then 0 < u f < u, 0 < ug < u and (u f) A (ug) = 0 in C[0.1]. This shows that u cannot be an atom, and so C[0,1] is an atomless Riesz space. Next, let 0 < f E L1 [0, 1]. Then the measurable set A = {x E [0,1] : f (x) > 01 has positive measure. Pick a measurable subset B of A such that B and C = A\ B both have positive measure, and note that the functions f XB, f Xc E L1 [0, 1) satisfy

0 < f Xe <- f, O < fXc <- f, and (fXB)A(fXc) = 0 in L1[0,1]. Therefore, f cannot be an atom, and so L 1 [0,11 is also an atomless Banach lattice. I

Problem 2.3.4. Show that in general the norm dual of an atomless Banach lattice need not be atomless by proving that the norm dual of the atomless Banach lattice E = C[0,1] has atoms. Solution: Fix an arbitrary point t E [0, 1] and consider the continuous linear functional be : CIO, 11 -. R defined by be (f) = f (t). We claim that bt is a discrete element, and so Problem 2.3.2 will guarantee that bi is an atom in CIO, I]*. To see that this is the case, pick some x' E C[0,1]' satisfying 0 < x' < bt. By the Riesz Representation Theorem, there exists a unique regular Borel measure µ on [0, 11 satisfying x'(f) = f f dµ for each f E C[0, 1]. We claim that Suppµ = {t}.

Since x' 0 0, we know that Suppp 0. We verify next that Supp µ = {t}. Indeed, if s E Suppµ and s 0 t, then by choosing a function 0 < f E C[0,11 satisfying f (s) = 1 and f (t)) = 0, we see that

0<J fdµ=x*(f)
x'(g) =

µ=J gdµ = g(t)µ({t}) = µ({t})be(g) . {e}

2. Basic Operator Theory

74

Therefore, x' = p({t})at. Thus, bt is a discrete element (and hence an atom) in the Banach lattice C[0,

I

Problem 2.3.5. If g, h : E - R are two disjoint non-zero positive linear functionals on a Riesz space, then show that there exist disjoint vectors u, v

in E+ such that g(u) > 0 and h(v) > 0. Solution: Since g and h are non-zero and positive, we can find some 0 < x E E such that g(x) > 0 and h(x) > 0. By scaling, we can assume that g(x) > 2 and h(x) > 2. Since g and h are disjoint, we have 0 = (g n h)(x) = inf{g(y) + h(x - y): 0 < y < x} . Pick any y E [0, x] such that g(y) + h(x - y) < 1. Then g(x - y) > 1 and h(y) > 1. Let u = (x - y) - y n (x - y) and v = y - y A (x - y). These two elements are disjoint, and

g(u) = g(x - y) - g(y n (x - y)) ? g(x - y) - g(y) > 0. Similarly,

h(v) = h(y) - h(y n (x - y)) ? h(y) - h((x - y)) > 0. Thus, the elements u and v satisfy the desired properties.

Problem 2.3.6. Let E be a Riesz space and let 0 < f E E-. Show that f is an atom in E- if and only if f is a lattice homomorphism, that is, if and only if f (x V y) = maxi f (x), f (y) } for all x, y E E. Solution: Assume first that f is an atom in the order dual E. To show that f is a lattice homomorphism, it is enough to show (according to Theorem 1.34) that f (u) f (v) = 0 whenever u n v = 0. Assume, contrary to our claim, that there are two disjoint positive elements u, v E E such that f (u) > 0 and f (v) > 0. For each x E E+ let g(x) = lim f (x n nu). A straightforward verification shows

that g is additive on E+ and hence the formula g(x) = g(x+) - g(x-) defines a linear functional on E. Clearly, 0 < g < f and g(u) = f (u) > 0. The Riesz space E^' is necessarily Archimedean and so, in view of Lemma 2.30,

the functional f is a discrete element in E-. Therefore, there exists a constant A > 0 such that g = A f . Clearly A = 1 because g(u) = f (u) > 0, and hence we also have g(v) = f (v) > 0. On the other hand, by the definition of the functional g, we obviously have g(x) = 0 for each x that is disjoint to u. In particular, g(v) = 0, a contradiction. For the converse, assume that f is a lattice homomorphism, and assume that g, h are two disjoint functionals satisfying 0 < g, h < f. We must show that at least one of them is zero. If not, then we can find two disjoint elements u, v E E+ such that g(u) > 0 and h(v) > 0; see Problem 2.3.5. Hence, f (u) f (v) > g(u)h(v) > 0.

On the other hand, since f is a lattice homomorphism, we have 0 = f (u n v) _ f (u) f (v), which is impossible.

Problem 2.3.7. Let S, T : E - F be two regular operators between Banach lattices with F Dedekind complete. Show that:

(a) S < T in f, (E, F) implies S* < T* in f, (F*, E*).

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75

(b) IT`I < ITI'. Give an example where I T * I < I T I ' holds true.

Solution: (a) Since E is a Banach lattice, we have Lr(E, F) C L(E. F). Assume that two operators S. T E Lr(E, F) satisfy S < T, i.e., Sx < Tx for each x E E+.

Let 0 < y' E F. If r E E+, then note that S'y*(x) = y*(Sr) < y'(Tx) =T'y'(x). This shows that S'y' < T'y' for each 0 < y' E F', i.e., S' < T` in L,.(F', E'). In other words, the linear isometry T - T' from Lr(E, F) to Lr(F', E') is a positive operator.

(b) From ±T < ITI and part (a). it follows that ±T' < ITI'. Consequently, we

have IT'I = T' V (-T') < ITI'. For an example of a regular operator T satisfying Ti I< ITI'. consider the f, defined by operator T: fl T(x) = (xi - 12,12 - r3, r3 - x4, ...) _ ( X I - X 2 - X 3 , ...) - (x2, x3, x4, ...)

for each r = (r t .12....) E f t L. Clearly, T is a regular operator, and an easy argument shows that ITIx = sup ITyI = (11 +12,x2+13.13+r4....) IYI<_r

for each r E f+. We shall establish that IT'I 0 ITI' by proving that IT'Ib # ITI00 for each Banach-Mazur limit 0 on fem.

R be a Banach-Mazur limit and let e = (1.1,1, ...), To see this, let p: f,, the constant sequence one. Clearly, (ITI'O, e) = (0, ITIe) = 0(2,2,2 ....) = 2, and so IT* 10 # 0. On the other hand. if t' E f;. satisfies 1,01 < 0 and y = (yt,y2,...) is a null sequence. then from IW(y)I < It''I(Iyl) S 0(IyI) = lim Iy,,I = 0. we see that y = Ti is a null sequence in f, we have T'ti'(x) = ti'(Tx) = 0 f o r each r E f I . In other words. T* 4, = 0 f o r all i' E f;, with I 'i < p. This implies

IT'Iq= sup IT'vi =00ITI*o. vl <_o

and so IT'I # ITI'. Therefore. IT'I < ITI'.

I

Problem 2.3.8. Show that a Banach lattice E is a KB-space if and only if E is a band in its double dual E**. Solution: Assume first that E is a KB-space. Clearly. E has order continuous norm, and so E is (by Theorem 2.26) an ideal in E. To see that E is a band in E". suppose that a net {ra } C E satisfies 0 < x,, j r" in E". Then IIx,, II < IIx" II < oc for each n, and we claim that {ra } is a Cauchy net. To see this, assume by way of contradiction that {x } is not a Cauchy net. This means that there exist a sequence of indices satisfying a j and some e > 0 such that r.,,,, II > E for each n. Since the sequence {x0,, } is norm bounded and increasing, and E is a KB-space, we obtain that (x,,,) is a Cauchy sequence. However, the latter conclusion contradicts our assumption that Ilxa.,+ - x,"., II ? e for each n. So. {x,, } is a Cauchy net. If X E E is its norm limit, then xQ j x in E:

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76

see Problem 1.2.12(c). Since E is an ideal in E", it follows that xa T x also holds

in E. This implies x" = x E E, and so E is a band in E. For the converse, suppose that E is a band in E", and let a norm bounded sequence {xn} in E satisfy 0 < xn T. We must show that the sequence {xn} is norm convergent. The formula

x"(x') = lim x'(xn) , x' E E' , n-»oo defines a positive linear functional on E', i.e., x" E E;7. An easy argument shows

that xn T x" in E**. Since, E is a band in E", it follows that x" E E. Now from xn(x') = x'(xn) j x"(x') for each x' E U. = UE fl E+ and the fact that U. is a w"-compact subset of E', it follows from Dini's classical theorem (see for instance [7. Theorem 9.4, p. 70]) that IIx. -

xnll = Sup Ixs.(x') - xn(x')I ! 0. x' E UE.

This shows that E is a KB-space, and the solution is finished.

Problem 2.3.9 (Dodds-Fremlin [251). Let E and F be two Banach lattices with F having order continuous norm. If X E E+, then show that the set

B = {T E 4(E, F) : T [O, x] is norm totally bounded) is a band in Gr(E, F). Solution: Clearly, B is a vector subspace of the Banach lattice G4(E, F). The verification that B is a band will be done by steps.

STEP 1: If T E B and R, S E G,.(E, F) satisfy T = R + S and IRI A ISI = 0, then

R,S E C,(E,F). To establish this claim, fix x E E+ and let e > 0. From

IRIxi A ISIxi: X. E E+ for each i and Exi = x) j. [IRI A ISI]x = 0. (see 16, Theorem 1.16, p. 15]) and the order continuity of the norm in F, there n IRIxi A ISIxiII < z

exist vectors x1, ... . xn E E+ with rls 1 xi = x and IIE

From T[O, xi] C T[0, x] and the norm total boundedness of T[O, x], we see that all sets T[0, xt] are norm totally bounded. So, for each 1 < i < n there exists a finite subset 4ii of [0,xi] such that T[0,xi] C T(4i) + 11UF for each i. where OF is the closed unit ball of F. Put it = I FM, 1 yi : yi E 4'i for each i = 1, ... , n}, and note that 4' is a finite set. Now if z E [0, x], then using the R.iesz Decomposition Property we can write

z = F 1 ti, where 0 <_ z; xi for each i. For each 1 < i < n pick some yi E 4ii such that Tzi E T(yi) + 2n UF. Let y = E 1 yi E 4) and for simplicity put wi = zi - yi. Clearly, Iwil < xi and IIR(wi)II < sn for each i. Next, taking into account the lattice inequality 0:5 1 u I - Jul A Ivl < lu + v) (valid in Riesz spaces),

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77

we get n

n

IIR(z - y)II

=

II

F'R(wi)il <- E IIR(wi)II i=1

i=1

n

n

IS(wi)III + III IR(wi)I A IS(wi)III

FIIIR(wi)I - IR(wi)I

n

n

5 EII(R+S)(wi)II + IIE IRIk, A ISlx+li i=1

i=1

n

n

< EIIT(wi)II + IIE IRIk, A ISIxill i=1

i=1

n

2n+2 =E.

< i=1

Thus, R(z) E R(4b) + EUF, and so R[0, x] C R(4) + EUF. Therefore, R[0, x] is norm totally bounded. That is, R E B. By symmetry, S E B is also true.

STEP 2: If T E B, then T+ E B, i.e., B is a Riesz subspace in C,(E, F).

This follows from the preceding part by observing that every T E f, (E, F) satisfies T = T+ + (-T-) and IT+I A I - T-I = 0. STEP 3: If {T0} C B satisfies Ta T T in C,. (E, F), then T E B.

Fix x E E+ and let E > 0. Since F has order continuous norm, there exists some Q with IITx - Tsx II < z . Next, pick an arbitrary finite subset A of F such that Ts[0, x] C A+ z UF. Now if 0 < y:5 x, choose some a E A with II Toy all < z

-

and note that

IIa - TyII<- IIa - TsyII+IITy - TsyII < Ila - TsyhI+IITx - TpxII <

z+Z=E.

That is, T [O, x] C A + EVF, and so T [O, x] is norm totally bounded, i.e., T E B.

STEP 4: If 0:5 S < T holds in C,.(E, F) With T E B, then S E B. This conclusion can be obtained from a result of H. H. Schaefer [70, p. 142] (see also [6, Theorem 11.17, p. 172]), a special case of which is stated below.

If E is a Dedekind complete Banach lattice and x E E+, then the order interval [0, x] is the closed convex hull of the set of all components of x. A short proof of this result goes as follows. Let C C [0, x] be the closed convex

hull of the set C of all components of x. If C 96 [0, x], then there exists some z E [0, x] that does not belong to C. Now, by the Separation Theorem, there exists some non-zero linear functional f E E' and some scalar a such that f (z) > a and f (y) < a for all y E C. But then, by Theorem 1.17, we have f +(x) = sup{ f (y) : y E [0, x] and y n (x - y) = 0} = sup f (y) < a. VEC

So, we have a < f (z) 5 f+(z) < f+ (x) < a, a contradiction. Thus, C = [0, x]. We now establish the claim in Step 4. Let .A denote the set of all convex combinations of components of T, and note that according to Step 1 we have

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78

.A C B. By the preceding property. we conclude that A is r-norm dense in [0. T]. Since IIRII < IIRIIr holds for each R E 4,(E,F), it follows that A is also dense in [0. T] with respect to the operator norm. Now it should be straightforward to verify that S (as being the norm limit of a sequence of A) belongs to B. This complete the solution.

2.4. Compact and Weakly Compact Positive Operators Problem 2.4.1. Show that a positive operator T : E -+ F between two Riesz

spaces dominates an operator S: E -' F if and only if -T < S < T. Solution: Assume first that 7' dominates S. Then for each x E E+ we have ±Sx < ISxI < TIxI =Tx. Thus. ±S < T or. equivalently, -T < S < T. For the converse, assume that -T < S < T. This implies that for each y E E+

we have -Ty < Sy < Ty or ±Sy < Ty, and so I Syl = (Sy) V (-Sy) < Ty. Consequently. if x E E, then

ISxI = ISx+ -Sx-I < ISx+I +ISx-I
i

Problem 2.4.2. Let T : E - F be a positive operator between two Banach lattices. Show that for an operator S: E - F the following statements are equivalent.

(a) T dominates S. (b) T* dominates S*

(c) T** dominates S* .

Solution: (a)

(b) Assume that T dominates S. By Problem 2.4.1 this is equivalent to -T < S < T. Since the adjoint mapping R' -. R` is a positive operator, it follows that -T' < S' < T', which (by Problem 2.4.1 again) is equivalent to saying that the positive operator T* dominates S*. (c) The validity of this implication follows immediately from the pre(b) vious implication (a) : (b). (c) (a) Assume that the positive operator T*: E" -. F" dominates S. By Problem 2.4.1 this is equivalent to -T" < S" < T". Since E is a Riesz

subspace of E" and T agrees with T", it follows that -Tx < Sx < Tx holds for each x E E-. Using Problem 2.4.1 once more. we see that T dominates S.

I

Problem 2.4.3. Prove the following basic properties of compact operators.

(a) Sums and scalar multiples of compact operators are compact operators.

2.4. Compact and Weakly Compact Positive Operators

79

(b) The composition of two (or more) bounded operators at least one of which is compact is likewise a compact operator. (c) Bounded finite-rank operators are compact. (Recall that an operator between two vector spaces is said to be finite-rank if its range is finite dimensional.) (d) A bounded operator is compact if and only if its adjoint is likewise a compact operator. (e) The norm limit of a sequence of compact operators is a compact operator. Solution: (a) That the sum of two compact operators is compact follows immediately from the fact that the sum of two totally bounded sets of a normed space is always totally bounded. (If C C Uk 1 B(x,, r) and D C Uj'j 1 B(y3, e), then C + D C Uk 1 U 1 B(x, + y).2e).) The fact that scalar multiples of compact operators are compact is obvious.

(b) Let X --- Y - Z be a scheme of bounded operators between Banach spaces. If either S or T is compact, then TS(Ux) = T(S(Ux)) must be a totally bounded subset of Z. (Keep in mind that every bounded linear operator carries norm totally bounded sets to norm totally bounded sets.) That is, TS is a compact operator.

(c) Let T : X -i Y be a bounded finite-rank operator between Banach spaces. This means that there exist xi. x2, .,X; E X' and J1, yz, ... , yk E Y such that Tx = 1 x, (x)yi holds for all x E X. Notice that each x; (Ux) is a bounded subset of R. So, if {xn} is an arbitrary sequence of Ux. then there exists a subsequence of {xn}, say {z, }. such that x; (zn) ri for each i. Now note that k

k

Tzn = Exi(zn)yi 1=1

Yaiyi 1=1

in Y, proving that T is a compact operator.

(d) Let T: X - Y be compact and consider Y as a closed vector subspace of C(U. ). where UU is the closed unit ball of Y' equipped with its w'-topology. Now fix y' E UY and let e > 0. Since T(Ux) is a norm totally bounded subset of Y, it follows from the classical Ascoli-Arzela theorem that T(U,x) is an equicontinuous Vy. of y' satisfying bounded subset of C(Uv ). So. there exists a

IT.-(y') -Tx(z*)I = Ix(T'y` -T'z')I < z for all z' E Vy. and X E Ux. So. IIT*y'-T'z'II = Supreer Ix(T'y'-T'z')I < z < E. for all z' E Vy . If the linear functionals yl ..... y;n E UU are chosen so that UY C Um 1 Vy,, then T' (Uy) C Um B(T'y; . e), proving that T* (U}) is a norm 1

totally bounded set. Now assume that T*: Y* - X' is a compact operator. By the preceding case,

we know that T**: X" - Y" is also a compact operator. In particular, the set

T(Ux)=T"(Ux)9T"(UT)

2. Basic Operator Theory

is a norm totally bounded subset of X". Since X is a subspace of X", it follows that T(Ux) is a norm totally bounded subset of X. Thus, T is a compact operator. (e) Let IT, } C L(X, Y) be an arbitrary sequence of compact operators such that IIT.a - TII -. 0 holds for some operator T E G(X,Y). Fix e > 0, and then choose some k such that IITk - TII < E. Since Tk is a compact operator, there exist yi, ... , y,,, E Y such that Tk(Ux) C U°_1 B(ys, E). Next, fix x E UX and then pick some I < i < m such that II Tkx - y, II < E. Now notice that

IITx-y,11 s II(T-Tk)xII+IITkx-y:ll <- IIT - TkII+IITkx - y.II <

E+E=2E.

Therefore, T(UX) C U; `-1 B(y;, 2E) so that T(Ux) is a norm totally bounded subset

of Y. This shows that T is a compact operator.

Problem 2.4.4 (Gantmacher [30]). For a bounded operator T: X - Y between Banach spaces show that the following statements are equivalent.

(a) T is weakly compact.

(b) T"(X") C Y. (c) T*: (Y', a(Y', Y)) - (X', a(X *, X")) is continuous. (d) T*: Y' --. X' is weakly compact.

Solution: (a)

(b) We know that UX is a(X", X')-dense in U. V. That is, if bar denotes the a(X", X' )-closure operation in X", we have Ux = UX. Taking

into account that T**: (X", a(X", X')) -. (Y", a(Y", Y*)) is continuous and that (by (a)) the a(Y", Y')-closure of T(U) in Y" lies in Y, we get

T"(UX)=T" UX CT"(Ux)=T(UX)CY. Hence,

T" (X") C Y.

(b)

(c) Let yo °1Y-Y' 0 in Y', and let x" E X". From T"x" E Y,

we see that

(T'ya,x**) = (y;, T"x ') -. 0. This shows that T'y; 0"X*-X"' , 0 in X. (c)

(d) By Alaoglu's theorem, we know that the closed unit ball of UY

is a(Y',Y)-compact. Therefore, T'(UY) is a a(X',X")-compact subset of X. Thus, T' is a weakly compact operator. (d) tor

(a) From the above established implications, it follows that the opera-

T.': (X..,a(X.. X'))

-

(Y..,a(Y..,Y."))

is continuous. Thus, T"(UX) is a weakly compact subset of Y. Since Y is norm dosed in Y it follows that Y is also weakly closed, i.e., (Y", a(Y", Y'**))dosed, in Y**. From a(Y", Y') C a(Y", Y'**), it follows that T" (U1.') fl Y is

also a(Y", Y')-compact. Now use that T(Ux) = T"(Ux) C T"(Uy) fl Y and the fact that a(Y", Y') and o(Y, Y') agree on Y to conclude that T(Ux) is a relatively weakly compact subset of Y. That is, T is weakly compact, and we are done.

2.4. Compact and Weakly Compact Positive Operators

81

Problem 2.4.5. Prove properties similar to the ones listed in Problem 2.4.3 for weakly compact operators. Solution: (a) Since the sum of two relatively weakly compact sets is relatively weakly compact, it follows that the sum of two weakly compact operators is weakly compact. Also, it should be clear that scalar multiples of weakly compact operators are weakly compact.

(b) The composition of two operators one of which is weakly compact is also weakly compact. This follows from the fact that an operator between two Banach spaces is bounded if and only if it is weakly continuous.

(c) From Problem 2.4.4, we know that a bounded operator between Banach spaces is weakly compact if and only if its adjoint is weakly compact. (d) Consider a sequence IT,, } C £(X. Y) of weakly compact operators satisfying

IITn - T II - 0 for some T E C(X, Y), where X and Y are Banach spaces. Fix X** E X. By Problem 2.4.4, we know that {T,;'x"} C Y, and from the relation

T"x"II -' 0, we see that T"X" E Y. That is, T"(X") C Y holds, and so (by Problem 2.4.4) the operator T is weakly compact.

Problem 2.4.6. Show that the range of an arbitrary compact operator is separable.

Solution: Let T : X Y be a compact operator between Banach spaces. Then T(Ux) is a norm totally bounded subset of Y and hence a separable metric space. Let A be a countable subset of T(Ux) that is dense in T (UX ), and let B = U 1 nA. We claim that the countable set B is dense in R(T). To see this, let y = Tx E R(T). Choose some natural number k such x E kUX. This implies y = Tx E kT(UX) C kA = kA C B. Hence, B is a countable dense subset of R(T), proving that the range of T is separable.

Problem 2.4.7. Show that a subset of a normed space is norm totally bounded if and only if it is contained in the closed convex hull of a sequence that converges to zero. Solution: Let A be a non-empty subset of a normed space X. Assume first that there exists a sequence {xn } such that IIxn II - 0 and A C co {xn}. Since the set (0, xl, x2, ...} is a compact subset of X. it follows (from Mazur's theorem) that co {xn } is a norm totally bounded set. This implies that A is also norm totally bounded.

For the converse assume that A is norm totally bounded. Let us denote the closed unit ball of X by U. The construction of the desired sequence will be done by induction. Start by letting A0 = A and ko = 0. Pick a finite subset F1 = {xl, x2, ... , xk, } of 2Ao such that 2Ao C F1 + 2-IU. (Clearly, such a set exists since 2Ao is norm totally bounded.) Put Al = (2Ao - Fl) f12-1U, and note that Al is a norm totally bounded set. Now for the induction hypothesis, assume that a pair (An-1, Fn) of subsets of X has been constructed such that is a norm totally bounded subset of 2-nU and the finite set Fn = {xk.,_1+1, xk.,_1+2, . , xk, } is a subset of the totally

2. Basic Operator Theory

82

bounded set 2An_1 satisfying 2An_1 C Fn+2-nU. Let An = (2An_1-Fn)f12-nU, and note that each An is a norm totally bounded set. Now choose a finite subset of 2An satisfying 2An C Fn+1 + 2-n-1U. Thus, the Fn+1 = {xkn+1, , induction step produces the pair (An, Fn+1) with the same properties. Notice also that An C 2-nU for each n > 1. We claim that the sequence (xn} produced by the above induction process satisfies A C co {xn }.

To see this, let n > 1 and choose some k such that kn+1 < k < kn. It follows that xk E Fn C 2An C 2 2-nU or IIxkII < 21-n. This implies IIxnII - 0. Now let x E A. Then an easy inductive argument shows that for each n > 1 there exist integers ml,..., mn with ki_ 1+1 < mi < k, for each 1 < i < n such that the vector z2 Eco{xn} satisfies

IIx - (z +

+...+)II<4 n

Thus, x E co {xn } and so A C co {xn }, as desired.

Problem 2.4.8 (Terzioglu [781). Show that an operator T : X - Y between two Banach spaces is compact if and only if there exists a sequence {x;,} in X' satisfying IIxnll - 0 and IITxII < sup Ixn(x)I for each x E X. Solution: Assume first that T is a compact operator. Then T` is also a compact operator, and so T'(UY) is a norm totally bounded subset of X'. By Problem 2.4.7 there exists a null sequence {xn} in Y' satisfying T(UU) C co{xn}. Now notice that for each x E X we have

IITxII = sup Iy'(Tx)I = sup IT'y'(x)I <- SuPlx;,(x)I y'EU4

y

V"

n

For the converse, assume that there exists a sequence {xn} of X' satisfying IITxII -+ 0 and IITxII <_ suPn Ix,,(x)I for all x E X. Now define the operator

S:X - +coby

S(x) = (xi (x), x2(x).... )

.

Let u = (Ilxi II, IIxsII, ...) and note that u E co. Moreover, S(Ux) 9 [-u, u] holds. Since the order interval [-u, u] is a compact subset of co, it follows that S(Ux) is a norm totally bounded subset of co, and, of course, a norm totally bounded subset of the vector subspace Z = S(X) of co. Next, define the operator R: Z - Y by R(Sx) = Tx.

To see that R is well defined, notice that if Sx = Sy, then x,,(x - y) = 0 holds for each n, and so from IITx - TyII = IIT(x - y)II < supra Ix,,(x - y)I = 0 we get Tx = Ty. Observe that the inequality IIR(Sx)II = IITxII S supIxx(x)I = IISxII. n

implies that R is a continuous operator. Now from T(Ux) = R(S(Ux)), it follows that T(UX) is a norm totally bounded subset of Y, and so T is compact. I

Problem 2.4.9. Show that an operator T : X -a Y between Banach spaces is compact if and only if T factors with compact factors through a closed subspace of co.

2.4. Compact and Weakly Compact Positive Operators

83

Solution: The "if" part is trivial. For the "only if" part assume that T : X - Y is a compact operator. By Problem 2.4.8, there exists a null sequence {x;,} of X' satisfying IITxII < sup Ix;,(x)I

(*)

n

1

for each x E X. We can assume that r;, 54 0 for each n. Put an = Ilxn 11 2 and y;, = 114 11 2x,,. Clearly, IIy,;II = an - 0. Now define the operator S: X - co by 1

S(x) = (yi (r) y2 (x)....) , and note that IISxll. = sup. Iy, (x)l for each x E X. By Problem 2.4.8, S is a compact operator. Let Z = S(X) be the norm closure of S(X) in co. Clearly, S : X --+ Z is a compact operator.

Next consider the operator R: S(X) - Y defined by R(Sx) = Tx. If Sx = Sy, then x,,(x - y) = 0 for each n, and so from (*). we get Tx = Ty. This shows that R is well defined. Moreover, if AT = sups lan I, then it follows from the relation Ixn(x)I = lanyn(x)I A1Iyn(x)I that IIR(Sx)II = IITxp < sup l xn(x)I < Al sup ly;,(x)I = A1IISxHI00 . n

n

This shows that R: S(X) Y is a continuous operator, and so R has a continuous extension to Z = S(X) (that will be denoted by R again). Clearly, T = RS. To complete the proof, we shall show that R is also a compact operator. T o this end, define f n E c o by fn (A1. A2, ...) = anAn. Then 11A II = an --' 0 and also fn(Sx) = anyn(x) = x,,(x). Since IIR(Sx)II = IITxII -5 sups Ifn(Sx)I for each x E X, it follows that IIR(z)II < sup. I fn(z)I for all z. Now Problem 2.4.8 I guarantees that R is indeed a compact operator.

Problem 2.4.10. Let r : Sl - s2 be a continuous mapping on a compact Hausdorff space S1, and let T : C(f2) -+ C(S2) be the composition operator defined by T f = f o r. When is T a compact operator? Solution: Assume that the composition operator is compact. This means that the set T([-1, 1]) is a totally bounded subset of C(fl). By the classical Ascoli-Arzeli theorem, the set T((-1.1]) is an equicontinuous subset of C(12). In particular, for each wo E 0, there exists an open neighborhood I' of wo such that

I f(r(w)) - f(r(wo))I <

2

(*)

for all wEVWOand all f EC(Q)with-1< f <1. We claim that r(w) = r(wo) for all w E V 0. To see this, assume by way of contradiction that for some wl E V 0 we have r(w1) T(wo). Pick a continuous

function satisfying 0 < f < 1, f (r(wl)) = 1, and f (r(wo)) = 0, and note that I f (r(wl)) - f (r(wo)) I = 1 contradicts (*). Thus, we have shown that for each w E fl there exists an open neighborhood V, of w on which r is constant. From f2 = U EO VW and the compactness of Q, it follows that there exist wl , ... , wk E Q such that f? = Uk 1 VW, . In particular, r has a finite range. Let t1, . . . , to E fI be the distinct values that r attains. From

W; = {w E Sl: r(w) = t,} = r-1({t,}) = r-1(fl\ {t1..... t:-1,t{+1,...,tn}),

2. Basic Operator Theory

we see that each W; is a clopen subset of Q. Moreover, fI = U', W, is a finite partition of S1 and r is constant on each W1. It is easy to see that these are the only types of continuous functions r : Q --I' fl whose composition operators are compact. Also, observe that if 1 is a connected topological space, then for a composition operator to be compact, it is necessary and sufficient that the continuous function r be constant.

Problem 2.4.11. Let T : 11 --+ Y be a bounded operator, where Y is a Banach space. Show that the operator T is weakly compact if and only if the set {T(en): n = 1, 2, ...} is a relatively weakly compact subset of Y (where en denotes the sequence whose nth entry is one and every other is zero).

Solution: Let T : 11 - Y be a bounded operator. Put A = {T(en) : n = 1, 2,...l and U = U. Clearly, A C T(U). So, if T is weakly compact, then obviously A is a relatively weakly compact subset of Y. For the converse, assume that A is a relatively weakly compact subset of Y. By the Krein-Smulian theorem (see, for instance, (6. Theorem 10.15, p. 158]) the

closed convex circled hull C of A is also a weakly compact subset of Y. Now assume that x = (XI, x2, ...) E 11 satisfies F_,°__1 1x1) < 1, i.e., x E U. For each

is let u, = F11 x;T(e;) = T(E, 1 x;e;). Clearly, un E C for each n and from u -+ Tx, we see that Tx E C. Thus, T(U)C C, proving that T is a weakly

I

compact operator.

Problem 2.4.12. For a Banach apace X prove the following. (a) An operator T : X tj is weakly compact if and only if T is compact.

(b) An operator T : co -' Y is weakly compact if and only if T is compact. Solution: For this problem we must use the fact that 11 has the Schur property. That is: A subset of 11 is relatively weakly compact if and only if it is norm totally bounded; see, for instance, [6, Theorem 13.1, p. 2001.

(a) This follows immediately from the fact that t1 has the Schur property.

(b) Notice that: T : co - Y is weakly compact

T * : Y' - co =11 is weak ly compact

e=: T': Y* -1

1

is comp act

T : co - Y is compact. So, T is weakly compact if and only if it is compact.

Problem 2.4.13. Let {yl, y2, ...} be a countable relatively weakly compact subset of a Banach space Y; in particular, this is so if {y1,} is a weakly convergent sequence in Y. Show that the operator T : i'1 -, Y, defined by 00

T(A1, A2, ...) =

Xn1/n , n=1

2.4. Compact and Weakly Compact Positive Operators

85

is weakly compact.

Solution: Notice first that the set {y', Y2 ...} (as being relatively weakly compact) is norm bounded. This guarantees that the series En is norm convergent in the Banach space X for each sequence A = (A 1, A2, ...) E f i . So, T is a well-defined bounded linear operator. Now invoke Problem 2.4.11 to conclude that T is a weakly compact operator. I

Chapter 3

Operators on AL- and AM-spaces

3.1. AL- and AM-spaces Problem 3.1.1. A vector u in a subset A of a vector space X is called an internal point of A if for each x E X there exists some Ao > 0 such that u+AxEA for all JAI
Solution: Assume first that e is an internal point of X+ and let x E X. Then there exists some a > 0 such that e + a(-x) > 0. This implies x < oe, so that e is an order unit. Conversely, suppose that e is an order unit and let x E X. Fix some A0 > 0 such that Aox < e. Now notice that for each 0 < A < A0 we have

Ax = (o)Aox5A0 -e<e. So, e - Ax > 0 for all 0 < A < Ao. Similarly, replacing x by -x, there exists some Al > 0 such that e + Ax > 0 for all 0 < A < A1. These show that e is an internal point of X+.

Problem 3.1.2. If E is an AM-space with a unit e, then show that the formula IIXII- = inf{A > 0: 1xI < Ae} = min{A > 0: IxI < Ae}

(*)

defines an equivalent lattice norm on E. Under this norm E remains an AM-space and its closed unit ball coincides with the order interval [-e, e]. 87

3. Operators on AL- and AM-spaces

88

Solution: Let us verify that

II

- I1x

is a norm. Since e is a unit, the set {A >

0: lxi < Ae} is non-empty for each x E E and so the function 11- II,. is well defined.

Clearly. IIxll,, > 0 and 110II,o = 0. If IIxII. = 0. then IxI < ne for each n. This implies that x = 0. For the triangle inequality note that if lxi < Ae and I yI < µe, then Ix+yl < IxI +1Y1 < (A+µ)e. and so IIx+y11, < A+u. This implies IIx+yI1x < 11x11x+11y11C.' The homogeneity IDAxll. = JAI - 11x1lx is obvious.

Since x E [-e, e] is equivalent to lx) _< e, it follows that every x in [-c. e] satisfies IIxIIx < 1, that is. I-e,el C B = {x E E: 11x11, < 1}. Now assume that x E B. This implies that for each n we have IxI < (1 + 1)e, whence IxI < e. That is, x E (-e, e.], and consequently B = [-e. e]. This equality implies immediately that for arbitrary positive elements x. V E B their supremum x V y also belongs to B. In particular, it follows that 11 I1 x is an Al-norm. It remains to show that the (original) norm II . II of E is equivalent to II IIxFrom Ixl < 1lxllxe, we see that -

Ilxll < 11e11.11xIix

for each x E E. Since (E, 11 11) is a Banach lattice, Theorem 1.31 guarantees that the identity operator I: (E, 11- II) - (E, 11 - IIx) is continuous. That is, there exists some constant c > 0 satisfying IIxII ? cllxll,, for each x E E. Consequently, cllxllx < Ilxll < 11ell - Ilxll,, holds for each x E E, and so the norms Il are equivalent.

.

II and 11

IIx U

Problem 3.1.3. If E is an AM-space with a unit e, F is an arbitrary Banach lattice and T : E -. F is a positive operator, then IITII = IITell. Solution: The positivity of T implies ITxl < TIxi for each x E E. Now notice that UE = [-e, e], i.e., x E UE if and only if lx1 < e. This implies IlTxll = IIlTx1II < IlTell 0 for each x E UE. Therefore, 11TI1 = SUPXEUe IITxII = IITeil.

Problem 3.1.4. Show that a Banach lattice E is an AL-space (resp. an AM-space) if and only if its norm dual E' is an AM-space (resp. an ALspace). Moreover, establish the following:

(a) If E is an AL space, then E' is a Dedekind complete AM-space with unit e', where e*(x) = IIxII - IIx-II for each x E E. (b) If E is an AM-space with unit e. then E** is also an AM-space with unit e. Solution: We show first that if E is an AL-space, then E' is an AM-space. To this end, assume that x' A y' = 0 in E*. Put m = max{llx`Ii, Ily'll}, and note that m < 11x' + y* 11. Fix e > 0 and then select some x E E+ with IIxII = 1 and 11x' +y'11 < (x'+y')(x)+e. From [x' Ay'j(x) = 0, it follows that there exist u,v in E+ satisfying u+v = x and x'(u)+y'(v) < e. Next, from (u-uAv)A(v-uAv) = 0, 0 < u + v - 2(u A v) < x, and the fact that E is an AL-space we get Ilu-vAuli+11v'-vAui!=1Iu+v-2(uAv+)II
89

3.1. AL- and AM-spaces

Therefore,

x'(x)+y'(x)+E =x'(V)+y'(u)+x (u)+y (v)+E

<

IIx' + y' ll

x'(v) + y'(u) +2c < x'(v - v A u) + y'(u - v A U)+&

m(lly-vnull+Ilu-vnull)+3E m+3E. Since e > 0 is arbitrary, it follows that IIx' + y' II <- m, and hence

Ilx'Vy'11=IIx'+y'11=max{Ilx'II,IIy II}, proving that E' is an AM-space. In order to establish that the norm dual of an AM-space is an AL-space, we need the following property.

(P) If E is a Banach lattice and x' A y' = 0 holds in E', then for each e > 0 them exist x, y E E such that IIxfi < 1, IlyII < 1, x n y = 0, and

lix'Il <x'(x)+E and IIy'II <-y'(y)+E. To see this, fix f > 0 put 6 = 3, and then pick two vectors u, v E UE satisfying <- x' (u) + 6 and 11y* 11 < y' (v) + 6. From [x' A y'I(u) = 0, we see that there exist ul, u2 E E+ with ul + u2 = u and x'(ul) + y'(u2) < 6. Similarly, there exist vi, v2 E UE satisfying v i + v2 = v and x' (vl) + y' (v2) < 6. Now consider the vectors 11x* 11

x=u2-vlAu2 andy=vl-vlnu2. Clearly, x,yEUE,xAy=0, and

= x'(u2)-x'(vi nut) !x'(u2)-6=x'(u)-x'(ul)-6

x'(x)

(11x'11-6)-6-6=Ilx'II-36 11x* 11 -'F

Similarly. y' (y) > 11 y* II - E

Next, we shall show that the norm dual of an AM-space is an AL-space. To this end, let E be an AM-space and let x' n y' = 0 hold in E'. Fix c > 0 and then use property (P) to choose two vectors x, y E UE satisfying

xny=0.

Ilx'll<-x'(x)+E,

and Ily'II<-b (y)+E.

Now note that IIx' + y' 11

<

Ilx' Il + lly' 11

< x'(x+y)+y'(x+y)+2E<-II= IIx'+y'il.ilxvyll+2E
llx'll+lly'll+2E.

Since e > 0 is arbitrary, we see that lix' + y' 11 = IIx' lI + ily' II. This shows that E' is an AL-space.

Next, note that if E' is an AL-space, then by the above conclusions E" is an AM-space, and hence E (as a closed subspace of E") is also an AM-space. Similarly, if E' is an AM-space, then E is an AL-space. Now we shall verify the claims (a) and (b). Assume first that E is an ALspace. So, the formula e'(x) = IIx+ii - IIx-11 defines a positive linear functional e*: E -, R. For this you must use Problems 3.1.7 (listed below) and 1.2.2. Clearly,

3. Operators on AL- and AM-spaces

90

[-e'.e'] C C'. Now let :r' E U'. Then Ir'I E L". and so for each x E E- we have Ix'I(x) < I1f.r'i . Ilxll < Slrll = c'(x) or Ix'I < c'. Thus. -e' < r' < e'. and Se) U' C [-e'. e']. Consequently. U' = [-e'. eel proving that E' is an AAI-space with unit e'. Finally. let E be an A,11-space with unit e. Put

Q-e,e] = {x" E E": - e <s" <e} and note that --e,e] C We. Now assume x." E L"'. Then Ix"I E U", and for each 0 < r' E E' we have lx..I(x.) < IIx..II e(x') - IIx'II < IIx'II = Therefore. Ix" I < e or x" E (-e. e]. This shows that U" C Q-e. e] is also true, and so U" = Q-e. e.]. Consequently. E" is also an AM-space with unit e.

Problem 3.1.5. Let E be a Banach lattice and let E., be the principal ideal generated by a vector x E E. Show that E. under the norm. Ilyllx = inf{A > 0: lyl < a[xf }

is an AM-space with unit jxf. Solution: Let x E E. The verification of the fact that . IJ X is an 111-norm on E= is straightforward and is left to the reader. We shall show that - IIx is a II

II

complete norm. To this end, notice first that flu - vllx < c in E= is equivalent to lu - vl < clxl. In particular. we have

UEr={yEE=: Ilyllx<1}={yEE.: lyl<-Ixl}=[-Ixl.lxl] To establish that the lattice norm II Ilx is complete, it suffices (according to Problem 1.2.16) to show that every monotone II I1,,,- Cauchy sequence in E., is Il , -convergent in E. So, let {un} be a II Jl,o Cauchy sequence in E. satisfying -

II

0 < u,a

T.

Fix e > 0, and then select some no such that IIu - u,nllx < e or

Iun -u,,, l < clxl for all n. rn > nfl. This implies Ilun - um II < clixll for all n, in > no. In particular, {un } is a Cauchy sequence with respect to the original norm II - II on E. Since E is I1- 11-complete. there exists some u r= E such that Ilu,, - ull -, 0. Now

a glance at Problem 1.2.12 guarantees that u,, j it in E. Since 0 < u,,, - un < clxl holds for all in > it > no, it follows that 0 < u - u,, < flxl for all n > no. From, 0 no. This shows that E. is l(- II,,-complete, and so E. under the norm II

Ilx. is an A111-space with unit lxl.

I

Problem 3.1.6. Show that every AL-space has order continuous norm. Solution: We know that every L1(µ.)-space has order continuous norm. Now if E is an AL-space. then (by Theorem 3.5) E is lattice. isometric to some L1(p)-space, and hence it must have order continuous norm. I

Problem 3.1.7. Show that a Banach lattice E is: (a) an AL-space if and only if IIx + yll = IIxII + Ilyll for x. y E E" and (b) an AM-space if and only if IlxVyll = max{Ilx[l. Ilyil} for x. y E E+.

3.1. AL- and AM-spaces

91

Solution: (a) If IIx + vii = IIxII + Ilyil holds for all x. y E E+, then E is clearly an AL-space. Now assume that E is an AL-space. Then (by Theorem 3.5) E is lattice isometric to some Ll (p)-space. Now notice that if 0 < x, y E Ll (p), then

IIx+ylli= f(x + y)djz =

fxdz + Jydi.L=ilxlll+Ilylli.

(b) If IIx V yll = IIxII + Ilyll for all x, y E E+. then E is clearly an AM-space. For the converse, assume that E is an AlMM-space. Then (by Theorem 3.6) E is lattice isometric to a closed Riesz subspace of some C, (Q)-space. Now notice that if 0 < x.y E C(fl), then [x V y](w) = x(w) V y(w) < max{IIxIloo, IIxII,.} for each w E Q. Therefore, IIx V yllx = sup [x V y](w) < max{ IIxII IIyIIx} < IIx V yllx, wEll

a

and so IIx V yllx = max{ IIxII<. Ilyllx } for all 0 < x, y E C(sl).

Problem 3.1.8. Show that the lattice operations in any AM-space are weakly sequentially continuous. Solution: Let E be an AM-space. By Theorem 3.6. we can assume that E is a closed subspace of a C(fl)-space for some compact Hausdorff space Q. By the Riesz

Representation Theorem, we know that a sequence (fn) in C(n) satisfies fn Jv f if and only if {fn} is norm bounded and fn (w) - f(w) for each w E Q. Therefore, if fn - f in C(fl), then Ifnl - If I also holds true in C(fl). Since fn - + f in E if and only if f,, JL f in C(fl), it follows that E has weakly sequentially continuous lattice operations. I

Problem 3.1.9. Show that the lattice operations in an AL-space need not be weakly sequentially continuous.

Solution: We shall show that the AL-space E = L1[0, 1] does not have weakly sequentially continuous lattice operations. Keep in mind that E' = L,o[0,1]. For each n consider the partition

{0.2-n. 2.2-".3.2'", ... (2n - 1) . 2-n, l } of the interval [0, 1] and define the nth- Rademacher function rn: [0, 1)

R by

ra(t) = (-1)k-1 for (k - 1)2-" < t < k2-" (k = 1.2.... , 2n) . The graphs of rl and r2 are shown in Figure 1. We claim that if f : [0,1]

If is

a Lebesgue integrable function, then lim

f

n-+oo 0

l rn(t)f (t) dA(t) = 0.

That is, r,, - 0 in E.1 Since Irnl = 1 -w-+ 1 also holds in E, this will establish that E does not have weakly sequentially continuous lattice operations. To show that limn_,, f0 ra(t) f (t) da(t) = 0 for each Lebesgue integrable func-

tion f, it suffices to establish the claim for the case f = X(a.b), where [a,b) is a 'This is the famous Riemann-Lebesgue lemma from the theory of integration. completeness we present its proof.

For

3. Operators on AL- and AM-spaces

92

dA(t) = fo it is enough to show that lim far(t) dt = 0 for each 0 < a < b<

dt. Therefore,

subinterval of (0, 1J. (Why?) Clearly. fo

yI

y =r1(t)

pl

1t0

1.

y=r2(t)

1 t---o -o

t

:2

-1 +

i----o

-14

:--o

:-o

Figure 1. The graphs of rl and r2 To this end, fix 0 < a < b !S 1 and e > 0. Fix no such that 2-no < mint C. 4 Next, choose any n > no and consider the part ition {0, 2!, . - .... , 22- 1.1 }; for simplicity, let to = 2 and note that the points a and b are related to the to as shown in Figure 2. Since for any three consecutive points t,-1. t;, t;+1 we have dt, where dt = 0, we conclude that fo r a(t) dt = f dt + f e = t,,,-1 or c = t.,,; see Figure 2.

tk-1 a

tk tk+1

tm-1 b tm tm+l

t

Figure 2 Consequently, for each n > no we have J0brn(t) l

dtl

<

Jn

r(t)l dt +

Jc

Jr(t)I dt

= (tk-a)+(b-c)<e+2e=3e. This means that lim

dt = 0, as desired. (For a different solution seethe

proof of Lemma 11.56.)

Problem 3.1.10. Let E be a Banach lattice. For X E E+ put

[-x, xl = {y E E: -x< y < x } and -x
Establish the following:

3.1. AL- and AM-spaces

93

(a) [-x. x] and Q-x. xD are both convex sets.

(b) [-x, x] is a(E, E')-closed and Q-x, xD is a(E", E')-compact. (c) [-x, x] is a(E", E')-dense in Q-x, xU. Solution: (a) Obvious. (b) This follows from the fact that the cones of E and E" are both weakly closed; see Problem 1.2.12.

(c) Assume by way of contradiction that [-x, x] is not a(E". E')-dense in [-x, x]. By the Separation Theorem, there exist some x' E E' and some y" in [-x.xj such that x'(y) = 0 for each y E [-x,x] and x'(y") 36 0. It follows that

Ix'1(x)=sup{x'(y): y E E and -x
Ix'I(x) = 0, and so x'(y") = 0,

which is impossible. Hence, [-.r, x] is a(E". E')-dense in [-x, x].

Problem 3.1.11. Show that every regular operator from an AM-space with unit to any lp-space, where 1 < p < oo, is compact.

Solution: Let T : E

tp be a positive operator, where E is an AM-space with unit a and 1 < p < oo. If U = [-e, e] is the unit ball of E, then T(U) C [-Te,Te]. To see that T is compact, notice that the order intervals in 1p (for 1 < p < oo) are norm compact. Hence, every regular operator (which is the difference of two positive operators) from E to 1p with 1 < p < oo is compact.

Problem 3.1.12. Let E be an AM-space. Extend the norm of E to its Dedekind completion E6 as in Problem 1.2.21. Show that the Banach lattice

E6 is also an AM-space. If E is an AM-space with unit e, then E6 is also an AM-space with unit e. Solution: Let E be an AM-space. We shall show that the standard extension of the norm 11 - 11 of E to the Dedekind completion E6 is also an M-norm.

To this end, let 0 < 1, y E E. Choose two sequences of {xn} and {yn} of E satisfying i < xn and y < yn for each n, IIxnII - 11ill, and Uyn1I -,11fl. Since E is an AM-space, it follows from Problem 3.1.7 that max{II±II, Il II} < 11i V yII <

IIx,, V ynll = max{IIxnII IIynUI} - max{11111, 11011}

Hence, IIi v uII = max{ hill, 11011 }, and so E6 is an AM-space.

Now assume that E is an AM-space with unit e. Let us denote the closed unit balls of E and E6 by U and U6. respectively. Also let

[-e.e]={yEE: -e
3. Operators on AL- and AM-spaces

94

into account that [-e, el is a norm dosed subset of Ed (see Problem 1.2.12), we get -e < i = limt1 Ei < e. Hence, U6 C 1-e, el. Consequently, U6 = I-e, el, and so E6 is an AM-space with unit e too. I

Problem 3.1.13. Assume that S and fl are two compact metric spaces, p is an arbitrary (finite) Borel measure on S, and that K : S x (Z -' IR C(l) denote the integral is a continuous function. Let also T : C(S) operator with kernel K, i.e., T is defined by

Ta(w) = is

w)x(s) dp(s)

for al l x E C(S) and each w E fl. Establish the following.

(a) T is a compact operator.

(b) The modulus ITI: C(S) - C(Il) of the operator T (which exists according to Theorem 3.14) is also an integral operator satisfying IT[a(w) = fsIK(s,w)Ix(s)dp(s) for all x E C(S) and all w E fl. (c) The norm of T is given by IITII,.= sup{IITxQI,,: IIxII. < 1} = sup IS, Solution: (a) Let d and p be the metrics on S and f2, respectively, and let e > 0. Since K is a uniformly continuous function, there exists some 6 > 0 such that d(si,s2) +P(w1,w2) < b implies CK(sj,wl) - K(82, U)2)1 < e. Now if x E C(S) satisfies IIxjj,o < 1 and w1iw2 in ft satisfy p(wl,w2) < a, then

ITx(w1) -Tx(w2)I < j IK(s,w1) s

-

dp(s) < E/(S)

This shows that the norm bounded set {Tx: llxjloo < 1) is an equicontinuous subset of C(1l), and hence it is a norm totally bounded set. This proves that T is a compact operator.

(b) Fix 0 < x E C(S) and w E fl. Clearly, `TIx(w) < fs JK(s,w)Jx(s)dp(s). w), i.e., a(s) = 1 if K(s,w) > 0 and a(s) = -1 if K(s,w) < 0, and then choose a sequence {xn} C(S) such that xn(s) -. a(s) Denote by o the sign function of

for p-almost all s. Replacing (if necessary) each function xn by [x, V (-1)] A 1, we can assume that J1x,,jI,,, < 1 for each n. Since jx,,(s)x(s)I < x(s) for each s E S and all n, it follows from the Lebesgue Dominated Convergence Theorem that

fs

I K(s, w) Ix(s) dp(s) = is K(s, w)a(s)x(s) dp(s) s

= n-oo K(s,w)xn(s)x(s)dp(s) urn s is <

[TIx(w).

Hence, ITIx(w) = f5I K(s,w)Ix(s)dp(a) holds for each w E 0. (c) Clearly, }JTIt,o < sup,,,En fsjK(s,w)I dp(s). For the converse inequality pick some wo E f2 such that f5IK(s,wo)j dp(s) = suP,,En fsIK(s,w)I dp(s). Again,

3.1. AL- and AM-spaces

95

denote by a the sign function of K(., wo), and then choose a sequence {xn} C C(S) satisfying Ilxnll,o <_ 1 for each n and x,, (s) -+ a(s) for p-almost alls. Now use the Lebesgue Dominated Convergence Theorem once more to see that

is

K(s,wo)I dp(s)

= =

Is

K(s, wo)a(s) dp(s)

lim

IK(s, wo)xn(S)diz(S)

s IITI I . Hence, IITIIoo = max.,EC1 fS I K(s, w) I dp(s)-

Problem 3.1.14. Show that every Markov operator carries order units to order units. Solution: Let T: E be a Markov operator between two AM-spaces with units u and v, respectively. That is, T is positive and Tu = v. Let e by an arbitrary order unit in E. Then there exists some A > 0 satisfying u < Ae, and so (by the positivity of T) we have v = Tu < ATe. Since v is an order unit in F, this implies that Te is also an order unit in F. I

Problem 3.1.15. Every Euclidean space Rn can be considered either as an AM-space with unit e = (1,1, ... .., or as an AL-space under the usual norms n

Ilxil = man Ixil

--

and

Ilxlli = > Ixtl , 1=1

respectively. An m x n matrix A = [ash] with non-negative real entries is called Markov (resp. stochastic) if E 1 aid = 1 for each 1 <_ i < m (resp.

Em1atj=1 for each 1 < j < n). Now let A be an m x n matrix with non-negative entries and consider A as an operator from Rn to Rm. Show the following: (a) The positive operator A: (Rn,11 . (R'n, it ll,o) is a Markov operator if and only if the matrix A is a Markov matrix. (b) The positive operator A: (Rn, it - 111) --' (Rm,11 - 111) is a stochastic operator if and only if the matrix A is a stochastic matrix.

Solution: (a) Notice that Ae = (En=1 all,

En= a23,..., E 1 a,,,.). Thus, the

positive operator A: (Rn, II . II,o) - (Rm, II is a Markov operator if and only if Ae = e or if and only if air = 1 for each 15 i < m. Clearly, this is equivalent to saying that A is a Markov matrix.

(b) Notice that for each x E R+, we have

mn

n

m

IIAxIIi =EEaijxj =E[Ea`,]xl. i=1 j=1

j=1 i=1

(*)

3. Operators on AL- and AM-spaces

96

Now if A: (R", II III) - (R', II . III) defines a stochastic operator, then we have

IIAxIII =IIxIII for each x E R'. Letting x = ej, the ph unit vector in R", we see that Em, a;., = 1 holds for all 1 < i < n, i.e., A is a stochastic matrix. For the converse, assume that A is a stochastic matrix. Then it follows from (*) that IIAxIII = IIxIII for all x E R'. That is, A defines a stochastic operator.

3.2. Complex Banach Lattices Problem 3.2.1. If E is a Banach lattice and z = x + ty E Ec, then show that

IzI = sup Ixcos9+ ysinOl =suplxcos9+ysin9l. OER

9E (0,2a]

Solution: Notice that for each angle 0 E R there exists an angle 9 E [0, 27r] such that cos 0 = cos 9 and sin 0 = sin 9. This implies

IzI = sup

[x cos 9 + y sin 9] = sup [x cos 9 + y sin 0] .

9E]0.2w)

OER

If the suprema below exist, then clearly sup Ix cos 0 + y sin 0l = sup lx cos 0 + y sin 0l . GE10,2x]

OER

Now for each 0 E [0.2a] we have

xcos9+ ysin9 < lxcos9+ ysin9l = [xcos0+ysinB] V [-xcosB - ysin9] = [x cos 0 + y sin 9] V [x cos(a + 9) + y sin(a + 9)] < I z I , and consequently

IzI = sup Ixcos9+ysin9l = suplxcos9+ysinol, BE 10,2,x]

9ER

as claimed.

Problem 3.2.2. If E is a Banach lattice, then for arbitrary z, zl, z2 E Ec and A E C show that:

(a) IzI > 0, and lzl = 0 if and only if z = 0. (b) IzI

IzI I + IZ21.

Use the above properties to verify that the function II - 11c: Ec - R, defined by Ilzllc = IIIzIII, is a norm. Solution: (a) We proved in Problem 3.2.1 that for each z = x + ay E Ec we have IzI = SUPOER Ix cos 9 + y sin 9I . This implies IzI > 0.

If z = 0, i.e., if x = y = 0, then IzI = 0 is trivially true. On the other hand, if I z I = 0, then Ix cos 9 + y sin 01 = 0 for each angle 9. Letting 0 = 0 and 0 = z , we get IxI = Iyi = 0 or x = y = 0. Therefore, IzI = 0 implies z = 0.

97

3.2. Complex Banach Lattices

Now notice that I l z l l c= I l l z l l l> 0 for each z E E. Moreover, Ilzllc = III=III = 0

z = 0.

IZI = 0

(b) Let A = 0+13 E C and z = x+zy E E. Then Az = (ax-3y)+z(Qx+ay). Also, we can assume that x. y belong to C(a), where (1 is some compact Hausdorff space. Therefore, Iazl

= = I (ax - oy) + z(,3x + ay) I = (a2 + 32)(x2 + y2) =

(ax - ,3y)2

a2 _+02

(,3r + ay)2

x2 -+y2

= IN H IZI For the homogeneity of II - llc note that: Ilazllc = IIIAzlII = II1A1. kill = IN-11141 = IAl IlzlIc.

(c) Let z1 = x1 + tyl and z2 = x2 + zy2. Then z1 + z2 = xl + z2 + z(y1 + y2), and so for each angle 0 we have

(xl+x2)cos0+(yl+y2)sin0= (x1cos0+ylsinO)+(x2cos0+y2sin9) < Izll+1221 This implies I Z1 + z21 = supOER [(X1 + x2) cos 0 + (y1 + y2) sin 0] < IZI I + I Z21- Thus, llzl + z211c =

Illzl + Z2111 <- II

Iz11 + Iz2111 <- IIIz1lhi + IlIz2111= IIz1IIc + Ilz2llc

I

holds for all zl, z2 E Ec.

Problem 3.2.3. Show that the complerification of CR(Sl) coincides with CE(1l). Moreover, show that for each pair of functions f,g E CR(n) we have If + zgI = f2 + g2 and hence Il f + igllc = II ./2 + 92I1. Solution: Recall that Cc(1) is the complex vector space consisting of all continuous complex-valued functions defined on 0. Now if f E Cc(f1), then there exist two (uniquely determined) functions Ref and Imf in CR(S2) such that f(w) = (Ref)(w) + z(Imf)(w) for each w E Q. This means that we can define a mapping T: CC(I) -e (CR(fl))c via the formula

Tf =Ref+zImf. It is a routine matter to verify that T is a surjective isometry that leaves CR(S2) invariant and such that IT f I = Ti f I for all f E Cc(Il). By means of this isometry we can identify Cc(I) with (CR(f ))c. The remaining properties follow from this identification.

Problem 3.2.4. Show that the complexzfication of R' coincides with C". Also, show that all norms on C" (or R") are equivalent. Solution: From Problem 3.2.3, it follows that (R")c = Cc({1,2,...,n}) = C". We shall prove that all norms on C" are equivalent. Let us denote by II . Ill the usual tl-norm, i.e., llzlll = Ek=1 izkl for each z = (zl,z2,...,z") E C". An easy argument shows that a sequence {zk}, where zk = (zi . x2, ... , zn), satisfies Ilzk - 2111 -. 0 if and only if z;` -+ z, holds in C for each 1 < i < n.

3. Operators on AL- and AM-spaces

98

C'. Assume that a sequence {zk} in C"

Now let II 1{ be an arbitrary norm

satisfies IIzk - zll1 -' 0. From n

2k =

k) _ [1

(Z1.ZZ,...,Zn k k

Zk

i=1

where e, is the standard i`h unit vector, the preceding remark and the inequality IIzk - all

n

-

Iz' 1=1

we see that flak - z[1 --+ 0. This shows that the identity I: (C",11 III) -. (C",11.11) is continuous. In particular, the II 111-compact set K = {z E C": II zll 1 = 11 is also a II 11-compact subset of C". The continuity of the norm function z [1zII guarantees that m = minZEK IIzII > 0 and M = max,EK IIzII < so. Now using the II, we obtain that mIIzIIi < IIzII <- IIzI1i for all z E C". identity IIzII = I1zIl1 . , 11, This proves that 11 II is equivalent to 11 111. Thus, all norms on C" are equivalent. I 11

Problem 3.2.5. Show that the complexification of the real Lp(µ) coincides with the complex Lp(p). Also, show that for any pair of real functions fig in Lp(i) we have If + tgI = /f2 + g2 and that Ilf + zgllc = II V1rf_2 + g2 II,. Solution: The solution is similar to that of Problem 3.2.3.

1

Problem 3.2.6. Recall that II .III- and II IIoo-norms on C" are defined via the formulas n

IIzIIi =

Izil

and

IIzIIx = max Izil.

i=i

If A = (aid] is an n x n matrix, then the II III- and the II 11"C-norms of A are defined by IIAIII = sup IIAzI11

and

IIAII,,. = sup IIAzIIx. Ilzllx<_I

11x11151

Show that: n

IJAlli = Ima n

n

Iaiil

and

I[AII> =

E Ia;I .

Solution: Let A = [a,.] be an n x n matrix with complex (or real) entries. Before proceeding with the solution, we need to recall a few definitions. If a = a + to is a complex number, then its conjugate is the complex number a = a - i;3. We have as = lal2. The conjugate of the matrix A is the matrix A' = A'. where A = ]a;;] and B' denotes the transpose of any matrix B = [b;,] and is defined by B` _ [bj,]. In C" the dot product of two vectors x and y is defined by (x, y)

:-1

3.2. Complex Banach Lattices

99

If A is any n x n matrix, then for all x, y E Cn we have the "duality" identity (Ax, y) = (x, A*y) = (x, Aty)

(*)

Moreover, if y is an arbitrary vector in Cn, then and

IIyIIi = sup I(y,x)I II=IIoo51

INIT.= sup I(y,x)I

(**)

II=il1<1

Let us establish these formulas. Consider the 11-norm of y first. If x E Cn satisfies En , then Ilyil=IIyll1.Onthe Iv <E 1IvIIzIsE other hand, if for each i we choose some angle O; such that y;e'0' = Iyi I (where t is the imaginary unit!), then the vector x = (e'*1, ... l e"'-) satisfies IIxlIoo = 1 and < 1, then l(y,x)I = E I Iv I = IIyII1. Now let us look at the 1,-norm of y. If:5IIxIIi we have I(y,x)I = IE Iv,zI 5 En I Iy.ll=.l <_ Ilvlloo[E" 1IxI] IIyII.. Choose some k with Iykl = Ilylloo, and then pick an angle 0 with yke'* = Iyki. If we let x = e'mek, then IIxIi1 = 1 and I(y,x)I = Iykl = Ilylloo We now return to our problem. For any n x n matrix B = [b;,] let 1

n

r(B) =

lm a c

n

and

Ib;; I

o(B) = Im ax r (a; I .

Clearly,

o(B) = r(Bt) = r(B*)

and r(B) = o(Bt) = o(B*). First, we shall establish the formula for the norm IIAIIoo If llzIlao 5 1 and y = Az, then n

n

n

lye l = I a;z, l < t=1

Ia;IIz;I <_ E Ia.;I <_ o(A) i=1

1=1

for each i, and so IIAIIoo <_ o(A). Next, choose some natural number m such that E I Ian; I = o(A) and then for each j select an angle 9; satisfying anje'B° = ia*n, 1. Clearly, the vector z = eie") satisfies IIzIIoo = 1. Now, if we let y = Az, then y n = Ei=1 a.nie"., = E; 1 la.,I = a(A), proving that IIAIIoo = o(A). We now compute the 11-norm of A. Using (*) and (**), we see that IIAIII

sup IIAzII1 = sup

=

114<1

sup

11-011 :51

[

II=II.51

sup

114-:51

sup

[

I(Az,x)I ]

II=IIe <1

sup I (Az, x) I ] IIzfl151

[

sup

I

(z, A*x) I ]

112111 <1

II=IIiPi

II= 51 o(A*) = r(A),

I(A x+z)l] =

IIyIIuP

1

IIA xlloo

and the proof is finished.

Problem 3.2.7. Let E be a real Banach lattice with complezsfication Ec. A vector subspace J of Ec is said to be an ideal whenever Izil 5 Iz2I and z2 E J imply z1 E J.

3. Operators on AL- and AM-spaces

100

(a) Show that the ideals in Ec are precisely the vector subspaces of Ec of the form J = Jo ED U0, where Jo is an ideal in E.

(b) A bounded operator T: F - F on a (real or complex) Banach lattice is called ideal irreducible if there is no non-trivial closed ideal in F which is T-invariant. That is, T is ideal irreducible if and only if T (J) C J and J a closed ideal in F imply either J = {0}

orJ=F.

If T : E E is an operator on a real Banach lattice, then show that T is ideal irreducible if and only if Tc : Ec -+ Ec is likewise ideal irreducible.

Solution: (a) Assume that Jo is an ideal of E, and let J = Jo ® tJo. Clearly, J is a vector subspace of E.. Now assume that Ixi + tyt I < Ix + tyl and that x + ty E J. From Ixi l <_ Ixi + tyl l <- Ix + tyl <_ IxI + Iyl E Jo,

it follows that xl E J0, and similarly yl E Jo. Hence, xt + tyt E J, and so J is an ideal in E. For the converse, assume that J is an ideal in E.. Let Jo = { x E E : There exists some y E E with x + ty E J) .

Clearly, Jo is a vector subspace of E. Now if x + ty E J, then from the equality

y + t(-x) = (-t)(x + ty) E J it follows that y E Jo. That is, x + ty E J implies x, y E Jo or J C Jo®tJo. Moreover, if x E Jo, then pick some y E E with x+ty E J,

and then note that Ix + tOl = IxI 5 Ix + tyl, which implies x = x + t0 E J. In particular, x E Jo implies tx E J. Thus, if x, y E Jo, then x + ty E J, and this shows that Jo ® tJo g J. Therefore, J = Jo ® tJo. (b) Start by observing that an ideal J = Jo ®zJo in E. is closed if and only if J0 is a closed ideal in E. Next, assume that T is ideal irreducible, and let J = Jo ®iJo be a closed Ta-invariant ideal in E. If x E Jo, then x + tO E J, and consequently

Tx + t0 = Tc(x + t0) E T(J) C J = Jo ®tJo. This implies Tx E Jo and so Jo is a closed T-invariant ideal in E. Since T is irreducible, either Jo = {0} or Jo = E. This implies either J = {0} or J = E., i.e., T,, is an irreducible operator. Finally, for the converse assume that T. is an irreducible operator, and let J0 be a closed T-invariant ideal in E. Then J = Jo ® zJo is a closed ideal in EE, and from

TT(J) 9 T(Jo) ®tT(Jo) 9 Jo 9 tJo = J,

we infer that either J = {0} or J = Ec. This implies either Jo = {0} or Jo = E, i.e., T is an irreducible operator.

Problem 3.2.8. Show that all conclusions regarding domination of operators stated in the results 2.34-2.40 are also true for complex Banach lattices. Solution: Recall (from Problem 1.1.9) that if X. and Yc are the complexifications of two real vector spaces, then every operator T: X. - Y. can be represented by

3.2. Complex Banach Lattices

101

linear operators. Now assume that T : EE -' F, is an operator between two complex Banach lat-

T -STJ

tices having the matrix representation T = I S properties.

.

Then we have the following

L

(1) T is a compact operator (resp. weakly compact) if and only if S and T are both compact (resp. weakly compact) operators. (2) If T is dominated by a positive operator R, then S and T are both dominated by R.

For (1) we shall consider the case of compact operators only. Assume first that

T is a compact operator and let {xn} C U. If we put zn = xn + s0 E E. then {z-n} C UE, and so (by passing to a subsequence and relabeling if necessary) we can assume without loss of generality that {Tzn} converges in E.. Now from

TZn = IS

-Si [xnl = [Txnl TJ l 0 J

= Txn + 1Sxn

,

lSxnJ

we see that both sequences {Txn} and {Sxn} converge in F. This proves that S and T are both compact operators. For the converse, assume that both S and T are compact operators, and let {zn} C U&, where Zn = xn + zyn. Then T Txn - Syn $ [x,,l = Txn - Slln + t(Sxn +Tyn) Tzn = (*) T yn = Sxn +Tyn S Moreover, from llx II < 1Iznllc < 1, 1IynII < IIznIIc < 1, and the compactness of the operators S and T, by passing to appropriate subsequences and relabeling, we can assume that the four sequences {Txn}, {Tyn}, {Sxn}, and {Syn} are all norm

convergent in F. From (*), we see that {Tzn} converges in F, and so T is a compact operator. For (2) assume that T is dominated by a positive operator R. If x E E. then let z = x + zO and note that 1r1 T, I OJ ITx + zSxl < Rjzj = RjxI. ITzI = I

LS

This implies ISxI < Rixi and ITxI < Rixj for each x E E, so that R dominates both S and T. To show how one can generalize the results 2.34-2.40 to complex Banach latticers, we shhall prove only Corollary 2.35. To this end, let us assume that -STJ : E, -. E,, is an operator on a complex Banach lattice and T is T=IT S dominated by a compact positive operator R. A direct computation shows that

T

2

_

T3 _

T2-S2 -ST-TS

ST+TS

T2-S2

and

T3-TS2-S2T-STS S3-ST2-T2S-TST -S3+ST2+T2S+TST T3-TS2-S2T-STS

3. Operators on AL- and AM-spaces

102

By part (2) above, both S and T are dominated by the compact positive operator R. So. by the (real) Theorem 2.34, all entries of the matrix T3 are compact operators from E to E. By part (1) above, T3 is a compact operator. I

Problem 3.2.9 (Wickstead). Let E = R2 be equipped with the sup norm - 11. (i.e.. II (x, y) II. = max { lxl, lyl }) and let F = R2 be equipped with the it-norm II 111 (i.e.. II(x,y)II1 = Ixl + IyD). Now consider the operator T : E -- F defined by 11

-

T(x, y) = (x + y, x - y) -

Show that T is a non-positive regular operator satisfying the inequalities

IITIIc>2f >2=11TII. Solution: To see that T is a regular operator note that if T1. T2 : E -+ F are defined via the formulas T1 (x, y) = (x + y. x) and T2(x. y) = (0. y), then T1 and T2

are both positive operators and T = T1 - T2. (The regularity of T is. of course, automatic since here E is a finite dimensional space.) Now from IIT(x.y)II1

II(x+y.r-y)II1=Ix+yl+Ix-yI = = 2max{IxI.lyl} = 2II(r,y)IIx .

we see that IITII = 2. On the other hand, we claim that IITII,, ? 2v/2. To see this, consider the vector

= (1.1) + z(l.-1). Then IzI = (f, f), and so IIzII,'..,, = IIIzIIIL = f. F om Tc(z) = T(1,1) + zT(1. -1) = (2, 0) + t(0.2)

.

we get ITc(z)I = (2,2). This implies IITc(z)II1.,. = IIITc(z)III1 = 4. Therefore, IIT,,zhl1,,, = IIITczi II1= 4 = 2.r2_11,11....

which proves that IITII,, >- 2f > 2 = IITII.

U

Problem 3.2.10. Let H be a real Hilbert space with inner product Extending the inner product to He via the formula

("I-z2) = (x1+1y1,x2+Zy2) = ((xl,x2)+(yl,y2)I+z[(yl,x2)-('I,y2)f, (*) we turn He into a (complex) Hilbert space. Show that the standard norm II

- I1c on Hc defined by

I1zhIc = Ilx+tyllc =sup1lxcase+ysinOll OER

is not in general an inner product norm--and hence II with the norm induced by the above inner product.

. IIe

does not coincide

Solution: We leave it for the reader to verify that the formula (*) for (x1,22) indeed defines an inner product on If.

3.2. Complex Banach Lattices

103

Now we shall show that II - IIc does not satisfy the Parallelogram Law. To this end, consider the real Hilbert space H = L2[0,1], and let x = Xlo,l) and y = X[},11' Put u = x and v = ay. Then fI cos 9 + X(,.1l sing ]2 dt Ilu + vI12 _ sup J I X[o.1) 2 OER o sup [ cost 8 + z sin201 = 1

,

OER 2

fExIo,)coso_xL,lIsmo12dt Ilu - vii= Sup BER

e

SUP OER [ 2

cost g + 1 sin20]

29

1

fixiO,,)]2 dt = 2 ,

IIull Ilvll2

=

j[xt,ij]2dt = I .

Hence, Ilu + vIIC + Ilu - vlIC = 1 # 2 = 2(Ilu11C + IIvII,2) This shows that the norm 11. llc does not satisfy the Parallelogram Law and so it cannot be the norm obtained from an inner product.

Problem 3.2.11. Let X be a real Banach space and define N : X x X - R by:

N(x, y) = (IIx1I2 + IIyII2) 2

Establish the following:

(1) The real vector space X ® X normed with N is a Banach space. (2) The same function N (as defined above) now considered on the complexification X. = X ®zX, that is, N(x+zy) = (IIxII2+IIyII2)', is a norm on X,, if and only if X is a real Hilbert space. Solution: (1) Note that IIxII, Ilyll : N(x, y) and N(x, y) < IIxII + IIxII Hence, I (IIxII + Ilyll) <_ N(x, y) < IIxII + Ilyll

(**)

holds for all (x, y) E X ®X.

Let us verify that N is a norm on X ® X. Clearly, N(x, y) > 0 for each x ® y E X ® X, and from (,*) it follows that N(x, y) = 0 if and only if x = y = 0. For the homogeneity of N notice that if A E R, then N(A(x, y)) = N(Ax, Ay) = [ llaxll2 + Ilayli2 ] _ [ A2(IIxII2 + 11yI12) I1 = I AI N(x, y).

For the triangle inequality observe that N((xl, yl) + (x2, y2))

= N(xl + x2, y1 + y2) = [ IIxI + y1112 + IIx2 + y2112 ] [ (11x111 + Ilyl11)2 + (11x211 +

11Y211)2 11

(11x1112 + 11x2112) + (Ily1112 + Ilyzll2)

= N(xl, X2) + N(yl, y2).

3. Operators on AL- and AM-spaces

104

that a sequence {xn `, y,,) in X a) X Finally, notice that it follows from converges to x te y with respect to the norm N if and only if Ilxn - xll 0 and Ilyn - yll - 0. This easily implies that X A X equipped with the norm N is a Banach space.

(2) Assume first. that X is a real Hilbert space with inner product Then, by Problem 3.2.10, the complexification Xe of X is also a Hilbert space whose inner and is given by product is the extension of

(xl +Zyl,x2+Zy2) = In particular, if z = x + zy, then llzll =

(x1,y2)]-

(111112+I1y112)"

=N(x.y).

This shows that N is a norm on X. For the converse, assume that N is a norm on X, and let x, y E X be arbitrary vectors. Then (1+z)(x+zy) = (x-y)+z(x+y), and from the homogeneity of N we get N(x-y,x+y) = N((1+z)(x+zy)) = or (IIx-y112+IIx+y112)l _ 21(11x112 + Ilyll2)1 That is,

111-y112+I1z+y112 =2(IIXI12+Ilyll2),

for all x. y E X. This shows that the norm of X satisfies the Parallelogram Law. and so X is a (real) Hilbert space.

Problem 3.2.12. Let Ec be the complexification of a real Banach lattice E and assume that I z I< u for some z E Ec and it E E+. Show that for each c > 0 there exist positive vectors ul, ... , un in E and complex scalars cu,.. . , an such that Iail <_ 1 holds for each i = 1, .... n, n I ui = 1, and

z -F la;uiI <EU. Solution: Assume Izl < u and let e > 0. By the Kakutani-Bohnenblust Krein Representation Theorem 3.6, we identify the complex Banach lattice Eu zEu with some Cc(S))-space in such a way that u corresponds to the constant function one.

E Q such that Since z(f2) is a compact set, there exist points z(11) C U", D(z(wi), E). Consequently, if Vi = z'1(D(z(wi), t)), then each V, is a non-empty open subset of 0 and Sl = U,"__1 Vi. Next, pick a partition of unity {u1, 112, ... , un } subordinate to the open cover {V1, .... V,' I. That is. each u, is a continuous function from Sl to [0, 11, each u, vanishes outside Vi and E 1 ui = 1; see [7. Theorem 10.9, p. 85]. Clearly, Iui((&,) [z(w) - z(wi)] I < ui(w)e holds for each

i and all w E Q. Therefore, if ai = z(w.). then Tail < 1 (and ai E JR if z is a real-valued function) and for each w E St, we have tt

'z(w)

-E i=1

n

1E ui(w)[z(w) - z(w01 1=1

n

n

<_ I u,(w) [z (W) - z()] I < i=1

This shows that Iz - Lei=1 a.u.I < Eu holds in Ec.

ui(w)E = F. s=1

3.3. The Center of a Banach Lattice

105

3.3. The Center of a Banach Lattice Problem 3.3.1. Show that a positive operator T : E - E on a Rie8z space is an orthomorphism if and only if I + T is a lattice homomorphism. Solution: Let T : E -. E be a positive operator on a Riesz space. Assume first that I + T is a lattice homomnorphisin and let x A y = 0 in E. Then we have

0 < x A Ty < (x + Tx)

A (y

+ Ty) = [(I + T)x] A [(I + T)y]

= (I+T)(xny)=0. This implies x A Ty = 0, and so T is an orthomorphism. For the converse, we need the following lattice inequality: If u, v and w are positive vectors in a Riesz space. then

uA(v+w)
x<(v+u)A(v+w)=v+uAw. On the other hand, wehavex
xATy=TxAy=TxATy=0. and hence

0<

[(I + T)x] A [(I + T)y] = (x + Tx) A (y + Ty) < (x + Tx) A y + (x + Tx) A Ty

< xAy+TxAy+xATy+TxATy=O. Therefore, [(I +T)x] A [(I +T)y] = 0, and thus I + T is a lattice homomorphism.

Problem 3.3.2. Use Theorem 1.17 to establish that if E is a Dedekind complete Banach lattice, then the lattice operations of Z(E) are given by

(S V T)x = S(x) V T(x)

and

(S AT)x = S(x) AT(x)

for allS,TEZ(E) andallxEE+. Solution: Assume that E is a Dedekind complete Banach lattice. We need the following property: If S and T are central operators, then so are Ti. jS V T, and

SAT. To see this. pick some A > 0 such that jTxj < AIxj for each x E E, and note that for each u E E+ we have jTIu = supl1i<, ITxj < suplrl
SvT=

2(S+T+jT-Sj) and SAT=

z(S+T-jT-Sj).

Now let S. T E Z(E). We shall establish the formula for the infimum only-the

other formula follows from the identity SAT = S + T -SVT. Assume first that S and T are also positive operators. Notice that if y A z = 0, then y A Tz = 0

3. Operators on AL- and AM-spaces

106

and Sy A Tz = 0 hold, in which case we have Sy + Tz = Sy V Tz. Thus, using Theorem 1.17, for each x E E+ we have

(SVT)x = sup{Sy+Tz: yAz=O and y+z=x} = sup{SyVTz: yAz=O and y+z=x}

< SxvTx. Since (S V T)x > Sx V Tx is trivially true, we infer that (S V T)x = Sx V Tx. 2(S+T-SVT)x= z(Sx+Tx-SxvTx)=SxATx Consequently, (SVT)x= also holds for each x E E+. We now consider the general case. By the property stated at the beginning of

the proof, we know that 0 < S - SATE 2(E) and 0 < T -SATE 2(E). So, from (S - S A T) A (T - S A T) = 0 and the preceding case, it follows that for each x E+ we have 0

[(S - SAT) A (T- S A T)]x

(S-SAT)xA(T-SVT)x [Sx - (S AT)x] A [Tx - (S A T)x] Sx ATx - (S AT)x

I

or (S AT)x = Sx ATx.

Problem 3.3.3. Show that every central operator T : E -+ E on a Banach lattice is principal ideal preserving, i.e., show that T(Ex) C Ex holds for each x E E+. Solution: Let T : E - E be a central operator on a Banach lattice. Pick some A > 0 such that ITxl < Alxl holds for all x E E. Next, fix some u E E+, and assume y E E. Choose a scalar a > 0 such that lyl < au, and note that ITyl < Alyl < aAu

holds. This implies Ty E Eu, and so T(Eu) C E,,. Therefore, T leaves invariant every principal ideal.

Problem 3.3.4. Show that in a Banach lattice E: (1) Its center Z(E) with the operator norm is an AM-space with unit the identity operator 1.

(2) If E is also Dedekind complete, then the center 2(E) (as a Riesz space) coincides with the band generated by the identity operator I

in 4(E); and so Gr(E) = 2(E) ( D

.

Solution: (1) Let E be a Banach lattice. We consider 2(E) as a subset of C(E). We claim that 2(E) is a closed subset of L(E). To see this, assume that a sequence {T,.} of 2(E) satisfies T,, -+ T in G(E). Let Ixi A lyl = 0 in E. Since each T,, is a central operator, we have lxl A IT,.xI = 0 for each n. Now using the fact that the lattice operations are norm continuous, we see that Ixl A ITyI = 0. This implies

that T is an orthomorphism, and so (by Theorem 3.29) T is a central operator too. Thus, 2(E) equipped with the operator norm is a Banach space-and a units] (sub)algebra of G(E).

3.3. The Center of a Banach Lattice

107

Next, notice that if T E 2(E), then it follows from Theorem 3.30 that IITII = sup IITxhI = sup IIITI(IxI)II = IIITIII = IITIIr IIxII<1

Ilxll<_1

This implies that the operator norm of T coincides with the r-norm of T, and so the operator norm is a lattice norm. Thus, 2(E) is a Banach lattice with the operator norm. Clearly, the identity operator 1 is a unit for Z(E) (and I is both an algebraic

unit and an order unit). So, in order to establish that 2(E) is a Banach lattice with unit I, we must show that Uz(E) = [-I, I] holds. Clearly, [-I, I] e UZ(E). So, what needs verification is the inclusion Uz(E) C [-1, I]. In order to establish this, we need the following lattice-algebraic property of the operators in 2(E).

(P) If two operators S.T E 2(E) satisfy S A T = 0. then ST = 0.

To see this, assume that for some A > 0 we have Sx < Ax and Tx < Ax for all x E E+. Now note that if x E E+, then 0:5 (ST)x = S(Tx) < S(Ax) = )S(x) and 0 < (ST)(x) = S(Tx) < ATx, and so 0 < (ST)x < A(Sx A Tx) = A(S A T)x = 0. This implies ST = 0.

Now let T E Uz(E), i.e., IITII < 1. To finish the proof, we must show that ITI < I. For this, it suffices to establish that ITI < al holds for all a > 1. If this is not the case, then (ITI - aI)+ > 0 holds in the Riesz space 2(E) for some a > I. So, there exists some x E E+ such that y = (ITI - a1)+r > 0. But then, from the above property (P), we see that

(ITI - aI)y = [(ITI - al)+ - (ITI - al) ]y (ITI - al)+y - [(ITI - aI)- (ITI - aI)+)x

(ITI - aI)'y > 0, and so ITy! = ITly ? ay > 0. This implies IITyII > aIIyII > 0 and so I1TII > a > 1, which is impossible. Hence, ITI < al holds for all of > 1, and consequently IT[ < I.

(2) Now assume that E is a Dedekind complete Banach lattice. Clearly 2(E) is an ideal of E. As a matter of fact 2(E) = El, the principal ideal generated by I in C ,(E). To see that El = B1, let 0 < T E B1. Then, by Problem 1.2.6, we have T A nI T T. Next, suppose x n y = 0. Then x A [(T A nf)y] = 0 for each n, and so (by the order continuity of the lattice operations) we get x A Ty = 0. Thus, T is an orthomorphism and so (by Theorem 3.29) T E 2(E). Therefore, 2(E) = Br holds true in this case.

Problem 3.3.5. If A is a finite measure, then show-with appropriate identifications-that S(Lp(p)) = L.(µ) for each 1 < p:5 oc. Solution: Note that if u = 1 (the constant function one). then the principal ideal E coincides with i.e. E,, = Li(p). Moreover, EEu = LD(µ). Thus, by Theorem 3.34, Z(Lp(µ)) can be identified with L,,,(µ), where as in the proof of Theorem 3.34 every function f E L.(p) is identified with the multiplication operator Af f defined by Al f (g) = f g for each g E L1(µ). (For a more general result see Problem 5.2.7.) 1

3. Operators on AL- and AM-spaces

108

Problem 3.3.6. Prove the second part of the Maeda-Ogasawara-Vulikh Representation Theorem 3.35 by establishing the following general result for Riesz spaces: If F is an order dense Riesz subspace of an Archimedean Riesz

space E and F is Dedekind complete in its own right, then F is an ideal in E. Solution: Let F be an order dense Riesz subspace of an Archimedean Riesz space E such that F is Dedekind complete in its own right. We claim that for each x E E+ the non-empty subset D,, = {y E F: 0 < y < x}

of F satisfies D= j x in E. To see this, fix some x E E+ and assume that y < z holds in E for all y E Ds and some z E E. Clearly, y < x n z for all y E D=. Now suppose by way of contradiction that x - x n z > 0. Pick some u E F satisfying

0
then s = sup A also holds in E. To see this, assume that s = sup A holds in F and let a < t hold in E for all a E A and some t E E. If s A t < s is true, then there exists some v E F satisfying 0 < v _< a - ant, and so for each a E A we have a < a n t < a - v < s. Since a - v E F, the latter contradicts the fact that s = sup A holds in F. Hence, s = s n t < t holds in E, proving that s = sup A is true in E. Our next task is to show that F is an ideal of E. To see this, assume 0 < x < y holds with x E E and y E F. By the first part of the proof, we know that D= j x. Since D= T:5 y holds in F and F is a Dedekind complete Riesz space in its own right, there exists some w E F such that D= T w holds in F. Now by the second

part of the above proof, we also have D= j w in E. This implies x = w E F, so that F is an ideal of E. Now consider the Maeda-Ogasawara-Vulikh Representation Theorem. We know that E is an order Riesz subspace of C°O(Q), the order unit e of E corresponds to the constant function 1 on Q, and E is Dedekind complete. By the property of Riesz spaces proven above, E is an ideal of C°O(Q) and, consequently,

C(Q) C- E. Next, notice that C(Q) is an order dense Riesz subspace of C°°(Q). Also, by virtue of Q being extremely disconnected, C(Q) is Dedekind complete in its own right. So, by the above discussion, C(Q) is an ideal of C°D(Q), and clearly contains E. This shows that E = C(Q).

Problem 3.3.7. Show that a central operator on a Banach lattice commutes with all band projections. Solution: Let T : E - E be a central operator on a Banach lattice and let P: E - E be a band projection. This means that B = P(E) is a projection band, i.e., B ® Bd = E. From Theorem 3.29 we know that T(B) C- B and T(Bd) C- Bd.

Now fix some xEEand write x=y®zwith yEBand zEBd. Then TyEB and Tz E Bd, and so

(TP)x = T(Px) = Ty = P(Ty) = P(Ty) + P(Tz) = P(Ty + Tz) = P(Tx) = (PT)x

3.3. The Center of a Banach Lattice

109

for each x E X. Consequently, TP = PT, and therefore T commutes with every band projection.

Problem 3.3.8. This problem justifies the name "diagonal projection" for the band projection P2 of 4(E) onto the center of the Banach lattice E. Let E = R". the Euclidean n-dimensional (Riesz) space. Then G(E) = G.(E), the vector space of all n x n matrices with real entries. If D is a diagonal matrix with diagonal entries, then we write D = diag (dl, ... , dn). Establish the following properties.

(a) If A = [a,2] and B = [b,,J are two n x n matrices, then A _> B holds

in C,. (E) if and only if aj > bil for all i and j. (b) If A = [a,,] and B = [bij] are two n x n matrices, then

IAI=[lajl], AVB=[aijVbil] and AAB=[aidAbij. (c) The center of E consists of all diagonal matrices. (d) If A = [a,, ] is an n x n matrix, then Pz(A) =diag (all, an,

, ann)

Solution: (a) It suffices to show that an n x n matrix A = Jai,] defines a positive operator on R" if and only if a,, > 0 holds for each i and j. It should be clear that if A has non-negative entries, then it defines a positive operator on R". For the converse, assume that A defines a positive operator on R". Notice that this is equivalent to saying that a,jx_? > 0 for all xi,...,x, > 0 and every i = 1, ... , n. Letting x., = 1 and xk = 0 for k & j, it yields ail > 0 for all i and j. (b) We establish the formula for the absolute value only. So, let A = [aij] be an n x n matrix and put B = (I a., (]. From part (a), it follows that ±A < B. Now assume that another matrix M = [m,j] satisfies A < M and -A < M. By part (a),

we get ±a,., < m,, or Ia,, I < m,, for all i and j. This implies B < M, and thus (Al = A V (-A) = B.

(c) Let D = diag (di...., d") be a diagonal matrix and put A = maxi
holds for each x E R". This shows that D is a central operator.

For the converse, assume that an n x n matrix A = [aij] defines a central operator. This means that there exists some A > 0 such that I AxI < A xl holds for haveI1:n=iai,x,1 <_AIx,Iforallxi,...,x"ER"and all allxER". That is, we

i = 1,...,n. Now assume that i 36 k. Letting xk = I and x," = 0 for m 96 k, we see that n

a,,x,

IakI = I

Alx,(= 0.

j=1

Thus, aik = 0 holds for all i 96 k, and this shows that A is a diagonal operator. Therefore, Z(Rn) consists of all diagonal matrices.

110

3. Operators on AL- and AM-spaces

(d) If A = [a1t], let D = diag (all.... a"n) and B = A- D. Then D E Z(R"), IBI A I = 0 (i.e., B E Id), and A = D + B. This guarantees that Pz(A) = D = diag (al I.... ann). U

Problem 3.3.9. Let E be a Dedekind complete Banach lattice. According to the Maeda-Ogasawara-Vulikh Theorem 3.35 we can consider E as an order dense Riesz subspace of COO (Q), where Q is the Stone space of E. Show that

in this case Z(E) = C(Q). Solution: Observe first that since E is Dedekind complete, E is an ideal of C°°(Q). To see this, assume that 0 < x < y holds in C°°(Q) with y E E. Since E is order dense in C' (Q), the set D = {z E E: 0 < z < x} satisfies D j x in C°°(Q). Since E is Dedekind complete and z < y holds in E for each z E D, it follows that there exists some u E E such that D j u holds in E. Now observe that the order denseness of E in C°°(Q) implies that D j u also holds in C°°(Q). Hence, x = u E E must be the case, and so E is an ideal of C°O(Q). Fix h E C(Q) and consider the multiplication operator T: C(Q) C°D(Q). If x E E, then from the inequality ITxI = IhIIxI < IIhLILIxI and the fact that E is an ideal of COD(Q), it follows that Tx E E. Thus, for every h E C(Q), the multiplication operator defined by h is a central operator on E. For the converse, assume that T : E - E is a central operator. By the first part of Theorem 3.36 there exists some function h E C°°(Q) such that Tx = hx holds for each x E E. We shall finish the proof by showing that h E C(fl). To see this, assume by way of contradiction that h 0 C(fl). This means that for each n the open set V = {q E Q: Ih(q)I > n} is non-empty. Thus, the clopen set W = V is non-empty, and so 0 < Xw E C°D(Q). Since E is order dense in C°°(Q), there exists some y E E such that 0 < x < Xw. This implies ITxI = Ihxl ? nx > 0 and so IITxII ? nIIxfl for each n. Hence IIZII > n for each n E N, which is impossible.

This contradiction guarantees h E C(Q), as desired.

Problem 3.3.10. Recall that the principal band Bx generated by a vector x in a Riesz space E is the smallest band that contains x. Prom Problem 1.2.6 we know that Bz = {y E E: IyI A nlxl j Iyi}. If E is Dedekind complete, then BB is a projection band, i.e., B= ® Bz = E. We denote the band projection of E onto Bx by P. Let E be a Dedekind complete Banach lattice and let T, Z : E -- E be two disjoint regular operators (i.e., ITI A IZI = 0) with Z 54 0 and Z E Z(E). Show that for each 0 < x E E and each 0 < e < IIZII there exists a non-zero

component a of x such that Pa(ITIa) < ca. Solution: A glance at Problem 3.3.4 guarantees that we can suppose that T and

Z are positive operators with Z satisfying 0 < Z < I. Fix 0 < e < IIZII and let x > 0. Replacing E by Br, we can assume that x is a weak unit. Let Q be the Stone space of E and identify E with the ideal C(Q) of CO°(Q), where x corresponds to the constant function 1 on Q. By Theorem 3.36, there exists some h E C(Q) satisfying Z(z) = hz for all z E E and IIZII = Iihq°o Now notice that

3.4. The Predual of a Principal Ideal

111

the set V = {q E Q: h(q) > E} is a clopen subset of Q and soy = Xv E E. From Theorem 1.17, we know that

0=(TAZ)y=inf{Tz+Z(y-z): zEE+ and zn(y-z)=0}. This implies that there exist a component Xw of y = Xv (where W is a clopen subset of Q) and some qo E V such that T(Xw)(go) + h(go)Xv\w(qo) < E.

In view of h(q) > E for each q E V. we see that qo E W. From the inequality T(Xw)(go) < E, we infer that there exists some clopen set W1 C W satisfying

0
3.4. The Predual of a Principal Ideal Problem 3.4.1. Establish the following useful result: If Y and Z are closed subspaces of a normed space X. one of which is finite dimensional, then Y + Z is a closed subspace of X. Solution: Assume that Y is a closed subspace of X, Z is a finite dimensional subspace of X, and let 7r: X X/Y denote the quotient map, i.e., 7r(x) = [x] for each x E X. Then rr(Z) is a finite dimensional subspace of the normed space X/Y, and so rr(Z) is a closed subspace of X/Y. Now note that Y+Z = ir-1(7r(Z)). Since >r is continuous, the latter guarantees that Y + Z is a closed subspace of X. I

Problem 3.4.2. Let E be a Riesz space and let J be an ideal in E. Establish the following:

(a) The set C = {[x]: x E E+} is a (convex) cone in E/J. (b) The vector space E/J ordered by the cone C is a Riesz space. (c) The quotient map x ' -, [x] is a lattice homomorphism from E onto

E/J. Solution: (a) It should be clear that C + C C C and aC C C for all a > 0. Now assume that [x] E C fl (-C). This implies that there exist x1 > 0 and x2 > 0

such that x1 E (x] and -x2 E [x]. Thus, x1 - x E J and-X2- x E J, and so x1 +x2 = (xl -x) - (-X2- x) E J. The latter, in view of 0 < xl < x1 +x2, implies [x] = [xl] = 0. This shows that C is a cone in E/J. The reader should notice that the inequality [x] < [y] holds in E/J if and only if there exist xl E [x] and yl E [y] such that xl < yl in E. This means that whenever [x] < [y] holds in E/J, we can assume without loss of generality that x < y.

(b) We claim that the lattice operations in E/J satisfy [x] V [y] = [x V y] [x] A [y] = [x n y], and I[x]l = [Ixl]

3. Operators on AL- and AM-spaces

112

In order to establish these formulas, it suffices to show that [x]+ = [x+] for each

x E E. So, let x E E. From x+ > x and x+ > 0, it follows that [x+] > [x] and [x+] > 0 are both true. To see that [x+] is the least upper bound of jx] and 0, assume that some [y] E E/J satisfies [y] > [x] and [y] > 0. Pick yl E [y]. -Ti E [x]. and z E [y] such that yl > xl and z > 0. This yields

z+(yl -z)+(x-xi) =x+(yt-xl)>x. and by letting u = (y1 - z) + (x - x1) E J we have x < z + u. Consequently, x+ <- (z + u)+ < z+ + u+ = z + u+. Taking into account that u+ E J (since J is an ideal), we see that [x+] < [z + u+] = [z] = [y]. This shows that [x]+ = [x+], and the validity of the formulas has been established.

(c) The formula [x] V [y] = [x V y] established in (b) demonstrates that the quotient map is indeed a lattice homomorphism. I

Problem 3.4.3. Let E be a nonmed Riesz space and let J be a closed ideal in E. Establish the following.

(a) The quotient norm is a lattice norm-and so E/J is a normed Riesz space.

(b) If E is a Banach lattice, then E/J is also a Banach lattice. Solution: (a) Assume I[z]l < 1[y11 or. equivalently, [IxI] 5 [IyIj. We must show 111 [x[111 < III [y] 111, where III 'III denotes the quotient norm of E/J.

Replacing IxI by IxI A lyi (and using the fact that the quotient map is a lattice

homomorphism). we see that it suffices to show that 0 _< x <_ y in E implies III [x] 111 <_ III [y] 111. To this end, assume 0 < x < y in E.

Fix some z E [y] and let u = IzIAy. From ly-uI = ly-Iz1Ayl : ly - zl E J, we get u E jy] and 0 < u < y. Moreover, from 0 <_ x - u A x.= (x - u) + <_ (y - u)+ E J, we obtain u A x E [x]. Therefore. 111[x1111 <- 11u A x11 < 1lull =111zI A y11 <- II Iz1 11=11z11,

holds for all z E [y]. So, li l[x] i Il < inf=E[yl I[ zll = 111[y]III. and we have established

that III -III is a lattice norm.

(b) This follows immediately from (a) and part (d) of Theorem 1.11.

Problem 3.4.4. Show that the norm completion of a nonmed Riesz space E is a Banach lattice containing E as a Riesz subspace. Solution: Let E be a normed Riesz space and let t denotes its norm completion. Also, let C = E+ be the norm closure of E+ in E. It should be clear that C+C C C and aC C C for all a > 0. Now assume that x E C n (-C). Pick two sequences

{x,j and { y } in E+ satisfying x,, - x and y - -x. Then from the relation

0<<x, <x,, +yn-.x-x=0 it follows that xn--.0. Thus, x=0, andsoCis a (convex) cone in E. We shall denote the order induced by the cone C on k by
3.4. The Predual of a Principal Ideal

113

Start by observing that from the inequality Ix+ - y+I < Ix - yI valid for all x. y E E, it follows that the function p: E -+ E C E. defined by p(x) = is x+.

uniformly continuous. Therefore, p has a (uniformly) continuous extension to all of

E, which we shall denote by p again. We claim that p(x) = r+ = x Vc 0 holds in E for each x E E. This will establish that (E. C) is a Riesz space. To see this, fix x E E. Pick a sequence {xn } C E such that xn -+ r. Then we have x+ = p(rn) -+ p(x). and so p(r) >c 0. On the other hand, from xn :5 x.+. it easily follows (by taking norm limits) that x
satisfying u -+ z - x and zn -+ z. Clearly. zn - u -+ z - (z - r) = x. Now. from zn - u < zn = z,+, . it follows that (zn - u,,)+ < zn for each n. and so by taking norm limits, we see that p(x) = lim(zn - un)+
Z VC y = lim .r.,, V yn

nix

and

x AC y = lim xn A yn . n- 'X

(*)

From the above formulas. it follows immediately that E is a Riesz subspace of E.

To finish the proof. we must show that the norm of k is a lattice norm. To this end, let Ivlc. Sc Iu'Ic in E. Choose two sequences {vn} and {u:n} in E such that vn -+ v and wn uw. From (*). we get Ivnl IvIc and lwnl -+ IwIc. Using (*) once more, we see that Ivnl A lu'ni -' IvIc Ac Iwlc = Inc. Finally, using that Ivnl A Iwnl < II Ivic II <-

implies II fvnl A IwnJ II

Il lw,,l II. by taking norm limits, we obtain

II Iwlc II. This completes the proof that (E, C) is a Banach lattice.

Problem 3.4.5. Let E be an arbitrary Al-space, that is, E is a normed E+. Show Riesz space such that IIxVyII = max{IIxII, IIxII} holds for all x, y E the following.

(a) The norm completion of E is an A.41-space.

(b) If E has a unit e and the closed unit ball of E coincides with the order interval [-e. e], then the norm completion of E is an AMspace with unit e. Solution: (a) Assume that E is an Al-space. and let k denote its norm completion. By Problem 3.4.4. E is a Banach lattice containing E as a Riesz subspace. Now assume that x A y = 0 holds in E. Pick two sequences {xn} and {yn} in E satisfying xn -i x and yn y. Since (xn - Tn A yn) A (yn - rn A y,) = 0 for each n and since E is an Al-space, we conclude that ll(xn - X. A y,,) V (y, - xn A yn)Il = coax{ Ilxn - xn A ynll Ilyn - xn A y.111. From xn A yn -+ r A y = 0 and (*-*), we get Ilx V y0I = max{IIxII IIxII}. This shows

that k is an AM-space. (b) Assume that E has a unit e. i.e., UE = (-e, el. Also, let

-'c.ej={xEE: -e.<x<e}.

3. Operators on AL- and AAJ-spaces

114

Clearly, Q-e. eQ C U. For the reverse inclusion, let 0 t- r E Ut, i.e.. 111II < 1. Pick a sequence {x,,} in E consisting of non-zero vectors such that X,, r. Let y = fit- E E and note that Ilyn II = 1 for each n. This implies that y E Q-e., e.D mid yn -- ;I=I. Since the order interval Q-e, eD is norm closed. we infer that

That is. -e < -II?IIe < r < IIxIle < e. Hence, Ut C Q-c.cf, and so UE = [-c. e]. Therefore. E is an AM-space with unit r.

-c < ' < e.

Problem 3.4.6. Let E be an arbitrary L-apace, that is, E is a nonmed Riesz space such that JIx + yII = IIxII + Ilyll holds for all X. y E E with r A y = 0. Show that the norm completion of E is an AL-space. Solution: Assume that E is an L-space. and let k denote its norm completion. By Problem 3.4.4. E is a Banach lattice containing E as a Riesz subspace. Now assume that xAy = 0 holds in E. Pick two sequences {rn} and {yn} in E satisfying

xn - x and yn -. y. Since (xn - xn A y,) A (y, - rn A yn) = 0 holds for each n mid since F. is an L-space, we conclude that

II`xn-rn/1yn)+(yn-rn/\yn)I1=11rn-xnAynll+llyn-xnnynll (***) F oni .r A yr, - r n y = 0 and (* * *), we see that IIr + yll = Aril + Ilyil. This shows that k is an AL-space.

Problem 3.4.7. Let X be a nonmed space and let X be its norm completion. For each f E X', let f denote the unique continuous (linear) extension of f to f(. Show the following.

(a) The mapping f p-+ f is a surjective linear isometry from X' onto X' (and so, subject to this linear isometry. X' = X'). is a surjectit'e lattice (b) If X is a norrned Riesz space., there f isometry from X' onto f('. Solution: (a) Straightforward. (b) To show that f - I is a lattice isometry, we need to verify only that this mapping is a lattice homomorphism. To see this, fix f E X' and let r E X+. Then the modulus of f in X' is given by

IfI(r) = sup{If(y)I: y E X and IyI Similarh; the modulus off in X- satisfies

r}

.

111(r) = sup{ If (y)I : y E X and IyI

r}

.

Clearly. If I (x) < If I (x)- For the reverse inequality, pick y E X such that I yI < r.. Choose a sequence {yn} in X satisfying yn -, y. and then let zn = Iy,,V(-x)IAx E X for each n. Clearly. z - y and IznI < x holds for each n. This implies

If(y)I = n-x lint If(zn)I = linl n-xIf(zn)I

IfI(x)

for all y E X With I yI < X. It follows that

IfI(x)=sup{If(y)I: yEX and Iyl <x}<-If I(x).

3.4. The Predual of a Principal Ideal

115

and so If I (x) = 111(x) for all x E X+. The latter guarantees If I = If I, which means

that f '-. f is a lattice homomorphism. This completes the solution.

Problem 3.4.8. Let p be an arbitrary lattice seminorm on a Riesz space E, and define its null ideal by Np = {x E E: p(x) = 0}. Establish the following.

(a) Np is indeed an ideal in E.

(b) The function p: E/Np - R defined by 0([x]) = p(x) is a lattice norm on E/Np and p([x]) = infyE[r) p(y).

(c) If p is an M-seminorm (resp. an L-seminonn), then p is an Mnorm (resp. an L-norm). Solution: (a) This is straightforward. (b) First, let us establish that p is well defined. To this end, let y E [x], that is, p(Ix - yI) = 0. Then from Ip(x) - P(y)I < P(Ix - yI) = 0, we see that p(x) = p(y), and so p is well defined. Since p(y) = p(x) holds for each y E [x], it follows that

P([x]) = P(x) = inf P(y) To see that p is a lattice seminorm, assume I[x]I < I[y]I, i.e., [IxI] < [IyI]. We can suppose IxI < IyI. This implies p([x]) = p(x) = P(IxI) P(Iyi) = p(y) = A([y]). Finally, to get that p is a lattice norm, notice that p([x]) = p(x) = 0 implies

xENpor [x)=0. (c) Assume first that p is a lattice M-seminorm, and let [x] A [y] = [x n y] = 0 in E/Np. Replacing x by x - x n y and y by y - x A y and using that the quotient map is a lattice homomorphism, we can assume without loss of generality that x A y = 0 holds in E. Now notice that P([x] V [y]) = P([x V y]) = p(x V y) = max{P(x),P(y)} = max{P([x]),P([y])) .

This shows that p is an M-norm. If p is an L-seminorm, then a similar argument shows that p is an L-norm.

Problem 3.4.9. Complete the details of the proof of Lemma 3.38. Solution: Let us recall the background associated with Lemma 3.38. We consider a Banach lattice E and we fix a positive linear functional 0 <'0 E E. As usual, EE denotes the ideal generated by 0 E E', i.e.,

EE={x'EE': 3A>0 such that Ix'I 0: Ix'I < A0}, is an AM-space with unit 0. The linear functional 0 defines a lattice seminorm I[ III on E via the formula IIxIIo = O(IxI)

3. Operators on AL- and AM-spaces

116

This is an L-seminorm in the sense that (besides being a lattice seminorm) it also satisfies fix + yll0 = IIxfl0 + IlyJ10 for all x, y E E. If we consider the null ideal of 0, i.e., the closed ideal NO = {x E E: 4(Ixl) = 0}, then the quotient seminorm induced on the quotient Riesz space E/N0 by II 110, i.e., the lattice seininorm II[x111m = Yinf i

11y110 = 004

is in fact an L-norm. The norm completion of the normed Riesz space (E/NQ.11. 110)

is an AL-space, which we shall denote E(¢). Also, we shall denote the norm dual of E(O) by E(q,)'. Keep in mind that E(¢)' is an AM-space with unit e' defined by the norm of E(p), i.e., e'([xJ) = (b(x+) - i(x-) for each x E E. There is a natural mapping 0: E; --+ EM* or x' 1-4 4'x , defined by 4_. (1x1) = x'(x),

(*)

for all x' E EQ and all x E E. To see that 4' is well defined, we argue as follows. Fix x' E EE, and then select a scalar A > 0 such that Ix' 1 < A0. Now assume that y E [x], i.e.. [y] = [x1, or 0(j x - yI) = 0. From

Ix'(x)-x'(y)I <- Ix'1(Ix-yl)
(t)

it follows that 4_. is a continuous linear functional on E/N0. Therefore. 4x has a unique continuous linear extension to the norm completion E(Q)' of E/N0, which we shall denote by 4x. again. Lemma 3.38 now can be stated as follows.

The mapping 4': EE --+ E(4p)' as defined by (*) is a surjective lattice isometry.

We shall verify the above claims by steps. It should be clear that 4' is a bounded (linear) operator.

STEP I: 0 is a surjective operator. Let %P E E(O)'. So, there is some A > 0 such that 1%Y(1x1)1 < Ait[x111m = A0(ixl)

holds for each x E E. Now if we consider the linear functional x*: E - R, defined then from 1x'(x)J < A0(jxj) for each x E E, we see that Ix'1 < A0 by x'(x) or x* E E. Moreover, note that +x. _ y, and so fi is surjective. STEP II: 4' is one-to-one.

Assume that (Dx = 0. This means that x'(x) = 0 for all x E E. so that x* = 0. Therefore, 0 is one-to-one. STEP III: 4' is an isometry.

Fix x' E E. Then from Ix'(x)I = 0.- QXD1 <- II4=-II.111x1110 = II4'x-114(Ixi)

3.4. The Predual of a Principal Ideal

I14: 110. This shows that IIx' II x = inf {A > 0: we get Ix' I On the other hand, from Ix* I S IIx' II o& , it follows that

117

Ix I < A0} < II'

[[.

0r-([x])1 = Ix*(x)I <_ Ix`I(Ixi) <_ IIx'II (IxI) = IIx'Iloo 11[x]114 II <_ 11x*IIx. is also true. Consequently,

for each x E X. This implies that

III:- II = IIx' II. holds for each x' E E. and so 4 is a linear isometry.

STEP N: t is a strjective lattice isomorphism.

Let x' > 0 and [x] > 0. If 0 < y E [x], then It.. ([x}) = t.- ([y]) = x'(y) > 0. This implies -Cz- > 0, and so 4) is a positive operator. On the other hand, if $= > 0, then for each x E E+ we have x'(x) = 4s- ([x]) > 0. That is, the inverse of 4) is also a positive operator. This guarantees that 4) is a (surjective) lattice isomorphism.2

2 Here we use the fact that a surjoctive one-to-one operator T : F - C between two Riesz spaces is a lattice isomorphism if and only if T and T-1 are both positive operators. To see this, let s, y E F. From x < x V y and y < x v y, we get Tx < T(x v y) and Ty < T(x V y) or Tx V Ty < T (:r v y). The latter conclusion applied to T-1 yields T- I u V T- I v < T-1(4 V v) for

all u, v E G. Letting u = Tx and v = Ty, we get x V y < T-1(Tx V Ty), and by applying the positive operator T, we see that T(x V y) < Tx V Ty. Hence. T(x V y) = Tx V Ty, and so T is a lattice isomorphism.

Chapter 4

Special Classes of Operators

4.1. Finite-rank Operators Problem 4.1.1. Let f. f, I ... , fk be linear functionals on a vector space. Show that f lies in the linear span of fl, ... , fk if and only if k

nKerf,CKerf. :=1

I n particular, a collection gi, ... , gn of linear functionals on a vector space is linearly dependent if and only if ni#JKerg; C Kergj for some j. Solution: Let f, fl..... fk be linear functionals defined on a real vector space V. (The arguments below show that the conclusion is true for vector spaces over any field.) Note first that if f = Ek 1 a, f then clearly n= 1 Ker f, C Ker f. For the converse, let nk 1 Ker f, C Ker f . Consider the vector subspace of Rk

W = {(fl(v)..... fk(v)) E Rk: v E V}

,

and then define the linear functional 0: W -- R by b(f 1(v), ... , fk(v)) = f (v). Note that if (f, (v),.... fk(v)) = (fl (u), ... , fk(u)), then v - u E nk 1 Ker f, and so by our hypothesis f (v) = f (u), which shows that ¢ is well defined. The verification of the linearity of 0 is straightforward. Now notice that 0 extends linearly to all of Rk. That is, there exist scalars A,-., Ak such that 0(t1, ... , tk) = Lk 1 A,t, holds for all (t1.... , t,) E Rk. This easily implies f = E A f,. (Notice that this proof is applicable to vector spaces over any field.)

Problem 4.1.2. Show that an operator T : V - W between two vector spaces is a finite-rank operator if and only if there exist vectors wi, ... , wn in W and linear functionals gi, ... , gn on V such that T = Fn g, ® w;. 119

4. Special Classes of Operators

120

Solution: Assume first that T =

gi 0 wi. If U is the vector subspace of W generated by wl...., w,,, then clearly Tv E U for all v E V, i.e., T(V) C U. Since U is finite dimensional, T(V) (the range of T) is likewise finite dimensional. For the converse, assume that T is a finite-rank operator. i.e., the range T(V) of T is finite dimensional. Fix a basis w1..... w of T(V) and for each v E V write Tv = F ,'t 9t(v)wt, where each gi is a scalar-valued function defined on V. From n

E9i{av1 + 3v2)wi = T(ovi + i3v2) = aT(v'1) + 3T(r2) ,=1 n

_ Efa9:(v'i) i_1

it follows that gi(av1 + 13v2) = ag,(vl) + :3g; (v2) for each i. all v1. 1,2 E V. and all scalars a and 3. Thus, each g, is a linear functional and T = 1 g, a lci.

Problem 4.1.3. Let T : V W be an operator of rank k and let wl, ... , wk be a basis of the range of T. Then there exist (uniquely determined) linear functionals fl, . . . , fk on V such that T f= 9. wi. Show that f1... . fk are linearly independent. Solution: The uniqueness of the linear functionals f, should be immediate from the linear independence of the vectors w1, .... wk. The existence and linearity of the ft follow from Problem 4.1.2. Now suppose by way of contradiction that f'.. .. , f are not linearly independent. By rearranging, we can assume without loss of generality that there exist scalars A2, A3, ... , Ak such that f1 = k

T=Ef,owi

; 2

A, fi. This implies

k

k

k

fi®wl+F, fixuri=E f,:9(A1wi)+1: f,5i:w, i=2

i=2

i=1

i=2

k

1: ft & (wi + \.W,). i=2

This shows that rank of T is at most k - 1. a contradiction. Consequently, the linear functionals f l ..... fk are linearly independent.

Problem 4.1.4. Let T : V - V be a finite-rank operator on a vector space and assume that T = f; 0 v; _ E?_1 gj 0 uj are two arbitrary representations of T, where the vectors vi and uj need not belong to the range of T. Show that m n f; (vi)

Egi(u;)

That is, the scalar tr(T) _ Es t fi (vi) (called the trace of T) is independent of the representation of T.

Solution: Fix a basis w1.... , wk of the range of T and write T

hr 0 Wr. where the linear functionals hr are uniquely determined. It suffices to show that for an arbitrary representation T = 1 fi 0 vi we have E 1 fi(vi) = Er=1 gr(wr).

4.1. Finite-rank Operators

121

To this end, let T = f, ® vi be a representation of T. Assume first that the linear functionals fl, ... , fn are linearly independent. Then for each 1 < i < n there exists some xi E V such that fj(xi) = bid; see Problem 4.1.1. This implies v, E T(V) for each i and so v,If, = Ek=1 )4Wr for appropriate (uniquely determined) scalars a; . Hence, T = E. ® v, = Er=1(E 1)4f,) ® W,_, and from this we infer that hr A . f, for each 1 < r < k. Consequently, 1

(E a*fi) (wr) = > hr(wr) .

fi (E \Twr) =

ft (v,) _

r=1

i=1

1=1

r=1

r=1

i=1

Next, suppose that fl,.. .. f are linearly dependent. Then there exist linearly for appropriate independent linear functionals 0k,.. . , 01 such that fi = scalars p,,. This implies F f, ®vi = Ev_1 0. (E 1 p,vi). By the preceding case, we have 1

Ehr(wr) = r=1

= Efv.), v eo Mµvi) = EMµev)(v.) i=1 v=1 t=1

v=1

8=1

and the solution of the problem is complete.

Problem 4.1.5. Show that the trace of a finite-rank operator on a Banach space is similarity invariant. That is, show that if T : X -+ X is a finiterank operator on a Banach space, then tr(T) = tr(S'1TS) for each invertible bounded operator S : X -. X. Solution: Let T : V -' V be a finite-rank operator on a vector space and let

T = En fi ® ui be a representation of T. If S: V - V is an invertible operator, 1

then for each v E V we have n

n

[S-1TS]v = S-1(T(Sv)) =S-1(Efi(Sv)ui) 1=1

_>fi(Sv)(S-1ui)

2=1

[E(fi o S) ® (S-1ui)]v. 1=1

This shows that S-'TS = En1(f, o S) ® (S-1u,). Now a glance at Problem 4.1.4 yields

=

tr(S-'TS)

n

n

n

E(f, o S)(S-'u,) = >fi(S(S-1u,)) = >fi(ui) = tr(T),

as desired.

Problem 4.1.6. Let X be a Banach space and let {xi, x2, ... , xn} C X' and {x1i x2, ... , x } C_ X satisfy x, (xj) = bid for all i and j. Show that the finite-rank operator P =En 1 x; (9xi has rank n, is a continuous projection, and has range the vector subspace generated by {x1, x2, ... , x } . Use this conclusion to show that every finite-dimensional subspace Y of X is complemented, i.e., there exists a closed subspace Z of X satisfying

YnZ={0} and YED Z=X.

4. Special Classes of Operators

122

Solution: For the first part notice that the condition x, (x,) = 8;, for all

i

and j easily implies that both collections {xl. x2, .... x,,) and {x 1. x.2..... xn } are linearly independent. This implies that the range of the continuous operator x; 0 xi is the n-dimensional vector subspace generated by the collection P of vectors {x1, x2, ... , x }. Thus, P is a finite-rank operator of rank n. To see that P is a projection, note that for each x E X we have n

n

P2(x) = P(Px) = P( j=1

i=1 n

n

i-1

tt

E E x, (x)x;(xi)xJ =1 ,=1

n

x; (E r; (x)x=)xJ xi (x)x, = P(x). ,=l

Thus. P2 = P. and so P is a projection. For the second part. let {X1, r2..... xn } he a basis of the vector subspace Y of X. Pick linear functionals xi . x2.... , x,, on Y such that x, (r,) = b,, holds for all i and j. Since the norm topology of X induces the Euclidean topology on Y. each x, is norm continuous. By the Hahn-Banach theorem, each x, has a continuous extension to all of X. which we shall denote by x, again. That is. we can assume that x; E X' for each i. Now consider the continuous projection P = >,n_i x, ox, and use the preceding conclusion to see that P(X) = Y. If we let Z = (I - P)(X). then Z is a closed subspace of X satisfying Y fl Z = {0} and Y :r Z = X.

Problem 4.1.7 (Grothendieck [361). For a Banach space X establish that the following statements are equivalent.

(a) X has the approximation property. (b) For every Banach space Y the finite-rank operators Y'

X are

,r,.(Y. X) -dense in C(Y, X).

(c) For every Banach space Y the finite-rank operators X* 0 Y are r,(X. Y) -dense in C(X, Y).

Solution: (a)

(b) Fix a Banach space Y and let T' E C(Y. X). Also, let C be a compact subset of Y, and fix e > 0. Then T(C) is a norm compact subset of X. Since X has the approximation property, there exists a finite-rank operator S E C(X) such that II Sx - xli < e holds for each x E T(C) or (I STy - Tyl( < e for each y E C. This (by taking into account that the operator TS E C(Y. X) is a finite-rank operator) shows that T can be approximated uniformly on every compact subset of Y by finite-rank operators of C(Y, X). That is. Y' & X is r,(Y. X)-dense in C(Y, X). (b)

(c) Let T E C(X, Y). We can assume that T ,4 0. Fix a norm compact

subset C of X and let e: > 0. By our assumption, the finite-rank operators in C(X) are rr(X)-dense in C(X) = C(X, X ). Consequently, there exists a finitefor all x E C. This implies rank operator S E C(X) satisfying IISx - xli < IITSx - TxlI 5 IITII - IISx - xii < e for all x E C. Since the operator TS E £(X. Y) has finite-rank, it follows that the vector space of all finite-rank operators X' Yr, Y is -r,(X. Y)-dense in C(X.Y).

4.1. Finite-rank Operators

123

(a) Letting Y = X, we see that the identity operator I: X - X

(c)

belongs to the r,(X )-closure of the finite-rank operators in L(X). Now a glance at Lemma 4.8 shows that X has the approximation property, and the solution is finished.

Problem 4.1.8. Let X be a Banach space with the approximation property. If Y is a Banach space and T : Y -, X is a compact operator, then show that T*: X' --+ Y* belongs to the norm closure of the finite-rank operators

from X' to Y*, i.e., T' E X" ® Y'. Solution: Assume that T : Y - X is a compact operator and fix e > 0. Since the Banach space X has the approximation property, there exists (by Theorem 4.12) a finite-rank operator S = 1 y,, ® x, in C(Y, X) such that JIT - SD] < e. Then S' x,, ® y,, E C(X *, Y*). where now x is considered as a vector in X ', is a finite-rank operator, and IIT' - S'II = JIT - SD] < e holds. This shows that

Ts EX"9 Problem 4.1.9. Consider the rank-two operator T = 10 u - u 0 1 on Li[0,11, where 1(s) = 1 and u(s) = s for each s E [0,1]. Also, recall that each function v E L,,.[0,1) is viewed as a continuous linear functional on L1 [0, 1] via the formula (x, v) = fo v(s)x(s) ds. Prove the following.

(a) Tx(t) = fo (t - s)x(s) ds for all x E L1 [0.1] and each t E [0,1]. (b) The modulus of the operator T satisfies IT]x(t) = fo Is - t]x(s) ds for all x E Ll [0, 1] and t E [0, 1].

(c) ]TI is not a finite-rank operator. Solution: (a) Let T = 10 u - u ® 1. If x E L1(0,1], then for each t E (0.11 we have

Tx(t) =

((1 ® u)(x))(t) - [(u ® 1)(x)](t) [

=

I

f x(s) ds]u(t) - [

oI u(s)x(s) 0

r1

[u(t) - u(s)]x(s) ds

I

1

(t - s)x(s) d. s.

(b) Repeat the solution of part (b) of Problem 3.1.13.

da] 1(t)

4. Special Classes of Operators

124

(c) For each k let xk(s) = (k + 1)(k + 2)sk, and note that ITIxk(t) _ (k + 1)(k + 2)

J0

1

Is - tlsk ds

= (k + 1)(k + 2) / t (t - s)sk ds + (k + 1)(k + 2)

J (s - t) sk d., t

o

1)S2 - (k + 2)tsk+1]

= [(k + 2)tsk+l - (k+ 1)sk+2]

sa=t

[(k+2)tk+2-(k+1)tk+2]+[(k+1)-(k+2)t-(k+l)tk+2+(k+2)tk+2] = = 2tk+2 - (k + 2)t + k + 1.

Since the countable collection of polynomials {ITIxi(t).ITIx2(t),ITIx3(t)....} is linearly independent, it follows that the range of ITI cannot be finite dimensional. Hence. the modulus of the 2-rank operator T is not a finite-rank operator.

Y between two Banach spaces is Problem 4.1.10. An operator T : X said to be a nuclear operator if there exist two sequences {x;,} C X' and {yn} C Y satisfying F°° 1 11x* 11 11y. 11 < oc and x T(x) = Y, xn(x)yn

n=1

n_ x; 0 yn in C(X.Y). Show that:

for each x E X, i.e., T =

(a) Every nuclear operator is compact. (b) If X andY are Banach lattices, E°D_lllx;,II
Solution: (a) Let Al =

En==1

11411

Ilynll < X. From

Ilxn(x)y, II <- E II6;6II - IlynII - Ilxll = "1114 < x. n=1

n=1

it follows that the series defining T(x) is norm convergent in Y for each x E X and that IIT(x)II Mfl=II holds. This implies that T is a bounded operator. Now for each k consider the finite-rank operator Tk : X - 1 defined by Tk(x) = Fn=1 x,,(x)yn. Then for each x E X with lid < 1 we have rx

00

IIT(x) -Tk(x)II = II

.

n=k+1

xn(--)ynll <

0.

IIxn1I ' IIyn11

n=k+1

This shows that the sequence {Tk - T} converges uniformly to zero on Ux, which T in C(X, Y). Since each Tk is a compact operator, it follows means that Tk that T is also a compact operator.

(b) Now assume that X and Y are Banach lattices, C =

K=

En

1

IIymII < oo. This implies

,

IIx,, II < oo and

II=nlf - IIynII < oo. and so the formula

T = >n 1 x;, S, yn defines a nuclear operator from X to Y.

4.2. Multiplication Operators

125

Letu=>' 1lynIEY+. IfxEX.then x x

x

ITxI = JE xn(x)Ynl E Irn(r)I - lynI : n=1

n=1

Ixn(x)I - Iuf = CDD4 - Iui n=1

and so T carries X into the principal ideal Yu. Moreover, if Tk is the finite-rank operator of part (a). then Tk is also a finite-rank operator from X to Yu, and for each r E X with lixit < 1 we have

IT(x) -Tk(x)I

x x I E rn(x)Ynl c [ n=k+1

Hx II]u.

n=ktl

En Thus, 11 T -- Tk , <11

k+1 114 for each k. and this shows that T: X Y,, (i.e.. T as an operator from X to the AM-space Y;,) is also compact. Now, by Theorem 3.14. the operator T: X -' Y has a compact modulus that is given by the Riesz-Kantorovich formula: JTIr. = step{Ty: y E X and iy) < x) for each x E X+. Since. Y. is an ideal in Y, the latter implies that JT) considered as an operator from X to Y is also the modulus of the operator T : X Y. Since the 11.1Ix-topology on Yu is finer than the norm topology on Yu, it follows that DTI: X Y is likewise a compact operator.

4.2. Multiplication Operators Problem 4.2.1. Prove the following implications: Quasi-interior Point Order Unit Weak Order Unit Also show that in general no converse implication is true. Solution: Let E be a Banach lattice and let u > 0. Assume first that u is an order unit. Then Eu = E, and so Eu = Eu = E. That is. u is a quasi-interior point. Now assume that u is a quasi-interior point of E. Then from Lemma 4.15(2), it follows that II Ixi A nu - Ixf II -. 0 for each r E E. Since JxJ A ne j, Problem 1.2.12(c) guarantees that IxI A nu 1 JxJ for each x E E. The latter (in view of Theorem 1.27)

shows that Bu = E, and so u is a weak unit of E. To see that no converse implication holds true. note that the constant function 1 on 10, 1] is a quasi-interior point in L1 [0.1] but fails to be an order unit. Finally, the function u(t) = t is a weak unit in C. [0, 1] but it is not a quasi-interior point. 0

Problem 4.2.2. Show that a Banach lattice E has a weak unit if and only if E has an at most countable maximal subset of pairwise disjoint vectors. Solution: Let E be a Banach lattice. Assume first that E has a weak unit, say e. Then the singleton {e} is the required maximal subset of E. For the converse, let D be an at most countable maximal subset of pairwise disjoint vectors in E. We can assume that D is countable, say D = {x1.x2.... }, and that flxnjj = 1 for

4. Special Classes of Operators

126

each n. Since E is a Banach lattice, the series e = n i 2 x converges in E and 0 < 2 x < e holds for each n. Now notice that if some vector x E E satisfies x 1 e, then x 1 x for each n. Since D is maximal, we infer that x = 0. This shows that e is a weak unit of E.

I

Problem 4.2.3. Show that every separable Banach lattice has a quasiinterior point-and hence it also has a weak unit. Solution: Let E be a separable Banach space and fix a countable dense set {x1, x2,. ..}. Choose a sequence {an} of positive weights such that the series u = En l anjxnj is norm convergent. For instance, for each n pick some an > 0 such that an jj xn p < 2L . Clearly, 0 < an jxn j < u for each n, and so xn E E. holds for each n. Therefore, {x3 , x2, ...} C_ Eu and from this we infer that Eu = E. That

is, u is a quasi-interior point.

I

Problem 4.2.4. In a Banach lattice with order continuous norm, show that a positive vector is a quasi-interior point if and only it is a weak unit. Solution: Assume that E is a Banach lattice with order continuous norm and let e> 0 be a weak unit. This means that x A ne T x holds in E for each x E E+, and so (by the order continuity of the norm) we have jjx A ne - xjj - 0 for each x E E+. Now a glance at Lemma 4.15 guarantees that e is a quasi-interior point. The converse was established in Problem 4.2.1.

1

Problem 4.2.5. Prove that the Banach lattice M[0,1], of all signed Borel measures on [0, 1] of bounded variation, does not have quasi-interior points. Solution: Notice first that M10, 1] as an AL-space has order continuous norm; see Corollary 3.7. So, by Problem 4.2.4, the notions of weak unit and quasi-interior point coincide. Now assume by way of contradiction that the Banach lattice M[0,1] has a weak

unit, say 0 < p E M[0.1]. This guarantees that p A v = 0 implies v = 0. This implies that p({x}) > 0 for each x E [0.1]. To see this, assume that µ({x}) = 0 for some x E [0.1]. Then p A o_ = 0, where b= is the Dirac measure supported at x. and so bT = 0. which is impossible. If we let En = {x E (0,1): p({x}) > },

then it should be clear that each E is a finite set. But then E = U', En mustnbe a countable set, a contradiction. This contradiction establishes that the AL-space .M[0.1] has neither a quasi-interior point nor a weak unit. (The same proof shows that if f is an uncountable compact Hausdorff space, then ca(0) is an AL-space without quasi-interior points.) I

Problem 4.2.6. Show that if E is a Banach lattice, then its set of quasiinterior points is either empty or else it is dense in E+. Solution: Let A C E+ be the collection of all quasi-interior points of a Banach

lattice E. Assume that A 0 0, and let x E E+. Fix some e E A and note that ae + x is a quasi-interior point for each a > 0, i.e., ae + x E A. Now notice that (oe + x) = x. This shows that x E A. and consequently A = E+.

I

4.2. Multiplication Operators

127

Problem 4.2.7. If T : E - F is a strictly positive operator between two Banach lattices, then show that the adjoint operator T*: F* - E' carries quasi-interior points to quasi-interior points. Solution: Assume that T : E - F is strictly positive, and let e` E F' be a quasiinterior point. Now let 0 < x E E. The strict positivity of T implies Tx > 0. Therefore, since e' is a quasi-interior point of F. we have T'e'(x) = e'(Tx) > 0. This implies that T'e' is a quasi-interior point of E'; see Lemma 4.15.

Problem 4.2.8. Fix a function 0 E C(S2) (with SZ compact and Hausdorff) and let A0 denote the closed unital algebra generated by 0 in C(Sl). That is, A0 = {p(¢) : p is a polynomial}-. Show

that A0 is a Riesz subspace of C(Sl).

Solution: Consider the square root function f (t) = f on [0, 1]. By the classical Stone-Weierstrass Approximation Theorem, the function f can be approximated uniformly by a sequence of polynomials, say {q,,}. That is, flgn - f Il,o -i 0.

Assume that x is a function of the form x = p(o), where p is a polynomial. We can suppose without loss of generality that IIxII, < 1 (otherwise, we replace x Clearly, rn(x) = gn(x2) = gn(p(o)2) E A0 for each n and by the function

IIrn(x)-IxIlL = (j gn(x2)-f(x2)jj. <- Ilgn-flloo -'0. Thus, IIrn(x)-IxIll. -0,

-

and so IxI E A0. From the inequality IIlyl jzjII S Icy - zII, for all y, z E C(1), we get IxI E Ae for each x E A0. This implies Ao is a Riesz subspace of C(Q). I

Problem 4.2.9. Let (6, E, p) be a or-finite measure space and let 0 be in

Li(p). Show that for the multiplication operator Me,: Lp(p) - Lp(p), where 1 < p < oc, the following statements are equivalent. (1) The function 0 is constant on some measurable set of positive measure. (2) NI0 commutes with a rank-one positive operator. (3) Mo commutes with a rank-one operator. (4) NO commutes with a non-zero finite-rank operator.

Solution: (1)

(2) Fix a measurable subset A of 0 satisfying 0 < p(A) < oc and 0(a) = c for each a E A. Then it is easy to verify that Mo commutes with the rank-one positive operator K: Lp(p) -. Lp(p) defined by Kx = (fAxdp)XA (2)

(3)

(4) Obvious.

(4) (1) The proof of this implication is almost identical to that of Theorem 4.20. The arguments in the proof of Theorem 4.20 remain the same up to the verification of formula (**). After that we need to conclude that there exists some measurable set A with 0 < p(A) < oe such that p(O(a)) = 0 for all a E A. Recall that p has a finite number of zeros, say al , ... , a,,,. So, if A, = {a E A: O(a) = A, I. then A = U;" A,. Hence p(A,) > 0 for some i, and thus 0 = A, on this A. 1

4. Special Classes of Operators

128

Problem 4.2.10. Let r :

1? be a continuous mapping on a compact Hausdorff space Q. and let C,: C(S2) -. C(St) be the composition operator defined by C,(x) = x o r. Establish the following. SZ

(a) C, is a contractive positive operator satisfying C,(1) = 1. (b) C, commutes with a rank-one positive operator. Solution: (a) Clearly, C, is a positive operator and I [CT(x))(w)I = Ix(r(w))I <_ Ilxllx

for each ,.1E Q. Therefore, IIC,xIIx = sup,,En1[C1'(x))(ua)1 < Ilrll"'. This shows that 11C,11 < 1. and so C, is a contraction. Moreover, note that [C,(1))(w) = 1(r(w)) = 1 for each w E Q. This shows that C,(1) = 1 (and that II C, II = 1).

(b) Let r: S2 - 11 be a continuous mapping. We claim that there exists

some 0 < r' E C'(S1) satisfying CTx' = x'. To see this, consider the set C = {x' E C+(S2): x'(1) = 1} and note that C is non-empty, w'-compact, and convex. Using (a), an easy argument shows that CT (C) C C, and moreover CT : (C, u'') - (C, uw') is continuous. Now invoke the Brouwer--Schauder-Tychonoff theorem (see, for instance [4, Corollary 14.51, p. 485)) to conclude that CT has a

fixed point. Thus, there exists some x' E C with CTx' = x', i.e., x* (x) = x'(xor) holds for each x E C(1). (Regarding this conclusion, see also Lemma 9.45.) Next, fix some x' > 0 such that CT r.' = r' and consider the rank-one operator K = x' ,S 1. We claim that K and C, commute. Indeed, for each x E C(st). on one hand we have

C,Kx = C,(x'(x)1) _ [x'(x)1] o r = x'(x)1, and on the other hand

KC,x = K(x or) = x'(x o r)1 = x'(x)1 .

I

Thus. KC, = C,K.

Problem 4.2.11. Show that for a continuous map r : S2 -p Q. where Il is compact and Hausdorff. the following statements are equivalent.

(1) The map r is surjective. (2) The composition operator C, is an isometry. (3) The composition operator C, is one-to-one.

Solution: (1) ; (2) Fix x E C(f) and let t E Q. Since r is surjective, there exists some w E 1? such that r(w1) = t. So,

Ik(t)I = Ix(r(w))I = I(CTx)(w)I <_ IICsiII,

holds for all t E 0. This implies Ilxllx < suptEQIx(t)I < IICTxIIa Since, by Problem 4.2.10. C, is a contraction, we infer that IIC,xDIx = Ilxllx. Thus. C, is an isometry. (2)

(3) Obvious.

4.3. Lattice and Algebraic Homomorphisms

129

(3) (1) Assume by way of contradiction that r is not surjective. This means that there exists some wo E S2 such that wo V r(Q). Since fl is compact and r is continuous, we know that r(S2) is compact, and hence closed. Now pick some

function x E C(Q) such that x(wo) = 1 and x = 0 on r(SZ). But then C.,(x) = 0, contradicting the fact that C, is one-to-one. Hence, r is surjective. I

4.3. Lattice and Algebraic Homomorphisms Problem 4.3.1. According to Theorem 4.25, for any given lattice homomorphism T : C(f2) C(Q) there exist a mapping r: Q - f2 and some w E C(Q) such that for each f E C(1l) and each q E Q we have Tf(q) = w(q)f(r(q))

Show that w = T10 and that the mapping r is uniquely determined and continuous on the set Qo = {q E Q: w(q) > 0}. Solution: Clearly, w = Tin E C(Q), and if w(q) > 0, then r(q) is uniquely determined. Now assume that q,, -b q holds in Qo. Then w(q0) -i w(q) > 0 and

w(9Q)f(r(9a)) =Tf(q.)

Tf(9) =w(q)f(r(9))

So. f (r(q.)) -y f (r(q)) for each f E C(SI). This implies r(q,,) -+ r(q), and hence r is continuous on the set Qo = {q E Q: w(q) > 0}. 1

Problem 4.3.2. Give an example of a lattice homomorphism which is not an algebraic homomorphism.

Solution: Consider the operator T : C[O,11 -- C[O,1] defined by T(f) = 2f . Notice that f Ag = 0 implies T(f) AT(g) _ (2f) n (2g) = 2(f Ag) = 0. and so T is a lattice homomorphism. However. T(12) = T(1) = 2. 134 (2. 1)(2. 1) = T(1)T(1), and this proves that T is not an algebraic homomorphism. 1

Problem 4.3.3. Show that for a Markov operator T : C(Q) - C(Q) the following statements are equivalent.

(1) T is an algebraic homomorphism. (2) T is a lattice homomorphism. (3) T is a composition operator.

Solution: (1) (2)

(2) This is Theorem 4.26.

(3) This follows from Theorem 4.25 and the fact that Tin = 1Q.

. (3) (1) Assume that T f = for for some continuous function r : Q -. Q. Then for each f, g E C(Q) and each q E Q we have

[T(fg)](q)

=

[(fg)o rI(q) = (fg)(r(9)) = f(r(9)).g(r(9))

=

[(Tf)(9)] [(Tg)(9)] = [T(f)T(g)](q)

4. Special Classes of Operators

130

That is. T(f q) = T(f)T(g). and so T is an algebraic homomorphism.

I

Problem 4.3.4. Let 11 and Q be two compact Hausdorff spaces, and consider the convex set C of all Markov operators from C(1l) to C(Q), i.e.,

C= {TE C(C(n).C(Q)): T>0 and Tln=1Q}. Show that a Markov operator T is a lattice homomorphism if and only if T is an extreme point of C. Solution: Assume first that T is an extreme point of C. Fix a function h E C(fl) such that In < h < 2 lit, and then define the operators S, R: C(Q) - C(Q) by

Sf=TT( 1)

and

Rf=2Tf-Sf.

Notice that S, R E C and T = 1 2S + .12 R. Since T is an extreme point of C, it follows that T = S. This implies T(hf) = T(h)T(f) for each f E C(1?) and all h E C(c) satisfying In < h < 2 In. Now consider any non-zero g E C(fl).

Then the function h = In + I- E C(f2) satisfies In < h < 2. 10, and so T((ln + -)f) = T(ln + Tj.- -)T(f) for all f E C(rl). This easily implies T(y f) = T(g)T(f) for all f E C(ft) and all g E C(fl). Thus, T is an algebraic homomorphism, and hence a lattice homomorphism.

For the converse, assume that T E C is a lattice homomorphism. By Theorem 4.25, there exists a continuous mapping r : Q - n such that T f = f o r for each f E C (Q). Assume that T = QS+ (1- (x)R holds with S, R E C and 0 < a < 1. Clearly, T. S'. R' : ca(Q) -+ ca(f2), where ca(Q) and ca(fl) are the AL-spaces of all signed Bore] measures on Q and S2, respectively. Also, it should be noticed that the norm dual of every Markov operator carries probability measures to probability measures. Next, observe that for each q E Q and each f E C(1l), we have

(f. T'bq) _ (Tf.6q) = (Tf)(q) = f (r(q)) = (f. bj(q)) . This shows that T'bg = bi(q). and consequently br(q) = T'bq = oS'bq + (1 - a)R'bg . Since the extreme points of the convex set of all (Borel) probability measures on 0 are the Dirac measures (why?). it follows that T'bq = S'bq = R'bq for each q E Q. Therefore.

(Tf)(q) = (Sf)(q) = (Rf)(q) for all f E C(S2) and each q E Q. Therefore. T = S = R, and so T is an extreme point of the convex set C.

Problem 4.3.5. Show that a positive operator T : E -+ F between two Banach lattices is a lattice homomorphism if and only if Tc : Ec - Fc is a complex lattice homomorphism, i.e., ITczI = Tizj for all z E Ec. Solution: Let z = .r + zy E Er. and put u = lxi + iyl E E and v = Taxi + Tiyl _ Tu E F. It is a routine matter to verify that C Fe.. and so T: Eu F,, is a Markov operator. Moreover. it is easy to see that in order to verify that

TIzI. it suffices to consider T as an operator from E to F,.. From the Kakutani Bohnenblust- Krein Representation Theorem, we can assume that

131

4.3. Lattice and Algebraic Homomorphisms

T : C(1l) - C(Q). Theorem 4.28 guarantees this T is a composition operator. That is, there exists a continuous function r: Q - Q such that

Tf(q) = f(r(q)) holds for all f E C(1l) and all q E Q. In particular, note that for each q E Q we have ITTzl(q)

= ITS + tTyj(q) =

Vlr(Tx)l + (Ty)2 (q)

[(Tx)(q)]2 + [(Ty)(q)]2 =

[x(r(q))]2 + [y(r(q))]2

x2 + y2(T(q)) = Ix + iyl(r(q))

_

[TIzMI(q)

That is, ITzl = TIz[, as desired.

I

Problem 4.3.6. For a multiplicative linear functional 0 on a Banach algebra A establish the following.

(a) If a E A is an idempotent element (i.e., a2 = a), then show that 0(a) = 0 or q5(a) = 1. (b) Let p and q be two idempotent elements in A. Then show that the linear functional V' defined on A by V,(a) = O(gap) is also a multiplicative functional--even though the mapping a - qap need not be an algebraic homomorphism. Solution: (a) If a2 = a, then multiplicativity of 0 yields ]0(a)]2 = O(a2) = 0(a). This implies either 0(a) = 0 or 0(a) = 1. (b) It should be clear that tl' is a linear functional. For the multiptica.tivity of 0 notice that for any a and b we have tG(ab) = o(gabp) = O(q)-O(a)¢(b)O(p)

Therefore, from 0(p) = [O(p)]' and ¢(q) _ [0(q)]2, we get = [O(q)]20(a)0(b)[0(P)]2 = [O(q)O(a)O(p)j[0(q)O(b)O(p)] _ O(gap)O(gbp) = t(i(a)ip(b). This shows that t& is also multiplicative. V,(ab)

Problem 4.3.7. Fix any Banach space X and let Y = X q) X % . ®X be the direct sum of n > 1 copies of X; in particular, let Y = C" or Y = R" where n > 1. Show that the only multiplicative linear functional on the Banach algebra £(Y) is the zero functional.

Solution: Let 0 be a multiplicative linear functional on L(Y). Since T = TI holds for each T E L(X), it follows from the multiplicative property of 0 that 0(T) _ cb(T)0(I). This shows that ¢ = 0 if and only if 0(I) = 0. So, we must show that 0(1) = 0. Define the bounded operator J: Y -0 Y by

J(T1,x2,...,xn) = (xn,x1,x2,...,x.-1) and note that J" = I. Therefore, 0(1) = [0(J)]", and so in order to establish that 0(I) = 0, it suffices to prove that 0(J) = 0.

4. Special Classes of Operators

132

To this end, for each 1 < i < n define the bounded operator T, E C(Y) by T. (XI , x2..... X.) = (0,...,0,x,,0-.,0), where the vector x, occupies- the (i + 1)position. Also. define T E C(Y) by T (xl. X2, ... , (r,,, 0, 0, ... , 0). Clearly, T,2 = 0 for all 1 < i < n. and therefore from [©(T,)J2 = o(T,2) = 0 we get o(T,) = 0 o(T,) = 0, for each i. Now note that J = ; 1 T, holds. This implies ¢(.I) as desired.

Problem 4.3.8. For a Banach lattice E. a majorizing vector subspace X of E. and an arbitrary nonmed Riesz space F establish the following automatic continuity property: Every positive operator T : X -- F is continuous. Solution: Let F be the norm completion of F. According to Problem 1.2.21 the Dedekind completion F6 of P is a Banach lattice that contains k as a Banach subF as an operator from X to F6, i.e.. T : X -* P. lattice. Now consider T : X Since F6 is Dedekind complete and X majorizes E, it follows from the Kantorovich Extension Theorem 4.32 that the operator T : X -+ F6 has a positive linear ex

tension to all of E, say t: E -+ P. By Theorem 1.31 the positive operator t is continuous. This implies that the operator T: X - F is likewise continuous.

Problem 4.3.9 (Arendt [13]). Let F be a Dedekind complete Riesz space and let R: E -' H be a positive operator between two Riesz spaces. Then the positive operator R gives rise to a positive operator T '-. TR from Cr(H, F) to Cr(E, F). Establish the following. (a) If R is a lattice homomorphism, then T

TR is interval preserv-

ing.

(b) If R is interval preserving, then T i--. TR is a lattice homomorphism.

Solution: (a) Assume that R is a lattice homomorphism. A straightforward verification shows that the mapping T --+ TR, from L,.(H. F) to Cr(E. F). is indeed a positive operator.' To verify that this positive operator is interval preserving.

fix 0 < T E C(H. F) and assume that some operator S E Cr(E, F) satisfies 0 < S < TR. We must show that there exists some operator Ti E Cr(H, F) satisfying 0 < Tl < T and T,R = S. We start by defining an operator T1 : R(E) F by T,(Rx) = Sx, where R(E) is the range of the operator R in H. Observe that Tl is a well-defined operator. Indeed, if Rx = Ry. then from

lSx - Syl <- Six - y! < TRIx - yE = TIRx - Ryl = 0. it follows that Sx = Sy. Next. notice that for each x E E we have

T (Rx) = Sx < Sr+ < TR(x+) = T(-x)'

(*)

In addition, it is easy to see that the mapping p: H F. defined by p(h) = T(h+). is sublinear. So. by (*) and Theorem 4.31. there exists a linear extension of Ts to 1 Operators between vector spaces of operators are called transformers.

4.3. Lattice and Algebraic Homomorphisms

133

all of H (which we shall denote by T1 again) satisfying T1h < p(h) = T(h+) for each h E H. This implies Ti < T, and moreover if h E H+, then

-T1(h) = T1(-h) < T((-h)+) = T(O) = 0.

and so T1h > 0. Therefore, T1 is a positive operator. This operator satisfies

0
(b) Assume that R is interval preserving, and let S, T E Cr(H, F). Now fix x E E+. From [0, Rx] = R[0. x], it follows that two vectors y, z E [0, Rx] satisfy

y + z = Rx if and only if there exist vectors u, v E [0, x] such that u + v = x, y = Ru, and z = Rv. This implies [SR V TR] (x) = sup {SR(u) + TR(v) : u, v E E+ and u + v = x}

= sup{Sy+Tz: y.zEH+ and y+z=Rx} = (SVT)(Rx) = [(SVT)R](x). Therefore, SR V TR = (S V T)R in Cr(E, F), and this shows that T '- TR is a lattice homomorphism.

Problem 4.3.10 (Aliprantis-Burkinshaw-Kranz [9]). Let R: H -' F be a positive operator between two Dedekind complete Riesz spaces and let E be an arbitrary Riesz space. For the positive operator T s-' TR, from C,(H, F) to C, (E, F), establish the following.

(a) If R is an order continuous lattice homomorphism, then T .- RT is a lattice homomorphism.

(b) If the operator T RT is a lattice homomorphism and the order dual F' is non-trivial, then R is a lattice homomorphism. Solution: (a) We shall use here the following characterization of the supremum of two operators:

If X and Y are two Riesz spaces with Y Dedekind complete, then for each pair of operators A, B E Cr(X. Y) and each x E X+ we have: n

n

{j:Ax, VBxi: xi E E+ and Ex, =x} j (AVB)(x). i=1

1=1

(For a proof see [6, Theorem 1.16, p. 15].) Now assume that R is an order continuous lattice homomorphism. and let S,T E Cr(E, H). Using the above formula, for each x E+ we have n

n

R(S V T)(x)

=

x, = x))

Sxi V Txi : xi E E+ and

R(sup

n

n

sup {R(j:Sx,

vTxi) : xi E E+ and

t=1

n

sup { E RSxi v RTxi : 2=1

=

[(RS) v (RT)] (x).

xi = x} i=1

n

xi E E+ and E xi = x} i=1

4. Special Classes of Operators

134

Thus, R(S V T) = (RS) V (RT), and so T - RT is a lattice homomorphism.

RT is a lattice homomorphism. We (b) Assume that the transformer T claim that R is also a lattice homomorphism. To see this, fix some 0 < p E F^' and then use Lemma 4.13 to obtain that for each h E H we have o Qv IRhI = to ®®Rhl = (o ®®Rh) v (-(d a Rh)) = R(,p®h) v R(p ®(-h))

= R[(a o h) v (o o (-h))] = R(p o lhl) = p ®Rlhl . This easily implies Rihi = IRhl, and so R is a lattice homomorphism.

I

Problem 4.3.11 (Kim [48]: Andd-Lotz [55]). For an arbitrary positive operator T : E - F between two Banach lattices establish the following.

(a) If T is interval preserving, then its adjoint T*: F'

E' is a lattice

homomorphism. (b) T is a lattice homomorphism if and only if T' is interval preserving. Solution: (a) Let T: E -, F be an interval preserving positive operator between two Banach lattices. Start by observing that T' y' = y' oT holds for each li E F. By Problem 4.3.9(b), the operator y' - y' oT = T* y* is a lattice homomorphism. That is, T*: F' -+ E' is a lattice homomorphism. (b) Assume first that T: E - F is interval preserving. By Problem 4.3.9(a), it

follows that the operator y' '- y' o T = T' o y' (i.e., the operator T*: F' - E') is interval preserving. For the converse, assume that T' is interval preserving. Then by part (a). the

operator T*': E" -' F" is a lattice homomorphism. Since E is a Riesz subspace of E" and T" = T on E. it easily follows that T : E - F itself is indeed a lattice

I

homomorphism.

Problem 4.3.12. Let E = e,, and let c denote the Riesz subspace of all convergent sequences. Show that the limit functional Cim: c -+ R, defined by Cim(x) = limn-,, xn for each x = (xt, x2, ...) E c, is a lattice homomorphism that has a lattice homomorphic extension to all of f,. Solution: It should be clear that c is a majorizing Riesz subspace of fx. In addition. for any x = (xl. x2, ...) E c we have

ICim(x)[ _ Irim

x, =

m

Cim(lxl)-

ri homomorphism. By Corollary 4.36, This shows that the limit functional is a lattice Lim has a lattice homomorphic extension to t,

4.4. Fredholm Operators Problem 4.4.1. Show that a bounded operator T : X --. Y between real Banach spaces is Fredholm if and only if the operator Tc : Xc -+ Yc is likewise Fhedholm--in which case we have i(T,,) = i(T).

4.4. Fredholm Operators

135

Solution: The solution is based upon the following simple algebraic property:

A real vector space V is finite dimensional if and only if its complexification V, = V ® zV is finite dimensional-in which case we have

dimV=dimV, To see this, assume first that V is finite dimensional. Let e1, es, ..., en be a basis of

V and fix v. u E V. Write v = =z a. e. and u = =1,3,e,, and then note that n

n

v+zu=>aje3+zEQjej =F (a;+z13,)e;. j=z

j=1

Since the collection of vectors e 1, e2.... , en is also linearly independent in V,, (why?),

the latter identity easily implies that let, e2, ... , en) is also a basis in V.. So, V,, is finite dimensional and dim VV = dim V. For the converse, assume that VV is finite dimensional and let {uj + be a basis in V. Now let v E V. Then there exist complex numbers {aj + 0, }j such that n

n

n

J=1

1=1

i=1

v+z0=E(aJ+z,3J)(u,+zw;)=Dajuj - 33w;)+z[E((I,w;+Q;u;)]. This implies v =

u, - ,3, w.), and so V is spanned by the finite collection

of vectors { u1, ... , nn } U { w1, ... , wn }. Therefore, V is finite dimensional.

Y between real Banach spaces. Now consider a bounded operator T: X Clearly, TT(x + zy) = Tx + zTy = 0 if and only if Tx = Ty = 0 or equivalently, if and only if x, y E N(T). This implies N(TT) = N(T) ® zN(T) = [N(T)],,. The identity R(TE) = R(T)E) zR(T) = [R(T)], is also straightforward. Next, assume that T : X -. Y is a Fredholm operator. Pick a closed subspace

V of X and a finite dimensional subspace W of Y with X = N(T) ® V and R(T) ® W = Y. It is easy to see that the vector subspace V. of X. is closed, X. = N(T.) ®V., and Y. = R(TE) ED W.. From the above property, we infer that the operator Tc is a Fredholm operator and that i(Tc) = n(TT) - d(TT) = dim N(TT) - dim We = dim N(T) - dim W = i(T). For the converse, suppose that Tc is a Fredholm operator. From the property

stated at the beginning of the solution, we have dim N(T) = dim N(TT) < oo. Next, pick a finite dimensional subspace W1 of Y. such that Y. = R(Tc) ®[4'1 = [R(T) ® zR(T)] ® Wl ,

(*)

and let U = {x E X : there exists some z E X such that x + zz E W1 }. Clearly, U is a vector subspace of X. Since W'1 is finite dimensional, an easy argument shows that U is also finite dimensional. Moreover, from (*), it easily follows that R(T) + U = Y. Now a glance at Problem 4.4.3 guarantees the existence of a finite dimensional subspace U1 of U such that Y = R(T) ® U1. This implies d(T) < oo. Therefore. T is a Fredholm operator and the solution is finished. I

Problem 4.4.2. Prove Rieszs lemma without using the Hahn-Banach Extension Theorem: If Y is a proper closed subspace of a (real or complex)

4. Special Classes of Operators

136

Banach space X, then for each 0 < e < 1 then exists some unit vector x E X such that

Ily-xll>1-e

holds for all y E Y. Also, present an example of an infinite dimensional proper closed subspace Y of a Banach space X such that for each unit vector x E X \ Y we have d(x, Y) < 1.

Solution: Fix some xo E X \ Y. and note that d = d(xo, Y) > 0. Next. select some yo E Y with llxo - yoll < ids and let x = 1z-V0 Clearly. Ilxll = 1 and if y E Y, then llx - yII

= 11 xo-yo

y= 11

I zo- Vo+ Izo-Va xo-Vo

I>

d - Izo-Vol > 1 - E .

Now we present an example of an infinite dimensional closed vector subspace Y of a Banach space X such that d(x. Y) < 1 holds for each unit vector x E X. We let X = c. the Banach space of all convergent sequences equipped with the norm defined by 11xII = 2IIxIIx +

Z-n+rIxnI n=1

for each x = (x1, x2, ...) E c. We also let Y = co, the closed vector subspace of c consisting of all sequences converging to zero.

Now take any unit vector x E X. We claim that d(x, co) < 1. To see this, fix some k with xk satisfies

0, and note that the vector y = (T.1, x2.... 12-k, 0, 0....) E co

x

Y11

n>k

x

Ix-1 + E 12,. Ixnl < 21IxIIx + E 2-1 Tjx"j = l . n=k+1

n=1

This shows that d(x. co) < Iix - yll < 1.

Problem 4.4.3. This problem presents some elementary algebraic properties that have been used in the proofs of the results in Section 4.4. Establish the following.

(a) If W is a vector subspace of a vector space V, then W has an algebraic complement, i.e., there exists a vector subspace U of V satisfying W n U = {0} and W + U = V. (b) If U and W are two vector subspaces of a vector space V, then there exists a vector subspace W1 of W satisfying V n W1 = {0}

andU+Wi =U+W. (c) If T : V - W is a linear operator between two vector spaces, then the quotient vector space V/N(T) is linearly isomorphic to R(T). Solution: (a) Let W be a vector subspace of a vector space V. Fix a Hamel basis {e=};El of W and then choose a family {f,}jEj of linearly independent vectors in V such that the collection { e; : i E 11 U { fi: j E J} is a Hamel basis of V. (Zorn's lemma guarantees that such a family of linearly independent vectors exists.) Now

4.4. Fledholm Operators

137

let U be the vector subspace generated by the family { f, We claim that U is an algebraic complement of W. To see this, assume first that x E W n U. Then we there exist a family of scalars {A1},EI with Ai = 0 for all but a finite number of i and a family of scalars {µ,},E with pj = 0 for all but a finite number of j satisfying x = E,EI A,ei = E,,E it jfj. This implies >iEI Aiei + 0. The linear independence of the set

lei: i E I} U { fj: j E J} implies A, = 0 for all i E I and µj = 0 for all j E J.

Therefore, x = 0, and so W fl U = {0}. T o see that W + U = V, pick any v E V. Since lei: i E I} U { fj: j E J} is a Hamel basis, there exist families of scalars {Ai},EI with Ai = 0 for all but a finite

number of i and {µI}.,E with pI = 0 for all but a finite number of j such that v = EiEI Aiei + `)E p.f . If we let w = E.EI Aiei E W and u = `)E µ,fi E U. then v = w + u holds. Therefore, V = W E) U.

(b) By part (a) there exist vector subspaces U1 of U and W1 of W such that

U = U1e(WnU) and W = Wie(WnU). It is easy to check that W1nU1 ={0}.

So, U+W =U1e(WnU)eW1 =UeW1. (c) Consider the mapping t: V/N(T) -+ R(T) defined by T([x]) = Tx. Since [x] = [y] is equivalent to x - y E N(T) or Tx = Ty, it follows that the mapping t is well defined. If T[x] = Tx = 0, then x E N(T) or [x] = 0. That is, t is one-to-one. Moreover, t is linear and surjective. Therefore, t is an algebraic isomorphism, and so the vector spaces V/N(T) and R(T) are linearly isomorphic. I

Problem 4.4.4. Let X be an n-dimensional vector space, let Y be an mdimensional vector space, and let T : X -i Y be an operator. If we let r = dim R(T), then show that n(T) = n - r and d(T) = m - r; conclude from this that i(T) = n - m. Solution: Choose a vector subspace V of X and W of Y such that X = N(T)®V and Y = R(T)eW. Since V and R(T) are isomorphic, it follows that n(T)+r = n. Thus, n(T) = n - r. On the other hand, we know that d(T) = dim [Y/R(T)] = dim Y - dim [R(T)] = m - r. Thus, i(T) = n(T) - d(T) = n - r - (m - r) = n - m. (See also Problem 2.1.8.)

Problem 4.4.5. For a bounded operator T : X -p Y between Banach spaces define T : X/N(T) -. Y via the formula t[x] = Tx. Let rr : X - X/N(T) be the quotient map and J: R(T) - Y be the natural embedding. Now consider the scheme of bounded operators X "-+ X/N(T) --T!-- R(T) J - Y,

and note that T = JTrr. If T is a Fredholm operator, then show that all operators in the above scheme are Fredholm operators and compute their indices. Also, verify directly that i(T) = i(J) + i(T) + i(ir). Solution: Assume that T : X -+ Y is a Fredholm operator. We shall compute the indices of the introduced operators individually.

4. Special Classes of Operators

138

The operator 7r: X - X/N(T) is surjective and its null space is N(T). So, i(7r) = n(7r) - d(7r) = n(T) - 0 = n(T). The operator T: X/N(T) -. R(T) is an isomorphism, and thus i(T) = 0. Finally, the natural embedding J: R(T) -. Y satisfies N(J) = (0) and R(J) = R(T). Therefore, i(J) = dim N(J) - dim Y/R(J) = 0 - dim Y/R(T) _ -d(T). Consequently, we have

i(J) + i(t) + i(7r) = -d(T) + 0 + n(T) = n(T) - d(T) = i(T) = i(JT7r), which, of course, is the assertion of the Index Theorem.

Problem 4.4.6. If K : X - X is a compact operator on a Banach space, then show that the operator I - K has a finite descent. Solution: Assume by way of contradiction that R((I - K)n+i) is a proper vector subspace of R((I - K)n) for each n. Since (I - K)n = I - Kl for some compact operator K1, it follows from the proof of Lemma 4.45 that R((1- K)n) is a closed subspace for each n. So, by Riesz's Lemma 4.44, there exists a sequence {yn} in X satisfying yn E R((I -K)n) for each n and IIyn - yII > 1 for each y E R((1-K)n+i)

Now if lm< n, then we have

Kym-Kyn=ym-[yn+(I-K)ym-(I-K)yn] and y = yn + (I - K)y,n - (I - K)yn E R((I - K)-+l). This implies that for n 0 m we have JjK ym - Ky.11 > I l ym - y ll >_ 1 Consequently, the sequence { Kyn }

cannot have any convergent subsequence, contrary to the compactness of K. This contradiction establishes that I - K has a finite descent.

Problem 4.4.7 (The Fredholm Alternative). Let X be a Banach space and K E Y (X ). Show that for each A # 0 exactly one of the following two exclusive statements is true:

(a) For each y E X the equation x - AKx = y has a unique solution. (b) The equation x - AKx = 0 has a non-zero solution.

When (b) is true, show also that the equation x - AKx = 0 has a finite number of linearly independent solutions. Solution: Let us verify first that statements (a) and (b) are mutually exclusive. Assume that (a) and (b) are true simultaneously. If some x 96 0 is a solution for the equation in (b), then letting y = 0 in (a), we see that 0 and x are two distinct solutions of (a), which is a contradiction. Next, we shall show that either (a) or (b) is true. We consider the following two cases. CASE I:

o(K).

In this case, the operator - K is invertible. So, the operator I - AK is likewise invertible. This implies that for each y E X the equation y = x - AKx = (I - AK)x has a unique solution, namely, x = (I - AK)-'y.

CASE II: z E a(K).

4.5. Strictly Singular Operators

139

Since K is a compact operator. it follows that a is an eigenvalue of K, and so the equation x - AKr = 0 has a non-trivial solution. Now Lemma 4.45 guarantees that the non-trivial vector subspace

NK(1)={xEX: Kx= ax}={rEX: r-AKx=0} is finite dimensional.

Problem 4.4.8. Let T : X -+ Y be a bounded operator between Banach spaces having a pseudoinverse S : Y equation

X. Fix some y E Y and consider the

Tx = Y.

(*)

Show that (*) has a solution if and only if Sy is a solution of (*). i.e, T(Sy) = y. Solution: If x = Sy is a solution of (*), then (*) has at least one solution. For the converse. assume that some x E X satisfies Tx = y. Then we have

y = Tr = (TST)x = TS(Tx) = TSy = T(Sy). That is. Sy is a solution of (*).

Problem 4.4.9. Show that a bounded operator between Hilbert spaces has a pseudoinverse if and only if it has a closed range. Solution: Let T : H1 - H2 be a bounded operator between Hilbert. spaces. If T has a pseudoinverse. then it follows from Lemma 4.52(2) that there exists a closed subspace li' of H1 such that R(T) -61 1V = H2. But then, by Theorem 2.16, R(T) is a closed subspace.

For the converse, assume that R(T) is closed. Since, N(T) and R(T) are closed subspaces of the Hilbert spaces H1 and H2, respectively, it follows that they

are both complemented-keep in mind that every closed subspace in a Hilbert space is complemented. This. coupled with Lemma 4.52, guarantees that T has a pseudoinverse.

4.5. Strictly Singular Operators Problem 4.5.1. Show that a bounded operator T : X -> Y between Banach spaces is strictly singular if and only if T is not bounded below on any infinite dimensional closed subspace of X. Solution: Since IITxII < IITII IIxik for each x E X. it easily follows that if Z is a closed subspace of X, then T : Z -+ T(Z) is an isomorphism if and only if there exists some r > 0 such that e1Iz11 <_ IITzII for each z E Z. That is, T: Z - T(Z) is an isomorphism if and only if T is hounded below on Z. Therefore, T is strictly singular if and only if I' is not hounded below on any infinite dimensional closed subspace of X.

i

4. Special Classes of Operators

140

Problem 4.5.2. Let X; (i = 1, 2, ... , n) and Y) (j = 1.2.... , in) be all real (or all complex) Banach spaces and consider a bounded operator gym. T: X1 T, As in Problem 1.1.11, it is easy to see that there exist (uniquely determined)

bounded operators Tai : Xi -+ Yj (i = 1, 2..... n; j = 1, 2..... in) such that T can be represented by the in x n matrix T1

T2

T= Tm1 Tm2

T,,,n

Show that T is strictly singular if and only if each T;, is strictly singular. Solution: For simplicity, let X = X1 X2 - - ri- X and Y = Y, :-- 1s -:- --- I'M. Assume first that the bounded operator T: X - Y is strictly singular, and assume -

by way of contradiction that some operator T,, is not strictly singular. This means that there exists an infinite dimensional vector subspace V of X, such that T,, is bounded below on V, i.e.. there exists some c > 0 such that ;1T,,tyll > rllrll holds {0},whereV for alli'E V. If we let ti'={0}G{0}e (0) i?-V-,D{0}! occupies the ith summand. then it is easy to see that V is an infinite dimensional vector subspace of X and IDTrll > cJJxll holds for each r E V. The latter contradicts the fact that. T is strictly singular. Therefore, each T,, is strictly singular.

For the converse, assume that each T,, is a strictly singular operator. Let

P,; X X he the natural projection of X with range X, and define R, : } Y 0 c:; y, 0 e 9 0. where y, occupies the jt1i coordinate. by Ry, = 0 e 0 e Now consider the scheme of bounded operators X A - X, T.. 1l +-e Y. and

0T,,X,:0x...:,0, let T,3 = R;T,,P,. Clearly. ?,(rltDr2tD--where the vector T,,x, occupies the jth component. By Corollary 4.62. each 7j, is a strictly singular operator. Now from T = F,n i E 1 T,, and Corollary 4.62. we easily infer that T is a strictly singular operator. I =040M..

Problem 4.5.3. Show that a bounded operator T : X - Y between two real Banach spaces is strictly singular if and only if TT : X,, -. Ye is strictly singular. Solution: Assume that T'C is strictly singular. If T is not. strictly singular. then there exists an infinite dimensional closed subspace Z of X such that T: Z -. T(Z) is all isomorphism. So. there exists some constant f > 0 satisfying Fllzll < IITzll for each z E Z. Clearly, Z. = Z C; zZ is an infinite dimensional closed subspace of Xe. Now if a + zb E Zr, then for each angle 0 we have IITc(a+zb)Il

=

IITa+zTbll > II(Ta)cost+(Tb)sinOII IIT(asiii9+bcose)Il > ellasinO+bcosoll -

That is. Te restricted to Z, is an isomorphisin, which is impossible. Therefore. T is strictly singular. This implies Ella+zbli <_

IITc(a+tb)II.

4.5. Strictly Singular Operators

141

For the converse, assume that T is strictly singular, and assume contrary to our claim that there exists an infinite dimensional vector subspace V of X. and some b > 0 such that IIT,,zII > 611z11 holds for all z E V. Now consider the vector subspace of X ® X defined by

W= {xl ®x2 : x1, x2 E X and x1 + 1x2 E V } . Note that W is an infinite dimensional vector subspace of X ® X and that

IIITeT](xl aX2)II

= >

IITx111+ IITx2II > IITT(xl+Z2)II 611x1 + =211 > 4

Ilxi ® x211

holds for each x1 ® x2 E W. This shows that the operator T ®T : X ®X -+ Y eY is strictly singular, contrary to the conclusion of Problem 4.5.2. Therefore, TT must be strictly singular. (The proof of this part was communicated to the authors by Julio Flores.)

Problem 4.5.4. Prove the following property of Lp-spaces that was used in Theorem 4.58: If X is an infinite dimensional vector subspace of any Pp-space, then for each m E N the vector subspace

Xm={x=(xl,x2.... )EX:

x1=x2=...=xm=0}

is also infinite dimensional. Solution: The proof is based on the following property: If Y is an infinite dimensional vector subspace of any ly-space, then for every m E N the vector subspace Ym = {y E Y : y,,, = 0} is also infinite dimensional. The proof of this claim is as follows. Consider two arbitrary linearly indepen-

dent vectors u, v E Y. If u,,, = 0 or Vm = 0, then either u E Ym or v E Ym. If u,,, # 0 and v,,, 96 0, then the vector w = umv - Vmu E Ym is non-zero since u and v are linearly independent. Therefore, we have shown that if u, v E Y are linearly

independent, then there exists a non-zero vector in the linear span of {u,v} that belongs to Y,,,. Now let { u 1, u2, ... } C Y be a countable linearly independent set.

For each n, let w be a non-zero vector in the linear span of {u_, u2.) that lies in Y, . It follows that {w1, w2, ...} is a countable linearly independent subset of Y,,,, and so Y,,, is infinite dimensional.

We are now ready to prove the result using induction on m. For m = 1, the conclusion follows immediately from the above established property. Now assume that X,,, is infinite dimensional for some m > 1. From

Xm+1 = {x = (xl. x2, ...) E X : XI=X2=

= xm = 01

= (xEXm: xm+1=0} and the above established result, we infer that X,,,+1 is infinite dimensional. This completes the induction, and the desired conclusion has been established.

Problem 4.5.5. Assume that for a pair of Banach spaces (X, Y) we have the following:

(a) Each infinite dimensional closed subspace of X contains a closed vector subspace that is isomorphic to X.

4. Special Classes of Operators

142

(b) No closed vector subspace of Y is isomorphic to X. Show that each operator in £(X, Y) is strictly singular. (It is known that the pair (El, Ep), where 1 < p < oc, satisfies properties (a) and (b); see 152, Proposition 2.a.2, p. 53].) Solution: Let (X, Y) be a pair of Banach spaces satisfying properties (a) and (b), and assume that contrary to our claim some operator T E £(X, Y) is not strictly singular. Then we can find an infinite dimensional closed vector subspace X0 of X such that the restriction T : Xo -+ Y is an isomorphism. By (a) there exists a closed vector subspace X1 of X0 that is isomorphic to X. Consequently, the closed vector subspace T(XI) of Y is isomorphic to X which contradicts (b). Hence, C(X, Y) consists of singular operators.

Problem 4.5.6. This problem presents an example of a strictly singular operator whose adjoint is not strictly singular. Establish the following.

(i) If Y is a separable Banach space, then there exists a bounded surjective operator T : fl -> Y whose adjoint T*: Y* fx is a linear isometry. (ii) If T : fl -+ fp, where 1 < p < oo, is any bounded operator satisfying (i), then its adjoint operator T*: Lq -, f., where p + 4 = 1, is not strictly singular. Solution: (i) Assume that Y is a separable Banach space. Fix a countable norm dense subset {yl, yZ.... } in the closed unit ball U. of Y. Now define the linear operator T: el -+ Y by T(A1. Az, ...) _ En 1 Anyn. Observe that for each vector

r.=(A1,A2,...)Eel we have x

oc

x

n=1

n=1

n=1

This implies T(U1) C U,., where U1 is the closed unit ball of 11. From Ten = yn, where en E U1 is the sequence having its nth coordinate equal to one and every other equal to zero, it follows that T(U1) = L',.. Therefore, in view of the Open Mapping Theorem, the operator T is an open mapping, and hence T is surjective. In particular, we have IITII = 1. To see that T*: Y' - t ,,v is a linear isometry, note that for each y' E Y' we have IIT'y'II = sup sup Iy*(Tx)I = sup Iy (y)) = IIy`h1 xEUi

xEUi

yEUy

(ii) Fix any 1 < p < oc. Then the pair of Banach spaces (el, Pp) satisfies properties (a) and (b) of Problem 4.5.5. So, if T: el -' ep is any operator described in part (i) above. then T must be strictly singular, while its adjoint operator T: Y' -+ e. being a linear isometry is not strictly singular.

Problem 4.5.7 (T. Oikhberg). Show that there is an infinite dimensional Banach space X with a non-zero multiplicative linear functional on the Banach algebra £(X ).

4.5. Strictly Singular Operators

143

Solution: Let X be an infinite dimensional hereditarily indecomposable Banach space, in brief X is an HI-space. This means that no infinite dimensional closed subspace of X is isomorphic to a direct sum of two infinite dimensional Banach spaces. The existence of the HI-spaces was established by W. T. Cowers and B. Maurey in [32]. Moreover, it was shown in [32, 33] that: If X is an HI -space and T E C(X ), then there exist a scalar and a strictly singular operator S E C(X) (both uniquely determined) such that T is of the form T = al + S, where I is the identity operator on X.

Now define a function 0 on C(X) by 0(T) = ©(aI + S) = a, where T = al + S is the unique decomposition of T as a sum of a scalar multiple of I and a strictly singular operator. (The uniqueness of a and S follows from Corollary 4.62.) Clearly,

0 is a non-zero function. Now let T1, T2 E C(X) so that T1 = a1I + S1 and T2 = a21 + S2, where S1 and S2 are strictly singular operators. Then we have

AT1=Aa1+AS1, T1+T2=(a1+a2)+(S) +S2), and T1T2 = 0102 +(a I SI +a2S2 +S1S2). According to Corollary 4.62 these decompositions are the unique decompositions of the above operators as sums of multiples of the identity operator plus singular operators. This implies

OAT,) = Aa1 = ad(T1) , o(Ti + T2) = a 1 +a I = q(T1) + 6(T2), and = 0102 = q(T1)4(T2), and so 0 is a non-zero multiplicative linear functional on C(X). o(T1T2)

Problem 4.5.8. Show that if 1 < q < p < oo, then the natural embedding J: LP[0j1] --+ Lq[O,1] is not a strictly singular operator.

Solution: As indicated in the hint of the problem, the solution will be based upon the following famous classical inequality, known as Khintchine's inequality, regarding the standard sequence {rn } of Rademacher functions on [0, 1] that were introduced in Problem 3.1.9.

For each fixed 1 < p < x there exist two constants Ay > 0 and By > 0 such that n v n 1i rf n r P / (t) dt] < B,, Ianl21 Apl Ianl2J t=1 =1 t=1 J holds for all choices of the complex numbers a1, a2, ..-,a,, . I

For a proof of this inequality see [23. Theorem 1.101 or [52, Theorem 2.b.31. This inequality implies the following. If X denotes the closed vector subspace generated by the sequence {rn} of Rademacher functions in L2[0,1], then for any

1 < p < oc the vector space X also coincides with the closed vector subspace generated by the sequence {rn} of Rademacher functions in LP[0,11 and that the two norms II - IIP and II - 112 are equivalent on X. In particular, if r, s E (0, oo), then the two norms 11

-

IIr and II 11, are equivalent on X.

144

4. Special Classes of Operators

Now assume that p and q satisfy 1 < q < p < oc and consider the natural embedding J: LP[0,1] - Lq[0,1]. By the above, J restricted to X is an isomorphism. This proves that the positive operator J: Lp[0,1] -, Lq[0,1] is not strictly singular.

Problem 4.5.9 (Hernandez-Rodriguez-Salinas [37]). A bounded operator T : E -. X, from a Banach lattice to a Banach space, is said to be disjointly strictly singular if for each disjoint sequence {xn } of non-zero vectors in E the restriction of T to the closed vector subspace generated by {xn } is not an isomorphism. Clearly, a strictly singular operator from a Banach lattice to a Banach space is disjointly strictly singular. Show that if p and q satisfy 1 < q < p < oc, then the natural embedding J: Lp[0,1] Lq[0,1] is a disjointly strictly singular operator. Solution: Let {xn} be an arbitrary disjoint sequence of non-zero vectors in Lp. Let Y be the closed linear span of this sequence in Lp[0,1]. A straightforward verification shows that Y is linearly isometric to tp. Similarly, the closed linear span Z of the sequence {xn} in L.10, 1) is linearly isometric to the Banach lattice Pq. Since 1 < q < p < oc, we know from Theorem 4.58 that the embedding of tp to fq is strictly singular, and so Jly cannot be an isomorphism. Therefore, the I natural embedding J: Lp[0,1) -- Lq[0,1] is disjointly strictly singular.

Chapter 5

Integral Operators

5.1. The Basics of Integral Operators Problem 5.1.1. Assume that p and v are or-finite measures. Show that if the kernels of two integral operators ST: : E - F satisfy S(s, t) = T (s, t) for p x v-almost all (s, t) E S x T, then Sx = Tx for each x E E. Solution: Assume that the kernels of two integral operators S, T : E - F satisfy S(s, t) = T(s, t) for p x v-almost all (s, t) E S x T, where µ and v are o-finite measures. That is, there exists a p x v-null set D such that S(8, t) = T(8, t) holds for all (s, t) 0 D. Fix some x E E and pick a v-null subset B1 of T such that for each t f B1 both functions belong to L1(µ) and satisfy and

Sx(t) =

J

S(s, t)x(s) dp(s) and Tx(t) = / T(s, t)x(s) dp(s).

From (p x v)'(D) = 0 and the o-finiteness of the measures, it follows from Fubini's Theorem 1.97 that there exists a v-null subset B2 such that for each t V B2

the subset At = {s E S: (s, t) E D} of S is a p-null set. Now note that if B = B1 U B2, then B is a v-null set and, moreover, for each t 0 B the functions and belong to L1(µ) and satisfy S(s, t)x(s) = T(s, t)x(s) for all s outside of the p-null set At. This implies Sx(t) =

r S(s, t)x(s) dµ(s) = s T(s, t)x(s) dµ(s) = Tx(t) s

for all t 0 B. Therefore, Sx = Tx for each x E E. This means that the operators S and T are identical, i.e. S = T. I

Problem 5.1.2. Let (S, E1, p) and (T, E2, v) be two a-finite measure spaces,

and consider a p x v-measurable function K: S x T - R. Assume that a function f E Lo(p) satisfies

t) f

E L1(µ) for v-almost all t E T. Show 145

5. Integral Operators

146

that the function g: T -a R defined v-almost everywhere by the formula

g(t) = is K(s, t) f (s) du(s) s

is v-measurable, i.e., g E Lo(v). t) f E L1(µ) for vSolution: Assume that for some f E Lo(p) we have almost all t E T. So, there exists a v-null subset Bo of T such that K(., t) f E

L, (p) for all t 0 Bo. Without loss of generality, we can assume that f (a) E R for each s E S. By Lemma 1.90 there exists a sequence {Kn} of p x v-integrable functions satisfying I KK(s, t)I < IK(s, t)I and Kn(s, t) -. K(s, t) for all (s, t) in S x T. Since I K (s, t) f I < I K (s, t) f I and since each Kn is p x v-integrable for each n E N, Fubini's Theorem 1.97 guarantees the existence of some v-null set B such that for each t 0 Bn the function s .- Kn (a, t) f (s) is p-integrable and the formula gn(t) = fs Kn(s, t) f (s) d14(s) defines a v-integrable function. If B = U O B, then B is a v-null set and from the Lebesgue Dominated Convergence Theorem it follows that gn(t) g(t) holds for each tit B. This shows that g is a measurable p function, i.e., g E Lo(v).

Problem 5.1.3. Let

be a p x v-measurable function such that.

(a) fs JT(s, t)I dp(s) < 1 for v-almost all t E T. (b) fT IT(s, t) I dv(t) < 1 for p-almost all s E S. Prove that the integral operator T with kernel T(., ) is a bounded operator from L2(p) to L2(v). Solution: Fix x E L2(p), and formally let y(t) = fs T(s. t)x(s) dp(s). We shall verify that the formula defining y indeed gives a function that belongs to L2(v). To this end, apply Holder's inequality in connection with the first condition on the kernel to see that for v-almost all t E T we have:

IT(s.t)x(s)I dp(s) = S <_

S

f[IT(s,t)IIx(8)I2J

f

f

s

IT(s.t)Idp(s)

IT(s, t)IIx(s)I2 dp(s)}

1

[ f fT(s, t)I dp(s),

(

s

IT(s. t)I Ix(s)I2 dµ(8)1

Therefore, we get Iy(t)I2 dv(t)

<

[f IT(s. t.)x(s)I dp(s)]2 dv(t)

Ts

IT

IT <_

f

{f IT(a,1)Iix(8)12 dp(s)] dv(t) .

Changing the order of integration (by virtue of Tonelli's theorem) and using the second condition on the kernel, we finally obtain T IT

ly(t)I2 dv(t) < f

s

[fT IT(s. t)I dv(t)] . Ix(8)12 dp(e) s fs I x(8)12 dµ(8) = IIxII2

5.1. The Basics of Integral Operators

147

This proves that the expression defining y is indeed a function that belongs to L2(v). Moreover, it follows that the integral operator with kernel defines a bounded operator from L2(µ) to L2(v) satisfying 1ITh < 1. Remark: General conditions on the kernel guaranteeing that the integral operator generated by this kernel belongs to C(Lp(µ), LQ(v)) for some p. q E [1, xJ can be found in [45, Chapter XIJ. See also Problem 5.1.10. 1

Problem 5.1.4. ([20, Section 4.1]) Let T: L2(p) - L2(v) be an integral operator. Assume that there exists a subset B E E2 such that v(B) = 0 and for each x E L2(µ) we have f, JT(s, t)x(s)I dµ(s) < oo for each t 0 B. That is. the exceptional set in the definition of the integral operator T is independent of the "input" function x. Prove that fS JT(s, t) I' dµ(s) < oc for v-almost all t E T. (Operators that satisfy this condition are known as Carleman operators.) Solution: Fix any point t E T \ B and define the positive linear functional Oe: L2(µ) - IIt via the formula qt(x) = fsIT(s,t)Ix(s)dµ(s). The hypothesis fs JT(s, t)x(s) I dµ(s) < oc for each t B and each x E L2(µ) implies that Ot is a well-defined positive linear functional. It follows that Ot is a continuous linear functional on L2(µ). Therefore, by the Riesz Representation Theorem, this linear functional is represented by a unique function in L2(µ). This implies that T(., t) E L2(µ) for each t V B. 1

Problem 5.1.5. Assume that (S, E1, µ) and (T, E2, v) are two a-finite measure spaces, and let 1 < p, q < oc satisfy p + 1 = 1. Show that an integral operator T : Lp(p) -+ Lo(v) maps the closed unit ball of Lp(p) to an order bounded subset of Lo(v) (i.e., there exists a function g E Lo(v) such that jTxl < g for each x E Lp(p) with JIxIIp < 1) if and only if there exists a v-null set B E E2 such that

fIT(s,t)IdIL(s) < oc for each t E T \ B. In particular, an integral operator T : L2(µ) -+ L2(v) is a Carleman operator if and only if T maps the closed unit ball of L2(µ) into an order bounded subset of Lo(v). Solution: We shall denote by U the closed unit ball of the Banach space Lp(µ), i.e., U = {x E Lp(p): JIxjjp < 1}. Suppose first that there exists a v-null set B E E2 such that is JT(s, t) 11 dµ(s) < oo

for each t E T \ B. In view of this condition, the function

g(t) _ [

Js

[T(s, t)]9 dµ(s) ]

148

5. Integral Operators

is finite for each t

B, and in view of Problem 5.1.2 the function g is v-measurable.

Thus, g E Lo(v). We claim that g satisfies the required property. To see this, fix x E U. Then there exists a v-null setlB1 such that

Tx(t) =

Js

T(s, t)x(s) dp(s)

for each t 0 B1. Therefore, by Holder's inequality, for each t 0 B U Bl we have ITx(t)I <_ [ f [T(s,t)]'1dµ(3)] °

.

[ f Ix(s)Ipdµ(s)]

g(t)IIxIIp
That is, ITxI < g holds for each x E U.

For the converse, assume that there exists a function g E Lo(v) such that ITxI < g for each x E U. Without loss of generality, we can assume that g(t) < 00

for each t E T. If we assume that the measure p is separable,' then the proof is considerably easier. We shall present this simpler proof first and then we shall provide the general proof. So, assume that the measure p is separable. In this case, Lp(p) is separable. So, there exists a countable dense subset {x1, x2,. ..} of U. Now for each n pick a v-null set Bn such that ITxn(t)I = I fS T(s,t)xn(s)dp(3)I

holds for each t

g(t)

Bn. Therefore, for each t that does not belong to the v-null set

U'l Bn we have [ I [T(s, t)]°dµ(s) ] q = sup I I T(s, t)xn(s) dp(s) I = sup ITxn(t)I < g(t) < 00. S

nEN

S

nEN

Thus, T satisfies the desired condition. For the general case, assume that p is an arbitrary a-finite measure. As before, the function T(s, t) is the kernel of the operator T. We shall say that a non-empty subset D of Lo(p) and the kernel T(s. t) are compatible if there exists a null set Bo E E2 such that for each t V Bo and for each x E D:

(1) The function s - T(s, t)x(s) is p-measurable and its extended Lebesgue integral exists. (2) The function xt, defined by xt (s) = Ix(s)I [Sgn T(s, t)] if T(s, t) # 0 and

xt(s) = 0 if T(s, t) = 0, belongs to D, i.e.. xt E D.

Note that conditions (1) and (2) are both independent of the representative chosen from the equivalence class [x] which, in accordance with our agreement, is

denoted by x rather than [x]. In addition, for each t V Bo the (extended) value of the integral fS T(s, t)x(s) dp(s) is also independent of the representative from the equivalence class [x]. Using Fubini's theorem, it is easy to see that if T(., ) is non-negative, A is a solid subset of Lo(p) and D = A n Lo(p)+, then (D,T) is an example of a compatible pair. 1 Recall that a measure p is separable if and only if Lp(p) is a separable Banach space; for details see [4, Section 12.51.

5.1. The Basics of Integral Operators

149

Now let (D, T) be an arbitrary compatible pair. Then an extended function y: T \ Bo - [-oc, oo) can be defined via the formula

y(t) = sup{J T(s, t)x(s) dp(s): x E D } .

(*)

It should be noticed that y is a function and not an equivalence class. If the set D is not countable, then the v-measurability of the function y, which is the pointwise supremum of v-measurable functions, is far from being obvious. Nevertheless, according to Y. I. Gribanov (see [45, Theorem XI.1.21) the following result is true.

(Gribanov's theorem) If (D, T) is a compatible pair, then the function y is v-measurable. Moreover, there exists a sequence {fn} in D such that

J

y(t) = sup T(s.t)fn(s)dp(s) nEN S for almost all v E T.

(**)

Now we are ready to prove the converse of our statement-without assuming that p is separable. Consider the p x v-measurable function IT(-,.) 1. As noted above, the pair (D, IT(., )I), where D = U+ is the positive part of the closed unit ball U in Lp(p), is compatible. Let Bo E E2 be the corresponding v-null set. By Gribanov's theorem, the function

J

y(t) = sup IT(s, t)Ix(s) dp(s) xED S is v-measurable and there is a sequence {xn } in D such that

y(t) = sup

J

IT(s, t)Ixn(s) dp(s) nEN S for all t E T \ B1, where B1 is some v-null set in E E2. This obviously guarantees

that y < g, and so y(t) < g(t) for all t E T \ B2, where B2 is some v-null set in E2. (As a matter of fact, it is easy to see that y is the supremum of the set {ITIx: x E D} in the Dedekind complete Riesz space Lo(v).) Since is a p x v-measurable function, Fubini's theorem guarantees that there exists a v-null set B3 E E2 such that for each t V B3 the function T(., t) is p-measurable. Finally, let us verify that the p-measurable function IT(-, t) I belongs to Lq (p) for each t E T \ (Bo U B1 U B2 U B3 ). To this end, note that a p-measurable

function f : S - R belongs to Lq(p) if and only if If IIq = sup

f f (s)x(s) dp(s) = xED sup f If (s)Ix(s) dp(s) < oo. T

zEU T

Applying this to the function IT(., t) 1, we get [JT [T(s, t)]' dp(s)]

ai

J = sup J IT(s, t)Ix(s) dp(s) T = sup IT(s, t)Ix(s) dp(s) xEU T xED

= y(t) < g(t) < oo . This establishes that T satisfies the desired condition.

U

150

5. Integral Operators

Problem 5.1.6. Give an example of an integral operator on L2[0,1] which is not a Carleman operator-where [0,1] is equipped with its Lebesgue measure. Solution: This example can be found in [20, Section 4.1.3]. Let g be an arbitrary non-negative function in Lr [0, 1] such that IIgfli = z and g tt L2[0, (for instance,

let g(s) =

and consider on (0,1] x (0,1] the function T(s,zJt) = g(js - t1). is a measurable function. To show that the integral operator T

a s ), 1

Clearly, with kernel T (s. t) belongs to C (L2 [0,11) , we will verify that the kernel satisfies the

conditions given in Problem 5.1.3. Indeed, for each t E [0, 1] we have rI

J0

ri

re

:

IT(s, t)I ds = / g(t - s) ds + / g(s - t) ds _< 2 o

ri

Jo

g(s) ds = 1.

By symmetry, fa IT(s, t) I dt < 1 also holds for each 8 E [0,1]. Finally, for each t E (0, 1] we have

fo IT(8,t)12ds =

/tg2(t-s)ds+

119

2(s-t)ds

o

i t

::

j92(u)du+

J

92(u) du

0

J0

1g2(u)du=oo.

This shows that the kernel T does not satisfy the Carleman condition at each point t E [0,1]. Remark: In view of Problem 5.1.4, the above example demonstrates that in general the exceptional set in the definition of an integral operator depends on the function the operator is acting upon. I

Problem 5.1.7. Let {Tn } be an increasing sequence of regular integral operators and let T be another regular integral operator. Show that Tn T T holds in Gr(E, F) if and only if Tn(s, t) T T(s, t) holds for p x v-almost all (s, t) in CE X CF. Solution: Without loss of generality we can assume that E and F are order dense

ideals in Lo(p) and Lo(v), respectively-and so we have CE = S and CF = T. Suppose first that T T T in 4(E, F). From Theorem 5.5, it follows that there exists a p x v-null set A such that Tn(s, t) T < T(s, t) holds for each (s, t) f A. by and by -)XA, respectively, we can assume that Tn(s, t) T:5 T(s, t) holds for each (s, t). Let K(s, t) = limn-,,. Tn(s, t) for each (s, t) E S x T. So, K is a p x v-measurable function, and since for each x E E+ we have Tnx(t) = fs T,, (s, t)x(s) dp(s) T fs K(s, t)x(s) du(s) and Tnx(t) T Tx(t) for v-almost all t E T, we infer that Tx(t) = f5 K(s, t)x(s) dp(s) for v-almost all t. From the uniqueness of the kernel of the operator T (see Corollary 5.7), we get T(s, t) = K(s, t) for p x v-almost all (s, t). Hence, T. (s, t) T T(s, t) for p x v-almost

Replacing

all (s,t) E S x T. For the converse, assume that Tn(s, t) T T(s, t) for pxv-almost all (s, t); without loss of generality we can assume that Tn (s, t) T T(s, t) for all (s, t). Now if x E E+,

5. 1. The Basics of Integral Operators

151

then for each t E T we have T,, (s, t)x(s) T T(s, t)x.(s) for p-almost all s E S. So. by the Lebesgue Dominated Convergence Theorem, we get

Tx(t)

f

T

that

t E T.

Problem a

If

T

the

T T in

an

that T

f T(s,

E

in Lo(p), E a

a

I= and

a

a

an

E

we

in

and

x T is also a 0 for a-finite measure, there exists a sequence {o } of p x p-step functions such that 0 < 0,,(s, t) T T(s, t) for all (s, t) in S x T.

In order to establish that I J. T, it suffices to show (according to Problem 5.1.7)

that I I T holds true when the kernel of T is of the form T(s,t) = XA(S,t) for some p x p-measurable subset A of S x S with (p x p)'(A) < oo. So, suppose that T(s,t) = XA (s, t) for some p x p-measurable subset A of S xS with (p x p)' (A) < 00 and all (s, t) E S x S. Since 1 = XS belongs to Lo(p), there exists (according to Corollary 1.91) a

disjoint sequence {V} in E satisfying Xy:, E E and p(,,) < oo for each n, and S = Un V,,. If we let S = U 1 V,, then Xs" E E and p(S) < oo hold for each n. Moreover, Xs" T 1 in Lo(p). In particular, note that if x E Lo (p), then the 1

sequence {x n nXs" } in E satisfies 0 < x n nXs" T x in Lo(p). Now fix some n. Since the measure p is non-atomic, for each natural number k there exist pairwise disjoint p-measurable subsets Bl , ... , Bk of S satisfying S = Uk B" and B") for each i = 1,...,k. For each 1 < i < k let

C, = S \ U,#; B.. Clearly, XB,', Xc, E E for each i = 1.... , k. If for an arbitrary subset D of S x S we let Dt = Is E S: (s. t) E D}, then TX B^ (t)

=

j\A(st)xB'dP(s) _ jXAtrB(8)dP(8)

for p-almost all t E S. This implies that

0 < [(T AI)XS"](t) < [TXB, +IXc; ](t) = "(k") +Xc; (t) for p-almost all t and all i = 1, ... , k. This implies 0 < [(T A 1) <Sn) (t) <

for all k and p-almost all t E S. Consequently, [(T A I)Xs"](t) = 0 holds for p-almost all t, or (T A I)ks,, = 0 for each n. Finally, let x E E+. From 0 < (TAI)(xAnX5") < (TAI)(nX,") = 0, we easily get (T n I)(x A nXs") = 0 for each n. Since 0 < x A nXs" T x in E and TAI is order continuous, we infer that (T A 1)(x) = 0 for each x E E+. That is. T A I = 0.

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152

Problem 5.1.9 (Abrainovich Aliprantis-Zame (31). Show that for an arbitrary Dedekind complete AAI-space E with unit e the following two statements are equivalent. (a) There exists a probability measure space (S2. E. 7r) such that E is lattice isometric to L,.,, (ir) with e corresponding to the constant func-

tion 1 on 0. (b) E admits a strictly positive order continuous linear functional. Proof. (a) = (b) If (it, E, ir) is a probability measure space, then L,,. (1r) is Dedekind complete with order unit 1, and integration against r yields a strictly positive order continuous linear functional on L, (7r). Moreover, if T: E -+ L,o(rr) is a surjective lattice isometry with T(e) = 1, then it should be clear that the mapping q: E ---+ R, defined by d(x) = f T(x) dir, is a strictly positive order continuous linear functional on E satisfying q(e) = 1. (b) (a) Let ¢: E -. R be an order continuous strictly positive linear functional. Replacing 0 by m we can suppose that 0(e) = 1. Next, define III E - R by the formula IlfylII = 0(lyl). It is easy to check that III III is an III L-norm on E, which (in view of the order continuity of o) is also order continuous. Let k be the completion of (E, III ' III)- From Problem 3.4.6 we know that E is an AL-space. Next, we claim that E is an ideal in E. Clearly, E is a Riesz subspace of E. We

must verify that whenever 0 < z < y holds with y e E and Z E E, then z E E. So, assume 0 < z < y E E with y E E. Since E is norm dense in its completion E, there C E converging to z; that is, lllz, - zlll - 0. Since z < y exists a sequence and the mapping u,.-+ u A y is III - 111-continuous, we can assume (by replacing

with { z A y}) that 0 < z < y holds for each n. Also, by passing to a subsequence (if necessary). we can suppose that 111z,, - zIII < nz for each n. Now let k+m

Wk.m = V Zi . 1=k

and note that 0 <_ wk,,,, < y for all k and m. Since E is Dedekind complete, there exists some wk E E satisfying wk.,,, T.,, wk in E, and since III 'III is order continuous on E, it follows that IIlwk.m - Wk III m-. a 0. From k+m

k+m

k+m

k+m

IWk.m - ZI = I V Zi - Z+ I = I V (Zi -- Z)I < V Iz. - it < E IZi - ZI , i_k i=k

111

we see that Il lwk.m-zl Il

<-

Ek±k

i=k

i=k

ll Iz, -zl II < k, and so by the order continuity of the

norm III - III, we get II link - zl II = limm--,,Q 11lwk.m - zI II

for each k. In particular,

limk-w II Iwk - z I ll = 0. Clearly. the sequence {wk} is decreasing and bounded from

below by zero. Using once more the Dedekind completeness of E. it follows that wk j w holds in E for some w E E. Now from the order continuity of the norm 111.111 on E, we infer that IIIwk - wl ll -+ 0. But then 111z - n%III = limk-.x II Iz - wkl ll = 0,

proving that z = w E E.

5.1. The Basics of Integral Operators

153

The preceding conclusion also implies that e is a weak unit in E. To see this,

assume that u n e = 0 holds in k for some u E E. Pick a sequence {u,,} of E+ such that II Iu,, - uIII -- 0. Now let v,, = u n u and note that (since E is an ideal of E) we have v E E and 0 < v,, < u for each n and that IIIvn - uIII -+ 0. Clearly, v A e = 0 for each it, and since e is an order unit of E, we infer that v = 0 for each it. This implies u = 0 and it confirms that e is a weak unit of E. Next, notice that by the Kakutani-BohnenbhLst-Nakano Representation Theorem 3.5, we can find a probability measure space (Q. E. 7r) and a surjective lattice isometry T: E -+ L1 (r) such that T(e) = 1. Since T is a surjective lattice isomorphism of k onto L1(7r), it sends norm dense ideals of k to norm dense ideals in L1(7r), and hence T(E) must be a norm dense ideal in Ll(7r). Furthermore. T(E) = L,,(7r) since T(e) = 1. Finally, from IIxII = inf{.1 > 0: IxI < ,1e} and the definition of the II lIx norm in L,o(r), it follows that T: (E. II - II) - (Lx(7r), II II:,,) is also a surjective lattice isometry.

Problem 5.1.10. Let (S, E1, µ) and (T, E2, v) be two arbitrary measure spaces, and assume that 1 < p, q < oc satisfy 1 + q = 1. If a function T(., ) E Lp(µ x v), then show that the formula Tx(t) = f T(s, t)x(s) dp(s)

is

defines a regular integral operator from Lq(µ) to Lp(v) such that its norm satisfies IITII < (When p = q = 2, such an operator T is called a

Hilbert-Schmidt operator.) Solution: From Fubini's Theorem 1.97 we know that T(., t) E Lp(µ) for v-almost all t E T. So, if x E La(te), then E L1(tc) for v-almost all t E T. This implies that the formula

Tx(t) = JT(s.t)x(s)di1(s) defines a measurable function, i.e., Tx E Lo(v). We claim that Tx E Lp(v). To see this, notice that Holder's inequalityf implies

fT ITx(i)Ip dv(t) =

j T(s, t)x(s) dp(s)I p dv(t)

< j ({j IT(s,t)I"dt.(s)JpIlxllq)"dv(t)

(JT IfS IT(s, t)I" diL(s)] dv(t))

IIxIIq .

IIxIIq This (according to Theorem 5.11) shows that T is a regular integral operator from Lq(p) to Lp(v) and that IITII 5 IIT( .)lip. I So. IITxIIp <

Problem 5.1.11. Show that Hilbert-Schmidt operators are Carleman operators.

5. Integral Operators

154

Solution: Let T : L2(p) L2(v) be a Hilbert-Schmidt operator. Then. according to the definition of a Hilbert- Schmidt operator (see Problem 5.1.10). its kernel belongs to L2(p x v). That is,

!IT(.. )Ilz = ff xT I T(s.t)12d(p x v)(s.t) < Applying Fubini's theorem to this integral, we obtain that fs IT(s. t)J2 dp(s) < xo for v-almost all t E T. This shows that T is a Carleman operator. see the definition of a Carleman operator in Problem 5.1.4. 1

Problem 5.1.12. Assume that (S, E1, p) are (T. E2. v) are two a-finite measure spaces, and assume that (S, E1, p) is atomic - which in this case means that S is a union of at most countably many atoms. Show that a linear operatorT: E - Lo(v). where E is an ideal in Lo(p), is an integral operator if and only if T is order continuous. Proof. The order continuity of each integral operator T: E Lt,(v) has been established in Corollary 5.12 and it is true for any measure space (S. El, p). It is the converse statement that requires the extra assumption that (S. E1. p) is atomic. Since p is a-finite, we can represent S as a union U,,E.\' An of atoms of positive measure. where H is at most countable. Clearly, each r E E can be written as r = >nEN An xq , where A,, are scalars and the series converges p-almost

everywhere. Since T is order continuous, it follows that Tx =

E,4- anl'ka

where again the series converges v-almost everywhere. For each s E An and t E T let T(s. t) = --L--TkA (t). It is obvious that the

function T(., -) is p x v-measurable. Let us verify that T(., ) is the kernel of the operator T. For each x E E we have J

T(s.t)x(s)dp(s) = F, (t) nEA'

T is an integral operator.

5.2. Abstract Integral Operators Problem 5.2.1. Let E and G be two Banach function spaces associated with Lo(ir). Show that E n G is also a Banach function space which is order dense if both E and F are order dense ideals. Solution: Observe first that if E and F are order dense ideals, then their intersection E n G is also order dense. Indeed, if 0 < x E L' (7r). pick some y E E with 0 < y < x, and then select some z E G such that 0 < z < y. Hence. z E E fl G and

0
5.2. Abstract Integral Operators

155

Let II ' IIE and II JIG denote the lattice norms of E and G, respectively, and define the function II JJ : E n G - [0, oo) via the formula IIxII = IIXIIE + IIxIIG

A straightforward verification shows that II . II is a lattice norm on the ideal E n G. We claim that (EnG, 11 11) is a Banach lattice. To see this, assume that {xn } C EnG is a II ' II-Cauchy sequence. This implies that Ix.) is both a II ' IIE-Cauchy sequence and a II ' JIG-Cauchy sequence. Since E and G are Banach function spaces, there exist x E E and y E G such that Jlxn-xJJE-+0and JJxn-yIIG--'0. Now, according to Problem 1.2.13, there exist a subsequence {xk. } of {xn}. a

vector u E E+, and a vector v E G+ such that

Jxk,,-xl
Ixk,,

- yI - nV

for each n. This implies x = y. Letting w = x = y, we obtain w E E n G and IIxn - wit =

IIxn - wllE +

IJxn - LIIG = IIxn - xlI E + IIxn - YIIG -0-

This shows that (E n G. II ' IJ) is a Banach function space.

U

Problem 5.2.2. Let G be an ideal in a Riesz space E. let F be a Dedekind complete Riesz space, and let T : G -+ F be a positive operator. Assuming that T has a regular extension to all of E, establish the following. (a) The operator T has a minimal positive extension to all of E. That is, there exists a positive extension TG : E -' F of T such that if S: E - F is any positive extension of T. then TG < S. (b) The minimal extension TG of T is given by the formula

Tu(x) = sup{Ty: y E G and 0 < y< x} for each x E E+. In particular, TG(x) = 0 for each x E Gd. (c) If T is order continuous, then TG is also order continuous. Proof. Let S: E --' F be a regular extension of the positive operator T: C The fact that G is an ideal in E implies that for each x E G+ we have

F.

S+(x) = sup{Sy: y E E and 0 < y < x} = sup{Sy: y E G and 0 < y:5 x} = sup{Ty: yEG and 0 < y < x} = Tx. This shows that S+ is a positive extension of T to all of E.

(a) Let C = {R E G,+. (E, F): R = T on G}, i.e., C is the set of all positive linear extensions of T from G to all of E. By the above, C 54 0. Now if R1, R2 E C, then for each x E G+ we have

(R1nR2)x = inf{R1y+R2(x-y): yEE and 0
= Tx.

156

5. Integral Operators

This shows that R1 A R2 E C, and so C is directed downward. Since C consists of positive operators and 4(E, F) is Dedekind complete, there exists a positive operator TG E C, (E, F) such that C J To. From {Rx: R E C} I TGx for each x E E+ and Rx = Tx for each x E G, it follows that TGx = Tx for all x E G, i.e., TG is a positive linear extension of T, and obviously TG = minC.

(b) Define the mapping T,,: E+ - F+ via the formula

T,,,x=sup{Ty: y E G and 0
Ty1 + Ty2 = T(yi +y2) < Tm(u+v). This implies Tmu+Tmv < T,,,(u+v). Now assume that y E G satisfies 0:5 y:5 u+v. By the Riesz Decomposition Property there exist vectors yl, y2 E E+ satisfying

0 < yl < u, 0 < y2 < v, and y = yl + y2. Since G is an ideal, the latter implies yl, y2 E G, and so Ty = Ty1+Ty2 < T,,,u+Tmv for all 0:5 y < u+v. Consequently, Tm(u + v) < T,,,u + Tmv. Thus, T,,, (u + v) = Tmu + Tmv, and so Tm is additive. By Theorem 1.15, the formula T,,, (x) = T,,,(x+)-T,,,(x") extends T,,, (uniquely) to a positive operator on all of E. We shall show that TG = Tm. To see this, notice first that since Tm is a positive extension of T, we have TG < Tm. On the other hand, if x E E+, then for each y E G with 0 < y < x we have Ty = TGy < TGX, and so

Tmx = sup{Ty: y E G and 0 < y < x} < TGx. Therefore, Tm < TG is also true, and thus To = T,..

If 0 < x E Gd, then y E G and 0 < y < x imply y = 0. This shows that TGx=0 for each xEGd. (c) Assume that T: G -+ F is also order continuous and let 0 < xa j x in E. Clearly, TGxa T:5 TGx. To see that TGx is the least upper bound of the net

{TGxa} assume that TGxa < u for all a. Fix y E G with 0 < y < x and note that 0 < y A xa j y in E (and, of course, in G). The order continuity of T implies T(y A xa) j Ty. So, from

T(yAxa) =TG(yAxa)
Problem 5.2.3. By means of Nakano's theorem (see Problem 1.2.22), show directly that if E is a Banach function space associated with a a -finite measure 7r, then its order continuous dual Ea has weak units. Solution: Let E be a Banach function space associated with a a-finite measure 7r. For each f E ED let Cf = Nf = {x E E: IfI(IxI) = 0}d be the carrier of f. We claim that each pairwise disjoint subset of En is countable. To see this, let C be a pairwise disjoint subset of En consisting of non zero linear functionals. According to Nakano's theorem (see Problem 1.2.22), for every pair f, g E C with

f # g we have CJ fl C. = {0}. Since Cf is a band of E, there exists a unique set D f E E with 7r(D f) > 0 such that C f = {x E E: x = 0 7r-a.e. on fl \ D f };

5.2. Abstract Integral Operators

157

see Theorem 1.92. Notice that the mapping f - D f is one-to-one and that the condition Cf fl C9 = {0} is equivalent to D f fl D. = 0. Since it is c-finite, the family { D f : f E C} is at most countable. This implies that C is at most countable.

Now assume that C = (fl, 12,. ..} is a countable maximal pairwise disjoint family in EA satisfying II fn II = 1 for each n; by Zorn's lemma such a family exists. Since E,; is a Banach lattice (in fact, it is a band in E'), it follows that the functional f = F,OD_1 2L Ifnl E En is a weak unit in E. (If C is finite, we let f = E9EC

Problem 5.2.4. Let E be a normed Riesz space. Show that En separates the points of E if and only if for each 0 < x E E there exists some 0 < 0 E En satisfying 0(x) > 0. Solution: If En* separates the points of E and 0 < x E E, then there exists some 0 E E,; such that O(x) # 0, and so either 01(x) > 0 or 0-(x) > 0. Conversely, assume that for each 0 < x E E there exists some 0 < 0 E En satisfying 0(x) > 0 and let y E E be a non-zero vector. Pick 0 < 0 E En such that 0(IyI) > 0, and then use the formula 0 < 0(IyI) = Iyi(0) = sup{iG(y): >G E E and ItPI <_ 01 to conclude that for some t
Problem 5.2.5. Assume that (fl, E, 7r) is a a -finite measure space and let E be an order dense function space associated with Lo(ir). Show that a function g E En satisfies g = 0 if and only if fn g(w)x(w) dir(w) = 0 for each x E E. (This guarantees the uniqueness of the function g in the Representation Theorem 5.26.) Solution: Assume that a function g E Eo satisfies fn g(w)x(w) d7r(w) = 0 for each x E E. If g 0 0, then there exists some e > 0 such that the ir-measurable set A = {w E Sl: Ig(w)I > E} has positive measure. Since E is order dense in Lo(a), there exists some y E E satisfying 0 < y < EXA < Igi. Now if x = (Signg)y, then

xEEand 0=

Jfl

g(w)x(w) d7r(w) =

f

Sl

I g(w)I y(w) &r(w) > e

J JA

y(w) d7r(w) > 0 ,

which is absurd. This contradiction establishes that g = 0 a-a.e.

Problem 5.2.6. Let (S2, E, a) be a or-finite measure space. Show that the order continuous dual of L... (7r) coincides with L1(a), i.e., Loo(a)*. = Li(a).

Solution: Let E =

By Theorem 5.26, E is an order dense Banach function space associated with Lo(a). Clearly, L1(a) C E. Now assume that f E E. This means that fn f (w)x(w) d7r(w) is a real number for each x in L,,,(a). In particular, if x = Sign f , then x E L (7r), and so fn Ill da = f n f (w)x(w) da(w) E R. This shows that f E Li(ar), and thus E C Li(ir). Hence, L1(a).

Problem 5.2.7. Let (11, E, a) be a a -finite measure space. If E is an order dense Banach function space in Lo(7r), then show that the center of E coincides with Le(a), i.e., 2(E) = La(ir).

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5. Integral Operators

Solution: Start by observing that the Maeda-Ogasawara -Vulikh completion of E can be identified with Lo(ir) equipped with its (7r-a.e.) pointwise multiplication:

see Theorem 3.35. Now fix some operator 0 < T E S(E). Then, according to Theorem 3.36, there exists a function 0 < f E Lo(7r) such that

T(9)=19 for each ,q E E. To finish the solution, we must show that f E Lx(a).

To this end. assume by way of contradiction that IIfllx = x This implies that for each n the measurable set A = {w E fl : f (w) > n} has positive measure. Replacing (if necessary) the set A by some of its subsets in E, we can assume that 0 < 1r(An) < oc and XA E E hold for each n. Then TXA,, = P kA. > nXA,,, and so IITII > n for each n, a contradiction. Therefore, f belongs to L.x(ir). and from this we easily conclude that Z(E) = L,,(ir).

Problem 5.2.8. For an order bounded sequence {xn} in a Banach function space with order continuous norm consider the following convergence properties:

(a) xn -° x

(b) xn - x

(c) xn "'.r

(d), and (b)

(d) 1Ixn - X11

0.

(c). Does (d) = (b) ?

Show that (a)

(b). (c)

Solution: (a)

(b) The equivalence of (a) and (b) is established in Prob-

lem 1.6.4. (This proof is also valid in any function space.) (b)

(c) This is Corollary 1.83.

(c) 4=* (d) Let {xn } be a sequence in a Banach function space E such that Iln I < u for each n and some u E E. Assume first that xn -" r. If IIxn - xII 74 0, then there exist a subsequence {yn} of {xn} and some e > 0 satisfying Ilyn -xII > F for each n. From xn --+ x, it follows that there exists a subsequence {zn } of {yn } satisfying z .x; see Theorem 1.82. By the equivalence of (a) and (b), we have z,, - -+x. But then the order continuity of the norm implies Ilzn - xII -+ 0, which contradicts Ilz,, - xll > e for each n. Hence. x,, 1-+x implies Ilrn - xII -' 0. For the converse, assume that IIxn - xII -* 0. and let {y } be a subsequence of {rn}. Since IIy -xII - 0, it follows from Lemma 5.21 that {yn} has a subsequence that converges 7r-a.e. to x. This implies that xn -yx is true; see Theorem 1.82. (d) = (b) Let E = Ll [0, 11. Now consider the sequence of closed intervals

la' 1J' !0' 131- - . Also, let {xn} be the sequence of the characteristic functions of these intervals. Then it is easy to see that 0 < x < 1 for each n, llxn III -. 0, but xn(w) -74 0 for (0, 1]. L0. 2J' [12-11, l0' 3J' 13' 31' 13' 1J' [0' 411414-241-124. 41-

each w E [0, 11.

1

Problem 5.2.9. If {xn} is an order bounded positive sequence in some Banach function space E in Lo(ir), then prove that xn -' 0 is equivalent to x.,

o (E .E-*

0 for each y E E.*. 0, i.e.. to the property: A, xn(w)y(w) dir(ty) Is this equivalence true if {xn} is not assumed to be a positive sequence?

5.2. Abstract Integral Operators

159

Solution: Assume that E is a Banach function space, and let {xn} be a positive sequence satisfying 0 < .rn < r for each n and some x E E+. It is not difficult to see that xn a E E' 0 is equivalent to fA xn(w) d7r(w) i 0 for each A E E with XA E E and ir(A) < c. Suppose first that xn -"-+0 is true, and assume by way of contradiction that

there is some A E E satisfying kA E E, a(A) < oo, and fAxn(w)dir(w) f 0. Therefore, there exists a subsequence {yn} of {xn} and some b > 0 such that fA xn(;.r) d7r(w) > 6 for each n. Since. xn --1-- 0, there exists a subsequence {Z,, I of {yn } satisfying zn -2- 0. But then the Lebesgue Dominated Convergence Theorem guarantees that fA zn(w) dir(ty) - 0, a contradiction. So, xn o E E' 0.

For the converse, assume that xn ° E E 0. Fix a set A E E of finite measure with XA E E, and for each f > 0 and n let An(E) = {w E A: xn(w) > E}. From xn, we get 0 < E7r(An(E)) < fAxn(w)d7r(w). Since fAxn(w)ddr(w) - 0, the latter implies 7r(An(E)) - 0. Thus, x,, ' 0. The equivalence is false if the sequence {xn} is not positive. For an example, let E = L,,[0,1] and consider the sequence of Rademacher functions {rn} as introduced in Problem 3.1.9. We have En = L1 [0, 1] and rn -W- 0; see the solution of Problem 3.1.9. However, the sequence {rn} does not converge in measure to zero since no subsequence of {rn} converges pointwise almost everywhere to zero.

Problem 5.2.10. If E and F are Banach function spaces, then show that (En ® F)dd is a principal band in Gr(E, F). Solution: We can assume that E and F are order dense ideals. Let e be a weak unit in F and 6 be a weak unit in E. (The existence of these weak units is guaranteed by Corollary 5.22 and Theorem 5.26.) Clearly, the operator p 0 e belongs to (E,; ®F)dd . Assume now that some elements 0 < x' E Eo and 0 _< x E E

satisfy (x* 0 x) A (0 0 e.) = 0. From 0 < (x' A 0) ®(x A e) < (x' ®x) A (p ®e), we see that (x` A b) ®(x A e) = 0. This implies that either x' A p = 0 or x A e = 0.

So, either x' = 0 or x = 0, i.e.. x' ®x = 0. This shows that 0;9 e is a weak unit in (E,; (9 F)dd, and the conclusion follows.

Problem 5.2.11. Show using Theorem 5.28 instead of Theorem 5.30 that if E is a Banach function space with order continuous norm associated with a v-finite measure p, then every continuous operator T : E - L,,.(v) is a regular integral operator. That is, £(E, L.. (v)) = GK(E, L.(v)). Solution: Since L,,,(v) is a Dedekind complete AM-space with unit, Theorem 3.9

guarantees that £(E, Lc (v)) = C,(E. L. (v)). We can assume that E is order dense in L,o(p). Fix any weak unit 6 E En; hence p(s) > 0 for p-almost all a E S. As in the solution of Problem 5.2.10. we see that the operator 0®1T is a weak unit in £,r (E, Therefore, to establish the identityC(E, La,(v)) = GK(E, L0,(v)),

it suffices (in view of Theorem 5.28) to show that the zero operator is the only operator T : E -. L.,, (v) that satisfies T A (p ®1T) = 0. So. let T : E - Lx(v) be a positive operator satisfying T A (6 ®1T) = 0. Since E has order continuous norm, it follows that T = 0 if and only if T xA = 0 for each measurable subset A C E1 of finite measure satisfying XA E E. Assume.

5. Integral Operators

160

by way of contradiction, that there exists some A as above and such that TXA > 0. Then there exists some e > 0 such that v = (TXA - e)+ > 0. By the continuity of T there exists some 5 > 0 such that:

f E E and 11f II < b imply I T f (s) I< e for /,almost all s E S.

(*)

The order continuity of the norm in E implies that there is some 0 < n < p(A) such that if B C A and p(B) < ii, then IIXBII < 5. To see this, assume by way of contradiction that this is not the case. This means that there exists some sequence { Bn } such that B C A for each n, µ(B,) --+ 0, and 11;(B. (I > 6 for each n. By passing to a subsequence, we can assume that p(Bn) < aL holds for each n. Now p(B;) < 21-n it follows that Xc 10 if C = U n Bt, then from p(Cn) < holds in E. Since E has order continuous norm, the latter implies IIXc,,II 1 0. But then from 0 < XB S XC we obtain IIXB II 0, contrary to IIXa II > 6 for each n. This establishes the existence of q. Since 0 is a weak unit in E.*, Problem 1.6.9 guarantees the existence of some

real number a > 0 such that C E E1, C C A and p(C) > rl imply fc Odp > a. Pick any B E E1 with B C A. If p(A \ B) < r), then IIXA\BII < 6, and (*) yields 0 < TXA\B = TXA - TXB < E1T,

whence TXB ? (TXA

- e)+ = v. On the other hand, if p(A \ B) > r), then we

have fA\Bodu>a. Thus, if w = VA (air), then 0 < w

F, and if B E E1 satisfies B C A, then TkB+(m®1T)XA\B =TXB+(fA\B¢dp)1T > w. But then Theorem 1.17 implies

0< w <

inf {TXB + (0 o 1T)XA\B : B E El and B C A}

= [TA(o(&1T)J(XA)=0, which is impossible. This contradiction shows that TXA = 0, and hence T = 0, as desired.

Problem 5.2.12. Show that Dunford's Theorem (Corollary 5.31) is not valid for continuous operators between LI -spaces. Solution: According to Problem 5.1.8 the identity operator I : L1 (0,11 -+ L1 [0,11

is disjoint from every regular integral operator. So, if I were a regular integral operator, then I would coincide with the zero operator, which is absurd. Hence, not every positive operator between L1-spaces is an integral operator. (The preceding argument establishes really the following result: If T: L1[0,11 -# L1 [0,11 is an integral operator, then I + T is not an integral operator.)

Problem 5.2.13. Assume that (S, E1 i µ), (T, E2, v), and (Q, E3, 9) are a-finite measure spaces and that E, F. and G are ideals in Lo(,u), Lo(v), and Lo (7r), respectively. Assume also that in the scheme E -- F T + G the operators S and T are both regular and order continuous. If either S or T is an integral operator, then show that TS is likewise an integral operator.

5.2. Abstract Integral Operators

161

Solution: Let {x } C E be an order bounded sequence satisfying x -9- 0. According to Theorem 5.30 it suffices to show that TSx s- 0. Consider first the case when S is an integral operator. Since S is regular, the sequence is order bounded in F and, since S is an integral operator, Sx -2- 0. Hence, by Corollary 1.86, Sx _ 0 in F, and so (by the order continuity of T) we have TSx -9- 0 in G. In particular, TSx L. 0. Thus, in this case, TS is indeed an integral operator. Assume now that T is an integral operator. Since S is regular, the sequence

{ Sx } is order bounded in F, and we claim that Sx -r 0 holds in F. If not, then there exist some B E E2 with v(B) < oo, some e > 0, and some subsequence of (x,,) such that for each n we have v({t E B:

E}) > e.

(**)

Now, according to Theorem 1.82, there exists a subsequence { z } of {y } such that

z -2- 0. But then, by Corollary 1.86, we have z -9+ 0, and so by the order continuity of S. we get Sz,1-- 0. That is, Sz R-0. This (according to Corollary 1.83) implies Sz - 0, contrary to (**). Consequently, Sx,, 0. Since T is an integral operator, it follows from Theorem 5.30 that TSx -F 0. Using Theorem 5.30 once more, we get that TS is an integral operator. Regarding this problem. two remarks are in order.

(1) When S is an integral operator, the assumption T E Gu (F, G) can be replaced by the weaker assumption T E LO(7r)), and when T is an integral operator we do not need the assumption that T : F -. G is regular.

(2) In general, the composition of integral operators may fail to be an integral operator. We refer to [49] for an example of two compact integral operators on L2[0,11 whose composition is not an integral operator.

Problem 5.2.14. Assume that (S, El, t) and (T, E2, v) are two a-finite measure spaces with (S, E1, p) non-atomic. Also, suppose that E and F are ideals in Lo(S, El, p) and Lo(T, E2, v), respectively, and let S: E -+ F be a regular integral operator. Let FO denote the order ideal in F generated by the range space of the

operator S and assume that there is an operator T : FO - E such that TS = IE, the identity operator on E (in a sense T is a right inverse of the operator S). Show that T cannot be an integral operator. Solution: Assume, contrary to our claim, that T is an integral operator. Therefore, T is order continuous as an operator from FO to Lo(µ). By the hypotheses S: E -+ F is a regular integral operator and, therefore, S considered as an operator from E to Fo is also a regular integral operator. Hence, by Problem 5.2.13 and the remark at the end of it, the composition TS is again an integral operator. However, TS is the identity operator and according to Problem 5.1.8 the identity operator cannot be an integral operator, since (S, E1 i p) is non-atomic.

5. Integral Operators

162

Problem 5.2.15. Assume that (S, El, p), (T, E2, V), and (Q, E3, 0) are three a -finite measure spaces and that E, F, and G are function spaces in Lo(p), Lo(p), and Lo(0), respectively. Suppose that in the scheme of operators E S F L G both S and T are regular integral operators with and T(., .), respectively. kernels Show that the composition operator R = TS is an integral operator whose and T(., ). i.e., kernel is given by the convolution of the two kernels R(s, q)

=fS(s, t)T(t, q) dv(t)

.

Solution: Since T and S are regular integral operators, they are both order continuous. So, by Problem 5.2.13, the operator R = TS is an integral operator. Let us calculate the kernel of R. Without loss of generality, we can assume that both operators S and T are positive and, hence, their kernels S(-, ) and T(., ) are non-negative functions. For each x ErE+ we have Sx(t) = t S(s, t)x(s) dy(s)

is

for v-almost all t E T, and therefore

f

f for 0-almost all q E Q. Since all functions are non-negative, an application of fTSx](q) =

Tonelli's theorem yields that

T(t, q) [ / S(s, t)x(s) dp(s)] dv(t)

t

[TSx](q) = J [4 S(s, t)T(t. q) dv(t)]x(s) dµ(s) S

T

and that the function s'--p l fT S(s, t)T(t, q) dv(t)] x(s) is integrable for 0-almost all q E Q. Consequently, R(,9. 9) = fT S(s. t)T(t. q) dv(t) is the required kernel for the operator R.

Problem 5.2.16. Let (S, E1 i p) and (T, E2, v) be two a-finite measure spaces with (S. E1, p) non-atomic, and let 0 < K(.,-) E Lo(p x v). Do there exist a function 0 < f E Lo(p) and a subset B E E2 such that: (a) v(B) > 0. (b) For each t E B we have K(s, t) > f (s) for p-almost all s E S?

Solution: We claim that the existence of a function f and a set B with the required properties is equivalent to the statement that there exists an e > 0 such that the set U, = {(s, t) : K(s. t) > e} includes a non-trivial rectangle A x B, that is, (p x v)(A x B \ Uj = 0 for some A E E1 with u(A) > O and some B E E2 with v(B) > 0. Indeed, assume first that there is an e > 0 such that (p x v)(A x B\U.) = 0 for some A E El with p(A) > 0 and some B E E2 with v(B) > 0. Assume also that p(A) < x. Let C = A x B \ UI, and let Ct denote the t-section of C fort E T. Since (p x v)(C) = 0, Fubini's theorem implies that p(Ct) = 0 for v-almost all t E T. We will assume that p(Ct) = 0 for each t E B: otherwise we will remove from B

5.2. Abstract Integral Operators

163

the v-null set of the exceptional points t. We claim that the function f = EXA is as required. Indeed, for each t E B, if s Ct, then clearly K(s, t) > EXA(s). Since p(Ct) = 0, this means that K(s, t) > f (s) for p-almost all 8 E S, as claimed.

For the converse, assume that there exists a function 0 < f E Lo(p) and a subset B E E2 such that v(B) > 0 and for each t E B the inequality K(s, t) > f (s) holds for p-almost all s E S. Let us denote this set by St. Since f > 0, we can find an E > 0 such that the set A = {s E S: f (s) _> E}

satisfies p(A) > 0. We will verify that (p x v)(C) = 0, where C = A x B \ U,. For each t E T let Ct denote the t-section of C. We claim that p(Ct) = 0 for each

t E T (and hence (p x v)(C) = 0 by Fubini's theorem). Indeed, if t 0 B, then Ct = 0. On the other hand, if t E B, then Ct C S \ St and so again p(Ct) = 0. Thus, indeed, the rectangle A x B is contained in U(. To answer our question in the negative, it remains to recall that in general there exist measurable subsets of S x T of positive measure that include no nontrivial rectangles. (See, for instance, Exercises 136.12 and 136.13 in [83], where

I

S=T=[a,b].)

<00

Problem 5.2.17 (Benedek-Panzone [16]). Let (S, E1, p) and (T, E2, v) be two a-finite measure spaces. For each p E [1, oo) we denote by Lp,.,,(µ x v) (or simply by Lp,OC) the collection of all functions f E Lo(p x v) such that

IIfiip,oo=esssup[L1f(5,t)11'(t)1"

usual, the equivalent functions are identified. Similarly, we denote by Lp,I (p x v) (or simply by Lp,I) the collection of all functions f E Lo(p x v) such that if IIp,1 =

[jIf(stw'dv(t)]dlL(s) < oo .

JS The non-negative real numbers IIf IIp,o and IIf Ilp,l are called the mixed

norms of f. Show that: (a) The function II' IIp,,,. is well defined, i.e., if f, g E Lp,". satisfy f = g it x v-a.e., then Ilfllp,c,,, = II9IIp,oo.

(b) The collections Lp,,, and Lp,1 (with their mixed norms) are both Banach function spaces associated with Lo(p x v). (c) If 1 < p, q < oo satisfy n + 1 = 1, then Lp,l = Lq,oo. Solution: (a) Let f, g E Lp,00 satisfy f = g It x v-a.e. This means that there exists a It x v-null set A such that f (s, t) = g(s, t) holds for all (s, t) 0 A. Since It and v are a-finite measures, for p-almost all s E S the function s i-4 v* (A.) is v-integrable,

where A. = {t E T : (s, t) E Al, and fs v'(A8) dp(s) = (p x v)'(A) = 0. This implies that v'(A.) = 0 for p-almost all s E S. So, for p-almost all s E S we have IT

[f (s, t)]° dv(t) = IT\A.'f (s, t)]° dv(t) = fT\A [g(s, t)]° dv(t) =

and from this we easily get if IIp.oo = II9ilp, 0.

j[g(s, t)]' dv(t),

5. Integral Operators

164

(b) We shall verify that L.,,, is a Banach function space; the case of Lp,1 can be done similarly. Obviously, if f E LP.,0 and A is a scalar. then A f E LP,.. To verify that LP,,,, is closed under addition, let f,g E Lp,,,. Using the triangle inequality of the norm in Lp(v), for p-almost all s E S we have [J if(s,t) + g(s, t) IP dv(t)] ° T

< <

[J If(s, t) IP dv(t)] ° + [J I9(s,t)IPdv(t)] °

T Ill IIP.x + II9IIP.x .

T

This implies Of +

ess sup aES

[IT if (s. t) + g(s. t)IP dv(t)] ° <_ II f IIP.x + H9IIP.x

Therefore, f +g E LP.,r, and II - Ilp.x satisfies the triangle inequality. The verification of the other two axioms of the norm is straightforward and is omitted. If If I < 1.91, then clearly Ill IIP.x _< IIgIIp,x. This implies that II . IIP.x is a lattice

norm on Lp, and that Lp,,,,, is an ideal in Lo(p x v). In particular, Lp,,, is a Dedekind complete Riesz space.

Next, we shall verify that Lp,,c is a Banach lattice. According to Problem 1.2.16, it suffices to show that every increasing norm Cauchy sequence in Lp is norm convergent. To this end, assume that {f.1 is a II - IIp, Cauchy sequence in Lp,,, satisfying 0 < f, j . Without I ms of generality, we can suppose that 0 < f,, (s, t) < fn+1(s, t) holds in IR for all n and each (s. t) E S x T. Let f (s, t) = limn_.,, f,, (s, t), and so f : S x T [0, oo]. Now fix some M > 0 such that 1 / IIfnIIP.oo = esssup [J [fn(s,t)]pdv(t)] ° < Al

,

aES

for each n.

T

This implies that we have f7[fn(s,t)]Pdv(t) 1< M for p-almost

all s E S. By Beppo Levi's classical theorem, we infer that for p-almost all s E S we have f (s, ) E Lp(v) and that fT[f (s, t)]P dv(t) < M. This implies s [ fT[f (s, t)]P dv(t)] P < M. Therefore, if we can prove that f (s. t) < x for estsup

S

p x v-almost all (s, t) E S x T, then it will follow that f E LP,,* To see that f (s, t) < oo for p x v-almost all (s. t) E S x T, assume by way of contradiction that this is not the case. Then there exists a p x v-measurable subset A of S x T with 0 < (p x v) *(A) < oo satisfying f (s, t) = oo for each (s, t) E A. By Fubini's Theorem 1.97, the function s F-+ v'(A,), where A, = {t E T : (s, f) E Al, is v-integrable and (p x v)'(A) = fs v' (A,) dp(s). In particular, the p-measurable set B = Is E S: v'(A,) > 0} satisfies juo(B) > 0. But then for each s E B we have

j [f(s, t)J dv(t) ? j.[f(s.t)]pdv(t) = oc. which is impossible. Therefore, f E Lp,,,. We shall finish the proof of the completeness of the normed Riesz space L.,,,

by proving that IIf,, - f IIP.x -. 0. To this end, let c > 0. Fix some k such that 11 f,,, - fn Ilp,,o < e for all m, n > k. In particular, for p-almost all s E S we have fTIfm(s.t)-fn(s,t)IPdv(t) <eforalln,m> k. From f,,,(s,t) f,,, f(s, t), it follows

5.2. Abstract Integral Operators

165

that for p-almost all s E S we have fT[f(s,t) - fn(s,t)]pdv(t) < e for all n > k. This implies II f - fnllp,. <- e for all n > k. That is. fn - f in Lp,,,,. + 1 = 1. We shall

(c) Let 1 < p < oc and let q be its conjugate index, i.e., prove that Lp,1 = Lq.W.

First, we shall establish that the norm on Lp,1 is order continuous. To this end, assume that xn j 0 holds in Lp,l. This implies xn(s, t) j 0 for p x v-almost all (s, t) E S x T. Therefore, for p-almost each a E S we have xn(s, -) j 0 in Lo(v), and so the Lebesgue Dominated Convergence Theorem yields Ilxn(s. )lIp =

[fT (xn(s. t)]p dv(t)} ° 10

for p-almost all a E S. This and the hypothesis that the functions Ilxn(s, -)I[p belong to L1(p) imply (again by the Lebesgue Dominated Convergence Theorem) that llxnllp,l =

j[j[xn(s.t)1"dv(t)] ° dp(s)

10.

=f

Therefore, 11 - llp,, is order continuous.

The order continuity of the norm II - llp,, guarantees that Lp.1 = (Lp,1)n. That is, each bounded linear functional o on Lp,1 is order continuous and hence allows an integral representation by Theorem 5.26.

Our next step is to establish the inclusion L.... C Lp,l. Take any function g E Lq,oo, and let us verify that xg E L1(p x v) for each x E Lp,1. So, fix x E Lp,1. Applying Holder's inequality, we see that Js

f I x(s. t)g(s. t)I dv(t)j dp(s)

<-

<-

f

Ilx(s, -)Ilp . lI9(s, -)IIq dp(s)

ess sup II9(s -) llq - is ll x(s, .) ll p dp(8)

s

SES

e

ll9llq.x -

IIXIIp,1

.

Now an easy application of Tonelli's Theorem 1.98 shows that xg E Ll (p x v). So, g defines a bounded linear functional on Lp.1, i.e., g E Lp.1, and 11911 < II9IIq,oc, where II9II denotes the norm of g as a bounded linear functional on Lp,1. We shall show now that in actuality we have equality. That is, we claim that II9II=II9Ilq,oc-

(*)

To this end, fix t > 0 and suppose that II9IIq,w = 1. Thus, J. Ig(s, t)l q dv(t) < I for p-almost all a E S. Without loss of generality we can assume that g(a, t) > 0 for all (s, t) E S x T. According to the definition of the space Lq,., we can find some A E E1 such that 0 < p(A) < x and llg(s, -) 119 = [fT gq(s, t) dv(t)] 4 > 1- e for each s E A. Consider the function x(s. t) = a a XA(s)[9(s, t)]4-1. From Js

LfT [s(s. t)]p dv(t)] P

dp = Tt1 j[j[g(8,t)]_1)idv(t)]

dp(s)

= pI f [ JT[9(s, t)I° dv(t)] ° dp(s) <

1.

5. Integral Operators

166

it follows that x E Lp,1 and that Ilxllp-1 < 1. Consequently,

IIgII >- g(x) =

v) = jxgd(x xT 1,(A7

(s)Ig(s,t)]d(µ x v)(s,t)

j[g(s , t)]Q dv(t) dµ(s) > (1 - e)Q. 1a

Since e > 0 is arbitrary, we get II9II ? 1 and (*) has been established. The inclusion Lq.,, C_ Lp,l = (L9,1);, guarantees that is an order dense

take any 0 < g E L;,1.

ideal in (Lp,1)n. To establish the equality Lq,a =

Since the measure µ x v is a-finite, there exists a sequence {gn} E Lq,,o satisfying 0 < g. (s, t) T g(s, t) for µ x v-almost all (s, t) E S x T. So, for µ-almost all s E S we have 0 < gn (s, t) T g(s, t) for v-almost all t E T. This implies [1T" n (s'

t)]4 dv(t)]

a

T [JT g(8, t)]Q

dv(t)] a

for µ-almost all s E S, and consequently (from Lemma 1.101) we get

IlgnIIQ.x = es sup [f [gn(s,t)1Qdv(t)J" T esssup [JT[g(s,t)I°dv(t)]"

.

The function g belongs to Lq.op if (and only if) the expression on the right-hand side is finite, or, equivalently, if the sequence { Ilgn Ilq,,o } is bounded. But gn T 9 in LP'1 implies that I Ign II T Ilgll and, as we proved above, I(gn IIq.. = Ilgn II . This proves that the sequence {Ilgn(Iq,.) is bounded from above by IIghI, and consequently g E Lq,,p. The solution is finished.

Problem 5.2.18. The result of this problem can be found in many books; see for instance [24, 28, 45, 46]. For the notation used here see Problem 5.2.17. Assume that (S, E1, µ) and (T. E2, v) are two a -finite measure spaces. T -- R defines, via the Show that a a x v-measurable function usual formula

Tx(t)

= fs T(s, t)x(s) dµ(s),

an integral operator from L1(µ) to Lp(v) for some 1 < p < oo if and only if belongs to Lr,,.. Moreover, prove that in this case the integral operator T determined by the kernel T(., ) is a regular operator and its norm is given by IITII = IIT(-, )IIp,.. Solution: Assume first that

E Lp,,o for some I < p < oo. We must show maps L1(µ) to Lp(v). that the integral operator T generated by the function

That is, we must show that for each x E L1(µ) the formula fs T(s, t)x(s) dµ(s) defines a function that belongs to Lp(v). Without loss of generality we can suppose that T(s, t) > 0 holds for all (s, t) E S x T. Pick a non-negative function x E L1(µ) and let

y(t) =

r T(s, t)a (s) dp(s) .

is

5.2. Abstract Integral Operators

167

Now let 1 < q < oc satisfy n + v = 1. From Holder's inequality, we get y(t)

= JT(s.t)x(s)d/1(s) = f [[T (s, t)]px(s)] ° [x(s)]1

[f [T(s, t)]px(s) dp(s)]

<

P

'

S

[J [T(s,t)]Dx(s)du(s)]

dp(s)

[ f [x(s)]q(l 0) dp(s)]

P

s

llxlli

Therefore, using Tonelli's theorem, we obtain that IlyllD

= L ly(t)Ip dv(t)

f

[f [T(s, t)]px(s) dp(s)]

llxl[

dv(t)

[ f \ f [T(s, t)]D dv(t))x(s) dp(s) ] Ilxlli

11

S

T

lIT(', )llD.. ' llxlli < 00. defines an This shows that y E Lp(v), and so the it x v-measurable function integral operator T: L1(p) - Lp(v) that satisfies IITII <_ IIT(', )hl,.,,. To establish the reverse inequality IITII > llT(',')IID.,,, consider the adjoint Since both Banach lattices L1(p) and Lp(v) have operator T': Lq(v) order continuous norms, Theorem 5.29 guarantees that T' is also an integral operator whose kernel T' (t, s) satisfies T' (t, s) = T(s, t).

Fix some E > 0 and then find some set A E E1 such that 0 < p(A) < 00 and [fT[T(s, t)]p dv(t) ] o > IIT(', )llp,,o - E for each s E A. In particular, we have eSsSUP8ES [fT[T(s, t)]D dv(t)]

o > IIT(', ) Ilp,,, -e. Now let U denote the closed unit ball of Lq(v) and let U+ be its positive part, i.e., U+ = U n Lq (v). Clearly, the set t)ti(t) dv(t) : 1P E U+ j is a norm bounded subset of L00+ (A), and T(U) = { fT hence it is also order bounded. In particular, since Li(p) is Dedekind complete, Now using part (2) of Lemma 1.101, we see that supT'(U+) exists in

=

IIT' lI

sup IIT'IGII = sup 11f LEU+

%LEU+

T

II sup IT T(-, t)'(t) dv(t)

su

T(', t)V,(t) d(t) II °Q

II =

II

j[T(., t)]p dv(t) II.

[fT[T(s,t)]pdv(t)]

"E

IIT(,S)IID. - E . Since IITII = IIT' II and e > 0 is arbitrary, we have IITII >

and so indeed

IITII =

For the converse, assume that a p x v-measurable function T(., ) defines an beintegral operator from L1(p) to Lp(v). We want to show that the kernel longs to L9,,o. This follows immediately from the previous estimate by considering the dual operator for which we have IITII = IIT' II > Hence,

E

Lp,.

11p..

5. Integral Operators

168

Problem 5.2.19 (Lozanovsky [54]). Dunford's theorem stated in part (2) of Corollary 5.33 asserts that: If (S. E1, µ) and (T, E2, v) are or-finite measure L,,(v) is spaces and I < p < oo, then every bounded operator T : L1(µ) a regular integral operator. Present an alternate proof of Dunford's theorem using Problem 5.2.18. Solution: Let T : L1(µ) -. Lp(v) be a bounded operator, where 1 < p < oc. As usual, we let 1 < q < oo satisfy v + 1 = 1. By virtue of Theorem 3.9, we can assume that T is a positive operator and assume also that IITII < 1. Let Ei be the collection of all subsets A of E1 with 0 < µ(A) < cc. For each set A E Ei let UA denote the collection of all non-negative functions K E Lp,a, = LQ 1 satisfying IIKII9.. <_ 1 and TXA(t) =

fK(s.t)A(s)dP(s)

(*)

for v-almost all t E T. that We shall prove that there exists some p x v-measurable function belongs to each UA. If this is established, then it will follow from (*) that for each

p-step function 0 we have TO(t) = fS Ko(s, t)di(s) dµ(s) for v-almost all t E T. Then, using the fact that the step functions are norm (and also order) dense in L1(µ), a continuity argument will guarantee that for each x E L1 (p) we must have

Tx(t) =

r Ko(s, t)--(s) dp(s) S

for v-almost all t E T. This proves that T is an integral operator with kernel K0. We claim that the family of sets {UA}AeEf has the finite intersection property. To see this, let A1, ... , A E Ei, and then pick pairwise disjoint µ-measurable sets C1.... , Ck E E{ such that each A, is a finite union of some subcollection of the

sets C1,....Ck. If we define the function K: S x T - R by k

K(s.t) _

Xc,(s)(TXc,)(t)

:-1

then it is not difficult to check that K E UAW for each j = 1.... , n. Therefore, the family {UA}AEE{ has the finite intersection property.

To finish the proof, it suffices to show that each UA is compact in the weak* topology or (L,,,x,Lq,1). To see this, assume that a net {Ka} in some UA satisfies Ka -IC- K in Lp x. Obviously K >_ 0 and IIKIIp,x < 1. It is easy to see that for each set B E E2 with v(B) < oo the function 0(s, t) = XA(s)XB(t) belongs to L., 1. This implies

L [IJA K(s, t) dp(s)] dv(t)

_

is XT 1 Qm

K4d(p x v) = lim a

f

KK¢d(p x v)

SxT

f. [ fA Kc(s. t) dp(s)] dv(t)

e JB TXA(t) dv(t)

5.3. Conditional Expectations and Positive Projections

169

for all B E E2 with v(B) < oo. But then Lemma 1.96 yields

TXA(t) = f K(s. t) d1(s) = fK(st)xA(s)dP(s) A

for v-almost all t E T. That is, K E UA, and the solution is finished.

Problem 5.2.20. Let E and F be two Banach function spaces with F having order continuous norm. Show that every regular integral operator from E to F carries order bounded subsets of E to norm totally bounded subsets of

F. That is, show that if x E E+, then for each regular integral operator T : E - F the set T[O, x] is norm totally bounded. Solution: Let E and F be as above and fix x E E+. According to Problem 2.3.9 the collection of operators

8 = {T E L,.(E, F) : T(0, x] is norm totally bounded }

is a band in C,(E, F). From Theorem 5.25 we know that E is a normal function space. But then, from Theorem 5.28, we have £,.(E, F) = (E ® F)dd. Since every finite-rank operator is a compact operator, it is easy to see that E; 'O F C S. Consequently, 1 G,.(E, F) = (E (9 F)dd C 8, and the solution is finished.

5.3. Conditional Expectations and Positive Projections Problem 5.3.1. Let (Sl, E, µ) be a probability space, and assume that A is a a-subalgebra of E. Show that for each function f E L.(A) we have Ilf IIA.oo = Ilf IIE.o

Solution: Assume that A is a o-subalgebra of E, and let f E Loo(A). Clearly, f E LS(E) and: IlfIIA.,r

= inf{M>O: A={wESl: If(w)I>M}EAandµ(A)=0},

IIfIIE,,,

= inf{M>O: A={wEQ: If(w)I>M}EEandµ(A)=0}.

Since for each scalar c we have {w E Q: If (w) I > c} E A, it follows that the collections where the above infima are taken coincide. Consequently, the two norms are the same, i.e., IIfIIA.x = IIlIIE,o-

Problem 5.3.2. This problem shows that every Banach lattice with order continuous norm and a weak unit can be represented by a Banach function space that "lives" between L,,. and L1. Let E be a Banach lattice with order continuous norm and a weak unit e. Show that there exists a probability measure space (fl, E, µ) and an ideal F in

L1(E) containing the constants, i.e., LS(E) C F C L1(µ), and a surjective lattice isomorphism T : E -- F such that T (e) = 1.

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Solution: For the solution we shall need the following property of Banach lattices with order continuous norms.

If a Banach lattice E has order continuous norm, then for each x > 0 there exists a positive linear functional on E that is strictly positive on E, the ideal generated by x. We begin by presenting a proof of this result. Fix x > 0. For each 0 < f E E', put Nf = {y E E: f(IyI) = 01 and CJ = Ni. Clearly, f is strictly positive on Cf and, since f is order continuous, NJ is a band in E. Let J= U f>0 CJ Since 0< f< g implies C f C C9, it easily follows that J is an ideal in E. We claim that J is order dense. To see this, let 0 < z E Jd, i.e., -

z-LCfforall0< f EE'. Thatis,zEC1=(Nf)dd=NfforallO< f EE'. Thus, f (z) = 0 holds for all 0 < f E E', and so f (z) = 0 for all f E E. This implies z = 0 or Jd = {0}, and so J is order dense in E. Since E has order continuous norm, J is also norm dense in E. Next, we claim that J is norm closed. To see this, assume that a sequence { yn } C J satisfies y,, - yin E. For each n pick some 0 < f, E E' with II fn II = 1

such that yn E Cf., and let f = rn 1 - fn Clearly, 0 < f E Es. From N f C Nf for each n, it follows that Cf. C Cf for each n. Since Cf (being a band) is norm closed, we see that y E Cf C J. Thus, J is norm closed and consequently (since

it is norm dense) J = E. In particular, there exists some 0 < g E E' such that x E C9, and from this we see that g is strictly positive on E.

Now let E be a Banach lattice with order continuous norm and a weak unit e. By the preceding part, there exists a positive linear functional 0: E -. R that is strictly positive on the ideal Ee. Since e is a weak unit, it follows that 0 is a strictly positive linear functional on E. We can assume that O(e) = 1. Next, we consider on E the L-norm defined by IIIxIII = o(IxI) and, as in the solution of (b) (a) of Problem 5.1.9, we can show that E is an ideal in the norm completion k of (E, I I I - III) and that e is a weak unit in E. Since (E, III - III) is an AL-space, it follows from the Kakutani-Bohnenblust-Nakano Representation Theorem 3.5 that we can identify k with an Li(p)-space over some probability measure space (ft, E, p) with e corresponding to the function 1. Since E is an ideal in E and e = 1 E E, it easily follows that L,. (p) S E C Li(p).

Problem 5.3.3. Show that for every a-subalgebra A of E the conditional L1(E) - L1(E) is a strictly positive order conexpectation operator tinuous contractive projection with range Ll (A). That is, show that:

(1) If f E Ll(E) and f > 0, then E(fIA) > 0 in L1(A). (2) If fn 10 in L1(E), then E(fnIA) 10 in Ll(A).2 has norm one, i.e., (3) The operator

1-

2 Recall (see Theorem 1.80 and the paragraph thereafter) that since we deal here with a finite measure space, each ideal E in Lo(E) satisfies the countable sup property. Accordingly, a positive

operator T : E - E is order continuous if and only if f 10 implies T f. j 0.

5.3. Conditional Expectations and Positive Projections

171

(4) If g E L1(A), then E(gIA) = g, and so and

(5) The operator

E(1IA) =1.

A) leaves LS(E) invariant and IIE('IA)Iloo = 1.

Solution: (1) Assume that a function f E L1(E) satisfies f > 0. Then for each A E A we have fA E(f IA) dp = fA f dp > 0 in L1(A). This implies (see Problem 1.6.14) that E(f I A) _> 0. Moreover, from fn E(f I A) dµ = fo f dµ > 0, we infer that E(fIA) > 0 in L1(A). Therefore, the operator is strictly positive.

(2) Let fn 10 in L1(E) and assume that some function g E L1(A) satisfies 0 < g< E(fn I A) in L1(A) for each n. Using the Lebesgue Dominated Convergence Theorem, we see that 0<

jgd< inf .6 (fIA)du = inf fads 1 0,

which is impossible. Hence, g = 0 and thus E(fnIA) 10. (3) Assume that f E L1(E) satisfies Ilf 111 < 1. The positivity of

IIE(fIA)II1 = j IE(fI-4)I d p <

j

yields

f

E(Ifl IA)dp = n IfI d p = Ilflll < 1.

The latter implies that the operator E(-IA): L1(E) LI(E) is bounded and that IIE(-IA)II1 < 1. Taking into account that E(1IA) = 1, we get 1. (4) and (5) These are obvious.

Problem 5.3.4. Assume that {fZi}iE1 is an at most countable partition of a set fl. That is, I is at most countable, fti 0 0 for each i E I, Il flftj = 0 for i # j, and UiE1 fli = Q. Establish the following. (a) The o-algebra generated by the family

{fai}iEI is given by

A={UlEJfti: JCI}. (b) A function f : 12 -i R is A-measurable if and only if it is of the

form f

= LiE I ai Na, .

(c) If 12 is at most countable, then every a-algebra of subsets of 12 is generated by a unique at most countable partition of 11. (d) Now suppose that (fl. E, p) is a probability measure space and that {fli}iE, is an at most countable measurable partition of f2, i.e, besides {fii}IEI being a partition of f2, we also have f2i E E for each i.

If it (1li) > 0 for each i E I, then for any f E LI(11,E,p) we have

E(f I A) _

N n.) fsz, f dp Xn, . iE 1

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Solution: (a) Since every element of A belongs to the a-algebra generated by the family {Sl,},El, it suffices to verify that A itself is a v-algebra. This claim follows easily from the identities

Q = U Sii

and

iEl

iEl\J

iEJ

and the fact that A is closed under arbitrary unions. (b) If a function f : S l R is of the f o r m Jr = L;E f then it should be clear that f is A-measurable. For the converse, suppose that f is A-measurable. To establish that f is of the form Jr = E,Ef aixi, , it suffice to show that f is constant on each Sl,. To see this, assume by way of contradiction that there exist some i E I and two distinct points wt, w2 E Sl, such that f (wi) 34 f (w2). Now notice that the sets a'-K',.

Ot =

{W E ili: f(w) = f(wt)}

and

O2

= {w E Sl;: f(w) = fPx)}

belong to A and they are both proper non-empty subsets of Sl which is impossible. This proves that f must be of the desired form.

(c) Assume that 1 is at most countable and that B is a a-algebra of subsets of Sl. Let us show first that there is at most one partition of Sl that generates B. To see this, assume that the two partitions {Sli}iE f and {O,},EJ of fl generate

Fix some io E I. Then there exists a non-empty subset Jt of J such that Sli = U,EJ, O,. Now fix any index j E J. There exists a non-empty subset It of I such that O) _ U,EI, fli C Slb. Since the family {flifjE f is pairwise disjoint, it follows that. Jr, = {io}. This implies that Jt must be a singleton too, say .It = {j}. Consequently, f1 j,, = Oj. Thus, each Sli must be some O; and similarly every 6, must equal some Sl,. In other words, the two partitions {Sl, };El and {O; }jEJ coincide, and so B is generated by at most one partition, which is necessarily at B.

most countable. Next, we shall verify that B is generated by a partition. For each w E Sl let

B,,,=n{UEB: WEU}. 0, then necessarily B.,, = B,,,,. Indeed, take any U E B such that wt E U. If w'2 0 U, then w2 E U`, and therefore (keeping in We claim that if B,,,, n B,,,

mind that the set U` belongs to B) we get that B,,,, n B,,,, g U n U` = 0, contrary to our assumption. This proves that w2 E U and consequently Bw,, C BW,. By

symmetry we have B,,,, C B,,,,, whence B,, = B12. This establishes that the collection {B1.,: w E Sl} is a partition of Q.

To finish the proof, it remains to be shown that each B, belongs to B. To we this, let A1, = Sl \ B,,,. Since Sl is at most countable. A1.; is likewise at most countable. Now notice that for each s E A,,, there exists some U, E B with W E U. i.e., and s f U,. In particular, we have B,, g (l,E 4,U,. Now let t E t E U, for each s E A,,,. If t B1.;, then t E A,,, and so t f UU, a contradiction. n,EA.,U,,

Hence, t E B1. and thus f,EAWU, C B,,,. Consequently. B1., = desired.

E B, as

5.3. Conditional Expectations and Positive Projections

173

(d) Assume that the function f belongs to L1(S2, E, p). By part (b) the function ' a, E$EI [ µs; n. dµ] Xa, is A-measurable. Now for each A = VIED fti E A we have

f

that

f >2[µIffl,f dp]Xn,dµ e

r

[ J >2 iEJ

fn` dµ] X.. dµ

>2f fdli=f iEJ n, = JA

4A

f dµ]X

AT? IF)

A iE!

fdµ

fdµ.

This shows that E(f I A) = EiE! [ µ(n,) fn. f d p] Xn, , as claimed.

Problem 5.3.5. For an arbitrary convex function tai : R -. R, any function f E L1(E) for which ip o f E L1(E), and any o-subalgebra A of E establish the following inequalities-known as Jensen's inequalities: (a) V) (ff dµ) < f i,i o f dµ.

(b)

tP(E(fIA)) < e('ofIA) in L1(A).

Solution: Assume that the function f E Ll (12, E, µ) satisfies 0 o f E LI (p). (a) It is well known that at each point a E R the convex function t/, has left and right derivatives and the inequality TV;.(a) < ty'r(a) holds. Moreover, if m satisfies t/,r(a) < m < Oe' (a), then the straight line with slope m passing through (a, t/i(a)) supports that graph of tG at a, i.e.,

m (x - a) + b(a) < 10(x)

(*)

holds for each x E R; see Figure 1.

1!

y = m(t - a) + iP(a)

=tL(t) a

t

Figure 1. The support property of a convex function

In particular, letting a = f f dp and choosing m as above, we have m [f (w)

- Ji dµ] + .& (f f dli) <- +G (f (w))

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174

for it-almost all w E Q. Integrating yields m

[I f dµ - I f dµ] + V (I f dµ) < f b(f) dµ

.

or w(f fdµ) < ftpo fdµ. (b) Assume that A is a a-subalgebra of E. Without loss of generality we can assume that both f (w) and .6(f I A)(w) are finite for each w E Q. Let

D={(a,8)ER2: at +3
sup {at +0: (a,13) E D} = tb(t) . Since R2 is separable. there exists a countable subset Do of D that is dense in D. In particular, notice that

sup {at + 8:

E Do} = 0(t).

(**)

Now fix (a,#) E Do. Substituting f (w) for t in at + /3 < tu(t), we get

af(w) +Q <- t(f(w)) for each w E Q. This inequality, in conjunction with E(1IA) = 1 and the monotonicity and linearity of the conditional expectation operator E(JA), yields

aE(fIA)+8=P(af +IIA) S£(tiofIA) in L1(A). So, for each (a,#) E Do there exists some p-null set A,,,a E A such that -'(of + 01A) (w) <- E(' o f IA)(w) for each w V AQ,a. If we let A = U(Q.0)EDAQ,y3, then A E A, µ(A) = 0 and

aE(fJA)(w)+B <E(t'ofIA)(.)) for all w 4 A and all (a_3) E De. Taking the supremum over Do and using (**), we obtain *(E(fJA)(w)) <_ E(, o f JA)(w) for each w 0 A. Therefore, V,(E(f JA)) < E(1 o f IA) in L1(A). It is worth pointing out that we can get the conclusion of part (a) from part (b). Indeed, if we take A = {O, f1}, then L1 (A) consists of the constant functions on ft. Moreover, it is easy to see that E (f I A) = (f f dµ)1 and E(' (f) I A) _ (f V )o fdµ)1. Consequently, part (b) implies ii(f f dµ) < f it- o f dµ.

Problem 5.3.6. Give an example of an ideal E in L1(E) such that E is a Banach lattice in its own right and E contains a non-trivial norm closed order dense ideal in E. Solution: Let ft = (0, 1] and let E be the a-algebra of all Borel subsets of [0,1]. Choose some u E L1(E) \ L,>(E) such that u > 1; for instance let u(t) = 71t. Now consider the principal ideal E in L1(E) generated by the function u and equip E with the uniform norm IJ pu. We know that (E, II JIu) is a Dedekind complete Banach lattice. Moreover, it should be clear that ideal in E.

is a non-trivial order dense

5.3. Conditional Expectations and Positive Projections

175

We claim that the II - Ilu-distance from u to L r(E) is strictly positive. To we this, let y E L,.: (E) and assume that some .1 > 0 satisfies Iu-yl < Au. This implies

(1 - A)u < y and so if 0 < A < 1 were true, then y could not belong to LS(E), a contradiction. Hence, A > 1. Consequently, Ilu - yllu = inf{A: Iu - yJ < Au} > 1. In particular. we have d(u, L,, (E)) = inf { JIu - yllu : y E Lx (E) 12: 1.

Finally, let F be the norm closure of L,,(E) in (E, II II.) Then F is a norm closed proper ideal in E. Obviously, this ideal is order dense in E. 0

Problem 5.3.7. Show that the conclusions of Lemma 5.42 do not hold in general for the remaining values of the indices. Solution: (a) We shall present an example of a contractive projection T on L,,[0.1] such that T1 = 1 but the adjoint operator T' does not fix 1. Pick a multiplicative rion-zero positive linear functional tt, on L,o[0.1]. (Since 11, as an AM-space with unit, can be represented as a C(K)-space in view of the Kakutani-Bohnenblust-Krein Theorem 3.6, non-zero multiplicative positive 1 and (by Theorem 4.30) we have linear functionals exist.) In particular, IIivII = 1. Consider next the rank-one operator T = i: 0 1 on L."'[0,11, that is.

Tx = v(x)1 for each x E L,j[0,1]. Clearly, the operator T is a positive contractive projection and T1 = 1. Now notice that the adjoint operator T' = 10 *. satisfies

T'(1) = 11 x :1(1) _ (1,1)v = 0 0 1. that is. T' does not fix the function 1. (b) We will present here a contractive operator T on L2[0,1] such that Ti = 1 but T is not positive. First consider the three-dimensional Euclidean space R3 and the straight line L passing through the origin in the direction of the vector (1, 1. 1). Let R be the rotation of R3 about L through a "small" positive angle 0, say 0 = 20°. Clearly, R is a linear isometry, R fixes the vector (1, 1. 1) and R is not positive. B = [3, 3). and C = [3, 1]. and let Next, consider the sets A = [0. 3).

Y = {e1, + C2k0 + C3x : Cr. C2i C3 E R } . It should be clear that Y is a closed vector sublattice of L2[0,1] such that 1 E Y.

Now equip R3 with the f2-norm and consider the lattice isometry J: R3 -' Y defined by J(xl, r2, r3) = 3x1 k,, +3x2 +3X3 V- Also, let Q denote the orthogonal projection from L2[0,11 onto Y. If we consider the scheme of bounded operators

R3 --B- R3 _L, y L2(0,11, then it should be clear that T is a and let T = JRJ-1Q: L2[O. 11 contractive operator such that T1 = 1 but T is not positive. For instance, notice L2[().1]

that Tom.

L2 [0, 1].

---I

U

Problem 5.3.8. Show that the conditional expectation operator E(. JA) satisfies the following property: If a function f E Li (E) has its support in some set A E A, then the function E(f [A) also has its support in A. Solution: Assume first that 0 < f < Y,. Then 0 < E(f IA) <- E(x IA) = X.. holds in L1(A). and so Stipp E(f IA) C A. For the general case, let a function f E Lr (E) satisfy Supp f 9 A. From If I A n XA I If I and the order continuity of the operator

5. Integral Operators

176

C(-IA), it follows that E(IfI A nXJIA) T E(IfI IA) in L1(A). So, by the preceding case, we see that SuppE(fIA) C SuppE(IfIIA) = Un 1 Supp(IfI Any,,) C A.

Problem 5.3.9 (Dodds-Huijsmans-de Pager [26]). Suppose that E is an ideal in L1(E) containing the constants, and let 0 < 40 E E. Assume also that for some o-subalgebra A of E the operator P: LI(E) LI(E), defined by Px = E(OxIA), leaves E invariant. Show that P is a projection if and only if there exists a subset A E A such that Supp o C A and C(,01 A) = x,, in LI (A). Solution: By the definition of the operator P we have P1 = E(oIA). Assume that P2 = P. Using the averaging property of the conditional expectation operator, we obtain

E(,PIA) = P1 = P21 = P(E(01A)) = E(oE(4IA)IA) £(&IA)E(1IA) = E(oIA)2.

This guarantees that. £(01A) is a characteristic function of some set A E A. To verify that Supp q C A, suppose by way of contradiction that Supp o is not a subset of A. This means that 0y,,, is a non-zero positive function. By Problem 5.3.8 we have Supp E(ox. I A) C Ac. Now from X,,, < 1 and the positivity of P we get

P(X,,.) < P(1) _ £(qIA) = 7r,,. That is, Suppe(oy",IA) c A. This implies that the set Supp £(qX,,.. IA) E A has s-measure zero, and so E(oX,,, IA) = 0.

In turn, the latter implies qj,,, = 0, which is a contradiction. For the converse, assume that Suppo C A and £(4IA) = X,. We must show that the operator P is a projection. Start by observing that for each x E E we have P2x = P(e(gxIA)) = £(9E(oxIA)1A) = E(oI A)E(OrI A) = X,E(oxI A) . Notice that for each x E E we have Supp ox C Supp b. Hence, from Problem 5.3.8, it follows that Stippe(4xIA) C A, and so XE(4xIA) = E(OxIA) = Px. This proves

that p2 = P.

I

Problem 5.3.10 (Moy [58]; Dodds-Huijsmans-de Pager [261). This problem is a modification of Theorem 5.45 for linear operators that are not necessarily projections. Let E be an ideal in L1(E) containing the constants, and let T : E - E be a linear operator such that: (a) If xn ---+ 0 in E, then f I Tx III -+ 0. (b) T leaves LS(E) invariant. (c) For each h E L.(E) and each g E E we have T(hTg) = ThTg.

Then there exists a o-subalgebra A of E and a function 0 E Eo so that Tx = E(OxIA) for each x E E.

Solution: Let Y = {hELx(E): T(hf) = hTf for all fE

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177

and note that Y is a vector subspace of L,,(E) such that 1 E Y. We claim that Y is an order closed subalgebra of L,, (E). If this claim is true, then by Corollary 5.41 there exists a a-subalgebra A of E such that Y = L,,(A). The fact that Y is a subalgebra follows immediately from condition (c). To C Y, h E L,, (E) and h -g'h show that Y is order closed in L., (E), assume We must verify that h E Y. To this end. fix any f E L,. : (E). Since in f -Q' hTf in E. In view of h -' h in L,o (E), we certainly have h f -' h f and -T(hf)[11 - 0, (a) we have T(hf)111 -. 0, and consequently f for each n. The latter shows that T (h f) = hTf as h E Y implies T (h f) = and thus h E Y. for each f E

Next, we claim that T(E) C Y. To see this, we shall establish first that C Y. So, let r E L,,(E). By (b), the function Tx belongs to Thus, in view of (c), for each f E L,,, ,(E) we have T ((Tx) f) = TxT f . This shows that Tx E Y, and thus T(Lx(E)) C Y. Now, since L,,. (E) is order dense in E and Y order closed in L,c (E), condition (a) easily implies that Tx E Y for each x E E.

To complete the solution, we shall consider two cases. Assume first that E = L,,, (E). Define the linear functional 4): L,,(E) --' IR by fi(x) = fnTxdp. L1(E). Therefore, there exists a Condition (a) guarantees that 4) E function 0 E L1 (E) such that 4)(x) = fo ox du for each x E L,,. (E). Using this and the fact that for each set A E A the function ',, belongs to Y, for each A E A and x E E we obtain

x dp = I mxXdp = 4i(x) =

fin T (xX) dit = Jr

Tx dp =

r

Tx dp .

In, JA a Since Tx is A-measurable, we can conclude that Tx = E(¢xIA). The general case of an arbitrary E can be easily reduced to the previous one. Indeed, consider the restriction of T to L,,(E). We denote this restriction by T,,.. It should be clear that T,, also satisfies conditions (a), (b) and (c), and consequently, by the previous part, there are a a-algebra A of E and a function -0 E L1(E) such that T,,.(x) = E(4xIA) for each x E L,o(E). This and condition (a) imply immediately (since L,,(E) is an order dense ideal in E) that the representation IA

Tx = E(OxIA)

is valid on all of E. In particular, the obtained representation of T implies that Ox E L1(E) for each x E E. This shows that the function 0 belongs to E,; .

Problem 5.3.11. Give an example of a Markov projection on L1 [0, 1] that is not bistochastic. Show that such a projection cannot be contractive. Solution: A Markov projection which is not bistochastic can be found in Example 5.56. We present another example below. Let Sl = [0, 11, let E be the a-algebra of all Lebesgue measurable subsets of [0, 1], and let p be the Lebesgue measure on [0, 1]. Also, we let A denote the a-subalgebra generated by the two sets A 1 = [0,12 ] and A2 = [ 1,1] . We now 2 consider the operator P: L1(E) -p L1(E) defined by the formula Px = E(bxJA). where 0 = 2X011 +2Xij,11. A glance at Problem 5.3.4 shows that E(OIA) = 1, and so in view of Problem 5.3.9 we know that P is a positive projection on L1 [0, 1] such

that P1 = 1.

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178

It retrains to show that the adjoint operator P^' does not fix the function 1. To this end, consider the function x = Pxdp and note that ox = 0. This implies Px. = E(,pxjA) = 0, and so we have f = 0 0 a = f x d1c. By Lemma 5.48. the adjoint operator P^' does not fix the constant function 1. This projection P cannot be contractive since, otherwise, Lemma 5.42(a) would guarantee that the adjoint operator P" must fix the function 1. a contradiction. Of course. it can also be shown directly that IIPIH > 1. 1 Corollary 5.52 was generalized to Lp(E)-spaces for 1 < p < oc with p 96 2

by T. Ando 1101. Under the assumption that the contractive projection is positive, the restriction p 0 2 is redundant. We present below a very simple proof of this result.

Problem 5.3.12. For a vector subspace Y of some Lp(E) with 1 < p < 00 establish the following.

(a) If Y is a closed vector sublattice, then Y has a weak unit. (b) Y is a closed vector sublattice if and only if it is the range of a contractive positive projection on Lp(E). Solution: (a) Let Y be a closed vector sublattice of an Lp(E) with 1 < p < oc. By Zorn's lemma there exists a maximal pairwise disjoint family {x,, hE. of functions in Y. If DA E E denotes the support of x),, i.e., D), = Jw E Sl: x,\(w) # 0}. then { Da } AEA

is a tt-a.e. pairwise disjoint family of subsets of E. Since p is a probability

measure, it follows that the set I is at most countable. We can assume that A = N and that llxn(Jp = 1 for each n. Let x = J:n t 2 E V. and we claim that x is a weak unit in Y. To see this, let y E Y satisfy x 1 y = 0. If y 54 0. then y 34 xR and y I xn holds for each n, and so the family {y, xt, x2, x3....} is pairwise disjoint and includes {x1..r2.... } properly, contrary to the maximality property of {x1,x2.... }. Hence, y = 0 and so x is a weak unit in Y. (b) Assume first that Y is the range of a contractive positive projection on LP(E)Since the norm of Lp(!l. E, u) is strictly monotone, it follows from Lemma 5.43(b) that Y is a closed vector sublattice of LP(E)-

For the converse, assume that Y is a closed vector sublattice of Lp(E). By part (a), we know that Y has weak units. Let U E Y be a weak unit in Y with IIuIIp = 1. Replacing 1 with Supp u, we can assume without loss of generality that Supp u = Q. Now define a new probability measure v on E via the formula

v(A) = ju'dit. Clearly, p and v are equivalent measures and the Randon-Nikodym derivative of

v with respect to µ equals up, i.e., = up. This implies that the mapping J: Lp(ft E. v) Lp(1l. E, µ), defined by

a

J(f)=f(dµ)A=fu, is a surjective lattice isometrv; see [8. Problem 37.11, p. 3581.

5.3. Conditional Expectations and Positive Projections

179

It follows that, Y1 = J-1(Y) is a closed vector sublattice of Lp(fl,E,v) and since J-1g = g/u for each g E Lp(1, E, it), we see that 1 E Y1. Therefore, by Lemma 5.40, there exists a a-subalgebra A of E such that Y1 = Lp(fl, A, v) and consequently, by Corollary 5.52, the conditional expectation operator E(JA) on L1(11, E, v) acts as a positive contractive projection from Lp(ll, E, v) onto Y1. To finish the proof, notice that the operator P = Lp(E, p) -+ Lp(E, it) is a positive contractive projection with range Y. It is interesting to point out that the above result almost characterizes the Lp spaces. Namely, T. Ando [11] proved that: If every closed vector sublattice of a Banach lattice E is the range of a positive contractive projection, then E is lattice isometric to an Lp- or a c0(r)-space.

Later, L. Tzafriri [81] and J. Lindenstrauss and L. Tzafriri [51] established an isomorphic version of this result.

Problem 5.3.13. According to Problem 5.3.12(b), if a Markov projection P on some Lp(E) with 1 < p < oo is contractive, then its range is a vector sublattice of Lp(E). Show that the converse is not true. That is, give an example of a Markov projection P on Lp(E) such that its range is a vector sublattice of Lp(E) but IIPII > 1. Solution: We shall exhibit an example of a Markov projection P on Lp(E) such that its range is a vector sublattice of Lp(E) but IIPII > 1. Let n = [0, 11, let E be the Borel a-algebra, let µ be the Lebesgue measure, and let A = {O, (1) be the trivial a-subalgebra of E. So, E(xlA)

fo x (w) dp(w) 1

holds for each x E L1(E). Now let 0 = 2Xfo,1/21 and note that E(OIA) = 1. This implies that the formula Px = E(OxIA) defines a Markov projection whose range Y is the one-dimensional subspace spanned by the constant function one. In particular, Y is a vector sublattice of Lp(E) for each 1 < p:5 oo.

However, for each 1 < p < oc the projection P is not a contraction, i.e., IIPII > 1. To see this, consider the function x = 2"Xfo,1/21 and note that IIxIIp = 1 An easy calculation shows that Px = 21,, . 1, whence IIPII ? IIPxlI = 20 > 1. 1

Problem 5.3.14. Show that a projection on a Hilbert space is an orthogonal projection if and only if it is a contraction. Solution: Let P: H - H be any projection on a (real or complex) Hilbert space. Assume first that P is also an orthogonal projection, that is, the decomposition H = Ker (P)®R(P) satisfies Ker (P) 1 R(P). Now for each x E H write x = x1+x2 with x1 E Ker (P) and X2 E R(P) and note that IIPxII2 =11x2112 <_ Ilxl 112 + 11x2112 = IIx1 + X2 112 = 11x112

Thus, IIPxII < llxll for each x E H and this proves that P is a contraction.

5. Integral Operators

180

For the converse, assume that P is a contraction. Let a E Ker (P) and 6 E R(P) and let (a, b) = a + 02 E C be the inner product of the vectors a and b. Then for each scalar A we have IIAbII2 = IIP(a + Ab) 112 < IIP1I2 Ila + Ab112 < IIaII2 + 2Aa + IlAbll2.

Therefore, IlaII2+2Aa > 0 for each A E R, and from this it easily follows that a = 0. Observing that (ta, b) = - j3 + at and applying the preceding conclusion to the

vectors to E Ker (P) and b E R(P), we get /3 = 0. Thus, (a, b) = 0 holds true for

all a E Ker (P) and b E R(P) , i.e., Ker (P) 1 R(P). This shows that P is an orthogonal projection.

Problem 5.3.15. Give an alternative proof of Lemma 5.38 by employing Theorem 5.57. That is, show that for each a-subalgebra A of E and for each 1 < p < oo the operator leaves the space Lp(E) invariant and is contractive on this space. Solution: We need to consider only 1 < p < oo. Let q be the index conjugate to p. Take any function x E Lp(E) and consider the function e(xIA). Recall that this function is A-measurable, and therefore its II . II-norm can be calculated as follows:

IIe(zlA)Ilp = sup{(e(xIA),y): y E Lq(A), IIyIlq 51} = sup{(e(xIA),y): y E L.(A), IlYIlq 5 1} = sup{(x,e(yIA)'): y E L.(A), IIyIIq <_ 1} = sup{(x, e(ylA)) : y E L.(A), IIYII9 < 1} <

sup{(x, Y): y E L.(E), IIyIIq <_ 1} = IIxI[p

This establishes both that e(xIA) E Lp(A) and that lit(-IA)Ilp <_ 1.

Remark. Formally speaking, there is a circular argument in proving Lemma 5.38 by using Theorem 5.57 since our proof of Theorem 5.57 depends upon this lemma, where we used that each conditional expectation operator acts as a contractive operator on L2(E). However, the reference to Lemma 5.38 can be avoided by invoking Theorem 5.54.

5.4. Positive Projections and Lattice-subspaces Problem 5.4.1. If P is a positive projection on an Lp(µ)-space for some 1 < p < oo, then show that its range is again order isomorphic (up to an equivalent norm) to an Lp(v)-space for some measure space (521, E1, v).

Solution: In view of Theorem 5.60 the range Y of the operator P is a closed lattice-subspace of Lp(p) which is lattice isomorphic to a closed vector sublattice F of Lp(fZ, E, µ). But each closed vector sublattice of an Lp-space is an abstract Lp-space and it is lattice isometric to an Lp(v)-space for an appropriate measure space (ill, El, v); see [5, Theorem 10.12, p. 711 . Some remarks are in order regarding this problem.

5.4. Positive Projections and Lattice-subspaces

181

(1) If we replace the assumption "a positive projection" by "a contractive projection," then a corresponding version of the above result is also true. For p = 1 this was established by A. Grothendieck [351 and for the remaining values of p by L. Tzafriri [801. (2) We do not claim in this problem that Y is order isomorphic to the original Lp(p)-space. This is not true in general. However, we have the following

result: If a positive projection P: Ip - Pp, 1 < p < oo, has an infinite dimensional range, then the range of P is order isomorphic to Pp.

Problem 5.4.2. Let (xij : i E I = {1, 2, ... , n}; j E J = {1, 2, ... , m}} be a finite subset of vectors in a vector lattice. Then we have n m m n m n and V A xrj)J = A V xij A V xi.T(i) = V A xij i=1 j=1 i=1j=1 rEIJj=1 rEJ'i=1

.

Solution: The solution will be based on the two (infinite) distributive laws that are valid in vector lattices. We state and prove them for later use in the solution. (Dl) If a non-empty subset D in a vector lattice E has a supremum, then for each vector a E E the set D A a= {x A a: x E D} also has a supremum and sup (D A a) = (sup D) A a.

(D2) If a non-empty subset D in a vector lattice E has an infimum, then for each vector a E E the set D V a = {x V a: x E D} also has an infimum and inf (D V a) = (inf D) V a.

To prove (Dl), let s = sup D and fix an arbitrary vector a E E. Then for each x E D we have x A a < s A a. This shows that s A a is an upper bound for the set D A a. To see that s A a is the least upper bound of D A a, assume that a vector y satisfies x A a< y for each x E D. We have

x+a-sVa<x+a-xVa=xAa
mm

m

)1=1)2=1

)n=1

n

A V xi,rci) = A A ... A

IEJ! i=1

1x1.,J V x,;,V ... V xnJn,

We claim that the right-hand side of this identity is equal to V=1 Aj 1 xij. This will be proved by induction on n. For n = 1 and any vectors {x11, x12, ... , x1,n} we have ATl=1 xja, = V;=1 A 1 xi), and so the desired identity is true for n = 1. Now assume that for some n > 1 and for all finite subsets of vectors of the form {xij: i = 1,2,...,n; j = 1,2,...,m} we have m nm mm A A ... A [x1,jI V x2.73V ... V xnan1 = V A xi)

ji=1j2=1

)n=1

i=1 j=1

(*)

182

5. Integral Operators

To prove our claim for n + 1 consider an arbitrary finite subset of vectors i = 1,...,n+ 1; j = 1, ... , m}. To simplify the computations below, let = A,+,=1 xn+l.jn+1 Using the induction hypothesis (*) and the distributive

{x,J :

x

laws, we see that

n m

n+1 m

VAxij

[V A x,,]vx t=13=1

1=1 J=1

in

e

m

A

m

[A J1=1j2=1 in

... A fx1.j,Vx2.J2V j..=1 m

m

A A ... A [x1.JIVx2.j2V-jn=1

J1=1J2=1

mm

AA

m

...

A [ A x1.j1 V X2-j2 V

V x, ,, V xn+l.jn+1]

jn=1 j_+1=1

j1=1j2=1

mm

m

m

A A ... A

[xi J1Vx2.12V

V xn.J.. V Zn+1.J..+1] .

j,,=1Je+1=1

J1=192=1

I

Thus, (*) is true for n+ 1, and the solution is finished.

Problem 5.4.3. If Y is a vector subspace of a vector lattice E, then the vector sublattice of E generated by Y coincides with Y^V and also with YV ^. That is,

R(1,)=yAV=YV^

Solution: A glance at Problem 5.4.2 shows that Y^11 = YV^. Also. it should be clear that Y^V C R(Y). Therefore, in order to establish the identity Y^V = R(Y), it suffices to show that Y^ V is a vector sublattice of E. To see this, let x = V i A j7-1 x, and y = up i Aai ypq be two vectors in Y^V. Then t n in k n k m 1

x+y= V Ax,J+p=1q=1 V A= V Vt=1p=1J=1q=1 A A(xij+ypq) EY"V s=1j=1

Now let a be a scalar. If a > 0, then n n ax=alrV A xi,] = V A ax1J EY^V i=1j=1 s=1j=1

and for a < 0 we have ax

nm

A =a [V i=1j=1

n m V axij E YV^ =YAV ] = i=1j=1

The above show that Y^ V is a vector space. Finally, from n

xVy= [V

Ax1v['/

i=1j=1

t

Ayiq]

EY^V,

P=1q=1

we conclude that Y^V is indeed a vector sublattice of E, as desired.

I

5.4. Positive Projections and Lattice-subspaces

183

Problem 5.4.4. Show that a vector subspace of a finite dimensional vector lattice is a lattice-subspace if and only if it is the range of a positive projection. Solution: Let Y be a vector subspace of some R". If Y is the range of a positive projection, then Y is a lattice-subspace by Theorem 5.59. For the converse, assume that Y is a lattice-subspace. Then the identity op-

erator I: Y -y Y is a positive operator and Y (as being finite dimensional) is Dedekind complete. By the Kantorovich Extension Theorem 4.32, the operator I has a positive linear extension P: B , Y, where B is the ideal generated by Y in R". Since B is closed. B is a band in the Dedekind complete Riesz space R". Write R" = B E) Bd, and let PB denote the band projection of R" onto B. Now notice that the operator PPB : R" - R" is a positive projection whose range is Y. I

Problem 5.4.5. For a vector subspace Y of some C(K)-space establish the following.

(a) V separates the points of K if and only if R(Y) separates the points of K. (b) Assume that the induced cone Y+ = Y fl [C(K)I+ is generating in Y, i.e., Y = Y+ -Y+. Then Y majorizes C(K) if and only if R(Y) majorizes C(K). (c) Assume that: (i) Y is a Dedekind complete lattice-subspace. (In particular, this is so if Y is a finite dimensional lattice-subspace.) (ii) Y separates the points of K and 1 E Y.

Then the Miyajima projection P: R(Y) --* R(Y) extends uniquely to a positive projection on C(K) with range Y. Solution: (a) If Y separates the points of K, then R(Y) automatically separates the points of K since Y C R(Y). For the converse, assume that R(Y) separates the points of K. If Y does not separate the points of K, then there exist two distinct points t1, t2 E K such that x(t1) = .r(t2) for each x E C(K).

Now let y E R(Y) and choose x;, E Y (i = 1, ... , n: j = 1, ... , m) such that y = \/; 1 A xe,. Since the lattice operations in C(K) are pointwise, we have 1

n

y(t1)=V

rn

.=1,=1

nm

/\xf,(t1)=Vn2i,(t2)_]/(t2).

3=1j=1

That is, y(t1) = y(t2) for each y E R(Y). In other words, R(Y) does not separate the points of K, a contradiction. This shows that Y separates the points of K. (b) If Y majorizes C(K), then 1Z(X) also majorizes C(K). For the converse, assume that R(Y) majorizes C(K). Let .r E C(K) and pick some y E R(Y) such that x < y. Now choose functions x1 E Y (i = 1, 2..... n; j = 1, 2, ... , m) such that y = V 1 A' )'= 1 xv . Since Y+ is generating, for each pair (i. j) there exists some z,, E Y+ such that yip < z,,. If we let z = E' 1 z,_, then z E Y+ and it is easy to see that x < y < z. This shows that Y is majorizing C(K). 1

F'

184

5. Integral Operators

(c) Since R(Y) includes the constant function 1 and separates the points of K, it follows from the classical Stone-Weierstrass theorem that R(Y) is norm dense

in C(K). In particular, if the operator P: R(Y) -, R(Y) extents to a positive C(K), then P (as a positive operator defined on a Banach operator P: C(K) lattice) is continuous. Since P agrees with P on R(Y), it is uniquely determined. To see that P: 1Z(Y) -. R(Y) does have a positive linear extension to all of C(K), note

that the operator P: 1Z(Y) -- Y satisfies all the assumptions of the Kantorovich Extension Theorem 4.32. So, according to this theorem, a positive linear extension

of P: R(Y) -. Y to a positive linear operator P: C(K) - Y exists. This extension is a positive projection on C(K) with range Y that extends the Miyajima projection

I

P: R(Y) -+ R(Y).

Problem 5.4.6. Let E = C[0,1] and let Y be the vector subspace generated by the functions 1 and s, where s(t) = t for each t E [0, 1]. That is, Y consists of all linear functions. Establish the following. (a) Y is a lattice-subspace of E. (b) R(Y) consists of all piecewise linear continuous functions. (c) The Miyajima projection P: R(Y) -+ R(Y) is given by

Px(t) = tx(1) + (1 - t)x(0) for each x E R(Y) and each t E [0,1]. (d) If we define Q: C[0,1] -. C[0, 1] by Qx(t) = tx(1) + (1 - t)x(0) for each x E C[O,1) and each t E [0, 1], then Q is the only extension of the Miyajima projection to a positive projection on all of C[0,1]. (e) There are infinitely many bounded projections on E with range Y. Solution: Recall that a function x E C[0,1) is said to be piecewise linear if < a,, = 1 of the interval [0,1] and there exist a partition 0 = ao < al < constants mi, bi (i = 1, 2, ... , n) such that x(t) = m,t + b, for each a, 1 < t < ai. If x(t) = mt + b for each t E 10, 1], then x is called a linear function.

(a) It should be clear that the vector space Y generated by the functions s and 1 is precisely the vector space of all linear functions. If z1(t) = m1t + b1 and x2(t) = met + b2 are two linear functions, then the linear function whose graph is the line joining the points (0,max{x1(0),.2(0)}) = (0,max{bi,b2}) and (1, max{x i (1), x2(1) }) = (1, max{m1 + bi, m2 + b2 }) is the least upper bound of x1

and x2 in Y. That is, the linear function y(t) = [max{m1 +b1,m2 +b2} - max{b1,b2}]t - max{bi.b2} satisfies y = x1Wx2 in Y. This shows that Y is a lattice-subspace of C[0,1). (b) We shall denote by L the collection of all piecewise linear functions. Clearly, L is a vector sublattice of C[0,1] and Y C L. Therefore, R(Y) C L. So, what needs verification is the inclusion L C R(Y).

5.4. Positive Projections and Lattice-subspaces

185

For each 0 < a < b < 1 and each m E R, we consider the continuous function ua,b,m : [0, 1] -+ R defined by 1

ua.b,m(t) =

0

if 0
m(t - a) if a < t < b tm(b - a) if b < t < 1

.

Now notice that if x is a typical piecewise linear function as defined at the beginning of the solution, then x = b1 + UO,a,,mi + ua1.as.m2 + Ua2,a3.m3 + ... + uan-i.1,mn

Therefore, in order to establish that L C R(Y), it suffices to show that each function Ua,b,m belongs to R(Y). To this end, consider the functions y1, y2 E C[0,1) defined by yl (t)

=

[m(s - a1)+J (t) _

y2(t)

=

[-m(s - ,01)+](t) _

if 0 < t < a

0

m(t-a) Jr a<(-<1, 0

if 0 < t < b

-m(t - a)

if b
Clearly, y1, y2 E 7Z(X) and they satisfy y1 + y2 = ua,b,,,,. Hence, each function of the form Ua,b,m belongs to R(X ), and thus R(Y) = L.

(c) The mapping P: R(Y)

R(Y) defined by

Px(t) = tx(1) + (1 - t)x(0) for each x E R(Y) and each t E [0, 1] is clearly a positive projection on R(Y) with range Y. By the uniqueness of the Miyajima projection, P must coincide with the Miyajima projection.

(d) The operator Q: C[0.1]

C[0,1], defined by Qx(t) = tx(1) + (1 - t)x(O)

for each x E C[0,1] and each t E [0, 1], is a positive linear extension of the Miyajima projection to all of C[0.1]. Since any such extension must be continuous and R(Y)

is norm dense in C[0,1], it follows that Q is the only extension of the Miyajima projection to a positive projection on all of C[0,1].

(e) Fix any two points a and 0 in [0,1] such that 0 < a < Q < 1. For each x E C[0,1] let Tx be the linear function whose graph is the line joining the points (a, x(a)) and ((3,x(,3)), i.e.,

Tx(t)

(t - a) + x(a).

Then T is a continuous linear projection on C[0,1] with range Y. However, it is easy to see that the projection T is not positive.

Problem 5.4.7. If Y is a majorizing lattice-subspace of a Banach lattice, then show that the Miyajima projection P: R(Y) - R(Y) is continuous. Solution: We consider ?Z(Y) equipped with the lattice norm of E. Then, according to Theorem 5.59(5). there exists a lattice norm 111-111 on Y satisfying IIxil : IIIxtII

for each x E Y.

186

5. Integral Operators

According to Problem 4.3.8 the operator P: (R(Y), II . II) - (Y, III . III) is continuous. Therefore, there exists some M > 0 such that IIIPxIII < MIIxII for each x E R(Y). This implies IIPxII < MIIxII for each x E R(Y). The obtained inequality shows that the Miyajima projection P: (R(Y), II II) - (R(Y), II . II) is

I

continuous.

Problem 5.4.8. This problem presents an application of lattice-subspaces to harmonic analysis. Recall that a C2 junction u : 0 -' R, where 0 is an open subset of R2, is said to be harmonic if it satisfies Laplace's equation = 0 at every point of O. O2u = + Now let D = { (x, y) : x2 + y2 < 11 be the open unit disk. We shall denote by N the vector space of all continuous functions defined on the closed

unit disk D = { (x, y) : x2 + y2 < 11 and harmonic on D. The classical Dirichlet theorem states that if f is a continuous function on the unit circle 8D = { (x, y) : x2 + y2 = 11, then there exists a unique continuous extension

f of f to all of D which is harmonic on D. The harmonic extension f in polar coordinates is given by Poisson's classical formula _

(e) - 12x JOr jr,

2a

coso,sinO 1-r dO(r 6) E [0, 1) x [0 27r] ' , '

1-2r sin B-Q)+r

Establish the following.

(a) A function f E C(D) satisfies fo > 0 if and only if f0D > 0, where feD denotes the restriction of the function f to 8D.

(b) The mapping Q: CA -+ C(D), defined byQf = f8D, is a positive projection whose range is N. (c) ?{ is a lattice-subspace of C(D) and Q is the unique extension of the Miyajima projection to C(D). (d) ?{ is an AM-space with unit 1, the constant function one, and the mapping f F-' f, from C(8D) to H, is a surjective lattice isometry. Solution: (a) Assume first that f (x, y) > 0 for all (x, y) E D. Since f extends continuously to the boundary OD, we have f (x, y) > 0 for each (x. y) E OD. Hence,

from f = f on OD, we see that feD > 0. For the converse, assume that feD > 0, i.e.. f (x, y) > 0 for each (x, y) E OD. Fix some (r, e) E D. From 2r cos(B - 0) < 2r, we obtain

2

1-2rcos(8-0)+r2> 1-2r+r2=(1-r)2>0.

q_2 For each 0 E [0, 2a] we have f (cos 0, sin 0) > 0, and so > 0 for r. each 0 E [0, 21r]. Now a glance at the Poisson formula guarantees that f (r, 0) > 0 for each (r, 0) E [0,1) x [0, 21r].

(b) The linearity and positivity of Q are easy consequences of the Poisson formula and part (a). The identity Q2f = Qf for each f E C(D) follows immediately

5.4. Positive Projections and Lattice-subspaces

187

from the uniqueness property of the Dirichlet theorem. The fact that Q is the identity on it (and hence its range is 7{) also follows easily from Dirichlet's theorem.

(c) Since it is the range of the positive projection Q. Theorem 5.59 guarantees that it is a lattice-subspace of C(D). To see that the positive projection Q is uniquely determined, notice first that 1 E ii and that 71 separates the points of D. For instance, let (xl, yl ), (x2.112) E D

satisfy (xi, yl) 0 (x2, y2). For instance, if xl 0 x2, then the function g(x, y) = x satisfies g E it and g(x1, ii) 0 g(x2, y2). From the Stone-Weierstrass theorem, we infer that the vector sublattice R(it) generated by 71 is norm dense in C(D). Recall

also that the Miyajima projection P: R(7i) -- R(7{) is a uniquely determined positive projection on R(7{) with range R. So, since every positive projection on Cam) is continuous, we infer that Q is the unique extension of P to a positive projection on C(D) with range 7{. (d) The proof will be based upon the following well-known maximum modulus principle for harmonic functions.

If u E it, then max Iu(x. y)I = nlax

I u(x. y)I.

(x.y)ED

Hence if f E CA satisfies 11f II. < 1, then II Q f II x < 1. Since Q1 = 1, it easily follows that the projection Q has norm one. From Theorem 5.59(6); we know that under the norm 1IIhII1 = IIQIhIIIx

the lattice-subspace it is an AM-space with unit, where the unit is the function Ql = 1. Now notice that for each h E 7 i we have IIhII. = IIIhIIL = IIIQhIII,,, 5 IIQIhIII, = IIIhI II <_ IIQII - IlhIt = IIhHI. This shows that IIIhIII = IIhIIx for each h E 7{. In particular, the lattice-subspace N with the induced norm from C(D) is an AM-space. Moreover, it should be clear that the closed unit ball of 71 coincides with the closed interval [-1, 1]. That is, (7{, II - II,o) is an AM-space with unit 1.

From the above conclusions it is easy to see that the mapping f - f, from C(c9D) to it, is a surjective lattice isometry.

Chapter 6

Spectral Properties 6.1. The Spectrum of an Operator Problem 6.1.1. Establish the following identity-known as the second resolvent identity. If S, T E £(X) and A E p(T) fl p(S), then R(A, S) - R(A, T) = R(A, S)(S - T)R(A,T) . Solution: For each A E p(T) fl p(S) we have R(A, S)(S - T)R(A,T)

= R(A. S) [(S - A) - (T - A)1 R(A, T) = R(A, S)(S - A)R(A, T) - R(A, S)(T - A)R(A,T)

= -(A - S)-1(A - S)R(A,T) + R(A, S)(A - T)(A - T)-1 = R(A. S) - R(A, T) ,

I

as desired.

Problem 6.1.2. Generalize Corollary 6.8 by proving that the derivatives of the resolvent function of an operator T E £(X) are given by the formulas d"R(A,T) dA"

for n = 1, 2, ...

_ (-1)"n![R(A,T)]"+1

.

Solution: By Theorem 6.6. we know that if Ao E p(T) satisfies IA- AoI < R( then 00

00

n=O

n=O

R(A,T) = E(-1)n(A - A,)nR(AO)n+i = E [(-1)nR(A0)n+1] (A - AO)n.

189

6. Spectral Properties

190

Now a glance at Theorem 1.73(3) shows that d"R(A0,T) = n! [(-1)'R(Ao)n+17 = (-1)'n!

[R(A0)]n+1

1

dA "

I

holds for each n = 1. 2_...

Problem 6.1.3. If X is a Banach space and T : X -+ X is a bounded operator, then show that lim AR(A, T) = I. A-.oo

Solution: Note that for each JAI > IITII, we have (*) AR(A,T)=A(a+a +T2+.__) =l+a(I+a+a + From lima-,,(I - a) = I and Problem 2.1.11, we get lima.,o(1 - a)-1 = I. I+a(Ia)`1

Hence. (*) implies lima,, AR(A, T) = I. An alternative solution can be obtained as follows. As above, for each A E C with JAI > IITII we have AR(A. T) = I + En 1 an . Now, a glance at Problem 1.5.6 shows that lima En00=1 An = 0. This implies AR(A. T) = I.

Problem 6.1.4. For any operator T E C(X) show that [R(A. T)]' = R(A, To) holds for each A E p(T). Solution: Let A E p(T). By Theorem 6.14 we have [R(A, T)]' = [(A - T)-1] 0 = [(A and we are done.

- T)'] -1 = (A - T')-1 = R(A, T') , U

Problem 6.1.5. Let T E £(X). If Ao E p(T), then show that R(Ao,T) cannot be quasinilpotent. Solution: Start by observing that an operator S E G(X) is invertible if and only if 0 a(S). In particular, an invertible operator cannot be quasinilpotent. Now assume that T E G(X) and let Ao E p(T). Therefore, AO - T is invertible. This implies that R(Ao, T) = (Ao - T)-1 is also invertible, and thus, by the above, R(Ao. T) cannot be a quasinilpotent operator.

Problem 6.1.6. Show directly that if T : X

X is a bounded operator I

on a (real or complex) Banach space, then the limit limn-00 IIT"!In always exists in R and I I Jim IIT"IJn =inffIT"IIn. I

1

Solution: Let a =1 info JIT"IJ n >0. From the inequality a< III- II n we see that lim inf"_,o JJTnIJ n >a.

Now let e > 0 and then fix some k such that IITkJI k < a + c. For any positive integer n write n = km + r, where m > 0 and 0 < r < k. Note that IIT"II

=

IITr(Tk)mJI :5 IITIir . IIT'`IIm = IITIJT . (JJTkhI k )km JITJIT . (a +

)n_T < 111(a

+ )n.

6.1. The Spectrum of an Operator

191

1

1

r IITn II n <_ 111 n (a + e), from which it follows o+So, where M = maxo<,
lim

supn_. IITn II n

< a + e for all e > 0, or lim supra-. IITn II n < a. This implies -1 1 1

-

a = IimSupra-.oo IITnIIn = liminfra-,o IITnlln. So, a = limn-,,. IITnIIn.

0

Problem 6.1.7. For any operator T E £(X) show that

r(T)=inf{a>0: Solution: Consider the set A = {a > 0:

Ilan 11 --0}.

- 0 } and let ao = inf A. We

11

must show that r(T) = ao. Take any A and 3 such that r(T) < Q < A. Since r(T) = limn-". IIT°" II ^ , it follows that there exists some k E N such that IIT"' II " < 13 for all n > k. This n - 0. Thus, A E A for implies < 1 for all n > k, and so each \ > r(T), and consequently ao = inf A < r(T). To see that ao = r(T), assume by way of contradiction that ao < r(T) and fix some 0 satisfying ao < 0 < r(T). Clearly, (3 E A. However, since 0 < r(T), it

_Y

follows that Q < IITn II ° for all n, and so 1 < T^

0, which is impossible. This

contradiction establishes that r(T) = ao.

Problem 6.1.8. If S, T : X X are two commuting bounded operators on a Banach space, then show that r(ST) < r(S)r(T). Solution: Assume that two operators S, T E C(X) commute, i.e., ST = TS. Then we have (ST)n = SnTn for each n, and so II(5T)nl11=

II-I1° . IIT-II°

holds for each n. Therefore,

r(ST) =

ll mII(ST)nII° < h [JISnll° IIrII" J

[,l

IIS"II° ]

. [n' IIT"II" l = r(S)r(T),

as claimed.

Problem 6.1.9. Let T : X X be a bounded operator and assume that for some non-zero complex number a the series 00 a-(n+l)Tnx converges in norm for each x E X. Show that a E p(T). Solution: Assume that the operator T and the scalar a satisfy the stated properties. Clearly, lim a-(n+1)Tn+lx = 0 for each x E X. n-»oc The series Sx = En°_o -(n+1)Tnx defines an operator from X to X. If for each n we consider the bounded operator Sn = Fn o a-i'+1>T' on X, then Snx -' Sx for all x E X. By the Uniform Boundedness Principle, S E £(X). Now for each

6. Spectral Properties

192

xEX we have n

lim [(a [(a - T)S]x = nix =

t=0

n

n

i=0

t=0

lim

J

lim [x. - a-(n+1)T"+lr] = X.

=

n -. )o

That is. (a - T)S = I. Similarly, S(a - T) = I. Therefore, a - T is an invertible operator or a E p(T).

Problem 6.1.10. Let T E C(X). Show that r(T) = limn, I1T"IIn = 0 if i

and only if limn , IIT"xll n = 0 for each x E X . Solution: If limn...

117" II

= 0 and x E X. then from .1

IITnXII

< (IITn11

.1

' Iixli)" = II7"`II n

Ilril° - 0.

it follows that IlT"xIl k -. 0. For the converse, assume that IIT"xll * - 0 holds for each x E X. Fix a complex number a 0 and let x E X. Since lIT"sll^ - 0, there exists some no E N such Ila-(n+1) T"x.ll ( 1 for all that lITnxII r < 2 for all n > no. This implies Fn > no. In particular, the series Eno a'("+1)T"x converges in X for each r EX.

By Problem 6.1.9. a E p(T) for each a 34 0. This and the fact that a(T) 0 0 imply that p(T) = C \ {0} or a(T) = {0}. So, 117' Ilk = r(T) = 0.

Problem 6.1.11. If X is a Banach space and T E £(X), then show that o,(Tk) = {Ak: A E a(T)} for each k (and conclude from this that r(Tk) = [r(T)]k). Also establish that an operator T E £(X) satisfies 117' II

0 if and only

ifr(T)<1.

Solution: In this proof, we shall employ the following simple algebraic property (P): (P) If A, B,C E G(X) satisfy CA = BC = I. then A = B = C-1. To see this, note that A = IA = (BC)A = B(CA) = BI = B. Now let T E £(X) and fix some natural number k. Assume first that A E a(T).

In this case, we claim that Ak E Q(7). Indeed, if Ak ¢ a(Tk), then there exists sonw S E £(X) such that (Ak - Tk)S = S(Ak - Tk) = I. Since T)(Ak'1 +,\k-2 T+ ... + ATk-2 + Tk-1) , Ak - Tk = (A

-

+ \7'k-2 +Tk-1)S and B = S(Ak-1+Ak-2T+ .+ATk-2+Tk-1) satisfies (A-T)A = B(,\-T) = I. it follows that the pair of operators A = (Ak-1 + Ak-2T +

Now property (P) implies A E p(T). which is a contradiction. Hence. Ak E a(Tk), and so { AL*: A E a(T) } 9 a(Tk).

6.1. The Spectrum of an Operator

193

For the converse inclusion, fix some p E a(Tk) and consider the polynomial p(z) = p - zk with factorization (_1)k k(rl

p - zk =

- z)(r2 - z)...(rk - z),

where r1. r2i ... , rk are the roots of the polynomial p counted with their multiplic-

ities. In particular, p = rk holds for each i. The above factorization yields p - Tk

=

(-1)k(r1 - T)(r2 - T)...(rk - T).

Now notice that if each operator r, - T is invertible, then it follows from the above factorization that p - Tk is also an invertible operator, which is impossible. Hence, for some 1 < i < k the operator ri - T is not invertible, i.e., ri E a(T). This implies

p = rk, and sop E {Ak: A E a(T)J. Thus, a(T) C {Ak: A E o(T)} is also true, and so

o(Tk) _ {'\k: AE o(T)}.

For the last part assume first that r(T) < 1. Fix some 5 > 0 such that r(T) < 5 < 1 and note that lim"_,,. IIT"II° = r(T) implies that there exists some no such that IIT" II < 5 or IITn II < 6" for each n > N. Since bn -. 0, we see that

117' 11 -+ 0. For the^ converse assume IITn II -. 0. Then IITk II < 1 for some k, and so 1 r(T) = info IIT"II* <- IITkIII < 1.

Problem 6.1.12. If T : X - X is an invertible bounded operator on a Banach space, then show that o(T-1) = {A-1: A E o(T) }. Solution: Start by observing that since both T and T-1 are invertible, the number zero is neither in a(T) nor in a(T-1). Now let p E a (T -1). Put A = a and assume by way of contradiction that A E p(T). This means that there exists a bounded operator S E C(X) satisfying S(A - T) _ (A - T)S = I. Then

I = S(A - T) = STT-1(A - T) =

ST(AT-1

- I) _

ST) (p - T-1)

,

TS) = (A - T)S = I. Now invoking property (P)

and similarly (p - T-')

stated at the beginning of the solution to Problem 6.1.11, we see that p V o(T-1), which is a contradiction. Hence, A = µ E a(T) and p = A-1. Therefore, we have

a(T-1) C {A'1: A E a(T)}. Now, by the symmetry of the situation, if A E a(T) = a((T-1)-1), then there exists some p E o (T -1) such that p-1 = A. Thus, A-1 = p E a (T -1) . That is, {A-1: A E a(T)} C a(T'1). This proves that o(T-1) _ {A'1: A E a(T)}. I

Problem 6.1.13. If an operator T E £(X) satisfies IITkjI < 1 for some k, then show that limn_ f1Tn99 = 0. Solution: If an operator T E C(X) satisfies IIT'II < 1 for some k, then

r(T)=infIIT"II°
0.

1

6. Spectral Properties

194

Problem 6.1.14. Let 11 be a compact Hausdorff space and let µ be a Borel

measure on Q. Assume that either E = C(fl) or E is a Banach function space that is order dense in Lo(u). If 0: f) -' C is a continuous function, then show that the spectrum of the multiplication operator M0: E -i E coincides with the range of 0, i.e., a(MO) = R, = 10(w): w E S2}. Solution: We start by observing the following properties: The inverse operator M;1 exists if and only if 0(w) i4 0 for each w E 12, in which case we have Mm 1 = Mo-'. The identity A - NO = MA_4, holds for each A E C. By the above properties, A - Mo is invertible if and only if 4)(w) 0 A for each w E Il.

Since 0 is continuous, the latter happens if and only if A ¢ Rb. This establishes the desired equality a(M4) = Rm. I

Problem 6.1.15. For a non-empty compact subset K of C establish the following properties.

(a) There exists a bounded operator T : e2 -+ e2 such that a(T) = K.

(b) If S: Cc(K) - Cc(K) is the multiplication operator defined by (Sx)(z) = zx(z), then a(S) = K. Solution: (a) Let K be a non-empty compact subset of complex numbers. Assume first that K is a finite set, say K = {Al, A2, ... , An}. Then the operator defined on Cn via the diagonal matrix A = diag(Al, A2, ... _A.) defines a bounded operator whose spectrum is precisely K. Now assume that K is an infinite set. Let (A1, A2, ...} be a countable dense subset of K, and define the operator T : e2 - e2 via the formula T(xI,x2.... ) _ (A1x1,A2x2.... ).

If en denotes the sequence whose n`h coordinate is equal to one and every other is zero, then Ten = Anen is obviously true for each n. This shows that An - T is not one-to-one, and so An - T is not invertible. Therefore, {A1, A2, ...} c a(T). Since a(T) is a compact set, it follows that K C a(T).

To see that K e a(T), let A 0 K and then choose some e > 0 such that IA - zI > e holds for all z E K. This implies 00 1A

II(A - T)xII = II ((A - Al)xl, (A - A2)x2,

- AnI2IxnI2,

> eIIxfl

n=1

for all x E e2. That is, A - T is a bounded below operator. On the other hand, if y = (yl,112.... ) E e2, then the vector z = ( , ...) also belongs to e2 and satisfies (A - T)z = y. This shows that A - T is also surjective, and hence an invertible operator. In other words, we have shown that A 0 K implies A 0 a(T). The above properties establish that a(T) = K. (b) If ip(z) = z, then according to Problem 6.1.14 the spectrum of the multiplication operator S = Mo satisfies a(S) _ {4)(z): z E K} = K. (See also Problem 7.1.11.)

1

6.1. The Spectrum of an Operator

195

Problem 6.1.16. If an invertible operator T : X -+ X on a Banach space satisfies IIT z fl < M < oo for all integers n (such an operator is called a

Cesaro operator or a double power bounded operator), then show that o(T) C {A E C: CAI = 1}. Solution: Here is a solution different than the one suggested by the hint. The given conditions imply

r(T) = lim llTnHH ° < lim Al' = 1, n-oo

n-oo

and similarly r (T -1) < 1. Now let A E or(T). Then JAI < r(T) < 1. On the other hand, by Problem 6.1.12, we know that E a(T-1). This implies 71 < r(T-1) < 1 or JAI > 1. Consequently,

IAI=1, andthus t(T)C{AEC:IAI=1}.

U

Problem 6.1.17. Let Ti: Xi - Xi (i = 1, ... , n) be bounded operators on Banach spaces. If X = X1 ® ®Xn is the direct sum Banach space, then show that:

(a) The direct sum operator T = T1® ®Tn : X T(x1®... ®xn) = Tlxl ®... ®Tnxn ,

X, defined by

is invertible if and only if each Ti is invertible-in which case the inverse operator T-1 satisfies T-1 E) (b) The spectrum of the direct sum operator T is the union of the spectra of the operators Ti, i.e., o(T1 ® . ® TO = U 1 o-(Ti). Solution: (a) If each T, is invertible, then the direct computation shows that the operator S = T 1 1 ® . . . ®T,,-1 satisfies ST = TS = I and so S = T-1. Conversely, assume that T-1 exists. Let yi E X, and consider the vector ®0 9yi 909 . ®0, where the yi occupies the ith position. To see that y = OED T, is one-to-one, assume T,y, = 0. Then Ty = 0, and so y = 0 or y, = 0, so that T, is one-to-one. To show that T, is surjective, let T-1y = x = x1 ® ® x,,. Then

get T,xi=yi. This shows that T. is indeed surjective-and so each T,: X; - X; is invertible. from

(b) A scalar A E a(T) if and only if A-T= (A-T1)ED is not invertible on X. In view of (a) this is equivalent to saying that A E U 1 o(T,).

Problem 6.1.18. Show that an operator T : X - X on a complex Banach space is nilpotent if and only if it is quasinilpotent and has finite descent. Solution: Clearly, if T is nilpotent, then T is also quasinilpotent and has finite descent. For the converse assume that T is quasinilpotent and has finite descent. Let p = b(T) < oo. To finish the proof, it suffices to show that R(T') = {0}. To this end, assume by way of contradiction that R(T') 96 {O}.

Suppose first that T has finite ascent. This implies that a(T) = p and that X = N(TP) e R(TP). In particular, R(TP) is a closed T-invariant subspace of X on which T is invertible. The latter conclusion guarantees that the spectrum

6. Spectral Properties

196

of the restriction of T to R(TP) has a non-zero point. Consequently, from the decomposition T = TI N(TV) ® TIR(TP), it follows that the spectrum of T has a nonzero point (see Problem 6.1.17), which contradicts the fact that T is quasinilpotent.

Now assume that T has infinite ascent. According to Theorem 2.14, we can renorm R(TP) so that it becomes a Banach space and, moreover, the restriction S of T to R(TP) is continuous. Since p is the descent of T, it follows that S is also surjective. The fact that T has infinite ascent implies N(T) fl R(TP) 0 {0}; see

part (d) of Problem 2.2.8. So, the surjective operator S: R(TP) - R(TP) is not invertible. But then, by Theorem 6.20, S (and hence T) has a non-zero eigenvalue. I Hence, the spectrum of T has a non-zero point, which is a contradiction.

Problem 6.1.19. If S, T E L(X), then show that a(ST) U {0} = a(TS) U 10). Conclude from this that:

(a) a(ST) = {0} if and only if a(TS) = 10}-

(b) r(ST) = r(TS). Give an example of two operators S, T E G(X) such that a(ST) # a(TS). Solution: The solution will be based upon the following algebraic property.

(*) If A, B, C E £(X) satisfy (I - AB)C = C(I - AB) = I, then (I + BCA)(I - BA) = (I - BA)(I + BCA) = I. We verify the first identity. From C(I - AB) = I, we get CAB = C - I, and so

(I + BCA)(I - BA) = I + BCA - BA - BCABA

= I+BCA-BA-B(C-I)A = I+BCA-BA-BCA+BA=I. (a) It follows from (*) that if 10 o(AB), then 10 o(BA). Now let A E o(ST) be non-zero. Assume by way of contradiction that A 0 o(TS). This means that there exists a bounded operator D: X - X such that (A - TS)D = D(A - TS) = I. Rewriting this identity, we get

(I - 1TS)(AD) = (AD)(I - 1TS) = I. Letting A = IT, B = S, and C = AD in (*), we see that

(I + SCT)(I - IST) = (I - 1ST)(I + SCT) = 1. Rewriting the latter identity, we get

['(I + SCT)](A- ST) = (A- ST) [,(I +SCT)] = I. Therefore, A V o(ST), which is a contradiction. Thus, o(ST) U {0} 9 o(TS) U {0}. By the symmetry of the situation, we have o(TS) U {0} C o(ST) U {0}. Hence, a(ST) U {0} = a(TS) U {0}.

(b) Note that

r(ST) = max

AEo(ST)

IAA =

max

AEa(ST)U{0}

IAA =

max

)LEO(TS)U{0}

IAA = max CAI = r(TS) . AEo(TS)

6.2. Special Points of the Spectrum

197

For a counterexample consider the operators S, T : el - el defined by

S(xl, x2, x3, ...) _ (x2, x3, ...) and T(xl, x2, x3, ...) _ (0, xl, x2, x3, ...) . Notice that ST =1, and so a(ST) = {1}. On the other hand, we have

TS(x,0,0,0.... ) = (0,0,0,...), and so TS is non-invertible, i.e., 0 E a(TS). So, a(ST) _ 1154 {0,1} = o(TS). (See also Problems 6.2.9 and 8.2.7.)

1

Problem 6.1.20. Let T E £(X) be an operator and let zo E C be a point such that every A E a(T) \ {zo} is an isolated point of a(T). Show that a(T) is at most countable and that if a(T) is countable, say a(T) = {Al, A2.... then An -' zo. Solution: Assume that a bounded operator T E C(X) satisfies the stated property. Clearly, a(T) must be at most countable. Next, suppose that a(T) is countable, say a(T) = {Al, A2,. ..}. Now for each e > O let C. _ { A E a(T) : I A - zoo > e }. If some CC is an infinite set, then it must

have an accumulation point, say p. It follows that p is an accumulation point of a(T) satisfying it E a(T) \ {zo}, contrary to our hypothesis. Therefore, each C, is a finite set and from this we easily infer that An - zo. I

6.2. Special Points of the Spectrum Problem 6.2.1. Show that an operator T EC(X) is not one-to-one if and only if zero is an eigenvalue of T. Solution: If 0 is an eigenvalue of T, then there exists some non-zero vector x such that Tx = Ox = 0. This shows that T is not one-to-one. Conversely, if T is not one-to-one, then there exists some non-zero vector x such that Tx = 0 = Ox. Therefore, 0 is an eigenvalue of T.

Problem 6.2.2. Let T : V

V be a linear operator on a vector space.

Show that a collection of eigenvectors of T any two of which correspond to distinct eigenvalues is a linearly independent set.

Solution: Let T : V - V be a linear operator on a vector space. The proof is by induction. For n = 1, the conclusion should be obvious. For the induction step, assume that the result is true for any collection of n eigenvectors of T corbe n +I eigenvectors responding to distinct eigenvalues. Now let vl,... , v,,, of T corresponding to the distinct eigenvalues Al, ... , An, A,,+1, respectively. To are linearly independent, we must show establish that the vectors v11. .. , vn,

that if alvl + ... + anVn + an+l Vn+1 = 0 holds, then necessarily al = a2 = ... = an = an+1 = 0.

(*)

6. Spectral Properties

198

So, assume that (*) is true. This implies aIA1 u1 + ... + anAnVn + Qn+iAn+lvn+1 = a1Tv1 + + anTvn + Qn+1Tvn+1

= T(alvl +

+ anxn + an+lvn+l)

=T(0)=0. Multiplying (*) by An}, and subtracting it from the preceding equation, we get

- An+i)vn = 0. Hence, our induction hypothesis yields a,(A, - An+1) = 0 for i = 1....,n. Since al (Al - An+1)v1 +a2(A2 -

An+1)v2+.-.+ an(A1

= an = 0. From (*) we A; - An+l # 0 for each i, we infer that al = a2 = get an+lvn+l = 0 or an+1 = 0. Therefore, ai = 0 for all i. and so the vectors v3, ... , vn, vn+1 are linearly independent.

Problem 6.2.3. Let T : X - X be a bounded operator on a Banach space. Show that if for some A E C there exists a sequence {xn} of unit vectors such that ll\xn - TxnfI - 0. then A E a(T), i.e., show that Qa(T) C a(T). Solution: If A E aa(T), then the operator A - T is not bounded below. and so it cannot be invertible. Hence, A E a(T). I

Problem 6.2.4. Show that a complex number A is an approximate eigenvalue of an operator T E C(X) if and only if there exists a sequence {xn} in X satisfying xn 74 0 and Axn - Txn ---+ 0. Solution: Assume that a sequence {x } in X satisfies r" f' 0 and Axn -Tx,, - 0. Since xn 7 0, there exist some e > 0 and a subsequence {yn} of {xn} such that Ilii

II

from

>_ a for all n. Now let zn = ; t g and note that II zn II = 1 for all n. Moreover, IIAzn - TznII =

it follows that Azn - Tzn

IIAyn - TY-II < IIAyn - TynII Ilynll

0. This shows that A E aa(T).

0.

I

Problem 6.2.5. Let T : X - X be a bounded below operator on a Banach space. If y > 0 satisfies IITxII ? yIlxtl for each x E X, then show that

Q,(T)n{AEC: IAI
X is an isometry, then aa(T) C {A E C: IAI = 1 }.

Solution: Assume that IITxII > yIIxll holds for each x E X and some y > 0. If JAI < y and x E X satisfies IIxII = 1, then IITz - AxII ? IITxll - IAI - IIxII ? y - JAI > 0. This shows that A cannot belong to the approximate point spectrum of the operator

T. That is,oQ(T)nJA EC: IAI
for each n. we have r(T) = 1, and thus aa(T) must lie on the unit circle. (Since T need not be invertible, this conclusion does not follow from Problem 6.1.16.)

6.2. Special Points of the Spectrum

199

Problem 6.2.6. Let T : X --+ X be a non-invertible bounded operator on a complex Banach space. If some y > 0 satisfies IITxII > yIIxII for each x E X, then show that D(0, y) = {A E C: IAI < y} C ar(T). Solution: First of all note that since T is not invertible, 0 E o(T). Moreover, since T is bounded below, 0 E ar(T). Let 0 = sup{a > 0: D(O,a) C ar(T)}. Clearly, D(0.03) C Cr(T) C a(T), and so D(0, Q) C a(T). To finish the proof, it suffices to show that 3 > y. To establish this claim, assume by way of contradiction that Q < y. From the definition of Q it follows that there exists a sequence of complex numbers {An}

satisfying 0 < IA,, < 0 + and An V ar(T) for each n. By passing to a subsequence, we can assume that An -+ A for some A E C. Clearly, CAI = 0, and from D(0, $) C a(T), we get A E a(T). Now notice that II(An - T)xII 2 IITxII - IIAnxII > yIIxII - IAnI - lixil >- ZaIIxI)

(*)

holds for each x E X. Since each operator An - T has dense range, it follows from Lemma 2.8 that each An - T is an invertible operator. But then, by Theorem 2.5, the operator A - T is itself invertible, a contradiction. This contradiction completes the solution. I

Problem 6.2.7. If P E £(X) is a non-trivial projection (that is, if P2 = P and P 36 0, 1), then show that o(P) = ap(P) = (0, 11. Solution: Assume that P E G(X) is a non-trivial bounded projection. Let Y = P(X) and Z = (I - P)(X). Then Y and Z are non-trivial closed vector subspaces of X and X Y e Z holds. If y E X and z E Z are both non-zero, then

Py=y

and

Pz=O=Oz.

Therefore, 0 and 1 are eigenvalues of P. Now fix A E C such that A 0 0 and A 0 1. Notice that if x = y+z E X = Y®Z, then

(A-P)x=Ax-Px=(A-1)y+Az. So, if (A-P)x=0. then (1-A)y=OandAz=0. SinceA96O,l,wegety=z=0 or x = 0. Thus, A - P is one-to-one. + z satisfies (A - P)u = x. Now if x = y + z E X, then the vector u = Hence, A - P is a surjective one-to-one operator, and is thus invertible. Therefore, p(P) = {A E C: A # 0,1} or ap(T) = a(T) = {0,1}. 1

Problem 6.2.8. Let S E £(X) and T E C(Y) be two bounded operators, where X and Y are two complex Banach spaces. The operator S and T are said to be similar whenever there exists an invertible operator R in £(X, Y) satisfying T = RSR-1. In other words, S and T are similar if there exists an isomorphism (i.e., a surjective, one-to-one, and continuous operator) R : X -+ Y such that the diagram shown in Figure 1 commutes.

6. Spectral Properties

200

If S and T are similar operators, then show that a(S) = a(T),

ap(S) = ap(T),

aa(S) = oa(T),

and ar(S) = a,.(T).

X

IT

S1

X RY F[gure 1

Solution: By the symmetry of the situation, it suffices to prove that

a(S) S a(T), ap(S) c op(T), and aa(S) c aa(T) hold. We shall do this in each case separately.

(a) ap(S) C ap(T).

Let A E ap(S). Pick some non-zero vector x E X such that Sx = Ax. Since R is an isomorphism, we have Rx 54 0, and from the commuting diagram we get T(Rx) = RSx = A(Rx). This shows that A E ap(T), and so ap(S) C ap(T).

(b) va(S) c aa(T). Let A E aa(S). Pick a sequence of unit vectors so that Ax, - Sx,, 0. Since the operator R (as being an isomorphism) is bounded below, there exists some 6 > 0 such that I!Rx II > 6 for each n. Now note that

(A - T )Rx = ARx - T

ARx - RSx = R(ar -

0.

This implies (see Problem 6.2.4) that A E a, (T), and so ca(S) C o,,, (T). (c) a(S) C i7(T)-

Let A E a(S). Assume A - S is a non-surjective one-to-one hounded operator. By part (a), we know that A - T is one-to-one. We claim that A - T is not surjective. To see this, assume by way of contradiction that A - T is surjective. and let x E X

be an arbitrary vector. Pick some yo E Y such that (A - T)yo = Rx. Since R is surjective. there exists some xo E X such that yo = Rxo. Therefore,

Rx=(A-T)yo=(A-T)Rxo=ARxo-TRxo=ARxo-RSxo=R(A-S)xo. Since R is an isomorphism, it follows that (A - S)xo = x. This shows that A - S is surjective, which is a contradiction. Hence, A - T is not surjective, and so A E a(T),

i.e., a(S) C a(T). (d) ar(S) C a,.(T). Let A E ar(S). This means that A - S is one-to-one and (A -- S)X 96 X. To see that A-T is one-to-one, assume that (A-T)y = 0 for some y E Y. Pick some x E X

such that y = .T.. and note that (A - S)x = [R-1(A - T)R[x = R-1(A - T)y = 0. Since A - S is one-to-one. it follows that x = 0. and so y = Rx = 0.

6.3. The Resolvent of a Positive Operator

201

On the other hand, if (A - T)Y = Y, then

(A - S)X = [R-'(A - T)]R(X) = R-' (A - T)Y

= R-1((A-T)Y)=R-1(Y)=X, which is a contradiction. Hence, (A - T)Y 96 Y, and so A E a,-(T).

Problem 6.2.9. For two operators S, T E C(X) establish the following. (a) A non-zero complex number is an eigenvalue of ST if and only if it is an eigenvalue of TS. (b) If X is finite dimensional, then show that or(ST) = o(TS); compare this with Problem 6.1.19.

Solution: (a) Assume that for some A 0 and some x 96 0 we have STx = Ax. This guarantees Tx 0 0. Now note that TS(Tx) = T(STx) = T(Ax) = ATx. Therefore, A is an eigenvalue of TS. By symmetry, the converse is also true.

(b) Since X is a finite dimensional complex Banach space, we can identify X with some On. Also, we can view S and T as n x n (complex) matrices. In this case, it should be noted that the spectrum of any operator consists only of eigenvalues.

From part (a), we know that the matrices ST and TS have the same nonzero eigenvalues. Now notice that 0 is an eigenvalue of the matrix ST if and only if det(ST) = det(S)det(T) = det(TS) = 0. Since det(ST) = 0 if and only if det(TS) = 0, we see that 0 E o(ST) if and only if 0 E o(TS). Therefore, a(ST) = a(TS) must be true in this case. (However, as we saw in Problem 6.1.19, the two sets o(ST) and a(TS) need not coincide in general.) I

6.3. The Resolvent of a Positive Operator Problem 6.3.1. Assume that En00=I JznI exists in a Banach lattice E, where zn = xn + tyn E Ec. for each n. Show that E n00=1 z exists in E. and E ° °_ 1 zn 1

<E

1

I zn I, where the series are assumed to converge in norm.

Solution: If k and 0 are fixed, then k

1 E xn n=1

k \1 Cos 0 + ( Ek ynl1 Cos B = F,(x,l COs 0 + yn sin 9) 1

/

n=1

/

n=1 k

00

E IZnI <_ F, IZnIn=1

Hence,

n=1

(E,,=IX,,) cos9+ (En 1 yn) sin0 < i°° 1 Iznl holds for all 9. This implies 00 00 00 \ IEZnl =sup [ ( xn 1 Cos 0 + (> Y-) sin 0 ] <

n=1

OER

n=1

/

n=1

/

00 I Zn I ,

n=1

6. Spectral Properties

202

as desired.

Problem 6.3.2. If S, T : E -+ E are positive operators, then show that r(S + T) > max{r(S), r(T)}. Solution: Let S, T : E - E be two positive operators. Then 0 < S < S + T holds, and so 0!5 S" < (S + T)" for each n. This implies II S" II 5 11(s + T)111 for each n, and so

r(S)= lim"IlS"II" < lim"II(S+T)"II°

=r(S+T).

Similarly, r(T) < r(S + T), and so max{r(S),r(T)) < r(S + T).

Problem 6.3.3. Let T : E -' E be a positive operator on a Banach lattice. If T is a-order continuous (resp. order continuous), then show that R(A, T) is also a-order continuous (resp. order continuous) for each A > r(T). Solution: We prove the a-order continuous case only. So, assume that T: E - E is a a-order continuous positive operator on a Banach lattice. If x" 10 in E, then Tx" j 0, and so 7-2x" = T(Tx") j 0. This shows that T2 is a a-order continuous operator. By induction, Tk is a a-order continuous operator for each k = 0, 1, ... . So. if A > r(T) is fixed, then each positive operator Sk = En=1 A ("+i)T'k is a-order continuous. Moreover, Sk I R(A, T) and IISk - R(A, T) II - 0 in C(E). Now let y" 10 in E, and assume that 0 < y < R(A, T) y" 1. For each k and n we have

0 < y:5 R(A. T)y" = [R(A. T) - Sk)y" + Skyk : [R(A, T) - Sk]yl + Skx" . Since Skx" In 0 for each k. the above inequality implies 0 < y < [R(A. T) - Sk] xl for each k. Hence, 0 < IIyII 5 II [R(A,T) - Sk] xlII -. 0, and so IIyII = 0 or y = 0. This shows that R(A, T)y" j 0, proving that R(A, T) is a a-order continuous positive 1 operator.

Problem 6.3.4. Let T : E -i E be a positive operator on a Banach lattice. If a scalar a > 0 satisfies ax < Tx for some x > 0 or of < T'x` for some 0 < x` E E*, then show that a < r(T)Solution: Assume that ax < Tx holds for some a > 0 and x > 0. Since T is a positive operator. an easy inductive argument shows that a"x < T'"x for each n. This implies a" 5 II T" II or a < I I T" I I " for each n. Consequently, we have a < lim"_.,o IIT" II n = r(T). A similar argument applies when ax' < T'x'.

i

Problem 6.3.5. Let T : X X be a bounded operator on a complex Banach space and let p,,,,(T) denote the unbounded (open) component in C of the resolvent p(T). Establish the following.

(a) If a(T) is at most countable (in particular, if every non-zero point in a(T) is an isolated point), then p,,,(T) = p(T). (b) A closed subspace Y of the Banach space X is T-invariant if and only if Y is R(A,T)-invariant for each A E p,,, (T).

6.3. The Resolvent of a Positive Operator

203

(c) If a closed subspace Y of X is T-invariant and Tly is the restriction

of T to Y, then px.(T) C p(Tly) and R(a,Tly) = R(A,T)ly for each A E p,,.(T).

(d) If poo(T) = p(T), then for every T-invariant closed subspace Y of X we have a(Tly) C a(T). (e) Assume that an open annulus AR1,R2(Ao) lies entirely in px(T) and that R(,\, T) = F,n__,. (a - \o)-A,, is the Laurent series expansion of the resolvent in AR1,R2(Ao) If Y is a T-invariant closed subspace of X, then Y is also An-invariant for each n E Z. Solution: (a) Assume that a(T) is at most countable. We shall show that p(T) is arcwise connected. To see this, let A, u E p(T) with A # p. We claim that there exists a straight line passing through A that lies entirely in p(T). Otherwise, if every straight line L1 passing through A intersects a(T), then a(T) would be uncountable, which is impossible. Similarly, there exists a straight line L2 passing through it that is not parallel to L1 and lies entirely in p(T). If z is the intersection of L1 and L2, then the continuous path consisting of the line segments from A to z and from z to it lies entirely in p(T) and joins A with p . This shows that p(T) is arcwise connected-and hence p(T) is connected. In particular, p,,.(T) = p(T).

(b) Let Y be a closed subspace of X. Assume first that Y is T-invariant. Fix C defined any y E Y and y' E Y1, and consider the analytic function f: p,,(T) by

.f (A) = y' (R(A. T) y) .

From the Neumann series, it is easy to see that for each A E C with IAA > r(T) the vector subspace Y is also R(A, T)-invariant. So, R(A, T)y E Y for all JAI > r(T).

Since y' E Y', we obtain f (A) = y' (R(A, T)y) = 0 for each JAI > r(T). The analyticity of f and the connectedness of pa (T) imply y'(R(A,T)y) = 0 for each A E p,,,,(T), all y` E Yl and all y E Y. Now an easy application of the HahnBanach theorem shows that R(A,T)y E Y for all A E p,, (T) and all y E Y. That is. Y is R(A. T)-invariant for each A E p .(T). For the converse, assume that Y is R(A. T)-invariant for every A E p,o(T). According to Problem 6.1.3 we have lima AR(A, T) = I. Therefore,

lim T(AR(A. T)) = T.

a-.x

(*)

Now note that if A E p(T), then

T(AR(A, T)) = ATR(A, T) = -A(A - T - A)R(A,T) = A2R(A,T) - A. So, for each y E Y we have [T(AR(A, T))] y = A2R(A, T)y - Ay E Y. Now a glance at (*) shows that Ty E Y for each y E Y. So, Y is T-invariant.

(c) Assume that Y is a closed T-invariant subspace of X, and let A E p,.(T). By

part (b), Y is also R(A,T)-invariant. From (A-T)R(A.T) = R(A,T)(A-T) = IX, it follows that (A - T)R(A, T)y = R(A, T)(A - T)y = y for each y E Y. This shows that R(A, T)Jy is the inverse of the operator A-TIy E C(Y). Therefore, A E p(Tly) and R(A,Tly) = R(A.T)Iy.

6. Spectral Properties

204

(d) From part (c) we have p(T) = pp(T) C p(TIy). Taking complements yields

a(TIY) s a(T). (e) Assume that AR,,R,(Ao) = {A E C: R1 < IA - Aol < R2} lies entirely in pp(T). Fix some R such that R1 < R < R2 and let CR denote the positively oriented circle centered at AO with radius R. According to Laurent's Theorem 1.78 for each n E Z we have

An -- 2vt f 1

CR

R A,T A-Ao F: IT dA.

Now fix y E Y. Part (b) guarantees that R(A,T)y E Y for each A E p,, (T), and so from Problem 1.5.3 we get Any

_L - 2vt

R(\,T)gl

CR (A-Ao

dAEY

for each n E Z. So, Y is An-invariant for each n E Z, and the solution is finished.

I

Problem 6.3.6. Give an example of a positive operator T : E -+ E on a Banach lattice and some T-invariant closed ideal J in E such that: (a) There is some A E p(T) for which J is not R(A, T)-invariant. (b) The spectrum a(TIj) is not included in a(T), where TJj denotes the restriction of T to J. Solution: Let E = t2(Z), the Banach lattice of all square summable sequences defined on the integers Z. We let T : E - E be the right-shift operator. That is,

T(...,x-2,x-1,

x1,x2,...) _ (...,x-3,x-2, ,x0,x1,x2,...)

1

0

0

for each (...,x_2, x_ 1, X0, x1, x2, ...) E 12 (Z). It is easy to see that T is a surjective lattice isometry; and so r(T) = IITII = 1. The inverse of T is the left-shift S: E - E defined by

(...,x-1,x0,`1,x2,x3,...).

S(. 0

0

We claim that a(T) = F = {A E C: JAI = 1}. To see this, consider first some A E a(T). Then on one hand JAI S IITII = 1, and on the other hand (from Problem 6.1.12) we have lz1 :5 IITII=1 orJAI>1. Thus, JAI= 1, and so a(T) C_ r. For the reverse inclusion, let A E r. We claim that the vector y E t2(Z) defined by yn = 0 if n 34 0 and yo = 1 does not belong to the range of the operator A - T. To establish this, assume by way of contradiction that there exists some vector x = (... , x-2, x-1, xo, x1, x2, ...) E t2(Z) such that (A - T)x = y. That is, Axn - xn_ 1 = yn for each n E Z. It follows that xn = a xo for each n > 1, and from

Ixol = Ixnl -, 0, we see that xo = 0. This implies x_1 = -1 and x_,, = -An for each n > 1. But then we have 1= Ix-n I - 0, which is impossible. This establishes that y f R(A -T), and so A E a(T). Thus, r g a(T) is also true, and so a(T) = F. Since T is invertible, 0 0 a(T). Now let

x_i=0 for all i>1}.

6.4. Functional Calculus

205

Clearly, J is T-invariant and the restriction of T to J is the familiar forward shift.

From Example 6.21 we know that o(Tij) = {z E C: IzI < 1}. This implies o(TIj) ¢ o(T). Finally, notice that 0 E p(T) and R(O,T) = -S. Consequently, the closed ideal J of £2(Z) is T-invariant but not R(O, T)-invariant. (Compare this problem with Problem 6.3.5.)

6.4. Functional Calculus Problem 6.4.1. Show directly (i.e.. by avoiding the Spectral Mapping The-

orem) that if p(A) = Fk-eakAk is a polynomial. then for every bounded operator T : X -+ X on a Banach space we have a(p(T)) = p(a(T)) = {p(A): A E o(T) 1. Solution: Let p(A) = Ek=o akAk be a polynomial-with real or complex coefficients. For each p E C we have the factorization

p -p(T) = (-1)nl'an(Al -T)(A2 -T)...(A,, -T),

(*)

where A1, A2, ... , An are the roots (counted with their multiplicities) of the polynomial q(A) = p - p(A). Assume that p E o(p(T)), i.e., the operator p - p(T) is non-invertible. In

view of (*), there exists some root A, of the polynomial q(A), i.e., p = p(A,), such that the operator A, - T is non-invertible, that is, A. E o(T). This implies p E {p(A): A E a(T) }, and so a(p(T)) C {p(A): A E o(T) }. Now let a E a(T), and put p = p(a). In this case, we claim that the operator p - p(T) cannot be invertible. Indeed, if p - p(T) is invertible. then it follows from (*) that each A, - T is invertible. Since a = A, for some i, we get a I a(T), which is a contradiction. Therefore, we have p = p(a) E o(p(T)), and consequently {p(A): A E o(T) } C a(p(T)). Thus, o(p(T)) = {p(A): A E o(T) } = p(a(T)).

Problem 6.4.2. Assume that f is an analytic function defined on a neighborhood V of oc, that is, the function f is defined and is analytic on an open set of the form V = {A E C: IAA > R} for some R > 0. Suppose also that f vanishes at infinity. If C is a Jordan contour in V surrounding the spectrum of an operator T E £(X), then show that fc f (A)R(A, T) dA = 0. Solution: Assume that f satisfies the stated properties. Then, according to Cauchy's Integral Theorem 1.70, there exists some ro > IITll such that A

=jf(A)R(AT)dA=

jf(A)R(A1T)dA

holds for all r > ro, where C,. is the positively oriented circle centered at zero with radius r.

6. Spectral Properties

206

Now let e > 0. Pick some rl > ro such that If (A)I < e for all A E C with Al I> rl. So, according to Theorem 6.3, for each r > r1 we have

IIAII = Il f f(A)R(A,T) dAll < e tar r-1T = 2ae7_7" C.

Letting r - oo, we get IIAII < 2ae for each e > 0. That is, IIAII = 0 or A = 0, as claimed.

Problem 6.4.3. Let T : X - X be a bounded operator on a Banach space. Fix any f E .F(T) and let C be a Jordan contour surrounding a(T) in the domain of analyticity of f. Show that

f(T)x= a: f f (A)R(A, T)x dA

holds for each x E X. In particular, if Tx = ax for some x E X and a scalar a E C, then show that f (T)x = f (a)x. Solution: Fix some f E .F(T) and let C be a Jordan contour in the domain of analyticity of f surrounding a(T). The formula

f(T)x = -L j f(A)R(A,T)xdA follows immediately from Problem 1.5.3.

Now assume that Tx = ax. From the identity (A - T)x = (A - a)x, it follows that R(A,T)x = (A - T)-lx = 1x for each A E C. So, from the preceding case and Theorem 1.73 we get

jf(A)R(AT)xdA=-

f(T)x E-i

[J x .

JcfxdA

dA]x = f(a)x,

as desired.

Problem 6.4.4. Let T E £(X), let f E.F(T), and let g E.F(f (T)). Establish that the composition go f belongs to.F(T) and that (gof)(T) = g(f (T)). Solution: If f is a constant function, then the conclusion is trivial. So, we can assume that f is non-constant. In this case, f is an open mapping. Now let V and U be open neighborhoods of a(T) and a(f(T)) on which f and g are analytic, respectively. According to the Spectral Mapping Theorem 6.31,

we have a(f (T)) = f (o(T)), and thus f (V) is an open neighborhood of a(f (T)). Replacing U by U1 = U n f (V) and V by f -1(Ul ), we can suppose without loss of generality that f (V) C U. Therefore, g o f E .F(T). Pick a Jordan contour C2 in U surrounding or (f (T)), and let W be the open set satisfying a(f (T)) C W C W C U whose boundary is C2. Clearly, a(T) C f-1(W). Replacing V by f -1(W ), we can also assume that f (V) n C2 = 0. Finally, fix any Jordan contour C1 in V surrounding o(T).

207

6.4. Flunctional Calculus

f_M is

Fix it E C2, and note that the function h: V --# C defined by h(A) = analytic, i.e., h E F(T). From h(A)[µ - f(A)] = 1 for each A E V, it follows that

h(T)[µ - f (T)] = [p - f (T)]h(T) = I. Therefore, h(T) = [p - f (T)]-1, and so

h(T) = * j v-I a' dA = [p - f(T)] ,

= R(µ, f(T))

for each µ E C2. This implies

zn f 9(µ)R(µ, f (T')) dµ

g(f(T))

2

r,

_ a"

c fC1

i

=J J 1

,

W,

p- a r dAdµ µ-f Ra,T

2

{t-f(A) dµdA

c d R A,T dA

2vt

J lg(f(A))R(A,T)dA (gof)(T), as claimed.

Problem 6.4.5. Let T : X X be a bounded operator on a Banach space, and let V be an open neighborhood of a(T). Assume that a sequence If"} of analytic functions defined on V converges uniformly to some function f : V C. Show that f E.F(T) and that II fn(T) - f(T)II --+ 0. Solution: Since the sequence {fn} of analytic functions converges uniformly to f on V, it follows that f is analytic on V and so f E .F(T). This can be easily seen from Morera's theorem; see Problem 1.5.2. Indeed, if C is any Jordan contour that lies in V, then from 0

jfn)d'__.jf(')dA, Morera's

we see th at f f (A) dA = 0, and so by theorem f is analytic on V. Now fix e > 0 and consider any Jordan contour C in V surrounding a(T). Put M = maxaEC IIR(A. T) II, and pick some no such that I fn(A) - f (A)I < 2i ` for all

n > no, where t(C) denotes the length of the contour C. Now note that for each n > no we have

Ilfn(T) - f(T)II = zn Il f [fn(a) - f(A)JR(A,T)dAll < _L. 21r This shows that

XTf

. Mt(C) = E.

II fn(T) - f(T)II = 0.

Problem 6.4.6. Let T : X -+ X be a bounded operator on a Banach space, and let a be a spectral set of T. Also, let (Y, Z) be the pair of reducing subspaces as determined in Theorem 6.34. If h E .F(T). then show that

h(T)Iy = h(TIy).

6. Spectral Properties

208

Solution: If o = 0 or a = a(T), the conclusion is trivial. So, assume that a is a non-trivial spectral set. Let us recall the "background" for this problem. Fix a non-empty spectral set a such that a' = a(T) \ a is also non-empty. If we let 26 = d(a, a') > 0, then the two non-empty open sets V. = UAEO B(A, 6) and V0' = UAE0' B(A, 6) satisfy

aCV a'CV,',

V,nV,'=(

and

.

That is, if a is a spectral set, then there exist disjoint open neighborhoods V. and V, of a and a', respectively.

Next, consider the functions f f,,: V, U V,, - C defined by

o(A)

1

0

if A E V, if A E V,-

and

f,, (A) =

0 1

if A E V, if A E V,, .

Clearly, f f,' E F(T), and (fo)2 = f"' (f ,)2 = f ,

fofo' = 0, and f, + f,' = 1.

So, by Theorem 6.29, we obtain

[fo(T)]2 = fo(T), [fo'(T)]2 = fo'(T), fo(T)fo'(T) = 0, and f., (T) + f,'(T)

= I.

It follows that f,(T) and f,, (T) are both projections on X satisfying

X = R(f,(T)) ® R(f,,(T)) Moreover, the Spectral Mapping Theorem implies

a(f,(T)) = f,(a(T)) = {0,1}

and

o(f,,(T)) =

{0,1}.

This shows that f,(T) and f,, (T) are both proper bounded projections. If we let Y = R(f,(T)) and Z = R(f,-(T)), then (Y, Z) is a reducing pair for T. Now fix some h E .F(T) . We must show that h(T) I y = h(T I y ). To this end, pick a Jordan contour C1 in V, surrounding a and a Jordan contour C2 in V,, surrounding a'. Let C be the union of C1 and C2. So, C is a Jordan contour surrounding a (T) . Now for each y E Y we have

h(T)(y)

_

[h(T)fo(T)]y

[zx jh(A)fc(A)R(AT)d)t]Y

f

h(A)R(A,T)dA]y

=

f

_

[zx.I h(A)R(A,TIy)dA]y

= h(TIy)yThis shows that h(T)Iy = h(TIy) holds.

I

209

6.4. Functional Calculus

Problem 6.4.7. Let X be a Banach space. Show that there do not exist operators S, T E £(X) such that ST - TS = I, the identity operator on X. 1 Solution: Fix S, T E L(X) and assume by way of contradiction that ST-TS = I. To see this, start by observing that from Problem 6.1.19 we know that (*) a(ST) \ {0} = a(TS) \ {0} .

The identity ST - TS = I implies ST = I + TS, and so using Problem 6.4.1, we see that a(ST) = 1 + a(TS). (**) From (*) and (**), it follows that a(ST) {0} and a(TS) # {0}. Now let A be a non-zero point in the spectrum of ST. We claim that A is a negative integer. To see this, notice that (*) implies A E a(TS) and so (**) yields 1 + A E a(ST). Similarly, if 1 + A j4 0, then 2 +A E a(ST). Since, a(ST) is a compact set, there must exist some k E N such that k + A = 0 E a(ST). This implies that A is a negative integer. Let AO = min [Z n a(ST)]. Consequently, by (**), the negative integer AO - 1 satisfies AO - 1 E a(TS), and so AO - 1 E a(ST), which is a contradiction. This establishes that the identity operator cannot be of

the form ST-TS for anyS,TEL(X). Problem 6.4.8. Let T : X -, X be a bounded operator and let a be a spectral set of T. Also, let (Y Z,) be the unique T-reducing pair of subspaces such

that X = Y,®Z a(TIy,) = a, anda(T1Z,) = a(T)\a; see Theorem 6.34. If we let T, = Tly%, then show that ap(T,) = a n ap(T)

,

ar(T,) = a n ar(T) , and ac(T,) = a n ac (T) .

Solution: We shall verify each identity separately. I: ap(T,) = or n ap(T).

Let A E ap(T,). Then there exists some non-zero vector y E Y, satisfying T,y = Ty = Ay. This implies A E ap(T), and so op(T,) g a n op(T). For the reverse inclusion, let A E a n ap(T). Pick a non-zero vector x E X

satisfying Tx = Ax, and write x = y + z with y E Y, and z E Z. Clearly, Ty + Tz = Ay + Az or Ty - Ay = Az - Tz E Y, n Z, = {0}. This implies Ty = Ay and Tz = Az. If y = 0, thenz 00, and so A E o(TI2,) =a(T)\a, which contradicts A E a. Hence, y 0 0 and Ty = Ty = Ay implies A E op(T,). Therefore, or n ap(T) C- ap(T,), and consequently ap(T,) = a n ap(T). II:

ar(T,) =an ar(T).

Let A E a, (T,), and so (A-T,)(Y,) 74 Y,. We claim that (A - T)(X) 94 X. To see this, assume by way of contradiction that (A - T)(X) = X holds true and fix y E Y. Pick a sequence {xn } of X such that (A - T)xn -- y. Then we have {(A-T)P,(T)xn} C (A-T,)(Y,) and (A-T)P,(T)xn P,(T)y = y. This implies 1 If S, T E C(X), then the operator IS, T] = ST - TS is called the commutator of S and T. In this terminology, the conclusion of this problem can be stated as follows: The identity operator on any Banach space cannot be a commutator.

6. Spectral Properties

210

y E (A - T,)(Y,), and consequently (A - T,)(Y,) = Y,, which is impossible. This establishes that (A - T)(X) 34 X, and so A E ar(T). Therefore, ar(T,) C afla,.(T). Now assume that A E a fl a,(T). This implies (A - T)(X) 96 X and that

the operator A - TZe : Z, - Z, is invertible. Now if (A - T,)(Y,) = Y then the latter conclusion implies that (A - T)(X) = X, which is impossible. Hence, (A - T,)(Y,) # Y. must be true, in which case we infer that ,\ E a,(T,). Therefore, an ar(T) C a,. (T,) is also valid, and thus or, (T,) = an a, (T). III: ac(T,) = a fl a,(T).

This can be proven in a similar manner as II above.

Problem 6.4.9. Let ).o be an isolated point in the spectrum of an operator T E L(X). Show that the resolvent of T has an essential singularity at AO if and only if either the ascent or the descent of A0 - T is infinite. Solution: By Theorem 6.39 we know that Ao is a pole if and only if both a(Ao - T) and 6(A0 - T) are finite. This implies the conclusion. I

Problem 6.4.10. If T E C(X) and Ao E Q(T) is a pole of order k of the resolvent function

T) around A0i then show that:

(1) Ao is an eigenvalue of T. (2) Yao = N((Ao - T)k) and Z.\O = R((Ao - T)k). Solution: (1) Since Ao is a pole of order k of the resolvent of T around A0, it follows from Theorem 6.39 that (T - 1\0)k-'A_1 0 0 and (T - Ao)kA_l = 0. A0)k-'A_1x 96 0, then the non-zero vector So, if x is any vector in X with (T y = (T - Ao)k-1A_1x satisfies (T - Ao)y = (T - Ao)kA_ix = 0. This implies Ty = Aoy, and so A0 is an eigenvalue of the operator T.

(2) According to Theorem 6.39 we have a(Ao - T) = 6(Ao - T) = k. If we let Y = N((Ao - T)k) and Z = R((Ao - T)k), then by Theorem 2.23 both Y and Z are T-invariant closed subspaces satisfying X = Y ® Z, and the operator Ao - T is quasinilpotent on Y and invertible on Z. The quasinilpotence of A0 - T on Y implies that a(Ao - TIy) = {0}. Invoking the Spectral Mapping Theorem, we get o(TJy) = {Ao}. Now from the identity

T = TIy ® Tjz, we obtain a(T) = a(TIy) U a(TIZ); see Problem 6.1.17. Since A0 0 o(TIZ) and o(TIy) = {Ao}, we see that a(TIZ) = a(T) \ {A0}. Now a glance at Theorem 6.34 guarantees that Y = Ya, and Z = Zao.

Problem 6.4.11. Let T E £(X) and let A0 be an isolated point in the spectrum a(T) such that every neighborhood of A0 contains points that are not eigenvalues of T. Show that A0 is a pole of the resolvent R(-, T) if and only if Ao - T has finite descent. Solution: If A0 is a pole of the resolvent T), then it follows from Theorem 6.39 that b(Ao - T) < oo. For the converse, let p = 6(Ao - T) < oo and assume by way of contradiction that a(Ao - T) = oo. In particular, we have N((A0 - T)P+l) 34 {0). This implies

6.4. Functional Calculus

211

that the surjective operator Ao - T: R((Ao-T)P) - R((Ao-T)P) is not invertible. Now a glance at Theorem 6.20 implies that there exists some e > 0 such that every A E D(A0, e) is an eigenvalue of T, which is a contradiction. Hence, a(Ao - T) < oo. T). I Now invoke Theorem 6.39 to conclude that Ao is a pole of

Problem 6.4.12. If a quasinilpotent operator T : X -. X on a Banach space satisfies p = b(T) < oo, then show that Tp = 0.

Solution: Assume that T E C(X) is quasinilpotent, i.e., o(T) = {0}. From T). But then, from Problem 6.4.11, we know that 0 is a pole of the resolvent Theorem 6.39, we obtain that 0 is a pole of order p. If we consider the spectral set a = {0}, then it follows from Problem 6.4.10 that Yo = N((-T)P) = N(TP) = X and Za = {0}. Consequently, TP = 0.

1

Problem 6.4.13. For a spectral set a of an operator T E C(X) establish the following.

(a) If 0 0 a, then there exists an operator S E C(X) such that

P,(T) = TS = ST. (b) If 0 E a, then there exists an operator S E C(X) such that

P,(T) = I + TS = I + ST. Solution: According to Theorem 6.34 there is a unique T-reducing pair (Y Z,) such that o (T I y,) = a, a (T (Z,) = a (T) \a, and P, (T) is the projection of X onto Y, along Z,.

(a) Assume 0 V a. This means that the operator TIy, : Y. - Y, is invertible. Let R : Y,

Y, be the inverse of the operator T I y, , and consider the operator

S = RP,(T) E L(X). Then for each x=y,+z,EY,ED Z,=X we have TSx = TRP,(T)(y, + z,) = TRy, = y, = P,(T)x, and

STx = RP,(T)T(y,+z,)=RP,(T)(Ty,+Tz,)=RTy,=y,=P0(T)x. This shows that P,(T) = ST = TS. (b) If 0 E o(T). then 0 ¢ a' = o(T) \ a. So, according to part (a), there exists an operator R E C(X) such that P,, (T) = I - P, (T) = RT = TR. If we let S = -R, then we have P,(T) = I + TS = I + ST.

Problem 6.4.14. Let a be a spectral set of an operator T E C(X). If there exists some r > 0 such that

aC {AEC: IAI r}, then show that the range of the spectral projection P,(T) is given by

R(P,(T)) _ {x E X : lmn' - 0) . Solution: Let a be a spectral set of an operator T E C(X) and let r > 0 be such that a C {A E C: IAA < r} and a(T) \ o C {A E C: JAJ > r}. Also let (Y Z,) be the T-reducing pair, where Y, is the range of the projection P,(T).

6. Spectral Properties

212

0 in From o(Tly,) = or, it follows that r(TIY,) < r. Consequently, (Tlr )" n (T rn)"x L(Y,); see Problem 6.1.11. In particular, - 0 for each x E Y, and so

Y,C{xEX: fnx-.0}. For the reverse inclusion, assume that some x E X satisfies T ;,r - 0 in X. If we choose an no such that II Tnx (I < 1 for all n _> no, then IITnxII < r for each n > no. This implies limsupn... " IITnxII S r, and from Theorem 1.77 we see that the series E°°_0 converges for each A E C with IAI > r. Next, choose some R > r such that o(T) \ o C {A E C: IAI > R} and let A = {A E C: r < IAI < R}. Clearly, A fl o(T) = 0 and for each A E A we have m

m

0o

n. (A-T) T"xTT _ (A - T) lim T lim (A-T)E T"x ___4T m~ n=0 = m~ n=0 n=0 Tm+lx m-noo(x-

Air') =x.

Since A C p(T), it follows that A - T is one-to-one for each A E A, and so for each A E A we have R(A, T)x = E,°_o Tn, . Using the formula for the residue of this series, we get x = T°x = sue:

R(A,T)xdA = [Z*s j R(A,T) dA] x = P,(T)x E Yo .

fcR Consequently, {x E X : Trx claimed.

R

- 0) C Y and so {x E X:

x

- 0) = Y, as I

Problem 6.4.15 (R. Melton). For an entire function f : C - C having the series expansion f (A) = F,' O anA" the following statements are equivalent.

(a) Each coefficient an of the expansion is non-negative.

(b) For each positive operator T on any Banach lattice the operator f (T) = EO_1 anTn is positive. (c) There exists an infinite dimensional Banach lattice E such that f (K) is positive for each positive compact operator K on E. Solution: It should be clear that (a)

(c). What needs verification is (b) the implication (c) (a). So, assume that (c) is true. The solution will be based upon the following three properties. (1) If every pairwise disjoint subset of an Archimedean Riesz space E is finite, then E is finite dimensional.

(2) Let {xn} be a disjoint sequence in a Banach lattice such that the series n=1 xn is norm convergent. Then x,,, > 0 holds for each n if and only if En==1xn>O. (3) If x is a non-zero positive vector in a Banach lattice E, then there exists a positive linear functional x' E E* satisfying x'(x) = 1 and x* (y) = 0 for ally .L x.

6.4. Functional Calculus

213

We shall provide proofs to these claims below.

(1) We start by observing that for any x > 0 there exists an atom u such that

0 < u < x. Indeed, if x is not an atom, then there exist two vectors 0 < ui < x and 0 < t4 < x such that ui n u2 = 0. If neither of them is an atom, then there exist two vectors 0 < ui < u1 and 0 < u2 < ui such that u21 n u2 = 0. Clearly, the vectors u2, ui , t4 are pairwise disjoint. If neither ui nor u2 is an atom, then we can continue this process. The process must terminate in a finite number of steps and produce a desired atom. Otherwise, if the process continues, then we can construct a countable disjoint sequence of non-zero vectors, contradicting our assumption. Since any two atoms are linearly dependent or disjoint, it follows from our assumption that there exists a maximal finite number of pairwise disjoint atoms, say u1, u2i ... , un. We claim that the linear span V of {ul, u2, ... , un} coincides with E. To see this, let x E E+ be in the ideal generated by {u1, u2i ... , un }; we can

assume that 0 < x < > 1 u,. For each 1 < i < n pick some Ai > 0 satisfying x n ui = Aiui, and note that i=1

i=1

i=1

i=1

This shows that V coincides with the ideal generated by {u1, u2, ... , un }. Now an easy argument shows that the band generated by {ul, u2, ... , un} also coincides with V; see also the solution to Problem 2.3.2. Finally, to see that V = E, it suffices to show that Vd = {0}. If there exists some 0 < z E Vd, then by the first part there exists an atom u such that 0 < u < z. However, the latter leads to a contradiction since the collection of pairwise disjoint atoms (U1, u2, ... , un, u} contradicts the maximality property of {u1, u2, ... , un }.

(2) If xn > 0 for each n E N, then clearly Fn '=l xn > 0. Now assume that En°=1 xn _> 0. From En 1 xn = IE00 1 xnI = F,;,°=1 Ixnl, it follows that (Ixn I - xn) = 0. Taking into account that Ixn I - xn > 0 for each n, the latter shows that Ixn I - xn = 0 or xn = Ixn I > 0 for each n.

(,'

1

(3) Pick any y' E E' with y'(x) = 1. According to Problem 5.2.2 there exists

a positive linear functional x' E E' such that x'(z) = y'(z) for all z E E,, and

x'(y)=0 for ally Ix. We now return to the solution of the problem. Since E is infinite dimensional, property (1) guarantees the existence of a pairwise disjoint sequence of positive unit vectors. By property (3) above, for each n there exists some positive linear functional x;, E E' satisfying x;, (xn) = 1 and x;, (x,n) = 0 for all n 96 m. Choose a sequence {an} of positive scalars satisfying 0 < An < 1 for each n and such that the formula

x

K = L.i An (xn (9 xn+1) n=1

defines an operator on E. Clearly, K is a positive compact operator. Moreover, note that Kx1 = A1x2 < x2. An easy inductive argument shows that Knxl < xn+1 for each n.

6. Spectral Properties

21.1

By our hypothesis, the operator

x

ff

I=Ea,lt:,l , =.n

is positive. In particular. 0 < f (h jr, =

i < at,.r 1

T,

1

Now according to property (2) above. we mast have anJ',i_1 0 for each 1t > 0. 1 This inlplics a,, > 0 for each n > 0. and the solution is finished.

Chapter 7

Some Special Spectra

7.1. The Spectrum of a Compact Operator Problem 7.1.1. Show that the only compact operators with closed range are the finite-rank operators. Solution: Let T : X -. Y be a compact operator between two Banach spaces, and assume that its range R(T) is closed. The surjective operator t: X/N(T) - R(T), defined by t[x] = Tx, is an isomorphism. In particular, from T (UX,N (T)) = T(U) and the compactness of T, it follows that the closed unit ball Ux,N(r) of X/N(T) is compact. Therefore, the vector space X/N(T) is finite dimensional. This implies I that R(T) is finite dimensional, and so T is a finite-rank operator.

Problem 7.1.2. For each 1 < p < oc define the integral positive operator T : Lp[0,1] - Lp[0,11 by T f (t) =

j(s2 1

+ t2)f (s) ds ,

where the interval [0,11 is equipped with its Lebesgue measure. Show that T is a finite-rank integral operator and find its spectrum. Solution: To see that T is a finite-rank roperator, note that

Tf(t) =

j

1

l +t2)f(s)ds = ILLJo ' f(s)ds]t2+s2f(s)ds. J

This shows that T is. in fact, an operator of rank 2. If we let u1(t) = t2 and u2(t) = 1 for each t E [0,1] and xi(f) = fo f(s)ds and X2* (f) = fo s2 f (s) ds for each f E Lp[0,11, then T = zi g,u1 +x2 Ou2. So, according to Theorem 7.6, the non-zero eigenvalues of T are the non-zero eigenvalues of the 215

7. Some Special Spectra

216

matrix

A=

i

si (ur) x (u2)

fo tz dt

fo dt

3

[x(1zj) x (uz)

Lft4dt

fo tz dt

l5

The characteristic polynomial of A is: P4(A) = det(A - Al) = Az - 3a - as Solving the equation PA(A) = 0 yields A = 513 5. So, according to Theorem 7.6, L the spectrum of the operator T is o(T) = {0.5 3V5-' s is 5 }.

The reader can verify that the eigenvalue Al = 53V5corresponds to the eigen15

function (unique up to a constant multiple) fl(s) = 5sz + 05- and the eigenvalue A2 = 5 is 5 corresponds to the eigenfunction f2(s) = 5sz - f.

Problem 7.1.3. Consider the Volterra operator V : LAO. 11 - LAO, 11 (1 < p < ac.) defined by

Vf(t) =

f f(s)ds. r

0

Show that V is quasinilpotent, i.e., a(V) = {0}, by proving that for each A # 0 and each g E Lp[0,1] the equation (A - V) f = g has a unique solution given by

f (s) _ gel

f

dt; + Zg(s)

0

Solution: Let 1 < p < oc. Fix a non-zero complex number A and define the operator S : Lp [0.1 ] - L[0, 1] via the formula

Sf(t)

eI

Jo

f(f)e-i d( + -if (t).

It should be clear that S is indeed a linear operator. What needs verification is that S maps L.10, 1] into Lp[0,1] and that it is a bounded operator. To see this, fix f E Lp[0,1), and define g: [0,1] - R by

g(t) _ lea rt f(e)e-a dt;. 0

Clearly, g is a continuous function (see, for instance, [8, Problem 22.6. p. 176]), and so g belongs to Lp [O, 11. Moreover, if Al = 1 e &' , then for each t E [0,1] it follows from Holder's inequality that Ig(t)# < M fo dd 5 MIIfIIP. Hence, IIYIIp < AIIIf IIp Now from Sf = g + f, we obtain a IIsflip < II9IIP+ atOflip <- Of+ fXl)IIIIIp-

This establishes that S is a bounded operator. Next. we claim that (A - V)Sf = S(A - V)f = f for each fimction f E Lp[0,1]. Since the continuous functions are norm dense in Lp[0.1] (here we use the fact that

7.1. The Spectrum of a Compact Operator

217

1 _< p < oc), it suffices to prove that (A - V)S f = S(A - V) f = f holds true for each function f E C[0,1]. So, let f E C[0,1]. Then for each t E [0, 11 we have

[(A-V)Sf](t) = -let f

f f

f

x

f

f f(e)e-ad1+f(t)- [let A

+ft?eff(s)e-fds-of 0

+a

f(e)e-Add, s-0 I

f(a)ds

de+f(t) - aet

aef f

fo

f(t;)e-A de 0

ftf(s)ds - a ftf(s)ds 0

0

f (t)

Similarly, for each f E C[0.1] and each t E [0.1] we have

[S(,\-V)f](t) = ASf(t)-S(Vf)(t) aek

4 +f(t)

10

-asef

ft{jf(s)ds]e_ aff(s)da 0

e aet

10

f(t)e a dd+f(t)

+zel ft [ ff(a)ds] d(e-f) 0

f

ae1 f e f(e)e A

f

f

0

e

t

j(a) da

f(t) + a f t f(s)ds 0

0

- ael = f(t)

f 0f(s)ds

f

ael f

f

-

f(#,)e-l A - a f f (a)ds 0

7. Some Special Spectra

218

The above equations show that S(A - V) f = (A - V)Sf = f holds for each f E Lp [0, 11. Therefore. S = (A - V)-', and the desired conclusion follows.

Problem 7.1.4. Show directly (i.e., without using Problem 7.1.3) that the Volterra operator has no eigenvalues-and conclude from this that the Volterra operator is quasinilpotent.

Solution: Consider the Volterra operator V : Lp[0,11 - Lp[0,1] (1 < p < oo) defined by

Vf(t) =

t f(s)ds.

J0

First, let us show that T does not have 0 as an eigenvalue. i.e., that T is one-to-one. To see this, assume T f = 0 or

M fo t

Then for each subinterval (closed, open, half-closed, or half-open) [a, b] of [0,1] we have f b f (t) dt = f o f (t) dt - fo f (t) dt = 0. Since every open subset of [0,1] is an at most countable disjoint union of intervals in 10, 1], it follows that fo f (t) dt = 0 for each open subset 0 of [0, 1]. Since the Lebesgue measure is regular, the latter implies fA f (t) dt = 0 for every Lebesgue measurable subset A of [0, 1]. Now consider the two Lebesgue measurable sets B = {x E [0,1] : f (x) > 0} and C = {x E [0,1] : f (x) < 0}. From fe f (t) dt = fc f (t) dt = 0, it follows that B and C are both Lebesgue null sets. This implies f = 0 a.e., and so T is one-to-one. Next, assume that for some A 0 0 and some f E Lp[0.1] we have t

10

f(s) ds = Af(t)

(**)

for almost all t E [0, 1]. Since the function F(t) = fo f (s) ds is (absolutely) continuous (see, for instance [8, Problem 22.6, p. 176]) and A j4 0, it follows from (**) that f can be assumed to be a continuous function. But then from (*-*) and the Fundamental Theorem of Calculus, it also follows that the function f is differentiable and Af'(t) = f (t) for each t E [0, 1]. Solving this differential equation yields f (t) = Ce t. Now one more glance at (**) shows that f (0) = 0. This implies C = 0 or f = 0. Therefore, V does not have any non-zero eigenvalues. Since the Volterra operator is a compact operator (see Example 7.8), it follows from Theorem 7.3 that o(V) = {0}. That is, the Volterra operator is quasinilpotent.

Problem 7.1.5. Let T : E -' E be a positive contraction on a Banach lattice. Show that 1 belongs to the approximate point spectrum of T if and only if r(T) = 1.

Solution: Let T : E

E be a positive operator which is also a contraction, i.e.,

IITII<<1. Then IIT"IIIITII"<-1,andsor(T)<-IIT"II^ <<1. Assume that 1 E a. (T). This implies 1 < r(T), and hence r(T) = 1. For the converse, suppose that r(T) = 1. By Theorem 7.9, we know that 1 belongs to the

7.1. The Spectrum of a Compact Operator

219

spectrum of T (and hence, as a boundary point, it also belongs to the approximate point spectrum of T). I

Problem 7.1.6. Show that the hypothesis r(T) > 0 cannot be dropped from the Krein--Rutman Theorem 7.10. L1 [0, 1] be the standard Volterra operator defined by V f (t) = fo f (s) ds for each f E L1 [0.1] and all t E 10, 1]. According to Problem 7.1.4, V is a compact quasinilpotent positive operator, i.e., r(T) = 0. We claim that 0 cannot be an eigenvalue of the operator V.

Solution: Let V : L1 [0,1]

To see this, assume that a function f E L1 [0.1] satisfies V f (t) = fo f (s) ds = 0 for all t E [0, 11. This implies fa f (s) ds = 0 for all a, b E [0,1], and from this we conclude that f (s) = 0 for almost all s E [0, 1]. Consequently, 0 cannot be an eigenvalue of V. (See also Problem 7.1.4.)

Problem 7.1.7. Show that the compactness hypothesis of the operator in the Krein-Rutman Theorem 7.10 cannot be dropped. Solution: Consider the shift operator T: f2 - 12 defined by T(x1,x2.... ) = (0,x1,x2,...). Then T is a non-compact positive operator-in fact, it is a lattice isometry. So, II7''j II = 1 holds for each n, and so r(T) =

IIT' II Al = 1.

Now note that if T(x1ix2,...) = (0,x1ix2.... ) = (x1,x2,...), then it is easy to see that xi = 0 for each i. Hence, x = (x1, x2, ...) = 0, and this shows that r(T) = I cannot be an eigenvalue of T. a

Problem 7.1.8. Let T : E - E be a positive operator on a Banach lattice and let p be a polynomial with non-negative Show that r(p(T)) = p(r(T)). Solution: Assume that a polynomial p has non-negative coefficients, and let T : E - E be a positive operator. Then the operator p(T) is also positive, and so, by Theorem 7.9, we know that r (p(T)) E a (p(T)). At the same time, from the Spec-

tral Mapping Theorem or Problem 6.4.1, we know that p(a(T)) = a(p(T)). Since the polynomial p has non-negative coefficients, we have Ip(A)I < p(IAI) < p(r(T)) for each A E a(T). Therefore, r(p(T)) = max{Iµj: p E a(p(T))} = max{Ip(A)I: A E a(T)} = p(r(T)) ,

I

as claimed.

Problem 7.1.9. Prove the following generalization of the Krein-Rutman Theorem 7.10. If a power compact positive operator T : E -* E satisfies r(T) > 0, then there exists some vector x > 0 such that Tx = r(T)x. Solution: Assume that r = r(T) > 0 and that Tk is compact for some k > 1. By Theorem 7.10, there exists some x > 0 such that Tkx = rkx. Now consider the

operator S = Ek o r'Tk-1-', and note that the vector y = Sx satisfies k-2

k-1

r`Tk-1-'x =

y = Sx = t=0

riTk-1-3 x + rk-1x > rk-1x > 0. i-0

7. Some Special Spectra

220

Now we have

k-1

Ty-ry = TS.x-rSx=(T-r)Sx=(T-r)[Er'Tk-1-',x :=o

= (Tk - rk)x = 0. That is, Ty = ry, and so r is an eigenvalue having a positive eigenvector. Problem 7.1.10. Prove Lemma 4.18 using the Krein Rutman theorem. That is, show by using the Krein-Rutman theorem that if Sl is a compact Hausdorf space without isolated points, then the only weakly compact multiplication operator on C(S1) is the zero operator. Solution: Assume that St does not have isolated points and suppose by way of contradiction that there exists a non-zero weakly compact multiplication operator AI,0 on C(1l). Then MOMO = Afp is a non-zero compact positive multiplication operator; see [6. p. 337]. So, replacing p by 02. we can suppose that p > 0 and that MO is compact. Clearly, 1

1

1

r = r(M.,) = lim IIAl 11n = lim IIMM..IIn = lim (IIo"IIx)" = IIpfl00 > 0. n--no n- Orn-= Consequently, by the Krein-Rutman Theorem 7.10 there exists some 0 < x E C(Q)

such that A10(x) = Ox = rx, or (0 - r)x = 0. So, if V = {w E 1: x(w) > 0}, then V is a non-empty open subset of Q and b(w) = r for all for all w E V. Now let { f, } C C(V) be any sequence in C(V) satisfying II fn IIx <_ 1 for each

n. By Tietze's extension theorem, we can assume that each f" is defined on all of 11. Now the compactness of AfO guarantees the existence of a subsequence {gn} of { fn } and a function f E C(n) such that II Afm(gn) - f II:x - 0. This implies that {g,,) converges uniformly to if on V or that g,, -+ i f holds in C(V ). This shows that the closed unit ball of C(V) is compact, and hence C(V) is a finite dimensional space. In turn, this implies that V (and hence the open set V) must be a finite set. But then fl has isolated points, which is a contradiction. Consequently, there are a no non-zero weakly compact multiplication operators on C(SI).

Problem 7.1.11. Show that there exist a Banach space X and a non-empty compact subset K of C such that o(T) K for each T E £(X). Solution: Let X be the HI-space introduced in Problem 4.5.7. As explained there, each operator T E C(X) is of the form T = Al + S. where A is a constant and

S i5 a strictly singular operator. Therefore, o(T) = A + o(S), and the spectrum of T is at most countable since by Theorem 7.11 the spectrum of S is at most countable. Consequently, if K is an arbitrary uncountable compact subset of C, then no operator in £(X) can have K as its spectrum. (In connection with this problem see also Problem 6.1.15.)

Problem 7.1.12. Assume that S E £(X) is strictly singular and that an operator T E £(X) has an at most countable spectrum. Show that the spectrum of S+T is at most countable and that zero and the points of a(T) are the only possible accumulation points of o(S + T).

7.I. The Spectrum of a Compact Operator

221

Solution: The solution is a modification of the proof of Theorem 7.11. For any A 0 0 that does not belong to o(T) the operator A - T is invertible and hence Fredholm of index i(A-T) = 0. Therefore, by Theorem 4.63, the operator A-T -S is also Fredholm of index zero. Now Lemma 4.55(3) guarantees that there exists some open disk D(A) centered at A such that for each p E D(A) the operator it - T - S is Fredholm and the functions n(p - T - S) and d(p - T - S) are constant on the set D(A) \ {A}. Fix a non-zero A ¢ o(T). We claim that there exists a half-ray that starts at A and does not meet the set o(T) U {0}. Indeed, otherwise, every half-ray starting at A and not containing zero would intersect the spectrum of T, and so o(T) would be uncountable, which is a contradiction. Consequently, we can find some Al such that JAI I > IIT + SII and the line segment L = [A, All joining A and A, contains neither zero nor any point from o(T).

Since the operator p - T - S is invertible for IAI > IIT + SII, it follows that n(p - T - S) = 0 and d(p - T - S) = 0 for each p that is close enough to A,, and hence, by our choice of the disk D(A1), we have n(p - T - S) = d(p - T - S) = 0 for every p E D(A,). The compactness and connectedness of L guarantee that there exists a finite number of disks D(tk) that cover L and satisfy D(l:,) fl D(t3}1) 0 for 1 C j < k - 1; see also Problem 1.5.7. Using this chain of disks connecting the point

A, with A and keeping in mind that the functions n(p - T - S) and d(p - T - S) are constant on each set D(&) \ {t';}, we can conclude that n(p - T - S) and d(p - T - S) are constant on the union D = Uk 1[D(t;) \ {t; }l. Moreover, since

n(p - T - S) = d(p - T - S) = 0 for each p E D(A1) \ {A1}, it follows that n(p - T - S) = d(p - T - S) = 0 for each p E D. This implies, in particular, that the operator p - T - S is invertible for each p E D(A) \ {A} and so if A belongs to o(T + S), then A is an isolated point of o(T + S). To finish the proof, it suffices to show that for every e > 0 the compact set

A,={AEo(S+T): IA - pl>e for all AEo(T)U{0}} is finite. Indeed, since AE is compact and disjoint from o(T) U {0}, the previous arguments guarantee the existence of a finite number of points pi, ... , p such that each of them is outside of a(T) and the disks D(p,) cover AF. This implies that AE e JAI,_, it.). That is, AE is finite, and the solution is finished.

Problem 7.1.13 (T. Oikhberg and V. Troitsky). Let 0 be a non-zero multiplicative functional on the Banach algebra £(X ), where X is an arbitrary Banach space with dim X > 1. Show that: (a) For each T E £(X) we have cb(T) E o(T). (b) If for T E C(X) there is a reducing pair of closed subspaces (V, W) such that dim V < oo and T vanishes on W, then q5(T) = 0.

(c) If T E £(X) is a strictly singular operator (in particular, if T is compact), then cb(T) = 0. Solution: (a) Since 0 is non-zero, we know that 40(I) = 1. Fix T E £(X) and pick any A E p(T). So, there exists an operator S E £(X) such that (A - T)S = I.

7. Some Special Spectra

222

This implies that [A - 0(T)]O(S) = 4(1) = 1, and hence A - O(T) 0 0. That is, 0(T) A for each A E p(T). Consequently, 0(T) E C \ p(T) = o(T). (b) If dim X < oo, then the result is a consequence of Problem 4.3.7. So, let dim X = oo. Replacing, if necessary, V by a larger subspace, we can assume without loss of generality that dim V > 1. Consider the collection B of all operators S E C(X) that leave V invariant and vanish on W. Clearly, the functional tb = 0Ie is multiplicative on B. Furthermore, the algebra B is obviously isomorphic to C(V) and T E B. Since V is finite dimensional and dim V > 1, it follows from Problem 4.3.7 that 0 = 0. Thus,

O(T) = (T)=0. (c) Let T E C(X) be a strictly singular operator. From Theorem 7.11 we know that the spectrum o(T) is at most countable and that every non-zero point in o(T) is isolated and is an eigenvalue of finite (geometric) multiplicity. Therefore for each e > 0 the set A,, = {A E off): IAI > e} is finite and consists of isolated points of the spectrum o(T). Thus, Af is a spectral set of T. Consider the spectral subspaces V and W of X corresponding to A,, and Be = o(T) \ & respectively. Theorem 6.34 guarantees that V ®W = X and also that o(T I v) = A, and o(T I w) = B. Consider the bounded operator Ti : X -- X defined by Tiv = Tv for v E V and Tlw = 0 for w E W. Clearly, the pair (V,W) reduces the operator T1. Since dim V < oo, part (b) applies and it follows that O(TI) = 0. Finally, consider the operator T2 = T - T1. In view of the properties of the pair of the spectral subspaces (V, W), the spectrum of T2 is contained in the set BB and therefore, by (a), we have 0(T2) E BB, whence I0(T2)I < e. At the same time we have 0(T2) = 0(T) - 0(T1) = 0(T). That is, we have established that I0(T)I < e, a and hence 0(T) = 0 since e > 0 is arbitrary.

7.2. Turning Approximate Eigenvalues into Eigenvalues Problem 7.2.1. Show that if X is an arbitrary Banachspace, then for each x = (XI, x2, ...) E &, (X) the norm of the vector [x] E X is given by II[x] II =

IIy)I.0 =1im sup I[xn[I

inf Solution: Let X be an arbitrary Banach space and let x = (rl,x2....) E t,,(X). Put a = lim sups-a Ilxn II If y = (yl, y2, ...) E [x], then z = xn - yn 0, and so from IlxnII = Ilvn + znll <- IIvul + IIZnII <- IIvII. + IIZnII, we get

a = limsup I(xnII S limsup(IIy(Ix + Ilznll] = IIvIIx n-w n-oc Hence, a < infvE(2i IIYIIOQ = II(xIIl.

Now let e > 0 be fixed. Then there exists some k such that Ilxnll < 0 + e holds for all n > k. If we consider the vector y = (yl,y2,...) E t.(X) defined by

7.2. Turning Approximate Eigenvalues into Eigenvalues

223

y;=0if1k,thenyEJxJand llylloo 5 ct + e. Since e > 0 is arbitrary. II (x] 11 5 Ilyll30 5 a. Hence,

I

II (x] II < Il yll. = a, as desired.

Problem 7.2.2. Assume that

is a sequence of real Banach spaces. Show that (with the obvious identifications) for each 1 < p < oo we have

fp((Xl)cED (X2)cED ...Pp(Xi®X2®...)ED 2fp(XI®X2q) ...). Also show that

c0(Xl ®X2(D ...)®2Cii(XI (D X2®...)

co((Xl)c®(X2)c®...)

Solution: We establish the result for the fp-sums when 1 < p < oo. To this end, (B... ). Put x = (r1,x2.... ) and let z = (x1 +1yi.x2 +2y2....) E fp((X1)c ( D y = (yi, y2....). and note that both sequences x and y belong to fp(XI tip X2 (i) ). This allows us to define a mapping T: fp((X i)c t1) (X2)c ffi ...) -' fp(X1 E}) X2 1 ... ) tD zep(Xi 0 X2 kD ...

via the formula T(z) = x + sy. We claim that T is a surjective (topological) isomorphism. F o r the additivity of T note that if z = (.r1 + zy1. x2 + y2 ' . . . ) ) and it = (a1 + zbi, a2 +02- ..), then we have T(z + it) = (xi + a1, x2 + a2, ...) + ?(y1 + b1, y2 + b2,...) = (x + zy) + (a + zb) =T(z)+T(u).

For the homogeneity of T let A = a + id, and note that T(Az)

= T ((a + zf3)(xl + M), (a +'1f3)(x2 + z'y2)....)

= Taxi - iyi + z(0xi + (M), axe - j3y2 + t(/3.1'2 + aye).... ) = (axi -,3y1, (tx2 - dye, ...) + z(/fxi + ayi. f3x2 + oy2....) = ((kx - fly) + 1(R3r + (ky) = (a + m13)(r + zy)

= ATz. We leave it to the reader to verify that. 7 is surjective and one-to-one. Finally, we shall show that T is continuous. To see this, let z = (xi + zyl, x2 + zy2.... ) E fp((X i )c +'B (X2)c kD ...)

and consider the vectors r = (xi,r2.... ) and y = (yi,y2.... ). Then for each 0 E IR we have oc

,

Ilxc(.KO+ySill 9llp

sin Hllp

n=1 00

1

Ilxn+tynllp)"

<-

II=llp

n=1

Hence, the norm of T (z) = x + zy in ep(X1®X2 m ) ®zfp(X i ® X2 (B... ) satisfies IIT(z)II

sup upilxcosO+ysinellp 511=11p O

On the other hand, we know that iIT(z)II ? 2IIzII1,5 IIT(z)II 5 IIziIp

Ilyllp) ?

ZIIzIIp. That is,

7. Some Special Spectra

224

holds for each z E Pp((X1)c®(X2)c® . . ), proving that T is a surjective topological isomorphism.

Problem 7.2.3. Let {Xn} be a sequence of real (or complex) Banach spaces, and assume that 1 < p, q < oo satisfy v + q = 1. Show that (X1®X2®...p(XI

®XZ®...)q,

where the equality is understood subject to the natural duality

(x,x*) = E(xn,xn) n=1

forx = (x1,x2,...) E (X1

®X2®... )p, x` = (xi,x2.... ) E (Xi

Similarly. show that

(X1®X2(D ...)1 = (Xi®XZ®... (X1®X2®...)o

and

= (Xi®X2®...)1,

Solution: We shall establish the result for 1 < p, q < oc only. This will be done by showing that the mapping

F: (Xi ®XZ ®...)q - (X1

GX2®...)p,

defined by Fx'(x1, x2, ...) _ F-n 1(xn, xn) = F_n 1 xn(xn), is a surjective linear isometry, where x' _ (xi, x2, ...) E (XI ED X2* ®

)q.

To this end, fix x' = (xi, x2, ...) E (XI- ® X2 ® . . . )q. Then for each vector x = (x1.x2,...) in (X1 ® X2 ®. )p we have Ixq(xn)I <_ IIxnII . IIxnII. and so for each k we get the estimate

(xn,xn) < n=1

Ilxnlip)

IIxnII . IIxnII n=1

n=1

=

IIz'lIq

-

ll

Ilp

n=1

This implies that the formula Fx' (x) = F_,°,° 1 (xn, xn) defines a continuous linear functional on (X1 ® X2 ® . )p and (*)

IIFx* ll <_ IIx' lIq

Moreover, it should be clear that Fx` + Fy' = F(x' + y') and F(ax') = \Fx', that is, F is a linear operator. Now let f E (X1 ®X2/® )p. For each n define x,*,: X,, -4 1R by

xn(xn) = PO, 0....,O,xn,O,O.... ), where the vector xn E Xn occupies the n`h coordinate. Then x,, E X. and moreover f (x1, x2, ...) = En=1 xn(xn) holds for each vector x = (x1i x2, ...) in (X1 ® X2 iD .. )p. Next, fix 0 < e < 1. For each n choose some yn E Xn satisfying

IIxnII = 1 and xn(yn) ? ellxnll put zn =

and note that for each k we

7.2. Turning Approximate Eigenvaiues into Eigenvalues

225

have k

k

k

E(E IIx

II4)

= E Ilxnll°-1EIIxn11 < F, IIxn114-lxn(1/n) n=1

n=1

n=1

f(Z1.Z2.....Zk.0,0.0....) k

IIfii(EIIzniIp)P

=llfll(IlxnllQ)°

n=1

n=1

< IIf1I holds for each k and each 0 < c < 1. It follows that x' = (xi, x2, ...) E (Xi e X2 e . . )q, f = Fx', and IIx'IIq : IIFx'II. Consequently, e(En=1 IIxnII4)

Now use (*) to conclude that II Fx' II = IIx' Ilq Therefore, F is a surjective linear isometry-which is also a lattice isometry if each X. is a Banach lattice. I

Problem 7.2.4. Let {X,,} be a sequence of all real or all complex Banach spaces, and consider the Banach space X = (X1 ® X2 (B .. )p for some 1 < p < oo. For each T = (T1,T2, ...) E (G(Xi) ®G(X2) ®... ),,. define the mapping.FT : X -b X by

,FT(x1,x2,...)

(T1x1,T2x2,...).

Establish the following.

(a) FT is a bounded operator on X, i.e.. FT E £(X). (b) The mapping T .FT, from (G(Xi) ®£(X2) (D ... ). to £(X), is a linear isometry-and therefore for each 1 < p < oo the Banach space (C(X1)®1C(X2)®

is a closed subspace of Z ((XI ED X2 ED

Solution: (a) Fix 1 < p < oc, pick T = (T1,T2,...) E (1(X1) ®1(X2) ® ... )00, and put Al = supra IITnII = IITII < oc. If x = (xl.x2....) E X, then note that

x

M

(IIFTxllp)p = E IITnxnllp < n=1

x

IITTIlpllxnllp < Mp y, IIxnIIp = Mnllxlip < oc n=1

n=1

This implies, II.FTxII p <_ AMIIxiIp < oc, and so FT is indeed a mapping from X

to X. It should be obvious that FT: X - X is a bounded operator satisfying IIFTII
also true. Consequently, IIFTII = IITII,, This shows that the mapping T - FT, from the Banach space (1(X1) ®1(X2) ® ). to 1(X), is a linear isometry. I

Problem 7.2.5. Verify that the mapping x H 1, from X to X defined by the formula x = [(x, x, x,..... )], is a linear isometry-which is also a lattice isometry if X is in addition a Banach lattice.

7. Some Special Spectra

226

Solution: Let J : X -y X be defined by J(x) = i = [(x. x, r.... )]. For the additivity of J note that if x, y E X, then J(x + y) = [(x + y, x + y, ...)] = [(x, x....) + (y, y.... = [(x, x, ...)] + [(y, y ...)] = J(x) + J(y) To see that J is homogeneous, let x E X and A be an arbitrary scalar. Then we have

J(Ax) = [(ax. Ax....)] = [A(x, X,...)] = A[(x, x.... )] = AJ(x). To see that J is an isometry, notice that for each x E X we have IIJ(x)Ii = 11((x,x.... )]11 =limsupIIxII = IIxII

n-oc

Finally, assume that X is a Banach lattice. Then the lattice operations of X are given by [x) V [y] = [(XI V yi, x2 V y2, ...)]

and

[x] n [y] = [(xl n yt, x2 A y2, ...)] .

Therefore, if x. y E X, then J(x V y) = [(x V y, x V Y'...)] = k V y = J(x) V J(y).

This proves that when X is a Banach lattice, J is a lattice isometry.

Problem 7.2.6. If X is a real Banach lattice, then show that a bounded operator T : X - X is positive if and only if the operator T : X -- X is positive.

Solution: Let T: X - X be a bounded operator on a (real) Banach lattice. Assume first that T is positive and let [x] > 0 in E. Then there exists some vector

y = (yl, y2,...) E (x] such that yn > 0 for each n. Since T is positive, Tyn > 0 likewise holds for each n. This implies

T[x] = [(Txi,Tx2....)] _ [(Tyi,Ty2.... )] > 0, and so f is a positive operator. For the converse, assume that T is a positive operator, and take any x E X. Then we have Ta = [(Tx,Tx,...)] 0. Consequently, there exists some sequence y = (yl, y2, ...) E [(Tx. Tx.... )] satisfying yn > 0 for each it It follows that yn Tx. Now a glance at Problem 1.2.12 guarantees that Tx > 0. This shows that T is a positive operator.

I

Problem 7.2.7. If_T : X - X is a surjective operator on a Banach space, then show that T : X - X is also surjective. Conversely, if T has a closed range and f is surjective, then show that T is also surjective. Solution: Assume that T : X X is surjective. By Corollary 2.15 there is a constant c > 0 such that for every y E Y there exists some x E X satisfying Ilxll S cllyll and Tx = y. Now let [y] = [(yl, y2....)] E X. For each n pick some xn E X satisfying llxnll < cllynli < cllyll« and Txn = y, . Therefore, if we let x = (xi,x2.... ), then x E t,,,,(X) and T[x] = [(Tx1,Tx2.... )] _ [y]

7.2. 7Lrning Approximate Eigenvalues into Eigenvalues

This shows that T is surjective. For the converse, assume that the operator T: X

227

X has a closed range and

that T is surjective. Fix some y E X. Since T- is surjective, there exists some

x = (xl,.r2....) E fx(X) such that T[x] = [(Tx1.Tx2.... )] = y. This implies y. Since the range of T is closed. we infer that y E R(T). Therefore, Tx,, R(T) = X, and so T is surjective.

Problem 7.2.8. Show that a positive operator T : E E on a real Banach lattice is a lattice homomorphism if and only if T : E - E is likewise a lattice homomorphism. Solution: Assume first that T is a lattice homomorphism and let x = (xl, x2....) and y = (yl, y2, ...) belong to t-,(E). Then we have T (Ix] V Iy])

= T(I(X1 V yl,x2 v y2....)]) = [(T (xl V yi ). T (x2 V y2), ...)] = [(Txi V Tyl. Tx2 V Ty2, ...)] = [(Tx1.Tx2, ...)] V [(Ty1.Ty2.... )I = T([ ]) V T(Iy])

This shows that f is a lattice homomorphism. For the converse, suppose that f is a lattice homomorphism. If X. Y E E, then [(TxVTy,Tx-

VTy,...)]

((Tx.Tx....)]V [(Ty. Ty.... )]

TrVTy=T(.i.V T([(xVy.xVy.... )] [(T(x V y). T(x V y)....)J

.

This implies T(x V y) = Tx V Ty. and so T is a lattice homomorphism.

U

Problem 7.2.9. Let X be an arbitrary real Banach space with complexifcation Xc = X C zX. Establish the following:

(a) A sequence z = (xl + zyl, x2 + 9y2, ...) in Xc is bounded, i.e.. z E t,,(Xc), if and only if the sequences x = (xl, x2, ...) and y = (yl, y2, ...) in X are both bounded. i.e., x. y E tx(X). In this case., write z = x + zy.

(b) Show that the mapping T : Xc , k e d, defined by T ([X + r y]) _ [x] + t[y]

for all x = (xl, 12, ...) and y = (yt, y2, ...) in tx,(X) r is a surjective (topological) isomorphism- which is a lattice isomorphism if X is also a Banach lattice. Solution: (a) This is a consequence of the following fact: If a+ lb E X E: z X . then (IIaII + IIbII) < Ila + zbll S hail + IIbhh

(b) First. let us establish that T is well defined. To this end. let the sequence = a + zb = (a, + tbl , a2 + tb2....) belong to [x -.- ty]. This means that the sequence a+tb is norm bounded, and Ila,, +zb,, II(an-x,,)+t(b -yn)II - 0.

7. Some Special Spectra

228

From the inequalities of part (a), it follows that the sequences a = (al,a2.... ) and b = (bl, b2....) are both bounded, i.e., a, b E f,r (X ). Ila,, - xn I] -+ 0, and I I bn - y.11 - 0. Thus. a E [x) and b E [y], and so [a] +,&[b] = [x] + t[y].

holds true, proving that T is well defined. We now verify the conditions needed to guarantee that T is a linear isomorphism. For the additivity of T, note that

T([x+ty]+[a+zb]) = T([(x+a)+z(y+b)])=[x+a]+t[y+b] ([x] + z[y]) + ([a] + t{b]) = T ([x + zy]) + T([a + tb]) .

The homogeneity of T can be proven as follows: = T([Ax + zAy]) = [Ax) + t[Ay] = A[x] + iA[y]

T(AIx + ty])

= A([x] + t[y]) = AT([x + zy]) To see that T is surjective, note that if [x] + z[y] E 19 zX. then r = (xz, xz....) and y = (yt,y2.... ) are both bounded sequences in X, and moreover the bounded sequence r + ty (see part (a)) in Xc satisfies T ([x + zy]) = [x] + z[y] .

To prove that T is one-to-one, let T([x + zy]) = [x] + t[y] = 0. This means that

[x] _ [y] = 0 or that x and y are null sequences in X. This implies that the sequence [r + zy] in f,(Xc) is also a null sequence, and so [x + ty] = 0 in Xc. So, T is one-to-one too. Finally, for the continuity of T note that: IIT ([x + ty]) II

=

II [x] + z[y]II = sup li [x] cos9 + [y] sin 9II BER

= sup II [(xi cos 9 + yl sin 9, x2 cos 9 + y2 sin 9....)] II BER

= sup( lim sup I I xn cos 9 + yn sin 9II ]

BER nix

< limSUP llxn +tynll = II[x+ty]II n- 3c

Therefore. T is continuous, and consequently by the Open Mapping Theorem, the operator T is a topological isomorphism.

Problem 7.2.10. Let T : X X be a bounded orrator on a real Banach space, and let T : X -p X be its extension to X. If Xc = X ED %X is the complexification of X, then in Problem 7.2.9 we defined the mapping T : Xc -- X ®zX by T ([x + zy]) = [x] + %[y]

for all x = (XI, x2, ...) and y = (yt, y2, ...) in f,,,(X) and proved that it was a surjective (topological) isomorphism. Establish the following properties. (a) The diagram shown in Figure 1 is commutative.

7.2. Turning Approximate Eigenvalues into Eigenvalues

229

(b) The spectra of the operators T, Tc, and T + zT coincide. That is,

a(T) = u(?e) = a(T + zT-),

ap(T) = ap(T e) = ap(T + d),

and

Qa(T) = sa(Tc) = aa.(T + t T-).

(c) If X is a Banach lattice and T is a lattice homomorphism, then I(T)c(z)l = TJzI

holds for all z E (X)c. Solution: (a) Note that if [z] = [x + zy] E Xc, then we have

(T + zT)T([z]) = (T + 2f )([ ] + z[y]) = T([x]) + zT([y]) = [(Txl,Tx2.... )] + z[(Tyl,Ty2, ... and

TTc([z]) = T[(Tc(xl + zyi), Tc(x2 + zy2)....)]

= T[(Txl + zTy1,Tx2 + zTy2, ...)l

= [(Tx1, Tx2, ...)] + z[(Tyl, Ty2,...)] So, (T + zT)T = TTc. That is, the diagram in Figure 1 is indeed commutative. (b) From Problem 6.2.8, we know that

a(Tc) = a(T + zT),

Qp(Tc) = ap(T + zT),

and Qa(Tc) = a4(T + zT) .

The other equalities follow from Corollary 7.17.

(c) Assume that T is a lattice homomorphism. Then, by Problem 7.2.8, the operator T : X - X is also a lattice homomorphism. The desired conclusion now follows from Theorem 4.29.

1

Xc-

T

(X)c

I

TO

X.

(X )c

Figure 1

Problem 7.2.11. If T : X - X is a bounded operator on a Banach space, then for every free ultrafilter U on N we have:

(1) a(T) = a(Tu) (2) aa(T) = Qa(Tu) = op(Tu) Solution: (1) Fix some A E C. According to part (c) of Problem 1.4.9 we know that the operator A - T is invertible if and only if the operator (A - T)u = A - Tu is invertible-in which case we have (A - Tu)-1 = [(A - T)-1]u' From the above conclusion, we infer immediately that p(T) = p(Tu). This is, of course, equivalent to a(T) = a(Tu).

7. Some Special Spectra

230

(2) We shall show first that a.(T) = ap(Tu). To this end, let A E CQ(T). Then there exists some sequence {x,,} in X satisfying IIx,,II = 1 for each n and Tx,, - Ax,, -+ 0. So, if we let x = (X I, X2, ...), then [x] # 0 and Tu[x] = A[x]. Thus, A E ap(Tu), and so u0(T) C ap(TU)

For the reverse inequality, let A E ap(Tu) and pick some x = (x1, x2, ...) in e,,,,(X) with I I [x] I I = limu Ilxn I I = I and Tu [xJ = A [x] or limu [Tx - Ax,,] = 0. Therefore, for each k the set

Ak= {nEN: IITx,, -AxnII <

and IIxnOI> 1}

belongs to U, and so it is an infinite subset of N. Moreover, we have Ak+1 S Ak for each k. Now if we choose a sequence of natural numbers {mk} satisfying Mk E Al, and Mk < mk+1 for each k, then x,,,,k ,4 0 and Tx,,,k - Ax,,,k -) 0. This shows that A E CQ(T), and so ap(Tu) C oa(T). Consequently, aa(T) = ap(Tu).

Since, ap(Tu) C u.(Tu) is always true, what remains to be proven is the inclusion a,,(Tu) C ap(Tj) = oa(T). To this end, let A E aa(Tu). This means that A - Tu = (A - T)u is not bounded below on X. Now if A does not belong to aa(T), then A - T is bounded below on X. But then it follows from part (c) of Problem 1.4.9 that (A-T)u must be bounded below on XU, a contradiction. Hence, A E as (T) and so as (Tu) C a,, (T) is true, as desired. I

7.3. The Spectrum of a Lattice Homomorphism Problem 7.3.1. Let G = {z,.. . , z,,} be a subset of the unit circle r consisting of p distinct numbers. If G is closed under multiplication, then show that 2xk

G={zEC: zp}={e P : k=0,

..,p-1)

Solution: Fix any z E G and consider p+ 1 elements z, Z2'. .. , zp+1. Each of them lies in G and so z'n = zE for some 1 < m < e < p+ 1. This implies zn = 1 for some 1 < n < p. In particular, 1 E G and from zzn-1 = zn = 1, we see that G is a subgroup of r. Next we will show that zp = 1. To this end let v = min{n E N : 1 < n < p and z" = 11. Clearly the set H = {1, z, ... , z'-,) is a subgroup of G and so p is divisible by v, and thus 0 = 1. The above proves that the solutions of the equation SP = 1 are precisely the p elements of the group G. Therefore G = { e' : k = 0,1, ... , p -1 }. 1

Problem 7.3.2. Fix an angle 9 E R and let G = {e"'9: n E N }. Establish the following.

(a) If 9 is a rational multiple of 27r, then G is a finite subgroup of the unit circle r (and so G consists of roots of unity). (b) If 9 is not a rational multiple of 27r, then G is dense in r. Solution: Clearly G is closed under multiplication.

7.3. The Spectrum of a Lattice Homomorphism

231

(a) Assume that the angle 0 is a rational multiple of 27r, that is, 6 = 2rr-,'- for some k E N and m. E Z. Consequently ernei = elk*' = 1. Now if n E N, then we can write n = ek+r, where 0 < r < k, and so enei = (ekee)ee''d' = erei. This shows that G is a finite subset of r. Since G is closed under multiplication, it follows from Problem 7.3.1 that G is a subgroup of r consisting of roots of unity.

(b) Suppose that the angle 8 is not a rational multiple of 27r. In this case, for any distinct n, rn E N we have cn8i # ernei. Indeed, otherwise, e"8' = em9' would imply that e("-'")B' = 1, and so (n - rn)O = 27rq for some q E Z or, equivalently. 0 = qr1t 27r, which is impossible. In particular, G is an infinite set. Next fix any k E N and for each 1 < j < k let F1

1'}={ex': 2z(j-1)
sonic n, in E N with n > in such that c"82, c"'8' E I'j. Pick xI, x2 E [ ( - 1), 2 ) eme' = e- r2'. Finally, let .r = XI - X2 and note that such that en8i = e l" and ex' = e(n-m)8 E G. Consequently, eexr E G for each e E N. This jxj < implies that for each I < j < k there exists some e E N such that eeri E I'). Since I k E N is arbitrary, we infer that G is dense in I'.

Problem 7.3.3 (Kronecker's theorem). Let G be an arbitrary subgroup of the unit circle r. Show that either G is dense in r or the it consists of roots 2vk of unity, i.e., G = {e P ' : k=0,11 ... , p -1 } for some p E N. Solution: The conclusion follows immediately from Problems 7.3.1 and 7.3.2.

1

Problem 7.3.4. Establish the following result that was used in the proof of Theorem 7.20. If p is a regular Borel probability measure on a compact Hausdorff space ft and some function z E Cc(1l) satisfies fsi z(t) d;i(t) = 1 and IzI = 1, then z(t) = 1 for each t in the support of p. Solution: Replacing S2 by Supple, we can assume without loss of generality that Supp tc = U. Now suppose that a function z = x + ty E Cc(51) satisfies Izi = 1 and fsz z(t) dµ(t) = fn x(t) dp(t) + z f f2 y(t) dp.(t) = 1. Clearly,

Jx(t)dli(t)=1

and

JsI

I

y(t) dta(t) = 0 .

From IzI = 1, it follows that x < 1. If r(to) < 1 holds for some to E 1), then there exists an open neighborhood V of to such that x(t) < I for each t E V. Taking into account that u(V) > 0, the latter implies

1=

f

U

.r(t) dtz(t) + f

x(t) dte(t) = IV

x(t) dll(t) < 11(v) + /A(st \ V) \V

which is impossible. Hence. x = 1, and from 1 = Iz12 = r2 + y2 = 1 + y2, we get

y= 0. Therefore, z=x+zy=x=1. Problem 7.3.5. Let T : E - E be an interval preserving positive operator. If a complex number A = IAIe'° belongs to the residual spectrum of T. then

7. Some Special Spectra

232

show that for each integer n the complex number I \Ie"'B is an eigenvalue of the adjoint operator T*. Solution: By Theorem 6.19 we know that A is an eigenvalue of T. Also, by TheE' is a lattice homomorphism. According to Corollary 7.22. orem 1.35. T*: E* the point spectrum of T' is cyclic, and the desired conclusion follows.

Problem 7.3.6. Assume that for some continuous function 0: fl -. R on a compact Hausdorff space the multiplication operator M0: C(f2) -, C(f2) is a lattice homomorphism. Show directly (i.e., without using Theorem 7.23) that the spectrum of the multiplication operator M, is cyclic. Solution: Note that the multiplication operator 14 is a lattice homomorphism if and only if 0 is a non-negative function. Now a glance at Problem 6.1.14 shows that the spectrum of MM coincides with the range of 0. a subset of 10, oo). This implies that the spectrum of A is cyclic. I

Problem 7.3.7. Let T : X - X be a bounded operator on a Banach space. If the approximate point spectrum of T is rotation invariant, then show that the spectrum of T is also rotation invariant. Solution: Assume k = IAIeie° E a(T), and let C. = {z E C: Izl = Ia1}. We must show that C C a(T). If C n aa(T) 54 0, then C C aa(T) by our hypothesis. So. assume that C fl as (T) _ 0. In this case, A E a(T) \Qa(T). By Theorem 6.18 we know that a(T) \aa(T) is an open subset of C. Therefore, there exists some f > 0 such that 9o - e < 0 < Oo + e implies lAle`e E a(T) \ aa(T). Now let A={6

[ 0 . 2 7 r] :

jAleie E a(T) \ aa(T) for each 9 E (90 - b. Bo + b) } .

Since e E A, we see that A 96 0. Let 60 = sup A. Since a(T) is closed, we have Ialei8o-6o), IJ1le`0e11+60) E a(T). From C n aa(T) = 0, it follows that IAIe`(6°-&,), Ialet(e"+6°) E a(T) \

aa(T)

We claim that 6o = 2a. To see this, assume that 60 < 2ir. Since a(T) \ a,, (T) is open, then-as above-we can find some q E A such that 60 < n < 27r, which is a contradiction. So, 60 = 2ir, which implies C C a(T), and the proof is finished. I

7.4. The Order Spectrum of an Order Bounded Operator Problem 7.4.1. Let T : V -, V be an invertible operator on a vector space, and let T-1: V -> V be its inverse. Assume that W is a T-invariant subspace of V such that T : W -. W is also invertible. Then show that T-1 leaves W invariant and that the inverse of Tlty (the restriction of T to W) = T-11w holds. is the restriction of T-1 to W, i.e., (TIw)-1

7.4. The Order Spectrum of an Order Bounded Operator

233

Solution: Let S: W -+ W be the inverse of T: W - W, i.e., S = (TIw)-. If w E W, then we have

T(T-1w) = T(Sw) = w. Since T is one-to-one, it follows that T'lw = Sw for each w E W. That is,

T-1

also leaves the vector subspace W invariant and T-11 w = (TI w) -1 holds.

Problem 7.4.2. Describe the Banach lattice E if E = R. Solution: Assume that E = R, so that EA = RA. Now notice that a function f E RA belongs to E if and only if there exists some x E R such that If (a)I < x holds for each a E A. This means that k consists of all bounded real-valued functions on a, that is, E = le(a). I

Problem 7.4.3. Let a be an arbitrary index set. Show that a Riesz space E is Dedekind complete if and only if the Riesz space EA is Dedekind complete.

Solution: Let E be a Riesz space. Assume first that E is Dedekind complete, and let D C EA be non-empty and bounded from above. That is, there exists some

g E EA such that f < g holds for each f E D. This is equivalent to f (a) S g(a) for all f E D and each a E A. So, if we fix a E A, then the non-empty subset (f (a): f E D} of E is bounded from above in E. Since E is Dedekind complete, for each a E A there exists some h(a) E E such that SUP/ED f (a) = h(a) holds in E. Thus, we have defined a function h: A - E (i.e., h E EA), and we claim that h = sup D holds in EA. Clearly, f < h holds for each f E D, i.e., h is an upper bound of the set D in EA. To see that h is the least upper bound of the set D, assume that some 0 E EA

satisfies f < 9 for each f E D. So, for each a E A we have f (a) < 0(a) in E for all f E D. This implies h(a) = supfED f (a) < 9(a) for each a E A, that is, h < 0. Therefore, h = sup D holds in EA, and consequently EA is a Dedekind complete Riesz space. For the converse, suppose that EA is a Dedekind complete Riesz space, and let

D be a non-empty bounded from above subset of E. Pick some x E E such that d:5 x holds for each d E D. It follows that the non-empty subset b = {d: d E D} of EA is bounded from above by i. So, according to our hypothesis, f = sup D exists in EA. In particular, if ao E A is any fixed index, then d = d(ao) < f(ao) holds in E for all d E D, i.e., f (ao) E E is an upper bound of D. To finish the solution, we shall show that f (ao) = sup D holds in E. To this end, assume that some vector y E E satisfies d < y for each d E D. Then g is an upper bound for the set D. This implies f < g in EA, and in particular f (ao) < g(ao) = y. Thus, f (ao) = sup D holds in E, proving that E is Dedekind

I

complete.

Problem 7.4.4. Let E be a nonned Riesz space, and define the function P. by

IIIfIII=inf{IIxII: xEE and IfI <x} for each f E E. Establish the following.

(*)

7. Some Special Spectra

234

(a) III 'III is a lattice norm that extends II

. II

to E.

(h) If E is a Banach lattice, then t equipped with the lattice norm given by (*) is a Banach lattice containing E as a Banach sublattice. Solution: (a) We shall verify first that the function III -III is a norm. Clearly. II If I II ? 0 holds for each f E E and 1110111 = 0. Now assume that 1111111 = 0. This means that there exists some sequence (x.1 C E+ satisfying

IfI <_ i,, for each n and l1r II -- 0. From I f (a)I < ±n(Q) = x,,. we see that [If (a) II 5 Ilxn II for all n and each a E A. This implies II f (a) II = 0 for each Q E A,

and so f =0. For the homogeneity of III III, let f E E and let A be a non-zero scalar. Clearly,

IAfl < s if and only if IfI S

IIIAf11I

, and so

= inf{IIxII: x E E and lAfI < r} =

lAI inf{IIxII :

x E E and 1tl < i }

IAI inf{IIyII: y E E and I f

=

IAI ' IIIfIIi

-

For t he triangle inequality, assume IfI < x and I9I < b with X. Y E E+. Then we have If +9I <- IfI+191 <_ r+il _ (x.+y). and so 11If+9I II <_ ilr+yii <_ IIrll+IIyII Taking infirna yields 11If + 9I II < 11 If 111 + 11I9Iii.

Next, we shall establish that III ' III is a lattice norm. To this end. assume that 191 holds in E. Now if z E E+ satisfies 191 < i. then If 1 < ! is also true, and IfI .x) 111f111 < IIyII holds. This implies 111f III 5 inf{1Jz11: z E E and 1915 z} = III9IIt. That is, the norm III . III is a lattice norm. Finally, let us show that. III ' is an extension of II - II To this end, let x E E. From I&I = 1x1 and IxI E E, we see that 11111115 11 IxI 11 = IIxII. On the other hand. 111

if y E E satisfies 1±1 < y, then IrI 5 y. and so IIyII = II I'I II S IIyII This implies IIxII : inf{IIyII: y E E and Iz1 < y} = IIIt11I. Hence. 1112111 = IIziI for each r. E E. and thus III - III extends the norm 11 II of E to E.

(b) We only need to show that III ' III is a complete norm. To this end, let { f } be a III ' I II-Cauchy sequence of E. By passing to a subsequence if necessary. we can assume without loss of generality that 111 fn11I < 2 r holds for each n. Now for each it pick some vector rn E Et satisfying If, +r - f 1 <- ±, and -2---r .

Since E is a Banach lattice. the series y _ k rk is norm convergent in E and 11 y 11 < 2 -

for each n.

fn(o)I < r for each a E A and all n. it easily follows that is a Cauchy sequence in E for each a E A. Since E is a Banach space, f,, (a) -, f (a) holds in E for each a E A. So. we have defined a function f : A - E. We shall finish the solution by proving that f E E and that 11 If, - fill 0. From

{

7.4. The Order Spectrum of an Order Bounded Operator

235

We first show that f E E. Start by observing that for each a E A we have n-1

Ifn(a)I = I fn(a) - f1(a)I + If1(a)I = IE(,fk+1 - fk)I + Ifl(a)I k=1

n-1

00

xk + Ifl(a)l

Ifk+1 - Al + Ifl(a)I k=1

k=1

y1 + If1(a)I

Letting n -+ oo, we get If (a) 1 < y1 + If, (a) I for each a E A, and then using that f1 E E, we easily infer that f E E. Finally, for m > n we have

x

m-1

m-1 7

Ifm(a) - fn(a)I = I E [fk+I(a) - fk(a)JI k=n

F, Ifk+1(a) - A(a)1 S > xk = yn k=n

k=n

Letting nz - oo, we obtain If (a) - fn(a)I < yn for each n and all indices a E A. This implies 111f,, - f III

is a Banach lattice.

Ilyn II < s

,

and so II If,, - f III

- 0.

Therefore, (E, III

III )

Problem 7.4.5. Let S, T : E -+ E be two order bounded operators on a Riesz space. If (S + zT )" : Ec - Ec is invertible, then show that the operator S + zT: Ec --+ Ec is also invertible and that its inverse is order bounded and satisfies [(S + zT)-1J' = [(S + zT)']-1 = (S + zT)-1. Solution: Let S, T : E --+ E be two order bounded operators on a Riesz space such that the operator (S + zT)' = S + zT : EE -+ EE is invertible. We shall show first that the operator S + zT: Ec -. Ec is also invertible.

To show S+zT is one-to-one, let (S+zT)(x+zy) = Sx-Ty+z(Tx+Sy) = 0. This easily implies

(S+zt)(i+zy) = [(S+zT)(x+zy)J'=0. and so (given that S + it: Ec -i EE is invertible) we get i + zy = 0. This implies x + zy = 0, proving that the operator S + zT: E,, - E. is one-to-one. To see that the operator S+ iT is surjective, let x+zy E Ec. Then i+zy E EC, and since S + it is surjective, there exists some f + zg E Ec such that

(S+zT)(f+zg)=Sf-Tg+z(Tf+Sg)+zy. Now if ao E A is any fixed index, then and T(f(ao)) +S(g(ao)) = y. So, if a = f (ao) and b = g(ao). then a + zb E Ec and

S(f(ao)) - T(g(ao)) = x

(S+zT)(a+zb) = Sa-Tb+z(Ta+Sb) =x+zy. This shows that S + zT : Ec Ec is also a surjective operator. The above show that S + zT : EE -+ E. is invertible. Let S1i T1: E - E be two linear operators such that S1 + zT1 is the inverse of S + zT: EC - E.. That is, we have (S + zT)(S1 + zT1) = (S1 + zT1)(S + iT) = I(= I + ii). This implies (S + zT)(S1 + zt1) = (S1 + zT1)(S + it) = Z (= I + if).

So, the operator S + it: (EA)c -- (EA)c is invertible and (S + it) -' = S1 + iT1.

7. Some Special Spectra

236

Since S + d' is invertible on the subspace E,, of (Ea ). it follows that A + zB leaves invariant Er: see Problem 7.4.1. This means that S1 and T1 leave k invariant, and so (by Theorem 7.31) Sl and T1 are both order bounded operators. Therefore, the inverse S1 + iT1 of the operator S + zT : EE -. EE is an order bounded operator. Finally, one more glance at Problem 7.4.1 guarantees that the inverse of the operator (S+zT)-: is the restriction of Si +&Ti to This implies

[(S+zT)"]-1 = (S+zT)-1 = Sl +&TI = (Si +iTi)`= [(S+zT)-']'. and the solution is finished.

Problem 7.4.6. If T is an order bounded operator on a Dedekind complete Banach lattice, then show that o0(T) = ao(T) = a(T). Solution: Let T: E E be an order bounded operator on a Dedekind complete Banach lattice, and let TT : EE --+ EE be its extension to the complexification of E. Then for any complex number A = a + ib we have

(A-T.)"=((a-T)+r(b-T)]'=(a-T)'+z(b-T)"=a-T+l(b-T). Now, from Problem 7.4.5, it follows that the operator A - TT : Er - E, has an order bounded inverse if and only if the operator A - Ta : Ev. Ec has likewise an order bounded inverse - and also if and only if A - Tom : Re -. -Pr is invertible. These equivalent statements simply show that ao(T) = ao(T) = a(T). I

Problem 7.4.7. Establish the following elementary properties that were cos

used in Example 7.36. Consider the matrix M = [sin © on R2, where 0 is an arbitrary angle. Then:

(a) For each n E N we have M' =

sn i nO

rs(i)n

sin m sill

d

acting

- sin no cos no

(sin nod (b) The modulus of M" is the matrix IM"I = I cos nol [Isinnol Icosn&l

'

(c) The symmetric matrix I M" I has eigenvalues I cos n0] f I sin nOI . (d) The Euclidean norm of lA' I is 111 Al" 111 = I cos nol + I sin nol.

Solution: (a) This proof is by induction. Assume that the formula is valid for some n > 1. Then we have - sin riot rcos o - sin dl cos no sm d cos d J cos o cos rap - sin p sin n© -(sin d cos no + cos o sin np) cos o sin no - sin o sin no sin p cos nip + cos p sin no cos[(n + 1)0] - sin[(n + 1)O] sin((n + 1)o] cos[(n + 1)o]

AI"A! =

cos no

[sin nO

(b) This is a special case of Problem 8.3.1. We have IM-1 = Icosnol sin ndl 11

I - sinn©l[I cos noIsinn'I I cos npl

I sin nol

I cos ndl

7.5. The Essential Spectrum of a Bounded Operator

237

(c) If we let a = l cos nil and b = I sin nOI, then the characteristic polynomial of IMnI is

det [AI - IMnl] = det

[Aa

-ba] A

= (A - a)2

- b2.

Therefore, the eigenvalues of IMnI are the solutions of the quadratic equation

(A-a)2 -b2 = (A - a + b)(A - a - b) =0. Solving this equation yields: A = a ± b = I cos n¢I ± I sin nOl.

(d) Since IAfnI is a real symmetric matrix, its Euclidean norm coincides with its spectral radius; see Problem 8.1.9. So, we have II IMn I ll = l cos ndl + I sin ntbl.

7.5. The Essential Spectrum of a Bounded Operator Problem 7.5.1. For the identity operator I on a Banach space X establish the following:

(a) X is infinite dimensional if and only if II [I]II = 1. (b) X is finite dimensional if and only if II[I] II = 0.

Solution: (a) If X is finite dimensional, then £(X) = K(X). Consequently, E(X) = {0}, the trivial Banach algebra. So. II[1] 11 = 1 implies that X must be infinite dimensional. For the converse, assume that X is infinite dimensional. Then I V K (X ), and so

11[j]II

> 0. From 0 < 11(1111 = 11(11211

!5

11(11112, we see that 11(1111 >- 1.

Since

II[I]II 5 I1111 < 1 is obviously true, we infer that 10111 = 1

(b) This equivalence follows immediately from the preceding part.

Problem 7.5.2. Let A be a unital Banach algebra, and let G(A) denote the non-empty collection of all invertible elements of A. Show that G(A) is an open subset of A and that the mapping x -+ x-1, from G(A) to G(A), is continuous-in particular. G(A) is a topological group. Solution: Assume first that some element a E A satisfies Ilall < 1. From the inequality Ilanll < Iialln it follows that the series b = E°°_oan is norm convergent in A. Now the continuity of the product map (x, y) 1-4 xy implies that

). an-Ear+1 =e, (e - a)b = (e - a)Ean =nL=O n=0

n=0

where e denotes the unit of A. Similarly, b(e - a) = e. This shows that the element

e - a is invertible and (e - a)-' =

°

an a".

Now let any a E G(A), i.e., assume that a is an invertible element of A. Pick some element x E A such that IIx - all < 11nl Then from

l

Ile - xa-'II = II(a - x)a-'II < llx - all Ila-'Il < I

7. Some Special Spectra

238

and the preceding conclusion, it follows that xa-1 = e - (e - xa-1) is an invertible

element. That is, there is some y E A such that (xa'1)y = y(xa-1) = e. This implies x(a-ly) = e, and thus x has a right inverse. Similarly, x has a left inverse, and this guarantees that x is invertible,' that is, x E G(A). Consequently, C G(A), and so G(A) is an open subset of A. Since e E G(A), the B(a, I open set G(A) is automatically non-empty. Finally, let us show that the mapping x - x-1, from G(A) to G(A), is continuous. To this end, fix a E G(A). Then by the above, B(a, T,-'-g) C G(A). Now let

y E B(a, I). Clearly, IIy - all IJa-111 < 1 or 1 - Ily - all . Ila-1II > 0. From the identity Y-1-a-1

= a-1(a a-' (a

y)y-'

(y-'

-

= a-1(a - a 1) +a-1J ') - a + a 1(a - y)a-1 , y)[(y-'

= - y)
Ila-'

II, and so

IIy-' - a-'11 < 1- a - a-y Ily - all holds for all y E B(a, Wig). This implies that the map x " x-1 is norm continuous at each element a E G(A).

Problem 7.5.3. Let S: 1p -- gyp, where 1 < p < oo, be the forward shift operator defined by

S(x1,x2,x3,...) = (09 x1, X21 x31...). Show that the essential spectrum of S coincides with the unit circle, that is,

a (S)=r={AEC: JAI=1}. Solution: Clearly, S is an isometry, and so JISIJ = 1. From Example 6.21 we know C {A E C. JAI 5 1}. We that a(S) = a(S') = {A E C: JAI 5 1}. Hence, claim that:

(a) For any A E C the operator A - S is one-to-one, i.e., N(A - S) = {0}. (b) For any A E C with JAI < 1 the operator A - S is bounded below, and hence A - S has closed range. (c) For each complex number A with JAI < 1 the null space of the adjoint

operator (A- S)'=A-S' is {au: aEC}, where u=(1,A,A2,...) E gyp. We verify each claim separately below.

(a) If x = (x, i x2, ...) E N(A - S), then we have Ax, = 0 and Axi+1- xi = 0 for each i > 1. If A = 0, then xi = 0 for each i > 1, i.e., x = 0. If A :A 0, then x, = 0 and x;+1 = Zxi for each i >>-1, and sox = 0 in this case too. Thus, N(A-S) = {0}. (b) If A E C satisfies JAI < 1, then for each x E en we have II(A - S)xII >- IISsII - IIAXII = llXlI - JAI . lixll = (1- IAI)IJxlI 1 If uv = wu = e in a unital algebra with unit e, then w = we = w(uv) = (wu)v = ev = v.

7.5. The Essential Spectrum of a Bounded Operator

(c) From Example 6.21 we know that S*: fq

239

fq is given by

S'(rl.x2,x3.... ) = (x2.x3.r4,...). So, if r E N(A - S'), then Ar, - .r.,+1 = 0 for each i > 1. This implies xj+1 = A`xt for each i > 1. and so x = xt (1, A. A2....). On the other hand, if x = a(1, A, A2....).

then it is easy to see that (A - S')x = 0. Therefore. N(A - S') = {au: a E C}. In particular. (b) and (c) show that for each A E C with IAI < 1 the operator A - S is a Fredholm operator (of index -1), and so the operator A - S is essentially invertible. This implies that a (S) C F.

Now let A E F = do(S) C ao(S). Since the operator A - S is not bounded below and N(A - S) _ {0}. it follows that A - S does not have closed range. This implies that A - S is not a Fredhohn operator. Consequently, the operator A - S is not essentially invertible, and so A E a.(S). Thus, 1' c a ,,.(s) is also true. and

hence i, (S) = F. Problem 7.5.4. Show that i f an operator T : X

X on an i n f i n i t e dimen-

sional Banach space s a t i s f i e s a(T) = {A0}. then a ,(T)

Solution: We know that apaq(T) C a(T). If Ao

a.,.A(T), then a,_'(T) = 0, a

contradiction.

Problem 7.5.5. Show that the power compact operators on a Banach space are precisely the essentially nilpotent operators. Deduce from this that every bounded operator on a Banach lattice that is dominated by a compact positive operator is essentially nilpotent. Solution: Let T : X X be a bounded operator on a Banach space. Then T" is a compact operator if and only if the equality [T]" = [T"] = 0 holds in the Calkin algebra it(X). That is, T is power compact if and only if [T] is essentially nilpotent. Now let T : E - E he a bounded operator on a Banach lattice that is dominated by a compact positive operator. By Corollary 2.35 we know that T3 is a compact operator. and so T is essentially nilpotent. 2

Problem 7.5.6. Let Ao be an isolated point in the spectrum of an operator T E G(X). Show without using Theorem 7.44 that if the spectral projection P,,0(T) is of finite-rank, then A0 is a pole of the resolvent of T. Solution: Let Xl = R(Pa(, (T)) and X2 = N(P),a(T)). According to our hypothesis. X 1 is finite dimensional. X = X 1 e X2. and the closed subspaces X 1 and X2

are both invariant under T. Moreover, if Ti = Ttx, (the restriction of T to X,). then a(T1) = {Ao} and a(T2) = a(T) \ {Ao}. Also, for brevity write P = Pao and

Q=I-P.

Notice that the resolvent operator satisfies R(A,T) = R(A.Ti) G R(A.T2) = R(A.T)Ix, E9R(A.T)Ix2.

From Ao if a(T2). it follows that R(A,T2) = R(A.T)Ixz = R(A.T)Q is analytic around A0. Furthermore, since X1 is finite dimensional. all singularities of R(A. T1) are poles; see Corollary 6.41. So, Ao is a pole of R(A.T1) = R(A.T)I,, = R(A.T)P. This implies that A0 is also a pole of R(A, T) = R(A. T)P + R(A, T )Q.

7. Some Special Spectra

240

Problem 7.5.7. Give an example of an operator T E C(X) and some point A E a(T) \ o (T) that is not an eigenvalue of T. Solution: Let S: f2 - t2 be the forward shift, i.e., S(xi, x2, ...) _ (0, xl, x2, ...). The spectrum of S coincides with the closed unit disk; see Example 6.21. According

to Problem 7.5.3, we have o,.,(S) = r, the unit circle. Clearly, 0 4 a,(S). t2 is the forward shift, then ir(T) is the inverse of a(S) in C(X). If T: f2 This implies that 0 E o(S) \ o,.,(S) = {A E C: Ian < 1}. Now notice that 0 is not an eigenvalue of S.

Problem 7.5.8. Show that every strictly singular operator on an infinite dimensional Banach space is essentially quasinilpotent.

Solution: Assume that T E C(X) is a strictly singular operator on an infinite dimensional Banach space. According to Theorem 7.48. in order to establish that T is essentially quasinilpotent, we must show that each non-zero point A E a(T) is an isolated point of o(T) and that the spectral projection PA(T) has finite-rank. To this end, let A E o(T) \ {0}. From Theorem 7.11 we know that A is an isolated point of a(T). Next, according to Problem 6.4.13, there exists an operator

S E C(X) such that Pao(T) = TS = ST. This implies that the operator Pao(T) in C(X) is strictly singular. Since Pao(T) is the identity operator on its range, it follows that P,\,, (T) has finite-rank, and the solution is finished.

I

Problem 7.5.9. Establish the following properties regarding the measure of non-compactness.

(a) For each subset A of a metric space we have X(A) = (b) A subset A of a metric space is relatively compact if and only if x(A) = 0. (c) A bounded operator T E C(X) is compact if and only if X(T) = 0. (d) If T E C(X, Y) and S E C(Y, Z), then X(ST) < X(S)X(T). Solution: (a) Let A be a subset of a metric space (M, d). If X(A) = oo, then x(A) = oc is also true. So, we can assume that X(A) < oo. Pick any r > X(A) and then choose points x1, ... , xn in M such that A C U, 1 B(x,, r). Since the set C(xi, r) y E M : d(xi, y) < r } is a closed set and satisfies B(x r) C C(xi, r), it follows that for each e > 0 we have n

n

n

A S Ui=1 B(xi, r) = U B(xi. r) S U C(xi, r) S U B(xi, r + e) . i=1

i=1

i=1

This implies X(A) < r + e for each r > X(A) and all e > 0. Thus, X(A) < X(A). Since -K (A) < X(A) is obviously true, we easily conclude that X(A) = X(A).

(b) It is easy to see that for a subset A of a metric space the condition X(A) = 0 is equivalent to saying that A is a totally bounded subset. That is, a subset A of a metric space A is relatively compact if and only if X(A) = 0. (c) This follows immediately from (b).

7.5. The Essential Spectrum of a Bounded Operator

241

(d) Assume T E C(X, Y) and S E C(Y Z). Fix r > x(T) and p > x(S). Pick yl.... , y, E Y such that T(Ux) C U 1(y, +rU y ). Similarly, choose z1..... z,,, E Z such that S(L'y) C U;"-1(z3 + pUz), and note that n

ST(Ux) =

n S(U[y,+rUy])

1=1

n

= US(y,+rUy) i=1 n

on

U [Sy, + rS((ry)] C U U [Sy, + rz1 + rpUz] . 1=13=1

1=1

This implies X(ST) < rp for all r > X(T) and p > X(S). So. x(ST) < X(S)X(T).

Problem 7.5.10. If X is a Banach space, then show that the measure of non-compactness x : C(X) - J is a seminorm. Also. show that this seminorm induces a norm on the Calkin algebra C(X). Solution: Recall that if T E C(X). then the measure k(T) of non-compactness of T is defined by

X(T) = inf I r > 0: ? x1..... zn such that T(Ux) C U. 1 B(x,. r) } . In particular. we have X(T) > 0 for each T E C(X ). We now verify the remaining desired properties of the function k : C(X) - R. 1. The homogeneity, i.e.. X(AT) = IAIX(T) for all T E C(X) and all scalars A.

The verification of the homogeneity property will be based upon the identity µB(0. r) = B(0, I pI r) for each r > 0 and p. Fix T E C(X) and a scalar A # 0. If r > X(T), pick vectors x1, ... , xn such that T(Ux) C U 1 [x1 +B(0, r)] and [Ax= + B(0. IAIr)]. This implies note that AT(Ux) C Un 1 [Ax= + AB(0. r)] = U X(AT) < IAIr for each r > X(T). and from this it follows that X(AT) < IAIx(T) Now let p > X(AT). Choose y1..... yk such that AT(Ux) C Uk,-, [y, +B(0. P)]. 1

This implies T(Ux) C Uk a + -1B(0, p)] = U`i [

a holds for each p > X(AT). and so k(T) <

+B(0. IXI

Consequently,

? or IAIX(T) < X(AT). <' X(T) < l i Therefore. X(AT) = JAI-X(T). For A = 0, the preceding identity should be obvious.

2. The triangle inequality. i.e., X(S+T) < X(S) +.,<(T) for all ST E C(X). Let S,T E C(X ). Fix r > X(S) and p > X(T). Pick vectors x1.... , x and y1..... yk such that S(Ux) C Un1 [x, + B(0. r)] and T(Ux) C Uk_1 [yj + B(0, p)]. It follows that n

k

(S+T)(Ux) C S(U.x)+T(Ux) C U U[x,+yl+B(0.r)+B(0.p)] ,=1.7=1 it

k

U i=1.7=1

U[r,+yj+B(O.r+p)].

This implies X(S + T) < r + p for all r > X(S) and all p > X(T). Consequently. X(S +T) < X(S) + X(T).

242

7. Some Special Spectra

3. If S - T is a compact operator, then X(S) = X(T). Let S - T = K, a compact operator. From S = T + K. the triangle inequality, and Problem 7.5.9(c), it follows that X(S) = X(T + K) < X(T) + X(K) = X(T). Similarly, from T = S+ (-K) we get X(T) S X(S). Therefore. X(S) = X(T). A consequence of (3) and the other properties above is that the function I X: £(X) - R when restricted to the Catkin algebra t(X) is a norm.

Chapter 8

Positive Matrices

8.1. The Banach Lattices A1n(1R) and AL(C) Problem 8.1.1. If A is an n x n matrix with real or complex entries, then show that

(Ax, y) = (x, V y)

for all x, y E C". In particular, show that A is Hermitian if and only if (Ax, y) = (x, Ay) for all x, y E C. Solution: Let A = (a,, ) be an n x n matrix with complex entries. Fix x, y E C", and note that n

n

n

Ax = (F aljxj, F a2jxj.... , E anjxj ). and At y

J=1

1=1

j=1

n

n

n

t=1

i=1

(Ea11Yz. t=1

Consequently, we have Y)

(Ax,

_1

i=1 )=1

1=1

i=1

1yi

] =(x Ay).

j=1

(Ae-A)y)

Finally, note that (Ax, y) - (x, Ay) = (x, holds for all x, y E C. So, if (Ax, y) = (r., Ay) or (Az. y) - (x, Ay) = 0 for all x, y E C", then (x, (A - A)y) = 0 for all x, y E C". Letting, r. A)y. we get ((A - A)y, (A - A)y) = 0 for

allyEC". This implies (.4`-A)y=0 for all y E C", and so

A.

Problem 8.1.2. Show that for an n x n matrix A the following statements are equivalent. 243

8. Positive Matrices

244

(a) A is unitary. (b) The rows of A form an orthonormal basis in C". (c) The columns of A form an orthonormal basis in C". (d) A is inner product preserving, i.e., (Ax, Ay)=(x, y) for x, yEC". (e) A is Euclidean norm preserving, i.e., IlAxII = IIxII for x, y E Cn. (f) A carries an orthonormal basis to an orthonormal basis. Solution: Clearly, (a), (b), and (c) are equivalent. (a)

(d) Note that (Ax, Ay) = (A Ax, y) = (x, y).

(d)

(e) We have IIAxII2 = (Ax,Ax) = (x,x) = 11x112, and so IlAxII = IIxII

for all xEC". (e) (a) The proof of this implication will be based upon the following property of Hermitian operators.

(P) A Hermitian n x n matrix H is equal to zero if and only if (Hx, x) = 0 forallxECn.

To establish (P), assume that a Hermitian n x n matrix H satisfies (Hx, x) = 0 for

all XE C'. Then for all x,yEC" we have 0 = (H(x + ty), x + zy) = (Hx, x) - t(Hx, y) + t(Hy, x) + (H(ty), ty) = t [(Hy, x) - (Hx, y)] = t [(Hy, x) - (x, Hy)] = t [(Hy, x) - (Hy, x) ]

= 2zIm(Hy, x). Consequently, Im(Hy, x) = 0 for all x, y E C". Replacing y by ty, we also have Im(H(ty), x) = Re(Hy, x) = 0. So, (Hy, x) = 0 for all x, y E C" and this implies that H = 0. To see that (e) implies (a), note that IlAxII = IIxII is equivalent to the identity

(Ax,Ax) = (x, x) or ((A`A - I)x,x) = 0 for each x E C". Since TA - I is a Hermitian matrix, it follows from (P) that A A -I = 0 (and also that AT -I = 0). That is, TA = AAA = I, and this shows that A is a unitary matrix. (f) Let {e1,...,e,,} be an orthonormal basis. Then IIAe1II = IIe.II = I (d) for each i, and (Aei, Ae,) = (ei, e,) = 0 for i # j. This implies that {Ael,... , Ae" } is an orthonormal basis. (f) (e) Assume that A carries an orthonormal basis {e1, ... , e,, } to an orthonormal basis {Ael,... , Aen}. Then for each x = E 1.\iei we have n IA,12 = IIAshI2 ,

IIxI12 = i=1

and so IlAxII = IIxII holds for each x E C".

Problem 8.1.3. As in the case of operators, two n x n matrices A and B are said to be similar if there exists an invertible matrix C such that A = C'1BC. Recall also that the truce tr(A) of an n x n matrix A = [a,]

8.1. The Banach Lattices M"(R) and M"(C)

245

is the sum of its diagonal entries, i.e., tr(A) = En, aii. Establish the following.

(i) Two similar matrices have the same characteristic polynomial. (ii) This part deals with the trace of matrices.

(a) Show that if A and B are two arbitrary n x n matrices, then we have tr(AB) = tr(BA). (b) Show that the trace of an n x n matrix A coincides with the sum o f its eigenvalues. That is, i f Al, ... , A" are the eigenvalues of

A (counted with their multiplicities), then tr(A) _ F,1 Ai. (c) Show that two similar n x n matrices have the same trace, i.e., the trace of an n x n matrix is similarity invariant.

(d) Consider an n x n matrix A with column vectors cl,... , c,,. Also, let e; denote the ith unit vector in Cn and consider ei as a linear functional acting on C" (or R") via the formula ei(x) = (x,ei) = xi. Show that A, as an operator on Cn (or R"), can be identified

with the finite-rank operator T = E l ei ® ci. (e) Using the notation of part (b), show that tr(A)

ei(ci).

Solution: (i) Assume A = C-1BC. Then we have

pA(A) = det(AI - A) = det(C-'(,\I - B)C) = det(C'1)det(AI - B)det(C) = [det(C)]-'det(AI - B)det(C) = det(AI - B) = pu(A) . (ii) (a) Observe that the (i, i)th entry of the matrix AB is Ek=1 aikbki and the (k, k)c1i entry of BA is n

1 bkiaik So, n

n

F'aikbki = E E bkiaik = tr(BA).

tr(AB) = E FIaikbki = i=1k=1

n

n

k=1i=1

k=1i=1

(b) From the expansion formula of the determinant, we see that PA (A)

= det(AI - A)

\n+an-lAn_1 + ...+a,A+ao = An - tr(A)An-1 + + a,A + (-1)ndet(A) .

On the other hand, we also have A1)(A

PA(A)

A1)...(A

n

An _

An)

+a,A+(-1)n[IAi.

\i=1

d=1

The above equations show that n

tr(A) _ E ,\i i=1

and

det(A) = H Ai . i=1

8. Positive Matrices

246

(c) Let A and B be two similar n x n matrices. By part (i), we know that PA(A) = pB(A) So. A and B have the same eigenvalues (together with their multiplicities), say A,...... n. Now (b) above implies tr(A) = En, A, = tr(B). (d) Note that if x = (xl. x2, ... , xn) E C", then n

T(x) =

n

e, 0c,(x) =

n

e, WC. =

x, c, = Ax,

where in the last expression x is considered as a column vector. (e) Clearly, En

= En

1(c,); = Ei= 1 e,(c,). (See also Problem 4.1.4.) Problem 8.1.4. Show that each eigenvalue of a unitary matrix has modulus 1 a=i

one.

Solution: Let U be a unitary n x n matrix, i.e.. U-1 = U`. Then with respect to the Euclidean norm we have

= (Us,Ur.) = (UtUx,x) = (x, x) = 11x112, or IlUxfl = Ilxll for each x E C". This shows that U defines a surjective isometry. By Problem 6.2.5. it follows that every eigenvalue of U lies on the unit circle. A direct proof goes as follows. Assume that a unit vector z E Cn Satisfies Uz = then IIL'x112

I'\ 12

= A3(z, z) _ (As, Az) = (Uz. Uz) = (U`Uz. Z) = (z. Z) = 1.

and so 1Al=1.

Problem 8.1.5. Show that every Hermitian matrix has real eigenvalues and that eigenvectors corresponding to distinct eigenvalues are orthogonal. Solution: If A is a Hermitian matrix and a unit vector z E C" satisfies Az = Az, then = (z. z) = (z, ,\z) = (z, Az) _ (Atz, z) = (Az, z) = (Az, z) = A(z. z) = A. That is, every eigenvalue of A is a real number. Assume that two non-zero vectors z,w E C" satisfy Az = A1z and Aw = A2w with Al 36 A2. From the preceding part. we know that Al and A2 are real numbers. Now note that.

(At - A2) (z. u) _ Al (z, w) - A2(z. w) = (A1 z. w) - (z, )i2w) (A1 w) - (z, A2w) _ (Az, w) - (z, Aw) (.4z, w) - (Az, w) = 0. This implies (z, w) = 0. and so z and w are orthogonal vectors.

Problem 8.1.6. Recall that a square matrix A is called diagonalizable if there exists an invertible n x n matrix R (called a diagonalizing matrix for A) such that the matrix R-1 AR is diagonal. Establish the following.

(a) A matrix A is diagonalizable if and only if A has n linearly independent eigenvectors.

8.1. The Banach Lattices M,,(R) and M,,(C)

247

(b) Every Hermitian matrix is diagonalizable by a unitary matrix. Solution: (a) Assume that an invertible matrix R diagonalizes the matrix A, that is, R-1AR = D or, equivalently, AR = RD. From Problem 8.1.3(i), we know that D = diag(A1i A2, ... , A"), where A1, A2, ... , A" are the eigenvalues of A (counted with their multiplicities). If Cl, c2, ... , c, denote the column vectors of the matrix R, then note that the equation AR = RD can be written in the form

I

I

Act

Ace

I

I

... ... Ac" = Alcl I

...

I

I

I

A2c2

I

... ...

c"

(*)

I

That is, Ac; = Aici for i = 1, 2, .. ., n. Since the (column) vectors c1, c2, ... , cn are linearly independent, it follows that A has n linearly independent eigenvectors. Conversely, assume that Ac; = Aici for each i and that the vectors Cl, c2, ... , cn are linearly independent. Then the matrix R with columns c1i c2, ... , c,+ is invertible and, in view of (*), it satisfies AR = RD or R-1 AR = diag(A1, A2, ... , A"). (b) In order to solve this part, we need the following property that is also useful in many other contexts.

(Closed Subspaces Have Orthogonal Complements) If V is a closed subspace of a Hilbert space 7{, then V e V1 = N.

To see this, note first that the Pythagorean theorem guarantees that V ® V -L is a closed subspace. Take any x E N and consider the unique nearest to x vector v E V. By Problem 1.1.18(f) this vector exists and x - v 1 V. Therefore, we have

x=v+(x-v)with vEVandx-vEV1. Hence,V®V1=?{.

We now return to our problem. So, let A be a Hermitian matrix. From Problem 8.1.5, we know that eigenvectors corresponding to distinct eigenvalues of A are orthogonal. We shall prove first that A is diagonalizable. To see this, let V be the vector subspace generated in C" by the collection of all eigenvectors of A. It should be clear that Av E V for all v E V, i,e., V is A-invariant. We claim that V = C". To see this, assume by way of contradiction that V 36 C". Then, by above,

we have V e VI = C", and so the vector subspace

V1={zEC": zlv=0 for all vEV} of C" is different from {O}. Observe that V1 is A-invariant. Indeed, if z E V1, then note that for each v E V we have (Az, v) = (z, Av) = 0, that is, Az E V1. But then, since V1 is finite dimensional, A must have a non-zero eigenvector lying

in V1, say z E V1. This implies, z E V and so z = 0, which is a contradiction. Hence, V = C". So, A has n linearly independent eigenvectors, and thus it is diagonalizable.

To complete the solution, we need to show that A has n orthonormal eigenTo this end, let A1, ... , Ak be the distinct eigenvalues of A, and let

vectors.

Vi = {z E C": Az = Aiz}. By Problem 8.1.5, we know that Vi 1 Vj if i 96 j and

(D

E)

If Ei = (ei, ... , e;,j,) is an orthonormal basis of Vi,

8. Positive Matrices

248

then E = U 1 E, is an orthonormal basis of C" consisting of eigenvectors of A. So, if U is the n x it matrix whose columns are the vectors of the orthonormal basis E. then U is a unitary matrix that diagonalizes A. I

Problem 8.1.7. An n x n matrix A is said to be positive semidefinite whenever (Az, z) > 0 for all z E Cn. Establish the following:

(a) If A is a positive semidefinite matrix, then A has non-negative eigenvalues.

(b) A Hermitian matrix A is positive semidefinite if and only if it has non-negative eigenvalues. Solution: (a) Let A be a positive semidefinite matrix. If A is an eigenvalue of A and z E C" is a unit vector satisfying Az = Az, then note that A = A(z, z) = (Az, z) = (Az, z) > 0. (b) Assume that A is a Hermitian matrix. Then. according to Problem 8.1.6(b),

the matrix A can be diagonalized by a unitary matrix. That is, there exists a unitary matrix U satisfying UtAU = diag(A1...... "). where Al, A2.... , A" are the eigenvalues of A counted with their multiplicities. Now letting z = U y. we see that ti

A,Iy.I2 > 0

(Az, z) = (AUy, Uy) = (UtAUy, y) = :=1

for all y E C'. From this, it should be easy to establish that A is positive sernidefinite if and only if A, > 0 for each i. I

Problem 8.1.8. For each k let Ak =

1 ] be

an n x n matrix. It is easy

to see that the sequence of matrices {Ak} converges to a matrix A = (aid] in the Banach lattice Mn(C) if and only if limk_x a;11 = air for each i and j.

Show that an n x n matrix A satisfies Ak - 0 in

if and only if

every eigenvalue of A has modulus less than one. Solution: Assume first that every eigenvalue of A has modulus less than one. Then we have

lim IpAkjJk = r(A) = max{CAI: A is an eigenvalue of A) < 1. kx So, if we fix some real number 6 satisfying r(A) < 6 < 1. then there exists some ko such that 11AklI'1 < b or IlAkil < bk for all k > ko. This implies Ak -. 0. For the converse, suppose that Ak - 0 and let A E C be an eigenvalue of A.

Pick some unit vector z E C" such that Ax = Az. and note that Akz = Akz for each k. In particular, we have IAIk = IjAkzlj = IIAkzII <_ IVAkjj

This implies limk_,p IAIk = 0, and from this we infer that IAI < 1 for each eigenvalue

A of the matrix A. (Compare this solution with that of Problem 6.1.11.)

8.1. The Banach Lattices M,,(]R) and M,,(C)

249

Problem 8.1.9. Consider a matrix A E M,,(C). If Al is the (non-negative) largest eigenvalue of the Hermitian matrix A A, then show that 11 All = VTI In particular, if A is Hermitian, then show that its Euclidean norm coincides with its spectral radius. Solution: Since (AAz, z) = (Az, Az) = IIAxII2 > 0 holds for all z E C", we infer that the Hermitian matrix AA is positive semidefinite. Thus, by Problem 8.1.7, the matrix AA has non-negative eigenvalues, say Al > A2 > . . . > An > 0. Moreover, since AA is Hermitian, it is (according to Problem 8.1.6) diagonalizable by some unitary matrix, say U. That is, there exists a unitary matrix U such that U'T AU = diag(A1, Az, ... , An).

Fix a unit vector x E C" and let y = U-1 x. Clearly, IIyll = 1. Then we have IIAxII2

= (Ax, Ax) _ (x,AAx) _ (Uy,AAUy) n

n

= (y, UtAtAUy) _ > A,Iy,I2 < F AiIy+I2 = Al This implies IIAII = sup11111=1 IIAxil

On the other hand, there exists some unit vector z E Cn satisfying A Az = Ajz, and so IIAII2 > IIAxII2 = (Az, Az) = (A`Az, z) _ (A1z, z) = A1(z, z) = A1. Therefore, IIAII ?

A 1.

For the last part, assume that A is Hermitian, i.e., A = A. So, in this case, we have AA = A2. Since o(A2) = {A2: A E Q(A)} by the Spectral Mapping Theorem, we see that IIAII = max{IAI : A E o(A)} = r(A).

Problem 8.1.10. Show that an n x n matrix is quasinilpotent if and only if it is nilpotent. Solution: Let A be an n x n matrix. If A is nilpotent, then Ak = 0 for some k, and so An = 0 for all n > k. Therefore, r(A) = limn IIA"II '" = 0. For the converse, assume that A is quasinilpotent. That is, r(A) = 0, or equivalently o(A) = {0}. This implies Al = A2 = . . . = An = 0, and hence the characteristic polynomial of A is pA(A) = A". By the Cayley-Hamilton Theorem 8.3, we have An = 0, so that A is a nilpotent matrix. (See also Problem 6.1.18.)

Problem 8.1.11. Let A = [a;j] be an n x n matrix with complex entries. If we let ate = b1j + zcij and consider the real matrices B = [b 3] and C = [c;j], then we can write A = B+zC and view A as a vector in the complexification M. (R) ® W. (R) = M. (C) of the real Banach lattice Mn(1R). Show that the modulus of A = B + zC in Mn (]R) ® zMn (R) is the positive

matrix JAI = [la;,l]. Solution: We know that if a, b E R. then supeER(a cos 9 + b sin 0) = vora- + ; see the solution to Problem 1.1.7. Also, we know that a non-empty family of n x n real

8. Positive Matrices

250

matrices where Da = [d ], satisfies SUPAEA Da = D = [dij] in 'in(R) if and only if sUPAC- ,j dv = dj for all i and j, i.e., supaen D,, _ [SIPAEA diXi ] Now the modulus of A in Aln(R) ® tMn(R) is given by IAI

= sup (B cos 9 + C sin 9) =sup [b,,, cos 9 + c,, sin 9 ] 9ER

OER

[supp(bijcos9+c jsin9)] =

(bij)2+(G))21

[Ia;jl], 0

as desired.

Problem 8.1.12. Prove the Cayley-Hamilton Theorem 8.3. That is, show that every square matrix satisfies its characteristic polynomial. Solution: Let PA(A) = An + an_iAn-1 +

+ alA + ao be the characteristic

polynomial of an n x n matrix A. The proof employs the well-known formula

C(cofC)t = (cofC)tC = det(C)l

for an n x n matrix C = [cj). The cofactor matrix cofC = [cofc.;j] is the n x n matrix whose (i, j)th entry is given by cofcij = (-1)i+jdet(C;j), where C,j is the (n - 1) x (n - 1) matrix that is obtained from C by deleting its ith row and jth column. So. for each A E C we have (AI - A)[cof(AI - A)]t = det(A - I)I = pA(A)I.

(*)

Now notice that each entry of cof(AI - A) is a polynomial of degree at most n - 1. This implies that the matrix [cof (AI -A)]' is of the form Ek=u A* At, where the Bk are fixed n x n matrices. Rewriting (*), we get

A"I + a,-1An-lI + an_2An-2I + a1Al + aol . Since this holds for each A E C, it follows that:

I = B.-I,

an-1I = Bn_2 -ABn-i , an-21 = Bn-a -ABn-2

all = Bo - AB1,

aoI = -ABo.

8.2. Operators on Finite Dimensional Spaces

Multiplying the above equations by An, get

An-1,

251

An-,,..., A, and I, respectively, we

An = AnBn-1 ,

an-lA n-1 n-2 an-2A

= An-1 Bn-2 - An B,-1 = A n-2 Bn-3 - A n-1 Bn-2 ,

a1A = ABo-A2B1,

aoI = -ABo, + a1 A + a0I = 0. For a and then adding yields A' + an_ 1 An-1 + an-2An-2 + different proof of the Cayley-Hamilton theorem see Problem 8.2.10.

8.2. Operators on Finite Dimensional Spaces Problem 8.2.1. Show that for an operator T : X - X on a finite dimensional vector space the following statements are equivalent.

(a) T is invertible. (b) T is one-to-one. (c) T is surjective.

Solution: (a)

(b) Obvious.

(b) : (c) If T is one-to-one and lei,..., en } is a basis of X, then Tel, ... , Ten are n linearly independent vectors, and so they must be a basis of X. This implies that T is a surjective operator. (c) . (a) Assume that X is n-dimensional and that the operator T : X -. X is surjective. If n = 1, then T is clearly invertible. So, assume that n > 1. To finish the solution, we must show that T is one-to-one.

To this end, assume by way of contradiction that Tx = 0 for some x 54 0. Choose vectors x2i ... , xn such that {x, x2, x3, ... , xn } is a basis of X. If V is the (n - 1)-dimensional vector space generated by the linearly independent vectors

x2, ... , xn, then the linear operator T : V - X is also surjective. This implies n = dim(X) _< dim(V) = n - 1, which is a contradiction. Hence, T is one-to-one, and consequently it is invertible. I

Problem 8.2.2. A square matrix A is said to be upper triangularizable if there exists an invertible matrix C (called an upper triangularizing matrix for A) such that C-'AC is an upper triangular matrix. Show that every matrix A is upper triangularizable and that if C is an upper triangularizing matrix for A, then the diagonal entries of C-1 AC consist of the eigenvalues of A (counted with their multiplicities).

8. Positive Matrices

252

Solution: Let A = [aij] be an arbitrary n x n matrix with complex entries. Also, let {e1, e2, ... , en} be the standard orthonormal basis of Cn. Then A defines an operator T : Cn - Cn via the formula T(x) = Ax for each x E C". So, for each k we have

n

T(ek) = Aek =

aikei.

(t)

i=1

By Theorem 8.4 there exist T-invariant subspaces V1 C V2 C C Vn = C" such that dim(V;) = i for each i. In particular, there exist linearly independent vectors fl, f2, ... , fn such that (fl, f2,. .., fi} is a basis for V for each i. If we consider the basis (fl, f2, ... , f, I of Cn, then we can write n

T(fj) _ Ebkjfk

(tt)

k=1

for all j = 1, 2, ... , n. Since T (f j) E Vj for each j and { fl, ... , f j } is a basis of Vj, it follows that bij = 0 for i > j. That is, the matrix B = [ bij ] is upper triangular. Let *

* *

b22

* *

* *

0

0

bn-ln-1

0

0

0

B= bnn

Next, f o r each j choose scalars c i j (i = 1, ... , n) such that n

fj = E cjjei , i=1

and note that the matrix C = [cij] is invertible. It follows from (t) that

T(fj)

ckjT'(ek) k=1

ckj k=1

aikei) =

( i=1

F'(E aikckj )ei i=1

,

k=1

and from (tt) we get

bkj(>2cikei) _E(Ec$kbkj)ei

T(fj)=Ebkjfk= k=1

k=1

i=1

i=1

k=1

Therefore, E1C=1 aikckj = Ek=1 c kbkj for all i and j. In matrix notation this is

equivalent to AC = CB or C-1 AC = B, which shows that the matrix A is indeed upper triangularizable! To finish the proof now look at part (i) of Problem 8.1.3. 1

Problem 8.2.3. Assume that a matrix A E Mn(C) has the distinct eigenvalues J11i ... , .\k with multiplicities ml,... , mk, respectively. If q(.) is a polynomial, then show that the characteristic polynomial of the n x n matrix q(A) is Pq(A) (A) = [A - q(,\,)]

m,

[A - q(1\2)]

12 ... [A - q(A,, )] mk

8.2. Operators on Finite Dimensional Spaces

253

Solution: Let A be an n x n matrix having distinct eigenvalues A1, ... , Ak with multiplicities ml, ... , mk, respectively. By Problem 8.2.2, we know that the matrix A is similar to an upper triangular matrix, and so we can suppose that A is itself an upper triangular matrix. In particular, the diagonal entries of A are the eigenvalues Ai each appearing mi times. It is easy to see that q(A) is also an upper triangular matrix having diagonal entries q(A1),... , q(Ak) with each element q(Ai) appearing mi times. This implies the desired formula. I

Problem 8.2.4. This problem describes the minimal polynomial of an operator T : X -> X on an n-dimensional complex Banach space. (a) Show that the spectrum of T is finite and consists of eigenvalues. Moreover, i f we let a(T) = {A1, ... , Ak}, then show that k < n. (b) Show that there is a non-zero polynomial p(A) such that p(T) = 0. (c) Prove that some root of the polynomial p(A) in (b) must belong to the spectrum of T; and so the spectrum of T is non-empty. (d) Assume that ir(A) is a polynomial of minimal degree and leading coefficient one satisfying 7r(T) = 0. Show that ir(A) = (A - A1)°' ... (A - Ak)ak

,

where, as in (a), a(T) _ {A1, ... , Ak} and each ai is the ascent of the operator Ai - T. (This establishes that the polynomial 7r(A) is uniquely determined; it is referred to as the minimal polynomial of the operator T.) (e) Show that a polynomial q(A) satisfies q(T) = 0 if and only if q(A) is divisible by the minimal polynomial 7r(A) of T. In particular, prove that two polynomials i,li(A) and 4'(A) satisfy tli(T) = 0(T) if and only if there exists a polynomial 0(A) such that q(A) = i(A) +0(A)7r(A). Solution: (a) Here we shall use the following simple fact: An operator on a finite dimensional vector space is invertible if and only if it is one-to-one. Thus, a complex number A belongs to the spectrum of T if and only if A - T is not one-to-one or, equivalently, if and only if A is an eigenvalue of T. Therefore, the spectrum of T consists of eigenvalues. Now notice that eigenvectors corresponding to distinct eigenvalues are linearly independent; see Problem 6.2.2. Since the maximum number of linearly independent vectors in X is n, it follows that a(T) consists of at most n complex numbers.

(b) Let {x, ... , x } be a basis of X. Since for each i the n+1 vectors xi, Txi,... , T"xi must be linearly dependent, there exists a non-zero polynomial pi (,\) such that pi(T)xi = 0. If p(A) = pi(A)p2(A) ...p,, (A), then p(A) is a non-zero polynomial satisfying p(T)xi = 0 for each i. Since {x1,. .. , x } is a basis, the latter implies p(T)x = 0 for each x E X, i.e., p(T) = 0. (c) Assume that p(,\) is a non-zero polynomial satisfying p(T) = 0, and write

p(A) =a(A-ri)kt ...(A-rm)km,

8. Positive Matrices

254

where r1..... r,,, are the distinct roots of p(A). If ri V a(T) for each I < i < m, then the operator T - ri = -(r, - T) is invertible for each 1 < i < m. This implies that the operator 0 = p(T) = a(T - r1)k1 ... (T - rm)k- is also invertible, which is impossible. Hence, r, E Q(T) must hold for some i, and consequently a(T) is non-empty. (Of course, we already know that the spectrum of T is non-empty, but the above proof is direct and elementary.) (d) Let ir(A) he a non-zero polynomial of minimal degree with leading coefficient

one satisfying 7r(T) = 0. Write 7r(A) _ (A - rl)k,...(

r")k-

where r1..... r,,, are the distinct roots of 7r(A). Also, let ai denote the ascent of the operator r, - T. Consequently, N((ri - T)O,) = N((ri - T)k) for all k > ai. We claim that ki < ai for each i. Indeed. if ai < ki holds for some i, then from (T-ri)k'[[J(T-rj)k., r] =7r(T)x=0.

poi

it follows that the non-zero polynomial q(A) =

has degree

(A-ri)Q,

less than the degree of 7r(A) and satisfies q(T)x = (T - ri)c - [fi#i(T - rr)k'x] = 0 for each x E X. That is, q(T) = 0. This contradicts the minimality property of 7r(A), and so ki < ai holds for each i. Next, we claim that k, = ai for each i. To see this, assume by way of contradiction that k, < a, for some i; without loss of generality we can suppose that 96 0. This i = in. Fix some x E N((r,,, - T)',-+') such that y = (r,,, implies (r,,, - T) y = 0 or T y = r,,, y. and consequently m-1

m-1

0 = 7r(T)x

(ri-T)k`!(r,,,-T)kmx= ]I (ri-T)k,y

[ i=1

i=1

M-1

[J1 (r,-rm)k.]y00, i=1

which is a contradiction. This contradiction establishes that k, = of for each i. It remains to he shown that a(T) = {r1.... , r.). Indeed, if r, 0 o(T) for some f,#,(A-rj)ki has degree i. then r; -T is invertible. and so the polynomial q1(A) = less than the degree of 7r(A) and satisfies q, (T) = 0, which is impossible. Thus. every

ri belongs to the spectrum of T. Finally, let r E a(T) and fix some nou-zero vector

-E X with Tz=rz. If r0 r, for each i. then we have 7r(T)z = ir(r)z A 0. which is also a contradiction. This establishes that o(T) = {r1, ... , r,}. (e) Let q be a non-zero polynomial. If q(A) = p(A)7r(,\), then q(T) = 0. For the be the distinct converse assume that q(T) = 0. Let {r1..... rk} U {rk+1 . C p(T). roots of the polynomial q(A). where {r1..... rk} C Q(T) and {rk+1 . Assume that each root r, has multiplicity rni. Thus, Ti

L.

r,)"`]

q(A) = a i=1

[

H (A - r,)"'.] i=k+1

= g1(A)g2(A)

8.2. Operators on Finite Dimensional Spaces

255

where a is the leading coefficient of the polynomial q, qi(A) = IlkI(A - ri)m, and

q2(A) = of k+i(A - ri)'".. Since the operator q2(T) = aH;'_k+l(T - ri)m, is invertible, it follows that qi (A) is a non-zero polynomial satisfying qi (T) = 0. Now

for each 1 < i < k let A = min{m a; }, where a; is the ascent of the operator lli=1(A - r;)a,. Clearly, r; - T, and consider the non-zero polynomial q3(A) = 7k ql (A) = g3(A)p(A) holds for some non-zero polynomial p(A). Since the degree of the polynomial q3(A) is less than or equal to the degree of the minimal polynomial rr(A)

and q3(T) = 0, it follows that q3(A) = ir(A). Thus, q(A) = g2(A)p(A)7r(A). This shows that q(A) is divisible by 7r(A).

Finally, notice that two polynomials O(A) and V)(A) satisfy 0(T) = 1/i(T) if and only if iJ'(T) - 0(T) = 0. By the preceding part, this is equivalent to having Vi(A) - 0(A) = 8(A)7r(A) for some polynomial 8(A). Therefore, 0(T) = (T) if and only if there exists a polynomial 8(A) satisfying ii(T) = 0(T) + 8(T)ir(T). I

Problem 8.2.5. Let T : X -+ X be an operator on a finite dimensional vector space. If a pair of vector subspaces (Y, Z) reduces the operator T, then show that pT(A) = pTj r (A)pTlz (A). Solution: Let T : X -+ X be an operator on a finite dimensional complex vector space and let (Y, Z) be a pair of vector subspaces that reduces T. Fix a basis

{yi,...,yk} of Y and a basis {z1,...,z,,,} of Z. Clearly, {yi,...,yk,zi,...,z,,,} is a basis of X. Moreover, we have m

k

Tyi = Eaiiyi (j = l.....k)

and Tz,, = EbrpZr (p r=1

i=1

Therefore, the matrices representing the operators T : Y -+ Y and T : Z --+ Z with

respect to the bases (yi,...,yk} of Y and {zi,...,z,,,} of Z are given by

A

=

.'.

aik

b11

b12

bi m

a22

a2k

b21

b22

b2m

ak2

akk

bml

bm2

bmm

all

a12

a21

akl

and

B=

It follows that the matrix representing the operator T: X X with respect to the basis 1{y1, ... , yk, z1, ... , zn, } is the block diagonal (k + rn) x (k + m) matrix

C=

[A

BJ . This implies that AI - C = I AIkO A

0

and so AI,,, - B, ,

pT(A) = det(AI - C) = det(AIk - A) det(AI,,, - B) = pTl,. (A)pr,. (A) , as claimed.

Problem 8.2.6. Let T : X -+ X be an operator on an n-dimensional vector space. If two n x n matrices A and B represent T with respect to two different

bases of X, then show that A and B are similar. Solution: Assume that the two n x n matrices A = [a12J and B = (bij) represent the operator T with respect to the bases

and {fi.f2,...,f},

8. Positive Matrices

256

respectively. That is, n

T(ek) = Aek =

a,ke,

for k = 1, 2.... , n,

(*)

bk, f k

for

j = 1,2, ... , n .

(**)

i=1

and

0

T (f,) = B fi = k=1

Next, for each j choose scalars c,., (i = 1, 2, ... , n) such that

fj =Ec,,ei. i=1

and note that the matrix C = (c;l) is invertible. Using (*), we get n

n

T(fl) = 1: ck,T(ek) _ k=1

ck,

n

(>aike,) = E(Eask%)ei, i=1

k=1

n k=1

1=1

and from (**) we obtain n

n

n

T(f,) k=1

n

n

i=1

k.1

bkj(>ckei) _

bkjfk =

i=1

k=1

Ckbkj)e,-

Therefore, Ek=1 a,kCkj = Ek=1 C,kbk, for all i and j. In matrix notation this is equivalent to AC = CB or A = CBC-1, which establishes that the matrices A and B are similar.

Problem 8.2.7. If S, T : X -+ X are two operators on a finite dimensional vector space, then show that ST and TS have the same characteristic polynomial-and conclude from this that if A and B are two n x n square matrices, then AB and BA have the same characteristic polynomial. Solution: Assume first that S is also a projection, i.e., S2 = S. If Y = S(X) and Z = (I - S)(X). then X = Y ® Z, S is the identity operator on Y, and S vanishes on Z. Assume that Y is a k-dimensional vector subspace. Then there exists a basis B of X such that S has the matrix representation S

_

[1k

0

0 0 '

where Ik is the k x k identity matrix and the "diagonal" 0 is the (n - k) x (n - k) zero square matrix. Now assume that T = [t,.,] is the matrix representation of T with respect to the basis B, and write the matrix T in the block form

_

CD

T- U V

'

where the size of C is k x krand that of V is (n - k) x (n - k). Now note that

ST= f 0 0]

and

[Co TS= IU 01

8.2. Operators on Finite Dimensional Spaces

This implies that for each A PST(A) = det

257

0 we have

AIk - C

-D

0

AIn_k

= det

Alk - C

0

0

_ An_kpC(A)

AIn-k An-kpC(A).

Similarly, for each A # 0, we have pTS(A) = So, PST(A) = prs(A) for each A. That is, ST and TS have the same characteristic polynomial.

We now consider the general case. Let W = S(X), the range of S. and let {Sxl .... , Sxk } be a basis of 14". Pick vectors yk+1, ... , yn such that the set of vectors {Sx1, ... , Sxk, yk+1. - -, yn } is a basis of X. Since the vectors Sxl,.... Sxk

are linearly independent, the vectors x1.....xk are also linearly independent. So, we can find vectors xk+1, ... , xn such that {xl, . . . , xk, x1,+1, - - -, xn } is a basis of X.

Next, consider the linear operator R: X - X defined by RSx, = x, for I < i < k and Ry, = x, for k + 1 < i < n. Clearly, R is an invertible operator, and an easy argument shows that (RS)2 = RS. i.e., RS is a projection. By the first part (RS)(TR-1) = RSTR-1 and (TR-1)(RS) = TS have the RSTR-1 and ST have the same characsame characteristic polynomial. Since teristic polynomial, we conclude that ST and TS have the same characteristic polynomial. (See also Problems 6.1.19 and 6.2.9.)

1

Problem 8.2.8. Let V be a finite dimensional vector space. If A is a nonempty subset of V.. then its annihilator Al- is the vector subspace of V defined by

A1={v'EV': v'(a)=0 for all aEA}. Similarly, the annihilator of any non-empty subset B of V* is the vector subspace of V defined by BI = {v E V : b*(v) = 0 for all b' E B}. Assume that T : X -i Y is an operator between two finite dimensional vector spaces. As usual, we define its adjoint operator T*: Y' -+ X' via the duality identity

(Tx,y') = (x,T*y*) = y'(Tx) for all x E X and all y' E Y. Establish the following:

(a) [R(T)]' = Ker(T'). (b) [Ker (T)] -L = R(T`).

(c) dim R(T') = dim R(T). Solution: (a) For this identity note that: y' E [R(T)]1

c-

(Tx, y') = (x,T'y') = 0 for each x E X

e T' y' = 0 y' E Ker (T') . (See also Theorem 2.13.)

(b) Fix X' E R(T'), and then pick some y' E Y' such that x' = T'y'. Now for each x E Ker(T) we have x* (x) = T'y'(x) = y'(Tx) = 0. This shows that x' E [Ker(T)J1, and so R(T') 9 [Ker(T)]1.

8. Positive Matrices

258

Now let x* E [Ker (T)]1, and assume by way of contradiction that x* V R(T*). Then, by the Separation Theorem, there exists some x E X satisfying x*(x) 0

and (x,T*y*) = (Tx,y*) = 0 for all y* E Y*. It follows that Tx = 0 or that x E Ker (T). But then x* E [Ker (T)]1 implies x* (x) = 0, which is a contradiction. Therefore, x* E [Ker(T)]-'- implies x' E R(T*) or [Ker(T)]1 C R(T*). Consequently, [Ker (T)]1 = R(T*).

(c) Let n = dim X = dim X*. Fix a basis {xl,... , xp} in Ker (T), and then choose vectors xp+1, .... X. such that {x1, ... , xp, xp+l, ... , xn} is a basis of X. Next, select vectors xj E X (j = 1, ... , n) satisfying xj* (xi) = o,, for all i and j. Clearly, {xi, x2, ... , xn } is a basis of X*.

If X = E , .fix,, then Tx = E,"_p+1A,Tx,. This shows that

dim R(T)
(*)

Furthermore, we have {xv+1,... , x;, } C [Ker (T)] 1 = R(T' ), where the last identity holds true by virtue of part (b). Consequently,

n - p < dim R(T*).

(**)

From (*) and (**), we see that dim R(T) < dim R(T*). Applying this inequality to the operator T*: Y* -e X' and taking into account that T** = T, we get

dim R(T) < dim R(T*) < dim R(T**) = dim R(T).

Therefore, dim R(T) = dim R(T*) holds true. (Regarding this problem, see also 1

Theorem 2.13.)

Problem 8.2.9. If T : X -4 Y is an operator between finite dimensional vector spaces, then show that

dim Ker(T) + dim R(T) = dim X. Solution: A glance at the inequalities (*) and (**) in the solution of the preceding problem yields dim R(T) = dim R(T*) = n - p, where n = dim X and p = dim Ker(T). Therefore,

dim Ker(T) + dim R(T) = p + (n - p) = n = dim X.

The result can be obtained directly from (*) (and thus avoiding the use of the Separation Theorem). The only thing that is needed here is to observe that the vectors Txp+1,...,Txn (introduced at the beginning of the solution of part (c) in the preceding problem) are linearly independent. To see this, notice that if Exp+1 A-Txi = T(E p+1 A,xi) = 0 holds true, then r," p+1 A,x, E Ker(T), and

so we must have En p+1 \ixi = ? 1 pixi for appropriate scalars pi. Since the

set of vectors {x1, ... , xp, xp+1, ... , xn } is a basis of X, the latter implies 1i = 0 for each i with p + 1 i S n. This guarantees that {Txp+ 1, ... , Tx,,} is a basis of I R(T) and n - p = dim R(T).

Problem 8.2.10. Use Theorem 8.9 to present an alternate proof of the Cayley-Hamilton Theorem 8.3.

8.2. Operators on Finite Dimensional Spaces

259

Solution: Let T : X -» X be an operator on a finite dimensional complex v e c t o r space. Also, let Al, ... , Ak be the distinct eigenvalues of T with multiplicities M I ,- .. , Mk, respectively. According to Theorem 8.9. we can decompose X in the form X = Xl e X2 e e Xk, where each vector subspace X, has dimension m,, is T-invariant, and the operator T - A,: X, -' X; is nilpotent. In particular. we have (T - A,)''- = 0 on X,.

Now if Si = 11r#,(T - A,)',, then pr(T) = flk=1(T -

S,(T - A,)'" then

for each i. k

k

k

S,(T - A)"'.x, = F S,(0) =0.

Pr(T )x = I: pr(T)x, _ ,=1

=1

t=1

Therefore, pr(T) = 0. In particular, if A is any matrix representing the operator T. then pr(A) = 0 is true, proving the Cayley-Hamilton theorem.

Problem 8.2.11. Show that every square matrix is similar to its transpose. Solution: Let A be an n x n square matrix, and let A1.... , Ak be the distinct eigenvalues of A with multiplicities ml, ... , mk, respectively. As usual, in terms of the basis {e1.... , e" } of the standard unit vectors. the matrix A defines an operator T: C" C" via the formula T(x) = Ax for all x E C". According to Theorem 8.9. there exist T-invariant subspaces X 1, ... , Xk of X such that:

(1) dim(X,) = m, for each i = 1.2,...,k. eX2b...,EXk. (2) Cn =X1 (3) The operator T - A,: X, -p X, is nilpotent for each i = 1.2.... , k. F o r each 1 < i < k let {fl',

B = {fI

.

.

. .

.f , ,)} be a basis for X,. Clearly, the set of vectors ,,,

f2,....frn,41 , f2,."' f"12,..., fk, fk,..., fm"}

is a basis for C". Since the operator T leaves each X, invariant, it is easy to see that the matrix representing the operator T with respect to the basis B is a bock-diagonal matrix of the form diag(A1, A2, ... , Ak), where each A, is an in, x m, square matrix. In particular, it follows that the n x n matrices A and B = diag(A1, A2, ... , Ak) are similar. This implies that At and Bt = [diag(A1. A2, ... , Ak)]t = diag(Ai, A2..... Ak)

are similar matrices. Next, let S: C"

C" be the operator defined via the matrix

Bt = diag(At. A2.... , Ak) with respect to the basis B. Then S leaves each X, invariant and S - A, : X, -. X. is quasinilpotent for each i = 1..... k. This follows from the fact each T - A, has the same property which guarantees that the matrix A, - A, (and hence A, - A,) is quasinilpotent. This implies that the matrices B and Bt have a common Jordan form, which guarantees that B and Bt are similar matrices. Thus, we have shown that A - B - Bt - At, and so the matrix A is similar to its transpose At. I

8. Positive Matrices

260

Problem 8.2.12. Let T : Y - Y be a bounded operator on a Banach space. A vector y E Y is said to be cyclic (or more precisely T-cyclic) if the vector subspace generated by {y,Ty,T2y,T3y,...} is dense in Y. For an operator T : X - X on an n-dimensional vector space establish the following.

(a) A vector x E X is T-cyclic if and only if {x, Tx,... ,T"-ix} is a basis of X.

(b) The operator T has a cyclic vector if and only if T has a matrix representation of the form:

000 1

00

0

1

00 00

0.00

-ao

-al -a2 (*)

0 0 0... 0 0 -an_3 000 0 0 0 ...

1

0

0 -an-2 1

-an-1

(c) If x is aT-cyclic vector and 71x = -aox-a1Tx-

-an_iTn-ix,

then the characteristic polynomial of T is given by pi'(A) = An +

an-1'kn-1 + an-2An-2 +... + aiA + ao.

Solution: (a) Since every subspace of a finite dimensional space is itself finite dimensional (and hence closed for the Euclidean topology), it follows that a vector x E X is T-cyclic if and only if the vector subspace generated by its orbit {x,Tx,T2x,...} is all of X. Now fix a vector x E X. If {x,Tx,...,T"-ix} is a basis of X, then it should be clear that the vector subspace generated by {x, Tx, T2x, ... } coincides with X. For the converse, assume that the vector subspace generated by {x, Tx, T2x, ... }

coincides with X. If {x,Tx,...,T"-1x} is not a basis of X, then it is easy to see that there exists some 0 < k < n - I such that Tkx lies in the span of the k vectors x, Tx, ... , Tk- lx. This implies that the vector subspace generated by {x, Tx, T2x, ... } has dimension at most k < n, and so it cannot coincide with X, which is a contradiction. Hence, if x is T-cyclic, then {x, Tx,... , T"- 1x} must be a basis of X. (b) Assume that the operator T has a cyclic vector, say x. By (a) this implies that {x, Tx,... , T"-lx} is a basis of X. If we let bi = x, b2 = Tx,... , bn = 7'-'X, then with respect to this basis we have: Tb1

= b2 = (0,1,0,0,...,0),

Tb2 = b3 = (0,0,1,0,...,0),

Tbn-1 = bn = (0, 0, 0, 0, ... ,1) , Tbn = (-ao, -ai, -a2, -a3, ... , -an-1) .

8.2. Operators on Finite Dimensional Spaces

261

These vectors written as columns yield the matrix representing the operator T with respect to the basis {bl, b2,. .., bn }. This matrix is of the desired form (*). For the converse, assume that the operator T has a matrix representation of the form (*). This means that there exists a basis {b1, b2, ... , bn } such that:

Tb1 = (0'1'0'0'...'0)=b2, T2b1

= Tb2 =

(0, 0,1, 0, ... , 0) = b3 ,

T3b1 = Tb3 = (0,0,0,1,...,0) = b4, Tn-lb1 =Tbn-1 = (0,0,0,0,...,1) = bv, Tnb1 = Tbn = (-ao, -a1, -a2, -a3, ... , -an-1)

So, if x = b1, then {x,Tx,...,Tn'1x} = {b1,b2,...,b, } is a basis of X. By part (a), this guarantees that x is a T-cyclic vector. (c) Assume that x is a cyclic vector for the operator T. Then, according to part

(a), the set {x,Tx,...,Tn-1x} is a basis of X, and so (by part (b)) the operator T has the matrix representation with respect to this basis given by (*), where Tnx = -aox - a1Tx - - an-17' 1x. So, what needs verification is the fact that the characteristic polynomial of an n x n square matrix A of the form (*) is given by pA(A) = An + an-1 An-1 + an-2An-2 + ... + al A + ao. The proof is by induction. It is straightforward to prove the validity of the claim for n = 1 and n = 2. So, for the induction step, we assume that the claim is true for all n x n square matrices of the form (*). Now assume that A E Mn+1(C) is of the form (*), and let -ao, -a1, ... , -an be the n + 1 scalars of the (n + 1)th column of A. Therefore, the characteristic polynomial of A is:

-1

A

0 0

0

-1

A

1A

PA(A) = det(AI - A) = dot 0 0 0

0

...

0 0 0

0 0 A-1 0 0

0 0

0

0 0 0

ao a1 a2

0

an-2

A

an_1

-1 A+an

If we expand this determinant with respect to the first row, then we have PA (A) = A det B - ao (-1)nt1 det U,

where det B is the characteristic polynomial of a matrix of the form (*) having last

column with scalars -al, -a2, ... , -an, and U is an n x n upper triangular matrix having each diagonal entry equal to -1. Clearly, det U = (-1)n and our induction hypothesis implies that det B = An + an A"-1 + an An-1 + + a2A + a1. Therefore, +an_1An-2+...+a2A+a1) -ao(-1)n+1(-1)n PA(A) = = An+1+anAn+an-An-1+...+alA+ao. A(An+anAn-l

This completes the induction, and the solution of the problem is finished.

I

8. Positive Matrices

262

8.3. Matrices with Non-negative Entries Problem 8.3.1. Let A = [ai,] be an n x n matrix with real entries. Show that A (as a regular operator on Rn) has a modulus and that I AI = []ai, I]

Solution: We have observed before that an n x n matrix C = [cij] with real entries defines a positive operator on Rn if and only if c,j > 0 for all i and j. This implies that if B = [b,j] and C = [ci1] are two n x n matrices with real entries, then B > C if and only if b,3 > ci3 for all i and j. Now let A = [a,3] be an n x n matrix with real entries, and let. B = [1a,j]]. By the above B > A and B > -A both hold. Now if another n x n matrix C = [c,3]

with real entries satisfies C > A and C > -A, then c,j > a,j and cj > -a,3 for all i and j. Therefore, c,j > 1a,3I for all i and j, and so C > B. Consequently, [AI = A V (-A) = []a,3[] in the Dedekind complete Riesz space M,,(lR). (See also Problem 8.1.11.)

Problem 8.3.2. Show (by means of an example) that although the spectral radius of a (non-zero) positive matrix is an eigenvalue, it need not have a strictly positive eigenvector. Solution: Consider the 2 x 2 positive matrix A = 10 OJ . Clearly, Q(A)

1, 0},

and so r(A) = 1. Now an easy computation shows that (up to a scalar multiple) the vector (1, 0) is the only positive eigenvector of A corresponding to the eigenvalue 1. This vector is not strictly positive. (Notice that this conclusion does not contradict the Perron-Robenius Theorem 8.26 since the matrix A is not irreducible.) I

Problem 8.3.3. Assume that a non-zero positive matrix A commutes with a strongly positive matrix. Show that if aA > 0 and Ax > .1Ax for some x > 0, then Ax = aAx. Solution: Assume that A is an n x n non-zero positive matrix that commutes with an n x n strongly positive matrix B. Also, suppose that AA > 0 and that Ax > )IAx holds for some x > 0. To see that Ax = AAX is true, assume by way of contradiction that Ax > aAx.

Let y = - (Ax - aAx) > 0. Then Ax = AA(x + y) > 0, and since B is strongly positive we have B(x + y) >> 0 and By >> 0. Since A is a non-zero positive matrix, we have BAy = ABy > 0, and so Ay > 0 is true. This in turn implies that ABy = BAy > 0. Now note that

AB(x + y) = ABx + ABy >> ABx = BAx = \AB(x + y)

.

In particular, there exists some e > 0 satisfying AB(x + y) >- (AA + e)B(x + y). This implies aA + e < AA, which is impossible. Hence, Ax = aAx. 1 aij (the sum of all Problem 8.3.4 (Frobenius). For each i let ri = elements in row i) and put p = mine
8.3. Matrices with Non-negative Entries

263

Solution: Pick some x > 0 such that Ax = r(A)x, and then choose some k such that xk = maxi <, 0. Then we have n

n

r(A)xk = (Ax)k = E akjxj < (E ak,)Xk < Rxk . 1

j=1

1j=1

This implies r(A) < R.

Nowife=(1,1,...,1),then foreach 1
Problem 8.3.5 (Gershgorin [311). This exercise presents a simple but very useful "approximation" of the spectrum of an arbitrary matrix. Let A = [aid] be an n x n matrix and let r; _ Ejn#.jatjj and Cb = {z EC: Iz - at;I _< ri} for each i. Show that a(A) C Ut` Cs. 1

Solution: Fix an eigenvalue A of A, and then pick some x = (xl,... , x,) 54 0 satisfying Ax = Ax. Choose some k such that Ixk I = maxi <: 0. From Ask akjxj, we get (A - akk)xk = Ej#k akjxj. This implies

IA-akkl=IFakjkl 5 F lakjljXk j#k

j#k

j0k

and so A E Ck. Consequently, a(A) C U", C.

Problem 8.3.6. If an arbitrary n x n matrix A = [a;3] E ..(C) satisfies laid > E j,6; Iaij l for each i, then show that A is invertible. Solution: Notice that Problem 8.3.5 implies that A = 0 does not belong to the spectrum of A. Thus, A is invertible.

I

Problem 8.3.7. A non-negative n x n matrix A = [a=j] is said to be a Markov matrix if the sum of the elements in each row equals one, i.e., if atj = 1 for each i. (This is equivalent, of course, to saying that A defines a Markov operator on R'2.) Show in four different ways that every Markov matrix A satisfies r(A) = 1 by using: (a) Problem 8.3.4, (b) Lemma 8.14(4),

(c) Gershgorin's theorem (see Problem 8.3.5). and (d) the spectral radius formula. Solution: (a) Following the notation of Problem 8.3.4, we have ri _ E;=1 a, = 1 for each i. Therefore, r = mini<=
(b) If e = (1, r(A) = 1.

1), then Ae = e. Now a glance at Lemma 8.14(4) yields

8. Positive Matrices

264

(c) We know that r(A) is an eigenvalue of A. Therefore, by Problem 8.3.5, there exists some 1 < i < n such that

r(A)-a,,

Ir(A)-a,

a,J = 1-a,,. i#=

This implies r(A) < 1. Since Ae = e, we conclude that r(A) = 1. (d) Equip R" with the sup norm II - II, i.e.. IIxII = maxi<,
Problem 8.3.8. Let A be an n x n non-negative matrix. Then: (a) A defines a lattice homomorphism on Rn if and only if each row has at most one non-zero entry. (b) A is interval preserving if and only if every column of A has at most one non-zero entry. Solution: Let A = [a,)] be a non-negative matrix. (a) Assume first that A defines a lattice homomorphism on R" equipped with its standard ordering. Since the (column) unit vectors el, .... e are pairwise disjoint, it follows that the non-negative (column) vectors Ael,.... Ae" are also pairwise disjoint. But Ac, is simply the ph column of A, and so the matrix A has pairwise disjoint columns. This implies that A has at most one non-zero entry in each row. For the converse, assume that A has at most one non-zero entry in each row. Now consider a row i of A. If A has a positive entry in row i, then let j, be the column where this happens, i.e., a,3, > 0. If row i does not have a positive entry, let j, = 1. Note that Ax = (al,, xi, , a2j,a,2, ... , a"i xi ). So. if x A y = 0, i.e., x, A y, = min{x,, y, } = 0 for each i. then Ax A Ay =2 (a1J, x, . a2)2 x,2 .... a.1. x)n ) A (a l7, y,,, , 0212 y}2 ..... a.,. yj. )

= ((((alnx,,) A (a13,y,,), (a212x,2) A (a2J2y12)...., (0.,i,.xj..) A (a"1..yi_)) ` \a19s (xi A yl ), x272 (x2 A y2), ... , a,,, (x A y")) = 0.

This shows that A defines a lattice homomorphism on R". (b) The direct proof of this claim is laborious. The easiest way to see its validity

is by using part (a) in conjunction with Theorem 1.35 (see also Problem 4.3.11) asserting that: The matrix A defines a lattice homomorphism if and only if its transpose At defines an interval preserving operator. 1

Problem 8.3.9. If an n x n positive matrix A = [a;j] satisfies o(A) = {1}, then show that at, = 1 for all i. Use this and the preceding exercise to show that if a positive matrix A with a(A) = {1} defines either a lattice homomorphism or an interval preserving operator, then A = I, the identity matrix.

8.4. Irreducible Matrices

265

Solution: Assume that a non-negative n x n matrix A = (a,3] satisfies o(A) = {1}.

Then the matrix B = (A - I)2 satisfies a(B) = {0} or \, = A2 = ... = an = 0,

where Al, a2, .... ,\n are the eigenvalues of the matrix B. In particular, we have

tr(B) = 1 ai = 0. Clearly, bi, = (aii - 1)2 + Ek#i aikaki for each i. Since A has non-negative entries and _i 1 b,i = tr(B) = 0, it follows that (a - 1)2 = 0 for each i. That is, a,i = 1 for i = 1, ... , n. (The same formula for the trace of B also shows that for each i we have akak, = 0 for all k # i.) Now assume that, in addition to a(A) = {1}, the matrix A defines either a lattice homomorphism or an interval preserving operator. Then, according to Problem 8.3.8, the matrix A either cannot have more than one non-zero entry in each row or else it has no more than one non-zero entry in each column. This shows that in both cases A must coincide with the identity matrix. I

Problem 8.3.10. Let A be an n x n positive matrix. If A is nilpotent, i.e., r(A) = 0, then show that there exists some vector x > 0 such that Ax = 0. Solution: Assume that A > 0 and A" = 0 holds for some p > 2. Now assume by way of contradiction that for each x > 0 we have Ax > 0. That is, A is a strictly positive matrix. This easily implies that A" = 0 is likewise a strictly positive matrix, which is impossible. Thus, there must exist some x > 0 such that Ax = 0. 1

8.4. Irreducible Matrices Problem 8.4.1. For a vector subspace J of C" show that the following statements are equivalent.

(a) J is an ideal, i.e., J = Jo ® 2Jo, where JO is an ideal in Rn. (b) J is a band, i.e., J = B E D where B is a band in R". (c) There exists some non-empty subset I of 11, 2, ... , n} such that

J={(zl,...,z")ECn: zi=0 for all iET}. Solution: The solution is based upon the structure of ideal and bands in R". We shall establish the following basic property.

A vector subspace V of R" is an ideal if and only if it is a band-and also, if and only if there exists a subset Z of {1,2,...,n} such that

V={xERn: x,=0 for all iE2}. To establish this basic property, let V be an ideal in R". If V = {0} or V = R", then take I = 11, 2, ... , n} or I = 0, respectively. So, assume that V is a proper subspace of R". We define the set I as follows:

I={iE{1,2,...,n}: xi=0 for all xEV}. We also let W = {x E R": x, = 0 for all i E Z}. It should be clear that W is an ideal in R" and that V C W. We claim that V = W.

8. Positive Matrices

266

To see this, let x E W. Notice that if i Z, then there exists some y E V such that y, 0 0. From the inequalities 0 < e; < i1 jyj and the fact that V is an ideal in R", we infer that each unit vector e; belongs to V for each i V Z. But then we have x = E 1 xie; _ E; ji x;e, E V. This shows that W C V, and so V = W. The claim that W is also a band follows from the simple fact x° -Q- x in R" implies x° - x; in R for each coordinate i. Now the equivalence of (a) and (b) should be obvious. For the equivalence of (a) with (c) notice that if JO = {x E R": x; = 0 for all i E Z}, then

J={(zl,...,zn)EC": z;=0 forall iEZ}=Jo®zJo. This completes the solution of the problem.

Problem 8.4.2. Show that an n x n positive matrix A is irreducible if and only if At is irreducible. Solution: Notice that for any n x n matrix A we have = Also, observe that a matrix B E Mn(R) is strongly positive if and only Bt is strongly positive. Now use statement (3) of Theorem 8.18 to see that (I+At)"'1

(I +

[(I+A)n-1]e

is strongly positive (I + At)n-1 is strongly positive At is irreducible,

A is irreducible

A)"-1

as claimed.

Problem 8.4.3. For positive n x n matrices A and B establish the following.

(a) If Ak is irreducible for some k > 1. then A is irreducible. (b) If A is irreducible, then A + B is irreducible. (c) If A and B are irreducible, then AB and BA need not be irreducible. Solution: (a) Assume that Ak is irreducible for some power k > 1. Now let J be an ideal in C" such that A(J) C J. Then Ak(J) J is also true. So, by the irreducibility of Ak, we have either J = {0} or J = C". Therefore, A is an irreducible matrix.

(b) Assume that A is irreducible and that B is a non-negative n x n matrix. Take any ideal J such that (A + B)(J) C J. Since 0 < A < A + B. it follows that A(J) C J. The irreducibility of A implies that either J = {0} or j = C". So. the matrix A + B is irreducible.

Then AB = BA = I 0 0J = I. This shows that A and B are both irreducible and that AB = BA (the identity matrix) is not an

(c) Let A = B = I 1

irreducible matrix.

0J

.

I

Problem 8.4.4. Show that a non-negative matrix A is irreducible if and only if the matrix I + A defines a Krein operator.

8.4. Irreducible Matrices

267

Solution: Recall that a positive operator T : E -. E on an AM-space with unit is said to be a Krein operator if for each x > 0 there exists some k > 1 such that Tkx is an order unit. Also, keep in mind that the order units of R" are simply the strictly positive vectors in R", and that R" with the sup norm is an AM-space

with unit e = (1, 1,...,1). Now assume that A is an irreducible matrix and let x > 0. According to statement (3) of Theorem 8.18 the matrix (I + A)n-' is strongly positive. This implies (1 + A)'-'x >> 0, and so I + A is a Krein operator. For the converse, suppose that I + A is a Krein operator, and let an ideal J

in C" satisfy A(J) C J. We claim that J must be a trivial ideal, that is, either J = {0} or J = C". To see this, assume that J 91{0}, and fix some 0 < x E J. It is easy to see that (I + A)kx E J for each k > 1. Since I + A is a Krein operator, there exists sonic k > 1 such that (I + A)kx. is an order unit. Therefore, .1 contains an order unit, and since J is an ideal, it follows that J = C". This shows that A is an irreducible matrix. lI

Problem 8.4.5. Use Problem 8.4.4 and Theorem 9.42 to present an alternate proof of the fact that the spectral radius of a positive irreducible matrix has a unique (up to scalar multiples) strictly positive eigenvector. Solution: Let A be an irreducible positive matrix and assume that Ax = Aox for some Ao E C and some x > 0. Since (according to Lemma 8.17) A does not have zero columns, it follows that the vector Ax has at least one positive coordinate, say the i`h one. This implies Aoxi E R and Aoxi > 0. Hence, A0 is a positive eigenvalue of A.

Next, notice that (I + A)x = (1 + Ao)x. Since I + A defines a Krein operator on R" (see Problem 8.4.4), it follows from Theorem 9.42 that:

(1) 1+Ao=r(I+A)=1+r(A). (2) The vector x is strictly positive and (apart from scalar multiples) is the only positive eigenvector of I + A. So, (1) implies Ao = r(A), and (2) shows that the vector x is strictly positive and (apart from scalar multiples) is the only positive eigenvector of A.

Problem 8.4.6. Consider a positive irreducible matrix A = (ai3] and assume that At = [a,j(t)] is a positive matrix-valued function defined for each

t in some subinterval (a, b) of R such that for some to E (a, b) we have aij(t) = aij for each (i, j). Establish that r(At) = r(A). Solution: To show that limt.t r(At) = r(A), we must show that if t" - to, then there is a subsequence {sn} of {tn } such that limn., r(A,,,) = r(A). So, fix a sequence {tn} C (a,b) such that to -' to. By Theorem 8.11, for each n there exists a positive unit vector xn such that r(At )xn. Since the closed unit ball of R" is compact. there exists a subsequence {xk,, } of {x,, } such that yn = xk,, -> y > 0. Let Sn = tk,,, and note that

lim r(A,,, )yn = lim A,,, y,, = Ay > 0, noo n-00

8. Positive Matrices

268

where Ay > 0 is true by Corollary 8.21. This implies that there is some AO > 0 such

that limn. , r(A,n) = AO. and so Ay = Aoy. By Corollary 8.23, Ao = AA = r(A).

That is, we have limn., r(A,,,) = r(A), and thus limit, r(At) = r(A).

8.5. The Perron-Frobenius Theorem Problem 8.5.1. Show that n arbitrary complex numbers 2b' . , z satisfy E i= , zi I = E i= 1 Izi l if and only if there exists some angle 0 such that zi =jzi Ieie for each i = 1,... , n.

Solution: If zi = Iz,leee for each i = 1,. .. , n, then it should be clear that

Et

We shall establish the converse by induction. First we I Et 1 zi I = 1 I zi 1. shall establish the conclusion for n = 2. To this end, assume that two complex numbers zi = Izlleie1 and 22 = Iz2le:s2, where 0 < 01, 02 < 21r, satisfy 1z1 + 221 = Iz1 I + Iz21. Squaring both sides, we get Izl + 2212 = (Z1 + z2)(11 + z2) = (fziI + Iz21)2 or 14112 + 12212 + 21 z2 + 21z2 = I Z112 + 12212 + 21 zi l

. 1x21

+ e'(e2`z) = 2. Letting z = or z2 - 2z + I = (z - 1)2 = 0. This implies z = a/(et-82)

expression yields

Simplifying the latter we get z + 2 = 1, and so 81 = 02.

e'(B1-ea), e'(8I'e2)

Therefore, our conclusion is true for n = 2. For the induction step, suppose the claim is true for some n, and assume that the non-zero complex numbers z 1, .... zn, zn+l satisfy I Ee 11 zi I = E - Iz,1. From the inequalities n+1

n+1

n+1

n

n

Elz'l=IEZiI=I(EZ) +Zn+il
i=1

i=1

we see that

zi) + zn+1l = IF-"I=1 zi4 + Ion+1l- So. according to the case

I

n = 2, there exists an angle 0 < 0 < 21r such that F-t 1 zi = j E 2n+1 =

(*)

i=1

loll

zi I eie and

Izn+lle'8.

A glance at (*) also shows that I _i 1 ziI = induction hypothesis, there exists an angle 0 < each i = 1, ... , n. Now note that Izil)e,e

z, le.e

So, according to our < 21r such that zi = Izile'-o for Et11zi1.

=

zi lzil)e'O. iTl i=1 =1 This implies e`9 = e'm. Hence, ch = 0, and the induction is complete. i=1

=I

i=1

Problem 8.5.2. Consider the polynomials: (i)

1, (ii) A3 - A, and (iii) A4 + a2 - 1.

Which of the above polynomials are the characteristic polynomials of positive irreducible matrices?

8.5. The Perron-F
269

010

Solution: (i) Consider the matrix A = 0

1

0

0

0

1+

1

0 0

1

A+A2+A3 = 0 1

0

0

0

1

0

0

1

0

0

, and notice that 0

0 0

1

1

1

1

1

1

1

0 0

1

1

1

1

1

0+0

By part (4) of Theorem 8.18 the positive matrix A is irreducible. Now a simple calculation shows that A

pA(A)=det

(ii)LetB=

0

2

2

0

0

-1

0 -1

0

-1 =A3-1.

A

A

0

0

0

_

01

1

1

0

0

2

1

B2=2 0 1

0

1

0

0

1

2 0

0

. Then we have

,

1

and so B + B2 is a strongly positive matrix. This implies that B is an irreducible matrix. Now notice that Ps(A) = det

-

-

A

-

0

-

A

0

=

A3

- A.

A

(iii) Let p(A) = A4 + A2 - 1. Solving the quadratic (A2)2 + A2 - 1 = 0 yields Consequently, the roots of the polynomial p(A) are A2 = -li Al=

-12

s,

A2=_

-12 s,

VW5s, and A4=- 125z.

Therefore, r(A) = 1A31= IA41= 1 s Since r(A) is not a root of p(A), it follows that the polynomial p(A) = A4 + A2 - I cannot be the characteristic polynomial of any irreducible positive matrix. (In fact, it cannot be the characteristic polynomial of any positive matrix!) I

Problem 8.5.3. Consider the following Markov matrix

00 2 A-_ 01 01 00 1

00

1

Show that A is irreducible and satisfies cr

1

2

0 0 0

(A) = (1}.

8. Positive Matrices

270

Solution: Since A is a Markov matrix, we know that r(A) = 1; see Problem 8.3.7 Now the patient reader should verify the following computation: A10 16

6 2 4

6 6

3

1

2

2

6 6

8

4

2

4 2

This, combined with condition (4) of Theorem 8.18, guarantees that A is a positive irreducible matrix. A glance at statement (2) of Theorem 8.29 shows that A is a primitive matrix. So, aper(A) = {r(A)} = {1}. Next, let us discuss the eigenvalues of A. Again. a direct computation yields

PA(A)=det

A

0

01

AI

0

0

-1z -1 a -1

A

This polynomial has the real roots Al = I and -1 < A2 < - 2 . The other two roots A3 and A4 are complex conjugates with modulus less than one. Indeed, if we look at the function f (A) = A3 +A2 +A+ 2, then f(A) = 3A2 + 2A + 1 > 0 for each A, and so (since its discriminant b2 - 4ac = 4 - 12 = -8 is negative) f (A) is strictly increasing. This implies that p(A) has only two real roots.

Problem 8.5.4. This exercise illustrates the Perron-Frobenius theorem. Consider the positive matrix

0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0

A=

00000001 1

1

1

0

1

1

0

1

1

0

1

1

0

1

1

1

0 0 0 0' 0 0 0 0 0 0 0 0 0 0 0 0

(a) Show that A is irreducible. (b) Show that the characteristic polynomial of A is PA(A) = (A2

- 3)(A

- 1)2(.1 + 1)2(A2 + 1).

Obtain from this that on the circle of radius one there are six eigenvalues: 1 with multiplicity two, -1 with multiplicity two, z and -i. (c) Show that the spectrum of A is rotation invariant under 180°. Solution: A direct computation shows that if x = (x1. x2, x3, x4, x5i x6, x7 . xs) is an arbitrary vector in R8, then

Ax= (x5,x6.x7,x8,x1+x2+x3,xi+x2+x4ix1+x3+x4.x2+x3+x4). (t)

8.5. The Perron-FFobenius Theorem

271

(a) Let J 34 C" be an ideal that is invariant under A, i.e., A(J) C J. There exists a non-empty subset I of (1, 2,..., 8) such that J consists of all vectors x E C"

satisfying xi = 0 for each i E Z. Now assume that 1 E I. From (f), we see that 5 E Z. Otherwise, if 5 V 2, then the vector e5 belongs to J while Ae5 = el V J. Once we are convinced that 5 belongs to Z, it follows from (t) that 2,3 E Z, and then 6,7 E Z, and eventually I = { 1, 2, ... , 81. This shows that in this case J = {0}.

Now notice if 2 E Z, then 6 E 2, and so 1 E 2, which (by the preceding argument) implies J = {0}. Similarly, it is easy to see that once I is assumed to be non-empty, then 1 E 2, and so J = (0). This shows that A is an irreducible positive matrix.

(b) The computation of the characteristic polynomial is left as an exercise for the reader.

(c) From the characteristic polynomial, we infer that r(A) = f. Therefore, QP,,(A) = {V3-, -f}. In other words, the eigenvalues on the peripheral spectrum of A are spaced apart by 180°. By statement (5) of the Perron-Frobenius Theorem 8.26, the spectrum of A is rotation invariant under the angle 180°.

Problem 8.5.5. Let A be a primitive matrix and let (A)" -+ S, where r = r(A). Show that each column of S is an eigenvector corresponding to the eigenvalue r. Solution: Let cj denote the jth column of the matrix S. Clearly, cj = Sej. From Corollary 8.30, we know that the vector

cj = Sej = limo(* n-o

"ej

is an eigenvector of A corresponding to the eigenvalue r, i.e., Acj = rcj.

Problem 8.5.6. Show that an n x n positive matrix is primitive if and only if it defines a Krein operator. Solution: Let A be an n x n positive matrix. If A is a primitive matrix, then (according to condition (2) of Theorem 8.29) there exists some m E N such that Am is strongly positive. So, Amx >> 0 holds for each x > 0. This proves that A defines a Krein operator. (Keep in mind that in R" the order units are precisely the strictly positive vectors.) For the converse, assume that A defines a Krein operator. Note first that A is irreducible. To see this, assume that an ideal J ,- {0} is A-invariant, i.e., A(J) C J. Fix some 0 < x E J, and then pick some k such that Akx >> 0. It follows that

A k x E J, and thus the ideal J contains order units. This implies J = C", and hence A is an irreducible matrix. Next, for each j pick up some m, E N such that Ammej >> 0.

To finish the solution, we shall prove that if m = ml + m2 + + m", then A' is strongly positive. Indeed, since A as an irreducible matrix has no zero rows, it follows that Amej = A'"-'".'(Am,ej) >> 0. Since Amej is the jth column of A'", we see that A' has positive entries. Now condition (2) of Theorem 8.29 guarantees that A is a primitive matrix. I

8. Positive Matrices

272

Problem 8.5.7. We have two glasses; call them 1 and 2. Glass 1 contains 100 grams of water and glass 2 contains 100 grams of wine. We transfer 10 grams from glass 1 into glass 2 and stir the mixture very well. Subsequently, we transfer 10 grams from glass 2 to glass 1 and again stir the mixtures in both glasses very well. We repeat this process ad infinitum. If wn and wn denote the amounts (in grams) of wine in glasses 1 and 2, respectively, after the nth stage of this process, show that

lim w,l, = lim wn = 50. n-oo TO Solution: We follow the notation given in the hint. So, at stage n - 1 we have w,l,_ 1 grams

of wine in glass 1 and wn_ 1 grams of wine in glass 2. (The water-wine

mixtures are stirred very well so that the mixtures in both glasses have uniform densities.) Now we perform the two parts required at the nth step. First, we transfer ten grams of the mixture from glass 1 to glass 2, and then stir the mixtures in both glasses very well-so that the mixtures again have uniform densities. After this transfer, the amounts of wine in both glasses are: and W2 = wn-1+1 00-W, I- I W1 = wn-1- 100wn-1 = Next, we transfer ten grams from the (well-stirred) 110-gram mixture of glass 2 to glass 1 (and afterwards stir both mixtures very well). Then the amounts of wine lown_1

in the two glasses are: wn 2

wn

= Wl + 110

--

W2

= Iown-1 + liven-1 + tiown-1 = iiwn-1 + i- wn-1 2 _ 10 _1 1 _ 101 1 1 10 2 110 W2 - 11 W2 - 11 1'Own-l +wn-1 I'iwn-1 + llwn-1

_ 10

These equations can be written in the following matrix form:

rn'] _

I1

111 nn

2 11

wn-1

11

Given that wa = 0 and wo = 100, the preceding recursive formula yields wn1 wn

10

1

11

I1

111

10 11

n

10

Now notice that the positive matrix A =

1

1 1

1o

11

11

is a Markov primitive

matrix. This implies that r(A) = L° + 11 = 1 and Ae = e, where e = (1, 1); see Problem 8.3.7. Therefore, according to Corollary 8.30, we have 10 1 " 0 1 1wW'1'

_

11

W.2

11

10

11

100

1

where c > 0 is a constant. This implies wn - c and wn -+ c. Since wn + wri = 100

holds at each stage n, it follows that 2c = 100 or c = 50. Hence, wn -. 50 and wn - 50.

Chapter 9

Irreducible Operators

9.1. Irreducible and Expanding Operators Problem 9.1.1. Verify the "Implication Scheme" of Theorem 9.6: Strongly Krein N14. 'W" Krein Strongly xpanding Expanding

Ideal Irreducible Weakly Expanding Band Irreducible

Solution: In the implications Strongly Krein

:

Krein Operator

Ideal Irreducible

Band Irreducible

only the implication "Krein Operator Ideal Irreducible" needs verification. To this end, let T : E -+ E be a Krein operator on an AM-space with unit. Also, let .1 96 {0} be a closed T-invariant ideal in E. Fix some 0 < x E J, and then pick some n E N such that T"x is a strong unit. From the T-invariance of J, it follows that T"x E J, and from this we infer that J = E. Therefore, T is ideal irreducible. To see that an ideal irreducible operator need not be a Krein operator, consider the operator T : R2 - IIt2 defined by T(x. y) = (y. r). In the implications Strongly Expanding

Expanding Operator Weakly Expanding

Band Irreducible, 273

9. Irreducible Operators

274

only the implication "Weakly Expanding Band Irreducible" needs verification. This can be established with an argument similar to the one in the preceding case. The implication "Strongly Krein Strongly Expanding" should be obvious. The implication "Krein Operator Weakly Expanding" follows from the fact that a strong unit is always a weak unit. Finally, we shall verify the implication: Strongly Expanding

Ideal Irreducible.

To see this, assume that T : E - E is a strongly expanding operator on a Banach lattice, and let a closed ideal J 34 {0} be T-invariant. Fix some 0 < x E J, and

note that the vector u = Tx E J is a quasi-interior point. So, from Eu C J, it follows that E = E C J or J = E. This shows that T is ideal irreducible. Problem 9.1.2. Show that every weakly expanding operator is strictly positive.

Solution: Let T : E

E be a weakly expanding operator, and let x > 0. Since T is weakly expanding, there exists some n E N such that T"x is a weak unit, and

so T"x > 0. Now from T"-n(Tx) = T"x > 0, it follows that Tx > 0. Thus, T is strictly positive.

Problem 9.1.3. Let S, T : E - E be two arbitrary positive operators on a Banach lattice. If 0 < S < T and S is band (resp. ideal) irreducible, then show that T is likewise band (resp. ideal) irreducible. Solution: Assume that S is ideal irreducible, and let T(J) C J for some closed ideal J. If 0 < x E J, then from 0 < Sx < Tx E J it follows that Sx E J. That is, S(J) C J. Since S is ideal irreducible, either J = {0} or J = E. This shows that T is ideal irreducible. Exactly the same arguments show that if S is band irreducible, then T is also a band irreducible operator.

Problem 9.1.4. If S and T are two positive operators on a Banach lattice and one of them is band (resp. ideal) irreducible, then show that S + T is band (resp. ideal) irreducible. Solution: It follows from Problem 9.1.3 by observing that 0 < S < S + T.

Problem 9.1.5. Give an example of a band irreducible positive operator T that is not a supremum of a family of ideal irreducible positive operators dominated by T. Solution: Let T be any band irreducible positive operator that is not ideal irreducible (for instance, take T in Example 9.7). Then each operator S satisfying 0 < S < T is not ideal irreducible-otherwise, by Problem 9.1.4, T itself would be ideal irreducible. This, of course, implies the desired conclusion.

0

Problem 9.1.6. We know that each central operator leaves every ideal invariant while an ideal irreducible operator does not leave invariant any nontrivial closed ideal. Therefore, the following conjecture seems plausible: If S is a central positive operator and T is an ideal irreducible positive operator, then S A T = 0. Disprove this conjecture.

9.1. Irreducible and Expanding Operators

275

Solution: Let R be any ideal irreducible positive operator and consider T = R+I. Then T is also ideal irreducible. If S = I, then S is a central operator, and obviously

SAT=S. Problem 9.1.7. If a positive operator T on a Banach lattice is either strongly expanding, or expanding, or strongly Krein and S is a strictly positive operator, then show that TS also possesses the corresponding property of T.

Solution: Let S, T : E

E be two positive operators on a Banach lattice with S strictly positive. Fix some x > 0. and note that Sx > 0. (a) Assume that T is strongly expanding. Then TSx = T(Sx) is a quasi-interior point. This shows that TS is strongly expanding.

(b) Suppose that T is expanding. Then TSx = T(Sx) is a weak unit. This shows that TS is expanding.

(c) Assume that T is a strong Krein operator. Then TSx = T(Sx) is a strong unit. This shows that TS is a strong Krein operator.

Problem 9.1.8. If a Banach lattice E has order continuous norm, then show that a positive operator on E is expanding if and only if it is strongly expanding.

Solution: The claim follows from the following fact: In a Banach lattice with order continuous norm a positwe vector is a quasi-interior point if and only if it is a weak unit. Indeed, if e > 0 is a weak unit, then x A ne 1 x holds for each x E E+. and so (by the order continuity of the norm) lix - xAnell - 0 for each x E E+. Therefore, e is a quasi-interior point. The converse is obvious. I

Problem 9.1.9. Show that a positive operator on a Banach lattice with order continuous norm is band irreducible if and only if it is ideal irreducible. Solution: This claim follows immediately from the following result: If a Banach lattice has order continuous norm, then every closed ideal is a band. I

Problem 9.1.10. If T : E -r E is a or-order continuous band irreducible positive operator, then show that:

(a) T is strictly positive. (b) T carries weak units to weak units. (c) E has weak units. Solution: (a) Let x > 0. and let B= denote the band generated by x in E. Now assume by way of contradiction that Tx = 0. Since every y E B+ satisfies yAnx j y and T(y A nx) = 0 holds for each n. it follows from the a-order continuity of T that Ty = 0 for each y E B,+,. Consequently, Ty = 0 for each y E B=. This implies T(B,x) = {0} C B=, i.e., that B= is T-invariant, and so the band irreducibility of

9. Irreducible Operators

276

T guarantees that Bx = E. But then T = 0 which is impossible. Hence, x > 0 implies Tx > 0, and so T is strictly positive.

(b) Let u > 0 be a weak unit in E, i.e., B = E. Since, by part (a), T is strictly positive, it follows that v = Tu > 0, and so B,, 0 {0}. Now fix some x > 0, and note that x A nu j x implies T(x A nu) ? Tx. From T(x A nu) < nv, it follows that Tx E B,,. So, B is T-invariant, and the band irreducibility of T guarantees that That is, v = Tu is a weak unit in E.

(c) Fix some y > 0 and consider the vector u = E. =0 2R T° > 0. We claim that the principal ideal Eu is T-invariant. Indeed, if IxI < \u, then 00

ITxI < Tjxj < aTu = a

2n Tir2 Tn

00

= 2AIITIJ

n=0

2

r

T 1,,

s 2.IIT11u,

n=0

and so Tx E Eu. Therefore, T(Eu) C Eu C Bu. Since T is a-order continuous, it easily follows that T(Bu) C Bu. Now taking into account that T is band irreducible and Bu 0 {0}, we see that Bu = E. This shows that u is a weak unit in E, and consequently E has weak units. I

Problem 9.1.11. For a positive operator T : E --+ E on a Banach lattice establish the following properties.

(a) If T' is ideal (resp. band) irreducible for some k > 1, then T is also ideal (resp. band) irreducible. (b) If T is ideal irreducible, then T2 need not be ideal irreducible. Solution: (a) Assume that Tk is ideal irreducible for some k > 1, and let J be a closed ideal in E such that T(J) C J. Then Tk(J) C J also holds, and so, by the ideal irreducibility of Tk, it follows that J = (0) or J = E. This shows that T is ideal irreducible. A similar argument shows that if Tk is band irreducible for some k > 1, then T itself must be band irreducible.

(b) Consider the positive operator T : R2 - R2 defined by T(x, y) = (y, x). Clearly, T2 = I, and so T2 is neither ideal nor band irreducible. We claim that T is ideal irreducible. To see this, let J 34 {0} be an ideal in

R2 which is T-invariant. Let 0 < (x, y) E J. Then either x > 0 or y > 0. From T(x, y) = (y, x) E J, we see that 0 « (x + y, x + y) = (x, y) + (y, x) E J. This implies J = E, and so T is ideal irreducible.

I

Problem 9.1.12. For a Markov operator T : E - E on an AM-space with unit e establish the following.

(a) There exists some 0 < 0 E E* satisfying T*cb = ¢. (Note that if E = C(Q) and e = 1, then 0 is a regular Bored measure on Sl, known as an invariant measure for T.) (b) If 0 < 0 E E* satisfies T*0 =,i, then the rank-one positive operator S = 0 0 e satisfies ST = TS = S.

9.1. Irreducible and Expanding Operators

277

Solution: (a) Consider the w"-compact convex set

C={ ,EE`: tai>Oandtb(e)=1). Then the adjoint operator T*: E' - E' is w`-continuous and satisfies T* (C) C C. (Indeed, if ) E C, then (T*1i,)(e) _ ?/i(Te) = 41(e) = 1, and so T"ii E C.) Thus, by the classical Tychonoff Fixed Point Theorem, there exists some 0 E C such that

T'O =0. (b) Notice that for each x E E we have:

STx = S(Tx) _ (O®e)(Tx) _ O(Tx)e = [(T`c&)x]e = O(x)e = Sx, and TSx = T(Sx) = T(cb(x)e) _ g(x)Te = ((x)e = Sx. Therefore, ST = TS = S. (For related results see Lemma 9.45 and Corollary 9.46.)

I

Problem 9.1.13. Consider the positive operator T : C[O,1] -+ C[0,1) defined by

Tf(t) = i

J f(s)ds,

(*)

for all 0 < t < 1 and T f (0) = f (0). Show that: (a) T defines indeed a positive operator on CIO, I]. (b) T is one-to-one (and hence strictly positive). (c) T is a Markov operator. (d) T is not band irreducible-and hence not ideal irreducible either. (e) The point spectrum of T is the set

op(T)_{AEC: IA-11<2}U{1}. (f) T is not a compact operator.

Solution: (a) We need to show that T f is continuous at zero for each f E C[0,1

To see this, fix f E C[0,1], let e > 0, and then choose some b > 0 such that u, v E [0,1] with Iu - v) < d imply If (u) - f (v) I < e. So, ifs E [0, 11 satisfies 0 < s < 6, then If (s) -1(0)1 < e. In particular, for each t E (0, b] we have

ITf(t)-Tf(O)I= iIJ'[f(s)-f(0)]dsI < i JIf(8) - f(0)1d8 < et=e. This shows that T f is continuous at zero. Now it should be clear that T as defined by (*) is a positive operator on C(0,1]. (b) Assume that some f E C[0,1] satisfies T f = 0. This implies J o f (s) ds = 0 for each 0 < t < 1. The Fundamental Theorem of Calculus shows that we have f (t) = It f a f (s) ds = 0 for all 0 < t < I. So, T is one-to-one-and consequently T is a strictly positive operator.

(c) Notice that 1 (the constant function one) is the unit of C[0,1]. It follows that for each 0 < t < 1 we have T1(t) = fo 1 ds = 1. Therefore, Ti = 1, and so T is a Markov operator. In particular, A =e 1 is an eigenvalue of T.

9. Irreducible Operators

278

(d) Consider the band B = { f E C[0,1]: f (s) = 0 for all s E [1,1] }. It is easy to see that T does not leave B invariant. Thus, T is not band irreducible. (e) To find the eigenvalues of T. we must find all complex numbers A = a + 3z for which there exists a non-zero continuous function f : [0, 1] - C such that

(a)

T f (t) = i 1 f (s) ds = A f (t) 0t

for each 0 < t < 1. Since T is one-to-one, we see that A = 0 cannot be an eigenvalue

of T. So, we can assume from the outset that A 0 0. In view of (c), we can also assume that A 0 1. Now a glance at (a) guarantees that for every solution f of (a) the derivative f'(t) exists for each 0 < t < 1. Rewriting (a) in the form fo f (s) ds = At f (t) and differentiating, we get f (t) = Af (t) + At f'(t) or Atf'(t) + (A - 1) f (t) = 0.

((3)

We are looking for eigenvalues different than 0 and 1. In this case, aside from scalar multiples, the solution of (0) is given by f(t) = ti-1 = eQ-1)Int (7)

for each 0 < t < 1. Letting t -+ 0+ in (a) yields f (0) = Af (0) or f (0) = 0. This shows that the function f(t) = tI-1 as defined in (-y) is an eigenfunction of T if and only if limt-o+ ti-1 = 0. Taking into account the formula

tI = e(s-1)Int = elnte-*_

Int

we see that limt.o+ t -1 = 0 if and only if a - a2 - $2 > 0. Now note that

a-Q2-Q2>0

(k 2+Q2a<0 2)2

(a -

+ Q2 < (1)2

JA-21 <2. Therefore, the point spectrum of T is op(T) = {A E C : IA - 1 < } U 11). z

2

(f) Since the spectrum of T is uncountable, T cannot be a compact operator. A direct proof that T is not compact goes as follows. Consider the sequence { fn} in C[0,1] defined by ns if 0:5 s < n fn (s)

if -1<s<1.

1

Clearly, 11 fnIj = 1 holds for each n. Moreover, a direct computation shows that

Tfn(t) =

2 1 - 2 nt

if 0
Notice that {Tfn } converges pointwise to the discontinuous function g : [0, 1] - R given by g(O) = 0 and g(t) = 1 for 0 < t < 1. This shows that no subsequence of {Tfn } can converge uniformly, and consequently T is not a compact operator.

9.1. Irreducible and Expanding Operators

279

Problem 9.1.14. Let 1 < p < oo. We consider the operator T of Problem 9.1.13 defined on Lp[0, oo) via the same formula

T f (t) = i

re

J0

f (s) ds,

(**)

for all t > 0 and T f (0) = f (0). Show that the formula (**) defines a one-to-one strictly positive operator from Lp[0, oo) into Lp[0, oo) such that IITIIp = p 11

Solution: Consider an arbitrary continuous function f : (0, oo) -+ R with compact

support, i.e., f E Cc (0, oo). Choose some M > 0 so that 0 < f (s) < M for all s > 0. If K = f °D f (t) dt, then the function

if 0<s<1 ifs>1 belongs to Lp(0, oo). Clearly, 0 < T f < g, and so T f belongs to Lp(0, oo) for each I < p < oo. Also, in view of the inequalities t f(s)ds}p = t[Tf(t)]p S t[g(t)]p+ 0: t[i

f 0

it follows that

ftf(s)ds] t[e

plo

=0.

Now integrating by parts and using Holder's inequality, we get

IITfllp =

f'[1f'f(s)ds]pdt= l1 j[jf(8)ds]"d(t1_1')

pp I

<

f

f(t)[Tf(t)]p-1

0

dt

[Tf(t)]a(p-1)dt)

pIlfllp(f p

=

pp1 Ilflip (IITfllp)Q

This easily implies (II Tf IIp)p

°<

IIf IIp or

IITfllp < PPI llfllp for all f E C,(0, oo). In other words, the mapping

T: C,(0, oo) - Lp(0, oo) defines a continuous operator such that IITII <_ p1 Since C,(0, oo) is norm dense in Lp(0, oo), the operator T has a unique continuous (linear) extension t to all of Lp(0,oo) and IITII < p 1. Our next objective is to show that t f (t) = 1 f f (s) ds holds for all f E Lp(0, oo) and all t > 0. o

9. Irreducible Operators

280

To this end, let 0 < 0 be a step function. Choose some C > 0 satisfying 4(s) < C for all s > 0, and then pick a sequence {fn} in CC(0,oo) with lim f I fn(s) - m(s)Ipds = 0. We can assume that lim f(s) = q(s) for almost all s; see Problem 1.2.13. In view of the inequality

IfnAC - ml = IfnAC -OnCI 0 and all n. Since lim IITfn - TmII p = 0, we can also assume (by passing to a subsequence) that T f n (t) - To(t) for almost all t. Next, observe that for each fixed t > 0 we have 0 E L1(0, t) and so, by the Lebesgue Dominated Convergence Theorem, we see that 7'0(t)

lim T fn(t)

= lim f

JU

e

jct(s)ds.

fn (s) ds =

Now let 0 < f E Lp(0, oo). Choose a sequence {mn } of step functions with 0 < bn T f In view of

- Tflip = 0, we can assume that TOn(t) -+ Tf(t) for almost all t. Taking into account that for each fixed t > 0, we have f E Lp(0, t) C L1(0, t), the Lebesgue Dominated lim I140n

Convergence Theorem yields

Tf(t)= limTon(t)= lim efo b (s)ds= n--oo n-oo

tf(s)ds

e

fo

for each t > 0. Thus, T = T. Next, we shall show that IITII = it must be established that IITII >

p 1 ' We already know that IITUU <

To this end, let

(s("-I-1)pfn(s)

if 0 < s < 1 if s > 1 .

0

SI

ds = n, and also

Then IIfnIIp = L Sn

TAM =

t-1)p_'

n

1+n p-1

t-1

{

if 0 < t < I if t > 1 .

Consequently, we have IITfn llp

=

JlTfn(t) fl+np-1 n

f1+np-1

I

]pff t' '-'dt+ f 0

x

1

t-pdtJ

1

]p(n+pl1),

and so [n + p11] n 1+n(p-1

n

= IlTfnlIp :5 IITII ' IIfnIIp = IITII ' n°

This implies IITII

1

[1+ n 1 1v __ D

p1.

So,

9.1. Irreducible and Expanding Operators

281

from which it follows immediately that 11TH > PP t .

Finally, we establish that T is one-to-one. Assume that T f = 0 for some function f E Lp(0. x). Then f f f (.s) ds = 0 for all t > 0. This implies f = 0 a.e., and so the operator T is one-to-one. (See also the solution to Problem 7.1.4.) 1

Problem 9.1.15. Prove the following converse of Corollary 9.14. If for a positive operator T : E -+ E on a Banach lattice there exists some A > r(T) such that R(A, T) is strongly expanding, then T is ideal irreducible. Solution: Let T : E - E be a positive operator on a Banach lattice. Also, assume that for some A > r(T) the operator R(A, T) is strongly expanding. By Theorem 9.6

(or by Problem 9.1.1), the operator R(XT) is ideal irreducible. Suppose that T(J) C J for some closed ideal J in E. Now fix some vector .r E J. and note that T"x E J f o r each n = 0, 1, ... . This itnplices R(A,T)x = rn a A`(n+l)Tnx E J. and therefore the closed ideal J is R(A, T)-invariant. Since R(A, T) is ideal irreducible, we get that either J = {0} or .J = E. This shows that T is ideal irreducible.

Problem 9.1.16. For a positive operator T : E -+ E on a Banach lattice show the following.

(a) The operator T is ideal irreducible if and only if {0} and E' are the only w'-closed bands in E' that are invariant under the adjoint operator T*: E* -, E*. (b) If the adjoint operator T' is band irreducible, then T is ideal irreducible. Give an example of an ideal irreducible operator whose adjoint is not band irreducible. (c) If E has order continuous norm, then T is ideal irreducible if and only if T' is band irreducible. Solution: We use the notation suggested in the hint of the exercise. For each ideal J in E and each band B in E' we define the polars

J° = B° =

{x' E E*: (x, x') = 0 for all x E J I and {x E E: (x, r') = 0 for all x.' E E'} .

Next, we shall present several properties of the polars. For the discussion below J

denotes an ideal in E and B a band in E. 1. J° is a w'-closed band in E' and B° is a norm closed ideal in E. Clearly, J° and B° are both vector subspaces. We verify that J° is a w'-closed

band. To see that J° is an ideal, assume that x. y' E E' satisfy lit I < Ix' l with x' E J. Then for each 0 < x E J. we have ly (x)I <_ ly'I(I=I) <_ lx'1(x) =sup{x'(z): z E E and IzI <x}

= sup{x'(z): zEJ and I:I<x}=0. So. y' (x) = 0 for all .r E J. That is, y' E J°, and hence J° is an ideal in E. Next, notice that if {x,*, 19 J° satisfies 0 < x. 1 x' in E. then for each 0 < x E J, we

9. Irreducible Operators

282

have x*(x) = sups x,(x) = 0. This implies x' E J°, and so J° is a band in E. The corresponding properties for the polars of B can be proven in a similar manner.

2. If J is a closed ideal in E, then J°° = (J°)° = J. Similarly, if B is a w'-closed band in E', then B°° = B. Assume that B is a w'-closed band in E. It should be obvious that B C B°°. Suppose by way of contradiction that B°° # B. This means that there exists some x' E B°° such that x' V B. Since B is w'-closed, there exists (by the Separation

Theorem) some x E X such that x* (x) # 0 and y* (x) = 0 for all y' E B°O. In particular, y' (x) = 0 for all y' E B. This implies x E B°, and therefore x' (x) = 0, which is impossible. Hence, B°° = B. The corresponding result for J can be proven in a similar manner.

We now return to our problem. Let T : E

E be a positive operator on a

Banach lattice.

(a) Assume that T is ideal irreducible, and let B be a T'-invariant w'-closed

band in E. If x E B°, then for each x' E B we have (Tx,x') = (x,T'x') = 0. This shows that Tx E B°, and so B° is T-invariant. Since T is ideal irreducible, it follows that either B° = {0} or B° = E. This implies either B = B°° = {0} or

B=E'.

For the converse, suppose that the only w'-closed T'-invariant bands in E' are {0} and E. Let J be a closed T-invariant ideal in E. If X0 E J° and x E J, then (x, T'x') = (Tx, x') = 0. This implies T'x' E J°, and so the w'-closed band J° is T'-invariant. From our hypothesis, J° = {0} or J° = E. This means J = JO° = E or J = J°° = {0}. Consequently, T is ideal irreducible.

(b) If T' is band irreducible, then To does not leave invariant any non-trivial w'-closed band in E. By part (a), T must be ideal irreducible. The adjoint of any positive operator on C[0,11 cannot be ideal irreducible since the norm dual of C[0,1] does not have any quasi-interior points; see Problem 4.2.5 and Theorem 9.3. Therefore, any ideal irreducible positive operator T on C[0,1] provides the desired example. (c) This follows from (a) and the fact that a Banach lattice has order continuous norm if and only if every band of E' is w'-closed; see [5, Theorem 9.1, p. 60].

Problem 9.1.17. Give examples of positive operators S and T on atomic and non-atomic Banach lattices such that T is ideal irreducible, ST < S, and S is a non-expanding strictly positive operator. (This shows that the inequality TS < AS in Lemma 9.13 cannot be replaced by ST :5 AS.)

Solution: For the atomic case let E = R2, and consider the pair of positive operators S, T : E

E given by the matrices

T = 1l

Ol]

and S = 10

U,

.

9.2. Ideal Irreducibility and the Spectral Radius

283

Clearly, T is ideal irreducible and S is strictly positive. Since Sel = el, the operator rrIa] for each (a, Q) E S is not expanding. Finally, note that we have

E

=S

=(/3+a,0)=SIo]

This shows that ST = S, and certainly ST < S is also true. For the non-atomic case, let E = Lp[0,1], where 1 < p < oo. Fix any strictly positive functional ¢ E E' such that 0(1) = 1, for instance let O(x) = f0 x(t)dt, and consider the positive rank-one operator T = 4®1. Clearly, the operator T is an ideal irreducible positive operator on E. Next, let A = X(0.11 and consider the operator

S = XAT on E. Clearly S is strictly positive but falls, to be expanding. In fact R(S), the range of S, is the one-dimensional vector space R(S) = {AXA: A E R}. Now note that if x E E+, then Sx = XATX = XAO(x)1 = O(x)XA,

and from 0(1) = 1 it follows that STx = S(O(x)1) = O(x)S(1) = c(x)cb(1)XA = O(x)XA

I

Therefore, ST = S, and certainly ST < S is also true.

9.2. Ideal Irreducibility and the Spectral Radius Problem 9.2.1. Let E be a Banach lattice and let T : E -+ E be a positive operator having a non-trivial invariant closed ideal. Show that there exists some 0 < u E E satisfying Eu 0 E and T(Eu) C Eu. Solution: Let J be a non-trivial T-invariant closed ideal and fix some 0 < x E J.

Now consider the element u = n0 s

T" w . Since J is T-invariant, we see that

u E J. If y E E satisfies IyI < Au, then

ITyi s Tlyl <_ ATu = A00E ,I

.

n+ Tu

n=0

00

ff x = 2AIITII E

Z

r

i T"1

< 2AIITIIu.

n=0

This implies Ty E E,,, i.e., Eu is T-invariant. Consequently, T(Eu) C E C J. Thus, the vector u > 0 satisfies the desired properties.

I

Problem 9.2.2. Let T : X -+ X be a bounded operator on a Banach space. If A is an eigenvalue of T, then show that the corresponding eigenspace NT(A) = {x E X : Tx = Ax} is a T-hyperinvariant closed subspace. Solution: Assume that a bounded operator S: X -+ X commutes with T, i.e.,

ST=TS. IfxENT(A),then T(Sx) = S(Tx) = S(Ax) = ASx,

and so Sx E NT(A). In other words, NT(A) is S-invariant, and thus the closed subspace NT(A) is T-hyperinvariant.

I

9. Irreducible Operators

284

Problem 9.2.3 (Ercan-Onal [29] ). Let T : X - X be a bounded operator on a Banach space and let x be a vector in X. Show that T is locally quasinilpotent at x if and only if Ix. (Tnx)I ^ 0 for each x` E X*. If X is also a Banach lattice, then show that T is locally quasinilpotent at x if and only if j x* (T"x) I n 0 for each 0 < x* E X' . Solution: If IITnxII ^ 0

0 and x' E X', then clearly Ix'(Tnx)1"< 11x* 11n

IITn4II°

0.

whence Ix'(Tnx)I °-+0.

For the converse assume that 1x* (Tnx) I ° ---. 0 for each r' E X'. We must establish that IITnxII" -+0. We shall present two solutions below. The first one is due to V. C. Troitsky. Let f > 0 and consider the sequence { T^x }. We claim that the sequence { T^x } is weakly bounded. To see this, fix x' E X'. From 1x* (Tnx) I "-+ 0, it follows that there exists some no such that Ix' (Tnx) I " < e

for all n > no. The latter is equivalent to Ix' < I for all n > no. This implies that the sequence of real numbers {x' (2_r): n E N} is bounded for each x' E X'. i.e., the sequence IT,.) } is weakly bounded. By the Principle *x } is also norm bounded. Pick some of Uniform Boundedness, the sequence { T (T%x) I

Al > 0 satisfying II T-r II < A! for each n E N or, equivalently. IITnrII 1 < FM * for each it. Consequently, we have 0 < hm inf II 2'x 11 < lim sup llrxlj" < lim sup (AI" = E . ^ n- oc n--= n-. c

Since e > 0 is arbitrary, we get lim infra-x flT' xf l" = lim 0. This shows that IITnxII

{Tnx11

i = 0.

The second solution of the converse is due to Z. Ercan and S. Onal [29]. Assume by way of contrauiiction that IITnxII " f+ 0. So, there exist some f > 0 and a strictly

increasing sequence {kn } of natural numbers such that f < IITk^ xfl Now consider the sequence {xn } of X defined by

for each it.

xn =

Tk"x. Ekn = 1 holds for each it. Moreover, we Clearly, llxnIl converges weakly to zero. To establish this. fix r' E X'. From claim that -» 0, it follows that there exists some t such that I x' (Tk^ x) I' < 2 Ix ' (Tkn x) I for all n > t. Therefore, for n > f we have ILW

Ix'(xn)I = e Ix'(Tk^x)I < This implies r' (xn) - 0. and thus xn 0-

/(

.(

)k..

_/

(2)A^'

A glance at Problem 1.3.5 reveals that the sequence {xn } has a basic subsequence. By passing to a subsequence and relabelling, we can assume without loss of generality that {xn} is itself a basic sequence. If {x;,} denotes the sequence of

9.2. Ideal Irreducibility and the Spectral Radius

285

biorthogal functionals of the basic sequence {xn} defined on the vector subspace Y generated by {xn}, then we can extend each x,, to X with preservation of its norm. From part (c) of Theorem 1.37 we know that Ilxnll < TI z < ti < oo for each n, where sc is the basis constant of the basic sequence {xn}. That is, the sequence {x,*,} is norm bounded and satisfies x,,(x,n) = Now consider the continuous linear functional g = EO°_1 2-x;, and note that Iecng(xn)II3^ = Ifk,. Ig(Tk"x)I

-= 2E f+0.

Imo"

2

However, the last conclusion contradicts our assumption. This contradiction establishes that lIT"x1I ^ -, 0.

Finally, let us assume that X is a Banach lattice and that lx* (T nx) I "

-' 0 for

each 0 < x' E X'. If f E X' and e > 0 are given, then choose some no > I such that I f + (T"x) I " < 2 and I f - (T" X) I " < hold for each n > no. So, if n > no, then we have

If (Tnx)I < I f+(Tnx)I + If (Tnx) I < 2^ + 2- = 2-f < En . Therefore, I f (Tnx) I " < e for each n > no. This shows that If (Tnx) I

each f E X. By the previous case, we get llTnxll I _ 0.

0 for U

Problem 9.2.4. This problem presents a one-to-one positive operator which is locally quasinilpotent at some positive vector but fails to be locally quasinilpotent. Consider the positive operator T : 12 - 82 defined by

T(xl, x2, x3, ...) _ (XI, xl, 2 , 3 , ...)

Show that T is one-to-one, is locally quasinilpotent at e2, and fails to be quasinilpotent.

Solution: Notice that Ten = nen+1 for all n > 2. Now using an inductive argument, we see that T"e2 = (n+1 en+2 Hence IITne2I1 = n+1 1 for each n. This implies

0.

nlim 00IITne21I" =

1

On the other hand, IIT"112: IIT"e1112! 1 implies 11n 11 n > 1 for each n. Hence,

T is locally quasinilpotent at e2 > 0 but it is not a quasinilpotent operator.

I

Problem 9.2.5. Let T : X - X be a bounded operator on a Banach space and denote by QT the set of all points where T is locally quasinilpotent, i.e.,

QT = {x E X :

i

lim 1ITnxIIn = 0}

.

Show that QT is a T-hyperinvariant vector subspace. Solution: We shall show first that QT is a vector space. To this end, let x. y E QT and fix e > 0. Pick some no such that llTnxll < e" and IIT"yll < en hold for all n > no. It follows that IIT"(x+y)II^ S (IIT"xll + IIT"yII)^ < (2fn)1 < 2F for all

9. Irreducible Operators

286

= 0, and so x + y E QT. Also note that

n > no. Therefore, limn-,, IIT"(x + y) 11 if a is a scalar, then

Um j17'(,\x)1I" = lim 1IAT"(x)II° = n-oo lim IAIl - lim 1I7'xIJ* =0. "oo n--x n--x and so ax E QT. Consequently, QT is a vector subspace of X. Finally, let us show that Qr is a T-hyperinvariant subspace. To see this, assume that an operator S E C(X) satisfies TS = ST and let xo E QT. Then we have IIT"(Sxo)II'

0.

= IIS(T"xo)II" 5 11S11" 11T xoII^

This implies Sxo E QT, and so QT is a T-hyperinvariant subspace.

Problem 9.2.6. For the notation used here see Problem 9.2.5. For a bounded operator T : X -+ X on a Banach space establish the following properties.

(a) QT = {0} is possible. (b) QT can be dense without being equal to X.

Solution: (a) Let T: X

X be an isometry. Then for each non-zero vector

x E X. we have jjT"xII = 11xil, and so limn_ tIT"xO * = limn-, IIxjI n = 1. This shows that T cannot be locally quasinilpotent at any non-zero vector, and hence QT = (01-

(b) Consider the "backward shift" operator S: f1 - f1 defined by the formula S(x1, x2, ...) = (x2, T.3, ...). Notice that if a sequence x c= fl is eventually zero. then S"x = 0 for all sufficiently large n. This implies linen-, UUS"xII n = 0 for all sequences x E t1 that are eventually zero. In other words, QT contains all sequences in 11 that are eventually zero, and hence QT is dense in f1.

Now consider the vector u = (1. a. 3. er, ...) E f1, and note that // 7°17L=t2n1

This implies T" u ll = quently,

IIT"ujl

,2 2

i

1r,2 1

-

i

1

1

=211 tt.

or, equivalently, JIT"ull = 22 for each n. Conse= a 0 0, and so u V QT. Hence, QT 16 f1.

Problem 9.2.7. This problem presents a pair of positive operators that satisfy the hypotheses of Corollary 9.21. T:

L,,,,

Consider the positive operator

defined by T(x 1, x2r x3 ,.

.. =

x1+x22

x'+x3 21+z4 2 , 2

Establish the following:

(a) The operator T is a Krein (and hence an ideal irreducible) noncompact positive operator.

(b) The spectral radius of T equals one, i.e., r(T) = 1. (c) There is some 0 < 0 E Fx satisfying 11011 = 4)(e) = 1 and T`p = 6. where e denotes the constant one sequence, i.e., e = (1,1,1, ...).

9.2. Ideal Irreducibility and the Spectral Radius

287

(d) The rank-one positive operator S =,0 0 e on Q. satisfies

TS=ST=0®e. Solution: (a) Clearly, T is a positive operator. Now, let x = (xl, x2, ...) > 0 be a bounded sequence, and let k = min{i E N: xi > 0}. Then it is easy to see that Tkx > 2# (1,1,1, ...). This shows that Tkx is a strong unit of e,,., and so T is a Krein operator. To see that T is not a compact operator, consider the norm bounded sequence {x,,} C eo,, defined by x,, = (1, 1, ... ,1, 0, 0,...), where the 1's occupy the first n coordinates. It follows that

Tx _ (1,1,...,1,222

where the 1's occupy the first n - 1 coordinates. This implies IITx,, - Tx,,, (I = z has no convergent subsequence in eo,,. for all n 94 m, and consequently Therefore, T is not a compact operator.

(b) From T(1,1,1.... ) = (1,1,1, ...), we see that T is a Markov operator. 1(1,1,1, 1 for each n, and so

This implies JIT"II = SIT"(1,1, II* = 1. r(T) = (IT

(c) This follows immediately from part (b) of Problem 9.1.12.

(d) This follows again from part (b) of Problem 9.1.12.

1

Problem 9.2.8. Consider the operator T : C[0,1] - CIO, 1] defined by the formula 1

Tx(t) = fo sin(st)x(s) ds. Show the following:

(a) T is one-to-one and strictly positive. (b) T is a compact operator. (c) If J = {x E C[0,1) : x(O) = 0}, then J is a non-trivial closed ideal and T(J) C_ J holds-and so T is not ideal irreducible.

(d) For each x > 0 the function Tx is a quasi-interior point of J. In particular, the operator T : J -i J (the restriction of T to J) is strongly expanding-and hence ideal irreducible. (e) The spectral radius of the operator T : CIO, 1] - CIO, 11 is positive (that is, r(T) > 0) and coincides with the spectral radius of the

operator T : J - J (the restriction of T to J). Solution: (a) Assume that Tx = 0, i.e., fo sin(st)x(s) ds = 0 for each t E [0,1]. This implies 02ft FEW

1

sin(st)x(s) ds

= /

2"

1

e

[sin(st)x(s)] ds

fo

= (-1)" J s2i sin(8t)x(s) ds = 0 1

0

9. Irreducible Operators

288

for each t E [0, 1] and all it = 0, 1, .... Consequently, f a p(s2) sin(st)x(s) ds = 0 for each polynomial p. Now observe that the algebra generated by {1, s2} is (by the Stone--Weierstrass theorem) norm dense in CEO, 1], and so f f (s) sin(st)x(s) ds = 0 for each continuous function f E C[0,1] and each t E [0, 1]. Letting f (s) = x(s), we get fo sin(st)[x(s)]2ds = 0 for each t E [0,1]. Since sin(st) > 0 for all 0 < s,t < 1,

we see that x(s) = 0 for all 0 < s < 1, and so x = 0. This proves that T is one-to-one.

It should be clear that 0 < x E C(0,1] implies Tx > 0, and thus T is a strictly positive operator. (b) If f E C(0,1] satisfies Il f Il,o < 1, then note that

ITf(t1)-Tf(t2)I =

j [sin(stj) - sin(st2)J f (s) ds

f Isin(stl) - sin(st2)I ds L'lti- st2fds=fti-t2fJ adS=2ft1-t21 holds for all t1, t2 E 10, 1]. This shows that T([-1, 1]) is an equicontinuous subset of CIO, 11. By the classical Ascoli- Arzela Compactness Theorem, T((-1, 1]) is a norm totally bounded subset of CEO, 1], and so T is a compact operator.

(c) Clearly, J is a non-trivial closed ideal of C(0,1]. Since T f (0) = 0 for each function f E C[0,1], it follows that J is T-invariant. This shows that T is not ideal irreducible.

(d) We claim that if some u E J satisfies u(t) > 0 for all 0 < t < 1, then u is a quasi-interior point of J. Indeed, if 0 < y E J, then (y A nu)(t) T y(t) for each 0. Thus, u is a quasi-interior t r= [0, 1], and so (by Dini's theorem) fly A nu - yll,o point of J. Now let 0 < x E CIO, 11. Since sin(st) > 0 for all 0 < s, t < 1, it follows that Tx(t) = ff sin(st)x(s) da > 0 for each 0 < t < 1. Therefore, Tx is a quasi-interior point of J. This shows that T : J -, J is strongly expanding, and consequently it is also ideal irreducible.

(e) The operator T : J - J is a compact ideal irreducible positive operator. By Corollary, 9.20, the spectral radius r of TI,, is positive, i.e., r = r(TJJ) > 0. Hence, by the Krein-Rutman Theorem 7.10, there exists some 0 < x E J such that

Tx = rx. This shows that r(T) > r > 0. Since T: C[0,1] -' C[0,1] is a compact positive operator, again the KreinRutman Theorem 7.10 guarantees that r(T) is an eigenvalue of T. So, there exists some positive function y E CIO, 11 satisfying Ty(t) = fo sin(st)y(s) ds = r(T)y(t) for each t E 10, 1]. This implies y(O) = 0, and so y E J. Consequently, r(T) < r is also true, and therefore r = r(Tf j) = r(T).

9.2. Ideal Irreducibility and the Spectral Radius

289

Problem 9.2.9. If an ideal irreducible positive operator T : E -+ E commutes with a compact positive operator S: E -> E, then neither T nor S is locally quasinilpotent at any positive vector.

Solution: Assume that S, T: E

E are two commuting positive operators on a Banach lattice such that T is ideal irreducible and S is compact. Suppose by way of contradiction that either T or S is locally quasinilpotent at a vector u > 0. Then from II(ST)null "

II(ST)"uII n

= IIS"T"ull ° <_ IISII =

IIT"'S"uIl ° < IITII

II7""ull

^ , and

IISuuhI

it follows that the compact positive operator ST is quasinilpotent at the positive u > 0. So, by de Pager's Theorem 9.19, there exists a non-trivial f-hyperinvariant closed ideal J in E. In particular, since T commutes with ST. we have T(.1) 9 .1, which contradicts the ideal irreducibility of T. Hence, neither T nor S is locally quasinilpotent at any positive vector. 1

Problem 9.2.10. Corollary 9.22 asserts that if T is a compactly dominated ideal irreducible positive operator on a Banach lattice, then every positive operator S in the commutant of T has positive spectral radius, i.e.. r(S) > 0. Generalize this result as follows: If T : E -+ E is a compactly dominated ideal irreducible positive operator on a Banach lattice, then every positive

operator on E that commutes with T (in particular, T itself) cannot be locally quasinilpotent at any positive vector of E.

Solution: Let T : E - E be an ideal irreducible positive operator on a Banach lattice that is dominated by a compact operator. By Corollary 2.35, T3 > 0 is a compact operator. Now let S: E - E be a positive operator that cornrmrtes with T, and assume by way of contradiction that S is locally quasinilpotent at some vector u > 0. Then the positive operator T3S is compact. commutes with T, and is locally quasinilpotent at u (why?). By Theorem 9.19, there exists a nontrivial closed 1-hyperinvariant ideal J. In particular, we have T(.1) C J. which is impossible. So, S cannot be locally quasinilpotent at any positive vector of E. 1

Problem 9.2.11. Show that in Corollary 9.22 the hypothesis "S commutes with T" cannot be replaced by "S commutes with some power of T" even when T is compact. Solution: Consider the pair of positive operators S, T : R2 R2 defined by S(x, y) = (y, 0) and T(x, y) = (y, x). It should be clear that T2 = I and so ST2 = T2S. Therefore, S commutes with a power of the compact operator T. Now note that S2 = 0 and this implies r(S) = 0. 1

Problem 9.2.12. If an ideal irreducible positive operator T : E - E on a Banach lattice has a compact power (in particular, if T is dominated by a compact operator), then show that r(T) > 0.

9. Irreducible Operators

290

Solution: Let T : E -+ E be an ideal irreducible positive operator on a Banach lattice such that the operator S = Tk is compact for some k. Clearly, S commutes with T. Now Corollary 9.21 implies that r(T) > 0. 1

Problem 9.2.13 (de Pager [61]). Show that if a power of an ideal irreducible positive operator T dominates a compactly dominated positive operator, then r(T) > 0. Solution: Let T : E - E be an ideal irreducible positive operator such that for some k the operator S = Tk dominates a compactly dominated positive operator. Now use Corollary 9.29 to conclude that r(T) > 0. 1

9.3. Band Irreducibility and the Spectral Radius Problem 9.3.1. Show that the conclusion of Lemma 9.30 is false if the "compactness" of T is replaced by "ideal irreducibility" of T. That is, give an example of two commuting positive operators T, S: E --+ E on a Banach lattice such that: (a) T is strictly positive, order continuous, and ideal irreducible. (b) S is not order continuous. Solution: Let E = C(r), where r is the unit circle. We denote by 1 the constant one function. Fix a non-zero rational angle 0 and let T be the operator defined by means of the rotation by the angle 0. That is, Tx('y) = x(ryete) for each x E C(F) and y E r. Clearly, T is a surjective lattice isometry. This easily implies that T is strictly positive and order continuous. To continue the solution, we need the following result: If 6 is non-zero and rational, then the set {eine : n E N} is dense in r.

To see this, fix a natural number k and let e > 0. By Kronecker's theorem (see Problem 7.3.3) the subgroup {e"e : n E Z} of r is dense. This implies that the set {eine: n E Z and 1n1 > k} is dense in r. So, there exists some m E Z with 1ml > k such that 1e"ne -11 < e. From Ietme -11 = 1elme(1 - e-:me)1 = 11 - e-11, we can assume that m E N (and so m > k). Now if we let n = m - k E 111, then we have lesne e-skeI = le-.ke(esme - 1)1 = le$ms - 11 < E.

-

The latter, coupled with Kronecker's theorem, shows that the set {eine: n E N} is dense in r. Next, we claim that the rationality of 0 implies that the operator T is ideal

irreducible. To see this, let J be a T-invariant closed ideal in c(r) such that J # {0}. Pick a function x E J+ and some yo E 1' such that x(yo) 96 0. Now let y E P be an arbitrary point. Since 0 is a rational angle, it follows from (.) that the set {'ye'ns: n E N} is dense in r. From T"x E J+ and T"x(y) = x(ye'"g) for each n E N (and yo E {ryeiie : n E N}), it follows that for some n E N we must have

9.3. Band Irreducibility and the Spectral Radius

291

T' x(y) > 0. Therefore, for each 7 E I' there exists a function z, E J+ such that z, (-y) > 0. In particular, for each -1 E I' we can find a neighborhood V, of ry such that z,(rt) > 0 for all q E V. The compactness of I' guarantees the existence of a finite subset {y1, .... 7k } of r such that 1' = J 1 V,, . Now let z = Ek 1 z,, E J+, and note that z(y) > 0 holds for each y E 1'. This implies that z is an order unit of E. and consequently J = E. Now, in view of Problem 9.1.12 (or in view of Corollary 9.46) there exists some 0 < 0 E E' satisfying T'p = b and II oil = o(1) = 1. It is well known that C(1') does not admit any non-trivial order continuous linear functionals; see for instance 16, Example 4.5, p. 45]. Hence, 4 is not order continuous and consequently the positive

operator S = o (9 1 on E is not order continuous either. The reader should also observe the following two properties:

(i) Both operators T and S are Markov operators-and so r(T) = r(S) = 1. (ii) The pair of closed subspaces (Ker o. RI). where Rl = {A 1: A E R}, is a reducing pair for the operator S. Finally, we shall verify that T and S commute. Indeed, for each x E E we have

T(Sx) = T(4(x)1) = o(x)l and S(Tx) = o(Tx)l = [T'o](x)1= 4(x)1.

Problem 9.3.2. Let J be an ideal in a Riesz space E. Show that J is a band if and only if the quotient map x o- (x], from E onto E/J, is order continuous.

Solution: We know that the quotient map x '- [x] is a surjective lattice homomorphism; see Problem 3.4.2.

Assume first that J is a band and let x,, j 0 in E. To see that [XQ] 10 in E/J, suppose that (x0] > [x] in E/J for all a and some x E E'. For each a put yQ = (x - x,,)t and note that 0 < yQ T x. Now using that the quotient map is a lattice homomorphism. we see that (ye] = [(x - xQ)+] = ([x] - [xe])+ = 0 for each a. This implies {yo} 9 J, and since J is a band, we get x E J or [x] = 0. Therefore. [x,,] j 0 holds in E/J, and so the quotient map is order continuous. For the converse, assume that the quotient map is order continuous, and let a

net {xo} C J satisfy 0 < x,, T x in E. Then 0 =

j ]x] holds in E/J. This

implies [x] = 0 or x E J. This shows that J is a band.

Problem 9.3.3. Let T : E - E be a a-order continuous positive operator on a Banach lattice. If the commutant of T contains a a-order continuous band irreducible positive operator and a a-order continuous compact positive operator, then show that r(T) > 0. Solution: Let TS = ST and TK = KT, where S > 0 is a a-order continuous band irreducible positive operator and K > 0 is a or-order continuous compact positive operator. Fix some A > r(S) and observe that T [R(A, S)K] = [R(A, S)K]T > 0. Corollary 9.14 guarantees that R(A, S) is a or-order continuous strongly expanding positive operator. Since TK is a or-order continuous positive operator. it follows from Theorem 9.35 that r(R(A. S)TK) > 0. Finally, from

0 < r(R(A,S)TK) = r(T[R(A, S)K]) < r(T)r(R(A, S)K) ,

9. Irreducible Operators

292

we can conclude that r(T) > 0.

1

Problem 9.3.4. Let S, T : E - E be two commuting positive operators on a Banach lattice. If T is a a-order continuous band irreducible operator and some power of S dominates a a-order continuous compactly dominated positive operator. then show that both operators have positive spectral radius,

i.e.. r(S) > 0 and r(T) > 0. Solution: Assume that S' dominates some a-order continuous compactly dominated positive operator. Fix some A > r(T) and let Q = R(A. T)T. By Corollary 9.14. the operator Q is strongly expanding. Now from Theorem 9.35 it follows

that

0 < r(QSm) = r(R(A.T)TS') < r(R(A.T))r(T)[r(S)]"', and so r(S) > 0 and r(T) > 0 both hold.

Problem 9.3.5. Let Q. S: E

1

E be two positive operators such that:

(1) Q is a-order continuous and expanding. (2) S dominates a compactly dominated a-order continuous positive operator. in particular this is so if S is itself a-order continuous and dominated by a compact operator. Show

that r(QS) > 0.

Solution: Let Q and S satisfy the properties stated in the problem. Pick a o-order continuous positive operator L and a compact operator K satisfying 0 < L < S

andL
Consider the operator C = QL and note that C > 0 and that if Ckx > 0 for some x > 0 and some k. then Ck.r is a weak unit. (This follows from the fact that Ck = QR for some positive operator R.) In particular. we have Ck > 0 for each

k. Since 0 < C = QL < QK. Corollary 2.35 implies that C3 > 0 is a compact operator. Notice also that (LQ)3 = L(QL)(QL)Q is a compact strictly positive operator. Therefore, since C3 is also strictly positive, it follows that E has the countable sup property (see Lemma 9.15). This guarantees that the operators Q and L are order continuous. Assume by way of contradiction r(C) = 0. This implies r(C3) = [r(C)13 = 0.

Now taking into account (C3)2 = C6 > 0 and the order continuity of C, Theorem 9.31 guarantees the existence of a non-trivial f-hyperinvariant band B for C3 such that C°(B) 96 {0} (and hence C3(B) 7A {01). This implies that there exists some 0 < r E B such that C3x > 0. The preceding discussion guarantees that C3x is a weak unit. Since the weak unit C3x belongs to the band B, we see that B = E, a contradiction. So, r(QS) > r(C) > 0 holds. 1

Problem 9.3.6 (Schaefer [71); Grobler [341). Let T: E - E be a a-order continuous band irreducible quasinilpotent positive operator on a Banach lattice. If S: E E is a compact positive operator, then for each k > 1 show that [0, S] f1 [0, T'k1 = {0}.

9..4. Firein Operators and C(Q) -spaces

293

Solution: Assume that the operators T and S satisfy the properties stated in the problem. Fix some A > r(T) and let Q = R(A,T). Then Q is a or-order continuous positive operator, it commutes with T, and (in view of Corollary 9.14) it is also

expanding. Now if 0 < A < S and 0 < A < Tk. then from Theorem 9.35 it follows that 0 < r(QTk) < r(Q)[r(T)?k and so r(T) > 0, a contradiction. Hence. [0, S] fl [0, Tk J = {0} for each k > 1.

Problem 9.3.7 (Caselles 1211). A Harris operator is a positive operator T : E -. E on a Dedekind complete. Banach lattice such that some power of T is not disjoint from the band of integral operators. Prove that every a-order continuous band irreducible Harris operator has a positive spectral radius. Solution: Assume that T : E - E is a a-order continuous hand irreducible Harris

operator. Then there exists an integral positive operator k : E - E such that T A K >0. Fix some A > r(T). Then T commutes with R(A. T). and by Corollary 9.14 the operator R(A. T) is a a-order continuous strongly expanding positive operator. Now observe that the operator T dominates the compactly dominated a-order continuous

positive operator T A K. Therefore, according to Theorem 9.35. we have 0 < r(TR(A.T)) < r(T)r(R(A.T)). This implies r(T) > 0. as desired. I

9.4. Krein Operators and C(cl)-spaces Problem 9.4.1. Let T : C. (it) - C(1l) be a positive operator. Show that for each unit u E C(1l) we have IITII,, = IlTullu

and

r(T) = lim (IITulln)n

Solution: Assume that u > 0 is a unit. By Problem 3.1.2 we know that the formula

IIrIIu =infJA >0: IrI
;Ixli < 1} = [-u.uJ.

Since T: C(1) C(1?) is a positive operator. the norm of T" with respect to the lattice norm II - 'I,, is given by IIT"II,, = IIT"ull,,. Therefore. r(T) = slim (117- 11,,) " = bin (IIT"ulI,,) and the solution is finished.

Problem 9.4.2. Consider a positive operator T : C(f 2) - C(St) and put in = minT1(,u) and M = rnaxTl(.,j). .eE ll

WED

9. Irreducible Operators

294

Show that m < r(T) < Al and conclude from this that if Ti >> 0, then the operator T has a positive spectral radius. Solution: From the definition of the real numbers m and M, it follows that ml < Ti < All. This implies mT1 < T21 < MT1. Consequently, we have m21 < T21 < A121. By induction, we get m"l < T11 < M"1 for each n. Since the sup norm of the positive operator T" satisfies fT"]l = {T"1]],,., it follows that m" < 11T"11 < At" or m < 11 T"JI < M for each n. Therefore,

m < r(T) = lim IIT"]I n-or,

< M.

Now if Tl >> 0 holds, then m > 0, and so from (*) it follows that r(T) > 0.

(*)

1

Problem 9.4.3. Let p be a non-zero regular Borel measure on SI with full support, and assume that T : C(fl) -+ C(ft) is a positive integral operator generated by a continuous strictly positive kernel. That is, there exists a continuous real-valued function K on ft x f2 satisfying K(t, s) > 0 for all

(t,s)Ef2xf2 and T f (t) =

JK(t,s)f(s)d(s)

for each f E C(fl) and all t E ft. Show that T is a compact strong Krein operator. Solution: The compactness of T follows immediately from Problem 3.1.13. We must verify that T is a strong Krein operator. To this end, let 0 < f E C(SI). Then the open set V = {w E is f(W) > 0} is non-empty. Since p has full support, we have p(V) > 0. In particular, there exists some e > 0 such that the Borel set B = {w E Q : f (w) > e} satisfies p(B) > 0. Now if we choose some 6 > 0 such that K(t, s) > b for all s, t E i, then

Tf(t) =

jK(t. s)f(s) dp(s) > J

a

K(t, s)f(s) dp(s) >&p(B) > 0

for each t E ft. This shows that T f is a unit of C(SI), and so T is a strong Krein operator.

Problem 9.4.4. Let K : [a, b] x [a, b] -- [0, oo) be a continuous function such that for some c > 0 we have K(t, s) > 0 for all Is - t] < e. Show that the integral operator with kernel K(t, s) is a compact Krein operator. Solution: We claim first that T2 is also an integral operator. Indeed, for each f E C[a, b] we have b (T2f)(t) = [T(Tf)](t) = Jb K(t,u)[JK(u,s)f(a)ds]du

fb r rb

a

K(t, u)K(u, s) du] f (s) ds.

9.4. Krein Operators and C(52)-spaces

295

This shows that T2 is a integral operator with kernel given by the formula

K2(t, s) =

JJ.n

K(t. u)K(u. s) du.

(For this conclusion see also Problem 5.2.15.) By induction, the operator T" is an integral operator. We shall denote the kernel of the operator T" by K"(t. s). Notice that K (t, s) = fQ K(t, t (u. s) du holds for n = 2.3.4.... , where, of course, K, (s. t) = K(s, t). Now observe that K2(t.s) > 0 for all Is - ti < 2E. To see this, let s < t and t - s = Is - tI < 2E. Let uo = 'z , the midpoint of the interval [s, t], and put 6 = E - °2=i. Clearly, 0 < 6 < E, and if (uo - uI < b. then

Is - uI 0 and K(u. s) > 0 for all it E (uo - 6, to + 6). This implies b

0+3

K(t . u)K (u, s) du > 0. p fuu-S Repeating the above process. we see that there exists some positive integer n such K2(t, s) = J K(t. u)K(u, s) du >

that the kernel K (t, s) of the operator T" satisfies K (t. s) > 0 for all (t, s) in [a, b) x [a, b). Now a glance at Problem 9.4.3 shows that T' is a strong Krein operator. and consequently T is a Krein operator. The compactness of T follows from Problem 3.1.13.

Problem 9.4.5. By Theorem 9.42 we know that a Krein operator can have only one (up to a scalar multiple) positive eigenvector corresponding to a positive eigenvalue. Give an example of a Krein operator whose eigenvalues are all positive.

Solution: 1Let a and 3 be two arbitrary positive numbers. Consider the operator T = I I iJ on R2. Clearly. T is a strong Krein operator. A straightforward computation shows that Al = I + at and A2 = 1 are eigenvalues of T. These eigenvalues are both positive provided o3 < 1. Two corresponding eigenvectors are ( . &) and (0/ . - f ). 6

Problem 9.4.6. Show that the product of two commuting Krein operators is also a Krein operator. Give an example of two non-commuting Krein operators whose product is not a Krein operator. Solution: Let S. T : C(Q) C(S2) be two arbitrary commuting Krein operators, and let 0 < x E C(1). Pick some k E N such that Tkx is a unit. Since T and S commute, we have (ST) *x = Sk(Tkx). By Lemma 9.40 each Krein operator carries units to units. and hence Sk(Tkx) is a unit.

To see that the preceding conclusion is not valid for non-commuting Krein

operators, consider the positive matrices A = to

1

I and B =

1.1

of . From

9. Irreducible Operators

296

A2

=

2] and B2 =

121

11, we we that both A and B define Krein operators

[1

on R2. Now notice that AB = [2

l # BA =

[01

1and that AB is not a Krein

J

operator.

Problem 9.4.7. Theorem 9.47 was originally proved by M. G. Krein under C_ U for each a, where U = Int[C(f1)+]. Thethe extra hypothesis orem 9.47 implies, of course, that version. Show that the original Krein theorem implies Theorem 9.47. Solution: For each 0 < t < 1 let TQ,, = TQ + EI, where I is the identity operator on C(S1). Clearly. T,,., (U) C U, and so for each e > 0 there exists some 0 < q, E C. where C = {>v E C(SZ)* :

u=(1)=l}.

and a family of non-negative scalars { aa,, }a<EA such that T, o, + c o, = AQ,e

c.

(*)

We consider the open interval (0, 1) directed by the relation >- defined by fl t E2 whenever tl < t2. Notice that the net (x,),E(o.j) of real munbers, defined by x, = e,

satisfies x, - 0. Next. observe that the net {o, }0<,« of C has a w'-convergent subnet. We can assume that 0, - w - : - + 0 holds in C. and so T p, - - Ta¢ for each a E A. Now fix t1. From (*) it follows that a E A and then pick some t > 0 with 0 < Aa., = Aa.,O,(I)= TTO,(1) +EOE(1) < t+ 1.

This implies that for each fixed a the net {AQ_,}o<, 0. Without loss of generality we can assume that AQ,, f A,,. Finally, taking w'-limits in (*), we get lim[Tno, + Fl,j = lim J1,,,OE _ "OQ

for each a E A, and we are finished.

Problem 9.4.8. Let {TQ}OEA be a family of pairwise commuting positive operators on a C(n) -space. If there exist some A > 0 and some unit u such Au for each a, then show that there is some 0 < fi E C(SZ)' such that that TQp = Am for all a. Solution: For each a E A. let S = IT,,. Then each S. is positive and Sau = u for each a. Moreover. SSS,3 = S, 3S. for all a, 3 E A. According to Theorem 9.48 there exists some function 0 < ¢ E C(Q)' satisfying

aTp0 =S;4=o foreachaEA. So,Tad=At foreach aEA.

I

Problem 9.4.9. For a pairwise commuting family {Ta}QEA of positive operators on a C(Q) -space consider the following statements.

(a) If for some non-zero vector v E C(1l) we have Tav = v for each a, then there exists some 0 < E C(SZ)' such that Tic _ 0 for all a.

9.4. Krein Operators and C(S2)-spaces

297

(b) If for some unit u E C(Q) we have TQu = u for all a, then there exists some 0 < 0 E C(Sl)' such that Tao = 0 for all a. Show that (a)

(b) but (b) does not imply (a).

Solution: The implication (a)

(b) is trivial. To see that (b) does not imply

(a), let C(f2)r = R2 and consider the positive projection

the matrix o OJ . A direct computation shows that I

T: C(Q) I

0

C(Q) defined by

0110==

[0, However,

it is easy to see that T does not have any unit as an eigenvector.

Problem 9.4.10. Show that the commutativity assumption regarding the family of operators in Theorems 9.48 and 9.50 cannot be dropped. Solution: Let P denote the family of all (Borel) probability measures on [0.1]. For each p E P consider the positive operator T,,: C[0, 11 ---4 C[0,1] defined by T,, f (t) = f. 1f (s) dµ(s), i.e., T,, = p®1. A direct computation shows that TT = T

for all p, v E P. In particular, it follows that IT.: p E P} consists of non1 for commuting positive projections on C[0,1]. In addition, notice that each p E P, i.e., each T. is a Markov operator. Assume by way of contradiction that there is some 0 < m E M[0,1] = [C(fl)]' such that T,; 0 = 0 for each p. Scaling appropriately, we can suppose 0([0,1]) = 1. Hence, 0 = T;4= (19p)0= (1.0)p = p for each µ E P, which is impossible.

Chapter 10

Invariant Subspaces

10.1. A Smorgasbord of Invariant Subspaces Problem 10.1.1 (Schaefer [69]). Show that every linear operator on an infinite dimensional vector space has a non-trivial invariant subspace. Solution: Let T : X -+ X be a linear operator on an infinite dimensional vector space. Fix some x 0 0 and consider the T-orbit space Or of x. If Or 96 X, then Or is a non-trivial T-invariant subspace. So, assume that Or = X. In T2x.... } is linearly independent. this case, we claim that the set of vectors {x, Tx, Indeed, if the set of vectors {x, Tx, T2x,... } is linearly dependent, then there ex-

ists some k E N and scalars ao, a1, ... , ak_ 1 such that Tkx = _o a,T'x, and so OT coincides with the vector subspace generated by the finite set of vectors {x,Tx,T2x.....Tk-1x}. But then. X = Or is finite dimensional, which is a contradiction. Hence, the set {x, Tx, T2x.... } is linearly independent. The above properties show that every vector y E X has a unique representation

of the form y = F'_0 A,T'x with the A, equal to zero for all but finitely many i. Now define the non-zero linear functional 0: X -. R by b(y) = Exo A,. Clearly, V = Ker 0 is a proper vector subspace of X. To finish the solution, we shall show that V is T-invariant. To see this, let v E V. So, if v = ;_o A,T'r, then 0(x) = F_;_o A, = 0. Now notice that Tv = E°_p A,T'+1x, and so Q(Tv) = E'0 A, = 0. This shows that Tv E V, and so V is T-invariant, as claimed.

Problem 10.1.2. Give an example of a non-zero operator on 1R4 without non-trivial hyperinvariant subspaces. Solution: For each 0 E [0, 27r) let Ra : R2 --. R2 be the rotation operator through

the angle 0. That is, R3 =

cos $ - sin 0 sin$ cosh 299

10. Invariant Subspaces

300

Fix a non-zero angle 0 that is an irrational multiple of 2a and consider the operator T: R4 R4 defined by T(xj,r2.r3,x4) = (Rg(ri,x2).Re(T3,x4)) We shall show that this operator T does not have any non-trivial hyperinvariant subspaces. To establish this, we need to describe first the non-trivial invariant subspaces of T. Two obvious non-trivial invariant subspaces are:

Lt_{(x1,x2.0.0): x1,r2ER} and Vr={(O,O,r3,x4): r3,x4ER}. These invariant subspaces are not T-hyperinvariant. To verify this. consider the operator S: R4 - R4 defined by S(x1, X2- X3, x4) = (x3 x4, x1, x2)

and note that S commutes with T. However, it is easy to see that S leaves neither Vt nor VV, invariant.

Now for each -y > 0 and each angle 0 E [0, 27r) let V 3.

, _ {(rl.x2. yR3(xi.x2)):

X1, X2 E R}.

(*)

Clearly. each l'p,, is a subspace of R4. Moreover, from

T(x1.x2,yR,3(xl,x2))

=

(Re(xl, x2) yR3Re(xi. x2)) , it follows that T(V3,,) C VB,,. That is, V,3,., is T-invariant. Our next goal is to prove that the T-invariant subspaces described in (*) are all invariant subspaces of T that are different from Vr and V1. To this end, assume that V is a non-trivial T-invariant subspace that is different from Vr and Vt. We will need the following fact.

(.) If V n Vt 0 {0}, then V = Vt. Similarly. if V n Vr 0 {0}, then V = Vr. Let us verify the first claim. To see this, assume that there exists a non-zero vector r = (XI-X2-0,O) E V. Consequently, the vector Tx = (Re(xl. x2), 0.0) also belongs to V and, since Re(x,,x2) and (xl,x2) are linearly independent, it follows immediately that VI C V. For the reverse inclusion, assume that there is some non-zero vector v = (v1, v2, V3, v4) E V that does not belong to Vt. Then the vector (0, 0. v3, v4) = (v1. v2, v3, v4) - (v1, v2.0, 0) is non-zero and belongs to V.

Hence, a similar to the above argument applies and yields that Vr C V. But then X = V. Vr C V, contrary to V j& X. The second statement can be proven in a similar manner. We now return to our proof that V should coincide with one of the T-invariant subspaces described in (*). Since V is different from Vt and V., there exists some vector v = (v1. v2, v3, v4) E V such that Iv, I + It+2I > 0 and Iv31 + 11741 > 0- Without

loss of generality we can assume that v, + v2 = 1. Let y =

v3 + v4 > 0 and

let d E [0, 21r) be the angle satisfying the equality (v3, v4) = yR,3(vl, v2). We shall establish that V = VB,,.

Assume that x = (x1, x2, x3, x4) E V. Clearly, Ix1I + Ix2I > 0 since otherwise x E Vr, and (.) would imply that V = Vr, a contradiction. Therefore.

10.1. A Smorgasbord of Invariant Subspaces

301

we can assume that xI + x2 = 1. By Kronecker's classical theorem (see Problem 7.3.3), there exists a strictly increasing sequence of natural numbers {k} such (Si. 52). By passing to a subsequence of {k,,) if necessary, we that Rk" (vi, 172) can also assume that RB" (v3 v4) w4). Since for each n we have (Rk" (t'1. V2). Re" (L'3 V4)) = Tk" ('vi, 1.2, v3, v4) E V .

it follows that the vector u = (xu. x2. w3. w4) belongs to V. This implies that we see that x - W = (0, 0, x3 - IV3. x4 - w4) E V. and from this (in view of

r-w=0.i.e.. £3=W3 and x4=w4. Now from (x3.54) = (w3. w4) = =

lim RB'(i'3. v4) =line nix nix

Re,.

[YR3(vi, v2)]

nx

YR;3[ lim RB'(v1.r2)}

we see that (X1, X2 - x3, X4) E V,3,.,. Consequently, we have shown that V C Vs,.y.

The reverse inclusion can be proven in a similar fashion. To this end, assume that x = (X1.52. -yR3(xl. X2)) is a non-zero vector in V3,.,. It should be clear that (x1. x2) # 0: otherwise x = 0. which is a contradiction. As before, we can assume

that rI + x2 = 1, and so there exists a strictly increasing sequence of natural

numbers {kn} such that RB(51.52). This implies yR0(rl. x2) = yR;3 lim` IRB" (ill. t:2)] = 1i111 Re" (v3, v4) .

Since (RB" (vi, v2). R8" (v3. v4 )) E V. it follows that (rl. x2, ,yR3.(x1, r.2)) E V. Therefore, Vp,i C V. and consequently V Finally, we shall prove that no subspace of the form V,3,, is T-hyperinvariant. To see this, fix any angle a E (0.2s) and consider the operator S: R4 - R4 defined by

S(51.52.53.14) = (Xi.x2.R0(x3.x4)) For each x = (51, 52, 53.54) E Ht4, we have

TSr = and similarly

p STx = S(Re(x1,x2).Re(x3.x4)) = (Re(xl.x2),RQRe(x3,x4))

= Hence, TSx = STr for each .r E R4. i.e.. S commutes with T. Now notice that fo each (51.12. 7R;3(r1, r2)) E U3,,. we have S(5i. s'2, 7R;3(xl. X2)) = (XI, X2.1Ra+.9(x1.52)) 0 VV..;. .

This implies that no V3,, is invariant under S, and the solution is finished.

Problem 10.1.3. Let T : X - X be a bounded operator on a Banach space. Denote by Lat (T) the set of all closed T-invariant subspaces of X. Show that, under the standard set inclusion, Lat (T) is an order complete lattice with smallest and largest elements.

10. Invariant Subspaces

302

Solution: We consider Lat (T) ordered by the inclusion relation C. Clearly, {0} and X belong to Lat (T). So, Lat (T) has a smallest ({0}) and a largest (X) element with respect to the introduced ordering C.

Let F be a non-empty subset of Lat (T). Put v = nFE, F and note that Visa closed subspace of X. Now if v E V, then v E F for each F E .F, and so Tv E T(F) C F. This implies Tv E V, and hence V is T-invariant, i.e., V E Lat (T). It is easy to see that V is the greatest lower bound of the subset F in Lat (T). That is, V = inf.F in Lat (T). Next, let us denote by W the vector subspace generated by UFO, F. We claim that W is T-invariant. To see this, let x E W. This means that there exist subspaces

Fl, .... F E F a n d vectors xi E F ; (i = 1, ... , n) such that x = E 1 xi. Since T(Fi) C Fi for each i, it follows that Tx = E I Tx, E F1 + Fz + - - - + Fn C W, i.e., the vector subspace W is T-invariant. This implies that its norm closure W is also T-invariant, i.e., W E Lat (T). We claim that W is the least upper bound of the family F in Lat (T). To see this, let U be a closed subspace satisfying F C U

for each F E F. This implies W C U, and thus W C U. Therefore, W = sup.F holds in Lat (T). The above show that, under the operations

YAZ=YnZ and YvZ=Y+Z, the partially ordered set (Lat (T), C) is an order complete lattice having {0} as its smallest element and X as its largest. I

Problem 10.1.4. For a bounded operator T : X -. X on an infinite dimensional real Banach space establish the following.

(a) The operator T has a non-zero finite dimensional invariant subspace if and only if the operator Tc : Xc -. + Xc likewise has a nonzero finite dimensional invariant subspace.

(b) The operator T has a non-zero finite dimensional hyperinvariant subspace if and only if the operator Tc : Xc - Xc likewise has a non-zero finite dimensional hyperinvariant subspace. Solution: It should be clear that if V is a non-zero finite dimensional T-invariant (resp. T-hyperinvariant) subspace of X, then V. = V e tVV is a non-zero finite dimensional Ta-invariant (reap. T,,-hyperinvariant) subspace of Xc. (a) Assume that V is a non-zero finite dimensional To-invariant subspace. Pick

a basis {z1, zz, .... zk} of V and for each 1 < n < k let zn = xn + tyn E X. Now let V be the non-zero vector subspace of X generated by the set of vectors {x1, ... , xk, yl, ... , yk}. Clearly, V is finite dimensional, and so (since X is infinite dimensional) V is a non-trivial closed subspace of X. Since V is Ta-invariant, for each 1 < n < k there exist scalars A1,n, A2,n, ... , Ak,n such that Tczn = E;=1 Aj,nzj. So, if Aj,n = aj,n + tOj,n E C, then it follows from k

Txn + zTyn = Tezn = F, Aj,nz) = j=1

k

k

Qj,nyj) + t E(Qj,nxj + aj,nyj) j=1

3=1

10.1. A Smorgasbord of Invariant Subspaces

303

that Txn = j=1(aj nxj - /33,,,yj) E V and Tyn = Ej=1(Qj.nxj + aj,nyj) E V. These show that v is T-invariant.

(b) Now assume that V is Tc-hyperinvariant. As in the preceding part, let {z1, z2,

... , zk} be a basis of V, where zn = xn + :yn E Xc, and let V be the

non-zero finite dimensional vector subspace of X generated by the set of vectors {xt, ..., xk, yl,... , yk }. We shall show that V is T-hyperinvariant. To this end, assume that an operator S E £(X) satisfies ST = TS. This implies SeTe= T,S,,, and so (since V is Tc-hyperinvariant) V is Sc-invariant. In particular, for each n = 1, 2,. . .. k there exist scalars uj,,, = -y,n + zbj,, E C (j = 1, 2,. . ., k) such that k

Sxn + t$yn = Sczn =

k

k

,Uj.nzj = 1:('Vj.nxj - bj.nYj) + t F,(1'j,nyj { bj,nxj) J=1

j=1

j=1

Therefore, Sx,, = Ej=1(-tj.nzj -bj,nyj) E V and Syn = Ej=1(_yj,nyj +bj,nxj) E V for each n = 1, 2,..., k. This implies that V is S-invariant, and consequently V is T-hyperinvariant.

Problem 10.1.5. Let T : X ---p X be a bounded operator on a Banach space.

If X is not reflexive, then show that the double adjoint T**: X" - X" has a non-trivial norm closed T"-invariant subspace. Solution: If X is not a reflexive Banach space, then X is a non-trivial closed subspace of X" and T" (X) = T(X) C X. That is, X is T"-invariant, and so T" has a non-trivial closed invariant subspace.

Problem 10.1.6. Let 1 be a compact Hausdorff space. Fix some function 0 E C(I) and consider the multiplication operator Mm on C(fI) defined by Mp (f) = Of. If ito is a non-empty subset of SI and V = {f E C(11): f (w) = 0 for all w E SIo} , then show that V is a closed M0-invariant subspace of C(SI). Solution: Clearly, V is a closed subspace of C(f2). Now if f E V and w E (o, then we have [M0(f)](w) = O(w) f (w) = 0(w) 0 = 0. This means M0(f) E V, and so V is Mb-invariant.

Problem 10.1.7. Let 11 be a compact Hausdorff space. Fix a continuous

mapping r: fI - 1 and consider the composition operator C, on C(Q) defined, as usual, by C, (f) = f or. If S?.a is a r-invariant non-empty subset of fI (that is, r(N) C 1o), then show that the vector subspace

V = If E C(SI): f (w) = 0 for all w E SIo} is a closed CT-invariant subspace of C(SI).

Solution: Clearly V is a closed subspace of C(Sl). Now if f E V and w E fto, then we have r(w) E flo, and so [C, f J(w) = f(r(w)) = 0. This implies C,(f) E V, and hence V is C,-invariant.

10. Invariant Subspaces

304

Problem 10.1.8. If T : X -ti X is a bounded operator on a Banach space, then show that:

(a) Its commutant {T }' = IS E L(X): ST = TS } is a closed unital subalgebra of C(X).

(b) T has a non-trivial closed hyperinvariant subspace if and only if {T }' is a non-transitive algebra. Solution: (a) Let S, R E {T}' and let A be a scalar. Then, we have

(S+R)T = ST+RT=TS+TR=T(R+S), (AS)T = A(ST) = A(TS) = T(AS) , (SR)T = S(RT) = S(TR) = (ST)R = (TS)R = T(SR) ,

IT = TI =T. Therefore, S+T, AS, SR, and I all belong to {T }', and so {T}' is a unital subalgebra of the Banach algebra C(X ). To see that {T}' is closed in C(X ), assume that a sequence {Sn} in {T}' satisfies

S. -. S in C(X ). Then TSn = SnT for each n, and by letting n - oo and using the joint continuity of the map (A, B) '-4 AB (see Problem 1.1.14), we get TS = ST, i.e., S E {T}'. Therefore, {T}' is a closed subalgebra of £(X).

(b) By definition, a vector subspace V is T-hyperinvariant if and only if it is invariant under every bounded operator that commutes with T. It remains to notice that the latter is equivalent to saying that V is a {T}'-invariant subspace.

I

Problem 10.1.9. If X is a Banach space, then show that the only 1-hyperinvariant subspaces are {0} and X. Solution: Assume that V is a non-zero 1-hyperinvariant subspace of X. Since {I}' = L(X), it follows that V is invariant under every bounded operator on X. Fix some no-zero vector v E V, and let x E X. Also, choose some x' E X* with x' (v) = 1. Now if we let T = x' ® x, then T is a bounded operator, and so it leaves V invariant. In particular, we have Tv = x E V. Since x E X is arbitrary, it follows that V = X. I

Problem 10.1.10. Let A be an algebra of operators on a vector space. Show that Al = {al + A: a scalar and A E A} is the unital algebra generated by A. (This implies that any algebra of operators A and the unital algebra generated by A have the same invariant subspaces.) Solution: Let A be an algebra of operators on a Banach space X. It should be clear that Al = {aI + A: a scalar and A E A} is the vector subspace of C(X) generated by A and the identity operator I. It remains to verify that Al is a subalgebra of C(X). To see this, take any two operators a1I + Al and a2I + A2 in

Al and note that

(all + Ai)(a21 +A2) = a1a21+ (ajA2 +a2A1 +AiA2). Since a1 A2 + a2A1 + Al A2 E A, the latter shows that (all + Al)(a21 + A2) E A1.

10.1. A Smorgasbord of Invariant Subspaces

305

The obtained representation of Al shows at once that A and AI have the same I invariant subspaces.

Problem 10.1.11. Show that an algebra of operators A C £(X) has no non-trivial closed invariant subspaces if and only if for any pair x, y E X with x # 0 and every c > 0 there exists some operator A E A such that IIAx - yII < E. Solution: Assume first that A has no non-trivial closed A-invariant subspaces. This implies that A , = X for each x # 0. and the validity of the desired condition follows.

For the converse, suppose that for any pair x, y E X with x 0 0 and every ( > 0 there exists some operator A E A such that IIAx - yII < f. Also, let V he a non-zero closed A-invariant subspace. Fix a non-zero vector xo E V. Now if y E X and f > 0 are given, choose some A E A such that IIAx - yII < E. Since V is A-invariant, it follows that Ax E V, and from this we infer that y E V = V. Therefore, X C V so that V = X. This shows that A has no non-trivial closed invariant subspaces. I

Problem 10.1.12. Let A be a unital algebra of continuous operators on a Banach space X and let Ax = {Ax: A E A}. Show that A is non-transitive if and only if A has a non-trivial closed subspace of the form Ax. Solution: Assume first that for some x E X the closed vector space

is nontrivial. Note that if A E A, then for each T E A, we have TA E A, and so

T(Ax) = TA(x) E Ax. This shows that Ax is A-invariant. By continuity, we obtain that Xx. is also A-invariant. Now suppose that there is a non-trivial closed A-invariant subspace V. Fix some non-zero vector v E V, and note that for each A E A we have Av E A(V) C V. is a This implies that Av C V = V # X. Since v = Iv E AV, it follows that non-trivial closed subspace of X. I

Problem 10.1.13. Let T : X - X be a bounded operator on a Banach space and consider the set

V = {xE X: nim IIT"xII =0}. -oo

Establish the following.

(a) V is a T-hyperinvariant subspace. (b) If T is power bounded, then V is a closed T-hyperinvariant subspace.

Solution: (a) We show first that V is a vector subspace. To this end, let X. y E V and let A be a scalar. Then, from IIT"(Ax)II = IAI - IIT"'xII -' 0, and

0, 0 < IIT"(x + y)II = IIT"x + T"yII 5 IIT"xhi + IIT"yII it easily follows that x + y E V and Ax E V. Thus, V is a vector subspace. To see that V is T-hyperinvariant, let S E £(X) satisfy ST = TS. Now if x E V, then IIT"(Sx)II = IIS(T"x)II <- IISII . IIT"iiI --. 0.

10. Invariant Subspaces

This implies Sx E V, and so V is S-invariant. Therefore, the vector subspace V is T-hyperinvariant. (b) Now assume that there exists some M > 0 such that 117' 11 < M for each

n, and let v E V. Fix e > 0, and then pick some w E V such that IIv - wlI < E. Next, choose some no such that IIT"wiI < e for all n > no. Consequently, for each n n0 we have IIT"vII

IIT"(v - w)II + IIT"w11 < MIIv - wfl + IIT"wli < Me + E = (M + 1)f.

This shows that IIT"vII - 0, and so v E V. Thus, V = V and hence V is a closed T-hyperinvariant subspace.

Problem 10.1.14. Let S, T : X - X be a pair of bounded operators on a Banach space. As usual, the direct sum of S and T is the operator S ® T on X ED X defined by (S ® T)(x S y) = Sx ® Ty.

If a bounded operator R: X -i X satisfies RS = TR, then show that its graph GR = {x (D Rx: X E X) is a closed S ® T-invariant subspace of

X ®X. Solution: It should be clear that GR is a vector subspace of X ® X. For each x(D RxEGRwehave (S ® T)(x ® Rx) = Sx e T(Rx) = Sx ® R(Sx) E GR . That is, GR is S ® T-invariant. To see that GR is a closed subspace, assume that a sequence {x" ® Rx" } C GR

satisfies x" ®Rx,, - x ®y in X ®X. This is equivalent to having x,, - x and Rx" - y in X. Since R is continuous, it follows that Rx,, -y Rx, and so y = Rx. Therefore, x ®y = x ®Rx E GR, and thus GR is a closed subspace of X ®X X.

I

Problem 10.1.15. Show that the shift operator on any ep-space has a nontrivial closed hyperinvariant subspace. Solution: Recall that the shift operator S: tp -, 1p is defined by S(x1i x2, x3, ...) = (O, x1, x2, x3, ...) .

Let V = {(x1ix2.... ) E tp: x1 = 0}. Clearly, V is a closed subspace of 1p, and we claim that V is S-hyperinvariant. To see this, notice that V is the range of S (that 1 is, S(tp) = V), and then invoke Lemma 10.2.

Problem 10.1.16. Show that the converse of Corollary 10.13 is not true. That is, show that the dual algebra A' of a transitive algebra of operators A may be non-transitive. Solution: According to C. J. Read (66] there is a bounded operator T: 11 -, t1 without non-trivial closed invariant subspaces. Let A be the unital algebra generated by T. That is, A = {p(T) : p is a polynomial} I. Clearly, A is transitive.

10.2. The Lomonosov Invariant Subspace Theorem

307

Next, notice that the adjoint T': e, - e has non-trivial closed invariant For instance, for any arbitrary non-zero vector x E e,, the closed

subspaces.

subspace V generated by {x, T'x, (T' )2x, ... } is a non-trivial closed T'-invariant subspace. This is a simple consequence of the fact that e is not separable. Now it should be clear that V is a non-trivial closed A'-invariant subspace. 0

Problem 10.1.17. Let 0: f2 - R be a continuous function separating the points of a compact Hausdorff space SZ and let M0: C(Q) - C(Q) denote the usual multiplication operator defined by Mo(f) = Of. Establish that

{M0}'_{M9: gEC(f2)}. Solution: Clearly M,Mg = M9M4, for each g E C(f2), i.e.,

(Mg: g E C(ft)} C {M4,}'. Let T E {M4,}' and put h = T(1). Since TM4, = MOT implies T(cbf) _ OT(f ) for each f E C(1l), it follows that T(O) = h4, and by induction T(01) = ho " for each n > 0. The latter implies T(.p(O)) = hp(O)

(*)

for each polynomial p. Since the function 0 separates the points of (l, it follows from the Stone-Weierstrass Approximation Theorem that the algebra A = IPM: p a polynomial } is norm dense in C(fl). Hence, in view of (*), we get T f = h f for each f E C(fl)This shows that T = Mh, and consequently {MO}' C {Mg : g E C(fl)} is also true. I We now conclude that {M4,}' = {Mg: g E C(fl)}.

10.2. The Lomonosov Invariant Subspace Theorem Problem 10.2.1. Present an example of a polynomially compact operator which is not compact. Solution: Any nilpotent operator which is not compact provides a desired example. We will produce such an operator. Let X be an infinite dimensional Banach space and consider the operator T : X ® X - X ® X defined by T(x ® y) = 0 ® x. Obviously, T is a bounded operator which is not compact. It remains to notice that

T2=0.

1

Problem 10.2.2. If fl is a compact Hausdorff space without isolated points and a function 0 E C(fl) is one-to-one, then show that the multiplication operator M4, does not commute with any non-zero weakly compact operator. Solution: Let MOT = TM4, for some weakly compact operator T : C(fl) - C(O). By Lemma 4.19(3), we have T = Mf for some function f E C(f2). Now a glance at Lemma 4.18 reveals that f must vanish at every accumulation point of Q. Since f2 has no isolated points, this implies that f = 0. Consequently, T = M f = 0. 1

10. Invariant Subspaces

308

Problem 10.2.3. Show that every multiplication operator on C(Q) is a Lomonosov operator.

Solution: If Il is finite, then C(f) = It" or C(fl) =

C". and so every operator on

C(Sl) is compact - and hence, a Lomonosov operator. Consequently, we can assume

that 1 is an infinite set. Now let 0 E C(fl). Fix wo E Il and then choose two open sets V and W such that wo E V C V C W and WC # 0. Next, select a function f E C(f2) with f = I on V and f = 0 on W`. Then the operator Alf is non-zero. is not a multiple of the identity operator and commutes with Afo. By Theorem 4.20 the operator Alf commutes with a rank-one operator. Therefore, AfO is a Lomonosov operator.

I

Problem 10.2.4. Let (f2, E, A) be a a-finite measure space such that there exist two disjoint sets A, B E E with 0 < µ(A) < oc and 0 < p(B) < oo. Show that every multiplication operator on Lp(1I, E, p), where 1 < p < oo, is a Lomonosov operator. Solution: We know that the multiplication operators on Lt,(p) are determined by the functions of L,,,.(u); see Problems 3.3.5 and 5.2.7. So, fix any 6 E Lx(p), and let Jr = v . Clearly, the multiplication operator Mt is not a multiple of the identity operator and commutes with the multiplication operator Mo. Now a glance at Problem 4.2.9 shows that Al f commutes with a rank-one positive operator. Consequently, AI, is a Lomonosov operator.

Problem 10.2.5. Show that every Markov operator T on an AM-space with unit commutes with a rank-one positive operator-and hence T is a Lomonosov operator.

Solution: Let E be an AAf-space with unit e and suppose that T: E -, E is a Markov operator, that is, T is positive and Te = e. Consider the non-empty u'-compact convex set

C={0 EE+: y:,(e)=1} Notice that if 0 E C. then T'tp(e) = U;(Te) = rli(e) = 1, that is, C is T'-invariant. Moreover, it is easy to see that the mapping TO: (C. w') -- (C, w' ) is continuous. Therefore, by Tychonoff's Fixed Point Theorem (see, for instance, [4, Corollary 14.51, p. 4831) there exists some 0 E C such that T'd = 4. To finish the solution, we shall show that the rank-one positive operator on E given by K = o ® e commutes with T. Indeed, for each x E Ewe have

KTx = ¢(Tx)e = [T'O](x)e = a(x)e = Kx, and TKx = T(4(x)e) = q(x)Te = Q(x)e = Kr. Consequently, KT = TK = K.

Problem 10.2.6. Consider the operator T: C[-1, 11 -; C[-1, 1] defined by Tx(t) = x(-t) for each x E C[-1, 11 and all t E [-1,1]. (a) Show that T is a non-scalar surjective lattice isometry which is also a Markov operator--and so, by Problem 10.2.6, T commutes with a rank-one positive operator and is a Lomonosov operator.

10.2. The Lomonosov Invariant Subspace Theorem

309

(b) Exhibit a rank-one operator on C[-1, 1] that commutes with T. (c) Show that the compact operator S: C[-1, 1] - C[-1, 1], defined by

Sx(t) = Jx(s)sin(st)ds, commutes with the operator T. (d) Show that 1 and -1 are the only eigenvalues of T.

(e) Exhibit non-trivial invariant subspaces different from the eigenspaces N1 and N_ 1. Solution: (a) Clearly, T is non-scalar, one-to-one and surjective-if y E C[-1, 11 consider the function x E C[-1, 11 defined by x(t) = y(-t) and note that Tx = y. From Tx > 0 if and only if x > 0, it follows that T is a lattice isomorphism, and TI = 1 guarantees that T is a Markov operator. It should be obvious that

IITxII. = sup ITx(t)I = sup Ix(-t)I = sup Ix(t)I = llxll,o, -1
-i
-1
and so T is an isometry. (b) Assume that rb E C[-1, 11 is an even non-zero function, i.e., ¢(-t) = ¢(t) for

each t E [-1.1], and let K: C[-1.1] - C[-1, 1] be the rank-one operator defined by Kx = x(0)0. Now note that for each x E C[-1, 1] and each t E [-1,1] we have TKx(t) = T(x(0)6)(t) = r(0)0(-t) = x(0)O(t) = Kx(t), and KTx(t) = x(0)m(t) = Kx(t) . Thus, TK = KT = K, and so T commutes with the rank-one operator K. (c) If x E C[-1. 1] and t E [-1,1], then we have

TSx(t) = STx(t) =

j J

1

x(s) sin(-st) ds = -

1

I

j ./

r1

x(-s) sin(st) ds =

l x(s) sin(st) ds , and 1

J-lx(u) sin(ut) du JJ1

x(s) sin(st) ds. .1

1

This shows that the compact operator S commutes with T.

(d) Assume that Tx = Ax for some non-zero x E C[-1, 1]. This implies that x(-t) = Ax(t) for each t E [-1, 1]. If A = 0, then it follows that x = 0, and so A = 0 cannot be an eigenvalue of T. Thus, we can assume that A 54 0. Fix some to E [-1, 1] with x(to) 96 0 and note that x(-to) = Ax(to) 76 0. From

r(-to)x(to) = x(-to)x(-(-to)) = Ax(to)Ax(-to) = A2x(-to)x(to), it follows that A2 = 1. Therefore, the only possible eigenvalues of T are 1 and -1. To see that these are indeed the eigenvalues of T observe that Ti = 1 and that if

y(t) = t, then Ty(t) = y(-t) = -t = -y(t), i.e., Ty = -y. (e) To exhibit non-trivial closed invariant subspaces different from the eigenspaces N1 and N_1, for each 0 < b < 1 let Va = {x E C[-1,1] : x = 0 on [-b, 611.

10. Invariant Subspaces

310

It is easy to check that each V6 is a non-trivial closed T-invariant subspace that is different from the eigenspaces of T. I

Problem 10.2.7. This exercise is a combination of results obtained by W. B. Arveson and J. Feldman [15], C. Pearcy and N. Salinas [62], and P. Meyer-Nieberg [56]. Let X be a complex Banach space and for an operator T E £(X) let R(T) denote the norm closure in £(X) of the collection of all operators q(T), where q(A) is a rational function (i.e., a quotient of two polynomials with complex coefficients) that is analytic in a neighborhood of the spectrum Q(T). Clearly, R(T) is a commutative unital algebra of operators.

Show that if an operator T E £(X) is non-scalar and R(T) contains a non-zero compact operator, then R(T) is non-transitive. Solution: Let X be a complex Banach space and fix an operator T E £(X). Clearly, R(T) C {T}' and R(T) is a commutative unital algebra. If K is a non-zero compact operator in R(T), then it follows from Lomonosov's Theorem 10.19 that T has a non-trivial closed hyperinvariant subspace V. Thus, V is R(T)-invariant, and so R(T) is a non-transitive algebra. I

10.3. Invariant Ideals for Positive Operators Problem 10.3.1. Let S: E - E be a positive operator on a Banach lattice such that Sx0 > yxo holds for some vector xo > 0 and some scalar y > 0. Show that S is not locally quasinilpotent at xo. Solution: It follows that for each n E N we have S"xo > y"x0. Therefore, IIS"xoII -> 'r"IIxoII for each n, and so liminf IIS"xoII

>_ yliminf IIxofl

This proves that S is not quasinilpotent at xo.

^

= 7 > 0.

I

Problem 10.3.2. Let J be an ideal in a Banach lattice E. If J is invariant under some positive operator B : E - E, then J is also invariant under every operator dominated by B.' Solution: Let J be an ideal in a Banach lattice E that is invariant under a positive operator B : E -' E. Assume that an operator T : E -' E is dominated by B. Then

for any x E J we have ITxI < BIxJ E J, and so Tx E J. This shows that J is

I

T-invariant.

Problem 10.3.3. Give an example of a pair of non-commuting positive operators A and B such that A E (B), that is, AB - BA > 0. Solution: Consider the positive operators A, B : t2 - Q2 defined by A(xl, x2, x3, ...) = (x2, x3, ...)

and

B(xI, x2, x3, ...) = (0, x1, x2, x3, ...) .

1 Recall that an operator T: E - E is dominated by B if ITxI < Bjxi for each x E E.

10.3. Invariant Ideals for Positive Operators

311

That is, A is the backward shift and B is the forward shift. Then for each vector x = (x1, x2, x3, ...) in e2 we have ABx = x = (XI, x2, x3, ...) and BAx = (0. x2, X3.... ) . This shows that AB > BA. that is, A E [B) and AB 96 BA, i.e.. A and B do not commute.

For another example, take for T the operator B from above and let S: t2 -+ t2

be the (projection) operator defined by S(xi, x2, ...) _ (X1.12,... , xn, 0, 0....). Then for each x = (x1. x2, ...). we have

TSx = (0,11,x2.....xn-1,xn,0,0....) and STx = (Qx1,x2,...,xn_1,0,0.... ). This implies TS > ST and TS # ST.

U

Problem 10.3.4. For a positive operator B: E - E on a Banach lattice establish the following.

(a) The range ideal RB of B (i.e., the ideal generated by the range of the operator B) is given by RB = { y E E : 3 .r E E+ such that I yI < Bx }

.

(b) The range ideal RB is f-hyperinvariant. (c) For each A E [B) the range ideal RA is B-invariant. (d) The null ideal NB is A-invariant for each A E [B). In particular, NB is E-hyperinvariant. (e) Present an example of an operator A E (B] for which the ideal NB fails to be A-invariant. (f) If B is non-zero and [B) contains an ideal irreducible operator, then B is strictly positive. Solution: (a) It should be clear that (y E E : 3 x E E+ such that I yI < Bx } C RB . Now let y E R8. This means that there exist positive scalars o1, ... , on and vectors

x1 ....,x, such that IyI < E 1 o,IBr;i < B(E

o;Ix;I). Therefore, if we put 1 x = Fn a, Ix, I E E+, then I yI < Bx. This shows that RB C l y E E : 3 x E E+ such that I yI < Bx } . Consequently, RB = {y E E: 3 x E E+ such that IyI < Bx }. (b) Take any positive operator A commuting with B and let y E RB. Pick some x E E+ such that. IyI < Bx. Therefore, IAyl < AIyI < ABx = B(Ax). This shows that Ay E RB, and thus RB is A-invariant.

(c) Let y E RA. By part (a) there exists some x E E+ such that IyI < Ax. This implies IByI < BIyI < BAx < A(Bx). and so By E RA. Therefore, RA is B-invariant.

(d) Fix A E [B) and let x E NB. The inequalities BIAxI < BAIxI < ABIxI = 0 imply BIAxI = 0, and so Ax E NB. That is, NB is A-invariant.

10. Invariant Subspaces

312

(e) For an example. let B be the backward shift and A be the forward shift on e2. By Problem 10.3.3 we know that A E (B], and an easy argument shows that

NB = {x = (x1,0,0,0....): x1 E R} Clearly, NB is not invariant under A. (f) Assume that an operator A E [B) is ideal irreducible. Clearly, A # 0. According to the conclusion of part (a), the closed ideal NB is A-invariant. So, by the ideal irreducibility of A, we have either NB = {0} or NB = E. Since B # 0, it follows that Na = {0}. and so B is strictly positive.

Problem 10.3.5. Assume that K: E -- E is a compact operator on a Banach lattice satisfying Kxo # 0 for some xo E E. Show that there exists a compact operator C such that Cxo > 0 and ]CxI < I Kxl for each x E E. Solution: Let J = RK be the range ideal of K, i.e.. J is the ideal generated by the range K(E) of the operator K. We claim that j has a quasi-interior point. To see this. start by observing that the compactness of K implies that the range of K is separable: see Problem 2.4.6. So. there exists a countable set (un) of

non-zero vectors in K(E) that is norm dense in K(E). Let u = E-

12_n

lu,.ll

It

should be clear that u E J and so Eu C J. where Eu is the principal ideal generated by u in E. Therefore EL C J. Conversely, from the inequalities Iunl < 2'

and the fact that {un) is dense in K(E) it follows that J C Eu. This proves that E and hence u is a quasi-interior point in J. Since the vector Kro belongs to J and is non-zero, Lemnia 4.16 guarantees

that there is an operator V: J - J such that V(Kxo) > 0 and IVyI < Iyj for each y E J. Finally, consider the operator C = VK. Clearly C is a compact operator and Cxo = V Kro > 0. To complete the solution, it remains to notice that ICrI = IVKrI < KIxI for each x E E. I

Problem 10.3.6. For a positive operator T on a Banach lattice E let 21T = {A E C(E): 3 B E {T}'.* such that IAxI < BIx3 for all x E E} , where {T}', denotes as usual the set of all positive operators in the commutant of T. Show that

(a) The center of E is included in T. i.e., .(E) C 21T. (b) 2tT is a unital subalgebra of C(E). (c) An ideal J is e-hyperinvariant forT if and only if J is 2tT-invariant. Solution: (a) Since every operator S E Z(E) satisfies ISxI < AIxI = AIIxI for some A > 0 and the identity operator 1 commutes with T, it follows that S E %T, i.e., Z(E) C 2tT. (b) A straightforward verification shows %T is a vector subspace of C(E). Now assume that Si, S2 E 2iT. Pick two positive operators B1. B2 E {T}' such that B, dominates S, for i = 1.2. Now note that for each x E E we have 1S1S2xI < B1IS2ri _5 B1B2IxL,

10.3. Invariant Ideals for Positive Operators

313

i.e., S, S2 is dominated by Bl B2. Since Bl B2 E {T}', we see that S, S2 E 21T. The above show that 21T is a unital subalgebra of C(E).

(c) Assume that an ideal J in a Banach lattice E is t-hyperinvariant for T. Also, let S E 21T. This means that there exists some positive operator B: E -y E that commutes with T and dominates S. Since J is £-hyperinvariant for T, it follows that J is also B-invariant. But then, by Problem 10.3.2. J is automatically S-invariant. Therefore, J is 2ST-invariant. For the converse, assume that an ideal J in E is 21T-invariant. Next notice that if B : E ---+ E is a positive operator that commutes with T, then B E 21T, and so J must be B-invariant. This shows that the ideal J is t-hyperinvariant for T.

Problem 10.3.7. Recall that an operator T : E - F between two Banach lattices is said to be AM-compact, provided that T maps order bounded sets to norm totally bounded sets, i.e., if T [a, b] is a norm totally bounded set for each order interval [a, b]. Give an example of an AM-compact operator which is not compact. Solution: We shall use the following fact: If 1 < p < oc, then every order interval in £p is a norm compact set.

To see this, let a = (al, a2, ...) be a positive vector in some £p-space, where 1 < p < oc, and consider the order interval 10, a]. Clearly [0, a] is a norm closed subset of £p. Pick some k such that (Enkan)" < e for all n > k. If we let b = (at, a2, ... , ak, 0, 0 ...), then the interval [0. b] is obviously compact, and [0, a] C [0, b] + W.

where U is the closed unit ball of 4. This implies that [0, a] is a norm totally bounded subset of £p, and thus [0, a] is a norm compact subset of £p. The preceding statement shows that every positive operator from any Banach

lattice to any £p space with 1 < p < oc is an AM-compact operator. However, not every positive operator on an £P space is a compact operator. For instance, the identity operator and the shift operator are positive and non-compact. Hence, every non-compact positive operator on an £p space (with 1 < p < oo) provides an example of an AM-compact positive operator which is not a compact operator.

Problem 10.3.8. Let B : E

E be a positive operator on a Banach lattice and assume that there exists a positive operator S: E -- E such that

(1) S E (B], that is. SB < BS. (2) There exists some xo > 0 satisfying lim inf IIS"xoII n = 0.

(3) S dominates a non-zero AM-compact operator. Show that the operator B has a non-trivial closed invariant ideal. Solution: The proof is similar to that of Theorem 10.24. Let B, S, and x0 satisfy the properties stated in the theorem, and let K be a non-zero AM-compact operator dominated by the operator S, i.e., I Kxl < S(IxI) for each x E E. We can assume

10. Invariant Subspaces

314

II B I I < 1.

Then the series A = E'00 B' defines a positive operator on E and

A E [S), i.e., AS > SA. To see this, use first the hypothesis that S E (B] to show that SBn < B"S for each n > 0, and then note that 00

00

00

00

SA=S(yBn) =>2SB" 2B')S=AS. n=0

n=0

n=0

n=0

Thus, A E [S). This also implies that SAk < AkS for all k > 0. For each x > 0 we denote by Jy the principal ideal generated by Ax. From 0 < x < Ax we see that x E J,, and so J= is non-zero. Moreover, we claim that J" is B-invariant. Indeed, if y E JZ, then Iyl < \Ax for some A > 0, and so 00

Bnx < \Ax,

I ByI 5 Blyl 5 AB(Ax) = A n=1

which implies By E J=. So. J2 is a non-zero closed B-invariant ideal. The proof will be finished if we show that j. 0 E for some x > 0. To establish this, assume by way of contradiction that:

Jx = E for each x > 0.

(*)

We claim that without loss of generality we can suppose Kxo 54 0. To see this,

consider the ideal J_-,. If Ky = 0 for each 0 < y E J 0, then K = 0 on J 0, and so by (*) we get K = 0, a contradiction. Therefore, Kyo 96 0 for some 0 < yo E J 0. We claim that lim infn-oo IIS"yoll " = 0. To prove this, pick some scalar A > 0 such

that 0 < yo < \Axo holds and then notice that 0 < S"yo < aS"Axo < \AS"xo also holds for each n. This implies IISnyoII" < a" IIAII" . IISnxoll °, from which it follows that liminfn-o0 IIS"y0II' = 0. So, replacing (if necessary) x0 by yo, we can assume Kxo 0. By scaling, we can assume IIKII = 1. Also, replacing xo by ax0 for an appropriate scalar a > 1. we can suppose that llxo II > I and Il Kxo II > 1. Now denote by UO the closed unit ball centered at x0 > 0. By our choice of xO, we have 0 l UO

and

0 V K(Uo).

(**)

From (*), we know that 7 = E for each x # 0. Hence, for each y E E+ the sequence { y n nA(l x I) } is norm convergent and moreover, in view of Lemma 4.15,

y A nA(I xl) = y. In particular, for each x 96 0 there exists some n such that IIxo - xo A nA(Ixl)II < 1. Since the function z ' x0 A nA(Izl) is continuous, the set {z E E: IIxo - xo A nA(Izl)ll < 1} is open for each n. In view of the fact that 0 K(U0), the previous arguments guarantee that 00

K(Uo n [O,xo]) C U {z E E: IIxo - xo n nA(Izl)ll < 1). n=1

Since the AM-compactness of K implies that K(Uo n [0, x0]) is compact and since {z E E: IIxo - x0 A nA(Izl)ll < 1} C {z E E: IIxo - x0 A (n + 1)A(Izl)II < 1) for each n, it follows that

K(Uon[0,xo]) C {z E E: IIxo-x0nmA(Izl)II < 1}

10.3. Invariant Ideals for Positive Operators

315

for some m. In other words, there exists some fixed natural number m such that x E K(Uo n [0, x0)) implies xo A mA(ITI) E Uo.

In particular, xl = xo A rA(IKxoI) E Lo n [0, xo]. Since Kx1 E K(Uo n [0, xo]),

it follows that x2 = xo A mA(IKx1I) E Uo n [0,x0]. Proceeding inductively, we obtain a sequence {x71 } of positive vectors in Uo n [0, xo] defined recursively by the formula x71+1 = x0 A mA(I Kx71I ). Now we claim that

0<xn<m"A"S"xo for each n. The proof is by induction. First recall that SAk < AkS holds for all k > 0. For n = 1, the desired inequality xl = xo A mA(IKxoI) < mA(Sxo) is trivially true. For the induction step, note that if 0 < x" < m"A"S"x0 for some n, then the inequality I Kx71I < Sxn implies

x0 A rA(IKx7I) < mA(IKx7I) < mASx71 mn+iAS(A"S"xo) = mn+'A(SAn)Snxo

< mn+IA(A"S)S"xo < mn+'An+lsn+lx0. Thus, 11x7111

5 mn

Since lim inf IIS"xo II

IIS"xoll * for each n.

IIAII"IISnx0II, and so 11x7111 ^ 5 mIIAII

0. it follows that lim inf Ilxn 11

0. Thus, there exists a

n-ac n-oc subsequence {xk,} of {x, } such that Ilxk II * - 0. In particular, xk -' 0. Since {xk } C Uo, this implies that 0 E Uo = Uo, contrary to (**), and the solution is finished.

Problem 10.3.9. Let B: E -. E be a positive operator on a Banach lattice and assume that there exists a positive operator S : E -+ E such that:

(1) S E [B). that is, SB > BS. (2) S is quasinilpotent. (3) S dominates a non-zero AM-compact operator.

Show that the operator B has a non-trivial closed invariant ideal. Solution: Let B. S satisfy the properties stated in the theorem, and let K be a non-zero AM-compact operator dominated by S, i.e., I KxI < S(I xI) for each x E E.

We can assume that IIBII < 1. Then. the series A = En o B" defines a positive operator on E. Using the condition S E [B), we can easily verify that the inequalities SBk > BkS and SkA > ASk hold for each k. For each x > 0 we denote by Jz the principal ideal in E generated by Ax. Since x < Ax E J, the ideal J= is non-zero. Moreover, we claim that J= is B-invariant. Indeed, if y E J=, then IyI < AAx for some A > 0. and so ao

B"x < AAx.

IByI 5 Blyl 5 AB(Ax) = A n=1

which implies By E J=. So, Jr is a non-zero closed B-invariant ideal. The proof will be finished, if we show that J,, E for some x > 0. To establish this, assume by way of contradiction that

J= = E for each x > 0.

(*)

10. Invariant Subspaces

316

Since K 0 0, we can find some xo > 0 for which Kxo # 0. By scaling, we can assume IIKII = 1. Also, replacing xo by axo for an appropriate scalar a > 1, we can suppose that 11x0 11 > 1 and II Kxo II > 1. Now let Uo = (z E E: Ilxo - zll <- 11 be the closed unit ball centered at xo > 0. By our choice of xo, we have O f Uo

and

O V K(Uo).

(**)

By (*), we know that ji = E for each x 96 0. Hence, for each y > 0 the sequence {ynnA(Ixl )} is norm convergent and limn-. ynnA(Ixl) = y. In particular, for each x 0 0 there exists some n such that Ilxo-xonnA(IxI)II < 1. Since the function z xonnA(Izl) is continuous, we see that the set {z: Ilxo-xonnA(IzI)II < 1) is open for each n. In view of 0 K(Uo), the above arguments guarantee that 00

K(Uon[0,xo)) S U{x E E: Ilxo-x0nnA(Izl)II < 11. n=1

The AM-compactness of the operator K guarantees that the set K(Uo n [0, xo]) is

compact. This and the fact that the sets {z E E: Ilxo - xo A nA(Izl)II < 1} are increasing as n increases imply that

K(Uo) S (z E E: Ilxo - xo n mA(Izl)Il < 1) for some m. In other words, there exists some fixed m such that x E K(U0 n [0, x0)) implies that xo A mA(Ixl) E Uo. In particular, we have x1 = xo A mA(I Kxol) E Uo n [O, xol. Consequently

Kxl E K(U0 n [O,xo]) and so X2 = xo A mA(lKxll) E Uo n [O,x0j. Proceeding inductively, we obtain a sequence {xn } of positive vectors in Uo n [0, xo) defined by xn+1 = xo A mA(l Kxnl ). We claim that

05 xn5 m"S"A"xo holds for every n. The proof is by induction in n. For n = 1 the inequality xl = xo A mA(lKxol) < mA(Sxo) < mSAxo is trivially true. For the induction

step, recall that AS' < SkA and note that if 0 < xn 5 m"S"A"xo for some n, then 0 < xn+1

= X o A mA(I Kxn l) < mA(I Kx I) < mA(Sxn )

< mnt1A(SS"A"xo) = m"+'A(Sn+'Anxo) 5 mn+lsn+lAn+lxo Thus Since Jim IIS" II

0, it follows that n imo llxn lI

SmIISnIl.IIAII.IIxoII-for eachn. 0, and hence Jim llxn ll = 0.

However, since {xn } C Uo, we can conclude that 0 E Uo = Uo, contrary to (**), and the solution of the problem is finished. I

Problem 10.3.10. Recall that a bounded operator T : X -- Y between Banach spaces is called a Dunford-Pettis operator if xn -W-4 0 in X implies that jITxnII - 0 in Y. Show that every positive Dunford-Pettis operator on a Banach lattice which is quasinilpotent at a non-zero positive vector has a non-trivial closed invariant ideal.

10.4. Invariant Subspaces of Families of Positive Operators

317

Solution: Let B : E -i E be a positive Dunford-Pettis operator on a Banach lattice and let B be quasinilpotent at a non-zero positive vector. We can assume that B is strictly positive; otherwise NB is a non-trivial closed invariant ideal. Since B is a Dunford-Pettis operator, it follows that B carries order intervals onto relatively weakly compact sets [6, Theorem 19.12, p. 339]. This guarantees that the strictly positive operator B2 carries order intervals to norm totally bounded sets. That is, B2 is a non-zero AM-compact operator. Since B commutes with B2, 1

the desired conclusion follows immediately from Corollary 10.28.

10.4. Invariant Subspaces of Families of Positive Operators Problem 10.4.1. Let S be a multiplicative semigroup of operators on a Banach space and let S1 = S U {I}, where I is the identity operator. Show that S1 is again a multiplicative semigroup. Solution: Take two arbitrary operators T1, T2 in S. If neither of then is I, then both of them belong to S and so T1T2 E S since S is a multiplicative semigroup. If one of them, say T1, equals I, then T1 T2 = T2 and therefore the product again I belongs to S.

Problem 10.4.2. Let S be an additive semigroup of operators on a Banach space and let So = S U {0}, where 0 is the zero operator. Show that So is again an additive semigroup. Solution: Take two arbitrary operators T1, T2 in So. If neither of them is the zero operator, then both of them belong to S and so T1 + T2 E S since S is an additive semigroup. If one of them, say T1, equals to the zero operator, then T, + T2 = T2 1 and therefore the sum again belongs to So.

Problem 10.4.3. Let C be a collection of bounded operators on a Banach space X and let Qf be the set of all vectors in X at which C is finitely (locally) quasinilpotent. Show that Qcf is a vector subspace of X which is invariant under C and C', the commutant of C. Solution: It is obvious that Qf is closed with respect to scalar multiplication. To show that Qf is closed with respect to addition, choose two arbitrary vectors x, y E Qf and let F be any finite subset of C. We have

'(x+y)II" <[II."xII+II-1-"yII] since both IIF"xII

<2"max{IIF"xII"IIF"yII

0,

and II."yII " converge to zero by the definition of Qf. There-

fore, IIF"(x + y)II " .- 0, and sox + y E Qc. To prove that Qf is invariant under C, pick any C E C and let us verify that Cx E QCf. Notice that for each finite subset 7 of C and for each n E N we have

10. Invariant Subspaces

318

,X,C C g"+', where C = ' U {C}. Consequently, from IIgn+1xll n

we see that IIPCxll °

= [Ilgn+'xII

-' 0,

1

`)

0. This shows that Cx E QCf. The fact that Tx E Qc

for each T E C' can be proved in a similar manner. Moreover, in Lemma 10.40 we proved the stronger statement that Qc is invariant under [C). I

Problem 10.4.4. Let C be a collection of positive operators on a Banach lattice E and let N c =nCECNC. Show that Nc is invariant under C and [C). Solution: Take any x E N. and two arbitrary operators Co E C and To E [C). We need to show that Cox E Nc and Tox E N,,, that is, CICoxI = 0 and CIToxI = 0 for each C E C. The first equality follows from CICoxI < CCoixI = CO = 0. For the second, note that CTo < TOC implies CIToxI < CTo l xi <_ TO CI xI = Too = 0, I and so CIToxI = 0, as desired.

Problem 10.4.5. Let C be a collection of positive operators on a Banach lattice and let Sc be the multiplicative semigroup generated by C. Establish the following identities: [C) = [Sc), (C] = (Sc], and C' = S1. Solution: We will verify the first equality only. The other two can be verified similarly. We begin by verifying the inclusion [C) C [Sc ). Take an operator T E [C),

and an arbitrary operator S E Sc. We want to show that TS > ST. From the definition of Sc we know that S = C1C2 C for some operators C1, C2,..., Cn in C. We are also given that TC >_ CT for each C E C. Using this, we obtain

TS=TCIC2...Cn>C1TC2...C, >... 2C1C2...C,T=ST, proving that T E [C). To verify the reverse inclusion [Se) C [C), note that if T E [Sc), then T automatically belongs to [C) because C C Sc.

Problem 10.4.6. Let C be a commutative collection of positive operators on a Banach lattice. If C is quasinilpotent at some vector xo > 0, then C is finitely quasinilpotent at xo. Solution: Let F = {C1,. .. , Ck } be an arbitrary finite subset of C. We claim that Il."xoll" = 0. To prove this, fix 0 < e < 1 and let M = 1 + maxino and each i=1,...,k. Recall that the set .71 consists of all the products of arbitrary n operators from

T. Therefore, in view of the commutativity of the operators in C, an arbitrary operator S E P has a representation of the form S = C' C2' ... Ck k for some + ak = n. The latter condition implies integers a; > 0 satisfying a1 + 02 + that for each n there exists some 1 < in < k such that ain > B. . We will assume that n > kno and hence ai > no. Using the commutativity of the operators once again, we see that

CaiC2* 2 ...kx = CQl 1

k

1

i.,-1

in+1

k

0 1n

,

10.4. Invariant Subspaces of Families of Positive Operators

319

and so without loss of generality we can assume that i(n) = k. Then we have IISxoII"

IIC1"C2..CkkxoU

= IIC1II

"

IICkkxoII 1.

< Mk-1. (ek)* =

(Ek)°

Mk-1E.

This shows that II f"xo II 'n < Mk-If for all n > kno, and so limn 11.7-"xo 11 " = 0. Therefore, the collection C is indeed finitely quasinilpotent. I

Problem 10.4.7. Let 0 be a locally compact Hausdorff space and let S be a non-zero multiplicative semigroup of positive operators on Co(0) which are quasinilpotent at some xo > 0. If S is also an additive semigroup, then show that S has a common non-trivial closed invariant ideal. Solution: If Sx0 = 0 for each S E S, then there is nothing to prove, since the intersection of the null ideals J = nsES{x E Co(0): SjxI = 0) is a non-trivial closed S-invariant ideal. (The ideal J is non-trivial since S is non-zero and xo E J.)

So, we assume that Soxo > 0 for some So E S. Without loss of generality we can assume that the support set F = {t E A: xo(t) > 0} of the function xo is compact. Indeed, each S E S is certainly quasinilpotent at each element from the order interval [0, xo]. Furthermore, the functions with a compact support in [0, xo] are norm dense in 10, xo] and this proves our claim. We claim next that there exists a point to E A such that [Sxo] (to) = 0,

(*)

for each S E S. To see this, assume to the contrary that (*) is not true. Then for each t E 0 there exists an St E S such that [Stxo] (t) 0 0, and so there exists an open neighborhood G(t) of t such that

[Stxo](s)>et>0 for each 8EG(t). Take a finite cover G(tl ), ... , G(tm) of the compact set F, and consider the operator

S=St,+St,+...+Stm. By our hypothesis S E S, and so S is quasinilpotent at xo. On the other hand, we clearly have

m

[Sxo](t) _

[St.xo](t) > min{et1 ,...,et,,,} > 0 i=1

for each t E F. This implies that there exists some y > 0 such that Sxo > -yxo .

But, as we know from Problem 10.3.1, the last inequality implies that S is not quasinilpotent at xo, a contradiction. Therefore, a point to with the required properties exists. Finally, consider the ideal of Co(A):

J=Ix ECo(z): (xI
10. Invariant Subspaces

320

Obviously J {0} as 0 Soxo E J. It is also obvious that J is S-invariant. Furthermore, notice that J cannot be dense in Go(A) since [Sxo] (to) = 0 for each operator S E S. Consequently, the norm closure J of the ideal J is a non-trivial closed S-invariant ideal.

Problem 10.4.8. Let C be a norm bounded collection of operators in £(X ).

The joint (or the Rota-Strang) spectral radius of C is defined by

r(C) = nEN inf n -11C 11 , where we let C' = {C1C2 C": C, E C) and ([C"1[ = sup{J[SJ : S E C"}. Show that r(C) = rlimm VI-IC-n11 .

Solution: The solution is absolutely similar to that of the case of a single operator considered in Problem 6.1.6. One should observe only that for a collection of operators C the inequality ]ICn"mII < llC"[] - ]]Cm11 still holds for all n. in E N.

Problem 10.4.9. Assume that a positive matrix [air] defines a continuous operator on some fp, where 1 < p < oc, and this operator is quasinilpotent at some non-zero positive vector. Consider also the collection W of all double sequences w = {w,2 : i, j = 1.2,.. .... } of scalars such that the operator T. defined by the weighted matrix [wijaij] is a regular operator on 1p. (Clearly, W contains all bounded double sequences.)

Show that the family of operators {T.: w E W} has a common nontrivial closed invariant ideal. Solution: We denote by A the operator defined by the matrix [a,j]. Let J be any non-trivial closed A-invariant ideal. Such an ideal always exists by virtue of Corollary 10.31 since A is quasinilpotent at a non-zero positive vector. Since the norm on fa is order continuous, the ideal J is a band.

Fix an arbitrary w = i, j = 1.2....} E W. It is should be clear that if w is bounded, that is, ]u,,, I < c for some constant c > 0, then ]T.,] < cA, and consequently the band J is also IT.,I-invariant. To consider the case of a general w E 1V, assume first that w consists of positive

numbers and let w = {w,2 A n}. Obviously w,, E W and w T w. Therefore, T,,, T T. and consequently for each x E J+ we have T,,, x T T.,,x. Since T.,, x E J and J is a band. it follows that Twr E J. That is. J is T.,,-invariant. The general case follows immediately from the previous one by observing that for each w E W we have ]T.,, L < T.,,,.

[i

10.5. Compact-friendly Operators Problem 10.5.1. Prove that the following classes of operators consist of compact friendly operators:

10.5. Compact-friendly Operators

321

(a) compact positive operators,

(b) positive operators commuting with non-zero positive compact operators, (c) positive operators that dominate non-zero compact positive operators. (d) positive E'-integral operators (in particular, abstract integral operators), where E is a Dedekind complete Banach lattice. (e) positive integral operators. Solution: Recall that a positive operator B: E E on a Banach lattice is said to be compact-friendly if there exist two non-zero positive operators R, K : E - E with K compact and a non-zero operator C: E -. E such that RB = BR and JCxJ < RjxJ

and

JCxl < KIxI

for each x E E.

(a) Assume that B: E -. E is a compact positive operator. To see that B is compact-friendly, we simply let R = C = K = B. (b) Assume that B: E -p E is a positive operator that commutes with a non-

zero compact positive operator T : E - E. If we let R = C = K = T, then it is easy to see that B is a compact-friendly operator. (c) Assume that a positive operator B: E -+ E dominates a non-zero compact

positive operator T : E -i E, i.e., 0 < T < B. If we let R = B and C = K = T, then we see that B is a compact-friendly operator.

(d) Let T : E -+ E be a positive E'-integral operator on a Dedekind complete Banach lattice. This means that there exists a net {To} of positive operators such that 0 < To T Tin G, (E) and each TQ is dominated by a finite-rank operator. Now pick any index 3 such that To > 0, and then choose a finite-rank operator S satisfying Tp < S. Clearly, S is a compact positive operator. Now if we let

R = T. C = T3, and K = S, then we easily infer that T is a compact-friendly operator.

(e) According to Theorem 5.28 each integral operator on a Banach function space is an abstract integral operator. The conclusion now follows immediately from (d).

I

Problem 10.5.2. Let B : E - E be a positive operator on a Banach lattice. If there exists a non-trivial B-invariant projection band 0 in E such that Bx = 0 for each x E Ad, then show that B is compact friendly. Solution: Pick any vector 0 < v E Ad and any non zero positive functional 0 such that 0(x) = 0 for each x E A. (Such v and 0 clearly exist because 0 is a nontrivial projection band.) Now consider the positive (non-zero) rank-one operator

T = 0 0 v. A straightforward verification shows that BT = TB = 0. By letting R = C = K = T, we obtain the three operators R, C, and K needed to guarantee that B is a compact-friendly operator.

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322

Problem 10.5.3 (Sirotkin). For a non-zero positive operator T : E - E on a Banach lattice establish the following.

(a) If K E G(E) is a non-zero compact positive operator, then T + K is compact friendly. (b) There exists a compact friendly operator B E G(E) such that T < B. (c) If E is Dedekind complete, then there exists a compact friendly op-

erator S such that 0 < S < T. Solution: (a) Let K E G(E) be a non-zero compact positive operator and let B = T + K. Letting R = B and C = K, we see that R commutes with B, and R dominates the non-zero operator C which is dominated by the compact positive operator K. Hence, B is compact-friendly.

(b) Pick any non-zero compact positive operator K on E. For instance, let K be a rank-one positive operator. By part (a), the operator B = T + K is compactfriendly and satisfies T < B. (c) Let E be Dedekind complete. We assume that dim E = oo, since otherwise there is nothing to prove. Choose three pairwise disjoint non-zero bands A,, A2 and A3 such that Al ® A2 (D A3 = E. This is always possible since E is infinite dimensional (how?). Let Pi denote the band-projection on the band Ai. Since P1 + P2 + P3 = I, it should be clear that we can find i, j E { 1, 2,3) such that the operator S = PiTPj is non-zero. Obviously, 0 < S < T, and we claim that S is compact-friendly.

To verify that S is compact-friendly, we will establish that S satisfies the conditions of Problem 10.5.2. Let A = Di ® Ap Clearly, A is a band in E which is also S-invariant. It remains to notice that Sx = 0 for each x E Ad = Ak, where k 96 i, j. Therefore, Problem 10.5.2 guarantees that S is a compact-friendly operator. The assumption of Dedekind completeness can be considerably relaxed. The same proof goes through whenever there exist three pairwise disjoint non-zero bands Al, A2, and A3 such that Al ® A2 ® A3 = E.

Problem 10.5.4. Let 0 be a positive one-to-one function in C(Sl), where 11 is a compact Hausdorff space without isolated points. Show directly, i.e., without using Theorem 10.65, that the multiplication operator M o is not compact friendly.

Solution: Suppose that fl and 0 satisfy the properties stated in the problem. Assume by way of contradiction that M,, is compact-friendly. Then, there exist non-zero operators R, K, A: C(l) -+ C(fl) with R and K positive and K compact such that

RM, = MoR, R dominates A and K dominates A. By Lemma 4.19(3), there exists some 0 < h E C((Z) such that R = Mh. From the domination of A by R we have JAxi < hx for each 0 < x E C(Sl). Therefore, A E S(C(SZ)) and so, by Theorem 3.33, there exists a non-zero function 10 E C(fl) such that Ax =10x and 1,01 < h. Since A is dominated by a compact operator K, it

10.5. Compact-friendly Operators

323

follows from Theorem 2.34 that the operator A3 is compact. Clearly A3x = tli3x and the function 03 is non-zero since vp is non-zero. However, according to Lemma 4.18, a multiplication operator with a non-zero multiplier cannot be compact if f has no isolated points. This contradiction shows that Mm cannot be compact-friendly. I

Problem 10.5.5. A positive operator B: E - E on a Dedekind complete Banach lattice is called a generalized Harris operator if some power of B is not disjoint from the band (E* ® E)dd of E`-integral operators on E, i.e., if some power of B dominates a non-zero positive E`-integral operator. Recall that if some power of B is not disjoint from the band (En (& E)dd of abstract integral operators, then B is called a Harris operator. Of course, each Harris operator is a generalized Harris operator. For a generalized Harris operator B: E -, E establish the following.

(a) B is compact-friendly. (b) If B is quasinilpotent at a non-zero positive vector, then B has a non-trivial closed 1-hyperinvariant ideal.

Solution: (a) Let B: E - E be a generalized Harris operator on a Dedekind complete Banach lattice. This means that there exist some natural number m and a positive E'-integral operator 0 < S E (E' (& E)dd such that S S B'". Therefore,

there exist a positive operator A: E -+ E and a finite-rank operator T: E -+ E

such that 0
Problem 10.5.6. Show that Theorems 10.55 and 10.60 remain true if we replace the assumption lim IIT"x011" = 0 with lim inf I1T"xoll n = 0. n-.oo

n-oo

Solution: The validity of the proofs of Theorems 10.55 and 10.60 under this new assumption can be established exactly as in the solution of Problem 10.3.8. 1

Problem 10.5.7. Let T: 8p -+ ep (1 < p < oo) be a continuous operator with modulus. Assume that there is a non-zero positive operator S : tp -+ fp such that:

(a) S commutes with the modulus of T. (b) S is quasinilpotent at a non-zero positive vector.

Show directly, i.e., without using Theorem 10.24, that T has a non-trivial invariant closed ideal. Solution: It is enough to prove that the modulus ITI has a non-trivial closed invariant ideal. Clearly, this ideal is T-invariant as well. Therefore, we can assume

without loss of generality that T > 0. So, suppose that S > 0 satisfies TS = ST and n mo S" To I I " = 0 for some xo > 0. We distinguish two cases: Sxo > 0 and

Sxo=0.

10. Invariant Subspaces

324

We begin by assuming that Sxo > 0. Scaling xo if necessary and using the continuity of S, we obtain some k such that xq > ek > 0 and Sek > 0. Now let P: Pp --+ ep denote the natural projection onto the vector subspace generated by ek. Clearly, 0 < Px < x for each 0 < x E gyp. We claim that P7'mSek = 0

(*)

for each m > 0. To this end, fix m > 0 and let PTmSek = aek for some a > 0. Clearly TmS = ST'", and hence 0 < a"ek = (PT mS)"ek < (T 'S)"ek = T""S"ek < T "S"x0 . Consequently,

0
J= {xEPp: 3A>0and r>0-such that IxI
for all x E J, and consequently (ek, x) = 0 for all x E J. The latter shows that J is a non-trivial closed ideal of Pp. We claim that J is T-invariant. Indeed, if x E J, then there exists some positive scalar A > 0 and some r > 0 such that IxI < AE= QTiSek. Thus, ITxI < T(IxI) < AEi+jT"Sek, and hence Tx E J. That is, J (and hence J) is T-invariant. Finally, consider the case Sxo = 0. Let A denote the unital algebra generated by T in £(tp). Also, let

J={xEtp: IxI 0 and a constant c > 0 such that IAI < c E '=o V. But then, from r r ISxI < SIxI < SIAIxo < S(cET=)xo = c1:V (Sxo) = 0,

it follows that Sx = 0 for all x E J. If J = Pp, then S = 0, which is a contradiction. Therefore, J is a non-trivial closed T-invariant ideal. I

Problem 10.5.8 (Sirotkin). Let B: E -> E be a compact friendly operator on a Banach lattice such that no positive operator commuting with B dominates a non-zero compact operator. Show that for each at most countable family of positive operators commuting with B there exists a non-trivial closed ideal that is B-invariant and .F-invariant.

10.5. Compact-friendly Operators

325

Solution: Let B: E - E be a non-zero compact-friendly operator on a Banach lattice such that no positive operator commuting with B dominates a non-zero compact operator. Also, let F = {T" } be a sequence of positive operators such that T. B = BT" holds for each n. The solution is an adaptation of the proof of Theorem 10.55. Fix three non-zero operators R, K, C: E -+ E with K, R positive, K compact, and satisfying

RB = BR,

ICxI < R(IxI)

and

ICx1 < K(Irj) for each x E E.

Without loss of generality we can assume that IIBII < 1.

For each n pick some a > 0 such that the series T = En_ R"T" is norm convergent-and so it defines a positive operator-such that II B + TIP < 1. Consequently, the series A = E°°_e(B + T)" also defines a positive operator. Clearly, T commutes with B, and A commutes with B and T. For each x > 0 we denote by J= the principal ideal generated by Ax, that is,

Jy={yEE: IyI 0}. This ideal is non-zero because x E JJ in view of the obvious inequality x < Ax. If the norm closure J., # E for some vector x > 0, then Jr is a non-trivial closed (B + T)-invariant ideal. From 0 < B, T < B + T, it follows that J , is invariant under both B and T. Clearly, this ideal is also Tn-invariant for each n. We claim now that there exists some x > 0 such that Jr j4 E. To prove this, assume to the contrary that

Jr = E

(*)

for each x > 0, that is, Ax is a quasi-interior point in E for each x > 0. Since C 34 0, there exists some xt > 0 such that Cx.1 36 0. Then, AICx1I is a quasi-interior point satisfying APCx1I > ICx1I, and it follows from Lemma 4.16(1) that there exists an operator Aft : E -, E dominated by the identity operator such that x2 = A11Cx1 > 0. Put 111 = All C, and note that the operator 111 is dominated by the compact positive operator K and by the operator R.

By (*) we have Js, = E and so, since C # 0, there exists 0 < y < Ax2 such that Cy 0 0. Hence (again by (*)) the element AICyI is a quasi-interior point. Since ICyI < AICyI, there exists, by Lemma 4.16(1) again, an operator A12: E -+ E dominated by the identity operator such that x3 = A12Cy > 0. Since IyI < Ax2 and Axe is a quasi-interior point, it follows from Lemma 4.16(2) that there exists an operator Af : E -+ E dominated by the identity operator such that AI Axe = y. So, X3 = Af2Cy = A12CMAx2. Let 112 = A12CAIA and note that 112 is dominated by the compact positive operator KA and by the operator RA. If we repeat the preceding arguments with the vector x2 replaced by x3, then we obtain one more operator 113: E - E that satisfies 113x3 > 0 and that is dominated by the compact positive operator KA and by the operator RA. Consider the operator 113112111. From n3112111x1=113x3 > 0, we see that the operator 113112111 is non-zero. Moreover, as shown above, each 11; is dominated by a compact operator and, therefore, Theorem 2.34 guarantees that the operator n3112n1 is necessarily compact. A straightforward verification shows also that 1113112111x1 5 RARARIXI

326

10. Invariant Subspaces

for each x E E. But this contradicts our assumption, since the operator RARAR clearly commutes with B, and the solution of the problem is complete.

Problem 10.5.9 (Troitsky [791). Show that every essentially quasinilpotent compact friendly operator on a Banach lattice of dimension greater than one has a non-trivial closed invariant subspace. Solution: Let B: E - E be an essentially quasinilpotent compact-friendly operator on a Banach lattice of dimension greater than two. If B is quasinilpotent, then by Theorem 10.55 there exists a non-trivial closed B-invariant ideal. So, we can assume that r(B) > 0. Since the operator B is positive, we know (in view of Theorem 7.9) that r(B) E o(B) and so, by Corollary 7.49, the spectral radius r(B) is an eigenvalue of B. This guarantees (see Problem 10.1.4) that B has a non-trivial closed invariant subspace in Ec and also in E.

Problem 10.5.10 (Troitsky [791). Let S: E -. E be a quasinilpotent integral operator on a Banach function space such that S- is compact. Then each of the operators ISI and S+ has a non-trivial closed invariant subspace. Solution: Let B = ISI. This is a positive integral operator and so it is compactfriendly. If B is quasinilpotent, then it has a non-trivial closed invariant ideal. So, we can assume that B is not quasinilpotent, i.e., r(B) > 0. Since B = S+2S-, it follows that r. (B) = r(S) = 0. That is, B is essentially quasinilpotent. But then. according to Problem 10.5.9, the operator B = ISI has a non-trivial closed invariant subspace. The proof for S+ is similar. I

Problem 10.5.11. Show that the set of all compact friendly operators on a Banach lattice E need not be closed or open in G(E). Solution: Let E = CIO, 11 and let 0(t) = t. By Theorem 10.65 (or by Problem 10.5.4) the multiplication operator Mo is not compact-friendly. It should be

clear that there exists a sequence of positive functions On E C[0,1] such that IIOn - 0110, - 0 and each On has a flat. This implies that each multiplication operator M4 is compact-friendly and that Mm -+ M, in C(C[O,11). This proves that the set of compact-friendly operators is not a closed subset of C(CIO,11). Now, for each n consider the function On E C[0,1] defined by On(t) = 1 + it, and note that Ia n - III,,. -. 0. This implies that the sequence of multiplication operators (M,P } converges to the operator Ml = I in G(C[O,11), which is compactfriendly. However, since no function On has a flat, it follows that each operator M,. is not compact-friendly. This proves that the set of compact-friendly operators is not an open subset of C(C[O.1]) either. I

Problem 10.5.12. This problem presents an example of a Hilbert-Schmidt operator on L2[0,11 which is one-to-one, positive, locally quasinilpotent at a positive vector, but fails to be quasinilpotent. The construction of such an operator will be done in three steps as follows.

10.5. Compact-friendly Operators

327

(a) Let 0 < a < b < 1 and 0 < c < d < 1 and consider the positive integral operator T : L2 [a, b] -+ L2 [c, d] defined by b

Tx(t) =

sin(st)x(s) ds , c < t < d .

j

Then the operator T is one-to-one and compact. (b) Assume that 0 < a < b < 1. Then the positive integral operator T : L2[a, b] - L2[a, b], defined by

Tx(t) =

sin(st)x(s) ds, a< t < b,

JJ.n

has a positive spectral radius-and so T is a one-to-one compact positive integral operator which is not quasinilpotent.

al .............. a2

if .................

a

b

(a)

(c)

a3

.............. .............

a3 a2 al 1

is

z

(b) Figure 1. The kernel of the operator K

Consider the function K: [0, 11 x (0, 1] --+ JR defined as follows: the value K(s, t) is equal to sin(st) if (s, t) lies in each one of the shaded rectangles shown in Figure 1(b) and is zero everywhere else. The points ao, al, a2.... shown in Figure 1(b) are defined by

n-i

a0=1 and an = 1 - 2e

j for n > 1. k=0

The length of the interval (an+1, an) is equal to an - an+l = nT As usual, we shall denote by K the operator on L2[0,1] defined be the kernel K(s, t). That is, the operator K: L2[0,11 -+ L2[0,1] is defined by

Kx(t) =

J0

1

K(s, t)x(s) ds, 0 < t < 1.

10. Invariant Subspaces

328

Then the integral operator K: L2[0,1] - L210, 11 is a one-to-one compact positive operator which is quasinilpotent at some positive vectors but fails to be quasinilpotent. Solution: (a) The compactness of T follows easily from the fact that T is a Hilbert-Schmidt operator; see also Theorem 7.7. We shall prove that T is oneto-one. The geometric situation is shown in Figure 1(a), where the kernel function K(s, t) = sin(st) is defined over the rectangle Q = [a, b] x [c, d]. It is easy to see that Tx is a continuous function for each x E L2[a,b]. As a matter of fact (see [7, Theorem 20.4]) the function Tx E Coo [c, d] for each x E L2 [a, b] and [Tx]

jb

(n) (t) =

[FF sin(st)]x(s) ds.

(*)

Assume that Tx = 0 for some function x E L2 [a, b]. Then for each n, it follows

from (*) that fa 92n sin(st)x(s) ds = 0 for almost all t E [c, d]. The subalgebra A of C[c, d] consisting of all even polynomials contains the constant function one and separates the points of [a, b]. (Notice that the polynomial p(s) = 8 2 is one-to-one on [a, b].) So, by the Stone-Weierstrass theorem, A is uniformly dense in C(a, b]. Consequently, A is - 112-dense in L2[a, bJ. From (*) we see immediately that fa p(s) sin(st)x(s) ds = 0 for each p E A and almost all t E (c, d]. Consequently, there exists a sequence {p,,} in A satisfying pn --+ x a.e. and Ipn1 5 y a.e. for all n and some y E L2(a,b]; see [7, p. 207]. So, for almost all s E [a, b] we have pn(s)sin(st)x(s) ----+ sin(st)[x(s)]2 and Ipn(s)sin(st)x(s)I < y(s)Ix(s)I. Since y]xI E Li[a,b], the Lebesgue Dominated Convergence Theorem implies II

b

sin(st)[x(s)]2 ds =slim

for almost all t. Therefore, x = 0 a.e.

j

bp

n(s) sin(st)x(s) de = 0

(b) Since the kernel of T is symmetric, it follows that T is a symmetric operator. Consequently, r(T) = IITII2 > 0; see also Problem 11.1.10. If for 1 < p!5 oo we consider the operator T : LAO, 11 -' Lp[0,11 defined by

Tx(t) = fo sin(st)x(s) ds, 0 < t < 1, 1

then we have the following norm estimates: I

IITII00

= IIT11100 = sup

tE 10,11

=

J

sin(st) ds

0

t = 1 - cos 1

sup

0 . 4597

tE 10,1j

IITII2

<-

[f

1

[sin(st)]2 dsdt] 5

ri [2-4J eit2tdt]

:0.7009.

10.6. Positive Operators on Banach Spaces with Bases

329

(c) The fact that K is one-to-one follows easily from part (a). Since the kernel K(s, t) is positive and belongs to L2([0,1] x [0,1{), it follows that the operator K is Hilbert-Schmidt. In particular, K is a compact positive operator. To see that K is not quasinilpotent, notice that K leaves L2 [0, z ] invariant. (As usual, here we identify L2 [0, with the closed subspace of L2{0,1) given by 2] {x E L2{0, 1): x = 0 a.e. on [2,1J }.) Moreover, observe that the restriction of K to L2 [0, Z] is the integral operator with kernel sin(st). By part (b), the restriction of K to L2 [0, 2 J has a positive spectral radius. That is, K is not quasinilpotent when restricted to L2 [0, 2 ] . Therefore, K cannot be quasinilpotent. Finally, consider the positive function x0 = Xla,.il > 0. An easy verification shows that 0 < K"xo <_ Consequently, [IIK"xoII2]

< 1

from which it follows that lim"--.00[jJK"xo112]" = 0. This shows that K is quasinilpotent at the non-zero positive function x0. I

10.6. Positive Operators on Banach Spaces with Bases Problem 10.6.1. Let {xn} be a basis in a Banach space X and let 00

anXn: an>0 for each n = 1,2,...}.

C = Ix n-1

Show that C is a closed cone of X (called the cone generated by {xn }) . Solution: Let x = r1n=1 anxn and y = FO' 1,3nxn. If x, y E C, then clearly

x+yECandaxECforeacha>0. That is,C+CgCandaCCCforeach a > 0. Now assume that x E C fl (-C). From the definition of the basis, we see

that an = -an and hence an = 0 for each n. This shows that C fl (-C) = {0}. Therefore, C is a cone in X.

To show that C is a closed cone, assume that xn = E00 i=1 an,ixi E C and xn - x E X. From Theorem 1.37 we know that each coordinate functional ck is continuous. Consequently, for each k E N we have 0 < an,k = ek(xn) - ck(x) = ak. This implies ak > 0 and hence x E C. Therefore, C is a closed cone in X. I

Problem 10.6.2. If a Banach space X has an unconditional basis {un}, then show that:

(a) The cone C generated by the basis {un} is a lattice cone. (b) The Riesz space (X, C) has weak units. (c) After an equivalent renorming the ordered Banach space (X, C) becomes a Banach lattice with order continuous norm.

10. Invariant Subspaces

330

Solution: (a) According to Problem 10.6.1 we know that the set 00

C = {x = E anon : an > 0 for each n = 1, 2.... } n=1

is a closed cone. In the usual way this cone induces the coordinatewise ordering > on X. That is, for x = E°n°__i anun and y = En 1 Qnun in X we have x > y if and only if an >,On for each n.

We shall show next that for each x = En 1 anus and y = En 1 Mnun in X the series E', max{an,Qn}un and L..n i min{an,On}un are norm convergent in X. If this is done, then clearly (X, C) is a Riesz space and 00

0c

x V y= F, max{an, Qn}un

and

x n y=

min{an, Qn}un . n=1

n=1

We shall consider the first series only. To this end, let A = In E N: an > I3 }

and B = In E N: an < On)- Since the basis {un} is unconditional, both series F,nEA maX{an,Qn}un = EnEAanon and EnEB maX{an,Qn}un = EnEB 6nun are norm convergent. This implies that the series n 1 max{an,On}un is norm convergent.

(b) Pick a sequence {An} of strictly positive real numbers such that the series F-O°_i Anun is norm convergent in X. We claim that w is a weak unit. W= To see this, assume that a vector x = E°n°__1 anon satisfies w A x = 0. From 00

w A x = 1: min{An,an}un =0, n=1

it follows that min{A,, an } = 0 for each n, and from this we infer that an = 0 for each n. Therefore, x = 0 and so w is a weak unit in (X, C). (This conclusion can also be obtained from Problem 4.2.3 by observing that every Banach space with a basis is necessarily separable.) (c) The equivalent norm I I I . I I I on X that turns (X, C) into a Banach lattice is defined for x = F,n°__i anxn E X by 00

IIIxIIl=sup {IIE$nxnII: (131./32.... )ERN and 113nI:5 Ianl for each n n=1

The formula defining 111 111 makes sense by virtue of Theorem 1.51. The verification

that I I I 'III is a lattice norm is straightforward. From Lemma 1.49, it follows that for each x = En_l anxn we have 11x11 :5 IIIxIIl < KuIIxII ,

where KL is the unconditional constant of the basis {un}. Therefore, the ordered vector space (X, C) equipped with the norm I I I - III is a Banach lattice.

To verify that the norm is order continuous, let x.y j 0. Fix any e > 0 and yo. Then find some n E N such that 111(1- Pn)x,,III < e, where Pn is the natural projection of X onto the vector subspace generated by {x1ix2,...,xn}. Applying Lemma 1.49 again, we see that 111(I - Pn )x, I I I < e for each y > yo. It remains

10.7. Non-transitive Algebras

331

to note that IIIPnx7Il1 10 since x., j 0 and since we are dealing here with a finite number of coordinates only. Thus, I I Ix., I I I < e for all 7 that are large enough.

Problem 10.6.3. Let X be a Banach space with an unconditional basis {u,a}. If a linear operator T: X - X is positive with respect to the cone generated by the basis

then show that T is a bounded operator.

Solution: Let C be the cone generated by the unconditional basis

By

Problem 10.6.2 we know that C is a lattice cone and that we can renorm X so that (X, C) becomes a Banach lattice. Now a glance at Problem 1.2.14 guarantees that the positive operator T is continuous. I

Problem 10.6.4 (Spalsbury [76]). Consider the Volterra operator defined in Example 7.8. Show that there is no basis in L2[0,1] with respect to which the Volterra operator is positive. Solution: Recall that the Volterra operator V: L2[0,1] -. L2[0.1] is defined via the formula

Vf(t) =

f

f(s)ds.

0

From Problems 7.1.3 and 7.1.4 we know that V is a quasinilpotent one-to-one compact positive operator. Assume by way of contradiction that there exists a basis in L2[0,1] such that the Volterra operator V is positive with respect to the cone generated by this basis {x,,}. If P1 is the projection of L2[0.1] onto the vector subspace generated by xl, then, as shown in the proof of Theorem 10.66, we have

P1 Vmx1 = 0 for m = 1.2, ... . (t) So, if M denotes the closed vector subspace generated by the set {Vmx1: m > 1}, then M is V-invariant, and in view of (t) we have P1 x = 0 for each x E Al . Since Plxl = x1 0 0, it follows that x1 V 111, and thus rl is a non-trivial closed V-invariant subspace. By the well-known characterization of the non-trivial closed invariant subspaces of V (see [44]), there exists some 0 < a < 1 such that Al = If E L2[0- 1] : f = 0 a.e. on [0, al) . Because Vx1 E 111, we must have Vx1 = 0 a.e. on [0. a]. Since the Volterra operator is one-to-one on L2 [0, a] (see Problem 7.1.4), it follows that x1 = 0 a.e. on [0, a], and

consequently x1 E M, a contradiction. This contradiction shows that the Volterra operator cannot be positive with respect to any basis for L210,1]. a

10.7. Non-transitive Algebras Problem 10.7.1. If S2 is a compact topological space and X is a Banach space, then show that the space C(Q, X) of all X -valued continuous functions

10. Invariant Subspaces

332

on fi equipped with the norm

Hill = max Ilf()II is a Banach space-which is a Banach lattice if X is a Banach lattice. Solution: The verification of the norm axioms of the indicated function f +-+ II f II is straightforward. We show the completeness of the norm. Notice that convergence with respect to the norm in C(St, X) is equivalent to uniform convergence. So, if

{ f,,) is a Cauchy sequence in C(i2, X), then { f } is convergent uniformly to a function f : Q --+ X. It follows that f is continuous and II f,, - f II -+ 0. Hence, C(1, X) is a Banach space. Now assume that X is a Banach lattice. It is not difficult to see that C(it, X) is a Riesz space under the pointwise lattice operations (f V g)(w) = f(w) V 9(w)

and

(f Ag)(w) = f(w) A9(w)

Moreover, if If I S IgI in C(Q, X), then If (w) I < Ig(w) I for each w E St, and so Ilf(w)II < Ilg(w)II, from which it follows that IIf1I 5 IIgII. That is, the norm on C(St, X) is a lattice norm, and so C(1, X) is a Banach lattice.

Problem 10.7.2. Let X be a Banach space, and let the closed unit ball U* of X* be equipped with the w'-topology. Establish the following.

(a) If T : X -+ X is a compact operator, then T*: U* - X' (the restriction of T' to U') is a completely continuous function. (b) The collection of all completely continuous functions IC(U', X') is a norm closed subspace of C(U', X'), and so it is a Banach space in its own right.

Solution: (a) Recall that a function f : U' -+ X' is said to be completely continuous if it is continuous for the weak* topology on U' and the norm topology

on X'. Now assume that T: X -' X is a compact operator, and let xa - -x' in U'. If x E X is fixed, then from

(x,T'xa -T'x') = (x,T'(x, -x')) = (Tx,xa -x') -' 0, it follows that

T'xa _W'. T'x'

(*)

in X'. Since T is compact, the operator T' is also compact, and hence T'(U') is a norm totally bounded subset of X'. It follows that every subnet of the net {Txa} has a norm convergent subnet. In view of (*), every subnet of the net {Txa} has a norm convergent subnet to T'x'. This implies that IIT'xa - T'x'II -+ 0, and so T*: U' -i X' is completely continuous. (b) Let {f, } C K (U0, X') satisfy 11f, - f II --+ 0 for some f E C(U*, X*), and let e > 0. Pick some k such that Ilfn - f II < e for all n > k. Now assume that a net {xa } C U' satisfies xa -w-L+ x' in U'. Since fk belongs to JC(U',X'), there exists some index ao such that II fk(xa) - fk(x')II < e for all

10.7. Non-transitive Algebras

333

a > ao. Consequently, for a > ao we have

IIf(xQ) - f(x*)II

<-

IIf(x.) - fk(X,)II + Ilfk(x.) - fk(x*)II + Ilfk(x*) - f(x*)II

< E+E+E=3E. This shows that II f (x.)-f (x*) II -' 0, and so f E K(U*, X'). Therefore, K(U*, X*) is a closed subspace of C(U*, X*).

Problem 10.7.3. Assume that for a subalgebra A of £(X) there exists a non-zero operator To E £(X) such that: (a) To commutes with each operator in A. (b) For some B E £(X) its adjoint operator B* is not in the r,-closure of the vector subspace spanned by {aT*Ta : a E C(U*) and T E Al in C(U*, X*). Show that A* is non-transitive. Solution: By Lemma 10.73, there exist vectors x*' E X'* and x* E U' satisfying (x**

x**(B*x*)

=I

and

(x** ® x*, aT*To) = (a(x* )T*Tox*, x**) = a(x*) (T'To x*, x'') = 0

for all T E A and all a E C(U*). Therefore, (T*Tox*, x**) = 0 for each T E A. If y' = Tox' 96 0, then the relation (T*y*,x**) = 0 for each T E A coupled with Theorem 10.12 shows that A* admits a non-trivial A'-invariant closed subspace. Now assume that Tox' = 0, and let N denote the null space of To*. Since To commutes with A, it follows that N is a closed A'-invariant subspace. From x**(B'x*) = 1, we infer that x* 0 0. Since x' E N, we see that N is non-zero. On the other hand, if N = X', then To = 0, and so To = 0, a contradiction. Therefore, in this case, N must be a non-trivial closed A*-invariant subspace. I

Problem 10.7.4. Recall that a Banach space X is said to have the Schur Property if xn w' 0 implies IIxnII - 0. If X is a Banach space with the Schur property and A is a commutative subalgebra of £(X), then show that the dual algebra A* of G(X *) is nontransitive. Solution: If A consists entirely of scalar operators, then the conclusion is trivial. So, we can assume that there exists some non-scalar operator T E A. Fix some Ao in the approximate point spectrum of T and consider the operator To = T - A0!. Also, select a sequence {xn} of unit vectors of X satisfying IITOxnII = IITxn - AoxnII - 0.

Clearly, To commutes with A. Next, consider the subspace M of C(U', X *) generated by the collection (aT'TT : a E C(U') and T E A). We claim that the

restriction to U' of the identity operator I*: X* -, X' does not belong to the T,-closure of M.

10. Invariant Subspaces

334

To see this, assume by way of contradiction that I' belongs to the r; closure of M. Therefore, if e > 0 and x' E U' are fixed, then there exist a; E C(U') and T; E A (i =1, ... , m) satisfying 11 [ I' - E:'_ i a;T = To ] (x*) II < e. This implies m

(X, X-) -

a;(x`)(x, Ti* To(x')) <

(*W

for all x E X with flxll < 1. Taking into account that A is a commutative algebra, it follows from (*) that I (x, x') - E__, aj(x*)(Tox,T; x*)I < c for each x E X with lixil < 1. In particular, we have I (xn, x*) - E', ai(x')(Tox,,,TT x') I < E for each n, which, in view of (**), yields I (x,,, x') I < E for all sufficiently large n. In other words, the sequence {xn} converges weakly to zero in X. Since X has the Schur property, it follows that IIxn11 -' 0, contrary to the assumption that I1xn11 = 1 for each n. Therefore, the identity operator I' does not belong to the r8-closure of M and the conclusion follows from Problem 10.7.3.

Problem 10.7.5 (Honor [381). If A is a commutative algebra of bounded operators on 11, then show that the dual algebra A* of operators on t is non-transitive. Solution: This follows from Problem 10.7.4 by recalling that el has the Schur property; see [6, Theorem 13.1, p. 2001.

Chapter 11

The Daugavet Equation

11.1. The Daugavet Equation and Uniform Convexity Problem 11.1.1. Show that i f the vectors x1, ... , xn in a nonmed space satisfy n

n

11F xkll = E IIxkII, k=1

k=1

then IIEk=1 akxkll = Ek=1 akllxkll holds for each choice of non-negative

scalars a1,.-.,an.

Solution: The proof is by induction. For n = 1, the claim is trivially true. So, assume the claim is true for some n and let the vectors X1 , xn, xn+1 satisfy II Ek=1 xkll = Ek=1 IIxkII Also, let al,. .. an, an+1 be non-negative real numbers; we can suppose al > ak for each k. From n+1

n+1 II

xk ll

=

xk II <_ 11x111 + k=1

k=1

k=2

xk II

IIxkII,

<_

k=1

it follows that II Ek=2 xkIl = Ek=z IIxkII Now notice that the identity n+1

n+1

n+1!

F, akxk = a1 (F xk) - Dal - ak)xk, k=1

k=1

k=2 335

11. The Daugavet Equation

336

in conjunction with the induction hypothesis, yields

k=1

I

n+1

n+1

alIIE=kII k=1

IiEakxkll

-

n+ IIDal -

ak)xkII

k=2

n+1

al (E II=kVI) - n+1 1(a1 - ak)IIxkII k=1

n+1

k=2

n+l

EakllXkll >- Ilrakxkll k=1

k=1

Therefore. IlEk=1 akxkll = Ek±; akllxkll, and the solution is finished.

Problem 11.1.2. This problem presents another version of Lemma 11.4. Let X be a normed space and assume that a vector u E X and some unit vector v satisfy Ilu + vII > 2 - E for some f > 0. Show that for all a, Q E R+ we have

I1au+13vI1 > (1-E)(a+Q). Solution: Assume first that a+0 = 1. Without loss of generality we can suppose that a >.3. Then we have Ilau + Qv1I

?

Ila(u + v)II - II(a - Q)vll ? a(2 - E) - (a - Q)

= a+(3-QE> 1-E. Therefore, using the preceding case, we see that for all a, $ E R+ we have

Ilau+livll = (a+13)11-' C, 7u+ a+a vII 2! (a+A)(1 -E), as claimed.

U

Problem 11.1.3. Let V be a vector subspace of a nonmed space X, and assume that for a vector u E X and some c > 0 we have IIu + vIl > 2 - E for each v E V with Ilvll = 1. Show that for each v E V and all scalars A we also have IIAu +vII > (1- E) [IAI + IIvII]. Solution: Let v E V be a non-zero vector. If A > 0, then Problem 11.1.2 implies IIAu + vII = JIAU + IIvII u' 11 2! (1- E) [IN + IIvII]

If A < 0, then IIAu + vII = II IAIu - V11 > (1 - E) [IAI + IIvU]

and the solution is finished.

,

U

Problem 11.1.4 (C.-S. Lin [50)). Assume that two sequences {un} and {vn} in a normed space satisfy limn-oo[I[un + vnit - (IIu II + IlvnII)) = 0. Show that limn-w [ Ilaun + /3vnll - (allunII + AIIvnII) ] = 0 for all a > 0 and 0 >_0.

11.1. The Daugavet Equation and Uniform Convexity

337

Solution: Without loss of generality we can assume that a > (i. Then note that 0

> =

Ilaun + Ovn II - (allunll + iIIvnII)

Ilaun + av - (a - a)vnll - (allunli + $Ilvnil) allure+unit - (a -a)IIvnII - (allun11 +0IIvnII)

= allure + vnll -

a(Ilunll + IIvnll)

That is, we have established that

a[Ilun+vntt-(IlunII+IIvnII)]

Ilaun+Ovnll-(allunll+OIlvnll)<-0

holds for each n. This implies limn....,, [ llaun +

(allunll + MIvnII) ] = 0.

t3vnll -

1

Problem 11.1.5 (C.-S. Lin [50]). Generalize Problem 11.1.4 as follows. Assume that in sequences {un} (k = 1, 2, ... , m) in a Wormed space satisfy fm

m jI

n-oo

k=1

= 0. " -F IIun II ] k=1

k

Then show that limn-,* [ Il 1 akunll -E of non-negative scalars a I, ... , am .

1

akllunII ] = 0 for any choice

Solution: Assume that in bounded sequences {un } (k = 1.2, ...,in) in a normed space satisfy ll

x

HIM

rm

k=1

k=1

=0,

and let a1, 02, ... , am be in non-negative scalars. Without loss of generality, we can suppose that a1 > ak for each k. Now note that m 0

m

IIEOkunll -1:akllunll k=1

k=1

m

m

Ilal (y, un) - (al - ak)unII - >akllunll k=1

k=1

M

m

> a1IIE Uk II k=1

k=1 m

- >(al - ak)IIuII - Ek=1akllunlI k=I

m

M

a1 [III u II - L., liunll ] . k=1

k-1

Therefore, we have shown that a1

[IIFunII

k=1

- k=1 Ilunll]

Eakllunll < 0 k=1

k=1

holds for each n. This implies limn_ [ lirk

1

akunll - Ek 1 akllunll ] = 0.

Problem 11.1.6 (C.-S. Lin [501). Generalize Theorem 11.10 by showing that for a uniformly convex Banach space X and two operators S, T E C(X) the following statements are equivalent.

11. The Daugavet Equation

338

(a) III +S+TII = 1 + IISII +IITII. (b) There exists a sequence {xn} of unit vectors such that

=1+IISII+IITII. Solution: (a) : (b) Assume that III + S + TII = 1 + IISII + IITII. From the inequalities

1+IISII+IITII=III+S+TII
III+S+TII ? II(I+T+S)xnll Ilxn + IIS + TIIxnII - II (S + T)x - IIS + Tllx II

= 1+ IIS+TIl - II(S+T)xn - IIS+TIIxnII . Since 1 + IIS+TII - II(S+T)xn - IIS+TIIx,,II - 1 +IIS+TII, it easily follows from the above inequalities that limn-,, II(I+S+T)xnIl = 1+IIS+TII =III +S+TII. (b)

(a) Assume that a sequence {xn} of unit vectors satisfies

lim II(I+S+T)xn11=1+IISII+IITII.

n-.oo

Letting n -- oc and using the inequalities

I + IISII + IITII >- 1 + IIS+TII >- III+S+TII > II(I+S+T)xnll, we get 1+IISII+IITII > 1+IIS+TII > III+S+TII >- 1+IISII+IITII. This implies III + S + TII =1 + IISII + IITII

Problem 11.1.7 (C: S. Lin [501). If T: X - X is an invertible operator on a uniformly convex or uniformly smooth Banach space, then show that

III-2TII=1 implies Ill - TII < 1. Solution: Assume that III - 2TII = 1. We claim that 1 = III - 2TII ¢ a(I - 2T). Indeed, if we had 1 E a(I - 2T), then the operator I - (I - 2T) = 2T would not be invertible, contradicting our assumption that T is invertible. Since the Banach space X is either uniformly convex or uniformly smooth, it follows from Theorem 11.10 that the operator I - 2T does not satisfy the Daugavet equation. Therefore,

2111-TII=1121-2TII=III+(I-2T)II<1+II1-2TII=2. This implies III - TII < 1, as desired.

Problem 11.1.8. Let T : X - X be a non-zero bounded operator on a uniformly convex or uniformly smooth Banach space. Show that IITIII - T is non-invertible if and only if T satisfies the Daugavet equation.

11.1. The Daugavet Equation and Uniform Convexity

339

Solution: Notice that IITIII -T is non-invertible if and only if IITII E a(T). Since X is either a uniformly convex or a uniformly smooth Banach space, it follows from Theorem 11.10 that the latter is equivalent to saying that the operator T satisfies the Daugavet equation. I

Problem 11.1.9. If a continuous operator T : X , X on a uniformly convex or uniformly smooth Banach space satisfies the Daugavet equation,

then show that for each a > 0 the continuous operator eaT satisfies the Daugavet equation and that IIe°TII = e°IITII holds.

Solution: Fix it > 0 and consider the formal series f (A) = ea = E°

0

A"

Clearly, this series has non-negative coefficients and f(IIaTII) = e°OTI1. The desired conclusion now follows from Theorem 11.14.

Problem 11.1.10. Let (Q, E, p) be an arbitrary a-finite measure space. A linear operator T : L2(p) L2(µ) is called symmetric if (Tf, g) = (f, Tg)

for all f, g E L2(µ). where

denotes the usual inner product in L2(µ)

defined by (f, g) = fn f (w)g(w) dp(w). Establish the following. (a) Symmetric operators are bounded.

(b) An integral operator T on L2(µ) with a kernel K E L2(µ x µ) is symmetric if and only if K(s, t) = K(t, s) holds for p x p-almost

all(s,t)E1 xQ. (c) The integral operator T : L2[0,1J - L2[0,11, defined by

Tx(t) =

n 0

sin(st)x(s) ds,

is symmetric.

(d) Every symmetric operator T : L2(µ) -+ L2(µ) satisfies r(T) = IITII-

(e) Every symmetric positive operator T: L2(1) -+ L2(p) satisfies the Daugavet equation, i.e., III + TII = I + IITII. Solution: (a) Let T: L2(µ) -- L2(µ) be an arbitrary symmetric operator. If f E L2(µ) = L2(µ). then for each g E U (the closed unit ball of L2(µ)) we have I(Tg,f)l = I (g,Tf)I <_ IITII - IITfII :5 IITfII

This shows that the set T(U) is a weakly bounded subset of L2(p), and hence T(U) is also a norm bounded subset of L2(µ). Therefore, T is a bounded operator.

11. The Daugavet Equation

340

(b) Assume first that K(s.t) = K(1, s) for p x p-almost. all (s, t) E fi x fi. If f.g E L2(p). then using Fubini's theorem. we see that

f[

f f

Jf2 x f!

K(s.t)f(t)9(s)d(p x p)(s.t)

= it, II, K(t. s)9(s) dp(s)] f (t) dp(t) 1

f f(t)[ f K(t.s)g(s)dp(s)]dp(t)_ f2

f2

For the converse. assume that (TI. g) = (f. Tg) holds for all f. g E L2(p). Define the function F E L2(p x p) by F(s. t) = K(s, t) - K(t. s). If A. B E E have finite measure. then the functions yA and tB belong to L2(p) and satisfy

foxf!

F(s. t) t.a(s)' B(t) d(p x p)(s. t)

t) x

Jftxf2

p)(s.t)

= f [ f K(s.t)YB(t)dp(t)]1.t(s)dp(s) f2

it

-f [ f K(t, s) %A(s) dla(s)] tB(t) dp(t) f

0.

_ This implies that for each p x p-step function o = A

lfixf!

1

atNA,xB, we have

F(s.t)o(s.t) d(p x p)(s,t) = 0.

(*)

Since F- belongs to LAW. there exists a sequence of p x p-step functions {o } such that 0 < o,, I F+. This implies Im FI < F+ F = (F')2 E L1(p) and

fmxf2

ot,F -. (F*)2 p x p-a.e. So. using (*) and the Lebesgue Dominated Convergence Theorem, we get 0=

ff

x p)(s.t) -- Jf

( F)2d(p x p).

Hence, f ffmx12(F+)2 d(p x p) = 0, and so F+ = 0 p x p-a.e. Similarly. F- = 0 p x p-a.c. Therefore. F = F+ - F- = 0. This shows that K(s, t) = K(t. a) holds for p x p-almost all (s. t) E fi x Q. (c) Use (a) and the obvious equality sin(st) = sin(ts) valid for all s, t E (O, 11.

(d) Lent T : L2 (µ) - L2 (p) be a symmetric operator. Then for each f E L2(p) we have IITfII2 = (Tf.Tf) = (T2f. f) < IIT21I . Ilfh2- So, IITII2 <- I171 5 IITII2. whence IIT211 = IITII2

11.1. The Daugavet Equation and Uniform Convexity

341

Since Tk is also a symmetric operator for each k, an inductive argument guarantees that IIT2" ll = IITII2" for each it. Hence, r(T) = limn, IIT2" II 2 = IITII. (e) To see that every symmetric positive operator satisfies the Daugavet equation, we need to observe that the spectral radius r(T) = IITII belongs to the spectrum of T (see Theorem 7.9) and that L2(µ) is uniformly convex. Now invoke Theorem 11.10.

Problem 11.1.11. Let {xn } and {yn } be two sequences of unit vectors in a uniformly convex Banach space X. and let {On } be a sequence of unit vectors

in X. If On(xn) - 1 and On(yn) - 1. then show that 11xn - yn[1 -. 0. Solution: It follows from the hypotheses that On(xn + yn) - 2. Hence, from On (xn +Y.) <- II an + yn II < 2 we get II xn + Y.11 --' 2. Now the uniform convexity of 1 X guarantees that Ilxn - yn II - 0.

Problem 11.1.12 (Milman [571). Show that uniformly convex Banach spaces are reflexive. Solution: The arguments below are due to B. J. Pettis [64]. Let X be a uniformly convex Banach space. Fix some x" E X" such that 11x" 11 = 1. Since the closed unit ball U of X is a(X", X')-dense in the closed unit ball U" of X", it follows that there exists a net {xQ}QEA in U such that xQ °rx.x-I-I . x". Consider the d 2x in X Moreover, net {x. +xa}(Q,3) EAXA and notice that xQ +xa of from IIxQ +xa 11 < 2 for all (Q, 0) E A x A, we see that lim sup.,$ IIxQ + xp II < 2. On the other hand, by Problem 1.1.17, the norm function y" '-' 11y"11 is a(X", X')norm lower semicontinuous. This implies 2 = 112x-11 < lim inf0,Q IIxQ + xi311. The above properties show that lim0,p IIxQ + xail = 2.

Now use the fact that X is a uniformly convex Banach space to infer that IIxQ -xQII' 0. (See Lemma 11.6.) This shows that the net {xQ} is a norm Cauchy

net. If x E X is its norm limit, then it should be immediate that x" = x E X. Thus, X** = X holds true, and so X is a reflexive Banach space.

There is another way of proving this result using Problem 11.1.11 and the following famous theorem of R. C. James [391:

A non-empty weakly closed bounded subset of a Banach space is weakly compact if and only if every continuous linear functional attains a maximum on the set. So, according to James' theorem it suffices to show that every continuous linear functional 0 on X attains a maximum on the closed unit ball of X, i.e., it attains its norm. To see this, assume that 11oll = 1. Pick a sequence {xn} of unit vectors such that O(xn) ' 1. We claim that {xn } is a Cauchy sequence. Indeed, if {xn } is not a norm Cauchy sequence, then there exist a subsequence {yn} of {xn} and some e > 0 such that Ilyn+i -yn11 > e holds for each n. Now notice that 0(yn+t) -' 1 and O(yn) - 1. But then Problem 11.1.11 guarantees that llyn+i - ynll 0, which is impossible. This contradiction shows that {xn } is a Cauchy sequence. If xn x, then lid = 1 and O(x) = 1 = 11011, as desired. I

11. The Daugavet Equation

342

Problem 11.1.13. Recall that a Banach space X is said to be uniformly smooth if for each e > 0 there exists some 6 = b(e) > 0 such that if IIxII > 1, IIvII >-1 and IIx - yli < 6, then IIx + YIl ? IIxII + IIvII - EIIx - y11

(*)

Show that a Banach space X is uniformly smooth if and only if for each e > 0 there exists some -y > 0 such that for each unit vector u and for each v E X with IIvII < -y we have IIu + vII + IIu - vII < 2 + EIIvII

(**)

Solution: Assume first that X is uniformly smooth, and let c > 0. Fix some b > 0 so that IIxII ? 1, Hull > 1 and IIx - yfl < 6 imply (t) Next, choose some 0 < y < 1 such that < 6. Now take any unit vector u E X and any v such that IIvII < -y. Let a = u_ and b Clearly, Ilall 1, IIbII ? 1 < 6, and so (t) yields and Ila - bll = i 2 IIvII < IIx + yll > IIxII + Ilyfl - 2IIx - vii

i

Ila + bit ? Ilall + IIbII -

2lia - bit.

(tt)

Substituting the values of a and b in (tt), we get: 1'ry 112uhl >

>

,

iiu + vII + 11-71lu - vii - z 1` 7 Il2vli .

This implies 2llull ? IIu + vii + iiu - vii - elivll or Ilu + vii + Ilu - vii S 2 + Ellvll

That is, the validity (**) has been established. For the converse, suppose that X satisfies (**). Fix e > 0 and then choose some

0 < 6 < 1 such that 16 < -t. Choose any x, y E X such that IIxII > 1, Iiyll > 1 and lix - yll < 6. Clearly, lix + yii = ii2x - (x - y)ii > 2 - 6 > 0. Now consider the vectors u = =+y and v = :+yl Then Ilull = 1 and IIvII = 71=+y < z < v. Therefore. (**) applies and yields IIu + vII + flu - vii < 2 + Elivii

Taking into consideration that u + v =

=+y IIxII + :2

+y

.

and u - v = +yI , we get

Ilyil 5 2 + e z+y iix - yii

Consequently,

IIx + vii ? IIxII + IIvII - 2IIx - yi1 >- IIxII + Ilyll - Eiix - vii which is (*), and the solution is finished.

Problem 11.1.14 (Smulian (74]). For a Banach space X establish the following duality properties:

(a) X is uniformly convex if and only if its dual X ` is uniformly smooth.

11.1. The Daugavet Equation and Uniform Convexity

343

(b) X is uniformly smooth if and only if its dual X* is uniformly convex.

Solution: We shall establish the following two properties: (a) If a Banach space X is uniformly convex, then X' is uniformly smooth. (,0) If a Banach space X is uniformly smooth, then X' is uniformly convex. Notice that if (a) and (Q) have been established, then the desired conclusions follow. For instance, if X' is uniformly smooth, then X** is (by (,0)) uniformly convex. Since X is a closed subspace of X", the latter implies that X is likewise uniformly convex. Similarly, if X' is uniformly convex, then X must be uniformly smooth. We now proceed to establish (a) and (Q). (a) Let X be a uniformly convex Banach space and assume by way of contradic-

tion that X' is not uniformly smooth. Then, according to Problem 11.1.13, there 0, exist some e > 0 and two sequences {4n} and {70n} in X' such that 11',,11

Il0nll=1,and

II0n+ nll+II

-0.II>2+eIIiPnll

(*)

for each n. Since X is reflexive (see Problem 11.1.12), we can find unit vectors xn and yn in X such that [On + ', J(xn) = 110n + roll

and (On -'', 1(yn) = Iimn - *nII

It is easy to check that III¢n + O II -1I 5 II+'nII, i.e., I[On +' n](xn) -1I <- 110-11Therefore, I0n(xn) -11 <- 2II0nII. Similarly, I0n(yn) -1) 5 2110.11. Since II,PnII -+

0, we get On(xn) -+ 1 and 4.n(yn) -- 1. But then the uniform convexity of X guarantees (see Problem 11.1.11) that Ilxn - yn II < e for all n > no. Now for any

nano we have II0n+1Pnli+llmn-1WI = [On+' n1(xn)+[On-1Onl(T/n) = [On(xn) + On(yn)) + *n(xn - yn)

< 2+II*nII-Iixn-ynll < 2+eIIOnll, which contradicts (*). Hence, X' must be uniformly smooth.

(Q) Assume that X is uniformly smooth and let e > 0. According to Problem 11.1.13 there exists some 0 < y < 1 such that for each unit vector u and for all

vEXwith llvII5

we have IIu + vII + IIu - vII 5 2 + 4IIvII

(**)

Now suppose that two linear functionals x', y' E Ux satisfy 11x* - y* II > e Pick some unit vector y E X such that (x' - y')(y) > 2 and let z = ryy. Clearly,

11. The Daugavet Equation

344

Ilzil = 7 and (x' - y')(z) > 2 . Finally, using (**). we obtain that 1Ix.+y*11

=

sup [x'+y'I(x) !1x!1=1

=

sup

1x'(x+z)+y'(x-z)-(x'(z)-y'(z))]

II=II=1

sup [11x+z1l+Ilx-zllII=II=1

< 2+fIIzII- 2

z]

2+ a z 2- a

That is, we have shown that if x', y' E Jx- and llx' -y'll > e, then where 5 = 4 . In other words, we have shown that X' is uniformly convex.

a

Problem 11.1.15. Show that a locally uniformly convex Banach space X satisfies the Kadete-Klee property. That is, show that in a locally uniformly convex Banach space a sequence {x,,} is norm convergent to some vector x if and only if x 'w x and 11xn 11---- 11xII. Solution: Let a sequence {xn} in a locally uniformly convex Banach space satisfy xn -u + x and IIxn 11 - luxll. If x = 0, then Ilxn - xli = 11x+.11 -' 0 is trivially true. So, we can assume that x 96 0 and xn 0 0 for each n. Put yn for each n and y = A.. Then 11yn1I = 1 for each n and 11y11 = 1. Moreover, it should be clear (from the linearity of the weak topology) that yn -k-+ y. From the identity

xn-x=llxnll(yn-y)+[IIF rZl -I]x. it follows that llxn - .r11 -. 0 if and only if Ilyn - ylI -' 0. Therefore, in order to complete the solution, we must show that Ilyn - y11 0. To this end. assume by way of contradiction that fly,, - yfl 74 0. This means that there exist a subsequence of {y,,} (which we shall denote by {yn } again) and some 0 < e < 2 such that llyn - yll ? e holds for all n. Since X is locally uniformly convex, it follows that Ilyn + yll f 2 (see Lemma 11.15). Since Ilyn + yll <_ 2 for each n, this means that there exists another subsequence of {y,,} (which we shall

denote by {y,,} once more!) and some 0 < b < 2 such that Ilyn + yll < 2 - b for each n. Now pick some x' E X' satisfying llx'll = 1 and x'(y) = 1 and note that 1 r' (yn + y) l -- Ix' (y) + x' (y)1 = 2. On the other hand, we obviously have lx'(yn+y)I 511x'11.11yn+y11 <-2-6 for each n, a contradiction. Hence,llvn-yll - 0 and consequently llxn - xli -. 0 is true. The converse is true in any Banach space. That is, if llx,, - xll -, 0 in any Banach space. then clearly xn -w x and llxnll -- lixll are automatically true. it

Problem 11.1.18. Show that each Lp-space with 1 < p < oo is uniformly convex and uniformly smooth. Also show that the infinite dimensional L3 and L,,-spaces are neither uniformly convex nor uniformly smooth.

Solution: We shall establish first that the Lp(p)-spaces with 1 < p < 00 are uniformly convex- So, let (Il. E, p) be a measure space, and fix some I < p < oo. We shall verify that for arbitrary functions f, g E Lp(p) the following equalities and inequalities hold.

11.1. The Daugavet Equation and UnifQrm Convexity

345

(1) (The Parallelogram Law) If p = 2, then IIf+91Ip+IIf -9IIP=2(IIf112+119IIp) (2) (Clarkson's inequality # 1; (22]) If p > 2, then

IIf+9IIP+IIf -9IIP <2P-'(IIfIIp+IIgIIp) (3) (Clarkson's inequality # 2; (22]) If 1 < p < 2 and 1 < q < oc satisfy v+q= 1, then

IIf +9IIP+IIf -9IIP _<2(IIfflP+II9IIP)' If these properties are established, then II fn + 9n IIP - 2 with llfn IIP < 1 and 119n1IP <- 1 imply Ilfn - 9n11p -. 0. That is, Lp(p) is a uniformly convex Banach space for each 1 < p < oc. First of all, we will show that the Parallelogram Law is true in any inner product space X. Indeed, if X is any inner product space and x, y E X, then I1x+y112+1Ix-y112

= (x+y,x+y)+(x - y,x - y) = IIxI12+(x,y)+(y,x)+IIvII2+IIXII2-(x,y)-(y,x)+11vII2 = 21IxI12 +2

11Y112.

Next, we shall verify Clarkson's inequality #1. To this end, assume that p > 2. In this case, we claim that for all a,,3 E R we have la + 01P + ja - 3IP < 2P-1(I(kIP + 1,31P).

(*)

If (*) is established and f,g E Lp(p), then If (w) + 9(w) I P + If (w) - 9(w) I P

< 2P-'(If

(w)IP + I9(w)IP)

for p-almost all w E Q. and the validity of Clarkson's inequality #1 follows by integrating.

Now let us prove (*). Observe first that if (*) is true for all a,,3 E R+, then (*) is also true for all a, 0 E R. If a = 0 or Q = 0, then (*) is trivially true. So, we can suppose that a > Q > 0. Letting t = > 1, we can rewrite (*) as (t + I)P + (t - 1)P < 2p-1(1 + tP).

(**)

To see that (**) is true, consider the function f(t) = 2P-1(I+tP)-(t+1)P-(t-1)P for t > 1. Clearly, (**) is true if and only if f (t) > 0 holds for each t > 1. Taking derivatives, we see that

f'(t) = f"(t) = P(p-

= p(p -

p[2P-1tP-'

- (t + 1)[2P-'tp-2

1) [

= p(p - 1) (

2(2t)P-2

[(2t)p-2

1)P-'

- (t + - (t + - (t +

- (t 1)P-2

1)P-2

1)P-1

J , and

- (t - (t -

1)P-27 +

1)P-2] 1)P-2 ] [(2t)p-2

- (t - 1)P-2]) > 0.

The latter implies that the derivative f' is a strictly-increasing function, and conse-

quently f'(t) > f'(1) = 0 for all t > 1. In turn, this implies that f is an increasing function. In particular. we have f (t) > f (1) = 0, and this establishes the validity of

11. The Daugavet Equation

346

(**). (It is interesting to note that (*) is also true for any pair of complex numbers

aand0.) Next, we shall prove Clarkson's inequality #2. To this end, let 1 < p < 2. Recall that q = P"I > 1. In this case, we can prove with an argument similar to the preceding one that for each a,,8 E R we have

Ia + 01" + Ia - 0111 < 2(Ialp + 1AIP)

(t)

.

Now let f, g E Lp(p). Then from (t) we get

-

9(w)I9

If (w) + 9(w)I9+I f(w) for p-almost all w E 11. This implies 11 If

+9111 +

if _ g19

11p_1

e

< 2(If(w)IP + I9(w)IP) ° r 91Q)P-1(1141 i

(If + 9111 + If [2P-1 j (If (w)IP

e

+ Ig(w)IP) dp(w)1

T

2( IIf IIp + IIgIIP )' T

Prom q(p - 1) = p, we see that If + g111, If - 9111 E Lp_1(,a). Now taking into

account that 0 < p - 1 < 1, it follows from the "reverse" Minkowski inequality'

that

II If

If -9IQIIp_1 < IIIf+914+If -9IQIIp_,. Since IIIf + gIQIIp-1 = IIf + 9IIp and If - 9I9IIp-1 = IIf - 9IIp, the preceding inequality can be written as +9I"Il,_,+II

IIf +9IIp+IIf -gIIp < IIIf +gI'+If -glpllp_1' and so Ilf + gilp + IIf - 91199 < 2(IIf IIP + IIgIIP) -

,

which is Clarkson's inequality # 2. If L, (p) and Lao (p) are not finite dimensional, then they are not reflexive. So, in this case, according to Problems 11.1.12 and 11.1.14 neither L,(p) nor L,o(p) is uniformly convex or uniformly smooth.

Problem 11.1.17. This problem discusses various differentiability properties of the norm. Let x be an arbitrary non-zero vector in a Banach space X. Establish the following.

(a) If 0 < tl < t2i then for each v E X we have IIx + tiyIl - Iixll < IIx + t2VII - IIzII -IIvII < t2 ti

-

< IIVII

This shows that the limit

x(v) = lira

lIx + tvll - llxlI

t

' It is well known that if 0 < r < 1, then for every pair of functions x, y E L,.(µ) we have IIx + yllr >- IIxlkr + IIYIIr, where IIZIIr = (fn Izlr dµ)

*

for each z E L,.(µ).

11.1. The Daugavet Equation and Uniform Convexity

347

exists and I0x(v)I <_ IIvII. Also, ¢x(x) = IIxII and Ox(-x) = -IIxII (b) The function v F- 0x(v) from X to R is sublinear.

(c) A linear functional x* E X* is said to be a 8upporting functional at x if IIx* II = 1 and x* (x) = IIxII We denote by Jx the non-empty

collection of all supporting functionals at x. Show that Jx consists of all linear functionals on X that are dominated by 4x(), i.e., Jx = {x* E X* : x*(v) < qx(v) for all v E X)

.

(*)

(d) For the function v H qx(v) the following statements are equivalent.

(i) 0x(') is linear. (ii) Ox(-v) = -,Ox(v) for each v E X. (iii) The norm of the Banach space X is Gateaux differentiable x+t t - x exists in R for each v E X. at x, i.e., (iv) There exists a unique supporting functional at x, that is, the set J. is a singleton. (v) Ox() is linear and is the only supporting functional at x. (e) The norm of a Banach space X is said to be Fr< chet differenx exists uniformly in v in at x if the limit limt,o x+tt the unit sphere Sx = {v E X : IIvII = 1}. Establish the following. a. If the norm is Fr< chet differentiable at x, then it is also Gateaux differentiable at x.

/3. The noun is Frechet differentiable at x if and only if there exists a (necessarily unique) linear functional x* E X* such that: for every e > 0 there is some S > 0 so that for each u E X with IIxII < S we have I IIx + ull - IIxII - x*(u) 1 < EIIuII

(**)

Solution: (a) Fix a vector v E X and consider the function 0: R - R defined by fi(t) = IIx + tvll - IIxII for t E R. Clearly, 1'(0) = 0. The triangle inequality implies that is a convex function satisfying -ItIIIvII 5 +'(t) 5 ItIIIvII The well-known properties of the convex functions guarantee that their difference ratio is monotone

and sofor0
t'(t1) - 0(0) = 0(t1) < 0(t2) = O(t2) -+G(0) < _ e2 t2 t1 t1

- IIvII

,

which is (a). This certainly implies that the limit

0x(v) = lim 0(t) = lim IIx + tvll - IIxII = Jim IIx + tvll - IIxII tlo t t-»o+ t t-.o+ t exists and satisfies lO=(v)I < IIvII.

11. The Daugavet Equation

348

Letting v = x, we see that '(t)=(11+tI-1)1lxll and sot;'(t)=tllxll for any t-> -1. Thus, 0=(x) = lim lix + txll - Ilxll = Ilxll , t-o+

t

and similarly

0r(-x) = tlm

Ilx - t X11 - II=II

= -IIxII -

(b) If .1 > 0, then we have IIZ+t(AV)II-11z11 = a[ _+ At FV11-11X11] for each v E X. Letting t - 0+, we get t ()v) = A4z(v). That is, o_(-) is positively homogeneous. To verify the subadditivity of 0_(-) pick any v, w E X and notice that for each t > 0 we have Ilx + tvll - IIxII + llx + twit - IIxII = Ilx + tvll + fix + twit - 21Ixil t t t > II2x + t(v + w)II - 211xll t

=

Ilx + 2 (v + W) 11 - IIxII t

z

.

So, letting t - 0+, we get 0=(v) + 0=(w) > O=(v + w). This establishes that is a sublinear function.

(c) Let J. = {x' E X*: 11x*II = 1 and x'(x) = IIxII} be the set of all supporting functionals at x. The Hahn-Banach theorem guarantees that J. is not empty. To verify (*) assume first that x' E J,,,.. Then for every scalar t and every v E X, we have

IIxII + tx'(v) = x'(x) + tx'(v) = x'(x + tv) < 11x + tv11, and consequently x'(v) < 11=+t `I-II=II for all t > 0 and v E X. This implies that

x'(v)<0=(v)foreachvEX,i.e.,J1C{x'EX': x'(v)
for all v E X, that is, 1Ix'11 < 1. However, -x'(x) = x'(-x) < 0=(-x) = -IIxII and so x'(x) > IIxII. This proves that x'(x) = IIxII and 11x'11 = 1, that is, x' E J. Hence {x' E X': x'(v) < tp=(v) for all v E X} C J=, and so

J= = {x' E X*: x' (v) < O=(v) for all v E X) . (d) The implication (i)

(ii) is trivial since if

is linear, then of course

O=(-v) _ -q=(v) for each v E X. Notice also that whenever follows from (a) that 0s(-) E J. (ii)

is linear, it

(iii) We have lim to-

lix + tvll - IIxII t

This proves that limt-0

=-

lim

t-.o+

Ill+tvtl-II=II

lix + t(-v)II - IIxII =

t exists and is equal to 0,,(v).

4:(v)

(iv) Assume contrary to our claim that the set J, is not a singleton. That is, suppose there are two distinct unit linear functionals x', y* E X' satisfying (iii)

11.1. The Daugavet Equation and Uniform Convexity

349

x'(x) = y'(x) = Ilxll. Since r' 34 y', the linear functionals x' and y' must be linearly independent. Otherwise, x' = Ay' implies Ilrll = x'(x) = Ay'(x) = Allrll. This yields A = 1, and so r' = y'. a contradiction. Since x' and y' are linearly independent, there is some u E X with x'(u) = 0 and y'(u) > 0. Similarly, there exists some v E X with y'(v) = 0 and x'(v) > 0. Now consider the vector z = v - u and note that x'(z) > 0 and y'(z) < 0. From llxll + tx'(z) = x* (x) + tx'(z) = x* (x + tz) < llx + tzll, it follows that 11=+tz 11-II=II > for each t > 0, and so e

lim

t-o

Ilx + tzll - Ilxll = lim llx + tzll - llrll > x'(z) > 0. t t t-.o+

-

Similarly, for each t < 0 we have

Ilrll + ty'(z) = y'(x) + ty'(z) = y'(x + tz) <- llx + tzll, and so

II=+tztl-II=II

lim t-o

< y(z) for each t < 0. This implies

llx + tzll - llxll = lim llx + tzll - Ilxll < Y *(Z) < 0. t t t-»o-

The above inequalities show that limt.o

II=+tzi1_11.1I

does not exist, which is a

contradiction. Hence, J= is a singleton.

(v) Assume that the set J= is a singleton, that is, J= = {x'} for some For a singleton J, the identity (*) established in (b) means that there exists exactly one linear functional which is (iv)

x' E X'. We want to prove that r' =

We claim then that 0= itself must be dominated by the sublinear functional linear. To see this, assume by way of contradiction that is not linear. Then

there exists some v E X such that -0=(-v) < 4=(v). (Here we use the facts that a sublinear functional p on X always satisfies the inequality -p(-z) < p(z) for

each z E X and that p is linear if and only if p(-z) = -p(z) for each z E X.) Consider the linear subspace 11.1 = {Av: A E IR} generated by v, and define the linear functionals g, h : Al - IR by

g(Av) = -A0=(-v)

and

h(Av) = AO:(v)

for Av E M. A moment's thought reveals that both g and h are dominated by the restriction of the sublinear functional to M. So, by the Hahn-Banach Extension Theorem, g and h have linear extensions to all of X that are dominated by o_. We denote these extensions again by g and h. It remains to notice that

g(v) =

< O,(v) = h(v) implies that g 0 h, and this contradicts the

preceding conclusion. Hence, so

= x', as desired.

(v)

(i) is trivial.

is a linear functional. Therefore 0,(-) E J=, and

(e) We first prove (o). So. assume that the norm is Frechet differentiable at x. That is, the limit o1(v) = limt.o II=+t°l1-II=II exists uniformly over v E Sx. Now notice that if v E X is an arbitrary non-zero vector, then we have II=+t» II-Il=il

ii=+(tllvll) IlVVl hj-II=II

11. The Daugavet Equation

350

In particular, the limit limt.o Ilx+e 11-11x11 exists in R for each v E X. This shows that the norm is Gateaux differentiable at x. Next, we shall prove (Q). Assume first that the norm is ft chet differentiable at x. By part (a) we know that the norm is Gateaux differentiable at x, and so

x' _

is a linear functional. In other words, we have lim Ilx+evll-11x11 = t

uniformly for v E SX. This means that for each e > 0 there exists S > 0 such that I1x+tvll-11x11

- x'(v) I < e

for each v E SX and each non-zero t E (-S, S). It follows that I Ilx + tell

- Ilxll - tx"(v) I < Eltl

for all Itl < S. Now if u E X satisfies 0 < Ilull < S, then we can write u = tv where

v =III E Sx and t = Ilull < S. This implies that Ilx + ull - Ilxll - X*(u) I <_ ellull, holds for all u E X with Ilull < S, and the validity of (**) has been established.

For the converse, assume that (**) is true. Fix e > 0 and then choose some S > 0 that satisfies (**). Next pick any v E Sx, any non-zero t E (-S, S), and let u = tv. By (**) we have I llx + tvll -11x11-

I < Elltvll = Eltl

or, equivalently, I

Ilx+evll-11x11

_ X*(v)I < C.

This proves that the limit limt-o Ilx+evtl-IlxlI = x* (v) exists uniformly over v E Sx. Therefore, the norm is FYechet differentiable at x, and the solution is finished.

Problem 11.1.18. Show that a Banach space X is uniformly smooth if and only if its norm is uniformly Frechet differentiable over the unit sphere of X, that is, the limit Ox(y) = aim

llx + tyll - 114

exists in IR uniformly for all unit vectors x, y E X X. Solution: As usual we denote the unit sphere of the Banach space X by SX, i.e., Sx = {x. E X: llxll = 1}. Assume first that the limit i'-.m

Ox(y)

In particular, this implies that for each unit

exists uniformly over x, y E Sx vector x the sublinear functional O consequence of this is that the limit lim [ Ilx+tvtl-II=II + L

Ilx + tyll - 114

Ilx-tvtl-11x11 1

j

is linear; see part (d) of Problem 11.1.17. A

= li o t

IIx+tvtl-11X11

+ eina 11-411-11=11

= MY) + Ox(-y) = O AY) - 4'x(y) = 0

1I.1. The Daugavet Equation and Uniform Convexity

351

exists uniformly over x, y E Sx. So, if E > 0 is given, then there exists some 0 < 'y < 1 such that Ilx+tyil-llx11 + 11x-tylll-11x11

I

<

for all x, y E Sx and all non-zero t E [-y, y]. This implies that for all unit vectors x and y and for all 0 < Itl <-y we have fix + tyll + lix - tyll < 2 + Eltl .

In particular, if u is any unit vector and v is any vector with 0 < llvll <_ y, then by letting x = it, t = llvll and y = Ilv I' we get Ilu + vll + hit - vii < 2 + Ellvll .

Clearly, this inequality is also true for v = 0. By Problem 11.1.13 the above condition is equivalent to the uniform smoothness of X.

For the converse, assume that X is uniformly smooth. We must show that there exists a function -t/': Sx x Sx -R such that for each c > 0 there exists some y > 0 such that I IIx+ty11-11x11

- ?/'(x, y) I < E

(*)

for all x, y E Sx and all t with 0 < Itl < y. The arguments below establishing this claim are due to J. Diestel. We start by observing that according to Problem 11.1.14 the norm dual X* of X is uniformly convex. So, given e > 0 there exists some 0 < 6 < 1 such that:

f,gESx and

11f+g11>2-J

IIf -gil<E.

(**)

Next, for each x E Sx select a functional fx E X' supporting at x, i.e., = llfxll 1 and fx(x) = 11x11 = 1. We claim that:

x,YESx and IIx-yll<6

Ilfx - fyll<E.

(***)

To see this, assume that two vectors x, y E Sx satisfy IIx - yll < 6. Then we have

Ilfx+fyll+IIx-yl1 ? (fx+fy)(y)+fx(x-y) f. (y) + fy(y) + f. (x)

- f. (y) = 2,

and so llfx + fy 11 > 2 - 6. Now a glance at (**) guarantees that 11 fx - fy 11 < E. We can extend the mapping x '-+ fx from Sx to all of X by letting fZ = llzllfe if z # 0 and fo = 0. Clearly, = f if z 34 0 and f=(z) = IIZI12 for all z E X. In addition, note that If,,(v)I <_ IIull Ilvll for all u,v E X. Now fix x, y E Sx, t E (0,1), and let zt = .+tt Then we have

11. The Daugavet Equation

352

f=(ty)

-

f=(x) + f=(tb) - 1

- f=(x + ty) - I < If=(x + ty)I - 1 -

t lix+tyll-11x11 t t t Ilx + tyll2 - Ilxll Ilx + tyll < Ilx + tyll2 - I f=+ty(x)I tllx + tyll tllx + tyll t

t

Ilx+tyll-1 =

t

-

f=+ty(x + ty) - l f=+ty(x)I = f=+ty(x) + t fz+ty(y) - I f=+ty(x)1

tlix+tyll <

tllx+tyll

tf=+ty(y) = f=+ty(y) = fz, (y) tiix + tyll ilx + tyll

In particular f=(y) < II=+tytl-II=II < f,., (y) for all t E (0,1). Similarly, for each II=+tvii-11x11 < f=(y) Therefore, for each non-zero t E (-1,0) we have fz,(y) < t E (- 1, 1) we have II=+tvll-11x11

t

_ f=(y) I

Ifze(y) - f=(y)I

11h, - f=1)

(t)

Next, put ry = min{ 4. } and choose any t satisfying 0 < Itl < 7. Clearly, lix + tyll ? Ilxll - Iti Ilyll =z 1 - Itl > 2, and an easy computation shows that the vectors x, zt E Sx satisfy tv+ 11=+e It = I < 2(11- Ilx Ilzt - x11 = + tyll I + It1) = =

2(I ilxil - lix + tyll I + itl) <_ 2(Ilx - (x + ty)II + Itl) 41t1 < S.

Consequently, it follows from (* * *) that 11f,, - f=II < c. Finally, a glance at (t) shows that for all x, y E Sx and all 0 < Itl < 1 we have II=+tv ll-II=II - f=(y) I < E, I

e

which is (*). This completes the solution.

11.2. The Daugavet Property in AL- and AM-spaces Problem 11.2.1. Show that a compact Hausdorff space 11 has an isolated point if and only if the Banach lattice C(11) has an atom. Solution: Assume first that wo is an isolated point of f2, i.e., assume that the set {wo} is a clopen subset of f2. We claim that the function f = X(,,,o) E C(f2) is an atom. To see this, assume that 0 <_ g, h < f satisfy g A h = 0. Since f (w) = 0 holds for all w 96 wo, it follows that g(w) = h(w) = 0 for all w 54 wo. Assume g 96 0,

i.e., g(wo) > 0. Then, from min{g(wo), h(wo)} = (g A h)(wo) = 0, it follows that

h(wo) = 0 or h = 0. That is, either g = 0 or h = 0 is true, proving that f is an atom of C(f2).

11.2. The Daugavet Property in AL- and AM-spaces

353

For the converse, suppose that 0 < u E C(Sl) is an atom. Fix some wo E Sl such that u(wo) > 0. We claim that u = X{o,o} If this is established, then {wo} is a clopen subset of 12, and so wo must be an isolated point of 12. To see that u = X{wo}, pick any w 34 wo. Since fZ is a compact Hausdorff space, there exist functions f, g: f1- [0, 1] satisfying f Ag = 0 and f (wo) = g(w) = 1. Now consider the continuous functions ul = fu and u2 = gu. Clearly, 0 5 U1, U2 5 u, ul A u2 = 0, and ul (wo) > 0. Since, u is an atom, it follows that u2 = gu = 0. In particular, we have u(w) = g(w)u(w) = U2(w) = 0 for all w A wo. This shows that u = X{0}, as desired.

Problem 11.2.2. This problem shows that a compact operator on an AMor an AL-space with atoms need not satisfy the Daugavet property. Let E = C(fl), where fl is a compact Hausdorff space having an isolated point, say wo. Consider the multiplication operator T : C(l) -> C(1l), defined by T f = -X{wo}f.. Establish that: (a) T is compact. (b) IITII = 1.

(c) III +TII<1<1+IITII=2. Solution: (a) Let { fn} be a sequence in C(Sl) satisfying II fnll,, < 1 for each n. Then {fn(wo)} is a bounded sequence of real numbers, and so it has a convergent subsequence, say fk., (wo) -' a. If f = -o X{.o}, then note that f E C(1Z) and IITfk,

-flloo =

-i 0.

This shows that T(U), where U is the closed unit ball of C(11), is a norm totally bounded subset of C(ft). Consequently, T is a compact operator.

(b) If f E U, then note that IITflloo =maxITf(w)I = max

If(wo)I < 1.

So, IITII = suPIEU IITf II o0 5 1. Since IIT1II0o = 1 and i E U, we see that IITII = 1.

(c) For the norm of I + T observe that if f E U, then f (w) + T f (w) = f (w) if w 96 wo and f (w) + T f (w) = 0 if w = wo. This implies I (f + T f)(w)I < 1 for each

wEQand each f EU,and so III +TII=supfEUIIf+TfIlooS1. Therefore, III + TII < 1 < 2 = 1 + IITII. This shows that the compact operator T does not satisfy the Daugavet equation.

Problem 11.2.3. If J : E - E is an isometry on an AL-space, then show that IIJ + TII = 1 + IITII for each operator T EC(E) that is disjoint from J. Solution: Recall that every bounded operator on an AL-space has a modulus and that its norm coincides with its r-norm; see Theorem 3.9. Now assume that an operator T E G(E) is disjoint from an isometry J E G(E), i.e.. 1.11 A ITI = 0. This implies IJ +TI = IJI + ITI. So, if x E E satisfies IIxII = 1, then using that E is an

11. The Daugavet Equation

354

AL-space, we obtain IIJ + TII

=

II IJ + TI II = II IJI + ITI II (I IJIIxI + ITIIxI II

=

II IJIIxI II + II ITIIrl II

IIJxII + IlTxll 1 + IITs.II

.

This implies l+IITII =1+sup11=11=1 IITxII <- [IJ+TII. and so IIJ+TII = 1+IITII I

Problem 11.2.4. Let f,g: [0,1] -+ R be two non-zero bounded real measurable functions, and consider the rank-one operator T = f 9- g acting on each Lp[0,1]-space. That is, the operator T : Lpj0,1] - Lp[0.1] is defined via the formula t

Th = (f 0 g)h = (h, f)g = [J h(s)f(s) ds] g 0

for each h E Lp[O,1].

(a) Verify directly (i.e., without using Theorem 11.20) that the operator T : L,,.[0, 1] -0 L,,,,[0,1 ] satisfies the Daugavet equation. (b) Show that T : L2 [0, 1] - Ll [0, 1] satisfies the Daugavet equation.

(c) If (g, f) = 0, then show that T: Lp[0,1] - Lpf0,1] does not satisfy the Daugavet equation for any I < p < oc. Solution: (a) We shall denote the Lebesgue measure on [0, 1] by A. Note that IITII = 11fII1- IIgIIx Thus, according to Corollary 11.5, we can assume without loss of generality that Ilf III = IIgIIx = 1. In particular, we have IITII = 1.

Now fix 0 < E < 1. From IIsOx = 1, it follows that there exists a measurable subset A of [0.1] of positive Lebesgue measure such that Ig(t)I > 1 - e holds for

eachtEA. Let Al={tEA: g(t)>1-c}andA2={tEA: -g(t)>1-e}. If A(A,) = 0, then A(A2) > 0. So, in this case, if we replace g by -g. f by -f and A by A2, we can assume that g(t) > 1 - E for each t E A.

Next, let x = Sgnf. That is, x(t) = 1 if f (t) > 0 and x(t) = -1 if f (t) < 0. Clearly, X E L[0,1] and Ix(t)I = I for each t. By the absolute continuity of the integral, there exists some 6 > 0 such that for each Lebesgue measurable subset B of [0, 1] with A(B) < 6 we have fe If(t)I dt < Z. Pick any Lebesgue measurable subset B of A with 0 < A(B) < 6 and consider the Lebesgue measurable function y = rXe- + Xe. Clearly, Ig(t)I = 1 for each t E [0, 1]. and so flyllx = 1. Now note

that fI

f(t)y(t)dt =

L1f(tt+Lf1t

f

=

f

>

1-2f If(t)Idt> 1 -E.

1

Ig(t)I dt +

o

B

e

f(t) dt - lB

11.2. The Daugavet Property in AL- and AM-spaces

355

Consequently. we have

III+TII ? Ily+Tyll. = Ily+ (f f(t)y(t)dt)9II x 0 r Il

r

ls+ (J' f(t)y(t)dt)9]

Rll x

.

Since y(f) = 1 and g(t) > 1 - f for each f E B. it follows that

III +TII>1+(1-f)2 for each 0 < f < 1. Letting a

0+. we get III + TII > 2. Therefore. III + TII = 2. and so T satisfies the Daugavet exluation.

(b) From part (a) we know that the operator S = g f : L, [0. 11 Lx [0.11 satisfies the Daugavet equation. It remains to note that S = T. and so 7' also satisfies the Daugavet equation.

(c) It is easy to see that T2 = 0. This implies r(T) = 0 or a(T) = {0}. Since T is non-zero, we have IITII > 0. Since for I < p < x the Banach space L (1:) is uniformly convex (see Problem 11.1.16). it follows from Theorem 11.10 that T

satisfies the Daugavet equation if and only if 0 < IITII E o(T) = {0). which is impossible. Hence, for each 1 < p < x the operator T: Lp[0. 1] - Lf,(0.11 does not satisfy the Daugavet equation. I

Problem 11.2.5. If p is a or-finite non-atomic measure. then show that every finite-rank operator on Lp(p), where 1 < p < x, is disjoint from the identity operator and so in this case, by Lemma 11.21. every finite-rankoperator on Li(p) satisfies the Daugavet equation. Solution: Assume first that (11, E. p) is a nun-atomic finite measure space. Fix f E L4 (it) and it E L; (it), where 1 < p < x and 1 < q < x satisfy 1 + 1 = I. To prove I A (f 0 u) = 0. it suffices to establish that I A (f s;, u)(1) = 0. To verify this. assume that some g E Ly (p) satisfies g < 1 A (f ti: u)(1). Fix f > 0. The absolute continuity of the integral guarantees the existence of some 6 > 0 such that fA f (t) dp(t) < f for each measurable set A with p(A) < 6. Next, choose some n with ',fl1 < f and then select pairwise disjoint measurable sets

A,-- A,, with p(A,) = '`'st) for each is this is possible since it is non-atomic. Since NA', + XA. = 1, it follows that g < I n (Jr S u)(1) < I(\.a;) + (f : u)(t.a. )

tA', + { f4. f (t ) dit(t) J u < YAK + fu

for each i. This implies g < as. Since f > 0 is arbitrary, the latter shows that g = 0. and from this it follows that I A (f : u) = 0. Now assume that (51. E. JA) is a v-finite measure space. Pick a sequence {52,,} in

E such that p(Q,) < x for each it and 0,, l Q. Also. fix f E LQ (p) and is E LP - (it), where 1 < p < x and I < q < x satisfy 1 + w = 1. By the preceAing case, we have

each n. So. if 0
11. The Daugavet Equation

356

I A (f ?, u) that I A (f 9 u)(g) = 0. This implies that I A (f le u) = 0 holds true in this CA.tie too.

Now let S =

u, be a finite rank-operator. Then

1_1 f,

k

Ofi$u,I
g=1

k

=

IA(EIf.ISIuil) I=t k

< EIA(IfilZ- IuJ)=0. t=1

(For the validity of the last inequality see the solution to Problem 3.3.1: see also Lemma 4.13.) So. I A ISI = 0, and thus the identity operator I is disjoint from the vector space of all finite-rank operators. I

Problem 11.2.6. Present an example of a bounded operator on some Lpspace with 1 < p < oc that is disjoint from the identity operator and fails to satisfy the I)augavet equation.

Solution: Let f = Y(u.i) u = xli,li, and consider the rank-one positive operator T = f L;, u: Lp[O,1I - L1,[O.11. By Problem 11.2.5. T is disjoint from the identity operator. and by Problem 11.2.4(c) T fails to satisfy the Daugavet equation. I

11.3. The Daugavet Property in Banach Spaces Problem 11.3.1. If T : X - Z and S: Y

Z are bounded operators

between nonmed spaces. then show that sup{ IITx + SyII

:

IIxII 5 1. IIyII < 1) =sup{ 11 TX + Syll : IIrII = IMI =1 } .

Solution: Assume that T : X

Z and S: Y -. Z are two bounded operators

between normed spaces. Let

0 = sup{IITx+SyII: IIxII 5 1. Ilyll < 1}. and .3 = sup{IITx+SyII Ilxll=llyll=l}. Clearly. 3 < ct. Pick a y r. E X and y E Y satisfti ing IIxf l < 1 and Ilyll 5 1. If r = 0 and y = 0. then IITx + SyII = 0 <0 is trivially true. Now consider the case where x Y6 0 and y = 0. Put it = , . and then pick a unit vector r E Y. Then, using Lemma 11.24. we sec that

(ITr +SyII = <

llxll

IITuii <_ IITuII = lllTu + OSvlI

3.

11.3. The Daugavet Property in Banachh Spaces

357

Finally, consider the case x 76 0 and y 0 0. Again using Lemma 11.24, we get IITx + Syli

=

IIIIxIIT(Il') + IIyIIS(;rvrl )II

< max{IIT(I=11)+S(

)II,IIT( TI-h)II}
So, we have shown that if x E X and y E Y satisfy IIxII <- 1 and IIyll 5 1, then II Tx + T y II < /3. This implies a < 0, and so a = 0. as desired. It should be clear that if T ® S: X ® Y -. Z denotes the direct sum operator,

defined by (T ® S)(x ® y) = Tx + Sy, and X ® Y is equipped with the norm I IIxeylI =maxi IIxII,IIyII}, then IITOSII =a=(3.

Problem 11.3.2 (Abramovich (1J). Let E be a o-Dedekind complete Banach lattice and let UE be its closed unit ball. Consider the set D of all vectors x for which there exists some collection { {XI, x , . . . , xn} of pairwise disjoint V xn, and let positive vectors in UE such that I XI = x1 V X2 V M(E) = sup{ IIxII : x E D 1.

Show that E is lattice isomorphic to an AM-space if and only if M(E) < oo. Solution: Assume first that E is lattice isomorphic to an AM-space. That is, assume that there exists an M-norm I I

I

I I I on E such that

alIxil < 11141 :5 /IIxII

(*)

for each x E E and some constants 0 < a <,3. Pick any collection {x1, x2, ... , of pairwise disjoint positive vectors in UE and let x = x1 V X2 V ... V xn. Then we have

II''IICkIIIxIII=a tmax IIIxiIII- Etmx Iix;llSE, whence M(E) < a < oo.

For the converse assume that M(E) < oc. Recall that in any o-Dedekind complete vector lattice every principal band is a projection band. For z E E+ the band projection on the principal band B. will be denoted by PZ; as is well known, P=(x) = supm x n mz for each x E E+. A direct verification shows that D is a solid subset of E satisfying the inclusions UE C D C AI(E)UE. We claim that D is also a convex set. To verify this, pick any x, y E D and a E [0,11. We must establish that ax + (1 - a)y E D. Since D is solid, we can assume without loss of generality that x and y are in E+. By the definition of D there are pairwise disjoint positive vectors x1,.. . , x in UE such that x = x 1 V V x and pairwise disjoint positive vectors yl,... , yk in LIE such

that y=yiV...Vyk. For each 1 < i _< n and each 1 < j < k let uj = Py, (xi) and vii = P,,, (yj). From the identity

uiJ Av,,,, =sup[x1Amy,] A

v,,,, = 0 for (i, j) 54 (µ, v). Similarly, (i, j)

U'J A u,,,, = 0 and V,_, A v,,,, = 0.

(µ, v) implies

11. The Daugavet Equation

358

Now let u; = x; - FA_ u;j and v., = vj - E" I nj. Since Ek-1 11;j = Py(r; )

and E ,_ 1 v,j = P . (y. ) , w e see that 0 < u, < x, and 0 < v , < y .

11u,11

IIv, II < 1, u, 1 y and v 1 x for all i and j, and hence u; 1 yy and u I x; for all i and j. In particular, we have u, 1 vj for all i and j. Also, from the definitions and v,, I v, if j # v. Now of the u; and v. it is easy to see that u; 1 u,, if i notice that n

k

i=1

J=1

k

x=E(u,+Eu,j) and

n

yE(vj +Ev,j

J=1

1=1

Letting w1 = au;, + (1 - a)v;j, we have 11w,_,11 1 for all i and j and n

k

ax+(1-a)yF,

n

n

=1j=1 1=1 J=1 where, as shown above, the vectors in this representation are pairwise disjoint and belong to UE. Hence, ax + (1 - a)y E D. and thus D is a convex set. Now we consider the Minkowski functional of D, i.e., the functional po : E -o R. defined by

po(x)=inf{A>O: rEAD}. From UE C D C M(E)UE, it follows that p,, (x) < IIxII < A1(E)p0 (x) for all x E E. That is, po is a lattice norm on E that is equivalent to the original norm II on E. Finally, we claim that po is an M-norm. To see this, assume that two II -

vectors x, y E E+ satisfy x n y = 0. We must show that pn (x V y) < max{p0 (r), pn (y) } .

Let A > max{p0 (r), pD (y) }. Clearly, X E AD and y E AD and from x n y = 0, it easily follows that x V y E AD. This implies pD (r V y) < A. and from this we obtain po(xV y) < max{p0(x),p0(y)}. Hence, p,, (x V y) = max{p0 (x), pD (y) } and so p,,, is an M-norm. as desired.

I

Problem 11.3.3. Let E = L1 [0.1] and F. = L,C[0. 1] for each n. For the Banach lattices E _ (r,' 16)En)x and F = (En 1®F,,)1 establish the following.

(a) Neither E nor F is lattice isomorphic to an AL- or an AM-space. (b) Neither (E(B F)1 nor (E® F),,, is lattice isomorphic to an AL- or an AM-space. (c) The Banach spaces (E (D F)1 and (E e F)"' satisfy the Daugavet property for weakly compact operators. Solution: (a) We use the notation introduced in Problem 11.3.2. We claim that M(E) = M(F) = ,c, and so neither E nor F is lattice isomorphic to an AM-space.

To see that M(E) = ac, start by letting xn = n(n + 1)x1

E L1[0. 1).

Clearly, Ilxn II1 = 1 for each it and if n 0 m, then xn A x,n = 0 holds in L1 [0, 1). Next, for each n let

xn = (xn,0,0.0.... ).

11.4. The Daugavet Property in C(Q)-spaces

Note that IIxn 11

359

= 1 for each n and xn A xm = 0 in E for all n 0 m. In particular,

we have IIX1 V X2 V ... V Xn lloo

II /(x1 V x2 V ... / xn, O, 0, o....) )II. +x2+...+xn,0,o,0,...)II

= II(xl

00

n

I1x1+x2+...+xnII1 =EIIxiIIl =it i=1

This implies M(E) > n for each n, and so M(E) = oo. For the Banach lattice F, for each n let yn = where the function 1 = Xlo,iI E LCO[0,1] occupies the nth coordinate. Clearly, yn A ym. = 0 holds in F for all n 56 m. Now notice that (0,0,...,0,0,1,0,0,...),

ill

IIy1Vy2v

_

..Vynlll=111,1,...,1,0,0,0,...)II1

n positions

a

-

1

This implies M(F) > n for each n, and so M(F) = oo. Rom F' = E, it follows that M(F') = oo. An easy argument also shows that M(E') = oo. Therefore, neither E' nor F' is lattice isomorphic to an AM-space, and hence neither E nor F is lattice isomorphic to an AL-space. (b) The above arguments applied to (E (D F)1 and (E (D F),o show that none of these Banach lattices is lattice isomorphic to an AL- or an AM-space.

(c) By Theorems 11.20 and 11.30 both Banach lattices E and F satisfy the Daugavet property with respect to weakly compact operators. Now, applying Theorem 11.30 once again, we see that (E(DF)1 and (EeF),,. also satisfy the Daugavet property with respect to weakly compact operators.

11.4. The Daugavet Property in C(Sl)-spaces Problem 11.4.1. Give an example of a continuous operator on C[0,1] that factors through co but that is not weakly compact. Solution: As the hint to the problem suggests, we can use here the following wellknown theorem of A. Sobczyk [75]: Every closed subspace of C[0,1] isomorphic to co is complemented. We begin by constructing a closed subspace X of C(0,1] which is isomorphic to co. To do so, for each n pick a non-negative function fn E CIO, 11 of norm one and with support in (n+1, J. Now consider the positive operator T: co-+CIO, 1] defined by T(al, a1, ...) = E' 1 an fn, where the convergence of the series is uniform, i.e.,

in the norm of C(0,1]. It is not difficult to see that T is a lattice isometry. So, if X = T(co), then X is a closed subspace of C[0,1] and T: co - X is a surjective

11. The Daugavet Equation

360

lattice isometry. Let J: X -. C[0,1] denote the natural embedding of X into CIO, 1].

By Sobczyk's theorem, X is complemented in C[0,1]. Let P: C[0,1] - C[0,1] be a continuous projection with range X. Now notice that if we consider the scheme of bounded operators

C[o,1] -e- X T_''co-z.X _L CIO, 1), then it is easy to see that P = (JT)(T'1P). This shows that P: CIO, 11 - C[0,1] factors through c0. Finally, we claim that P: C[0,1] - CIO, 1] is not a weakly compact operator. To see this, assume by way of contradiction that P is a weakly compact operator.

Since the restriction of P to X is the identity operator on X, it follows that IX (= the identity operator on X) is weakly compact. In particular, the closed unit ball of X is a weakly compact set. This implies that X (and hence co) must be a reflexive Banach space, a contradiction. So, P is an example of a non-weakly compact operator on C[0,1] that factors through co.

Problem 11.4.2. For a series F,', xn in a (real or complex) Banach space X the following statements are equivalent.

(a) The series En'=1 x is weakly unconditionally Cauchy. (b) The supremum SUP fr

1<1 SUP neN

llEn==1 Enxnll is finite.

(c) The supremum sup,,,,,=1 SUPmEN llrn

1

Enxnll is finite.

Solution: (a)

. (b) Assume that the series E°_i xn is weakly unconditionally Cauchy and fix x' E X*. This implies that s = E 1 I x' (xn) I < oo. Now for each 1 for each n we have

m E N and each choice of scalars E3, ... , cm satisfying m

m

m

00

/

I x+(L.. Enxn) I = IE Enx`(xn)I s E IEn1 - lx*(xn)l = > Ix'(xn)I <_ s < 00. n=1

n=1

n=1

n=1

Consequently, the set {En 1 Enxn : m E N and [Enl < 1 for each n} is weakly bounded. By the Uniform Boundedness Principle this set is also norm bounded, and the validity of (b) is established. (b)

(c) This is obvious.

(c)

(a) Assume that M = sup11.1=1 SUPMEN IIEn=1 Enxnll < oo and let

x` E X'. For each n put en = s, yam if x'(xn) 96 0 and fn = 1 if x'(xn) = 0. Clearly, [Enl = 1 for each n and for any m E N we have m m m

E [x'(xn)[ _ EEnx'(xn) = x'/1 EEnxn) < /

n=1

n=1

n=1

8x-11

rm

- Ilu Enxnll < n=1

Mllx'll < 00.

This implies that En '=i lx' (xn) I < M 11x' 11 < oo for each x* E X', and so the series F,°D=1 xn is weakly unconditionally Cauchy.

Problem 11.4.3. Prove that a series En'=1 xn in a Banach space X is weakly unconditionally Cauchy if and only if for each A = (1\1, 112, ...) E co

11.4. The Daugavet Property in C(1)-spaces

the series E

361

A xn is unconditionally norm convergent. Furthermore, show that the mapping T : co - X, defined by 1

00

T(A1, A2, ...) _ E Anxn , n=1

is a bounded operator that need not be weakly compact. Solution: If F_', .r,, is weakly unconditionally Cauchy, then according to Problem 11.4.2 there exists a constant Al > 0 such that for each finite subset a of N and for each collection of scalars {C,, : n E o} satisfying IEnI < 1, we have n11 <

M.

nEo

k +sup{IAn1: n > k} > 0. (1\1,,\2, Clearly, bk . 0. Now note that for any finite subset A of {k. k + 1,.. .} we have

11E nEA

Anxn II

=

Ilbk

bk

nEA

xn II < Albk

0.

Since A is an arbitrary subset of {k. k+ 1....}. it follows that the series n Anxn is unconditionally norm convergent. (See Problem 1.3.8.) For the converse, assume that the series F_' j Anxn is unconditionally norm convergent for each A = (A1, A2, ...) E co. This implies that the series of scalars

E°°_1 x'(Anxn) = E 1 Anx'(xn) converges for each x' E X'. Since A E co is arbitrary and co = PI, it follows that {x'(xn)} E 11, i.e., E'l Ix'(xn)I < oo holds for each x' E X. This shows that the series E;= 1 xn is weakly unconditionally Cauchy. (See also the footnote in the proof of Theorem 4.9.) The fact that T as defined above is a bounded operator follows immediately from statement (b) of Problem 11.4.2. To see that T might fail to be weakly compact. let X = co and {xn } be the standard basis in co. A direct verification shows that the series 1 x,, is weakly unconditionally Cauchy. To finish the solution notice that in this case T = I, the identity operator on co.

En

Problem 11.4.4. Let S be a topological space without isolated points and let G be an arbitrary open subset of S. Show that neither G nor its closure G has isolated points-where, of course, G and G are equipped with the induced topology.

Solution: Assume that S has no isolated points and let G be an open subset of S. Consider first an arbitrary point s E G. If s is an isolated point in G, then {s} is an open subset of C. This means that there exists an open subset V of S such that {s} = G fl V. Since G is an open subset of S, it follows that {s} is also an open subset of S, i.e., s is an isolated point of S, which is impossible. Thus, G has no isolated points. Now pick an arbitrary point p E C. If p is an isolated point of G, then there

exists an open subset W of S such that {p} = W fl G. This implies that W is an open neighborhood of p. As p E G, it follows that W fl C 0 0. Since W n G C W n G = {p}, we infer that {p} = W n G. That is, {p} is an open subset

11. The Daugavet Equation

362

of S, i.e., p is an isolated point of S, which is a contradiction. This shows that has no isolated points, and the solution is finished.

Problem 11.4.5. For a compact Hausdorff space 11 show that.

(a) If p is a regular Borel measure with full support (i.e., Suppµ=ll), then the natural embedding J.: C(12) --f L00(p) is a lattice isometry.

(b) If fl is also metrizable and has no isolated points, then there exists a non-atomic regular Borel probability measure on 11 with full support. Solution: (a) Assume that a regular Borel measure p has full support. It should be clear that the natural embedding Ju : C(1l) - Li(p) is a lattice homomorphism.

Now let f E C(fl). We must show that IIfII = IIfSince IIf11,o ? II!is obviously true, it suffices to establish that IIf IIu,,, ? IIf II To this end, let e > 0. Notice that the set V = j w E f : if (w) l > IIf 1 1,,. e) is a non-empty open set. Therefore, from Suppp = 1, it follows that p(V) > 0. This implies IIf II,..,, ? IIf II - e. Since c > 0 is arbitrary, the latter inequality yields

-

IIf 1Iµ.,, > If II,o, as desired.

(b) Assume that ) is a metrizable compact space without isolated points. In this case there exists a countable collection {01,02,...} of open sets that form a base of the topology on fl. Since 11 has no isolated points, it follows from Problem 11.4.4 that each On has no isolated points either. But then, according to Theorem 11.34, for each n there exists a non-atomic regular Borel probability mea-

sure pn on U. . Now consider the measure it = n 2-nµ,,. Then p is a non-atomic regular Borel probability measure on ). Moreover, we claim that it has full support. To see this, assume by way of contradiction that the open set Il\Suppp is non-empty. This implies that there exists some k such that Ok C fl \Suppµ. Consequently, 2 npn (On) > 2-kpk(Ok) = 2-k > 0,

0 = p(Ok) _ n=1

which is impossible. Therefore, p has full support.

I

Problem 11.4.6. Establish the following separability properties.

(a) If the adjoint of a bounded operator T : X -, Y between Banach spaces has a separable range, then T likewise has a a separable range. (b) If a Banach space X has a separable dual, then X itself is separable. Also, give an example of a separable Banach space whose norm dual is not separable. Solution: (a) Let T : X --+ Y be a bounded operator between Banach spaces such that the range R(T*) = {T*y*: y* E Y*} of its adjoint operator T* is separable. Pick a countable collection {yl, y2, ... } of linear functionals in Y* such that the set {T*yl, T*y2,...} lies in the unit sphere of R(T*) and is norm dense there. That

11.4. The Daugavet Property in C(n)-spaces

363

is, IIT`y,,I1 = 1 for each n and if z` E Y' satisfies IIT*z*II = 1, then for each e > 0 there exists some m satisfying IIT*z* - T*ymp < e. Next, for each n pick a unit vector xn E X such that I (T*yn)(xn)I > 21 and consider the closed linear subspace Yo generated in Y by the countable set (Txj,Txz.... }. Obviously Yo is separable, and so it suffices to verify that the range of T satisfies R(T) C Yo. If this is not the case, then there exists some x E X such that Tx Yo. But then, by the Hahn-Banach theorem, there exists some non-zero y' E Y' such that y'(Tx) = T*y*(x) 0 and y*(y) = 0 for all y E Yo. Scaling appropriately, we can assume that 1[T' y* 11 = 1. Now note that for each n we have IIT.y* - T'ynIl

? I(T'y' -T`yn)(xn)I = Iy'(Txn) - (T'y;,)(xn)I = I (T*yn)(xn)I > 2

which is impossible. This contradiction establishes that T has a separable range. (b) Assume that the norm dual X' of a Banach space X is separable. Consider the identity operator I: X X, and note that the adjoint operator I*: X' - X' coincides with the identity operator on X. Since, by our hypothesis, R(I') = X' is separable, it follows from part (a) that X = R(I), the range of 1, is also separable. For the last part notice that if we let X = C[O,1], then X is separable while C[O,1]' = ca[0,1] (the Banach lattice of all finite Borel signed measures on [0, 1]) is non-separable.

Problem 11.4.7 (Weis-Werner [82]). If n is a metrizable compact space without isolated points, then show that every weakly compact operator on C(n) satisfies the Daugavet equation. Solution: Assume that fZ is a metrizable compact space without isolated points and let T E C(C(fl)) be a weakly compact operator. Since fZ is metrizable, it follows that C(fZ) is separable. So, by Gantmacher's theorem (see Problem 2.4.4), T'" maps C(f1)** into C(fl). This shows that T" has a separable range. Therefore, by Problem 11.4.6, the adjoint operator T" (and hence also the operator T) has a separable range. Now a glance at Theorem 11.43 guarantees that T satisfies the Daugavet equation.

I

Problem 11.4.8 (Weis-Werner (82]). Modify the arguments in the proof of Theorem 11.43 using the Bartle-Dunford--Schwartz theorem [28, p. 306) to show that the assumption of metrizability made in Problem 11.4.7 is not necessary. That is, show that if n is a compact Hausdorff space without

isolated points, then every weakly compact operator on C(n) satisfies the Daugavet equation. Solution: The Bartle-Dunford-Schwartz theorem is the following result. A non-empty subset A of C(fZ)' = ca(fl) is weakly compact if and only if it is norm bounded and there exists a measure a E ca(Q) such that v(E) - 0 uniformly in v E A when µ(E) -' 0.

We shall use this theorem for the solution of our problem. Let T E G(C(fl)) be a weakly compact operator, where n is a compact Hausdorff space. It follows

11. The Daugavet Equation

364

that the representing kernel {pr,: w E s1} _ {T'&,: w E Q} of T is a relatively weakly compact set. So. by the Bartle Dunford-Schwartz theorem there exists a regular Borel measure p E ea(fl) such that each p,,, is absolutely eontmuous with respect to it. Since p is a finite measure, it follows that the set

B = {y E S1: p({..}) # 0} is at most countable. The absolute continuity of each if,,, with respect to is implies

that the set A = (w

2:

p.,({..:}) # o}

is a subset of B. Consequently. A is at most countable.

Now for each a E A let Q. = {w E Q : w -A a) = 1a)". Since St has no isolated points, it follows that each Q. is an open dense subset of Q. From the Baire Category Theorem (see. for instance [59. p. 294]), we conclude that the set

ni1a=A`={wEf1: p_({.;})=0} aE.a

is also dense in ft. This implies that condition (t) of Lemma 11.42 is satisfied. and so T satisfies the Daugavet equation. I

Problem 11.4.9 (Ansari (121). Fix: q E [0.1] and consider the rank-one positive operator T = 5q 0 1 on C[0,1]. That is. Ti = r(q)1 for each s E C[O,11. Establish the following properties of T.

(a) T factors through co (and hence T satisfies the Daugavet equation).

(b) The operator jT: C[0.11 -, C[0,1]" is not disjoint to jI, where j : C[O, 1] -- C[0.1]" is the canonical embedding and I is the identity operator on C[0,1].

(c) The adjoint operator T*: C[0.11' - C[0,1]' is not disjoint from I', the identity operator on C[0,11*. Solution: (a) Since T is of finite-rank. it easily follows that T factors through co. Consequently. T must satisfy the Daugavet equation. (b) To see that j1 and jT are not disjoint in C(C[0. II.C. [0. I]'*). note that [jI A jT] (1)

= inf {u + jTv: 0 < u, L E CIO, 1] and u + r= 1}

inf{u+v(q)1: 0
u = xt91. This is a bounded Bore] function. and so w E CIO. 1]". Obviously (w. bq) = 1. We shall show that it + v(q)1 > u for all 0 < u. r E CIO, 1] with u + v = 1. Indeed, for each 0 < u, v E C[0,1] satisfying u + v = 1. we necessarily have that [u + r(q)11(q) > 1, i.e.. u+L(q)1 > uw. Therefore. [jl A jT] (1) > tr > 0. (c) To show that 1* and T" are not disjoint, we will verify that [I'AT';(bq) > 0.

Note that T' = 1

bq and so T' (49) = b9 and that if 0 < it.v E C[0.1]' and

11.5. Slices and the Daugavet Property

365

p + v = bq, then p = abq, v = (1 - a)bq for some scalar a E 10, 1]. Now we have

[I'AT'](bq) = inf{µ+T'(v): 0 0.

0

11.5. Slices and the Daugavet Property Problem 11.5.1. Let X be a linear subspace of a Banach space Y and let J : X , Y denote the inclusion operator. Show that the following statements are equivalent.

(1) Every rank-one operator T : X tion. i.e., IIJ + TAI = 1 + 1ITh

Y satisfies the Daugavet equa-

.

(2) If yo E Y and .ro E X' are unit vectors, then for each 0 < c < 1 there exist a unit vector y E Y and some 0 < b < 1 such that: (a) Slice(Uy., y, b) C Slice(Uy., yo, E).

(b) y' E

y. J) implies jjxo + J'y'11 > 2 - e.

(2) Assume that yo E Y and xo E X' are unit vectors, and let Solution: (1) 0 < E < 1. Start by observing that the rank-one operator T = xo ® yo E C(X, Y) satisfies IITII = 1. So, by our hypothesis, IIJ + TII = 1 + IITII = 2. Fix some unit vector x E X satisfying Ilx + TxjI > 2 - E > 0 and xo(x) > 0, and then let II=X+Tz 2-, and b - 1 - FIX-+T:11

y-

+T=11

Clearly, Ilyll = 1 and 0 < b < 1. Next, fix y' E Slice.(Uy.. y, b). and note that

y*(x)+xo(x)y*(yo) = (.r+Tx,y*) = IIx+Txll(y,y') > From this, y'(x) < I and 0 < xo(x) < 1, we get

IIx+TxII(1-b)=2-E.

x0(r)y*(yo) > 2 - E - y'(x) > 1 - f > 0. This implies 0 < x; (x) < 1 and 0 < y* (yo) < 1, and therefore.

y'(Yo) > xo(x)y,(yo) > 1 - E > 0. That is. y' E Slice(Uy.. yo, c). and so Slice(Uy., y, b) g Slice(Uy-, yo, E). Finally, taking into account that y' E Slice(Uy-, y. 6) implies 0 < y'(yo) < 1, we get IIxo + J'y* ll

>

2

(x. xo + J'y') = J'y,(x) + xo(x) y*(x) + y'(yo)xo(x) > 2 - E.

11. The Daugavet Equation

366

(1) Let T = xa ® yo E L(X, Y) be a rank-one operator. We must (2) verify that IIJ + TII > 1 + IITII. Without loss in generality we can assume that IIxoII=IIyo1l=1 (and so 11Th Fix any 0 < E < 1. According to our assumption, there exist a unit vector y E Y and some 0 < b < 1 such that y' E y, 6) implies y' E Slice(Uy., yo, e)

and Ilxo + J'y' II > 2 - E. Clearly, Slice(Uy-, y, b) 96 0, and thus if we fix any y, 6), then we have y' (yo) > 1 - E and 0 < 1 - y* (yo) < E. So, y' E

IIJ+TII = IIJ'+T'II>II(J'+yo®xo)v II=IIJ'y+y(yo)xoll =

Ily'(yo)[xo

-Joy]

y*(yo)IIxo+J*y'll >

- [y(yo) -1]J'y'II

- [1- y'(yo)]IIJ'y'lI

(1 - E)(2 - E) - E.

Letting c -. 0+, we get IIJ + TII > 2, and the solution is finished.

Problem 11.5.2 Q43]). Let X be a linear subspace of a Banach space Y and let J: X - Y denote the inclusion operator. Show that the following statements are equivalent.

(a) Every rank-one operator T : X - Y satisfies the Daugavet equation, i.e., IIJ + TII = 1 + IITII. (b) For any pair of unit vectors yo E Y and xo E X* and each 0 < E < 1 there is x E X satisfying IIxII = 1, xo(x) > 1- e, and IIx + yoll -> 2 - e. (c) For each pair of unit vectors yo E Y and xo E X' and each 0 < E < 1 there is some unit linear functional y` E Y* satisfying y*(yo) > 1-E and llxo + J*y' ll > 2 - e.

Solution: (a) e=o (b) Assume that (a) is true. Let yo E Y and xa E X' be unit vectors, and let 0 < E < 1. According to statement (2) of Lemma 11.46 there exist a linear functional x' E X' of norm one and some 0 < b < 1 such that: (1) Slice(Uy, x', 6) C Slice(Ux, xo, c). (2) For each x E Slice(Uy, x', b) we have IIx + yo II >- 2 - E.

Clearly, any x E Slice(Ux, x', 6) satisfies the desired properties.

Now suppose that (b) is true. Let T = xo ® yo E C(X, Y) be a rank-one operator. We must verify that IIJ + TII > 1 + IITII. Without loss of generality we can assume that IIxoII = Ilyoll = 1 (and so IITII = 1). Fix 0 < e < 1. By our assumption there exists a unit vector x E X satisfying xo(x) > 1 - E and IIx + yo 11 > 2 - E. The latter implies 0 < 1 - xo(x) < e, and so

IIJ+TII > IIx+Txll = IIx+xo(x)yoll = Il [xo(x)x + xo(x)yoJ + [1 - xo(x)]xll >-

II xo(x)(x + yo) ll - 11 11 - xa(x)Ixll

> (1-E)(2-E)-E for each 0 < E < 1. Letting e 10 yields IIJ + TII > 2. Therefore, IIJ + TII = 2, and so T satisfies the Daugavet equation.

11.5. Slices and the Daugavet Property

367

(a) (c) Suppose that (a) is true. Then statement (1) of Lemma 11.46 is true and the validity of (c) follows immediately from statement (3) of this lemma. For the converse assume that (c) is true. Let T = xo ®yo E £(X, Y) be a rankone operator. We must verify that IIJ + TII > 1 + IITII. Without loss in generality we can assume that IIxoII = Ilyoll = 1 (and so IITII = IIx5II IIyoII = 1). Now fix any 0 < E < 1. According to our assumption, there exists a unit linear

functional y' E Y* satisfying Ilxa + J'y' II > 2 - E and y'(yo) > I - c. The latter implies 0< 1 - y'(yo) < e, and so

IIJ+TII = IIJ'+T'll >_ II(J'+yo ®xo)y1I = IIJ'y'+y'(yo)xall

= Ily'(yo)[xo-J*y*]-[y(Yo)-1]J'yII > y*(yo)llxo+J'y'11-[1-y'(yo)]IIJ*y*II

(1-E)(2-E)-E. Letting e -. 0+, we get IIJ + TII > 2, and the solution is finished.

1

Problem 11.5.3. If X is a reflexive Banach space, then X does not satisfy the Daugavet property for rank-one operators. Solution: If X has the Daugavet property for rank-one operators, then Theorem 11.50 implies that X satisfies the Daugavet property for weakly compact operators. Since X is reflexive, the operator T = -I is weakly compact and hence T satisfies the Daugavet equation. However, this is impossible since III + TII = 0. 1

Problem 11.5.4. Let C = 12', the positive cone of e2. Show that 0 is an exposed point of C that fails to be strongly exposed.

Solution: Consider the Banach space X = e2 and let C = t2+. Clearly, C is a norm closed convex subset of e2. We claim that the point 0 E C is an exposed point of C that fails to be strongly exposed. To prove this, let u = (1, 2, 3, 4, ...) E e2 . If we consider the linear functional z` = -u E e2 = t2, then it is not difficult to

see that z'(0) = 0 > z*(x) for each x = (x1,x2,...) > 0. That is, the continuous linear functional z" exposes 0, and so 0 is an exposed point. Now for each n let yn= 0,0,...,0,1, 2, 13,4,...), 1

i

1

where the zeros occupy the first n coordinates, and note that IIyn1I2 = lIu1I2 = 76 for each n. Moreover, for any v = (v1, v2, ...) E t2 = e2 we have (yn, V) I

-

I

i=n+l

7

i-n

0

I

E v2]

i=n+l n

(i-n

E vi, 2

1

i=n+l

0.

i=n+l

That is, (yn, v) --+ (0, v) for each v E t2 = e2. Since Ilyn II = that 0 is not a strongly exposed point of C.

765

74 0, this implies 1

Problem 11.5.5. Use Theorem 11.50 to present an alternative proof of the second part of Theorem 11.20: If E is an atomless AL-space, then E satisfies the Daugavet property for weakly compact operators.

11. The Daugavet Equation

368

Solution: Let E be an atomless AL-space. Without loss of generality we can assume that E = L1(it, E, p), where (St, E, p) is a Lion-atomic measure space. To establish that E satisfies the Daugavet property for weakly compact operators, we must show (according to Theorem 11.50) that E satisfies the Daugavet property for rank-one operators.

To this end, let 0 E Li(p) and u E L1(µ). Here it should be pointed out that we do not make any assumptions regarding the measure space (St, E, p), and It is exactly this therefore we cannot claim that Li(p) coincides with circumstance that makes the proof more complicated than that of Problem 11.2.5. To establish that O®u satisfies the Daugavet equation, it suffices to show (according to Lemma 11.21) that 0®u is disjoint from the identity operator, i.e., IA IO®ul = 0 holds in G,, (LI (p), Li (p)). Since 10 ®ul = 1010 Jul, we can assume without loss of generality that 0 > 0 and u > 0. Next, let 0 < x E L1(µ), and put

0<w=[IA(p(&u)](x)=inf{z+0(x-z)u: 0
(*)

Now fix e > 0. Since 0 is a continuous linear functional, there exists some b > 0 such that IIuiii < 6 implies IO(u)I < e. The absolute continuity of the norm of the vector x guarantees the existence of some ?i > 0 such that if A E E satisfies p(A) < ti, then IIXXA III < b (and so ()(XXA) < e also holds).

Since the function x is integrable, its support is a-finite. That is, we can write

Suppx = {w E it: x(w) > 0} = U,O=1 Stn, where it,, E E and p(S2,) < oo for each n. Next, fix some n and consider the measurable set Stn. Pick some k such that < q, and then select k pairwise disjoint measurable subsets S1, ... , Sk 11^

of it such that p(Si) =

p(k

for each 1 < i < k and Uk 1 Si = Stn. (This

selection is possible since the measure p is non-atomic.) If for each 1 < i < k we

let fi = xXs, E L1(p) and gi = x - fi = Ej#i ff, then we have 0 < gi < x, and so from (*), we see that w < gi + O(fi)u < fi + eu for each 0 < i < k. This implies w(w) < cu(w) for p-almost all w E Si and for each 0 < i < k. Therefore, w(w) < eu(w) for p-almost all w E Stn. In turn, this shows that w(w) < eu(w) for p-almost all w E Supp x. Since clearly w vanishes outside of the support of

x, it follows that 0 < w < eu for each e > 0. This shows that w = 0, and so [I A (0 (9 u)](x) = 0 for each x E L' (p). Therefore, I A (0 ® u) = 0 holds in G(L1(p),L1(14)), as desired. (See also Problem 5.1.8.)

Problem 11.5.6 ([43, 47]). Let

1

be a non-atomic measure space

and let Z be a Banach space. Show that the Banach space X = L1(µ, Z) of Z-valued Bochner integrable functions satisfies the Daugavet property for weakly compact operators. Solution: It is enough to show that condition (b) of Problem 11.5.2 is fulfilled. To this end, let y E X and X* E X' be two unit vectors and fix some 0 < e < 1. It is well known that the linear functional x' can be identified with a weak* measurable function taking values in Z*; see [24].

Choose some B E E such that IIXayiIx 5 z and IIXex'Ilx > 1 - 2. Next, choose any unit vector x E X with support in B and such that x"(x) > 1 - e. This easily implies IIx + yII x > 2 - e, and hence condition (b) of Problem 11.5.2 is fulfilled, as desired.

11.6. Narrow Operators

360

Problem 11.5.7 (Kadets [40]: Khalil [47]). Let 0 be a compact Hausdorff space without isolated points and let Z be an arbitrary Banach space. Show that the Banach space X = C(i2. Z) of Z-valued continuous functions satisfies the Daugavet property for weakly compact operators. Solution: It suffices to establish that condition (b) of Problem 11.5.2 is fulfilled.

So. fix 0 < c < 1 and consider two unit vectors y E X and x' E X. Also pick a unit vector it E X such that x'(u) > 1 - c. It is well known that the linear functional x' can be identified with a Z'-valued regular Borel measure of bounded variation: see [24]. In particular, x' has at most countably many atoms. Therefore (since St has no isolated points and IIyII = 1), we can find a point ...b E !1 and an open neighborhood V of w:o such that:

(a) IIy(wo)Ilz > 1 - f. (b) Ix'I(V) < f, that is, the total variation of r' on V is less than f. Now the Urysohn-Tietze extension theorem guarantees the existence of a func-

tion v E X such that v(w) = u(w) for each w E 11 \ V, v(wo) = y(wo) and IIv(w)IIz <- IIy(w)IIz for each w E V. In particular. 1 - c < IIv'IIx < 1. Let x = l = 11x u'. C!earlv. IIxiiX = 1 and Ilx + yIIX

?

IIx(wo) + y(wo)Ilz = II

>

Ti i

(wo) + y(wo)11z

t a Y(WO) + y(wo)IIz = (I =t r + 1) IIy(wo)IIz i

2(1-f).

A straightforward calculation also shows that

x'(x) = r'(xln,ti,) + x'(xk,.) > 1 - 4f. Thus. the unit vector x satisfies condition (b) of Problem 11.5.2, and so the Banach space X has the Daugavet property for weakly compact operators.

11.6. Narrow Operators Problem 11.6.1. Give an example of a weakly compact operator that is not narrow. Solution: Let E = Lr,[O, 1], where I < p < -1c. Since E is reflexive, the identity operator I on E is weakly compact. However. I is not a narrow operator. Indeed, if we let A = [0.1]. then for any measurable subset B of A with A(B) > (where A 2 Ixi = ka denotes the Lebesgue measure on [0.1]) and any function z E Lp[0.1] with we have IIIxllp=IIxllp>2_1P . This shows that the identity I is not a narrow operator.

Problem 11.6.2. Show that each weakly compact operator on L1(p) is a narrow operator.

11. The Daugavet Equation

370

Solution: It is well known that each weakly compact operator T: LI (A) - Y is a Dunford -Pettis operator; see for instance [6, pp. 335. 3361. Therefore, by Theorem 11.57, the operator T is narrow.

Problem 11.6.3. If in a scheme of bounded operators E S Y -T Z the operator S is narrow, then show that the operator TS: E

Z is narrow.

Solution: If A E E and e > 0 are given, then there exists some x E E such that IxI = XA and IISxdd <,E. But then we have II(TS)xfl = IIT(Sx)II < efITII, and this shows that TS is a narrow operator. I

Problem 11.6.4. Give an example of a narrow operator that is not regular. Solution: It is well known that there exists a compact non-regular operator T: L2[0,1] -+ L2[0.1]. (See for instance [6, Example 16.6].) By Theorem 11.57 the operator T is automatically narrow. I

Problem 11.6.5. Let T : E - Y be a narrow operator. Show that for each A E E there exists a sequence of Rademacher functions and such that 0.

supported by A

Solution: The proof is by induction. Fix A E E. By Theorem 11.53 there exists a measurable subset B1 of A such that p(B1) = 2µ(A) and IITr1II < 2-1. where rl = xa, - XA\B, For the induction step assume that the Rademacher function r satisfying 2^

k=1

has been constructed, where Bi" n Bj' _ 0 for i # j. p(Bk) = 2-"p(A), and II Tr" II < 2-". Applying Theorem 11.53 again, we can write each Bk as a disjoint

union Bk = CAI:' U DA: with p(C') = p(Dk) = 2 p(Bk) such that the function xk ='(C,; - XDk satisfies 11T.411 < 2'(2"+1) Now let r"+1 = Ek=1 xk It is easy to verify that the above inductive argument produces a sequence {r" } i1 of Rademacher functions with the desired properties.

Problem 11.6.6 (Plichko--Popov [65]). Show that the sum of a narrow and a compact operator is narrow. Solution: Let T : E --+ Y be a narrow operator and let K : E -. Y be a compact operator. We want to prove that T + K is a narrow operator. Fix A E E. By Problem 11.6.5 there exists a sequence of Rademacher functions satisfying Ir"I = to for each n and IITr"II - 0. By Lemma 11.56, we know that r -- 0. Since the operator K is compact, it follows that IIKr II -. 0 and hence II (T + K) r,, II --r 0. This, coupled with the condition Jr,, I = Xa for each n. implies that T + K satisfies the definition of a narrow operator. (Note that, in fact. this solution shows that the sum of a narrow operator and a Dunford-Pettis operator is narrow.) I

11.6. Narrow Operators

371

Problem 11.6.7 (Plichko-Popov [65]). Let E = Lp[O,1], where 1
From the definition of the Haar functions we know that the support of each Haar function is an interval of the form [-L,, 1] for some k, n E N. Next, let N = N1 U N2 be a splitting of N into disjoint infinite subsets. For j = 1, 2 we let E3

be the closed linear span of the Haar functions hi such that 2m-1 < i < 2' holds for some in E N,. Since the sequence {hl, h2, ...} is an unconditional basis, the closed subspaces E1 and E2 are complemented and E = El ® E2. We will show now that the natural projection P1 of E onto El along the subspace E2 is a narrow operator. To this end, fix A E E and e > 0, and then pick some 6 > 0 such that II XB II < E whenever B E E satisfies p(B) < 5; see Lemma 11.51. Let us also choose a finite collection {A1, A2,..., A,,} of pairwise disjoint intervals that are support sets of the Haar functions such that A(AAA) < 5, where A = Uk=1 Ak. We will also assume that 5 is small enough so that µ(A n A) > 2p.(A). Next, choose m1 E N2 large enough so that for each hi with 2tte1-1 < i < 2-1 its support Supp hi has measure at most µ(A1) and some of the supports of these functions cover A1. That is,

Al = U Supp hi , iEA.,,,

for an appropriate subset 0",, of [2"'' -1, 2m') n N. At the next step, let us choose m2 E N2 and a subset such that m2 > m1 and

of [2-2-1,2m2) nN

A2 = U Supp hi . iEAm2

We repeat the same procedure inductively for the sets A3, ... , A. and obtain the

sets of integers Omk for k = 3,. .., n. Let i be the sum of the Haar functions corresponding to the indices from the above sets O,,,k, i.e., By construction the function x E E2 and ixl = XA.

Ek-1 EiEAmk h;.

Finally, let B = An A and x = 1XB. As mentioned above, µ(B) > Zµ(A) and clearly IxI = XB. Also note that Ix - fl = XAAA, and hence IIx - III < since µ(AiA) < 5. Clearly, x E E2, and therefore P1x = 0. This implies that IIP1xII = IIP1(x - x)II < EIIPIII, and consequently the projection P1 is a narrow operator. By symmetry, the natural projection P2 of E onto E2 along the subspace E1 is also a narrow operator, and thus the identity operator I is the sum of two narrow operators, I = P1 + P2. (It should be noted that the above proof shows that our

11. The Daugavet Equation

372

previous conclusion is also true for each rearrangement invariant Banach function space on 10.11 with an unconditional basis.) I

Problem 11.6.8 (Plichko -Popov [65]). Show that each operator on Lp(0,11, where 1 < p < oc . is the sum of two narrow operators. Solution: Let E = Lp10.11. According to Problem 11.6.7 there exist two narrow operators P1 and P2 such that I = P1 + P2. Therefore, for each operator T on E we have T = TI = TP1 + TP2. Now a glance at Problem 11.6.3 shows that TP1 and TP2 are both narrow operators, and the solution is finished.

I

Problem 11.6.9. Let (1, E, µ) be a non-atomic probability space. Show that every generalized sequence of Rademacher functions {r,(A)} satisfies the conclusion of the Riemann-Lebesgue lemma, i.e.. fo xr dµ - 0 for each x E L1(µ). Solution: We begin by noting that the desired conclusion does not follow directly from Lemma 11.56 since the norm on Lr,(p) is not order continuous. However, a simple approximation does the job. Given X E L1(µ), fix any e > 0 and pick any function y E L x (µ) such that llx - yll, < c. By Lemma 11.56 there is no E N such that Ifs, yr dpl < e for all n > no. Also we have I fn(x - y)rn dpl < Ilx - yll, < e. This yields that I ff, xr,, dµ4 < 2( for each n > no.

Problem 11.6.10. Let (Q. E. µ) be a non-atomic probability space, and assume that E is a Banach function space with order continuous norm satisfying Lx(p) C E C L1(µ). Also, let {rn} be a sequence of Rademacher functions supported by a measurable subset A of S2. Establish the following.

0 in E.

(a) For each x E E we have xr

(b) For each x E L1(µ) we have IIx+xr,,111 - llxIll. Solution: (a) Fix x E E and let f E E*. Since E has an order continuous norm, we know that E' = E. that f can be identified with a measurable function. and that fx E L1(p). We must show that f (xrn) = fn f xrn dp -+ 0. Since fx E L1(µ), this follows immediately from Problem 11.6.9 in view of the obvious identity

Jf.zrndP = J fx t

f,

0 for each (b) Letting E = L1(p). it follows from part (a) that f0 Ixlrn dp x E L1(µ). Hence f(1(IxI + Ixlrn) dp - fn Ixl do = llxll, It remains to note that

Ilx+xrn14i =fs,I(l+rn)xldp= fsl(Ixl+Ixlr,,)dµsince 1+rn>0. Thus indeed I Ill+xrnll, - Ilxlli

11.7. Some Applications of the Daugavet Equation Problem 11.7.1. Let X be a Banach space and assume that there exists an increasing continuous function W : 1R R and a scalar y > 0 such that

11.7. Some Applications of the Daugavet Equation

373

for every finite-rank projection P we have L'(III - PII) > ti +'.t(IIPII) (*) Then show that X does not have an unconditional basis. Solution: Assume by way of contradiction that the Banach space X has an unconditional basis {en }.

Let A be the collection of all finite subsets of N. For each A E A we denote by P4 the natural projection on the closed subspace generated by {ek: k E Al and by QA the complementary projection. i.e.. QA = I - P.t. Since the basis is unconditional, Lenima 1.53 guarantees that 01 =sup{11P411:

-3c

Consequently

3=sup{I1QAII: AEAl < x. By the hypothesis 4t(11Q,4II) > '1 + for each A E A. Since t is increasing. we get tt(3) > + z'(IIP.41) for each A E A. Taking the supremum over A E A. we obtain t.'(a). (**)

On the other hand. for each A E A and for each vector x E X we obviously have

Q.4'II - 0. where An = {1.....n} n A`, and this implies that

IIQA11 <_ sUPAEA IIP411 = a. Taking the suprerum over A on the left side of the last inequality. we get .3:5 a. whence t; (;3) < v(a) in contradiction to (**). If X satisfies the Daugavet property for rank-one operators. then the function Wi(t) = t and the constant 1 satisfy (*). Therefore, Theorem 11.60 is a special case of this problem.

Problem 11.7.2. Show that every Banach space can be renormed so that in the new norm it fails the Daugavet property for rank-one operators. Solution: Let X be a Banach space. It suffices to establish that X can be renorined in such a way that its closed unit ball has strongly exposed points; see Theorem 11.63.

To see that this is possible, take any ro that dots not belong to the original closed unit ball U and then renorm X by taking for the new closed unit ball V the closure of the convex circled hull of U U {xo}. Note that

V={au+.3ro: n>0 and a+1.31<1}. Clearly. U C V C (1 - IIxuI')U. This implies that the Minkowski functional of V defines an equivalent norm with closed unit ball V. To finish the proof. we shall establish that x0 is a strongly exposed point of 1'. Since xo 0 U. it follows that there exists some O E X` with Iloli = I such that o(xo) > 1. Now assume that a sequence {anon + .3nr0} C V satisfies

o(anu +.3nro) --i ©(ru).

(t) We insist show that roll 0. For this, it suffices to establish that an - 0 and ;3n - 1. In view of an > 0. the inequality on + 1.3.1 < 1 and (t). it is enough to prove that 13nI

1.

11. The Daugavet Equation

374

If jF3n 1

that 1;3,,l

f+ 1. then by passing to an appropriate subsequence, we can assume ,3 < 1. Now note that for each n we have

JO(Qntin + 3nxo)I :5 CznIO(tLn)J + f3nlm(xo)

1 - j j3.1 + Ii3n1O(x0)

Taking limits and using (t) once more, we see that ,p(xo) < 1 - r3 +:?m(zo) < (1 - 3)0(xo) + f3¢(xo) = a(xo) , which is impossible. This contradiction completes the solution.

U

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Index AL-space, 81, 90 AM-compact operator, 313 AM-space, 87 90 represented as Lc(x ), 152 with unit, 8Z abstract integral operator, 154 additive semigroup of operators, 319 adjoint of resolvent, 130 adjoint of strictly singular operator, 142 adjoint operator, 66 algebra of operators, 304 non-transitive, 304, 305. 331 transitive, 306

unital, 304 algebraic complement of subspace, 136 algebraic homomorphism, 123 analytic function, 50.205 annihilator, 19 of a set, 252 of a vector subspace, 1.. 25Z antisymmetry property. 20 approximate eigemalue, 198.222 approximate point spectrum, 198. 232 approximation of vectors in E. 109 approximation property. 122 Archimedean Riesz space, 22. 25 72 108

Banach function space, 154 Banach lattice, 332 atomic, 282 complex, 96 with mixed norm, 163

with order continuous norm. ZL Z6. 91. 126. 329 Banach space direct sum. Z 145 quotient. 18 reflexive, 11.44.341. 367 renormed, 313. separable, 362

uniformly convex, 1L 338.341, 342.344 uniformly smooth, 338.342-4d band, 24. 12. 15. 76. IM 159.265.291 generated by a set. 24 generated by a vector, 24 generated by an ideal, 24 principal, 110 projection, 25 321 band irreducibility, 290 band irreducible operator, 273. 276, 281 quasinilpotent. 292 band projection, 108

ascent of operator, 66. 131. 210

Bart le-Dunford-Schwartz theorem, 363 basic sequence, 39

atom. I2. 14.352

basis, 31, 329

212

in C(fl)-space, 352 atomic Banach lattice, 282 atomic measure space, 154 atomless Riesz space, 23 automatic continuity, 132 backward shift. 69. 286.311

in CEO. 11. 37

in Lp[0,1. 34

in a B-space, 31.33 orthogonal, 244 orthonornal, 243 unconditional, 36, 40. 329 biorthogonal sequence, 40 379

Index

380

Bochner integrable function. 368 Borel measure, 9_4 126. 231. 362 bounded below operator. fi3

composition operator. 83- 128. 303 conditional expectation operator. 169 124 cone. 15. 22_0 2L 329

Calkin algebra, 23.1

Carleman operator. 141 carrier. 31 of functional. 156 of ideal, 31 Cayley-Hamilton theorem. 254. 258 center. 103 of Lp(p). 107. 152 of a Banach function space. 152 of Banach lattice. L051 ID6 central operator, IM 27A Cesaro operator, 1.95 chain of open sets. 51 characteristic polynomial, 245,

252. 255.

256,, 268

Clarkson's inequality. 345 clopen set. 26 closed ideal, 112 closest point to a set. 1 l cofinal set. 51 collection of operators, 317 finitely quasinilpotcnt, 318 locally quasinilpotent, 317 commutant of an operator, 304 commutator of two operators. 209 commuting Krein operators. 295 commuting operators. 191 296 compact operator. Z& 82 compact-friendly operator, 321 compactly dominated operator. 289. 292 complement algebraic of subspace. 136 of closed subspace. 12L 359 complemented subspace. 12L.3,59 complete measure. 52 completely continuous function. 332 completeness of normed Riesz spaces. 28 completion Dedekind, 32 93 Maeda- Ogasawara- Vulikh. 148 norm. 112 complex Banach lattice. 96 243 complex ideal. 99 complex lattice homomorphism. 134 complexification. 3. 96.92

of R'. 97 of Ca(f2), 97

of a B-lattice. 94 of a B-space. 22Z of a real normed space, 4 of a real vector space. 3. 5 component, 26. ZZ

composition of analytic functions, 206

generated by a Schauder basis, 329 generating. 2L 1.83 in a B-space. 329 in a normed Riesz space. 22 conjugate, 99 of complex number, 98 of matrix. 98 continuity of positive operators. 2Z, 35.1 contour, 205 Jordan. 20,5

contraction. 64.218 contractive projection, 17SL 125

convergence

,55

in measure. 52 notions in Lo. 158

order, 22 53 ultrafilter, 35 convex hull of components. ZZ convolution of kernels. 162 countable spectrum. 197 countable sup property. 54 curve in a topological space, 50 cyclic vector, 260

Daugavet equation. 335.363.365 Daugavet property, 352.353, 35 .359. 367 Dedekind complete Riesz space. 26. 3.. 54. 10 108. 155. 233 Dedokind completion, 3_0. 93

derivatives of the resolvent function. 189 descent of operator. 68. 13Z. 210 finite, 138 diagonal matrix. 103 diagonal projection. 109 diagoualizable matrix. 246 diagonalizing matrix. 246 direct sum Banach space. L 195 direct sum of operators. 306

direct sum operator. 195 Dirichlet's theorem, 186 discrete vector. 72

disjoint linear functionals. 24 disjoint sequence, 212 disjoint vectors. Z4. 123 disjointly strictly singular operator. 134 disjointness off from integral operators. 151 domination of operators, 100. 290, 310 double adjoint. 303 double power bounded operator. 1.95 dual of (p-sum of B-spaces, 224 duality of Abf- and AL-spaces. 88 Dunford's theorem. 160 Dunford- Pettis operator. 316

381

Index

dyadic point. 38

homomorphism algebraic, 129

E,;, order continuous dual of E, 31

lattice, 19. 105, 117 129.222.264 hyperinvariant closed subspace, 283

eigenspace, 283 eigenvalue. 241. 206. 240. 262 approximate, 19., 222 eigenvector of matrix, 262. 267 equivalent norms, 91 essential singularity. 210 essential spectrum, 231 of forward shift, 238 essentially nilpotent operator, 239 expanding operator. 273 exposed point of a set. 367 extension of additive function, 21 extension of linear functional, 1!1

factorization of compact operators. 82 filter of sets. 44 Finite Dimensional Separation Theorem, 1fi finite-rank operator, 79, 139,. 123. 127 215.

35

finitely quasinilpotent collection, 318 forward shift, 69.238, 311 Frechet differentiability of the norm, 347 Fredholm alternative. 138 Fredholm operator. 134 Ftobenius' theorem, 262 function linear, 184 analytic. 54. 205 F.ochner integrable, 368 completely continuous, 332 harmonic, 1.86

Z(X, Y), set of isomorphisms from X to Y. 67 ideal. 99, 265 complex, 22 invariant, 310 null, 116.311 null of an t-seminorm. L15 null of functional, 31 order dense, 58

order dense in Lo(s), 58 principal. 90 range of operator. 311 ideal irreducibility, 283 ideal irreducible operator, 1.04. 273. 276

identity operator, 160.356 implication scheme. 223 independence of eigenvectors, 197 of linear functionals. 114 index of operator. 134. L37 Index Theorem, 131 inequality Clarkson's, 346 Jensen's. 123 Khintchine's. 143 triangle, 96 inner product preserving matrix, 244 integral operator. 94. 145. 21& 294 internal point of a set, 82

matrix-valued, 267 piecewise linear, 189

interval preserving operator. 46. 132 231,

Rademacher. 91., 3211

invariant ideal, 310 invariant measure, 276 invariant subspace, 5. 213, 299 inverse of operator, 1 T 193, 232.235 invertible elements in a B-algebra, 23Z invertible matrix. 263

retraction, 13 functional multiplicative. 221

supporting set at a point. 347 functional calculus, 205

264

invertible operator. Z 133.232 235.251 Gantmacher's theorem, SO Gateaux differentiability of the norm. 347 generalized Harris operator, 323 generating cone. 21. 183 Gershgorin's theorem, 263 Gribanov's theorem. 199 Haar system, 34 harmonic function, 186 Harris operator. 293, 323 hereditarily indecomposable B-space, 143

irreducibility band. 290 ideal, 283 irreducible matrix, 255 266 irreducible operator. 213

isolated point. 352361 isolated point of a(T), 197, 210, 239 isometry. 8.353 lattice, 225

Hermitian matrix, 243.246.249

linear. 8. 142. 225 isomorphism lattice. 1.12

Hilbert space, 142 Hilbert-Schmidt operator. 153

James' theorem. 341

Index

382

Jensen's inequality, 1T3 joint continuity of composition, lil joint spectral radius, 320 Jordan contour, 205

KB-space. 15 Kadeta-Klee property. 334 Khintchine's inequality, 143 Krein operator. 266. 267. 271, 223.293 compact. 294

with positive eigenvalues. 295 Krein-Rutman theorem, 21.9 Kronecker's theorem, 231 L-space, 114

LO(p). space of measurable functions, 54

L,a, B-lattice of mixed norms, 163 Lp,.. B-lattice of mixed norms, 163 lp-sum of B-spaces. 223 C(X.Y), the B-space of bounded operators,

210

C. (E, F), the B-lattice of regular operators, 28 Lat (T), algebra of T-invariant closed subspaces, 301 Laplace's equation, 1.86 lattice homomorphism, 46, 74. 105 117, 129. 132.. 227, 229

complex. 130 order continuous, 133

spectrum, 230 lattice isometry. 225 lattice isomorphism, 117 lattice operations in G4(E. F). 26 weakly o-continuous. 91 lattice seminorm, 115 lattice-subspace, 1$1 finite dimensional. 183 law

Parallelogram. 115 limit functional, 134 linear function. 184 linear functional limit. 114 multiplicative, 131. 142 linear isometry, 8 114, 14 225 linearly independent functionals, 119 locally quasinilpotent collection. 317 locally quasinilpotent operator. 285 Lomonosov operator. 308 Lomonosov's Invariant Subspace Theorem, 307 M-space, 1.13

Maeda-Ogasawara-Vulikh Representation Theorem, 108. 114

majorizing vector subspace, 183 Markov matrix, 9 ., 263, 269 Markov operator, 95. 130, 276, 308 Markov projection, 172 matrix diagonal, 109 diagonalizable, 246 diagonalizing. 246

Hermitian, 233 20, 249 inner product preserving, 244 invertible, 263 irreducible, 265, 266 Markov. 95, 263. 269 nilpotent, 249. 265 non-negative. 262 norm preserving, 244 positive, 262 positive semidefinite. 248 primitive. 271 quasinilpotent, 249 stochastic, 95 strongly positive, 262 unitary. 244. 246 upper triangularizable, 251 upper triangularizing, 251 matrix representing an operator, 6 255 matrix-valued function. 267 maximum modulus principle, J U measure, 107. 126 atomic, 1114

Borel, 128. 362 complete. 52 invariant, 276

of bounded variation, 126 on a semiring, 59 separable, 148 measure convergence. 52 measure of non-compactness. 240 of operator, 241 of set, 244

minimal extension, 155 minimal polynomial. 253 mixed norm, 163 Miyajima projection, 183 continuous, 185 modulus, 121 262 of finite-rank operator, 123 of integral operator. 94 of matrix. 262 of regular operator, 24 Morera's theorem, 48 multiplication operator, 125. 127, 303, 307 weakly compact, 224 multiplicative linear functional, 131. 142.221 multiplicative semigroup, 319 multiplicity of eigenvalues, 252

Index

Nakano's theorem, 156 narrow operator, 369 nearest point mapping, 11 nearest point to a set, 11 Neumann series, 191 nilpotent matrix, 249.265 nilpotent operator, 70, 195 non-negative matrix. 262 non-transitive algebra, 304, 305. 331 norm, 1 Ftdchet differentiable, 347 Gateaux differentiable, 347

mixed, lM of integral operator, 94 of operator. 1 quotient, 19. 112 uniformly convex, U uniqueness in a B-Lattice, 28 norm completion, 112, 114 of Airspace, 114 of AM-space. 113 norm extension, 233 norm of integral operator, 94 norm preserving matrix, 244 norm totally bounded set. 81 norming subspace, 49

norms on C, 98 nuclear operator, 124

null ideal, k ¢9, 311 of an t-seminorm. 115 of functional, 116 one-to-one operator, L 251 operator AM-compact, 313 abstract integral, 154 adjoint, 66 band irreducible, 273, 276, 281 bounded below, 63 Carleman, 142 central, 106. 27A Cesaro, 195 compact, 26.82 compact-friendly. 320, 321 compactly dominated, 289, 292 composition, 83,3 128. 303

conditional expectation, 169.120 contraction, 04 direct sum, 195 disjointly strictly singular, 144 double power bounded, 195 Dunford-Pettis, 316 essentially nilpotent, 239 expanding, 273 finite-rank, 79. 119, 123. 12L 215. 355 F>redholm, 134

generalized Harris, 323

383

Harris, 293. 323 Hilbert-Schmidt, 153 ideal irreducible, 100 273, 276 identity, 151, 160, 356 integral. 994145. 215, 294 interval preserving, 46, 132. 231, 264

invertible, 7 193 232 235 251 irreducible, 223. isometry, 8

Krein, 266, 267, 271, 273. 293 locally quasinilpotent, 285 locally quasinilpotent but not quasinilpotent, 285 Lomonosov. 308 Markov, 95. 1130. 276, 3.08

multiplication, 125. 127, 3303 307 narrow. 369, 332

nilpotent, 70 195 non-strictly singular, 143 nuclear, 124 one-to-one, 7 251 order continuous, 25, 202 polynomially compact, 307 power compact, 219. 239 principal ideal preserving, 106 projection. L76 quasinilpotent, 192, 195 rank-one, 127. 354.365 regular, 93 shift, 219. 306

strictly positive. 127 311 strictly singular, 139.220. 240 strong Krein, 233 strongly expanding, 274. 275. 281 surjective, L 226, 251 symmetric. 328, 339 Volterra. 216. 219. 3.1 weakly compact, 78. 800. 84. 220. 367

weakly expanding. 274 with closed range, 65-67. 139 with countable spectrum, 197 with separable range, 81. 362 operator domination, 100. 310 operator norm. 1 orbit of vector, 260 order closed set, 23 order complete lattice of invariant subspaces, 301 order complete Riesz space, 22 order completeness of Lo(p), 54 order continuity of the norm in an AL-space, 90 order continuous dual, 31

of L,,, 152 order continuous operator, 22 , 202 order convergence, 22. 53 order dense ideal, 58

Index

384

order dense Riesz subspace, 54 order interval. 92 weakly compact. 22 order spectrum. 232 order unit, 8-7 125 ordered vector space Archimedean. 21 orthogonal basis. 244 orthogonal complement. 247 orthomorphism. 105 orthonornal basis, 244 Parallelogram Law. 345 partition of set, 131 Perron-Flobenius theorem, 268, 270 piecewise linear function, 184 point closest to a set, 11 dyadic. 38 exposed. 367 internal of set, 87 isolated. 302 361 isolated of o (T ). 239 nearest to a set. 11 quasi-interior, 125. 126 strongly exposed. 367 support of a set, 14 Poisson's formula, 1.86 pole of the resolvent. 210 polynomial

Kadets-Klee. 344 reflexivity. 20 Schur, 88.4 333

transitivity, 20 pseudoinverse of operator, 132 quasi-interior point, 125. 128 quasinilpotent matrix, 249 quasinilpotent operator. 19.2 195 quotient Banach space, 18 quotient map, 113 111 quotient norm, 18, 112

Rademacher functions, 91.370 range ideal of operator, 311 range of operator, 98, 139 rank of a matrix, 258 rank-one operator, 127. 354. 365 reducing subspace, 202 reflexive B-space, 11. 440 341, 367 reflexivity property, 20

regular operator, 92 representation Kakutani-Bohnenblust-Krein, 130

minimal, 253 polynomially compact operator, 307 positive matrix. 262 positive operator on a B-space, 329 positive projection, 169, 180 positive semidefinite matrix, 248 power compact operator, 219, 238 primitive matrix, 271 principal band, 110, 159

Maeda-Ogasawara-Vulikh. 1f18 of AM-space as 152 of finite-rank operator, 120 of operator by matrix. 255 resolvent of positive operator, 201 resolvent identity second, 181 retract of a space, 13 retraction, 13 Riemann-Lebesgue lemma, 91 Riesz space, 133, 155 Archimedean, 2 24, 52, 72 108 212 atomless, 73 Dedekind complete, 3¢, 30, 54. 105, 108, 155.233 order complete, 22

principal ideal. 881

Riesa subspace, 11 112

principal ideal preserving operator, 106 product rule for derivatives, 48 119, 199 projection, 22, 121., 1I.. > hand, 108. 110 contractive, 170. 175 diagonal. 109 finite-rank operator, 121 Markov. 117 orthogonal, 119 positive, 180 spectral. 211, 239 projection band. 321 property antisymmetry, 20 Daugavet, 352

order dense. 54 Riesx's lemma. 135 Riesz-Kantorovich formula, 26 125

characteristic, 245.250, 252. 255, 256, 268

*-convergence, 55

S-invariant subspace, 318 Schauder system. 37 Schur property. 84. 333 second resolvent identity, 188 semigroup of operators, 319 additive, 319 multiplicative. 319 semiring of sets, 52 separable B-space, 362 separable measure. 198

Index

separable range, 8L 362 separation of points, 157. 1133

separation of sets, 16 Separation Theorem Finite Dimensional, 16 sequence basic, 32 biorthogonal, 40 disjoint. 212 weakly unconditionally Cauchy. 360 series Neumann, 191 unconditionally convergent. 41, 361 weakly unconditionally Cauchy. 360 set clopen. 26 cofinal, 5.1

solid. 23 spectral, 20% 211

totally bounded, 169 weakly compact, 84 shift, 63 shift operator, 219,306 backward. 69, 286.311 forward, 5i9.. 238, 311

similar matrices, 243. 255 similar operators, 189 similarity invariance. 121 singularity essential, 210 slice of a set, 365 solid set, 23 space

385

of similar operators. 200 order. 232 spectrum of order bounded operator. 236 stochastic matrix, 95 strictly positive operator. 12L 311 strictly singular operator. 139. 220. 240 strong Krein operator. 273 strong unit. SZ strongly expanding operator. 273. 225.281 strongly exposed point. 367 strongly positive matrix. 262 subgroup of the unit circle. 230. 231 dense.231 finite. 230 subspace complemented, 121 invariant. 5.203 norming, 422

reducing. 207 subspace invariant under a collection. 318 subspace invariant under an operator. 5 subspace of an ultrapower. All sum of B-spaces. 223 sum of narrow operators. 370. 372 sum of unconditionally convergent series, 42 sum of vector spaces closed. 111 support point of a set. 14 supporting functional. 347

surjective operator, 2. a 226. 251 symmetric operator, 328, 339 system Haar. 34 Schauder, 3Z

L. 114 Al. 11,3.

Hilbert, 102 special points of the spectrum. 197 Spectral Mapping Theorem for polynomials. 205 spectral projection, 211. '1.39 spectral radius. 1990 192. 283, 289, 290, 282. 299 joint, 320

of sum of operators. 202 Rota-Strang, 320 spectral set, 209 211 spectrum, 189.220 approximate. 198. 232 essential, 237 of a direct sum operator. 195 of compact operator. 215 of composition of operators, 196. 201 of lattice homomorphism. 230 of multiplication operator, 194 of operator, 183 of order bounded operator, 236 of projection. 199

totally bounded set. 76.81. 169 trace. 121)

of finite-rank operator. 120 of matrix. 234 transformer, 132 transitive algebra. 306 transitivity property. 21) transpose of matrix. 259 triangle inequality. 96 U-limit of sequence. 35 ultrafilter. A1. 229 free, 229 ultrafilter convergence. AL ultrapower of a B-space. 44 unbounded component of p(T), 202 unconditional basis, 36. 40, 329 unconditionally convergent series, 41. 361 uniformly convex B-space. 11311, 342.344 uniformly convex norm, 11 uniformly smooth B-space. 342-344 uniqueness of kernel, 145 unit. 8L 330

Index

386

order, 87. 125 strong, 87 weak, 2M5 125. 126, IN. 169. 275 330 unital algebra, 122

unital algebra of operators, 304 unitary matrix, 244.246 upper triangularizable matrix, 251 upper triangularizing matrix, 251 vector cyclic, 260 vector sublattice. 118 closed, 178 generated by a vector space, 182

Volterra operator, 22 212. 331 weak unit, 25. 125. 12_6 11 275.330

in Eo, 156 weakly compact interval, 92 weakly compact operator, Z& $ l

884

220.

367

weakly compact set, 84 weakly expanding operator, 274 weakly sequentially continuous lattice operations, 91 weakly unconditionally Cauchy sequence, 360 weakly unconditionally Cauchy series, 360

Titles in This Series 51 Y. A. Abramovich and C. D. Aliprantis, Problems in operator theory. 2002 51Z Y. A. Abramovich and C. D. Aliprantis, An imitation to operator theory. 2002 49 John R. Harper, Secondary cohomology operations, 2002 48- Y. Ellashberg and N. Mishachev, Introduction to the h-principle. 2002 4Z A. Yu. Kitaev, A. H. Shen, and M. N. Vyalyi, Classical and quantum computation, 2002

46 Joseph L. Taylor, Several complex variables with connections to algebraic geometry and Lie groups, 2002

45 Inder K. Rana, An introduction to measure and integration, second edition, 2002 44 Jim Agler and John E. McCarthy, Pick interpolation and Hilbert function spaces, 2002 43 N. V. Krylov, Introduction to the theory of random processes, 2002 42 Jin Hong and Seok-Jin Kang, Introduction to quantum groups and crystal bases, 2002 41 Georgi V. Smirnov, Introduction to the theory of differential inclusions, 2002 40 Robert E. Greene and Steven G. Krantz, Function theory of one complex variable. 2002

39 Larry C. Grove, Classical groups and geometric algebra, 2002 38 Elton P. Hsu, Stochastic analysis on manifolds, 2002 32 Hershel M. Farkas and Irwin Kra, Theta constants, Riemann surfaces and the modular group, 2001

36 Martin Schechter, Principles of functional analysis. second edition. 2002 35 James F. Davis and Paul Kirk, Lecture notes in algebraic topology. 2001 34 Sigurdur Helgason, Differential geometry. Lie groups. and symmetric spaces. 2001

33 Dmitri Burago, Yuri Burago, and Sergei Ivanov, A course in metric geometry, 2001 32 Robert G. Bertle, A modern theory of integration, 2001 31 Ralf Korn and Elite Korn, Option pricing and portfolio optimization: Modern methods of financial mathematics. 2001

30 J. C. McConnell and J. C. Robson, Noncommutative Noetherian rings. 2001 28 Javler Duoandikoetxea, Fourier analysis, 2001 28 Llvlu L Nicolaeecu, Notes on Seiberg-V4itten theory. 2000 22 Thlerry Aubin, A course in differential geometry, 2001 26 Rolf Berndt, An introduction to symplectic geometry. 2001 25 Thomas Friedrich, Dirac operators in Riemannian geometry. 2000 24 Helmut Koch, Number theory: Algebraic numbers and functions. 2000 23 Alberto Candel and Lawrence Conlon, Foliations 1, 2000 22 Giinter R. Krause and Thomas H. Lenagan, Growth of algebras and Gelfand-Kirillov dimension. 2000

21 John B. Conway, A course in operator theory, 2000 20 Robert E. Gompf and Andrds L Stipsicz, 4-manifolds and Kirby calculus. 1999 12 Lawrence C. Evans, Partial differential equations. 1998 18 Winfried Just and Martin Weese, Discovering modern set theory. 1I: Set-theoretic tools for every mathematician, 1997

1Z Henryk lwaniec, Topics in classical automorphic forms, 1997

16 Richard V. Kadison and John R. Ringrose. Fundamentals of the theory of operator algebras. Volume 11: Advanced theory, 1997

15 Richard V. Kadison and John R. Ringrose, Fundamentals of the theory of operator algebras. Volume I: Elementary theory. 1997

14 Elliott H. Lieb and Michael Loss, Analysis, 1997 13 Paul C. Shields, The ergodic theory of discrete sample paths. 1996

TITLES IN THIS SERIES

12 N. V. Krylov, Lectures on elliptic and parabolic equations in Holder spaces, 1996 11 Jacques Dixmier, Enveloping algebras, 1996 Printing 10 Barry Simon, Representations of finite and compact groups, 1996 9 Dino Lorenzini, An invitation to arithmetic geometry, 1996

8 Winfried Just and Martin Weese, Discovering modern at theory. I: The basics, 1996 7 Gerald J. Janusz, Algebraic number fields, second edition, 1996 6 Jens Carsten Jantzen, Lectures on quantum groups, 1996 5 Rick Miranda, Algebraic curves and Riemann surfaces. 1995 4 Russell A. Gordon, The integrals of Lebesgue. Degjoy, Perron, and Heostock, 1994 3 William W. Adams and Philippe Loustaunau, An introduction to Grobner bases, 1994

2 Jack Graver, Brigitte Servatius, and Herman Servatlus, Combinatorial rigidity, 1993

1 Ethan Akin, The general topology of dynamical systems, 1993

This is one of the few books available in the literature that contains problems

devoted entirely to the theory of operators on Banach spaces and Banach lattices.The book contains complete solutions to the more than 600 exercises in the companion volume, An Invitation to Operator Theory.Volume 50 in the AMS series Graduate Studies in Mathematics, also by Abramovich and Aliprantis.

The exercises and solutions contained in this volume serve many purposes. First, they provide an opportunity to the readers to test their understanding of the theory. Second, they are used to demonstrate explicitly technical details in the proofs of many results in operator theory, providing the reader with rigorous and complete accounts of such details. Third, the exercises include many well-known results whose proofs are not readily available elsewhere. Finally, the book contains a considerable amount of additional material and further developments. By adding extra material to many exercises, the authors have managed to keep the presentation as self-contained as possible. The book can be very useful as a supplementary text to graduate courses in operator theory, real analysis, function theory. integration theory, measure theory, and functional analysis. It will also make a nice reference tool for researchers in physics, engineering, economics, and finance.

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www.ams.org