Principles of functional analysis (Graduate Studies in Mathematics)

Let H be a Hilbert space. For any elements f, g E H, II/+ 9112 =

11!112+ 2(/, g )+ 119112

we

have

1.4. Fourier series

17

f(x)

g(x) 0

1

2

0

3

1

3

2

Figure 1.1

and giVIng

II!- gll2 = 11!112 - 2(!, g)+ llgll2,

( 1 . 38) This is known as the parallelogram law, The name comes from the special case which states that the sum of the squares of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of the diagonals. Thus, it follows that in any Hilbert space, ( 1 .38) must hold for all elements f, g. This gives us a convenient way of checking whether a given Banach space is a Hilbert space as well. If one can exhibit two elements J, g of the Banach space which violate ( 1 .38) , then it clearly cannot be a Hilbert space. Recall the spaces C and B from Section 1 .2 and Example 4. Let us show Define f that they are not Hilbert spaces. For simplicity take a = 0, b to be 1 from 0 to 1 , to vanish from 2 to 3 and to be linear from 1 to 2 (see Figure 1 . 1 ) . Similarly, let g vanish from 0 to 1 , equal 1 from 2 to 3 and be linear from l to 2. Both f and g are continuous in the closed interval [0, 3] and hence are elements of B and C. But =

IIIII

=

llgll

=

�·

II!+ gil = II!- gil = 1 ,

violating ( 1 .38) . Thus, B and C are not Hilbert spaces. In Section 9. 1 we shall show that every Banach space whose elements satisfy ( 1 .38) is also a Hilbert space. 1.4. Fourier series

Let f(x) be a function having a continuous derivative in the closed interval (0, 21r] and such that / (27r) = f(O) . We can extend f(x) to the whole ofiR by making it periodic with period 21r. It will be continuous in all of JR, but its

1. BASIC NOTIONS

18

derivative may have jumps at the endpoints of the periods . Then according to the theory of Fourier series

0 f(x) = � + L a n cos nx + L bn sin nx,

( 1 .39) where

( 1 .40)

00

00

1

1

2 1 7r an 7r 0 f ( x) cos nx dx, 1 == -

bn

==

27r f(x) sin nx dx, 1 7r 0

1 -

and the convergence of ( 1 .39) is uniform. Now our first reaction to all this is that formulas ( 1 . 39) and ( 1 .40) are very complicated. So we recommend the following simplifications. Set

( 21r ) - 2 'P2k (x ) == 7r - l /2 cos1 kx, n2k == 7r 12 ak ,

1r/ 2)
1

==

1

2 ao ,

no == (

,

==

0

An important property of the functions 'Pn which was used in deriv.ing ( 1 .39) and ( 1 .40) is

10 27r (x )
( 1 .42)

==

{

0' 1,

if m # n, if m == n.

Now consider the sequence (no, n1, ···)of coefficients in ( 1 . 4 1 ) . A fact which may or 1nay not be surprising is that it is an element of For, n n 21r 21r f(x)
1 a L ] L0 0

10

0

where

l2 •

n fn(x) . L n1((Jj(x).

( 1 .43)

0

Since the Fourier series converges uniformly to f(x) and fn(x) is just a partial sun1, \Ve have n

Thus,

( cto, c� l· · · · )

L() aJ is

an

�

2r 7r J(x)2dx as n

�{)

� oo.

elerncnt of l2. and its norn1 is

1 . 4. Fourier series

19

I Figure 1.2

f C1

Hence for each E ( the set of functions with period 21r having continuous ) in such that first derivatives ) , there is a unique sequence 00 ( 1 .44) =

( o:o, o:1 , l2 7r (x) 2 1 2 I: o:] f dx, and ( 1 .41) holds. ( In general, the symbol C 1 refers to all continuously differentiable functions. At this point we use it to refer only to periodic ·

0

·

·

0

functions having period 27r. ) One might ask if we can go back. If we are given a sequence in l 2 , does there exist a function E such that ( 1 .41) and ( 1 .44) hold? The answer is a resounding no. The reason is simple. Since l2 is complete, every Cauchy sequence in But does every sequence { / } of functions has a limit in j such that in

f C1

l2

l . 2 C1 2 ( 1 .45) 1 rr [/j (x) - fk (x) ] 2 dx � 0 as j, k � have a limit in C1? We can see quite easily that this is not the case. In fact one can easily find a discontinuous function g(x) which can be approximated by a smooth function h(x) in such a way that {J 27r [g (x) - h(x)] 2 dx o is as small as we like ( see Figure 1 .2). Another way of looking at it is that ( ) / 1 7r 2 2 ( 1 .46) 11!11 = 1 f (x) 2 dx is a norm on the vector space C 1 , but C 1 ( or even C ) is not complete witl1 oo

respect to this norm. What does one do in the situation when one has a normed vector space V which is not complete? In general one can complete the space by inventing fictitious or "ideal" elements and adding them to the space. This may be done as follows. Consider any Cauchy sequence of elements of V. If it has a limit in V, fine. Otherwise, we plug the "hole" by

1 . BASIC NOTIONS

20

inventing a limit for it. One has to check that the resulting enlarged space satisfies all of the stipulations of a Banach space. This can be done as follows: Let V denote the set of all Cauchy sequences in V. If -

-

are members of V, we define

li m l 9k l · G + H {gk + hk }, o:G {ngk }, I G I k-+oo ==

==

==

We consider a Cauchy sequence equal to 0 if it converges to 0. With these definitions, V becomes a normed vector space. Moreover, it is complete. To see this, let == be a Cauchy sequence in V . Then for each n, there is an such that

( {9nk }) G n Nn 1 < l 9nk 9n zl l , k, l > Nn . Pick an element 9n l with l > Nn and call it fn· Then 1 , < Nn . 9 11/n- nkl l Let Fn be the element in V given by the constant sequence ( fn , fn , · · · ) . ( It qualifies as a Cauchy sequence. ) Then GI n - Fn l < -1 . Finally, let F { fn }. Then I Gn - F l < I Gn - Fn l + I Fn - F l 0 as n

-

n

k >

n

==

�

n � 00.

We note that in order to be logically precise, we had to consider equiv­ alence classes of Cauchy sequences. If this sounds artificial, just remember that it is precisely the way we obtain the real numbers from the rationals. In Section 8.6 we shall give an "easy" proof of the fact that we can always complete a normed vector space to make it into a Banach space ( without considering equivalence classes of Cauchy sequences ). In our present case , however , it turns out that we do not have to invent ideal elements. In fact , for each sequence satisfying (1.45) there is a bona fide, function (x) such that

f

( 1 .47)

{27r o [fj(x)- f(x)]2 dx � 0 as j J

� oo.

It may be that this function is very discontinuous, but its square is integrable in a generalized sense. No clain1 concerning pointwise convergence of the sequence is intended , but just that ( 1 .4 7) holds. To summarize: the completion of C1 with respect to the norm ( 1 .46) consists of those functions having squares integrable in a generalized sense. We denote this space by

{ fi }

21

1 . 4. Fourier series

L2 L2 ( [0 21r] ) . (When no confusion will result , we do not indicate the =

,

region over which the integration is taken. At the present time we are integrating over the interval [0, 21r ] . At other times we shall use different regions. ) Set Now suppose ( o:o , o: I , · · · ) is a sequence in

l . 2 n fn ( x ) 2: O:j
( 1 .48)

f E C 1 , and for < n n ( 1 .49) 1 /n - fml l 2 I m2:+ l appj l 2 m-2:+ l aJ Thus . there is an f E L 2 such that 1 /n - / I 0 as 0

Then

m

n

=

=

�

By ( 1 .42) , we see for

m

<

n,

----+

n �

0 as

m,

n ----+

oo .

oo .

{27r

am J fn (x) cpm (x) dx. We claim that 7r 2 ( 1 .50) am J{ f ( x ) cpm (x) dx . This follows from the fact that 2 ( !,g ) 1 1r f (x ) g(x) dx =

o

=

o

=

is a scalar product corresponding to ( 1 .46) . Hence,

l ( fn ,
�

n.

=

0

Letting

n

--4

oo ,

n. � 00 .

=

0

we get

( 1 . 51)

f C1

0

L2 . 2 f L. fn j""'=O Qj(n)
{ fn }

Then there is a sequence Conversely, let be any function in qf can be expanded elements in which converges to in Now each in a Fourier series . Thus,

( 1 . 52)

=

00

�

fn

1 . BASIC NOTIONS

22

Since it is a Cauchy sequence, we have 00

m n 2 j l a )aj I 1 /n - fm l j =O - l) 2 Since l 2 is complete, there is sequence ( ao, a 1 , nl - aj ) 2 0 as I:(aj ( 1 .54) j =O =

( 1 53 ) .

---+

a

00

j=

( 1 .55)

=

-

00

· · ) in

l2 such that

00

I: Qj 'Pj · 0

£2 . Moreover n 2 (a 1 /n - fl l jI:=O j l - aj ) 2 0

By what we have just shown, j

m , n ---+ oo .

n � <X' .

�

Set

·

0 as

E

�

as n � , oo ,

f f. Thus, we have Theorem 1 .4. Th ere is a on e-to-one correspondence between l 2 such th at if (no, a 1 , ) corresponds to J, then n ( 1 . 56) I L ai'Pi - /I I � 0

Hence,

=

·

·

and

L2

·

as n � oo ,

0

and ( 1. 57)

00

1 ! 1 2 = L a] , ai

=

0

(J,

'Pi ) ·

Note that this correspondence is linear and one-to-one in both directions. We shall have more to say about this in Chapter 3. As usual, we are not sat­ · We want to know if similar isfied with merely proving statements about statements hold in other Hilbert spaces. So we examine the assumptions we have made. One property of the sequence is

( 1 . 58)

(

'Pm , 'Pn )

_

-

{0,

1,

£2 . { 'Pn } m

# n, _

m - n.

Such a sequence is called orthonormal. So suppose we have an arbitrary of elements in H. If f Hilbert space H and an orthonormal sequence Then = (J, is an arbitrary element of H, set

( 1 . 59)

{ 'Pi } On 'Pn ) · n 2 2 n n n I / - L1 Oi 'Pil l 1 / 1 - 2 L1 ai ( f, 'Pi ) + L1 a7 1 / 1 2 - L1 nr . =

=

1 .4. Fourier series

23

n L 0T < 1 ! 1 2 ·

Hence,

Letting

n

� oo ,

1

we have

00

L al < 1 ! 1 � .

(1 .60)

1

Equation ( 1 .59) is called in equ ality. We also have

B ess el's identity,

··

while ( 1 . 60) is called

B ess el 's

Let ( 0:1 , 0: 2 , ) be a s equ ence of real numb ers, and let { be an orthono rm al s equ ence in H. Th en

Theorem 1 . 5 .

·

n L O:i
conv erg es in H as

Proof. For

m

<


1

n �

oo

if and only if ,

,

n,

2 0


An orthonormal sequence { sums of the form

1

c.

E >

00

and ( 1 . 62)

1 ! 1 2 == L ( J,
1

1 . BASIC NOTIONS

24 Proof. Let f be any element of H, and let

In

L )n
=

o:

)

be a sequence of sums of the form ( 1 . 6 1 ) which converges to f in H. In ) in l such particular, ( 1 .53) holds . Thus, there is a sequence ( o:o , o: 1 , _ _ 2 that ( 1 . 54) hold_:; . If we define f by ( 1 .55 2 , we know that f E H (Theorem 1 . 5) and fn --+ f as n --+ oo . Hence , f == f. This proves the first statement. To prove ( 1 .62) , we have by Bessel's identity ( 1 . 59) , ·

n

n I ! - L O:i
1

Letting

n --+

oo ,

·

;

o: .

0

.

Equation ( 1 .62) is known as

·

P ars ev al 's equ ality.

Theorem 1 . 6 has a trivial converse. If ( 1 .62) holds for all f in a Hilbert is complete. This follows space H, then the orthonormal sequence { immediately from ( 1 .59) . We also have



Theorem 1 . 7. If {

==

00

1

==

we have

n n lim ( O:j
1. 5 .

' 1

'

==

1

0

Problems

( 1 ) Show that statement ( 1 5) of Section 1 .2 is implied by statements ( 1 ) - ( 13 of that section , but not by- statements ( 1 - ( 9) .

)

)

25

1 . 5. Problems

, I ! - l:(J, 'Pk ) 'Pkl l for all f E H and all scalars a k . (3) Show that an orthonormal sequence {

(2) Let

·

·

·

1

1

( 4)

{ an }

a )

is Let c denote the set of all elements ( 1 , E l00 such that a convergent sequence , and let co be the set of all such elements for which � 0 as n � oo . Show that c and co are Banach spaces. ·

·

·

an

(5) Show that the operator given at the end of Section 1 . 1 satisfies the hypotheses of Theorem 1 . 1 . (6) Carry out the details of completing a nor1ned vector space by the method described in Section 1 .4. (7 ) Prove Theorem 1 .4 with

£2 replaced by

complete orthonormal sequence.

any

Hilbert space with a

(8) Show that the norm of an element is never negative. are elements of a Hilbert space and satisfy l x + Yl l == l x l + x, yshow I Y I , that either x == cy or y == ex for some scalar > 0. ( 10) If X k � x, Yk � y in a Hilbert space, show that (x k , Yk ) � (x, y). ( 1 1 ) If x , Xn are elements of a Hilbert space, show that

(9) If

c

1, · · ·

( 1 2) Show that in a normed vector space

n V j. L Vj = L V j + L j=l j=l j=n+l oo

oo

1 . BASIC NOTIONS

26

'Pv}

be any collection of or­ ( 13) Let H be a Hilbert space, and let { thonormal elements in H ( not necessary denumerable ) . If x E H, show that there is at most a denumerable number of the

'Pv such

'Pv) ( 14) Show that B [a , b] is a normed vector space. ( 15) For f ( n 1 , nn ) E lRn , show that 1 / l o == maxk I nk I and n n kl l f l 1 == L l k= 1 are norms. Is lRn complete with respect to either of them? ( 1 6) If { 'P n } is an orthonormal sequence in a Hilbert space H, and 1 / 1 2 L1 ( J, 'Pn ) 2 holds for each f E H, show that { 'Pn } is complete. ( 1 7) Prove: If { 'Pn } is an orthonormal sequence in a Hilbert space H and ( n 1 , ) E l2 , then there is an f E H such that nk = ( f, 'Pk ) , k 1 , · · , and 1 / 1 2 == L a� . k1 that

(x ,

=/=-

==

·

0.

·

·

,

00

==

·

·

·

==

00

·

=

( 18) Calculate

n n ( n )2 L1 a� L b� - L1 ak bk 1

and use the answer to prove the Cauchy-Schwarz inequality ( 1 .26) .

( 19) Show that

l cf l oo == l e i · 1 / l oo for f E Zoo .

(20) Show that the shortest distance between two points in a Hilbert space is along a straight line. Is this true in all Banach spaces?

1 . 5.

27

Problems

(21) Define

I fI *

=

t

inf { >

Is this a norm on IRn ?

0

:

I f I < t } , f E lRn .

I I

on IR2 such that the set {22) Does there exist a norm. B = { x E IR2 : ll x l l < 1 } ·

is a rectangle? an ellipse?

(23) Show that (1.29) can be solved in IRn for arbitrary b if 1 < lajk l n , 1 < j, k < n . -

(24) Show that ( 1 . 32) can be solved in l 00 for arbitrary b when 1 < I aJ·k l 2k 2 ' J. ' k - 1 ' 2 ' . ·

-

-

·

·

-

(25) Show that ( 1 .32) can be solved in l 2 for arbitrary b if 1 < i aik i 2jk , j , k = 1 , 2 , ·

·

·

.

Chapter

2

DUALITY

2 . 1 . The Riesz representat ion theorem

y)

y,

Let H be a Hilbert space and let ( x , denote its scalar product. If we fix then the expression assigns to each E H a number. An ass:_ gnment F of a number to each element x of a vector space is called a functional and denoted by F . The scalar product is not the first functional we have encountered. In any normed vector space, the norm is also a functional. The functional F has some very interesting and surprising == ( features. For instance it satisfies

(x, y) (x) (x) x, y)

x

(2 . 1 )

)

for 0: 1 , 0: 2 scalars. A functional satisfying ( 2. 1 is called linear. Another property is (2 . 2)

I F (x ) l < M l x l ,

which follows immediately from Schwarz ' s inequality ( cf. ( 1 . 28 ) ) . A func­ tional satisfying ( 2.2 ) is called bounded. Thus for fixed, is a == bounded linear functional in the Hilbert space H. "What is so surprising about this?" you are probably thinking to your­ self; "I am sure that we can find many other examples of bounded linear functionals." Well, you may be surprised to learn that there aren ' t any oth­ ers. In fact we have the following

y

Theorem 2 . 1 . For every bounded linear functional there is a unzque element y E H such that

( 2 . 3)

F(x) (x, y)

F on a Hilbert space H

F(x) ( x , y) for all x E H. ==

29

3

0

2.

Moreover,

IYI

(2 .4)

=

sup

x E H, x :fO

DUALITY

I FX(x) l . I I

Theorem 2. 1 is known as the Riesz representatiort theorem. In order to get an idea how to go about proving it , let us examine (2.3) a bit more closely. If assigns to each element the value zero, then we can take 0, and the theorem is trivial. Otherwise the we are searching for cannot vanish. However, it must be "orthogonal" to every for which == i.e. , we must have 0 for all such Let N denote the set of those satisfying 0. Suppose we can find a # 0 which is orthogonal to each E N. Then I claim that the theorem is proved. For clearly is not in N ( otherwise == we would have # Moreover, for each and hence E H, we have

F

x

y

( x, y) ==

x

y == F(x) 0; x

x. y F(x) == x y 2 == (y, y) 0), F( y) 0. I Y I x F(F(y)x - F(x)y) == F(y)F(x) - F(x)F(y) 0, showing that F(y)x - F(x)y is in N. Hence (F( y ) x - F(x)y, y) == 0, or y) F(x) (x, F(y) . Y I I 2 This gives (2.3) if we use F(y) y / I Y I 2 in place of y. � This is to be expected since we made no stipulation on y other than that it be orthogonal to N. ) We also note that the uniqueness and (2.4) are trivial. For if Y l were another element of H satisfying (2.3) , we would have ( x, y - y 1 ) == 0 for all x E H. l , showing that I Y - Y1 l 0. Thus, In particular this holds for x y Y Yl y . Now by Schwarz's inequality, I F (x) l == l (x, Y) l < l x i i Y I · Hence, F (x) l I sup I YI xl l _ However, we can obtain equality by taking x == y . In fact , I Y I F (y) / I I Y I I · This gives ( 2.4) . ==

=

==

==

==

>

x E H, x :fO

=

All that is now needed to complete the proof of Theorem 2. 1 is for us to find an element # which is orthogonal to N ( i.e. , to every element of N ) . In order to do this we examine N a little more closely. What kind of set is it? First of all, we notice that if are elements of N, so is and for any scalars For, by the linearity of +

y 0

x x 1 2 0:1 x 1 o:2 x 2 0:1 , 0:2 . F, F (n 1 x 1 +n2 x2 ) == o:1 F(x 1 ) +o:2 F(x2 ) = 0.

2. 1 .

The Riesz representation theorem

31

o: 1 x 1 + o:2 x2 (2. 2 ) { Xn } x

A subset U of a vector space V is called a subspace of V if is in U whenever are in U and 0: 1 , 0: 2 are scalars. Thus N is a subspace of H. There is another property of N which comes from and is not so obvious. This is the fact that it is a closed subspace. A subset U of a normed vector space X is called closed if for every sequence of elements in U having a limit in X, the limit is actually in U. In our particular case, if is a sequence of elements in N which approaches a limit in H, then by

x 1 , x2

{ xn } (2. 2 ) I F (x) l = I F (x) - F(xn ) l I F (x - Xn ) l < M l x - Xn l 0 Since x does not depend on we have F(x) 0. Thus, x E N, showing that N is closed in H. ==

----+

==

n,

as n ----+ oo .

Thus, we have a closed subspace N of H which is not the whole of H. We are interested in obtaining an element y #- 0 of H which is orthogonal to N. For the special case of two-dimensional Euclidean space, we recall from our plane geometry that this can be done by drawing a perpendicular. We also recal l that the shortest distance from a point ( element ) to a line ( subspace ) is along the perpendicular. The same thing is true in Hilbert space. We have Theorem 2 . 2. Let N be a closed subspace of a Hilbert space H! and let be an element of H which is not in N. Set

(2. 5 )

x

l x - z l· Then there is an element z E N such that l x - z l = d . Proof. By the definition of d, there is a sequence { Zn } of elements of N such that l x - Zn l � d. We apply the parallelogram law ( cf. ( 1 .38 ) ) to x - Zn and x - Zm· Thus I (x - Zn ) + (x - Zm ) 1 2 + I (x - Zn ) - (x - Zm ) 1 2 2 l x - Zn 1 2 + 2 l x - Zm 1 2 , or (2. 6 ) 4 l x - [( zn + Zm )/2) 1 2 + l zm - Zn l 2 2 l x - Zn l 2 + 2 l x - Zml l 2 . Since N is a subspace, ( zn + zm )/2 is in N. Hence, the left-hand side of (2.6) not less than d = inf

zEN

==

==

is

This implies

l zm - Zn l 2 2l l x - Zn l 2 + 2 l x - Zm l 2 - 4d2 0 as Thus, {zn } is a Cauchy sequence in H. Using the fact that a Hilbert space is complete, we let z be the limit of this sequence. But N is closed in H. Hence, z E N, and d = lim l x - zn l D l x - zl l · �

<

==

m, n

_____, oo .

32

2.

D UALITY

Theorem 2 . 3 . Let N be a closed subspace of a Hilbert space H. Then for each E H, there are a v E N and a orthogonal to N such that == v + . This decomposition is unique.

x

x

w

x, w

x l x - zl w

x

w

== 0. If fj. N, let z - E N be such that Proof. If E N, put v == and must == == d, where d is given by (2.5) . We set v == show that is orthogonal to N. Let #- 0 be any element of N and o: any scalar. Then

z, w x - z

u d 2 < l w + o:u l 2 == l w l 2 +2o:(w, -u ) o: 2 l u l 2 2 2 ( ) (w, u) (w, u) w u [ , ] 2 2 = I u l a + 2a ul l 2 + l u l 4 d - l u l 2 2 2 ( [ ) w, u (w, u) ] = l ul 2 a + + d2 l ul 2 l ul 2 ' where we completed the square with respect to o: . Take o: = -(w, u) / l u l 2 . Thus, ( w, u) 2 < 0, which can only happen if w is orthogonal to u . Since u was any arbitrary element of N, the first statement is proved. If x = v 1 where E N and is orthogonal to N, then v - v1 is both in N w and orthogonal to N. In particular, it is orthogonal to itself and thus must +

2

+

-

VI

== W I -

WI

vanish. This completes the proof.

+ WI ,

0

.

The proof of Theorem 2 . 1 (the Riesz Representation Theorem) is now com­ plete. Corollary 2 .4. If N is a closed subspace of a Hilbert space H but is not the whole of H, then there is an element y #- 0 in H which is orthogonal to N.

x

Proof. Let be any element of H which is not in N. By Theorem 2.3 , is orthogonal to N. Clearly, iv #- 0, for = v + w, where v E N and 0 otherwise would be in N. We can take as the element y sought.

x

x

w

w

Theorem 2.3 is called the projection theorem because of its obvious geomet­ rical interpretation. If is a bounded linear functional on a normed vector space X, the norm of is defined by

F

F

F (x) I ( 2 . 7) I F I = sup I I l . It is equal to the smallest number NI satisfying (2.8 ) I F (x) l < M l x l for all x E X. In this terminology (2.4) becomes (2.9) I YI I FI . x EX, x #O

==

X

2. 2.

33

The Hahn-Banach theorem

2 . 2 . The Hahn-Banach theorem

Now that we have shown that all of the bounded linear functionals on a Hilbert space are just the scalar products , you might begin to wonder whether Banach s paces which are not Hilbert spaces have any non.. zero bounded linear functionals at all. (In fact , one might be tempted to reason as follows : if the only nontrivial bounded linear functionals on a Banach space that has a scalar product are precisely the scalar pro_ducts , then Ba­ nach spaces which do not have scalar produ c ts should have none. ) How can we go about trying to find any that might exist? As we always do in such cases , we consider the . simplest cas e possible, Let X be a Banach space, and let # 0 be a fixed element of X. The set of all elements of the form forms a subspace Xo of X. Do there exist bounded linear functionals on X0 ? One candidate is ·

xo

Clearly this is

a

axo F(axo ) = a. linear fu nctionaL It is also bounded , since a xo l ) I F (nxo l = l n l = l xo l l .

So there are bounded linear functionals on such subspaces. If we could only extend them to the whole of X we would have what we want . However, difficulties immediately present themselves . Besides the fact that it is not obvious how to extend a bounded linear functional to larger subspaces , will the norm of the functional be .increased in doing so? And what happens if one needs an infinite number of steps to complete the procedure? The answers to these questions are given by the celebrated Hahn-Banach theorem ( Theorem 2 . 5 below) . L.et V be a vector space. A functional p( on V is called sublinear if \

x)

( 2 . 10)

(2. 1 1 )

p(x + y ) p (x) p(y) , V, p (o:x ) o:p(x), x E V, a > 0. x,

+

<

y E

=

Note that the norm in a normed vector space is a sublinear functional.

Theorem 2 . 5 . {The Hahn-Banach Theorem) Let V be a vector space, and be a sublinear functional on V. Let M be a subspace of V, and let let p be a linear functional on M satisfying

f (x)

(x)

(2. 12)

f (x )

<

p(x) , x E M.

Then there is a linear functional F (x) on the whole of V such that

(2. 13) (2. 14 )

F(x)

=

f (x) ,

x E M,

F( x) < p(x) , x E

V.

2. D UALITY

34

Before attempting to prove the Hahn-Banach theorem, we shall show how it applies to our case. We have

M

Theorem 2.6. Let be a subspace of a normed vector space that is a bounded linear functional on Set

X , and suppose

M. (x) f l l sup . = 11! 11 I I Then there is a bounded linear functional F(x) on the whole of X such that (2 . 15 ) F(x) = f (x), x E M, (2. 1 6) I F I = l fl · Proof. Set p(x) = I I I · l x l , E X. Then p(x ) is a sublinear functional and l (x) < p (x), x E M. Then by the Hahn-Banach theorem there is a functional F(x) defined on the whole of X such that (2. 15) holds and F(x) < p(x) = I I I · l x l , x E X. Since - F(x ) = F ( - x) < I I I · I - x l , x E X, we have I F (x) l < l f l · l x f l , x E X. Thus, I FI I II · Since F is an extension of f , we must have I I I I FI · f(x)

x E M, x=;fO

X

X

<

<

Hence, (2. 16) holds , and the proof is complete.

0

Since we have shown that every normed vector space having nonzero elements has a subspace having a nonzero bounded linear functional, it follows that every':normed vector space having nonzero elements has nonzero bounded linear functionals. Now we tackle the Hahn-Banach theorem. We first note that it says nothing if = V. So we assume that there is an element 1· 1 of �l which is be the set of elements of V of the forn1 not in Let

M

M. M1

(2. 17)

O:X }

+

x, 0: E IR, E M. X

2.2. The Hahn-Banach theorem

35

MI

Then one checks easily that is a subspace of V and that the represen­ tation (2. 1 7) is unique. To feel our way, let us consider the less ambitious task of extending to so as to preserve (2. 12) . If such an extension exists on it must satisfy

MI F f MI, F(ax i + x ) aF(xi) + F (x) aF(x i) + f (x ) . Therefore, F is completely determined by the choice of F(x i) · Moreover, we must have (2 . 18) aF (x i) + f (x ) < p (o:x i + x ) for all scalars and x E M. If o: > 0, this -means X X 1 F(x i) < -0: [p (o:x i + x ) - f (x ) ] p (x i + 0: ) - / ( -0: ) p(x i + z ) - f (z ) , where xja. If a < 0, we have 1 , y ( x (x x ] i ) (ax f(y) F(x i ) > -[p ) ) f p + i a where y - xjo: . Thus we need (2. 19) f (y ) - p (y - XI ) < F(x i) < p (x i + z ) - f (z ) for all y, z E M. Conversely, if we can pick F(x i) to satisfy (2. 1 9) , then it will satisfy (2. 18) , and F will satisfy (2. 14) on MI . For if F (x i) satisfies (2. 19) , then for a > 0, we have a.E (x t ) + f (x ) a[F (x i) + ! ( -0:X ) ] < o:p(x i + -0:X ) p(ax 1 + x ) , while for a < 0 we have aF(x i) + f (x) .- o:[- F(x i) + / ( - -o:X ) ] - ap( - -aX - XI ) p(ax i + x ) . So we have now reduced the problem to finding a value of F(x i) to satisfy (2. 19) . In order for such a value to exist , we must have (2. 20) f (y) - p(y - XI ) < p(x i + z)1 - f (z) for all y, z E M. In other words we need f (y + z) < p(x i + z) + p (y - XI ) · This is indeed true by (2. 1 2) and property ( 2 1 0 ) of a sublinear functional. we Hence, ( 2 . 2 0 ) holds. If we fix y and let z run through all elements of have J (y ) - p (y - X t ) < inf (p (x i + z ) - f ( z ) ] - C. Since this is true for any y E M, we have sup[f (y ) - p(y - XI )] < =

=

a

=

=

-

z =

=

+

=

=

=

=

=

<

.

!vf,

zEM

c =

yEM

C.

36

2.

DUALITY

We now merely pick F (x 1 ) to satisfy c

< F (x1 ) < C.

Note that the extension F is unique only when c = C. Thus we have been able to extend f from M to in the desired way. If M1 == V, we are finished. Otherwise there is an element x2 of V not in l'v1. Let M2 be the space "spanned" by x2 and By repeating the process we can extend .f to in the desired way. If M2 =I= V, we keep a stiff upper lip and continue. We get a sequence of subspaces each containing the preceding and such that f can be extended from one to the next . If, finally, we reach a k such that l'v1k == V, we are finished. Even if

l\J1

M1 .

M2

Mk

00

(2. 2 1 )

k= l

then we are through because each x E V is in so1ne Mk , and we can define F by induction. But what if ( 2 . 2 1 ) does not hold? We can complete the proof easily when V is a Hilbert space and p(x) == T' ll x l l for some positive constant !' · For then one can extend f to the closure 1\1 of by continuity. By this we mean that if { X n } is a sequence of el�ments in M which converges to \ x E V, then { f (xn ) } is a Cauchy sequence of real numbers and hence has a limit . We then define F(x ) = lim f (xn ) · The limit is \ ndependent of the sequence chosen. One checks easily that F(x) is a bounded linear functional on the set l'v1 and coincides with f ( x) on A1 . Since is a Hilbert space, there is an element E M such that F(x) = (x, for all x E ( Theorem == < !' · But F(x) can be defined as ( x , y) on the 2. 1 ) . Moreover whole of V, and its norm will not be increased . What do we do when the space V is not a Hilbert sp�ce and (2 .21 ) does not �old? This is not a trivial situation. In this case we need a statement known as Zorn's lemma concerning maximal elements of chains in partially ordered sets. This lemma is equivalent to the axiom of choice, the validity of which we cannot discuss here. We shall present Zorn's· lemma in Section 9 . 5 . There we shall complete the proof of the Hahn-Banach theorem for those spaces that require Zorn's lemma in the proof.

M

y)

y

I YI I II

M

M

2 . 3 . Consequences of the Hahn-Banach t heorem

The Hahn-Banach theore1n is one of the most important theorems in func­ tional analysis and has n1any far-reaching consequences. One of them is Theorem 2 . 7. Let X be q normed vector space and let x0 =I= 0 be an element of �X" . Then there 'is a bounded linear functional F(x) on such that

(2.22 )

IFI

=

X

1,

F (x o ) = l l x o l l ·

2. 3.

Consequences of the Hahn-Banach theorem

37

X such that f (xi )

Corollary 2 .8 . If x1 is an element of bounded linear functional f on then X I

X,

=

0.

=

0 for every

First we prove Theorem 2 . 7.

M

Proof. Let be the set of all vectors of the form o: xo . Then of by Define f on .f ( o: x o ) o: l l x o ll · Then f is linear, and

X.

M

M is a subspace

=

l f ( o: x o ) l == l o: l · l l xo l l

=

l l o: x o l l ·

Thus , f is bounded on M, and l l f l l == 1 . By the Hahn-Banach Theorem, there is a bounded linear functional F on such that I I F I I 1 , and F (axo) n ll x o ll · This is exactly what we want . 0

X

=

=

Corollary 2.8 is an immediate consequence of Theorem 2.7. If x 1 -# 0, there would be a bounded linear functional F on such that F (x 1 ) l l x1 l l · Thus, x 1 0 .

X

=

=

Another consequence of Theorem 2 . 6 is Theorem 2 . 9 . Let M be a subspace of a normed vector space X, and suppose x o is an element of satisfying

X d

(2.23)

==

d (x o ,

M)

==

inf l l x o - x l l

xEM

Then there �s a bounded linear functional F on F ( x o ) == d, and F(x) 0 for x E =

M.

>

0.

X such that I I F I I

1,

Proof. Let M1 be the set of all elements z E X of the form z ==

(2 . 2 4)

o:x o + x ,

0:

E

IR ,

X

E M.

Define the functional f on J\;/1 by f ( z) == o:d. Now the representation ( 2 . 24) x1 - x E is unique, for if z 0: 1 x o + x 1 , we have ( a - 0: 1 )x o which contradicts ( 2 . 23) unless n 1 == a and x 1 == x . Thus , f is well defined and linear on A/1 . It also vanishes on ltf. Is it bounded on Af1 ? Yes, since ==

=

M,

X

II == l l n�r o + x ll . l f ( n x o + x ) l == l n l d < l n l · l l x o + 0'

1 !1

Hence, f is a bounded linear functional on M1 with < 1 . Howeve r, for any E > 0 we can find an :r 1 E AI such t hat l l x o - x i I I < d + E . Th e n f ( xo - X I ) == d, and hence, d l f (:r o - x i ) I == 1 - c > d+ c, d+c l l xo - x i i i

2. DUALITY

38

1 /1

1 . We now apply Theorem which is as close to one as we like. Hence, 2.6 to conclude that there is a bounded linear functional F on X such that on M1 . This completes the proof. D I I F I I 1 and F =

=

=

f

Theorem 2.9 is a weak substitute in general Banach spaces for the pro­ jection theorem (Theorem 2 .4) in Hilbert space. Sometimes it can be used as a replacement . For any normed vector space X, let X ' denote the set of bounded linear E X ' , we say that functionals on X. If if

/, g

f g f (x ) g(x) for all x E X. =

=

The "zero" functional is the one assigning zero to all + by E X, +

h f g and o:f by =

g

x E X. We define

h(x ) f (x ) g(x) , x g (x) n f (x ) , x E X. =

=

=

Under these definitions , X ' becomes a vector space. We have been employing the expression

(x ) f l 1 /1 1 ��� l x l l ' f E X' . This is easily seen to be a norm. In fact ) (x + g(x ) l g (x) f (x f ) l l l l l sup < sup sup + l xl - l xl l xl · (2.25)

=

Thus X ' is a normed vector space. It is therefore natural to ask when X ' will be complete. A rather surprising answer is given by Theorem 2 . 10 . X ' is a Banach space whether or not X is.

{ fn } be a Cauchy seq.uence in X' . Thus for any c > 0, there is l fn - fm l < € for > N, or, equivalently, (2 .26) l fn (x ) - fm (x ) l < c-l l x l for > N, x E X, x # 0. Thus for each x # 0, { fn (x ) } is a Cauchy sequence of real numbers, and hence has a limit depending on x . Define

Proof. Let an N such that

ex

f (x)

== Cx .

m,

n

m,

n

2. 4 .

39

Examples of dual spaces

Clearly

f is a functional on X. It is linear� since f ( a: I X I + 0:2 X2 )

= =

n

It is also bounded. For let have (2.27)

l fn ( x ) -

=

lim fn ( O:I X I + 0:2X2 ) lim{ a:I fn (X I ) + 0: 2 /n (X2) } a: I f( x i ) + a:2/( x2 ) .

be fixed in (2.26) , and let

f (x) l < c ll x ll ,

n

m � oo.

Then v1e

> N, x E X.

Hence, l f ( x ) l < c l l x l l + l fn ( x ) l < ( c + 11 /n l l ) ll x ll for n > N, x E X. Hence, E X ' . But we are not finished. We must show that fn approaches in X'. For this we use (2.27) . It gives

f

f

€

ll fn - / II <

for

n

> N.

D

Since c was arbitrary, the result follows. We now give an interesting counterpart of (2. 25) . From it we see that l f (x ) l <

11/ 11 · l l x l l ,

and hence, ll x ll ;::: But by Theorem 2 .7, for each x and ( x ) l l x ll · Hence,

f

sup /EX ' , f-:/=0 E

l l x ll

=

II/II .

X there is an

=

(2.28)

l f (x ) l

max

/EX ' , f-:/= 0

f E X' such that 11/11

=

1

l f (x ) l

II/ II .

2 .4 . Examples of dual spaces

The space X ' is called the dual (or conjugate) space of X. We consider some examples. If H is a Hilbert space,. we know that every E H ' can be represented in the form

f

f (x)

=

(x , y) ,

x E H.

The correspondence f y is one-to-one, and 11/ 11 II Y II · Hence , we may identify H' with H itself. We next consider the space lp , where p is a real number satisfying 1 < p < oo (note that we have already mentioned l2 and l00 ) . It is the set of all infinite sequences x (x 1 , , xj , · · · ) such that �

=

=

·

·

·

00

40

2.

D UALITY

Set (2 . 29) The first question that comes to mind is whether lp is a vector space. In particular, is the sum of two elements in lp als o in lp? I n showing this we might as well show that ( 2 . 29) is a norm , i.e. , that (2.30) This will not only sh ow that lp i s a vector space , but that it is a normed vector space. Inequality (2.30) is kn ow n as Minkowski 's inequality. To prove (2.30) we first note that it is trivial for p == 1 , so we assume p > 1 . To proceed one would expect to raise both �ides to the p th power and evaluate , as we did in the case p == 2. However , experience has shown that this leads to quite a complicated state of affairs , especially when p is not an integer. Since this course is devoted to the proposition that only eaey th i n gs should be done, we follow an age-old trick discovered by our forefathers which notes that it is mu ch si mpl e r to try to prove the eq uival e n t state1nent -

(2 . 3 1 ) Next , we note that we m ight as well assume t h at the cotnponcnts ..r1 and of x and y are all no nn egati ve For ( 2 . 30) is s upp osed to hold for s uch cases anyway, and once (2.30) is prove d for such cases , it follows im 1n e di at c ly that it holds in general . This assumption saves us the need to write absolute value signs all the way th ro u gh the argurnent . Now (2.3 1 ) is e q u ival e n t to

YJ

.

00

00

:L x i ( x i + Yi ) P - 1 + LY ( x

1

1

t

t

+

Y ) pt

l

< l l x l l p l l .r + Y l l � - 1 + l l u l l p l l :r + Y l l � - 1

so it suffices to show that 00

:L Xi (x1 + Yi ) P- 1 <

(2 .32)

1

l l x + Y l l� Set

q

==

pj (p - 1 ) .

Then

=

l l x l l p l l x + Y l l �- 1 .

00

:L z;/p� l .

1

�

2. 4.

Examples of dual spaces

where

z ==

41

( z 1 , · · · , ZJ , · · · ) . Thus we want to prove 00

L

(2.33)

X i Zi

< llxllp ll z llq ·

1 This is known as Holder 's inequality. It implies ( 2. 32) , which in turn in1plies (2. 3 1 ) , which in turn /ilnplies Minkowski 's inequality (2.30) . Hence , if we prove (2.33) , we can conclude that lp is a normed vector space. To prove (2 .33) , we note that it suffices to prove (2.34)

00

L 1

X i Zi

< 1

llxiiP

when

==

ll zllq

==

1.

In fact , once we have proved (2.34) , we can prove ( 2 . 33) for arbitrary x E lp and z E l q by applying (2 .34) to the vectors z

x' = x / l l x i i P ,

'

==

z / l l z ll q , which satisfy ll x' I I P == ll z ' ll q == 1 . l\1ultiplying t h rough by l l x l l p l l z1l q gives (2.33) . Now (2.34) is an immediate consequence of a P bq (2.35) ab < - + p

for positive numbers a, b . In fact , we have 00

L X i Zi < L 1

xl? �

p

-

+

L

q

z9 �

q

-

1 1 == - + p

q

==

1.

To prove (2.35) , we graph the function y == .rP - 1 and note that the rectangle with sides a , b is contained in the sun1 of the areas bounded by the curve and the x and y axes ( cf. Figur� 2. 1 ) . Since the curve is also given by .r == y q - 1 , we have q aP b q ab < xP- 1 dx + y - 1 dy == +-.

ia 0

b i 0

-

p

q

0 A nat ural question to ask This con1pletes the proof. now is whether lp is complete. We leave this question aside for t he 1noin�nt and discuss the dual space l� of lp . Again we assun1e p > 1 . A hint is givcu by Holder's inequality (2. 33) . If z E l q , and we �et

f ( x ) ==

(2.36)

00

L 1

�ri Z i ,

then f is a linear functional on lp, and by ( 2 . 33) it is a lso bounded . \\�c show that all bounded linear functional� on lp are of t he forn1 ( 2 . 36 ) . Theorem 2 . 1 1 . l� ==

lq ,

where 1 /p + 1 / q == 1 .

no,,·

42

2. D UALITY

y

a

Figure 2.1

x = (x 1 , x2 , · · · ) E lp , f

·· )

Proof. Suppose and E l� . Set e 1 = ( 1 , 0 , , e2 = (0, 1 , 0, ) , and in general ej the vector having the j-th entry equal to one and all other entries equal to zero. Set ·

Then

·

·

Sn E lp , and

l x - sn l � = I: l xil p

·

00

�

0.

n+ l

Thus,

and

l f (x ) - f (sn ) l l f (x - sn) l < 1 1 ! 11 l x - sni i P =

·

� 0 as

n �

oo .

Hence,

00

f (x ) l: xj f(ej ) · Set ej) and z = (z1 , z2 , · · · ) . We must show that z E lq . Up until ( f now, x has been completely arbitrary. Now we take a particular x. We set q 2 when -# 0; , z z { i l il Xi = 0, when = 0. =

Zj

1

=

Zi

Zi

2. 4.

43

Examples of dual spaces

For this case

n P n pq l n q l sn l � = L1 l xii = L1 l zil ( - ) = L l zil , since = qj(q - 1 ) . Moreover , n q n � (sn ) = L1 Xi Zi = L1 l zil , and n q1P l f (sn) l < l f i ( L1 l ii l ) I Hence , n qlp n q L1 l zi l < 11! 11 (2: l zi l ) f , or n q1q ( L l zil ) f < 11 / 11 · Hence , z E l q , and l z l q ::; 11 !1 1· B u t n l f ( x ) l = I L Xi Zi l < l x l p l z l q , and thus , 11/11 < l z l q · This shows that 11 ! 11 = l z l q · 1

p

1

1

1

Thus , we have proved

Theorem 2 . 12. If f E

D

l�, then there zs a z E lq such that f ( x ) = L X i Zi , x E lp , 1

and ( 2 . 37)

11 ! 11

�

l zJiq·

By the way� we now have the answer about completeness . For if we corn­

pute l� , we find that it is j ust lp . Thus lp is the dual space o f a normed vector space and, hence , is cornplete by Theorem 2 . 1 0. It is for this reason that we waited until now to introduce t he spaces lp \vhen p =I= 2 , oc . Remember that we have not proved completeness for the case p = 1 . S i n c e it is very easy, we leave it as an exercise . '

44

2. D UALITY

b

a

Figure 2 . 2

£P P

consisting We wish to remark that the same ideas apply to the space on some interval a < < b such that of functions is integrable on this interval. The norm is given by

l x (t) !

t

x(t)

) l x i P (1b 1 x (t) I Pdt 1,he same reasoning shows that £P is a normed vector space and that ' l jp

=

Holder s ineq u ality (2.38)

F E (V)' there is a y E Lq such that F(x) 1b x(t) y (t) dt , x E £P ,

holds. Moreover, for each (2. 39)

=

and ( 2 . 40)

As another example, we consider the space C C [a , b] of continuous functions in the interval a < < b. Let f be a bounded linear functional on C and be a function in C. Now it is well known that we can approximate any continuous function as closely as desired by a step function, i.e. , a function that is constant on a finite number of subintervals covering [ a , b] cf. Figure 2.2) . If f were also defined for step functions, we could compute it for a sequence ap p r oaching and take the limit. "Even so," you say, "what makes you believe that it would be easier to compute f for a step function than for a continuous function?" Well, just

x(lett) x(t)

(

=

t

x(t)

2. 4.

Examples of dual spaces

45

look. If f were defined on step functions, we would let

{ 1, , aa << ts << st << b,b,

ks (t) ==

(2.41 )

0,

[ s] s

be the characteristic funct io n for the interval a , for > a , and define there is a set of numbers g(s) == f . Then for each step function

(k8 )

y (t) a == tq < t < · · · < tn == b ( called a partition of [a , b] ) such that n y (t) == L ai [ktt (t) - ktt- 1 (t) J. 1

1

By linearity, we would have

n 1 ( y ) == L a i [g ( t i ) - g ( t i - 1 ) J , 1 function g (s). But this is all wishful and f is determined once we know ' the

thinking, since f is not defined for step functions. "Well," you say, "can t we extend it to be defined for step functions?" That is an idea! After all, C is contained in the space B of bounded functions, which contains among other things the step functions. Thus by Theorem 2 . 6 , there' is a bounded linear functional on B which coincides with f on C and such that == In particular, is defined on step functions so we can set '

F

\

I F I 1 /1 1 ·

F

g (s) == F(k8 ).

(2 .42)

x(t) be any element of C. Then for any > 0, there is a 6 > 0 such l x(t ' ) - x (t) l < E whenever I t ' - t l 6 (in case you do not recognize it , this is just uniform continuity ) . Let a == to < t 1 < tn == b be any partition of [ a , b] such that rJ == m �x l ti - t i - 1 l < 6 .

Now let that

E

<

·

·

·

<

't

Then if t� is any point satisfying we have

< for ti - l < < ti . (t)] . y( t ) == L x(t � )[ kt l (t) -

l x (t ) - x (t� ) l

Let

.

n,

1

E

t

kt l _ 1

46

2.

Then

D UALITY

n

F (y) L x ( t� ) [g ( ti ) - g ( ti- 1 ) ] , ==

and we know that

1

I F (x) - F ( y ) l < I F I · l l x - Y l l < c-I I F I ·

In other words , n

(2.43)

lim

1] �0

L x ( t � ) [g ( ti ) - g ( t i- 1 )]

F(x ) .

1

exists and equals This limit , when it- exists , is better known Riemann-Stieltjes in t egral

as

the

1b x (t ) dg ( t ) .

( 2 . 44)

Thus , we have shown that the integral (2.44) exists and

F (x) = 1b x ( t ) dg( t ) ,

(2 .45)

x E C.

What kind of function is g ? To answer this, let n

y (t) =

L ai [kt; ( t ) - kt i - I ( t )]

1 be any step function on [a, b] . Then n

F ( y) ==

L O:i [g ( ti ) - g ( ti

-1

1

)] '

showing that n

< I F I · I Y I == 11 ! 1 1 m�x l ai l ·

i :L ai [g ( t i ) - g ( t i - 1 ) ] 1

1 This is true for all choices of the o: i . Take o: i ai == - 1 , otherwise. This gives n

L l g ( ti ) - g ( ti - 1 ) 1 <

( 2 .46)

1

t

=

lJ.

if g ( t i ) > g ( t i - 1 ) and

I II ·

Since y was any step function on [a , b] , (2 . 46 ) is true for any partition a == t0 < t 1 fn == b. Functions having this property are said to be of bo'll n ded varza tzon. The to tal varza tzon of g is defined as

< ··· <

n

V ( g ) = sup L l g ( ti ) - g ( ti- I ) I . 1

2. 4.

47

Examples of dual spaces

where the supremum ( least upper bound) is taken over all partitions of [a, b] . To summarize, we see that for each bounded linear functional f on C[a , b] , there is a fu nction g ( t ) of bounded variation on [a, b] such that

1b x ( t) dg (t) , x V ( g) < 1 / 1 · �---

(2.47)

f (x)

and Since

=

E

C [a, b] ,

n

I L X ( t � ) [g ( t i ) -- g ( t i - 1 )] I < I I X I I v ( g ) for all partitions a that

(2 .48) Hence,

(2.49)

=

1

to < t 1 <

·

·

·

< tn

1b x (t) dg ( t ) V (g)

==

b and all choices of t � , it follows

< l l x i i V (g ) .

==

1 !1 ·

It should be pointed out that the converse is also true. If x E C and g is of bounded variation, then. the Riemann-Stieltjes integral (2.44) exists and clearly gives a linear functional on C. By (2 .48) , this functional is bounded and its norm is < V ( g ) . There is a question concerning the uniqueness of the function g . We know that each bounded linear functional f can be represented in the fortn ( 2.47) , where g is of bounded variation. But suppose f can be represented Se t g 9 1 Th e n in this way by means of two such fun c tions 91 and clearly g is also of bounded variation and

(2.50)

1b x (t ) dg (t)

92 ·

=

0,

X E

==

..- 92 ·

C.

What does this imply co ncernin g g ? In order to answer this question we must recall two well-known properties of functions of bounded variation. The first is that the set of points of discontinuity of such a function is at most denumerable. The second is tha.t at any point t of discontinuity of such a function, the right- and left-hand limits g ( t + ) lim g ( t + 8) , =

g(t- )

=

8'\, 0

lim g ( t - 6)

8�0

both exist (i.e. , the d iscont i nu ity is merely a ju n1 p) . Now let c be auy point of continuity of g , a < c < b. Let 6 > 0 be given , and let z ( t ) be

48

2.

D UALITY

z(t) c

a

c+6

b

Figure 2 . 3

the continuous function which is identically one in [ a , c) , identically zero in [c + b", b] and linear in [c, c + b"] ( cf. Figure 2. 3) . Then

1b z ( t ) dg (t ) =g (c) - g (a) + 1c+O z ( t ) dg (t) .

By (2 .48) this gives

l g (c) - g (a) l < vcc + 8 ( g ) , the total variation of g in the interval [ c, c + 6] .

Now we make use of the fact ( which we shall prove in a moment ) that

(2. 5 1 )

�c+ «5 ( g )

-------+

0

as

6

�

0,

whenever g is continuous at c. Hence , g( c) == g ( a ) for all points c of continuity of g . Reasoning from the other end , we see that a necessary condition for (2.50) to hold is that

(2.52)

g (c)

==

g (a)

=

g (b)

for all points c of continuity of g . This condition is also sufficient. For if it is fulfilled, we have n

� x (t � ) [g ( ti ) - g ( ti - 1 )] = 0 1

for all partitions which avoid the discontinuities of g . Since the integral exists, it is the limit of this expression no matter how the partitions are taken provided rJ � 0. Since the discontinuities of g form at most a denumerable set , this can always be done while avoiding the discontinuities of g . Hence, (2.50 ) holds . Thus, we have proved

Lemma 2. 13. A necessary and sufficient condition for {2. 50) to hold is that g (c) = g (a) == g (b) at all poin ts c of continuity of g .

49

2. 4. Examples of dual spac es

This shows us how to "normalize" the function g in ( 2 . 47) to rnake it unique . If we define

t == a ' g (a) , g (a) + g (t ) - g (t+) , a < t < b, g(a) , t == b '

w ( t ) == then by Lemma 2 . 1 3 ,

1b x (t) dw( t )

If we now set

=

0,

x E G.

/g ( t) == g ( t) - w ( t ) '

then g is of bounded variation and satisfies

g (a) == 0 ,

(2. 53)

g(t) == g (t + ) ,

( 2 . 54)

a < t < b.

We call a function satisfying ( 2 . 53 ) and ( 2 . 54) normahz ed. By Lemma 2. 1 3 , there is at most one normalized function of bounded variation satisfying (2.4 7) . There is one 1nore thing we rnust investigate . By (2.48) and (2 .49) ,

V ( g)

=

1 1 ! 1 1 < V (g) .

We now show that V (g) < V (g) . To see this , let a == to < t 1 < · · · < tn b be any partition of [a , b] , and let E > 0 be given . Then there are points Ci such that ti < ci < ti+ l and :::;

Thus , if we take co == a.

l g ( ti + ) - g (c{) l < c/ 2 n .

en

n

==

b,

n-1

\VC

have n

,L l g ( t i ) - g (t i - J ) I < ,L l g ( f ; + ) - g ( r; ) l + ,L lg (c i ) - g ( ci - 1 ) 1 1

1

1

Tl

+ L l g ( ci- 1 ) - g ( t i 1 + ) I < V ( g) + f . 2

Thus . Since

E

To

was

arbitrary.

surnrnarize .

\\re

\\"e

-

V (.lj )

<

ll ( g) + E .

havP t l H' result .

havp

50

2.

DUALITY

C [a , b] there is a

Theorem 2 . 14. For each bounded linear functional f on unique normaliz ed function g of bounded variation such that

(2.55) and

b f (x) 1 x(t) dfJ(t), x E C[a , b] , =

( 2 . 56)

V ( g) =

l fl ·

Conversely, every such normaliz ed fJ gives C[a, b] satisfying {2. 55) and (2. 56).

a

bounded linear functional on

Proof. In case you have forgotten, it remains to prove (2. 5 1 ) . Let c > 0 be given. Then there is a l5 > 0 so small and a partition C

+

6=

TQ

<

71 < · · · <

Tm

=b

[c + 6, b] such that �b ( g ) - c < l g ( c + 6) - g (c) l + L1 l g ( rk ) - g ( rk - 1 ) 1 , and l g (c + 6 ) - g ( c) l < E . Now let c = t o < t 1 < · · · < tn = c + 6 be any partition of [ c , c + 6] . Then n L l g (ti ) - g (ti - 1 ) 1 L i g ( rk ) - g ( rk - 1 ) 1 vcb (g) 1 1 of the interval

m

+

m

<

m

. ) ( ( ) g g r r k 1 k l 1 1 Hence, n L1 l g (ti ) - g( ti 1 ) 1 < 2c . Since this is true for any partition of [ c , c + 6], we have vc +8 (g) < 2c . < 2c + L

-

This co1npletes the proof. Theorem 2. 14 is due to F. Riesz.

0

2. 5.

Problems

51

2 . 5 . Problems

( 1 ) Prove the statement following ( 2 . 7) .

F

is a bounded linear functional on a normed ve�tor space X, (2) If show that

I FI

==

sup

llxll < l

I F (x) l

==

sup

llxll = l

I F (x) l .

F(x) is called additive if F(x + y ) F(x) + F(y). If F is additive, show that F(o:x) aF( x) for all rational o: . ==

(3) A functional

=

( 4) Show that an additive functional is continuous everywhere if it is continuous at one point .

( 5) Prove that l 1 is complete. (6) Prove the Hahn-Banach theorem for a Hilbert space following the procedure outlined at the end of Section 2.2. ( 7) Let

M be a subspace of a normed vector space X which is not

dense. Show that there is a sequence � 1. and d(xn ,

{ xn }

C

X such that

M) (-8) If x E lp , y E l q , E lr, where 1 /p + 1 / q + 1 /r L1 l xi yi zil < l x i P I Y I q l z l r · z

00

==

l xn l

= 1

1 , show that

(9) If f is a bounded linear functional on X, let N be the set of all

x E X such that f( x )

x0

==

E X such that 0 . Show that there is an + every element x E X can be expressed in the form x == where x 1 E N. ( 1 0) Show that c�

==

[f (see Problem 4 of Chapter 1 ) .

( 1 1 ) Show that (2 . 1 1 ) implies p(O)

==

0.

o:x0 x 1 ,

52

2.

DUALITY

(12)

Prove the Hahn-Banach theorem for the case when V satisfies ( 2 . 2 1 ) without using Zorn ' s lemma.

(13)

Show that Theorem

(14)

Let F, G be linear functionals on a vector space V , and assume that F (v) == 0 ==} G(v) == 0, v E V. Show that there is a scalar such that G ( v ) == CF(v) , v E V.

2. 1

implies Theorem 2 . 3 .

C

( 1 5) Let F be a linear functional on a Hilbert space X with D (F) == X. Show that F is bounded if and only if N (F) is closed .

( 16)

A subset U of a normed vector space is called convex if o:x + ( 1 ---- n )y is in U for all x , y E U and 0 < o: < 1. Show that Theorem 2.2

holds when N is a closed convex set .

(17)

A set U is called midpoznt convex if (x + y) / 2 is in U for all

x , y E U.

Show that a closed, midpoint convex subset of a normed vector space is convex, but if it is not closed, this need not be the case.

( 18) For X a Banach space, show that a sequence {xk } converges in norm if and only if x ' (xk) converges uniformly for each x' E X' satisfying II x' II == 1 .

(19)

a

{xk } is sequence of elements in a normed vector space X and { n k } is a sequence of scalars, show that a necessary and sufficient condition for the existence of an x ' E X' satisfying x' (x k ) a k and ll x' ll A1 , If

==

==

is that

11

rt

1

1

I L 11k ak l < M il L 11kxk l l holds for each

n

and scalars fJ 1 , ·

· ·

, f3n .

( 20 ) If p ( x) is a sublinear functional on a vector space V and xo is any element of V, show that there is a linear functional F on V such that F(xo )

==

xo

and

- p ( - x)

<

F(x) < p(x) .

x

E \ '.

2. 5.

Problems

53

(21) Show that there is a functional f E l� satisfying

f

lim inf X k < ( x ) $ lim sup X k , k -+oo

k -+oo

x

==

( x 1 , x2 ,

·

·

·

) E l00 •

(22) If X is a Hilbert space, show that the functional F given by Theo­ rem 2.6 is unique. (23) Show that lp is not a Banach space when 0 < wrong? (24) If f E lp for some p <

oo ,

show that

I / I l P 1 / l oo as �

P

�

oo .

p

< 1 . What goes

-

4 ... • •

•'

• '

.

'

·-·

. ....

'

-

·-

... .

Chapter

,.· ...

3

LINEAR OPERATORS

3 . 1 . Basic properties

Let X, Y be normed vector spaces. A mapping A which assigns to each element x of a s et D (A) C X a unique element y E Y is called an operator ( or transformation) . The set D (A) on which A acts is called the domain of .A. . The o perator A is called linear if a ) D (A) is a subspace of X and b ) A (a1 x 1 + o:2x2 ) ments x 1 , x2 E D (A) .

=

o: 1 Ax 1

+

n2Ax2 for all scalars 0: 1 , 0: 2 and all ele­

Until further notice we shall only consider operators A with D(A)

=

X.

An operator A is called bounded if there is a constant !VI such that

(3 . 1 )

I I Ax l l < M l l x ll ,

x E X.

55

3. LINEAR OPERATORS

56 The norm of such an operator is defined by

I I A ll

(3.2)

M

= ���

I I Ax i l . ll x ll

which works in (3. 1 ) . An operator A is called Again, it is the smallest implies A x n � A x in Y. A if X n � x in continuous at a point xo E bounded linear operator is continuous at each point . For- if X n � x in X, then I I Ax n - A x i l < I I A II · l � x; � ·x ll � 0 .. We also have

X

X

Theorem 3 . 1 . If a linear operator A is continuous at one point xo E then it is bounded, and hence con tinuous at every point.

X,

Proof. If A were not bounded, then for each n we could find an element X n E X such that I I A x n ll > n l l x n ll · Set

Then Zn

�

xo . Since A is continuous at xo , we must have Azn Axn Azn= + Axo . n ll x n l l

Hence,

Ax n n l l x n ll

--

But providing a contradiction.

----+

�

Axo . But

0.

I I Ax n ll > 1' n l l x n ll D

B(X,

X

to Y. Y ) be the set of bounded linear operators from We let Y) is a normed vector Under the norm (3 . 2 ) , one easily checks that space . As a generalization of Theorem 2 . 10, we have

B( X , Theorem 3 . 2 . If Y is a Banach space, so is B(X, Y) .

Proof. Suppose {A n } is a Cauchy sequence of operators in for each E > 0 there is an integer N such that

I I A n - Am l l < Thus

( 3 . 3)

E

for m, n > N .

for each x #- 0,

II A n x - A m x l l < c l l x l l ,

m, n

> N.

B( X, Y) . Then

57

3. 2. The adjoint operator

This shows that { Anx } is a Cauchy sequence in Y. Since Y is complete, there is a Yx E Y such that A n x � Yx in Y. Define the operator A from X to Y by A x == Yx · Then A is linear tsee the proof of Theorem 2. 10) . Let m � oo in (3.3) . Then II Anx � Ax il < c llxll , n > N. Hence,

I I A xll < c llxl l + I I Anx ll

( c + II A n ll ) llxl l ,

<

n

This shows that A is bounded. Moreover,

I I An - A ll

< E,

Hence,

II An - All This completes the proof.

---+

0 as

n

n

> N.

> N. ---+

oo . D

3 . 2 . The adjoint operator

Suppose X, Y are nor1ned vector spaces and A E B(X, Y) . For each y ' E -Y' , the expression y ' ( A x ) assigns a scalar to each x E X. Thus, it is a functional F ( x ). Clearly F is linear. It is also bounded since

' ' ' I F ( x ) I == I y (A x ) I < I I y I I II A x I I < II y I I · I I A II · I I x I I · Thus, there is an x ' E X ' such that (3.4) y' (A x ) == x ' ( x ) , x E X . This functional x ' is unique , for any other functional satisfying (3.4) would ·

have to coincide with x ' on each x E X. Thus, to each y ' E Y ' we have assigned a unique x ' E X ' . We designate this assignment by A' and note that it is a linear operator from Y' to X ' . Thus , (3.4) can be written in the forn1 y

(3.5)

'

(Ax )

== A' y ' ( x ) .

The operator A' is called the adJoznt (or conjugate) of A. depending on the mood one is in . But no n1atter what the mood , the following is true. Theorem 3 . 3 .

A' E B ( Y' , X ' ) ,

and I I A' II

== I I A ll .

Proof. We have by ( 3 . 5 ) ,

I Hence.

A' y

'

' ( x) I == I y ' (A x ) I < II y II · I I A II · I I x I I ·

3. LINEAR OPERATORS

58

I A' I < I A l · To prove the reverse I Ax l < IJ A' I · l x l , x E X.

A'

This shows that E B (Y' , X') and that inequality we must show that (3 .6)

Again by (3.5) , we have

I y' (Ax) I < I A'y' l I x I < I A' l I y' l I x I , '(Ax) l < I A, I l x l , E X. sup -I Y I y'I I .

showing that

·

·

·

X

·

y' =f; O

We now appeal to (2.28) to obtain (3 .6) .

D

The adjoint has the following easily verified properties:

(A + B )' = A' + B' . ( o:A) ' = o:A'. (AB) ' = B'A'.

(3 . 7) (3.8) (3.9)

Why should we consider adjoints? One reason is as follows. Many problems in mathematics and its applications can be put in the form: given normed vector spaces X, Y and an operator Y) , one wishes to E solve

A B( X,

Ax == y . The set of all y for which one can solve (3 . 10) is called the rang e of A and is denoted by R ( A). The set of all x for which Ax = 0 is called the null space of A and is denoted by N (A). Since A is linear, it is easily checked that N (A) and R(A) are subspaces of X and Y, respectively. If y E R ( A), there is an x E X satisfying (3 . 10) . For any y ' E Y' we have y' ( Ax) y' ( y) . Taking adjoints we get A'y' (x) y' (y) . If y' E N (A') , this gives y'(y ) 0. Thus, a necessary condition that y E R(A) E N (A' ) . bviously. it is t hat y' (y) 0 for al be of great interest to know when this condition is sufficient . We h h answer in (3. 10)

==

=

==

l y

'

==

also

0

the next section. In doing so we shall find it picturesque ( if not gro tesque ) t ert n i n ol ogy.

Vv"ould s

a ll

convenif'nt

fi nd t e

to i n tr od u ce

s o 1n e

3. 3.

Annihilators

59

3 . 3. Annihilators

Let S be a subset of a normed vector space X. A functional x' E X ' is called an annzhilator of S if x ' (x) == 0 , x E S . The set of all annihilators of S is denoted by s o . To be fair, this should not be a one-way proposition. So for any subset r of X' , we call x E X an annihilator of r if x ' ( x) == 0, x ' E r. We denote the set of such annihilators of r by 0'T. Now we can state our necessary condition of the last section in terms of annihilators . In fact , it merely states (3. 1 1 )

c

R( A)

o N (A ' ) .

We aTe interested in determining when the two sets are the same. We first prove some statements about annihilators. Lemma 3 .4.

so

and

or

are closed subspaces.

Proof. We consider s o ; the proof for or is 'siinilar. Clearly, s o is a subspace , for if x 1 , x 2 annihilate S, so does n 1 x 1 +n 2 x 2 . Suppose �� E s o and x� � x' in X ' . Then x � (x) � x ' (x) , x E X. In particular , this holds for all x E S, showing that x ' E s o . D Lemma 3 . 5 . If l'vf is a closed subspace of X, then °(A1° ) == M.

Proof. Clearly, x E 0( M0 ) if, and only if, x ' ( �: ) == 0 for all x' E 1\![ 0 • But this is satisfied by all �; E M. Hence, A1 C 0( A1° ) . Now suppose x 1 is an element of X which is not in M. Since M is a closed subspace,

d ( x 1 , A!) == inf l l x 1 - z l l > 0 . z E Al

By Theorem 2 . 10, there is an x� E X' such that x � ( x 1 ) == d(x 1 , J\;f ) , ll x� ll == 1 , and x� ( x) == 0 for all x E Af ( i.e. , x' E AI0 ) . Since x 1 does not annihilate D x� , it is not in °(Af0 ) . Hence, 0(M0 ) C Af, and the proof is complete.

Let W be a subset of X. The subspace of X spanned ( or generated) by H' consists of the set of finite linear co111binations of elerr1cnts of VV, i . e. , sums of the for1n n

'""""' o: · x · L J .7 ' 1

where the a:1 are scalars and the Xj arc in lif/. The closure 'WT of W consists of those points of X which are the lirrlits of sequences of elen1ents of W. One

3. LINEAR OPERATORS

60

checks easily that W is a closed set in X. The closed subspace spanned by W is the closure of the subspace spanned by W. We have Lemma 3.6. If S is a subset of X, and M is the closed subspace spanned by S, then M0 = so and M = 0(S0) . Proof. The second statement follows from the first, since by Lemma 3. 5, M = 0 (M0) = 0( S0 ) . As for the first, since S C M, we have clearly M° C 5° . Moreover, if xj E S and x ' E so , then x' (

n

n

L aj Xj ) = L O:jX1 ( xj ) 1

1

=

0,

showing that x ' annihilates the subspace spanned by S. Moreover, if {zn } is a sequence of elements of this subspace, and Zn --+ z in X, then

x' (zn )

�

x'(z) . '

Hence, x' annihilates M, and the proof is complete.

Now returning to our operator A E B ( X, Y) , we note that

(3. 12)

R(A) 0

==

0

N (A') .

ForY y' E R(A)0 if, and only if, y' (A x ) 0 for all x E X . This in turn is true if, and only if, A'y' (x) == 0 for all x , i.e. , if A'y' == 0. Now applying Lemma 3.6, we have =

(3. 13) since R(A) itself is a subspace. Thus we have Theorem 3. 7. A necessary and sufficient condition that

(3. 14)

R(A)

==

0N(A ' ) ,

is that R(A) be closed in Y. 3 .4 . The inverse operator

Suppose we are interested in solving the equation

(3. 15)

Ax

==

y,

where A E B (X, Y) and X, Y are normed vector spaces. If R(A) == Y, we know that we can solve ( 3. 1 5 ) for each y E Y. If N(A) consists only of the vector 0, we know that the solution is unique. However, if the problem arises from applications , it may happen that y was determined by experimental means , which are subject to a certain amount of error. Of course, it is hoped that the error will be small. Thus, in solving (3. 15) it is desirable to know that when the value of y is close to its "correct" value, the sarnc will be true for the solution x . Mathematically, the question is: if Yl is close to Y2 and

3. 4 .

The inverse operator

6_1

A x i = Yi , i == 1 , 2 , does it follow that x 1 is close to x 2 ? This question can be expressed very conveniently in terms of the inverse operator of A. If R(A) == Y and N(A) = {0} (i.e. , consists only of the vector 0) , we can assign to each y E Y the unique solution of (3. 15) . This assignment is an operator from Y to and is usually denoted by A- 1 and called the inverse operator of A. It is linear because of the linearity of A. Our question of the last paragraph is equivalent to: when is A- 1 continuous? By Theorem 3. 1 , this is equivalent to when is it bounded. A very important answer to this question is given by

X

X,

Theorem 3.8. If Y are Banach spaces, and A E Y, N(A) = {0} , then A - 1 E (Y,

B X) .

B(X, Y) with R(A)

=

This theorem is sometimes referred to as the bounded inverse theorem. Its proof is not difficult , but requires some preparation. The main tool is the Baire category theorem (Theorem 3.9 below) . The Let X be a normed vector space and let W be a set of vectors in set W is called nowhere dense in if every sphere of the form

X.

X

l l x - xo ll < r, r > 0, contains a vector not in the closure W of W (i.e. , W contains no sphere) . A set W c is of the first category in if it is the denumerable union of nowhere dense sets, i.e. , if

X

X

=

00

w u wk ,

1 where each Wk is nowhere dense. Otherwise, W is said to be in the second category. The following is known as Baire 's category theorem. Theorem 3 . 9 . If X is complete, then it is of the second category. Proof. Suppose

(3. 16)

X were of the first category. Then 00

X ==

U wk , 1

where each wk is nowhere dense. Thus there is a point X1 not in w 1 · Since x 1 is not a limit point of W1 , there is an r1 satisfying 0 < r 1 < 1 such that the closure of the sphere does not intersect w1 . This sphere contains a point X 2 not in contains a sphere of the form

w 2 ' and hence

62

3. LINEAR OPERATORS

such that the closure of S2 does not intersect W�2 . Inductively, there is a sequence of spheres sk c sk 1 of the form:

Sk == { x : l l x - X k ll < rk } , 0 < rk < 1 j k , such that the clos�re of Sk does not intersect Wk . Now for j Xj E Sk , and hence ,

> k , we have

(3. 1 7) This shows that { x k } forms a Cauchy sequence in X. Since X is complete, this sequence has a limit xo E X. Letting j � oo in (3. 17) , we get

l l xo - X k I I < rk . Thus Xo is in the closure of sk for each k , showing that Xo is not in any of the wk , and hence not in 00

0

This contradicts (3. 16) .

Let X, Y be normed vector spaces , and let A be a linear operator from X to Y. We now officially lift our restriction that D(A) == X. However, if A E B (X, Y) , it is still to be assumed that D (A) == X. The operator A is called closed if whenever { X n } C D (A) is a sequence satisfying

Xn then x E D(A) and A x (3. 18)

----t

==

x in X,

A xn

y.

-----+

y

in Y,

Clearly, all operators in B (X, Y) are closed. Another obvious statement is that if A is closed , then N (A) is a closed subspace of X. A statement which is not so obvious is Theorem 3 . 10. If X, Y ate Banach spaces, and A zs a closed linear operator from X to Y, with D (A) == X, then

a) There are positive constants M, l l x l l < r.

r

such that I I A x l l < M whenever

b) A E B (X, Y) . Theorem 3. ib is called the closed graph theorem. The geometrical sig­ nificance of the terminology will be discussed later. We note here that the theorem immediately implies Theorem 3.8. For if A - l exists, it is obviously a closed operator from Y to X, and by hypothesis D(A - 1 ) == Y. Hence, we can apply Theorem 3. 10 to conclude that A - 1 is bounded. In making these observations , we note that it was even unnecessary to assume in Theorem

3.4. The inverse operator

63

3.8 that A E B(X, Y) . This hypothesis was only used to show that A- 1 is closed. But from the definition· of a closed operator we see that A - 1 is closed if, and only if, A is. So we might as well have assumed in the first place that A is merely closed. We can therefore replace Theorem 3.8 by the seemingly stronger Theorem 3 . 1 1 . If X , Y are Banach spaces and A is a closed linear operator from X to Y with R(A) = Y, N (A) = {0} , then A- 1 E B( Y, X) .

The reason we said "seemingly" is that Theorem 3.8 actually implies Theorem 3. 1 1 . To see this we use a little trick. Suppose A is a closed linear operator from X to Y. As we noted, D(A) is a subspace of X. If we use the norm of X, then D(A) is a normed vector space. The only trouble is that it is not complete ( unless A E B(X, Y) ) . So we use another norm, namely

+

l l x ii A == ll x ll II Ax l l · Now don't laugh. This is a norm on D(A) . But that is not all. If X and Y are Banach spaces and A is closed, then D (A) is complete with respect to this norm. For if { x n } is a Cauchy sequence with respect to this norm, then { x n } is a Cauchy sequence in X and {A x n } is a Cauchy sequence in Y. Hence, there are x E X , y E Y such that ( 3. 18 ) holds . Since A is closed, x E D(A) and A x == y. Thus, l l xn - x ! I A � 0. Now we forget about X and consider A as an operator from the Banach space D(A) with its new norm (3. 19) to Y. Moreover, A E B(D (A) , Y) , since (3. 19)

I I Ax ll < ll x i i A ·

Thus we can apply Theorem 3.8 to conclude that A- 1 E B(Y, D (A) ) , i.e. , that or

II A - 1 Y I I x

+ II Y I I Y < C IIY I I Y ·

This gives Theorem 3 . 1 1 . A similar trick can be used to show that rfheorem 3.8 also implies the closed graph theorem ( Theorem 3. 10 ) . This is done by introducing the cartesian product X x Y of X and Y. This is defined as the set of all ordered pairs (x , y ) of elements x E X, y E Y. They are added by means of the formula Under the norm

II (x , y) II == ll x l l x + II Y II Y ,

X x Y becomes a normed vector space provided X and Y are normed vector spaces. J\tioreover, it is a Banach space when X and Y are Banach spaces.

3. LINEAR OPERATORS

64

If A is an operator from X to Y, the graph G A of A is the subset of X consisting of pairs of the form

x

Y

(x , Ax ) . It is clearly a subspace of X x Y. Moreover , GA is a - closed subspace if, and only if, A is a closed operator (this is the reason for the terminology) . Equipped with this knowledge, we now show how Theorem 3. 10 is a conse­ quence of Theorem 3.8. Let A be a closed operator from X to Y defined on the whole of X, where X and Y are Banach spaces. Let E be the linear operator from G A to X defined by E( ( x , A x ) )

==

x.

By what we have j ust seen , GA is a Banach space . Moreover, E E B ( GA , X ) , and R(E) == X. Also N(E) == (0, 0) . Hence , we can apply Theore1n 3.8 to conclude that E - 1 is bounded , i . e . that ,

l l x l l + I I A.r l l < C l l x l l ·

from which we conclude that A E B ( X, Y) . We have shown that Theorems 3.8, 3. 10, and 3. 1 1 are equivalent. It therefore suffices to prove only one of then1. We choose Theorem 3. 10. Proof of Theorem 3. 1 0. We first note that (b) follows frotn (a) . For if x # 0 is any element of X, set z == r x /2 l l x l l . Then l l z ll < r , so t hat I I A z l l < Al. But A z == rA x /2 l l x l l , sho\ving th a t

I I A x ll < 2 Afr - 1 l l x l l .

Thus , (a) implies (b) . In order to prove (a) Un == { .z·

e1nploy Thcoren1 3 . 9 . Set

I I A x II < n } .

:

Then X ==

we

ex:;

U

[in .

1

Since X is con1plete. it 1nust be of t he second category ( T he o r e n 1 3 . 9 ) . a nd hence. at least one of the Un . say Uk . is not no,vhere dense ( note t he d o u b l e negative) . This n1eans t hat Uk contains a sphere o f t he forn1 v( ==

{

.z ·

:

l l .r

.1· o

-

II

< t}.

t > 0.

i.e uk is dense in �� . \\'"e n1ay t ake t h0 cent er .r o o f \ I t o b e i n uh· b�,. s h i ft i n g i t slightly. Fro1n this it follo\YS t hat t he s e t of Y<'ct o rs of t he for111 : - J'o . \Vhere z E uh' • is dense i l l t he sphcr0 . .

st ==

{

1·

:

l l .r I I < t } .

3. 4. The inverse operator

65

X E St , then X + xo E and for any > 0 there is a z E uk such l x + xo - z l < c . Now, all vectors of the form z - xo, z E Uk , are contained in U2k , since I A ( z - xo) l < I Az l + I Axo l < 2k. Hence , u2 k is dense in St . Since X E Um if, and only if, xjm E UI , UI is dense in Sr { x : l x l < r t /2k} , 0, Uo: is dense in So:r · Let 8 be any number satisfying and for each n 0 < < 1 . I claim that Sr UI /( I - 8 ) · (3.20) This means that l x l < r implies I Ax l < 1 / ( 1 - 8) , which is exactly what we want to prove. To prove (3.20) , let X be any point in Sr . Since UI is dense in Sr , there is an X I E UI n Sr such that l x i - x l < 8r, i.e. , X I - X E S8r · Now u8 is dense in S8r · Hence, there is an X 2 u8 n S8r such that l x2 + XI - x l < 82 r, i.e. , X 2 + X I - X E s8 2 r · But U82 is dense in s82 r , and there is an X 3 E U82 n S82 r such that l x 3 + x 2 + X I - x l < 83 r, i.e. , X3 + x 2 + x 1 - x E S8 3 r · Continuing in this manner, there is an Xn + I E U8n n S8nr such that n+ I I I:I Xk - x l < 8n+ I r. Since X n+ I E U8n , we have €

vt ,

For, if that

==

>

8

=

c

E

so that

k

k

J

J

k -j + I 1 8 l i I A � xn l < � I Axn l < o - 1 - 8 0 as k Thus, A L1 Xn is a Cauchy sequence in Y. Since Y is complete, this converges to some E Y. But y

k

I I:I Xn - x l < 8kr

�

�

0 as

k

� oo .

j,

� 00 .

66

3.

Since A is closed, Ax

==

IIYII <

y,

00

and

LINEAR OPERATORS

00

L1 I I Axn l l < L1 bn - 1 == 1 / ( 1 - 6) .

This completes the proof.

D

3 . 5 . Operators with closed ranges

Let X, Y be normed vector spaces and let A be an operator in B ( X, Y ) . The­ orem 3 . 7 tells us that R(A) consists precisely of the annihilators of N (A ' ) , provided R(A) is closed in Y. It would therefore be of interest to know when this is so. There is a simple answer when X, Y are complete. Suppose that N (A) == { 0} , i .e. , that A is one-to-one ( we shall remove this assumption later ) . Then the inverse A - l of A exists and is defined on R(A) . If Y is complete and R(A) is closed in Y, then R(A) itself is a Banach space with the same norm. If X is co1nplete as well, we can apply Theorem 3.8 with Y replaced by R(A) to conclude that A - 1 E B ( R(A) , X ) . Thus there is a constant C such that (3. 2 1 ) Another way of writing (3. 2 1 ) is (3.22)

l l x l l < C I I Ax l l ,

x E X.

Thus, if X, Y are complete, then a necessary condition that R(A) be closed in Y is that (3.22) holds . A moment 's reflection shows that it is also suf­ ficient . For if Yn � y in Y, where Yn E R(A) , set Xn == A - l Yn · Then, by (3 . 22) , l l xn - Xm l f < C I I Yn - Ym l l � 0 as 1n , n � oo . Thus, {xn } is a Cauchy sequence in X, and therefore approaches a limit x E X. Since A is bounded , Yn == Axn � Ax , showing that Ax == y . Thus, we have shown that when X, Y are complete and A is a one-to-one operator in B (X, Y ) , then a necessary and sufficient condition that R(A ) be closed in Y is that (3.22) hold. As you know, we have a great weakness for trying to "generalize" statements in hopes that they may apply to other situations. At this point we give way to ourselves and ask whether the same conclusion can be drawn without assuming that A E B ( X, y·) . True, we used Theorem 3.8, which assumes A E B (X, Y ) . But we could have used Theorem 3. 1 1 , \vhich comes to the same conclusion with only assuming that A is a closed operator from X to Y. So let us try to get away with assuming only this. Checking the rest of the proof, we find that the only other place we used the boundedness of A is in concluding that Axn � Ax . But I claim this is true even when A is only closed. For, remember that Axn � y in Y

3. 5. Operators with closed rang·es

67

while X n � x in X. By the definition of a closed operator, x E D(A) and Ax == y , showing that y E R(A) . Thus , we have proved Theorem 3 . 1 2 . Let X, Y be Banach spaces, and let A be a one- to-one closed linear operator from X to Y. Then a necessary and sufficient condition that R(A) be closed in Y �s that (3. 22) hold.

Of course this is all well and good if N(A) is j ust the zero element . But what does one do in case it is not? Do not despair. There is a method, but it needs a bit of preparation. We first examine N (A) a bit more closely. As we have remarked several times, it is a subspace of X. But more than that , when A is a closed operator, N(A) is a closed subspace of X. For if Xn E N(A) and Xn � x in X, then 0 == Axn � 0 . Since A is closed, x E D (A) and Ax == 0, i.e. , x E N (A) . Now let M be any closed subspace of a normed vector space X. Suppose we say that x E X is equivalent to u E X and write x r-..J u whenever x - u E M. Then one checks that x r-..J x, and x r-..J u is the same as u r-..J x. Moreover, x r-..J u, u r-..J v implies x r-..J v . A relationship having these properties is called an equivalence relation. Next, for each x E X, let [x] denote the set of all u E X such that u r-..J x . Such a set is called a coset. Clearly [x] == [u] if, and only if, x r-..J u. Moreover, if x is not equivalent to u, then [x] and [u] have no elements in common. We define the "sum" of two cosets by [x] + [u] == [x + u] .

However, we have to ascertain whether this definition makes sense. For suppose [x 1 ] == [x] and [u 1 ] == [u] . Then x 1 r-..J x , u 1 r-..J u, and hence, x 1 + u 1 r-..J x + u, showing that [x 1 + u 1 ] == [x + u] . We also define multiplication of a coset by a scalar by n [x] == [o: x] . Under these definitions, one can verify easily that the collection of cosets forms a vector space with zero element [0] == M. You may think that it is rather strange to form a vector space in which each element is a set . But there is nothing in the rules that prevents us from doing it . We are going to do something even crazier. We are going to give this vector space a norm by setting II [x] I I == d ( x , M) == inf I I x - z I I · zEM

O f course, we must check to see if this satisfies all of the properties of a

norm. For instance, if I I [x] I I == 0, then d (x, !vi) == 0. Here we need the fact that M is closed in X to conclude that x E l'vf so that [x] == [OJ . The triangle inequality is also easily verified. In fact, for x , u E X, there are sequences { zn } and { Wn } in lvf such that l l x - zn l l � d (x , 1\1 ) , l l u - wn l l � d('u , AI) .

3. LINEAR OPERATORS

68

Hence, d (x + u , M)

==

inf l l x + u - z l l

zEM

+

< l l x U - Zn - Wn l l < II X - Zn II II U - Wn II �

+

d (x , M) + d (u, M) .

It follows that the collection of cosets forms a normed vector space, which inappropriately is called a quotient ( or factor) space and is denoted by X/M. A very important property of such spaces is given by Theorem 3 . 13. If M is a closed subspace of a Banach space X, then XjM

is a Banach space.

Before proving Theorem 3 . 13, we show how quotient spaces help us find a counterpart for Theorem 3 . 12 when A is not one-to-one. Assume that X, Y are Banach spaces and that A is a closed linear operator from X to Y. Define the operator A from XIN (A) to Y as follows. D( A ) is to consist of those cosets [x] C D(A) . ( Note that [x] E D(A) if, and only if, x E D(A) . For, if x E D (A) and u E [x] , then x - u E N (A) , showing that u E D (A) . ) For [x] E D ( A ) , we define A [x] to be Ax. This definition makes sense, since Au == Ax whenever u E [x] . Moreover, A is a closed operator from XIN (A) to Y.' For if [ x n ] --t [x ] in XIN ( A ) and A [xn] --t y. in Y, then there is a sequence { zn } C N ( A) such that X n + Zn --t x in X and A (xn + Zn) --t y in Y. Since A is closed, x E D(A) and Ax == y . Hence, [ x] E D( A ) and

A [x]

==

y.

What is N (A) ? If A [ x ] == 0, then A x == 0, and hence, x E N (A) . Thus, [x] == [0) . The accomplishment of all this madness is that A is a one-to-one, closed linear operator from XIN (A) to Y. Moreover, XIN (A) is a Banach space by Theorem 3. 13. Thus, by Theorem (3. 12) , R(A) = R ( A ) is closed in Y if, and only if,

( 3.23 )

ll [x] ll

< C I I A [xJII ,

[x ] E D ( A ) ,

for some constant C. Rewriting ( 3 . 23 ) in our old terminology, we get ( 3.24 )

d (x , N(A) ) < C II A x ll ,

x E D (A) .

Hence, we have Theorem 3. 14. If X, Y are Banach spaces, and A is a closed linear operator from X to Y, then R (A) is closed in Y if, and only if, there is a constant C

such that (3. 24) holds.

It remains to prove Theorem 3 . 13. To that end, let {xn } be a Cauchy sequence in XI M. Then for each k , there is a number N ( k ) > N ( k 1 ) such --

Operators with closed ranges

3. 5.

that Set

Uk

==

69

l l [x m] - [xn ] l l <:2 - k ,

X N( k ) · Then

m, n

> N ( k) .

I I [uk + l] - [u k] II < 2 - k .

Thus, there is a Zk E M such that Set

lluk + l - U k + Zk li < 2 - A; .

==

Vk u k + l - uk + Zk , and n Wn � V k 1 ==

==

n Un+ l - 'lL I + � Zk . 1

Then and consequently for

n

>

mJ ,

n

n

0 as llwn - Wm ll < � l l vk l l < � 2 - k rn + l nt + l { wn} is a Cauchy sequence in X, and since X is complete, Wn m � oo .

�

Hence, in X. Thus,

n- 1 l l [un ] - [w + u i ] II < llun - W - U I + � zk l l

Hence,

[un ]

1

==

l lwn - 1 - w ll

�

---=+

n �

.

0 as

w

oo .

converges to a limit in XjN (A) , and the proof is cornplete.

"Wait a n1inute!" you cry. "It is explicitly stated in the definition of completeness that the whole Cauchy sequence is to converge to a lirr1it . Here { [u n l } is n1erely a subsequence of the original Cauchy sequence { [ xn] } .', Relax! There is nothing to worry about . A Cauchy sequence cannot go off in different directions. In fact , we have Lemma 3 . 1 5 . If a 8ubseq,u ence of a Cauchy sequence co n v e rg e s then the w

.

h o le sequence

converges .

Proof. Let {xn } be a Cauchy sequence in a nor1ned vector let E > 0 be given . Then there is an N so large that

l l .r n

.r 1 n

II

<

E,

m

space

, n > N.

Now if {.1·n } has a subsequence converging to x in such that

-'X" ,

then there is an

X. and

m

>

N

3. LINEAR OPERATORS

70

for all

n

> N. Now the proof is complete.

0

We now use Theorem 3. 14 to prove a result related to Theorem 3:7. Theorem 3. 16. Let X, Y be Banach spaces, and assume that A E B (X, Y) . If R(A) is closed in Y, then

R(A' )

(3. 25)

=

N(A) 0 ,

and hence, R(A ' ) is closed in X' (Lemma 3 .4). Proof. If x' E R( A') , then there is a y ' E Y' such that A ' y ' == x ' . For X E N (A) , x ' (x) == A' y ' (x) = y' (Ax) == 0.

Hence, x ' E N (A) 0 • Conversely, assume that x� E N(A) 0 • Let y be any element of R(A) , and let x be any element of X such that Ax == y . Set f (y)

=

x' (x) .

This , believe it or not , is a functional on R(A) . For, if we had chosen another X I E X such that Axi = y , then XI - x E N(A) , and hence, 1 X ( XI )

= X

1

(X) ,

showing that f depends only on y and not on the particular x chosen. Clearly f is a linear functional. But it is also bounded. For, if z is any element in N (A) , .f ( y ) == x' ( x - z ) , and hence, I J ( y ) l < ll x ' ll · ll x - z l l . Thus, l f ( y ) l < l l x' l l d (x, N(A) ) < C l l x' l l · I I Y I I by Theorem 3. 14. By the Hahn-Banach Theorem ( Theorem 2.6) , f can be extended to the whole of Y. Thus there is a y ' E Y' such that f (Y)

==

y' ( Y ) ,

y E Y.

In particular, this holds for y E R (A) . Hence, x' (x) == y ' (Ax) ,

x E X.

By the definition of A' , this becomes x' (x)

Since this is true for all

x

=

A' y ' (x ) ,

x E X.

E X, we have x'

==

A' y ' .

Hence, x' E R(A ' ) , and the proof is complete.

0

71

3. 7. The open mapping theorem

3 . 6 . The uniform boundedness principle

As an application of the closed graph theorem, we shall prove an important result know either as the Uniform Boundedness Principle or the Banach­

Steinhaus Theorem.

Theorem 3. 1 7. Let X be a Banach space, and let Y be a normed vector space. Let W be any subset of B (X, Y) such that for each x E X,

sup II A x i l < · oo .

AEW

Then there is a finite constant M such that I I A l l < M fo r all A E W. Proof. For each positive integer n, let Sn denote the set of all x E X such that II A x l l < n for all A E W. Clearly, Sn is closed. For if { xk } is a sequence of elements in Sn , and X k � x , then for each A E W we have II A x H == lim II A xk ll < n . Thus , x E Sn Now, by hypothesis , each x E X is in some Sn . Therefore, .

00

U sn .

X ==

1

Since X is of the second category (Theorem 3.9) , at }east one of the Sn , say SN , contains a sphere (note that each Sn is closed ) . Thus , there is an xo E X and an r > 0 such that x E SN when ll x - xo ll < r. Let x be an element of X, and set z == xo + r x/ 2 ll x ll · Then l l z - xo l l < r. Thus, II Az l l < N for all A E W. Since xo E SN , this implies

I I Ax i J <

4N II x l / r

,

A E W, x E X,

which is precisely what we want to show. The proof is complete.

0

3 . 7. The open mapping theorem

Another useful consequence of the previous theorems can be stated as fol­ lows . Theorem 3. 18. Let A be a closed operator from a Banach space X to a Banach space Y such that R(A) == Y . If Q is any open subset of D(A) , then

the zmage A ( Q) of Q is open in Y .

Proof. Let D == D(A) /N(A) , and define the operator A from D to Y by A [x ] == A x . Then A is a 9ne-to-one operator fron1 the Banach space D onto the Banach space Y. Consequently, it has a bounded inverse :

3. LINEAR OPERATORS

72

Let Q be an open set in D ( A ) , and let Yo E A ( Q) Then there is an x o E Q such that A x o == YO · Let E > 0 be such that ll x - xo ii A < E implies that implies x E Q. Take == c/ Co . Then II Y - Yo I I < .

6

6

II AA - 1 ( y - Yo ) I IA < Co l lY - Yo II < E. Consequently, there is a [ z] E D such that I I [ z] I IA < E an d A[ z] == y - YO· This implies that there is a z E D ( A ) such that ll z ii A < E and A z == y - Yo · Let x == z + x o . Then x E D (A) and l l x - x o i iA == l l z ii A < E. Thus, x E Q and A x == y . This means that I IY - Yo l l < impl ies that y E A (Q) . Henc e ,

6

A( Q) is open.

0

Theorem 3. 1 8 is known as the Open Mapping Theorem. A simple con. sequence IS

A be a closed operator from a Banach space X to a Banach space Y such that R(A ) == Y . If Q any open subset of X , then the image A(Q n D (A) ) oj Q n D ( A ) is open ir� Y .

Corollary 3 . 1 9 . Let

zs

This follows from the fact that Q n D (A) is open in D (A) . 3 . 8 . Problems

( 1 ) Prove that a normed vector space X is complete if and only if every series in X satisfying 00

l:1 l l xj II <

oo

converges to a limit in X .

(2) Let

A

be a linear operator from a normed vector space X o_nto a normed vector space Y. Show that A - l exists and is bounded if and only if there is a number M > 0 such that

ll x ll < M II Ax l l ,

x

J

E D (A) .

(3) If X, Y a1:e Hilbert spaces, is B ( X, Y) a Hilbert space?

(4) Prove : ( a )

N ( A ) == 0R ( A ' ) ; ( b ) R ( A ')

C

N (A) 0 •

( 5 ) Let X, Y be two normed vector - spaces, and let x #- 0 be any element of X and y any element of Y. Show that there ·is an operator A E B ( X, Y) such that Ax == y .

73

3. 8. Problems

(6) Let X, Y be normed vector spaces having the property that there is a one-to-one operator A E B (X, Y) such that A- 1 E B (Y, X) . Show that X is complete if and only if Y is. (7) If a normed vector space X has a subspace M such that M and Xj M are complete, show that X is a Banach space. (8) Let X, Y be Banach spaces, and assu1ne that A E B (X, Y) , B E B(Y, X) are such that BA = I. If E B (X, Y ) is such that A I I · I I B I I < 1 , show that there is an S E B ( Y, X ) such that ST = I

liT ­

T

.

(9) Let M be a closed subspace of a normed vector space X . Show that

M' =

��' (�)' = M0

Why did we assume M closed?

( 10) If X , Y are normed vector spaces and B (X, Y) is complete, show that Y is complete.

(11)

Show that a subset G of X x Y is the graph of an operator fro 1n X to Y if and only if G is a subspace such that (0, y ) E G implies y = 0.

( 1 2) If X, Y are nor1ned vector spaces , are the nor n1 s of (X X ' x Y' equivalent?

( 1 3) Prove that if { Tn } is a sequence in B (X, Y) such that liru for each x E X , then there is a E B (X. Y) such that for all E X .

T

x

( 1 4) Let X, Y be normed vector spaces. Show th a t X to Y is bounded if and only if sup

x E X , II x ll = l y 1 E y I , II Y 1 l l = 1

an

I Y ' (A .r ) I < oo.

( 1 5 ) If W is a subspace of X ' , show that Yf'

c

(0 T1l)0 •

x

Y) ' and

T11 x exists Tn x � Tx

operator

A fro

n1

74

3. LINEAR OPERATORS

( 1 6) If X, Y are normed vector spaces and A is a linear operator from X to Y which is bounded on any nonempty open set , show that A is bounded. ( 1 7) Let X, Y be Banach spaces, and let A be a linear operator from X

to Y such that N ( A ) and R(A) are closed and (3.24) holds . Show that A is a closed operator.

( 18) For X , Y normed vector spaces and A

show that

II A x l l f xEX d ( x , N ( A ) ) and that R ( A' ) is closed. .

Ill

E

B(X, Y) satisfying (3 .24) ,

I I A' y' ll y' EY' d ( y ' , N ( A ' )) . 1nf

( 1 9) Assume that X, Y are Banach spaces and that A is a closed linear

operator from X to Y which maps bounded closed sets onto closed sets. Show that R ( A ) is closed.

( 20) If X, Y are normed vector spaces and { Ak } is a sequence of oper­ ators in B ( X, Y) such that Ak x � A x for each X E X , show that A E B (X, Y) and II A II < lim inf i i Ak l l · (21) Let X, Y be Banach spaces, and assume that A is a closed operator from X to Y having an inverse A- 1 E B (Y, X ) . Let B be an operator from X to Y such that D( A) C D(B) and I I BA- 1 11 < 1 . Show that A + B has an inverse ( A + B) - 1 in B ( Y, X) . (22) Under the above hypothesef?, show that

(23) If X is a Hilbert space, if and only if A' x =

x.

A E B (X)

and

==

II A I I < 1 , show that A x x

(24) Let X, Y be Banach spaces, and let F ( x , y ) be a linear functional on X x Y which is continuous in x for each fixed y and continuous in y for each fixed x. Show that F E (X x Y)'.

3. 8. Problems

75

A is a linear operator on a Hilbert space X such that D(A) == X and ( x, Ay) == (Ax, y), x, y E X. Show that A E B (X ) . (26) Let {D k } be a sequence of dense open subsets of a Banach space X. Show that (25) Suppose

00

is dense in X .

X, Y be Banach spaces, and let A be an operator in B( X, Y) in Y . If R(A) is of the second category in Y, show R(A)Q ) dense A( is open in R(A) for each open set Q X. (28) Let be a subset of a Banach space X such that sup l x ' (x) l < oo , x ' E X' . Show that sup l x l < oo .

(27) Let with that

C

S

xES

xES

Chapter 4

THE RIE S Z T HE O RY FO R C O MPAC T O P ERAT O RS

4. 1 . A type of integral equation In Chapter I , we found a method for solving an integral equation of the form

x(t) = y (t) +

(4 . 1 ) We

1t k (t, s)x( s ) ds .

now want to examine another integral equation of a slightly different forrn, na1nely

x( t ) = y ( t ) + v(t)

( 4 .2)

.lb w ( s) x( s ) ds ,

which also co1nes up frequently in applications . Of course, now we are more sophisticated and can recognize t hat ( 4.2) is of the form

x

( 4 .3 ) where

f{

is an operator on

a

-

Kx

==

y,

nor1ned vector space X defined by

(-1. 4) where x 1 is a given elen1ent of X� and x � is a given element of X' . Clearly, J( is a linear operator defined e ver y vv h er e on X. It is also bounded since

77

78

4. THE RIESZ THEORY FOR COMPACT OPERATORS

Now clearly in order to solve (4.3) , it suffices to find

x == y + Kx,

Kx ,

i.e. ,

x� (x) .

Since

or

(4.5 ) Now if x� ( x1 ) # 1 , we can solve for

x� ( x) and substitute into ( 4.3) to obtain x = y + 1 X� (y) x1 ' ) - x1 (Xt .

(4.6)

By substituting ( 4.6) into ( 4.3) , we see that it is indeed a solution. Hence a solution of ( 4.2) is

x ( t ) == y ( t ) + provided

I: w ( s ) y( s) ds v ( t ) 1 - I: w(s)v(s) ds

1 b w ( s ) v ( s ) ds f. 1 .

(I ' ll bet you never would have dreamt of solving ( 4.2) in this way without first changing it to ( 4.3) . ) Concerning uniqueness , we see from (4.5) that if y == 0 , then x]_ ( x) rr1ust vanish, and hence, so must x. Now, suppose x]_ (x1 ) == 1 . By (4. 5) , we see that we must have

xi (y ) == 0

( 4. 7)

in order that ( 4.3) have a solution. So let us assume that ( 4. 7) holds. On the other hand, any solution of ( 4.3) must be of the form X

( 4.8) where

o: is a scalar.

== y + O:X l ,

x we have K x == x ]_ (y )x t + o: x ]_ ( X t ) x 1 For such

==

o:x 1 ,

showing that any element of the form ( 4.8) is a solution. How about unique­ ness? If x == K x == x ]_ ( x )x 1 , then x is of the form x == o: x 1 . Consequently, K x ==

o:x]_ ( X t ) x1

==

o:x1 ,

showing that there is no uniqueness. Set A == I - K , where I is the identity operator, i.e. , x E X. Then ( 4.3) is the equation

Ax == y. In ter1ns of A we have shown that if x ]_ ( xi ) # 1, of ( 4.9) for each y E �X". If J:·� (x 1 ) == 1 , one can

I x == x

for all

( 4.9)

there is a unique solution solve ( 4.9) only for those

4. 1 .

79

A type of integral equation

y

satisfying ( 4. 7) , i.e. , those that annihilate x� . In this case there is no uniqueness, and N(A) consists of all vectors of the form o: x 1 . Let us take a look at A'. Clearly, I' is the identity operator on We shall denote it by I as well. By definition,

X'.

x ' (K x ) -so that

==

x � (x)x ' ( x l )

K' x ' ( x ) , ==

K' x ' ( x ) .

Hence,

( 4 . 10)

K' x' x' ( x 1 ) x � . K'x' == j3x � . Thus , K' x ' = j3x � ( x 1 ) x � , =

If x' E N(A') , then x'

==

showing that

(3 ( 1 - X � (X 1 ) ) X� == 0. If x � (x 1 ) =f. 1 , we must have (3 = 0, i.e. , N(A') = {0} . If x� ( x 1 ) N (A') consists of all functionals of the form f3x i .

=

1 , then

We can put our results in the following form Theorem 4. 1 . Let X be a normed vector space, and let A == I - K, where K is of the form (4 -4). If N (A) == {0} , then R(A) Otherwise R ( A) is

closed in

N(A' ) .

X, and N

X. == (A) is finite dimensional, having the same dimension as

Yes, I shall explain everything. When x � ( x 1 ) == 1 , then R(A) consists of the annihilators of x� , and hence, is closed ( Lemmjli�-4 ) . Moreover, N(A) is the subspace spanned by x 1 , while N(A') is the subspace spanned by x� . As we shall see in a moment , both of these spaces are of dimension one. To introduce the concept of dimension, let V be a vector space. The ele­ ments v 1 , · · · , Vn are called linearly independent if the o nly scalars o: 1 , · · · , O:n for which

( 4 . 11) are n 1 = · · · = O:n = 0 . Otherwise, they are called linearly dependent. V is said to be of dimenszon n > 0 if

( a) There are n linearly independent vectors in V and

4.

80

THE RIESZ THEORY FOR COMPACT OPERATORS

( b) Every set of n + 1 elements of V are linearly dependent . If there are no independent vectors, V consists of j ust the zero element and is said to be of dimension zero. If V is not of dirnension n for any finite n , we say that it is infinite dimensional. Now suppose diln V == n ( i.e. , V is of dirncnsion n ) , and let V I , · · · , Vn be n linearly independent elements. Then every v E V can be expressed uniquely in the forn1

(4. 12) To see this , note that the set v , VI , · · · , Vn of n + 1 vectors must be linearly dependent . Thus, there are scalars (3, f3I , · · · , f3n , not all zero, such that (3 v + f3 IV I +

· · · + f3n vn == 0 . Now (3 cannot vanish, for otherwise the v 1 , · , Vn would be dependent. Dividing by (3, we get · an expression of the form ( 4. 12) . This expression is ·

·

unique. For if

then

( n1 - n� ) v1 + · · · + ( nn - n � ) vn == 0,

showing that n � == ai for each i. If dim V = n , we call any set of n linearly independent vectors in V a basis for V. We are now going to attempt to extend Theorem 4. 1 to a more gen­ eral class of operators . In doing so we shall make use of a few elementary properties of finite dimensional normed vector spaces. Let be a normed vector space, and suppose that it has two norms II · II I , II · l l 2 · We call them equivalent and write I I · lh I I · ll2 if there is a positive number a such that

X

"-J

( 4. 13)

Clearly, this is an equivalence relation ( See Section 3.5) , and a sequence {x n } converges in one norm if and only if it converges in the other. We shall prove Theorem 4.2.

If X is finzte

dzmensional, all norms are equivalent.

Before proving Theorem 4.2 we state some important consequences. Corollary 4.3. A finzte dimensional normed vector space is always com­ plete.

81

4. 1 . A type of integral equation

x1 , · · ,

(

Xn be a basis for X i.e a set Proof. Suppose dim X == n , and let of n linearly independent elements) . Then each E X can be written in the form ·

Set

x

.,

l x l 1 == (L1 l ni l 2 ) 1 12 . n

This is a norm on X, and by Theorem 4 . 2 , it is equivalent to the given norm of X. Thus if (4. 14)

sequence in IR n . (a �k) , · · · , a0:�k) )· ·is a Cauchy ( 1 , · , an ) whi h is the limit of

is a Cauchy sequence in X, then Since IRn is complete there is an element ' n in IRn . Set

,

(n(1k) · · · a( k) )

c

'

( 4. 15) Then

x(k ) x in X. Hence X is complete.

0

---+

Corollary 4.4. If M is a finite dimensional subspace of a normed vector space, then M is closed.

{ x( k) }

x ( k) x

Proof. If is a sequence of elements of M such that in X, it is a Cauchy sequence in M. Since M is complete with respect to any norm, 0 has a limit in M which must coincide with

x(k)

x.

---+

A subset W of a normed vector space is called bounded if there is a number b such that E W. It is called compact if each < b for all sequence of elements of W has a subsequence which converges to an element of W.

l xl

{ x ( k) }

x

Corollary 4.5 . If X is a finite dimensional normed vector space, then every bounded closed set T in X is compact.

x1 , · ,

Proof. Let Xn be a basis for X. Then every element expressed in the form (4. 1 5) . Set ·

·

x E X can be

n

(4. 16)

l x l o == I:1 l ai l ·

Let

This is a norm on X, and consequently it is equivalent to all oth e rs be a sequence of elements of T. They can be written in the form

{ x(k) }

.

82

4. THE RIESZ THEORY FOR COMPACT OPERATORS

(4. 14) . Since T is bounded, there is a constant Hence, there is a constant c such that

b

such that

l x( k) l < b.

l x ( k) l o = L l a� k) I < c. By the Balzano-Weierstrass theorem, { a �k ) } has subsequence which con­ verges to a number n 1 . Let us discard the x ( k ) not in this subsequence. k For the remaining indices k, { a � ) } has a convergent subsequence. Again, discard the rest of the sequence. Continuing in this manner, we have a subk sequence such that n � ) � n i for each i . Set x == n 1 x 1 + · · · O:n Xn . Since ( 4. 16) is equivalent to the norm of X, this subsequence converges to x in X. D Since T is closed, x E T. n

i= l

a

+

We now give the proof of Theorem 4.2.

x 1 , · · · , Xn

x

Proof. Let be a basis for X. Then every E X can be written in the form ( 4. 15) . We shall show that any norm I I I I on X is equivalent to the norm given by (4. 16) . In one direction we have ·

l x l == l n 1 x 1 + · · · + O:n Xn l < l o: 1 x 1 l + · · · + l nnxn l < m�x l xi l · l x l o ­ Conversely, there is a constant C such that l xl o < Cl xl , J

I.e. ' .

( 4. 1 7) In proving ( 4. 1 7) , it clearly suffices to assume ( 4. 18)

l ni

O:i

by this value, apply For if 2:: I is not equal to one, we can divide each satisfying (4. 1 7) and then multiply out . Now, if (4. 17) were not true for (4. 18) , there would be a sequence of the form (4. 14) satisfying

o:i

l x (k) l o = 1 and such that l x ( k ) I I � 0. As we just showed, there is a subsequence of { x( k) } for which n� k) � O:i for each i. Let x = o:1 x 1 + + O:nXn . Then n l x l o = L l o:i l == 1 , and l x ( k) - x l = II Li (a� k) - ai )xil l < Li l a� k) - ail · l xil l ·

1

·

·

83

4. 1 . A type of integral equation

Thus

x == 0, contradicting the fact that ll x ll o == 1 . This completes the proof.

0

Corollary 4.5 has a converse. Theorem 4.6. If X is a normed vector space and the surface of its unit sphere {i. e. , the set ll x ll == 1} zs compact, then X is finite dimenszonal.

In proving Theorem 4.6 we shall make lise of a simple lemma due to F. Riesz . Lemma 4. 7. Let M be a closed subspace of a normed vector space X. If M is not the whole of X, then for each number () satisfying 0 < () < 1 there is an element x o E X such that

ll xo II

(4. 19)

== 1 an d

d(xo, M )

> e.

Note the difference between Le m ma 4.7 and Theorem 2 . 2. The reasun we cannot take () == 1 in Lemma 4.7 is that in general, for x not in M, there may not be a z M satisfying

E

(4.20)

ll x - z ll

==

d(x, M ) .

This may be due to the fact that X is not complete. But even if X is a Banach space, ( 4. 20) may fail because of the geometry of X. We leave it as an exercise to show that if M is finite dimensional, we may take () == 1 in Lemma 4.7. We can now give the proof of Theorem 4.6. Proof. Let x1 be any element of X satisfying l lx1 ll == 1, and let M1 be the subspace spanned by x 1 . If M1 == X, then X is finite dimensional and we are through. Otherwise, by Lemma 4. 7, there is an x 2 E X such that ll x 2 ll == 1 and d ( x2, M1 ) > � · (Note that si � ce M1 is finite dimensional, it is closed. By the remark at the end of the preceding paragraph, we could have taken x 2 such that d ( x2 , M1 ) == 1 , but we do not need this fact here.) Let M2 be the subspace of X spanned by x1 and x2 . Then M2 is a closed subspace, and if M2 # X, there is an X 3 E X such that ll x 3 ll == 1 , d(x3 , M2 ) > � · We continue in this manner. Inductively, if the subspace Mn spanned by x 1 , · : · , X n is not the whole' of X, there is an X n+ l such that

ll xn + I II == 1 ,

and

d (xn +l , Mn ) ,

1 2

> -.

84

4. THE RIESZ THEORY FOR COMPA CT OPERATORS

We cannot continue indefinitely, for then there would be an infinite sequence { X n} of elements such that II Xn I I == 1 while

ll xn - X m ll

>

1

2,

n # m.

This sey_uence clearly has no convergent subsequence showing that the sur­ face of the unit sphere in X is not compact . Thus there is a k such that D Mk == X, and the proof is complete. It remains to give the proof of Lemma 4 . 7 . Proof. Since M # X, there is an X I E X \ M ( i.e. , in X but not in M) . . Since M is closed, d == d( X I , M) > 0. For any E > . 0 there is an x o E M such

that

I I x i - xo ll < d + E. Take == d( l - B) / B. Then d + == d/ B. Set I X X o o == x ll x i - xo l l ' Then ll xo ll == 1 , and for any x E M we have d x l - x o ll x + x o ) - x l l l > ( ll ll x xo ll - ll ll x i - xo ll l l x i - xo l l since ll x i - x ol l x + x o is in M. This completes the proof.

E

E

=

> (), 0

Another simple but useful statement about finite-dimensional spaces is Lemma 4.8. If V is an n-dimensional vector space, then every subspace of V is of dimenszon < n. Proof. Let W be a subspace of V. If W consists only of the element 0,

then dim W == 0. Otherwise, there is an element W I # 0 in W. If there does not exist a w E W such that WI and w are linearly independent, then dim W == 1 . Otherwise, let w2 be a vector in W such that W I and w2 are linearly independent . Continue in this way until we have obtained , Wm are linearly independent elements W I , · · · , Wm in W such that w , WI , dependent for all w E W. This must happen for some m < n , for otherwise the dimension of V would be more than n. This means that dim W == m < n, and the proof is complete. 0 ·

·

·

4. 2. Operators of finite rank

85

4 . 2 . Operators of finite rank

(

(

Encouraged by our success in solving 4.3) when K is given by 4.4) , we attempt to extend these results to more general operators. As we always do when we are unsure of ourselves , we consider an operator only slightly more difficult . The next logical step would be to takA of the form

K n Kx jL=l xj(x)xj , x1 E X, xj E X' . ( 4.2 1 ) Note that K is bounded , since n I Kx I I < ( L I I xj I I I I xj I I ) I x I · Moreover, R (K) is clearly seen to be finite dimensional. Conversely, any operat9r K E B ( X ) B (X, X ) such that R(K) is finite dimensional must be of the form (4 . 2 1 ) . For, let x , · · · , X n be a basis for R(K). Then for each X E X, n Kx L Oj (x)xj, where the coefficients o:j (x) are clearly seen to be linear functionals . They are also bounded, since by Theorem 4 . 2 , there is a constitnt C such that n < n L l o:j l C l L O:J Xj " · Hence, n L l o:j (x) l < C I Kx l < C I K I · l x l . Thus there are functionals xj E X' such that O:j ( x) xj ( x), and K is of the form ( 4 . 2 1 ) . An operator of the form ( 4 . 2 1 ) is said to be of finite rank. As a counter­ /

==

·

1

==

1

==

1

1

1

1

==

part of Theorem 4. 1 , we have ,_

'

Theorem 4.9 . Let X be a normed vector space, and let K be an operator of finite rank on X. Set A == I - K. Then R (A) is closed in X, and the dimensions of N(A) and N(A ' ) are finite and equal. Proof. Clearly, we may assume that the pendent in ( 4. 21 ) . If is a solution of

x

( 4.22)

Ax n x - L1 x� (x)xk

Xj and the xj are linearly inde­

==

y,

==

then

y.

86

4. THE RIESZ THEORY FOR COMPACT OPERATORS

x, it suffices to find x� ( x), , x� ( x ). Operating_ on

In order to determine we have 4. 22) with

(

xj,

·

n

or

xj(x) - k=L x�(x)xj(x k ) = xj (y), 1 ), ) == xj( ) y xj(x x�(x ] [ 6 L k jk k= 1 n

( 4 . 23 ) k

where §j is the Kronecker delta

l)j k

==

·

·

1 < j < -n, 1 �j<

n,

{ 0,

j :-1 k, 1 , j == k.

Now, (4.23) is a system of n linear equations in n unknowns. From the theory of such equations we know that if the determinant

�=

! 6j k - xj(xk ) l

x� (x),

does not vanish, then one can always solve (4 . 23) for the and the solution is unique. (If you are unfamiliar with this theory, do not despair. We shall prove more general results shortly. ) On the other hand, if � 0, there is no uniqueness , and ( 4 . 23 can be solved only for those which satisfy

)

y

==

n

(4 . 24 ) whenever n

(4.25)

L= l [6j k - xj(xk )]

j

Oj =

0,

1 < -k<

-

n.

Moreover, the number of linearly independent solutions of (4.25) is the same as the number of linearly independent solutions of

)] xj(x ,B 6 [ = 0, k k jk L k= 1 n

(4. 26)

1 <j<

The only fact nee'ded to complete the proof is that

(4. 27)

n.

x1 E N (A' ) if and only if

I I X == � � Oj Xj , n

1

( ) K1x1(x) = x' (Kx) == )1 : x�(x)x' ( xA.)

where the Oj satisfy 4 . 25 . For, by definition, li

87

4.2. Operators of finite rank

x' E X'. Hence, n (4. 28) K'x' = I: x' (xk )x�. This shows that any solution of A ' x' = x' - K ' x' = 0 must be of the form (4. 27) . Moreover, if x' is of this form, then n n K'x' = I: o:j I: xj(xk)x k , showing that ( 4. 27) is in N (A') if, and only if, the satisfy ( 4.25) . Hence, (4. 23) can be solved for those y which annihilate N ( A ') , and dim N ( A ) = D dim N ( A ' ) < This completes the proof.

for any

x

E

X and

1

j=1

k= 1

O:j

n.

Now that we can handle operators of finite rank, where can we go from here? It is natural to think about operators which are "close" to operators of finite rank such that (4 . 29) Then for

I I Kn n

( 4.30) If A = I -

sufficiently large,

K, we have

K l � 0 as

n � oo .

I l K - Kn l < 1 .

( K - Kn ) - Kn . Set Bn = I - (K - Kn ) · If X is a Banach space, then, by Theorem 1 . 1 , Bn has an inverse B;; 1 defined on the whole of X. One can verify directly or use Theorem 3.8 to show that A=I-

Bn 1 is a bounded operator. Hence ,

A = Bn (I - Bn 1 K�) . We leave as an exercise the trivial fact that the product of a bounded oper­ ator and an operator of finite rank ( in either order ) is an operator of finite = y is equivalent to rank. Since the equation

Ax

( 4.31 ) Theorem 4.9 applies. We obtain Theorem 4. 10. Let X be a Banach space, and assume that K E B(X) �s the limit in norm of a sequence of operators of finite rank. If A I K� then R (A) is closed in X, and dim N (A) = dim N ( A' ) < oo . ==

-

88

4. THE RIESZ THEORY FOR COMPA CT OPERATORS

Proof. From (4 . 3 1 ) we see that N ( A) == N ( I - Bn 1 K ) · Since A' == ( I ­ Bn 1 Kn )' B� , it follows that dim N ( A' ) == dim N [ ( I - Bn 1 Kn )'] . Hence, y E X, then ( I - Bn 1 Kn )xk dim N ( A) == dim N (A') < oo . If Axk Bn 1 y , showing that there is an x E X such that (I - Bn 1 Kn ):J.; == Bn 1 y , or equivalently, Ax == y . Hence, R(A) is closed and the proof is complete. 0 n

-4

-4

What kind of operators are the limits in norm of operators of finite rank? We shall describe such operators in the next section. 4 . 3 . Compact operators

Let X, Y be normed vector spaces . A linear operator K from X to Y is called compact ( or completely continuous) if D(K) == X and for every sequence {x n } C X such that l l x n l l < C, the sequence { Kx n } has a subsequence which converges in Y. The set of all compact operators from X to Y is denoted by K ( X, Y ) . A compact operator is bounded . Otherwise , there would be a sequence oo . Then { Kx n } could not have {x n } such that l l xn l l < C, while I I Kx n l l a convergent subsequence. The sum of two corr1pact operators is compact , and the same is true of the product of a scalar and a compact operator. Hence, K ( X, Y ) is a subspace of B ( X, Y ) . An operator of finite rank is clearly compact . For it takes a bounded sequence into a bounded sequence in a finite-dimensional Banach space. Hence, the image of the sequence has a convergent subsequence. If A E B ( X Y) and K E K ( Y, Z) , then K A E K ( X Z) . Similarly, if L E K ( X, Y�) and B E B ( Y, Z) , then BL E K ( X Z) . Suppose K E B ( X, Y) , and there is a sequence { Fn } of operators of finite rank such that -4

,

,

,

0 as n -----+ oo . I l K - Fn l l We claim that if Y is a Banach space, then K is compact . ( 4.3 2)

-----+

Proof. Suppose {xk } is a bound ed sequence in X, say l l xk l l < C. Since F1 is a compact operator, there is a subsequence { Xk 1 } for which { F1 xk1 } converges. Since F2 is compact , t here is a subsequence { Xk2 } of { Xk 1 } for which { F2 xk2 } converges. Continuing, we see that there is a subsequence {xkn } of {xkn _ 1 } for which { FnX kn } converges. Set Z k == Xkk · Thus , { Fnz k } converges as k oo for each Fn . Now -4

I I K zj - K zk l l < I I K zj - Fn zj l l + I I Fn Zj - Fn z k ll + II FnZ k - K z k l l < 2C I I K - Fn ll + II Fn Zj - Fnz k l l · Now let E > 0 be given. We take n so large that 2C I I K - Fn l l <

c /2 .

4. 3.

89

Compact operators

Since { Fn zk } converges , we can take j, k so large that I I Fn z� - Fnzk l l < c /2 . Hence, { K Zk } is a Cauchy sequence. Since Y is a Banach space, the proof is complete. D Note that in this proof we only made use of the fact that the Fn are compact . Hence, we have proved Theorem 4. 1 1 . Let X be a normed vector space and Y a Banach space. If L is �n B(X, Y) and there is a sequence { Kn } C K(X, Y) such that

I l L - Kn l l

�

0 as n

�

0,

then L is in K(X, Y) . To illustrate Theorem 4. 1 1 , consider the operator K on lp given by

K(x 1 , x 2 ,

·

·

·

, Xk ,

·

·

·

)

==

(x 1 , x 2 j2,

·

·

·

, xk / k ,

·

·

·

).

We shall see that K is a con1pact operator. To that end , set

Fn (X l , X 2 ,

·

·

·

)

==

(x 1 , X 2 / 2,

·

·

·

,

Xn / n , 0 ,

·

·

·

)

,

n

==

1 , 2,

·

·

·

.

For each n the operator Fn is of finite rank . The verification is sin1ple ; we shall leave it as an exercise. Moreover, for 1 < p < oo ,

I I ( Ii - Fn - l ) x l l � = sho,ving that For the case

p ==

oo ,

fn � �kJP

I I K - Fn - 1 I I p we have

<

<

ll

�� � -

1 / 1� .

�

llx ll= l l . I I ( K - Fn - d .r l l = = sup k < n k> n -. Thus, (4 . 32) holds for 1 < p < oo . By Theoren1 4. 1 1 , we can now conclude that K is a compact operator. Con1pact integral operators are considered in Section 1 1 .4. vVe see fro1n Theore1n 4 . 1 1 that K (X. Y) is a closed subspace of B ( X, Y) . Now that we know t hat . i n a Banach space, the li1nit in norm of operators of finite rank is con1pact . we wonder whether the converse is true. If K is compact , can \VC find a sequence of operat ors of finite rank that converges to K in norn1? .A.. s \\Te shall see later. t he answer is yes . if \Ve are in a Hilbert space. ( See Section 1 2 . 3 . ) It is also affirrnative for nutny well-known Banach spaces. For 1n any years it \vas helieverl to be true for a general Banach space.

4. THE RIESZ 1 ,HEORY FOR COMPA CT OPERATORS

90

However, examples of Banach spaces were found for which the assertion is false. Thus, if X is a Hilbert space and K E K(X) == K (X, X) , then Theorem 4. 1 0 applies. If X is a Banach space, the hypotheses of that theorem may not be fulfilled for some K E K( X ) . However, we are going to show that , nevertheless , the conclusion is true. We shall prove Theorem 4. 1 2 . Let X be a Banach space and let K be an operator in K(X ) . Set A == I - K. Then, R(A) is closed in X and dim N(A) == dim N (A') is finite. In part�cular, either R(A) == X and N (A) == {0} , or R(A) -1 X and N (A) # {0} .

The last statement of Theorem 4. 1 2 is known as the Fredholm alternat�ve. To show that R(A) is closed we make use of the trivial Lemma 4. 13. Let X, Y be normed vector spaces, and let A be a linear operator from X to Y. Then for each x in D (A) and E > 0 there is an element xo E D (A) such that

Axo == Ax ,

d(xo , N (A) )

==

d(x, N (A) ) ,

d(x, N(A) ) < l l xo l l < d (x , N(A) ) + E . Proof. There is an x 1 E N(A) such that l l x - x 1 l l < d (x , N(A) ) + E . Set 0 Xo == X - X 1 .

Returning to A == I - K, we show that R(A) is closed by proving that for some

C,

(4.33)

d(x , N(A) ) < C I I Ax l l ,

x E X,

and then applying Theorerr1 3 . 14. If ( 4.33) did not hold , there would be a sequence {x n } E X such that

d(xn , N(A) ) == 1 ,

I I Ax n ll

---t

0 as

n ---t oo .

By Lemma 4. 13 , there is a sequence { zn } such that

d ( Zn , N ( A) ) == 1 ,

1 < I I Zn I I < 2 ,

I I A zn II

-----+

0 as

n ---t oo .

Since K is compact , there is a subsequence {u n } of { zn } such that Ku n � w E X. Now Un == Kun + Aun � w, and, hence, Aw == 0. But this is impossible, since l l un - w l l > d(un , N(A) ) == 1 for each n . The proof that dim N (A ) is finite is simple. We employ Theorem 4.6. N (A) is a normed vector space. If we can show that the surface of the unit sphere in N (A) is compact , it will follow from Theorem 4.6 that N(A) is finite-dimensional. So suppose {x n } is a sequence in N (A) such that l l xn ll == 1 . Then {Kxn } has a convergent subsequence. Since X n == K xn ,

91

4. 3. Compact operators

the same is true for {x n } · Hence, the surface of the unit sphere in N ( A) is compact . How about N ( A') ? Well, A' == I - K' . If it turns out that K' is also compact , then we have dim N ( A') < oo as well. The compactness of K' will be shown in the next section. We assume it here. To show that d i m N ( A ) == dim N (A') , we first consider the case dim N ( A') == 0, i.e. , N (A' ) == {0} . _Since R(A) is closed , we see, by Theorem 3 . 7, that R(A) == 0N ( A') == X. Now suppose there is an x 1 -# 0 in X such that Ax 1 == 0. Since R(A) == X, there is an x 2 E X such that Ax 2 == x 1 and an X3 such that Ax3 == x 2 . Continuing, we can always find an X n such that Ax n == Xn - 1 · Now A n is a bounded operator with I I A n l l < I I A I I n (se e Se ction 1 .2) . This implies that N(A n ) is a closed subspace of X (cf. Section 3.5}. In addition , we have n N (A) c N ( A 2 ) c · · · c N (A ) c · · · , /'

and what is more, these spaces are actually increasing, because A n x n == A n - 1 (Ax n ) == � n - 1 X n - 1 == · · · == Ax 1 == 0, and A n - 1 X n == A n - 2 (Ax n ) == A n - 2 X n - 1 == · · · == A x2 == X 1 "# 0. Hence, we can apply Lemma 4.7 to find al Zn E N(A n ) slich that n I I Zn I I == 1 , d ( Zn , N (A - 1 ) ) > 1 / 2.

Since K is cornpact , {K zn } should have a convergent subsequence. But for

n

>

m,

I I K zn - K zm l l == l l zn - A Zn - Z m + A zm l l == ll zn - ( Zm - A zm + A zn ) l l > 1 / 2, since A n - 1 ( z m - A zm + A zn ) == 0 . This shows that { Kzn } cannot have a convergent subsequence. There must be something wrong. What caused the contradiction? Obviously, it was the assumption that x 1 -:1 0. Hence, dim N(A) == 0. We have just shown that dim N (A' ) == 0 implies dim N (A) == 0 . Con­ versely, assume that dim N( A) == 0. Then R(A') N(A) 0 == X' by Theorem 3. 16. Since A' == I - K' and we know that K' is compact , the argurr1ent j ust given implies that dim N (A') == 0. Hence, we have shown that dim N (A) == 0 if, and only if, dim 1\T (A') == 0. Next , suppose ==

dim N( A) ==

n

> 0,

dim N (A ' )

== m

> 0.

Let x 1 · · · , Xn � x� , · · , T� span N (A) and N (A') , respectively. I claim there are a functional x � and an element xo such that �

( 4. 34)

·

x � ( Xj )

==

0,

1 <j <

n,

x � (x n ) "# 0.

92

4. THE RIESZ THEORY FOR COMPACT OPERATORS

xj(xo) == 0,

( 4.35)

1 <j<

m,

x� (xo) -1 0 .

Assume this for the moment and put

Kox x� (x)xo, x E X. Then Kol X == X' ( XQ ) x0 , x' E X'. Set A 1 == A - Ko. We shall show that dimN (A 1 ) == 1 , dim N(A� ) == 1 . ( 4.37) But A 1 = 1-(K + Ko), and K + Ko is a compact operator. Thus , we have an operator A 1 of the same form as A with the dimensions of its null space and that of its adjoint exactly one less than those A, A', respectively. If and are both greater than one, we can repeat the process and reduce dim N(A 1 ) and dim N (A� ) each by one. Continuing in this way we eventually reach an oper ator A I k, w·h ere k is compact and either dim N (A) 0 or dim N(A' ) == 0. It then follows from what we have proved that they mpst both be zero. Hence, and the proof is complete. To prove (4. 37) , suppose x E N(A1). Then Ax = Kox x�( x)xo. Now N( A') , since it does not annihilate x�. Hence� we must xohaveis not in R(A)and== 0consequently A x = 0. Since x E N(A), x� (x) 0, (4.38) Therefore, by ( 4.34) , n x� (x) == I: ojx� (xj) == O:n x� (xn ) == 0, 1 showing that O:n == 0. Hence, x is of the form n- 1 X == 2:1 O:jXj . ( 4.39 ) Conversely, every element of the form (4. 39) In N(AI) · This follows from the fact th'at it is in N(A) and satisfies x � (x) == 0. This shows that 1. dim N(A 1 ) Suppose x' E N(A�), i.e. , A'x' bx' x'(xo)x� . But x� is not i:q.d K R(A') == N (A) 0 , since it does not annihilate Xn · Hence, x'(xo ) == 0 , an consequently, A' x ' 0. Thus, x' == 2: /JJ xj 1 (4.36)

==

I

I

m -

n -

of

m

n

=

=

-

m == n,

==

=

IS

=

n -

==

==

rn

==

4. 3. Compact operators

93

and

m

x' ( xo ) = L f1jxj ( xo ) = f3m x � ( xo ) == 0, showing that f3m

=

1 0. Hence, x ' is of the form m- 1

x' =

( 4.40 )

L1 (3j xj .

Conversely, every functional of the form (4.40) is in N (A� ) , since it is -in N(A') and N(Kb) . This proves (4.37) . It remains to show that we can find an x� satisfying ( 4.34) and an x o satisfying (4.35) . The functional x� can be found easily. To see this, let M be the subspace spanned by x 1 , · · · , X n - 1 [i.e. , the set of all x of the form (4.39)] . The element Xn is not in this finite dimensional (and, hence, closed) subspace. This implies that d( x n , M) > 0. Then, by Theorem 2 . 9 , there is an x � E X' such that ll x � l l = 1 , x� ( xn ) = d ( x n , M ) , and x� ( x ) = 0 for x E M. This more than suffices for our needs . We find x o by induction. In fact , we are going to prove Lemma 4. 14. If x � , · · · , x � are linearly independent vectors in x: then there are vectors x 1 , · · · , X m in X such that

j = k, j # k,

( 4.4 1 )

1 < j, k <

m.

Moreover, if x 1 , · · · , X m are linearly independent vectors in X , then there are vectors x � , · · · , x� zn X' such that (4 . 4 1} holds. Proof. Assume that the lemma is true for m == l linearly independent vectors in X' . Then, for any

xj ( x -

( 4.42)

l-1

L x� ( x ) xk )

==

1 > 1 . Let x � x E X,

-

xj ( x ) - xj ( x ) = 0,

,

· · ·

, x� be

1 < j < l.

1

Now if x� vani shed on o [ x � · · · , x z 1 ] (the set of annihilators of x� , then it would have to vanish on l- 1 ,

_

x - L x� (x)x k

for any

x

1 E X by (4.42) . This would mean that l-1

x� (x ) =

L1 x� ( x ) x� (xk) ,

x E X,

· · ·

, x�

_

1

)

,

4. THE RIESZ THEORY FOR COMPJ.. CT OPERATORS

94

or

l-1

==

x�

L x � (xk)x� . 1

But this is impossible, since x � , , x� are linearly independent. Hence, x� does not vanish on the whole of 0 [x� , , x�_ 1 ] . There must be at least one element xz in this set which does not annihilate x� . Thus , we can choose x z so that ·

·

·

·

·

·

x � ( x z) == 1 . Now, for each k #- l we can rearrange the functionals x� , , x� so that x � replaces x� as the last one. Since the xj with j #- k form a set of l - 1 linearly xj ( x z) == 0,

1 < j < l,

·

·

·

independent functionals, we can again use our induction hypothesis and the argument given above to conclude that there is an X k E X such that

xj (X k)

==

j #- k , X� (X k)

0,

==

1.

This implies that ( 4.41 ) holds for m == l . Since it is trivial for m == 1 , the proof is complete for the first part. For the second part , we use the argument given above. We let !vfj denote the subspace of X spanned by x 1 , , Xm with Xj missing. The vector Xj is not in Mj , and consequently, d ( xj , Mj) > 0. Then by Theorern 2.9, there is a x� E MJ such that xj (xj) == 1 . This proves the second part of the len1ma. D ·

·

·

We still must prove the fact that K' E K(X') whenever K E K(X) . This proof will be given in the next section. In Section 4 . 2 , when we considered operators of finite rank, we employed the theory of n, linear equations in n unknowns. We also remarked that if you are unfan1iliar with this theory, we shall prove it later from more general considerations. Our 1nore general result is Theorem 4. 1 2 . Suppose we have a system of the form n

( 4.43)

"""" a't)· · xJ �

j ==l

·

-

-

y· •

'l '

1
n.

can consider this as an equation of the form A x == y in IR� where x == ( .T J , x 11 ) , y == ( YI , , Yn ) . The operator A is clearly linear and bounded (with II A I I < Inaxi,J l aij l ) . l\1 oreover, if we set K == I - A, then K is an operator of finite rank (since R( K) C IRn ) , and, hence, compact. To find A' , WP h e-lVE \ by definition , vVe

·

·

·

,

·

·

·

(A' u , x ) == ( A x ) , u , x E IR� 'U ,

95

4. 4. The adjoint of a cornpact operator (Here, we are using the fact that IRn is a Hilbert space. ) Thus, if A'u == we see that n

L1 Vj Xj

n

n

n

( 4 .44)

n

L Ui L aij Xj = L Xj L aijUi ·

=

j =1

i=1 j =1 Since this is true for all x E JR� we have /'

v,

i=1

n

L aij Ui == Vj ,

1 <j <

n.

i=1 Now the range of A is evidently closed, being finite-dimensional. Hence, y is a solution of (4.43) if, and only if, it annihilates (or is orthogonal to) N(A') . This is the same as· saying n

L1 UiYi

( 4.45)

==

0

for all vectors u satisfying n

( 4.46 )

a ij U i L i=1

== 0,

1<j<

n.

Moreover, by Theorem 4. 12, dim N ( A ) == dim .lV(A') . This means that the number of linearly independent solutions of n

( 4.4 7)

L a ij Xj = 0,

1
n,

j=1

is the same as the number of linearly independent solutions of (4.46) . In particular, if the only solution of ( 4.4 7) is x == 0, then ( 4.43) can be solved for all y. We should remark that we have used much more than was necessary (we used a mountain to crush a flea) . 4.4. The adjoint of a compact operator

In this section, we shall prove Theorem 4 . 1 5 . -,Let X, Y be normed vector spaces, and assume that K is in K(X, Y) . Then K' is in K(Y' , X') .

In order to prove Theorem 4. 1 5 , we shall develop a few concepts . Let X be a normed vector space. A set V c X is called r-e latzvely compact if every sequence of elements of V has a convergent subsequence. The lin1it of this subsequence, however, need not be in V. Let E > 0 be given . A set of points W C X is called an E-net for a set U C X if for every x E U there is a z E W such that ! l x - z ll < E . A subset U C X is called totally bounded if for every E > 0 there is a finite set of points, W C X which is an c-net for

96

4. THE RIESZ THEORY FOR COMPACT OPERATORS

U. We leave it as an exercise to show that if U is totally bounded, then for each E > 0 it has a finite E-net W c U. If A is a map from X to Y , then the image A(U) of a set U C X is the set of those y E Y for which there is an x E U satisfying A x == y . We have Lemma 4. 1 6 . Let X, Y be normed vector spaces. A linear operator K from X to Y is compact if, and only if, the image K(U) of a bounded set U c X is relatively compact in Y. Proof. Suppose K E K (X, Y) . Since U is bounded, there is a constant C such that ll x l l < C for all x E U. Let {K xn } be a sequence in K( U ) . Then . ll xn l l < C. Since K is compact , {Kxn} has a co;nvergent subsequence. This means that K ( TJ) is relatively compact . Conversely, assume that K (U) is relatively compact for each bounded set U. Let { xn} be a bounded sequence in X. Then {K xn} has a convergent subsequence. Hence, K E K (X, Y) . 0 Theorem 4. 1 7. If a set U

C

X is relat�vely compact, then it is totally bounded. If X is complete and U is totally bounded, then U is relatively compact.

Before proving Theorem 4. 17 we shall show how it can be used to prove Theorem 4. 15. Proof. Since X' is complete ( Theorem 2. 1 0) , it suffices by Theorem 4. 1 7 to show that for any bounded set W C Y', the image K' (W) is totally bounded in X' . Let c > 0 be given , and let S1 denote the closed unit sphere ll x l l < 1 in X. Since K is a compact operator, K ( S1 ) is totally boundeq by Lemma 4. 16 and Theorem 4. 17. Thus, there are elements x 1 , · · · , Xn of S1 such that each X E sl satisfies

( 4.48)

IR n defined by y' ( K X ) )

for some i. Let B be the mapping frorn Y' to B y'

==

( y' ( K X 1 )

,

· · ·

,

n

.

--

Since R(B) is finite di1nensional, B is an operator of finite rank. Let W be a bounded set in Y' , say I I Y' I I < Co for y ' E H'. Then B ( �t" ) is totally bounded. Thus, there are functionals y � · · · , y�1 such that every y ' E W satisfies I I By' - By; II < c �

for so1ne j. Hence ,

( 4.49)

4.4. The adjoint of a compact operator

97

(

y'

for each i. Now let be any element of W. Then there is a j for which 4.49) holds for each i . Let be any element of Then there is an i such that 4 . 48 holds. Hence,

si· x ( ) I Y' (Kx) - yj (Kx) l < l y' (Kx) - y' (Kxi ) l + I Y'(Kxi ) - yj (Kxi ) l + l yj (Kx i ) - yj (Kx) l < I Y' I · I K x - Kxi l + c + I Yjl l · I l Kx i - K xl l < 2 Co c + E . Since this is true for any X E SI ,we have I K'y' - K'yjl l < ( 2Co + 1 ) c . This means that K' (W) is totally bounded, and the proof is complete. /'

0

It remains to give the proof of Theorem 4 . 1 7. Proof. Assume that U is relatively compact , and let c > 0 be given. We

e

shall show that U has an c-net consisting of a finite numb r of points of U. ( We actually do not need to show that the points belong to U. ) Let XI be any point of U. If U is contained in the set < c , we are through. > If every E U satisfies Otherwise, let E U be such that

x - xi I l x2 x l x2 - X II I E. (4. 50) l x - XI I c or l x - x2 l < c , we are also through. Otherwise, let X3 be a point of U not s;:ttisfying ( 4.50) . Inductively, if · · , Xn are chosen and no point of U satisfies l x - x i i E, · · · , l x - xn l > E, then the first statement is proved . Otherwise, let X n + I be such a point . We eventually stop after a finite number of steps. Otherwise_, { xj } would be a sequence of elements of U sat isfying l t xj - x k l > for j -1 k . <

Xl ,

·

I

>

£

This sequence would have no convergent subsequence, contradicting the fact that U is relatively compact . Hence, U is totally bounded. Conversely, assume now that U is totally bounded and that X is com­ be any sequence of points of U. Since U is totally bounded, plete. Let it is covered by (i.e. , contained in the union of a finite number of spheres of radius 1. At least one of those spheres contains an infinite number of elements of ( ounting repetitions . Let s be any one of the spheres containing an infinite number of elements of Throw out all elements of not in SI , and denote the resulting subsequence by There is a finite number of spheres of radius 1 2 which cover U. At least one of these contains an infinite number of points of Choose one, and denote it and denote the resulting by Throw out all points of not in

{ xn }

I

{ Xn } c

{ xn } 82 .

) ) I {xn } · / {X } . i n { xn i } S2 ,

{ xn l} ·

98

4. THE RIESZ THEORY FOR COMPACT OPERATORS

sequence by {x n 2 } · Continue inductively, and set Zn = Xnn · Then for each > 0, there is an N so large that all Z k are in a sphere of radius less than E for k > N. In other words, { Zn } is a Cauchy sequence. We now use the completeness of X to conclude that it has a limit in X. This completes the proof of Theorem 4. 1 7. 0

E

4 . 5 . Problems

( 1 ) Show that the set Iw consisting of all elements f == ( 0: 1 , · · · ) in [00 such that l o: n l < 1 /n is compact in l00 • The set Iw is called the

Hilbert cube.

f2) Let A1 be a totally bounded subset of a normed vector space X. Show that for each E > 0, M has a finite c--net N C M (not j ust merely N c X ) . (3 ) Prove that if M is finite dimensional, then we can take () Lemma 4.7.

=

1 in

( 4)- Show that if X is infinite dimensional and K is a one-to-one oper­ ator in K(X ) , then K - I cannot be in K(X ) . (5) Let X be a vector space which can be made into a Banach space by introducing either of two different norms . Suppose that con­ vergence with respect to the first norm always implies conv&gence with respect to the second norm. Show that the norms are equiva­ lent . (6) Suppose A E B (X, Y ) , K E K (X, Y) , \\rhere X, Y are Banach spaces. If R(A) c R(K) , show that A E K (X, Y) .

( 7) Suppose X, Y are Banach spaces and K E K (X, }r) . If R(K) show that Y is finite dimensional.

=

Y.

(8) Show that if X is an infinite dhnensional nol llH'd \'( 'c t or space . t ht'u there is a sequence {xn } such that ll .1·n I I == 1 . i l .r n - .r , n I I > 1 fo r m # n.

4. 5. Problems

99

(9) A normed vector space X is called strictly convex if ll x + Yll < ll x ll II Y I I whenever x , y E X are linearly independent . Show that every finite-dimensional vector space can be made into a strictly convex normed vector space.

+

/\

( 10) Let V be a vector space having n linearly independent elements v 1 , · · · , Vn such that every element v E V can be expressed in the form (4. 12) . Show that dim V == n . ( 1 1) Let V, W be subspaces of a Hilbert space. If dim V < oo and dim V < dim W, show that there is a u E W such that l !ull == 1 and ( u, v ) == 0 for all v E V. ( 1 2) For X, Y Banach spaces, let A be an operator in B (X, Y) such that R(A) is closed and infinite-dimensional. Show that A is not compact . ( 13) Suppose X is a' Banach space consisting of finite linear combinations of a denumerable set of elements. Show that dim X < oo . ( 14) Show that every linear functional on vector space is bounded.

a

finite dimensional normed

( 15) If M is a subspace of a Banach space X , then we define codim-'J\;/ == dim X/A1. If M is closed , show that codim M == din1 A/0 and codirn AJ0 dim M. ( 16) If M, N are subspaces of a Banach space X and codimN < din1 A/ , show that M n N # {0} . (If dim l\1 == oo , it is assun1ed that codimN < oo . ) ( 1 7) If A E B(X, Y) ,. R(A) is dense in Y and D is dense in X � show that A maps D onto a dense subset of Y. ( 18) Show that every infinite dimensional nor1ned vector space unbounded linear functional.

has

an

100

4. THE RIESZ THEORY FOR COMPACT OPERATORS

A

( 19) If X, Y are Banach spaces and is a closed linear operator from X to Y such that R(A) is not closed in Y, show that for each c > 0 there is an infinite dimensional closed subspace M C such that the restriction of to M is compact with norm < c.

D(A)

A

(20) Let { x� } be a sequence of functionals in X' , where X is a finite dimensional normed vector space, and assume that x � ( x ) � 0 as k � oo for each x E X. Show that l x/c � 0.

l I

A E B (X) , let I A I K KEKinf(X) I A + Kl l . Show that I A I K < 1 implies that R(I - A) is closed in X and dim N (I - A) = dim N (I - A') < oo.

(2 1) If X is a Banach space and =

(22) Show that the product of a bounded operator and an operator of finite rank (in either order) is an operator of. finite rank.

Chapter

5

FRED H O LM O P ERAT O RS

5. 1

..

Orientation

If X is a Banach space and K E K(X ) , we saw in the last chapter that A I - K has closed range and that both N(A) and N(A') are finite dimensional. Operators having these properties form a very interesting class and arise very frequently in applications. In this chapter, we shall . investigate some of their properties. Let X, Y be Banach spaces. An operator A E B (X, Y) is said to be a Fredholm operator from X to Y if ) ==

( 1 ) o:(A) == dim N(A) is finite, (2) R(A) is closed in Y,

(

(3) f1(A) == dim N (A' ) is finite� The set of Fredholn1 operators from X to Y is denoted by

(5. 1 )

i (A)

==

n (A) - f) (A) .

For K E K(X ) we have shown that i ( I - K) == 0 (Theorem 4. 1 2) . ,

101

102

5.

FREDHOLM OPERATORS

X, Y, N(A)

For convenience, we shall assume throughout this chapter that Z ar e Banach spaces ( unless we state otherwise ) . If E == with = {0} and in ( Theorem 3 . 1 1 ) . then has an inverse Is there any corresponding statement that can be made for an arbitrary E If "I 0 , cannot have an inverse. There is, however, a subspace of such that t he restriction of to has an inverse ( i.e. , is one-to-one ) . This can be effected easily by means of

(X, Y) A A- 1 B(Y, X)

R(A) Y , A A
A Xo

Lemma 5 . 1 . Let N be a finite dimensional subspace of a normed vector space Then there is a closed subspace of X such that

X. (a) Xo N n

Xo

=

{0}

x X

Xo and an X I E N such that xo x xo x 1 . If V1 and V2 are subspaces of a vector space V such that V1 V2 {0} , then the set of all sums VI + v2 , Vi E Vi , is called the direct sum of V1 and v2 and is denoted by VI EB v2 . Thus , the conclusion of Lemma 5 . 1 can be stated as X Xo EB N. The proof of Lemma 5. 1 will be given at the end of this section. Now, by Lemma 5. 1 , there is a closed subspace Xo such that X == Xo EB N(A). (5.2) Clearly, A is one-to-one on Xo. Let A be the restriction of A to Xo . Then To see this, note that if E A E B(Xo, R(A)). Moreover, R( A ) R(A). R(A), then there is an x E X such that Ax But x xo + XI , where xo E Xo , x 1 E N(A). Hence,I Axo Ax - Ax i showing thatl E R(A). Thus A has an inverse A- in B(R(A), Xo) ( here, we make use of the fact that R(A) is closed and, hence, a Banach space ) . We have found a sort of "inverse" for A. The only trouble with it is that A - 1 is defined only on R(A) and not on the whole of Y. This fact will make it extremely awkward in using the inverse. It would be more useful if there would be an operator Ao E B(Y, X) that equals A- I on R(A ). Such an operator can be found if there is a finite dimensional subspace Yo such that Y R(A) EB Yo . (5.3) Then we can define A0 to equal A. - 1 on R(A) and to vanish ( or do something else ) on Yo. That Ao E B(Y, X) would follow from Lemma 5 . 2 . Let X1 be a closed subspace of a normed vector space X, and {0} . Then let M be a finite dimensional subspace such that M X 1 {b) For each E there is an E This decomposition is unique. = +

==

n

=

=

==

=

=

y.

=

y

y,

y

==

n

==

103

Orientation

5. 1 .

X2 == X 1 EB M is a closed subspace of X. Moreover, the operator P defined by (5.4)

Px ==

{

x,

0,

x

E M, X E X1

is in B (X2 ) . Applying Lemma 5.2 to our situation , we see that there is an operator P E B (Y) such that y, y E Yo , p - y 0, y E R(A ) . Thus, the operator I - P is in B (Y, R(A ) ) . Moreover, Ao = A. - 1 (I - P) , and , since A. - 1 E B (R(A) , we have Ao E B (Y, That we can decompose Y inro the form (5.3) follows from _

{

X ).

X),

Lemma 5 . 3 . Let X be a normed vector space, and let R be a closed sub­ space such that R0 is of finite dimension n. Then there is an n-dimensional subspace M of X such that X= R EBM.

Note that in our case R(A) 0 = N(A') [see (3. 1 2)] , �hich is finite dimen­ sional. Hence, (5.3) is , indeed , possible. The proofs of Lemmas 5.2 and 5.3 , will be given at the end of this section. We now summarize our progress so far. Theorem 5 .4. If A E

(a) N ( Ao ) == Yo , {b) R(Ao) ==

Xo,

(c) Ao A = I on

Xo ,

{d) AAo = I on R(A) . In addition, (e) Ao A == I

-

Ft on

X,

104

5. FREDHOLM OPERATORS

AAo I - F2 on Y, where F1 E B(X) with R(F1 ) N(A) and F2 E B ( Y) with R(F2 ) Yo. Consequently, the operators F and F are of finite rank. (f)

=

=

=

2

1

To prove statement ( e ) , we note that the operator F1 == I - AoA is equal to on and vanishes on Xo . Hence, it is in B(X) by Lemma 5 . 2 . Similar reasoning gives ( f) .

I N(A)

We conclude this section by giving the - proofs of Lemmas 5 . 1-5 . 3. First , we give the proof of Lemma 5. 1 .

x 1 , Xn � � N x, x (5.5) xj(xk ) = 6jk , 1 < j, k < Let X be the set of those x E X such that xj ( x) = 0 for each j. Clearly, Xo is a subspace of X. It is also closed. For if Zn E Xo and Zn in X , then z 0 xj(zn ) xj( z ) for each j . Hence, z E Xo . Moreover, Xo n N = { 0 } . For if x E N, n X = L1 O'.k Xk , and henee, xj ( x) O:j for each j . If x is to be in Xo as well, we must have each O:j 0 . Finally, we must show that every x E X can be written in the form x xo + z, where xo E Xo and z E N. Let x be any element in X. Set n z L1 x�(x ) xk . Clearly, z E N. Moreover, we have xj (x - z ) xj(x) - xj (x) 0 for each j . Hence, x - z E X0 . This gives the desired decomposition. The uniqueness is obvious. D

Proof. Let be a basis for ( see Section 4. 1 ) . By Lemma 4. 14, , we can find functionals , in X' such that ·

·

·

·

·

·

n.

0

�

�

==

==

==

=

==

==

==

Next , we give the proof of Lemma 5 . 2 . Proof. We first prove

I Px l < C l x l , X E x2 . If (5.6) did not hold, there would be a sequence {x n } of elements of X2 0 in X ._ Since M is finite dimensionaL, 1 , while X n such that I Px nl l {Pxn } has a subsequence converging in M ( Corollary 4.5) . For convenience, we assume that the whole sequence converges. Thus Px n in Thert z (I - P)xn - z in X1 . Since both M and X1 are closed, we have z E MnX1 . (5 .6)

=

�

-4

�

M.

1 05

Further properties

5. 2.

Hence, by hypothesis, we must have z = 0. But ll z l l = lim II P x n ll = 1 . T his provides the contradiction that proves (5.6) . To prove that X2 is closed, we let { x n } be a sequence of elements of X2 converging to an element x E X. By (5.6) , { Pxn } is a Cauchy sequence in M and, hence, converges to an element z E M . Consequently, { (I - P) xn } converges to an element w E X1 . Hence, Xn = Px n P ) x n converges to z w E x2 . This shows that X E X2 , and the proof is complete. D

+ (I -

+

Finally, we give the proof of Lemma 5.3. Proof. Let x� , · · · , x � be a basis for R 0 • Since R == 0( R 0 ) (Lemma 3.6) , we see that x E R if, and only if, xj ( x ) = 0 for each j . Now, by Lemma 4. 14, there are elements X I , · · · , Xn of X such that (5.5) holds. The Xk are clearly linearly independent. For, if

then for each .i;

n

xj ( L ak x k ) = CXj = 0. 1 Let M be the n-dimensiop.al subspace of X spanned by the Xk . Then RnM == {0} . For, if x E M, then x is of the form n x = L ak x k , 1 and hence, xj ( x ) = aj for each] . If x is also in R, then we must have O:j = 0 for each j . Now, let x be any element in X . Set n

w

=

L1 x� ( x ) xk .

Then w E M and xj (x - w) = 0 for each j . Hence, x - w E R and we have the desired decomposition. This completes the proof. D 5 . 2 . Further properties

From the definition, it may not be easy to recognize a Fredholm operator when one sees one. A useful tool in this connection is the following converse of Theorem 5.4. Theorem 5 . 5 . Let A be an operator in B(X., Y) and assume that there are operators A 1 , A2 E B ( Y, X) , K1 E K (X ) , K2 E K (Y) such that

(5.7)

A 1 A = I - K1 on X

106

5.

FREDHOLM OPERATORS

and (5.8) Then A E (X, Y) . Proof. Since N(A) C N ( A1 A) , we have o:(A) < n (I - K1 ) < oo (Theorem 4. 12) . Likewise, R(A) :J R(AA2 ) = R (I - K2 ) . Hence, N(A ' ) c N (I - K� ) and j](A) < o:(I - K� ) < oo (here, we have made use of the fact that the adjoint of a compact operator is compact ; see Theorem 4. 15) . It remains to prove that R(A) is closed. This follows easily from Lemma 5 . 6 . Let X be a normed vector space, and suppose that X N EBXo , where Xo is a closed subspace and N is finite dimensional. If X 1 is a subspace of X containing Xo , then X 1 is closed. ==

Applying Lemma 5.6 to our situation, we know that there is a finite­ dimensional subspace Y1 of Y such that Y == R(I - K2 ) EB Y1 (Lemma 5.3) . Since R(A) :J R(I - K2 ) , we see by Lemma 5.6 that R ( A) is closed in Y. 0 Now we give the proof of Lemma 5.6.

+

Proof. Set M == N n XJ . Then x l == Xo EB M. For if X E xl , X == xo z , where Xo E Xo and z E N. Since Xo E xl , the same is true for z . Hence, z E M . We now apply Lemma 5 . 2 . 0

We now come to a very important property of Fredholm operators. Theorem 5 . 7. If A E (X, Y) and B E {Y, Z), then BA E {X, Z) and

i (BA) == i ( B ) + i (A) .

(5.9)

Proof. By Theorem 5.4 , there are operators

Ao E B(Y, X) , B0 E B(Z, Y) , F1 E K(X) , F2 , F3 E K(Y) , F4 E K ( Z) such that (5. 10 ) and (5. 1 1 )

AoA == I - F1 on X,

AAo == I - F2 on Y

Bo B == I - F3 on Y,

BBo

==

I - F4 on Z

(actually, the Fi are of finite rank, but we do not need to know this here) . Hence , AoBoBA == Ao (I - F3 ) A == I - F1 - Ao F3A == I - Fs on X, and

5. 2.

107

Further properties

where Fs E K(X) and F6 E K( Z) . Hence, by Theorem 5.5, we see immedi­ ately that BA E (X, Z) . To prove (5.9) , let Y1 that

==

R(A) n N (B) . We find subspaces Y2 , Y3 , Y4 such

Y1 EB Y2 N(B) == Y1 EB Y3 , Y = R(A) EB Y3 EB Y4 , ( see Figure 5 . 1 ) . This can be done by Lemn1as 5. 1 and 5.3 . Note that Y1 , Y3 , Y4 are finite dimensional and that Y2 is closed. Let di == dim Yi , i == 1 ' 3, 4. Now, N (BA) = N (A) EB X 1 , R(B) = R(BA) EB Z4 , where X 1 c Xo is such that A (X 1 ) = Y1 and Z4 == B ( Y4 ) . We claim that R(A)

=

(5. 12) Assuming this for the moment , we have o:(BA)

==

o: ( A) + d 1

j3(BA) == j3(B)

+

,

d4 ,

d 1 + d3 , ,B (A) == d3 + d4 . These relationships immediately give (5.9) . o: ( B )

==

0

To prove (5. 1 2 ) , we employ Lemma 5 .8. Let V, W be finite-dimensional vector spaces, and let L be a llnear operator from V to W. Then dim R ( L) < dim D( L) . If L is one-to­ one, then dim R( L ) == dim D ( L ) .

In our case, A is a one-to-one linear map of X 1 onto Y1 . Hence , dim X1 == dim Y1 . Similarly, B is one-to-one on Y4 , and its range on Y4 is ( by definition ) Z4 . Hence, dim Z4 == dim Y4 . Thus, we complete the proof of Theorem 5. 7 by giving the simple proof of Lemma 5. 8 . Proof. The second statement follows from the first, since L - 1 exists and

D ( L- 1 ) ==

R( L - 1 ) == D( L ) .

R( L ) ,

, Wn To prove the first statement , assurne that dim D ( L) < n , and let 'W J , , Vn be vectors in D ( L) such that be any n vectors in R ( L) . Let v 1 , ·

·

·

·

·

·

108

5.

FREDHOLM OPERATORS

Y1

X

A

y

R(BA}

8

z

Figure 5 .1

Lvi == Wi , 1 < i < n . Since dim D ( L) < n , the Vi are linearly dependent. Hence , there are scalars n 1 , , nn , not all zero, such that ·

·

·

Consequently, L( n1v1 +

·

·

·

+ On Vn )

==

n1w1 +

· · · +

On Wn ==

0.

This shows that the Wi are linearly dependent. Since the Wi were any n vectors in R ( L) , we see that dim R( L) < n. Since this is true for any n , we must have dim R ( !..� ) < dim D( L) . The proof is complete. 0 We end this section with a simple consequence of Theorems 5 . 5 and 5 . 7. Lemma 5 . 9 . Suppose A E

(Y, X) and i (Ao ) == -i (A) . Proof. By hypothesis , (5. 10) holds with F1 E K(X) and F2 E K(Y) . Thus, by Theorem 5 . 5 (with X and Y interchanged) , we see that Ao E (Y, X) .

5. 3.

109

Perturbation theory

Moreover, by Theorem 5 . 7, i (Ao ) + i (A) == i (I - F1 ) == 0. 0

Hence, i (Ao ) == - i (A) , and the proof is complete. 5 . 3 . Perturbation theory

If K E K (X) , then A == I - K is a Fredholm operator. If we add any compact operator to A, it remains a Fredholm operator. Is this true for any Fredholm operator? An affirmative answer is given by Theorem 5. 10. If A E and

( X, Y) and K E K ( X, Y ) , then A + K E (X, Y ) i (A + K)

(5. 1 3)

==

i (A) .

Proof. By Theorem 5.4, there are Ao E B (Y, X) , F1 E K(X) , F2 E K (Y) such that

Ao A == I - F1 on X,

(5. 14)

AAo == I - F2 on Y.

Hence, Ao (A + K) == I - F1 + Ao K == I - K]_ on X, and ( A + K) Ao == I - F2 + KAo == I - K2 on Y, where K1 E K (X ) , K2 E K(Y) . Hence, A + K E (X, Y ) ( Theorem 5 . 5) . Moreover, by Theorem 5 . 7, i [ Ao (A +-,K ) ] == i (A0 ) + i (A + K) == i (I - K1 ) == 0. But i (Ao ) complete.

==

-i (A) ( Lemma 5.9) . Hence, (5. 13) holds , and the proof is ·

0

Another result along these lines is given by Theorem 5 . 1 1 . Assume that A E (X, Y ) . Then there is an 17 > 0 such that for any T E B (X, Y) satisfying I I T I I < ry , one has A + T E (X, Y ) ,

i (A + T ) == i (A ) ,

(5 . 1 5) and

(5 . 16)

o:(A + T ) < o:(A) .

Proof. By (5. 14) , we have

Ao (A + T ) == I - F1 + Ao T on X and (A + T ) Ao == I - F2 + T Ao on Y.

1 10

5.

FREDHOLM OPERATORS

We take ry == I I Ao l l - 1 . Then I I Ao T I I < I I Ao i i · I I T I I < 1 , and similarly, I I TAo l l < 1 . Thus , the operators I + Ao T and I + T Ao have bounded inverses (Theorem 1 . 1 ) . Consequently, (I + Ao T ) - 1 Ao ( A + T ) == I - (I + Ao T ) - 1 F1 --on X (5 . 1 7) (A + T ) Ao ( I + T Ao ) - 1

(5. 18)

==

I

-

F2 (I + TAo ) - 1 on Y.

We can now appeal to Theorem 5 . 5 to conclude that A + T E 4> (X, Y) . Moreover, by Theorem 5 . 7 applied to ( 5 . 1 7) , we have i [ (I + A 0T ) - 1 ] + i (Ao ) + i ( A + T ) == 0. Since i [ ( I + A0T ) - 1 ] == 0, we see that ( 5 .-15) holds. It remains to prove (5 . 16) . To do this , we refer back to Theorem 5.4 (c) . From it , we see that Ao (A + T) == I + Ao T on Xo ,

+

and hence, this operator is one-to-one on Xo . Thus, N (A T) n Xo == {0} . Since X == N ( A) EB X0 , we see that dim N ( A + T ) < dim N(� ) . This follows from the next lemma ( Lemma 5. 12) . 0

-

Lemma 5 . 1 2 . Let X be a vector space, and assume that X == N EB Xo , where N zs finite dzmensional. If M is a subspace of X such that !Yf n Xo == {0} , then dim lv! < dim N. Proof. Suppose dim N < hypothesis� Xk

==

n,

Since dim N <

n,

and let x 1 ,

·

·

·

, X n be any

n

vectors in M . By

X k o + X k i , X k o E Xo , x k 1 E N , 1 < k < n . there are scalars n 1 , a n not all zero such that ·

Hence,

n

·

·

,

n

L (Xk:l:k == L Qk X k() E ..tX"() .

Since the .r k are in

l'vf ,

1 1 this can happen only i f n

L Ct k .T h· == 0 . 1

showing that the x· k are li nearly dependent . Thus dirn < true for any n , the proof is con1plete . \\le now no t e

a

n.

Since this is 0

converse of Theorcn1 5 . 7.

Theoren1 5 . 1 3 . A ssume that A E B(X� Y ) and B E [J ( } /: ) o r( " u ch that B A E 4> (X, Z) Then A E 4> ( ..X" . rr) 1}·. an d only 1,f. B E
.

·.

5. 3.

111

Perturbation theory

Proof. First assume that A E

BAAo == B - BF2 on Y, where F2 E K(Y) . Now Ao E ( Y, X ) by Lemma 5.9, while BA E ( X, Z) by hypothesis. Hence, BAA0 E ( Y, Z) ( Theorem 5 . 7) . Since BF2 E K (Y, Z) , it follows that B E ( Y, Z) ( Theorem 5. 10 ) . Next, assume that B E (Y, Z) , and let Bo E B ( Z, Y) satisfy (5. 1 1 ) . Then, BoBA == A - F3A on X. Now BA E ( X, Z) by hypothesis, while Bo E ( Z, Y) . Hence, Bo BA E 0 ( X, Y) , and the same must be true of A. This completes the proof. The following seemingly stronger statement is a simple consequence of Theorem 5 . 13. Theorem 5 . 14. A .ssume that A E B (X� Y) and B is zn B(Y, Z) ttre such that BA E ( X, Z) . If a: ( B) < oo , then A E

We shall see in the next section that we can substitute ,6(A) < place of o:(B) < oo in the hypothesis of Theorern 5 . 14.

oo

in

Before we continue, let us give some examples. Consider t he space lp, 1 < p < oo . If x == (x1 , x2 , · · · ) is any elernent in lp , define L1x

==

(x2 , X3 , · · · ) .

We can check that N (£ 1 ) consists of those elements of the forrn (x1 , 0, · · · ) , and hence, is of dimension 1 . Let Xo be the subspace consisting of elcn1ents of the form ( 0, X2 , X3 , · · · ) It is easily seen that Xo is a closed subspace of lp , and that .

lp

==

N ( LI ) EB Xo .

Moreover, R(£ 1 ) == lp , so that i (£ 1 ) Another example is given by L2 X

==

==

1.

(0, X I ' X 2

•

.

.

.

).

5. FREDHOLM OPERATORS

1 12

In this case, we have N(L2) = {0} , R(L2 ) = Xo . Thus, i(£2) == - 1 . Finally, consider the operator L3 x = ( O, x3 , x2 , X5 , x 4 , x 7 , x 6 , · · · ) .

Here, one has N(£3) == N(£1 ) , R(L3) = Xo , i (L3) = 0. Note that L 1 L2 = I, while L2L1 is I on Xo and vanishes on N ( £1 ) . In Section 4.3 we showed that the operator K given by K x = (x1 , x2j2 , XJ/3 , was compact on lp . It follows therefore, that Li i = 1 , 2 , 3. Examine these operators.

·

·

·

)

+ K is a Fredholm operator,

5 . 4 . The adjoint operator

If A E B (X, Y) , then we know that A ' E B (Y' , X') (Theorem 3.3) . If A E (X, Y) , what can be said about A'? This is easily handled by means of Theorems 5.4 and 5.5. In fact , by taking adjoints in ( e ) and (f) of Theorem 5.4, we have

A ' A� == I - F{ on X ' ,

A� A' == I -

F�

on Y ' .

Since A� E B(X ' , Y') and the Ff are also of finite rank (see Section 4.2) , we see, _by Theorem 5.5, that A ' E ( Y', X') . What about i ( A ' ) ? Is it related in any way to i ( A ) ? To answer this question, we must examine the operator A' a little more closely. Since o: ( A ') = dim N( A ' ) , we have o: (A ' ) = ,B(A) . To determine ,B(A ' ) , we set A" = (A ' ) '. Let X " be the set of bounded linear functionals on X' . We know that X " is a Banach space (Theorem 2. 10) . Moreover, A " is the operator in B (X " , Y " ) defined by (5. 1 9)

A"x" ( y ') = x" (A ' y') ,

x" E X " , y' E Y' .

Now, the set of elements in X " that annihilate R(A') is precisely N(A ") [(3. 12) applied to A ' ] . Hence, by Lemma 5.3, there is a subspace W C X' of dimension o: (A " ) such that X ' == R(A') EB W.

( 5 . 20 )

On the other hand, if o: (A) = n and x1 , , Xn form a basis for N(A) , then there are functionals x� , , x� such that ( 5 . 5) holds (Lemma 4. 14) . Let Z be the n-dimensional subspace spanned by the x� . Then Z n R(A' ) == {0} . For if z ' E Z, then ·

·

·

·

·

·

and hence, 1 < -k< - n.

5.4. The adjoint operator

z'

113

z'(xk )

If E R(A' ) as well, then == 0 for each k, since R(A ' ) == N (A)0 ( Theorem 3 . 16) . Hence, == 0. Now for any E X' , set

z'

U

' xk )

Then u ( we have

I

n

x'

I I "'""' == X - � X ( Xj ) Xj .

1

'

u'

k by (5 .5) . Hence,

== 0 for each

X'

(5.2 1)

==

E R(A ' ) . Since

x' -

u'

E Z,

R(A ' ) EB Z,

and d i m Z == o:(A) . We now apply Lemma 5. 12 to conclude that dim Z == _ dim W. But dim Z == o:(A) , and dim W == o: (A" ) == (3 ( A') . Hence, we have proved Theorem 5 . 1 5 . If A E (X, Y) , then A ' E (Y' , X') and

i (A ' ) == - i ( A) .

(5.22) We have just seen that

o:(A " ) . o:(A)

(5.23)

whenever A E (X, Y) . It should be noted that � in general, n ( A") > o:(A)

(5.24)

for arbitrary A E B(X, Y) . This can be seen as follows. Let element of X , and set

xo be any

x' (xo), x' E X' . I F( x ' ) I < l l x o II · l x ' l ·

F (x' ) == Then, This means that F(x ' ). is is an E X " such that

x�

a

bounded linear functional on X i . Hence, there

x�(x') x' (xo), x' E X' . The element x� is unique. For if x1 also satisfies ( 5 . 25) , then x� (x' ) - x� ( x' ) 0, x' E X' , sho,ving that x� x � . \Ve set x � Jxo. Clearly, J is a linear mapping of X into X" defined on the whole of Moreover, it is one-to-one. For if Jxo 0, we see by ( 5 . 25) that x'(xo ) 0 for all x' E X'. Hence, x0 0 . In our new notation , ( 5 . 25) becomes (5 . 26) Jl-· (x' ) ( ) x E X, x ' E X ' . ==

(5 .25)

==

=

==

X,

==

We also note that

( 5 . 2 7)

==

==

tJ is

a

==

x

'

x ,

bounded operator. In fact ,

I .J ( x.l ) I p su = II .J .r l l l l .r ' ll .r

'

I )I = sup .r( .1· l l ' ll x

,

we

=

have

l l:r ll .

1 14

5. FREDHOLM OPERATORS

where the least upper bound is taken over all nonvanishing x' E X' , and we have used (2.28) . In particular, we note that if x E N(A) , then Jx E N(A") . This follows from (5.28) A" Jx (y' ) = Jx (A ' u') = A' y ' (x) = y' (Ax ) , y , E Y ' . Since J is one-to-one, we have by Lemma 5.8 that

o:(A) = dim N(A) = dim J[N(A) ] < dim N(A") = o:(A") . This gives (5.24) . We now can make good our promise made at the end of the last section. We shall prove Theorem 5 . 1 6 . Assume that A in B(X, Y) and ·n in B{Y, Z) are such that BA E

(X, Y) and B E (Y, Z) . Proof. Taking adjoints , we have A' B' E

(Y' , X') and B' E ( Z' , Y') . In particular, o:(B") < oo . But by ( 5 . 24 ) , we know that o:(B) < o:(B " ) < oo .

Hence, the hypotheses of Theorem 5. 14 are satisfied, and the conclusion follows. 0 5 . 5 . A special case

Suppose K E K(X) and A = I - K. Then, for each positive integer n, the operator An is in ( X ) = ( X, X) and, hence, o: (An ) is fi}\lite. Since N(An ) C N(An + l ) , we have o:(A n ) < o:(A n + l ) , and we see that o:(A n ) approaches either a finite limit or oo . Which is the case? The answer is given by Theorem 5 . 17. If K is in K{X) and A == I - K, then there is an integer k n n > 1 such that N(A ) == N(A ) for all k > n . Proof. First we note that the theorem is true if there is an integer k such that N (Ak ) == N( A k + l ) . For if j > k and x E N(AJ + l ) , then AJ- k x E N(A k + l ) == N (Ak ) , showing that x E N (AJ ) . Hence, N (AJ ) = N (Aj + l ) for each j > k. This means that if the theorem were not true, then N(A k ) would be a proper subspace of N ( A k + l ) for each k > 1 . This would mean, by Len1rna 4.7, that , for each k, there would be a Z k E N (A k ) such that

ll zk ll

== 1 ,

d(zk � N( A k - 1 ) )

> 1/2.

1 15

5. 5. A special case

Since K is a compact operator in X, { Kzk } would have a convergent sub­ sequence. But fo r j > k ,

II K zj - Kzk ll == ll zj - Azj - Zk + Azk ll == ll zj - (z k - Azk + Azj ) ll > 1 / 2 , since AJ- 1 (z k - Az k + Azj ) == 0. This shows that { Kzk } cannot have a conver­ gent subsequence. Whence, a contradiction. This co1npletes the proof. I

Note

the

0

similarity to the proof of Theorem 4. 1 2 of the -p revious chapter.

For A E

r(A) == lim o: ( An ) , n---+ oo

as well. Are there other operators A E (X) with both r (A) and r' (A) finite? Let us examine/such an operator a bit more closely. If r ( A) < oo , then there must be an integer n > 1 for which the--conclusion . of Theorem 5. 1 7 must hold . To see this , note that o:(A k ) is a nondecreasing 1 sequence of integers bounded from above. Similarly, if r' (A) < oo , there must be an integer k m > 1 such that N(A' ) == N(A' m ) for k > m . If both r (A) < oo and r' (A) < oo , let j == 1nax [m , n ] . Then o:(A k ) == o:(AJ ) , (3 (A k ) == (3 (Ai ) for k > j . Hence, i (A k ) == n(A k ) - ,B(A k ) == o:(AJ ) - j](AJ ) == i (AJ ) for k > j . But i (A k ) == k i (A) . Hence, (k - j ) i (A) == 0 for all k > j , showing that we must have i (A) == 0. But if this is the case, we must also have m == n . To recapitulate, if A E (X ) , r (A) < oo and r' (A) -< oo , then i (A) == 0, and- there is an n > 1 such that ( 5 . 29) Moreover, we claim that

(5.30) This follows from the fact that if x is in this intersection, we have, on one hand , that An x == 0 and , on the other, that there is an x 1 E X such that x == A n x 1 . Thus , A 2 n x 1 == 0 , or X I E N(A 2 n ) == N (A n ) . Hence, , x 8 be a basis for N (A n ) . Then there are x == An x 1 == 0. Next , let x 1 , bounded linear functionals x � , · · · , x � , which annihilate R(An ) and satisfy ·

(5 .31 )

·

·

x� ( xk ) == §Jk ,

1 < j, k <

s.

To see this, we note that for each j. the subspace spanned by R( A n ) and the X h· \\�ith k =!= j is closed ( sec Len1ma 5 . 2 ) . Hence, by Theorem 2 . 9 , there

1 16

5. FREDHOLM OPERATORS

is an xj E X' which satisfies xj (xj ) == 1 and annihilates R(An ) and all the X k such that k -:1 j . Next , set

n

(5 .32 )

Vx

==

I:1 x� (x)xk .

Since V is of finite rank (and , hence, compact) , we see that A n + V E ( X) and i (A n + V) == 0 (Theorem 5. 10 ) . We note that o:(A n + V) == 0. If (An + V)x == 0, then Vx E R(A n ) n N(An ) . Thus , Vx == 0. But then A n x = 0, which means that x E N(An ) . On the other hand, Vx == 0 implies x� ( x) - == 0 for each k , and this can only happen if x == 0. The upshot of all this is that An + V has a bounded inverse E. Moreover, since R(V) C N (A n ) and R(An ) C N(V) , we have A n v == VA n == 0. Hence, (An + V) V = V (An + V) == V 2 , showing that or

VE == EV 2 E = EV

'

whence,

EA n == A n E == I - EV. Since E is bounded, EV E K(X) . We have proved the "necessary" part of

Theorem 5 . 1 8. A necessary and sufficient condition that A E (X ) with r (A) < oo and r' (A) < oo is that there exist an integer n > 1 and operators E in B(X) and K in K(X) such that

( 5 . 33 )

EA n == A n E = I - K.

Proof. To prove the sufficiency of (5.33 ) , set W = A n . Then by Theorem 5.5, W E (X) . Now by Theore m 5. 1 7, there is an integer m such that

N [ (I - , K) J ] == N [ (I - K) m ] , R [ (I - K) i ] == R [ (I - K) m ] , I

j>

m.

Thus,

N(WJ ) c N (EJ WJ ) = N [ (EW)J] == N [ (I - K) i ] = N[(I - K) m ] , since E and W commute by (5.33 ) . Hence, o:(Wi ) is bounded from above, and r (A) == r(W) < oo . Similarly, R(Wi ) :J R(Wj EJ ) == R [ (W E)J ] R [ (I - K) J ] == R[(I - K) m ] , .

showing that

N ( W '1 )

C

JV[(I - K ' ) m] ,

and hence, (3(Wi ) is bounded from above. This gives r' (A) = r' (W) < oo . The proof is complete. D

5.6. Semi-Fredholm operators

1 17

"Not so fast !" you exclaim. "We have shown that W E (X) , but we have not shown that A E (X ) ." Right you are! This we do by proving Lemma 5 . 19. Let A 1 , · · · , A n be operators in B(X) which commute, and suppose that their product A == A 1 · · · A n is in 4>(X) . Then each A k is in ( X) .

a

Proof. Clearly, N (A k ) C N (A) and R(A k ) :J R (A ) . Thus , ( A k ) and (3 (A k ) are both finite. Moreover, by (5 .3 ) , X == R(A) EB Yo , where Yo is a finite-dimensional subspace of X. Since R ( A k ) :J R( A ) , we see by Lemma 0 5.6 that R( A k ) is closed.

In connection with Theorem 5 . 18, we wish to make the observation that if A E 4>( X ) , i (A) > 0 and r ( A ) < oo , then the conclusions of Theorem 5 . 18 hold. In fact , since i ( A k ) == ki (A) > 0, we always have (A k ) > (3 ( A k ) , showing that r ' ( A ) < oo , and Theorem 5 . 18 applies directly. A similar statement holds if we assume i (A) < 0, r� ( A) <

a

00.

5 . 6 . Semi-Fredholm operators

One might wonder whether something can be said when some of the stip­ ulations made f? r Fredholm operators are relaxed . The answer is yes , and we give some examples in this section. Let 4> + (X, Y) denote the set of all A E B(X, Y) such that R (A ) is closed in Y and ( A) < oo . Such operators are called semi-Fredholm. If A E + (X, Y) , then (5.2) still holds . Let P be the ( projection ) operator defined by

a

(5.34) By Lemma 5 . 2 ,

p ==

{

I on N (A) , 0 on Xo.

P E B (X) . We also have

a

Lemma 5 . 20. Assume that A is in B(X, Y) and that ( A) < oo . Let P be defined by (5. 34). Then R (A) is closed in Y if and only if

(5.35)

II (! - P) x l l

<

II A x ll , x E X.

Proof. If R(A) is closed, then the restriction of A to Xo is one-to-one and has closed range. Hence, by Theorem 3. 1 2 ,

l l x l l < CII A x l l , x E Xo . But for any x E X, (I - P) x E Xo and A ( I - P)x == Ax. (5 . 36 )

This proves ( 5 . 35) . Conversely, assume (5.35) holds. If y E Y and Ax n � y , then { A x } is a Cauchy sequence in Y . By (5. 35 ) , { (I - P) x n } is a Cauchy sequence in X. n

118

5. FREDHOLM OPERATORS

Since X is complete, there is an x E: X such that ( J - P) x n � x. Thus, A(I - P) x n � Ax . But A(I - P) xn == Ax n � y . Hence , Ax == y , and y E R( A) . Accordingly, R( A) is closed in Y, and the proof is complete. D Suppose A E + (X, Y) . T hen by (5.35) �

l l x l l < C II Ax l l + II Px l l · Set

!x l

==

II Px ll ·

l nx l

==

lnl · lxl ,

(5.37) Then we have (5.38) (5 .39)

lx + Y l < lxl + IYL 0 only if x == 0. Thus, l x l is not a norm. A

but we do not have l x l == functional satisfying (5.38) and (5.39) is called a seminorm. The seminorm given by (5.37) also has the following property. If { x n } is a sequence of elements of X such that ll xn ll < C , then it has a subsquence which is a Cauchy sequence in the semi norm I · I (this follows from the compactness of P) . A seminorm having this property is said to be compact relative to (or completely continuous with respect to) the norm of X . Thus, we have shown that if A E + (X, Y) , then there is a seminorm compact relative to the norm of X such that (5.40)

ll x l l < C II Ax ll + l x l ,

x E X.

The converse is also true. In fact , we have Theorem 5 . 2 1 . Suppose A is in B{X, Y). Then A E + (X, Y) if and only if there is a semznorm 1 · 1 compact relative to the norm of X such that (5.4 0) holds. Proof. We have proved the "only if' part. To prove the "if ' part, assume that ( 5.40) holds. Then n (A) < oo. To see this , let {x n } be a sequence of elements in N(A) such that ll x n ll == 1 . Then there is a subsequence (which we assume is the whole sequence) such that

l x n - xm l Hence , by (5.40) ,

�

o as

m , n � oo .

ll xn - Xm l l < l xn - Xm l � 0. Thus, { x n } converges. But this shows that N(A) is finite dimensional (The­ orem 4.6) . Next , let P be defined by (5.34) . Then we claim that (5.35) holds. For if it did not , there would be a sequence { X n } such that II ( J - P)x n ll == 1 , Axn � 0 as n � oo .

1 19

5. 6. Semi-Fredholm operators Set Zn

==

(I - P) xn . Then

Zn

E Xo , and

il zn l l == 1 ,

Azn

�

0.

Since the sequence {zn } is bounded, it has a subsequence (assumed to be the whole sequence) such that � l zn - Zm l � 0 as m, n � oo . Hence, by (5 .40) ,

I I Zn - Zm I I < C II A ( Zn - Zm ) II + I Zn - Zm I � 0. Since Xo is closed, there is a z E Xo such that Zn � z. Hence, Azn � Az. But A zn � 0. This shows that z E N(A) . On the other hand , z E Xo . The only way these statements can be reconciled is if z == 0. But this gives further trouble, since II z II == lim I I Zn II == 1 . Our only way out is to conclude

that the assumption that (5.35) does not hold cannot be true. We now make 0 use of Lemma 5 . 20. Now we can prove the counterparts of Theorems 5 . 10 and 5 . 1 1 . Theorem 5 . 22. If A E + (X, Y) and K zs in K{X, Y), then A + K E + (X, Y) and (5. 1 3) holds. Proof. By The o rem 5 . 2 1 ,

ll x ll <

C I I ( A + K ) x ll + l x l + C II Kx ll , x E X.

Set

lxlo

==

l x l + C I I Kx l l ·

Then, l x l o is a seminorm which is compact relative to the norm of X . Thus , A + K E + (X, Y) by Theorem 5 . 2 1 . If A + K E 4>(X, Y) , then A == (A + K) - K E (X, Y) by Theorem 5. 1 0 , and ( 5 . 1 3 ) holds . If A + K tJ. (X, Y) , i.e. , if i ( A + K) == - oo , then the same is true of A by Theorem 5 . 10. In this D case, both sides of ( 5 . 13) are equal to - oo . We also have Theorem 5 .23. Suppose A E 4> + (X, Y) . Then there zs a number ry > 0 such that for each T in B(X, Y) satisfying II T II < 1] , we have A + T E 4>+ (X, Y) ,

a(A + T ) < a (A) ,

( 5 . 41 )

f)( A + T) < (3(A) ,

and {5. 1 5) holds. Proof.

By

Lenuna G . 20.

l l .r ll < C II ( A + T ) x l l + C II T.r ll +, II Px l l .

xE

\'" .

..

120 Take

5. FREDHOLM OPERATORS 17

= 1 / 2 C . Then

ll x ll < G I I (A + T ) x l l

+

or

1 I I Px l l + 1 1 x l l , 2

l l x l l < 2 C II (A + T) x l l + 2 I I Px ll · This immediately shows that A + T E 4> + (X, Y) (Theorem 5.2 1 ) . Moreover, since Px = 0 for x E Xo , we have ll x ll < 2 C II (A + T ) x l l , x E Xo , which shows that N(A + T ) n Xo = {0} . Since X = Xo EB N (A) , we see by Lemma 5. 12 that the first inequality in (5 .41 ) holds. If (3 (A) = oo , then the second inequality holds trivially. Otherwise, A E (X, Y) , and there is an 17 > 0 such that A + T (X, Y) and (5. 1 5) ho�ds (Theorem 5. 1 1) . Thus, (3(A + T) = ( A + T ) - n (A) + (3 (A) < (3 (A) . I

E

a

It remains to prove (5. 1 5) for the general case. Assume first that N(A) = { 0} . Then there is a constant Co such that

l l x ll < Co i i A x l l , x E X (Theorem 3. 12) . Take 17 < 1/3 Co . Then II x II < Co II Ax + Tx I I + Co I I T I I · I I x II , and consequently,

3 x < l l l l 2 Go i i (A + T) x l l , x E X. This implies that R(A + T) is closed and that (5.41) holds (in particular, ( A + T ) = 0) . Reversing the procedure, we have 3 < ll x l l 2 Co ( I I (A + T - T) x ll + I I T i l · ll x ll ) , x E X,

a

which gives

ll x l l < 3 Co i i A x ll ,

x E X.

Of course, we knew that this inequality held before (even without the factor 3 ) , but we wanted to see if the value of 17 that we picked allowed us to work back from A + T to A. As we see, it does , and we can conclude that

f)( A ) < j3(A + T ) , fro1n which we conclude in conj unction with ( 5.4 1 ) that a ( A + T) == a ( A ) 0 , !1 (A + T) (3 ( A ) . Thus, in this case (5. 15) holds. In the general case, we can find a closed a(A ) < a( A + T) , ==

subspace Xo of X such that

X == N (A) EB Xo

==

121

5. 6. Semi-Fredholm operators

( cf. Lemma 5 . 1 ) . We let Ao be the restriction of A to Xo . Then Ao E + ( Xo, Y ) , N ( Ao ) == { 0} , o: ( Ao) == 0, and

R ( Ao) == R(A) , {J (Ao) == {J (A) , i (Ao) = i (A) - o: ( A) . If To is the restriction of T to Xo, then we see that o:(Ao + To) == 0 , i (Ao + To ) = i (Ao ) . Since

A (z + x ) = A x , (A + T ) (z + x )

==

Tz + (A + T ) x ,

z E N(A) , x E Xo ,

the following lemma will allow us to conclude that

o:(A + T ) < o:(Ao + To ) + o:(A) ==_a(A) , i (A) == i (Ao ) + o:(A) , i (A + T ) == i(Ao + To) + o:(A)

==

i (A) .

This will complete the proof.

D

Lemma 5.24. Assume A E + ( X, Y ) , X .:z:: N EB X, B ( X , Y ) satisfies -

A (z + x)

==

C z + Ax ,

n

� dim N <

-

oo ,

AE

z E N, x E X,

where C is in B(N, Y) . Then A E + ( X , Y ) , and -

-

o:(A) < o:(A) + n ,

Proof. We may assume that n

Case 1 . C zo

==

i (A) = i ( A) + n .

1 , i.e. , N = {zo } . There are three cases . -

N(A) EB {zo} , o:(A) == _ = R(A , (A == o:(A) + 1 , R(A) ) (3 ) (3 (A) , i (A) == i (A) + 1 . Case 2. C zo # 0, C zo fl. R(A) . For this case we have N( A ) == N (A) , o: ( A ) == o: (A) , R ( A ) == R (A) EB { C z0 } , (3 ( A ) == (3 (A) - 1 , i ( A ) == i (A) + 1 . Case 3 . C zo # 0 , C zo E R(A) . In this case there is an element xo E X such that A x o == C zo . Thus , ==

0. In this_ case we have_ N(A)

-

==

A(,.\ zo + x) == ,.\ Czo + A x . Consequently, N( A ) == - N(A) EB {z0 - xo } , o:( A ) == o:(A) + 1 , R(A) R(A ) , (3(A ) == (3(A) , i (A) == 'i (A) + 1 . We see that in all cases the conclusions

of the lemma hold .

0

We end this section by making the following trivial but sometimes useful observations.

5. FREDHOLM OPERATORS

1 22

Proposition 5 . 2 5 . A E (X, Y) if and only if A E (X, Y) implies A' E _(Y' , X ' ) , so that the "only if ' part is immediate. Moreover, if A E + (X, Y) , then R(A) is closed in Y and o:(A) < oo. I f, in addition , A' E + (Y ' , X ' ) , then ,B(A) = o: ( A' ) < oo. Thus , A E ( X, Y) . 0 Theorem 5 . 26. If A E + (X, Y) and B E + ( Y, Z) , then BA E + (X, Z) , and {5. 9) holds.

Proof. By Theorem 5 . 2 1 ,

l l x ll < C1 I I A x l l + l x l 1 ,

x E X,

I I Y I I < C2 I I B Y I I + I Y I 2 , 1j E Y, where the seminorms I · l 1 and I · l 2 are compact relative to the norms of X and Y , respectively. Thus

and hence, It is easily checked that the seminorm

C1 I A x l 2 + l x l 1 is compact relative to the norm of X . Thus , BA E + (X, Z) by Theorem 5 . 2 1 . Finally, we note that if i (BA) is finite, i.e. , if BA E ( X, Z) , then we must have A E (X, Y) and B E (Y, Z) by Theorem 5 . 14 with (5 .9) holding by Theorem 5 . 7. On the other hand, if i ( BA) == -oo, then either A fl. (X, Y ) or B fl. ( Y, Z) ( or both ) . This means that either i (A) == - oo or i (B) == - oo ( or both ) . In this case, both sides of (5.9) are equal to lxl3

==

_ .,.,

0

- 00 .

We note that the proof of ( 5 . 23) depends only on the fact that n(A) < oo . Thus, we can ext�nd Theorem 5 . 1 5 to + (X, Y) , then

i (A ' ) In particular, i (A')

==

==

o:(A ' ) - o:(A " )

oo �� A ¢ ( X, Y) .

==

- i (A )

We define _ (X, Y) to be the set of those A A' E + (Y ' , X ' ) . We have

.

E

n ( .�, Y) SGCh

t h at

1 23

5. 7. Products of operators

Theorem 5 . 28. If A E _ (X, Y) and {5. 13} holds. Proof. Now, A' E + (Y ' , X ' ) and i (A ' + K' ) = i (A ' )

( Theorem 5.22 ) . Hence, A + K

by Theorem 5. 1 5 .

E

E

K(Y ' , X ' ) by The­

_ (X, Y) by definition , and ( 5 . 1 3 ) holds D

Theorem 5 . 29. If A E _ (X, Y) , then there is an 17 > 0 such that II T II < rJ for T in B{X, Y) �mplzes that A + T E _ (X, Y) , and {5. 1 5}, (5.4 1} hold. Proof. We know that T' E B (Y ' , X ' ) and that li T' II < 17 ( Theorem 3.3 ) . Hence, A' + T' E + ( Y ' , X ' ) for 17 sufficiently small, and i ( A' + T') == i (A ' ) , o:(A ' + T') < a ( A ' ) , ,B(A ' + T') < ,B(A ' )

( Theorem 5.23 ) . This translates into ( 5. 1 5 ) , ( 5.41 ) , and the conclus·i on fol­

lows.

D

As a counterpart of Theorem 5. 26 , we have Theorem 5 .30. If A E _ (X, Y) and B and {5. 9} holds.

E

_ (Y, Z ) , then BA E _ (X, Z ) ,

Proof. By definition, A ' E + (Y ' , X ' ) and B ' E + ( Z' , Y ' ) . Thus , ( BA) ' A'B ' E
by Theorem 5.26. Hence, BA E

==

D

5.7. Products of operators

We also have the following Theorem 5 .31 . If A is �n B (X, Y) , B is in B (Y, Z ) and BA then B E- _ (Y, Z ) .

E


Proof. Since dim R(BA)0 < oo , there is a subspace Zo such that dim Z0 < oo and Z = R(BA) EB Zo ( Lemma 5 . 3 ) . Since R(B) =:) R(BA) , we know that R(B) is closed ( Lemma 5.6 ) and R(B)° C R(BA ) 0 • Thus , dim R(B)0 < oo . D Consequently, B E _ (Y, Z ) .

In contrast , we have

5. FREDHOLM OPERATORS

1 24

Theorem 5 .32. If A is in B (X, Y) , B is in B(Y, Z) and BA E

_ ( Z ' , X ' ) . Theorem 5 . 3 1 now implies that A' E <J>_ (Y' , X') , which means that A E + (X, Y) . D

Two closed subspaces XI , X2 of a Banach space X are called comple­ mentary when XI n X2 = {0} and X == XI EB X2 . Either subspace is called a complemertt of the other. We say that a subspace XI c X is complemented if it has a complement . Some Banach spaces contain subspaces which are not complemented. We note Lemma 5 . 33 . If X1 is a closed complemented subspace of a Banach space X, then there is a bounded projection P on X with R(P) = X 1 . Proof. Let x2 be a closed complement of XI in X. Then each X E X is of the form X = X I + X2 , X k E Xk . Define Px == X I · Then P is a projection, and it is easily verified that it is a closed operator. Hence, it is bounded by the closed graph theorem (Theorem D 3. 10) . -

Theorem 5 . 34. If A E + (X, Y) and R (A) is complemented in Y, then there is an A0 E B (Y, X) such that AoA E (X) . Proof. Let P be a bounded projection from Y to R(A) . There is a closed subspace Xo E X such that

X == Xo EB N(A) . Then A has a bounded inverse A (rom R(A) to Xo . Let Ao Ao E B (Y, X ) , and AoA ==

{

I 0

=

AP. Then

on Xo , on N(A) .

Thus, Ao A E (X ) .

D

Theorem 5 .35 . If A E _ (X, Y) with N(A) complemented, then there is an Ao E B ( Y, X) such that AAo E ( Y) . Proof. There is a finite dimensional subspace Yo E Y such that

Y == R(A) EB Yo .

125

5. 7. Products of operators

Let P be the bounded projection onto R(A) which vanishes on Yo , and define Ao as above. Then on R(A) , AAo == 0 on Yo .

{I

Thus , AA0 E