Unbounded Linear Operators - Theory And Applications

=

V :neg) qE:1

H(x)

= g(x)

for any

g

E:3 with x E:neg)

Since:3 is totally ordered, it follows that H is in
G(ax

+ m)

=

aG(x)

+

G(m)

= aG(x)

+ F(m)

mE:D = :n(F)

with G(x) chosen so that (2)

IG(ax

+

m)1 ~ p(ax

+ m)

Tho proof of the theorem is now reduced to showing that a number, which we call G(x), existH tmdl t.hat. (2) holdH.

UNBOUNDED LINEAR OPERATORS

18

We first suppose that X is a real vector space. equivalent to (3)

Note that (2) is

+ m) :s; p(ax + m)

G(ax

for if (3), then -G(ax

+ m) =

G( -aX - m)

:s;

+ m)

p( -aX - m) = p(ax

As particular cases of (3), we want G(x) to be defined so that (4)

G(x

+ z) :s;

+ z)

p(x

Z

E:D

and (5)

-G(x

+ y)

= G( -x -

y)

:s;

p( -x - y)

Y E:D

It turns out, as will be shown presently, that the validity of (4) and (5) is enough to guarantee (3). From (1) we see that (4) and (5) are equiva~ lent to (6)

G(x)

:s;

p(x

+ z)

- F(z)

Z

E:D

and (7)

G(x)

~

-pC -x - y) - F(y)

Y E:D

respectively. Now, the right side of (6) dominates the right side of (7) for all z and y E:D, since p(x

+ z)

- F(z)

+ p( -x -

+ F(y) {p(x + z) y)

~

p(z - y) - F(z - y)

~

0

Hence, defining G(x) = inf F(z) I z E:D}, (6) and (7) and therefore (4) and (5) hold. To show that (4) together with (5) implies (3), three cases are considered. i.

G(ax

ii. (}(ax

+

For a

+ m)

>

0, (4) implies

= aG(x

For a

<

+ :):s; ap(x + :) =

p(ax

+ m)

mE:D

0, (5) implies

m) -, --'Iva ( -x -

~) ~

-up ( -x -

S) . .

p(ax

+ m) me:O

INTRODUCTION TO NORMED LINEAR SPACES

iii.

19

For a = 0 G(m) = F(m) ::; p(m)

Thus the theorem for real vector spaces is proved. Suppose X has complex scalars. The following proof is due to Bohnenblust and Sobczyk. The idea of the proof is to reduce the complex case to the real case and apply the above result. Write f(m) = Re f(m)

+ i 1m f(m)

meM

where Re f(m) and 1m f(m) denote the real and imaginary parts, respectively, of f(m). Since f(im) = if(m), we see that Ref(im)

+ i Imf(im) =

-Imf(m)

+ i Ref(m)

Thus 1m f(m) = - Re f(im) and (8)

f(m) = Re f(m) - iRe f(im)

Let X r be X considered as a vector space over the reals. As sets, X = X r • Now Re f is a linear functional on M considered as a subspace of X r and

IRe f(m) I ::; If(m) I ::; p(m)

meM

Hence, by what has already been proved, there exists a linear extension G of Re f to all of X r such that

IG(x)1 ::; p(x)

X

e Xr

In view of Eq. (8), define F on X by F(x) = G(x) - iG(ix)

x eX

A simple calculation shows that F is a linear extension of f to all of X. Given x e X, write F(x) in polar form F(x) = IF(x)le i8 • Then IF(x) I

= F(e- i8 x) = Re F(e- i8 x) = G(e-i8 x) ::; p(e-i8 x) = p(x)

ThuR t.he t.heorem is proved. 1.5.3 Uf~mark. In the proof of the- above theorem, the following rOHult. hlLloi booll showlI.

20

UNBOUNDED LINEAR OPERATORS

Suppose X is a vector space over the reals. Let M be a subspace of X and let p be a real-valued function on X with the following properties. p(x

+ y) :s; p(x) + p(y)

p(ax)

=

a~O

apex)

If f is a linear functional on M such that f(m)

:s;

p(m)

then there exists a linear functional F which is an extension of f to all of X such that F(x) :s; p(x) X E X 1.5.4 Corollary. Let m' be a continuous linear functional on a subspace M of normed linear space X. There exists an x' E X' such that x' = m' on M and IIx'll = IIm'lI. Proof. Define p on X by p(x) = Ilm'lIllxll. Since p satisfies C't) and (ii) of Theorem 1.5.2 and Im'ml :s; IIm'lIl1mll = p(m), mE M, there exists a linear functional x' on X such that x' = m' on M and

Ix'xl :s; Thus x' is in X' and

Ilx'lI

=

IIx'll :s; Ilm'lI.

sup

11"'11=1

Ix'xl

:XEX

Hence

Ilx'll

=

Ilm'llllxll

p(x) =

x

EX

On the other hand,

~ sup 11"'11=1

Ix'xl

=

xEM

sup

11"'11=1

Im'xl

=

Ilm'll

xEM

IIm'll·

1.5.5 Corollary. Let M be a subspace of normed linear space X. x E X with d = d(x, M) > 0, there exists an x' E X' such that

Ilx'll = M.

x'M = 0

1

and

Given

x'x = d(x, M)

Proof. Let M 1 be the subspace spanned by x and the elements of Define linear functional v' on M 1 by v'(ax

Then v'M = 0 and v'x For a ,..s 0 and m eM,

=

d.

+ m)

= ad

We assert t.hat. v' is in M~ with

Illlx + mil - lalll x + ~ II

~ lal d

Ilv'll =

1.

INTRODUCTION TO NORMED LINEAl SPACES

21

Thus for all a,

Iv'(ax

+ m)\ = lal d ::;

Hence v' E M~ and IIv'lI ::; 1. Ilx - mkll-4 d. Since

d

=

lIax

+ mil

There exists a sequence {mk} in M such that

v'(x - mit) ::; llv'lIl1x - mkll-4 Ilv'lI d

it follows that Ilv'll ~ 1. Thus Ilv'll = 1. The corollary follows upon taking x' E X' to be an extension of v' so that Ilx'lI = Ilv'll = 1. 1.5.6

Ilxli. yll =

Proof.

1.5.7

Given x E X, there exists an x' E X' such that IIx'll = 1 In particular, if x ~ y, there exists an x' E X' such that x'x - x'y.

Corollary.

and x'x = o ~ IIx -

Take M = (0) in Corollary 1.5.5.

Corollary.

For any x in normed linear space X,

Ilxll Proof. (1)

=

sup Ix'xl """11=1 X/EX'

For x' in the I-sphere of X'

Ix'xl ::; Ilx'll Ilxli ::; Ilxll

By Corollary 1.5.6, there exists a z' in the I-sphere of X' such that (2)

z'x =

Ilxll

The corollary follows from (1) and (2). As a simple application of Corollary 1.5.6, we prove the converse to Theorem 1.3.4. 1.5.8 Corollary. Let X and Y be normed linear spaces. complete, then Y is complete.

If [X, Yj is

Proof. Let {Yn} be a Cauchy sequence in Y. Choose Xo E X such that Ilxoli = 1. (Recall that X ~ (0) unless stated otherwise.) There exists an x' E X' such that x'xo = IIxoll = 1. Define Tn E [X, Yj by

T,.z • x'(x)y" Now

22

UNBOUNDED LINEAR OPERATORS

Hence liT.. - T mil ::; II x' I II y" - Ymll which implies that {T.. } is a Cauchy sequence in [X, Y]o Thus, by hypothesis, {T,,} converges in [X, Yj to some T. Since

Ily" -

Txoll

= IIT"xo - Txoll ::; liT" - Tlllixoll

{Y.. } converges to Txo and therefore Y is complete.

The next theorem is very useful and is needed in the proof of Theorem II.4.3. 1.5.9 Definition. A subset K of a vector space over the real or complex numbers is called convex if for every x and Y in K, the set {ax

+ (1

- a)y

10 ::; a

::; I}

is contained in K. 1.5.10 Theorem. Let K be a closed convex subset of normed linear space Xo Given x E X but not in K, there exists an f ~ 0 E X' such that Ref(x)

~

Ref(k)

kEK

The conclusion in the above theorem has the following geometric interpretation when X is the plane. Consider :n(f) as a line through the origin. Choosexosothatf(xo) = 1. Now, x lies on thelinef(x)xo + :n(f) and each k E K lies on the line f(k)xo + :n(f). Since f(x)

~

f(k)

k

K

E

it follows that K lies on one side of the line f(x)xo + :n(f) containing x. The proof of the theorem depends on the following lemma. 1.5.11 Definition. Let 0 be an interior point of a convex subset K of normed linear space X For each x E X let 0

A(x) = {a I a> 0, x where aK "'" {ak IkE K}.

E

aK}

Define the functional p on X by

p(x),= inf A (x) We shall call p the M inlcowBlci functional of K. 1(, A (x) r,I t/l and 0 'S p(:r:) < 00

'd

Since 0 is an interior point

INTRODUCTION TO NORMED LINEAR SPACES

1.5.12

Lemma.

23

Let K and p be as in the above definition.

Then for

x and y in X,

i. ii. iii.

p(ax) = ap(x) , a ~ 0 p(x + y) ~ p(x) + p(y) p(z) ~ 1 for all Z f K

Proof of (i). It is clear that p(O) a E A(ax), x is in a-1aK and therefore ~

p(x)

inf a- 1a

=

= O. Suppose

a

>

O.

Given

a- 1p(ax)

a.A(a,,)

Thus ap(x)

~

p(ax), a

~

O.

This result implies a>O

or ap(x)

~

p(ax).

Hence p(ax) = ap(x), a

~

O.

Proof of (ii). We first observe that if a and b are nonnegative real numbers, then (a + b)K = aK + bK. Indeed, if x and yare in K, then the convexity of K implies a b a+bx+a+b YEK

Thus ax bK C (a

+ by

E

+ b)K.

(a

a+b~O

+ b)K. Since x and yare arbitrary in K, aK + Obviously, (a + b)K C aK + bK. Hence (a

+ b)K =

aK

+ bK

To conclude the proof of (ii) , we suppose a E A (x) and bE A(y). Then x + y E aK + bK = (a + b)K. Hence p(x + y) ~ a + b. Since a and b are arbitrary in A (x) and A(y), respectively, it follows that p(x + y) ~ p(x) + p(y). Proof of (iii). Suppose Z f K but p(z) < 1. Then there exists an r, such that Z E rK. Since 0 E K and K is convex, it follows that rk = rk + (1 - r)O E K, k E K. Thus Z E rK C K, which is a contradiction.

o ~ r < 1,

Proof of theorem 1.5.10. .Since x , K and K is closed, there exist!;! an r-ball 8, r> 0, such that x + 8 and K are disjoint. Thus k o E K implies x - k o ' -ko K - 8 = K 1• Furthermore, 8 = -8 C K 1, showing that 0 is an interior point of K 1. Since K and 8 are convex, it follows thl1t [(1 is also convex. Hence tj)e Minkowski functional p of K 1 if! do{inn!l. W (I first provo tho thcore~ for tho cl1se when X is 11 real

+

UNBOUNDED LINEAR OPERATORS

24

normed linear space. On the one-dimensional space sp {xo I, where Xo = x - k o, define linear functional f by

Now, f(axo) ::; p(axo); for if a ~ 0, then f(axo) = p(axo) by Lemma 1.5.12. If a < 0, then f(axo) = ap(xo) ::; 0 ::; p(axo). Hence, by Remark 1.5.3 and Lemma 1.5.12, there exists a linear functional F which is an extension of f to all of X such that F(y) ::; p(y), Y E X. Moreover, F is bounded; for if Ilyll = 1, then ±ry ESC K 1• Hence ±F(y) = F(±y) ::; p(±y) ::; r

1

Since Xo = x - k o ¢ K 1 and -k o + k E -k o + K - S = K 1 for all k E K, it follows from (iii) of Lemma 1.5.12 and the definition of p that for each kEK, F(x) - F(k o) = f(xo) = p(xo)

~

1 ;:::: p( -k o

+ k)

;:::: -F(k o)

+ F(k)

Thus F;eO

and

F(x) ;:::: F(k),

kEK

proving the theorem for real normed linear spaces. If X is complex, then as in the proof of the Hahn-Banach extension theorem, let X T be X considered as a real normed linear space. Then by what has just been shown, there exists an fT ;e 0 in such that fT(X) ;:::: fT(k), k E K. Define f on X by

X;

f(x) = fr(x) - ifT(ix) It is easy to verify that f satisfies the demands of the theorem.

For a treatment of separation theorems for convex sets the reader is referred to Dunford and Schwartz [1], pages 409-418.

1.6

CONJUGATE SPACES

1.6.1 Definition. Normed linear spaces X and Yare called equivalent if there exists a linear isometry from X onto Y.

A particular caee of the next theorem ie eSl!lential for the etudy of <.liffcrontinl operators in Chnpe. VI nnd VII. The proof may be found in Dunforclnnd Sohwnrt,z Ill, pnlt0R 28ll-200.

25

INTRODUCTION TO NORMED LINEAR SPACES

We say that p' is conjugate to real number p if lip understanding that p' = 00 when p = 1.

+ lip'

= 1, with the

1.6.2 Theorem. Let (8, 2:, p,) be a positive measure space. If 1 < P < 00, then £~(8, 2:, p,) is equivalent to £p,(8, 2:, p,), p' conjugate to p, in which x' E £~(8, 2:, p,) is related to the corresponding g E £p,(8, 2:, p,) by x'f =

Is gf dp,

If, in addition, 8 is the union of a countable number of sets of finite p,-measure, that is, 8 is u-finite, then the theorem holds for p = 1. For convenience we write x' = g.

As noted in Example 1.1.7, lp = £p(8, 2:, p,), where (8,2:, p,) is a u-finite positive measure space. Thus we have the following result. If 1 ~ p < 00, then l~ is equivalent to lp" p' conjugate to p; y' E l~ is related to the corresponding y = (Yl, Y2, ' ..) E lp' by y'(a) =

L"" a,Yi

i-1

For convenience we write y'

=

(ai, a2, . . .).

1.6.3 Definition. A set K in normed linear space X is called orthogonal to a set F C X' if x'k = 0 for all k E K and x' E F.

The reason for the terminology orthogonal will be made clear in Sec. 1.7. The orthogonal complement in X' of K, denoted by K 1., is the set of elements in X' which is orthogonal to K. Even if K is not a subspace, K1. is a closed subspace of X'. 1.6.4

Theorem.

i. ii.

Let M be a subspace of normed linear space X.

Then

X'I M 1. is equivalent to M' under the map U defined by U[x ' ] = x~ where [x'] is in X'I M 1. and x~ is the restriction of x' to M. If M is closed (so that X 1M is a normed linear space), then (XI M) I is equivalent to M 1. under the map V defined by (Vz')x

= z'[x]

Proof of (i). Note that U is unambiguously defined, since [y'] = [x'] implios 0 = y'm - x'm, mE M. Cloarly, U is linear with range in M'. Givon m' , M', I.hol'O OXiH[,H, hy 0orollltry T,JiA, an x' e X' which iR nn

26

UNBOUNDED LINEAR OPERATORS

extension of m'. Hence U[x'] = x~ = m' which shows that
IIU[x'lll ::; inf

y'.[x']

Ily'll

=

= M'.

II[x'l II

On the other hand, there exists a v' E X' which is an extension of x~ such that Ilv'll = Ilx~lI. Therefore v' is in [x'] and

II U[x'] II

(2)

By (1) and (2),

1\

Proof of (ii).

U[x'lll

=

Ilx:V1I

=

=

Ilv'll

~

lI[x'lll

lI[x'JlI.

For z' E (X/ M)',

I(Vz')xl =

Iz'[xll ::;

liz' II

II [xlII ::; IIz'lllIxll

xEX

and

(V z')m = z'[m] = z'[O] = 0

mEM

Thus V z' is in M 1. with (3)

IIVz'\\ ::;

IIz'll

Since

\z'[xll

=

\(Vz')y\ ::;

IlVz'lIllyll

y E [x]

it follows that

Iz'[xl \ ::; II Vz' II II [x] II Thus

Ilz'll ::; IIVz'll which, together with (3), proves that V is an isometry. Given x' E MJ., let z' be the linear functional on X/M defined by z'[xl = x'x. Since

Iz'[xll

=

\x'y\ ::;

Ilx'llllyll

y E [xl

it follows that Iz'[xl\ ::; Ilx'lIl1[xlll. Hence z' is in (X/M)'. Vz' = x', proving that
Furthermore,

1.6.5 Definition. A set A in a normed linear space is called separable if there exists a countable subset of A which is dense in A. 1.6.6 Example. l", 1 ::; p < co, is Rl1Jlnrahlo. A count.ahlo Bubsot whic-,h ifl I1nlll'l" ill l" iF! tho !'l(,t, of 1~11 "'''11111'1\('111'\ of tl1l\ form (al, 0'1, . . • ,

INTRODUCTION TO NORMED LINEAR SPACES

27

ak, 0, 0, . . .), where the ak are rational (a complex number is rational if

its real and imaginary parts are rational). ,,,, is not separable. For suppose {Xk I is a sequence of elements in loo' We show that there exists an x E ,,,, such that IIx - xkll ;::: 1. Let Xl X2

Define

X

=

= =

(Xl!, X21, • • •) (XI 2, X2 2, • • • )

(aI, a2, . . .) by if IXkkl ;::: 1 if IXkkl < 1

Then x is in ,,,, and IIx - xjll ;::: laj - x/I;::: 1. It is shown in Zaanen [1], page 75, that £p(S, :3, p,), 1 ~ p < 00, is separable, where p, is a separable measure. In particular, £p(Q), 1 ~ p < 00 , is separable, where Q is a Lebesgue-measurable subset of En. Taylor [1], page 91, shows that £",(1) is not separable, where I is any interval on the line. 1.6.7 Theorem. If the conjugate of normed linear space X is separable, then X is also separable.

Proof. Let {x~l be dense in X'. Choose sequence {xd in X such that Ilxkll = 1 and IX~Xkl ;::: Ilx /II/2. We claim that the space Xl spanned by the Xk is dense in X. Suppose this is not the case. Then, by Corollary 1.5.5, there exists an x' :;z£ 0 in X' such that x' X 1 = O. Since {x~ I is dense in X', there exists a subsequence {x~.l which converges to x'. Since

Il xk1' . -

x'II> I(xkt' . - x/)x k~.1 -

=

Ix'.x .1 kt k t

> Ilx~,11 2 -

{x~, I converges to O. But this is impossible since x~, ~ x' :;z£ o. Thus Xl is dense in X. Since the set A, consisting of elements of the form n

L aiXi,

ai rational, is countable and dense in Xl, it follows that A IS

i-l

dense in X.

Hence X is separable.

Since loo = l~ is not separable but II is, the converse to the above theorem does not hold. 1.6.8 Df!finition. The natural map., denoted by J x, of normed linear Ilpnce X info nil Rrrond ron.i1loatr Rpnrr X" (the Banach flpace of bounded

UNBOUNDED LINEAR OPERATORS

28

linear functionals on X') is defined by (J xx)x'

=

x' EX'

x'x

If the range of J x is all of X", then X is called reflexive.

Remarks

1.6.9

i.

The natural map J from X into X" is a linear isometry. The linearity of J is clear, while from Corollary 1.5.7 we have

IIJxl1 = 'l,'l,.

iii.

I(Jx)x'\ =

sup

Ilx'l\ =

1

sup

I\x'~= 1

Ix'xl = Ilxli

Every reflexive space is complete, since a conjugate space is complete and isomorphisms preserve completeness. £p(S,~, p.), 1 < P < 00, is reflexive, as may easily be seen from the definition of the natural map and the representation of £~(S, ~, p.) given in Theorem 1.6.2.

A word of caution. Sometimes the argument given to show, for example, that £p = £p(S, ~, p.), 1 < P < 00, is reflexive is the following. £~ is equivalent to £~" which in turn is equivalent to £pu = £p. Hence £p is reflexive. The flaw in the argument is that the equivalence of a normed linear space with its second conjugate does not guarantee the reflexivity of the space. James [1] gave an ingenious construction of a Banach space X which is equivalent to X", yet the dimension of X"/ J xX is 1. 1.6.10 Theorem. separable.

The conjugate space of a separable reflexive space is

Proof. Suppose X is reflexive and separable. Then X" = J xX is separable. Hence X' is separable by :Theorem 1.6.7.

Since l~ = l", and II is separable but l", is not, i.t follows from the above theorem that II is not reflexive. 1.6.11 Theorem. is reflexive.

A Banach space is reflexive if and only ~f its conjugate

Proof. Let J be the natural map from X into X". Suppose X is reflexive. We must show that given x'" EX"', there exists an x' EX' such that

x"'Jx = (Jx)x'

(1) J)Clfinl1

x' -

:r"'''.

Thl1l1

x'x

xeX

x' e X' "'1Ir1 (1) iM

Ml\.ti",(jCld.

lEI

29

INTRODUCTION TO NORMED LINEAR SPACES

Assume X' is reflexive but X is not reflexive. There exists an x" E X" such that x" ~ JX. Since X is complete and J is a linear isometry, we know from Lemma 1.4.8 that J X is also complete and therefore is closed in X". Thus there exists an x'" E X'" such that x"'x" =;t. 0 and x'" J X = O. Since X' is reflexive, there exists an x' E X' such that

o =;t. x"'x"

(2)

=

x"x'

and

o=

(3)

x'''Jx = (Jx)x' = x'x

XE

X

Equation (2) shows that x' =;t. 0, while Eq. (3) shows that x' = 0, which is absurd. Hence X is reflexive.

1.6.12

Theorem.

A closed subspace of a reflexive space is reflexive.

Proof. Let M be a closed subspace of reflexive space X. m" EM", define x" E X" by

Given

x"x' = m"x~ where x~ is the restriction of x' E X' to M. Let m = J X-IX". It will be shown that m is in M and J Mm = m", proving that M is reflexive. Suppose m ~ M. Then there exists an x' E X' such that x'm =;t. 0 while x'M = O. Thus x~ = 0 and

o =;t. x'm

= x'J X-IX" = x"x' =

m"x~

= m"O = 0

which is impossible. Hence m is in M. For each m' EM', let m: be an element in X', which is an extension of m'. Then

Thus J Mm = m", completing the proof of the theorem. The next theorem is useful in proving convergence in norm of certain sequences in oCp(Q), 1 < P < 00.

1.6.13 D~finition. A sequence {x n } in normed linear space X is said to converge weakly to x EX if for every x' EX', X'X n -+ x'x. This is written 1.6.14

w X n -+ X.

UflnUlr/c.

Banach [1],

(IOIlVOJ'jl;llIWIl ill 1101'111 iH ,,}all HUIlIO

pa~owl~7-139,

nH wonk

has 8hown that

eOllvorl/:OIWO.

III

tl,

30

UNBOUNDED LINEAR OPERATORS

1.6.15 Theorem. Every bounded sequence in a reflexive space contains a weakly convergent subsequence. Proof. Let {Xk} be a bounded sequence in reflexive space X and let Xl = sp {Xk}. Then X I is separable, since each x E X I is the limit of n

elements of the form

L aiXi,

a.

Thus X~ is separable by

rational.

.=1

Theorems 1.6.12 and 1.6.10. Let {x~} be dense in X~. Since {X~Xk} is a bounded sequence, there exists a subsequence {X~Xlk} which converges. Since IX~X1k I is bounded, there exists a subsequence {X~X2k} which converges. Thus {X~X2k} and IX~X2d converge. By induction, one obtains sequences {Xjk} such that (1) (2)

1< _3.

IXu+I)kl is a subsequence of IXjk} {X;Xjk} converges

Setting Vk = Xkk, it is easy to see from (1) and (2) that {x;vd converges for eachj. Since Ix;} is dense in X~, it follows that {x'vd converges for each x' E X~. Define x" on X~ by X"X' = lim X'VIc /c-> ..

Clearly, x" is linear. Since Ix'vkl ~ Ilx'lllIvkll and {vd is a subsequence of bounded sequence Ixd, it follows that x" is bounded; that is, x" E X~'. Therefore the reflexivity of X I implies the existence of an x E Xl such that (3)

X'X = XliX' = lim X'Vk /c-> ..

x' E X~

Given z' E X', let z~ be the restrictio~ of z' to Xl.

By (3),

z'X = z~x = lim Z'Vk /c->

Hence {Vic} converges weakly to x, which proves the theorem. Convergence in norm is not the same as weak convergence in an infinite-dimensional reflexive space X. Otherwise, by the above theorenl, every sequence in the I-sphere of X contains a convergent (in norm) subsequence. Thus the I-sphere is compact, and therefore X is finitedimensional. The propcrl,ills of the nlttuml mo.p from X into X" lon.d to (Jl'Oof of tho followillK tllllorolll.

11.

flimple

INTRODUOION TO NORMED LINEAR SPACES

1.6.16 Theorem. Banach space.

Every normed linear space is a dense subspace of a

Proof. Given normed linear space X, let X consist of the elements of X and the elements of the Banach space Z = J xX C X" which are not in J xX. Define addition and scalar multiplication in X by the corresponding operations in Z. The norm of an element in X is the norm of the corresponding element in Z.

We shall call

X the completion of X.

HILBERT SPACES

1.7

In this section some of the basic theorems concerning Hilbert spaces are proved. The study of these spaces has been the subject of several books. The reader is referred to Akhieser and Glazman [1], Berberian [1], Halmos [1], and Stone [1]. 1.7.1 Definition. Let X be a vector space over the real or complex numbers. An inner product on X is a scalar-valued function ( , ) defined on the Cartesian product X X X with the following properties. t.

n. m. iv.

(ax, y) = a(x, y) (x, y) = (y, x)j that is, (x, y) is the complex conjugate of (y, x) (x + y, z) = (x, z) + (y, z) (x, x) > 0 whenever x ¢ 0

X, together with an inner product, is called an inner-product space. It follows from the above axioms that (x, ay)

(x, y

+ z)

=

a(x, y)

= (x, y)

(0, x) = 0

1.7.2

Example.

(x, y) =

+ (x, z) X E

X

For unitary n-space an inner product is given by

L" x,y.

x = (Xl, . . . ,x,,),

y

=

(yt, . . . ,y,,)

i-I

1.7••'1

Example.

An inner product on

.c~(O)

is given by

32

UNBOUNDED LINEAR OPERATORS

In Examples 1.7.2 and 1.7.3, (x, x)i gives the norm of x. We shall show that for any inner-product space, Ilxll = (x, x)t is indeed a norm.

Let X be an inner-product space.

1.7.4 Schwarz's inequality. for all x and y E X

\(x, y)1 ~ Ilxllllyll Equality holds if and only

~f

Then

Ilxll = (x, x)i

x and yare linearly dependent.

Two proofs are given. The first one is the usual short proof, while the second, though longer, is geometrically motivated. Proof. (1)

For any scalar X,

= IIxl1 2 - X(x, y) - X(y, x)

0 ~ (x - Xy, x - Xy)

+ \X12 IIyl12

For (y, x) = 0, the inequality is trivial and equality holds if and only if at least x or y is zero. Suppose (y, x) ~ O. Substituting X = IlxI1 2/(y, x) in (1) yields (2)

o<

-llxl12

-

+

4

IIxl1 lIyll2 \(x, y)12

whence (3)

Equality holds in (3) if and only if equality holds in (1), in which case x = (1IxI1 2/(y, x»y. In the plane, Schwarz's inequality follows from the fact that (x, y) = IIxllllyll cos 0, where 0 is th~ angle between x and y, 0 ~ 0 ~ 7r. The first step of the proof is to give meaning to cos 0 for an arbitrary inner-product space. To avoid trivialities, it is assumed in the sequel that x ~ 0 and y ~ O.Choose X so that x - Xy is perpendicular to y; that is, 0 = (x - Xy, y), or equivalently,

Thus we may think of a right triangle with hypotenuse x and sides Xyand x - Xy. Define cos 0 (1)

=

. lUll

1\r~11 Ilx - AyJ]

0 -

.

IlxII'

INTRODUCTION TO NORMED LINEAR SPACES

33

(Envision 8 as being the acute angle between x and y.) putation shows that

I(x, y)1 = IIxll IJyl1

(2)

From

IIxl1 2=

(x -

AY + AY, x

A simple com-

cos 8

Ay + AY)

-

and the fact that A was chosen so that (x - AY, Y) = 0, it follows that the "Pythagorean theorem" holds; Le., (3)

Thus (1) and (3) imply (4)

cos 2 8

Hence 0

+ sin

2

=

8

1

=:; cos 8 =:; 1 which, together with (2), implies

lex, y)1 =:; IIxlillYIl Suppose

I(x, y)I

=

IIxlillyli.

o= or x =

Ay.

1.7.5 Theorem. norm on X. Proof.

Then (2) and (4) show that

Ayll I[xll

sin 8 = [[x -

Given inner-product space X,

IIxII

= (x,

x)l

defines a

For x and y E X, Schwarz's inequality implies

IIx + Yll2 = (x + y,'x + y) = IIxl1 2+ (x, y) + (x, y) + IIYl12 =:; l[xl[2 + 21(x, y)1 + IIyI12 =:; IIxl1 2+ 211xlillYII + l[y112 = (11xll + Ilyll)2

+

+

Hence Ilx yll =:; Ilxll Ilyll· The other properties required of II II in order that it be a norm follow immediately from the definition of an inner product. An inner-product space will always be considered as a normed linear space with the norm Ilxll = (x, x)l. 1.7.6 Theorem. Suppose x" -+ x and 1/" -+ y in inner-product space X. Tlwn (x", 1/..) -+ (x, 1/). In other wortls, the inner product i.~ continuo'u8 O'TI. tht, mdr'lc Hl)(Lt~(' X X X.

UNBOUNDED LINEAR OPERATORS

Proof.

We write

+

(x n, Yn) - (x, y) = (X n, Yn) - (x, Yn) (x, Yn) - (x, y) = (X n - x, Yn) + (x, Yn - y)

By Schwarz's inequality I(xn, Yn) - (x, y)1 ~

Ilxn- xllllYnll + IlxllllYn

- yll

Since IIYnl1 ~ IIYII, Ilxn - xii ~ 0 and llYn - yll ~ 0, it follows that (x n , Yn) - (x, y) ~ o. 1.7.7 Definition. A Hilbert space is an inner-product space which is also a Banach space with norm Ilxll = (x, x)l. Thus £2(S, ~,JL) is a Hilbert space, but any nonclosed subspace of £2(S, ~, JL) is an inner-product space which is not a Hilbert space.

In the course of solving the following problem, the theorems which we desire to include in this section make their appearance. 1.7.8 Problem. Suppose M is a closed subspace of Hilbert space Given x E X, does there exist an element Px E M such that

Ilx -

x.

PxlJ = d(x, M)

In other words, does there exist a best approximation to x among all the elements in M-best in the sense that Ilx - Pxll :::; Ilx - zll for all z E M? M must be closed, for if x E if but x II M, then for all mE M, d(x, M) = 0

< Ilx - mil

If M is a line through the origin and X is the plane, then Px is the foot of the perpendicular from x to M or, what is the same thing, x - Px is perpendicular to M. The next lemma is motivated by this observation.

1.7.9 Definition. Elements x and Y in an inner-product space are called perpendicular or orthogonal, written x ..L Y, when (x, y) = O. Element x is perpendicular to a set M, written x ..L M, if (x, m) = 0 for all m E M 1.7.10 Lemma. Let X be an inner-product space (not necessarily complete) and let M be a subspace of x. Suppose that given x E X, there exists a Px EM such that x - Px ..L M. Then Px is the unique element in M for which d(x, M) = Ilx - Pxll. More specifically, the "Pythagorean theorem" holds; that is,

IIx - mill - 11;( - Pxll' + II/'x - m,1I 1

m.M

INTRODUCTION TO NORMED LINEAR SPACES

Proof.

35

It is clear that the Pythagorean theorem implies that

Ilx - mil> Ilx -

Pxll

m

= d(x, M)

~

Px,

mEM

Given m EM,

Ilx - ml1 2 =

(1)

(x - m, x - m)

= «x - Px)

Ilx -

Pxl/ 2

+ (Px - m), (x - Px) + (Px - m» + (x - Px, Px - m) + (Px - m, x - Px) + I/Px -

mll~

Since x - Px .1 M and Px - mE M, it follows from (1) that

Ilx Px.

mll 2 =

Ilx -

Pxl1 2

+ I/Px

- ml/ 2

Problem 1. 7.8 is now reduced to finding the "foot of the perpendicular" We start out by finding Px for certain M.

1.7.11

Definition.

A set in an inner-product space is called orthogonal

i] distinct elements in the set are orthogonal. An orthogonal set is called orthonormal if each element in the set has norm 1.

In l2 the set of unit vectors Uk

=

(0, 0, . . . , 1, 0, . . .) is orthok

normal. For the £2([0, 211"]) space of complex-valued functions, {(l/V211")eint l, n = 0, ±1, ±2, . . . ,is an orthonormal set. In order to motivate how to determine Px in the next lemma, take (1,0,0) and (0, 1,0). For x = (Xl, X2, xs) the foot of the perpendicular is

X

= ES with M the plane spanned by the vectors

Also, the length of Px is dominated by the length of x (Bessel's inequality). 1.7.12 Lemma. Let {UI, U2, . . . } be an orthonormal set in Hilbert space X and let M = sp ({Uk}). Then for each x E X there exists a Px in M such that x - Px ..L M. The formula for Px is given by Px =

L'" (x, Ui)U; ;=1

The serieB converges to Px independent of rearrangement.

..

L I(x, 14.)1 I-I

2

-

I/Pxl/ 2 ~

Ilxll»

Furthermore,

(BeBsel' B inequality)

UNBOUNDED LINEAR OPERATQRS

Proof. The convergence of the series will be established first. Since {Ul, 11,2, . • .} is orthonormal, it follows that

o ~ (x

n

n

i=l

i=l

L (x, Ui)Ui, X - L (X,Ui)Ui) =

-

IIxll 2 -

n

L l(x,ui)12 i=l

Hence .,

L I(x, ui)12 ~ IIxll

(1)

2

i=1 n

Thus for Sn =

L (X,Ui)Ui i=1

n+k

snll

IISn+k -

= (

n+k

L (X,Ui)Ui, i=n+l L (X,Ui)Ui) i=n+l n+k

L

I(x, u;)I2.-~ 0

as n

~

00

i=n+l

Therefore {Sn} converges to an element Px in :fe. Note that the argument also shows that the series converges independent of any rearrangement. Since Sn is in M and M is closed, Px is also in M. Now, by Theorem 1.7.6, n

(x - Px, Uj) = lim (x -

L (x, Ui)Ui, Uj) = (x, Uj) -

1~j

(x, Uj) = 0

i=l

n~oo

Hence x - Px ..L sp (lUi}) and therefore, by the continuity of the inner product, x - Px ..L M. By (1), n

IIPxl1 2 = lim n-+oo

(L

n

.,

i=l

i=l

L (X,Ui)Ui) = L I(x, ui)12 ~

(X,Ui)Ui,

i=l

IIxl1 2

.,

Suppose that

PIX

=

L (x, Vi)Vi, where

Ivd

is a rearrangement of

Iud·

i=1

Then by the above argument,

Ilx -

x -

Pxll = d(x, M) =

PIX

Ilx -

..L M.

P 1xll

Hence, by Lemma 1.7.10, and

Problem 1.7.8 is therefore solved for M = sp IUk}. The next step is to solve the prohloffi for M = Ap 11/,,.}, whol'O l?t" I is 110(, llo(Jossll.rily (~O\lll t,lLblll.

INTRODUCTION TO NORMED LINEAR SPACES

37

1.7.13 Lemma. Let M = sp {u,,}, where {u,,} is an orthonormal set in Hilbert space X. Thenfor each x e X, (x, u,,) ~ ofor at most a countable number of U", and Px = (x, u")u,,

L

(x,ua>",o

T he series converges to Px e M independent of rearrangement. x - Px 1.. M.

Moreover,

Proof. Once it is shown that (x, u a ) ~ 0 for at most a countable number of u a , the rest of the lemma follows from Lemma 1.7.12. For each sequence of elements {u",} in {u a }, we have, by Bessel's inequality, that '" I(x, U a )12 :::; Ilx11 2 • Consequently, there exist, for each positive integer

L

i=1

n, at most n of the (x, u a) such that

En =

I(x, ua)1 2:: IlxI1 2/n.

{ua I I(x, ua )12

2::

Define

11~12}

Then

{u

a

II(x, 1I a )1 > o} = n~1 En

is countable, since each En is finite. The class of those M for which there exists a solution to Problem 1.7.8 has now been enlarged to those M = sp {u,,}, where {u a } is an arbitrary orthonormal set. Thus, to solve the problem for general closed subspaces M, it remains to show that M is the closure of the span of some orthonormal set. 1.7.14 Lemma. Given any closed supspace M ~ 0 of Hilbert space X, there exists an orthonormal set {u a } such that M = sp {u,,}, Proof. The proof utilizes Zorn's lemma in order to prove the existence of a maximal orthonorm~l set 0 contained in M, maximal in the sense that 0 is not a proper subset of any orthonormal set contained in M. Let (p be the class of all orthonormal subsets of M. (p is not empty since it contains the I-point set {m}, where m eM and Ilmil = 1. Partially order (p by set inclusion. Let 3 be a totally ordered subclass of (p. Since V S is an upper bound in (p for 3, Zorn's lemma asserts that

tains a maximal element, say the orthonormal set {u a }. We claim that M = sp {u a }. Assume the contrary. Then there exists an x ~ 0 in M but not in M 1 = sp {u,,}, By Lemma 1.7.13, there exists a Px e M 1 such that x - Px 1.. MI. Thus the set consisting of (x - Px)/llx - Pxll and the 11" is all OIthollorma} set r,ontaininp; {u a } properly. But this is impolil,Hiblo, HiIlC'C' [11,,1 iH TlIl\.xinml. Hrmc'o /If = Flp rn"l.

38

UNBOUNDED LINEAR OPERATORS

From the preceding lemmas we have the following theorem. 1.7.15 Theorem. Let M be a closed subspace of Hilbert space X.' Given x EX, there exists a unique element Px E M such that x - Px ..L M. Furthermore, Ilx - Pxll = d(x, M) < Ilx - mil, mE M, m ;;e Px. 1.7.16 Definition. Subspaces M 1 and M 2 of vector space V are called linearly independent if M 1 ( \ M 2 = (0) or, equivalently, each x E M 1 + M 2 has a unique representation x = Xl + X2, Xl E M 1, X2 E M 2. The symbol M 1 EI1 M 2, called the direct sum of M 1 and M 2, is used only if M 1 and M 2 are linearly independent and denotes the set M 1 + M 2. 1.7.17

If M is a closed subspace of Hilbert space X, then

Corollary.

X

=M

EI1 M1.

where M 1. is the closed subspace of elements perpendicular to M.

Proof. If y EM (\ M 1., then 0 = (y, y) and therefore y = O. Given x E X, there exists, by Theorem 1.7.15, a Px E M such that x Px ..L M. Since x = Px

+ (x -

Px) EM

+ Ml.

the corollary is proved. For a fixed x in Hilbert space X the functional x' defined by The question arises whether all elements in X' are determined in this manner. The answer is in the affirmative, as is now shown. x'(z) = (z, x) is in X'.

1.7.18 Riesz representation theorem. Let X be a Hilbert space. Given x' E X', there exists a unique x E X such that x'(z) = (z, x) Moreover,

X

Ilxll = Ilx'll·

Proof. If x' = 0, choose x = O. exist as in the theorem, then (1)

Z E

o=

x'(z)

= (x, z)

Assume x' ;;e O.

If x were to

Z E m(x')

The first step then is to find an element v ~ 0 orthop;onnl to m(x'). Since m(x') is closed and is 1I0t all of :le, Rueh U I' CJxiHt.s by 'I'hoorolll 1.7,15, For

INTRODUCTION TO NORMED LINEAR SPACES

39

any scalar a, x = av is also orthogonal to m(x'). We need another condition to determine which a to choose. This condition is given by (2)

x'(x) = (x, x)

or ax'(v)

= (av, av) = aallvl1 2

Hence, choose a = x'(v)/llvI1 2 • We assert that x = av = (x'(v)/lIvI1 2 )v is the desired x. Since the map [u] ---t x'u is a 1-1 linear map from x/m(x') onto the scalars, x/m(x') is one-dimensional and therefore [x] ,t. [0] is a basis. Hence X = sp {x} EEl m(x'). Given z E X, z has the representation z = aX + w, for some w E m(x'). By the choice of x, (1) and (2) are satisfied. Thus x'(z) = ax'(x) = a(x, x) = (ax

+ W, x)

= (z, x)

Suppose x'(z) = (z, y) for all z E X. Then 0 = (z, x - y), Z E X, and, in particular,O = (x - y, x - y). Hence x = y. It remains to prove that Ilxll = Ilx'll· Since

Ix'(z)1 =

I(z, x)1 ~

Ilzllllxll

zEX

and

Ilx'll Ilxll it follows that

Ilxll

~ x'x

= (x, x) = IIxl1 2

Ilx'll·

1.7.19 Definition. For X a Hilbert space, let Ex be the map from X onto X' defined by Exx = x', where x'z = (z, x), Z E X.

It is clear that Ex is additive and that Ex(ax) = aExx. By the Riesz representation theorem, we know that Ex is an isometry with CR(E x ) = X'. Thus if X is a real Hilbert space, then JC is equivalent to X'. 1.7.20 Definition. A Banach space X with norm II 111 is called a Hilbert space if there exists an inner product on X such that (x, x)! = Ilxlll' 1.7.21 Theorem. X is reflexive. Proof.

1.7.19.

Let E

If X is a Hilbert space, then so is X'.

= E 3C

Furthermore,

be the map from X onto X', as in Definition

Define an inner product ( , )1 o_n X' by (:r', 11') I

-

(Ji]- 111', F:- IX')

UNBOUNDED LINEAR OPERATORS

40

where ( , > is the inner product on X. From the properties of E, it is easy to verify that ( , >1 is indeed an inner product. Since E-l is an isometry,

Thus x' is a Hilbert space. Let E 1 = Ex' be the map from x' onto X", as in Definition 1.7.19. Suppose x" e X". For each x' EX',

Thus X is reflexive. We conclude this section with the proof that all separable Hilbert spaces are equivalent to l2. 1.7.22 Lem,m,a. Let 0 be an orthonormal set in Hilbert Space X. following statements are equivalent. 'to

n. iii.

The

0 is a maximal orthonormal set in X sp (0) = X x = Px = ~ (x, u)u u.0

iv.

!lxll = IIPxl1

(Parseval's equality)

Proof. (i) implies (ii). If sp (0) ~ X, then it follows from Theorem 1.7.15 that there exists ayE X such that Ilyll = 1, y ..L sp (0). Thus 0 is a proper subset of the orthonormal set y V 0, which contradicts (i). (ii) implies (iii). By Lemma 1.7.13, x - Pz ..L sp (0). Since sp (0) = ac, x - Px ..L x - Px, whence x = Px. (iii) implies (iv) trivially. (iv) implies (i). Let 0 be a subset of orthonormal set 01. Assume there exists an x E 0 1 but not in 0.' Then (x, u> = 0 for each u E O. Hence, by (iv), 1 = IIxl1 2 = IIPxl1 2 = I(x, u)12 = O. Therefore 0 1 = O.

L

m0

Since 01 was an arbitrary orthonormal set containing 0, 0 is a maximal orthonormal set. In the literature, a maximal orthonormal set in X is called complete. In the £2([0, 211"]) space of real-valued functions, the orthonormal set {1/V211", cos nt/v;, sin nt/V;, n = 1,2, . . . } is a maximal orthonormal set. This statement is proved in McShane and Botts [1], pages

230-233. 1.7.23 Lem,ma. set 1:11 rnuntaNr..

In a separable Hilbert space, every orthonormal

INTRODUCTION TO NORMED LINEAR SPACES

Proof.

41

Let e be an orthonormal set.

Ilu -

v!l 2

= (u - v, U

v)

-

For U and VEe

= Ilu!l 2+ IIvl1 2 =

2

Thus e is countable, since a set of isolated points in a separable metric space is countable. 1.7.24

Theorem.

A separable Hilbert space is equivalent to l2.

Proof. Given a separable I-Iilbert space X, there exists, by Lemmas 1.7.14 and 1.7.23, a maximal orthonormal set Uh U2, • . . in X. By

L (x, Ui)Ui. 00

Lemma 1.7.22, each x E X has the representation x

=

Since

i=l 00

IIx!l2 =

(1)

L I(x, ui)I2 i= 1

the sequence {(x, Ui)} is in l2. Let T be the linear map from X into l2 defined by Tx = {(x, Ui)} By (1), T is an isometry.

It remains to prove that m(T) = l2. n

{04}

E

l2 be given.

Defining s"

=

L aiUi, it follows that (s,,} is a Cauchy i=l

sequence and therefore converges to some

VEX.

Hence

00

Tv

= {(

L a"Uk, Ui)}

k=l

Let

=

{ad

chapter II

LINEAR OPERATORS AND THEIR

CON~UGATES

There are essentially two methods of dealing with differential operators in the usual function space setting. The first is to define a new topology on the space so that the differential operators are continuous and then to develop and apply a general theory of continuous operators on a nonnormable topological linear space. This is known as L. Schwartz's theory of distributions (cf. Schwartz [1], Hormander [1], and Friedman [1]). The second method is to retain the Banach space structure while developing and applying a general theory of unbounded linear operators (cf. Browder [3.1 and Visik [1]). We use the second method for the following two reasons: 1,.

1,1,.

The linear differential operators usually encountered are closed or at least have closed linear extensions. Many of the important theorems which hold for continuous linear operators on a Banach space also hold for closed linear operators.

In this chapter, as well as in Chaps. IV and V, the second assertion is substantiated.

Throughout thi8 chapler, X and Yare normed linear 8paces over the 8ame 8calar8, and T 1:11 a lirl(~al' 0lJeralOT luwina domain a 8ubspace of X

LINEAR OPERATORS AND THEIR CONJUGATES

43

and range a subspace of Y. X and Yare assumed complete only when specifically stated. For any set M C X, TM denotes the set I Tm I mE M (\ 5J(T)}. CLOSED LINEAR OPERATORS

11.1

II.I.I Definition. X X Y is defined as the normed linear space of all ordered pairs (x, y), x E X, Y E Y, with the usual definitions of addition and scalar multiplication and with norm given by II (x, y)" = max Illxll, IlyII}. II.I.2 Definition. The graph G(T) of T is the set {(x, Tx) I x E D(T) }. Since T is linear, G(T) is a subspace of X X Y. If the graph of T is closed in X X Y, then T is said to be closed in X. When there is no ambiguity concerning the space X, we say that T is closed.

II.I.3

Remarks 1,.

ii. m. w. v.

T is closed if and only if Ix n } in D(T), Xn ~ x, TX n ~ y, imply x E D(T) and Tx = y. If Tis 1-1 and closed, then T-l is closed. The null manifold of a closed operator is closed. If :D(T) is closed and T is continuous, then T is closed. The continuity of T does not necessarily imply that T is closed. Conversely, T closed does not necessarily imply that T is continuous. This statement can be verified by the following examples.

Let 5J(T) be any proper dense subspace of X = Y and let T be the identity map. T is obviously continuous but not closed. II.I.4 Example. Let X = Y = C([O, 1]) and let C'([O, 1)] be the subspace of X consisting of the functions with continuous first derivatives. Define the linear differential operator T mapping C'([O, 1]) into Y by (Tx)(t) = x'(t), t E [0, 1]. T is clo13ed; for if Xn ~ x and TX n ~ y, then I x n } converges uniformly to x and I x~} converges uniformly to y on [0, 1]. It follows from taking antiderivatives of x~ and y that x is in C'([O, 1]) and that Tx = x' = yon [0, 1]. Thus T is closed. However, T is unbounded, since the sequence Ixn(t)} = It n } has the properties IITxnl1 = nand Ilxnll = 1.

It is shown in Chaps. VI and VII that large classes of differential operators are closed.

The closed-graph theorem, which is another fundamental theorem in functional I1l\alysis, shows when a closed linear operator is continuous. WhilA thiR thllorom is valid in mom p;OnC1rl1] topololl:i(~nl Iinrnr ArnOOA (cf.

44

UNBOUNDED LINEAR OPERATORS

Robertson and Robertson [1]), we present the proof for Banach spaces as an application of basic lemma II.1.7. The lemma is also applied later to prove the important Theorem II.4.3. II.I.5 define

For Z a normed linear space and r a positive number,

Definition.

Sz(r) = {z I z E Z,

Ilzll ::;

SZO(r) = {z I z E Z,

r}

Ilzll < r}

The following lemma is a simple consequence of the properties of a norm. II.I.6

Lemma ~.

ii. nt.

Given x E X, the translation map fx from X onto X defined by fx(z) = x + z is a homeomorphism. For any nonzero scalar a, the map ga from X onto X, defined by ga(X) = ax, is a homeomorphism. For any open set V C X and any x E X, fx(V) = x + V is open. Therefore A + V = U x + V is open, where A is an arbitrary x.A

w.

set in X. Given a set K C X, aK = ga(K)

= ga(K) = aKfor any scalar a.

II.I.7 Basic Lemma. Let X be complete and let T be closed. SyO(r) C TS x (l), then SyO(r) C TS x (l).

If

Proof. To prove the lemma, it suffices to prove that SyO(r) C TS x (l/l - a) whenever 0 < a < 1; for if this is the case, then given y E SyO(r), there exists an a, 0 < a < 1, such that y/1 - a is also in SyO(r). Hence there exists an x E Sx(1/1 - a) such that Tx = y/1 - a or T((l- a)x) = y. Sincell(l- a)xll::; 1,yisinTS x (1). LetO < a < 1 and let y E SyO(r) be given. By hypothesis and (iv) of Lemma II.1.6, it follows that for each nonnegative integer n, SyO(ra n ) C TSx(a n ). Taking n = 0, there exists an Xo E Sx(l) such that Ily - Txoll < ra j that is, y - Txo E SyO(ra). Hence, taking n = 1, there exists an Xl E Sx(a) such that Ily - Txo - TXIII < ra 2 ; that is, y - Txo - TXI E SyO(ra 2). Proceeding in this manner, a sequence {x n } is obtained with the properties that n

(1)

and

Ily -

< ran+!

i=Q

.

L Ilx.. 1/1 ,,-0 " defined by z" ... L Xi is Cl\.uehy ,-0 Consequently,

L Tx;1I

11 :::;

n,

e and therefore the sequence {ZII}

t-mrpWTWO

in Bl1nn,oh apfl.co X.

Thus

LINEAR OPERATORS AND THEIR CONJUGATES

45

ther;e exists an x E X such that [[xii::; 1/1 - e and Zn ~ x. From (1) it is clear that TZ n ~ y. Since T is closed, x must be in :n(T) (\ Sx(l/l - e) and Tx = y, whence y E TS x (l/l - e). II.l.B Open-mapping theorem. Let X be complete and let Y be of the second category. If T is closed and 0, the set Tx TSx(e) contains an open set Tx + SyO(r) for some r > O. By the basic lemma II.1.7, it suffices to show the existence of some r > 0 such that SyO(r) C TSx(e). Since

+

Y = \.J nTS x (l) and Y is of the second category, there exists a positive n=l

integer p such that p TS x (l) = pTSx (l) contains a nonempty open set. It follows from (ii) of Lemma 11.1.6 that TS x (e/2) must also contain a nonempty open set V. Thus

oE V

- V C TSx

(~) -

TSx

(~)

C TSx(e)

Since V - V is an open set about 0, there exists an r > 0 such that SyO(r) C V - V C TSx(e). Thus the proof of the theorem is complete. For an example of an incomplete normed linear space of the second category, the reader is referred to Bourbaki [1], Exercise 6, page 3. II.l.9 Closed-graph theorem. A closed linear operator mapping a Banach space into a Banach space is continuous. Proof. Suppose the domain of T is all of X with X and Y complete. The graph G(T) may be considered as a Banach space, since it is a closed subspace of Banach space X X Y. Define linear maps PI and P 2 from G(T) into X and Y, respectively, by PI(x, Tx) = x and P 2 (x, Tx) = Tx. Clearly, PI is 1-1 and satisfies the hypotheses of tbe open-mapping theorem. Consequently, PI has a continuous inverse. Since P 2 is continuous, T = P 2P I - I is also continuous.

In the closed-graph theorem it is essential that both X and Y be complete, as may be seen in the following two examples. The first-order differential operator in Example 11.1.4 is closed but not continuous. The domain of the operator is C'([O, 1]), which is not complete, while the range of the ope~ator is Y = C([O, 1]), which is complete.

11.1.10 IUl(.l Int,

Example.

II ho

l\

Lot X ho any illfii~ite-dimom;ional Banach space X. IT is usually roforred (;0 I1S 11

Hot, of lmHitl dnmCJIlI,H for

UNBOUNDED LINEAR OPERATORS

46

Hamel basis. It may be assumed that the elements in H are of norm 1. Define Xl as the vector space X with norm II III given by hiEH

Clearly,

Ilxlll ~ Ilxll. Xl is not complete.

Let Y1, Y2, . . . be any infinite

n

countable subset of H and let Sn

=

L

k- 2Yk.

It is easy to verify that

k=l

{Sn} is a Cauchy sequence in Xl which does not converge in Xl. Let T be the identity map from X onto Xl. Then T is closed and has a continuous inverse. However, T is not continuous; otherwise T would be an isomorphism, which would imply that Xl is complete.

As an application of the closed-graph theorem, the following fundamental theorem is obtained. It is also referred to in the literature as the Banach-Steinhaus theorem. II.I.ll Uniform-boundedness principle. Let X be a Banach space and let F be a set of continuous linear operators mapping X into Y such that for each x E X, sup II Txll < 00. Then sup II Til < 00 j that is, the operators TEF

TEF

in F, when restricted to a bounded set in X, are uniformly bounded.

Before proving the theorem, we indicate (at the expense of being somewhat verbose) how one might motivate the proof. It is required to find some constant M so that for all x E X, sup IITxl1 ~ Mllxll. A technique TEF

which is often used is to fix x and thereby induce a vector-valued function Ax: F ---+ Y defined by (Ax)T = Tx. In this context, we seek a constant M so that (1)

sup I (Ax) Til ~ TEF

By hypothesis, sup T,F

II(Ax)TII <

Mllxll 00,

XEX

for each x E X.

Thus, Ax is a

member of B(F, Y), the normed linear space of bounded functions from F into Y with norm defined by Ilhll = sup Ilh(T)II. Hence, by (1), an TEF

M is required so that IIAxl1 ~ Mllxll for all x E X. This, in turn, suggests defining the operator A: X ---+ B(F, Y), with the hope that A is bounded. The proof of the theorem consists of showing that A is continuous by applying the closed-graph theorem. Since B(F, Y) is not necessarily complete, B(F, ?) is considered instead, where :P' is the completion of Y.

Proof. Lot A bo t.ho linoo,r lllltp from X in (,0 llCF, 1') dn!illOd hy (Ax)'l' - 'l'x, 1'. /t', His oUHy to verify Lhat A it! dOIlOU, WhOlWO, lJy Ulll

LINEAR OPERATORS AND THEIR CONJUGATES

47

closed-graph theorem, A is continuous. sup II Txll = T.F

IIAxl1

Therefore

IIAIII!xll

~

xEX

Consequently sup T.F

IITII

~

IIAII

In the above theorem it is essential that X be complete.

r

Let X be

k

the subspace of l2 consisting of all elements of the form

a,u" where

i=1

U1 = (1,0,0, . . .), U2 = (0, 1,0,0, . . .), etc. For each positive integer n, let Tn: X ~ l2 be the linear operator defined by setting

if i ~ n ifi = n Then II Tnll

= n while for each x E X, Tnx ~ 0.

By taking Y to be the space of scalars in Theorem 11.1.11, we obtain the result that if X is complete and F is a subset of X' such that sup Ix'xl < 00, X E X, then F is bounded in X'. The following theorem x'di'

is what one might call the dual result. II.I.12

Theorem.

Suppose K is a subset of X such that

sup !x'kl k,K

<

00

x' EX'

Then K is bounded. Proof.

Let J be the natural map from X into X". sup I(Jk)x'j = sup k,K

kEK

Ix'kl <

00

By hypothesis,

x' EX'

Since X' is complete and J is a linear isometry, the uniform-boundedness principle assures us that sup Ilkll = sup IIJk!1 < 00. k,K

k,K

II.I.13 Definition. Let M be a subspace of X. An operator P is called a projection from X onto M if P is a bounded linear map from X onto M such that p2 = P.

As fi further I1pplication of the closoet-graph theorem we show that thero ill a 1-1 cOrrCElpOIIUUIICO betweon tho sct of projections onto M and cortltill e!oHou IlU!JSPlW(l!! of X.

48

UNBOUNDED LINEAR OPERATORS

II.I.I4 Theorem. Let M be a closed subspace of Banach space X. There exists a projection from X onto M if and only if there exists a closed subspace N of X such that X = M E9 N (M (\ N = (0) and X = M + N). In this case, there exists a c > 0 such that

11m + nil;::: cllmll

meM,neN

Proof. Suppose P is a projection from X onto M. Let N = m-(P). Since P is continuous, N is closed. Furthermore, N (\ M = (0), since v e N (\ M implies v = Pv = O. Since any x e X may be written x = Px + (x - Px) and x - Px is in N, X = M + N. Conversely, suppose N satisfies the hypotheses of the theorem. Then each x e X has a unique representation of the form x

= m+n

meM,neN

Define the operator P from X onto M by P(m + n) = m. Clearly, Pis linear and p2 = P. Moreover, P is closed. Indeed, suppose

Since M is closed, y is in M. Therefore y = Py and nk --t x - y. N is also closed, x - y is in N. Hence 0 = P(x - y) = Px - y. P is closed. By the closed-graph theorem, P is continuous and

IIPII 11m + nil ;:::

IIP(m

+ n) I

= Ilmil

Since Thus

meM, neN

The above proof shows that the 1-1 correspondence between projections Ponto M and the closed subspaces N such tha t X = M E9 N is given by P --t m-(P). II.I.I5 Theorem. There is always a projection from a Hilbert space onto anyone of its closed subspaces. Proof.

Corollary I.7.17 and Theorem II.1.14.

The above theorem is instrumental in obtaining a number of important results in Hilbert space. There are examples which show that the theorem does not hold in general if the Banach space is not required to be a Hilbert space. For a discussion of projections and corresponding references, the reader is referred to Dunford nnd Schwartz [11, pnges .1.13 and .'i!i4, ltlld to I'o!c:zyllski [J I. For lilliLo-diHlOIIHiollltlllllbllprwoH Wll lmvo tho rollow illjl; I'lll'IUlt.

49

LINEAR OPERATORS AND THEIR CONJUGATES

II.l.l6 Theorem. There is always a projection from a normed linear space onto anyone of its finite-dimensional subspaces.

Suppose M is a finite-dimensional subspace of X with basis Xl, X2, • . • , Xn . Let x~, x~, . . . , x~ be elements in X' such that X~Xj = 5i;, where 5ij is the Kronecker delta. The may be constructed as follows. Let M i = sp I Xl, X2, • . • ,Xi-I, Xi+l, . • • ,xn }. Since M i is finite-dimensional and therefore closed, there exists a E X' such that V;Xi ~ 0 and Choose = V;/V;Xi. It is easy to verify that i = O. Proof.

x;

v;M

v;

x;

r n

the operator P defined by Px

=

X;(X)Xi

is a projection from X onto M.

i=l

II.l.l7 Remarks. If X is an n-dimensional normed linear space, then so is X'. To see this, let Xl, . . . ,Xn and x~, . . . ,x~ be as in the proof of the above theorem, where now X = M. It is easy to verify that the set Ix~, . . . ,x~} is linearly independent and that each x' E X' has the

r n

representation x'

=

'

X' (Xi)X;.

Thus every finite-dimensional normed

i= 1

linear space is reflexive, since JxX, X, X', and X" have the same dimension and therefore J xX = X". 11.2

CONJUGATE OF A LINEAR OPERATOR

As we shall see in subsequent chapters, the concept of the conjugate of a linear operator is very useful in obtaining information about the range and inverse of T. In the remaining sections of this chapter, theorems are presented which show how certain properties of an operator and its conjugate are related. When T is bounded on all of X, the conjugate of T is the map T' which takes each y' E Y' to y'T E X'. For example, suppose X = Y is an n-dimensional normed linear space over the complex numbers with basis Xl, X2, • . . ,Xn • Let M = «(Xij) be an n X n matrix whose elements are complex numbers. Then M, together with the basis Ix;}, determines T

r n

by setting

=

TXi

(X;jXj.

Let x~, x;, . . . ,x~ be elements in X' such

j=l

that X;Xj = 5i;, the Kronecker delta. Then T' is the operator determined by the adjoint matrix M*, together with I x~}, defined by setting n

~ T "Xi = i.I

,

(XjiXj'

i-I

Now if T is no longer continuous, then y'T need not be in X'. To define in a similar manner the conjugate of such an operator, the natural proeodure is to single out those 1/' E Y' f,9J' which y'T is continuous. Lett.illlJ: thiR RllhRPIW(1 of V' bo 1,ho
UNBOUNDED LINEAR OPERATORS

50

especially in cases when T is a differential operator whose domain, considered as a subspace of £2(12), consists of certain smooth functions. In most applications, ~(T) ,t. X but ~(T) = X, where X is complete. Before defining T' formally, we prove the following theorem. 1l.2.1 Theorem. Let M be a subspace dense in X and let Y be complete. If A is a bounded linear map from Minto Y, then there exists a unique continuouslinear extension A of A to all of X and IIAII = IIAII. The conjugate space M' is equivalent to X'. Proof. Given x EX, there exists a sequence {Xk} in M which converges to x. Since IIAxk - AXil1 :s:; IIAllllxk - xiii, {Axd is a Cauchy sequence which therefore converges to some y in Banach space Y. Let Ax = y. In order to show that A is unambiguously defined, suppose that {zd is a sequence in M which also converges to x. Then by what has just been shown, the sequence Ax!, Az 1, AX2' Az 2, . . . converges to some WE Y. Thus subsequences {Axd and {Az k } both converge to w. Since AXk ~ y, Y = w = lim Azk. Clearly, A is a linear extension of A to all k.... .,

of X.

Now IIAxl1

= n.... lim IIAxnl1 :s:; IIAIIlim Ilxnll = IIAllllxll ., n.... .,

Hence IIAII :s:; IIAII. Since A is an extension of A, IIAII ~ IIAII. Thus IIAII = IIAII. If Y is taken to be the scalars, then the map from M' onto X' which takes A in M' to A E X' is a linear isometry. 1l.2.2 Definition. Let the domain of T be dense in X. The conjugate T' of T is defined as follows. ~(T') = {y' I y' E V', y'T continuous on ~(T)}. For y' E ~(T'), let T' be the operator which takes y' E ~(T') to iTT, where iTT is the unique continuous linear extension of y'T to all of X. ~(T') is a subspace of V', and T' 2S linear. T'y' is taken to be y'T rather than y'T in order that CR(T') be contained in X'.

The requirement that the domain of the operator be dense in X is not as restrictive as might appear at first glance. Suppose ~(T) is not dense in X. Then one can define Xl as the closure of the domain of T and obtain the conjugate T' as a mapping from a subspace of Y' into X~. By inspecting the theorems which require that ~(T) = X, one can usually remove this restriction by considering Xl in place of X. This is done, for example, in the proofs of Theorems V.l.6 and V.l.8. The following simple examples are presented in detail in order to give somo "fr.clilll/;" for I,ho dofinition of It eOlljul/;ItLc opomj,or. Tho conjugl\toF! of (~ort.ltill diITorollt.iul opornl,orfl Itrll ddlll'lllillllll ill Chup. VI.

LINEAR OPERATORS AND THEIR CONJUGATES

Il.2.3

Example. U1

Let X = Y =

= (1,0,0, . . .),

be the unit vectors in

[p.

51

1

[p,

~ p

<

00,

and let

= (0, 1, 0, . . .), etc.

U2

Define T by

n

T(X1, X2, . . . ,xn , 0, 0, . . .) =

(I jxj, X2, Xa, . . . ,x"' 0, 0, . . . ) j=1

Suppose y' = (a1' a2, . . .) E :D(T').

~

Then for k

1,

Since Ilukll = 1 and y'T is bounded on :D(T), a1 = 0. Also, any element (0, b1, b2, . . .) E [p' = [~ is in :D(T'). Hence the domain of T' consists of all the elements in [p' which have zero as their first term. Suppose T'y' = (C1, C2, . . .), where y' = (0, a2, aa,. .) E :D(T'). Then k~2

and

C1

= O. Thus T'y' = y'.

Il.2.4 Example. Let X = Y = £2([0, 1]) and let :D(T) be the subspace of X consisting of those f such that (1)

f is absolutely continuous on [0, 1] with f' f(O) = f(l) =

(2)

= :

E £2([0,

1])

°

Define T by Tf = f'. Strictly speaking, :D(T) is the subspace of cosets [f] E X, where f satisfies (1) and (2) and Tf is the coset [1']. Recall that an absolutely continuous function is differentiable almost everywhere. Since C;(I) is contained in :D(T), :D(T) is dense in X by Theorem 0.9. Suppose y E :D(T') C Y' = £2([0, 1]) and T'y = x. Then for f E :D(T), (3)

f y(t)f'(t) dt

= yTf

= (T'y)f

=

101 x(t)f(t) dt

Since f satisfies (1) and (2) and x is in £2([0, 1]) C £1 ([0, 1]), we may inte!/;rate by parts and get (4)

f

x(t)f(t) dt -

-

f

(ll(t)

+ C)I'(t) dt

52

'"

UNBOUNDED LINEAR OPERATORS

where C is an arbitrary constant and H(t) = and (4),

o = 10

(5)

1

(y(t)

"t

10

+ H(t) + C)f'(t) dt

xes) ds.

f

Thus, by (3)

5)(T)

E

Letfo(t) = fot (y(s) + H(s) + Co) ds, where Co is chosen so thatfo(l) = Thenfo is in 5)(T), and from (5) it follows that

o = 10

1

Iy(t)

o.

+ H(t) + Col2 dt

Hence, as elements in X, y = -H - Co (considered as functions, y = -H - Co a.e.). It follows from the definition of H that y satisfies (1) and y' = -x = - T'y. On the other hand, if y satisfies (1), then considering y as an element of Y', Holder's inequality and integration by parts give lyTfl =

110

1

y(t)f'(t) dt I =

1

110

y'(t)f(t) dt

I~

f

Ily'lIllfll

E

5)(T)

Thus y is in 5)(T'), since IlyT11 ::; lIy'll. We have shown that 5)(T') is the subspace of £2([0, 1]) consisting of those elements y which satisfy (1) and for which T'y = -y'. Hence T' is a proper extension of - T. Il.2.5 Example. Let X = £p(I), Y = £q{I), 1 ::; p, q < 00, where I is a compact interval. Let k be a bounded measurable function on I X I. Define the linear map K from X into Y by (Kf)(t) =

j; k(s, t)f(s) ds

It follows from Holder's inequality that K is in [X, Y]. y E Y' = £q,(I) and K'y = x EX' = £p,(I). Then for f EX,

Suppose

£x(s)f(s) ds = (K'y)f = yKf = £yet) [£ k(s, t)f(s) dsJ dt Hence, by Fubini's theorem, ;; x(s)f(s) ds =

£f(s) [!r k(s, t)y(t) dt] ds

f EX

Thus, we may infer from Theorem 1.6.2 that (1('1/)(11) - X(II) - [/c(II, t)1I(l) (ll

11

f

Y'

53

LINEAR OPERATORS AND THEIR CONJUGATES

Unless otherwise specified, we shall assume throughout the remainder of this chapter that the domain of T is dense in X.

II. 2. 6

Theorem.

T' is a closed linear operator in V'.

Proof. Suppose~ y: - t y' and T'y: - t x'. Then for each x E 'J)(T), y:Tx - t y'Tx and y:Tx = T'y:x - t x'x. Thus y'T = x' on 'J)(T). Hence, by the definition of T', y' E 'J)(T') and T'y' = x'. Therefore T'is

closed. The above theorem is used in Chap. VI to prove that certain differential operators are closed. For instance, the operator T l defined by

= {f I f absolutely continuous on TJ = f'

'J)(T 1)

[0, 1], f' E£ 2([0, I])}

is closed in £2([0,1]), since it was shown that - T l is the conjugate of the operator T defined in Example II.2.4. Next, some topological properties as well as the "size" of 'J)(T') are investigated. Even though 'J)(T) may be all of X, it is still possible for 'J)(T') to consist solely of the zero vector. The following example which demonstrates this is due to Berberian.

II. 2. 7 Example. Take X = Y = l2 and take 'J)(T) to be the span of the unit vectors Uk = (0, 0, . . . , 1, 0, 0, ..) E l2. Let k

{Uk; I k,j

be any double indexing of {Uk}. TUk = Uk,

=

1,2, . . . }

For each k, define j = 1,2, . . .

and extend T linearly to X. Suppose y' = (aI, a2, . . .) E 'J)(T'). for each k, T'y'Uk; = y'Uk = ak, j = 1,2, . . .. Now .,

L IT'y'Uk;12 ~

IIT'y'112

i-I

Hence

Th(1r(\forp 1/' - O.

o ...

lim T,'y'Ukl ""' at j-+-a

1

~

k

Then

54

UNBOUNDED LINEAR OPERATORS

For an example of a differential operator T with 'neT') = (0), the reader is referred to Stone [1], Theorem 10.10, page 447. II.2.8 Theorem. 'neT') = Y' if and only if T is continuous. is the case, then T' is also continuous and IIT'II = IITII.

If that

Proof. Clearly, if T is continuous, then y'T is continuous for each y'eY'. Thus'n(T') = Y'. Suppose'n(T') = Y'. LetSbethel-sphere of 'neT). For each y' e V', sup ly'Txl ::; IIT'y'll. Hence, by Theorem 11.1.12, IITII

=

sup II Txll XES

<

:XES 00.

Now, for each xeS, IT'y'xl ::; Ily'IIIIT11.

Thus IIT'y'll ::; IITIIIIY'II, and therefore IIT'II 5: IITII. IITxl1

=

sup ly'Txl

lIy'li =

1

=

sup IT'y'xl ::; IIT'lIllxll

I!y'li =

By Corollary 1.5.7, X

e 'neT)

1

Hence IITII ::; IIT'II and the theorem follows. The linear operators which one usually encounters have the property that they are restrictions of closed operators. Theorem II.2.11 characterizes such operators. II.2.9 Definition. A set F of linear functionals on a vector space V is said to be total if given any v ;:e 0 in V, there exists an f eF such that f(v) ;:e o.

In Examples II.2.3 and 11.2.7, the domain of T' is not total. lary 1.5.6 shows that a conjugate space is total.

Corol-

II.2.1O Definition. T is called closable if there exists a linear extension of T which is closed in X. The domain of T is not required to be dense in X. II.2.11 Theorem. Statements (i),. (ii) and (iii) are equivalent. is not required to be dense in X.)

('n(T)

T is closable. T has a minimal closed linear extension; i.e., there exists a closed linear extension '1' of T such that any closed linear extension of T is a closed linear extension ofT. m. For any y ;:e 0 in Y, (0, y) is not in the closure of the graph of T. If 'neT) = X, then the statement that 'neT') is total is equivalent to the above three statements. In this case, T' = ('1')', where '1' is the minimal closed extension of T. 2.

ii.

Proof. (i) implies (ii~). Let l' be a closed linear extension of T. If 11 e Yand 11 ¢ 0, then (0, y) ~ G(1') :::> G(T). Hence (0, y) _ (J['l'), since (J(t) iA (~loR()(l in X X Y.

LINEAR OPERATORS AND THEIR CONJUGATES

55

Suppose (0, y) ~ G(T) for any y r5- 0 in Y.

(iii) implies (ii).

'1' as the operator whose graph is G(T); that is, :D('1') = {x I (x, z)

E

G(T) for some

Z E

Y}

Define

'1'x = z

Then '1' is unambiguously defined and is a closed linear extension of T. Furthermore, '1' is the minimal closed linear extension of T; for if T 1 is a closed linear extension of T, then G(T 1) ~ G(T) = G('1'). (ii) implies (i) trivially. Suppose that :D(T) = X. We proceed to show that :D(T') is total if and only if statement (iii) is valid. Let :D(T') be total and let (0, y) be in G(T). Then there exists a sequence {x n } in :D(T) such that Xn ~ 0 and TX n ~ y. Thus, for each y' E :D(T'), y'Tx n ~ 0 and y'Tx n ~ y'y. Since :D(T') is total, it follows that y = O. Assume (iii). Let y r5- 0 be an element in Y. Then (0, y) ~ G(T) and therefore there exists a z' E (X X V)' such thatz'(O, y) r5- 0 and z'G(T) = O. Defining x' E X' and y' E Y' by x'x = z'(x, 0) and y'y = z'(O, y), we obtain

o=

z'(x, Tx) = x'x

o r5- z'(O, y)

+ y'Tx

X E :D(T)

= y'y

From these two equations, we have y' E :D(T') and y'y r5- O. Thus :D(T') is total. Suppose y' E :D«'1')'). Then y''1' is continuous on :D('1') and, in particular, continuous on :D(T). Thus y' E :D(T'). Suppose y' E :D(T') and x E :D('1'). Since (x, '1'x) E G('1') = G(T), there exists a sequence {x n } in :D(T) such that Xn ~ x and TX n ~ '1'x. Hence

ly''1'xl =

lim ly'Txnl ~ IIT'y'll lim

n->oo

n->oo

Ilxnll = IIT'y'lIllxll

Therefore y''1' is bounded, which means y' E :D«'1')').

Hence

:D(T') = :D«'1')')

Since T'y'

=

('1')'y' on the dense subspace :D(T), it follows that T' =

('1')'.

The theorem just proved shows that the operator in Example II.2.3 is not closable.

11.2.12 Corollary.

A linear-operator which maps a Banach space into a Banach space is continuous if and only if the domain of its conjugate operator 1:H total.

56

UNBOUNDED LINEAR OPERATORS

Proof. Let X and Y be complete and let XJ(T) = X. Suppose XJ(T') is total. Then by Theorem II.2.11, '1' is closable. Since XJ(T) = X, '1' must be its own closed extension. Hence '1' is continuous by the closed-graph theorem. If '1' is continuous, then XJ(T') = V', which is total. II.2.13 Lemma. If X is reflexive and F is a subspace of X', then F is total if and only if F is dense in X'.

Proof. If F is total but not dense in X', there exists an x" ~ 0 in X" such that x"F = O. By the reflexivity of X, there exists an x ~ 0 such that 0 = x"x' = x'x for all x' E F. But this is impossible since F is total. If F is dense in X', then F is total, since any set dense in a total set is also total. II.2.I4 Theorem. Suppose Y is reflexive. Then '1' is closable if and only if XJ( '1") is dense in Y'. In that case, the minimal closed extension of '1' is J y-lT"Jx, where J x and J yare the natural maps from X into X" and Y onto Y", respectively.

Proof. The first assertion of the theorem is a consequence of Lemma 11.2.13 and Theorem 11.2.11. Thus the conjugate '1''' ofT' is defined, and by Theorem 11.2.6, '1''' is closed. Since J y and J x are isomorphisms, it follows that E = J y-IT" J x is also closed. Given x E XJ(T"J x ) and y' E XJ(T'), (1)

y'(Jy-lT".Jxx) = (T"Jxx)y' = (Jxx)T'y' = T'y'x

Now XJ(T) C XJ(T"J x ); for if x

E

XJ(T), then

[(Jxx)T'y'l = IT'y'xl :::;

l!y'IIIITxll

y'

E

XJ(T')

Thus II (Jxx) '1"11 :::; IITxll, and therefore, by the definition of XJ(T'), J xX E XJ(T"). Consequently, we obtain from (1) y'(Ex) = T'y'x = y'Tx

X E XJ(T), y'

E

XJ(T')

Since XJ(T') is total, Ex = Tx for all .r E XJ(T). Thus E is a closed linear extension of T. To prove that E is the minimal closed linear extension of '1', it suffices to prove that G(E) C G(T) = G(T). Suppose (u, v) E G(E) but (u, v) ¢ G(T). Then there exists a z' E (X X V)' such that

z'(u, v)

¢

0

,'(x, 7'x) - 0

x,5)('1')

57

LINEAR OPERATORS AND THEIR CONJUGATES

Let x' E X' and y'

E

Y' be defined by x'x

= z'(x, 0)

y'y

= z'(O, y)

Then (2)

x'u

(3)

x'x

+ y'v = z'(u, v)

+ y'Tx

~

0

= z'(x, Tx) = 0

Equation (3) implies y' E 5)(T') and T'y' = -x'.

X E

5)(T)

Since

a substitution of u for x in (1) gives y'v = T'y'u = -x'u

which contradicts (2).

1l.2.15

Theorem.

Hence G(E) C G(T).

1" is continuous if and only if 5)(T') is closed in Y'.

Proof. Suppose T' is continuous and y: -----+ y', y: E 5)(T'). Since T' and {y~} are bounded, there exists some constant M such that IIT'y~11 ~ M. Therefore ly'Txl

= lim ly~Txl = lim IT'y~xl ~ Mllxll n-HO

As a result, y' E 5)(T') whence 5)(T') is closed. Conversely, if 5)(T') is closed, then T' is a closed linear map from Banach space 5)(T') into Banach space X'. By the closed-graph theorem, T' is continuous. Under the assumptions that X and Yare complete and that T is closed, it is shown in Corollary 11.4.8 that the statements "T continuous," "T' continuous," and "5)(T') closed" are all equivalent.

Il.2.16 Theorem. Every operator in [Y', X'] is a conjugate of an operator in [X, Y] if and only if Y is reflexive. Proof. Suppose Y is reflexive and A E [Y', X']. Define T E [X, Y] by T = Jy-lA'J x . Then T' = A, since for all y' E Y' and x E X, T'y'x

=

y'(Jy-lA'Jxx)

=

(A'Jxx)y' = (Jxx)Ay' = Ay'x

Suppose every operator in [Y', X'] is a conjugate operator. Let x~ E X' and Xo e X be chosen so tha t x~x~ = 1. Given y" E Y", define A E [Y', X'l (-~

by A 1/

-

Y" (Y')x~

58

UNBOUNDED LINEAR OPERATORS

By hypothesis, there exists aTE [X, Y] such that A y'Txo = T'y'xo = Ay'xo = y"(Y')x~xo = y"y'

=

T'.

y'

Hence, E

Y'

Therefore Y is reflexive. 11.3

STATES OF LINEAR OPERATORS

We shall first explain the notion of a state diagram. The diagram is a bookkeeping device for keeping track of some theorems concerning the range and inverse of T as well as T'. Motivated by the known theorems relating T and T', we classify various possibilities of meT) and T-l by

I: meT) II: III: 1: 2: 3:

=

Y

m(T);;e Y but meT) m(T);;e Y

= Y

T-l exists and is continuous. T-l exists but is not continuous. T has no inverse.

= Y, we say that T is in state I or that T is surjective, written TEL Similarly, we say that T is in state 3, written T E3, if T has no inverse. If TEll and TEl, we write T EIll. The same notation is used for T'. If T E111 1 and T' Ela, we write (T, T') E (111 1 , la). Thus, (T, T') has 81 classifications, which are described in the "checkerboard" 11.3.14, which is referred to as a state diagram.

If meT)

In this section, theorems are proved which show the impossibility of certain states for (T, T'). These eliminated states are shown by shading the corresponding squares. Some of the squares can be eliminated if additional assumptions are put on X and Y, for example, X reflexive or Y complete. In these instances the letters X-R or Yare put in the corresponding squares. The state diagrams 11.3.14 and 11.4.11 were obtained by Goldberg [1]. II.3.1

Theorem.

If T' has a bounded inverse, then meT') is closed.

Proof. Suppose T'y: ----t x' E X'. there exists an m > 0 such that

Since T' has a bounded inverse,

Thus {y:J. is a Cauchy sequence which converges to Bome y' in Banach space Y'. Rinn(~ 7" iA (']OAP
LINEAR OPERATORS AND THEIR CONJUGATES

59

The theorem shows that state III is impossible for T'. we write T' ¢ lIt.

For brevity,

II.3.2 Definition. If C is a subset of X', then the orthogonal complement of C in X is the set .LC = {x I x E X, x'x = 0 for all x' E C}. II.3.3 Remarks. K.L and .LC are closed subspaces of X' and X, respectively. AlsoK.L = K.L and.LC = .LC.

The proofs of Theorems 11.3.4 to 11.3.8 are left to the reader. II.3.4

Theorem.

If M is a subspace of X, then .L(M.L) =

M.

II.3.5 Theorem. If N is a subspace of X', then (.LN).L :J N. is reflexive, then (.LN).L = N. II.3.6 Remarks. Dieudonne [1] has shown that subspace of X', then (.LN).L = N. Since we later (.LN).L = N, for N finite-dimensional, we shall prove Dieudonne's theorem. Since N C (.LN).t, it follows from Theorem 1.6.4 (1)

dim

.L~

=

dim

(.L~)' =

If X

if N is a reflexive use the fact that this special case of that

dim (.LN).L ;::: dim N

If x~ ,x~, ., x: is a basis for N, then the map A from Xj.LN into Un defined by A[x] =- (x;x, x~x, . . . ,x:x) is 1-1 and linear. Thus (2)

dim

.L~ ~

n = dim N

It follows from (1) and (2) that N = (.LN).L.

II.3.7 t.

ii.

Theorem (R(T).L = (R(T).L = ;neT') (R(T) = .L;n(T')

In particular, T has a dense range if and only if T' is 1-1.

By interchanging the roles of T and T' in the above theorem, we do not quite obtain the dual type theorems. 11.3.8

. i. ii.

£?:i.

Theorem J.(R(T') ~ ;neT) If ~(T') is total, then ;neT) = .L(R(T') (', :nCT) ~ C ;n(T).L

In pn.rI1'I'lIlnr,

1/
1'R lolnl, Own 'I' 1'S 1-1.

UNBOUNDED LINEAR OPERATORS

60

11.3.9

Theorem.

If l' and 1" each has an inverse, then (1'-1)' = (1")-1.

Proof. By Theorem II.3.7, :£)(1'-1) =
x'1'-1(1'x) = 1"y'x = y'1'x

X E

:£)(1')

Thus (1'-1)'x' = y' on
Ilzn/l =

1 and

if 1'z~ ~ 0 if 1'Zn = 0 II.3.lI

Theorem.


Proof. Suppose «(1") I) X' if fllld ollly if ('I"l 0'

and II.3.1l,
61

LINEAR OPERATORS AND THEIR CONJUGATES

The next theorem shows that additional states for (1', 1") fail to exist when Y is complete. 11.3.13 Theorem. bounded inverse.

Let Y be complete.

If meT) = Y, then 1" has a

Proof. The argument is analogous to the one in Theorem II.3.11, with use being made of Theorem 11.1.11.

As a summary of the preceding theorems, the following state diagram is obtained. For example, we see from an inspection of the diagram that T has a bounded inverse if and only if the range of 1" is all of X'. In regard to the squares which remain open, we shall present in Sec. II.5 examples which exhibit the existence of the corresponding states when both X and Yare reflexive. 1l.3.14

State diagram for linear operators

Y: Cannot occur If Y II compl.t•.

UNBOUNDED LINEAR OPERATORS

62

STATES OF CLOSED LINEAR OPERATORS

Il.4

This section is devoted to the construction of a state diagram for closed operators. Actually, the only "deep" theorem is 11.4.3, which plays a very important role in later sections. The theorem was first proved for bounded, instead of closed, operators by Banach [1], Theorem 1, page 146. For the generalization to closed operators, see Kato [1], Lemmas 324, 335, Goldberg [1], Theorem 5.1, and Rota [1], Theorem 1.1. Browder [2], Theorem 1.2, gave similar results for closed linear operators in Frechet spaces. 1l.4.1

and S SyO(r)

If T' has a bounded inverse (T not necessarily closed) the I-ball in X, then TS ~ SyO(r), where r = 1/11(T')-111 and {y Illyll < rl·

Lemma.

is =

Proof. Suppose y E SyO(r) but y tI TS. Since TS is closed and convex, there exists, by Theorem 1.5.10, a nonzero y' E Y' such that Re y' (y) ~ Re y'(TS). Assert that Re y'(y) ~ ly'Txl for all XES n :D(T) = SI. Indeed, if x E SI and y'Tx is written in polar form ly'Txle i9 , then e- i9 x E SI. Hence Re y'(y) ~ Re y'T(e-i9 x) = ly'Txl

Consequently, y' E :D(T') and

Ily'llllyll

~ ly'YI ~ sup ly'Txl =

IIT'y'll

~ rlly'll

XESl

Thus 1l.4.2

r

> O.

Ilyll

~

r, which contradicts the supposition that y E SyO(r).

Suppose TS ~ SyO(r) , where S is the I-ball in X and If T-l exists, then it is continuous with II T- 1 11 ~ 1/r.

Lemma.

Proof. If T-l exists, then for x ,e 0 and 0 < c < 1, (1 - c)rTxlllTxll is in SyO(r). By hypothesis, there "exists some Z E S n :D(T) such that Tz = (1 - c)rTx/IITxll. Since T is 1-1, z = (1 - c)rx/IITxll. Hence IITxl1 ~ (1 - c)rllxll, which implies that T-l is continuous with liT-III ~ 1/r. Theorem. Suppose X is complete. If T is closed and T' has a bounded inverse, then TS ~ SyO(r), where r = 1/11(T')-111 and S is the I-ball of X. Thus ffi(T) = Y and T is an open map. If T-l exists, it is continuous and SyO(1/1IT- 111) is the largest open a-ball which is contained in TS. II.4.3

Proof.

All but the last statement of the theorem are immediate

conscqllenncs of LCmmltH ll.4.l, 11.1.7, alld TJ .4.2. RIIJlpmlO T--I ()xists ILlld SyO(a) C '/'S. Tlwll fl'OllI LOIllIl\ll. 11.1.~ ltlill TlwOI'OIllH 11.:1.11 ILIIlI

63

LINEAR OPERATORS AND THEIR CONJUGATES

11.2.8, it follows that =r As a summary of Theorems II.3.1l, II.3.13, and 11.4.3, we have the following dual results. Note that neither X nor Y need be complete, nor do we require that T be closed in (i). II. 4.4 't.

ii.

Theorem m(T') = X' if and only if T has a bounded inverse. Suppose that X and Yare complete and that T is closed. m(T) = Y if and only if T' has a bounded inverse.

Then

Il.4.5 Corollary (Banach-Mazur). If Y is a separable Banach space, then there exists a continuous linear operator mapping II onto Y. 1\£oreover, the conjugate of Y is equivalent to a subspace of l",. Proof.

Let {yd be a sequence of elements of norm 1 which is 00

dense in the unit sphere of Y.

Define T e [ll' Yj by T({ ad) =

L akYk.

k=l

Let

Ul

= (1, 0, 0, . . .),

IIT'y'll =

U2

= (0, 1, 0, 0, . . .), etc. Then for y' e Y'

sup IT'y'(Uk) I = sup ly'Ykl = sup k

lIull =

k

1

IY'YI = Ily'll

Hence T' is an isometry and the range of T is Y by Theorem 11.4.4. 1I.4.6 Definition. The 1-1 operator '1' induced by T is the operator from 'J)(T) j'Jt(T) into Y defined by

'1'[xj = Tx Note that 1'is 1-1 and linear with the same range as T. The importance of considering'!' is that certain results which hold for 1-1 linear operators may be applied to '1' in order to obtain information about T. The proof of Corollary 11.4.8 is a case in point. The next lemma shows that l' has some of the essential properties of T. 11.4.7 Lemma. Suppose 'Jt(T) is closed and '1' is the 1-1 operator induced by T. Then i. £?:.

T is closed if and Only if l' is closed. 7' 1:S continuou,q if an,d only if _1' is continuous, in which case

IITII - II til,

t·it'.

:1)«1')') ""

:1)('1") and

'

11(1')' /1'11 "" 11'I"y'll

64

UNBOUNDED LINEAR OPERATORS

Proof of (i). Let T be closed. Suppose [xnl ---+ [x], [xnl E'D(T)/m(T), ---+ y. Then there exists a sequence Vn E m( T) such that X n Vn ---+ x. Since T(x n - v n) = T[xnl ---+ y and T is closed, x is in 'D(T) and Tx = y. Thus [xl E'D(T) and T[xl = Y, which proves that l' is closed. Conversely, let l' be closed. Suppose X n ---+ x and TX n ---+ y. Then [xnl ---+ [xl and T[xnl = TX n ---+ y. Hence [xl E'D(T) and l'[xl = y, or, equivalently, x E 'D(T) and Tx = y. Therefore T is closed. Proof of (iz). Suppose T is continuous. Then

and T[xnl

IIT[xlll = IITzl1

~

IITllllzl1

inf

Ilzll

z E [x]

Hence IIT[x]1I ~

Thus

111'11

~

IITII.

IITII

,,[xl

=

I!TIIII[xlll

On the other hand, if Tis continuous, then

IITxl1 =

IIT[xlll

~

IITIIII[xlll

~

IITllllxl1

Therefore IITII ~ IITII· Combining these results, we obtain IITII = 111'11. The proof of (iii) follows easily upon replacing T by y'T and l' by y'T in the proof of (ii). Il.4.8 Corollary. Let Y be complete and let T be closed. If T' is continuous, then 'D(T) = X. If both X and Yare complete and T is closed, then the following three statements are equivalent.

i. '/,'/,. t'/,'/,.

T' is continuous. T is continuous on all of X. 'D(T') is closed.

Proof. Suppose T' is continuous. Assume, first, that T is 1-1. Let Y 1 = CR(T) and let T 1 be the operator T mapping 'D(T) into Y 1 • Then, by Theorems II.3.7 and II.3.9, (T1-I)' has an inverse and (1)

Since T' is continuous, it follows that T~ is also continuous. Thus (1) shows that (T1-I)' has a bounded inverse. Since T is closed, it follows that T1-I is also closed. Thus we may apply Theorem IIA.3 to T1-I and obtain X = CR(T1-I) = 'D(T). If T is not 1-1, then the 1-1 operator l' induced by T is closed and its conjugate is bounded by Lemma II.4.7. Thus, by what has just been shown, X/meT) = 'D(f') = <J)(T)/m(T). Hence X = 'D(T). If, in addition, X is complete, then T is continuous by the dosedgraph theorem. Hence 'D(T') = Y'. This shows that (i) implies (ii) and (i1:) implies (i#). Assuming (iii), (i) is n. consequence of the dosedgraph theorom aJlpliod

(,0 '/".

LINEAR OPERATORS AND THEIR CONJUGATES

65

li.4.9 Corollary. Suppose X is reflexive. If there exists a bounded linear operator which maps X onto Banach space Y, then Y is reflexive. If M is a closed subspace of X, then XjM is reflexive.

Proof. Assume that T is an isomorphism from X onto Y. It follows from Theorem II.4.4 that Til is also an isomorphism. (One can also easily prove directly that Til is an isomorphism.) Since TilJ x = J yT and both J x and Til are surjective, it follows that

Y" = rst(T"J x) = rst(J yT) C rst(J t) Thus Y is reflexive. Let us only assume that rst(T) = Y. Then by Theorem 11.3.13, T'Y' is isomorphic to Y'. Since Y' is complete, T'Y' is also complete and therefore a closed subspace of reflexive space X'. Hence T'Y' is reflexive by Theorem 1.6.12. Thus, by what was just proved, Y' is reflexive and therefore so is Y. The last statement of the theorem now follows, since the map from X onto X/M defined by x ~ [x] is bounded and linear. One of the most important consequences of Theorem 11.4.4 is Theorem IV.1.2, which is proved when closed operators with closed range are discussed. We come to the final theorem which is needed to complete the state diagram for closed operators. Il.4.10 Theorem. Suppose X is reflexive. then rst(T') is dense in X'. Proof.

If T is closed and 1-1,

Let T 1 be the operator T cbnsidered as a map onto

It is easy to see that rst(T') = rst(T~). By Theorems 11.3.7 and 11.3.9, T~ has an inverse and (T~)-l = (T1-l)'. Since T is closed, it is clear that T 1 is closed and therefore so is T1-l. Applying Theorem II.2.14 to T1-r, we have that ~«TI-l)') is dense in X'. Since

the theorem follows. At times it will be convenient for us, throughout the remainder of the book, to refer to the two state diap;rums rather than to the various theorems ill SO(:s. II.a n.nd IIA. For examplo, (iii inspection of Diagram 11.3.14 shows t.hal. 7' fIJI I if alld only if' '/" fIn. If T is dosed, X reflexive, and Y

66

UNBOUNDED LINEAR OPERATOR$

complete, then an inspection of II.4.l1 shows that T E II s if and only if T' E III 2• 1I.4.11

State diagram for closed linear operators

i T_ Y: Cannot occur if Y is complete.

X-c: Cannot occur if X is complete and T is closed. X - R-c: Cannot occur if X is reflexive and T is closed.

11.5

EXAMPLES OF STATES

In this section examples are given which exhibit those states of (T, T') which can occur even when both X and Yare reflexive. Examples are also given which show that the states of (T,T') corresponding to the squares with entries can occur when the corresponding hypotheses on X or Y arc removed. Silwo somp of Lbo ILrgUlnUllts dopolld Oil propt'rtioM of (lOlIIpU(Jt OpOl'lL-

6-7

LINEAR OPERATORS AND THEIR CONJUGATES

tors, it is suggested that the reader who is unfamiliar with the theory of compact operators return to this section after Chap. III. Where possible, compact linear operators are constructed. T continuous on X

(11, 11): Let T be the identity operator on X. (la, III l ): Let T be the left-shift operator defined by T( {Xk}) Obviously, TEla and by II.3.14, T' EIII l . (1111, la): Let T be the right-shift operator defined by

T({xd) = {xk-d

1

<

k

where Xo = O. T E 111 1 and by 11.4.11, T' E Ia. It is easy to see that the left-shift and right-shift operators are conjugates of each other.

T compact on X (11 2 , 11 2): Let T be defined by T( {Xk}) = {xk/k}. T is compact since Tn ---t T, n = 1, 2, . . . ,where Tn is the compact operator from X into Y defined by Tn({xd) = (yd, Yk = xk/k, 1 ~ k ~ n, and Yk = 0, k > n. Tn is compact since it is bounded and its range is finite-dimensional. Clearly, m(T) is dense in Y, and T has an unbounded inverse. Since a compact operator cannot have an infinite-dimensional range which is complete, T E 11 2• From 11.4.11, T' E 11 2 • (II a, 111 2): Let L be the left-shift operator on l2' and let A be the compact operator in example (11 2, 11 2). Define T = AL. Then T is compact and in lI a. From II.4.11, T' E 111 2•

(111 2 , II a): Let T be the conjugate of the operator in example (lI a, 111 2). Since the conjugate of a compact operator is compact, T is compact and from 11.4.11, T E111 2 and T' Ella. (IlIa, lila): Let T be the zero operator. Less trivially, let T l be the compact operator in example (11 2, 11 2) and let Land R be the left-shift and right-shift operators, respectively. Then T = T1RL is a compact operator in IlIa. By 11.4.11, T' E IlIa. Y not complete

The next eXflmplo dcp~nds on the following; observation, I\, e'_()Inpll'(~j, lillPIU' op~mj,ol' from mflexive space X into Le.1. '/'I Ill' UIf' O)H·"/l.l.or' '/' ('ollFlidl'l'l'd /l.A II. 1l1llJl f!'Om X Ollt,O VI = (!l(7'),

SUP!'OHO 7' iA

y,

Tcompact on X

UNBOUNDED LINEAR OPERATORS

68

Then T 1 is also compact. To see this, suppose Ix n } is a bounded sequence in X. Since T is compact, there exists a subsequence Iv n } of Ix n } such that TV n ----+ y E Y. By Theorem 1.6.15, there exists a subsequence IZn} of Iv n } which converges weakly to some x E X. The continuity of T implies U'

'W

that TZ n ----+ Tx. Since TZ n ----+ y, it is clear that TZ n ----+ y. Hence y = Tx, since y'y = y'Tx for all y' E Y'. Thus Tlv n -> Tx E Y I , showing that T 1 is compact.

(1 2 , 1I 2) : Let To be the compact operator in example (11 2 , 11 2) and let T be the operator To considered as a map from X onto R(T o). Then T is a compact operator in 12 , and by 11.4.11, T' E Ih (1 3 , 111 2): Let L be the left-shift operator on l2. Choose K as the operator in the preceding example (1 2 , 11 2), Defining T = KL, T is a compact operator in h Hence T' is also compact. Since a compact operator on an infinite-dimensional normed linear space cannot have a bounded inverse, T' ¢ 1. Thus T E 111 2 , by 11.4.11. X not complete

T continuous on X

(1 2 , III I ): Let X be the normed linear space obtained by renorming l2 as in Example II. 1.10. Define T as the identity map from X onto l2. Then T ¢ 1; otherwise the incomplete space X would be isomorphic to Banach space Y. Thus T E 12 and T' E I1I I by I1.4.11. (III, 11): Choose X as any proper subspace dense in Y = l2 with T as the identity map on X. (I1 2 , III 1) : Let A be the operator in example (1 2 , 111 1). Setting Xl as the domain of A and Y 1 as any proper subspace dense in l2' define T:X I X Y I ----+ l2 X l2 by T(x, y) = (Ax, y). Take the norm on l2 X l2 to be given by II(x, y)11 = (11x11 2 + IlyI12)~. With respect to the inner product «x, y), (u, v» = (x, u) + (y, v) l2 X l2 is a separable Hilbert space which, by Theorem 1.7.24, is equivalent to l2. T is easily seen to be in 11 2• Given z' E (l2 X l2), define x' and y' E l~ by x'(x) = z'(x, 0) and y'(y) = z'(O, y). Then

Iz'(x, y)1 ::; Hence

Ilx'llllxll + Ily'llllyll ::; (11x'll + IIY'II)II(x,

Ilz'll ::; Ilx'll + Ily'll· IIT'z'll

For

Ilxll =

~ IT'z'(x, 0)1

1,

= jCA'x')xl

Thus (1)

IIA 'x'il ~. te:'1L 11 7"1,'11 > c.. Ir(A')-'TI

y)11

LINEAR OPERATORS AND THEIR CONJUGATES

69

Similarly,

IIT'z'll

~

/T'z'(O, y)1

=

IY'Y/

I/T'z'll

~

Ily'll

lIyl/ =

1

Thus (2)

Hence, by (1) and (2), there exists some m

IIT'z'lI This shows 1"

E

1.

~

>

m(llx'l/ + Ily'lI)

0 such that

~

ml/z'll

It follows from II.4.11 that 1"

E

II1 1 •

(II a, III 1): Let A, Xl, and Y 1 be as above. Define 1': X 1 X Y 1 ~ 72 X l2 by T(x, y) = (Ax, Ly), where L is the left-shift operator on l2. T is easily seen to be in II a. Arguing as in the above example, we obtain

IIT'z'll

~

IIL'y'll = lIy'll

Recall that L' is the right-shift operator and, in particular, is an isometry. Thus 1" E 111 1, by II.4.11. l' compact on X

X complete but not reflexive

(II 2 , III 2 ) : If 0 and

Ilxll

p

~

Ilxll

q•


:::; q:::; oco, then x

Indeed, if q

<

oco,

=

(Xl, X2, . . • )

then Ilxll~ =

E

lp implies:r E lq

'"

L /xklplxkl q- p <

oco,

k=l

since {XkJ is bounded. Now Ilxll q ~ !xkl, 1 :::; k. Thus IIxll~ :::; IIxll~-Pllxll; whence Ilxll q :::; Ilxll p • If q = oco, then clearly Ilxll oo :::; Ilxli p • Let A be the identity map from II into l2. Then by what was just proved, A is continuous. Define T:l 1 ~ l2 by l' = KA, where K is the compact operator in example (11 2 , 11 2). Since 1" is compact, it is not in 1 and its range is separable. However, loo is not separable. Thus 1" E 111 2 by 11.4.11. (111 2 , Ilia): Let 1'1 be the compact operator l' in the preceding example and let R be the right-shift operator on ll. Define 1': II ~ l2 as the operator T 1R. Then l' is clearly in III 2• Since compact operator 1" has a separable range, 1" is in III. Thus 1" E IlIa, by 11.4.11. The next example is for the only square in 11.4.11 which has not been accounted for.

X complete but not reflexive

Y not complete

l' compact on X

7'0.

UNBOUNDED LINEAR OPERATORS

Then by the argument given in example (11 2,11 2),1'1 is a compact operator with an unbounded inverse. Define l' to be the operator 1'1 considered as a map from li onto Y = oo M

M

L IkYkl :::; L IkYk k=I k=I

vknl

+

M

L IVknl

k=I

M

:::; k=I L IkYk -

vknl

+ 1 ----} 1

as n ----}

00

M

Thus

L IkYkl

:::; 1 for all M, and therefore x

=

(kYk I is an element in li'

k=I

Hence TV n ----} Y = Tx, which shows that l' is compact. Since 1" is compact, its range is a separable subspace of inseparable space loo. Thus 1" E III. It follows from II.4.l1 that 1" is in 111 2 • Since the operators in the above examples are all continuous on X, we have also shown that the state diagram for closed linear operators is the same as the state diagram for continuous linear operators defined on all of X. In order to gain some insight into the reason for this, we mention the following theorem which was proved by Goldberg in (2). Roughly speaking, it is shown that by letting E be 'neT') with norm defined by Ily'll + IIT'Y'II, E becomes a conjugate space. Moreover, T~, which is the bounded operator 1" with respect to E, becomes the conjugate of a bounded operator 1'1. Finally, 1'1 and T~ have the same states as l' and 1", respectively. Thus the state diagram for (1', 1") is the same as the state diagram for (1'1, T~).

II.5.1 Theorem. Suppose l' is closed. Let E denote the space 'neT') with norm Ily'IIr = Ily'll + IIT'y'll and let T~ be the operator 1" mapping E into X'. Define J:Y----}E' by (Jy)y' = y'y and A:E----} (JY)' by (Ay')Jy = (Jy)y'. Then 1,.

n. m. iv.

E is linearly isometric to (JY)' under the map A. JT is a continuous linear map from 'neT) into JY. T~ = (JT)' A. l' and T' are in the same states as bounded operators JT and (JT)', respectively.

Suppose that X is reflexive and Y is com.plete. Let Z be the closure of J Y in R' and let .J '1': X --) 7. 111: t/if' ('(jilt i'll1/li71.~ Uncar l:;rJ(m.~ion of .J l' to a.1l IIf X. '/'hI'11 '/' (lnd '/" (1'/'1' in thl' HII/III' ,~/IlII'H (/.q I'l' lInd (,rr)', 1'1"~Jlf'I·lilll't!l.

LINEAR OPERATORS AND THEIR CONJUGATES

71

A state diagram for linear operators in a topological linear space setting was constructed by Krishnamurthy [1]. Six examples which demonstrate the possibility of states corresponding to the blank squares still unaccounted for in Diagram 11.3.14 may be found in Goldberg [1]. II. 5.2 Definition. A scalar A belongs to 'to n.

Let X = Y and let I be the identity operator on X.

p(T), called the resolvent of 1', if l' - AI has a dense range and a bounded inverse. u(T), called the spectrum of 1', if A , p(T).

The set u(T) is decomposed into the following disjoint sets by letting A belong to n't. Pu(T), called the point spectrum of 1', if l' - AI E 3. iv. Ru(T) , called the residual spectrum of 1', if l' - AI E III but not in 3. v. Cu(T) , called the continuous spectrum of 1', if l' - AI E 2 but not in III. Since 1" - AI = (1' - AI)' and l' - AI is closed when l' is closed, an appeal to the state diagrams for (1' - AI, (1' - AI)') verifies the following relations. Il.5.3

Corollary

'to n. m. w.

v. v't. vii. viii.

u(T) = u(T') Pu(T) C Ru(T') U Pu(T') C Ru(T) U Cu(T) C Ru(T') U Cu(T') C Cu(T) Ru(T) C Pu(T') Ru(T') C Cu(T) U If l' is closed and Ru(T') C Pu(T)

Pu(T') Pu(,-T) Cu(T')

Pu(T) X is reflexive, then Cu(T)

= CI1(T') and

11.5.4 Remark. The same argument used to prove the above corollary can be used to prove the following generalization. Let B be a continuous linear operator with X ::) X)(B) ::) x)(T) and range in Y. (Y not necessarily X.) Then (1' - AB)' = 1" - AB'. Define pn(T) as in (i) of Definition 11.5.2, with B replacing I. Similarly, define un(T), etc. Then (i) through (vii) of Corollary II.5.3 hold, where t.he Hylnbol UII'('1") ,'opllt(1(1A u('l") wherever it appears; for example, (ii) now rUlLdl'l I'ul/('I') C UUII'('1") U l'ulJ'('1").

72

UNBOUNDED LINEAR OPERATORS

11.6

CHARACTERIZATIONS OF STATES OF OPERATORS

Motivated by certain important concepts in the study of Banach algebras, we show that the states of closed linear operators can be characterized in terms of these concepts. II.6.1

Definition ~.

n.

iii.

T is called left (right) regular provided there exists an operator A E [V, X] such that AT = I, (TA = I), where I is the identity map on ~(T) (Y). The set of all left (n'ght) regular elements will be denoted by R z (R r ). T is called a left (right) divisor oj zero if there exists an A ~ 0 in [V, X] such that TA = 0 (AT = 0) on Y (~(T». The set of all left (right) divisors of zero is denoted by D z (Dr). T is called a leJt (right) topological divisor oj zero if there exists a sequence (An} in [V, X] such that IIAnl1 = 1 and TAn -+ 0 (AnT -+ 0) in [V] ([~(T), Xl). The set of all left (right) topological d1~visors of zero is denoted by Zz (Zr).

For convenience, we shall use notations such as III = Dr to mean that T is in state III if and only if T E Dr and 1 = ezz to mean that TEl if and only if T is in the complement of the set Zzj that is, if T is not a left topological divisor of zero. II. 6.2

i. ii. ~n.

iv.

Theorem

III = Dr 1 = ~.zz 3 = Dz 2

= Zz (\ eD z

Proof of (i). If T E III, there exists soine y' E Y' such that y' ~ 0 and y'CR(T) = O. Define A E [V, X] by Ay = y'(y)xo, where Xo ~ 0 is fixed in ~(T). Then A ~ 0 and ATx = 0 for all x E ~(T). Thus T E Dr. Conversely, suppose T E Dr. Let A ~ 0 be in [V, X] and let AT = O. Then CR(T) C ;n(A) ~ Y, whence T E III. Proof oj (ii). Suppose TEl and suppose there exists a sequence (An} in [V, X] such that An ~ 0 and TAn -+ 0 in [V]. Now there exists an m > 0 such that for each y in the l-sphere of Y,

Hence An -+ 0 in [V, Xl and therefore T ~ Zz. On the other hand, suppose ']' ~ 1. Then there oxists I1soquencc I x" I in ~(']') BliCh thnt Ilxnll = 1 and 'l'x n -+ O. Fix 1/ e Y' with Ilu'll - 1. 1+'01' OIwh n,
73

LINEAR OPERATORS AND THEIR CONJUGATES

by AnY = y'(y)xn. Then IIAnl1 = 1 and IITA nll ~ II Txnll ~ O. Hence T EZ!. Proof of (iii). Suppose there exists and x r5- 0 such that Tx = O. Define A E [V, X] by Ay = y'(y)x, where y' is some nonzero element in V'. Then A r5- 0 but T A = O. Hence TED!. If TED!, there exists an A r5- 0 E [V, X] such that T A = O. Consequently, (0) r5- CR(A) C m:(T). (iv) is a trivial consequence of (ii) and (iii). II. 6.3 Theorem. Then I = eZr • Proof. {An} in [V,

Suppose X and Yare complete and T is closed.

Suppose TEl and also in Zr. {Ynl E Y such that

Then there exist sequences

Xl and

IIAnl1 = 1

1 IIAnYnll ~ 1 - n

IIYnl1 = 1

in

[~(T),

X]

Let T be the 1-1 operator induced by T. The proof that IIA nT/I = /IAn Til is the same as the proof that /IT/I = /lTII in Lemma IIA.7. Moreover, T is closed. Since TEl, there exists a sequence {x n} such that TX n = Yn. The sequence ([xnJl in ~(T)/;n(T) is unbounded since /lAnTII ~ 0 and

Thus, for Zn = [x nJ/II[xn] II,

II Tznll =

ll~:j,\'

=

1I1!~:JIII

=\II[;n]1I ~ 0

This implies that l' does not have a bounded inverse, which contradicts Theorem II.1.9. Hence TEl implies T E eZr • Conversely, suppose T ¢ Zr. To show TEl, it suffices to show that T' E 1 by 11.4.11. Assume T' does not have a bounded inverse. Then there exists a sequence {y:1 E ~(T') such that Ily:/I = 1 and T'y~ -+ O. Let Xo be an element of norm 1 in ~(T). Define An E [V, Xl by

Then IIAnl1 = 1, and for all x in the I-sphere of ~(T),

Boueo

IIA" 7'/1 --. 0,

whieh

(lolltrndic~tll

tho Iluppositioll 7' ¢ Zr.

74

UNBOUNDED LINEAR OPERATORS

II. 6.4 Theorem. If X and Yare complete and T is closed, then the following statements are equivalent.

TEll T ¢ZIV Zr

't.

n. iii.

T E Rl

(\

Rr

Proof. (i) is equivalent to (ii) by Theorems 11.6.2 and 11.6.3. We show that (i) is equivalent to (iii). If TEll, then T-1 E [Y, X], TT-1 = I on Y, and T-1T = I on XJ(T). Thus, by definition, T E R l ( \ R r. If T E R l ( \ Rr, then clearly T is 1-1 and CR(T) = Y. Hence from 11.4.11, TEll.

Combining Theorems 11.6.2 to 11.6.4 we obtain the following result. II.6.5 Theorem. Suppose X and Yare complete. Then the states for a closed linear operator with domain dense in X and range contained in Y are characterized by 't.

n. n't.

iv. v. vi. vii. viii. ix.

II.7

11 = R l (\ R r = e(Zr V Zl) 12 cannot exist I a = D z ( \ eZ r 111 cannot exist 11 2 = Zr (\ Zz (\ e(D r V D l )

lI a = Zr (\ D! (\ eDr 111 1 = Dr (\ eZI 111 2 = Dr (\ Zz (\ eD! IlIa = Dr (\ D z

ADJOINT OF AN OPERATOR IN HILBERT SPACE

The Riesz representation theorem -shows how a Hilbert space can be identified with its conjugate. Consequently, if X and Y are Hilbert spaces, then T' can be identified with an operator mapping a subspace of Y into X as follows. II.7.1 Definition. Let X and Y be Hilbert spaces. Let Ex and E y be the isometries from X onto X' and Y onto Y', respectively, as in Definition 1.7.19. The adjoint of T, written T*, is defined by T* = E X -1T'E y . II.7.2

i.

Remarks The usual definition of T*, which is equivalent to the one above, is: ~(T*) iF! the subspace of Y consist,illll; of thoso 11 for which fv(:1:) = ('l';r, 7/) iH (;ollt.inIlO\lR Oil :0('1'). For' Hllnh 7/lni. '1'*11 = z, whorl' .l'u(:r) - (x, z) for' nil :1' ~ :1)('1'). '1'1\(11'1'1'01'(\ \.I1lI o(lOl'lLl.or

LINEAR OPERATORS AND THEIR CONJUGATES

75

T* is characterized by the equation (Tx, y) = (x, z) = (x, T*y) ii.

iii.

From the definitions of Ex and E y , it is clear that T* is T' when the spaces are identified with their conjugates. In particular, the state diagrams for (T, T') and (T, T*) are the same. If X = Y and T = T*, then T is called self-adfoint. Thus, a self-adjoint operator is closed, and Diagram II.4.11 shows that II, II 2 and 1II 3 are its only possible states. For a treatment of self-adjoint ordinary differential operators, the reader is referred to Neumark [ll and Dunford and Schwartz [1], part II.

(

chapter III

STRICTLV SINGULAR OPERATORS

The concept of a strictly singular operator was introduced by Kato [1] in his treatment of perturbation theory. It was shown that certain properties of T are shared by operators of the form T + B, where B is strictly singular. In this chapter, properties of strictly singular operators and, in particular, compact operators are obtained, the results of which are used in Chap. V. Throughout this chapter, X and Y denote normed linear spaces. Completeness is assumed only when specifically stated. III.1

COMPACT OPERATORS

111.1.1 Definition. Let B be a bounded linear operator with domain in X and range in Y. B is called strictly singular if it does not have a bounded inverse on any infinite-dimensional subspace contained in its domain. If B is defined on all of X, it suffices to consider only closed infinitedimensional subspaces. The most important examples of strictly singular operators are the compact operators. These playa major role in the study of differential and integral oquations. In eases whoro the normod lineal' SpfWOS ILro not Ilssuml'd (~omJlhl\'l', it. is oonvuniollt. t.o (\oIlHidOl' ,LlHO pr'Il!:OlllplL<:L OpOI'Il,(,orH.

77

STRICTLY SINGULAR OPERATORS

IIl.l.2 Definition. Let K be a linear operator with domain in X and range in Y. If KS is totally bounded in Y, where S is the I-ball in X, then K is called precompact. If KS is compact in Y, then K is called compact. IlI.l.3

Theorem.

Every precompact operator is strictly singular.

Proof. Let B be a precompact operator with domain in X and range in Y. B is bounded since a totally bounded set is bounded. Suppose B has a bounded inverse on a subspace M C 5)(B). Then BS M is totally bounded, where SM is the I-ball in M. Since B has a bounded inverse on M, it follows that SM is totally bounded in M. Hence M is finite-dimensional by Theorem 1.4.6.

Examples of strictly singular operators which are not compact are given in Sec. II1.3. IlI.l.4

i.

ii.

'm.

Remarks

Every bounded linear operator with finite-dimensional range is compact. For if K: X ~ Y is such an operator, then letting S be the I-ball of X, KS is a bounded set in the finite-dimensional subspace CR(K) of Y. The compactness of KS is therefore a consequence of Corollary 1.4.3. An operator is compact if and only if it takes every bounded sequence into a sequence which has a convergent subsequence. An operator is precompact if and only if it takes every bounded sequence into a sequence which has a Cauchy subsequence. If K is precompact as a mapping from X into Y and 9" is the completion of Y, then K, when considered as a map into 9", is compact by Theorem 0.3. Thus K ~ [X, Yj is compact if and only if it is precompact, provided Y is complete.

III.l.S Lemma. Suppose /K N } is a sequence of precompact operators in [X, Yj. If K N ~ K in [X, V], then K is precompact. Proof. Let e N such that

> 0 be given.

By hypothesis, there exists an integer

(1)

Since K N is precompact, there exist elements Xl, X2, .'1:; for which

S of X such that given X ~ S, t.hcrn is nn

(2)

., Xm

in the I-ball

78

UNBOUNDED LINEAR OPERATORS

Hence, from: (1) and (2), IIKx - KXil1 ::; IIKx - KNxl1

+ IIKNx -

KNxi/1

+ IIKNXi

-

sse

KXil1 < "3 + "3 + "3

which implies that K is precompact. As an application of the lemma, the following example of a compact operator is given. 1I1.1.6 Example. Let I = [0, 1] and let 1 < p, q < 00, with p' and q' conjugate to p and q, respectively. Suppose k(s, t) is in £r(l X I), where r is the larger of p' and q. Then the linear operator K defined by (Kx)(t) =

10

1

k(s, t)x(s) ds

is compact as a map from £,,(1) into £q(1). Proof. To show, first of all, that K maps £,,(1) into £q(I), suppose x is in £,,(I). Since r' ::; p, we have, hy Holder's inequality, that x is in £r,(I), Ilxll r ' ::; /Ixll", and (1)

IIKxll qq = fa1 dt I fa1 k(s, t)x(s) dslq ::; fa1 dt [( fa1 Ik(s, t)lr ds )q1rllxllr,q] ::; /Ixll"q fa1dt (fa1 I(k, t)/r ds)q1r

Since 0

<

q/r ::; 1, Holder's inequality implies that for any g E£1(I),

(2)

Taking get) = fa1 Ik(s, t)i' ds, we obtain from (1) and (2) that (3)

Thus K is continuous as a map from £,,(I) into £q(I). The operator K will next be shown to be compact under the additional assumption that k is continuous on I X I. Let Ix,,} be a bounded sequence in £,,(1). Then for y.. = Kx.., we have from the inequality I/x"lh ::; Ilx"l/" that 17I"(t) I

~

ill

1":(8, t)X.(H)I dR ~ 1l~~~~1 IIc(R, t)lllx"l/p

STRICTLY SINGULAR OPERATORS

79

Hence {Yn} is uniformly bounded on I. Since k is uniformly continuous on I X I, there exists for each c > 0 a fJ = fJ(c) > 0 such that for all SE

I,

For such t 1 and t 2 ,

Since {x n } is bounded in "cp(I) , (4) shows that {Yn} is equicontinuous on I. Hence, by the Ascoli-Arzela theorem, {Yn} contains a subsequence converging in C([O, 1]) which in turn implies convergence in"cp([O, 1]). Thus K is compact. Removing the restriction that k be continuous, we nevertheless know by Theorem 0.9 that there exists a sequence {k n } of functions continuous on [0, 1] X [O,l]whichconvergesin"c,(I X l)tok. LetKnd"cp(I),,,cq(I)] be defined by

Then by what was just proved, K n is compact. k in (3) gives, for each x E "cp(I) ,

II(Kn - K)xllq ::; I[xll p (

10[l 10(l [kn(s, t)

Thus K n ~ K in ["cp(I) , "cq(I)]. III. 1.5.

Substituting k n

k(s, t)I'dsdt

-

)1

1,

~0

-

k for

as n -

eX)

Consequently, K is compact by Lemma

In general, if Y is not complete, the limit in [X, Y] of compact operators need not be compact. This may be seen in the following example. IIl.I.7 Example. Let Co be the subspace of l", consisting of the sequences which converge to O. Let To be the operator mapping Co into l2 defined by

T o( (

ak})

=

{~k}

1 ::; k

Define Tto be the operator To considered as amapfromcoonto Y =
-k

Ilxkll = 1 in Co, but TXk ~ Y in l2, where y = {I, ,!, j-, . . . j. Now, y is not in Y, Rince 1'x = 11 implies x = /1,),1, . . . j, which is not in Co. TTol1c'o 1'l':I'k! hilt:! 110 Huhscquol1no whi<:h(r,onvcrg;cs in Y; that is, T is not c'omplll'l, lIoWI'VnI', Ull' !'4l1C!II11I1(\(! or oplll'/l,I,orR 'I'n :rll _.. ~ Y, ddinod hy

80

UNBOUNDED LINEAR OPERATORS

Tn ({ak\) = {lSd,wherelSk = ak/kfor1 ~ k ~ n and ISk = O,k > n,converges in [co, Y] to T. Moreover, (R ( Tn) is finite-dimensional and therefore Tn is compact. Corollary III.1.l0 shows how compact operators may arise from closed operators which do not have a closed range. The following theorem, excluding the assertion that the operator is precompact, is due to Kato [1]. III.I.B Definition. A subspace M of a vector space V is said to have finite deficiency in V if the dimension of V / M is finite. This is written . V d1m M

<

00

Even though M is not contained in 'J)(T), a restriction of T to M will mean a restriction of T to M (\ 'J) (T) . III.I.9 Theorem. Let T be a linear map from a subspace of X into Y. Suppose that T does not have a bounded inverse when restricted to any closed subspace having finite deficiency in X. Then given e > 0, there exists an infinite-dimensional subspace M(e) contained in 'J)(T) such that T restricted to M(e) is precompact and has norm not exceeding e. Proof. The hypothesis implies the existence of an Xl E X such that 1 and IITxll1 < 3- le. There is an x~ E X' such that Ilx~11 = 1 and X~Xl = Ilxlll = 1. Since ;n(x~) has deficiency 1 in X, there exists an X2 E ;n(x~) such that IIx211 = 1 and IITx211 < 3- 2 e. There exists an x; E X' such that Ilx;11 = 1 and X;X2 = IIx211 = 1. Since ;n(x~) (\ ;n(x;) has finite deficiency in X, there exists an Xa E ;n(x~) (\ ;n(x;) such that Ilxall = 1 and IITxal1 < 3- ae. Inductively, sequences {xd and {x~} are constructed in X and X', respectively, with the following properties.

IIxIII =

(1) k-l

(2)

Xk E ( \

;n(x~)

or, equivalently, X~Xk = 0

1

~ i


i=l

It is easy to verify that the set of Xk is linearly independent, whence M = sp IXl, X2, • • • } is an infinite-dimensional subspace of 'J)(T). It will now be shown that the restriction T M of T to M has norm not exceedm

ing e.

Suppose

X

=

L

,-I

aiXi.

Then from (1) and (2),

STRICTLY SINGULAR OPERATORS

81

In fact,

l::::;k::::;m

(3)

< m. Then from

For suppose (3) is true for k ::::; j

(1) and (2),

i

X;+l X =

(4)

L aiX:+1 (Xi) + ai+1 ,=1

Hence, by (4) and the induction hypothesis, i

L lad IX:+lxil

la;+ll ::::; Ix:+ 1xl +

::::;

IIxll +

i=1

i

L 2i-lll xll

=

2i /lxlI

,=1

Thus (3) follows by induction and m

IITxl1 ::::;

L 10:;1 II Txil1

m

::::;

,=1

L2

i

-

3-i Ellxll

1

::::;

EllxU

i=l

Hence IIT M II ::::; E. To prove that T M is precompact, it suffices to show that T M is the limit in [M, Y] of a sequence of precompact operators. For each positive integern, define TnM:M - ? Y to be Ton sp (Xl, X2, . . . , xn } and 0 on sp (X n+l, Xn+2, • • .}. Clearly TnM is linear and has finiten+k dimensional range. Moreover, TnM is bounded on M; for if X = a,Xi,

L

,=1

then by (1) and (3), n

II TnMx11

::::;

L \a,IIITxill ,=1

Thus TnM is precompact by III.1.4.

n

::::;

L 2i- 13-iEllxll ,=1

Since

00

IITMx - TnMxll::::;

L

10:;1 II TXili

i=n+1 00

::::; Ellxll

L

2i-l3-i

-?

0

as n

-?

00

,=n+l

it follows that TnM converges to T M in [M, Y], completing the proof of the theorem. Ill.l.LO Corollary. Let X and Y be complete. If T is closed but 0 there exists an infinite-dimensional closed Rubspaw M(e) contained in 5)(1') such lI{al T restricted to M(e) is compact with norm not ('xrr"',f1'n(l e.

82

UNBOUNDED LINEAR OPERATORS

Proof. Let W be a closed subspace having finite deficiency in X. Assume T has a bounded inverse on W. Then TW is closed; for if TX n ~ y, Xn E W, then (xnl is a Cauchy sequence and therefore converges to some x in Banach space W. Since T is closed, x is in ~(T) and Tx = y. By hypothesis, there exists a finite-dimensional subspace N of X such that X = W + N. Hence TX = TW + TN. Since TW is closed and TN is finite-dimensional, TX is also closed by Theorem 1.4.12. But this contradicts the hypothesis that
It is shown in Example IIL3.7 that the conjugate of a strictly singular operator is not necessarily strictly singular and conversely, a bounded operator which has a strictly singular conjugate need not be strictly singular. This lack of duality, fortunately, does not carryover to precompact operators, as the next theorem shows. III.I.II Theorem. A bounded linear operator is precompact if and only if its conjugate is compact. Proof. Let K be a precompact linear operator which maps X into Y. We shall prove that K' is precompact and therefore compact since X' is complete. Given e > 0, there exist Xl. X2, • • • , X n in the I-ball Sx of X such that for each x E Sx there is an Xi for which (1)

IIKx - KXil1

e

< "3

Let A be the map from Y' to unitary n-space given by Ay'

=

(y' KXI' . . . ,y'Kx n)

Since A is clearly bounded and linear, we know by Remark (i) of IILIA that A is compact. Hence there exist y~, . . . , y~ in the I-ball Sy, of Y' such that for each y' E SY', there is a y; for which IIAy' - Ayjll

<;

In particular, 1 ~

i $

11

STRICTLY SINGULAR OPERATORS

83

From (1) and (2),

- yjKx;! + lyiKx; - YiKxl z ::;; IIKx - Kx;11 + 3 + I/Kx; - Kxll < z

IK'y'x - K'yjxl ::;; ly'Kx - y'Kx;1

+ ly'Kx;

Thus I/K'y' - K'y;11 ~ e whence K' is precompact. Conversely, suppose K' is compact. Then by what has just been shown, K" is compact. Now, K"J x = JyK, where J x and Jy are the natural maps from X into X" and Y into ylI, respectively. Since J x is bounded, it is clear that K"J x is compact and, in particular, J yK is precompact. Since J y has a bounded inverse, it is easy to see that K is precompact. If the conjugate of a bounded linear operator is compact, it does not necessarily follow that the operator is compact. Indeed, in Example III. 1.7, Tn -> T in [co, Y]. Thus T~ -> T' in [Y', c~]. Since Tn is compact, so is T~. Hence T' is precompact and therefore compact, since Y' is complete. However, T is not compact.

111.1.12 Corollary. If the range of a precompact operator is complete, then the range is finite-dimensional. Proof. Suppose B E [X, Y] is precompact and its range Y l is complete. Then the operator B l , considered as the map B from X onto Y l , is precompact. Furthermore, B~ has a bounded inverse by Theorem II.3.I3. Since B~ is compact, Theorem III.1.3 asserts that Y~ is finitedimensional. Thus Y l is finite-dimensional. 111.1.13 Remark. Suppose B is a precompact operator in [X, Y]. Since a totally bounded set in a metric space is separable, BS and therefore ., Cll(B) = U nBS is separable, where S is the I-ball in X. Thus, in light n=l

of Corollary III.l.I2, CR(B) is no larger than a proper subspace dense in Y, provided Y is an infinite-dimensional separable Banach space and X is infinite-dimensional. Goldberg and Kruse [3] show that for such X and Y, there always exists a compact operator in [X, Y] whose range is dense in Y. We have also seen in Theorem III.1.3 that B cannot have a bounded inverse on X whenever X is infinite-dimensional. However, in the paper just cited, it is shown that there exists a 1-1 compact operator in [X, Y] if and only if X' contains a countable total subset. 111.2

RELATIONSHIP BETWEEN STRICTLY SINGULAR AND COMPACT OPERATORS

Thoro iA It cloAo connection hotwrM t,ho Cfn,AS of strictly singular operators lind "Ito I'IIIHH or PI'l'('"I11Plld npomf,ori'l. FoJ' l'xlI.mplo, Kn,t,o [n, pn.Jl:o 2RIl,

UNBOUNDED LINEAR OPERATORS

84

has shown that these two classes coincide in the space of bounded operators mapping a Hilbert space into a Hilbert space. Feldman, Gokhberg, and Marcus [1] prove that the only nontrivial closed two-sided ideal in [Xl, where X = lp, 1 :::; p < 00, or X = Co, is the space xlX] of compact operators. Since it will be shown that the strictly singular operators S[X] constitute a closed two-sided ideal in [Xl, it follows that S[X] = x[X]. The following two theorems give insight into the connection between the two classes of operators. III.2.1 Theorem. are equivalent. '/,. ii.

m.

Suppose B

E

[X, Y].

The following three statements

B is strictly singular. For every infinite-dimensional subspace M C X, there exists an infinite-dimensional subspace N C M such that B is precompact onN. Given e > 0 and given M an '/,'nfinite-dimensional subspace of X, there exists an infinite-dimensional subspace N C M such that B restricted to N has norm not exceeding e.

Proof. (i) implies (ii). Suppose B is strictly singular and M is an infinite-dimensional subspace of X. Then B M , the restriction of B to M, is strictly singular. Hence the existence of an N, with the properties asserted in (ii), is a consequence of Theorem III.1.9 applied to B M • (ii) implies (iii). If M is an infinite-dimensional subspace of X, then B does not have a bounded inverse on any subspace having finite deficiency in M; otherwise, B would be precompact and have a bounded inverse on an infinite-dimensional subspace. But this contradicts Theorem III.1.3. Hence (iii) follows by applying Theorem III.1.9 to B M . (i) follows easily from (iii).

The next theorem, due to Lacey [1], implies that if one adds to the above theorem that N has finite deficiency in M, then a characterization of the precompact operators is obtained. lII.2.2 Lemma. If S is a bounded set in X and x~, x~, , x: are in X', then for any e > 0, there exist Xl, X2, . . . ,Xm in S such that given XES, one can find an Xk such that

Proof.

The map A from X to unitnry n-Apo.co definod by A:I' -

(.r;;r, :r~,r, ' , , ,.r;,:r)

85

STRICTLY SINGULAR OPERATORS

is bounded and linear. Hence A is compact. Thus there exist , Xm in S such that given XES, there is an Xk such that

Xl,

X2,

lII.2.3 Theorem. Suppose B E [X, Y]. Then B is precompac~ if and only if for every e > 0 there exists a subspace N having finite deficiency in X such that B restricted to N has norm not exceeding e. Proof. Let B be precompact and let e > 0 be given. There exist X2, . . . ,Xn in the i-ball S of X such that given XES, there is an Xk such that Xl,

(1)

Let y~,

, y: be in the i-sphere of Y' such that y:BXi = flBxill.

Now

n

N = (\ ;n(y:B) has finite deficiency in X.

Suppose X E N (\ Sand Xk

;=1

Since X is in ;n(y~B),

is such that (1) holds.

Thus, from (1) and (2), (3)

Since X was arbitrary in N (\ S, (3) implies that B restricted to N has norm not exceeding e. Conversely, let e > 0 be given. By hypothesis, there exists an N of finite deficiency in X such that B N , the restriction of B to N, has norm not exceeding e. Since IIBNII = IIBNII, we may assume that N is closed. Hence, there exist VI, V2, . • • ,Vn in X such that

X

= N

E9 sp

lVI, . . .

,vnl

From the construction of the projection in Theorem II.1.16, it follows that there exist x~, x~, ... ,x: in X' such that every x E X has a unique representation of the form n

(4)

:r

= 'II -I-

_

L X;(X)li; ,-I

86

Since

UNBOUNDED LINEAR OPERATORS

IIBNII

~

€, we have IIBxl1

(5)

~

€llu\l +

~

Ilxll +

n

L Ix:(x)IIIBvi\l

;=1

By (4),

\lull

(6)

n

L Ix:(x) I Ilv,ll

;=1

Combining (5) and (6) and choosing K = max it follows that

IIBxl1

(7)

~

€llxll + €K

n

{IIBv,ll, \lv,ll; 1

~ i ~

n},

n

L Ix:(x)! + K .=1 L Ix~(x)1 .=1

XeX

Given 17 > 0, there exist, by Lemma III.2.2, elements Xl, S such that given v e S, there is an Xk such that

,xm in

X2,

n

L Ix~v -

x:xkl < 17

.=1

Thus, substituting obtain

Since

€

>

0 and 17

v-

>

Xk

for

X

in (7) and noting that

IIv -

Xkl! ~

2, we

0 are arbitrary, it follows that BS is totally bounded.

III.2.4 Theorem. The set SIX, Y] of strictly singular operators and the set of precompaet operators (PX[X,. Y] in [X, Y] are closed subspaces. The set of compact operators xIX, Y] is a subspace (not necessarily closed). Proof. It is obvious that the sets are closed under scalar multi~ plication. Suppose K and L are in xIX, V]. Let {x n } be a bounded sequence in X. Since K is compact, there exists a subsequence {Yn} of {x n } such that {KYn} converges. Since L is compact, there exists a subsequence {Zn} of {Yn} such that {Lz n } converges. Hence {(K + L)z,,} converges, which shows that K + L is compact. Suppose K and L are in (pxlX, Y]. Then K and L are in xIX, f], where f is the completion of Y. Hence, by the above result, K + L is in xlX, f], or, equivalently, K + L is in (pxIX, Y]. Rupposc K find L are in SIX, Y]. Let. M ho ILIl illfinito-climollHionnl !:luhelllWe of X. Them, hy Th(1orom III.2.1, t,horo OXiHI,H nil infiniLo-dilll(\ll~

STRlcnv SINGULAR OPERATORS

87

sional subspace N 1 C M such that K is precompact on N 1 and there exists an infinite-dimensional subspace N C N 1 such that L is precompact on N. Since K is also precompact on N, K + L is precompact on N by what we have just shown. Hence K + L is strictly singular by Theorem 111.2.1. (PX[X, Y) was shown to be closed in Lemma 111.1.5. Let K n ~ K in [X, Y), where each K n is strictly singular. Suppose K has a bounded inverse on subspace M C X. Then there exists a c > 0 such that for all meM,

IIKml1 :2 cllmll

(1)

Choose p so that (2)

Then from (1) and (2),

Thus K p has a bounded inverse on M. Since K p is strictly singular, M is finite-dimensional, whence K is strictly singular. If Y is complete, then the space of precompact operators coincides with the space of compact operators and therefore X[X, Y) is closed. Example 111.1.7 showed that if Y is not required to be complete, then X[X, Y) need not be closed. lII.2.5 Theorem. Let Z be a normed linear space. If K is in X[X, YJ, (PX[X, Y), or S[X, Y), then for A e [Z, X), KA is in X[Z, Y), (PX[Z, Yj, or S[Z, Y), respectively. Similarly, if B is in [V, Z), then BK is in X[X, Z], (PX[X, Z), or S[X, Z), respectively. Pro~f. We shall only prove the theorem for K e S[X, Y). The proofs for K in X[X, Y) or (PX[X, Y) are also easy. Suppose KA has a bounded inverse on a subspace M of Z. Then there exists a c > 0 such that for all x e M

IIKAxl1

:2 clJxll :2 II~ II IIAxll

Thus K has 11 bounded inverso on AM,jl.nd therefore AM is finite-dimenBut A iH 1-1 liIirwo [(A is 1-( Honce ll! is finite-dimensional. COlll:loqllontly, K A is strictly sinll:ullLl'.

MioJla!.

UNBOUNDED LINEAR OPERATORS

88

Suppose BK has a bounded inverse on a subspace N of X. there exists a c > 0 such that for all x E N, IIBllllKxll ~

IIBKxl1

~

Then

cllxll

Thus K has a bounded inverse on N, whence N is finite-dimensional. Therefore BK is strictly singular. III.3

EXAMPLES OF NONCOMPACT, STRICTLY SINGULAR OPERATORS

For the purpose of exhibiting some strictly singular operators, we prove Theorem IlL3.4 and Corollary IlL3,6, which are due to Whitley [1]. Ill.3.1 Definition. A linear operator which takes bounded sequences onto sequences which have a weakly convergent subsequence is called weakly compact.

III.3.2 Remark. Every weakly compact operator is bounded. For suppose B is a weakly compact operator which is unbounded. Then there exists a sequence Ix n } such that I Bx n } converges weakly and II BXn II ---t 00. But this contradicts Theorem Il.1.12. III. 3.3 Theorem. If at least X or Y is reflexive, then every operator in [X, Y] is weakly compact. Proof. The proof follows easily from Theorem 1.6.15 and the continuity of the operator. III.3.4 Theorem. A weakly compact operator on a Banach space which maps weakly convergent sequences onto norm convergent sequencies is strictly singular. Proof. Let BE [X, Y] satisfy the hypotheses of the theorem. Suppose B has a bounded inverse on a closed subspace M eX. Let lmd be a sequence in the I-ball of M. Then there exists a subsequence IVk} of lmd such that lBvd converges weakly to some y E Y. Since BM is closed, y is in BM; otherwise there exists a y' E Y' such that y'y = 1 and y'(BM) = 0, implying that lBvk} does not converge weakly to y. From the assumption that B has a bounded inverse on M, it follows that IVk} converges weakly to BM-ly. From the hypothesis, lBvd converges in norm to y and therefore lvd converges in norm to BM-ly. Hence the I-ball of M is compact and M is finite-dimensional. Therefore B is strictly singular.

The proof of the next theorem requiroR quite fl, bit of intep;mtion theory and will 1I0t ho ilwludod 111'1'0, Thl' rlmdor iH l'efer'l'OlI to J)ulll'Ol'd

STRICTLY SINGULAR OPERATORS

89

and Schwartz [1], Theorem VI.8.12, the remark preceding the theorem on page 508, the remark following Theorem V1.8.14 on pages 510-511, and the representation of the conjugate spaces in tables on pages 374-379. We cite only a few spaces X for which the following theorem holds. 111.3.5 Theorem. Every weakly compact map from X or X' into a Banach space takes weakly convergent sequences onto norm convergent sequences whenever X is anyone of the following Banach spaces. t.

ii. tn.

£1(8,~, fJ.)

and £00(8, ~, fJ.), where (8, ~, fJ.) is a positive measure space. C(S), where 8 is a compact Hausdorff space. B(S), where 8 is an arbitrary set.

Thus every weakly compact operator from X or X' into a Banach space is strictly singular.

Pelczynski [2] proved that the strictly singular operators and the weakly compact operators are the same in the following spaces: t.

ii.

[C(8), Y], Y an arbitrary Banach space and 8 a compact Hausdorff space. [£1(8,~, fJ.1), £1(8, ~, fJ.2)], 8 a topological space, ~ the field of all Borel subsets of 8, and fJ.1 and fJ.2 nontrivial measures (for details of these spaces, see Dunford and Schwartz [1], Chap. IV).

111.3.6 Corollary. Let X be anyone of the spaces in the above theorem. Then every bounded linear map from X or x' into a reflexive space is strictly singular. Proof.

Theorems III.3.3 and III.3.5.

The rest of this section gives examples of strictly singular operators which are not compact. 111.3.7 Example. Let X = l1 and let Y be any infinite-dimensional separable reflexive Banach space. By Corollary 11.4.5, there exists a bounded linear operator B which maps X onto Y. Hence B is strictly singular but not compact by Corollaries II1.3.6 and III.1.12. Without referring to Corollary III.3.6, B can be shown to be strictly singular based on the fact that weak convergence is the same as norm convergence in l1, as remarked in 1.6.14. B' is not strictly singular, since it has a bounded inverse by Theorem IIAA. Now B" maps X" into reflexive space Y", and X" may be identified with the conjugate of loo. Thus, by Corollary III.3.fi, B" is strictly singular. This example shows, in particular, that t,he eonjup;nte of 11 strictly singular operator need not be strictly singular lLll
90

UNBOUNDED LINEAR OPERATORS

The following theorem, which we do not prove, is needed for the next two examples of integral operators which are strictly singular but not compact. The proof of the theorem appears in Dunford and Schwartz [1], Corollary IV.8.U, and in Zaanen [1], pages 324-325. III. 3. 8 JL(8) <

Theorem. Let (8,~, JL) be a positive measure space with A sequence {Yn} in £1(8, ~,JL) has a weakly convergent sub-

00.

sequence if and only if it is bounded in £1(8, zero uniformly for all Yn as JL(E) ~ 0.

~, JL) and JE Yn dJL converges to

III. 3.9 Corollary. Let k be bounded and measurable on [0, 1] X [0, 1]. Define K as a map from £1([0, 1]) into £p([O, 1]), 1 ~ P < 00, by

(Kf) (0 =

10

1

k(s, Of(s) ds

Then K is weakly compact. For f E £1([0, 1]) and E a measurable subset of [0, 1],

Proof.

[.I:: I(Kf)(O Ip dtr

(1)

lP

~ (JL(E»1Ip sup Ik(s, 0110 If(s)1 ds 1

05;8,t5;1

Taking E = [0, 1] and 1 < P < 00, (1) shows that K is bounded and therefore weakly compact by Theorem III.3.3. For p = 1, let {in} be a bounded sequence in £1([0, 1]). Then, by (1), {Kfn} is bounded in

JL(E)

~

0.

IE I(Kfn)(OI dt

converges to zero uniformly for all fn as It follows from Theorem III.3.8 that K is weakly compact.

£1([0, 1]) and

The next two examples exhibit integral operators which are strictly singular but not compact. III.3.10 Example. [0,1] by

k(s, t)

°

(s, t)

2

(8 t) ,

°

For n = 1, 2, . . . , let k be defined on [0, 1] X

E

1 1 ( 2n' 2n- 1

J X (22n'j 2j--zn+ 1 J' j

when

E

8

(1_, _1_ J X (~j -t}, = °or t = ° 2 n 2 n-

1

Cloarly k iH houndounnullloltHllflthlo.

2n

= 0, 1, . . .

,2 n-

1 -

1

1

1

3J'

2j + 2n j = 0, 1, . . . ,2 n -

-

Defillo tho oorrospolldinp; intogral

91

STRICTLY SINGULAR OPERATORS

operator K:£1([0, 1])

~ £p([O,

1]), 1

f

(Kf)(t) =

~ P

<

00,

by

k(s, Of(s) ds

Then K is strictly singular by Corollary III.3.9 and Theorem III.3.5, However, K is not compact. To see this, define sequence Un} by

S E

(in' 2:-

1 ].

n = 1,2,

otherwise Then IIfnl11 = 1 and a simple calculation shows that

where I lip is the norm on £p(I). sequence in £p(I).

Hence (Kfn} has no convergent sub-

III.3.11 Remarks. It is shown in Dunford and Schwartz [1], Theorem VI.4.8, that the conjugate of a weakly compact operator is weakly compact. Hence, taking the operator K in the above example as a map from £1([0, 1]) to £1([0, 1]), K' is also strictly singular by Theorem III.3.5. III.3.12 Example. Let k be' bounded and measurable on [0,1] X [0, 1J and let rp be in £1([ -1, 1]). It follows from Theorem 0.11 that the operator K defined by

(Kf)(t) =

101 k(s, Orp(t -

8)f(s) ds

is bounded as a map from £1(1) into £1(1), I = [0, 1]. To show that K is strictly singular, it suffices, by Theorem III.3.5, to show that K is weakly compact. Let E be a measurable subset of I. Then for f E £1(1) and M = sup Ik(s, t)l, Fubini's theorem gives 0:;;',1:;;1

(1)

k I(Kf)(OI dt ~ M k dt £Irp(t -

s)llf(s)1 =M

£(h_.!rp(O! dt) If(s)1 ds

Since rp is integrable on [-1, IJ, there exists a 8 tluch that whenever E has me.aure less than 8,

f/C " 1.p(l)1 elt < I

ds

He

> 0, depending on e > I

0,

92

UNBOUNDED LINEAR OPERATORS

For such E, (1) implies that

k I(Kf) (t)1 dt ::; Me

Ilfll ::;

1

Hence K is weakly compact by Theorem III.3.8. As a particular case of this example we see that for g bounded and measurable on [0, 1] X [0, 1], the operator K 1 defined by (Kd)(t)

= [l g(s, t) f(s) ds

O
Jo It - sl"

is strictly singular as a map from £1([0,1]) to £1([0,1]).

g(s, t) = k(s, t) It - sl"

Upon choosing

O
where k is given in Example 111.3.10, K 1 = K, which is not compact. Feldman, Gokhberg, and Marcus [1] show that the compact operators and the strictly singular operators on £'11([0, 1]), 1 ::; P < 00, P r£ 2, are not the same. For additional examples of noncompact, strictly singular operators, the reader is referred to Goldberg and Thorp ll] and Lacey and Whitley [1].

III.4

CONJUGATES OF STRICTLY SINGULAR OPERATORS

In this section we consider when the conjugate of a strictly singular operator is strictly singular and, conversely, when the strict singularity of the conjugate implies the strict singularity of the operator. The results given here are due to Whitley [1].

III.4.1 Definition. A normed linear space Y is called subprojective if for every closed infinite-dimensional subspace Wof Y, there exists a closed infinite-dimensional subspace W 1 of W and a projection from Y onto W 1. A Hilbert space is subprojective. The proof of the next theorem may be found in Pelczynski [1].

III.4.2 Theorem. The spaces lp, 1 ::; P and Co are subprojective.

<

00,

£'11([0, 1]), 2 ::; p

<

00,

lII.4.3 Theorem. Let X and Y be cmn]Jlele and lei Y be 8uuprojective. If the conJuaate of 7' t lX, YI?:II IIlr'idly lIinrrnla'l', fhe//. '/' ill IItrict.lll win(/ular

93

STRICTLY SINGULAR OPERATORS

Proof. Assume T is not strictly singular. Then there exists an infinite-dimensional closed subspace M of X such that the restriction T M of T to M has a bounded inverse. Since TM is a closed infinite-dimensional subspace of Y, the subprojectivity of Y implies the existence of an infinite-dimensional closed subspace Vof TM and a projection P from Y onto V = TN, where N = TM-IV eM. Since P is surjective, P' has a bounded inverse on the conjugate space (TN)' by Theorem II.3.13. We show that the strictly singular operator T' has a bounded inverse on the infinite-dimensional subspace P'«TN)') , which is a contradiction resulting from the assumption that T is not strictly singular. Suppose v'e(TN)' and x is in the I-sphere of N. Then (1)

II T'(P'v') II

~ IT'(P'v')xl

= Iv'PTxl = Iv'Txl

Since T M has a bounded inverse and N is contained in M, it follows that for Ilxll = 1 and x eN, (2)

As a consequence of (1) and (2), we get

v' e (TN)' Thus T' has a bounded inverse on P'«TN)'). lIlA. 4 Corollary. Suppose that X is reflexive and that X' is subprojective. If T e [X, Y] is strictly singular, then T' is also strictly singular. Proof. Since Til = J yTJ x-I, where J x and J yare the natural maps from X into X" and Y into yII, respectively, Til is strictly singular by Theorem III.2.5. Since T' is in [Y', X'] and X' is subprojective, T' is strictly singular. lIlA. 5 Corollary. Let T be in [X, Y]. If X is a Hilbert space and T is strictly singular, then so is T'. If Y is a Hilbert space and T' is strictly singular, then so is T. Proof. By Theorem 1.7.21, the conjugate of a Hilbert space is a Hilbert space. Since a Hilbert space is subprojective, the corollary follows from Corollary II1.4.4 and Theorem III.4.3.

chapter IV

OPERATORS WITH CLOSED RANGE

Theorem IV.1.2 provides the motivation for the study of closed operators with closed range. There are many important applications of the theorem. For example, given a differential operator T defined on some subspace of £p(!J), one is interested in determining the family of functions y E £q(!J) for which Tf = y has a solution. If it is known that T is closed and has a closed range (examples of such differential operators are given in Chaps. VI and VII), then the space of such y is the orthogonal complement of the solutions to the homogeneous equation T' g = O. This information is very useful since oftentimes 'Jt(T') is finite-dimensional. We shall also see the vital role that closed operators with closed range play in perturbation theory. As before, T is a linear operator with domain in normed linear space X (not necessarily dense in X) and range in normed linear space Y. The spaces are assumed complete and the operators are assumed closed only when specifically stated.

IV.1

MINIMUM MODULUS OF AN OPERATOR

lV.l.l Lemma. Let X and Y be complete and let T be closed. Then T haR n, II01Jndr,r! h,."r,TRr. 7} and onl1/ 'Ii 'I' iR 1-1 nnd haR (J cloRrr! rarl(/(~.

OPERATORS WITH CLOSED RANGE

95

Proof. Suppose l' has a bounded inverse. in the proof of Theorem 11.3.1 shows that CR(T) is suppose l' is 1-1 and CR(T) is closed. Then 1'-1 is a Banach space CR(T) into Banach space X. Hence the closed-graph theorem. IV.l.2

Theorem. domain dense in X.

i. ii. iii.

Replacing 1" by l' closed. Conversely, closed operator from 1'-1 is continuous by

Let X and Y be complete. Suppose l' is closed with Then the following statements are equivalent.

CR(T) is closed. CR(T')is closed. CR(T') is the orthogonal complement of ;neT); that is, CR(T') = ;n(T).1

iv.

CR(T) is the orthogonal complement of ;n(T'); that is, CR(T) = .1;n(T')

Proof. Theorem II.3.7 implies that (i) is equivalent to (iv). (i) implies (iii). Let T be the 1-1 operator induced by T. Then T has a bounded inverse by Lemma IV.I.l and therefore CR«T)') = (Xj;n(T»' by Theorem II.4.4. Now, by (ii) of Theorem 1.6.4, (Xj;n(T»' is equivalent to ;n(T).1 under the map V given by (Vz')x = z'[x]. Thus (1)

V(CR(T)')

= ;n(T).1

Since '.D(T') = '.D«T)') by Lemma II.4.7 and since (T)'y'[x] y' E '.D(T'), it follows from (1) and the definition of V that

T'y'x,

CR(T') = V(CR(T)') = ;n(T).1 (ii) is an immediate consequence of (iii). (ii) implies (i). Let Y 1 = CR(T) and let 1'1 be the operator l' considered as a map into Banach space Y 1. Obviously, 1'1 is closed. To prove (i), it need only be shown that CR(T 1) = Y 1 • By Theorem II.4.4, it sufficies to prove that T~ has a bounded inverse. Since CR(T 1) = Y 1, T~ has an inverse by Theorem II.3.7. Furthermore, CR(T~) is easily seen to be CR(T'). Hence, T~ has a bounded inverse by Lemma IV. I. I.

To each linear operator l' having a closed kernel we associate a number 'Y(T), which plays a vital role in subsequent sections. The

rnunher arises from the following conRiderations. SUppOAO T iA dOAod nnel X nnd Y f~ro-complcte. Let T be the 1-1 operntor illdlll'('(l hy '1'. Them, by Lommn TV.1.t, meT) = m(l') iA <'loAnel

96

UNBOUNDED LINEAR OPERATORS

if and only if l' has a bounded inverse or, equivalently,

o < .mf {II1'[X]111 lTIXTIr [x] ~ ~(1n; ), x ~ meT) }

I

. { d(x,IITxl1 = mf meT)) x ~ ~(T), x ~ meT) IV.1.3 Definition. Let meT) be closed. written 'Y(T), is defined by 'Y(T)

inf

=

x.:D(T)

where %

is defined to be

}

1

The minimum modulus of T,

II Txll d(x, meT))

00.

11'.1.4, Example. Let X = £p(I) and Y = £q(I), where I compact and 1 :::; p, q :::; 00. Define T as follows. ~(T)

=

[a,

bl

is

= {f I j
Tf = j
(Recall that an absolutely continuous function is differentiable almost everywhere.) meT) is the space of polynomials of degree at most n - 1. Successive integration by parts shows that givenf ~ ~(T), there exists a z ~ meT) such that (1)

Thus, for 1

< q :::;

(2)

IJ(t) - z(t)\:::; IIj
00,

Holder's inequality gives

IIj
a)n-I+l/q'

1) ![(n - 1)q'

+ 1]W

where II Ilq is the norm on £q(I) and q' is conjugate to q. (1) implies that (3)

For q = 1,

IJ(t) - z(t)\ < IIf
If WI1 1\lI;rl'l' 1.0 writ.l' 1/00 - 0,

-

00 0 ...

(n - 1)1

I, ILllci

00 /00

....

I, t.holl it, fol1owil

97

OPERATORS WITH CLOSED RANGE

from (2) and (3) that for 1 (4)

Ilf - zllv ~

~

p, q

~

00,

Ilf(n) Ilq(b - a)n-I+l/q'+l/p (n - l)![(n - l)q' + l]W[(n - l)p + p/q'

+ 1j1'P

Since z is in ;neT), we may conclude from (4) that 'Y(T) ~

(n -

l)![(n -

l)q'

+ l]1lq'[(n

(b _ a)n

-

l)p

+ 1 + p/q'j1/V

I+W+1/v

where 1 ~ p, q ~ 00,1/ 00 == 0, 00 0 == 1, and 00/00 == 1. In particular, if p = q = 2, then (T) 'Y

> -

(n - 1) !(2n - l)!(2n)! (b _ a)n

An estimate for 'Y(T) is given in Chap. VI for a rather large class of ordinary differential operators T. IV.I.S Example. Take X = Y = lv, 1 ~ P ~ 00. Let ! Ad be a bounded sequence of numbers and let T be defined on all of X by

For each scalar A we compute 'Y(T),.) , where '1\ = AI - T and I is the identity operator. Define E),. to be the set of integers k for which A ¥ Ak. Then 'Jt(T),.) is the set of elements of the form ((31, (32, . . .) E lv, where (3j is arbitrary if j ¢ E),. and (3k = 0 if k E E),.. For simplicity we compute 'Y(T),.) when X = ll. The computation for 1 < P ~ 00 is essentially the same. Let x = !17d be in X. Then for y = !(3d E ;n(T),.) ,

Since (3k can be chosen to be 17k for k ¢ E),., it follows that

II [x] I =

d(x, ;n(T),.»

=

L 117kl hE),.

Therefore,

(1)

98

UNBOUNDED LINEAR OPERATORS

which implies that "(T)..) ~ inf

(2)

IA - Akl

= m

k,EX

On the other hand, given e > 0, there exists some i E Ex such that IA - Ail < m e. Let Ui be the element in II all of whose terms are zero except the ith term which is 1. Then from (1) and (2),

+

Since e

>

0 was arbitrary,

inf

IA - Akl =

"((AI - T)

k,Ex

The next theorem was proved in the discussion preceding Definition IV.1.3.

IV.l.6 Theorem. Let X and Y be complete and let T be closed. T has a closed range if and only if "(T) > O.

Then

Theorem IV.1.6, together with the next theorem, shows that if X and Yare complete, then any two of the following three statements imply the other. T closed CR(T) closed

IV.l.7

Theorem.

If "(T)

>0

and CR(T) is closed, then T is closed.

Proof. LetT be the 1-1 operator induced by T. Then ('1')-1 is continuous with domain a closed subspace of Y. Thus ('1')-1 is closed, and therefore '1' is also closed. Hence T is closed by Lemma 11.4.7. IV.l.8 Theorem. Let 'JL(T) be closed and let 'J)(T) be dense in X. "(T) > 0, then "(T) = "(T') and T' has a closed range.

If

Proof. Let Y 1 = CR(T) and let T 1 be the operator T considered as a map onto Y 1• The 1-1 operator '1'1 is surjective and has a bounded inverse, since "(T 1) = "(T) > O. Hence, by the State diagram 11.3.14, ('1'1)' is also in 11' Before evaluating "(T'), we need the following observations.

i.

By Theorems 11.3.7 and 1.6.4, Y' /'JL(T') = Y' /CR(T)J. and Y' /ffi(T)J. is equivalent 1.0 Y: under the map [V'] - y~, whore 11;/ ifil thC1 TI1Al.ridion of 11'10 VI. III plI.l'lil'1I111r, 11111'111 ... 111I~IIl.

OPERATORS WITH CLOSED RANGE

ii. iii.

99

It is clear that II(T)'y'll = II(T1)'y~ll. Thus, for y' E D(T'), 11<'1\)'y~11 = IIT'y'll by Lemma 11.4.7. By Theorems 11.3.9 and 11.2.8, 11«1'1)')-111 = I (Te 1),11 = II Te 1/1.

From (i), (ii), and (iii) it follows that 1

II «1'1)')-111

Thus T' has a closed range since "{(T')

>

°

and T' is closed.

The above theorem does not hold if "{(T) is not required to be positive. Indeed, Diagram 11.3.14 shows that state (1 2 , 111 1 ) can exist. However, if T is assumed to be closed, then the following corollary due to Kato [1] is obtained. IV.l.9

Corollary.

If X and Yare complete and T is closed, then

"{(T) = "{(T').

Proof. By Theorems IV.1.6 and IV.1.2, the statements "{(T) = 0, CR(T) is not closed, CR(T') is not closed, and "{(T') = are all equivalent. If "{(T) > 0, then by the theorem just proved, "{(T) = "{(T').

°

The remaining portion of the section gives some conditions under which a closed operator has a closed range. IV.l.10 Theorem. Let X and Y be complete and let T be closed. A sufficient condition that T have a closed range is that T maps bounded closed sets onto closed sets. If ;n(T) is finite-dimensional, then the condition is also necessary. Proof. Suppose T maps bounded closed sets onto closed sets. If CR(T) were not closed, then "{(T) = 0, which implies the existence of a sequence {[xnll in XJ(T) j;n(T) such that (1)

II[xn]11

=

1

and

We show that the existence of such a sequence leads to a contradiction. Consequently, CR(T) has to be closed. Let {znl be a sequence such that Zn E [x n] and Ilznll ~ 2. Then TZ n = TX n -... O. If IZnl has no convergent subsequence, it is a closed bounded set. Thus, by hypothesis, I TZ n I is a closed set, whence TZ N = 0 for !:lome N. nut tlwn III;I'NIII = II[ZN]/I--= 0, which contradicts (1). If .:,,' -. t for !'IOm(1 HII1IHII(\I\(1I1I'1' Iz", I, 1.111'11 'I'z ~ 0, HiIH'1' 'I'z", ) 0 al1d '/' iii

100

UNBOUNDED LINEAR OPERATORS

closed.

Hence

which again contradicts (1). Suppose ;)"[(1') is finite-dimensional and (1') and '1', the 1-1 operator induced by 1', has a bounded inverse. Thus

Consequently, there exists a sequence {Zn} in ;)"[(1') such that Xn + Zn ~ x. Since {x n } is a sequence in the bounded set S, {Zn} is also bounded. The finite dimensionality of ;)"[(1') assures the existence of a subsequence {Zn'} which converges to an element Z E ;)"[(1'). Hence X n ' ~ x - z. Since S is closed and Tis a closed operator, it follows that x - Z is in 1>(1') n 8 and y = Tx = T(x - z) E 1'8. Thus TS is closed. IV.l.ll Lemma. Let Y be complete. Suppose that M and N are linearly independent closed subspaces of Y such that Y = M EEl N (cf. Definition 1.7.16). Then Y' = M1. EEl N1.. Proof. Clearly, M1. n N1. = (0). By Theorem I1.1.14, there exists a projection P from Y onto M with ;)"[(P) = N. Giveny' E Y', let y~ = y'P and let y; = y'(I - P). Then y' = y~ + y;, y~ E N1. and y~ E M 1.. IV.l.12 Theorem. Let X and Y be complete and let l' be closed. Suppose N is a closed subspace of Y such that (1') is dense in X, then T'Y' = T'N1.. Proof.

Define To by

1>(1'0) = 1>(1') X N C X X y

To(x, n)

=

Tx

+n

The linear operator To is closed. Suppose (Xlr, nk) ~ (x, y) E X X Yand To(xk, nk) ----> z. Then Xk ~ x, nk ~ y and TXk + nk ~ z. Hence yEN and TXk ----> Z - y. Since Tis closed, x is in 1>(1') and Tx = Z - y. Thus (x, y) E 1>(1') X N = 1>(1'0) and To(x, y) = Tx + y = z. By hypothesis, O. To prove that O. Since (1'),

IITxll = II 1'o(x, 0)11

~ ')'(1'0)

d(x, 0), ~R(To»

= ')'(1'0) d(x, ;)"[(1'»

OPERATORS WITH CLOSED RANGE

101

Suppose :D(T) is dense in X and y' E :D(T') is given.

Let

Y 1 = CR(T) E9 N

Then, by Lemma IV.1.11 ,

where CR(T)O and N° are the orthogonal complements in Y~ of CR(T) and N, respectively. Thus, if y~ is y' restricted to Y 1, there exist y~ E CR(T)O and y~ E N° such that y~ = y~ y~. Let v' be a continuous linear extension of y~ to all of Y. Then v' is in CR(T).L, and therefore v' is in :n(T') by Theorem IV.1.2. Hence T'y' = T'(y' - v'). Moreover, y' - v' is in Nl., since y' - v' is y~ on N.

+

IV.l.13 Corollary. Let X and Y be complete and let T be closed. CR(T) has finite deficiency in Y, then CR(T) is closed. Proof. space of Y.

If

Y = CR(T) E9 N, where N is some finite-dimensional sub-

IV.l.14 Remarks. The above corollary need not hold if T is not closed. To show this, we first make the following observations.

'/,.

ii.

m.

Every linear operator defined on a finite-dimensional normed linear space V is continuous. This can be shown by first taking V to be unitary n-space Un and then using the fact that Un is isomorphic to V. Every linear functionalf on X with closed kernel is continuous. This follows from the fact that the 1-1 operator J induced by f maps the one-dimensional space X/:n(f) (assumingf ~ 0) onto the scalars. Hence jis continuous and therefore f is continuous by Lemma 11.4.7. There exists an unbounded linear functional on X, provided X is infinite-dimensional. Take {VI, V2, . . .} to be an infinite subset of a Hamel basis {xa } of X, where Ilxall = 1. Let f be the linear functional on X defined by f(x a ) = 0 when Xa ~ Vk, 1 =:; k, and f(Vk) = k. Then f is clearly unbounded.

By the above results, :n(f) is not closed and X = :n(f) + sp {x}, x ¢ :n(f). The linear operator T defined on X by T(u + ax) = u, U E :n(f), has range :n(f). IV.2

INDEX OF A LINEAR OPERATOR

Tho not.ion of t.ho indox of It linOlir opcfn1;or plltyS lUI ovor-incrcusinp; role ill tho At.udy of
UNBOUNDED LINEAR OPERATORS

102

to Gokhberg and Krein (1] for a discussion of the development oftheorems relating to an index. Recently, Atiyah and Singer [1] gave a general formula for the index of an elliptic operator on any compact oriented differentiable manifold. IV.2.1 Definition. The dimension of ;neT), written aCT), will be called the kernel index of T and the deficiency of
In Example IV.1.4, aCT) = nand (3(T) = O. In Example IV.L5, aCT) is the number of Ak which are O. (3(T) = 0 if {1/Ak I is a bounded sequence. (3(T) = 00 if infinitely many of the Ak are O. IV.2.2 Definition. If aCT) and (3(T) are not both infinite, we say that T has an index. The index K(T) is defined by K(T)

= aCT) - (3(T)

with the understanding, as in the extended real number system, that for any real number r,

oo-r=oo

r-oo=-oo

and

The indices of certain differential operators are determined in Chap.

VI. IV.2.3 i.

Proof of (i).

dim

Y
Proof of (ii). (~)

Then

aCT') is the deficiency of
n.

(1)

Let the domain of T be dense in X.

Theorem.

'JI

.

By Theorems 1.6.4 and 11.3.7,

=

dim (

Y )'
= dim
By Theorems IV.1.2 and 1.6.4, ",'Y'/

(:J(1) "'" dllll (\1('1") = dim

m('/') \'"

I

"-=

dilll ~ll('l')' •• dim ~1l('1') .'"-" a('/')

103

OPERATORS WITH CLOSED RANGE

Hence (1) and (2) give K(1')

=

a(1') - (3(1') = (3(1") - a(1") = - K(1")

IV.2.4 Definition. normally solvable.

A closed linear operator with closed range is called

A bounded linear operator which has a finite index and which is defined on a Banach space is oftentimes referred to in the literature as a Fredholm operator. Fredholm [1] studied certain integral equations which gave rise to such operators. In view of the development of the theory of unbounded operators, the following definition is frequently used. IV.2.5 Definition. A closed linear operator which has a finite index is called a Fredholm operator. IV.2.6 Remark. Corollary IV.1.13 shows that, in a Banach space, a Fredholm operator is normally solvable.

Examples of differential operators which are Fredholm operators are given in Chap. VI. The following theorem was proved by Gokhberg and Krein [1] for l' and B Fredholm operators. IV.2.7 Theorem. Let X and Y be complete and let l' be normally solvable with finite kernel index; that is, a(1') < 00. Suppose B is a linear operator with domain a subspace of a normed linear space Z and range in X. ~.

n. nt.

If B is closed, then l'B is closed. If B is normally solvable, then l' B is normally solvable. Let the domain of l' be dense in X and let Z be complete. If l' and B are Fredholm operators, then 1'B is a Fredholm operator and K(1'B)

iv.

=

K(1')

+ K(B)

Let Z be complete and let the domains of l' and B be dense in X and Z, respectively. If B is closed with finite deficiency index, that is, (3(B) < 00, then 'J)(1'B) is dense in Z.

We first prove the following lemmas. IV.2.8 Lemma. X. Then

i.

Let M be a closed subspace having finite deficiency in

For any subspace V of X there exists a finite-dimensional subspace N contained in V Buch that

104

UNBOUNDED LINEAR OPERATORS

n.

If V is dense in X, then V (\ M is dense in M.

Proof of (i). The dimension of V IV (\ M does not exceed the dimension of the finite-dimensional space X 1M, since the linear map 'II from V /V (\ M to XI M defined by 'II(x + V (\ M) = x + M is 1-1. Hence there exists a finite-dimensional subspace W of V such that

(1)

By Theorem II.1.14, there exists a projection P from V onto W with = O. Since PV is finite-dimensional and P is continuous, it follows that P(V (\ M)

W = PV CPV = PV

(2)

,Wn be a basis for W. Then, by (2), there exist elements in V such that PVi = Wi, 1 ~ i ~ n. Since the set (wd is linearly independent, the space N = sp (VI, V2, • • • ,vn } is an n-dimensional subspace of V and (0) = N (\ (M (\ V). By (1),

Let WI, W2,

VI, V2,

• • • ,V n

dim Hence it follows that

V _=

M(\V

dim W

=

dim N

V = M (\ V E9 N.

Proof of (ii). Suppose V = X. Then from (i), X = M E9 N, where N is a finite-dimensional subspace of V. Therefore V = M (\ V E9 N. It follows from Theorem II.1.14 that M is isomorphic to XIN under the map'll, where 'II(m) = m N. For clarity we write

+

(3)

M ~! N

:::) !

= (M (\ V) E9 N :: M (\ V

N

N

Since V is dense in X, V IN is dense in X/No

Noting that

V

'II(M (\ V) = N

it follows from (3) and the continuity of

'II

and

'II-I

that

is dense in M.

Lemma. Let X and Y be complete and let T be normally 80lvable If M is a sub8pace (not nece88ar£ly cl08ed) of X 8uch that M + meT) i8 closcd, then 'l'M 1:,~ closcd. In parl'iI:'IIlar, 'if M is closed and meT) is jinitcd'im,I'nx£onal, Ihen 'I'M 'ix rll/xI:d.

W.2.9

OPERATORS WITH CLOSED RANGE

105

Proof. Let T 1 be the operator T restricted to '.D(T) 1\ (M + ;neT»~. Then T 1 is closed and ;n(T 1) = ;neT). Hence 'Y(T I ) ~ 'Y(T) > O. Therefore T 1 has a closed range; that is, TM = T1(M + ;neT»~ is closed. Proof of Theorem lV.2.7

Proof of (i). Suppose Zn ---+ Z and TBz n ---+ y. Since 'Y(T) is positive, it follows that the sequence {[Bz n]} in '.D(T)j;n(T) is a Cauchy sequence. Thus there exists and [x] E Xj;n(T) such that [Bz n] ---+ [x]. This implies the existence of a sequence {xnl in ;neT) such that BZn + Xn ---+ x. We first show that {x n I is bounded. Suppose {x n I is unbounded. Then there exists a subsequence IX n ' I of {x n I such that Bzn ,

and

+ Xn'

11- ---+ 0

----;-;-11x-n ---';-,

Since Ix n,jllxn,1I1 is a bounded sequence in the finite-dimensional space ;neT), there exists a subsequence Ixn"l of IXn'l and a v E;neT) such that xn"jllxn"l1---+ v. It follows that and Since B is closed, v = B(O) = 0, which cannot be, since Ilvll = 1. Therefore (xnl is a bounded sequence in ;neT). Consequently, there exists a subsequence I Xn' I of {X nI and aWE ;neT) such that Xn' ---+ w. Hence Bzn, ---+ x - w. Since Zn' ---+ Z and B is closed, Z is in '.D(B) and

Bz = x -

W

= lim

n'....... ce

Bzn'

Since T is closed and TBz n, ---+ y, Bz is in '.D(T) and T Bz = y, showing that T B is closed.

Proof of (ii). IV.2.9.

= TBZis closed by Lemma

If BZis closed, then (R(TB)

Proof of (iii). The linear map 1] from ;n(TB)j;n(B) to (R(B) 1\ ;neT), defined by 1][x] = Bx, is 1-1 and surjective. Hence, for N 1 = (R(B) 1\ ;neT) (1)

a(TB = a(B)

+ nl

nl

= dim N l

Let N 2 be a subspace of ;neT) such that ;neT) (2)

n2

= N I El1

N 2•

Then

= dim N 2

ThcsuhspuCCB meR) /Llld N 2nro lincn:rly inaopendcnt, since Bx EN 2 C impJioHl/;reN,I\N v '" (0).

meT}

106

UNBOUNDED LINEAR OPERATORS

Since (j(B) and dim N 2 are finite, CR(B) and CR(B) EB N 2 are closed by Corollary IV.1.13 and Theorem 1.4.12. By hypothesis, 5)(T) is dense in X. Thus, by Lemma IV.2.8, (3)

where N s is some finite-dimensional subspace of 5)(T). (4)

ns

Now 'JL(T) = N 1 EB N 2 C CR(B) EB N 2• that Tis I-Ion all of N s and

=

Hence

dim N s

This, together with (3), implies

TX= TCR(B) EB TN s

(5)

(In general, if X = U EB V, where 'JL(T) C U and V is a subspace of 5)(T), then Tis I-Ion V and TX = TU EB TV.) It follows from (5) and the fact that T is I-Ion all of N s that (6)

(j(TB)

= (jeT)

+ dim TN s = (jeT) + ns

From (1), (2), (4), and (6) we obtain K(TB)

=

a(B)

= a(B)

+ nl -

(jeT) - ns

+ aCT)

- (j(B) - (jeT) = K(T)

=

a(B)

+ aCT)

- n2 - (jeT) - ns

+ K(B)

Proof of (iv). By Corollaries IV.1.13 and IV.1.6, the 1-1 operator f) induced by B has a bounded inverse on the closed subspace CR(B). Since 5)(T) is dense in X, 5)(T) (\ CR(B) is dense in CR(B) by Lemma IV.2.8. Thus, by the continuity of f)-I, it follows that (7)

5)(B) = f)-tCR(B) = f)-l5)(T) 'JL(B)

n

CR(B) C f)-l5)(T)

n =

CR(B) :D(TB)/'JL(B)

Since 5)(B) is dense in Z, (7) implies that :D(TB) is dense in Z. Multiplying a Fredholm operator on the left by a normally solvable operator does not guarantee that the product is even a closed operator. This is shown in the following example.

IV. 2. 10 Example. Let T be the derivative operator in Example ILIA. Let B: X ---+ X = C([O, 1]) ho dofined ItR tho linenr opomtol' whic:h tukoR (\Iwh x e X into t.ho c:onl:lt.ltIIt. flllll:l.ion :r(O). H iA ohviollAly 1\ pro-

OPERATORS WITH CLOSED RANGE

107

jection. However, BT is not closed. Let xn(t) = e-nt/n. Then X n ~ 0 and BTx n is the constant function -1, but lim BT:c n = -1 ~ BTO = O. n-HO

IV.2.11

Definition.

r

Suppose X = Y.

Given a polynomial

n

define peT)

=

aSk, where TO is the identity operator I defined on all of

k=O

X.

The domain of peT) is the domain of Tn.

IV.2.12 Corollary. Let X = Y and let X be complete. If there exists a scalar Ao such that AoI - Tis normally solvable and has finite kernel index, then for any polynomial p, the operator peT) is closed. Proof. Let the degree of p be n. The proof is by induction. For n = 0 the corollary is trivial. Assume the corollary to be valid for n = k and let v be a polynomial of degree k + 1. We write

V(A)

=

(Ao - A)q(A)

+c

where q is a polynomial of degree k and c is a constant. veT) = (AoI - T)q(T)

Then

+ cI

It suffices to prove that (AoI - T)q(T) is closed. By the induction hypothesis, q(T) is closed. Hence, by (i) of Theorem IV.2.7, (AoI - T)q(T) is closed. IV.2.13 Definition. Let X = Y. The set of scalars A for which AI - l' is a Fredholm operator is called the Fredholm resolvent of 1', denoted by F p(T). The set of A for which A ¢ F peT) is called the Fredholm spectrum of 1', denoted by Fu(T).

The next theorem was proved by Rota [1], Theorem 3.2, for differential operators and later generalized by Balslev and Gamelin [1] to Fredholm operators. The proof makes use of Theorem IV.2.7. IV.2.14 Theorem. Let X = Y be a Banach space over the complex numbers and let ']' be closed. For any polynomial p, /. '/1.

p(Fu('l')) C Pu(p('/')). lCqual7'iiJ holds if ~(1') is dense in X. 8U.1JPflH/· :JJ('J') iH d/'I/HI' i'f/ X. Tf f,hrrr r:tiHfH (l. Il ~ p(Fp('!')), tJum

UNBOUNDED LINEAR OPERATORS

108

peT) is closed and has domain dense in X.

Moreover

n

K(IJ.I - peT»~ =

2: K(AJ -

T)

;=1

where AI, A2, . . . , An are the zeros of the polynomial counted according to their multiplicity. Proof of (i). (1)

Given scalar v, let AI, A2' ..

v - peA)

=

(AI - A)(A2 - A) ..

IJ. -

peA)

, An be such that

(An - A)

Then (2)

vI - peT)

= (Ad -

T)(Ad - T) . . . (AnI - T)

Suppose v is in p(Fu(T». Then, by (1), one of the Ai is in Fu(T), say Ak. Since AkI - T commutes with AJ - T, 1 ~i ~ n, it is obvious from (2) that (3)

'.R(Ad - T) C '.R(vI - p(T»

and

ffi(vI -

peT»~

C ffi(Ad - T)

Since a(AkI - T) = 00 or (3(Ad - T) = 00, it follows from (3) that vI - peT) is not a Fredholm operator; that is, v E Fu(p(T»). Suppose 'J)(T) is dense in X. Then for 1 ~ i ~ n, AJ - T has domain dense in X. If each AJ - T is a Fredholm operator, then it follows from (2) and Theorem IV.2.7 that vI - peT) is a Fredholm operator. Hence, if v is in Fu(p(T», then some Ad - T is not a Fredholm operator. Thus, Fu(p(T» C p(Fu(T» which, when combined with what was shown, proves Fu(p(T» = p(Fu(T».

Proof of (ii). Suppose we hav.e the representations given in (1) and (2) with IJ. in place of v. It was shown in the proof of (i) that IJ.I - peT) is a Fredholm operator if and only if each AJ - T is a Fredholm operator. Thus (ii) follows from Theorem IV.2.7.

chapter V

PERTURBATION THEORY

One of the prime motivations for the study of perturbation theory can, perhaps, be best described by the following situation. An operator A is given about which certain properties are to be determined; e.g., find K(A). If A is "complicated," one tries to express A in the form A = T + B, where T is a "simple" operator and B is so related to T that knowledge about properties of T is enough to gain information about the corresponding properties of A. .For example, if A n

is a differential operator

L akDk, T might be chosen to be anDn, with B k=O

as the remaining lower-order terms. A theory which tells us under what conditions B can be disregarded in the determination of K(A), for example, is very useful. Some of the numerous applications of perturbation theory appear in Chap. VI. Most of the material presented in this chapter is based on portions of the deep results obtained by Gokhberg and Krein [1) and Kato [1). The first paper also contains an extensive bibliography which is supplemented by Bartle in the American Mathematical Society Translations, ser. 2, vol. 13. Thro'Uoh01lt this rhapler, B is a linear operator with domain a subspace X and mno(! a I/uill/pac() of norm(!d Unear I/paCI' Y. A 1/

(~f nnrml'd lin(!(1.r I/pa.('('

UNBOUNDED LINEAR OPERATORS

110

in the previous chapters, T is a linear operator with domain in X and range in Y. The spaces are assumed complete and T is assumed closed only when specifically stated. ;neT) is not required to be dense in X. V.1

PERTURBATION BY BOUNDED OPERATORS

The following lemma, which is due to Krein, Krasnoselskii, and Milman [1], is needed throughout this chapter. The proof depends on Borsuk's antipodal-mapping theorem [1] and will be omitted. We refer the reader to Gokhberg and Krein [1], Theorem 1.1, and to Day [1]. 17.1.1

Let M and N be subspaces of X with dim M > dim N Then there exists an m ;;e 0 in M such that

Lemma.

(thus dim N

<

00).

Ilmil

=

d(m, N)

The lemma need not hold if dim M = dim N < 00, as may be seen by taking X to be the plane, with M and N nonperpendicular lines through the origin. When X is a Hilbert space, the lemma has the following easy proof. We first note that M n Nol ;;e (0); otherwise ' N = d'1m (N Nol E9 N d1m

ol) = d'1m Nol X

which contradicts the hypothesis.

> EEl Nol = d'1m M _ d'1m·M Nol

Take x ;;e 0 in M

Ilx - nl1 2 = IIxl1 2 + IInl1 2 ~ IIxl1 2 Thus d(x, N) =

n EN

Ilxl[.

17.1.2

Theorem.

;neT).

If

i. ii.

n Nol. Then

Suppose 'Y(T) then

>

O.

Let B be bounded with ;nCB)

~

IIBII < 'Y(T),

aCT + B) :::; aCT) dim Y j
Proof of (i).

For x ;;e 0 in 'Jt(T

+ B) and IIBII

< 'Y = 'Y(T),

'Y1I[x]1I :::; IITxl1 = I[Bxl[ :::; IIBlll1x11 < 'Yllxll where [x] E Xj'Jt(T). Thus Ilzll > Lemma V.1.1, aCT + B) :::; aCT).

II[x]l\ = d(x,

meT)), and therefore, by

Proof oj (ii). Let XI = ~!)('1') Itncl let B I bo B reHtl'ictou to ~D(T). ConRidclI'ili1/: '/' Illid HI /1.101 OpPI'/I(,OI'H with dOT1millfol c\PIHI(1 in XI. t,hn 0011-

PERTURBATION THEORY

111

jugates T' and B~ exist with domains in Y' and ranges in X~. Theorem IV.1.8. -yeT')

=

-yeT)

> IIBII 2': IIBIII =

By

IIB~II

Hence, it follows from (i) applied to T' and B~ and from Theorem IV.2.3, that dim ----:::=TiiFy=;=~ = dim
-I-

--=:=~~y~~ =
B)

aCT'

+ BD ~

aCT') y

= dim-==c-
V.l.3 Corollary. Suppose T has a bounded inverse. If B is bounded with IIBII < -y = 111fT-III, then T -I- B has a bounded inverse. If, in addition, ~(B) :J ~(T), then

dim

_J__ = dim -=;::;;:;=Y==i~
For x

Proof.

f

~(T


-I- B) choose a positive integer N so that

m < -y N N ow for 0 S k

I (T -I- B

-

~

IIBII

N,

~ B) x 112': IITxll

-II

(1 - ;)

112':

Bx

(-y -

IIBII) IIxll

Thus T -I- B has a bounded inverse and (1)

Since 11(lIN)BII < -y - IIBII, Theorem V.1.2, together with that for 0 ~ k ~ Nand !J(T) = dim Y /
(2)

tJ

(1' -I-

B - Ie

t

1 B)

= tJ ( T +

B -_; B /-

(1),

implies

NB)

~tJ(T+B-~B)

I 12

UNBOUNDED LINEAR OPERATORS

It follows from (2) and Theorem V.1.2 that ~(T)

:::; (3(T

+ B)

:::; ~(T)

which proves the corollary. As a consequence of the corollary we obtain the following result. V.l.4 Corollary. Let B be bounded with ~(B) :> ~(T). Then the B-resolvent of T = {X I T - XB has a dense range and a bounded inverse} is an open set. V.l.S 00

>n i. ii. iii.

Lemma. Suppose that T 1 is a linear extension of T such that = dim ~(Tl)/~(T).

If T is closed, then T 1 is closed. If T has a closed range, then T 1 has a closed range. If T has an index, then K(T 1) = K(T) + n.

Proof of (i). By hypothesis, ~(Tl) = ~(T) ED N, where N is a finitedimensional subspace. Hence, G(T 1) = G(T) + Z, where G(T) and G(T 1) are the graphs of T and T 1 , respectively, and Z = {(n, TIn) In EN}. Thus, if G(T) is closed, then G(T 1) is closed, since Z is finite-dimensional. Proof of (ii).

If (R(T) is closed, then (R(T 1) is closed, since

and TIN is finite-dimensional.

Proof of (iii). It is easy to see that it suffices to prove (iii) for the case when n = 1. Suppose that. ~(Tl) = ~(T) ED sp {x}, for some xE~(Tl)' ThenT 1X = TX ED V, where V = sp {T 1xl when T 1x¢(R(T) or V = (0) when T 1x E (R(T). If T 1x ¢ (R(T), then it is easy to verify that (jeT) = (j(T 1) + 1 and that '.R(T) = '.R(T 1). Therefore, K(T 1 ) = K(T) + 1. If T 1x E (R(T), then (R(T 1) = (R(T) and there exists a z E ~(T) such that Tz = T 1x. Hence, '.R(T 1 ) = '.R(T) ED sp Ix - d. Thus a(T 1) = aCT) + 1 and K(T 1) = K(T) + 1. The following theorem is due to Kato. give a simplified proof of (iii).

Using Corollary V.1.3, we

V.l.6 'fheorem. Let X and Y be complete and let ~J)(T) C ~l)(n), SUppORt! that 7' I'll nor'J'llallll HIIINI,!I/(, nnd l/.nll an -ind,':/', If II HII <' "((7'), then

PERTURBATION THEORY

1,.

n. m.

+

113

l' B is normally solvable. aCT B) ~ a(T); ~(T B) ~ ~(T) K(T + B) = K(T)

+

+

Proofs of (i) and (ii). By Theorem V.L2, aCT + B) ~ aCT). It is easy to verify that l' + B is closed. Suppose aCT) < 00. If m(T + B) is not closed, there exists, by Corollary III.L1O, a closed infinite-dimensional subspace M C X)(T + B) such that for 'Y = 'Y(T) and all x EM,

11(1' + B)xll <

IIBIDllxl1

('Y -

Hence, for each x -;e 0 in M 'Yd(x, :neT»~ ~ IITxl1 ~

IIBxl1 + 11(1' + B)xll < (IIBII + 'Y

-

IIBI!) Ilxll

Therefore d(x, :n(T» < Ilxll, 0 -;e x EM"whence dim M ~ aCT) < 00 by Lemma V.LL But this contradicts dim M = 00. Hence l' + B is normally solvable. Suppose ~(T) < 00. Let Xl and B 1 be defined as in the proof of (ii) of Theorem V.L2, where it was shown that IIBIII < 'Y(T'). Since aCT') = ~(T) < 00 by Theorem IV.2.3, we may apply the result just proved to 1" and B~ to conclude that 1" + B~ is normally solvable and

+ B~) Since 1" + B~ is normally solvable and 1" + B~ = (1' + B Theorems IV.L2 and IV.2.3 show that
(1)

1 )',

1)

(2)

Thus (J(T)

aCT' ~ ~(1'

+ B~)

+ B) by (1)

= ~(T

+ B)

and (2).

Proof of (iii). Let I denote the closed interval [0, 1) and let Z be the set of integers together with the "ideal" elements 00 and - 00 • Define 'P:I ~ Z by 'P(A) = K(T + AB). Let I have the usual topology and let Z have the discrete topology; Le., points are open sets. To prove (iii), it suffices to show that 'P is continuous. Indeed, if 'P is continuous, then 'P(l) is a connected set which therefore consists of only one point. In particular, K(T) = 'P(O) = 'P(1) = K(T + B). In order to prove the continuity of 'P, we first show that K(1'

+ B)

=

K(1')

for IIBII sufficiently small. Suppose aCT) < closed subspace M of X sueh t,hat

x-

M ED ~n('J')

00.

Then there exists a

UN80UNDED LINEAR OPERATORS

114

Let T M be the restriction of T to M (\ ~(T). Then T M has a bounded inverse since it is closed, 1.1, and al(T M ) =
(J(T M

provided

IIBII < "'(T M ). ~(T

K(T) = K(T M )

= (leT)

Now

+ B) = ~(T) = ~(T) (\ M

Hence (1) and Lemma V.1.5 imply that for (2)

+ B)

+ aCT)

= K(T M

EB meT)

IIBII < "'(T M ),

+ B) + aCT)

= K(T

+ B)

Suppose (J(T) < 00. Then, as in the proof of (ii), we apply (2) to T' and B~, and thereby obtain (3)

for

K(T

+ B)

= K(T

+ B t)

= -K(T'

+ B~)

= -K(T') = K(T)

liB II sufficiently small.

Given AO E [0, 1], T + AoB is normally solvable and has an index by (i) and (ii). Thus, by the results obtained above, it follows that cp(A)

=

K(T

+ AoB + (A -

for A sufficiently close to AO' the proof of (iii).

Ao)B) = K(T

+ AoB) = cp(AO)

Hence cp is continuous.

This completes

V.I.7 Corollary. Let X, Y, T, and B satisfy the hypotheses in Theorem V.1.6. There exists a number p > 0 such that aCT + AB) and (3(T + AB) are constant in the annulus 0 < [AI < p. Proof. The proof of the corollary is a modification of Lemma 8.1 in Gokhberg and Krein [1], where B was taken to be the identity operator. We first assume aCT) < 00. For x E m(T + AB) and A ~ 0,

Tx = -ABx

whence Bx E
and

Thus /LlId

PERTURBATION THEORY

115

It follows that (1)

m(T

+ AE)

C

n

D Ic

10=1

where Rl

=

(R(T)

and It is easy to see that and We show by induction that R n and D n are closed subspaces of Y and X, respectively. R l is closed by hypothesis, and D I is closed since B is continuous. Suppose Ric and D Ic are closed. By Lemma IV.2.9, Rk+l = TD k is closed and therefore Dk+l = B-lRk+l is closed since B is continuous. Hence D n and B n are closed by induction. Define Xl =

n D 10=1

k

Yl =

and

.

n

Rk

10=1

It follows from the definition of DIc and Ric that TX I C Y l and BX I C Y I . Let T l and B I be the operators T and B, respectively, restricted to i)(T) n Xl with ranges in Y I . Since T is a closed operator and Xl is closed, T l is also closed. We show that (R(T l ) = Y l . Let y be an element in Y I =

.

n

TD n •

Then for each n

~

1 there exists an

Xn E

D n such

n=l

that TX n = y. Since ;neT) is finite-dimensional and D n :J D n +l , there exists an integer k o such that for k

~ ko

From the way the sequence {Xk I was chosen, together with the fact that Dk C D k ., k ~ k o, it follows that k

~

ko

Hence Xk o E n D k = Xl and TXko = y, which shows that T I is surjective. k':?:.ko Thus there exists, by Theorem V.1.6, a number p > 0 such that for

IAI < p, (2)

K(T

+ AB)

= K(T)

anrt

a(T I

+ AB

I)

... K(T I ) "" a('l'I)

116

UNBOUNDED LINEAR OPERATORS

Now, (1) implies that ;neT particular,

+ XB)

= ;n(Tl

+ XB 1)

for X ~ O.

In

(4)

It is clear from (2), (3) and (4) that OI(T + XB) and (j(T + XB) are constant in the annulus 0 < 171.1 < p under the assumpt'on OI(T) < 00. If OI(T) = 00, then (j(T) < 00 by hypothesis. As in the proof of Theorem V.1.5, we apply the result just proved to the appropriate conjugate operators in order to prove the corollary when OI(T) = 00. V.I.B Theorem. Let X and Y be complete and let B be continuous with S:>(T) C S:>(B). Define U to be the set of X for which T + XB is normally solvable and has an index. Then i. ii.

U is an open set. If C is a component of U, that is, a largest connected subset of U, then on C, with the possible exception of isolated points, OI(T + XB) and (j(T + XB) have constant values nl and n2, respectively. At the isolated points,

and Proof of (i). of T.

For X E U, apply Theorem V.1.5 to T

+ XB

in place

Proof of (ii). The component C is open, since any component of an open set in the space of scalars is open. Let 01(71. 0) = nl be the smallest integer which is attained by 01(71.) = OI(T + XB) on C. Suppose 01(71.') ~ nl. Owing to the connectivity of C, there exists an arc r lying in C with endpoints 71. 0 and X'. It follows from Corollary V.1.7 and the fact that C is open, that about each IL E r there exists an open ball S(IL) contained in C such that 01(71.) is constant on the set S(IL) with the point IL deleted. Since r is compact and connected; three exist points

on r such that

, S(X n ) cover rand SeX,)

n

S(X;+I) ~ cJ> O:::;i:::;n-1

We assert that 01(71.) = 01(71. 0) on all of S(X o). Indeed, it follows from Theorem V.1.5 that 01(71.) :::; 01(71. 0) for Xsufficiently close to 71.0. Therefore, since 01(71. 0) is the minimum of 01(71.) on C, 01(71.) = 01(71. 0) for X sufficiently close to 71. 0• Since 01(71.) is constant for all X ~ 71.0 in 8(71.0), thiA COIIHtl1nt must be OI(XO). Now OI(X) is I:OIIH(,ILII t on tho Aot 8 (X,) wi (,h tho poi 11 I, X,

117

PERTURBATION THEORY

deleted, 1 ~ i ~ n. Hence, it follows from (1) and the observation a(X) = a(X o) for all X E S(X o), that a(X) = a(X o) for all X ~ X' in S(X') and a(X') > nl. The result just obtained can be applied to T' + XB~, as in Theorem V.1.6, in order to prove the analogous results for (j(T

+ XB) = a(T' + XB~)

V.l.9 Remarks. The following examples show that Theorem V.1.6 does not hold in general even though IIBII = 'Y(T). If we take T to be the identity operator mapping an infinite-dimensional Banach space onto itself, then 'Y(T)

=

1

=

and

IITII

a(T) = (j(T) = 0

However, T perturbed by - T is the zero operator which has 00 as its kernel index and deficiency index. Less trivially, in Example IV.1.5, choose Xl, X2, • • • to be a strictly decreasing sequence of numbers converging to a positive number r. Then a(T) = (j(T) = 0, 'Y(T) = r, but 'Y(T - rI) = O. Hence, T - rI does not have a closed range, and therefore (j(T - rI) = 00 by Corollary IV. 1.6. V.2

PERTURBATION BY STRICTLY SINGULAR OPERATORS

If B is strictly singular with no restriction on its norm, we obtain the following important stability theorem due to Kato. V.2.1 Theorem. Let X and Y be complete and let T be normally solvable with a(T) < 00. If B is strictly singular and 5)(T) C 5)(B), then i. ii. m.

T + B is normally solvable. K(T B) = K(T) a(T XB) and (j(T XB) have constant values nl and n2, respectively, except perhaps for isolated points. At the isolated points,

+ +

00

>

a(T

+

+ XB) >

nl

and

(3(T

+ XB) > n2

Proof of (i). Since a(T) < 00, there exists a closed subspace M of X such that X = M EEl 'J'L(T). Let T M be the operator T restricted to M n 5)(T). Then T M is closed with m(T M ) = m(T). Suppose that T + B does not have a closed range. Then it follows from Lemma V.1.5 that the e1mlP.d oprmtor T M dool(not have a closed l'fLnge. Hence thC1TO l"XiRtR, hy Corollary TTT.l.lO, n. dORnel infinit.n-dimoTlRinnfLI BuhApnr.O

+n

118

UNBOUNDED LINEAR OPERATORS

M 0 contained in :n(TM )

= :n(TM

+ B) such that

II(T M + B)xll < 'Y(;M)

IIxll

xfiM o

Thus, since T M is 1-1, it follows that for all x in M o,

which shows that B has a bounded inverse on the infinite-dimensional space Mo. This, however, contradicts the hypothesis that B is strictly singular. We now prove a(T + B) < 00. There exists a closed subspace N I such that

;n(T

(1)

+ B)

= ;n(T

+ B) (\ ;neT)

E9 N I

Let T I be the operator Trestricted toNI. By Lemma IV.2.9, CR(T I ) = TN! is closed. Moreover, T I is 1-1. Hence its inverse is bounded. Since B = - T I on N I and B is strictly singular, N I must be finite-dimensional. Therefore (1) implies that ;n(T + B) is finite-dimensional.

Proof of (i,t). We have shown above that for all X, T + XB is normally solvable and a(T + XB) < 00. Thus, applying Theorem V.1.6 to T + XB in place of T, it follows that cp(X) = K(T + XB) is continuous from [0,1) into Z, where [0,1) has the usual topology and Z is the topological space consisting of the integers and - 00 with the discrete topology. Hence cp is a constant function. In particular, K(T) = cp(O) = cp(l) = K(T

+ B)

Proof of (iii). Since T + XB is normally solvable for all X, (iii) is an immediate consequence of Theore~ V.1.8. V.2.2 Corollary. Let X and Y be complete with :neT) dense in X. Let T be normally solvable and have an index. Suppose :neT) C :nCB) and both Band B' are strictly singular; for example, B is compact. Then ~.

n. m.

T + B is normally solvable. K(T + B) = K(T) a(T + XB) and (J(T + XB) have constant values nl and n2, respectively, except perhaps for isolated points. At the isolated points, and

Proof.

Apply Theorolll V.2.1 {,o 7"

(J(T R,lld

+ XB) > n2

11'

Whllll

(J(T)

<

00.

119

PERTURBATION THEORY

V.2.3 Corollary. Let X = Y be complete. 8uppose there exists a scalar Xo such that (XoI - T)-1 is strictly singular on all of Y. Then for all X, AI - T is a Fredholm operator with zero index.

We write

Proof.

(1)

AI - T = (I

+ (X

- Xo)(XoI - T)-l)(X oI - T)

Since (XoI - T)-1 is assumed to be strictly singular on X, we have from Theorem V.2.1 that K(I

(2)

+

(X - Xo) (X 01 - T)-l) = K(I) = 0

Furthermore K(X o1 - T) = 0

(3)

Thus (1), (2), (3), and Theorem IV.2.7 imply K(Al - T) = O. V.2.4 Corollary. LetX = Y be any of the spaces £1(8, ~, !J.), £.,(8, ~, !J.), C(8), B(S), or their conjugates. Suppose there exists a scalar Xo such that (XoI - T)-l is weakly compact on all of X. Then for all X, AI - T is a Fredholm operator with index zero. Proof.

Theorem 111.3.5 and Corollary V.2.3.

Goldman [1] proved that to any normally solvable operator T which does not have an index, it is always possible to add a compact operator B E [X, Y] such that for all X 7'" 0, T + XB does not have a closed range. To show this, we present a similar proof given by Whitley, which does not depend on a theorem of Mackey's used by Goldman. V.2.5 Lemma. The conjugate space of a separable normed linear space contains a countable total subset. ProQ,f. Let {Xk I be a countable set dense in normed linear space X. For each k there exists an x~ E X' such that IIx~1I = 1 and X~Xk = I!xkll. We show that {x~l is total. Assume there exists an x 7'" 0 in X such that x~x = 0 for each k. Since {Xk I is dense in X, there is an XN such that Ilx - xN11 < Ilxll/2. Hence

IIxNl1 ~ !lxl! - IIx - xNII > II~II But

which iH n cOIlf;rndict.io!1.

TllIlA {x; I iR t.ot.al.

UNBOUNDED LINEAR OPERATORS

120

V.2.6 Theorem. Let X and Y be complete and let T be normally solvable with no index. Then there exists a compact operator BE [X, Y] such that T + XB does not have a closed range for all X ~ O. If'St(T) is separable, then T + XB is I-I. Proof. Since dim
in (\ 'St(y;) = .L sp I y~, . . . ,y~} but not in 'St(Y~+I)' The existence of i=l

Yk+l is assured by Remark II.3.6 and the linear independence of Iy;}. Thus y;
.

L X~(X)Yi

(2)

~:lix~IIIIYill

Bx =

The completeness of Y, together with the absolute convergence of the series, implies that Bx exists, B E [X, Y], and B is the limit in [X, Y] of the operators B n defined by n

L X~(X)Yi

Bnx

=

~:lix~III1Yill

Since B n is bounded with finite-dimensional range, B n and, in turn, Bare compact. Now
x~x

= 0

Continuing in this manner, we see that 0 = x~x, 1 ~ i, and consequently Bx = O. In particular, B is I-Ion Xl = sp IXi} C 'St(T), for Bx = 0 E
+

(3)

\1('1' + AR)xll

~ 'Y d(x, m('l'

+ XB))

'Y - 'Y(T

+ AB) > 0

pERTURBATION THEORY

121

'St(T + "AB) = 'St(B) (\ 'St(T), since (T + "AB)x = 0 implies CR(T) (\ CR(B) = (0) and Tx = O. Hence, defining B 1 to be the restriction of B to 'St(T), 'St(B 1) = 'St(T) (\ 'St(B) = 'St(T + "AB). Thus, for x E 'St(T) , we know from (3) that

Now Bx

E

liB 1x l

=

1 TXlII (T + "AB l)xll

~

TXl7 d(x, 'St(B 1))

Hence B l has a closed range which must be finite-dimensional, since B 1 is compact. But this is impossible, since dim Xl = co and B is I-Ion Xl C'St(T). Hence CR.(T + "AB) is not closed. If 'St(T) is separable, then {Xi} may be chosen so that Xl = 'St(T), in which case 'St(T + "AB) = 'St(B) (\ XI = (0). V.3

PERTURBATION BY UNBOUNDED OPERATORS

Sz.-Nagy [1] observed that certain perturbation theorems remain valid if the perturbing operator B is not necessarily bounded but satisfies some less restrictive conditions. The conditions are that B be bounded or compact on ~(T) under certain renormings of ~(T). V.3.l

Definition.

Let the T-norm

IIxllT

=

II liT be defined on ~(T) by

IIxli + IITxl1

B is called T-bounded if ~(T) C ~(B) and B is bounded on ~(T) with respect to the T-norm; i.e., there exists a number M such thatfor all x E ~(T),

IIBxl1

~

M(llxll + IITxlD

B is called T-strictly singular (T-compact) if ~(T) C ~(B) and B is strictly singular (compact) on ~(T) with respect to the T-norm.

Obviously, if B is bounded and ~(T) C ~(B), then B is T-bounded. Similarly, B strictly singular (compact) implies that B is T-strictly singular (T-compact). V.3.2 Remark. It is very easy to see that if X and Yare Banach spaces and T is closed, then ~(T) with the T-norm is a Banach space. V.3.3 Theorem. Let X and Y be complete and let T be closed. :oCT) C :D(B) and B is closable, then B is T-bounded.

If

Proof. Let D T be :D(T) with the T-norm and let fJ be a closed linear extension of B. Then lJ is closed- and therefore continuous on Bnnnl'h AP/WO l)7" TlOlwo n iA ILIRO continuouR on D7••

UNBOUNDED LINEAR OPERATORS

122

V.3.4 Corollary. Let X and Y be complete. Suppose there exists a scalar Ao such that AoI - T is normally solvable and has finite kernel index. Then for any polynomial p of degree n, Tk is p(T)-bounded for 0 :::; k :::; n. Proof.

Corollary IV.2.12 and Theorem V.3.3.

The method of generalizing Theorem V.1.6 stems from the following result due to Sz.-Nagy [1].

Suppose '.D(T) C

V.3.5 Lemma. Let Y be complete and let T be closed. '.D(B). If there exist numbers a and b such that b

<

and

1

for all x E '.D(T) , then T Proof.

II(T

(1)

and

II(T

IIBxl1 :::;

allxll + bllTxl1

+ B is closed.

It follows from the hypotheses that for all x E'.D(T) ,

+ B)xll + B)xll

:::; ~

IITxl1 +

IIBx\1 :::;

IITxl1 - IIBxl1

~

allx\l + (1 + b)IITxll -allxll + (1

- b)IITxll

or (2)

IITxl1 :::; (1 -

b)-l(allxll + II(T + B)xID

Suppose Xn ~ x and ('1' + B)x n ~ y. Then (2) shows that {Tx n I is a Cauchy sequence. Hence there exists aYE Y such that TX n ~ y. Since T is closed, x E'.D(T) = '.D(T + B) and TX n ~ Tx. Thus, by (1),

II(T

+ B)(x

Hence (T

+ B)x

V.3.6

n -

n -t

Theorem. by

x)11 :::; (T

allxn -

xII + (1 + b)IITxn

-

Txll ~ 0

+ B)x, anli the lemma is proved.

Theorem V.1.6 holds if we replace the requirement

IIBII < 'Y(T)

IIBxl1 :::;

allxll + bllTxl1

X E '.D(T)

where a and b are nonnegative numbers such that a

+ h(T) < 'Y(T).

Proof. By Lemma V.3.5, T + B is closed. We first assume that Let T 1 and B 1 be the operators T and B, respectively, considered as mappings from Banach space D 1 into Y, where D 1 is '.D(T) with norm II III defined by

a

> O.

Ilxlll "'" /1,II.r.11 + bllTxl1

PERTURBATION THEORY

123

The idea of the proof is to apply Theorem V.1.6 to T 1 and B I in order to obtain the assertions about T + B. We therefore want to know how 'Y(T 1) compares with 'Y(T). This can easily be obtained in the following manner for a > 0 and b > O. Given x in D I , d(x, ;n(TI)) where

=

inf

nE~(T)

Ilx - nlll = a

[xl E <J)(T)/;n(T).

inf

nE~(T)

Thus = inf XED.

Ilx - nil + bllTxll = all[xlll + bllTxl1

IITxl1 all[xlll + bllTxl1

IITxll/ll[xlll

= inf XED.

a

+ bIITxll/ll[xlll

'Y(T)

a

+ b'Y(T) >

1

Since IIBIII ~ 1 < 'Y(T I) and T I is bounded on the Banach space D I, it follows that the conclusions of Theorem V.1.6 hold for T I and B I in place of T and B, respectively. However, m(T I + B I) = m(T + B), and ;n(T I + B I) = ;neT + B). Thus Theorem V.1.6 holds for T and B, where a> O. If a = 0, choose E > 0 so that E + b"{(T) < 'Y(T). Then IIBx!1 ~ Ellxll + bllTxll, and the theorem is proved by what was just shown. V.3.7

Theorem.

Theorem V.2.1 holds if B is T-strictly singular.

Proof. Let D I , T 1, and B I be as in the proof of the preceding theorem with a = b = 1, and apply Theorem V.2.1. To see that T + B is closed, let E be the identity map from <J)(T) onto D I. Then E is closed and T + B = (T 1 + BI)E. Thus T + B is closed by Theorem IV.2.7. V.3.B

Corollary.

Suppose T and B are closed and B is T-compact.

Then t.

Given E > 0, there exists a constant K f E <J)(T) ,

IIBfl1 n. m.

~ Kllfll

= K(s) such that for all

+ Ell Tfll

If Y is complete, then T + B is closed. If X and Yare complete, then T is normally solvable with finite kernel index if and only if T + B is normally solvable with finite kernel index. In this case, K(T) = K(T + B).

Proof of (i). Assume (i) is false. Then for some positive illtol/;cr n, thoro 0xiRtA '!til i . E ~D(T) su(~h thltt ~-

(I)

II/U.. II > /I. II!.. I + ell'I:t'"II

E

>

0 and each

124

UNBOUNDED LINEAR OPERATORS

Let gn = fn/llfnIlT, where

II

liT is the T-norm.

Then from (1),

(2)

Now, IlgnllT = 1 and B is T-compact. Hence there exists a subsequence {v n } of {gn} such that {Bvn} converges in Y. Since Ilvnll < IIBvnll/n and {IIBvnll} is bounded, Vn ~ 0 in X. From the assumption that B is closed it follows that BV n ~ O. Thus TUn ~ 0 by (2). But this is impossible, since 1 = IlvnllT = Ilvnll + IITvnl1 ~ O. Proof of (ii).

+ B is closed by (i) and Lemma V.3.5. Any set which is T + B-bounded is T-bounded, since

T

Proof of (iii). (i) implies the existence of a constant K such that for all

f IITfl1 S II(T

E '.D(T)

=

+ B)fll + IIBfl1 s

'.D(T

II(T

+ B) + B)fll + Kllfll + ill Tfll

whence IlfilT S 2(K

+ 1)llfllr+B

Therefore the T-compactness of B implies the T + B-compactness of B. Thus if T + B is normally solvable with OI(T + B) < 00, then T = (T + B) - B is also normally solvable and K(T + B) = K(T) by Theorem V.3.7. In particular, OI(T) < 00. The converse statement is a particular case of Theorem V.3.7. The next corollary generalizes a result due to Beals [2]. V.3.9 Corollary. Let X and Y be complete and let T be a Fredholm operator with domain dense in X. If B is T-strictly singular and B' is T'-strictly singular, then (T + B)' = T' + B'. Proof. We first note that if iJ and V are linear operators with equal, finite indices and V is an extension of U, then U = V. This follows from the relations OI( U) S OI(V)

{3(U)

~

{3(V)

and

which imply OI(U) = OI(V) and (3(U) = (3(V). Now, (T + B)' is clearly an extension of T' follows from Theorems V.3.7 and IV.2.3 that K«T

Thus ('I'

+ B)')

+ B)'

=

-K(T

= 7"

+ R'.

+ B) =

-K(T)

=

K(T')

K(U) = K(V)

+ B'. =

K(T'

Moreover, it

+ B')

chapter VI

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

In this chapter we shall consider properties of certain linear operators which arise from ordinary differential expressions of the form

where D = d/dt and the coefficients ak are complex-valued functions of a real variable, subject to certain conditions on an interval I. A differential expression T may give rise to many operators T which have their domains in ,cp(l) and ranges in ,cq(l), where 1 ~ p, q ~ 00 For example, T may be defined as follows:

n

Tf = Tf =

L akDkf k-O

Thus the operators determined by T are distinguished by means of their dOlHl1ins. The ones which we. present are those which are of recognized importlUwo und to whioh the gCIl{)rn.1 theory developed in the preceding (\hnptors npplioH.

UNBOUNDED LINEAR OPERATORS

126

When there is no possibility of confusion, a function in £1' °(1) and the element in £1'(1) which it determines are used interchangeably. Unles8 mention is made regarding the nature of I, the interval is arbitrary.

The main results, with somewhat different proofs, in Sees. VI.I to VI.5, are due to Rota [1] for 1 ::; p = q ::; 00.

VI.I

CONJUGATES AND PRECONJUGATES OF DIFFERENTIAL OPERATORS

In order to determine some properties of a differential operator l' with domain a subspace of, say, £",(I) and range in £q(1), 1 < q ::; OCJ, we introduce what we shall call the preconjugate of T. Since the domains of the operators l' which we consider are not dense in £",(1), the preconjugate '1' of l' is introduced to compensate for the nonexistence of 1". The idea is to consider l' as a map from a subspace of the conjugate space £",(1) = £~(I) into the conjugate space £q(I) = £;,(1) and to define '1' in such a way that, in certain cases, Tis the conjugate of '1'. VI.I.I Definition. Let X and Y be normed linear spaces and let l' be a linear operator with domain a total subspace of Y' and range in X'. The preconjugate '1' of Tis defined as follows. '£)('T) is the set of those x E X for which there exists ayE Y such that Ty'x = y'y for all y' in '£)(1'). Let 'Tx = y. Since '£)(1') is total, '1' is unambiguously defined. It follows that '1' is a linear map from a subspace of X into Y and Ty'x = y'('Tx)

X E '£)('1'),

y' E '£)(1')

VI.I.2 Lemma. Let l' be a linear operator with domain a total subspace of Y' and range in X'. Then the preconjugate of l' is closed. Proof. Suppose X n -----+ x and ''Fx n -----+ y. Then for each y' E '£)(1'), Ty'x n -----+ Ty'x and Ty'x n = y'('Tx n ) -----+ y'y. Thus Ty'x = y'y, y' E '£)(1'), which means that x E '£)('1') and 'Tx = y. VI.I.3 Lemma. Let l' be a linear operator with domain a total subspace of Y' and range in X'. Suppose '£)('1') is dense in X. Then ('1')' is an extension of 1', and therefore l' is closable. If, in addition, X and Yare reflexive, then '1' = ('1')', where '1' is the minimal closed linear extension of T. Proof. If y' E '£)(1') and x E '£)('1'), then y'('Tx) = Ty'x. Thus y' E '£)«'1')') and ('T)'y' = Ty', y' E '£)(1'). Since the conjugate operator ('1')' is closed, Tis closable. Let X and Y be reflexive. Suppose '1' ¢ ('1')'. Then therp. exists some (v', ('T)'v') j! G(T) = (J('J'). Thus there exists !1 z" E (Y' X X')'

127

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

such that z"(v', ('T)'v') r!' 0, and z"(y', Ty') = 0, y' E ~(T). Define y" E Y" and x" E X" by y"y' = z"(y', 0) and x"x' = z"(O, x'). Then (1)

y"y'

+ x"Ty'

(2)

= z"(y', Ty') = 0

y"v'

+ x"('T)'v'

y'

E

~(T)

r!' 0

Thus, since X and Yare reflexive, there exist x E X and y (3)

y'y

(4)

+ Ty'x v'y

= 0

+ ('T)'v'x

y' ¢'

E

E

Y such that

~(T)

0

Now, (3) shows that x is in ~('T) and 'Tx = -v. But then (4) implies that v'y + v'('Tx) = v'y - v'y r!' 0, which is absurd. Hence 'i' = ('T)'. VI.l.4 Lemma. Let A be a closable linear operator with domain dense in X and range in Y. Then A = '(A').

Proof. Since A is closable, ~(A') is total by Theorem II.2.11. Hence the preconjugate of A' is defined. Now, '(A') is an extension of A; for if x E ~(A), then A'y'x = y'Ax, y' E ~(A'). Thus, by the definition of '(A'), x is in ~('(A')) and '(A')x = Ax. Since '(A') is closed by Lemma VI.1.2, it remains to prove that G('(A')) C G(A) = G(A). By an argument analogous to the one given in Lemma VI.1.3, it follows that G('(A'))

C G(A).

VI.l.S Lemma. T = ('T)'.

If T is the conjugate of a closable operator, then

Proof. Suppose T = A', where A is closable. VI. 1.4, A = '(A') = 'T. Hence

Then, by Lemma

T = A' = (A)' = ('T)' by Theorem II.2.11. VI.l.6

Definition.

Let T be a differential expression of the form

where each ak is a complex-valued function on I. For each positive integer n, define A,.(l) to be the 8et of' complex-valued functions f on I for which !<,.-I) ... nil-I! existfl and i8 absol:utrlll continu01t8 on c11Cry compact 8ubinterllnl o! I. I,,!, A u(l) .... r!(l).

128

UNBOUNDED LINEAR OPERATORS

The maximal operator T.,p,q, 1 ::; p, q ::; and I, is defined as follows.

r

00,

corresponding to T, p, q

n

T. ,p,qf = Tf =

akDkf

k=O

f

E

The operator T~p,q is defined to be the restriction of T.,p,q to those CJ)(T.,p,q) which have compact support in the interior of I.

Since an absolutely continuous function is differentiable a.e., it follows that Tf is defined a.e. on I for f E An(I). Before proving Theorem VI. 1.9, we make the following observations which indicate what one might expect (T~p,q)' and '(T~p,q) to be.

r n

Suppose T =

ak Dk , ak E Ck(I), 0 ::; k ::; n.

Since Co""(I) is con-

k~O

tained in 5)(T~p,q), we know, by Theorem 0.9, that 5)(T~p,q) is dense in .£p(I) when 1 ::; p < 00 and is total in £""(I). Thus T:'p,q has a conjugate (T:'p,q)', defined with domain a subspace of £~(I) = £q,(I) and range in £;(I) = £p,(I), where 1 ::; p, q < 00, p' and q' conjugate to p and q, respectively. For 1 < p, q ::; 00, T~p,q may be considered as a map with domain a total subspace of the conjugate space £p(l) = £~,(I) and range in the conjugate space £q(I) = £~,(I). Hence the preconjugate 'T:'p,q of T~p,q is defined as a map from a subspace of £q,(I) to £p,(I). To simplify the discussion, we assume that I = [a, b] is compact. Writing Tn = T~p,q, let g be in 5)(T~) (g E CJ)('T R )) and let

Then for f E 5)(TR ) , f(g*))

Thus, regardless of whether g is in CJ)(T~) or in CJ)('T R ), (1)

In order to obtain an expression for g*, equation (1) suggests suc(akg)J
t

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

129

Since ak is in Ck(I), 1 =:; k =:; n, (akg)(k-l) is absolutely continuous. integration by parts gives for 1 =:; k =:; n,

t

(2)

J: - t

(akg)JCkl = (akg)JCk-ll

(akg)'JCk-ll

= (akg)JCk-ll - (akg)'JC k-2l k-l

=

l

Hence

J: + t

(-l)i(a kg)(JlJCk-l-ilJ:

(akg)C2l JC k-2l

+ lab (-l)k(akg)(klf

i=O b

•

where v Ja denotes v(b) - v(a) for any functIOn v defined at a and b. (1) and (2),

(3)

t

Since f

n

gTf =

k-l

l l

n

(-l)i(akg)(JlJCk-l- il

k=l i=O E 'JJ(T B )

J: + t l

From

(-l)k(akg)( klf

k=O

has compact support in the interior of I,

JCkl(a)

=

=

JCkl(b)

0

O=:;k=:;n-l

Hence, from (1) and (3),

t g*f = t

(4)

(T*g)f

where T*g is the function defined a.e. by

r n

T*g =

(-l)k(akg)(kl

k=O

Since 'JJ(T B ) is densein£p(I),1 =:; p equation (4) implies g* = T*g a.e. on I.

<

oo,andtotalin£~(l)

= £JI),

Hence

Thus T* appears to determine the conjugate and preconjugate of T B in a manner similar to that in which T determines T B •

r n

akD\ ak E Ck(I), 0 =:; k k=O Lagrange, or formal, adjoint T* of T is the expression defined by VI.l.7

Definition.

Let T

"'·0 -

=

r"

Ar-O

(-l)hDk(akU)

=:; n.

The

130

UNBOUNDED LINEAR OPERATORS n

L (-l)k(akg)(k)(t) exists for almost

If g(n) exists a.e., then (r*g)(t)

k=O

all t. The following lemma was proved in the above discussion. n

VI.l.B

Lemma.

Let r =

L akDk, ak

E

Ck(I) , 0 ::; k ::; n.

If f and g

k=O

are in An(l), where I

= [a, b] is compact, then n

k-l

f grf = l l

J: + f

(-l)l(akg)(j) j
fr*g

k= 1 j=O

The next theorem was proved by Halperin [1] for p = q = 2 and by Rota [1] for 1 ::; p = q ::; 00 n

VI.l.9

Theorem.

Let r be the differential expression r

=

L akDk,

k=O

where ak E Ck(!), 0 ::; k ::; n, and an(t) ¢ 0, t E I.

<

00

p, q ::;

00

1 ::; p, q R 'TT,p,q =. TT*.q',p'

1

<

Then

Proof. Write T R = T~p,q and T* = TT*,q',P" Suppose g E 5)(T*) and f E 5)(TR). Then f vanishes outside of some compact interval [a, b] contained in the interior of I. Therefore we have, by Lemma VI.1.8,

£grf = t grf = t

(1)

If 1 ::; p, q

<

00,

(r*g)f =

t

(r*g)f

then (1) shows th~t

Hence g E 5)(T~) and T~g = T*g. If 1 < p, q ::; 00, then considering T R as an operatorfrom £~. = £p(l) into £~. = £q(I) and f as a functional in £~" it follows from (1) that

Hence g E 5)('T n ) and 'TRg = T*g. show 5)(T~) C 5)(T*), 1 ::; p, q

q::;

00,

<

To prove the theorem, it remains to 00, and 5)('T R ) C :D(T*), 1 < p,

131

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Suppose g E :D(T~) (g f :D('T R)) and T~g = g* ('TRg = g*). for every f E :D(T R ),

Then

£(Tf)g !r fg*

(2)

=

To show that g is in :D(T*), it suffices to show that for every compact interval 1 0 = [a, b] C I, g, considered as a function, is equal a.e. to a function in An(Io) and T*g = g* a.e. on 1 0 • The remark follows from the observations that any interval is the union of an ascending sequence of compact subintervals and that continuous functions which are equal a.e. are identical. Let Do = {f If E :D(T R ), f has support in 1 0 1. Since elements in :D(TR ) have support in the interior of I, it follows that, given f E Do, j
(3)

=

(at (t - S)n-k-l j(n)(S) ds (n - k - I)!

0::;;k::;;n-1

ia

Now, f vanishes outside of Io.

Hence (2) and (3) imply

n-l (4)

lab an(s)g(S)j
L t dt l

ak(t)g(t)

(~

k=O

=

=~)=k~;!

j
(b dt (t g*(t) (t - S)n-l j
ia

ia

(n - I)!

All the integrands in (4) which involve both sand t are in £1(1 0 X 1 0 ), since ak is continuous on 1 0 , g E £q,(Io) C £1(/ 0), and j
(5)

0

=

fa

b

j(n)(s)

[

an(s)g(s)

+

n-I \'

~

k=O

J.

b

(t (n

=

s)n-k-l k _ I)! ak(t)g(t) dt

-

J.•

b

(t - s)n-l * ] (n _ I)! g (t) dt ds

for all f e Do. Lot F(s) be the expression ineide the square brackets in (5). We now show thnt F is equivalent on lo·to a polynomial of degree at most n - 1. OiVPIl Q E £q{T II ) Aunh l,hnt q iA Ol'fJlOP;OIlI1J to the subspace

UNBOUNDED LINEAR OPERATORS

132

h defined by

h(t) =

(t (t - S)"-l

Ja

h(t) = 0

t ¢ 10

is easily seen to be in Do with hCn)(t) hCn) in place of jCn), whence

o=

(6)

tE1 0

(n _ I)! Q(s) ds

f

Q(t) a.e. on Io.

=

Thus (5) holds for

Q(s)F(s) ds

for all Q E £q(l o) orthogonal to @ C £q,(Io); that is, (6) holds for all Q E @ol when 1 < q ::; 00 or for all Q E ol@ when;q = 1. Hence, Theorem II.3.4, Remark 11.3.6, and the finite dimensionality of @ imply FE ol(@ol) =

l
@

q= 1

Thus F is equivalent on 1 0 to a polynomial v of degree at most n - 1. Therefore, except for a set of Lebesgue measure zero, ,,-1

(7)

an(s)g(s)

= v(s)

-

\'

(b (t (n

L. I,

k=O

=

s)n-k-1 k _ I)! ak(t)g(t) dt

+

(b (t - S)"-l g*(t) dt

Js

(n - I)!

Now the right hand side of (7) and l/a" are absolutely continuous on 1 0 • Thus we can redefine g on a set of measure zero to be absolutely continuous on 1 0 so that (7) holds on all of 1 0 • Differentiating both sides of (7), we obtain for almost all s E 1 0 (8)

(a"g)'(s)

=

an(s)g'(s)

= v'(s)

+ g(s)a~(s)

+ an-1(s)g(S) +

L 1 (t -

n-2

b

s)n-k-2

(n - k - 2)! ak(t)g(t) dt

k=O

_

(b (t - s)n-2 g*(t) dt (n - 2)!

Js

Since g, a~ and l/an are absolutely continuous, it follows from (8) g' is equivalent to an absolutely continuous function. But then absolutely continuous on 1 0 ; for if the derivative g' of an absolutely tinuous funct.ion g is r-f)uivnlr-nt t.o n (~ontilluoIIS fundion h, thon

that g' is cong' is

133

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

identically h.

This follows from g(x) - g(a) =

lax g'(t) dt = lax h(t) dt

Repeated differentiation of both sides of the equalities in (8), together with a repetition of the argument just used to prove g' is absolutely continuous, shows that g(n-l) is absolutely continuous and T*g = g* a.e. on 1 0 • (Recall that v is a polynomial of degree at most n - 1.) Thus the proof of the theorem is concluded. Before proving the next corollary, we take a closer look at T*. n

VI.l.1O

L akDk,

Let T =

Lemma.

ak E Ck(I) , 0 :::;; k :::;; n.

Then for

k=O n

L bkg(k) a.e.

T*g =

k=O

where

O:::;;k:::;;n

For 0 :::;; k :::;; n, bk is in Ck(I). Proof.

Let t be such that g(n) (t) exists.

Then

n

L (-I)k(akg)(k)(t)

(T*g)(t)

(1)

k=O

By Leibnitz's rule, (ajg) (j)(t)

(2)

Setting i = k and j (1), that

l

=

k, k

+ 1,

. . . , n in (2) implies, together with

n

(T*g)(t)

bk(t)g(k)(t)

k-O

Hilwa DkaP-k)

iH ill

e (!), 0 k

= ap) :::;; k :::;;

IUlO aj is ill CiE/) , 0 ~ Ie :::;; fl.

.i :::;;

n, it follows that b k

134

UNBOUNDED LINEAR OPERATORS n

VI.I.ll

Remark.

If T* is identified with

L

bkD/c as given in the

/c=0

preceding lemma, then the Lagrange adjoint T** of T* is defined.

In

n

the same way, T** can be identified with

L c/cD/c, where k=O

C/c

=

.f

(-1)J

J=k

(f)

b/ J- k )

VI.I.12 Lemma. Let C/c be as in Remark VI. 1.11. ak on I for 0 ~ k ~ n. Thus T** = T. of I.

Then

C/c

is identically

Proof. Let II = [a, b] be any compact subinterval of the interior Suppose g is in C""(11) and! is in C;(11). Upon replacing T by n

T* =

o

L bkDk

k=O ~ k ~

in Lemma VI.1.8 and noting that 0 = f
n - 1, it follows that

Since C; (11) is dense in £1(11) and both Tg and T**g are continuous on 1 1, it follows that Tg = T**g. By taking g to be the functions g/t) = t}

O~j~n

we obtain a/c = C/c on II, 0 :::; k :::; n. Since 1 1 was an arbitrary compact subinterval of the interior of I, the lemma follows. The lemma could also have been proved by computing the derivatives of c} in terms of the ai. VI.I.13

Corollary

TT,P.q = 'T';.,
1 ~ p, q 1

<

<

00

p, q:::;

TT.P.q is closed when 1 :::; p, q

<

00

00

or 1

<

p, q :::;

00

Proof. Since T** = T, p" = p, and q" = q, the corollary follows from Theorem VI.1.9 and the fact that conjugate and preconjugate operators are closed.

VI.I.H Corollary. The conjugate of TT.P,O is the minimal closed extenS1:on of 'p;i,o'.p' when 1 < p, q < 00,

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Proof.

135

By Corollary VI.1.13 and Lemma VI.1.3, T''T,P.t1.

= ('TIl, ,q,, I,)'' . = TIl,.,. ,Cl,,I', T

MINIMAL AND MAXIMAL OPERATORS

VI.2

n

Throughout this section,

T

=

I

akDk, where

k=O

O::;k~n

an(t) ;= 0, tEl

VI.2.1 Definition. The minimal operator corresponding to (T, p, q) denoted by TO,T,P,q, is defined to be the minimal closed extension of T~p,q when 1 ::; ,p, q < 00 When 1 < p, q ~ 00, the minimal operator is defined to be TT',q',p" VI.2.2 t.

Remarks

There is no ambiguity in the definition of the minimal operator since it follows from Corollary VI.1.14 that R T'. T ,q,,p, = T T.P,q

ii.

1

< p, q <

00

The reason for the seemingly unnatural definition of the minimal operator for the case when p and q are not both finite will be made clear in subsequent theorems. The maximal operator is an extension of the minimal operator; for when 1 ::; p, q < 00, then TO,T,P,q = T~p,q C TT,P,q, since TT,p,q is closed. When 1 < p, q ::; 00, it follows from Theorem VI. 1.9 that

Now TT,P,q = (T~,q"p')' by Corollary V1.1.13. Hence we can conclude from Lemma VI.1.5 that ('TT,P,q)' = TT,P,q. Thus TO,T,P,q C TT,P,q by (1). From the definition of the minimal operator, Theorem VI.1.9, and Corollary VI.1.13, the following relationships are obtained. JlI.2.3

Theorem.

For 1 ::; p, q

<

00,

136

UNBOUNDED LINEAR OPERATORS

For 1

< P, q $

iii.

iv. VI.2.4

<Xl"

T O,T,P,tl = T'.T ,11,.1'. T'O,T•• , ,)' ,12.p, = (TR" T ,q ,p

o=

Xl

T T, p''Il n

Suppose y is in A n(!), ry

Lemma.

or some a E I.

=

0 a.e., and

=

yea) = y'(a) = ... = y(n-ll(a)

Then y

=

0 on I.

Proof. Suppose y(XI) r!' 0 for some Xl E I. On la, Xl] define

For definiteness, assume

> a.

n-l

sex) =

2:

Iy(k) (X) I

k=O

and Xo =

sup {x Is(x)

Since s is continuous, s(xo)

=

=

0

X E

la,

Xl]}

0, or equivalently,

O:S;k:S;n-l Thus (1)

y(k)(X)

=

y(kl(X) - y(k)(xo)

O:S;k:S;n-l

and n-l

(2)

()

(k) ( )

y(n-l)(x) = (X y(n)(t) dt = _ (x \' ak t y t dt lxo lxo ~ an(t) k=O

Letting M(x)

=

sup s(t), (1) and (2) imply xo:S;t:S;x

(3)

o<

M(x)

:s;

M(x)

Ix: :~: (I ~~~~~ I+ 1) dt

X

> Xo

But this cannot be, since the integral in (3) is less than 1 when X is sufficiently close to xo. Thus y = 0 on I. The proof for Xl < a is similar. Vl.2.5

Theorem.

dim ff'{,(TT,P,q)

:s; n,

1

:s;

P, q $

00.

Proof. Suppose IYI, Y2, .. , ,Yn} is a linearly independent subset of ff'{,(7'T,'P,a)' For a E I, tho clot.orminallt of tho lIluLrix (l// k ) (a)), 0 $

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

137

k :$ n - 1, is not zero; otherwise there exists a set of numbers

C1, C2,

. . . , en, not all zero, such that n

°= L CiYi(k)(a)

= 0, 1, . . . ,n - 1

k

i=l n

Hence Y

=

L ciYisatisfies the hypotheses of Lemma VI.2.4 and is therefore ;= 1

°on 1.

This contradicts the assumption that {Yl, Y2, . . . , Yn I is linearly independent. Thus, given g E 'Jt(TT,P,q), we know from Cramer's rule that the system of equations

L n

g(k)(a) =

k = 0, 1, . . . , n - 1

Yi(k)(a)Xi

;=1

n

has a solution Xi = bi , i = 1, 2,

. , n.

Therefore g -

L biYi satisfies ;= 1

n

the hypotheses of Lemma VI.2.4.

Hence g =

L biYi, which proves that 1=1

dim 'Jt(TT,p,q) cannot exceed n. Note that the proofs of Lemma VI.2.4 and Theorem VI.2.5 also hold even when ak is locally integrable, ~ k ~ n - I, and l/a n is locally in

°

£",(1).

VI.2.6

1

~

p, q

Definition. A pair of numbers (p, q) is called admissible if < 00 or 1 < p, q :$ 00.

VI.2.7 Theorem. Let (p, q) be admissible. If anyone of the minimal or maximal operators corresponding to (T, p, q) or to (T*, q', p') has a closed range, then all four of the operators are Fredholm operators and

dim 'J)(TT,p,q) = K(T

) (T ) T,P,q - K O,T,p,q

'J)(TO,T,P,q)

Proof.

For convenience we write T

=

TT,p,q

To

=

TO,T,p,q

T.

= T.o,q'.'P'

UNBOUNDED LINEAR OPERATORS

138

If either '1' or '1'* has a closed range, it follows from Theorems VI.2.3 and IV.1.2 that either '1'*0 or To has a closed range. Thus we need only consider the cases when To or '1'*0 have closed range. Suppose To has a closed range. If 1 ::; p, q < 00, then by Theorems IV.2.3, VI.2.3, and VI.2.5,l

Thus '1' and To are Fredholm operators.

Similarly, by Theorem VI.2.3,

Hence we have shown that To, '1', '1'*0, and '1'* are Fredholm operators when To has a closed range and 1 ::; p, q < 00. If 1 < p, q ::; 00, then from Theorem VI.2.3 it follows that

Thus To, '1', '1'*0, and '1'* are Fredholm operators when CR(To) is closed and (p, q) is admissible. If CR(T*o) is closed, then the above results applied to '1'*0 in place of To and '1'* in place of '1' prove that '1'*0, '1'*, To, and '1' are Fredholm operators when (p, q) is admissible. Finally, we note that (1)

· 'J)(T) < d' Y d1m 'J)(To) ;rt(T) - 1m CR(To)

+

=

(j

('1' ) 0

<

00

since the linear mapping v:'J)(T)j('J)(To) + ;rt(T» -----+ Y jCR(To), defined by v[x] = [TxJ, is 1-1. Since 01.('1') < OCJ, (1) implies dim 'J)(T)j'J)(To) < OCJ and therefore, by Lemma V.1.5,

VI.2.8 Lelnlnu. Given compact interval [a, b] and numbers Co, CI, . . . , Cn-l, there exists a function g which is infinitely differentiable on the line such that g(k)(a)

=

Ck

g(x) ""' 0

k

= 0,

1, . . . ,n - 1

139

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Proof.

Let P be any polynomial such that p(kl(a) = Ck, k =0, 1, n-l

. , n -

1; for example, P(x) =

l

~~

(x - a)k.

By Theorem 0.8

k=O

there exists a rp which is infinitely differentiable on the line, has support in [a - 1, b), and is identically 1 on an interval containing a as an interior point. Define g(x) = p(x)rp(x). Then by Leibnitz's rule, k = 0, 1, . . . ,n - 1

from which the lemma follows.

VI.2.9 Lemma. If I contains one of its endpoints a, then every f in the domain of the minimal operator corresponding to (T, p, q), (p, q) admissible, satisfies the conditions

o=

. . . = j
f(a) = f'(a)

Proof. For convenience we write To = To,T,p,q, T R = T~p,q, '1\ = TT',q',p" Since T; = T* for 1 ~ p, q < 00 and To = T~ for 1 < p, q ~ 00, it follows that for f E XJ(To), g E XJ(T",)

(1)

Let [a, b] be a compact subinterval of I. For any g in XJ(T*) such that g(x) = 0 for all x 2':: b in I, we have from (1) and Lemma VI, 1.8 that n

k-l

l l

o=

(-I)iDi(akg)Dk-i-lf

J:

k= 1 j=O

Since, by Lemma VI,2.8, g(kl(a) may be chosen to have any prescribed values and g(kl (b) = 0, it is easy to show that

o= VI.2.1O Theorem. admissible. Then ~.

n.

k = 0, 1, . . . ,n - 1

Let I contain one of its endpoints and let (p, q) be

The minimal operator co'rresponding to (T, p, q) is 1-1. The maximal operator corresponding to (T, p, q) has range dense in £q(l), 1 ~ p, q < 00,.

"roo!. ThuH

j
'J'U,T,lJ'V

If

'J'O,T,IJ,UY

is 1-].

= 0, t,h6n

y

..fa by Lommafol VI.2.0 and VI.2.4.

140

UNBOUNDED LINEAR OPERATORS

Since T;,p,q = To,T.,q',p" 1 ~ p, q < 00, it follows from what was just proved and Theorem 11.3.7 that TT,p,q has range dense in £q(I). VI.2.11 Theorem. Suppose I contains one of its endpoints and (p, q) is admissible. If anyone of the minimal or maximal operators corresponding to (T, p, q) or (T*, q', p') has a closed range, then TT,P,q is surjective and TO,T,P,q has a bounded inverse. Proof. The theorem follows from Theorems V1.2.1O and V1.2.7 and Lemma IV. 1. 1.

MAXIMAL OPERATORS CORRESPONDING TO COMPACT INTERVALS

VI.3

VI.3.1 Theorem. Let I = [a, b] be compact and let T be the maximal operator corresponding to (T, p, q), 1 ~ p, q ~ 00, where T is the differential expression

Then 2.

n.

T is surjective and has kernel index n. Iff E £q(I), 1 ~ q ~ 00, then Ty = f, where yet)

= It ia

G(s, t) fees)) ds an s

and G is the continuous function on I X 1 expressed as the following quotient of determinants.

G(s, t) =

Yl(S) yi(s)

Y2(S) yHs)

Yn(S)

Yl(n-2)(s) Yl(t) Yl(S) y£(s)

Y2(n-2) (s) Y2(t) Y2(S)

Yn(n-2)(s) Yn(t) Yn(S)

y~(s)

y~(s)

Yl(n-l)(s)

Y2(n-l)(s)

Yn(n-l)(s)

y~(s)

with Yl, Y2, .. , ,y,. any basis for meT).

iii,

~(T) ~ II ;,.

II:l!![ l.t

IO(B, t)lq' ds

rqt l

l~p~oo

I
APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS 1

"{(T) ;:::

111-11II an

max

a~8~t

00

IG(s, t) 111;1

141

1~p~oo

q = 1

where "{(T) is the minimum modulus of T, £p(I), and q' is conjugate to q.

II lip is

the norm on

Proof of (i). From the condition put on an, we may assume, without loss of generality, that an = 1. We first prove that not only is T surjective, but that given f E£1(1) and constants co, C1, . . . , Cn-1, there exists ayE :D(T) such that Ty

=f

(1) O::;k~n-1

This is equivalent to finding y such that the following system of equations is satisfied. y(t) =

Co

y'(t) =

C1

+l +

l

y'(s) ds y(2)(S) ds

(2)

yCn-1)(t) = Cn _1

+

l

n-1

+

yCn)(s) ds = Cn -1

l L -ak(s)yCk)(s) k=O

+ f(s)

Letting Y, C and F be the 1 X n matrices,

Y(t) = [

:,~t~ J

C = [

yCn-1) (t)

~: J

F(t) = [.

Cn-1

~ .J

f(t)

and letting A(t) be the n X n matrix (defined for almost every tEl),

o

1

o

1

o

A(t)

we may write (2) in the form. (3)

Y(t) - C

+

t

A(s) Y(s)

+ 11'(8) d8

ds

UNBOUNDED LINEAR OPERATORS

142

with the understanding that an integral of a matrix B is the matrix of integrals of the entries of B. The problem now is to solve (3). To do this, the method of successive approximations is used. Define Yo(t) =

c

(4)

Y k+1(t) = C

+ fal A(s) Yk(s) + F(s) ds

k = 0,1, . . .

Let n

IIYk(t)11 =

L max IYik(S) I

i=1 a::;.:,>t

where the y/'(s) are the entries in Yk(s). IA(s)1 =

Defining

L /bii(s)l i,i

where the bii(S) are the entries in A(s), it follows easily from (4) that k = 1,2, .. The integral in (5) exists, since IA(s)1 is integrable and 1\ Yk(s) - Yk-1(S) II is continuous as a consequence of the continuity of the entries of Yk(s) Yk-1(S). It will now be shown, by induction, that for all tEl, O:::;k

(6)

where J(t) = lIA(s)1 ds and M = IIY1(b) - Yo(b)ll. When k = 0, (6) is trivial. Suppose (6) holds for k = j - 1. Since J'(s) = IA(s)1 a.e. and J(a) = 0, (5) and the induction hypothesis imply IIYi +1(t) - Yi(t) II

:::; fat

IA(s)IIIYi(s) ~ Yi - 1(S) II ds :::; (j

~ I)! fat J' (s)Ji- 1(S) ds

MJi(t)

=-.-,J. Thus (6) holds by induction. Now, IIYmet) - Yk(t) II converges uniformly to zero as m, k ~ 00; for if m > k, then (6) implies

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Since

eJ(b)

=

I

j=O

P\b) , the assertion follows.

143

Hence each entry y/.(t) of

J.

Yk(t) converges uniformly on I as k ---+ 00 to a continuous function Yi(t). Letting Yet) be the column matrix with entries Yi(t), it follows from letting k ---+ 00 in (4) that Y satisfies (3). Consequently a solution to (1) exists. By what was just proved, there exist Yo, Yl, . . . ,Yn-l in ;neT) such that

o ~ i, k

~

n - 1,

aik

the Kronecker delta n-l

The set {Yo, . . . ,Yn-d is linearly independent; for if 0

=Y=

L

aiYi,

i=O

then 0 = y(k)(a) = ak, 0 ~ k ~ n - 1. Thus aCT) by the remark preceding Definition VI.2.6.

= n, since aCT)

~

n

Proof of (ii). The proof uses the method of variation of constants. Suppose f E£q(I) C £1(1) is given. We seek aYE :neT) so that Ty = f. Let Yl, Y2, . . . , Yn be a basis for ;neT). Express Y in the form n

yet)

=

L Ck(t)Yk(t) , where the Ck(t) are to be determined. k=l

The conditions

which are initially put on the Ck are that they behave like constants when we differentiate Y successively. That is to say, we want n

y'(t) =

L Ck(t)y~(t) k=l n

y(2)(t) =

L Ck(t)Yk(2)(t) k=l

(7)

n

y(n-l)(t) =

L Ck(t)Yk(n-l)(t) k=l

The system of Eqs. (7) holds, provided that when we differentiate Y successively, we obtain

(8)

C~(t)Yl(t)

+ C~(t)Y2(t) +

+ C:(t)Yn(t) = 0

c~ (t)y~ (t)

+ c~(t)y~(t) +

+ c:(t)y:(t)

= 0

UNBOUNDED LINEAR OPERATORS

144

= 0, 1 ::=; j ::=; n, it follows

Assuming that (7) holds and recalling that rYi in a straightforward manner, that n

(ry)(t) = an(t)

(2:

CU t )Yk(n- ll (t)

a.e.

k=l

Thus if the Clc can be determined so that the system of Eqs. (8), together with the equation

holds, then Ty = f. An inspection of the proof of Theorem VI.2.5 verifies that the determinant Yn(t) y~ (t)

Yl(t) y~ (t)

is not zero for any t € I. Thus, for almost every t € I, the system of Eqs. (8) and (9) can be solved for c~(t) by Cramer's rule. Choosing Ck(t)

(t

I

•

= ia Ck(S) ds, we obtam

(10)

where G is given in (ii). Proof of (iii).

It follows from (10) and Holder's inequality that

l
(11)

and (12)

ly(t)1 ::=;

I~ II an

max IG(s,

00

a:;;':9

t)IIITylll

q = 1

Since dey, meT)) ::=; IIYllp, the proof of (iii) follows from (11) and (12). Note that since G is continuous, max IG(s, t)/ is continuous as EL function as.:5t ~

of f nnrl, in pnrtir,uln,r, ill in £,,( 1), 1

P ::=;

00,

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

VI.3.2

Theorem.

145

Let I be an arbitrary interval and let

T

be the differ-

ential expression ak locally integrable on I,

0

~ Ii; ~ n -

1

llocally in £",,(1) an Then the maximal operator corresponding to (T, p, q), 1 closed.

Proof. Suppose

~

p, q

~

00,

is

Let T be the maximal operator corresponding to (T, p, q).

(1) Tfn~

g E£q(I)

For J a compact subinterval of I, define T J to be the maximal operator corresponding to (T, p, q, J). Considering fn and f as elements of £p(J) and g as an element of £q(J), it is clear from (1) that

(2)

Now, by Theorem VI.3.1, T J is surjective and 'Y(T J) > O. Hence T J is closed by Theorem IV.1.7. Therefore we may conclude from (2) that f E 'J)(T J) and T Jf = g. Since J was an arbitrary compact subinterval of I andfis in £p(I), it follows thatf E 'J)(T) and Tf = g. Thus T is closed. VI.3.3 Corollary. Let I = [a, b] be compact and let T be as in Theorem VI.3.1. Suppose T is a 1-1 closed operator which is a restdction of the maximal operator corresponding to (T, p, q), 1 ~ p ~ 00, 1 < q ~ 00. Then T-1 is compact. Proof. Suppose {fk} is a bounded sequence in CR(T) C £q(I). By Theorem VI.3.1, there exists a G, continuous on I X I, and a sequence {Zk} in ~(T 1), where T 1 is the maximal operator corresponding to (T, p, q), such that (1)

Now H6lder's inequality implies

(t G(s, t) fk(S) ds

Ja

an(s)

146

UNBOUNDED LINEAR OPERATORS

Ill/a"lloo

where M =

sup IG(s, t)1.

Hence {yd is uniformly bounded

a<.,t
on I and, in particular~ isbounded in £p(I). Moreover, (zd is bounded in £p(I). If this is not the case, then there exists a subsequence {zk'\ such that for Xk' = T-l!k' and

(2)

Since ;n,(T l ) is finite-dimensional, there exists a subsequence {Zk"} of {Zk'} such that zk"/lIzk,,lI---+ v E ;n,(T l ). Thus, by (2) and Hence v E ~(T) and Tv = 0, which is impossible since T is 1-1 and Ilvll = 1. Therefore (zd is bounded. Consequently (zd has a convergent subsequence. Hence in order to prove that T-l is compact, it suffices to prove that {Yk'} has a convergent subsequence, where (Yk'} is any subsequence of (Yk}. For convenience, take (Yk'} to be {Yk} . We show that {Yk} is equicontinuous. Let e > 0 be given. Since G is uniformly continuous on I X I, there exists a 0 = O(E) > 0 such that for all s E I, (3)

Now

Thus (3) and Holder's inequality intply

IYk(t 2)

-

Yk(t l )!

~

q Mlt l - t 2 I l/ 'llfkllq

+ e II~JL (b

- a)1/Q'llfkllq

Since (11!kIIQ} is bounded, it follows that (yd is equicontinuous. Therefore, by the Ascoli-Arzela theorem, (Yk} has a subsequence converging in C(I) which, in turn, must converge in £p(I). Hence T-l is compact. The following example exhibits very simple maximal operators which do not have closed range.

VI.3.4 Example. Let I = [0, 00) and let n = D - ~e-u, where>. is a SCalftf. We show thnt fOf 7\ = 'r'A,P,f/' (p, q) admiRRihlo, nnd nil >., 'fA

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

147

does not have a closed range, or equivalently, by Theorem VI.2.11, T), is not surjective. Givenflocally integrable on I, the general solution in A1(I) for which TAy(t) = y'(t) - Ae-ity(t) = f(t) a.e. is given by y(t) = ee-
(1)

+ e-
fat e
where cp(t) -£Ae-it and e is an arbitrary constant. Since Icp(t) I = IAI, (2)

tEl

Let f(8) = e-
2.

00.

Hence Iy(t) I 2: Ie + tle-'A' , which is not in £p(I) , 1 ~ p ~ Thus f is not in CR(T A) when 1 ~ p ~ 00 and q = 00. 1 < q < 00. Define f on I by

££.

f(t)

k - 1

~

t

< k,

00.

k = 1,2, 00

Then f is in £q(I), since le-
elAI and

L l/kq <

00.

Let

Ie=l

y be as in (1). Given a positive integer N, it follows from (2) and the definition of f that for N + 1 2: t ~ N,

I

Iy(t) I = ee-
+ e-
E

e
I ~ e- 1e + let ~ 1- 1 1Aj

00

L 11k =

Since

00,

y is not in £p(I), 1 ~ p ~

00.

Hence f is

1e=1

not in CR(T'A,P,q) when 1 :::;; p ~ 00 and 1 < q < 00. To summarize, it has been shown that for all A (3)

T'A,P,q is not surjective when 1 ~ p ~

00

and 1

<

q ~

00.

The last case to consider for (p, q) admissible is

iii.

1 ~ p < 00, Si nee q' = co, 1 < p' ~ co, and (TA)· ... - T --A, it fol1owHfroffi (c:l) that T('A). ,V',l" is not surjectivCl ""Id 1,1 II 11'0 fOl'(1 , hy ThoOl'(lIn VU!.ll, 7"A,IJ'Q iR 1101; H\II'jO(~tivo. q - 1,

UNBOUNDED LINEAR OPERATORS

148

The above example also shows that the essential resolvent: {x I Xl - T v,p,p has closed range}

is empty when v = eitD and 1 S p SOC>. observation that Tv,p,p - XI = eitTTA,P,P' VIA

This is a consequence of the

EXTENSION THEOREMS

It is seen from Theorems VI.3.1 and VI.2.11 that minimal operators determined by certain differential expressions on a compact interval have a bounded inverse, while the maximal operators are surjective. This section treats the following problem. Let T 1 be a linear operator mapping a subspace of a Banach space onto a Banach space and let To be a restriction of T I such that To has a bounded inverse. Under what conditions does there exist a linear operator T which is a restriction of T I and an extension of To, written To C T C Tr, such that T is surjective and has a bounded inverse? The problem was considered by Visik [1) in his investigation of partial differential operators in a Hilbert space setting and by Browder [2) for more general spaces. VI.4.1 Definition. Let T I be a linear operator from one normed linear space into another and let To be a restriction of T I . The pair (To, T I ) is called potentially solvable if T I is surjective and To has a bounded inverse. The pair of operators (To, T I ) is called solvable (f there exists a linear operator T such that To eTC T I , T is surjective, and T has a bounded inverse. We shall also say that T solves (To, T I ). VI.4.2 Theorem. Let X and Y be Banach spaces. Suppose that (To, T I ) is potentially solvable with 5:>(T I ) C X and m(T I ) = Y. A sufficient condition that (To, T 1) be solvable is that T 1 be closed and there exist projections from X onto :n(T I ) and Y onto m(T o). An operator T which solves (To, T I ) may be constructed as follows. Suppose x = X o E9 :n(T I ) Y = Yo E9 m(T o) where X 0 and Yo are closed subspaces.

T(u

Proof.

Define T by

+ v) = Tou + T1v In a very natural way, an operator T which solvos (To, T 1)

will be determinod aftnr (1onFlidonnp; Borne proportios T must have,

149

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Now,

Y =
(1)

where Yo is the closed subspace which is the kernel of a projection onto
('1'1-1 is used in the set theoretic sense.)

We note that

for if z is in ~(To) and Tlz is in Yo, then Toz = Tlz €
and on

= (0).

~(T),

T(u

+ v) = Tou + Tlv

If ~(T) is chosen to be ~(To) ED Tl-lY O, then '1' might not be 1-1; for if v ~ 0 is in ;)[('1'1), then v is in Tl-lY O. Thus Tv = Tlv = O. To avoid this situation, we make use of the hypothesis that (2) where X 0 is a closed subspace of X by defining 'P by (3) (4)

From the way '1' was determined it is clear that To eTC '1'1 and that '1' is 1-1. We now show that '1' is surjective. Suppose y € Y. By (1), there exist u € ~(To) and w € Yo such that y = Tou + w. Since '1' 1 is surjective, there exists a v € ~(Tl) such that TID = w. Thus v is in '1'1-1 Yo. By (2), thereexistsavo€;)[(Tl) C TClYosuchthatv - voisinX o. Hencev - Vo is in X o n TI-lY a, and u + (v - vo) is in ~(T). Furthermore, 7'(u

+ (11 -

lill))

= J'-fJU + W

-

y

UNBOUNDED LINEAR OPERATORS

150

To prove that T-l is continuous, it suffices, by the theorem, to prove that T is closed. Suppose

closed~graph

.

(5)

Since 1'1 is a closed extension of 1', y = T 1x. The existence of a projection P from Yonto
Since 1'0- 1 is continuous, (7)

Hence from (5), (6), and (7),

(8)

T 1vn ..--+ y - w = T 1x -

W

But Vn is in X on T I-IY o, and both X oand Yo are closed. by (8), X - To-1w is in X o and T1x - w is in Yo. Since

Therefore,

we have

Hence X is in ~(To) EB X o n T I-IY o = l' is closed and the theorem is proved.

~(T),

and Tx = T 1x = y.

Thus

VI.4.3 Corollary. Suppose X and Yare Hilbert spaces. If (1'0,1'1) is potentially solvable and 1'1 is closed, then (1'0,1'1) is solvable.

Proof. Since To has a bounded inverse, the minimal closed extension To also has a bounded inverse. Hence
VI.4.4

Corollary.

Let

T

=

L akD\ ak

E

Ck(I), 0 ~ k ~ n, an(t) ,;e 0,

k=O

t E I, and let (p, q) be admissible. then the pair is solvable.

Proof.

If (7'O,T,V.q, TT.v.q) is potentially solvable,

ThoorOlllH VI.2.7, II. I. HI

lLlld

Vr.4.2.

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

15 1

For the remainder of this chapter, r is the differential expression

and (p, q) is admissible. VI.4.5 Corollary. If I is compact, then (To,..,p,q, T..,p,q) is solvable. If 1 < p =::;; <Xl and 1 < q ::; <Xl, any linear operator which solves (To,..,p,q, T..,p,q) has a compact inverse.

Theorems VI.3.1 and VI.2,l1 and Corollaries VI.4.4 and

Proof. VI.3.3.

The construction of T in Theorem VI.4.2 leads to the following characterization of all the differential operators which solve a pair of minimal and maximal operators. VI.4.6 Theorem. Suppose (To, T l) is solvable, where To = To,..,p,q and T l = T..,p,q. Then an operator T which solves the pair can be constructed as follows. i,

Letvl,v2, . . . ,vmbeabasisfor'J[(T..*,q',p.)andletYl,Y2, . . . ,Ym be in .£q(I) such that i,j

ii.

Let

Xl,

X2,

=

1,2, . . . ,m;

Oij the Kronecker delta

. ,Xm be elements in ::O(T l ) such that TlXi = Yi

If {Zl' Z2, . . . ,Zm} is a set of elements in 'J[(T I) (the elements need not be distinct), define T as follows. ::OCT) T(u

+ v)

= Tou

=

::OCTo) E9 sp {Xi

+ Tlv

U f

+ z;}

::0 (To) , v f sp {Xi

+ z;\

Then T solves (To, T 1). Conversely, any T which solves (To, T 1 ) is of the form given by (ii) , where {111, lh, . . . , 1/", I is l'inearly independent and

Y - 01('1'11)

(j) Hp

1111, Ih, . . . ,Ym}

152

UNBOUNDED LINEAR OPERATORS

Proof. Condition (i) is used to obtain a Yo as in Theorem VI.4.2, while (ii) is used for the purpose of exhibiting X o n TI-IY O• Writing T* = Tr*,q',p" we note that CR(T*) is closed by Theorem VI.2.7. Hence Theorems IV.1.2 and VI.2.3 show that

Thus if lVI,

1

~ p, q

1

<

<

p, q ~

00

00

, vm} and fYI, . . . ,Ym} are as in (i), then Yo = sp fYI, . . . ,Ym}

and CR(To) are linearly independent.

Since

1

<

p, q ~

00

it follows that Y = CR(T o) E9 Yo. Let both {Xl, X2, . . . , x m} and IZl, ... , Zm} be as in (ii). Then sp {Xi + z;} is a subspace of TI-IY O• We obtain next an X o with the properties required in Theorem VI.4.2. Let TV be a closed subspace of X such that (1)

x

= m(T t )

E9 sp IXi

+ z;}

E9 TV

The existence of TV is assured since ;R(T I) + sp {Xi sional. Define X 0 = sp {Xi + z;} E9 TV. Since

+ z;}

is finite-dimen-

it follows from (1) that

Hence the operator T solves (To, T I) by Theorem VI.4.2. Conversely, suppose T solves (To, T I), and Y = CR(T o) E9 Yo, Yo = sp IYI, Y2, . . . , Yml, dim Yo = m. Then :neT) C :n(T o) E9 TI-IY O, as was shown in the proof of Theorem VI.4.2. Since T is surjective, we are assured of element8 Ut, U2, . . . , Um in :neT) such that TUi = Tlxi = Yi, 1 ~ i ~ m. Thus there exist Zl, Z2, . . . ,Zm in ;R(T) slwh that Ui = Xi + z,.
APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

VE

153.

Tl-IY o, we have that v is in '.D(T) and Tv =

m

m

m

i=l

i=l

i=l

L a,Yi = L a;Tui = T L a;(xi + Zi) m

Since Tis 1-1, v =

L a;(xi + Zi).

Thus T has the form given in (ii).

i=l

VI.5

BOUNDARY VALUE FUNCTIONALS

VI.5.1 Definition. A closed linear operator which is an extension of the minimal operator and a restriction of the maximal operator corresponding to (T, p, q) is called a differential operator corresponding to (T, p, q).

Writing T 1 = TT,p,q and To = TO,T,p,q, we let D I be the space :,o(T l ) with the T I-norm II III defined by

Since T 1 is closed, D l is a Banach space. Suppose T is a differential operator corresponding to (T, p, q). If Do = '.D(T o) and Dr = '.D(T) are equipped with the Tl-norm, then Do and Dr are closed subspaces of D I with Do C Dr C D l . On the other hand, given a closed subspace M of D l such that Do C M, the operator T M which is defined to be T l restricted to M is a differential operator. Thus there is a 1-1 correspondence between the differential operators and the closed subspaces M of D I which contain Do, the correspondence being M - t T M • Suppose M is a closed subspace of D I and contains Do. Then by Theorem 11.3.4, M = .l(M.l), where M.l C D~, the conjugate space of D l . Since Do is contained in M, it is obvious that M.l C Do.l. Thus M is the intersection of the null manifolds of a set of linear functionals which are continuous on D l and which annihilate Do, the set of functionals being M.l.

Conversely, given any set {B",l of linear functionals which are continuous on D l and which annihilate Do, the space M = (\ ;n(B",) is a a

closed subspace of D 1 which contains Do. The discussion leads to the following definition and theorem.

VI.5.2 D(1inition. Let D 1' and Do be as in the above discussion. A junctional U whir.h iR r.ontinwmR on D1"and whid! annihilat{~iI /)0, that iR, IJ e 1>11.1 C 1>;, iii rall(l!l a Immultrry vtrl"" juuf·tIOlWI.

154

UNBOUNDED LINEAR OPERATORS

'VI. 5.3 Theorem. There is a 1-1 correspondence between the differential operators corresponding to (r, p, q) and the set of subspaces of the form

(1)

where {Bal is a set of boundary value functionals. The differential operators are obtained by restricting T 1 to sets of the form (1) ; i.e., differential operator T is the restriction of T 1 to those y E ~(T1) which satisfy the equations BaY = 0

all

a

VI.5.4 Definition. If T is the restriction of the maximal operator to (\ :n(Ba), where {Bal is a set of boundary value functionals, T is said to a

be determined by {Bal.

VI.5.5 Example. The notation used in this example will be that given in the discussion preceding Definition VI.5.2. Suppose that I contains one of its endpoints a. Let functionals B k be defined on D 1 by n-1

Bk(y) =

L bk;,y(j)(a)

k = 1,2, . . . ,n

;'=0

where the bk ;, are constants. Then each B k is a boundary value functional. To see this, let 1 0 be any compact subinterval of I which contains a- as an endpoint. Define D 2 to be the space ~(T1) with norm defined by

IIyl12 = IIyll1 +

n-1

L max ly(j)(t)1

i=O

ld.

It is not difficult to verify that D 2 is a Banach space. Since lIyl11 ::; IIyll2 for all y E D 1, we have that D 1and D 2 are isomorphic by the open-mapping principle. Each B k is clearly bounded on D 2 and therefore bounded on D 1. By Lemma VI.2.9, y E ~(To) implies y(j)(a) = 0, 0 ::; j ::; n - 1. Hence BkD o = 0, 1 ::; k ::; n. Thus {Bd is a set of boundary value functionals, and the restriction of the maximal operator to those y E~(T 1) which satisfy the "boundary conditions" Bky

=

n-1

L bkjy(jl(a) =

0

k

= 1,2, . . . ,n

i-O

is a differential operator. If the determinant of the matrix (b ki ) is not zero, it follows from Lemmn VI,2.4 ihut tho differel\t.iLti opemtor is 1-1.

t55

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Assuming that the minimal operator has a closed range, the next theorem characterizes, in terms of boundary value functionals, the differential operators. VI.5.6 Theorem. Let TO.T,p,q have a closed range and let T be a differential operator corresponding to (T, p, q). If IBa} is a set of boundary value functionals which determine T, then

dim sp IBa} = K(T I) - K(T) = dim

~~~~ <

Thus T is the restriction of the maximal operator to those y satisfy the conditions

where m = K(T I) - K(1') and IB I, B 2 , pendent subset of I B a }.

... ,

00

E ~(TI)

which

B m } is any linearly inde-

Proof. By Theorem VI.2.7, To = TO,T,p,q and T I = TT,p,q are Fredholm operators. Since-sp IBa} = DTl. C D~, where DT and D I are ~(T) and ~(TI), respectively, with the TI-norm, Theorem VI.2.7 implies

dim sp IB a } = dim DTl. = dim

(Z;)' = dim Z; ~ dim ~f~

0)

<

00

Thus dim sp IBa} = K(T I) - K(T) by Lemma V.1.5. VI.5.7 Corollary. Suppose the minimal operator TO,T,p,q has a closed range. If I B a } is a set of boundary value functionals which determines TO,T,p,q and I V~} is a set of boundary value functionals which determines TO,T*,q',p', then

dim sp IBa} = dim sp

I V~} =

<

K(TT,p,q) - K(To,T,P,q)

00

Proof. We write T 1 = TT,p,q, T* = TT*,q',p" To = TO,T,p,q, and T*o = TO,T*,q',p" Let CB = sp IBa} and'l) = sp IV~}. If 1 ~ p, q < 00, then T~ = T*o and T~ = T*. Hence it follows from Theorems IV.2.3 and VI.5.6 that 00

>

dim CB = K(T I) - K(To) = -K(T~)

+ K(T~)

= K(T*) - K(T*o)

If 1 00

>

<

p, q ~

00,

then T~.

dim ClI '"" ,,('I'.) - K('1'o)

=

= T 1 and

T~ = To.

=

dim'l)

Hence

K{7'~.) C K('I'~) = K('I'.) - K(T..) 0=

di III 'l)

156

UNBOUNDED LINEAR OPERATORS

The differential operators which solve the pair consisting of the minimal and maximal operators are characterized in terms of boundary value functionals as follows. 17.1.5.8 Theorem. Suppose (To.•.p.q, T.,p,q) is solvable. Let {B a } be a set of boundary value functionals which determines the differential operator T. Then T solves (To,.,p,q, T•.Q,q) if and only if the following two conditions are satisfied.

The dimension of the space CB spanned by the set {B a I is dim 'Jt(T.,p,q). If Y1, Y2, . . . ,Ym is a basis for 'Jt(T•. p,q) and B 1, B 2, . . . ,Bm is a basis for CB, then the determinant of the matrix (B;Yi) is not zero.

'/,. n.

Proof. Suppose T solves (To, T 1), where To = To,.,p,q and T 1 = T.,p,q. Then by Theorem VI.5.6

To prove (ii), assume the determinant II (BiYi) I is zero. exist scalars C1, C2, . . . ,cm, not all zero, such that

+ cmB 1Ym

m

L C,Yi.

Let Y =

Then there

= 0

ThenO ~ Y E 'Jt(T 1) and Bky = 0,1 ::; k ::; m.

Hence,

;=1

by definition, Y is in :n(T) and 0 = T 1y = Ty, which is impossible, since T is 1-1. Conversely, assume (i) and (ii). Suppose 0 = Ty = T 1y. Then m

Y =

L b;Yi for some b1, b2,

.. , bm. Since Y is in :n(T) and {Bd deter-

;=1

mines T, (1)

o=

m

Bky =

L biBkYi

k = 1,2, . . . ,m

;=1

The assumption that 1\ (BiYi) II ~ 0, together with (1), implies b; = 0, 1 ::; i ::; m. Thus Y = 0, whence Tis 1-1. Now, by (ii) and Theorem VI.5.6, (2)

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

157

Since T 1 is surjective and Tis 1-1, it follows from (2) that 0 = (3(T); that is, T is surjective. Furthermore, the closed-graph theorem applied to T-l proves that T-l is continuous. Hence T solves (To, T 1). For further properties of boundary value functionals, the reader is referred to Rota [1], Lemmas 5.3 to 5.5. Kemp [1) considers the effect on the minimal and maximal operators corresponding to (T, p, p), 1 ~ P ~ 00, when the leading coefficient an of T vanishes in the interior of I. The conjugates of the operators are determined, and the domains of differential operators are described in terms of boundary value functionals.

VI.6

SOME A PRIORI ESTIMATES

In this section, some inequalities, called a priori estimates, which are used to determine properties of certain differential operators, are derived. As we shall see in the next chapter, these estimates have their counterparts in the theory of partial differential operators.

VI.6.1

Lemma. For n a positive integer and 1 be the linear manifold defined by

t.

n.

p

~

00,

let Wn,p(I)

If f is in Wnp (1), then all the derivatives f(k), 0 ~ k ~ n, are in £p(1). For each S > 0, there exists a constant K depending only on s, p, and the length of I (l may be unbounded), such thatforallf E Wn,p(l),

IIJ
~

~ Kllfl!""l

+ sllJ
o~

k

~

n - 1, 1

o~

k

~

n - 1, p =

~

p

<

00

00

where II l!p,l denotes the norm on £p(I). Considered as a function of the length of I, K may be chosen to be nonincreasing.

Proof. It suffices to assume that g is real-valued. The lemma will first be proved for n = 2. For n = 1, the lemma is trivial. Suppose f is in W2,p(I), 1 ~ p ~ 00. Let I I j } be a sequence of nonoverlapping bounded intervals of equal length L whose union is I. Let J be any It. Subdivide J into throe Bubintm'vals of equal length and choose J 1 and J 2 to ho t.he two RubintorvnlR which nro 'soparnted hy the middle third Mil hi II !.prvnl.

158

~

=

UNBOUNDED LINEAR OPERATORS

For x E J 1 and Y E J 2 there exists, by the mean-value theorem, a ~Z.II in J such that f(x) - fey) = f'W

(1)

X-Y

Since Ix - yl ~ L/3, it follows from (1) and the absolute continuity of I' that for each t € J, (2)

If'(OI =

Thus, for p =

If'W + l 00

!"(s) ds

I ~ 3L-l(lf(x) I + If(Y)!)

+ I£f"(s) ds I

and t E J, II'(t) I ~ 6L- 111fIl00.1

+ LIIf"1100.1

Since J was an arbitrary 1;, I' is in £00(1) and

111'1100.1 ::; 6L- 1 1IfII00.1 + LIIf"1I00.1

(3)

Suppose 1 ~ p < 00. Integrating both sides of (2), first with respect to x on J 1 and then with respect to Y on J 2, gives 3- 2L

2

+ £, If(y)1 dy + 3::; [ If(s)1 ds + 3- 2L2 £ If"(s) I ds

11'(01 ::; [,

If(x) I dx

2L2

1[

f"(s) ds I

Thus, from Holder's inequality, II'(t) I ::; 3 2L-2£1, p'llfllp,J

+ £1, p'IIf"llp.J ::; L1/p'2(32pL-2pllfll~.J + 11f"1\~.J)1/p

(5)

Consequently,

Since J was an arbitrary 1;, we obtain

111'11~.1 = (6)

=

p p L 1\1'1I~.1j ::; 18 L-p L \\fll~.1j + 2 Lp L 11f"11~.1; i i i 18PL-pllfll~.1

+ 2pLpllf"\I~.1

Let E > 0 be given. Suppose p = 00. If the length of I, written leI), does not exceed E, then taking L = leI) in (3) gives

(7)

11f'11...1 ~ l(~) Ilfll".1 + 5111"11.. ,J

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

159

00 > leI) > e, then I is the union of nonoverlapping intervals II and I 2, where l(11) = ke/2 for some positive integer k and e ~ l(1 2) ~ e/2. Applying (3) to f on the interval II with L = e/2 yields

If

(8)

Similarly, taking L

=

l(12) in (3) gives

(9)

1I!'11"'.1, ~

I; Ilfll"'.1, + ellf"II"'.1.

Thus, from (8) and (9), (10)

For e >

Ilf'II",.1 ~

I; Ilfll"'.1 + ell!"I1",.1

°fixed, define 12 K(l(1»

=

leI) 12

1e

leI) ~ l(1)

e

>e

Then K is a nonincreasing function of the length of I, and for (11)

11!'11",.1

~

00

>

leI),

Kllfll",.1 + ell!"II",.1

by (7) and (10). If leI) = 00, then (11) still holds upon choosing L = e in (3). Hence the lemma is proved for n = 2 and p = 00. By making use of (6), a similar argument proves the lemma for n = 2 and 1 ~ p < 00. Assume the lemma to be true for n = j. Suppose f is in W i +l. p (1), 1 ~ p < 00. Then for any compact subinterval J of I, f(j-ll is in W(2.V) (1). Now, the lemma holds for n = 2. Hence, given 1/ > 0, there exists a K o depending only on 1/, p, and the length of J such that f(jl is in £p(J) and (12)

Given 1/1 > 0, there exists, by the induction hypothesis, a K 1 depending only on 1/1, p and the length of J such that (13) Thus, from (12) and (13),

UNBOUNDED LINEAR OPERATORS

160

Choosing '71 so'that 1 - K 0'71

i, it follows from (14) that

;::::

(15)

Since '7 and '71 are arbitrarily small and K o and K 1 can be chosen to be nonincreasing functions of the length of J, the lemma for 11. = j + 1 follows from (13) and (15). By induction, the lemma is proved. The proof when p = 00 is the same, with I II", replacing II II~. VI.6.2

Theorem.

Let r be the differential expression r =

L" cl
where 1/cn and ~.

ii.

Ck,

0 :::; k :::;

11. -

1, are in £",(1).

If f is in 'J)(T"p,p) = (g I g f A n (1) n £p(I), rg f £p(I) I, 1 :::; p:::; 00, then jCk), 0:::; k:::; 11., is in£p(I). There exists a K, depending only on p, 11., the length of I, and the largest of the numbers 111/cn ll",,r and Ilckll""r, 0 :::; k :::; 11. - 1, such that for all f f 'J)(T"p,p) ,

o :::; k o :::; iii.

:::;

11.,

1 :::; p

k :::;

11.,

p =

<

00

00

Suppose 1/c n and Ck, 0 :::; k :::; 11. - 1, are bounded on the line. Then K, considered as a function of the length of I, can be chosen to be nonincreasing.

Proof. We shall prove the theorem for 1 :::; p ment for p = 00 is the same, with II II", replacing II arbitrary compact subinterval of I. Since

< 00. The arguII:. Let J be an

n-1

rf (1)

j
and jCk), 0 :::; k :::; is in £p(J) and

11. -

=

L

CkjCkl

k=O Cn

a.e.

1, is continuous on J, it follows from (1) that .. - I

L IIf(k)II",J) :::; CTr(11. + 1) (1Irfll:,J + Cr" L Ilrlll:,J )1/"]

IIf(")II",J :::; Cr (II rfll",J + Cr

k-O

n-l

1<-0

jCn)

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

where CI is the largest of

161

Ill/cnll oo ,I and Ilck\loo,I,

0 ::; k ::;

71. -

1.

Hence

n-l

IIJ
(2)

K o (1ITfll~,J

+

L

k=O

IIJ
where K o depends on CI , p, and n. Given e > 0, there exists, by Lemma VI.6.1, a constant K r depending only on e, p, and the length of J such that (3)

Thus, from (2) and (3), (4)

(1 -

Kone) IIJ
KoIITfll~,J

::;

KoIITfll~,I

In particular, for e chosen so that 1 -

+ KoKrnllfll:,J + KoKrnllfll~,I

Kone ;;::: i,

(5)

Now I may be expressed as the union of compact subintervals J r, J 2 , . . . , where J r C J 2 C . . .. Furthermore, we know by Lemma VI.6.1 that K r may be chosen to be a nonincreasing function of the length of J. Hence, letting J = J r in (5), it follows that i = 1,2, . . .

Therefore

f(n)

is in £p(I) and

(6)

where K depends only on p, 71., C I, and the length of I. Finally, there exists, by Lemma VI.6.1, an M depending only on p and the length of I such that (7)

Since K rand M, considered as functions of the length of I, can be chosen to be nonincreasing, the theorem follows from (6) and (7).

VI.6.S

Corollary.

Let r =

L" akD\

where each ak is a constant.

At-a

Then for the differential expression D, T',1J,l' "" P(TD,tI,JI) , where P is the n

polynomial

L a~z~ and t M-II

~ p :$

<Xl.

UNBOUNDED LINEAR OPERATORS

162

Proof. For convenience we write TT,P,P as TT' and TD,p,p as T D. Supposej E 'J)(P(T D)) = 'J)(Ti). Thenj E An(I) n £p(I) , and j
o ::; k

::; n.

Thus Tj =

L ak!(k) is in £p(I); that is,! e 'J)(TT)'

Furtherk=O more, (P(TD»j = T<J. Since Theorem VI.6.2 implies 'J)(TT) C 'J)(T'lJ) , we may conclude that TT = P(T D).

VI.6.4 i.

Remarks

As seen in Lemma VI.6.1, Wn,p(I), 1 ::; p ::; 00, concides with {g I g E An(I), g(i) e £p(I), 0 ::; i ::; n}. If we define a norm II \\n,p,l on Wn,p(I) by Iljlln,p,l =

f

11j
k=O

where II IIp,l is the norm on £p(I), then Wn,p(I) is a Banach space. To see this, suppose (jj I is a Cauchy sequence in Wn,p(I). Then obviously (f/k)} is a Cauchy sequence in £p(I) for 0 ::; k ::; n. Hence (fP)} converges in £p(I) to some hk • By Theorem VI.3.2, Tk,p = TDk,p,p is closed. Since eachJi is in 'J)(Tk,p), o ::; k ::; n, and Tk,pjj = f/k) ~ hk, it follows that h o = lim Ji is in 'J)(Tk,p) and hO(k) = Tk,ph o = lim Tk,pjj = lim jP) i~

j-+ 00

ii.

co

Thus {fj} converges in Wn,p(I) to h o, proving that Wn,p(I) is complete. If X n+ 1 is the Banach space £p(I) X £p(I) X . . . X £p(I)

---.

with norm

II (go,

gl, . . . , gn) II = •

'./

n

~

n+1

L Ilgillp,l,

then Wn,p(I) is

i=O

equivalent to the subspace M n+l of Xn+l consisting of those elements (i, 1', . . . ,j(n» where j is in Wn,p(I). Since Wn,p(I) is complete, M n+l is also complete and hence a closed subspace of Xn+l' Thus if 1 < P < 00, then X n +1 is reflexive, implying that Wn,p(I) is reflexive. If 1 ::; p < 00, then Xn+l is separable and therefore Wn,p(I) is separable.

VI.7

THE CONSTANT COEFFICIENT AND THE EULER DIFFERENTIAL OPERATORS

VI.7.1 Definition. Let T be a linear operator with domain and range conta1:ned in normed linear space X. 7'he (JRRential Rpectrl/.m of T,

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

163

written 0-.( T), is defined by ue(T) = {X I
VI.7.2

Theorem.

Let

T

L akDk, where each ak is a constant, and = L akz The maximal operator TT,P cor-

=

k=O

n

let P be the polynomial P(z)

k

•

k=O

responding to (T, p, p) and the interval [0, i.

ii. iii.

Ue(TT,P)

00)

has the following properties.

= {P(z) I Re z = OJ

For X e Ue(TT,p),
is the number of zeros of X - P(z), counted according to their multiplicity, which lie in the half plane Re z < O. Proof. Since XI - TT,p = T"-T,p, Theorem VI.2.11 implies X e Pe(TT,p) if and only if A! - TT,P is surjective, 1 ~ P ~ 00. Suppose 1 ~ P < 00. Now, by Theorems VI.2.7 and IV.2.14 and Corollary VI.6.3,

Thus the problem reduces to finding Fu(TD,p). the equation (AI - D)y = j has a solution yet)

= -

lot eMt -

o)

Forje£p(J),J = [0, 00),

f(s) ds

in AI(J O) for every compact subinterval J o of J. SUppORO R{l X < O. Thon?J 18 in J;p(J) by theorem 0.11. ThUB fJ(X! _. '1'11",) - O. Now, (i"l ill ill m(>.! - TII,Il), 1\IIl)lv(XT - TII,I') ~ 1.

164

UNBOUNDED LINEAR OPERATORS

Hence a("l>.I - TD,p) = 1 and Re"l>. < 0

(2)

Suppose Re "l>. > O. By Theorem V1.2.11, H - TD,p = TX-D,v is surjective if and only if THD,pl is surjective. Given g E £p,(J)

is a solution in AI(J) (\ £p,(J) to the equation "l>.y + y' = g by Theorem 0.11. Thus TX+D,pl is surjective and consequently (3(H - TD,p) = O. Since eXt is a solution to "l>.y - y' = 0 on every bounded subinterval of J and eXt is not in £p(J), it follows that a(H - TD,p) = O. Thus (3)

Re"l>.

>0

Suppose Re"l>. = O. Then"l>. is not in Pe(TD,p) = Fp(TD,p); otherwise it would follow from Theorem V.1.6 that K(p,[ - Tv,p) = K(H - Tv,p) for all p, sufficiently close to"l>.. But this contradicts the fact that both (2) and (3) hold. If p = 00, then by Theorem V1.2.7, (T.(TT,«» = (Te(T,-,l), which was just shown to be the set (p*(z) IRe z = OJ, where p* is the polyn

nomial

1 (-I)kakzk

n

1 (-I)k akDk.

corresponding to T* =

k=O

Since

k=O

P*(z) = P( -z) for all complex numbers z, it follows that U.(TT,«» = (P(z) IRe z = O}

Since (H - TT'P)' = TO,X-T-,P" 1 ~ p < 00, Theorem 11.3.7 implies that CR(H - TT,P)iS dense in £p([O, 00», 1 ~ p < 00. Thus (i) is proved. The rest of the theorem follows from (2), (3), and Theorem 1V.2.14. The classical method of solving the Euler differential equation n

1 bktkDkf = g, where each bk is a constant, is to introduce a change of k=O

variable, thereby transforming the given equation into one of the form vft = gl, where v is a differential expression with constant coefficients. A similar procedure is used in the proof of the next theorem. n

VI.7.3

Theorem. Let T =

1 bktkDk, where each bk is a constant, and k-O

let P be the polynomial

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

165

with lip = 0 when p => 00. The maximal operator TT,p corresponding to (T, p, p) and the interval [1, 00) has the following properties. i,

U-.(TT'P) = (P(z) IRe z = O} For A E U.(TT,p), CR(Al - TT'P) is a proper dense subspace of £p([I, 00».

ii. iii.

For A E P.(TT,P)' AI - TT,P is surjective, 1 If 1 ~ p < 00 and A is in P.(TT,P)' then

~

p

~

00.

is the number of zeros of A - P(z), counted according to their multiplicity, which lie in the half plane Re z < O.

Proof. Let J 0 = [0, 00) and J 1 = [1, 00). Then £p(J 1) is equivalent to £p(J 0) under the map 7J: £p(J 1) - 7 £p(J 0) defined by O~s<

(1)

00

with the understanding that e'/ p = 1 when p = 00. The statement follows from the well-known theorem concerning integration by substitution (cf. McShane [1], 38.4, page 214). The inverse map '17- 1 is given by (2)

f(t) = ('I7- 1g)(t) = t- 1 / Pg(log t)

l~t
Since e' and log t are monotone increasing and infinitely differentiable on the line and J o, respectively, it follows from (1) and (2) that (3)

Suppose that f is in A n (J 1).

(cf. McShane [1], 9.3, page 51). 'l7f = g is in An(Jo), and by (2),

f'(t)

=

t- 1-

1/ p

=

t- 1-

1 p /

g'(s) ].=log t

[(~ . ds

It follows by induction that for 1

-

~

-

1 1 1p t- - / g(s)].=log t p

-

!)p g(S)] k ~ n,

.=logt

Then

166

UNBOUNDED LINEAR OPERATORS

Hence

Let v be the differential expression

and let T ',p,P be the maximal operator corresponding to (v, p, p) with respect to the interval J o. Then from (4) and the definition of 1/ (5)

Thus, by (3) and (5), T.,p,p

=

1/-1T.,p,p 1/, or equivalently, for any

scalar~,

Since 1/ is a linear isometry, T.,p,p and T.,p,p have the same essential spectrum and K(Al - T.,p,p) = K(Al - T ',P,p) for ~ E p.(T.,p,p). The theorem now follows from Theorem VI.7.2. The results in the remaining sections of this chapter generalize those due to Balslev and Gamelin [1] for the constant coefficient and the Euler differential operators.

VI.8

PERTURBATIONS OF THE BOUNDED COEFFICIENT AND THE EULER OPERATORS

VI.B.l Theorem. Let J be the interval [a, 00) and let T be the maximal operator corresponding to (7, p, p), 1 < P < 00, where 1 an

-

ak

E

E

£~(J)

£~(J)

O=::;k=::;n Let B be the maximal operator corresponding to (v, p, p), 1 each bk measurable on J


00,

where

167

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Then i.

B is T-bounded if and only if each bk is locally in £p(J) and 0=::;k=::;n-1

In this case, given s > 0, there exists a K, depending only on and p, such that for all f E 'J:J(T), IIBfllp,J =::; Kllfllp,J

ii.

S

+ sllTfllp,J

Furthermore, T r +. = T + B, where T r +. is the maximal operator corresponding to (7 + V, p, p). B is T-compact if and only if each bk is locally in £p(J) and lim ..... .,

j 8+

1

Ibk(t) Ip dt

=

0

8

In this case, T and T r +. have the same essential spectrum. X E p.(T), K(Xl - T) = K(XI - T r +.).

For

Before proving the theorem, we need the following lemma.

VI.8.2 Lemma. Given interval I with length greater than 1 and E > 0, there exists a K, depending only on S, the length of I, and p, 1 < P < 00, such that for all b locally in £p(I) and all f in the domain of the maximal operator T corresponding to (D, p, p),

Ilbfll~.r =::; (sllf'II:.r

+ Kllfll:,r) 18,8+1] sup j8+1 Ib (t)I P dt CI 8

K may be chosen to be nonincreasing as a function of the length of I. Proof. Let II and I 2 be nonoverlapping subintervals of I such that I = II U h with II "to the left" of h For 7J > 0 such that t 7J and t - 7J are in I for all t E II and I 2 , respectively, choose cp E C' ([0, 7JJ) so that 0 =::; cp(t) =::; 1 on [0, 7Jl, cp(O) = 1, and cp(7J) = O. For f E 'J:J(T) and t E II,

+

f(t) - -

[(~

d (cp(s)f(t

jll (s l

+ H»

dB ... - jo,r.~ cp(s)f'(t

+ s) ds

- Iu~ cp' (8)f(t + 8) dB

168

UNBOUNDED LINEAR OPERATORS

Letting M = max 1~'(8)1, we obtain from Holder's inequality o:::;.:::;~

(1)

IJ(t)!:::; fo~ If'(t

+ s)1 ds + M fo~ IJ(t + s)1 ds :::; 7/1/P' [( fo~ If'(t + s)!p ds t + M (fo~ IJ(t + s)lp ds t :::; 27/I/p' (fo~ If' (t + s) Ip ds + Mp fo~ If(t + s) Ip ds t P

P ]

P

Taking 7/ sufficiently small, it follows that there exists a K I depending only on e, p, and the length of I such that If(t)lp :::; ~ fo~ !f'(t

(2)

+ s)lp ds + K

e {t+~ = 2 }t 1f'(s)lp ds

+K

I

fo~ IJ(t

(t+~

I

}t

+ s)lp ds

If(s)lp ds

From the conditions put on 7/ we see that as the length of I increases, 7/ can be kept fixed so that (1) still holds. Thus K 1 can be chosen to be nonincreasing as a function of the length of I. Letting a be the left endpoint of I I, Fubini's theorem and (2) imply that for tEl I and 7/ sufficiently small, (3)

h, Ib(t)f(t)!p dt :::; h, f+~ Ib(t)lp G[f'(s)[p + KtIf(s)[p) ds dt :::;

;;~ 1f'(s)lp + Kt\f(s)lpds J~aX(8-~,a) Ib(t)lpdt

: :; (-2e 11f'11~.I + KII\f!I~'I)

sup

[8,8+11CI

(8+1 Ib(t)lp dt

}8

For t E h J(t)

= -

(~

d

}o ds (~(s)J(t - s)) ds

Thus, by the argument used to establish (2),

As in (3), (4)

h. Ib(t)J(t)lp dt :::; £~ IJ'(s)lp + KIIJ(s)lp ds t+~ Ib(t)lp dt : :; (-2e 1\f'II~,I + Kl[lfl[~'I) [•.•sup /."+1 Ib(t)lp dt +IICI

The lemma now follows from (3) nod (4).

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

169

Suppose bk is locally in £p(J) and

Proof of the theorem.

O.::s;k.::s;n-l

Let J 1 be any subinterval of J of length greater than 1. Since any nonnegative numbers e and d satisfy (e P + d p )1lp .::s; e + d, it follows from Lemmas VI.8.2 and VI.6.1 and Theorem VI.6.2 that given J.l > 0 and 1/ > 0, there exist constants K o and K 1, depending only on p and J.l and p and 1/, respectively, such that for all f E 'neT),

Using equations (1) and (2), a simple computation shows the existence of a K 2, depending only on p and e, such that for all f E 'neT), (3)

It follows from Theorem VI.6.2 that there exists a constant C, depending only on p, such that for all f E 'neT), (4)

We may derive from (3) and (4) an inequality of the form n-l

(5)

IllIfl!p,J,.::s;

lllbk!(kl l\p.J ~ 1

Kllfllp.J,

+ ellTfllp,J,

f

E

'neT)

k=O

where K depends only on p and e. It is clear from (5) that 'neT) is contained in 'n(B) and that B is T-bounded. To prove TT+. = T + B, it suffices to show that 'n(TT+.) = 'neT). If f E 'neT), Tf and IIf are in £p(J) by (5), which implies f E 'n(TT+.)' On the other hand, if f is in 'n(TT+.), then (T + lI)f is in £p(J). Now, f(j), 0 .::s; j .::s; n - 1, is continuous on each compact subinterval J 1 of J. Hence IIf and Tf = (T + lI)f - IIf are in £p(J 1). By considering only those J 1 of length greater than 1 we have from (5) and the triangular inequality that

whence (0)

O<e
170

UNBOUNDED LINEAR OPERATORS

Since (6) holds for all subintervals J 1 of length greater than 1, it follows that vf is in £p(J). Consequently, rf = (r + v)f - viis in £p(J), implying that f is in :oCT). Thus T + B = TT+' and (5) implies

f E :OCT) where K depends only on p and c. Suppose B is T-bounded. Let 'P be a function in C; (El) such that 'P = 1 on [0, 1]. For each 8 ;::: 0, define 'Ps to be the function 'P translated so as to be 1 on [8,8 + 1]; that is, 'Ps(t) = 'P(t - 8). Now n-l

L bk'Ps(k) = bo

B'Ps =

on

[8,8

+ 1]

k=O

Hence (7)

(1'+1 Ibol Y'P ~ IIB'Psllp,J ~ IIBIITII'PsIIT =

where

p

II liT is the

1

00 -00

T-norm on :oCT).

dk 'Ps(t) dt = dtk

1

00

-00

IIBIIT(iI'Psllp,J

+ IIT'P.llp,J)

Since

k d 'P(t - 8) dt = dt k

1

00

_00

'P(k)(t) dt

it follows from (7) that (/.'+1 Ibo(t)IPdtY'P

~

00

IIBIIT [(/_

00

1'P(t)\PdtY'P n

+ lilakilco (!-ooool'P(k)(t)IPdtr'P] k=O

Thus sup ass
{s+l

J8

Ibo(t)lp dt

<

00

8

Suppose (8)

sup ass
(s+1

J8

IMt) Ip dt

<

00

O~i~k-l
8

°

We show that (8) holds for ~ i ~ k. Define h to be in C; (El) so that h(k) is 1 on [0, 1]. This can be done, for example, by defining h to be 'Pg, where g = tk/k!. Let hs(t) = h(t - 8). Then h.(k)(t) = 1 on [8,8+ 1], and on [s,

8

+ 1]

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

171

Hence (9)

(/.'+1 Ibk(Olp dtYIP ::; \lBIITllh.IIT + M

k-1

I

(/.'+1 \b.(t)lp dt)l/P

.=0

where M

~

Ih(i) (t) I, 0

sup

::; i ::; n - 1.

Since

-"'<1<'" n

Ilh.IIT::; (/_"'",

jh(t)lp dt)l/P

+ Illa;ll", (j_"'", Ih(i'(Olp dtYIP i=O

equation (9), together with (8), implies O::;i::;k Thus (8) holds for 0 ::; i ::; n by induction.

Proof of (ii).

Hence (i) is proved.

Suppose bk is locally in £p(J) and

(10)

For each positive integer N

> a define B N on :oCT) by on [a, N] on (N, 00)

We show that B N converges to B in the space of bounded operators on :oCT) with the T-norm. It follows from Lemma VI.8.2 and Theorem VI.6.2 that there exist constants K and C depending only on p such that for all f E :OCT), .. -1

(11)

L (tv'" Ibk(t)j
IIBf - BNfl1 ::;

k=O

n-1

Hl

,

k=O

+ Kllj
(sup /.8+1\bk(t)!Pdt)l/P N5,8<'" 8

(12)

Considering :.D(T) to he the Bnrlach HPILcc~ith t.he T-norm, it is clear from (10), (11), (Lilt! (12) Llmt UN - t JJ ill [:'0(7'), £,,(.I)J. To prove 1,111Lj, JJ iH 'l'-eolll\lILd, it, HllrJi('('H j,o Hhow t.Il1LL IllLdl llN iH 'J'-eomplLld.. HUPPOHI\ 1)'11

UNBOUNDED LINEAR OPERATORS

172

is T-bounded. Then UP) \, 0 ~ k ~ n - 1, is uniformly bounded on J, since equation (2) in the proof of lemma VI.8.2 implies the existence of a constant K depending only on p such that

which, together with (12), shows that UP)} is uniformly bounded on J. Furthermore, UP)} is equicontinuous on la, N], 0 ~ k ~ n - 1, since, again by (12), IfP)(t) - fP)(s) I =

If

fP+1)W

I

d~ ~

It - S\1 Ip 'IIf/k+1)llp,J

~ Cit - S\1IP'(llhllp,J

+

IITJillp,J)

Hence, by the Ascoli-Arzelli theorem, {h} has a subsequence {hot which converges uniformly on la, N], and U;,o} has a subsequence U;,d which converges uniformly on la, N]. Thus {hI! and U;,d converge uniformly on la, N]. Continuing in this manner, a subsequence {gj} of {h} is obtained such that for 0 ~ k ~ n - 1, {gP)} converges uniformly on la, N]. Thus "-1

l ~ l

I!BNg; - BNgjllp,J ~

(faN Ibk(t)lp\g;Ck)(t) - gP)(t)\P )11P

k=O

n-1 k=O

(sup Ig;Ck)(t) - gP)(t)I) (faN Ibk(t)!p dt)1 lP a$!$N

implying that {BNgj} converges in £p(J) as j ~ 00. Hence B N is T-compact and therefore B is T-compact. Clearly, any set which is AI - Tbounded is T-bounded. Moreover, it follows from (i) that any set which is AI - T - B-bounded is also T-bounded. Hence B is both AI - Tand AI - T - B-compact. Thus it .follows from Theorem V.3.7 that p.(T) = p.(T + B) = Pe(T,+v) and A E p.(T) implies

S

Conversely, suppose B is T-compact. Assume there exists some {sd of positive numbers converging to 00 such that

> 0 and a sequence

Let {!Ps} be the functions defined in the proof of (i). (13)

Then

173

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

Now

which means that {CP'k I is T-bounded. Thus, by passing to a subsequence if necessary, we may assume that (Bcp'kl converges in £p(J) to some y. Given a finite interval Joe J, CP'k = 0 on J 0 for all Sk sufficiently large, since cP has compact support. For such Sk, as

Sk

~

00

Since J 0 was an arbitrary finite subinterval of I, y must be the zero element in £p(J); that is, Bcp'k ~ O. This, however, contradicts (13). Hence lim

......... /.'

,+1

Ibo(t)lp = O.

The proof that

proceeds by induction as in the proof of (i). the theorem.

This concludes the proof of

For results related to Theorem VI.8.1, the reader Birman [1].

IS

referred to

VI.8.3 Corollary. Given J = [a, 00) and 1 < p < 00, let T and L be the maximal operators corresponding to (T, p, p) and (l, p, p), respectively, where

bn and (an

~ bn) in £ ... (J)

bk locally in £p(J) ~ Ibkl p Ell 0, 0

. f+l Then :neT) - :D(l,) , O',(T) = 0'.(1,), and>.

f

~

p,(T) implies

k

~

n

174

UNBOUNDED LINEAR OPERATORS

Proof. In view of Theorem VI.8.1, it suffices to prove the corollary for l = T + bJ)n. Theorem VI.6.2 implies that :oCT) = :0 (L). For any 13calar X, n-l

X-

(1)

l =

X-

= (1

T

+ b (X - + n

T

an

bn) +a (X

- T)

.

L akDk -

k=O

+ n~ ~

1

k=O

n

bn

ak -

an

X)

Dk - bn -

X

an

Let AA be the maximal operator corresponding to (1 + bn/an)(X - T). It follows from (1) and Theorem VI.8.1 that
(Tf)(t) = ~ aktkjCk)(t)

each ak constant, an

;C

0

k~O

(If)(t) = (Tf)(t)

+

n

~ bk(t)tkf(k)(t) k=O

the coefficients bk being subject to the following conditions: i. ii.

bn and l/(an + bn) are in £",(J). bk is locally in £p(J) with lim· s-t

00

,:.+1 -t1 Ibk(t)lp =

Je

0

o ~ k ~ n.

Then :OCT) = :O(L), u.(T) = u.(L), and lor X E p.(T), K(X! - T) = K(X! - L)

Proof. As in the proof of Theorem VI.7.3, we let 1] be the linear isometry from£p([I, 00)) onto £p([O, 00)) defined by (1]1)(8) = eo/pl(e') , o ::;; s < 00. Then, as was shown, for I E :oCT) and g = 1]f, k-l

(1)

tkJ
{n,-0. (i!-ds - (!P + f))

(/(8)]

} 0_10Kl

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

175

and (Tf)(t) = t-llp

tao +

I ak kfl (~ - (~ + j))

k=1

j-O

P

g(S)]

} .-logt

Furthermore, (1) implies that

As in the proof of Theorem VI.7.3, it follows that (2)

and (3)

where M and V are the maximal operators with domains contained in corresponding to the differential expressions

£p([O,

00»

and

respectively.

Noting that for x

~

0,

the theorem follows from (2) to (5) and Corollary VI.8.3. As an application of the above theorem, the following result, due to Rota [2], is obtained..

rt,

VI.S ..') I~Xff,m",1(l. For.! 00) arull < p < 00, letL be the maximal opornj,or eorrllMpollclilllC 1.0 (l, p, p), whoro l if! t,hn Riemann diffClfollt.ia,1

176

UNBOUNDED LINEAR OPERATORS

expression (If)(t)

(1)

=

t(t

+ I)j<2)(t) + (at + b)f'(t) + ct 2t(t++dt 1)+ e j(t)

with a, b, c, d and e constants.

We write (1) in the form

(If) (t) = t 2f(2) (t)

+ at.f' (t) + cf(t) + (II f) (t)

(2)

where (3)

with b2(t) = lit, b1(t) = bit and bo(t) = (d - c)/(t + 1) + elt(t + 1). Clearly, bo, b1 and b2 satisfy the conditions of Corollary VI.8A. Hence ue(L) = ue(T), where T is the maximal operator corresponding to t2j(2)(t) + atf'(t) + cf(t) = (Tf) (t). Furthermore, if i\ is in Pe(T), then K(i\I - T) = K(i\I - L). Thus, by Theorem VI.7.3, (4)

ue(L)

=

{-r 2 + -l p2

+ (1

- a) P

+ c + ir (a

- 1-

~) P -oo
For i\ f PeeL), K(i\I - L) is the number of zeros of the polynomial (5)

i\ -

c -

a(z - ~) - (z - ~) (z - 1 -

~)

counted according to their multiplicity, which lie in the halfplane Re z < O. In particular, if a = c = 0, then (4) a.nd (5) imply that 0 is in PeeL) and K(L) = 0, since the zeros of the polynomial in (5) with i\ = 0 are lip and 1 + lip.

VI.8.6 Theorem. Let J be compact and let T be the maximal operator corresponding to (T, p, p), 1 < P < 00, where

O~k~n-I

b" and :" in .£.,(J)

APPLICATIONS TO ORDINARY DIFFERENTIAL OPERATORS

177

Suppose B: £p(J) - t £p(J) is a bounded linear operator with XJ(B) ::) :D(A), where A is the maximal operator corresponding to (Dn, p, p) and J. Then T B is a Fredholm operator with index n whenever

+

IIBII <

(n -

l)![(n -

l)P'Z; 1jI/p'(np)1/pllbn ll""

where l is the length of J and p' is conjugate to p.

Proof.

Let K be the maximal operator corresponding to (v, p, p),

n-l

where v

L (bkjbn)Dk.

=

An argument similar to that given in the proof

k=O

of Theorem VI.8.1 shows that K is A-compact and that T = bn(A + K). Since B is bounded, it follows that K is A + (ljb n)B-compact. Hence

(1)

K(T

+ B)

= K

(A + K + :n B) = (A + :n B) K

whenever the indices exist.

Now, for

(2)

+ :n B)

K(A

II :n B II <

'Y(A),

= K(A) = n

The theorem now follows from (1), (2) and Example IV.1.4. As another application of perturbation theory, the following result, stated somewhat classically, is obtained. VI.B.7 Theorem. Given J a compact interval, let k be a bounded measurable function on J X J, aj € £l(J), 0 ~ j S n - 1, and 1jan € £,.,(J). Then, except perhaps for isolated scalars A, the equation n

L

aj(t)x(j) (t)

+ A £k(s, t)x(s) ds

= yet)

a.e.

Y € £l(J)

3=0

has precisely n linearly independent solutions in An(J).

Proof.

Define T to be the maximal operator corresponding to

n

(L k-O

ajDi, 1, 1) and J.

Define K on £l(J) by

UNBOUNDED LINEAR OPERATORS

178

Then, by Theorems VI.3.1 and VI.3.2 and Example III.3.12, T is normally solvable, aCT) = n, (J(T) = 0, and K is strictly singular. Hence, except perhaps for isolated A, (J(T

+ AK)

=0

by Theorem V.2.1.

and

a(T

+ AK)

= K(T

+ AK)

=

K(T) = n

chapter VII

THE DIRICHLET OPERATOR

In this chapter, a very small portion of the theory of partial differential operators is presented. A Dirichlet boundary value problem in a Banach space setting is considered, and the vital role played by a fundamental inequality is discussed. Few proofs are given, since the arguments are very involved and rely on some deep theorems in classical analysis. For further study and additional references, we refer the reader to Agmoll, Douglis and Nirenberg [1], Bers and Schechter [1], Browder [3], [4], Hormander [1], Friedman [1], and Sobolev [2].

VII.1

THE SOBOLEV SPACE wm,p(n)

This section is concerned with the definition and properties of the space Wm,p(Q), which is referred to as a Sobolev space. The definition and discovery of its basic properties are due to Sobolev [1] and [2]. See also Friedrichs [1]. Throughout this chapter, Q is an open subset of En, n ~ 1. The points of En are denoted by x =' (Xl, X2, • • • ,xn), and integration is with respect to Lebesgue measure. For D, = a/ax, and a = (aI, a2, . , an), ILlI n-tup\o of nOilIlCll;lltivo integers, we set D" .;, D l",D 2". . . . D~a •. Thn ordm' of f)" iH

lui'

n

L i-I

IYi·

180

UNBOUNDED LINEAR OPERATORS

So as to motivate the definition of Wm,p(Q) , let us consider the onedimensional case where Q is an open interval I. It was shown in Theorem VI.6.2 that for 1 ~ p ~ 00, Wm,p(I)

=

If I j
e £p(l),

~

0

k

~

m,

pm-I) absolutely continuous I

is the domain of the maximal operator corresponding to (T, p, p) and I, m

where

T

=

L akDk with Ija

and ak e £",(1), 0 ~ k ~ m.

m

k=O

all

Now for f e Wm,p(I), successive integration by parts shows that for C0"'(I),

ep e

In fact, Wm·p(I) has the following characterization. VII.I.I Theorem. A necessary and sufficient condition that f e £p(I) be in Wm,p(I), 1 ~ p ~ co, is that for each integer k, 1 ~ k ~ m, there exists a gk e £p(I) such that ep e Co"'(I)

We first prove the following lemma. VII.I.2 Lemma. Let g be locally integrable on I. integer k,

£gep(k)

If for some positive

= 0

then g is, almost everywhere, a polynom2al of degree at most k - 1. Proof. Let V be the vector space Co"'(I) and let V* be the vector space of linear functionals on V. Define D to be the ordinary derivative operator mapping V into V and define the transpose D* from V* into V* by (D*F)ep = FDep, Fe V*, ep e V. Let Fg e V* be defined by

Fgep Then, by hypothesis,

=

£yep

ep e V

THE DIRICHLET OPERATOR

181

that is, F g is in ~«D*)k). Assert that F the representation

E

~«D*)k)

implies that F has

where P is a polynomial of degree at most k - 1. Define FIe V* by

Then

~(FI) =

DV.

Indeed, suppose cp

=

Assume k = 1.

Dif> for some if> e V.

Then

where if> has support in compact interval [a, b] C I. Thus DV C ~(FI). On the other hand, if cp is in ~(FI) and has support in compact interval [a, b], define if> on I by

if>(x) =

f

cp(t) dt

h

Then if> is infinitely differentiable on I, and since cp = 0, if> has support in [a, b]. Hence ~(FI) C DV, and therefore ~(FI) = DV. Given F e ~(D*),O = (D*F)cp = FDcpforallcp e V. Thus~(FI) = DV C ~(F). Since dim V /~(FI) = 1, F = cF 1 for some constant c. Indeed,

where CPo is chosen so that FIcpo

Fcp

1.

=

=

cF1CP

This means that for all cp e V, =

£ccp

Thus the assertion about F e ~«D*)k) is proved for k = 1. Assume the assertion holds for k = J, and 0 = (D*)j+lF = (D*)i(D*F). Thep. there exists a polynomial Pi-l of degree at most J - 1 such that (1)

Choose Pi to bo defined by

(D*F)cp Ii,

polynomiflJ

=

1'10

£Pi-ICP

cpeV

that DPi = -Pi-I.

Then for Pi e V*

182

UNBOUNDED LINEAR OPERATORS

integration by parts shows that for all '" E V, (2)

(D*Pi)", =

£Pi"" £Pi-I'" =

Thus by (1) and (2), D*(Pi - F) = O. Hence, by what has been shown, there exists a constant c such that Pi - F = cF 1, or equivalently, F", =

£(Pi -

c)",

Since Pi - c is a polynomial of degree at most j, the assertion about FE ;n«D*)k) follows by induction. In particular, since F g is in ;n«D*)k), there exists a polynomial P of degree at most k - 1 such that for all '" E V,

o= F

g", -

£P'" = £(g -

p)",

Hence g = P a.e., which proves the lemma. Proof of the theorem. The necessity was proved in the discussion preceding the theorem. Assume the existence of the gk as in the theorem. Successive integration by parts implies the existence of a Gk such that G.(k-I) is absolutely continuous, Gk(k) = gk a.e., and '" E

Co'" (I)

Hence for all '" E Co'" (I),

Thus f - Gk is a polynomial, almost everywhere, of degree at most k - 1 by Lemma VII.1.2. In particular, there exists a constant c such that fCm-I) is Gm(m-I) + c a.e. and is absolutely continuous, and fCk) = Gk(k) = gk aselementsof£p(1),1 ::; k::; m. Hencefisin Wm,p(1). Based on Theorem VII.1.I, we obtain the following characterization of Wm,p(1) for 1 < P ::; 00. Let D be the ordinary derivative considered as a map from Co'" (I) C £p,(1) into £p,(1), where p' is conjugate to P and 1 < P ::; 00. Let f and gk be as in Theorem VII.1.I considered as elements of the conjugate space £p(1) = £~,(1). Then '" E

~(D)

Thus f is in Wm,p(I) if ILnd only if f iH in :D«J)k)') ILnd (Dk)'f "'" (_l)k g!.

THE DIRICHLET OP&RATOR

183

In light of the above discussion, the Sobolev space is defined as follows.

VII.l.3 Definition. Let n be an open subset of En. Given a nonnegative integer m, define Wm,p(n), 1 ~ p ~ 00, as the linear space of those f E .cp(n) having the property that for lal ~ m, there exists a ga E £p(n) such that

The norm on Wm,p(n) is given by

The generalized a-derivative of f is defined to be ga which is unique by Theorem 0.10. We write Daf = gao

VII.l.4 '/,.

ii.

Remarks For f E Cm(n) (\ Wm,p(n), integration by parts shows that the generalized a-derivative of f, lal ~ m, coincides with the a-derivative of f in the classical sense. For 1 < p :::; 00, the definition of the space Wm,p(n) is equivalent to the following definition. Consider Da as a linear map from Co" (n) C £p,(n) into .cp.(n), p' conjugate to p. Then

where (Det)' is the conjugate of Da. (-l)lal(Da)'f

iii.

=

Moreover,

ga

where ga is the generalized a-derivative of f. Thus the generalized a-derivative is a conjugate operator. Meyers and Berrin [1] proved that for 1 ~ p < 00 and n any open set in En, Coo(n) (\ Wm,p(n) is dense in Wm,p(n) with respect to the Wm,p(Q) norm.

For the remainder of the chapter, we assulIle 1

Vll.l.5

Theorem.