Introduction to Operator Theory and Invariant Subspaces

tr (NT) = tr (N 1 / 2T N 1/ 2 )

L (N

with

Xi

1 2T N 1 / 2 ei , ei) /

= Nl/2 ei .

But then:

= tr N

<

00,

and our claim follows from Lemma 2.5. On L«, we can define a norm, the norm of uniform convergence on the ball of £(H), that is, if E L; :

e

)Iell

=

sup{le(T))

j

T E £(H),

IITII

~

I},

(5)

and from Lemma 2.4 follows that £'(H) is also the dual of £. equipped with this norm. We now consider the application:

it> : )I (H) ---. defined by it>(N)

= 4>N, where
t:

is given by formula (4).

Topologies on the Space 0/ Operators

105

Lemma 2.6. - 4.> is a surjective isometry from ),/(H) equipped with its norm onto L; equipped with the norm (5). Proof of Lemma 2.6. - Indeed, we have, for N E ),/ (H)

Il¢NlI

=

sup{\tr (NT)I

=

IINIIN'

j

T E £(H),

IITII :S

I}

by Lemma 2.2, and this proves that 4.> is an isometry. Let's now show that it is surjective. All we have to show is that the image of ~ is dense in L; : the image of an isometry is always closed. Also, since this image is a vector subspace, it is dense in £. for the norm defined by formula (5) if and only if it is dense for the weak topology o(£.,£(H)) (B.B. [1], p. 34). Assume on the contrary that 4.>()'/) is not dense in L; for 0(£., £(H)). Since (again) it is a vector subspace, it must be contained in a closed hyperplane: there exists U i- 0 in £(H) such that, for every N in )I, ¢N(U) = o. We show that this is impossible. Since U i- 0, lUI i- O. Let Zl, ... ,Zn, with Ilzill = 1, mutually orthogonal, such that IU I i- 0 on F = span {Zl' ... , zn}, and let P be the orthogonal projection onto F. Then IUIP i- 0 on F, and \UIP = 0 on FJ... If x E H , we have:

(Px, x) and so P is a positive operator. Let N = PV·, where V· is given by Proposition 1.1, b), and satisfies lUI = V·U. We have:

¢N(U) = tr (NU) = tr (PV·U) = tr (PIUI) = tr (IUIP)

i-

0,

since IUIP is a positive non-zero operator. Since P is nuclear (as any finiterank operator), we get a contradiction, which proves our Lemma. We now finish the proof of Theorem 2.1 : 4.> is a surjective isometry, so we can use it to identify £(H) with the dual of ),/ (H), since it was the dual of L«, by Lemma 2.4. This proves our Theorem. From this Theorem follows that T

T

--+

sup{ltr(NT)1

j

--+

II T II t:(H)

N E ),/(H),

and

IINIIN:S

I}

Chapter V

106

are two equivalent norms on £(H). But we can show now that they are equal. Indeed, put E = JI(H), with its norm, E· the dual space, which is £(H), endowed with the norm IITII = sup{ltr (NT)I i N E J/(H), IINII.w ~ I}, and E;, the space £(H) with the usual operator norm. The dual norm of Ei is IINII(E;). = sup{ltr (NT)! i T E £(H), 11TIIop ~ I}, and this norm induces a norm on E by the same formula, the dual of which is Ei. But by Lemma 2.2, this induced norm is exactly the nuclear norm; therefore Ei = E+, and we obtain the formula: IITllop = sup{ltr (NT)j ; N E JI(H), IINII.w

< I}.

This may look like a complicated proof for a formula which is obvious anyway. Indeed, let's give a direct proof of it. One one hand, if IINII.w

< 1,

jtr (NT)l

<

IINlI.w·IITI\ ::; IITII.

On the other hand, take e > O. We can find x E H, Ilxll = 1, with IITxl1 > (1 - e) IITII, and then y, with Ilyll = 1, such that (Tx, y) = IITxll. Consider now the rank one operator x 0 y. It is nuclear, and has nuclear norm at most 1. Furthermore, To (x 0 y) = Tx 0 y , so by formula (8), Section1 : tr (T

0

(x 0 y)) = tr (Tx 0 y)

(Tx, y) > (1 - c) IITII,

and our formula is proved. From now on, we make the (isometric) identification corresponding to and we just say that £(H) "is" the dual of JI(H).

~,

Since £(H) is a dual, it has a weak-* topology, namely a(£(H), J/(H)), which we now describe. We call ultraweak topology on £(H) the topology defined by the family of semi-norms :

(6) where (x,k2:0, (yil,~o

are two sequences of points in H, satisfying:

(7)

107

Topologies on the Space of Operators

Theorem 2.7. - The topology G(£(H), N(H)) coincides with the ultraweak topology.

e

Proof. - 1) Let be a linear functional on £(H), continuous with respect to the semi-norms (6). Then, there are sequences (Xdi~O, (Yi)i~O' satisfying (7), such that:

I€(T)I ~ P(Xi),(Yi)(T) , for T E £(H). By Lemma 2.5, € belongs to L«, which means that € is continuous for G(£(H), N(H)). 2) Conversely, we wish now to show that the elements of

£. are continuous

for the ultraweak topology. We start with elements of £"".

Lemma 2.8. - IE € E £"", there exists n eI, ... ,en, e~, ... , e~

~

1, and two orthonormal sequences

in H, a sequence of positive real numbers AI, ... , An,

such that

n

€

L AiWei.< .

=

I

Proof of Lemma 2.8. - We know by (2) that

€ can

be written:

p

LW

€ =

x i • yi

,

I

for some finite sequences xl, ... , X p i YI, ... , Yp in H. Consider two families (Xi)i~q, (y;k~q, such that the applications: p

It : Y ~ L(Y, Yi)Xi 1 q

h :

Y ~

L(y,y;·)x~. I

e

are equal. Let's put = 2:1 Wx'_.y'.. J J If u., v are in H, with the rank 1 mapping u. ® v, we get :

€' (u ® v) =

L (u ® v (xj), y;. ) i

L (xi, v) (u, y;.) i

= (v, L (u, yj)xi) i = (v,h(u)) = (v, It (u)) =

€(u®v),

108

Chapter V

by a similar argument. So € = €' on all rank one operators, and therefore on all finite rank operators. But if P is the orthogonal projection onto

obviously €(TP)

= €(T) , €,(TP)

= €,(T) , so €(T) = €,(T) for every T E

£(H). Now, we consider the operator: p

U

Y

-+

L(Y,Yi)Xi I

and write U = V lUI (Proposition 1.1). lUI is positive, self-adjoint, has finite rank, so there is an orthonormal basis el, ... ,en, made of eigenvectors, associated to the eigenvalues AI, ... , An, with Ai ~ o. Therefore:

i = 1, .. . ,n, n

Uy =

L

Ai(Y, ei)e~

,

I

with e~ = Vei. The sequence (e~) on ImlUI), and

is orthonormal (because V is an isometry n

L Ai(y,ei)e~

,

1

so

€ = L~=l

AiW e i .<' and the Lemma is proved.

We prove now the second part of Theorem 2.7. Let 4> E £" : we wish to show that it is ultraweakly continuous. We may assume 114>11 ::; 1. We write

4> = L.k>O 4>k, with 4>k E £"", II4>k II < 1/2 k , for k >

o.

By Lemma 2.8, each

¢>k can be written : nk

L A~k)w

4>k =

i=l

elk) ,/(k)

,

.,

where for each k > 0, (e~k))i' (flk))i are two orthonormal sequences. Let's k show that L.~===l A~k) ::; 1/2 , for k > o. Indeed, let u, be the operator defined by :

U e~k) J

=

f~k) J'

i = 1, ... , nk

,

Topologies on the Space of Operators

on span{e~k)

L:~l

).~k).

109

; i ~ nk}, and 0 on the orthogonal. Then IIUkl1 = 1, tPk(Uk) =

So,

Now, we can write :

with "It

Xk

L().~k»)1/2e~k)

,

i=1 "It

Yk

1/2 fi(k)

L().~k»)

.

i=l

Therefore: "It

L L Ij().~k») 00

1/2e~k)

11 2

k=l i=l 00

"It

< L1/2 k,

LL).~k) k=l i=l

and the same for E~ tinuous.

IIYkI1 2 ,

k

and we have proved that ¢ is ultraweakly con-

Important Remark. - The ultraweak topology is stronger than the weak topology. Indeed, let To. be a net, converging to T for the ultraweak topology. This means that for every semi-norm P(z,.),(y;), P(zil,(y;) (To.

- T)

-j.

0,

or

But if we take the families (xd, (Yi) reduced to a single element, x, Y respectively, we get that for all x, Y in H,

((To. - T)x, y)

-j.

0,

which means that To. -+ T weakly. The terminology "weak" and "ultraweak" is confusing, and therefore is rather bad. It's now commonly adopted, so we have used it here, but something like "pointwise-weak" (to recall the action on the points of H), and "nuclearweak" (to recall the duality with JI) would probably have been better.

110

Chapter V

Proposition 2.9. - On the bounded subsets of £(H), the weak and ultraweak topologies coincide. Proof. - Of course, we need only to consider balls, and, in fact, the unit ball. The unit ball of £(H) is compact for 0(£, N) (since £ is the dual of N), and it is also compact for the weak topology (lemma 2.3). Since these topologies are comparable (remark above), they are identical on the balls. We now turn to strong topologies. We say that a net To converges to T strongly if, for every x in H, Tax ---. Tx. In other words, this is the topology of pointwise convergence on elements of H. It is defined by the family of semi-norms :

Px(T) =

IITxll,

xE H,

and a fundamental system of neighborhoods of 0 is :

where z i ,

•.• , X

n

are points in H, and e >

o.

We say that a net To converges to T ultrastrongly if, for every sequence of elements in H with

(Xi)i~O

L IIxil12 < 00,

In other words, this topology is defined by the family of semi-norms :

P(xd(T) =

L II TxiI1 2, i

where

(8) A fundamental system of neighborhoods of 0 is :

where e

> 0 and (X{)i~O

are sequences satisfying (8), for j

= 1, ... ,n.

This time, the terminology goes in the right direction: the ultrastrong topology is stronger than the strong topology.

Topologies on the Space of Operators

111

Proposition 2.10. - On the bounded subsets of £(H), the strong and ultrastrong topologies coincide. Proof. - Let W be a neighborhood of 0 for the ultra-strong topology. There exists e > 0 and a sequence (Xi)i~O, satisfying (8)' such that: {T E £(H) i

L IITxil12 < e}

C W.

i

Let i o be such that strong topology :

Li>io IIxill2 < e/2, and

V = {T E .c(H) ; Then, if T E V

n B(H),

IITxill <

V the neighborhood of 0, for the

Ve/(2io) , i = 1. .. , i o }.

T is in W, and our Proposition is proved.

Since the reader might be ultra-tired of these notions, we summarize the results. We have introduced the following topologies: 1. Norm topology in

Tn

-+

T if IITn -

.c (H) : Til -+ o.

2. Ultra-Strong topology : Defined by the family of semi-norms:

E II xil1 2 <

P(z;) (T)

=

Li IITxiI12,

where

00.

A net TOt converges to T ultra-strongly if, for every sequence elements in H with L IIxil12 < 00,

(Xdi~O

of

3. Strong topology:

Defined by the family of semi-norms: pz(T) = IITxll, where x E H. A net TOt converges to T strongly if, for every x in H, I (TOt - T)xll -+ O. 4. Ultraweak topology: Defined by the family of semi-norms:

where L: II xil1 2< 00, L: IIYil1 2 < 00. A net TOt converges to T ultraweakly if, for all such sequences (Xi), (yd

L I(TOt - T)Xi,Yi)1 i

-+

o.

lle

Chapter V

5. Weak topology

](Tx,Y)I, where z , Y

Defined by the family of semi-norms: Pz,II(T} are points in H. A net TOt converges to T weakly if :

((TOt - T)x, y)

--+

0,

for all z , Y in H.

1) is stronger than 2), 2} than 3), 3) than 4), 4) than 5). 2) and 3) coincide on balls, 4) and 5) also coincide on balls. We finally turn to the description of the linear functionals, continuous for these various topologies.

Theorem 2.11. - Let M be an ultraweakly closed subspace of .c(H), M· its dual, 8 M its unit ball, ¢ a linear functional on M.

1) The following conditions are equivalent: a) ¢ is weakly continuous, b) ¢ is strongly continuous, c) ¢ can be written as a finite sum ¢ = E~

WZi,Yi •

2) The following conditions are equivalent:

a) ¢ is ultraweakly continuous, b) ¢ is ultra-strongly continuous, c) ¢ can be written as an infinite sum ¢ = E~ 00,

E IIYil1

2

<

Wxi,y, ,

where

E II xil1 2 <

00.

d) the restriction of ¢ to 8M is ultraweakly continuous,

e) the restriction of ¢ to 8M is weakly continuous, f) the restriction of ¢ to 8M is ultra-strongly continuous,

g) the restriction of ¢ to 8 M is strongly continuous. Proof. - 1) Clearly, c) implies a), which implies b). We prove that b) implies

c). The strong topology on .c(H) is the topology of pointwise convergence on H : TOt --+ T if, for every z , TOtx --+ Tx. If we identify (as we already did) T with the collection (TxhEH' £(H) appears as a closed linear subspace of H H , each H being equipped with the norm, and the product with the product topology. By Hahn-Banach Theorem, each ¢ on £(H), strongly continuous,

119

Topologies on the Space of Operators

can be extended to a linear functional on HH, continuous for this topology. But, by definition of the product topology, every 4> in (HH). is a finite sum of functionals of the form u 0 f r. , where f r. (T) = T z , and u is a linear functional on H, that is u(z) = (z, y), for some y in H. Therefore, there are vectors Xx, ••• , x n , Yx, ... , Yn in H, such that: n

n

n

L Wr.;

4>(T) = LUiOfr.;(T) = L(Txi,Yi) 1

1

,Yi (T).

1

2) Clearly, c) implies a), which implies b). We now prove that b) implies

c). Let 4> be an ultra-strongly continuous functional. Therefore, there exists with 2: IIxil12 < 00, such that: a sequence (xik::~:o, 00

14>(T) I ~

(L I T x iI12) 1/ 2 ,

(1)

T EM.

i=O We consider the Hilbert space 12(H) = {(Zi)i~O 00 } , equipped with the norm :

j

z,

E H 'Vi,

Then the sequence x = (Xdi>O is an element of 12(H). Let on 12(H) defined by T((zi)i~o) = (TZi)i~O. Then:

14>(T) I ~

l'

(2: IlziI12)l/2 <

be the operator

111';;11·

Thus 4> extends on 12 (H) to a functional still satisfying:

We come back to the identification (here for 1') : l' --+ (Tz)zEH' where iI is equipped with the norm. The previous formula shows that 4> can be identified to a linear functional on it , continuous for the norm: therefore, there exists an element y in it , such that:

(Tx, jj) H '

4>(1')

00

4>(T)

=

L(Txi,Yi), o

Chapter V

114

which proves c). The fact that d) and e) are equivalent follows from Proposition 2.9, the fact that f) and g) are equivalent follows from Proposition 2.10. Clearly, a) implies d) which implies f). To show that f) implies b), we observe that if the restriction 4>ISM is ultra-strongly continuous, condition (1) above holds for T E BM, and therefore for all T EM: so 4> is ultra-strongly continuous on M.

Proposition 2.12. - Let M be an ultraweakly closed subspace of £(H) and 1. M its pre-annihilator (or pre-orthogonal), that is :

1.M

=

{N E J/(H)

j

tr (NT)

= 0,

for all T EM}.

Then the dual of the quotient J/(H)j1. M is M, and the weak topology u(M, JI j 1. M) is the topology induced on M by u(.c, J/).

Proof. - This is a general result in Functional Analysis: since M is closed in = (1. M)1. , and this result can be found in B.B. [1], Proposition

0(£, JI), M 3, p. 43.

In Part VI, we will apply the results of this chapter to the subalgebra of .c(H) generated by a single operator.

Topologies on the Space

0/ Operators

115

Exerclaes on Chapter V.

Exereise 1. - Let (en)n>O be the canonical basis of '2(1N), and Un be the operator defined by :

= (x, en)eo .

U« (x) a) Prove that

U~

is given by : U~

(x) = (x, eo)en ,

that the sequence (Un)n~O tends to 0 ultra-strongly, that the sequence U~ does not tend to 0 strongly, and that Un does not tend to 0 in norm. b) Deduce from a) that the norm topology is strictly stronger than the ultra-strong topology. c) Deduce from a) that the mapping T -+- T· is not continuous for the strong topology, nor for the ultra-strong topology. d) Show that the sequence U~ tends to 0 weakly. Deduce that the strong topology is strictly stronger than the ultraweak topology (and than the weak topology).

Exercise 2. - Let W be a neighborhood of 0 in £(H), for the ultra-strong topology. Show that there is a point z such that : sup

IITzll = +00.

TEW

(Hint: assume the converse, apply Banach-Steinhaus theorem and use exercise 1, b), to get a contradiction.)

Exercise 3. - Let (en)n~O be the canonical basis of 12(IN) , and Pn be the orthogonal projection onto F n = {Aen ; A E et}. Let Tm,n = Pm + mPn. a) Let (Xi)i>O be a sequence of points in H = '2(1N), with L IIxil12 < 00, and c > o. Show that one can choose m and then n large enough so that :

L IlT

2

m ,n X i 11

< e.

i

Deduce that the set .M = {Tm,n; m, n E IN} has 0 as an accumulation point, for the ultra-strong topology. (nkh~o be two increasing sequences of integers. Show b) Let (mk)k~O, that there exist x, y, such that (T mA: ,nA: x, y) does not tend to O. Deduce that no sequence (TrnA: ,nA:)k in .M converges weakly to o. c) Deduce from a), b) that £(H) is not metrizable for any of the weak, ultraweak, strong, ultra-strong topologies.

116

Chapter V

Exercise 4. - a) Let N be a finite rank operator, P the projection onto ImN. Show that: sup Itr (UPN)! = IINIIJI . 11U1I~1

Deduce that:

= IINIIJI .

Itr (KN)I

sup K compact, IIKII$1

b) Deduce the same formula for any N nuclear. c) Show that the dual of K (H) can be identified to )I (H), by means of the duality (K, N) ~ tr (K N).

Exercise 5. - Let Pn be the orthogonal projection, in l2(IN), on {Afn ; A E ~} . Show that, for any bounded sequence of scalars (an)n~o, the series L anPn converges for the strong operator topology, to an operator U, and that:

IIUII =

sup lanl· n

Deduce that t.(H) , equipped with its norm, contains a subspace isometric to loo(IN) (see B.B. [1], p. 107 for the terminology). Deduce that t. (H) is not a separable Banach space.

Exercise 6. -a) Let Bt. the unit ball of t.(H). Show that the application (U,T) ~ UT is continuous from Bt. x t.(H) into t.(H) , when they are both equipped with the strong topology. b) Deduce that the composition of operators is separately continuous for the strong topology. and (Tk)k~o of operators, which c) Deduce that for all sequences (UJc)k~O are strongly convergent to U and T, the sequence (UkTk) converges strongly to UT. Exercise 7. - We now show that on the product t. (H) x t. (H) , the composition of operators is not continuous for the strong topologies. Let z , Y be two points in H, with norm 1, and:

v

= {T E t.(H) ; I(Tx,Y)1 ::; I}.

Let also xl, ... , x n be a finite sequence of points in H, and

W = {T E t.(H) ;

IITxil1 ::;

We will find U and T in W such that TU

~

1, i

= 1, ... ,n}.

V : this will prove the assertion.

Topologies on the Space 01 Operators

117

a) Using exercise 2, show that there is a point z E H, that: sup

IITzl1

=

IIzl1 =

1 , such

+00.

TEW

b) Construct U E W , such that U x = AZ for some ..\ > c) Choose R E W such that IIRUxl! > 1. d) Construct an operator U', IIU'I! ~ 1, such that:

o.

(U'RUx, y) ~ 1.

e) Put T = U' R and conclude.

Exercise 8. - We equip £(H) with the weak topology. Show that the mappings T -+ T·, T -+ UT, T -+ TU (U fixed) are continuous.

Notes and Comments :

We have followed the presentation of J. Dixmier [1], with a few notational changes. We are greatly indebted to Francoise Piquard and Gilles Godefroy for their help in writing the Banach space version, which now follows.

Complements on Chapter V :

A. The space £(E) as a dual space. Let E be a Banch space, with dual E·. We consider the rank-one operators: for x in E, z" in E· , and z in E. On the space l(E·), we put the topology of point-wise convergence on the rank-one operators:

Ta

-+

T if and only if «(Ta

-

T)x·, x)

-+

o.

118

Chapter V

Then the unit ball B.c(E-) is compact for this topology. So .c(E*) is the dual of some space Z, which we now describe. First, we observe that by a computation analogous to the one we did for the proof of Theorem 2.1,

IITII =

sup{tr (T

0

z"

x ® x*) ; x E E,

E E*,

lIxll.llx*11 = I}.

Hence, by Hahn-Banach Theorem,

Bz = conv {x ®

z"

i x E E,

z"

E E*,

Ilxll.llx*11 =

I},

where the closure is taken for the norm of Z. Formally, this norm is given from the duality with £(E*), but it can be expressed more explicitly. The set of linear combinations of rank-one operators x ® z" is called the algebraic tensor product of E and E* , and denoted by E ® E*. On this space, the norm is by definition the gauge functional of the unit ball, that is the gauge of:

conv {x ®

z" ; x E E, z"

E E*,

Ilxll.llx*11

= I},

that is : Since the algebraic tensor product E ® E* is dense in Z, every u in Z can be written: 00

U

=

LU

n '

1

where, for all n, Un is in E ® E* and satisfies U can be written:

lIunll

(1)

u

with:

= I/2 n. This shows that

00

L

Ilxnll·l x~1I

<

00.

1

Formula (I) means that : 00

(u,T)

L(x~,Txn), 1

and moreover :

The space Z is called the projective completed tensor product of E and E* , and is denoted by E ®'Ir E* or E®E*.

119

Topologies on the Space 01 Operators

One checks directly that the duality is given by the trace: this comes from formula (7), Section 2, for rank-one operators. One also has (LindenstraussTzafriri [I], vol. 1, p. 31) :

Proposition. - On £(E), let TK be the topology of uniform convergence on compact subsets of E. Then the dual of £(E) is a quotient of the space E®E*.

B. The Approximation Property.

Definition. - We say that E has the Approximation Property if the set of finite rank operators is dense in £ (E) for TK . The following Theorem, due to A. Grothendieck [I], shows the links between the Approximation Property and the approximation by compact operators:

Theorem. - 1) Let X be a Banach space. The following are equivalent:

a) X has the Approximation Property, b) For every Banach space Y, the finite rank operators are dense in the space £(Y, X), for the topology of uniform convergence on compact sets,

c) For every Banach space Y, the finite rank operators are dense in the space £(X, Y), for the topology of uniform convergence on compact sets,

d) For every Banach space Y, every compact operator in £ (Y, X), every c > 0, there exists a finite rank operator T I in £(Y,X), with

liT - Till < c.

2) If X* has the Approximation Property, d) holds also when the roles of X and Yare reversed. We have a mapping j from E®E* into £ (E), defined by : 00

T(x) =

L x~(x)

Xn .

I

The image of this mapping is the space of nuclear operators on E.

Proposition. - The mapping j is injective if and only if E has the Approximation Property.

When it is the case, j(E®E*) is identified to the predual of £(E*), and we have a duality between the nuclear operators on E and the space £(E*).

Chapter V

120

We refer the reader to Lindenstrauss-Tzafriri [I] (vol.1, p. 31 and seq.) and to A. Grothendieck [I] for a detailed presentation of these matters. Further results concerning the space of compact operators and its duals can be found in G. Godefroy-P. Saphar [I]. A very important fact should also be noted: A. Szankowski [1] proved that the whole space £(H) does not have the Approximation Property. This is the first "natural" example of a space without the Approximation Property. The consequences of this fact upon Operator Theory have never been investigated, but obviously they should exist. C. Strong and Uniform Convergences. In general, quite obviously, a sequence of operators may converge strongly without converging in operator norm. However, on the space L oo , the following result holds (due to Thierry Coulhon [1]) :

Theorem. - Let Tn be a sequence of operators on L oo , with Tn -----+ I strongly, then Tn - t I in operator norm.

IITnll = 1.

If

A result of a similar nature holds for a class of operators defined on £ (H) : A mapping

is a positive operator. The following result was obtained by Th. Coulhon-Y. Raynaud [I] :

Theorem. - Let
PART III

BANACH ALGEBRAS

Notre arne est un trois-mats eherchant son !carie; Une voix retentit sur le pont: "Ouvre l'oeil !" Une voix de la hune, ardente et folie, erie: "Amour...gloire ...bonheur!" Enfer! e'est un ecueil !

In this Third Part, we introduce the reader to elementary Banach algebra techniques. First, from a general point of view: Gelfand's Theory and its consequences. Then, we turn to Operator Algebras: the algebra generated by a single operator, and the C· -algebra generated by a normal operator and its adjoint.

Chapter VI

Banach Algebras

In this chapter, we give some outline of Gelfand's theory about Banach algebras. We then show how to apply it to the algebra generated in £(H) by a single operator, in order to deduce some representation theorems and Invariant Subspaces. In the next Chapter, we will apply the C· -algebra techniques to normal operators. 1. Spectral Theory.

A Banach Algebra A is a complex Banach space, equipped with a multiplication (x, y) ---+ xy, from A x A into A, satisfying the following properties, for all x, y, z E A, all A E ~ :

a) (xy)z = x(yz), b) x(y+z) = xy+xz, (y+z)x

yx+zx,

c) A(XY) = (AX)Y = X(AY) , d) Ilxyll ~ Ilxl/·llyll. If moreover xy = yx, the algebra is said to be commutative. If there is a unit for the multiplication, that is an element denoted by 1, such that: x.l = l.x = x, for all x E A, the algebra is said to be unitary. A sub-algebra B is a vector subspace, which is stable for the multiplication: if x, y E B, xy E B. A sub-algebra B of A is called an ideal if it satisfies the following property : For all x E A, all y E B, xy E Band yx E B.

Examples: a) The space C(X) of continuous functions on a compact set X, with complex values, is a commutative algebra with unit, the multiplication being pointwise multiplication of functions. b) The space Ll~), equipped with the convolution of functions, is a commutative algebra, with no unit.

Chapter VI c) The space £(E) of all operators on a Banach space E is an algebra, for the composition of operators. This algebra has a unit (the identity), but is not commutative. The set K (E) of all compact operators is an ideal of £, (E) . In this Chapter, we consider only algebras with unit. The next propositions are extensions to this larger frame of results which were proved in Part I, Chapter II, for £(E). The proofs are identical and are left to the reader. Instead of )'.1, we just write A, as we already did for operators.

Proposition 1.1. - If Ilxll < 1, 1 + x is invertible. The set of invertible elements is open in A, and, on this open set, the mapping x - X-I is continuous.

IT x E A, the resolvent set p(x) is :

p(x) = {). E cE ; ). - x is invertible}. The spectrum u(x) of x is the complement of p(x) in cE •

Proposition 1.2. - For x E A, u(x) is a non-empty compact subset of cE . The mapping). - (). - X)-l is analytic on p(x). The spectral radius of x is :

r(x) = sup{I).1 ; ). E u(x)}, it satisfies :

r(x) =

lim Ilxnll 1jn. n---+oo

Theorem 1.3 (Gelfand-Mazur). - Let A be a Banach algebra with unit, such that every non-zero element in A is invertible. Then A is isometric to cE • Proof. - Since the spectrum of each element is non-empty, for each x E A, there is a ). E o(x) such that ). - x is not invertible. The assumption implies x =)., and A is the set of multiples of 1. Since 11).11 = 1).1 = 1, this correspondence is an isometry. Definition. - Divisors of 0: An element x in A is called a divisor of 0 if there exist two sequences (zn)n"2:0, (z~)n;: >:O, for all n, such that : XZ n

-

0,

of elements of A, with

z~ x -

0,

n - + 00.

For instance, I(t) = t is a divisor of 0 in C([O, 1]).

Ilznll

= Ilz~

II =

1

Banach Algebras

Proposition 1.4. - Every point in the boundary elements in A is a divisor of O.

aG

the group of invertible

Remark: Such a point is not itself in G, since G is open. Proof. - There exists a sequence X;I X

tt.

G, since

Xn

E G and x 1

:s: XZ

The same way,

ZnX ----t

:s:

Ilx~lx-I11

o. Set

which implies 1/llx~III-Jo

in G, with X n -Jo z , We know that G. So, by Proposition 1.1,

(xn)n~O

tt.

Zn

Ilx~11111x-xnll,

= x;I/llx~lll.

n = (x - xn)zn + 1/llx~111

Then

Ilznll = 1, and

-Jo o.

0, and the Proposition is proved.

Let now B be a closed subalgebra of A, containing 1. For x E B, we distinguish between: UB (x), spectrum of x considered as an element of B, and 0' A (z}, spectrum of x considered as an element of A. The comparison is as follows:

Theorem 1.5. - Let B be a closed subalgebra of A, containing 1. For every x E B, 0' A(x) e UB (x), and the boundary aUB (x) is contained in auA(x). Proof. - The first assertion is clear. For the second, let A E aUB (x). Then A- x is in aG B , boundary of the invertible elements in B. So A- x is a divisor of 0 in B, therefore in A. This implies that A- x is not invertible in A, and so A E UA(x). But A E aUB (x), and so there exists a sequence (An)n~O, An -----+ A, An E PB(x). Then An EPA(X), so A E auA(x) . We observe, however, that UA(X) and UB(X) need not be the same. As an example, consider the right shift S on 12(7L), and let 8 be the closed su balgebra of (H) generated by S, that is the closure in (H) of finite sums Lk~O Clk Sk. Trivial computations show that S-1 is not in this algebra, so o E UB (S). From Theorem 1.5 follows that UB (S) = D, though 0' C(H) (S) =

.c

.c

c.

Corollary 1.6. - IE PA(X) is connected, UA(X) = UB(X). Indeed, Us (x)\u A(x) is open, by Theorem 1.5 (and equal to PA (x)nUB (x)), and: which is the reunion of two disjoint open sets. An application of this Corollary is the following : if we have an operator T on H, and if we know that u(T) elR, then, in order to determine u(T) , we may work in any subalgebra of £(H) containing T and I.

Chapter VI

11!6

2. Ideals and Homomorphisms. Let 1 be an ideal in A, the quotient space Allis an algebra, where the multiplication is defined by multiplication of elements in each class. If x is the class of z , we also recall that the quotient norm is : IIzll

=

inf{llxli j x EX}.

The quotient mapping q, defined by :

q(x) = ±,

x E A,

from A into All, is an homomorphism. One checks easily that if 1 is closed, the quotient AlI is complete, therefore a Banach algebra, and that it has a unit, as soon as A does. An ideal 1 is said to be proper if 1 ¥= A. A proper ideal is said to be maximal if, for every proper ideal 1', the inclusion 1 c l' implies 1 = l' (in other words, there is no proper ideal which is strictly bigger). A proper ideal cannot contain any invertible element: otherwise, it would contain 1, and therefore the whole algebra.

Proposition 2.1. - Every maximal ideal is closed. contained in some maximal ideal.

Every proper ideal is

Proof. - Let 1 be a maximal ideal. The closure I is also an ideal, containing 1. But, since 1 does not contain any invertible element, it does not intersect the open ball {x E A jill - z II < I}, and neither does I. This proves that I is proper, and maximality implies 1 = I. Let now 1 be a proper ideal. We consider the family of all proper ideals which contain 1, ordered by inclusion. This set is inductive: if (Ii) is a subfamily which is totally ordered, the reunion uili is a proper ideal (since, again, none of them intersects the ball {z E A j 111 - z] < I}), which is a majorant for 1. Therefore, by Zorn's axiom (see for instance B.B. [1], p. 20), there are maximal elements: they are maximal ideals containing 1.

Example. - In the algebra C ([0,1]), the set of functions which vanish on a prescribed set A C [0, 1] is a closed ideal:

FA

= {f

E C([O, 1]) ; f(t)

= 0,

Vt E A}.

A maximal ideal is obtained when A is reduced to a single point: F{a}

=

{f E C([O, 1])

j

f(a) = O}.

127

Banach Algebras

An homomorphism mapping satisfying :

t/J

between two Banach Algebras A and B is a linear

t/J(xy) = t/J(x)t/J(y) ,

for all x, yEA

The multiplicativity property provides automatic continuity:

Proposition 2.2. - Every homomorphism t/J from A into G:: is continuous, and if t/J :j:. 0, then t/J(1) = 1, Iit/JII = 1 , and t/J(x) E u(x) , for all x EA. Proof. - a) The kernel K er t/J is an ideal in A. Since it is an hyperplane, this ideal is maximal, therefore closed, by Proposition 2.1. b) Let x, Ilxli = 1 . Then:

so It/J(x) I ::; 11t/J111/n. Letting n ~ +00, we get It/J(x) I ::; 1, so IIt/JII < 1. But if 1/; :j:. 0, there is a point x such that t/J(x) :j:. 0. Then t/J(x) = t/J(l).t/J(x), and 1/;(1) = 1. This proves also that IIt/JII = 1. c) For x E A, t/J(x) - x E K er 1/;, which is a proper ideal, and therefore contains no invertible elements. So t/J(x) E u(x). A non-zero homomorphism from A into

(V is

called a character. The spec-

trum of A, denoted by X(A), is the set of all characters on A. It is of course a

subset of A*, the dual of A. We put on A* the topology u(A*, A), that is the topology of pointwise convergence on elements of A. The theory which follows should remind the reader about similar results in Banach spaces: one shows that the unit ball of E* is weakly compact, and one considers the space E as a closed subspace of E**, identifying each x to the functional on E* : € ~ €(x) (see B.B. [11, Part I, chap. II, Sections 2, 3).

Proposition 2.3. - As a subset of the unit ball of A*, endowed with the topology u(A*, A), the spectrum X(A) is compact. Proof. - All we have to show is of course that x(A) is closed for u(A*, A). Let € E BA- , be an accumulation point of x(A) for u(A*, A). We know that, for all z , y in A, all e > 0, the set : VZ,Y;l:

= {" E A* ; I€(x) - ,,(x)1

< e , I€(y) - ,,(y) I < s , I€(xy) - " (xy) I < s}

is a neighborhood of € for u(A*, A). Since € is an accumulation point of X(A), this neighborhood contains a character f/J. But since:

€(xy) - €(x)E(Y) = E(xy) - fjJ(xy) + t/J(x)(fjJ(y) - €(y))

+ (t/J(x)

- €(x))€(y),

Chapter VI

128

this implies :

+ ItP(x) I + I€(y) I) ~ e(1 + IIxll + Ilyll)· E: > 0, we get €(xy) = €(x)€(y). The same way,

I€(xy) - €(x)€(y)I

~ e(1

Since this is true for every obtain €(I) = 1. So € is in X(A), and our Proposition is proved. For x in A, we define its Gelfand's Transform

x from A into C(x(A)). The correspondence x

--+

II x llC(x(A»

:

{tP --+ tP(x)},

--+

x is called the Gelfand's morphism. .If = {x; x

We have:

x by

we

We put:

E A}.

sup{lx(tP)1 i tP E x(A)} sup{ltP(x)1 ; 1/J E x(A)}

< r(x)

by Proposition 2.2

< Ilxll and this last formula proves that the application x

--+

x is continuous, from A

into C(X(A))So far, we have not exhibited any non-trivial ideals or characters, and in some cases there are none (see exercises). But further information can be obtained for the case of a commutative Banach algebra, to which we now turn. Though the whole algebra f,(H) is not commutative, commutative Banach algebras come naturally when one looks at the algebra generated by a single operator and identity, or by I, T and T'" in the case of a normal operator. 3. Commutative Banach Algebras.

Theorem 3.1. - Let A be a commutative Banach algebra. For every x in A,

u(x) = {x(tP) ; tP E x(A)}, and therefore

r(x) = sup{x(tP)

i

t/J

E x(A)}

=

II xllc(x(A»-

Proof. - IT A E u(x), x' = A - x is contained in a proper ideal, namely:

Ax' = {ux';

u E

A}.

This is an ideal, since A is commutative, and it does not contain I, otherwise x' would be invertible. This ideal is contained in a maximal ideal M. We now show that M is the kernel of some character.

Banach Algebras

1£9

Lemma 3.2. - An ideal is maximal if and only if it is the kernel of some character. Proof. - We have already seen that the kernel of a character was a maximal ideal. Conversely, let .M be a maximal ideal. Then the quotient AIM is a Banach algebra; let q be the quotient map. If some proper ideal I was contained in AIM, q-l(I) would be a proper ideal, containing M, with q-l(I) i- M. This is impossible, since M is maximal. So AI M contains no non-trivial ideal. Thus, every element is invertible (if an element x is not invertible, Ax' is a proper ideal), and by Theorem 1.3, AI.M is isometric to ~. Let 4> be the isometry from AI.M onto 0 q is a character on A with K er t/J = M.

The Radical of A, denoted by Rad(A), is the intersection of all maximal ideals contained in A. The algebra is said to be semi-simple if Rad(A) = {O}. By Theorem 3.1, we have: Corollary 3.3. - The following statements are equivalent:

a) x E Rad(A) , b) z E n,pEX(A){Y ; t/J(Y)

=

O},

c) X(t/J) = 0, for all t/J E X ( A),

d) r(x) = 0, e) lim IIx nll l / n

= o.

Theorem 3.4. - Let A be a commutative Banach algebra. morphism x ~ x has the following properties :

a) i(t/J)

= 1,

The Gelfand's

for all t/J E x(A) ,

b) A contains the constant functions, and separates points in A : that is, for all x and Y in A, there is a t/J in A, with t/1(x) i- t/J(y) .

c)

x is invertible in

d)

IIxllc(x(A» =

C(x(A)) if and only if x is invertible in A.

lim IlxnIl 1/ n,

e) Gelfand's morphism is bijective from A onto semi-simple.

A if and

only if A is

Proof. - a) follows from Proposition 2.2, since for every character, t/1(I) = 1.

b) We have A(t/J) = t/1(oX) = oX, and therefore A contains the constant functions. Moreover, if for every z E A, x(t/Jd = X(t/12), t/Jdx) = tP2(X) , and so

t/Jl

=

t/J2'

Chapter VI

190

c) % is invertible in C(x(A)) if and only if %(1/1) 1= 0, for all 1/1 in X(A). This is equivalent to 0 fI. 0' (x) , by Theorem 3.1, or to x invertible in A. d) This computation was made during the proof of Proposition 2.2, b). e) The mapping x --+ is injective if and only if Rad(A) = {O}.

x

We now apply this theory to the subalgebra of £(H) generated by a single operator. Let T be an operator on a Hilbert space H. Let's consider M, the largest commutative subset of £(H) containing T. The existence of M is ensured by Zorn's axiom. Of course, M contains I ,T, T2 ... , and R(A, T) for A E p(T). Moreover, M is an algebra: if AI, A 2 are in M, so are Al + A 2 , A IA 2 , and M is closed in operator norm. IT an operator A is in M, it commutes with every M in M. IT A is invertible in £(H), its inverse also commutes with every M in M, which proves that A-I is also in M. This shows that for every A in M, O'(A) is the same in £(H) and in M. If T is a normal operator, .M also contains T+ .

We may apply Gelfand's Theory to .M, which is a commutative Banach algebra, and we obtain:

Proposition 3.5. - We have:

O'(T) = {T(1/1)

j

1/1 E x(M)}.

In other words, for every A E X(.M), there is a character 'r/J of the algebra

.M such that : 1/1(T)

=

A.

The proof is an obvious consequence of Theorem 3.1 and of the fact that:

O'M(T) = O'(T). This Proposition will be used later (Chapter XIII) in order to show that some functions of an operator are not O. Indeed, if f(T) E .M,

1/1(f(T)) = f(t/J(T)) = f(A), and if we can choose A such that f(A) =I- 0, we obtain that f(T) is non-zero.

Banach Algebras

191

4. Functional Calculus. Let A be a commutative Banach algebra with unit. Let x E A, and 1 (x) be the set of functions which are analytic on o(x). For f E 1(x) , we define:

f(x) =

~ f 2,,1'"

l-

f(A)(A - x)-ldA,

where r is a curve around o(x) contained in the set of analyticity of f, and positively oriented. The properties of this functional calculus are the same as for the functional calculus described in Part I, Chap. II, 2 ; one gets, for instance:

f(o(x)) = o(f(x)). 5. C*-algebras. A mapping x --+ x" from A into itself, is an involution if, for every x, y in A, every A in «:,

(x*)* = x,

(x + y)* = z"

+ v' ,

(AX)* = Xx*, (xy)*

=

y*x·.

A C· -algebra is an algebra equipped with an involution, such that:

(1) We observe that 1· = 1.

Examples. - £(H) and K (H) are non-commutative C· -algebras, the involution being T --+ T* , the adjoint of T. - H X is a compact topological space, the space C(X) of continuous functions on X is a commutative C· -algebra, the involution being f --+ 7. In a C* -algebra, an element x is said to be normal if xx* = x" x, self-

adjoint if z"

= x.

Proposition 5.1. - Let A be a commutative C· -algebra. Then:

a) IIx*1I

=

Ilxll, for all

x E A,

b) x* is invertible if and only if x is,

d) If x is normal,

IIxl12 = Ilx 211,

e) u(x*) = conj(o(x)) ,

r(x) =

Ilxll,

Chapter VI

lSI!

where conj(n(x)) denotes {X; A E n(x)}.

Proof. - a) We have, by (1) :

and therefore

IIxll ::; Ilx*II, and the converse inequality follows,

replacing x by

z" . b) If x is invertible, xx- 1 = 1, so (x-1)*x* = l.

IIx 211 2 = II(x 2)*x211 = II(x*px211 II(x*x)*(x*x)11 = IIx*x11 = IIxll 4 , and so c) IT x is normal,

2

for n = 2 k , k = 1,2, ... , which implies Ilxnll 1/ n = Ilxll, and r(x) = d) and e) have the same proof as in Part I, Chapter II.

Ilxll.

Thus, by c), we have a new proof of Theorem 3.2, Chapter IV : a normal compact operator has an eigenvalue A with IAI = IITII. Indeed, we see here that for a normal operator, r(T) = IITII, so there is a point A E n(T) , with IAI = 11TH. But for a compact operator, such a A must be an eigenvalue, by Theorem 2.4, Chapter IV. H A, Bare C* -algebras, an homomorphism 4J is a "
(2)

Theorem 5.2 (Gelfand-Naimark). - Let A be a commutative C* -algebra. The mapping x --+ i is an *-isometry From A onto C(x(A)). Proof. - We have seen that r(x) = Ilxll, for every normal z , therefore for every x in a commutative algebra. In such an algebra, by Theorem 3.1, r(x) = lIillC(x(A». Therefore, the application x --+ x is an isometry. We now show that for any character 'f/J, 'f/J(x+) = 'f/J(x). We need a lemma: Lemma 5.3. - IF x is selF-adjoint, n(x) eIR. Proof of Lemma 5.3. - We first observe that for x self-adjoint, for all yEA, all A E a:: ,

11('\ - x)YII = II(X - x)yll,

(3)

Banach Algebras

199

Indeed,

11(.\ - x)y11 2

x)yll

=

11((.\ - x)yt(.\ -

=

Ily·(X -

x)(.\ - x)yll

= lIy·(.\ -

x)(X - x)yJl

=

II((X - x)y)· (X - x)yll

= II(X- x)yI12. Now, if .\ E 8o(x) , there is a sequence (Zn)n~O, Ilznll = 1, such that (.\x)zn ~ 0, which implies (X - x)zn ~ 0, and .\ = X. For x E A, we put u = (x+x·)/2, v = (x-x·)/2i. Then x = u+iv, z" =u-iv,and u, v are self-adjoint. So tP(u) , tP(v) are real, by Proposition 5.1, e). Therefore,

f/J(x)

= f/J(u) + if/J(v) = f/J(u) - if/J(v)

= f/J(x·).

Finally, we have to show that Gelfand's morphism is surjective. We have seen (Theorem 3.4) that A is a subalgebra of C(x(A)) which separates the points in A, is stable by taking conjugates, and contains the constant functions. Therefore, by Stone-Weierstrass Theorem, A is dense in C(x(A)). But J. is closed, being the image of an isometry, as we just showed. Thus A = C(X(A)) , and Theorem 5.2 is proved.

It can be shown that any C· -algebra (non necessarily commutative) is *isometric to some closed C· -subalgebra of £(H) (see for instance S. Sakai [1]). Theorem 5.4. - Let A be a commutative C· -algebra. We assume that A is generated by a single element a and its adjoint a· : this means that the finite sums 1: cxk"aka·' are dense in A. Then

a is an

homeomorphism from X(A) onto o(a).

Proof. - We know that a is continuous from X(A) onto o(a). If tPl, tP2 E X(A) satisfy ti(1/JI) = a(tP2), then tPda) = tP2(a), tPI(a·) = tP2(a·), and tPl, tP2 coincide on the whole algebra. This proves that a is injective. We know it is surjective, by Theorem 3.1. Since x(A) is compact, a is an homeomorphism. Proposition 5.5. Let A be a C· -algebra, B be a sub-C" -algebra. An element x E B is invertible in B if and only if it is invertible in A' Consequently, for x E B, 0 A (x) = 0 B (x) . Proof. - An element which is invertible in B is also invertible in A. Conversely, if x E B and X-I exists in A, z" is invertible in A, so x·x is invertible in A.

Chapter VI

19.4

But O'A(X·X) eIR, since x·x is self-adjoint. So, by Corollary 1.6, O'A(X·X) C Us (x·x). Therefore 0 fI. O's(x·x), (X·X)-l E B. But 1 = (X·X)-lx·x, X-I =

(x·X)-lx· E B. If one takes A- x instead of z , one gets

0'

A(x)

= Us (x), as we announced.

6. An Invariant Subspace Theorem.

In this Section, we show how the Banach Algebra techniques can be applied to Operator Theory, in order to find Invariant Subspaces.

Theorem 6.1. - Let E be a Banach space, T an operator on it, with:

IITII Then

~

1/2,

liT - III

~

1/2,

IIT2

-

Til

< 1/4.

T has Hyperinvariant Subspaces.

Remark. - The assumptions are given in their simplest form, which is not the optimal one. They essentially mean that the product T(T - I) has a norm which is substantially smaller than the product of norms IITII·/IT - III. The practical advantage of such an assumption is obvious: it is easy to check, much easier than any assumption on the spectrum, the resolvent, or the rate of growth of the iterates. Proof. - We consider the series : I +

L

n~l

(- 1)n . -1. 1 (- - 1) .. ' (1- -

~-

n!

2

2

2

n

+ 1)( -4T 2 + 4T )n .

This series converges in operator norm. To see this, we observe that, on one hand,

I -

4T + 4TII < 1, 2

and, on the other hand, that the coefficients have modulus < 1. Indeed, the product! . (! - 1) ... n + 1) is made of n terms, all of them < 0 except the first one. This product is of the sign of (-1) n-l. So, for n ~ 1,

(! -

I

(-I)n .!.(~-l) n!

2

2

... (~-n+l)1 2

=

(_l)n-l .~'(~-I) n!

2

~ . 2 - 3/2

2

and all terms are

< 1.

2

... (!-n+l) 2

2

k - 3/2 k

n - 3/2

n

Banach Algebras

195

Since the series converges in operator norm, let U be its sum. Of course, U commutes with T. Then :

L

U- I =

(-:l" .~.

(~

+ 1)(-4T 2 + 4T)",

(1)

- n + 1)11- 4T 2 + 4TII".

(2)

(~-

-1)'"

n

"~l

IIU -

III -< ~ L-

.!. (! 2 2

(-1)" n!

1) ... (!

2

"~l

But elementary computations (using Taylor coefficients) show that for 0 1,

vI=t

=

1 + ~ (-1)" . ! L- n! 2

.(! 2

1) ... (!

2

- n + l)t"

'


"~l

so:

L

(-1)" 1 1 1 - - 1)··· (- - n n! 2 2 2

~--(

+ l)t "

l-vI=t <

1.

"~l

With t =

II -

4T2

+ 4TII,

we get by (1) :

IIU - III <

(4)

1.

But, by formulae (1) and (3) :

U2

=

I - (-4T 2 + 4T)

which implies:

(U - 1+ 2T)(U + I - 2T) = O. We observe now that the operator (U U

+ I - 2T) is not zero. Indeed, if

+ 1- 2T

= 0,

then:

U - I = 2(T - I). But this is impossible, since liT - III ~ 1/2, and So there is a point x =j:. 0 such that :

(U - I

+ 2T)x

=

IIU - III < l.

O.

Set Xl = (U - 1+ 2T)x, and consider F = K er (U - 1+ 2T). This is a closed linear subspace of E, invariant by any operator which commutes with T, and F =j:. O. Moreover, F :f E, since U - I + 2T is not identically zero. Indeed,

IIU - III < 1, and 112TII

~

1/2.

This proves the Theorem.

196

Chapter VI

Remark. - There are obvious examples of operators satisfying our assumptions : first, any projection P with IIPI1 ~ 1/2 and liP - III ~ 1/2 (but a projection has obvious invariant subspaces, anyway), also any operator of the form P + W, where P is an orthogonal projection on a Hilbert space, and IIWII < 1/12. In this construction, Banach Algebra techniques come the following way: they allow us to exhibit U as (formally) I - J I - T.

Banach Algebras

197

Exercises on Chapter VI.

Exerc ise 1. - a) Let K(H) be the space of compact operators on H , and I be a closed ideal of K(H). Show that I contains a rank-one projection; deduce that I = K(H). b) Assume now that I is a closed ideal, but not contained in K(H). Show that I contains the projection onto some infinite dimensional closed subspace. Deduce that I = £(H). c) What are the closed ideals of £(H)? What are the characters in £(H)? (see also Exercise 10 below.)

- Let E = {I E C([O, 1]), I' E C([O,l])}, with the norm 11II1 = 11/1100 + 111'1100' Show that E is a Banachalgebra, for the usual multiplication of functions, and that Gelfand's transformation is neither isometric, nor surjective. Show that the closed ideals in E are not the intersection of maximal ideals containing them. Exercise 2.

Exercise 3. - Let E be the space of functions I on [0, i}, of class the norm:

11111 =

sup

i: 1/(~,(t)

tElo,l] k=O

en,

with

I. .

Show that E is a Banach algebra for the usual operations. Find the maximal ideals.

Exercise 4. - a) Let 1t{7L) = {(a n )nE7L ;

if a

= (a n )nE7L '

L lanl < co}, with the norm:

We define the multiplication by a.b

= c , with: (1)

Show that the set of maximal ideals is homeomorphic to the Torus II. b) Let:

+00

A(Il)

{/=Lc]'e ii 8

-00

;

Llcil
Chapter VI

198

with the norm:

IlfllA

=

L leil·

Show that A(Il) is isometric to I.(Z)' and that it is an algebra for pointwise multiplication of functions. Deduce from a) that the spectrum x(A(n)) is Il , and that the Gelfand's transform of each function f is f itself. Show that u(J) is the image of f. Deduce that if f E A(Il), and f((J) i 0 for all (J, then 1/ f is also in A(n) (Wiener-Levy Theorem).

Exercise 5. - Show that if A is a commutative Banach algebra, the function x --+ r(x) (spectral radius of x) is continuous. Exercise 6. - In 11 (IN) , we define the multiplication by a.b = c, with

a) Show that 11 is a commutative Banach algebra with no unit. b) Show that there is a bijective correspondence between the closed ideals and the subsets of IN. c) Show that there is a bijective correspondence between the maximal ideals and the integers. d) Characterize the subalgebras of II .

Exercise 7. - Let A = co(IN) , with the usual multiplication of sequences, and let I be the set of sequences with only a finite number of non-zero terms. Show that I is an ideal, dense in A, and that I is contained in no maximal ideal. Exercise 8. - Let A = lp( IN), 1 :::; p < 00, equipped with the usual termwise product. a) Show that if a, b are in A, a.b E I p / 2 • Deduce that A is a Banach algebra, and that A2 is an ideal of A, dense in A. b) Show that A2 is contained in a maximal ideal. c) Describe the Gelfand's transformation, and show that it is injective, that it has dense range but that this range is not closed. Exercise 9. - Let A = L 1 ([0,1]) , with the multiplication:

(f.g)(x) =

fax f(x -

t)g(t)dt.

199

Banach Algebras

a) Show that .A is a commutative Banach algebra. b) Let w be the constant function: w(x) = 1, for all x E [0,1]. Compute the n-th power w n of w in .A, deduce that A is the closed algebra generated by w, that the spectral radius of w is 0, and that 0 is the only character on .A. Exercise 10. - The only closed ideal of £(H) is K (H) (this exercise uses some material from Chapter VII). a) Let t/J be character on £(H). Show that for any projection P, t/J(P)

is

°or

1.

b) Show that for any projection of finite rank, t/J(P) = 0, and for any projection of infinite rank, t/J(P) = 1.

Jo+ oo

c) Let A be a normal positive operator, and A = >"dE). be its spectral decomposition (see Chapter VII). Then, for any >"0 > 0,

( A

~

B means that A - B is a positive operator.)

Show that:

d) Assume that 1/J(A) = 0. Deduce that 1/J(E(>..o)) = 0, so E(>..o) has finite rank. Deduce that, for every >"0,

where

IIB).o II < >"0

and C).o has finite rank. Deduce that A is compact.

e) Let A be any operator with 1/J(A) = 0. Writing A = UIAI (Chapter IV, Lemma 3.5), deduce that A is compact. f) Deduce that K ert/J = K (H) . Exercise 11. - Let T be an operator on a Banach space E. For>" E p(T), we denote by A(>") the algebra generated by I, T, (>.. - T)~l , closed in £(E). a) Let>.. E p(T). Show that for any J.L close enough to >.., A(>" + J.L) c A(>"). b) Show that A(>"

+ J.L)

= A(>") (consider the segment between>"

+ J.L

and

>") . c) Deduce that A(>") is constant in each connected component of p(T).

Chapter VI

d) Consider the right shift on 12 (71) . Show that A(O) is isometric to IAI > 1, is isometric to H?" (see Chapter VIII).

Loo{IT) and that A(A), for

Notes and Comments :

For further developments, the reader may consult for instance Sakai [1] or Kadison-Ringrose [11. Many examples of applications of Banach Algebra techniques to Operator Theory can be found in the excellent book by R. G. Douglas [1]. The Theorem in Section 6 is due to the author (unpublished). However, it is a mere rephrasing (in the language of Operator Theory and Invariant Subspaces) of an idea of J. Esterle [2]. Exercises 1 to 9 were indicated to us by G. Bourdaud. Exercise 10 gives a simple proof of the fact that £(ll) has no other closed ideal than K(H). It was communicated to us by Thierry Coulhon and Yves Raynaud.

Chapter VII

Normal Operators

1. Algebra generated by a normal operator.

Let H be an infinite dimensional complex Hilbert space, T a normal operator on it : T and T· commute. We call AT,T. the closed subalgebra of l(H) generated by T and T+ : it is the closure, for the operator norm, of all finite sums E ak,lTkT· l . This algebra is a commutative C+ -algebra, which we call C· -alqebra generated by T. We also recall that, for a normal operator T, r(T) = IITII (Chapter VI, Proposition 5.1).

Theorem 1.1. - Let T be a normal operator. The Gelfand's morphism is an *- isometry from AT,T. onto C(o(T)). The image of the operator T by this application is the identity function f(A) = A, A E o(T). If f) : C(o(T)) ----+ AT,T" is the inverse of this morphism, of the analytic functional calculus (defined for f E T(T)).

q)

is an extension

Proof. - Applying Theorem VI.5.2 (Gelfand-Naimark), we see that Gelfand's morphism G is an *-isometry from AT,T· onto C(X(AT,T")) : if p(x, y) is a polynomial in two variables,

G :

----+

{4> E X(AT,T" )

----+

p(4J(T), 4J(T+ ))},

with 4>(T+) = 4>(T) . By Proposition VI.5.5, the spectrum of T is the same in £(H) and AT,T" . By Theorem VI.5.4, t is an homeomorphism from X(AT,T·) onto u(T). Using this homeomorphism, we can identify these two sets, namely replace :

by:

{A E o(T)

----+

p(A,I)},

and we know that this last set is dense in C(o(T)). Considering G on the whole algebra AT,T- (and not only on polynomials), we get an *-isometry : G : AT,T.

----+

C(o(T)).

(1)

Chapter VII The image of T is the function {¢ ---+ ¢(T)} , which has been replaced, in our identification, by the function {-X ---+ -X}, -X E u(T). Let ~ be the inverse mapping of G, from C(u(T)) onto AT,T. : to each I E C(u(T)) we associate an operator I(T) in the algebra AT,T· . Of course, if I = p is a polynomial, we get the usual p(T). Now, if I is analytic on u(T), we write:

I(T) =

~

( f(-X)(-X - T)-ld-X. 2~1r Jr By Proposition VI.5.5, we know that if -X E p(T), (-X - T)-l E AT,T•. Therefore, the whole integral is in AT,T•. Finally, since ~ = c:», we get:

ITT)

=

I

I,

E C(u(T)),

and

(f(T)r = J(T*), by the *-properties of G. This new functional calculus is much more general than the one we defined in Chapter II : f needs only to be continuous on u(T), instead of being analytic on a neighborhood of u(T). We immediately give an application of this new functional calculus : Proposition 1.2. - A positive self-adjoint operator T has a square root: there is a positive self-adjoint operator U satisfying U 2 = T. Proof. - We know that since T is self-adjoint, u(T) cIR (Proposition V.5.3). Since T is positive, a(T) cIR+ (since every -X E u(T) is in aa(T) , there is a normalized sequence (xn)n~O such that TX n - -Xx n ---+ 0, so (Tx n, x n) ---+ -X, and the numbers (Txn,x n) are positive). We consider the function f : -X ---+ -X 1/4 on u(T) : it is continuous on IR+. Set Y = I(T). Then y* = f(T)* = J(T*) = f(T), since f is real and T self-adjoint. So Y is self-adjoint. Set U = v>. Then U is self-adjoint, and positive, since:

(UX,X) Finally, U2

= y 4 = T.

=

(y 2x , x )

=

Indeed, set g(-X)

(Yx,Yx)

= -X 4 •

= IIYxI1 2 •

Then go f(-X)

= -X

on u(T), so

g(Y) = T. The formulas proved in Chapter II, Section 2 (Proposition I, 2, Theorem 3, Proposition 4) are still valid here. But there is no need to discuss them now, since we are going to extend this new functional calculus further.

Normal Operators

2. Spectral Measures. We begin with some general considerations about vector valued integration ; in the next Section, we will apply them to the case of normal operators. Let Ii. be a locally compact space. A spectral measure E is a function defined on the Borel subsets of Ii. , with values in £(H) satisfying: a) for every Borel subset 6 b) E(0)

C

ti., E(6) is a self-adjoint projection in H,

= 0, E(ti.) = I,

c) for all z , y in E, if we define E z ,1I by :

E Z'1I(6) = (E(6)x, y), we obtain a regular measure on ti., equipped with the Borel a- algebra. (We recall that a measure JJ. is said to be regular if for all Borel subsets:

J.L(A)

inf{JJ.(O) ; 0 open containing A} sup{JJ.(K) ; Kcompact contained inA}.)

d) IT 61 , 62 are two Borel subsets of Ii. , with 61 n 62

= 0, then:

e)

This last property implies :

E(6) = E(6)2

E(6t E(6).

So:

E z,z(6)

=

(E(6)x,x)

=

IIE(6)xfl2,

and for every x in H, Ez,z is therefore a positive measure on Ii. , of total mass:

Chapter VII 3. Integration with respect to a spectral measure. We now develop an integration theory with respect to a spectral measure. By many respects, it is analogous to the usual integration theory. We first consider step-junctions : n

f =

L

Ci 16. ,

i:::::l

where Cl, ••• , Cn are complex numbers, 61 , ••• ,6,., Borel subsets of ~, and 16 is the characteristic function of 6 : it has value 1 in 6, 0 outside. Then, by definition, the integral of f with respect to E will be the operator:

/ IdE

= /

tCil"dE

=

tCiElii.

i:::::l

i:::::l

For z , y in H, we get : n

((/ fdE)x, y)

L ci(E(6i)X, y) I::::: 1

,.,

L Ci Ez,y(6i) 1:::::1

/ fdEz,y. Therefore,

1((/ fdE)x,x)1 And thus, sup IIzll~1

1((/ fdE)x,x)1 ~

sup If(t)l· t

But:

and so sup

1((/ IfldE)x, y) I < sup If(t) I, t

IIzll~I,lIyll~l

which means that:

II/ IfldEl1 ~ s~p If(t)l·

(2)

Normal Operators Applying (1) to

I

instead of

III, we get also:

11/ IdEIl

~ s~p

(3)

I/(t)l·

We call 800 (.6.) the algebra of bounded Borel functions on .6., equipped with the norm

11/1100 = Step functions are dense in BOO(~). of fIdE to functions I E Boo (~),

sup tEA

I/(t)l·

Therefore, we may extend the definition by posing :

/ IdE =

li~

IndE,

/

where (In)n~o is a sequence of step-functions, converging to I in BOO(~). This definition does not depend on the sequence (/n)n~o, This way, we get (2) and (3) for IE BOO(.6.) , and

((/ldE)X,y)

=

IdEx,y,

/

I

E

BOO(~),

x,y E

H.

4. Spectral Representation of a Normal Operator. We now use spectral measures to extend the functional calculus exhibited in Section 1. We first come back to the general setting of a commutative C·algebra, and then we apply the result to the algebra generated by a single normal operator.

Theorem 4.1. - Let if be a commutative C· -algebra contained in £(H), and let ~ = X(A). There exists a unique spectral measure E on ~ such that:

/A iue ,

U =

'rf U E A.

Moreover: a) The *-isometry

fI

--+

fA ius, from

an *-homomorphism :

I

--+

i

onto A can be extended as

C(~)

IdE,

from BOO(.6.) into £(H), with:

II

i

IdEl1

~ 11/1100,

I

E

BOO(~)

Chapter VII b) For every Borel subset fJ c

~,

every U in A, U and E(fJ) commute.

c) An operator V in £(H) commutes with all U in A if and only i6t commutes with all the projections E(fJ) , fJ E ~. Proof. - We know (Theorem 5.2, Chap. VI) that the application U ~ U is an *-isometry from C(~) onto A. Therefore, for all x, y in H, the inequality :

I(Ux,y)1 <

IIUII·llxll·llyll

lIUlloo·llxl!·!Iyll

=

shows that the mapping :

U

~

(Ux,y)

is a continuous linear functional on C(~), space of the continuous functions on the compact set ~, with norm::; Ilxll.llyll. By F. Riesz Representation Theorem (see Rudin [1]), there exists a regular Borel measure on ~, J.Lx,y, such that:

(Ux,y) = /Ud J1 X,y,

(1)

VUEA,

and

(2) For every

Q

E
/ UdJ1(u,y

= (U(Qx),y) = Q(Ux,y) =

But since the mapping U ~ in Boo(~),

U is surjective, we

/ /dJ1(u,y = which implies that J1a:r;,y Fix now / E Boo (~).

=

Q /

Q /

UdJ1x,1I .

get that for every function /

/dJ1x,y ,

QJ1:r;,y. The same way, we show that J1:r;,ay

= iiJ.L:r;,II'

The application :

(x, y)

~

/ /dJ.L:r;,1I

is linear in z , antilinear in y, and continuous, since:

1/ /dJ.L:r;.!l1 < 11/lloo·lIxll·llyll· Therefore, there exists an operator U/ in l (H) such that :

(U/x,y) = / /dJ1:r;,y

x,y E H.

This makes sense for every / E Boo (~). But if / E C(.6.), / = U for some U E A, and therefore U/ = U. This shows that the mapping f ~ U/ is an extension of . , inverse of Gelfand's morphism, from C(.6.) onto A. Let's call it

i.

Clearly • is linear, and we now show that it is multiplicative.

Normal Operators Let VI, V in A. Then

Vl V = @ , and so, for

I Vl VdILz,y = and consequently, for any

I

(VI V x, y)

x, y in H,

= I Vl dILv Z,JI

,

E C(~),

I IV dILz,JI

I I dILv z,y

=

(3)

•

This implies that the two regular Borel measures V dIJ-z,y and dIJ-Vz,y are equal, and so equation (3) holds also for every I E BOO(~). For y E H, I E BOO(~), we put z = Uj(y). Then, for any V in A,

II VdIJ-Z'71 = IldILVZ'71 = = (Vx,z) and this implies, for

I, 9

in

=

(UfVx,y)

I VdIJ-z,z ,

BOO(~),

I I gdILz,y

=

I gdIJ-z,z .

Therefore, (Ugx,z)

(Ugx,Ujy)

and finally,

(4) which shows that

i

preserves multiplication: it is an algebra homomorphism.

We now show that it preserves involution. First, we take U E A, such that fJ takes its values inIR, Then U is self-adjoint, since Gelfand's morphism preserves involution, Thus,

I fJ

dIJ-z,7I =

I fJ

(Uz, y)

dfiy,z ,

which shows that : IJ-z,y

Now, for

I

E BOO(~),

fiy,z .

(5)

Chapter VII

which shows that

U·J

(6)

Finally, we have:

by formula (2), and so :

(7) We now define the spectral measure announced in the statement of the Theorem. For every Borel subset 0 C ~, we set :

Since 16 takes only real values, U 1" is self-adjoint (formula (6)). Since 1~ = 16, E(0)2 = E(6) (formula (4)), and E(6) is a projection. If 0 = 0,16 = 0, and E(6) = O. If 6 =~, 16 = 1, and E(6) = I. Ifx,yEH,

Ez,y(o)

=

(E(6)x, y)

=

115d#LZ,y = #Lz,y(O),

which shows that Ez,y is a regular Borel measure on

~.

If 61 , 62 are two Borel subsets of 6., 151n62 = 151.152, which implies:

If 01, 02 are disjoint, 151u6 2 = 151 + 152, and therefore:

Therefore E is a spectral measure on 6.. If

I

E

Boo (6.), x, y E H , then : ((lldE)X,y)

= 1 IdEz,y = Ild#LZ,y =

Consequently, formulae (4), (6), (7) show that I ~ preserves involution, and is continuous.

(UJx,y).

J IdE

is multiplicative,

Normal Operators b) is of course a special case of c). Let V be an operator which commutes with all U in A. Then V commutes with all E(e5). Conversely, if V commutes with all E(e5) , we get, for any z , y E H :

I with z

= V·y,

UdEvz,JI

=

I

(UVxty)

=

(VUx,y).

UdEz". ,

and so :

Finally, the spectral measure is obviously unique. If E' was another one, with U

then:

I and so EZ,JI

=

E~,JI

t

UdEz"JI = A

I iJdE~'JI

=

t

I'"

UdEz,JI '

for all iJ,

for all x t Y E H. This finishes the proof of Theorem 4.1.

We now come back to the setting of Section 1. T is a normal operator, A = AT,T- is the algebra generated by T and T·, A = u(T) = X(AT,T- ), and t is the identity function. Therefore, Theorem 4.1 gives now:

Theorem 4.2. - Let T be a normal operator. There exists a unique spectral measure E on u(T) , such that:

f.

T =

>"dE.

u(T)

With this measure, we define, for I E BOO (u(T)) ,

I(T)

f.

=

IdE.

u(T)

This application extends the functional celculus, previously defined for I E l(T) , and then for I E C(u(T)). We now list some properties of this functional calculus:

Theorem 4.3. - Let T be a normal operator. We have:

a) If 11,12 E BOO(u(T)) , a,f3 E

~,

(all + f3h)(T) = ah(T) + f3h(T),

150

Chapter VII

b) (/(T))*

!(T*), for IE BOO (u(T)) ,

II/(T)II < 11/1100' for I E Boo(u(T))' d) 11/(T)xI1 2 = JU(T) 1J1 2 dE z ,z , x E H, IE Boo(u(T)) ,

c)

e) If U commutes with T and T* , it commutes with all operators I(T),

I E Boo(u(T)) , f) If In, I E 8 OO (u (T )), and if In converges to I uniformly on u(T)' then In(T) -+ I(T), in operator norm. If In(x) converges to I(x), Vx E u(T) , then In(T) ---+ I(T) for the strong operator topology. Proof. - a) to e) are just rephrasing of Theorem 4.2, f) follows from c) and

d). Remark. - One can show that if U commutes with T, it commutes with T· (Fuglede's Theorem, see Douglas [1] and Exercise 7). The spectral measure E is called a resolution of Identity. Indeed,

I = E(o(T)) =

fIdE. JU(T)

We have seen that the mapping I ---+ I(T) transformed characteristic functions into projections. Since step functions are dense in 8 OO (u (T )), we see that every normal operator is the limit (in operator norm), of a sequence of linear combinations of projections.

L2 •

The simplest example of normal operator is a multiplication operator on Fix a function 4> in L oo and consider the mapping: M~:

Then M; = Mj"

and

M~

I

---+

¢f.

is normal. We will show that, conversely, every

normal operator can be represented this way. We say that a point z E H is cyclic for (T, T·) if the set {U z; U E is dense in H. This means also that the set of finite combinations is dense in H.

L

AT,T*}

ak,ITkT· 1z

Theorem 4.4. - Let T be a normal operator. We assume that there is a point z E H which is cyclic for (T, T·). Then, there is a positive regular Borel measure J.L on u(T), and a unitary transformation A from H onto L 2 (u (T ), J,L ), such that for every g E L 2 (o(T ), J.L),

A E o(T).

Normal Operators

151

If 1 E BOO(u(T))' and if I(T) is the previously defined operator, then:

where MJ is the multiplication by

1

on L 2 (u(T ),Il ) .

Proof. - Let E be the spectral measure given by Theorem 4.2. Put Il Then, for 1 E BOO(u(T)),

(/(T)z, z) = Let H o = {/(T)z ;

Ho

=:)

1 E 8°O(u(T))}.

1

u(T)

=

Ez,z'

Idll.

Then:

{/(T)z; 1 E C(u(T))} = {Uz; U E AT,T·},

and since z is cyclic for (T,T·), H o is dense in H. We define A o , from H o into L 2 (u(T ),Il ) , by :

Ao(J(T)z) = 1 , If I(T)z

= g(T)z, then: o

=

11(1 -

g)(T)zI12 =

1

u(T)

If - gl2dEz ,z ,

which shows that A o is well-defined. It is an isometry, which can be extended into an isometry A from H into L 2 (u(T ),Il )' The image is the closure of BOO(u(T)) in L2 (u(T), Ill, which is L 2 (u(T ),Il) : the operator A is surjective. If g E B00 (u(T)) ,

Aof(T)g(T)z

Ao(Jg)(T)z

fg ,

which shows that:

and:

and our Theorem is proved. If there is no vector cyclic for (T, T·), one may split the space H into pieces in which there are such vectors, and one gets:

lSf

Chapter VII

Theorem 4.5. - Let T be a normal operator. There is a locally compact metric space X, with a Borel a -algebra T, a regular positive measure IL on T, a bounded continuous function h from X into (C, a unitary transformation A, from H onto L 2(X, T ,IL), such that:

We won't use this result in the sequel, and we refer the reader to DunfordSchwartz [I} for its proof. We now turn to the study of a simple example:

Example. - The operator M of multiplication by ei 8 on L 2 (TI, dO/ 21r ) (we just write M instead of M e i8).

Proposition 4.6. - The spectral measure E oE M is defined by :

where MIll is the multiplication by 16, characteristic function of 6, Borel subset oE TI, in the space L 2(TI,dO/21r).

Proof. - First, this operator is the same as the right shift on 12(71) , if we identify a function in L 2 (TI, dO /211") with its Fourier series. Therefore, its spectrum is the unit circle (see Chapter II). Now, let 9 E L oo (TI , dO/ 21r ) . The operator g(M) is the operator of multiplication by g, on L 2 (TI, dO /21r). Therefore: Mg = and if we take 9

= 16,

!

gdE,

we get the Proposition.

The scalar spectral measures E f,f are given by :

and more generally, for 9 E Boo (IT),

which gives : E

f,f

=

1112

dO . 21r

The invariant subspaces of this operator will be studied in Chapter IX.

Normal Operators

159

Remark 4.7. - An operator T satisfying, for every x E H :

IITxl1 = liT· xII , is normal.

Proof. - We write:

IIT*x1l2 = (x,TT*x) ,

and, with B, = T*T, B 2 = TT* , we get:

But B l

-

B 2 is self-adjoint, and:

IIB l

-

B 2 11 =

sup

112:11=1

1{(Bl

-

B 2 )x,x)1

0.

This remark applies to a surjective isometry, for instance the right shift on

l2(7.1). But it does not apply to an isometry which is not surjective: for instance, the right shift on l2(IN) is not a normal operator. 5. The spectrum of a multiplication operator. Let q, be a function in L oo ([0, 1], dt), and let M rfJ be the operator of multiplication by q, on L 2([0, I],dt). As we already said, this is a normal operator. We now describe its spectrum. Assume first q, to be continuous. Then A - MrfJ is invertible if and only if A - q, is an invertible function in L oo ([0,1], dt) , which means that A - r:/> does not take the value 0. So we get immediately that o(Mr/Jo) = q,([0, 1]). If q, is in Loo([O,I],dt) but not necessarily continuous, things are not so easy. Indeed, q, is only defined a.e., though l1(MrfJ) is a well-defined set. So, we introduce a new notion, the essential range of r:/>, denoted by RrfJ :

RrfJ

=

{A E e

j

W, neighborhood of A, p(q,-l (V)) > O},

where P is the Lebesgue measure on [0,1]. Then we get:

154

Chapter VII

Theorem 5.1. - u(M¢)

=

R¢.

Proof. - 1) Let A E R¢. Let Vn be the disk, centered at A, with radius lin, and let Un = ¢-I(Vn). Choose functions In in L 2([0,1]), with II In II = 1, and In = on U~ : this is possible since P(U n ) > 0. Then:

°

II(A - ¢)/nl ~

= / IA - ¢(t)121In(t)1 2dt =

1 IA -

¢(t) 121In(t)1 2dt

Un

~ supesstEunlA - ¢(t) 12 < I/n 2

,

by the choice of Un. So A - ¢ is invertible.

2) Conversely, let A f/; R¢. There exists a neighborhood V of A which satisfies P(¢-I(V)) = 0. Therefore, there exists an e > such that IIA¢lloo ~ e. So 11 (A - ¢) is in L oo , and is the required inverse for A - M¢.

°

6. Hyperinvariant Subspaces for Normal Operators. Theorem 4.2 gives obviously that every normal operator has Hyperinvariant Subspaces. Indeed, u(T) is not reduced to a single point: if u(T) = {a}, C(u(T)) is (C, so AT,T. is (C, which implies H = (C , contradicting our original assumption according to which H was infinite-dimensional (another argument is : if u(T) = {a}, r(T - a) = 0, so liT - all = 0). Therefore, we can find two disjoint Borel subsets 61 and 62 , with:

Put PI = E(c5d, P2 = E(c52) . Then PI, P2 are two projections, and F I = K er PI is the required Hyperinvariant Subspace. It is hyper invariant by Theorem 4.3, e). We observe, however, that the restriction of a normal operator to an Invariant Subspace is not a normal operator in general (we have already met the easy example of the right shift on 12(IN), restriction to 12(IN) of the right shift on 12(1L)). The reason is that the adjoint of TIF is not T*IF in general, unless F is itself invariant by T* : indeed, this adjoint is P(T*IF), where P is the orthogonal projection onto F. Such an operator: restriction of a normal operator to one of its invariant subspaces, is called subnormal. The question of the existence of Invariant Subspaces for subnormal operators remained open for a long time. Though one

Normal Operators

155

may think that the techniques developed in this chapter give strong information on the operator, they are unsufficient to solve this problem, which was settled only recently (1978) by Scott Brown [1], who gave a positive answer. His approach does not rely at all on Spectral Measures j it will be presented (in a slighly different context), in Chapter XIII. The question of existence of Hyperinvariant Subspaces for subnormal operators remains open. Another class, which is also an extension of normality, has received a positive solution to the Invariant Subspace problem. We say that an operator T is hyponormal if the commutant T·T - TT· is a positive operator. This is of course the case of any normal or subnormal operator. Very recently (1987), S. Brown [2] has shown that every hyponormal operator has Invariant Subspaces.

156

Chapter VII

Exercises on Chapter VII.

Exercise 1. - Let T be a normal operator on H, and E be its spectral measure. Show that A is an eigenvalue of T if and only if E ({A}) -I- O. If .\ is an eigenvalue, show that E({.\}) is the orthogonal projection onto Ker (A-T). Exercise 2. - Let T be normal, and put ITI = VT*T. Let V be such that T = VITI (cf. Chapter IV, Lemma 3.5). Show that there exists ¢, Borel function on o(T) , such that V = ¢(T). Deduce that V and ITI commute. Exercise 3. - Define T, from L 2 l1R) into itself, by : T f(t)

= f(t + 1).

Show that T is normal and find its spectral decomposition.

Exercise 4. - Let A be a self-adjoint operator. Show that: for all t EIR.

Exercise 5. - Show that for a normal operator, o(T) = oa(T). Exercise 6. Let T be a normal operator. Show that T is unitary (resp. selfadjoint) (reap. positive) if and only if o(T) C C (resp. o(T) C IR), (resp. o(T) cIR+ ). Exercise 7. - (Fuglede's Theorem). Let T be a normal operator, and A commuting with T. a) We put U = (T + T*)/2, V = (T - T*)/2i. Show that U and V are self-adjoint and that, for any A E ~ ,

b) Deduce that:

c) Show that AT* - AT can be written as iW>., where W>. is self-adjoint. Show that:

Normal Operators

157

Ded uce that :

d) Consider the mapping:

A

-+

eAT- Ae- AT- .

Show that this is a bounded entire function, with values in £(H). Deduce from Liouville's Theorem that it is constant, and conclude that: eAT- A = Ae>.T-.

e) Deduce that T+ A

= AT+

(Fuglede's Theorem).

Notes and Comments.

Our presentation can be found in many sources. Further information can be found in Dunford-Schwartz [1], J. Conway [1], and many others. A detailed study of essential ranges and multiplication operators, as well as generalizations (Toeplitz operators) can be found in the book of R. G. Douglas [1]. For vector valued integration, the canonical reference is Diestel-Uhl [11. Exercises 1 to 4 come from the book by J. Conway [11. Exercise 7 (a short proof of Fuglede's Theorem) is due to M. Rosenblum and is reproduced from H.R. Dowson [11.

Complements on Chapter VII.

As it is written in the above pages, the previous theory of spectral decompositions for normal operators does not make sense in a Banach space, since T and T+ do not operate on the same space, and so we cannot wonder whether they commute or not. However, spectral decompositions do make sense, and it is a legitimate question to ask in which Banach spaces and for what operators they hold.

158

Chapter VII

A nice theory in this direction has been built recently by E. Berkson, T .A. Gillespie, P. Muhly [1], [2], for invertible operators with bounded iterates :

After equivalent renorming, such an operator becomes a surjective isometry. It's quite natural to investigate first surjective isometries, since they are normal in Hilbert spaces and make sense in Banach spaces. Moreover, one knows (Proposition 1.12, Chapter II), that the spectrum of an isometry is always contained in the unit circle. Let X be a Banach space. A spectral family of projections is an application

t E IR

--+

E (t) E

.c (X) ,

where E(t) is a projection from X into itself, satisfying the following properties :

a) SUPtaR IIE(t) II < 00, b) E(s)E(t)

= E(t)E(s) = E(s), if s :S t.

c) For each to in IR, E(t) is right continuous at t = to, for the strong operator topology, and has a left limit for this topology. This left limit is denoted by E(t d) E(t)

--+

o)' I, when

t --+ +00, and

E(t)

--+

0, when t

--+

-00.

In the sequel, the above family will be concentrated on [0,211"1, which means that: for t :S 0, E(t) = 0 ,

E(t)

=I ,

for t

~

211".

An operator U is said to be trigonometrically well bounded if it can be written:

u =

fG>

eilJdE(s),

J[O,2'1f]

where E(.) is a spectral family of projections concentrated on [0,211"]. Such an operator admits a BV -functional calculus, where BV is the algebra of functions with bounded variation on the Torus. If we denote by L~ a summation with n t- 0, their theorem says:

159

Normal Operators

Theorem (E. Berkson, T.A. Gillespie, P. Muhly). - Let X be a reflexive Banach space, U an operator on it with bounded iterates. The following properties are equivalent:

a) U is trigonometrically well bounded, b) sup II N,t

N

1

L' -en

.

m t

U"1I <

00,

-N

c) for every t E [0, 21r],

converges strongly to an operator B(t), and : sup IfB(t)11

<

00.

tE[O,2",j

They prove also that, under the previous assumptions, there is an operator A, with u(A) C [0,21r], such that U = ei A • The second result, very important in the theory, is as follows: if the space X is a D.M.D. space, every operator U with bounded iterates is trigonometrically well bounded. The D.M.D. spaces are defined by the fact that differences of martingales form an inconditional basic sequence in L 2(X) (see D. Burkholder [1]). There are other definitions, in terms of the Hilbert transform (see D. Burholder [1], J. Bourgain [1]). As examples of these spaces, one finds the classical L p spaces (1 < P < +(0). Therefore, one gets for surjective isometries on L p spaces a theory of spectral decomposition which extends that of unitary operators on Hilbert spaces. A point which should be emphasized is that the notion of "spectral family of projections" is weaker than that of "spectral measure" (additivity disappears). Therefore, it can handle more operators, and in particular some translation operators on Lp(G) (p t=- 2, G locally compact abelian group), which arise naturally in Harmonic Analysis and are not spectral in Dunford's sense (see H. Benzinger, E. Berkson, T.A. Gillespie [1]).

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PART IV

ANALYTIC FUNCTIONS

Chaque Ilot signale par I'homme de vigie Est un Eldorado promis par Ie Destin ; L'Imagination qui dresse son orgie Ne trouve qu'un recif aux clartes du matin.

The purpose of this Fourth Part is to give a complete description of the invariant subspaces of the right shift on 12(IN) and 12(Z). Quite obviously, this operator has Invariant Subspaces, but to describe all of them is not easy at all. We will consider the operator as acting on the Hardy space H2 or on L 2 (II) . The first one is a space of analytic functions, and the description of the Invariant Subspaces of our operator requires the Beurling decomposition of these functions into inner and outer parts. This decomposition, together with its corollaries: Jensen's Inequality, Szego's Theorem, is of course of independent interest.

Chapter VIII

Banach Spaces of Analytic Functions

First, we recall our notation: D is the open disk {A E
Let n be an open set in the plane. A function u , defined in in ill. , is said to be subharmonic if : a)

n, with values

-00 ::; u(z) < +00, for all zEn,

b) u is upper semi-continuous on n, that is : if X n ~ x, u(x) 2: limsuPn_oo u(x n ) . Equivalently, this property can be described by the fact that for every a EIR, the set {x ; u(x) < a} is open. c) If the closed disk D(a, r) is contained in n,

u(a) <

+71" /

dO

u(a + rei 8 ) -

211'"

-71"

,

(I)

d) No integral in (1) is ~oo.

remarks. - 1) If u is continuous, a), b), d) are satisfied. 2) Properties a) and b) imply that u is bounded from above on every compact Ken. Indeed, if we set K n = {z E K j u(z) ~ n}, K n is compact, and, by a), one of the Kn's must be empty. So, every integral in (I) is < +00. 3) The meaning of Condition c) is that the value of u at the center of a circle is smaller than the mean of the values on this circle. A function u is said to be harmonic if the equality holds in (1), for all a, r , such that D(a, r) en.

Lemma 1.1. - If u is subharmonic in onIR then 4> 0 u is subharmonic.

n

and 4> monotone, increasing, convex

Proof. - Since 4> is increasing, convex, it is continuous on IR. Therefore, 4> 0 u

Chapter VIII

16.1

is upper semi-continuous. Moreover, if D(a, r) c ~

0

n~

we get:

u(a)

since ~ is convex and dO /21r is a positive measure on [-1r, 1r 1, of total mass 1.

Proposition 1.2. - Let I be holomorphic in continuous in n, for 1 ~ p < 00.

n.

Then

I/IP

is upper semi-

Proof. - By virtue of Lemma 1.1, all we have to do is to show the result for p = 1. But:

f

+W"

dO

I/(a + rei 9)1-

21r

-W"

In the sequel, rRl and of the complex function I.

Proposition 1.3. - If

I

~I

I

~

f+W"

I(a + reiO)

-W"

dO

-I

=

21r

I/(a)l·

are respectively the real and imaginary parts

is holomorphic in

n, rRl

and

are harmonic.

~I

Proof. - We write once more the equality :

I(a), and take real and imaginary parts.

Proposition 1.4. - Let u be a continuous subharmonic function in n, K a compact contained in n, h a real continuous function on K, harmonic in KO, such that u(z) ~ h(z), if z E 8K. Then u(z) ~ h(z), z E K. Proof. - Put Ul = u - h. Then Ul is subharmonic in KO. Let's assume that for some z E KO, uI(z) > O. Since Ul is continuous on K, it attains its maximum, m, on K, and since Ul ~ 0 on 8K, the set: A

=

{z

E K i uI(z)

= m}

is a compact subset of KO. Let Zo be a boundary point of A. Since Zo E KO , one can find an r > 0 such that the closed disk D(zo~ r) is also contained in KO. But some arc of 8D(zo~ r) is contained in AC. Therefore, Ul(ZO)

f

= m ~

and this contradicts the fact that

Ul

+W" -W"

"0

Ul(za

+ reI)

dO

-

21r

,

is subharmonic in KO.

Analytic Functions

165

Lemma 1.5. - Let C; be the circle of center 0, radius r , 0 < r < 1, and f a continuous function on Cr , with real values. There exists a function g, harmonic in D(O, r), with g = f on Cr.

Proof. - We define the Poisson Kernel: This is the family of functions Pr(O), O::;r
Pr(O)

1 - 2r cos 0 + r 2

(1) '

and one checks immediately that if z = re i 8 ,

Pr(O - t) =

eit ~(-.t-)

+Z

e' - z

(2)

.

We recall moreover that the family (Pr)r is an approximate unit in the space Lp(II, dO /211"), when 1 ::; p < 00. This means that if f E Lp(ll, dO /211")'

f * P;

~

f,

(3)

r ~ 1- ,

in Lp(ll, dO/211"). This property is an easy consequence of the following elementary facts about the Poisson Kernel:

Pr(O) and

1

181<e

~

0, 0::; r

dO

Pr(O) -

~

<1

1,

r

~

1-,

for all e

211"

1,

(4)

> o.

(5)

The reader may find a proof of these facts in any book dealing with Harmonic Analysis (see for instance Hoffman [11, Katznelson [11, etc). However, in Proposition 2.2 below, we will prove a stronger fact, namely that the convergence in (3) holds almost everywhere (in short: a.e.). Now, to prove our Lemma, we put, for r' < r, z

= r' ei 8 ,

Chapter VIII

166

.

with, for n E 1L Cn

=

!

~

t

't

e- I n f(re l

_~

dt 21r

) -

•

Since f is in L 2 , the series E~oo cne i n 8 is convergent, and, taking r' get g(re i 8 ) = f(re i D) , so f = 9 on Cr. Moreover, by the properties of the Poisson Kernel,

= r , we

and 9 is the real part of an holomorphic function, and so is harmonic in D(O,r).

Proposition 1.6. - If u is a continuous, subharmonic function in D, and if:

is the mean value of u on C«, then m(r) is an increasing function of r , T

<

°

~

1.

Proof. - Let rl < T2. Let h, given by Lemma 1.5, the continuous function on D(O,T2), equal to u on Cr 2 , and harmonic in D(O,T2)' Then, by Proposition 1.4, u ~ h in D(O, T2)' Therefore,

h(O) and our Proposition is proved. 2. Basic facts about Fourier Series. In this paragraph, we recall a few basic facts about Fourier series. Any function f E Lt{II, d8/21r) (and, a fortiori, in Lp(II, d8/21r) has a Fourier series:

(1) where:

The coefficients

Cn

tend to

°when

n

--+

±oo (Riemann-Lebesgue). We put

167

Analytic Functions

and:

11/1100 =

sup essl/(eiB)I. BE"

We define the partial sums of the Fourier series :

n

sn(/) =

L

Cje

ij B

,

-n

and the Cesaro averages of these partial sums : 1 Un

(f)

= -

n

n-I

L sk(/). 0

The following Theorem can be found for instance in Hoffman [11 : Theorem 2.1. - a) The function 1 is in Lp(II), 1 < p < the sequence (un)n~O is bounded in L p •

00,

if and only if

b) The function verges in L 1 •

1

is in L 1 (II) if and only jf the sequence (un)n~o

c) The function verges uniformly.

1

is continuous if and only if the sequence (un)n>O con-

con-

d) The function 1 is in Loo(II) if and only jf the sequence (un)n~O converges to 1 in the topology u(L oo , LI). This last fact implies that (un)n~O is bounded in L oo . For l(e i O) E LI(II,dO/211'), we define:

l(re i O)

=

(Pr * l)(fJ)

= IW"

Pr(O - t)/(e i t )!!:!.- . 2~

-W"

Then: Proposition 2.2. - For almost every 8, when r Proof. - Fix fJ Ell. We have:

Let

g(t) =

I

t

-W"

l(e i 8 )

dO -

211'

--+-

1- .

Chapter VIII

168

Let

A =

2~ [Pr(O - t)(g(t) - tf(ei8))n~~ll"

= ~ (Pr(9 - 1I")(g(1I") - 11" f(e i8)) - Pr(O + 1I")(g(-11") + 11" f(e i8»)) 211"

=

2~ (Pr(O - 1I")g(1I")) - Pr(O - 1I")f(ei8) ,

°

= 11".

and this last quantity tends to when r ~ 1- , except if 0 that for every 0 "f:. 0, Pr(9) ~ 0, when r ~ 1- .) Let now

For 6

> 0, we set :

B~

(

1It- 81>6

B~

=

P:(9 - t)(g(t) - tf(ei8)) dt , 211"

P:(9 - t)(g(t) - tf(ei8» dt .

(

1I

t - 8 15: 6

IT

(We recall

Irl > 6 > 0, a

direct computation shows that:

IP r' (r ) I ~

°

B~,

2r(1 - r (1 _ 2r cos 6

when r

and this quantity tends to For

211"

~

2 )

+ r 2 )2 ,

1- . Therefore,

B~

0, when r

~

we write :

B~

=

-

(

11'15: 0

dr P:(T)(g(O - r) - (9 - r)f(e i8)) 2 . 11"

1

But P:(T) is an odd function, since P; is even. Therefore,

B~

=

(O P:(T)(g(O _ r) _ (8 _ T)f(ei8)) dr

-

10

211"

-

f

o

»-211"dr

"0

P:(T)(g(9 - r) - (8 - r)f(e'

-0

(O P:(r)(g(9 + r) _ g(8 _ r) _ 2r f(ei8)) dr

10

211"

= (O P'(T)T(g(8 + T) - g(8 - T) _ f(e i8)) dT .

10

r

2r

211"

~

1- .

169

A nalytic Functions But :

g(O + r) - g(O - r) = 2r

f.8+t 8-t

-+

1 ~.

-+

f(e i 8 ) ,

2~

f E LI(II,dB/21r). Therefore, for every

when T -+ 0, for almost every 0, since e > 0, if 6 is small enough, we have:

and then we let r

f(e i t ) dt

This proves our Proposition.

The convolution with the Poisson Kernel can be extended to measures : Proposition 2.3. - Let JL be a finite Baire measure on the unit circle, and define:

f(re i . ) Then

=

i:

f is harmonic in the open disk dJLr =

P,(O - t)d!,(t) .

and the measures :

f(re l

'8

dO

) -

2~

converge to JL in the weak-* topology of measures, when r

-+ 1- .

The proof of Harmonicity is as in Lemma 1.5 above, and if 9 is a continuous function, one checks immediately (using the fact that the Poisson Kernel is an approximate unit) that:

when r

-+ 1- .

3. HP Spaces. For p, 1 ~ P ~ 00, we denote by HP (II) the space of (equivalence classes) of functions f(e i 8 ) in Lp(II,dO/2~), whose Fourier coefficients cn(J) are 0 for n < 0. In other words,

(1) For f(e i8 ) E HP, we define:

f(z) =

L

cnz n

,

zED.

(2)

n~O

Since (Cn)n>O D.

-+

0, this series defines an analytic function inside the open disk

Chapter VIII

170

The reader will observe that we keep the same notation (/) for the function defined on C and its analytic continuation inside the disk. For p, 1 < p :S

00,

We now set, for 1 :S p <

we have:

0 :S r < 1 :

00,

(3) and from Proposition 1.2 and Proposition 1.6 follows that for given p, /, M p ( / , r) is an increasing function of r, 0 :S r < 1. Moreover, if z = re i 8 ,

and we have seen that /

* P;

M p ( / , r)

in L p , when r

--+ /

--+

II/lIp,

Finally, we can write, for 1 :S p <

--+

when r

1- . This implies:

--+

1- .

00 :

(4) This is also true for p =

00,

with:

by the maximum modulus principle for analytic functions. On HP, we put the norm induced by L p • Clearly, HP is a closed linear subspace of L p , since for each n, the mapping / --+ cn ( /) is continuous

(Icn (/ )! :S

II/lip)·

We also observe that for p ~ 1, Hoo c HP C HI C £1. Formula (2) above allows us to extend /(e i 8 ) as an analytic function /(z) inside the unit disk. Conversely, one can recapture /(e i 8 ) from /(z) :

A naif/tic Functions

171

Proposition 3.1. - Let IE HP. For almost every fJ, when r

-+

1- .

This is indeed a trivial consequence of Proposition 2.1, stated for functions in L 1 . This convergence is called radial, since re i 8 moves along a radius, when r -+ 1-. In fact, one can show that this convergence holds non-tangentially: in any angular sector centered at fJ, which does not contain the tangent at fJ to the circle (see for instance W. Rudin [1], p. 243). Before continuing our study, we list a few easy facts about HI, H2, H?".

Proposition 3.2. - If I E HI and l(e i 9 ) EIR a.e., then I is constant: there is a a Em. such that l(e i 9 ) = a a.e. Proof. - By definition, for n > 0,

Taking conjugates,

f

'll"

9

•

'9

dfJ

e- m I(e' ) -

271"

-11"

= o.

Therefore, I has all Fourier coefficients equal to 0, except co(f) : this shows that I = co(f) a.e .. One could also say that I(z) is analytic and real-valued, therefore constant.

Corollary 3.3. - If both

I

and

f

are in HI,

I

is constant.

Indeed, (I + /)/2 and (I - i1)/2i are both in HI, and are real-valued. Therefore, they are constant, by the previous Proposition, and their sum is

(1 - i)I/2. Proposition 3.4. - The space H2 is a Hilbert space. If

g(ei 8 )

=

L

I,

g are in H2 , and :

d n ei n 9

,

n~O

we have:

(L Ic n~O

(f,g) =

f

'll'

_'II"

'9

'9

dO 271"

I(e' )g(e' ) -

Proof. - This is obvious on the definitions.

nI

2)1/ 2 ,

Chapter VIII

112

Notation : To avoid ambiguity, we distinguish between:

L:=o a"z", P : the set of trigonometric polynomials L:=-M a"zn.

P+ : the set of polynomials

Proposition 3.5. - Let 4> in Loo(TI). The space H2 is invariant by the multiplication by 4> (denoted by M4J) if and only if 4> is in HOO. Proof. - a) Assume that M",(H2) HOO. b) Conversely, if a polynomial,

Then

4> E H?", M",(P+) C H2 : '11" /

for n

C H2.

'fJ

'8

e- l n 4>(e l )p(e l

4>.1 E H2,

)

4> E H2 n L oo =

indeed, if p(z)

dO

'8

so

-

= L~

akz k is

= 0,

2~

_'II"

< O. But P+ is dense in H2 for its norm, so M",H2

C

H2.

Corollary 3.6. - H'" is a Banach algebra. Indeed, if 4>, '¢ are in H?",

so 4>'¢ E HOO . As a Corollary to Proposition 3.1 above, we give the following:

Theorem 3.'1 (F. and M. Riesz). - H JI. is a complex measure on Il , and if :

i:

.in"dIL(8) = 0,

for n = 1,2, ... ,

then JL is absolutely continuous with respect to Lebesgue measure: there is a function f in HI such that:

f dO . 2~

Proof. - For zED, we define :

j(z)

= /'11" _'II"

dJL(t).

1 - ze- l t

00

Pr(O - t) =

LTlnlein(fJ-t) , -00

Analytic Functions we have also :

f(z) =

r:

179

Pr(9 - t)dl'(t) .

(I)

Hence:

So

I

is in HI. But then we have: -·8 I(re' ) =

fir

dt . 211"

Pr(S - t)/(t) -

-1("

(2)

and the conclusion now follows if we compare (I) and (2) and apply Proposition

2.3. 4. Jensen's Formula. Jensen's Inequality.

The following Proposition gives a useful example of an harmonic function: Proposition 4.1. - Let n be a simply connected domain, and assume that I is analytic in n and does not vanish in this region. Then there is a function F analytic in n, such that eF = I. As a corollary, log I/(z)1 is harmonic in n. Proof. - Fix Zo in

n and put: F(z)

=

l I(~)d~

,

where r is any path from Zo to z, contained in n. Then F is analytic in n, and F'(z) = I(z), z E o. Since I does not vanish in n, I' /I is analytic in n, and we can apply this to f' / I instead of I : we get a function F, such that F' = 1'/1. We can add a constant to F, so eF(zo) = I(zo). Put h = le- F : then h' = 0, and h(zo) = I, so h is identically equal to I, and I = eF • Take now 9 = ~F. Then 9 is harmonic, III = eg , and log III = g. Proposition 4.2. - J~1r

log 11

-

ei81dS /21r =

o.

Proof. - Let n = {z ; ~z < I}. Since n is simply connected, and 1 - z f. 0 in n, there is a F analytic in n, such that eF(z) = 1- z, and F(O) = O. Since ~(I - z) > 0 in n, we have, for zEn,

UlF(z) = log 11 For small 6 > 0, let

r

- zl ,

I~F(z)1

< 11"/2.

be the path :

f(t) =

6~ t

~

211" - 6,

Chapter VIII and 1 the circular arc whose center is at 1, and passes from ei 5 to e- i 5 within the unit disk. Then :

27f -6 log 11 - et81_ . dO =

1 a

21r

1

ri-.-

I.

r

2Z1r

dz z

F(z)-

=

1 !R-.-

1

2Z1r,..,

dz

F(z)-, Z

by Cauchy's Theorem, and since F(O) = O. But the last integral is smaller than Co log 1/0, where C is a constant, since the length of 1 is smaller than 1rO. Letting 0 -+ 0, this gives the result.

Theorem 4.3 (Jensen's Formula). - Let n = D(O,R) , 1 be analytic in n, with 1(0) -I- O. Let 0 < r < R, and al,"" aN be the zeros of 1 in D(O, r), listed according to their multiplicity. Then:

IJ(O)!

ITN -) r I

n=l an

=

exp

17f

~

log I/(re i 8)1- . 21r

-7("

Proof. - We order the roots so that aIt ... , am are in D(O, r), lam+ll ···laNI = r . Put m

g(z)

=

I( ) z

2

_

N

II r(an-z) r - anz II

n=l

n=m+l

(an-z)'

so g(z) is analytic on some disk D(O, r + E:), E: > 0, and has no zero in this disk. Therefore, by Proposition 4.1, log Ig(z)I is harmonic in this disk, which implies: dO log Ig(rei 8 ) I- = log Ig(O) I .

1-7f 7("

But g(O)

1(0) n~

21r

I:nl . To obtain the formula, we observe finally that:

Indeed, each term r2 -

r(a n

iinz - z) ,

has modulus 1, and for n > m, each term

satisfies:

by Lemma 4.2.

n~m,

Analytic Functions

175

Corollary 4.4 (Jensen's Inequality). - If / E HI and /(0) =I 0,

"1 _,,"

dO log 1/(eSo)l- ~ 21r

log 1/(0)1.

Proof. - We cannot take immediately r = 1 in the previous theorem, since is not analytic in a larger disk D(O, R), with R > 1. For any r, 0 < r < 1, by Jensen's Formula:

"1 1"

_,,"

Fix e >

dO log I/(reSo)l- ~

I

log 1/(0)1.

21r

o. A fortiori,

_,,"

dO log(l/(reSO)1 + c) > log 1/(0)1. 21r

When r - 1-,

a.e. and in L I

•

Thus,

"1

_,,"

dO (log(l/(eSo)1 + c) ~

log 1/(0)1.

21r

Now, when c - 0, the sequence log(l/(eSO)1

+ c)

is decreasing, and thus:

Corollary 4.5. - For every function / in HI, not identically 0, the function log 1/( fSO) I is integrable. Proof. - First, we observe that:

and so all we have to show is that:

"1

_,,"

dO log I/(eSo) I > 21r

-00.

IT 1(0) =I 0, this follows from Corollary 4.4. IT 1(0) zng(z), with g(O) =I 0, and apply the Theorem to g.

=

0, we write I(z)

=

Chapter VIII

176

We observe that, as stated here, Jensen's inequality is a discontinuous estimate, in the following sense : assume that we have a sequence of functions (In)n~o, with In(O) --+ 0, and In --+ I (say: uniformly on II). Then, for each n, we apply Jensen's inequality to In, thus getting estimates which become worse and worse, when n --+ 00, since log I/n(O)1 --+ -00, though at the limit we apply Jensen's inequality to g = 1/zlc. This phenomenon is partly due to the fact that Jensen's inequality takes into account only the first coefficient of the Taylor expansion of I. A stronger version, taking into account a prescribed number k of coefficients in this Taylor expansion has been obtained by P. Enflo and the author (see the "Complements", at the end of this Chapter, in which the general problem of continuity of the functional :

will also be studied).

Corollary 4.6. - If a function I in HI is not identically 0, it cannot vanish identically on a set of positive measure. Indeed, since log III is integrable, the measure of the set where it takes the value -00 must be O. From Jensen's inequality, we deduce the following one, which can be interpreted as a similar inequality, involving the measure Pr(t - O)dO /21r instead of the measure dO /21r :

Proposition 4.7. - For any function

I

'll"

_'II"

I E HI, for 0 ~ r < 1, t E II,

"0 dO log I/(e' )IPr(t - 0) ~ 21r

. log I/(re't)l.

Proof. - Fix zo, Zo = rei! , and apply Jensen's inequality to the function:

4>(z) = I( z +

:0 ).

1 + zoz

We get:

I

'll"

_'II"

log 14>(f

dO

io)l-

21r

~ log 14>(0)1 = log If(zo) I·

Let A be the left-hand side of the above equation. Observing that (f i O+z)/(l +

A nalytic Functions

177

ze i 8 ) has modulus 1, we get, after a change of variables :

A

=

! !

i8(

~

2)

I I/( i8)1 e 1 - ,. _~ og e -zoe 2i8 + (1 + ,.2)e i8 ~ i8 1 _,.2 dB log I/(e )1 .812 _~ 11 - Zoe' 21r

L _~

Zo

dO 21r

log I/(ei 8 )IPr (t - 0) -dO , 21r

as we announced. 5. Factorization of HP functions: Inner and Outer functions.

We recall our terminology: l(e i 8 ) is the value of the function on the unit circle, obtained as a radial limit a.e., and I(z) is the value inside the unit disk. An inner function is a function m in H?", satisfying Im(ei8)1 = 1 a.e., and thus Im(z)1 ~ 1 for every zED, since the numbers Moo(m,r) are increasing with v (see Section 1). For instance, m(e i 8 ) = fin8 (n ~ 0) is an inner function. We will see other examples later, with Blaschke products and singular functions. An outer function F is a function in HI such that:

F(z) = where

IC

IC

exp

!

~ _~

eit + -.-t-

z dt k(t) - , e' - z 21r

(1)

is a complex number of modulus 1, and k(t) a real integrable function.

The meaning of formula (1) is not clear at this point. The following Proposition clarifies the links between k and F :

Proposition 5.1. - For a function F in HI,

k(O) = log IF(e i 8 )I , Proof. - For z

a.e.

= re i8 , we decompose:

where Pv , Qr are real: P; is the Poisson Kernel and Qr the conjugate Kernel:

Q (0) _ r

-

2,. sin 8 1 - 2r cos 9 + ,.2

Chapter VIII

178

Therefore, (1) can be written, for z

F(z) =

= rei 8 ,

exp(P,. * k + iQ,.

It

* k).

So we get:

=

IF(z) 1

exp(P,. * k) .

(2)

Since k E £1, P,. * k -----. k a.e., when r -----. 1- (see Section 2). Since also F(re i 8 ) -----. F(e i 8 ) (Proposition 3.1), our Proposition is proved. We observe that by formula (1), F cannot vanish inside the open unit disk.

Proposition 5.2. ~ Let F be a function in H 1 , not identically O. The following are equivalent: a) F is an outer function,

b) 1£ 1 is a function in H1 such that I/(e i8)1= IF(ei8)1 a.e., then:

IF(z)l

I/(z) I,

~

c) log IF(O)I

\jz E D,

I '" '" + 1

=

_'"

Proof. - 1) a) implies b). Let F be outer, and IF(ei8)\ on II. Then by Proposition 5.1,

F(z)

=

It

exp

dO

log IF(ei8)1- . 27r

1

in H1 such that I/(e i8)1=

ei8 z . dO -'-8- log I/(e'8)1 - .

_'" e' - z

27r

By formula (2) above, with k replaced by log 1/(ei8)1, log IF(re i8)1 =

f'" P,.(O - t) log 1/(eit)1 dt .

1-",

27r

Now, by Proposition 4.7, the right-hand side is larger than log I/(re i8)1. This proves our claim. 2) b) implies a). We consider the outer function: G(z) =

It

exp

I

'"

eit + z

-'-t-log

_'" e' - z

. dt IF(ed)l- . 27r

Since F is integrable, G is in HI, and:

lG(ei8)1

=

IF(ei8)1,

c.e.

Therefore, lG(z)1 ;::: IF(z) 1 in D, by b). Using again the fact that G cannot vanish in D, we find that F / G is analytic in D, has modulus 1 on C, and modulus ~ 1 in D : this is possible only if F /G is constant, and this constant must be of modulus 1. This proves that F is an outer function.

A nalytic Functions

179

3) a) implies c) is obvious on the definition of an outer function. 4) c) implies a). We define G as above. Then ]FI/IGI ~ 1 in D, = 1 on C, and IF(O) I/IG(O) I = 1. Therefore, F /G is constant. We now turn to a first decomposition of functions in HI

Proposition 5.3. - Let f be a function in HI, non identically O. Then f can be factored into m.F', where m is inner and F an outer function. This factorization is unique, up to a constant of modulus 1. Proof. - We put :

F(z)

=

exp

"1

eit + z "t dt eit _ z log If(e' )1 211" •

-,,"

This is an outer function. Put now:

f(z) m(z) = F(z) , one obtains a function analytic in the open disk, and Im(eit)1 = 1, by Proposition 5.1. Therefore, m is an inner function. If f = mlFl is another decomposition, IFI = IFll, so IF(z)1 = IFt{z) I in D by Proposition 5.2, and F = KFl , for some KE ~ ,with IKI = 1. 6. Factorization of Inner Functions: Blaschke products. Singular Functions.

In this paragraph, we study inner functions and get for them further factorizations.

Theorem 6.1. - Let

(an)n~o

be a sequence of complex numbers, with

n

Then the infinite product:

converges uniformly on each disk Izi ~ r < 1. Each an is a zero of B(z), with multiplicity equal to the number of times it is repeated in the sequence, and B(z) has no other zero in D. Finally, IB(z)1 < 1 in D, and IB(ei 8)1 = 1 a.e.

Proof. - Put ; bn

=

Ian I an - z an 1 - iinz

,

Bn

B'n

Chapter VI/I

180 Fix r < 1. For

Izl < r,

which shows that B(z) converges uniformly in the disk of radius r, and that B(z) is analytic in D(O, r}. Furthermore, each an is a zero of B with correct multiplicity, and B(z) =I- 0 otherwise. That IB(z) I < 1 if Izi < 1 is obvious, since it is true for partial products. The radial limit B(e i8) therefore exists a.e., and IB(e i 8 )1 ~ 1. It remains to show that IB(ei8)1 = 1 a.e. But for any function f in the monotonicity of M, (r, J) (Section 3, formula (4» gives :

n»,

We apply this to B~ = B/B n, and we observe that Ibk(e i8)1 lB n (ei8)1= 1 a.e., thus:

1 a.e., so

So:

I

""

IB~(reiO)1

-11"

dfJ 21r

~

111" IB~(eio)1 _,,"

dO = 21r

I""

IB(eiO)1 dO .

-11"

21r

But Bn(z) --+ B(z) uniformly on {lzl = r}, so B~ --+ 1 uniformly on this circle. This implies that f~,,"IB( ei8)IdfJ /21r ~ 1, and since IB(ei8)I ~ 1, we get i8 IB(e )1= 1 a.e. A function of the form :

with Ln(1 - Ian I) < 00, is called a Blaschke product. M is a positive integer, the set of an's may be infinite, finite or empty.

Proposition 6.2. - Let m be an inner function. Then the zeros an of m inside the open disk D satisfy Ln(1 - Ian I) < 00. Proof. Dividing if necessary by a power of z, we may assume m(O) =I- O. Let laol ~ latl ~ ... be the enumeration of the zeros of m inside the unit disk. Fix an integer N and choose r, with IaN I ~ r < 1. By Jensen's Formula,

" I

-,,"

log Im(re

iO) I -~

21r

~ log Im(O) I +

N

L log -anrII ' 0

Analytic Functions

181

which implies : N

2:o IOg -anr s 1

-log]m(O)I·

1

Letting r

-+

1- , we get : N 1 Llog-,-, < C, o an

with C = - log Im(O) I. But for 0 < x < 1, log l~:t

2: z . So we obtain:

and the series converges.

Theorem 6.3. - Every inner function m can be factored as m = B.G, where B is a Blaschke product and G is an inner function with no zero in the open disk. Proof. - We may assume that m has infinitely many zeros an, otherwise this is trivial. Dividing by zM if necessary, we may also assume that m(O) =f O. Put, as before, bn (z )

= -lanl

an -_ z , B(z ) an 1 - anz

= n°O bn () z

.

0

We know that this product converges, by Proposition 6.2 and Theorem 6.1. Put B n = n~bk, 9n = mlBn. For fixed n, e , IBn(z)1 > 1 ~ e, if ]zl is sufficiently close to 1. Therefore,

s~p Ign(re

i 8)1

s

1

1_ c '

and by monotonicity, this holds for all r < 1. So, letting e -+ 0, we get:

But gn -+ G has no zeros.

= m] B,

uniformly on each circle

Izl = r ,

So G is in H'" and

Chapter VIII

182

11], Duren [1]) that the function G, inner

It can be shown (cf. Hoffman

with no zeros inside the disk, can be written under the form :

G(z) = exp

I

'"

+z

ei8 -.-8-

_'" e' - z

dJl(6) ,

where Jl is a positive measure, singular with respect to Lebesgue measure. Such a function is called a singular inner function. The simplest example (obtained with the Dirac measure at 6 = 0) is the function:

z+l G(z) = exp - - . z-l 7. The Disk Algebra A(D). The Disk Algebra A(D) is the space of functions which are analytic inside the open disk D and continuous on the closed disk D : therefore, it is C(TI) n BOO , equipped with the norm :

IIflloo =

sup 8En

1/(ei 8 )f =

sup

1/(z)1 .

zED

This is of course a separable Banach space: the polynomials P+ are dense for the norm. This Banach space has an interesting property, connected with the extension of functions defined on a compact of 0 measure. This property will be used, in an Operator Theory context, in the next Part.

Theorem 7.1 (Fatou). - Let K be a compact in TI, with 0 Lebesgue measure. There exists a function 4> in A(D), such that:

Remark. - By virtue of Corollary 4.6, K must be of measure 0 for such a function to exist.

Proof. - Since K is compact, I1\K is the reunion of a countable family (In)n~o of open intervals, pairwise disjoint. Let en = P(In). For every n, let Yn be a positive function, of class Cion In, satisfying Yn < lie, Yn = 0 at the end-points of In, and:

A nalytic Functions

189

We put y = 0 on K, Y = Yn on In. The function y has the following properties : - 0 ~

- K

Y ~

lie,

= {O j y(O) = O},

- y is of class Cion IT\K and continuous on IT,

- log y is integrable. Put now w = log y. Then w satisfies : a) w = -00 on K, and if dist(O, K) --+ 0, w(8) --+ -00, b) w ~ -Ion IT, c) w is continuous on IT and of class Cion IT\K, d) w is integrable. We now set:

h(z) = =

I

'l " _'II"

eiB

+z

eiB _

dO

z w(O) 271"

(Pr * w)(O) + i(Qr * w)(O) ,

if z = re i O , where P; is the Poisson Kernel and Qr the conjugate Kernel (see Section 2). Since P; ~ 0, from b) follows that rRh :os: -1 in D. Set ur(O) = (Pr * w)(O) ; then U r --+ w, when r --+ 1- , uniformly on IT, since w is continuous. We now show that, since w is Cion IT\K, the functions v r (8) = (Qr * w)(O) converge uniformly on every closed interval contained in IT\K. This will imply that h(ei O) is continuous on IT\K.

Lemma 7.2. - If w is of class Cion a closed interval 10 - 00 1 ~ a, on this interval, V r converges uniformly to :

v (0) = _

Ill"

W

(0 + t) - w (0 - t) dt 2tan(t/2)

_'II"

271"

Proof. - Take 00 = O. The function

ePO(t)

=

w(O + t) - w(O - t) 2 tan(tI2)

is integrable on 1= {t ; 0 ~ t ~ min(O + a, a - O)}, and:

vr(O) - v(O)

=

=

I'll" cPB(t)(l- 2rsinttan(t/2)) dt -ll" 1 - 2r cos t + r 271"

,I .

_,..

2

r) 2

eP B(t)

(1 1 _ 2r cos t

+ r2

dt 21r .

Chapter VIII

184

And one checks for the Kernel gr(t)

(1 - r)2

=

1 - 2rcost + r 2

the following properties: 0 < gr (t) < 1, and gr (t)

--+

0, r

--+

1- , except for

t = O. The Lemma follows by a computation similar to the one made for the

Poisson Kernel (Proposition 3.1). We now come back to the proof of the Theorem. We set 9 = 1/h. Then, since h does not vanish in D, 9 is analytic in D, continuous in D, by Lemma 7.2, and the zeros of 9 on C are exactly the points of K. This proves our Theorem. We observe that ~g < 0 on D\K.

Corollary '1.3. - Let K a compact contained in the unit circle, with 0 Lebesgue measure. There is a function ¢ in A(D), with ¢(z) = 1 for every z E K, and I¢(z) I < 1 if z E D\K. Proof. - We take ¢ = erJ , where 9 is given by the previous theorem. Theorem 7.4 (Rudin, Carleson). - Let K a compact contained in the unit circle, with 0 Lebesgue measure, and let / be a continuous function on K, with complex values. There exists a function F in A(D), such that the restriction of F to K is f, and such that : IIFlloo = sup 1/(z)l· zEK

Proof. - Let ¢ be the function given by Corollary 7.3. For every h in A(D), we have:

Write ¢nh

= h + v«, where v«

is in A(D) and

v« = 0

on K. We get:

inf{llh + -rP1100 ; -rP E A(D), tP = 0 on K} inf II h n

~

+ tPlloo

inf II¢nhll n

lim II¢nhll n

sup Ih(z)l. zEK

But conversely, for every -rP in A(D), with .,p = 0 on K : Ilh + t/Jlloo

=

sup Ih(z) + tP(z) I > sup Ih(z) + ,p(z)[ zEC

zEK

sup Ih(z)l, zEK

Analytic Functions

185

and so : inf{llh +

1/11100 ; 1/1 E A(D), 1/1 = 0 on K} = sup Ih(z)l.

(1)

zEK

Call AK the set of restrictions to K of functions in A(D), equipped with the norm induced by C(K), and call S the subspace of A(D) consisting of functions 1/1, 1/1 = 0 on K. Finally, let R be the restriction operator, from A(D) into C(K). Formula (1) means exactly that for every h in A(D), IlhIIA(D)/s

=

(2)

IIRhllC(K) .

This means that R is an isometry from the Banach space A(D)/ S, into C(K), and therefore has closed range in C(K). The restriction operator from A(D) into C(K) has obviously the same range. We now show that this range is dense in C(K) : the restriction operator from A(D) into C(K)will therefore be surjective. Assume not. Since the range is a vector space, it is contained in a closed hyperplane: there exists a measure J.I. on K such that :

VI E A(D).

/ R(f)dJ.l. = 0,

By Hahn-Banach Theorem, J.I. may be extended to a measure on C(II) , also denoted by It. Therefore :

/ 1 lK In particular taking

1 = ei n 8 , / e

in8

dJ.l.

= 0,

n ~ 0,

VI E A(D).

we find:

lK dlt = 0,

Vn

~

O.

By the F. and M. Riesz Theorem (Theorem 3.7), this implies that l KdJ.l. is absolutely continuous with respect to Lebesgue measure. But since P(K) = 0, this implies It = 0, and the image of R is dense. So, R, from A(D)/S onto C(K) is a surjective isometry. This does not quite finish the proof of our theorem: given I in C(K), we want to find F in

A(D) with IIFlloo

= 11/1100'

We may of course assume that 11/1100 = 1. The fact that R is a surjective isometry implies that there are functions ¢ E A(D) and 1/1 E S, with:

¢

= 1 on

K,

14>1 < 1 on D\K,

(1)

186

Chapter VIII

,p Put 9

= 0 on K,

= + ,p. We have Igi ~ 0

There is an Cl such that:

1

II + ,plloo ~

(2)

2.

1 on K, Igl ~ 2 on C. Let 0

= {z E

1

be the set:

1

C ; Ig(z)! < 1 + 4"}'

> 0 such that II < 1 - Cion Oi, so we can find an rnl in IN on O~ .

On 0

1,

we get :

For k = 1,2, ... , let: 1

Ok

= {zEC; Ig(z)1 < 1+ 4k } '

Then the Ok'S are decreasing and contain K. Assume ml been chosen so that : !jm1gl + ... 2

+ _1_ mk- 1g l < k 1I 2 -

1-

~k 2

< ... <

rnk

have

on 0k-l ,

and on Ok-l :

1

21 m1gl +

1 ... + 2k_ll4>mk-lgl

1

< (2 + ... + < 1-

Using the fact that there is an ck a rnk such that:

+

1

4k- 1 )

1 2k •

> 0 such that I I < 1 - Ck on Ok we find

~1mlgl+"'+2~I4>mkgl and on

1

2 k - 1)(1

I

<

1-2k~1

onOk'

o,

Finally, the function:

F is in A(D) and satisfies :

F=f

onK,

and IIFlloo ~ 1. This finishes the proof of the Theorem.

187

Analytic Functions

Exercises on Chapter VIII.

Exercise 1. - Show that the function I(z)

=

zn In is not in HOO.

En~1

Exercise 2. - Show that if 1 is harmonic and z/(z) is harmonic, 1 is analytic. Exercise 3. - Let 1 in HI. Show that there exist g, h in H2 such that

1=

gh.

Exercise 4. - Let h be a positive integrable function on IT. A necessary and sufficient condition in order that h = 1/1 2 , for some function f in H2 is that J logh > -00. Exercise 5. - Let 1 E L p , 1 < P < 9 (z) =

00,

~ 2&71'"

and: (

1(A) dA .

Jc >. -

z

Is 9 in HP ?

Exercise 6 (Hardy). - Let 1 E HI, I(z) = En~o

L

n~1

(Hint: first assume that

an 11"

/ -11"

In the general case, write

.!.Ianl ~

anz n. Show that:

71'"·11/111 .

n

~ 0 for all

n, and use the formula:

dO 1 (71'" - 0) sin(nO) = -. 271'" n

1=

gh, with g, h in H2 (see exercise 3), with:

where B is the Blaschke product of the zeros of f.)

Exercise 7. - Let F be an outer function in HI. Assume that 1 is in HI, and that f IF is in L 1 • Show that f IF is in HI. Show that this property characterizes outer functions. Exercise 8. - Let 1 in HI. Show that a) or b) imply that function: a) 1I f is in HI, b) ~f(z) > 0, in D. If f is inner, show that 1 + f is outer.

1

is an outer

Chapter VIII

188

Exercise 9. - Let HJ be the subspace of HI consisting of functions 1(0) = O. Show that H'" can be identified to the dual of L 1 / HJ. Show that P+ is dense in HOO for u( HCXJ , L 1 / HJ). Exercise 10. - Let I be analytic in the open disk. Prove that product if and only if :

I

with

I is a Blaschke

a) I/(z)1 ~ 1 in D,

b) lim

r-+l-

I

"

_,,"

.

d8

log I/(re lo)1- = 2w

o.

Exercise 11 (Maximal ideals of A(D)). - Let lo(z) be the identity function: lo(z) = z . Let t/J be a character on A(D). Let 0: = t/J(/o). Show that for every polynomial p, t/J(p) = p(o:). Deduce the same relation for every function I in A(D). Deduce that the maximal ideals are of the form I a t;

= {I

E

A(D) ; 1(0:)

=

o).

Notes and Comments.

Our presentation is a mixture between many authors. For further information the reader should consult one of the excellent books which deal with HP spaces: Hoffman (11, Garnett (1], Duren 11], Rudin [1]. The exercises also come from these books, except Exercise 11, which was communicated to us by Richard Aron.

Analytic Functions

189

Complements on Chapter VIII :

One may study the continuity of the decomposition:

f = B·S·F

(1)

which we have found during the previous sections. The function f is in HI, B is the Blaschke product made with its zeros, S is a singular function and F an outer function in HI. So the question is : if a sequence (In)n~o converges to I in HI, what about the corresponding terms B n , S«, F n ? Do they converge to the terms B, S, F respectively ? The answer is "no" in general. Indeed, if:

l+z S(z) = exp - - , l-z this is a singular function, as we have already seen. But if we set Sr(z) = S(rz) , o < r < 1, we get a sequence of functions in HI, which converges to S in HI. These functions are outer (to see this, one checks easily that they give an equality in Jensen's inequality, so we apply Proposition 5.2). The following theorem was obtained by L. Bonvalot [1] :

Theorem 1. - The decomposition (1) is continuous at every function has no non-trivial singular factor S.

I

which

Since this decomposition is obtained by means of Jensen's Inequality, its continuity is related to that of the "Jensen's Functional" :

J(f) =

Ill"-ll" log I/(ei9)1 21("dO .

(the quantity exp J(J) is known as Mahler's measure, and it is used in Number Theory.) The continuity of Jensen's Functional has been studied by the same author, and the result is similar:

Theorem 2. - Jensen's Functional is continuous on HI at every function which has no non-trivial singular factor in the decomposition (1). n

.

f

When I is a polynomial of degree n 1 say I = p = Eo a; z1 , estimates on J(/) have been intensively studied by many authors. For instance, Kurt Mahler [1] proved that : logLla;1 > J(J) > logLla;l-nlog2.

190

Chapter VIII

(see also Arestov [1], Beller [1], Beller-Newman [1] Kurt Mahler [2] for other aspects of these estimates.) Recently, P. Enflo and the present author (B. Beauzamy- P. Enflo [1]) studied Jensen's Inequality for polynomials (or HI functions) satisfying: Ie

Llajl ~

dLlajl. j?:.O

°

(2)

We say that a polynomial satisfying (2) has concentration d at degree k. For such a polynomial, P. Enflo and the author proved that there is a constant C(d, k) such that:

f"

log

-,,"

ii1)I

Ip(e dO ~ C(d,k). Lj?:.o laj I 271"

(3)

Numerical estimates where computed by the present author [13] for C(d, k) it is the largest value (for t > 1) of the function:

f d,k (t)

2d

=

t. log -(t---l-)-((-:-~- -:~-)

Ie-+-I---l)

(4)

Let C(d, k) be the best constant satisfying (3) ; the precise value of C(d, k) is unknown. However, it was shown in [13] that, for d = 1/2,

C(I/2,k)

-2k log 2 ,

~

(5)

and that, asymptotically, when k --. +00,

C(d, k)

~

-2k.

(6)

The precise value of C(d, k) has been computed by A. K. Rigler, S. Y. Trimble, R. S. Varga [1] for the class of Hurwitz polynomials: these polynomials have real positive coefficients, and their roots are either real negative, or pairwise conjugate, with negative real part. For this class, the best constant they find (in the case d = 1/2) is -2k log 2, the upper bound in (5), and it is obtained for the polynomial (z + 1)2k /2 21e • The problem is also solved for other values of d and the extremal polynomials are given; we refer the reader to the above mentioned paper for an exposition of the complete results.

Chapter IX.

The Multiplication by ei 9 on H2(ll) and L 2(ll ) .

We consider in this Chapter the operator M of multiplication by ei 9 : f ---+ ei 9 f . Clearly, this operator is an isometry, both on H2 and on £2. The difference is that it is surjective in £2, and not in H 2 • IT we identify each function in L 2 with the sequence of Fourier coefficients:

/ =

L

C1c

ei k9

---+

(c1ch E z

,

1cEZ the operator M becomes the right shift on , 2 (1l ) : indeed, (c1chEZ is replaced by (ck-dkEZ when / is replaced by ei 9 f. The same holds on H2

/

L

C1c

ei 1c9

---+

(ckh E IN '

1cEIN and: S : (co, Cl, C2, ..• )

---+

(0, co, Cl, C2,' •• ),

is the right shift on ' 2(IN). In both cases, we use the words "right shift" to indicate that this operator pushes to the right. Some people use "bilateral shift" to designate the shift on 12(1l). We find this terminology quite confusing: it is not the operator which is "bilateral", it is the space !

The operators M or S have obvious invariant subspaces. In ' 2 (IN), the sets:

are clearly closed and invariant for S. The definition of Fn also makes sense in 12(1l). However, if we look at M on £2 [Il}, we see immediately another type of Invariant Subspace: Fix a

1ge

Chapter IX

measurable subset A elI, with 0 < peA) < 1 (P(A) is the Lebesgue measure of A), and consider :

This is a closed subspace of L2(II,d6/21r}, invariant by M, not equal to {a} since peA) > 0, and not equal to the whole space since peA) < 1. Let's call Invariant Subspaces of the form F n "of type 1"', and of the form G A "of type II" (we will give precise definitions later). We already see that the Invariant Subspaces depend on the frame: in H2 and L 2 the situation does not look the same : only type I seems to exist in H2, and both types exist in L 2 • We now investigate this question more in detail. 1. The multiplication by ei O in H2 .

Theorem 1.1. - Let F be a closed subspace of H2, F M. There exists an inner function m such that :

t-

{O}, invariant by

(1) Conversely, any subspace of this form is closed and invariant for M.

Proof. - Obviously, a subspace F given by (1) is invariant for M. It is also closed, because the multiplication by an inner function m is an isometry from H2 into itself (since Im(eiO)1 = 1 a.e.) and the image of an isometry is always closed. Let now F be a closed subspace, invariant by M. If all functions f(z), EM, vanish at 0, we may write, for some k > 0, F = zk Fa, where Fa is invariant, and contains a function f which does not vanish at o. This way, we need only to consider the case when F contains such a function, say fa. We denote by 9 the orthogonal projection of the constant function 1 onto F. This means :

f

1

=

9

+1-

g,

where 9 E F, 1 - 9 E FJ.... The function 1 is not in FJ..., since:

Therefore, 9

f:. o.

We will show that

Igl

is constant on II.

Multiplication by ei8

Igl 2 nt-O, But

199

takes only real values. Taking conjugates, we get, for all n E 1l,

!

'8

ll'"

e,n

'8 2 dO Ig(e')1 -

21r

-ll'"

= 0,

which implies that Igi is constant : there is a A E

Ig(e i 8)1

=

such that :

~

A a.e.

We deduce that Ig(z)1 ~ A in D, and g/A is an inner function. Let's show that F = g.H2. First, we observe that, since 9 E F and F is invariant, the functions g, e i8 g, e 2i8 g, . . . , are a 11·In F ,an d so are th e fi rnit e sums '" L..,..k~O a/ce ik8 g, tha t is the products p.g by polynomials. Let now f be in H 2 , and (Jn) n~O be a sequence of polynomials in P+ , converging to f in H2 (for instance the partial sums of the Fourier series). Then, since 9 is in H?", ot« --+ gf in H2. Therefore, gf E F and gH2 C F. Assume this inclusion to be strict. Then, there exists a function f in F, f .1 9 H2. In particular, for n 2:: 0,

! L

ll'"

. dO e- 1n8fg - = 0.

(2)

21r

-ll'"

But 1 - 9 .1 ein 8 f , since 1 - 9 .1 F :::> ein8F. So, for n e i n 8 (1 -

g)f -dO

21r

-ll'"

Therefore,

! So:

l '" ein8f -dO -ll'"

21r

=

=

2:: 0,

0.

!ll'" ein8f-gdO- . 21r

-ll'"

l '" e,.n 8f g -dO ! -ll'" 2,..-

=

0,

and by (2) we get that this equality holds for every n E ~.

°

°

This proves that fg = 0. Since Igi = A t- a.e., we must have f = a.e., and F = gH2. As we already said, we get an inner function if we replace 9 by m

= g/A.

Chapter IX

19-4

Corollary 1.2 (Benrling's Theorem) - Let f be in H2. The vector space spanned by functions ei n 8f , n ~ 0, is dense in H2 if and only if f is an outer function. Proof. - We put

Ff

=

{ f, ei8f , e2i9f , ... } span

Clearly, Ff is invariant by M. a) Assume Ff = H2. We write the decomposition:

f = m.F, into inner and outer parts. Then :

which implies that m is constant, and that f is outer. b) Assume f to be outer. Since Ff is a closed invariant subspace for M, by Theorem 1.1, there is an inner function m such that :

This implies in particular that there is a function h in H2 such that :

f We write h

= hi.ho, where

=

m.h.

hi is inner, and h o is outer. Then:

and mh, is inner. Since f itself is outer, the uniqueness of the decomposition into inner and outer parts implies that mk; is constant, which in turn implies that both m and hi are constant. Therefore, F] = H 2 • We observe that the above proof gives in fact the more precise result:

Corollary 1.3. - If f is a function in H2, and f = /i.fo is its decomposition into inner and outer parts, we have:

Multiplication by ei 8

195

2. Szego's Theorem. We now derive Szego's Theorem from Beurling's Theory. There are, however, direct proofs (see Hoffman [1]). Let A0 be the subspace of A(D) consisting of the functions / which satisfy /(0) = O.

Theorem 2.1 [Ssegd]. - Let h be a real positive integrable function on II. Then: dO (1) inf 11 - g(e18 )12 h(O) -dO = exp log h(O) - . gEAo

j lf

.

jlf

271"

-If

211"

-If

Proof. - Assume first that log h is integrable, that is J~lf log h dO/211" > Put: 1 ei 8 + z . dO G(z) = exp -'-8log h(eI 8 ) - • l

-00.

jlf

2

-If

e

-

Z

21T'

Then G is an outer function,

and G 2(0)

= exp

I.

-If

log h(O) -dO . 27r

If 9 E Ao, we have:

But the Oth Fourier coefficient of (1- g)G is G (0) (since / (0) = 0). Therefore:

This proves one inequality in (1). Conversely, since G is outer, by Beurling's Theorem, there are polynomials Pn(z) such that PnG converges to the constant function G(O) in H2. Then Pn(O)G(O) --+ G(O) =j:. 0, and Pn(O) --+ 1. We can assume Pn(O) = 1. Thus we can write Pn = 1 - gn, and gn is in Ao . But then:

which proves the converse inequality.

196

Chapter IX

Now, if we assume f~ff logh(e i 8 )d8/ 21r = - 00, we observe that for every e > 0, log(h + e) is integrable, and by the preceding computations, ~

inf

I I

gEAo

= exp

ff

-ff

ff

log(h(8)

~

d8

+ E:) - •

-ff

When e

d8

.

11 - g(e ' 8 )12(h(8) + e) -211" 211"

0, this last expression tends to 0, and we get :

Corollary 2.2. - Let / be in L 2 (TI, d8/211"). Then

jf and only jf:

I

ff

log 1/(ei8)1

-ff

d8

=

-

2'1['

-00.

This follows immediately from Szegd's Theorem, applied to h =

1/1 2 •

3. Multiplication by ei B on L 2 (TI, d8/ 21r ) . Let now J.L be a positive Borel measure on TI. The operator M : / ~ eie/ is a surjective isometry from L 2 (TI, dJ.L) onto itself. As we already said at the beginning of this Chapter, in the case where J.L is the Lebesgue measure, M can be identified to the right shift on 12(1L). We now describe its Invariant Subspaces.

Proposition 3.1. - A closed subspace F contained in L 2 (#-, ) satisfies MF = F if and only if there exists a measurable subset A C TI, such that: F = {/ E L 2

j

/(6)

0 c.e. on A C }

=

Proof. - Set L 2(A) = {/ E L 2 j /(8) = 0 a.e. on A C } . Clearly, L 2(A) is closed, and ML 2(A) =L 2(A). Conversely, let F be a closed subspace of L 2 (#-, ), and assume ei BF = F. Then also e- i BF = F, and, more generally, einB F = F, \In E 1L. We first study the case when F is generated by a single function [; For / E L 2 (#-, ), we set: Ff = span {e in8/

j

n E

Z},

Multiplication by ei B

197

where the closure is taken in L 2 (IL) . By Stone- Weierstrasa Theorem, every function gEe (II) is the uniform limit of trigonometric polynomials, that is of finite sums L~K ateik8. Therefore, for every function 9 E C(II), 9 I E Ff. Let h be a function in Fj. Then:

In

gIlt

for all 9 E C(II),

= 0,

and {h E L 1 (IL). So Ih = 0 a.e., and Ih = O. Let B = {8 j 1(8) = O}. Then all functions in Ff are 0 a.e. on B. Conversely, if tP = 0 a.e. on B, and h is orthogonal to F, tPh = 0, by what we just proved. Therefore, tP E F.L.L = F, and Ff = L2(A) , with A = BC. The Proposition is therefore proved for Fj . We now turn to the general case: F satisfies M F = F. Let (In)n>o be a dense sequence of functions in F. For each n, let :

An = {O

j

fn(O) f O}.

Then:

with A

= UA n .

Proposition 3.2. - A closed subspace F

C

L 2(II, IL) satisfies :

if and only if there exists a Borel function tP such that:

(1) Proof. - a) If ¢ satisfies (1), ei B¢ H 2 C H2, so F is invariant by M, and:

b) Conversely, assume that F satisfies nn~oein8 F = {O}. If G c F is a closed subspace, we denote by F e G the orthogonal complement of G in F. So we put:

Chapter IX

198

Then, for n

~

0,

and therefore : '\J7n~Oe

F

F

in8L C

In.

e (EBn~oein8L)

,

= {O},

= nn~oein8F

which means :

=

F

in8L

In.

'\J7n~Oe

=

Let 4> be any function in L, with 114>112

4> ..1 Therefore,

e

in 8

•

1. We get, for n ~ 1

4>.

[1r 14>1 2

e i n 8 dJ,t

=

O.

ff

Taking conjugates, we obtain the same equality, this time for all n E Z, n =j:. O. This implies that the measure 14>1 2 dJ,t has all Fourier coefficients equal to 0, except the Oth, which is 1. So we get:

(2) The space L has dimension 1 : assume that there is a function 4>' in L with 4>' ..1 4>, in L 2 ( ,.,.) . Since 4> ..L e i n 8 4>' , for n > 0, we get:

[~~ t/>~'e-'n8dll

=

> 0,

0,

n

0 ,

n>

(3)

and since 4>' ..1 e i n 84> , for n > 0, we get :

[~~ t/>'~e-'nodll

o.

Taking conjugates,

[~~ t/>~'' n8dll

=

0,

n > O.

(4)

Using finally the fact that 4> ..1 4>' , we obtain from (3) and (4) :

!_~~ t/>~'

e'nodll

=

0,

n E :1:,

which means that 4>4>',.,. = 0, or 4>'14>1 2 ,.,. = O. But 14>12d,.,. = ~:, so 4>' = 0 a.e. L, each element f in F can be written : Since F = EBn~oein8

f

=

La

ne

in8

4> ,

n~O

with

L lanl 2 < 00.

So F

= 4>.H 2 ,

and our Proposition is proved.

Multiplication by ei 8

199

This Proposition can be strenghtened :

Proposition 3.3. - If M F c F and M F ¢ such that:

Proof. - As previously, we set L

=

G

i- F, there exists a Borel function

= F e ei 8 F.

Put:

L EB ei 8 L EB e2i 8 L EB ••• ,

and L o = FeG. The operator M is surjective on L o : if we take a function h in L o such that h ...L ML o , Mh EMF, so h ...L Mh, and thus h E L. Therefore, h E L o n L implies h = O. From Proposition 3.1, we deduce that there is a Borel subset A C II such that L o = L 2(A). We now take ¢ E L, and we repeat on G the proof of Proposition 3.2 : we find J¢1 2 dJ,t = dO/21r. But ei n 8 ¢ ...L 1A.¢, since Lo...L ei n 8 L , for n> 0, and 1A4> E L o = L 2(A). This means:

[ffff e

in8

j-fff

e

o,

1A 1¢1 2 dJ,t

in8

dO 1A = 0,

n> O.

n> O.

21r

Taking conjugates, we get that, for every n E 7l, n

I.-ff

ei n 8 1 -dO = 0 A

2

1r

=f. 0,

'

which means that 1A is constant. This is possible only if A has measure 0, and we are back to the previous case.

Proposition 3.4. - Let I E L 2(II,dO/21r). The functions ei n 8 1, n > 0, generate L 2 if and only if we have simultaneously:

a) P{O ; 1(0) = o) = 0, b) f~ff

log 1/(0)IdO /211" =

Proof. - As before, we set :

-00.

Chapter IX

200

1) Assume first F, = L2. Then, by Proposition 3.1, a) holds. We now show that: inf jW" 11 _ (ale itl + ... + akeik8)121f12 dO = O. (1) kjah···,ak

27r

-Ir

Lemma 3.5. - Let I E L 2 , such that P{9 ; 1(9) functions ei n 8 I, n ~ 0, generate L 2 if and only if :

=

O}

= o.

Then the

- { ei8,e 2i8 , . .. } , 1 E span

in the space L 2(II, 1/1 2 d9/ 27r) .

Proof of Lemma 3.5. - Assume first that the functions ein8I, n ~ 0, generate L 2 . For every E > 0, we can find a finite sequence of scalars ao, ... , ak , such that: W" "8 ~ " "8 2 d9 [e-' I - L" aje" II -W" 27r j=o

j

and so,

which means that : - { ei8,e 2i8, ... } , span 1 E -

in the space L 2(II, 1/1 2d9/ 27r) . Conversely, one checks immediately that, if this holds, also e- i 8 , e- 2i8, ... , belong to span{ei8,e2i8, ...}, in the space L 2 (II,1/ 12d9/27r). Therefore, in this space: span {e in8 ; n ~ I}

=

span {e in8 ; n E 1l}.

This equality implies that in the space L 2 (II, dO /27r), span {e in8I

;

n ~ I} = span {e in8I

I

By Proposition 3.1, this last space is L 2 , since Lemma and therefore (1).

;

n E :IL}.

satisfies a). This proves the

By Szegd's Theorem (Section 2), we have: inf gEA o

L -W"

and this means that f~1r

11 -

gl21/1 2-d9 = 27r

log I/ld9/27r =

exp

L -W"

log 1/1 2 -dO ,

27r

-00.

2) Conversely, let's assume that a) and b) hold. Once again, by Szegd's Theorem, (I) is satisfied, and by Lemma 3.5, we get F] = L2.

eoi

Multiplication by ei fJ 4. Multiplication by x on L 2[O,I].

We now turn to the study of the multiplication by x on the space L 2[O,I]. Obviously, if A is a measurable subset of [0,1], and if we set, as before,

we obtain a closed invariant subspace of Mr. in L2[0,1]. It is non-trivial if and only if < P(A) < 1 (of course, P is now the Lebesgue measure on 10,1]).

°

Conversely, we have:

Proposition 4.1. - Any closed subspace of L 2[0, 1], invariant for Mr., is of the form L 2(A), for some measurable set A C [0,1]. Proof. - Let F c 10,IJ with Mr.F C F. We may of course assume that F is not the whole space. Take a function f E F, f 1= 0. First, we consider:

Since Ff is not the whole space, there is a function 9 in L 2[0, 1], 9 =f. 0, with 9 1.. Ff. Therefore,

i

1

x· f(x)g(x)dx

=

0,

k

~

o.

This implies that for every polynomial p,

f

p(x)f(x)g(x)dx

O.

Since polynomials are dense in L 2 [0, I], we get:

° a,e. = ° a.e. on

f(x)g(x) =

°

AC , < P(A) < 1, and Set A = {x ; I(x) 1= O}. Then f Ff C L 2(A). If this inclusion was strict, we could find a function h in L 2(A), h 1.. Ff. 1 Then, as before, we would have f0 pfhdx = 0, for every polynomial p, which implies I.h = o. So I would be 0 on the set where h is =f. 0, which is a subset of A of positive measure. This contradicts the definition of A, and proves our assertion for F]: For F itself, we do as in the proof of Proposition 3.1 : we take a dense sequence (/n)n:;:::o in F, and we obtain F = L 2(A), where A = uA n , each An being the set {In 1= O}.

eoe

Chapter IX

Corollary 4.2. - For every function

where A = {x ; I(x)

1

in L 2[0, 1],

i- O}.

Corollary 4.3. - For every polynomial p,

Therefore, the operator M z on L 2[0, 1] has an interesting property, in terms of cyclic vectors. There is a cyclic vector 10 (namely the polynomial 1) such that for every polynomial p, p(Mz)/o is also cyclic. Let's understand what this means for the functional calculuses and the Invariant Subspace problem. We have seen (Proposition 4.1) that the non-cyclic elements were the functions 1 with non-trivial support: P({/(x) = O}) > 0. We may wonder: starting with a cyclic element, like 10 = 1, how can we get to a non-cyclic one, 1, using a functional calculus ? This question makes sense in general (that is : not only for this operator), and is precisely at the origin of the attempts to use functional calculuses to solve the Invariant Subspace Problem: given an operator T, we start with a cyclic point xo, and we try to find a function 1 such that I(T) is well-defined and I(T)xo is non-cyclic. Such attempts will be described in the next chapters. Here, the answer is fairly easy to give: if f is a polynomial, I(Mz ) 1 = f will still be cyclic, and the same remains true if f is in the Disk Algebra or in H?? . The HOO functional calculus has not been introduced yet (see Chapter XI), so let's accept temporarily the following fact (which is easy to admit) : for 1 E H?", the operator I(Mz ) is the multiplication by f(x) on [0, 1[. Of course, since f is analytic on a neighborhood of the segment [0,11, we get that 1 still satisfies P({/(x) = O}) = 0. If we want to get to a non-cyclic function, we have to take 1 in L oo , vanishing on a set of positive measure on [0,1]. In the present case, f(Mz ) makes sense when f E L oo ' because the operator M z is normal, and u(Mz ) = 10,11 (see Chapter VII, Section 5), and f is bounded on u(Mz ) . But such a functional calculus, admitting functions which are not analytic, just bounded on u(T), does not exist for other classes than normal operators. This is why, in my opinion, the idea of solving the Invariant Subspace Problem by means of a functional calculus does not look very promising.

Multiplication by ei 8

e09

We have studied in detail three operators: multiplication by ei 8 on H2, and on L 2 , and multiplication by x on L 2[0,1]. We observe that, quite interestingly, the types of Invariant Subspaces encountered are quite different. The first one has Invariant Subspaces of type m.H2, the second one has Invariant Subspaces of type m.H2 and L 2(A), and the third one, of type L 2(A) only. All separable Hilbert spaces are of course isometric, and these three operators look "more or less" the same. But their Invariant Subspaces are quite different conceptually. This indicate that the type of Invariant Subspaces (and therefore the means which are used to find them) must heavily depend on the underlying space and on the specific representation of the operator. This also casts some doubts upon the possible success of methods like Banach Algebra methods, which do not use the structure of the underlying Hilbert space. They may be quite suitable for some classes of operators, under some specific assumptions (we have met such a situation in Chapter VI, Section 6), but I don't see how they could handle the whole problem. All this explains why the solution of the Invariant Subspace Problem is not easy: we have to find general means of exhibiting Invariant Subspaces of various natures. One can imagine that a given method may produce Invariant Subspaces of one type, for some class of operators, but it does not look simple to find a method which would work in all cases. Moreover, the three examples we have studied are very trivial ones : they are just multiplication operators. But if things look already bad at such an early stage, the future does not look very promising !

Chapter IX

Exercises on Chapter IX

Exercise 1. - Let H be a Hilbert space, 12 (IN, H) the space

Let S be the right shift on 12 (IN, H) , and let G be a closed invariant subspace of S in this space. Show that there is a Hilbert space HI and a function F, defined on D, with values in £(H1,H), such that: a) F is analytic on D and if zED, IIF(z)11 ~ 1, b) F(e i 8 ) is defined a.e. and is an isometry, c) the subspace G is of the form F.l2(IN, HI).

Exercise 2. - Let / be in H?", Show that 1, I, /2, ... form an orthonormal basis of H2 if and only if I(z) = Az, for some A, IAI = 1. Exercise 3. - Let F be the subspace of H2 : F

=

{I E H 2 i there is N(J) such that TI n ~ N(J) , /(1 - l/n 2 )

= a}.

Determine the closure of F in H2.

Exercise 4. - Show that the functions {e i n 8 I ; n ~ o) span L 2(Il) if and only if the functions {ei n 8lin ~ o} span L 2(Il). Exercise 5. - Let m be a non-constant inner function, M m the operator of multiplication by m on H2. a) Show that M m is a c.n.u, isometry (see definition of c.n.u. in the next chapter). b) Show that M m is unitarily equivalent to the multiplication by ei 8 if and only if there exist A, IA\ = 1, and a, [o] < 1, such that: m () z =

A z-a 1- ii:z '

zE D.

Exercise 6. - Let 4>, t/J be in L oo , with 14>1 = t/JI 1 a.e. 4>.H2 = t/J.H2 if and only if 4> = At/J, for some A, IAI = 1.

Show that

Multiplication by ei 8 Exercise 7.

~

I

Let

205

be an outer function. For A > 0, set :

UA

=

min(A, -log lJ(e i 8 )I) .

(1)

1) Let IA be the outer function such that log IIAI = UA and IA(O)/(O) > O. Prove that IIA I ~ eA , and that UA (ei B) + log I/(e i 8 ) I s O.

2) Show that IluA+logl/llll--+O,when A--+ +oo,andthat IA(O)I(O)--+ 1.

3) Choose a sequence (An)n>o, increasing fast enough, so that:

L (1 -

IAJO)I(O)) <

00.

n~O

Let In

= I An

be the corresponding functions. Show that:

L III - In.f112

<

00.

n

4) Deduce from 1) that:

I/n(z).I(z)1 < 1,

zE

D,

and from 3) that:

In(e i 8).f(e i 8 ) --+ 1,

n --+

00,

a.e.

Compare with Beurling's Theorem (Corollary 1.2).

Exercise 8. - Prove that the set of cyclic vectors, for the right shift on '2(Z) is dense in this space. Exercise 9. - The set of cyclic vectors for the right shift on 12(IN) is not dense. We will show that no sequence of outer functions can converge in H2 to

I(z)

= z. Assume that a sequence (In)n~O a) Show that In(O) --+ /(0) =

of outer functions converges to

I.

o.

b) Show that : 71" / -'If

dO log Iln(ei 8 )1~ 21r

(Use the Theorem 2, in the "Complements" of Chapter VIII.) c) Obtain a contradiction and conclude. This applies as well to any Blaschke product. Conversely, singular functions are limits of sequences of outer functions : if I is singular, I (r z) , o < r < 1, is outer.

Chapter IX

206

Exercise 10 (Szego's Theorem, general case). - Let JI. be a positive measure on n, Jl.a. and Jl.a its absolutely continuous and singular parts, with respect to Lebesgue measure. Show that: inf gEA o

{II - fl 2 d Jl. in

= inf {II - fl 2 d Jl.a. gEA o

in

(If F is the projection of 1 onto Ao , closure in L 2 (JI. ), then Exercise 11. - Let a We define d = a * b by : dn

= (akhEz

=

L

be in co(71), b

aibn-i ,

•

1- F = 0, Jl.a-a.e.)

= (bi)iEZ

be in ld71).

n E '/l.

iE'll Then (dn ) is in co('ll). Show that a is cyclic for the right shift S on co(71) if and only if, for every b E ld71) , d has infinitely many non-zero coefficients di'

i < o.

Notes and Comments:

Our presentation is a mixture between several authors. For further acquaintance, we recommend our usual references: Hoffman [1], Garnett [I], Duren [1], Douglas [1], Rudin [1]. Exercise 7 is taken from Garnett [1].

Complements on Chapter IX

1. Approximation of a non-cyclic vector of S on

12(IN) by cyclic ones.

These results are closely related to the approach to the Invariant Subspace problem described in Chapter XIV. Take /(z) = exp((z + l)j(z - 1)). This is a singular inner function, and /r(z) = /(TZ) , 0 < r < 1, are outer functions (as easily checked), which converge to / a.e. and in L 2 , when circle.

T -+

1-. They do not vanish on the unit

Multiplication by ei B

207

Take a sequence (Fn)n>o of outer functions with:

Then

111/Fnll oo

cannot be bounded. Indeed, if:

then:

But: 1-

f / r; ,

and this implies that : 1 E span {f, z[, z2 t,

.. .},

and since 1 is cyclic for S in 12(IN), 1 would be cyclic. This is not true. So we have the following quantitative problem: How fast does Ill/Fnil oo ---+ +00 ? The following theorem is due to L. Kahanpaa [1] :

Theorem. - There is a C > 0 such that, for every e > 0, every outer function F with:

II F

-

1112

<

C,

(1)

then:

There is a function F satisfying (1) and such that:

General results, concerning the approximation of non-cyclic elements by cyclic ones for a given contraction operator were obtained by Rene Martel [1].

eOB

Chapter IX

2. Convolution operators. We have studied multiplication operators on HP spaces (and further work will be done in Chapter XII), we may consider similar questions for convolution operators. In what space should a distribution T be taken so that the operator:

(1) is continuous from HP into itself? H we write Fourier expansions :

f =

Ea

ne

in8

,

n~O

(1) means that (tna n) is the sequence of Fourier coefficients of an HP function. For this reason, the required space is called the space of multipliers of 1 BP, and is denoted by M 1 HP (people often speak about "multipliers of HP ", in short). In H2, one sees immediately that a necessary and sufficient condition for T to be in M1 H2 is that the sequence (t n ) should be bounded. On L 1 , defining M1 L 1 the same way, one checks easily that T is in M 1 L 1 if and only if T is a measure on II. On HI, things are not so simple, and no complete description is known. We now describe a space, introduced by N. Varopoulos, which was thought for some time to be a reasonable candidate for M 1 HI. First, we consider the space (see the "Complements" at the end of Chapter

V) Its dual Vo• is the space of all operators from Co into 11, equipped with the operator norm. Let V be the bidual of Vo ; it is a dense subspace of 1(X){IN x IN). Now we define: JlV

=

{c = (Ci,k)i,kEIN ;

Set m n = Ci,k when i +k to the sequence (m n ) .

=n

C

E V , Ci,k

= Ci',k' if i + k = i' + k'}

; we can identify the double sequence (Cj,k)j,kEIN

Multiplication by eiO

209

Let now .M be the space of measures on II, 1'.M the set of Fourier transforms. We finally introduce : 1'M(IN)

=

{(an)n;?:O; there is a measure JJ , such that ,l(n)

= an

, 'Vn ~ a},

equipped with the quotient norm:

Then, 1.M(IN) is contained in )IV. Indeed, let JJ be a measure. It defines a continuous linear functional on Vo• (and thus an element of V) by means of the formula: (ai,k)i,kEIN

--+

f " (7"""""L..: _,,"

"0 e''U )dJJ(8) ai,k e'1

(an operator from Co into 11 is here represented by its matrix (ai,k)j,kEIN)' E 12, b =

Also, JlV is contained in M1 HI. Indeed, if a = (an)n~o (bn)n>o E 12, we set: n

(a * b)n = L an-kbk , k=O which is the sequence of Fourier coefficients of the function f 9 ,

f

= Lane

inO,

n~O

9 = Lbne n;?:O

inB.

Define T, from 12 ® 12 into l'HI, by : a®b

--+

a

* b.

The operator T is surjective and has norm 1, if the norm on 1 H I is defined by the formula :

IlaiiTHl = IlfllHl. Then if 0: = (O:j,k) E '2®'2, and fJ = (fJi,k) E co®co, the (fJi,kO:i,k) and is in '2®'2, and this remains true if fJ is in V.

product fJo: is

If fJ E JlV and (m n ) is associated with it, one checks that:

T(fJa)

=

mT(a),

and this shows that (mn ) is in M1 HI, with norm K I!fJIINv. Both inclusions 1.M(IN) c JlV c M1'H I are strict. This was proved by G. Bennett [1] for the first, and by Francoise Lust-Piquard [1] for the second.

This page intentionally left blank

PART V

DILATIONS and EXTENSIONS

o Ie pauvre amoureux des

pays chimeriques ! Faut-il le mettre aux fers, le jeter a. la mer, Ce matelot ivrogne, inventeur d' Ameriques Dont le mirage rend le gouffre plus amer ?

In this Fifth Part, we have gathered results inspired by the following idea: instead of studying the original operator T defined on a Hilbert space H, one builds a bigger Hilbert space }I, and an isometry f on it, which "extends" T in some sense. If the links between T and f are strong enough, one can derive interesting informations for T from well-known facts about isometries. If the new space }I contains H as a closed subspace, such a construction is called a dilation, and if it contains it as a dense subspace, the construction is an extension.

Chapter X

Minimal Dilation of a Contraction

Let H be a Hilbert space, and T a contraction on it, that is an operator satisfying

IITII

~ 1.

As previously, we denote by T· the adjoint of T. Since is also a contraction.

IIT·lI

=

IITII, it

Let's first give or recall a few definitions : a) T is an isometry if IITxl1 = equivalent to T· T = I , or to :

Ilxll,

(Tx,Ty)

for all x in H. This property is

(x,y),

for all z , y in H. b) T is unitary if it is a surjective isometry. This is equivalent to the fact that both T· T = I and TT· = I. As a special case, a unitary operator is normal. c) A closed linear subspace F is called reducing for T if T F = F. d) T is said to be completely non-unitary (in short: c.n.u.] if there is no reducing subspace F such that TIF is unitary. . 1. Construction of the isometric and unitary dilations.

The following Proposition, valid for isometries only, provides a decomposition of the space into two parts: one on which T is surjective, and the other on which T is c.n.u.

Proposition 1.1 (Wold's Decomposition). - Let T be an isometry. Set

Then H = H o E:B HI, and moreover: a) The space H 0 is reducing for T,

Chapter X

b) The space HI is invariant for T, reducing for T·, and Tln 1 is c.n.u. Of course H o or HI may be {o}. Proof. - Set L = (1m T).L = H e T H (we recall that this notation means that L is the orthogonal complement of T H). Then : HI

since (ImT).L Therefore:

=

=

EBn~oTn

(1)

L,

KerT· (see B.B. !1], p. 39). But TnL 1.. TmL, if n

i-

m.

(2) From this, we deduce that, when n

-+ 00,

HI = HeHo.

Now, we observe that THo = H o, so T is unitary on H o. The fact that T HI c HI is obvious from formula (1). It implies T·T HI c T· HI, and since T·T = I, Let y E HI. Then, for x E H o ,

(T·y, x)

= (y, Tx) =

0,

so T·y E HI, and T· HI C HI. So finally HI is reducing for T· . Clearly, Tln 1 is c.n.u. : if there is a subspace F C HI with TF = F, then F C H«. This implies that F = {O}, and our Proposition is proved.

Corollary 1.2. - Let T be an isometry on H. There exists a Hilbert space H as a closed subspace, and a unitary operator l' on it, such that Tin = T.

it , containing

We then say that T may be extended into a unitary operator.

Proof. - We will use the previous decomposition of H. On H o , T is already unitary, so we care only about HIOn HI, we consider once again :

We denote by l2(71,L) (or (TI Lh, see B.B. [1] p. 102), the space:

l2(71, L) = {(x n )nE71 ; X n E L 'tin,

L nE71

IIxnl1 2 < oo},

£15

Minimal Dilation equipped with the norm:

I (x n ) 1I1~

(L)

=

(L

Ilx nl1 2 ) 1/2.

nE?L Take now HI = 12(Z,L). This is a Hilbert space, and HI can be identified to a closed subspace of HI. Indeed, by formula (1) above, every point h in HI can be written:

with

Xk

E L for all k

Ilhll

~

=

0, and : 00

00

o

o

(2::: IIT kx kI12) 1/ 2 (L Il xkI12P / 2.

To h , we associate the po in t in

if 1

where we underline the Oth coordinate. We get: 00

(L Il xkI1 2)1/ 2, o

and the mapping h ---t it is an isometry. The point Th becomes (... ,0, Q, xo, x}, ... ). If S is the right shift on l2(?L, L), that is :

S is a unitary operator and T = SIH 1 • To construct iI and i , we finally take ti«, S on HI.

iI

= Hi, EEl

HI, and T is T on

Let now H, )/ be two Hilbert spaces, H being a closed subspace of )/. Let T be an operator on H, and U an operator on )/. We say that T is the projection of U, and we write T = Pr (U) if T = PU, where P is the orthogonal projection of )/ onto H. This is clearly equivalent to :

(Tx,y) = (Ux,y) ,

X

E )/, Y E H.

We also say that the couple ()/, U) is a dilation of the couple (H, T) if: for all n

~

O.

We say that this dilation is isometric if U is an isometry, and that it is unitary if U is unitary.

Chapter X

e16

An isometric dilation is said to be minimal if : )I

= span

U"(H)},

{U"~o

An unitary dilation is said to be minimal if :

This terminology is justified the following way: for an isometry (not invertible), we are interested in the images T"(H), n ~ 0, and the smallest dilation should contain only these spaces, therefore is the closed linear space they generate. For a surjective isometry, we are interested in the images T"(H), for n E 1l.

Proposition 1.3. - Let T be a contraction on H. There exists a minimal isometric dilation (U 1 , )lI) of (H, T). If P is the orthogonal projection from )11 onto HI, we have: TP = PU 1

and H is invariant for

Proof.

~

,

Ui .

We consider the space '2( IN, H), defined as above:

equipped with the norm :

Then H can be embedded in l2(IN,H) by x We consider the operator :

--+

(x,O,O, ... ).

which is self-adjoint and positive: (V x, x) ~ 0, for all x E H. By Chapter VII, Proposition 1.2, it has a square root, D, which is also positive, self-adjoint, and satisfies, for every x E H :

IITxll2 +

IIDxll2

(Tx, Tx) + (Dx, Dx) = (T*Tx,x) + (D 2x,x) =

= (x, x)

=

Ilx112.

(1)

Minimal Dilation IT i: = (Xn)n~O,

217

we now define:

and we get, by (1) :

II U l xllr2 ( H )

=

Il T x o1l 2 + II Dxoll2 +

L Ilx,1I ,~

= II xol1 + L Ilx,11 2

2

I

2

i~l

=

Il xl r ( H ) 2

,

which proves that U 1 is an isometry. Let P be defined as the projection from 12(IN, H) onto its Oth coordinate: this is a projection from 12(IN,H) onto H. We identify x with (x,D, ...). Then, for x E H, n ~ 0,

which proves that (12(IN, H), Ud is an isometric dilation. This dilation does not need to be minimal, since 12( IN, H) is too big a space in general. So we have to restrict it. We simply put:

which is a closed subspace of '2(lN, H). Clearly, U1 operates from }II onto itself, and the couple (}II, Ut) is the required minimal isometric dilation. The projection P from }II onto H is of course the restriction to }II of the projection from 12(lN, H) onto H. So we get, for every z E H, every n ~ 0,

which shows that, on }II, TP

PU 1

•

IT x E H, z E }II, we have: (T* z, z)

=

(T* z , pz)

=

(x, T pz)

=

(x, PU1z) = (x, U1z)

=

(U; z, z),

which proves that Ui = T* on H, and this implies of course that H is invariant under U;. One checks easily that :

This finishes the proof of our Proposition.

Chapter X

£18

Corollary 1.4. - For every x, y in H, every n, m

and if n

~

~

0, if n

~

m,

m,

From this Corollary follows obviously that the scalar products (Uix, Urny) do not depend on the construction of (Nb Ud. So, if (NL UD is another minimal dilation of (H, T) 7 there exists a surjective isometry A, from Nl onto i/ { 7 such that :

In this sense, the minimal isometric dilation is unique. We will now extend it further into a unitary dilation.

Proposition 1.5. - Let T be a contraction on H. There exists a minimal unitary dilation (JI 7 U) of (H, T). Proof. - We take (i/hUd given by Proposition 1.3. We extend (i/l,Ud into (Jl1 7 Ud by Corollary 1.2. We then take U = and:

o..

Corollary 1.6. - For every x, y in H, every n, m (Unx, Umy)

= (Tm-nx,y)

~

,

n~m,

= (x, Tm-ny) ,

n~m.

0, we have:

Once again, this implies that the minimal unitary dilation is unique in the sense that, if (N', U ') is another one, there is a surjective isometry A from )I onto )II, such that : U'A = AU. In the sequel of this Chapter, we will refer to U as "the" minimal unitary dilation of T (in short: m.u.d.), and we write simply U = UT. The following properties are obvious: - If T is unitary, UT = T, - If T

= T' e Til,

UT

= Ut EB U!J.,

- If T is an isometry and T = To then

UT

= To EB UTI'

e T1

is the Wold decomposition of T,

Minimal Dilation

219

Though the above construction is quite elementary and involves no deep tool, it allows us to give a very elegant proof of Von Neumann's Inequality, which, in its original form, was not so easily obtained.

Proposition 1.7 (Von Neumann's Inequality). - For every contraction T on H and every polynomial p, IIp(T) II ~

max Ip(z) I·

Izl$1

Proof. - Let (JI, U) be the m.u.d. of (H, T). We have peT) = Pp(U), so IIp(T)11 ~ IIp(U)lI· But since U is unitary, we have, by Chapter VII, theorem 4.3, c),

IIp(U)lI

<

max{lp(z)I ; z E u(U)} ~

max Ip(z)l.

Izl=1

We can now come back to the general study of (JI,U) :

Proposition 1.8. - Let (JI, U) be the m.u.d. of (H, T). Every eigenvalue of T of modulus 1 is an eigenvalue of U, and conversely. The corresponding eigenvectors are the same for both operators. Proof. - a) First, we have, for x E H, if IAI = 1 :

IlUx - AxII 2

= IIUx[l2 = 211xIl

2

= 2~(Ax

-

AX) + [IxII 2

2~(Ux,

AX)

2~(Tx,

- Tx, AX)

~ 211AX -

Tx[l.lIxll,

and this implies UX = AX if Tx = AX.

x

x

b ) Conversely, let's assume that U = AX, IA1 = 1, for some E }I. It's clear on the construction made in Corollary 1.2 that U on jj I has no eigenvector, so x must be in H o . Therefore, if U I is the isometric dilation of

T, we have Uli

=

AX. With, as before, X

=

(xn)n>O' we deduce from this

equation:

and for n

~

Txo

AXo,

Dxo

AXI ,

1,

L IIxn l 2 < 00, = X2 = ... =

from which we get, since

XI

Tx« = AXo,

Xn

=

...

Dxo =

=

0,

o.

This last property follows from the previous, since (I - T·T)xo = O. Indeed,

eeo

Chapter X

Lemma 1.9. - If T is a contraction and Xo is an eigenvector of T for the eigenvalue A, IAI = 1, Xo is also an eigenvector of T· , for the eigenvalue X. J

Ilxoll = 1. Then: I(T'*xo,xo)1 = l(xo, Txo)1 =

Proof of Lemma 1.9. - Assume

1,

which implies that T'* Xo is proportional to Xo : T'* Xo

a

=

(T'* xo, xo)

=

(xo, Txo)

= axo.

But then:

= X,

which proves our Lemma. Coming back to the Proposition, we see finally that UX = Ax implies x = (xo,O, ...), and Txo = AXo, and our Proposition is proved. 2. Decompositions of the space )( . We now study the decomposition of )I into a Hilbert sum of pairwise orthogonal subspaces. When we write an expression like: )I

= ... EEl L_ n EEl L- n+ l

we mean that every

x in

)I

EEl ••• EEl L o EEl L 1 EEl· •• EEl L n EEl •••

has a decomposition

x =

LXIc' IcEZ

where Xlc is in Llc for all k E Z. The LIc's are pairwise orthogonal, and therefore, IIxII II xlc1l 2P / 2.

(L

icE'lL

We now set

L

=

(U - T)( H) ,

where the closures are taken in

L'*

(U'* - T'*)(H) ,

)I.

Proposition 2.1. - For every m, n E IN, U'" L is orthogonal to U" L, and U": L'* is orthogonal to U" L'* . The space )( can be decomposed into: )I

= .'. EEl u,*nL '* EEl ••• EEl U'* L'* EEl L· EEl HEEl L EEl UL EEl ••• EEl UnL EEl ...

Proof. - a) We only need to show that U" L .1 L, for n 2: 1, and the same for L'*. But, for x, y E H,

(Un(U - T)x,(U - T)y)

=

= iU?», y) - (Un- 1 Tx, y) - (Un+ 1 x, Ty) + (UnTx, Ty) = (Tnx,y) - (Tnx,y) - (T n+ 1x,Ty) + (T n+ 1x,Ty) =0 and the same for (u·n (U'* - T'*)x, (U'* - T'*)y).

eel

Minimal Dilation

b) Let's show first that all terms in the decomposition are orthogonal. We still have to show that:

U"L 1. H ,

But one checks immediately that, for n, m

~ 0,

z , Y E H,

(U"(U - T)x, y) = 0,

(U+m(U+ - T+)x,y)

=

0,

which implies our claim. Let's now call )I' the expansion, and let's show that N = )I'. First, we observe that:

(1) Indeed, to prove (1), it's enough to see that:

U(U+ - T+)H

(B

UH = U (B (U - T)H.

(2)

But if:

x with

x', z" E

x' + (U - T)x",

H, then:

with:

T+x' + (I - T+T)x" ,

x' - Tx" .

Conversely,

x' = TXl

+ (I -

TT")X2 ,

x"

and this proves (2). Using (1), we get that UN ' = N'. But )II contains H, and U is the minimal unitary dilation, so N' = )I, and our Proposition is proved.

Chapter X We now put:

Proposition 2.2. - Let Ho denote the largest subspace of H on which T is unitary (cf Proposition 1.1). Then:

Consequently, if T is c.n.u.,

Proof. - Set Hb = span {M,M-}.L. Let z E and M-. Thus x 1- un L, x .1 U:" L-, for n

Hb. Then x is orthogonal to M ~ o. So x E H, by Proposition

2.1.

But also x 1- U- L, and so :

o

=

(x,U-(U-T)x)

=

(x,x)-(Ux,Tx),

which implies (Ux,Tx) = Ilx11 2 • This formula proves that Ux and Tx are proportional, and therefore U x = Tx for all x E H . Since Hb is invariant by U, it is also invariant by T. The formula T = UIHIo shows moreover that T is an isometry on Hb. Furthermore, T is surjective on Hb. Indeed, take x E Hb, and y = U· x. Then y E Hb, and Ty = z , So Hb c Hn , the largest subspace on which T is unitary. b) Conversely, if x E H o , U":» = Tn x , for n ~ 0, and U":x = T'":":»; for n:S o. So U":z E H, for all n E 'fl. Thus unx.l L, z L U-nL, so z E M. the same way, we show that x E M-. Thus x E Hb, and our Proposition is proved. We observe the formula: Ho

= span {M,M·}l=

nn>O (Ker (I - T·T)T n- 1 ) n (Ker (I - TT-)T· n- 1 ) .

The spaces M, M· are reducing subspaces for U, and therefore the spaces:

R.. = )leM are also reducing for U. We set:

R = UL~

,

R. = UIR••

The operator R is called the residual part of U, and R. the *-residual part. They are unitary operators.

Minimal Dilation

1!1!9

We also denote by PR., PR. the orthogonal projections from

onto R.,

}I

R. •. Proposition 2.3. - For every x in

PRX =

lim unT·nx ,

n-+oo

}I,

the following limits exist in

PR.X

=

}I

and:

lim u-nTnx.

n-+oo

Proof. - We only prove the second relation, the first one being similar. Since T is a contraction, the sequence IITnxlln~o is decreasing, therefore convergent. So, if 0 ~ m ~ n,

Now,

lIu-nTnxIl2+llu-mTmxIl2 - 2!R(u- nTnx, u-mTmx), which proves that limit. In order to x-yEM. The first fact H .L u m + n L. For

(u-nTnx)n>o is a Cauchy sequence in }I. Let's call y its show that y = PR. x, we have to show that y .L M, and is clear, since u-nTn x .L U'" L, for n the second, we write ;

~

-m, because

and this expression is in M. This proves our Proposition. IT T is not unitary, Land L. are not {O}, and ;

are reducing subspaces for U. On them, U acts like the right shift on 12 (Z,)/) ;

x = (xn)nEZ' with

X n E U" L for all n, then Uii: = (xn-d. We may write }I = M EB R.., and on M, U acts like the right shift. This means just, in fact, that for z EM, Ule x .L z , for all k ~ O. IT, instead of the whole space, we start with a point x E H and set

if

M(x) = span {Unx i n E Z}, we can do the same reasoning. Put

M_(x) = span {Unx ; n

~

oj.

Then M+ (x) nM_ (x) c H, and if T is c.n.u., one of them, say M+ (x), cannot reduce T. IT we take F = M+ (x) e U M + (x) , we get F =I- {O}, and U IeF .L F, for all k > O. So we have obtained ;

Chapter X

Proposition 2.4. - Let T be a c.n.u. contraction on H, and let (JI, U) be its minimal unitary dilation. For any x E H , set

M(x) = span {U"x ; n

E Z}.

In M(x), there exists a point a such that una is orthogonal to a, for all nEZ,ni=0.

Proof. - By what we just said, this is indeed satisfied for any a in M or M •. We are now going to study the spectral measure E of U. We recall that E BOO (II) , any z ,

o(U) C C, since U is unitary. By Chapter VII, for any f y E JI, (f(U)x, y)

IIf(U)xIl 2

In fdEz,J/(B), = In IfI dE z(B). =

2

z•

Theorem 2.5 (Nagy-Folas}. - If T is c.n.u., the spectral measure E of its minimal unitary dilation U is absolutely continuous with respect to Lebesgue measure on II, and so are the scalar spectral measures Ez,z, for x in H. Proof. - Take any a given by Proposition 2.4. Then:

and since (U"a,a)

=

° if n i= °and = lIal1

For every measurable subset 0

IIE(6)aI1 2 =

2

en, we get

(E(o)a, a)

=

I.

for n

= 0, we obtain:

:

dEa,a(B)

= IIal1 2 P(6),

(1)

where P is, as before, the normalized Lebesgue measure on II. If P(o) = 0, E(o)a = 0, and since E(o) commutes with Ur , E(6)un a = 0, for all n E 'lL, and so E(o)x = 0, for all x in M or M., and therefore on JI , by Proposition 2.2. Conversely, let 0 be such that E(6) = 0. We have E(o)a = 0, and by (1), P(6) = 0. Therefore, E is equivalent to the Lebesgue measure: for every measurable set 6, E(o) = if and only if P(O) = 0.

°

22S

Minimal Dilation

Let's now turn to the scalar spectral measures E,;,,;, x E H. Of course, if P(6) = 0, E,;,,;(c5) = 0, since E(6) = O. Conversely, let x E H, 6 en, such that E,;,,;(6) = O. This implies :

since 16(U) is a self-adjoint projection. So 0 = 1116(U)xll = IIE(c5)xll, and E(6)x = O. This relation holds, a fortiori, for any point a in M(x). Applying it to the point a given by Proposition 2.4, we get, by formula (I), P(6) = 0, and this proves the Theorem. We will see a direct proof of this result in Chapter XII. Take any point a given by Proposition 2.4. Write N = M(a) EB N'. Then, as we already saw, on M(a) U acts like the right shift. Therefore,

o(U) :) O(UIM(ll)) = C. But since U is unitary, o(U)

C

C. So we get:

Proposition 2.6. - If T is c.n.u., o(U) = C. We now come back to the scalar spectral measures E,;,,;. By Radon Nikodym's Theorem, they have a density with respect to Lebesgue measure: there exists a positive function f,; in L 1 [Il, d8/27r), such that:

The following result strengthens Proposition 2.6 :

Proposition 2.'1. - For every x =I 0 in H, log f,; is integrable. Proof. - For every polynomial p, we have:

Assume the conclusion to be false. Then, since flogf,; rem implies : inf Ilx - p(U)xll = 0, p,p(O)=o

which means:

= -00, Szego's

Theo-

ee6

Chapter X

The same way,

These equalities show that both M+(x} and M_(x} reduce U, therefore the subspace M+ (x) n M_ (x) C H reduces T, which contradicts the assumption according to which T is c.n. u. and proves the Theorem.

Minimal Dilation

Exercises on Chapter X.

Exercise 1. - Let T be an isometry on a Hilbert space H. Show that the following three properties are equivalent : a) T is c.n.u., b) For no point x in H, there exists a sequence (Xk)k~O z for all k,

such that Tk xk =

c) There exists a Hilbert space HI such that T is unitarily equivalent to the right shift on 12(IN,HI)o

Exercise 2. - An isometry U on H is said to be pure if The multiplicity of U is dim K er U· .

nn~O

un(H)

= {a}.

a) Show that a pure isometry of multiplicity 1 is unitarily equivalent to the multiplication by ei tJ on H2. b) Show that a pure isometry of multiplicity N is unitarily equivalent to the sum

Ear Me's

on

Ear H

2

•

Exercise 3. - For a E
Exercise 4. - Show that M(L) = H if and only if T" M(L·) = H if and only if T'": - t 0 strongly.

-t

0 strongly, and that

Exercise 5. - Let T be a c.n.u. contraction. Suppose that the intersection o(T) n C has Lebesgue measure O. Prove that H = M(L) = M(L·) Conclude that both sequences T" and T"": converge to 0 strongly. 0

Exercise 6. - For a polynomial p = Li>o ai zi , put a contraction on 11 (IN) ; show that :

Ipll

= Li lai I. Let T be

IIp(T) II < Ipll' and that this trivial inequality is best possible on 11. (About extensions of Von Neumann Inequality, see "Complements" below.)

Chapter X

228

Exercise 7. - A result about convergence in norm of the sequence in L p •

IITnll

to 0,

We say that an operator T on L p (or, more generally, on a Banach lattice) is positive if for every positive function f, T f is also a positive function. We set, for n ~ 1, An = (I + T + ... + Tn-l )/n. We assume that there is an> 1 such that II An II < 1, and that T is a contraction. We want to show that the sequence IITkll tends to 0, exponentially fast. 1) Show that I - An is invertible, and that its inverse B is a positive operator. 2) Set: C

=

.!:.((n-l)I+(n-2)T+ ... +T n- 1 ) n

Prove that I - An = (I - T)C. 3) Set D = BC. Show that D is positive, and that it is the inverse of

I-T. 4) Prove that for all m, D(I - Tm+l) = 1+ T

D

~

+ ... + T m,

and that

D(I - Tm+l). (the statement D

~

D' means that D - D' is positive.)

5) Deduce that D + DT + .,. + DTm -:; D2. 6) Show that the sequence DTm is decreasing, and deduce that:

mT m -:; mlrl?" -:; D + DT + ... + DT m . 7) Deduce that a {) < 1 such that :

IlTm ll -:; IID2 11/m, for

all

m. k

Deduce that the sequence

IITmll

~

Fix mo, show that there is

o.

tends to 0 exponentially fast.

8) Give an example showing that the result is false if the positivity assumption is removed.

Notes and Comments:

All results presented in this Chapter can be found in Nagy-Foias [1]. We have simplified the presentation and the notations. Exercises 1 to 6 also comes

Minimal Dilation

eeg

from Nagy-Foia., [11. Exercise 7 is a presentation of a nice result of A. Brunei and L. Sucheston [11.

Chapter X

290

Complements on Chapter X :

1. Von Neumann Inequality. We have met (Proposition 1.7) Von Neumann's Inequality:

IIp(T)11

~

max Ip(z)1

(1)

Izl~l

for any contraction T on a Hilbert space and any polynomial p. Two questions can be asked: if (1) holds for every contraction on a space E, must E be (isometric to ) a Hilbert space? and does (1) characterize contractions? For the first question, the answer is yes. The next theorem is due to C. Foiaa [11:

Theorem. - 1£ (l) holds for every contraction on a Banach space E, E is isometric to a Hilbert space. Proof. - Take y in E, lIyll = 1, and Xo in E·, with the rank-one operator T'x = xQ(x)y. Then IITII = 1. Fix consider:

IAI =

IIxoll = 1. Consider a, with lal < 1 and

1.

Then IBa(A) I = 1 (see Chapter VIII, Section 6). By (1), for all x E E,

Replacing x by (1 ~ aT)x, we get:

II (T ~ a)xll < II (1 -

aT)xll,

that is :

The point z being fixed, we choose Then:

for all

z, y E E, a E
with

xo, with Ilxan = 1, such that xQ(x) = 1.

lal < 1,

and also, obviously, for any

a E
Minimal Dilation

£91

Changing x into y, a into -0:, we obtain:

II v -

ax II =

II x - o:yll·

Take now a real 2: 1. Then :

Ilax - yll

=

Ilx - ayJl 1

2(llax - yl! + Ilx - ayll a a+y 1 II > Ii - +- 1x - -

=

-

2: 2:

2

2

a+l

-2-!Ix - yll IIx - yll

This implies that the radial projection: x

x

--+

W

is a contraction. This property characterizes Hilbert spaces( De FigueiredoKarlovitz [1]). To consider the second question, we must first rephrase (1). Say that T is similar to a contraction if there exists an invertible operator U such that T I = U-ITU is a contraction. If T is similar to a contraction, (1) implies:

IIp(T) II

s

K.

max Ip(z)I·

(2)

Izl~1

with K = IIUII.IIU-III. The question is now: if T is an operator on a Hilbert space such that (2) holds for any polynomial p, is T similar to a contraction? This question, due to P. Halmos, has been studied extensively, but is still open. An interesting result is due to V. Paulsen [11 : the conclusion holds under a slightly stronger assumption on T. We also mention a quantitative result of J. Bourgain [11 in this direction. IT T is defined on a finite-dimensional Hilbert space H, with dim H = n, and if T satisfies (2), there exists an operator U such that U-ITU is a contraction, and:

2S2

Chapter X

2. Dilations of positive operators on L p • The construction presented in this chapter applies only to Hilbert spaces. On L p spaces ( 1 ~ p < 00), a theory due to M. Akcoglu - L. Sucheston [11, [2] provides a dilation theory for positive contractions. An operator T on L p is positive if, for every function / which is positive, T / is also positive (this word has therefore a meaning here which is different from the one attributed to it in Chapter IV ; see also Exercise 7). A contraction, as before, satisfies IITII ~ 1. Theorem. - Let L be a L p space ( 1 ~ p < 00), and let T be a positive contraction on it. There exists another L p space B, a positive invertible isometry Q, from B into itself, a positive isometric embedding D of L into B, a positive projection P from B into itself, such that, for all n E IN

DT" = PQ"D. Moreover, the projection P can be chosen so that P(B) = D(L). If T is not a contraction, a similar result holds under some domination

conditions; see M. Akcoglu, L. Sucheston [2]. The proofs of these results rely on techniques which are quite different from those presented here for the L 2 case, and are much harder.

Chapter XI

The HOO Functional Calculus

1. Construction of the Calculus.

The construction of the minimal dilation, which we explained in the previous Chapter, will now allow us to construct an H'" functional calculus for contractions on a Hilbert space. So, as before, let H be a Hilbert space, T a contraction on H. Throughout this Chapter, we assume T to be completely non-unitary, which, as we recall, means that there is no subspace F with T F = F and the restriction TIF being unitary. Let now 4> in H?" (II). We will give a meaning to the operator 4>(T). For this, we first write the Fourier series of 4> :

4>(e i B) =

L

ak

ei k B

,

(1)

k~O

and, for 0 <

T

< 1, we set : (2)

So we may define :

4>r (T)

=

L

ak Tk T k ,

(3)

k~O

and this series converges in

.c (H) .

Proposition 1.1. - For every x in H, the limit lim 4>r(T)x

r--+l-

exists in H. The operator 4>(T) th us defined satisfies :

114>(T)\1

< 114>1\00'

(4)

Chapter Xl

Proof. - Let ()I, U) be the m.u.d. of (H, T). We first show that the limit lim,...... l - 4>r (U) x exists in )I. Since 4> E Hoo, it is bounded on the unit circle, and 4>(U) can be defined by means of the integral (see Chapter VII, Theorem 4.2) :

¢(U) = /.

u(U)

where E is the spectral measure of U. satisfies:

¢dE , By Theorem 4.3, Chapter VII, it

/ 1¢(e i 8 )

-

4>r(e i 8 )12 dE x, x(8)

/ 14>(e i 8 )

-

¢r(ei 8 )12 /%(8) :: '

since the measures Ez,z are absolutely continuous with respect to Lebesgue measure (see Chapter X, Theorem 2.5). Of course, the function /z is in £1. Furthermore, we know that 4>r -. 4> a.e, (Chapter VIII, Section 2), and that i8

sup l¢r(e )1 r,8

<

114>1100

(Chapter VIII, Section 1). From Lebesgue's Theorem, we deduce that, in )I,

¢r(U)x - 4>(U)x -. 0, and that:

114>(U)11

~

114>1100'

(5)

We observe that the convergence obtained is a strong convergence, not a convergence in operator norm. Since T = PU, we get:

(6) and this implies that 4>r(T)x converges, when r -. 1- , to a limit in H. We denote this limit by 4>(T)x, and from (5) and (6) we get:

¢(T)x = P¢(U)x, and

II¢(T) II < 114>1100, as we announced.

HOO Functional Calculus

1195

We now determine the adjoint of the operator ¢'(T). If ¢' E H oo, we denote by ¢'d the function defined by (see Chapter II, Proposition 2.2) :

A ED. This function is in H?", and if ¢'(eiB) = Ek~O akeiU , the Taylor series of ¢'d is : ¢'d(ei 8 ) = (ik eik8 .

L

k~O

Proposition 1.2. - The adjoint of ¢'(T) is given by :

Proof. - First, quite obviously, if T is c.n.u., so is T·. For r < 1, one checks immediately that :

When r ---+ 1- , the right-hand side converges, at every point z , to ¢'d(T'")x, and the left-hand side, to ¢'(T)'" z . This proves our Proposition. The mapping ¢' phism:

---+

¢'(T) , from H'" into £ (H), is an algebra homomor-

(4)?/J)(T) = ¢(T).p(T) ,

t/>,?/J E H oo

•

It defines a functional calculus. Obviously, if t/> is a polynomial, ¢'(T) coincides with the usual definition.

If 4>

= Lk~O

akeik8,with L lak 1<

00,

then t/>(T) = Lk~O

akT k .

Finally, if T is normal, we also get:

¢'(T)

=!.

¢'dE ,

O'(T)

where E is the spectral measure of T. In short, the Hoo functional calculus coincides with the previously known ones when both make sense. We now study the continuity properties of the H'" functional calculus.

Theorem 1.3. - Let T be a c.n.u. contraction. Then:

a) If ¢'n

---+

¢' in H?", then ¢'n(T)

b) If sup., II4>nlloo < 00 and ¢'n strongly and ultrastrongly.

---+ ---+

¢'(T) in £(H). ¢' a.e. on C, then t/>n(T)

---+

t/>(T),

296

Chapter XI

c) If
by Lebesgue Theorem. This proves b). For c), we have the same way, for every z , y in H, when n

--+ 00,

and our Proposition is proved. Let AT be the subalgebra of £(H) generated by T and I : this is the space of polynomials p(T) , pEP+ .

Corollary 1.4. - For every ,p in H?", the operator
Proposition 1.5. - Let v in H?", with Ivl < 1 in D. If T is c.n.u., so is

v(T), and, for every

u E H?",

u(v(T))

=

u

0

v(T).

Proof. - We consider the following transformation of the complex plane (Moebius transformation) :

z

--+

z-a

Ba(z) = --_-, 1- az

where a is a fixed point in D. We already met this transformation in Chapter VIII. Then IBa(z)1 = 1 if and only if operator on H, the operator V

Izi

1. Therefore, if U is a unitary

V = (U - a)(I -liU)-l is unitary, and conversely. V is called the Cayley Transform of U.

eS7

H'" Functional Calculus Here, we take v E HOO with

w(z)

Ivl < 1 =

on D. Thus

Iv(o)l < 1, and we set:

v(z) - v(O) , 1 -v(O)v(z)

which is a function in H?", satisfying Iw(z)1 < 1 on D. Therefore w(T) is a contraction. Set T' = v(T). Let H o be the space where T' is unitary. From what we said follows that Ho is also the space where w(T) is unitary, since w(T) is the Cayley transform of T'. We now show that T itself has to be unitary on H«. By Schwarz Lemma (see Cartan [1]), since w(O) = 0 and Iw(z)j < 1 on D, we can find a function a, analytic in D, such that w(z) = za(z), and lal ~ 1 on D. So a(T) is a contraction and :

w(T)

(1)

Ta(T).

For x E Howe get :

Ilxll =

IIw(T)xll = lI T a (T )x ll ~

IITxll,

which proves that T is an isometry on H o . Formula (1) implies that T is surjective on H o , since w(T) is. So TIHo is unitary, which implies H o = {O}, since we assumed T to be c.n.u. We have proved that v(T) was c.n.u. So w(v(T)) makes sense. For every polynomial p , we get :

For w E H?", let Pn be a sequence of polynomials converging to w for o(L oo , L.) (for example the Cesaro means of the Fourier series of w, see Chapter VIII, Section 2). Then, by Theorem 1.3, c), Pn(T') converges to w(T'} weakly. But also Pn ov converges to wov in o(L oo , Ld, so Pn ov(T) ~ wov(T) weakly. Therefore, w(T') = w 0 v(T). The outer functions playa special role in this functional calculus:

Proposition 1.6. Let T be a c.n.u. contraction, and F be an outer function, in HOO. The operator F(T} is injective and has dense range. Proof. - Let z be such that F(T)x =

o=

(Tn F(T)x, x) =

o.

For n

ur: F(U)x, x)

=

~

0, we have:

!W" ei n S F(e i S )/ %(8) dO . -W" 21r

Chapter XI

t98

This implies that the function G = F.fz is in HJ, the subspace of Hl consisting of functions which vanish at O. We set:

H(z)

=

exp

j

W"

eit + z -'-t-

-W"

e' - z

so H is an outer function, and lH(ei8)1 Proposition 5.1).

dt log fz(t) 2.... ' II

= Ifz(tJ)l, a.e. (see Chapter VIII,

Set G = Gi.Go, where G, is inner and Go outer. Then IHI = IGI/IFI on C. The two functions FH and Go are outer, and, on the unit circle: IFHI =

lFI·lfzl = IGI = IGol,

so they are equal in the unit disk (up to a constant of modulus 1), by Chapter VIII, Proposition 5.2. So F H = Go. This implies that: G F

Go . G" = H. G," F '

is in n» . So we see that fz is in u>. But fz is real valued, so, by Chapter VIII, Proposition 3.2, must be constant, that is fz = C, a.e. We deduce from the fact that G was in HJ that C . F is in HJ, meaning that C> F(O) = O. Since an outer function never vanishes inside the unit disk, C = 0, and fz = 0 a.e. But:

and so z =

o.

Let's show finally that F(T) has dense range. The operator T* is c.n.u. and F" is also outer (check on the definition) and in Boo. So F"(T*) = F(T)* is injective, and F(T) has dense range. Conversely, we have: Proposition 1.6. - For every function u which is not outer, there exists a completely non-unitary contraction T on a Hilbert space, such that u(T) = O. Proof. - We decompose u = m.F, with m inner, non constant, and we build T such that m(T) = O. We know (cf. Chapter VIII) that m.H2 is strictly contained in H2. Set H = H2 e m.H2 : this will be the required Hilbert space. Let M be the multiplication by z, on H2. One checks easily that the adjoint M* is given by : M*f

1

- (f(z) - f(O)) , z

HOD Functional Calculus

£99

which implies that M?" f ----t 0 when n ----t 00, for every function f in H2. The space m.H2 is invariant by M, so H is invariant by M*. We put:

We obtain a contraction T on H, with T':" = M*nI H n ----t 00. This implies that M and Tare c.n.u.

----t

0, strongly, when

By definition, if P is the projection of H2 onto H T

for n

~

P(MIH),

1. Therefore, for every w in HOD,

w(T) = P(w(M)IH)'

But we know that w(M) is the operator of multiplication by w on H2. In particular, for f in H, we have:

=

Pm(M)f

But m.] is in m.H 2 , so P(m.J)

= 0, and

m(T)f

=

P(m.J).

this implies m(T)

= O.

2. The Spectral Theorem for the HOD functional calculus. For the analytic functional calculus, we met (Chapter II, Theorem 2.3), the theorem : f(u(T)) = a(f(T)),

where

(1)

f is analytic on a(T).

We are now dealing with contractions, so a(T) c D. But f is not analytic on a neighborhood of D, only inside D. So formula (1) does not make perfect sense: if u(T) n C is not empty, how should f(a(T) n C) be understood? The function I is only defined a.e. on C, by means of a radial limit. So the left-hand side of (1) is not well defined, though the right-hand side is. For points inside the open disk, there is no difficulty:

Chapter XI

Proposition 2.1. - Let Then:

f

be a function in Hoo, and T a c.n. u. contraction.

f(o(T) n D)

o(f(T)).

C

(2)

Proof. - Set, for -X ED:

g(JL) = {(f(JL) - f(>..))/(I-£ - -X) /'(-X)

if JL E D\{A} if JL = -X

Then g is in HCXJ. Moreover,

g(T)(T - A) = (T - A)g(T) = f(T) - /(A). So, if A E o(T), A - T is not invertible, and /(T) - f(A) is not invertible either. This proves our claim. If / is in A(D), formula (2) can be extended to :

f(o(T))

(3)

o(/(T)).

C

More precisely, we have :

Theorem 2.2. - Let T be a c.n.u, contraction with spectral radius 1, and let A E o(T), with [AI = 1. Assume that / E H'" is continuous at A, that is :

f(A) =

lim

z-A,lzl<1

f(z)

exists. Then o(/(T)) contains I(-X). Proof. - Replacing T by T / A, we may assume A = 1, for simplicity. Put h(z) = I(z) - 1(1). Then h(l) = 0, and: lim

z-A,lzl<1

h(z)

=

(1)

0.

We consider the following sequence of functions:

Izi :S 1,

(1 - z)l/n ,

n ~ 1,

with, for instance, arg un(z) E

[

- '11'" 11"] ,.

2n 2n

We have un(l ) = 0, llu nlloo :S 2, and un(z) subset of D\l (2).

--+

1, uniformly on every compact

H OO Functional Calculus Lemma 2.3. - If the function h in H'" satisfies (1), then: when n

-+ 00.

Proof. - Let e > 0 be given. By (1), we can find a such that the neighborhood of 1 : VIl = {z; zED, Iz - 11 < a} has the property: if z E VIl , then Ih(z)1 < g. Since radial limits exist a.e., we also get that if Iz - 11 < a , and Iz I :s: 1.

Ih (z) I < e , Put CIl

= C n {Iz -11 < a}, and sup Iun(z)h(z) -

= C\C Il . h(z) I -+ 0, C~

(3)

Then, on C~, when n

-+ 00 ,

zEC~

by property (2) of the sequence (un)n~o, So, we can find no such that, for n

~

Iun(z)h(z) - h(z) I < e ,

no, n 2:: no, z E C~ .

On CIl , we have, by (3) : sup lun(z)h(z) - h(z)1 < 3g, zEC ..

therefore

Ilunh - hll < 4g, for

n 2:: no, and this proves the Lemma.

The properties of the HOO functional calculus (Proposition 1.1) imply:

h(T)un(T) - h(T)

-+

0,

in £(H).

(4)

We observe that un(T) - I cannot tend to 0 in £(H). Otherwise, for n large enough, un(T) would be invertible and so would be un(T)n = 1- T. But we have assumed 1 E o(T). Therefore, h(T) cannot be invertible, which means that f(T) - f(l) is not invertible, so f(l) E o(f(T)). The converse formula:

f(o(T))

:J

o(f(T))

is false in general: there are examples with o(f(T)) = D, and oCT) reduced to a point. See Foias, Nagy, Pearcy [1], and J. Esterle [11. To finish this Section, we wish to point out that if the spectral radius reT) is strictly smaller than one, the analytic functional calculus should be preferred to the HOO functional calculus. Indeed, it allows us to consider functions which are analytic on a neighborhood of {izi :s: r(T)} , a class which is much larger than HOO.

Chapter XI 3. The HOO functional calculus on an algebra generated by a single operator. Let T be an operator on a Banach space E. We have already met AT, the algebra generated by T in £ (H) : this is just the set of finite sums E,,~o a"T" . We consider now AT, closure of AT in £(B) for the ultraweak topology. Then we know (Chapter V, Proposition 2.12) that AT is a dual space, namely the dual of the quotient of the nuclear operators, JJ (H), by .1 AT, the pre-orthogonal of AT, that is : .1 AT

= {N E JJ(H) ; tr(Np(T)) = 0, for every polynomial p}

Since the trace is linear, we have also: .1 AT

= {N

E JJ(H) ; tr(NT")

=

0, for all k ~ a}.

To simplify our notation, we just put:

and AT is the dual of JJT. There is a weak topology on AT, which is o(AT, JlT), and, by Proposition 2.12, Chapter V, this topology is the trace on AT of o(£(H), JJ(H)). We observe that JJT' being a quotient of a separable space, is itself separable. So the balls of AT are metrizable for the weak topology (B.B. 11], p.

57). The algebra AT will be called ultraweakly dosed algebra generated by T. It contains the closure of AT in operator norm, and is contained in the closure of AT for the weak topology of £ (H) . The behavior of the HOO functional calculus with respect to this algebra is given by the following theorem :

Theorem 3.1. - Let T be a c.n.u. contraction on B and let AT be the ultraweakly closed algebra generated by T. Then: a) If (hn)n~o is a bounded sequence in Boo, converging to 0 at every point of the open disk D, hn(T) converges to 0 in AT for O(AT' JJT). H the convergence is uniform in D, then hn(T) converges to 0 in operator norm.

b) H (hn)n>o is a bounded sequence in Boo, converging to 0 a.e. on C, then hn(T) converges to 0 for the ultrastrong topology.

H'" Functional Calculu8

c) The mapping \If : h

--+

h(T) is continuous from HOO u(HOO, L 1/ HJ)

into AT, U(AT' )IT).

d) H \If is an isometry, \If is also an homeomorphism for the above weak topologies.

Proof. - a) Since the sequence (hn)n~o is bounded in Hoo, (hn(T» is bounded in l(H). Since weak and ultraweak topologies coincide on bounded subsets of l(H) (Proposition 2.9, Chapter V), we just need to show that hn(T) --+ 0 weakly, that is :

(hn (T)x, y)

--+

0,

for all x,y E H.

(hn(U)x, y)

--+

0,

for all x,y E H.

This follows from :

But since T is c.n.u. :

(hn(U)x, y)

1('

= /

l

.8

"hn (e )fz,JI(e

l

"8 )

dO -2 '

-pI

(1)

7r

where 12:,11 is in L 1 (see Chapter IX). Since h n (z) --+ 0, for all zED, this convergence is uniform in any disk of radius r < 1. We deduce that the derivatives h', h", ... , satisfy : for all k E IN. From this follows obviously by induction that for every k coefficient of h n , hn(k), satisfies : n

~

0, the k-th Fourier

--+ 00.

Consequently, for any polynomial p,

But polynomials are dense in L 1 , and the sequence (hn)n~o L oo. Therefore, for every 1 in L 1 ,

is bounded in

Chapter XI

and this implies our claim. The second part of a) is just Theorem 1.3, a). b) was proved in Theorem 1.3, b). c) We need to show that if h n --+ 0 for u( HOO , Ltl H J), then h n (T) --+ 0 weakly. But the assumption implies that for every k ~ 0, hn(k) --+ 0, when n --+ 00, and we proceed as in the proof of a). If
AT.

Indeed, let (Pn)n~o be a sequence of polynomials, bounded in L oo and converging to
£(H), and \II is surjective from HOO onto L. Moreover, L

Lemma 3.2. - A subspace L of a dual E· is u(E*, E) closed if and only if L n BE- is u(E·, E) closed.

Proof. - The reader may consult S. Banach [1] or Bourbaki [11 for more general versions of this result. However, we include a proof, which follows rather closely Banach's original ideas ([11, pp. 121-125). It's enough to show that for any 10 f/:. L, there exists x E E with lo(x) = 1 and I(x) = for all I E L ( a subspace L with this property is called regularly closed in Banach's terminology). Indeed, if L is not u(E·, E) closed, take 10 E F, 10 f/:. F. With the above point z , the neighborhood of fo

°

v

=

{g; Ig(x) - 10(x)1 < 1/2}

should intersect L : a contradiction. So, let 10 rf:. L, with 11/011 = 1. Of course, 10 rf:. nB L, for any n ~ 0. Since nBL is compact, we can strictly separate fo and nBL : there is a point X n in E, with Ilxnll = 1, and reals an > 0, en > 0, such that:

(1) for all

I

E nB L •

(2)

BOO Functional Calculus Next, we observe that :

(3) Indeed, assume not. Then, there is a sequence (In)n~o, with In E nBL for all n, such that In --+ 10 in E"'. But the In's are bounded in norm, so all of them belong to a given nOBL. We have:

Il/n - loll

2:: H/n - lo)(x noI > e no ,

a contradiction. Let e = inf n distE.(/o,nBL). Let: e

x;

2BE. + nBL ,

=

n 2:: O.

This is a compact set, for u(E"', E), which does not contain 10, by (3). So there is a point Yn in E, llYn II = 1, and a real Pn > 0, such that:

(4)

10(Yn) > Pn e

1 + nl(Yn)1 < Pn , 9 E BE·, IE BL, n 2:: o. 2g(Yn) From (5), taking f = 0, 9 such that g(Yn) = 1, we deduce:

e; and from (4),

IPnl < 1.

(5)

(6)

2:: ~,

From (5), we also deduce:

I!(Yn) I

s

(7)

Pn .

n Extracting a subsequence from the Yn's if necessary (we keep the same notation), we can assume that there is a I such that: and I 2:: e/2 .

(8)

We consider: This is a vector subspace of Co, by (7). In the space (c) of convergent sequences, condition (8) implies that the distance of the sequence (/0 (Yn))n to G is 2:: e/2. The dual of (c) is 11, so there is a sequence of complex numbers (Cn ) such that:

LICnl < 00, n

LCnl(Yn) = 0,

f

E

BL ,

n

L Cnlo(Yn) =1. n

Take z =

Ln CnYn

: this is the required point, and our Lemma is proved.

2.. 6

Chapter XI

Here, since \If is an isometry, L n Bl(H) is the image of the ball of Hoo, which is u(L oo , Ld compact. Therefore, B,f(H) is u(£(H), J/ (H)) compact, and therefore closed. By Lemma 3.2, L itself is closed for u(!(H), J/(H)). This proves that L = AT, and that \If is surjective. But then, 'If is an injective and continuous map from the compact BHao, equipped with u(L oo,L 1 ) , onto &:AT' equipped with u(AT,J/T). Therefore, 'Ill in an homeomorphism between these two spaces, equipped with their weak topologies, and our Theorem is proved.

H'" Functional Calculus

Exercises on Chapter XI.

Exercise 1. - Let T be a c.n.u. contraction on H t and let h in Boo be a non-zero function. Show that for every x in H such that h(T)x = 0, we have Tn x --+ 0 when n --+ 00. Exercise 2. - Let B a be the inner function:

Ba(z) =

z-a 1- az

--~,

Construct an operator T such that Ba(T)

lal < 1. = O.

Exercise 3. - (Mean ergodic Theorem). Let T be a contraction on a Hilbert space. Prove that the Cesaaro averages ~ L~-l Tk converge strongly in H, when n --+ 00. Exercise 4. - Let T be an invertible isometry on a Banach space E. Let

A(II) be the algebra of absolutely convergent Fourier series: / is in A(II) if / = L:~oo ci ei j 8 , with L:~oo ICil < 00. Show that f(T) can be defined by means of a normally convergent series, and that the usual properties of a functional calculus hold.

Exercise 5. - Let 8 be the right shift on 12(Z) and (Pn)n~O

be a sequence

of polynomials. a) Show that Pn(8)

--+

0 in £(12 ) if and only if Pn --+ 0 in L oo(I1).

b) Show that Pn(8) --+ 0 strongly if and only if (Pn)n>O is bounded in L oo (Il) and Pn --+ 0 in L 2 (II) . c) Show that Pn (8) --+ 0 weakly if and only if (Pn)n~O is bounded in L oo (II) and, for every j 2:: 0, the j th coefficient of Pn --+ 0 when n --+ 00.

Exercise 6. - Let 8 be the right shift on 12(Z), To every element of the algebra generated by 8 t that is to every finite sum Lk>O akSk, we associate the polynomial Lk~O

akz k•

-

a) Show that the closure of As for the operator norm can be identified to the algebra A(D) . b) Show that the closure of As for the strong topology can be identified to Hoo. c) Show that the closure of As for the weak topology can also be identified to BOO.

Chapter XI

Exercise 7. - Let S be the right shift on 12 (1L). To every element of the algebra generated by S and S-1 t that is to every finite sum E:=-M alcS k , we associate the trigonometric polynomial E~=-M alczlc. a) Show that the closure of the algebra AS,S-l generated by Sand S-I, for the operator norm, can be identified to C(IT) . b) Show that the closure of to L CXJ •

AS,S-l

c) Show that the closure of identified to L CXJ •

for the strong topology can be identified

AS,S-l

for the weak topology can also be

Notes and Comments: Section 1 comes from Nagy-Foias [1]. In Section [1] ; our proof follows J. Esterle [2]. 2, Theorem 2.2 is due to Foi~-Mlak

Complements on Chapter XI :

On the algebra generated by the right shift S, the weak and ultraweak topologies coincide. This property holds for a large class of operators, as described by Bercovici-Foiac-Pearcy [1]. A first example of an operator for which this property does not hold was built by D. Westwood [1]. Another was built by G. Cassier [11. Cassier's example has some interesting properties. Let T be the operator. Then the algebra AT is equal to the commutant {T}/, and also to the bicommutant {T}". Moreover, there is a nuclear operator A with the property that no finite rank operator B satisfies : tr (AU)

=

tr

(BU)

t

for all BEAT .

Chapter XII

C1-Contractions

Let E be a Banach space, and T a contraction on E. We say that T is a C 1-contraction if : For every x =j:. 0 in E, T" 0, when n ---l' +00. This is obviously the case for isometries. Here are two other examples:

r

Example 1. - On l2(lL), the weighted right shift : n E 'IL, is a contraction if

Iwnl

~ 1,

for all n E Z. It is a C 1- contraction if:

n

II

IWkl

f+ 0,

n

---l'

+00,

o

and Wk =I 0, for every k E 7L. For instance, for n ~ 0 gives a C 1-contraction.

Wn

= 1/4 for n

< 0 and

Wn

= 1

Example 2. - On the Hardy space J[2, the multiplication operator MtjJ, by a function 4> in Hoo, is a contraction as soon as 114>1100 ~ 1. It is a C 1-contraction when: P{O ; 14>(e i 8)1 = 1} > O. Indeed, put A

= {O; 14>( ei8) 1= 1}. 11 M ;

1112

Then, for

1 =I 0,

(/11" l4>n(ei8)/(eiO) 12 -11" ~ I/(eiO)1 2 dO)I/2

=

(!

A

dO) 1/2 2~

2~

> 0, since we know that a non-zero function in H 2 cannot vanish identically on a set of positive measure. In both examples above, the operator is c.n.u. except if n E 7L in the first case, or if 4> is identically 1, in the second.

W

These examples will be studied in great detail in Section 6.

n

= 1, for all

Ghapter XII

250

1. The extension of a Gt-contraction.

e,

In order to study T, CI-contraction on E, we will build a new space and extend T as an isometry on it. However, e here contains E as a dense subspace: this is quite different from the situation we encountered when we constructed the unitary dilation. The connections between both constructions will be studied in Section 3. We define a new norm on E, denoted by ~.~,

by the formula:

This is a norm because T is a GI-contraction. Obviously, for x E E,

(1) and

~xl

s Ilxll·

(2)

We denote by E the completion of E for the norm [. ~. The operator T can be extended, by formula (1), into an isometry from into itself, denoted by

e

T. From formula (2) follows that the canonical embedding from E into con tinuous and has norm 1. If E is a Hilbert space, so is

e, with the scalar product (for

e is

z , y E E),

(3) The limit on the right-hand side exists, since (Tm x, Tmy) is the sum of four quantities, decreasing with m : IITm(x ± y)lI, IITm(x ± iy)ll· One easily sees under what conditions the norms 11.11 and ~.~ lent:

are equiva-

Proposition 1.1. - The norms 11.11 and ~.~ are equivalent if and only if there exists on E a norm 11.111, equivalent to 11.11, for which T is an isometry. Proof. -a) Assume first that ~.~ is equivalent to 11.11 Then just take 11·111 since for ~.~, T is an isometry. b) Assume that there exists a norm 11.111, with :

ell·III ~

11·11 ~ GII·III .

= ~.~,

fSl

C I -coniraetions Then, for all x in E,

cllxlll >

C

C Ilxll·

So: ~x~

which implies that ~.~

~

c C IIxll,

and 11.11 are equivalent.

Therefore, in general (that is : except if T is an isometry for some equivalent norm on E), and E do not coincide. E is not a closed subspace of but a dense subspace. We now study the question: when is T surjective ? on

e,

e

e

t

Proposition 1.2. is a surjective isometry on of T is dense in E for the norm [. ~ .

e if and only if the image t is dense e. is a dense subspace of e,

Proof. - 1) Assume first 1m T to be dense in E for ~.~. Then Irn and since it is closed (T being an isometry), we get 1m f = in

e,

2) If f is a surjective isometry, T(E) = 1m T which means that Im T is dense in E for ~.~.

Corollary 1.3. - If T has dense range in E (for the original norm of E), is a surjective isometry. Indeed, a fortiori, this image is dense for

f

~.~.

2. Representing Functions.

In this Section and the next ones, we assume E to be a Hilbert space, and T to be a Cl-contraction, with dense range. The assumption "T has dense range" is a natural one when one is looking for Invariant Subspaces or Hyperinvariant Subspaces, since the closure of the image is hyperinvariant by T.

Being a surjective isometry, f is therefore normal. We denote by J-Lx,1I the scalar spectral measures (see Chapter VII), for z , y E So we may write, for i , g in BOO(II),

e.

(1) As a special case, this holds for

f,

g in

A(D).

Chapter XII

Let's compute the Fourier coefficients of JJz,,,. For k E 'Il, we have:

(2) Put simply CA:(x,y) = [T-kx,y]. If x, Y E l ,we can write:

CA:(x,y) = lim (Tm-A: x, Tmy). n-oo

(3)

We now show:

Proposition 2.1. - If T is c.n.u., the measures JJz,y, for x , y E l, are absolutely continuous with respect to Lebesgue measure. Proof. - We could deduce this from the results of Chapter X and the connections with the m.u.d. which we will study in Section 3. However, we prefer to give a direct proof. First, we observe that :

CA:(x,y)

=

1

4(CA:(x + y,x + Y)-Ck(X - y,x - y)+ + iCk(X + iy, x + iy) - iCk(X - iy, x - iy)X-4)

So we need only to establish the result for the measures J.Lz,z, which we just denote by JJz. We have to do it for every x E l , but we start with the case

xEE (x#O). Assume that, for some x E E, J.Lz is not absolutely continuous with respect to Lebesgue measure. Then there is a compact K c TI, with P(K) = 0, and

J.Lz(K) >

o.

By Fatou's Theorem (see Chapter VIII, Section 7), there is a function h in A(D), with h = 1 on K, and Ihl < 1 on D \ K. By the HOO functional calculus, h(T) can be defined. We consider the sequence (hn(T)x)n~o. This is a bounded sequence, since :

So there exists a subsequence h n " (T)x which converges weakly in E to a point z in E. Since the canonical injection from E into l is continuous, it is also weakly continuous, and this sequence also converges weakly to z in l .

259

C 1 -eoniractions For any g in A(D), we have, by (1) : lim [h nk (T)x,g(T)x]

[z,g(T)x]

k--+oo

=

lim k--+oo

111" hnk(eiB)g(eiB)d#lz (8) -II"

IK g(ei8)d#lz(8) . In particular, {z, xl = IJ,z(K) > 0, and so z 1= o. Using now the Rudin-Carleson Theorem (Chapter VIII, Section 7), we take a function ¢, with ¢(e iB) = e- i8 , for 8 E K, and 114>1100 ~ 1. We are going to show that T¢(T)z have:

= z.

Indeed, for every g in A(D), we

[(T4>(T) - I)z,g(T)xl = lim [h nk(T)(T4>(T) - I)x,g(T)x] k--+oo

=

lim k--+oo

=

111" (ei8¢(ei8) -

1) h nk (ei8) g(e i8) dlJ,z(tJ)

-II"

IK (eiB4>(eiB) -

1) g(ei8) dlJ,z (6)

= 0, by the definition of ¢. So we obtain, for every k

~

0 :

[(T¢(T) - I)z, (T¢(T) - I)hnkx] and, letting k

0,

-+ 00,

II (T¢(T)

- I)zll

0,

as we announced. Set V = 4>(T). Then V is a contraction on E, which commutes with T. For every finite sequence of complex numbers, (ak)k>o, we have:

(5) and thus:

IIT(L akTkz)11 k~O

This proves that T is an isometry on

Furthermore, T is surjective on Fz , since ¢(T)z E Fz , and z = T¢(T)z. This contradicts the fact that T is c.n.u., and proves our claim, when x E E.

Chapter XII

Since the measure "'z, x E E, is absolutely continuous with respect to Lebesgue measure, the series LkEZ ck(x)e i k8 is the Fourier series of an integrable function, which we denote by Az (B), and this function is the density of J.'z with respect to Lebesgue measure. So formula (1) can be written:

- - =I

[f(T)x, g(T)x]

dB , i 8)A f(e'°8)g(e z(B) -

(6)

271"

for x E E, f, 9 in Boo [Il] . We deduce from (4) :

(7) The functions Az are real valued, and more precisely positive, but the functions >'z,y are not real valued in general. To finish the proof of Proposition 2.1, we now take z E e. Since E is dense in we can find a sequence (xn.)n~O of points in E with X n --+ Z in We have, for every 9 in Loo(ll) :

e,

e.

In particular, for every Borel subset A ell,

If P(A) = 0, then J.'zJA) = 0, so "'z(A) = 0. This proves that J.'z is also absolutely continuous with respect to Lebesgue measure. Its density, denoted by >'%(8), has the Ck(Z) for Fourier coefficients, and:

Az(B)

"-J

L ck(z)e i k8 . kEZ

Our Proposition is proved; formula (6) extends without modifications to the case where z , y E e. From now on, we assume T to be c.n.u. We now investigate the transformation formulae for the functions >'z,II' x, YE

e.

Proposition 2.2. - Let f, 9 be in L oo (ll , d8/ 271" ) , and z , y E x' = f(T)x, y' = g(T)y. Then:

e.

We set

ess

C 1 -eontractions Proof. - We compute the kth Fourier coefficient of

" f = =

e

-ik8

dB 27r

Az'~'-

-W'

Az',~'

•

[1'-kx',y'j = !1'- k/(1')x, g(1')yJ

=

L:' -W'

e- 1k 8 / g A

z,1I

and this is the k th Fourier coefficient of

_dO11'" 2

by formula (1),

/g Az ,1I.

We observe that, since l' is a surjective isometry, 00(1') c C (Chapter II, Proposition 1.13). We have: Corollary 2.3. - All functions AZ ' 1I' x, Y E l , are 0 a.e. if eie

t. 00(1').

Proof. - Let / E Loo(II) , with /(e i 8 ) = 1 on 00(1'), and 0 on the complement. We know that /(1') = I (Chapter VII, Theorem 4.2). Applying Proposition 2.2 with 9 = /, x' = z , y' = y, we get :

which proves the claim. Proposition 2.4. - The spectrum of l' has positive measure, and :

00(1')

C

oo(T) n C.

Proof. - The first part follows obviously from the previous Corollary : for every x E l,

~X12

=

I

Az dfJ

211'"

imply that if x =F 0, Az is not identically o. This shows that 00(1') has positive measure. For the second part, we already observed that 00(1') C C. Let now A E ~ , A t. oo(T). Then A- T has an inverse, R(A, T). Since R(A, T) and T commute, we get, for m E IN, z E E,

Therefore,

Chapter XII

256

This shows that R(A, T) is continuous on E for the norm extended to into an operator which satisfies :

e,

0.1.

So it can be

OR(A, TH < IIR(A, T) II, and:

(A -

T)Rp., T)

R(A,T)(A - T)

I,

which shows that A i 0(1').

Remark 2.5. - During the above proof, we made the following observation: if an operator A from E into E commutes with T, it can be extended into an operator .A, from into satisfying :

e

e,

OA~

~

IIAII·

We also observe that the previous Proposition implies quite obviously that for any c.n.u. Cl-contraction, the set o(T) n C has positive measure. The function Az will be called "representing function" of the operator T at the point x. We now study several applications of these functions. The first one is an Invariant Subspaces Theorem, obtained by means of an estimate of the distance between a point and the space generated by its iterates.

Proposition 2.6.

~

eXP!IOg A%(O) dO 21r

Let x E E. We have:

=

inf {disfE(Tmx,span {T m+ 1x,T m+ 2x, .. .})} 2. (a) m~O

When T is invertible, this last quantity is equal to : inf {distE(Tmx,span{Tm-Ix,Tm-2x, ... })}2,

(b)

m~O

and both infima may be taken for mE'll.

Proof. - For every polynomial p(T) = Lk~O inf IITm(x - p(T)x) 11 2 m~O

= Ox =

akTk, we can write:

p(T)x0 2

j'" 11 _~

p(e i 8 )12 Az(6) d6 . 21r

If now we take the infimum of both sides upon all polynomials, we get a), by Szegd's Theorem (Chapter IX, Theorem 2.1). If T is invertible, we consider the polynomials q(T) = Lk$O bkTk, and we obtain b). The fact that the infimum may be taken for mE'll comes from the following observation: since T is a contraction, the infimum is actually a limit, when m ~ 00.

C t -contractions

e57

Corollary 2.7. - If T is invertible, log Az is integrable if and only if:

and this occurs if and only jf :

By means of the functions Az , we obtain a kind of formalization (or generalization), valid for the Cl-contractions, of the description of the Invariant Subspaces of the right shift on 12(Z) (see Chapter IX). Indeed, we see:

Theorem 2.8. - Let T be a c.n.u. C l -contraction, and let Az be the representing functions. If for some x E E, log Az is integrable, the operator T has Invariant Subspaces. This condition is obviously satisfied for the right shift, with, for example, x = el, and Az = 1. The Invariant Subspaces we obtain correspond to the Invariant Subspaces of type F n (or, more generally, of m.H2) we saw in Chapter IX, Section 1. The condition

is the abstract generalization of their definition (by "abstract" , we mean "valid for a larger class of operators"). However, it should be pointed out that the tools used to obtain this generalization are the same, namely Szego's Theorem, and no new concept has been introduced. What we proved in fact is that Szego's Theorem has a wider range of application than we originally mentioned. We will see later the corresponding generalization for the second type of Invariant Subspaces of the usual shift: the type L 2(A)' and we will come back to the meaning of both results. We now turn to a second application of the representing functions : the construction of a functional calculus, for convergent series. 3. Functional Representation of Convergent Series. In this Section, we assume T to be invertible and c.n.u. A functional calculus, as we met several times in the previous Chapters,

allows us to give a meaning to the operator f(T) , when f is taken in some class of functions. For instance, if we have a Fourier series, EkEZ akeik8, in

eS8

Chapter XII

some cases we can give a meaning to the operator E akTk. In this Section, we consider the inverse problem: a convergent series E akTkxo being given, what can we say about the properties of the "function" defined (formally) by the Fourier series EkEZ ake ik8 ? The following Proposition answers this question:

Proposition 3.1. - Let Xo E E, with Xo =1= O. Let

(1)

L akTkxo kEZ be a convergent series in E, with sum z . The Fourier series

L ake kEZ

ik8

(2)

defines a function 4J in the space £2 [Il, ..\zo dB /21r) , in the following sense: the partial sums E~N ake ik8 converge in this space, when M, N - 00. This function 4J satisfies moreover:

(3)

a.e.

Proof. - First, we observe that akTkxo - 0, when Ikl- 00. Since IITkxoll ~ 1 for k < 0 and lITkxol1 -f+ 0 when k --+ +00, we get in both cases that ak - 0 when Ikl - 00. The Fourier series EkEZ a"e ik8 therefore defines a distribution on Il , and more precisely a pseudo-fundion. We refer the reader to J.P. Kahane 11] for this concept, which we are not going to use here. For m, N E IN, we set : M

XM,N

LakTkxo, -N M

4JM,N(B) = Since the series (1) converges, XM,N therefore in t . But: M

~XM,N-XM/,N/~

= ~Lakfkxo-N ~

=

1

ik8 L a"e . -N -+ s:

in E, when M, N

--+ 00,

and

M'

Lakfkxo~2 -N' 2

~

_~ I4JM,N(B) - 4JM',N' (B) 1 ..\ zo (B) 21r '

which shows that (4JM,N) is a Cauchy sequence in L2(..\ z odB /27r), and therefore proves our first claim.

eS9

C 1 -eoniractions

1 E L oo (II) , we have

Furthermore, for every function

~/(f)x~2

=

lim

~f(f)XM,N

:

~2

M,N~oo

lim

M,N~oo

I""

_)("

1/(0)12 14>M,N(0) 12 .\%0 (0) dO 211"

(1/(8)1'1<1>(8)1' x•• (8) :: ' and this proves our Proposition. We observe that we don't need the convergence of the series (1) to be in E, but only in C, which is strictly weaker. We even don't need this convergence to hold in the usual sense: if the series is Abel summable (and Abel summabiiity is weaker than convergence of partial sums), the same result holds :

Proposition 3.2. - Let o < r < 1, the series :

XQ

E l, with

Xo

=F O. Assume that for every r,

are convergent, and Jet's denote their sums by X,.. If, when r ----4c 1-, x,. ----4c x in l, then the Fourier series (2) defines a function in L 2 (.\%0 dO /211"), in the following way : if we set, for 0 < r < 1,

4>,.(0)

=

L akrlkleik8 , kEZ,

there exists a function 4> in L 2 (.\:1:0 dO /211") such that :

when r

----4c

1-.

from which the first claim follows. The second is established as in the previous Proposition, replacing 4>M,N by 4>,..

Chapter XII

e60

4. Connections with Nagy-Foi as Dilation Theory. From T, c.n.u. contraction on H, we can build the minimal unitary dilation (J/, U) , as we explained in Chapter X. The connections between f on and ()I, U) are as follows :

e

Proposition 4.1. - For z , y E E, the numbers Ck(X, y) are the Fourier coefficients of the scalar spectral measure of the operator R., *-residual part of (J/, U). Proof. - It relies on a way of building the m.u.d. slightly different from the one presented in Chapter X. We only give the outlines and refer to M. Rome [11 and B. Beauzamy, M. Rome [11 for details. Let N be the closed subspace of 11 (1l, H) generated by sequences of the form (... ,0, -Ty,y, ...) ,

yE H,

and let ZH be the quotient of 11(Z,H) by N. If (YkhE 71 is in 11(1l,H), and fj is its class in Z H ,

The space Z H is a Hilbert space, for the scalar product :

H can be identified to a closed subspace of 1.(1l, H) by means of

u(x) =

(oo.,O,~,o,oo.),

where the Oth coordinate is underlined. Let u be the quotient operator, and 8 be the right shift on It (71, H), S the quotient operator. Then:

uT Set D

= (I

=

- T·TP/2, D = ImD, K l

s «.

= 12(1l, D),

K

= K;

ffi zH, and put:

4>(y) = (oo.,0,Dy,DTy,DT 2Y,oo.)ffiU(Y). Then 4> is an isometry. Define V on K by V

= 8- 1 on

Ks , S on Z H. Then:

261

Cl -eoniractions

for all Yl, Y2 E H , all n E IN. V is a unitary dilation of T, and one checks it is minimal. To identify the *-residual part of the dilation, we put: L = (V

0

4> - 4> 0 T)( H) ,

M = span{V n L; n E 7L}, and we know that R. = K eM. Here, M restriction R. = V Ifl.. is here S on Z H .

= K 1,

and so R.

=

Z

H. The

Finally, one obtains: Proposition 4.2. - If T has dense range, the injection extends to an isometry from onto Z H. Moreover,

e

tioT

ti

from H into Z H

SoU.

=

5. The Algebra A (11) . In order to continue our study of the C1-contractions, we need a few properties of the algebra of functions with absolutely sumrnable Fourier series. We denote by A(11) the algebra of functions on 11, the Fourier series of which is absolutely convergent:

L

1(8)

ak eikB,

kEZ

1I/11A

=

L

lakl,

kEZ

and A(11) is an algebra for the usual pointwise multiplication of functions. The reader is advised to carefully distinguish between A(I1) and A(D), the disk algebra, studied in Chapter VIII. Due to the definition of the norm, A(11) is isometric to the Banach space by the correspondence I -4 (akhEz. We refer the reader to J.P. Kahane [1] for a detailed study of A(I1).

11C~),

Functions in A(11) are obviously continuous, since they are uniforn limits of the partial sums of the Fourier series. We say that a function I in A(11) has strict support if its support (which is the closure of the set {8 ; 1(8) 1= O}) is a strict subset of 11 (this terminology is not extremely good, but it is better, anyway, than the one we used in some papers, with "compact" support. Since 11 itself is compact, every support has to be compact !).

e6e

Chapter XII

Therefore, a function with strict support will vanish identically on some interval of II. There is a restriction on the rate of decrease of the Fourier coefficients (when k -+ ±oo), in order that the function f may have strict support. These coefficients cannot decrease too quickly, when Ikj -+ 00. For instance, if lakI ~ 1/2 k , for k ~ 0, the function

will be analytic in the corona {1/2 < Izl < I}, continuous and therefore integrable on the unit circle. We have not seen, for the corona, any theory like the one developed in Chapter VIn for HI, but the same result holds: such a function cannot vanish on a set of positive measure of C without being identically zero. The way of proving this is as follows: we cover the corona by two simply connected sets DI and D 2 , the first one has in its boundary {Izl = 1 j I arg zl ~ 1r /2} and {lzl = 1/2 j I arg zl ~ 'If /2}, the second one {je] = 1 j [arg zl ~ 1r/2} and {izi = 1/2 j [arg zl >-: 'If/2}. We then make a conformal mapping to transform D 1 into the unit disk. This way, we see that the function f cannot vanish on a set of positive measure of the half-circle {izi = 1 ; Iarg zl ~ 'If/2}, and we repeat the argument for the second half. We now consider the sub-algebras of A(II) of the following form:

Ap = {f E A(II) ; f =

L

akeik8,

kEZ

L

laklpk < oo},

kEZ

where P = (Pk) satisfies: Pn ~

1

j

Pm+n ~

Pm·Pn

for all m,n E IN.

(1)

Such algebras are called Beurling Algebras. For them, a necessary and sufficient condition is known, concerning the weights (Pk), in order that Ap contains functions with strict support. If this is the case, the algebra is said to be non-Quasi-analytic (the converse, quasianalytic, refers to the above property of analytic functions: they cannot vanish on a set of positive measure without being identically zero). This condition comes from Paley-Wiener theorem, concerning the Fourier transform of functions with compact support. It can be expressed the following way:

C 1 -eontractions

fJ69

Theorem 5.1. - Let Ap be a Beurling Algebra in A(JI). It contains functions with strict support if and only if : logPn L l+n nE'll

<

The reader is referred to Y. Domar this result.

111,

2

00.

Y. Katznelson [1] for a proof of

H we forget about the condition Pm+n ~ Pm.Pn, the class Ap is not an algebra anymore. However, similar results hold, though more technical. The chief reference is Beurling-Malliavin 111, which is not easy to read. The following Lemma, due to A. Atzmon {ll, establishes a connection with the present situation.

Lemma 5.2. - Let (Pn)nE'll be a sequence of numbers with: 1) Pn ~ 1, for all n E Z,

2)

L

nE'll

logPn l+n 2

<

00,

3) there exists a constan t C > 0 such that Pn+ 1 ~ C Pn, for all n E Z, Then, for every arc in Ap ,

f

r = [60,6 1 ] ,

= LkEZ ake'ke,

0 ~ 60 < 6 1 ~ 271'", there is a function with support contained in r, satisfying:

L

jaklpk

<

f

00.

kE1l

Proof. - Let /3n = log Pn, n E 7l, and consider the piecewise linear function onffi, satisfying (n) = f3n for all n. Condition 2) implies : 00 / -00

(x) dx <

00,

(1)

1 + x2

and condition 3) implies that for any t inIR, sup 14>(t + x) - (x) I

<

00.

(2)

xEffi.

Therefore, it follows from Beurling-Malliavin [1], Theorem 1, that there exists a continuous, non identically 0 function 9 on ffi, with support contained in [60 10 1 ] , the Fourier transform 9 of which satisfies :

L:

Ig(zllezp(q,(z))dz <

00.

(3)

Chapter XII Using the fact that:

{ L

Ig(n - tllexp(?(n - t))dt =

onE'lL

[0 Ig(xllexp(?(x))dx ,

(4)

-00

we deduce from (3), using Fubini's theorem, that for some a, 0 < a < 1,

L

Ig(n - a)lexp(cP(n - a)) <

(5)

00.

nE'll From (2) and the definition of cP follows that there exists ad> 0 such that: (jn = cP(n) ~

and therefore, by (5) :

L

cP(n - a) + d

for all n E 'lL,

Ig(n - a) IPn <

00.

nE'lL

We define

fEe (TI) by : t E TI.

Then

f has the required properties. Indeed, ..

1 21["

f(n) = -g(n - a)

n E

1l,

and therefore :

E nE'lL

li(n)IPn = ~ 21["

L

19(n - a)IPn <

00.

nE'lL

Moreover, by construction, f is not identically 0 and is supported by the arc [80,8 1 1. This proves the Lemma. ing

A sequence (Pn)nE'lL satisfying the above conditions is often called a Beurlsequence. Any sequence of the form Pn = e1nlQ, for 0 < a < 1, is a Beurling

sequence. For us, the condition 3) in the above Lemma will automatically be satisfied, because we will take Pn = IITnxll, for some given point x. It seems unknown, however, if the Lemma holds without this (weak) regularity condition. We need two more results concerning the algebra .A (TI). For both results, detailed proofs and further comments can be found in J.P. Kahane [11, which is the chief reference for this algebra.

f65

C I -coniroctions

Theorem 5.3 (Wiener-Ditkin). - Let I in A(ll) with 1(0) = O. There is a sequence (/n)n~o of functions in A(ll), converging to f in A(ll) , each of them being identically 0 on some neighborhood of O. Proof. - For e > 0, set 6.~ = sup(O, l-Itl/£)' for It I < 7r, and V~ = 6.2~ Then V~ = 1 on 1-£,£1,0 on the complement of [-2£,2£1, and IIV~IIA One checks that (1 - V~)I -+ I, in A(ll) , when e --+ O. Theorem 5.4 (Wiener-Levy). - If f E A(ll) and 0, then 1/1 is in A(ll).

- 6.~. ~ 3.

I does not take the value

Proof. - The proof uses Gelfand's transformation, and was given in Exercise 4, Chapter VI. Before continuing the general study of Cl-contractions, we now develop two examples, which are quite characteristic and will motivate further results. 6. Two examples. A. A weighted shift on 12('Il). On

h ('Ill, we consider the weighted shift

S defined by : n E

with sn x

'Il,

W n = 1 for n ~ 0, and W n = 1/4 for n < O. Clearly, for every z =I- 0, 1+ 0, when n -+ +00, and IISII = 1. This operator is a Cl-contraction.

Furthermore, for every x =I- 0, we have: for all n E IN, with:

(L 41 I

C(x) =

k

k>O

Xkl

2

+

L

IXkI 2 ) 1/ 2 >

O.

k$O

This property obviously implies that there is no subspace where S is unitary, so S is c.n.u. Let's now determine what are Sand t . If x =

L

xne n, easy computations show that, for k ~ 0, Sk X

L

=

Xn-kWn-kWn-k+I·· .Wn-Ien ,

nE'Il

and if Y = LYnfn

E nEZ

Xn-m-kWn-m-k··· Wn-lYn-mWn-m •.. Wn-l

e66

Chapter XlI

If we set n = n' + m :

(s m+kX , smy) =

~ Xn'-kWn'-k··· L..J

Wn'+m-lYn'Wn, ... Wn'+m-l

nEZ If k ~ 0, when m --. ~ Xn'-kYn' L..J n'~k

+00,

+

(sm+k x, smy) has the limit:

~ L..J

Xn'-kYn'

/4 k - n ' +

~ Xn'-kYn' - /4 L..J

k - 2n '

n'
O~n'
And with the notation of Section 2,

We set xk

= Xk

for k ~ 0,

= 4kxk

for k < 0, and consider the function:

¢(e i 8 )

=

L

x~.eii8

.

iE'lL Then (fJ

= 1¢1 2 =

L xjxteiU-I)8 , ;,1

and the kth Fourier coefficient of ¢(fJ is that:

LI Xt+kXt,

that is Ck(X). So we see

(2) The same way, we obtain that if

L

y;.e

ii 8

,

iE'lL with yj = Yi for j ~ 0, and 4;y; for j < 0,

The function ¢(z) = L;EZ a;z; is analytic in the corona {1/4 < Izi < I}, and ¢(e i 8 ) is in L 2(TI,dO/2?r). By the argument we explained in the previous section (bringing the problem back to the unit disk by conformal maps), we see that: log 1¢(e i 8 ) 1-dO > -00. -'If 2?r

I.

C 1 -contractions

267

This implies by (2) that:

I"

log Az

-W"

dO

-

21r

>

(3)

-00.

Applying Corollary 2.7, we see that, for every x

t= 0,

which means that no point belongs to the closed linear span of its iterates. The operator has no cyclic vector : for every x t= 0,

T-1x

f/.

span{x,Tx, ...},

and therefore has a lot of Invariant Subspaces, B. A Multiplication Operator on H2(II). Let K be a simply connected subset of the closed unit disk, with nonempty interior, not containing 0, and such that the boundary r of K contains an interval of the unit disk. Let 4> be a conformal map from D onto K O • Let's assume furthermore that r is regular enough, so that cP extends to an homeomorphism from D onto K (this assumption is for simplicity i it is not essential). The function cP is in H?", and, more precisely, in A(D). It satisfies of course 114>1100 = 1. Let's call :

and I'

= 4>(1).

We consider the operator M~ of multiplication by cP, from H2 into itself. This is obviously a contraction. For any function f E H2, not identically 0,

which shows that

11M; /11 f+

°:

M~

Since K does not contain 0, M~ The function 1/4> is also in A(D) .

is a C 1 - contraction. is invertible, and its inverse is

Ml/~.

268

Chapter XII

For some 0: > 0, we can find an interval /2 where 1¢(ei8)1 < 1 - 0:. Therefore, for n 2 0,

I Minl ~

=

{1",n(~

/k·)12 ::

..

2 (_I_)2n /, I/(ei8)12 dO 1 - 0:

> (_I_)2n 0 -

where 0 = 0(/)

=

fl~

1-0:

211"

1~

'

1/1 2 dO/ 211" > O. So we get:

Lemma 6.B.l. - For every

I in H 2 , there are

6 > 0, C > 1, such that:

for all n 2

o.

This proves that for every l . the increase of the sequence IIMin III is exponential. Here again, the operator must be c.n.u. We know moreover (cf. Chapter VII) that the spectrum of M¢ is K. This operator has some very interesting properties which cast some light on the range of application of theories like unitary dilations. In order to develop them, we need two results about approximation of continuous functions by polynomials: Runge's Theorem and Mergelya s Theorem (the second is stronger than the first). The reader will find these two results in W. Rudin [I], p. 271 and p. 390.

Runge's Theorem. - Let K be a compact in the plane, with connected complement. Let f be a function analytic on a neighborhood of K. Then there is a sequence of polynomials converging to f uniformly on K. Mergelyan's Theorem. - Let K be a compact in the plane, with connected complement. Let f be a function continuous on K, analytic in KO. Then there is a sequence of polynomials converging to f uniformly on K. Proposition 6.B.2. - For every

f

f in H2,

E span{MtPI,M~f,

...}.

Proof. - Let VI be a neighborhood of K, V2 a neighborhood of 0, with VI n V2 = 0, and let g = 1 on VI, 0 on V 2 • This function is analytic on

269

C 1 -coniractions

a neighborhood of K U {O}, and, by Runge's Theorem, can be uniformly approximated by polynomials : for every e > 0, we can find a polynomial p(z) with: 11 - p(z)1 < E:/2, zE K. Ip(O)1 < ~/2, Replacing p by p - p(O) , we obtain:

11 -

Ip(O)1 = 0, We write p(z)

= Lk~l

p( z) I < e ,

zEK.

(1)

akz k. So we get, for every 8 ElI,

11 -

L ak
E:,

k~l

and, for every function f in H 2

,

which shows that: di st

(f, span {M~ I, MJ I, ... }) <

E:,

and proves our Proposition. The situation, in this respect, is therefore radically different from what we saw in example A, where no point (except 0) was in the closed linear span of its iterates. In Section 3, we investigated convergent series, involving the iterates of T. One may wonder: What is the difference between being the sum of a series Lk~l akT k and being in span {x, Tx, ... } ? The first is formally stronger than the second. Our present example shows that the difference is essential.

Proposition 6.B.3. - Except if f = 0, there is no series Lk~l convergent in H2, with / as its sum.

a/cM3f,

Proof. - Assume conversely that:

(1) Then

Chapter XII

270

when n

--+

+00.

Since the series (1) converges in H2, we have: 00

L ak4>k(re

i 8)f(rei 8

) --+

f(e i 8 )

,

a.e ,

when r

--+

1- .

1

Since f(e i 8 ) =j:. 0 a.e., this implies : 00

L ak4>k(re

i8

) --+

1,

a.e. ,

when r

--+

1- .

(2)

1

Set ¢(z) = ~~ akzk, for z E K. Then the property (2) shows that ¢ is in HOO(K) , this space being defined in a way similar to the usual HOO : the "radial" limits exist a.e, and are bounded. But then ¢ = 1 a.e. on r = 8K, and one sees that this is impossible, the same way as we did for H?", For further study of this operator, we need two Lemmas :

Lemma 6.B.4. - Every continuous function on I is the uniform limit, in I, of elements of F = span {I, 4>, 4>2 ...}. Proof. - Let I be continuous on I. We know that 4> is bijective from I onto I' c C, and so f 04>-1 is continuous on I'. Since I' is not the whole unit circle, we may, by Runge's theorem, find for every e > 0 a polynomial

q(z)

= ~k~O

qkzk , such that

1/0 4>-I(z) - q(z)1 <

s ,

for all z

« r.

Therefore, for all z E I,

lJ(z) -

L qk4>k(Z) I

I/(z) - q(4)(z)) I < e,

k~O

which proves the Lemma. We recall that L 2 (I) is the subspace of L 2 (II) made of the functions which are 0 a.e. on the complement of I. We obtain immediately:

Corollary 6.B.5. - The vector space F is dense in the space L 2(I). Indeed, from the Lemma follows that the closure of F in this space contains all trigonometric polynomials.

C I -contraetions

271

Lemma 6.B.6. - Let 9 in L 2(I) and

H a function h E L 2 (1) is orthogonal to Ff/' it satisfies h.g = 0 a.e.

Proof. - The assumption of orthogonality means that, for all k E IN,

Put f = g.h. Then f E LdI), and, in the duality (L I (I), L(X)(I)), is orthogonal to F. By Lemma 5.B.4, it is orthogonal to every continuous function on I , so is 0 a.e, This proves the Lemma.

Corollary 6.B.7. - Let 9 E L 2(I) and let A C I be the set where 9 Then the closure of Ff/ in L 2(I) is L 2(A).

i= O.

Proof. - By the previous Lemma, we see that every element of Fg , and therefore of Ff/' is in L 2(A). Assume that the inclusion of Ff/ in L 2(A) was strict. Then we could find hI in L 2 (A), hI E F/ ' and not identically o. But then g.h I = 0 by Lemma 6.B.6, which is contradictory. As a special case, we see that if 9

Proposition 6.B.8. - For every f

1

i=

0 a.e. on 1 , Ff/ is dense in £2 (I) .

9 E H2,

Proof. - We have:

by the previous Corollary, 9 being

i=

0 a.e,

This Proposition shows that, if Ff/ is the closure of Fg in H2, for every / E H2 , the iterates M~f, M~f, ... get closer and closer to Fg • Of course, f itself does not need to be in F g •

Chapter XII

e

The space and the operator T for this example are now easy to find : Corollary 6.B.5 shows that e is L 2(I) , and M~ id the multiplication by ~ on I, which is indeed a unitary operator. We denote it simply by M •. It is quite easy to find the Invariant Subspaces of M~ on L 2(I)

Proposition 6.B.9. - If F is an Invariant Subspace of M~ on L 2(1), there exists a Borel subset A c I, with P(A) > 0, P(I \ A) > 0, such that:

Proof. - We may find a family (gi), finite or countable, such that F is the closed su bspace (closed in L 2 (I) ) generated by the sets Ffli • Let Ai = {gi 1= O} . So F is the space generated by the spaces L 2(Ai), and therefore is L2(UiAi), and our Proposition is proved. We deduce:

Corollary 6.B.IO. - If F is an Invariant Subspace of Fn H2 = {O}.

M~

on L 2 (I} , then

This Corollary means that no Invariant Subspace of M~ is produced by its unitary extension M~. Since M~ is related to the dilation theory, we see an example, no invariant subspace of which comes from its unitary dilation. We now determine the Invariant Subspaces of M~

in H2.

Proposition 6.B.ll. - Every Invariant Subspace of M~ in H2 is of the form m.H2, where m is an inner function. Proof. - A subspace of the form m.H2 is clearly invariant. Let's show the converse.

Lemma 6.B.12. - Every function f in the disk algebra A(D) is the uniform ...}. limit of elements of F = span{I,~,~2, Proof of Lemma 6.B.12. - The function f 0 ~~I is continuous on K, analytic on KO. By Mergelyan's Theorem, for every e > 0, there is a polynomial p such that, for all u E K,

If 0 ~

- I ( u)

- p(u) I < c.

So, for every zED,

If(z) - p(4)(z)) I < s , which proves the Lemma.

C 1 -eontractions

Lemma 6.B.13. - For every outer function dense in H2.

219

I,

F] = span {I, 4>1, 4>2 I, ...} is

Proof. - By the previous Lemma, the closure Ff in H2 contains all functions f.h, for h in A(D), and, in particular, all functions zn I, for n ~ O. But these functions span H2 when I is outer (Beurling's Theorem, Chapter IX, Corollary 1.2), and this proves our Lemma. Let now 9 be a function in H2, 9 = m.] its decomposition into inner and outer parts. By Lemma 6.B.13, Fg = m.H2. This subspace is also invariant by u,», Finally, let G be any Invariant Subspace for Mq, on H2. From what we just said follows that G is also invariant by Mei' , so is of the form m.H2, by Chapter IX, Section 1. Our proof is complete. In this example, our operator is indeed a function of the usual shift Mq, = 4>(Me ifl ) . It is not surprising therefore that they should have the same Invariant Subspaces. 7. The spectrum of T and the spectrum of

T.

We have already seen that o(T) C C n o(T). We now study this inclusion more closely. We assume that T is invertible.

Proposition 7.1. - Let A E «: with IAI = 1, and A A - T contains all points x such that:

L k~O

log lI T - k xll < 1 + k2

fI. o(T). The image

of

+00.

Proof. - We may assume ,\ = 1 (just replace T by T / A). For a function 4>, of class Cion IT, we define q (0) by : 4>(0) - 4>(0) = (t i9 - l)q(O).

(1)

Let 4> = LkEZ akeik9, q = LjEZ bje ij 9 be the Fourier series. One checks immediately the following relations between the Fourier coefficients of q and those of 4> : if j ~ 0, bj I>j

Lal

bj

~Lal l~j

if j < O.

Chapter XII

Take x E E,

II xii

= 1, and put:

Then the sequence (ukhez is decreasing for k < 0, increasing for k ~ 0, and satisfies the assumptions of Lemma 5.2. Let's define (ukhez, by uk = Ikluk, k E 1L. Again, (uk) satisfies the assumptions of Lemma 5.2.

Lemma 1.2. - If Ekez lakluk <

00,

then

E;eZ IbjlUj < 00.

Proof. - We have:

Llbilu; = LLlad u; ;~o

i~OI>j

~

<

lall uo + la21(uo + Ul) + ... + lakl(uo + ... + uk-I) + ...

L klakluk

<

00,

k~O

and, the same way:

LLlallUj ;
~

la-llu-l + ... + la-kl(u-l + ... + U-k)

s L

kla-klu-k <

+ ...

00,

k~O

and our Lemma is proved. Since 1 rt. u(1'), we can find an arc r = [80,6 1 ] contained in u(1')c. By Lemma 5.2, there is a function 4> E A(TI) with support in r, such that LkeZ lak IU k < 00. We may assume moreover that q.(0) = -1. This function is of class C 1 • The function q defined by (1) satisfies, by Lemma 7.2,

L Ib;lo;

<

00,

ie'lL

and so the series get:

E i e 1L bjT;

converges in E. Let y be its sum. On C, we

4>(1') + I But q.(1')

= 0, since

supp 4>

=

c u(1')c. I

(1' - I)q(1'). So :

= (1' - I)q(T).

C 1 -eontraetions

e75

This implies :

x

=

(1' - I)q(1')x.

But since the series LjEZ bjTi converges in E, we have:

q(1')x

=

y E E,

and so : x

= (1' - I)y = (T - I) y,

which shows that s: is in the image of T - I, and proves our Proposition. Corollary 7.3. - IE, For every x E E,

then:

0(1')

= o(T) n C .

Indeed, for every A with IAI = 1, A - T is surjective by Proposition 7.1, therefore invertible. We now produce an example showing that the inclusion:

0(1') C o(T) n C can be strict. This example is a modification of example B in the previous Section. First, assume that the compact K, still simply connected, has the following property: 8K n C consists in a interval I' and an isolated point a'. Let, as before, 4> be a conformal map from D onto KO. Let 1= 4>-1(1'). Computations similar to those we made in Section 6 show that here again = L 2 (I) , o(M
e

This simple modification may give the impression that the difference between the two sets, (o(T) n C) \ 0(1') must be of measure o. This is false, as we now show. be a Let I', J' be two disjoint closed intervals in C, and let (a~)n~o dense sequence in J'. For every n, let K n be a simply connected compact, not containing 0, and such that 8 K n n C = I' u{a~}. We may moreover assume that 8Kn is regular: a rectifiable Jordan curve. In short, each K n is of the type just discussed.

Chapter XII

e76

Let 4Jn a conformal map from D onto K~. We put an = 4J;I(a~), 4J;I(I~). Of course, an ¢ In. Let En be the space H2, Tn be the operator M fln , and put:

In

=

=

(IT

(f Il/nl ~Jl/2

{(In)n~o

En) = j In E En for all n, o 2 0 which is obviously a Hilbert space. On E, we define T by : E

< oo},

Since each Tn is a contraction, obviously T is a contraction. This operator is C 1 • Indeed, if F

= (In)n , IITmFII~

=

LIIT:lnll~

n

i:

n

In

n

L > L /,

14J:' In(e

i9

2

)1

~:

In(e i 9 )12 dO > 211"

o.

The spectrum of each Tn is K«, The spectrum of T is uKn . Indeed, u(T) contains each K n , so it contains uKn . Conversely, if A ~ uKn , there is a 6 > 0 such that l4Jn(z) - AI > 6, for all z E D, and all n ~ 0, and the inverse of A - T is the multiplication operator, on En, by 1/(A - 4Jn). Therefore,

u(T) n C We now describe

T and

=

I'

U

J' .

! .

Lemma 7.4. - The space ! is

t Proof. - For F = (In)n~o,

~F~2

=

=

(~L.

(I.) ) •

we have:

liT: Inll~

~~ooL n

by Lebesgue Theorem

from which the Lemma follows.

e77

C 1 -eontractions

f is the multiplication by tPn in each L2 (In)' Precisely, if F E (n~ L 2(In )h t then TF = (tPnfn)n. Finally, we see that 0"(1') = I'. Indeed, a) O"(T) c r . HAt. lit with IAI = 1, and if ei S E In, then IA-4>n(e i S) / ~ = dist(A,I'), so: The operator

F

{)

= (fn)n~O,

and A b)

l'

is surjective. Since it is clearly injective, our claim follows.

aCT)

:J I'. Indeed, this is the case, since

aCT)

contains each O"(Mq,n) =

I'. This way, we see that 0"(1') n C contains I' and J' , though 0"(1') is reduced to I'. We observe that we are allowed to take as lithe complement arc of J' , though they have the same end-points. We need just to avoid to take these end-points among the (a~). For the operator T obtained this way, we will get

aCT) :J C, aCT) = I'. The sets that are spectra of C1-contractions have been determined by L. Kerchy [1]. 8. An Invariant Subspace Theorem for the C,-contractions. We are going to prove the following result, due to the author [31 :

Theorem 8.1. - Let E be a Banach space, T a C 1 , invertible contraction which is not a multiple of the identity. We assume that there exists a point x¥-O such that :

(1) Then T has Hyperinvariant Subspaces.

Remark. - The invertibility assumption is not essential, and condition (1) may be replaced by an assumption on a chain of approximate inverses of a single point. See [3].

Proof. - We may assume that T has no eigenvalue. Indeed, if A is an eigenvalue, the space K ere A - T) is closed, invariant under every operator which commutes with T, and is not the whole space since T is not a multiple of the identity.

Chapter XlI

278

The point x being given by (I), we put Ok = IIT-I k lx ll ~ for k E 1l. This sequence obvioulsy satisfies the assumptions of Lemma 5.2. Therefore, for every arc r = [8o~81] ~ there exists a function tP~ with support in r ~ with, if

¢

=

L

lej 10j <

00.

(2)

JEZ

We may of course assume that ¢ takes only real and positive values. Set TO¢(t) = ¢(t - 8). There is an angle 8 small enough so that the functions ¢~ TO¢ ~ T20¢ ~ ... , TNO¢ satisfy, if N is large enough: N

LTkO¢(t) > 0 k=O

for all t E TI.

Set u = 2:.:=0 TkOtP(t). All these functions are in A(IT), and u does not vanish I/u is also in A(TI). on TI, so, by Wiener-Levy Theorem (Theorem 5.2)~ The operator u(T) is well-defined, by a norm-convergent series, since u E A(TI) (recall that T is a surjective isometry). Moreover, it is invertible, and its inverse is (l/u)(T). Therefore, it is injective, and in particular, u(T)x i= O. Thus, for some k ~ 0 ::; k

s N,

This construction can be realized for every arc r. Let's take successively r N = [0, 27r / N] ~ for N = 1, 2, ... , and let's denote by tPN,kN the function obtained (or one of them) by the previous construction: this is a translate of 4> N ~ which satisfies : The supports of ¢N,k N have a width which tends to O. Let a = eiOo be an accumulation point of the middles of these supports. By construction, this means that, for every V, neighborhood of a, there is a function ¢N,kN' for N large enough, the support of which is entirely contained in V. We now consider the function 10(8) = ei 8 - ei 80 • It vanishes at 8 = 80 . By Wiener-Ditkin's Theorem (Theorem 5.3)~ there exists a sequence of functions (fn)n"2:0 in A(ll), convergent to 10 in A(ll), and each In is identically 0 on some neighborhood of 80 •

C t -eontractions

Since [«

--+

£19

fo in A(TI) , we get: IIfn(T) - fo(T)

II

--+

o.

(3)

But:

fo(T)x since

~Tx - ei8ox~

=

lim IITm(Tx - ei 8o x)1 1= m~oo

lim

IITmYII,

m~oo

with y = Tx - ei 90x, and y =I 0 since T has no eigenvector. Since moreover T is C t , we see that limm~oo IITmy11 > o. By (3), we may find no such that jno(T)x =I o. Put simply f = fno' for no chosen this way. The function j is identically 0 on some neighborhood Vo of (Jo. Choose No large enough so that <jJ = <jJNo,kNo has support in Vo. The functions f and <jJ are therefore disjointly supported. They both are in A (II) , satisfy f(f)x =I 0, <jJ(T)x =I 0, and <jJ moreover satisfies (2). Since the functions are disjointly supported, we have j(T)<jJ(f) = o. Set z = <jJ(f)x. Then we know that z =I 0, j(f)z = 0 (4).

Lemma 8.2.

~

The point z is in E.

Proof. - We consider the series EkEZ ckTkx. It converges in E, by (2). Let z' be its sum : then z' E E. The series converges also in <jJ(f)x = z, so z' = z is in E. This proves the Lemma. We now consider :

F = {y E E

j

j(f)y = o}

e, and its sum is

En Kerj(T).

This is a closed subspace of E. Indeed, if Yn --+ y, Yn E F and Y E E, then Yn --+ Y in and f(T)Yn = 0, so f(f)y = o. This subspace is invariant by T, and, more generally, by any operator U which commutes with T. Indeed, as we already observed, U can be extended into an operator U, which commutes with i , therefore with j(f). So :

e,

f(f)uy

=

j(f)Uy

=

Uf(f)y

= 0,

and U y is in E. The subspace F is not reduced to {O}, since it contains z, by (4) and Lemma 8.2. It is not the whole space, since f(f)x =I o. This proves our theorem.

Chapter XII

280

We see that we don't need the invertibility of T. All what is needed is that, for some point x , there is a sequence (xn)n>O of points in E, with:

T" X n

L

n~O

=

x,

logllxnll < 1 + n2

n ~ 0,

(1)

00.

(2)

We obtain:

Corollary 8.3 (Colojoara - Foiag}, - If E is reflexive and if both T and T* are C 1 -contractions, then T has Hyperinvariant Subspaces. Proof. - We are going to show that there are points z with a bounded sequence X n satisfying (1) and (2). Take € in E*, with II€II = 1. Then Ilr*n€11 o. Put 6 = lim IIT*n€ll. Take Yn E E, llYn II = 1, such that T*n €(Yn) = IIT*nell. Then €(Tn yn) = IIT*nell ~ 6. So the sequence Tn Yn does not tend to 0 weakly. Some subsequence Tnle Ynle has a weak limit x =j:. O. Fix m =j:. O. The sequence (Tn le -mYnle)k is bounded, has a weak accumulation point X m . But then TmTnle -m Ynle has Tm x m as accumulation point, so Tm x m = z , Since Ilxm II = 1 , our Corollary is proved.

r

In the course of this Chapter, we have met two Invariant Subspaces Theorems for the C 1 - contractions. The last one (Theorem 8.1) applies when the backward iterates do not grow too fast, and, if one applies it to the usual right shift on 12(1l) , gives subspaces of the type L 2(A) = {f j / = 0 on A C } . The first one (Theorem 2.8) shows that if log >':Z: is integrable, the point x does not belong to the closed linear span of its iterates, which is formally analogous ; ak = to the Invariant Subspaces, for the right shift, of type Fn = {(ak)k~o o for k < n} (corresponding in fact to m.H 2 ) . So we have found for C 1 contractions a formal description of both types of Invariant Subspaces of the right shift on 12(1l). Unfortunately, the conjonction of both types does not provide an answer to the question : does every C 1 - contraction have Invariant Subspaces ? Indeed, Theorem 8.1 does not apply if the backward iterates of each point grow too fast (and moreover we proved in [31 that the condition given on backward iterates was best possible to obtain this type of Invariant Subspaces). In order to be able to apply Theorem 2.8, we need log >':Z: to be integrable for some x . But the functions >':Z: are supported by a(T) (Corollary 2.3) and therefore the statement requires a(T) = C. But the example 6.B has backward iterates exponentially growing, for every point, and a(T) is a proper

C 1 -contractions

281

interval of C. So none of the results apply to it. As we said, this example has Invariant Subspaces for a trivial reason : it is just a function of the usual shift: M; = 4> (Mei' ). But how to recognize in general that an operator is a function of a given one? Moreover, we now observe a clear weakness in the above results: the assumptions are not invariant if T is replaced by a function f(T) , though the conclusion is. For instance, T may have slowly increasing backward iterates, and f(T) quickly increasing ones. So these results should by no means be considered as satisfactory.

282

Chapter XII

Exercises on Chapter XII.

Exercise 1. - Let T be a C1-contraction, and U be an operator such that: Unx - Tnx

--+

0,

n

+00

--+

for all x in E. Show that U = T .

Exercise 2. - Let T be a C1-contraction on a Hilbert space. Assume that T· is also a C1-contraction. Show that there exists a unitary operator U, and operators 8 1, 8 2 , continuous, injective, with dense range, such that:

Exercise 3. - Let T be a C 1-contraction on a Hilbert space E. Let defined as in Section 1.

i , e be

1) Show that the formula (Ax, y) = [z, yj,

e into

for x, y in E defines an operator A from adjoint, and has dense range F. Show that: F =

{x E E

; sup

E which is injective, self-

IIT·- n xll < oo}.

n~O

2) Can you give a meaning to these properties if E is only a Banach space ? 3) Determine the operator A in the special case of the operator Mq, on H 2 which was studied in Section 6.B. 4) We come back to the case where E is a Hilbert space. For x E F (defined in 1)), let LkE71 akT·-kx a series, convergent in E, with sum z. If z = 0, what can be said about the Fourier series LkE7[, akeik8 ?

Exercise 4. - Let A E u(1'). We want to show that there exists a sequence (xn)n~O of points in E, with Ilxnll --+ 1, ~Xn~ --+ 1, II(A - T)xnll --+ 0, ~(A

-

T)xn~

--+

O.

a) Show that there exists a sequence (Zn)n~O

~ Zn~ = 1,

~ (A - f) Zn ~

of points in E such that: --+

o.

C 1 -contractions

£89

b) Show that there exists a sequence (kn)n~o with k~ ~ k n for all n, every sequence (k~)

of integers, such that, for

c) Show that there exists a sequence (hn)n>o of integers, such that, for every sequence (h~) with h~ ~ h n for all n,

d) Put properties.

Un

Show that

max(kn,k~).

TU" Zn has the required

Xn

Exercise 5. - Let T be a C.-contraction on a Hilbert space E. We assume that T has a cyclic vector Xo ; let /0 be the representing function for this point, and let Jl.o = /0 dO /21r . Let m

1. We assume that Xo is cyclic for T'":

~

1) Prove that there exists a set Am C II, with Jl.o(A m) = 1, and a sequence (Pn)n~O

of polynomials, with: n

2) Let Bk = Am

--+ 00,

n [2k1rJm,2(k + 1}1r/m[, B ,k

0 E Am .

0 ~ k

< m, and

= {2k1r 0- ; 0 E Bk } m

Show that the sets B k are pairwise disjoint; deduce that P(A m } ~ l/m, where P is the Lebesgue measure.

3) Show that this is impossible for m large enough. Deduce that there exists a N such that TN has no cyclic vector. 4) Is the integer N universal? In other words, is there a N which fits all C 1 -contractions?

5) Find N in the case of the operator M fJ , Section 5.B. Exercise 6. of 8 2 ?

~

Let 8 be the right shift on 12(7L). What are the cyclic vectors

Chapter XII

Exercise 7. - Assumme T to be invertible, and satisfying : " logiiTRII LJ -1----=..:+-n-2~ nE?L

< 00.

Put Pn = IITnll, n E ?L, and let Ap be the corresponding Beurling algebra. We want to show that for every lEAp, l(a(T)) c a(I(T)) . a) Show that a(T) c C. b) let I = Lj aje ij fJ E Ap , and put, for N ~ 0, PN Let .\ = ei80 E a(T). Define s» by :

PN(e i8) _ PN(e i8o)

=

= L:!N aje ij fJ.

(ei8 _ ei8o)qN(ei8).

Show that qN E Ap , and that:

(e i8 _ ei8o)qN(ei8) ~

l(e i8) - l(e i8o) ,

when N

--+ 00.

c) Deduce that: (.\ - T)qN(T) ~

f(.\) - I(T)

in £(E). Deduce that f(.\) E a(I(T)).

Exercise 8. - Let

M~

be the multiplication operator studied in Section 6.

a) Show that the closure in operator norm of the algebra AM. can be identified to the space A(K) of functions which are continuous on K, analytic in KO. b) Show that the closure of this algebra for the strong operator topology or for the weak topology can be identified to BOO : every operator in these algebras is of the form :

f

--+

t/J 4> I,

for some function t/J E BOO .

Exercise 9. - Let T be a CI-contraction on a reflexive Banach space E. We assume that there exists a bounded sequence (Xj)j;::-o such that Tj Xj --h 0 weakly. a sequence of a) Show that there exists a sequence of integers (nk)J.:~O' positive numbers (o.j), with, for all k ~ 0, L::~ll (}:j = I, and a point z in E, with z i= 0, such that: --+

z,

in norm, when k

--+

+00.

eS5

C 1 -coniraciions

b) Deduce that there is a bounded sequence (Yk)k~O that: in norm, when k --+

of points in E, such

+00.

c) Using Theorem 8.1 and Corollary 8.3, show that T has Hyperinvariant Subspaces.

Exercise 10. - Let T be a C 1 -cont ract ion on a Banach space E. We assume that there exists a point Xo in E, and E: > 0, fJ > 0, such that :

We want to show that if T has no eigenvector, and if a series converges in E :

L "IkT-kxo

=

0,

k~O

then all "Ik'S are O. a) Show that there exists a C

> 0 such that: for all k

~

O.

b) Let g = ECkeik8 E A(JI). For m ~ 0, we set:

L

tPm(g) =

ckT lc+ m,

lc~-m

Let 4>(8) = Ek~O"lke-ik8.

Show that tPm(4))XO

--+

0, when m

--+

+00.

c) Let f(z) = Ek~O "Ikzk. Deduce from a) that this function is analytic in the disk D(O,l + g). Assume that f is not identically O. Show that it can be factored : L

f(z) =

II (z -

zl)h(z),

1=1

where ZI ••• ZL are the zeros of f on the unit circle, repeated according to their multiplicity, and h a function with no zero on the unit circle. d) Deduce that 1/h(e- i8) is in the algebra A(JI). Put hi = 1/ h(e- i8),

B

L = Il,=l (ea'8 - Zl).

e) Show that if u, v are in A(JI), and if tPm(u)XO then tPm (uv )xo --+ O. f) Deduce that tPm(4)' hi) eigenvec tor.

--+

--+

O. Show that if tPm(B)

0 when m --+

--+

+00,

0, then T has an

286

Chapter XII

Notes and Comments.

Results of Sections 1, 2 are due to M. Rome and the author [I] j the proof of Proposition 2.1 was communicated to us by A. Atzmon. Section 3 is due to M. Rome [I]. Section 4 is due to M. Rome and the author [I]. Section 5 : For further results about non-quasi-analytic classes, see Y. Domar [1], Beurling - Malliavin [1], J.P. Kahane [1], Y. Katznelson [1]. Lemma 5.2 is in A. Atzmon [2]. Section 6 is due to the author [5], [8]. Section 7, Proposition 7.1 is due to M. Rome and the author [I]. The rest of the Section is due to the author- [8]. Section 8 is due to the author [3]. Exercise 4 is due to the author [7]. Exercise 5 comes from Nagy - Foias [2]. Exercise 7 is due to A. Atzmon [1]. Exercise 9 is due to the author, and so is exercise 10 [2\.

Complements on Chapter XII.

1) The Theorem of Section 8 was extended by A.Atzmon [4], who gave results containing both this theorem and Wermer's Theorem mentioned in Chapter III. One of them is as follows :

Theorem. - Let T be an operator on a Banach space E. Assume there exist sequences (x n )nE1l in E and (Yn)nE1l in E·, such that, for all n :

Assume moreover that there exists a Beurling sequence (Pn) such that: for all n E 1l,

(a)

and that there is an integer k such that:

n

----+

±oo.

(b)

C l -eontraetions

e87

Then either T is a multiple of identity or it has Hyperinvariant Subspaces.

A similar result holds if the roles of X n and lin are reversed, but it's unknown if we can omit condition b), that is, assume only : log IIT"xoll L 1+ k kEZ 2

for some point

Xo

< 00,

in E.

2) There are links between the cyclic vectors for the usual shift on co(Z) and the Invariant Subspaces for the Cl-contractions (see B. Beauzamy [4]). Let H be a Hilbert space and T be an invertible Cl-contraction on H. Let A be the operator defined by the formula (see exercise 3) :

Let

f = L"EZ Ckt~ik8

E

A(II). As in exercise 10 above, for m ~ 0, we set:

tPm (I)

=

L

clcTIc+ m •

Ic~-m

Then:

Proposition. - At least one of the following holds:

a) There exists a function f in A(II), not identically 0, which satisfies tPm(J)x --+ 0, when m --+ +00, for all x E H (we say that T satisfies an asymptotic functional equation).

b) For every z , Y in H, the sequence ((ATlcx,Y)hEZ is in co(Z) and is cyclic in this space for the right shift. c) T has Invariant Subspaces.

The case a) is excluded if the set of the zeros of the function f is a set of harmonic synthesis of type Z A+ (the reader is referred to [4) for the definitions).

This page intentionally left blank

PART VI

INVARIANT SUBSPACES

Tel Ie vieux vagabond, pietinant dans Ia boue, Reve, Ie nez en l'air, de brillants paradis j Son oeil ensorcele decouvre une Capoue Partout oil la chandelle illumine un taudis. Charles Baudelaire ("Le Voyage").

During the course of the previous chapters, we have encountered many results about the Invariant Subspace problem, for various classes of operators. We will now study this problem specifically, and, during this Sixth Part, give two recent results: a positive one, for contractions on a Hilbert space, and a negative one: an example of an operator on the space ll, with no Invariant Subspaces.

Chapter XIII

Positive results

In this Chapter, we consider T, c.n.u. contraction on a Hilbert space. We have seen in Chapter XI how the HOO functional calculus was defined. In a first Section, we are going to investigate the following question: When is this functional calculus isometric ? 1. When is the HOO functional calculus isometric?

To say that the functional calculus is isometric means that:

114>(T) II = 114>1100

,¢ E

u»,

To say that it is isomorphic means that there exists a constant 0 such that:

First, we study this problem for ¢ E A(D), the disk algebra.

Proposition 1.1. - Let T be a c.n.u. contraction on a Hilbert space. The functional calculus is isometric on A(D) if and only if the spectrum of T contains the unit circle. Proof. - We know that the inequality :

114>(T)11

<

114>1100

always holds, by Chapter XI, Proposition 1.1. So we study the converse inequality. a) Let's assume that u(T) contains C. Then every point A of C is in ao(T), so in ol1(T). Therefore, there is a sequence (xn)n~O, IIxnll = 1, such that TX n - AXn --+ O. Take now ¢ in A(D) with II¢II 00 , and A E C such that

I¢(A)1 = 1. Then, with the above sequence

(Xn)n~O,

n--+oo,

Chapter XIII

which implies :

114>(T)X n II

--+

14>(A) I =

1,

and so :

114>(T) II

=

1.

b) Assume conversely that u(T) does not contain some point >'0 E C. Then some arc r around Ao is in p(T). We may of course assume (for simplicity) that Ao = 1. So I - T is invertible, and so is I - rT , for r close enough to 1. Moreover, there exist 0 < ro < 1 and C such that : r

~

ro .

This means: r

~

ro .

Consider the function :

which is in A(D). Then

l14>r(T)1I

II4>r 1100 =

~

C, though:

1

1_ r

--+

+00,

which proves that the functional calculus is not isometric (and, more precisely, is not isomorphic). We now turn to the study of the same question for the HOO functional calculus. The Theorem we now prove is due to C. Apostol [11. The answer provided is not as satisfactory as in Proposition 1.1, but it says that, for a contraction the spectrum of which contains C, we can always assume the functional calculus to be isometric if we are looking for Hyperinvariant Subspaces. Theorem 1.2 (C. Apostol). - Let T be a contraction on a Hilbert space, such that u(T) contains the unit circle. If the H'" functional calculus is not isometric, the operator has Hyperinvariant Subspaces. Proof. - First, we observe that, by Chapter XI, Proposition 1.2, this functional calculus is isometric for T if and only if it is for T*. Furthermore, we know that if both T and T* are Cl-contractions, they both have Hyperinvariant Subspaces, by Chapter XII, Corollary 8.3. Therefore, we can assume that one of them is not a CI-contraction. Next, we observe that if F is an Hyperinvariant

e99

Positive Results

Subspace for T, Flo is an Hyperinvariant Subspace for T·. So we can assume that T, rather than T·, is not Ct. This means that:

= {z ; T" X

F

-.

0, n -. +oo}

is not {o}. But this set is a closed hyperinvariant subspace for T, so it must be the whole space, otherwise our problem is solved. In other words, we may assume: Tnx -. 0, n -. +00 , for all x E H. A contraction satisfying this property will be called a Co-contraction. It is convenient to represent such an operator by means of a left shift on the space 12(IN,H) (in short: 12(H)). We recall that:

n

Proposition 1.3. - Let H be a Hilbert space, and T be a C o - contraction on it. Let S be the right shift on 12(H) :

There exists a subspace X of 12(H), invariant by the adjoint S· of S, and a surjective isometry V from H onto X, such that: T = V-I (S·lx)V.

Proof. - As we already did in Chapter X, we introduce: D

= JI-T*T,

which is positive, self-adjoint, and satisfies (see Chapter X, Proposition 1.3) :

from which follows clearly: 00

xEH.

We set:

Vx

(Dx,DTx,DT 2x, ...),

(1)

Chapter XIII which is, by (1), an isometry from H into '2(H). It satisfies :

therefore:

(2)

on H.

From this formula follows that the image of V is invariant under S* in 12(H). Set X = 1m V. This is a closed subspace of 12(H) and V is a surjective isometry from H onto X. Moreover, (2) indicates that:

(3) and our Proposition is proved. To simplify the notation, we write A

= S*'x.

We observe that, instead of 12(IN, H), we might have taken l2 (IN, F) , with F = ImD. The Proposition can be rephrased as follows: any Co-contraction on a Hilbert space is unitarily equivalent to the restriction of a left shift to one of its invariant subspaces. But the reader should not expect us to deduce from this result that any Co-contraction has Invariant Subspaces (this is unknown). The shift involved here is not the usual left shift on l2 (IN) : it is a shift on a product space l2 (IN, H) (some people say: a. shift of infinite multiplicity, but we will avoid this terminology). There is no deep statement behind Proposition 1.3 (and we have seen how easy the proof was), but this representation of the operator will be convenient for what follows. For any A,

IAI <

1, we define:

which is a subspace of '2(H) : it is the space of sequences (xn)n~O such that: Xn

Of course, E). is {O} if

IAI

= Anxo, ~

for all n

~

in l2(H)

O.

1.

We also denote by P). the orthogonal projection from '2(H) onto E)., and, finally, P the orthogonal projection from '2(H) onto X. We first state a few easy consequences of the definitions.

Positive Results

295

Lemma 1.4. - For any A in D, the operator

UI = (A - S*)(X - S) is invertible, and :

If we set:

U(A)

(X - S) ((A - S*)(X - S)) - I

,

then we have:

(A - S*)U(A) = I, U(A)(A - S*)

(1)

I-P>. , 1

IIU(A)J1 ~

(2) (3)

1 - IAI .

Proof. - For our first claim, we observe that:

(A - S*)(X - S)

IAI 2 - XS* - AS + S*S = (1 + IAI 2) - XS* - AS =

= (1 + IAI 2 )( /

-

U2 ) ,

with:

So:

We deduce:

I!Utl1l = (1 + IAI 2)-III(/ - U2 )- 11 1

< (1+IAI 2 ) -

<

1

1 1 ~ IIU2 11

1

- (1 - IAI)2' as we announced. The relation (1) is obvious on the definition. For (2), we observe that U(A)(A - S*) is a projection. Indeed, U(A)(A - S*)U(A)(A - S*) = U(A)(A - S*),

Chapter XIII

e96

by (1), and from the definition of U(,\) follows that:

So U(,\)(,X - S·) is the orthogonal projection onto (Ker (,X - S·)).1, and (2) follows. For (3), we write:

(X - S)U1 1 ,

U(..\) and, for all x in 12 (H), IIU(..\)xI1 2

= (U('x)· U(..\)x, x) = (U1 1 (..\ - S·)(X = (U 1 1 x , x ),

- S)U1 1 x, x)

So: IIU(..\)xI1

2

(1 _1 ,\1 )2 l1 x I12 ,

s

1

which implies (3) and finishes the proof of the Lemma.

Lemma 1.5. - For every ..\ E D, IIPP,,112 ~

l-IIU(..\)1I2 inf{II(..\ - A)x1l 2 i x EX, IIxll = I}.

Proof. - The projections P and P A are positive and self-adjoint, and so is their product. We have:

IIpPA Il =

=

sup I(PP"x,x)1

11:1:11=1 sup

I (x,

P"Px) 1

11:1:11=1

= II P"PII·

Therefore, IIPP" II = IIP"PII

=

sup{IIP"xI1

= 1-

2

i x E

X, Ilxll

inf{II(I - P,,)xI1

2

i

= I}

x E X, Ilxll 2

= I}

A)x1l i x E X, lIxll = I} 2 2 ~ 1 - IIU(..\)11 inf{Il(..\ - A)x11 ; x E X, Ilxll = I},

= 1 - inf{IIU(..\)(..\ -

as we announced.

Positive Results For a, 0

297

< a < 1, we put :

and: 1

Za = (D n u(T)) U (D n {A E p(T) ; all(A - T)-lll ~ l-IAI})' We have:

Proposition 1.6. - Let T be a Co-contraction. Then, for any a, 0 < a < 1, the set r a is contained in Za. Proof. - Take A in Za. Then : - either A E u(T)

n D.

Then:

inf{II(A - A)-lxll ; x EX, Ilxll = I}

0,

and by Lemma 1.5,

>

IIPPAII

1,

- or oX E p(A), and then: inf{IIP - A)-lxll ; x EX, IIxll = I}

1

and by Lemma 1.5, IIPPAI!2 ~ 1 -IIU(A)11 2 a2 (1 -IAI)2 ~ 1 - a2

,

by Lemma 1.4, 3), and the result follows. A subset G of the open disk D is said to be dominating (for the HOO

functions) if, for any ¢ in H?", sup{I¢(z)1 ; z E G} = lI¢lIoo. This is obviously the case for any corona {a < Izl < I}, with 0 < a < 1, since we know that the numbers

Moo(¢,r)

=

sup{I¢(z)1 ; Izl = r}

are increasing with r (Chapter VIII, Section 3).

e98

Chapter XIII

A set G is dominating if and only if almost every point in C is the nontangential limit of a sequence of points in G.

Proposition 1.7. - Let T be a Co -contraction, such that u(T) contains the unit circle. If, for some a, 0 < a < 1, the set Za is not dominating, the operator T has Hyperinvariant Subspaces. Proof. - If Za is not dominating, some subset C1 of the unit circle, of stricly positive Lebesgue measure, has the following property: for every 8 in C1, ts is not in Za if t < 1 is close enough to 1 : this means indeed that 8 is not the radial limit of points in Za. More formally, for every 1, ts i Za. Set

8

Eel, there exists

R, = {t8;

E: a

E: a

> 0 such that, if

ea


< t < I}.

But D \ Za has at most countably many connected components. For one of them, say G, we can find 4 points, 81, 82, 83, 84 in C such that the 4 open segments R 1 , R 2 , R 3 , R 4 are all in G. We connect 81 to 82 outside D and connect R 1 to R 2 inside D. We obtain a closed L curve such that :

(1) and L n D c G. The same way, we connect 83 to 8", and R 3 to R 4 , and we obtain L'. The domains limited by both curves are disjoint. If A E G, then A E p(T), and: 1

< 1-

IAI .

(2)

We now consider the integrals :

B' =

~ f 2&1r

JL'

(A - 83)(A - 84)(A - T)-ldA .

From properties (1) and (2) follows that these integrals are well-defined, and norm-convergent. Computations exactly similar to those made in Chapter II, Proposition 3.1, b) (using the resolvent equation and the fact that the domains bounded by Land L' are disjoint) show that B B' = O.

Positive Results

299

Moreover, since inside Land L' there are points of the unit circle, and since a(T) contains C, neither B nor B' is O. Let's explain this: take any s in the open arc ]SI,82[' This point is in the spectrum of T. By Gelfand's Theory (Chapter VI, Proposition 3.5), there is a character t/J on }r{, maximal commutative algebra generated by T, such that t/J(T) = 8. Of course, a fortiori, t/J is a character on .AT' We consider :

t/J(B) =

~

2&11'

/, (). - 8d(). - 82)(). - t/J(T))-ld>' L

= (t/J(T) - 81)(t/J(T) - 82) = (s - sd(s -

S2) i- 0,

and this proves our claim. Obviously, B commutes with every operator which commutes with T, so K er B is the required Hyperinvariant Subspace, and our Proposition is proved. So, if T has no Hyperinvariant Subspaces, for every a, 0 < a < 1, Za is dominating. Since Za era (Proposition 1.6), we see that r a is also dominating. From this assumption, we are now going to show that the H OCI functional calculus is isometric. We recall our notation from Chapter XI : AT is the ultraweakly closed algebra generated by T j this algebra is the dual of JlT = JI(H)jJ. AT, in the duality: (U,N) ~

tr (UN) ,

If K is a set in a Banach space, we denote by aconv(K) the absolutely convex hull of K, that is the set of all finite sums L Ctiki, where Cti E a:: , k; E K, and L: ICti I :s; 1. The closure of this set is denoted by aconv(K). We also recall that x ® y is the rank-one operator:

x ® y (z) = (z, y)x ,

zEH.

For the operator S" , the H OCI functional calculus is isometric:

1I¢(S")l1 = 1I¢1I0C1' The mapping \II, ¢ ~ ¢(S") is therefore weakly continuous, from the space HOCI, equipped with U(HOCI, LI/ HJ) onto As-, u(As-, Jls-) (Chapter XI, Theorem 3.1, d)).

900

Chapter XIII

For every .\ ED, the "evaluation" map

is continuous on Hoo for o(Hoo, Ltl HJ) (Chapter XI, Theorem 3.1, a)). So its compose with tw-I is a linear functional on As- ~ continuous for o(As-, Jls-). Therefore, this linear functional is given by an element of Ns«, which we denote by C>.. We have obtained:

Lemma 1.8. - For the operator S· (or, more generally, for any contraction for which the functional calculus is isometric), for every .\ ED, there exists an element C'A in Jls-, such that:

4>(.\), [or every function ¢ E H?" . The set of the C'A, .\ Era is large enough to generate the unit ball of

Jls- : Lemma 1.9. -If

ra

is dominating,

Proof. - For every ¢ E HOO , we have:

11¢(S*)11

=

II¢lIoo sup 14>(.\)1 'AED

sup 14>(.\) I 'AEr ..

The set aeonv { C>. j .\ Era} is therefore a convex su bset of BMs _ which norms all elements of As-. The equality follows from Hahn-Banach Theorem.

Lemma 1.10. - Let NEB Ms -

Then:

.

dist Ms - (N, aeonv(Bx

Proof. - Let e > 0 be given. C>'I' .. ' ,C'A ... , such that:

18)

Bx))

< a.

By Lemma 1.9, we can find al, ... , am,

m

liN -

EaiC>'iIlMs 1

<

c.

Positive Results Since Ai E fa for i

So there is a point

901

= 1, ... ,m, we have:

ei

in '2(H), with:

and

Therefore, ei can be written :

for some Xi E H, We set:

Ilxil! = 1. m

= liN -

A

L Qi(Pei ® Pei)l!)/s•. 1

Then:

m

A =

liN -

L Oi(Pei e ei) II )Is-. 1

Indeed, for every 4> E H'" ,

tr (4)(S·)(Pei ® Pei)) = (4)(S·)Pei, Pei) =

(4)(S·)Pei, ei),

because X is an invariant subspace for S· : Pe, is in X, so 4>(S·)Pei is in X also. So we obtain :

A ~

liN -

m

m

1

1

L oiC,\;II)ls_ + ilL Oi(C,\; -

Pei ® fi)ll)ls_

Lemma 1.11. - In Jls-, i = 1, ... ,m.

Proof of Lemma 11. - We have, for every function 4> E HOO :

90e

Chapter XIII

and:

All we have to show is therefore :

and it's enough to show it for ¢(z) = zn, n ;::: O. By the definition (1) of ei, since S· is the left shift, S'":«, f, -_

2 j (I-I\'1 \ n X". A, ) l 2 ( Ai

\n+l. Ai X" • . •) ,

and therefore,

(S·n fi , fi)

= (1 -1'xiI 2 ) L (,X?+k Xi , 'x~Xi) k~O

=

(1 - ]'xiI

L l'xil

2)'xf

2k

k~O

and this proves Lemma 1.11. We come back to the proof of Lemma 1.10. We have, by the above Lemma: m

A

m

< liN - LaiC).; II Jl s • + II I

L ai (ei -

Pei)

e fill Jl s •

I

m

< E: + L

lailll(I - P)eill

I

~

E:

+ a,

since:

This proves Lemma 1.10. To simplify the notation, we put, in )/s·

B=

aconv(Bx

e

Bx).

Lemma 1.12. - For every n ;::: 1, there exists N I , ... , N« in )/s·, with: k = l, ... ,n,

Positive Results such that:

909

n

liN -

LNkIINs. ~ an . 1

Proof. - We prove the result by induction on n. For n = 1, this is Lemma 1.10. Assume the result has been proven for n = m. Let Nk E a k - 1 B, k = 1, ... , m. By Lemma 1.10,

So there exists N m + 1 in IlN -

E;' Nkll.w s • B such that:

m+l

L

lIN-

m

NkllJl s· < alIN- LNkll)ls.,

1

1

and this proves the Lemma. We can now state :

Proposition 1.13. - Let a, 0 < a < 1, and T be a C o - contraction. If is dominating, then: 1

8)1s. c - - aconv(8x I-a

®

ra

8x) .

Proof. - Let N E 8)1s.' By Lemma 1.12, for every n 2: 1, there are N1, ... ,Nn , with Nk E ak-1B, such that: n

liN - L Nkll.v s•

< an ,

1

and:

ak-l ,

k = 1, ... ,n.

So: 1

< I-a' and: n

LN

k

1

Letting n -.

00,

E

1_ 8 , I a

n2:l.

we get that : I

-

N E -1-8,

-a

which proves the Proposition.

(1)

Chapter XIII

Lemma 1.14. - Let

0 < a < 1. H 1

4,

B)l5then,

C

--

I-a

4conv(B x 18> Bx) ,

for every 4> E Hoo, 114>(T)lI

Proof. - Let

4> E H?".

(1 - a) 114>1100'

~

We know that:

Using the duality between As- and )ls-, we deduce that for every e > 0, there is a N in B)I,. such that:

Using the assumption, we can find n ~ 1, aiJ ... , an in
tr But:

(1 ~

a

~

ni (Xi e Yi)4>(S.)) > 114>1100 -

lail

~ 1,

E.

= tr (4)(S*)x.: 18> Y.:)

tr ((x.: 18> Yi) 4>(S*»

= (4)(S*)x.:, Y.:)

= (4)(A)x.:, Yi) , since the points x.:, So we obtain :

Yi

are in X.

114>1100 -

1 e < 1_ a

n

L

lailll4>(A)II·ll xiH·IIYill

1

< 1 ~ a 114>(A)1I , and the Lemma follows. We can now prove the Theorem. As we already said, we can restrict ourselves to the case where T is a Co-contraction. Assume that T has no Hyperinvariant Subspaces. By Proposition 1.7, Za is dominating for every a, o < a < 1, and so is r a' By Proposition 1.13 and Lemma 1.14, for every 4> E Hoo,

114>(A)1I = 114>(T) II

~

(1 -

and so:

114>(T)II and the Theorem is proved.

=

114>1100,

a)II4>lloo ,

Positive Results

905

2. Invariant Subspaces. We are now going to prove:

Theorem 2.1. - Let T be a Co-contraction on a Hilbert space H. If for every a, 0 < a < 1, the set Z~ of points ..\ E D such that there exists a normalized weakly null sequence (zi")),,~o with:

is dominating, T has Invariant Subspaces.

Proof. - We use the notations of Section 1. The Theorem will be proved if we can find two points z , y in H such that: Co

=

x®

v,

(1)

Indeed, for every h E HOC) ,

h(O) = (h(T)x, y). Taking first h = 1, we see that (z, y) = 1, so x and y are non-zero. Furthermore, taking h(z) = zk, k ~ 1, we obtain (Tkx,y) = 0, so span {Tx, T 2x, ... } is orthogonal to y and is the required Invariant Subspace. To obtain x and y, it is clearly enough to use an approximation procedure:

Lemma 2.2. - If there exist two sequences x" ® y"

--+

Ilx" - x"+lll < IIY" - y,,+Ii1 <

Co,

(x,,),,~o,

(Y,,),,~o

in H such that:

(2)

in JlT

1/2",

for all n E IN

(3)

1/2",

for all n E IN,

(4)

then one can find x and y satisfying (lJ.

Proof of Lemma 2.2. - Conditions (3), (4) ensure that the sequences (Y,,),,~o are Cauchy in H. Let x, y be their limits. Then:

IIx" ® y" -

x®

yllJlT

~ ~

~

(x,,),,~o,

IIx" ® y" - x ® yllJl Ilx" ® (y" - y)IIJI + II (x" ~ x) ® yllJl Ilx"II·IIY" - yll + Ilx" - xll·IIYII,

and this last quantity tends to 0 when n

--+

+00.

This shows that x ® Y = Co.

906

Chapter XIII

We now turn to the construction of the sequences (xn)n>O - and (Yn)n>O. We start with Xo = Yo = o. Assume that we have built the sequences Xo, ... , Xn-l, Yo, ... ,Yn-l, such that: IIXk - xk-ll! ~ I/2 k

IIYk - Yk-lll ~ I/2 lIXk @Yk

-

k

,

,

k

COllJlT ~ I/4 +

1

k = 1,

,n - 1

(5)

k = 1,

,n-I

(6)

k = 0, ... ,n - 1.

,

We wish to build Xn, Yn satisfying (5), (6), (7) for k We know that :

We want to build

Xn

= Xn-l + Un,

Yn

= n.

= Yn-l + V n , such that:

Ilunl! < I/2

n

(9)

llvn II < I/2 n II(xn -

l

+ un)

We write e instead of 1/4

n

@ ,

(10)

n 1 (Yn-l + v n ) - ColIJlT < 1/4 +

and u, v instead of Un,

Lemma 2.3. - For every a, 0

(7)

.

(11)

Vn .

< a < 1,

Proof. - This is Lemma 1.9 above, with S· .

instead of

Z~

r a, and

T instead of

Let now fJ > o. For every a > 0, by Lemma 2.3 applied to e BJl T , we can find a finite sum L~ aiC>.. , with Ai E Z~ for all i, such that: N

II LaiC>.. - (Xn-l ® Yn~l

- CO}lIJiT < fJ

(12)

1

and: N

Llail < c.

(I3)

1

We now have to compare the norms of such quantities in )IT and )Is·. We recall that V is the embedding of H into 12(H) :

Vx

=

(Dx,DTx,DT 2x, .. .).

Positive Resu.lts

L

Lemma 2.4. - For any finite sum

II x ® y -

L aiCAJ..vT

aiG Ai, every points x, y in H, IlVx

=

907

e Vy -

L aiGAJ..v

S ••

Proof of Lemma 2.4. - For every function h in Hoo, tr (h(T)(x ® y -

L aiCAJ)

= (h(T)x, y) -

L aih(.Ai) L aih(Ai)

= (Vh(T)x, VY)h(H) -

= (h(S*)Vx, VY)'2(H)

= tr

-

L aih(Ai)

(h(S*)(Vx®Vy- LaiGAJ),

IIhll oo = 1, we

and if we take the supremum of both sides upon h in H'" with obtain the Lemma.

,AN are now given by (12) and (13):

The numbers N, a1, ... ,aN,A1 they depend upon a, 11. For i = 1 PiP: = -ai. We set:

, N, we choose

Pi

and

Pi

such that

N

L Piztd

U

(14)

1 V

=

N

~fJ~z(v;)

LJ

J

A;

(15)

,

1

and we will show that if a, 11 are small enough, and if III < ... < liN are conveniently chosen, the points u and v satisfy our requirements (9), (10), (11). To simplify our notation, we put:

A

=

(X n-1

+ u) ® (Yn-1 + v).

We want to compute IIA - Coll..vT ; according to Lemma 2.4, we can compute instead IIA - Goll.N s • . We write an expansion: N

A - Go

= X n-1 ® Yn-1

Co +

-

L PiP: (ztd ® zl7 ») i

(a)

i=l N

+ x ® LfJj zi;;)

(b)

i=1 N

+ LfJi zti )

®Y

(e)

i=1

+ LJ ~ p.fJ-~

I

J

z(vd

x,

e z(v;) A;'

it:i

and we will take care of each line separately.

(d)

Chapter XIII

908

zi

v

) is "close" (by definition) to being an eigenvector for T (or First, each V ) , for S·). So it is convenient to consider a point in 12(H) which is really an eigenvector for S·, for the eigenvalue A :

zi

v

Lemma 2.5. - Let A E D, 0 < a < 1, and that:

Then, there exists

Furthermore, if v~

IlzA11 =

E H, with

II (A - T)ZAII < a(1 - lAD· K er (A - S*) c 12(H) such that ileA I

eA E

VzAl1 < v'2a.

ZA

zi

v

)

~ 0 weakly in H, so does

ei

v

)

1, such

= 1 and

IleA -

in '2(H), when

00.

Proof of Lemma 2.5. - We observe that :

(16) Indeed,

II (S·

- A)V ZA I = IIVTz A

-

AZA II <

a(1 -

lAD,

and:

ll(I ~ AS·)-lll ~

1+

IAI + IAI 2 + ... =

_1_.

1-IAI

We set:

This is an isometry, and:

We need a sub-lemma: Lemma 2.6. - Let U be an isometry on a Hilbert space H. For every x in H,

dist(KerU"x)

= IIU·xll.

Proof of Lemma 2.6. - We know that U· = ir:' on 1m U and U· = 0 on (1m U)l. = K er U*. We decompose: Xl

and we have:

tr«

E ImU,

X2

E (ImU)l. ,

Positive Results

90Y

So we obtain:

Ilx - yll = dist (x, K er U·).

Conversely, let y E K er U· , with

Then:

and this proves Lemma 2.6. We come back to the proof of Lemma 2.5. We have :

Ker U"

=

Ker (S· - A),

since (I - AS·) -1 is invertible. IT we denote (as we already did in Section 1) by P>. the orthogonal projection from 12(H) onto Ker (A - S·), we deduce from Lemma 2.6 : IIp>.Vz -

Vzll = IIU·zll <

a,

and we take: P>.Vz>.

IIP>.Vz>.1I This proves Lemma 2.5. In the terms (a), (b), (c), (d) appearing in the decomposition of A - Co, d instead of d , and then the we consider the corresponding terms with differences.

et

Lemma 2.7. - The term (a) has a norm in Proof of Lemma 2.7. - In

)Is·,

zt

)IT

at most

fJ +

V2ea.

we write: N

(a) = VXn-l ® VYn-l - Co -

L aiei:» ® ei:d i=l

N

+ '" (e(lId L....J 0:'' > 'i

® e(II.-) - V Z(lIo} ® V z(lId) . Ai

i=l

Let (a') be the first line, (a") the second. By Lemma 1.11,

Ai

x,

Chapter XIII

910

So (a') can be rewritten : N

(a')

=

%"-1

~ Y"-l - Co -

L

aiC).i ,

i=l

which implies by (12) that II (a') liNT < ,.,. For (a"), we write simply : N

L laililetil e (etd - V ztd ) IIN

II (a") IINs • ~

s•

i=l N

+

L lailll(ei:d -

Vzi~d)

~ VztdllN s •

i=l

~ 2V2ea, by the same computation as in the proof of Lemma 2.2. We now turn to the term (b) :

Lemma 2.8. - There is an infinite subset IN 1 of IN , such that if VI < ... < VN are in IN 1, the term (b) has norm in )IT at most n + 4,.,y'€ (1 + ,.,). Proofof Lemma 2.8. - For every V1 < ... < with Ilh v1,...,vNlloo ~ 1, such that:

VN,

there is a function hV1, ...,VN ,

N

IIx ~

L /3jzl;;) liNT

N

(h Vl ,...,VN (T)x,

=

,=1

L /3jzl;;)) .

,=1

We can find an infinite subset of IN, IN'l' such that, if IN~ , the limit of h V 1 , ••• ,VN exists weakly in H?", when Let h o be this limit. We have:

VI VI

< <

< VN are in < VN ---+ 00.

N

(ho(T)x,

L /3j zt ; ))

0,

---+

,=1 when

VI

< ... <

VN ---+ 00,

since each sequence zi~;)

J

is weakly null. Set:

g = hV1,...,VN - ho .

Choose K such that IITK xII < ,." and write the Taylor series of g :

Positive Results where VI VI

< <

911

PK

is a polynomial of degree at most K - 1. Since 9 -+ 0 weakly when < VN E IN~ -+ 00, IlpKlloo -+ O. We choose IN~ C IN~ such that if

< VN

E IN~,

Since IIglloo ~ we have:

IIPKlloo < '7 < 1. Il h ... 1100 + IIholl oo lt .......N

~ 2, this implies

N

(g(T)x,

IIIKlloo ~ 3. Now,

N

L fJjzt ») = (PK (T)x, L Pjzt ») j

j

;=1

;=1

N

+ (/K(T}T K x, LPjzi;j»), ;=1 and: N

N

I(PK(T}x, LPjztj»)1 <

IlpKlloo ·11 LPjzi;j)ll,

;=1

;=1

N

1(/K(T}TKx, LPjzi;j»)1

< II/Klloo' IITKxll'

;=1

N

II L Pj zt

j )II·

;=1

To achieve the proof, we observe that, the sequences (zi~j)} being weakly null, J we can extract further subsequences which are almost orthogonal. This means that we can find IN 1 C IN"}, such that if VI < ... < VN are in IN1, N

N

;=1

;=1

IILP;zi;j)1I < (1+'1)(LIP;1 2)1/ 2 s ~(I+'7). This proves Lemma 2.8. The estimate of (c) is simpler : Lemma 2.9. - There is an infinite subset IN 2 C IN 1 such that if VI VN

are in

IN 2, the term (c) has norm in

)IT

at most

'7 + (I + '1) ~

Proof of Lemma 2.9. - We write: N

LVzi~d

N

L fJ.ei~d

® Vy

.=1

® Vy

i=1 N

+

L P.(Vztd • =1

ei~;))

e Vy

.

< ... < a IIyll.

Chapter XIII

91e

Let (e') be the term on the first line, (e") the term on the second. We have, for some function s e Boo, IIhll oo = 1, N

1I(!=')IINs* = (h(S*), EPiei:') 0

Vy)

,=1 N

=

E P,h(Ai)(ei:

i

) ,

Vy) .

• =1

Since the sequences (ei~»)v < ... < VN are in IN~,

VI

--+ 0 weakly, we can find IN~ for every i :

c INI such that if

So we obtain : N

N

• =1

i=1

lI(e')IIJls * < (EIPiI 2)1/2'(E71 2/ 4i)I/2 < 71 . For (e") , we write: N

II(e")IINs *

s II EPi(Vzi~d

- ei~d)II'lIVYII

1=1 N

:$

(1 + 71)

(E IP.1 211 Vzt o> - el:

i

)

11 2 ) 1/ 2

·IIVYII,

i=1

if the sequence

VI

< ... < VN is chosen in IN2

C IN~,

and therefore :

To finish, we have to compute the norm of (d). The computations are similar to the previous ones. Lemma 2.10. - There is a subset IN3 c IN2 such that if VI < ... < in IN3, the term (d) has norm in )IT at most 71 + (1 + tJ) 2 e y'2 a .

Proof of Lemma 2.10. - We write, in

(d)

)ls* :

VN

are

Positive Results and call (d') the first line, (d") the second. For (d'), we find h in HOD, with lIhll oo

II (d') II .vs • =

919

= 1, such that:

L fJiPih(.xi){ei~o),

V z1;;»),

ioFi

and we find a subset IN~

of IN2 such that if

VI

< ... < VN are in IN;,

if i,j= 1, ... ,N,with ifj. We get:

L IPillfJil/4i+i

II(d')II.vs • s

< " ,

i,j

and for (d") N

N

II(d")II.vs • s I LfJi(Vzto) - ei';il)ll·lI LfJ'jVzi;i)1I i=1

i=1

~ (1 +,,) if

VI

N

N

i=1

i=1

(L IPi1 211 Vzt a> - e1:;) 11 2)1/ 2 (1 +,,) (L IPiI 2) 1/ 2 ,

< ... < VN belong to a subset INa of

and our Lemma is proved. We now choose We observe also that:

IN~.

a

So we get:

and" small enough to get (11).

N

Ilull s

(1 +

")(L IPiI 2)1/ 2

<

Vi

= 1/21'6,

i=1

if " is small enough and 2.1.

VI

< ... < V N lacunary enough. This proves Theorem

Corollary 2.11. - Let T be a contraction on a Hilbert space such that u(T) is dominating. Then T has Invariant Subspaces.

Proof. - We may assume that T is Co (see Section 1), that u(T) = ua(T) , and that any sequence of almost eigenvectors is weakly null (see Chapter II). v Fix a, 0 < a < 1, and take .x E ua(T) with l.xl > a. A sequence (z1 » 11 sa tisfying : when n --+ 00, satisfies a fortiori the condition defining

Z~,

and we can apply Theorem 2.1.

Chapter XIII

91.1

Remark. -If ). E Za (defined in Section 1) and is not in O'(T) , we know that:

(1) So there is a point y,

Set x Then

Ilyll =

1, such that:

= ().-T)-ly, so ().-T)x = y, IIxll > l/a(I-I).I). Set finally z = x/llxll.

IlzlI

= 1, and :

II(). - T)zll

lIyll < = ~

a (1 - 1).1)·

(2)

Assume that the spectrum of T contains the unit circle. By Section 1, if we are looking for Invariant Subspaces, we may assume that for every a, the set of ). 's satisfying (1) is dominating. This ensures that there exists a z satisfying (2). But Theorem 2.1 requires not only a point z, but a weakly null sequence. So we cannot conclude immediately from these arguments that any operator with spectrum containing the unit circle has Invariant Subspaces. However, this is true, and was proved by S. Brown - B. Chevreau - C. Pearcy [2] by some refinements of the techniques presented here. The question of the existence of Hyperinvariant Subspaces for such operators is still open.

Positive Results

915

Exercises on Chapter XIII.

Exercise 1. - Let T be a Co-contraction on H and let U be its minimal unitary dilation. Show that, for every x, y in H,

Exercise 2. - Let that:

z).

as in Lemma 2.5, and

f).

IIPsn f ). 112 >

lim

given by this Lemma. Prove 1 - 2a 2

•

).Ez~,I).I-l-

Exercise 3. - Prove that for any x in H, the function :

is weakly sequentially continuous from H into )Is· (that is : if a sequence (Yn)n~O converges weakly in H to a point y, the sequence x ® Yn converges weakly in )Is. to x ® y). Exercise 4. - Show that in

)Is. :

v x ® Vy = x ® Py ,

x,yE H.

Exercise 5. - Assume that (xn)n>O is a sequence of points in H such that 0 in )IT, for all y in H. Show that V X n ® Z ~ 0, for every z in

Xn ® y ~

l2(H) . Exercise 6. - Let T be a contraction with dominating spectrum. Let A E o(T) n D, and let (xn)n~O be a weakly null sequence such that (A - T)x n ~ o. Show that X n ® y ~ 0, in )IT, for any y E H.

Notes and Comments:

The basic ideas developed in this Chapter come from Scott Brown's work [1], [2], [3]. Theorem 1.2 is due to C. Apostol [II. Corollary 2.11 appears in S. Brown - B. Chevreau - C. Pearcy [1], and the result mentioned in the Remark in S. Brown - B. Chevreau - C. Pearcy [2].

916

Chapter XIII

Our terminology differs slightly from the one adopted by some authors : they call "Co-contraction" a contraction T such that there is a function / in HOO with f(T) = o. In Nagy-Foiaa's terminology, an operator satisfying Tn x ~ 0, when n ~ 00 is called Co. (Coo if the adjoint has the same property, CO,I if the adjoint is a CI-contraction). Here, we don't use the adjoint at all, so we could simplify the terminology. Moreover, the use of Co to indicate that the operator is annihilated by an analytic function does not sound quite natural. The most natural question suggested by this work is : does any operator with r(T) = IITII have Invariant Subspaces? This question was stated long ago. We will also mention it at the end of next Chapter, since the results presented there also support this conjecture.

Chapter XIV

A Counter-Example to the Invariant Subspace Problem in Banach Spaces

The major progress in Modern Operator Theory is obviously P. Enflo's counter-example to the Invariant Subspace problem in Banach spaces (P. Enflo [1]), which opened new lines of thought. The very idea of building an example of an operator enjoying given properties was new in this field, and no previous attempt had ever been made to solve the Invariant Subspace problem in the negative direction. The ideas introduced in Enflo [1] had many applications, including to the positive direction on Hilbert spaces. Two other examples followed the lines initiated by P. Enflo: C. Read

[1] and the present author (B. Beauzamy [9]), the latter enjoying a stronger property, called "supercyclicity". C. Read's example was later strenghtened and simplified by himself (C. Read [2], [3]), so as to produce an operator on II with no Invariant Subspaces (See the "Notes and Comments" at the end of the Chapter for historical comments). C. Read's latest example was then simplified by A.M. Davie [unpublished], who communicated it to us. Though this construction is not the strongest (it does not possess any property like supercyclicity or hypercyclicity), it is by far the simplest, and we present it here. It produces a very elementary example of an operator on II with no Invariant Subspace. In the "Complements" at the end of this Chapter, we refer to more sophisticated examples and consider the problem in Hilbert spaces. Before turning to the construction of the example, we start with a few basic observations. Let T be an operator on a Banach space E. 1) To say that the operator T has no Invariant Subspaces is obviously equivalent to the fact that every non-zero point is cyclic. Indeed, if x is non-cyclic,

is a non-trivial Invariant Subspace.

Chapter XIV

918

Conversely, if T has a non-trivial Invariant Subspace, F, taking x E F, we see that Fr. C F, so x is not cyclic. are linearly 2) Take a cyclic point Xo for T. Then the iterates (Tkxoh:~o independent (since the space is infinite dimensional). Consider a finite sum Lk>O akTkxo. To such a sum, associate the polynomial p(z) = Lk>O akzk (thi; is well defined, by what we just said). On the space P+ of polynomials, pu t the norm : IIpll = II ak z k II·

L

k~O

Then the completion of P+ under this norm is a Banach space, which is obviously isometric to E, since the finite sums Lk~O akTlexO were dense. We keep for it the notation E. Moreover, to the sum :

we associate the polynomial :

L k~O

a/czk+l

=

z(L a/czle) . k~O

In other terms, T becomes the multiplication by z on the space P+, equipped with the norm 11.11. This can be expressed as follows:

Lemma. - Every operator with a cyclic vector can be represented as the multiplication by z on the completion of the polynomials in z , under some norm.

There is obviously nothing deep in this Lemma : it is a mere rephrasing of the original situation. But we understand that there is no restriction of generality if we try to find T as the multiplication by z on a space of polynomials. Moreover, it allows us to simplify the notations: if Xo is a cyclic point, and if p, I are polynomials, instead of p(T)xo we just write p , and instead of l(T)p(T)xo, we write l.p. The norm coming from the operator norm is written as lI'llop, that is : II l 1i op = IIl(T) II· 3) Under this representation, the point Xo becomes the polynomial 1, which is cyclic in E for the multiplication by z (by definition). To say that any point x in E is cyclic means that:

A Counter-Example

919

To obtain this property, since 1 itself is cyclic, it is enough to ensure that 1 E span {x, zx, z2x, .. .}.

This can be written: For every e > 0, there exists a polynomial I such that :

(1)

III x-III < c.

(In the original space, and with the original notation, this property would be written: 1I/(T)x - xoll < c.)

4) In order to produce an operator with no Invariant Subspace, we have to satisfy (1) for every non-zero x, and not just for polynomials. Indeed, satisfying it just for polynomials would ensure them to be cyclic : we would have a dense set of cyclic vectors. But this is by no means uncommon: many operators do have a dense set of cyclic vectors, such as the right shift of 12(71) (see Chapter IX, Exercise 8), or even of hypercyclic vectors (see Chapter III), and they still have Invariant Subspaces. H all polynomials are cyclic, take x E E, and a sequence (Pn)n~O of polynomials converging to x in E. Then, for every e > 0, there is a polynomial In such that:

II/nPn -

111 < c.

If we have

suPll/nllop <

00,

n

then: II/nx -

111

< Il/nx -

InPn II

+ II/nPn -

111

~ Il/nli op ·llx - Pnll + IllnPn ~

111

s ,

for n large enough, and this shows that x itself will be cyclic. So the problem of constructing an operator with no Invariant Subspaces requires the following two steps : a) build a norm on the space of polynomials such that every polynomial is cyclic for the multiplication by z, that is : For every e > 0, there exists a polynomial I such that : II/p

-111 <

e ,

(2)

b) Ensure that for every x in the completed space, for every c > 0, there is a sequence of polynomials (Pn)n>O, Pn --+ x, with bounded operator norms.

Chapter XIV

S20

5) There is certainly no uniqueness in the choice of the I's satisfying (2). Indeed, if by some II we can move p close to some XI, that is

then XI close to some X2 by 12 , ••• , Xn-I close to X n by In, and finally X n close to 1 by 'n+l, then the product In+l.ln ... 1t will move p close to 1, within C if CI, ••• Cn+1 were chosen small enough. Moreover, this product is likely to have considerable operator-norm. Fix now C > O. Say that I "acts on p" if it satisfies (2) and if moreover it has small operator norm, that is for instance : 1I'l1op ~

inf{II"lIop i I' satisfies (2)} + 1.

In what set around p can we expect the operator norms of the I's acting on this set to remain uniformly bounded? The best would of course be the whole unit sphere. However, this is not possible: this would contradict the fact that the spectrum of T is non-empty (see Exercise 3 below). So we cannot expect to find a constant C such that 1111lop ~ C, for all l's acting on the unit sphere. Therefore, we have to bound the norm 1I.lIop "locally", that is near each point z , with a bound depending on x. The type of the set, containing z , on which Illliop is bounded will determine the structure of the construction. It is a neighborhood of x in P. Enfto [11, B. Beauzamy [91, but on the present construction, it is computed using the coordinates on some finite-dimensional subspaces. So far, we have given only statements which were either trivial or too vague, but were required to explain on what lines an example can be found. We now turn to the construction itself. Theorem. - There exists an operator on

II

with no Invariant Subspace.

Proof. - We first recall or introduce some notation.

n

~

Let, as before, P+ be the set of polynomials with complex coefficients. For 0, let Pn be the set of polynomials with degree ~ n. If p

= E;~o

a;z; , we put: Ipl

=

Lla;l. ;~o

This is the 11 norm, which we denoted by Ipll. For simplicity, we drop the subscript 1.

SHl

A Counter-Example Moreover, for n

~

0, we put: n

E ajz j

=

Pn(p)

;=0 and val(p)

= min{j

; aj -1O}. This is the valuation of p.

Lemma 1. - Let n ~ 0, E:, s, M > o. There exists K > 0 such that, for every m, 0 ~ m ~ n, every polynomial 9 E P«, satisfying Igi ~ M and IPm(g)1 ~ s, there exists q E P«, with:

and

Iql

~ K.

Remark. - The reader will observe that the assumptions on 9 are of "concentration at low degrees", as we discussed at the end of Chapter VIII.

= 0, ... , n, we set

Proof. - For m

Om =

{p E

Pn ;

:

Ipl

~

M and IPm(p) I ~

()}.

This is a compact subset of Pn • For 9 E Om, we write the Euclidean division of zm by g, according to increasing powers of z : zm

with val(r)

=

gq

+r

,

> n. So we get: zm - r ,

gq

Therefore, there exists an open neighborhood U of 9 in Pn , such that, if g'E U,

We cover Om by U lJ • • • , UK.,. , corresponding to polynomials glJ ... ,gK... , and, for k = 1, ... ,Km , if 9 E Uk, we have

Finally, we take:

K and our Lemma is proved.

max max m

k

Iqkl ,

se»

Chapter XIV

Let now (an)n~o and (bn)n~o with ao = bo = I, and :

<

1

Let (Vn)n~-I

be two sequences of positive real numbers,

< bl < ... < an < bn < -..

al

be a sequence of positive real numbers, defined by : V-I

=

Vn

=

Vo

=

0,

(n - I)(an + bn )

n>

,

1.

of P+ as follows:

We define a basis (/n)n~o

10 = I, -If n 2: 2 and 1

~

r

~

n - I, or n = r

if ran ~ k ~ ran +

= I,

Vn-r-I ,

he

if (r - I)a n + V n- r < k < ran,

Ik

-If n 2: 2 and 1

~

r

~

(a)

Zk-I1.. ),

21(r- t )11.. -kllb..-

1

zk

,

(h)

,

(c)

n - I,

if r(a n + bn) < k ~ (n - I)a n + rbn , if (n - I)a n + (r - I)b n < k < r(a n + hn) , We also define formula (d) with n

an_r(Zk -

fA:

he = lie

zk - bn zle-bn

= 21(r- i >b..-kl/nl1.. zA: , (d)

by formula (c) with n = 1 when k = al < k < al + hi-

al

+ bl

,

and by

= 1 when

For fixed n 2: I, we have defined fie for V n-l < k ~ V n . Indeed, the sequence (b), (a), (b), (a), _.. allows us to go from V n-l to (n - I)a n : V n-l -+

an

-+

an +V n-2

-+

2a n

-+

2a n +V n-3

-+ ... -+

(n-I)a n

-+

(n-I)a n

•

Then the sequence (d), (c), (d), (c), ... allows us to go from (n - I)a n to (n - I)(an + bn ) = V n :

(n - I)a n -+

-+

an + bn

(n - l)a n + 2hn

-+

-+

(n - l)a n + bn

...

-+

-+

2(a n + hn)

(n - I)(a n + bn)

-+

(n - I)(a n + hn) .

In the first case, to ensure that the numbers involved are in the right order, we

need:

(r - I)a n + V n- r < ran,

A Oounter-Ezampl« which means V n -

r

<

9f!9

an, that is :

(n - 2)(a n -

l

+ bn-d < an ,

n

(Cl)

1.

~

In the second, we need:

(n - 1)4n

+ (r -

l)b n < r(a n

+ bn),

or (n - r - 1)an < bn, that is :

(n - 2)an < bn ,

(C2)

n 2: 1.

We see that for every k 2:: 0, he is of degree k: Therefore 10, Ii.. .. lie span Pk. We define a norm on P+ by the formula:

II Laklkll

Llakl,

=

k~O

for any finite sequence of complex numbers (ak)k~o, Let X be the completion of P+ for this norm. The space X is a Banach space, which is obviously isometric to 11. Let T be the multiplication by z . We will show that T is a continuous operator on X with no Invariant Subspaces. First, we establish a few consequences of the definitions: - If 1 ~ r ~ n - 1 and ran ~ k ~ ran + V n - r-l, by (a) : 1

1

---Ik-a n

--Ik' a n

a n - r +l

r

,

and therefore :

So:

Ilzk - zk-ranll

=

-

1

an -

+ r

1

a n - r +l

+ ... + -

1

~

2/a n- r

•

(1)

4 n-l

- The same way, from (c), we deduce that, if r(a n + bn) ~ k ~ (n - 1)a n + rb n 1

(2) Lemma 2. - If the sequences k 2:: 0,

(4n)n~O,

(bn)n~o

grow fast enough, for every

We postpone the proof of this Lemma. So (admitting Lemma 2), T is a continuous operator on X, with IITII ~ 2.

Chapter XIV

In the course of the proof of Lemma 2, we will moreover obtain the following estimates, which we admit temporarily:

lI T f ro + v :5 1Ia n - r II T f r4 - dl :5 1la n , liT/r(On +bn)-lll :5 l/b n . n - r-all

n

(3)

,

(4)

n

(5)

For m > 1, we define : Om

= {k E

=

IN i 3n > m , (n - m)a n :5 k :5 (n - m)a n

Un>m [(n - m)a n

,

(n - m)a n

+ vm-d

+ vm-t].

Lemma 3. -If the sequences (an)n~o, (bn)n~o grow fast enough, for m > 2, k ;» (m - l)a m and bm + am :5 s :5 bm + (m - l)a m, we have: -H k

i

Om,

liT- t, II :5 -IE k E Om, decomposing k = (n - m)a n

4.

+ i,

liT- fA: + a mz +-II j

with 0:5 i :5 Vm-l, :5 1.

We also postpone the proof of this technical Lemma. For m > 2, we define a linear mapping Qm, from P+ into QmfA:

=

f A: 0 { -amzj

if 0 :5 k :5 (m - l)a m , if k > (m - l)a m , k tI- Om , if k E Om , k = (n - m)a n + i

P(m-l)a m

,

by :

, j :5 Vm-l

n > m, so k ~ an > (m - l)am.)

(we observe that if k E Om, k ~ (n - m)a n

,

Corollary of Lemma 3. - IE m > 2, bm 9 E P+, then:

+ am

:5

.9

:5 bm

+ (m -

l)a m and

(6) Proof. - We first show the result for 9 = fk. - If k :5 (m - l)a m, Qmfk = fA:, and the left-hand side of (6) is O. - If k ~ (m - l)a m, k rI- Om, Qmfk = 0, and, by Lemma 3 : ~~T-

fA:" :5 4"Jk ~ = 4.

- If k 2:: (m -l)a m , k E Om, Qmfk = -amzj T-Qmfk

,

= -amz j +-,

and the left-hand side of (6) is, by Lemma 3 :

liT- t. + amzi +-II :5 1, which proves the Corollary when 9 = fk. To obtain it for every show that Qm is continuous on X.

g, we have to

925

A Counter-Example We recall that

i

- For

~ Vm-

Vm-

1 < (m - l)a m < V m , by (Cd.

1, the sequence

Ii

uses only a1, ... , a m-1 , b1, ... , bm- 1.

- For V m - 1 < i ~ (m-l)a m, we are in the situation to apply consecutively (b) and (a), and Ii is given by these formulas. But (a) uses a m - 2 , a m-1, am, and (b) uses am, bm- 1 . Therefore, there exists a constant C m a1,bl, ... ,am-1,bm - 1,a m , such that

i

- For

~

V m-1,

llil

,

depending only on the numbers

i=O, ... ,(m-l)am •

~ Cm,for

zi is computed using

- For V m-1 < i ~ (m - l)a m does not appear. So :

,

h, ... , Ii, and so

zi is computed using

:

h, ... , Ii,

and bm

and Qm is continuous on X, which proves (6). Moreover, we get:

and therefore : gE X.

Lemma 4. - Let g EX, with Ilgll = 1. We assume that for some m > 2, some r, 1 ~ r ~ m - 2, we have:

Then there is a polynomial

Ip

such that :

11cp(T)g -

III

~

31 a m - r - 1

.

Proof of Lemma 4. - Let h = Qmg, so Ihl ~ em. Applying Lemma 1, we find K = K(I/(a mCm ) , l/a m , C m , (m - 2)am ) , and a polynomial q in P(m-2)a m

,

such that:

ge6

Chapter XIV

and

Set f/J

= ZBm q.

Then f/J E

P(m-l)B m ,

and:

l ) Bm (f/Jh)

z(r+l)B m

and therefore :

II P (m Set rp

= _1

-

l/a m

1

f/J =

zb m

bm

II <

_zBm+b m bm

(7)

.

q.

From c), we deduce that if : r(a m

then:

+ bm )

1 k II-z

bm

-

~

< (m - l)a m + rb m

k

zk-b m

1 bm

II

,

•

If we put k' = k - bm , we have am ~

(m - 1lam ,

k' ~

and

II z k'

- -1z k' +bm

bm

II

(8)

Let's put: A

=

II P (m - l ) Bm+bm (rph)

1

= IIP(m-l)a m+bm(b

-

P(m-l)a m(tbh)

zbmtbh) -

m

If we write:

tbh =

II

P(m-l)a m(f/Jh)!I.

L li zi , i?O

since val(1/J) ~ am, we deduce from (8) :

(9) if we choose bm large enough.

St7

A Counter-Example Furthermore, denoting by dO the degree, we have:

if we choose bm large enough. So (d) gives :

Ilcph -

P(m-l)a...+b ... (cp

h) II ~

<

2(2(m-l)0 ... - i b_)/ma ...

Icphl

2(2(m-l)0_-i b_)/mo_

KC m bm

-

< I/am

(10)

again if bm is large enough. Moreover, since

cp can be written: (m-I)a... +b ...

cp

L

=

A_z- ,

o... +b...

and since Iql ~ K, we have We also have:

Icpl

~

K/b m

.

and by (6) :

Il r - g -

IIA_z-g -

r-Qmgll A_z·hll

~ ~

411gll, 4IA_lllgll, 4K < I/a m bm

II

,

if bm is large enough. This means :

(11)

By (I), if 1 ~

r ~ n -

I,

ran ~ k ~ ran

+ Vn-,.-I,

9tS

Chapter XIV

and since r we get:

~ 1'1 -

2, replacing n by m and r by r

+1

in the above estimates,

(12) Finally,

IIIp(T)g - 111 ~

+

IIIp(T)g - cphll

+ IIcph -

II P ( m - l )a m + b... (lph)

+ IIP(m-l)am (1/Jh) -

-

P(m-l)a".+b". (lph)

11+

P(m-l)a m (1/Jh) II

z(r+l)a m

II + lIz(r+l)a... - 111

1 111 2 < -+-+-+-+--am am am am am - r - l < 3/am - r - l , by (11), (10), (9), (7) and (12), and this proves Lemma 4. We now prove the Theorem. We want to show that, for every e there is a polynomial cp such that :

> 0, for every

9 E X with

Ilgl1

= 1,

IIIp(T)g-III < c. So, we fix e > o. Let k > 1 such that I/alr, < c/3. By Lemma 4, we need only to find r ~ 1 and m, with

r+k+l,

m ~ such that:

Indeed, then r ~ m - 2, and 3/am - r - l ~ 3/a1c < c. Let us assume conversely that for every r, every m with m - r - 1 we have:

~

k,

(13) We will obtain a contradiction. For each m, we can find D m such that:

>

0, depending only on

if 0 ~ Now, we write : 00

9

LCLi/i, k=O

i

~ Vm •

aI, b1 , ...

,am, bm ,

St9

A Counter-Example with

E;:o lail = 1, and for

> 1, we can write:

n

(n-l)a ..

(n-l)4..

L

L

ail;

i=o

fJn,i zi

.

i=o

We have also, for n > 2

Qn(

\1.. -1

L

L

ail;)

An,iZ;

i=O

;>(n-l)a..

with: -an

L

(14)

ai+(m-n)a....

m>n So we obtain : (n-l)a..

L

\1 .. -1

L

+

fJn,;zi

;=0

(15)

An,jZ;

;=0

From (13), (14), (15), we deduce that if r

~

1 and m - r - 1

~ k,

ra ...

L

;=\1... -1

IfJm,i1 <

l/ a m

.

(16)

+1

If ram + Vm- r-2 < j ~ ram + Vm- r-l, we are inside the range of application of formula (a) for the definition of 1;. So we get :

and applying (16) with r replaced by r+l, we obtain that, if r ~ 1, m-r-2 ~ k,

L

1

jail <

Combining this with (14), we get, for n

~ k

V.. -1

1

(17)

+ 2,

\1.. -1

L

<

la;+(m-n)a...1

L

lla m m>n < 2la n +l .

(18)

Chapter XIV

990

Taking now r

= 1 in

(13), we obtain, for n 2:: k + 2 : IPQIl(Qng)1

s -.!... an

and we deduce from (15) :

L IPn,; + An,;1 ;=0

<

1

(19)

an

from which follows, by (18), that :

Now, we observe that if Vn-l < i ~ (n - l)a n , we are in the range of application of formulas (a) and (b) for f;, and I; has no term zi, with Vn~2 < i ~ Vn-l. Thus: tln-l

2:

2:

a; I; -

{3n,; z; E

Ptln-'J

•

tlj=n-'J+ 1

i=tln-'J+1

So: tln-l

tln-l

< 2D n _t/an <1/~,

if an has been chosen large enough. We take m = n - 1 and obtain, when m 2:: k + 1 tI...

L

lail <

I/Jam+ l

(20)

.

;=tI...-l+l

Using (20) in (14), we find, when n 2:: k + 1 : tln-l

2: IAn,il s an 2: 2: lai+(m-n)Q...

i=O

m>n i=O tI...

~ an

2: 2:

m>ni=tI..._l+l

lail

1

A Counter-Example

and, using (19), for n

~ k

991

+2 : Vn-l

L

2

IPn,il <

an

j=O

that is : (21) and from (20), we get: (n-l)a n

L and noting that if

V n-l

<j

~

(n - l)a n

,

we conclude that : (n-l)a n

IPv n -

1 (

L

Vn-l

aj/j)1 < an-dan.

(22)

+1

Combining this with (21), we obtain, for n

~ k

+ 2,

This implies: tJn-l

Lo

1

lajl ~ D n - 1 -(a n - l + 2) , an

and this tends to 0, when n -+ 00. We obtain a contradiction with the assumption 2:~ lajl = 1, and this proves the Theorem. We still have to prove Lemmas 2 and 3.

Proof of Lemma 2. - We look separately, for k , at the cases (a), (b), (c), (d).

a) ran Then:

~ k ~ ran

+ Vn-r-l.

992

Chapter XIV

- If k = ran

+ Vn-r-l,

we deduce from (b) :

and using (1) and (b) : Ilzk+1-Onll

= < <

Ilz(r-l)On+Vn-r-l+11l 2

+lIzVn-r-l+lll

an-r-l 2 an-r-l

+ 2(1+vn-r-l-tOn-r)/bn-r-l

and therefore, liTfron+vn-r-l

II

if the sequences (an) and (6 n ) increase fast enough. So we get:

b) (r - l)a n Then:

+ V n - r < k < ran_

- If k < ran - 1,

- If k = ran - 1,

IlzrOn II ~

So:

if an is large enough.

1+

-

2

an -

by (1). r

,

A Counter-Example

c) r(a n + bn )

~ k ~

(n - l)a n + rb n

999

•

Then:

- If k < (n - 1) an + rb n , T I k = I k + 1 ,so II T I k II - If k = (n - l)a n + rb«, by (d) :

IITlkl1

=

(1 + bn )

= 1.

2((n-l)a n+l-i bn)/na ..

if bn is large enough. d) (n - l)a n + (r - l)b n < k < r(a n + bn ) Then:

•

- If k < r (an + bn ) - 1,

- If k

= r (an + bn) -

1,

and, by (2) :

IIzr(a.. +b II n )

Ilzra II

~ 2b~-1

+

~ 2b~-1

+ - - +1,

n

2

an -

and so, if b« is large enough,

liT /k II < This proves Lemma 2.

I/b n

.

r

~

1,

Chapter XIV

Proof of Lemma 3. Here again, we consider the same cases :

a) ran

~ k ~ ran

+ Vn-r-l

.

Then:

Since k > (m-l)a m , we have n > m. We distinguish between three sub-cases: 1) r

=n -

m (that is k E Om).

+ i, with 0

Then we write k = ran

+s

bm < j

Vm-l

~

j

~

Vm-l, and

~

+ (m - l)a m + bm <

2bm < an .

So ran

+ Vm-l <

k

+s <

(r

+ l)a n

,

and, by (b) : II zA:+tI II = ~

2(j+tI~tan)/bn-l 2(2bm~tan)/bn-l

< l/a n

•

if an is large enough. Also, since : j

+s

2bm <

~

Vm ,

we have by (1) : IlzA:+s-an - z;+tllI ~ 2/a m+l liT"'fA:

+ a mz;+"'11

~ a m llz A:+ "' lI ~

am

1 (-

an

+ amllzk+",-a n - z;+tllI

2 +--) am+l

<1.

2) r>n-m. Then s

> bm > V n ran

r ,

and therefore:

+ Vn-r-l <

(r - l)a n

k

+s <

(r

+ l)a n

,

+ V n- r < k + s - an < ran,

A Counter-Example

+ j t with 0

and if we write k = ran

~ j ~

995

Vn-r-l, we get, by (b) :

Ilz k +. !! = IIzk +. - a"ll = 2(;+.- t a n )/ 6.. -1 ~

2(26m-tan)/6n-l

t

since: j

+s

~

Vn-r-l

+ (m -

l)a m

+ 6m

+ (m -

l)a m

+ 6m <

~

Vm-l

~

Vn-r-l, and we get:

Therefore,

if an is large enough. 3)r
i + s ~ Vn-r-l, T· /k = - H i + S > Vn-r-l ,

- If

~

i

"c+., liT· /kll =

1.

and thus:

by (3)

< 2.ja n -

<

2

26m

r

jan -

r

< 1, if a n -

r

is taken large enough.

b) (r - l)a n + V n -

r

<

k

< ran'

Here again, n > m. - H k + s < ran, T. /k

= 2((r-t )a .. -k)/6 = 2./ 6n -

and therefore :

since s

s 26m •

1

/k+. ,

n - 1

zk+.

26m

•

996

Chapter XIV - If k

+s

~

ran,

T'"],k -liT" fkll ~

2(ra n - l - k)/ bn-

Tk+a-ran Tfran-I,

1

2 a/ bn- 1 2k+a-ran jan,

by (4) and Lemma 2

11

~ 4.2 jan

s 4.2 2bm jan

~

I,

if an is taken large enough.

c) r(a n Then:

+ bn ) ~

k ~ (n - I)an

T" fk = We have n

+ rb«,

zk+a -

bn

zk+a-b n

•

m. We consider first the case :

~

I) n = m. Then: and so :

k - If k

+e

~

(n - I)a n

+ s ~ (r + I) (an + bn ) . + (r + I)b n (which implies

IITafk11

- If k + s > (n - I)a n - if r < n - 2 :

+ (r + I)b n ,

r

~

n - 2),

= 1. k + 8 < (r + 2)(a n + bn ) , we have:

Ilzk+all = 2(k+a-(r+!)b n)/nan ::; 2(2(n-l)a n

- if r

~

-

!bn)/nan .

n- 2 :

IIz k + all = 2(k+B- !an+I )/b n 2(b n+2(n-l)a n -!an+ al/b n .

~

and the same way, if r < n - I,

and if r

=n -

I, Ilzk+a-b n

II

= 2(k+a-b n- !an+d/bn ::; 2(2(n-l)a n-!a n+d/ bn ,

and we get IITa /kll ~ I in all cases, choosing bn

,

an+l large enough.

A Counter-Example

997

2) We now consider the case n > m. - If k

+s

~

(n - 1)an + rb n

,

- If k + s > (n - 1)an + rb n , - if r < n - 1, we have k + s < (r

IIz k+ 6 11

+ 1)(an + bn ) ,

= 2«k+6)-(r+!)b.. )/na.. 2«n-l)a.. +2b m

~

-

!b.. )/na n

,

~ l/b~

and the same estimate holds for IIz k +6 - b.. II. - if r = n - 1, we have the same estimate for

Ilz k+611 = 2(k+6-!a ~

2(v.. +bm-

n

Ilzk +6- bn II, and

+ 1 )/b ..

i a .. +d/ b..

and we get:

IIT

6

/kll ~ l.

d) (n - 1)an + (r - 1)bn < k < r(a n + bn ) . Then:

T 6/ k =

2((r-i )b.. -k)/na.. Zk+6.

The hypothesis k > (m-1)a m again implies n 2:: m, and we consider separately the cases n = m and n > m.

1) n = m. Since s > bn , k + s > (n - 1)an + rb n • We distinguish between four different subcases :

a) k + s > (n - 1)(an + bn ) •

II Z k + 611 = 2(k+6- !a..+d/bn ~ 2(nb.. -!a .. +d/ b.. ,

IlT

6

/kll

~ 2bn/an+(nbn-!an+d/bn ~

if

an+1

is large enough.

1,

Chapter XIV

998

{l) k + 8 < (r + 1) (an + bn ) , r < n - 1. Then:

since

8 -

bn ~ (n - 1) an .

'"Y) (r + 1)(a n + bn) ~ k + Then, by (2) :

~ (n - 1)a n

8

+ (r +

1)b n, with r


1.

But k + s - bn (r + 1) ~ (n - l)a n , and so Ilzk+"-b..(r+l) II is bounded by a constant which depends only on an, and :

if bn is large enough. Furthermore, k

> (r + 1)(an + bn)

and thus:

liT" fkll

- s

>

rb n - nan ,

~ 2(r-!)b..-k)/na..

< -

2(na ..-!b..)/na.. br+ 2

~

1,

n

if bn is large enough.

6) k + s > (n - l)a n + (r + l)b n with r < n - 2. Since 8 < 2b n, we have k + 8 < (r + 2)(a n + bn) I and so : I

IIz k+"11 = 2(k+.-(r+!)b..)/na.. ~

if bn is large enough.

2(k+(n-l)a..-(r+!)b..)/na.. ,

A Counter-Example

999

2) n> m. - IT k

+8 <

r(a n

+ bn ) , Z A:+-

-- 2(A:+_-(r-t»/nIlB! 11:+.

T- IA:

= 2-/ nIlB fA:+_

liT- h;1I = 2-/

,

n llft

:::; 22b... /nIlB

since m < n,

:::; 2 ,

for b« large enough. - IT k + 8 ?: r(a n + bn ), then, using the fact that that, by (5),

IITII :::;

2, and the fact

we get:

liT- IA:II

::; 2- Ibn

s 2 2bm Ibn :::; 1,

if bn is large enough, since m < n. This finishes the proof of Lemma 3, and achieves the construction. The most noticeable difference between the present construction and the one given by P. EnOo [I] or the author [I] is that here the norm is defined at once, and not by an induction procedure. Moreover, the expression is rather simple, since each fA: depends only on two zi at most (in this respect, one cannot expect simpler !). But otherwise, most of the main features of the construction : the use of the 11 norm, concentration at low degrees for polynomials, use of very fast growing sequences, defined inductively, were already present in P. Enlio [11. So far, all existing examples of an operator with no Invariant Subspaces are built on a space which contains 11. It is unknown whether this condition is required or not. The only thing which is clear is that the "type of uniformity" used by P.Enflo or the author cannot be achieved in a Hilbert space. But this is only a technical remark. Also, all examples so far, for which the spectrum has been computed (C. Read II], B. Beauzamy [9]) satisfy :

r(T) <

IITII.

This supports the conjecture according to which an operator with r(T) might have Invariant Subspaces.

= IITII

Chapter XIV

Exercises on Chapter XIV.

Exercise 1. - Let T be an operator on a Banach space E, with a cyclic point Xo. Prove that T has no Invariant Subspaces if and only if : (PI) - For every e > 0, there exists a countable family of open balls (Bm)m~o, such that uBm = E \ {O}, and a family (Cm)m~O of positive reals, of points in E \ {O}, dense in E, there such that for every sequence (Yk)k~O exists, for each k ~ 0, a polynomial I Ie with :

and, if Yle E B m

,

Exercise 2. - Let T be as before. Prove that T has no Invariant Subspaces if and only if : (P 2) - For every e > 0, every family (Yk)Ie~O of points in E \ {O}, dense in E, there exists a family (B:")m~o of open balls, all centered on points Y;, such that uBm = E \ {O}, a family (Cm)m~O of positive reals, such that if Y; E B k, there is a polynomial ';,k with:

Exercise 3. ~ Let Xo be a cyclic point for T. Let oX E O'a(T), and (Yn)n>O, llYn II = 1, be a sequence of corresponding almost eigenvectors, that is (see Chapter II) : n -+ 00. Take: e

<

Assume that there is, for every n

IITxo - oXxoll IAI + IITII . ~

0, a polynomial

Show that: sup IIln(T) lIop n~O

00 •

'n such that:

94 1

A Counter-Example (Hint: Assume the converse, and write: II Ax o - Txol! ~ IIAxo - AI,,{T)Ynll

+

1I'"{T)(Ay,, - Ty,,)11

+ 1I'"(T)Ty,, - Txoll·) Deduce that if T has no Invariant Subspaces, that is if for every z , Ilxll = 1, every e > 0, there is a polynomial lz such that:

then: sup z,lIzll=1

II/ z (T )lI op = +00.

Exercise 4. - Dependence upon e of the Invariant Subspace property. To say that T has no Invariant Subspaces means: For every e > 0, for every x, y, with IIxli polynomial I such that :

II/(T)X -

yll <

= 1,

lIyll = 1, there exists a

e.

One may wonder: is it enough to have this property for some s , 0 < e < 1 ? The answer is "Yes" in a Hilbert space: Assume that there exists a a, 0 < a < 1, such that for every z , y, with Ilxll = 1, Ilyll = 1, there exists a polynomial I such that:

III(T)x - yll < a. 1) Let z , y, with Ilxll

= 1,

IlylI = 1. Let 11 be a polynomial, such that:

Set Zl = IdT)x. We may assume: !Iy-zlll !Iy-zlil

=a, = inf{lly-Azlll;

A Effi}.

Set: Yl = 2y - Zl, y~

z'1 Show that: Ily - y~1I

VI + 3a

2 -

1.

Chapter XIV

2) Let 12 be a polynomial such that:

Set

Z2

= 12 (T )x . Show

that:

3) Set:

and:

Show that:

III(T)x -

yll

~

al .

4) Define ind uctively an by :

Prove that the sequence an tends to 0 and conclude. 5) Study the same question in uniformly convex Banach spaces. 6) Conversely, in 11, take y

= el, and assume that

z

= el

- ae2. Then

Assume: Z2

Show that

IIz2 -

y{1I

= a,

=

+ ae2 _ a (e 1 + e2) . 1+a 2

el

and that:

II Z1 + Z2 2

_

yll

a,

so the above procedure fails.

Exercise 5. - Let H be a Hilbert space. Assume that there is an operator T on H with no Invariant Subspaces. We want to show that for every xo, every x in H, there is an operator A in AT such that II Ax - Xo II ~ s , and such that: IIAII = min{IIBII j IIBx - xoll ~ e, BEAT}.

A Counter-Example

Moreover, A satisfies

II Ax -

Xo

II = e and is unique.

a) Consider the set :

C = {I E P+ ; II/(T)x - xoll ~ e}, and let m = inf{lllll ; I E C}. Let (In) be a sequence in C with, for all n, Il/nli ~ m + lin. Show that there is a subsequence of (In) which converges for the ultraweak topology, to an element A E AT, and that IIAx - xoll ~ e , IIAII = m. b) Let I be a polynomial such that Ill(T)x - xoll < e. Show that there exists a polynomial I' such that :

1I/'(T) II ~ II/(T)II, and:

1I/'(T)x ~ xoll

e.

Deduce that IIAx - xoll = e. c) Assume that Al and A 2 satisfy the conclusion, with Al sider the operator (AI + A2)/2, and get a contradiction.

=1= A2.

Con-

Exercise 6. - Invariant subspaces and Ergodic Theory.

Let T be a contraction on a Banach space, with no Invariant Subspaces. We show that for every z , I n-I

.

;; LTJ x

-t

0,

a

in norm, when n

- t 00.

a) Show that for every e > 0, every z , there exists a polynomial I such that:

II/(T)(T - I)x -

xii <

e.

b) Fix e > 0, x E E. Take I given by a). Show that:

~ II/(T)(T _ I)x _ xii + 11/(T)1111 x - Tn x II.

II x + ... + Tn-Ix II n

n

(Hint: introduce the term I(T)(T - 1) :t+T:t+.~±Tn-l:t

.)

c) Deduce that, for n large enough, 11

x

+ ... + Tn-Ix n

il

< 2e.

Chapter XIV

d) Deduce that if T is a contraction on a Banach space, such that there are points x, Xo and ex < 1 such that, for all n, IITAx - xoll < ex, then T has Invariant Subspaces. (This is clear from Exercise 9, Chapter XII, when E is reflexive, but does not follow otherwise.)

Exercise 7. - Closed invariant convex subset of the unit ball (see also Chapter II). Let T be a non-zero contraction on a Banach space. We want to show that there is a closed convex subset C of the unit ball, with C =j:. 8, C =j:. 0, such that TC C C. a) Show that if T has no convex invariant set, for any x E B, with

= 8. z, y in E

Ilxll =

1,

conv{x,Tx,T 2x, ... }

b) Deduce that for every E: > 0, every there exists a polynomial I = L:j~O aj zj , with that: 1I (T )x - yll < g.

aj

with ~ 0 and

IIxii = lIyll = 1, L: aj = 1, such

'

c) Use Exercise 3 above to obtain a contradiction.

Notes and Comments:

Before the Banach spaces examples were found, A. Atzmon [2] produced an example of an operator on a nuclear Frechet space with no Invariant Subspaces. This example relies on properties of some spaces of analytic functions, and is quite interesting. There is no link, however, between his approach and the one which was used later in the Banach space case. P. Enflo's paper [1] was submitted to Acta Mathematica in february 1981. The paper was not well-written, and was quite complicated and not easy to understand, so refereeing took some time, as well as the minor changes which were made, so the paper was definitely accepted in 85, and appeared in 87. Using the ideas of P. Enflo [1), but simplifying and strengthening them, the present author also produced a Banach space example B. Beauzamy [9). This example had a stronger property, called 8upercyclicity : for every x =j:. 0, the set {CTn x ; C > 0 , n E IN} was dense in the whole space.

A Counter-Example This example was presented at the Functional Analysis Seminar, University of Paris VI - Paris VII, in february 84. It appeared in Integral Equations and Operator Theory in june 85. At the same time, C. Read also produced a counter-example (C. Read [1]). His paper benefitted of quick refereeing and "jumped over" the waiting line for publication in the Bulletin 01 the London Mathematical SocietU, so it appeared in july 84. Similar facilities were also offered, by the editors of this Journal, to the present author, who declined. To give such precision in uncommon, and would be of no interest, if C. Read had not, several times, unelegantly and unsuccessfully, tried to claim priority towards the solution of the problem. This behavior might be condemned with stronger words, but we remember we are presently writing for posterity. We have insisted upon several ideas introduced by P. Enflo, and which were used in all later counter-examples. But we would not like to give the impression that the other counter-examples are trivial modifications of his. C. Read's example on II (C. Read [3]), presented here in a simplified version is a major achievement in the theory (I am glad to point this out, despite the previous paragraph), and his latest result of an operator in 11 with all orbits dense (all points hypercyclic) is the "state of the art" in the Banach space setting. In the frame of Hilbert spaces, things are not so advanced. The operator which is "closest" to a counter-example" is the one built by the present author [12] : it has one hypercyclic point %0, and for every polynomial p with complex coefficients, p(T)xo is also hypercyclic. Therefore, the operator has a vector space of hypercyclic points (thus solving a question raised by P. Halmos [2]), but it may still have points which are not cyclic at all, thus having Invariant Subspaces. In my opinion, though being the "closest", this operator is still quite "far" from having no Invariant Subspaces, so far, even, that I have no idea whether such a counter-example exists or not. IT one compares the positive results of the previous chapter and the negative ones of this one, one sees the very large gap in between, and one understands better, perhaps, the opinion stated in the introduction about the state of development of the theory. A lot is still to be done, on a problem which is rather challenging, if not rewarding.

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INDEX

absolutely convex hull algebraic tensor product almost eigenvectors approximate unit approximation property

118 24 165 119

Beurling Algebras Beurling sequence Blaschke product

262 264 180

Calkin Algebra Cayley Transform Cesaro averages of the partial sums character (in an algebra) characteristic polynomial commutative algebra commuting with T (operator) compact completely non-unitary operator concentration d at degree k contraction convolution operators cyclic for (T, T*) cyclic point C* -algebra generated by a normal operator Co-contraction

95 236 167 127 6 123 32 79 213

dilation of an operator divisor of 0 dominating

212 124 297

eigenvalue eigenvector elementary hyperinvariant subspace elementary invariant subspace essential norm

6 6 33 33 94

299

190

91 208 150

32 141 293

Index

9.48

essential range essential spectrum extension into a unitary operator

153 95 214

finite support (vector with)

26

Gelfand's Transform

128

harmonic function hypercyclic point hyperinvariant subspace hyponormal operator

163 71 32 155

ideal inner function Invariant Subspace Invariant Convex subset involution isolated point of the spectrum isometric dilation isometry

123 177 32 32 131 30 215 213

Jensen's functional

189

Mahler's measure maximal ideal minimal isometric dilation minimal unitary dilation minimal polynomial multiplicity of an isometry multipliers

189 126 216 216 7 227 208

non-quasi-analytic algebra non-tangential convergence normal element in an algebra normal operator norming sequence nuclear norm nuclear operator nuclear operators on a Banach space

262 171 131 84 63 97 97 119

ou ter function

177

Index

partial isometry partial sums of a Fourier series Poisson Kernel pole of an operator positive operator on H positive operator on L p projection of an operator projective completed tensor product proper ideal pseudo-function pure isometry

88

167 165 30 86

228 215 118

126 258 227

quasi-nilpotent operator quasi-triangular

84 95

radial convergence radical (of an algebra) reducing subspace regular measure regularly closed subspace residual part of a dilation resolution of Identity resolvent of an operator resolvent set of an operator resolvent set (in an algebra) right shift operator

171 129 213 143 244 222 150 21 21 124 20

Schwarz map self-adjoint element in an algebra semi-simple algebra similar to a contraction (operator) singular inner function spectral family of projections spectral measure spectral radius (in an algebra) spectral radius of T spectrum of an operator spectrum (in an algebra) spectrum (of an algebra) spectrum in a finite dimensional space

120 131 129 231 182 158 143 124 22 21 124 127 6

Index

950

*-homomorphism *-residual part of a dilation step-functions strict support (function with) strong topologies sub-algebra subharmonic function subnormal operator supercyclicity support of a finite sequence

132 222 144 261 110 123 163 154 344 45

trace of an operator trigonometrically well bounded operator

98 158

ultrastrong topology ultraweak topology ultraweakly closed algebra unit in an algebra unitary dilation unitary operator unitary algebra

110 106 242 123 211 209 123

valuation of a polynomial

321

weak topology weighted shift

101 84

References

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