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(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
(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
_ ( 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
K contains all scalars A -# 0.
Before proceeding further, let us take the ingenious step of, hence-forth, denoting the operator A! by A. I am sure you will appreciate the great efficie? cies effected by this bold move. Throughout this chapter, we shall assume that X is a Banach space. We can also say something about
a (K - A) for A E K .
Theorem 6 . 2 . Under the hypothesis of Theorem 6. 1 , n (K - A) == 0 except for, at most, a denumerable set S of values of A . The set S depends on K and has 0 as its only possible limit point. Moreover, if A -# 0 and A rl. S, then n (K - A) = 0, R(K - A) == X and K - A has an inverse in B(X) .
E
Proof. In order to prove the first part we show that for any > 0 , there is at most a finite number of values of A for which I A I > E and n (K - A) i- 0. · To see this , suppose { A n } were an infinite sequence of distinct scalars such
1 29
6.
1 30
I An l > ( K - An)Xn
that
SPECTRAL THEORY
E and0. nWe( Kclaim - An) "# 0 . Let Xn "# 0 be any element such that that , for any the elements X , · , are n,
==
m
linearly independent . Otherwise , there would be an
I
·
·
such that
Xn
(6. 2) X I , · · · , X m are linearly independent . By (6.2) , Kx m+I == O} K X I + . . . + n m KX m == Q}AI X I + . . . + O m Am X m ·
while the
On the other hand , whence,
QI (AI - Am +I ) X ! + . . . + nm ( A m - Am+I ) X m == 0 . Since X I , · · , X m are independent and the An are distinct , (6.3) implies ni == · · · == Om == 0 , which is impossible because X m + I "# 0. Next , let Mn be the subspace spanned by x 1 · , Xn . Then Mn is a proper subset of Mn + l for each n. Hence, by Riesz 's Lemma ( Lemma 4.7-) , there is a Zn E Mn such that (6.4) l l zn ll == 1 and d (zn, Mn - 1 ) > 1 / 2 . Clearly, }{ maps Mn into itself. If (6.3)
·
,
·
·
then
n Kx == L nj Aj Xj . I Moreover, n -I (K - An)X == L Oj ( Aj - An ) Xj . 1 we have Hence, K - A n maps Mn into Mn - 1 · Now if > K Zn - K Zrn == ( - An) Zn + An Zn - K Zm == A n [Zn - A n 1 ( K Zm - ( K - An ) Zn) ] . But Kzm E Mm 0- Mn - 1 , while (K - An)Zn E Mn -I · Hence,
K
n
m,
showing that { K Zn } can have no convergent subsequence. This contradicts the fact that K is a compact operator. Thus, there cannot exist an infinite sequence { A n } of distinct values such that n(K - A n ) # 0 a nd I A n i > E . Now suppose A "# 0 and A fl. S. Then a ( K - ,.\) == 0. Fro n1 ( G . 1 ) and Theoren1 4 . 1 2 , we see that R ( K - A ) == X. All we need now i� t he bounded inverse
The spectrum and resolvent sets
6. 1 .
131
theorem ( Theorem 3.8) to conclude that K - A has an inverse in B (X) . This D completes the proof. For any operator A E B (X) , a scalar A for which n (A - A) # 0 is called an eigenvalue of A. Any element x # 0 of X such that (A - A) x == 0 is called an eigenvector ( or eigenelement) . The points A for which (A - A) has a bounded inverse in B (X) comprise the resolvent set p(A) of A. It is the set of those A such that n (A - A) == 0 and R(A - A) X ( Theorem 3.8) . The spectrum a(A) of A consists of all scalars not in p(A) . The set of eigenvalues of A is sometimes called the point spectrum of A and is denoted by Pa(A) . In terms of the above definitions , Theorem 6 . 2 states that =
( 1 ) The point spectrum of K consists of, at most , a denumerable set S having 0 as its only possible limit point . /
I
(2) All other points A # 0 belong to the resolvent set of K. We are now going to examine the sets A , p(A) and a(A) for arbitrary A E B (X ) . Theorem 6 . 3 . A and p(A) are open sets. Hence, a(A) is a closed set. Proof. If Ao E A , then by definition A - Ao E (X) . By Theorem 5. 1 1 , there is a constant 17 > 0 such that I JL I < 17 implies that A - Ao - JL E (X) and i (A - Ao - JL ) i (A - Ao) , n(A - Ao - JL) < n (A - Ao ) . Thus , if A is any scalar such that l A - A o l < ry , then A - A E (X) and =
( 6.5 )
i (A - A) = i (A - Ao ) , n(A - A) < a(A - Ao ) . In particular, this shows that A E A . Thus, A is an open set . To show the same for p(A) , we make the trivial observation that A E p(A) if and only if A E A , i (A - A) = O and n(A - A) == O. Thus, if A o E p(A) and l A - A o l < ry , then by ( 6.5 ) , i (A - A) == a(A - A) = 0. 0 Hence A E p(A) , and the proof is complete. Does every operator A E B ( X ) have points in its resolvent set? Yes; in fact , we have Theor �m 6 .4. For A in B(X), set
( 6.6 )
ra (A) == inf II A n l l 1 /n . n
Then p(A) yontains all scalars A such that I A I > ra (A) .
132
6. SPECTRAL THEORY This theorem is an immediate consequence of the following two lemmas:
Lemma 6.5 . If I -X I > II A I I , then ,\ E p(A)
Lemma 6 . 6 . If ,\ E a( A) , then for each n we have _x n E a(A n ) .
Using these lemmas, we give the proof of Theorem 6.4. Proof. If ,\ E a(A) , then for each n, we have _xn E a(An ) ( Lemma 6.6) . Hence, we must have I ,\ n I < II A n II , for otherwise, we would have ,\ n E p( A n ) ( Lemma 6.5) . Thus,
I -X I < II A n l l 1 /n , A E a (A) , n ;::: 1 , 2 , Consequently, I -X I < ra (A) for all ,\ E a(A) . Hence, all ,\ such that I -X I > 0 ra (A) must be in p( A ) . (6. 7)
·
·
·
.
Now we give the proof of Lemma 6.5. Proof. We have
(6.8)
Now if I -X I > II A ll , then the norm of the operator ,\ - 1 A is less than one. Hence, I - .x- 1 A has an inverse in B (X) ( Theorem 1 . 1 ) . The same is, ther�fore, true for A - ,\. 0 Now we prove Lemma 6.6. Proof. Suppose _x n E p(A n ) . Now,
An - ,\n
(6.9)
=
(A - A ) B
=
B(A - .X) ,
where
B = A n - 1 + ,\ An - 2 + . . . + _x n - 2 A + _x n - 1 . Thus, n(A - .X) = 0 and R(A - .X) = X . This me ans that ,\ E p(A) , contrary 0 to the assumption. This completes the proof. Lemma 6.6 has an interesting generalization. Let p(t) be a polynomial of the form p(t)
=
n
l:0 ak tk .
Then for any operator A E B (X) , we define the operator
where we take A 0
=
I. We have
Theorem 6 . 7. If ,\ E a(A) , then p (.X) E a(p(A) ) for any polynomial p(t).
1 33
6. 2. The spectral mapping theorerr1 Proof. Since A is a root of p( t ) - p(.X) , we have
p(t) :-- p(A) ==
(t - A)q(t ) ,
where q(t) is a polynomial with real coefficients. Hence,
p(A) - p(A) == (A - A)q(A) == q(A) (A - A) . Now, if p(A) is in p(p (A) ) , then (6. 10 ) shows that o:(A - A) == 0 and R(A 0 A) == X. This means that A E p(A) , and the theorem is proved . (6. 10)
A symbolic way of writing Theorem 6 . 7 is
( 6. 1 1 )
p(a(A) )
c
a(p(A) ) . Note that, in general, ther,e may be points in a ( p (A) ) which may not be of the form p(A) for some A E a(A) . As an example, consider the operator on JR2 given by A ( o: 1 , o: 2 ) == ( -o: 2 , o: 1 ) . A has no spectrum; A - A is invertible for all real A. However, A 2 has - 1 as an eigenvalue. What is the reason for this? It is simply that our scalars are real. Consequently, imaginary numbers cannot be considered as eigenvalues. We shall see later that in order to obtain a more complete theory, we shall have to consider Banach spaces with complex scalars . Another question is whether every operator A E B (X ) has points in its spectrum. An affirmative answer will be given in Section 9 . 1 . 6 . 2 . The spectral mapping theorem
Suppose we want to solve an equation of the form y,
x , y E X, where p ( t ) is a polynomial and A E B (X ) . If 0 is not in the spectrum of p( A) , then p (A) has an inverse in B (X) and , hence, (6. 12) can be solved for all y E X. So a natural question to ask is: What is the spectrum of p (A) ? By Theorem 6.7 we see that it contains p(a (A) ) , but by the remark at the (6. 12)
p(A) x ==
end of the preceding section it can contain other points . What a pity! If it were true that
p ( a(A) ) == a ( p ( A) ) , then we could say that (6. 1 2) can be solved uniquely for all y E X if and only if p(A) # 0 for all A E a ( A ) . Do not despair . There are spaces where (6. 1 3) holds . They are called � 6. 13 �
complex Banach 8paces. They arc the same as the Banach spaces that we
have been considering with the modification that their scalars consist of all complex nun1bers. By co1nparison, the spaces we have been considering are
134
6. SPECTRAL THEORY
called real Banach spaces .. The same distinctions are made for vector spaces and normed vector spaces. For a complex Banach space we have Theorem 6.8. If X is a complex Banach space, then JL E a(p(A) ) if and only if JL = p(A) for some A E o-(A) , i. e. , if {6. 1 3) holds.-
Proof. We have proved it in one direction already ( Theorem 6.7) . To prove it in the other, let /' 1 , , f'n be the ( cornplex ) roots of p( - JL . For a complex Banach space they are all scalars. Thus, ·
·
t)
·
p(A) - JL
c (A - !'1 ) · · · (A - !'n ) , c # 0. Now suppose that all of the /'j are in p(A) . Then each A - !'j has an inverse in B (X) . Hence, the same is true for p(A) - JL . In other words, JL E p(p(A) ) . Thus, if JL E a(p(A)) , then at least one of the 'Yj . must be in a ( A) , say l'k · Hence, JL == P(!'k ) , where /'k E a (A) . This completes the proof. D =
"Hold it a minute!" you exclaim. "You are using results for a complex Banach space that were proved only for real Banach spaces." Yes, you are right . However, if you go through all of the proofs given so far for real spaces , you will find that they all hold equally well for complex spaces with one important exception, and even in that case the proof can be fixed up to apply to complex spaces as well. For all other cases , all you have to do is to substitute "complex number" in place of "real number" whenever you see the word "scalar." What is the exception? I ' ll let you think a bit about it. In any event , if you do not wish to take my word for it , you are invited to go through the material so far and check for yourself. In short, all of the theorems proved so far for real spaces are true for complex spaces as well. Theorem 6.8 is called the spectral mapping theorem for polynC)mials. As mentioned before, it has the useful consequence Corollary 6.9. If X is a complex Banach space, then equation {6. 12) has a unique solution for every y in X if and only if p(>.. ) # 0 for all A E a( A) . 6 . 3 . Operational calculus
Other things can be done in a complex Banach space that cannot be done in a real Banach space. For instance, we can get a formula for p(A) - 1 when it exists. To obtain this formula, we first note Theorem 6 . 10 . If X is a complex Banach space, then (z - A) - 1 is a complex analytic function of z for z E p(A) .
By this , we mean that in a neighborhood of each zo E p(A) , the operator ( z - A) - 1 can be expanded in a "Taylor series ," which converges in norm to
6.3. Operational calculus
135
1
(z - A ) - , j ust like analytic functions of a complex variable. Yes, I promised
that you did not need to know anything but advanced calculus in order to read this book. But after all, you know that every rule has its exceptions. Anyone who is unfamiliar with the theory of complex variables should skip this and the next section and go on to Section 6.!1. The proof of Theorem 6. 10 will be given at the end of this section. Now, by Lemma 6.5, contains the set > II A II · We can expand 1 (z in powers of z - on this set . In fact , we have
A) - 1
Lemma 6. 1 1 .
(6. 14)
p(A)
lzl
If l z � > lim sup I An l 1 /n , then (z - A ) - 1 == L z - n An - 1 ' 00
1
.../
where the convergence is �n the norm of B{X). Proof. By hypothesis , there is a number 6 < 1 such that
(6 . 1 5 ) for n sufficiently large. Set
B == z - 1 A . Then, by (6.'1 5) , we have
(6. 16) Now,
n-1
n- 1
I - Bn == (I - B) L0 Bk == (L0 B k )(I - B), where, as always , we set B 0 == I . By (6. 16) , we see that the operators ( 6. 1 7)
converge in norm to an operator, which by (6. 1 7) mu st
z-
A == z ( I - B ) ,
be (I - B ) - 1 . Now
and hence,
(z - A) - 1
00
00
== (I - B) - 1 == L0 z - k - A k == L1 z -k Ak - 1 . z-1
This cqmpletes the proof.
1
D
1 36
6.
SPECTRAL THEORY
Let C be any circle with center at the origin and radius greater than, say, I I A I I · Then , by Lemma 6 . 1 1 , (6. 18) or
1
!c
00
z n (z - A) - 1 dz
=
L Ak- l 1
!c
k= l
z n - k dz
=
2 1f i1ln ,
(6. 19) where the line integral is taken in the right direction. Note that the line integrals are defined in the same way as is done in the theory of functions of a complex variable. The existence of the integrals and their independence of path (so long as the integrands remain analytic) are proved in the same way. Since (z - A) - 1 is analytic on p(A) , we have Theorem 6 . 1 2 . Let C be any closed curve containing a(A) in its interior. Then {6. 1 9) holds.
As a direct consequence of this, we have Theorem 6 . 1 3 . ru ( A)
==
max.-\E u ( A ) 1 ,.\ 1 and I I A n l l 1 / n
�
ru (A) as
n � oo .
E
Proof. Set m == max.-\ E u ( A ) 1 ,.\ 1 , and let > 0 be given . If C is a circle about the origin of radius a == m + E , we have by Theorem 6. 1 2 , 1 n n a M (2 1r a) = M a n + t , IIA ll < 2 1f where M == max I I ( z - A) - 1 1 1 · c 1 This maximum exists because (z - A) - is a continuous function on C. Thus, lim sup I I A n l l 1 / n < a == m + E . \
Since this is true for any E > 0, we have by ( 6. 7) , I 1 1 /n / n n < lim sup I I A l l n < ru (A) == inf I I A l l which gives the theorem .
m
< ru (A) , D
We can now put Lemma 6. 1 1 in the form Theorem 6 . 14. lf l z l > ru (A) then {6. 14) holds with convergence zn B(X). ,
Now let b be any number greater than ru (A) , and let f ( z ) be a complex valued function that is analytic in l z l < b . Thus, 00
(6.20)
f (z)
==
L ak z k , 0
J z l < b.
6. 3.
Operational calculus
We can define
137
f (A) as follows: The operators .
converge 1n norm, since .
L l akl · I Ak l < 00 0
oo .
This last statement follows from the fact that if < < b, then
ra (A) c
1 Ak l 1 1k
for k sufficiently large , and the series
:::;
c
is any number satisfying
c
L l akl ck 00
is convergent . We define
0
f (A) to be 00
(6. 2 1 ) By Theorem 6. 1 2 , this gives
(6. 22) ==
=
k 1 1 a f A) dz ( z z k 2 7r z !c 1 � 1 f(z)( dz, A) z 21rz !c 1
.
o
ra(A) f( z ) lzl
where C is any circle about the origin with radius greater than and less than b. We can now give the formula that we promised. Suppose does not Then is analytic in vanish for == 1/ < b. Set < b, and hence, is defined. Moreover, we shall prove at the end of this section that
g(z) f ( z ) . g ( z ) z l l g (A) 1 1 1 1 ) ( A) � d dz = I . (z g(z)(z f (A)g(A) � f A) z z 21rz !c 21rz !c Since f(A) and g (A) clearly commute, we see that f (A) - 1 exists and equals g (A). Hence, 1 1 (6.23) f (A ) 21ri J1c f (1z) (z - A ) - 1 dz. =
=
=
6. SPECTRAL THEORY
138
In particular , if
g(z) = 1 / f( z ) then
:2: ck z k , 00
.
0
l z l < b,
f (A) - 1 = L ck Ak . 00
(6. 24)
0
f(z)
a(A),
Now, suppose but not is analytic in an open set n containing analytic in a disk of radius greater than In this case, we cannot say that the series (6 . 2 1 ) converges in norm to an operator in B (X) . However, in the following way: There exists an open set w we can still define whose closure w c n and whose boundary 8w consists of a finite number of simple closed curves that do not intersect, and such that C w. ( That such a set always exists is left as an exercise. ) We now define by
ru (A) .
f (A)
1 j A) )( dz, ( z f z f (A) = � la 2 7r 2 w
( 6.25)
a(A) f (A)
where the line integrals are to be taken in the proper directions. It is easily checked that E B (X) and is independent of the choice of the set w. By (6.22) , this definition agrees with the one given above for the case when n contains a disk of radius greater than r A Note that if n is not connected, need not be the same function on different components of n. Now suppose does not vanish on Then we can choose w so that does not vanish on w ( this is also an exercise ) . Thus, == is analytic on an open set containing w so that is defined. Since 1/ == 1 , one would expect that == = I, in which case, it would follow that At the end of exists and is equal to this section , we shall prove
f (A)
f(z)
f( ) ( z ) z) f( ) g(z fz
u( )
a(A).
f (z)
(A) g (A) (A)g(A) g(A) f f g(A). f (A) - 1
Lemma 6 . 1 5 . If and
a(A)
.
g( z)
f(z) and g (z) are analytic in an open set n containing h(z) = f ( z)g( z ),
then h{A) -f(A)g(A). Therefore, it follows that we have Theorem 6 . 1 6 . If A is in B{X) and f(z) is a function analytic in an open exists and set n containing # 0 on then such that is given by 1 1 1 = d , la 7f 2 i w where w is any open set such that
a (A) f (A) - 1
f (z)
a(A), f (A) - 1 1 ) ( A z z f (z)
1 39
6. 3. Operational calculus (a) a (A)
c
c
w, w
n,
{b) ow consists of a finite number of s'lmple closed curves, {c) f (z) # 0 on w. Now that we have defined f (A) for functions analytic in a neighborhood of a(A) , we can show that the spectral mapping theorem holds for such functions as well ( see Theorem 6.8) . We have Theorem 6. 17. If f(z) is analytic in a neighborhood of a (A) , then
(6. 26) 'l. e. ,
J-L
a(f(A) )
_/
==
E a ( f ( A ) ) 'lf and only 'lf f.-L
==
J (a (A) ) ,
j ( A) for some A E
a(A) .
Pro'?f. If f ( A) # f.-L for all A E a (A) , then the function f ( z) - f.-L is analytic in a neighborhood of a(A) and does not vanish there. Hence , f (A) - f.-L has an inverse in B ( X ) , i.e. , J-L E p ( f (A) ) . Conversely, if f.-L == j ( A) for some A E a (A) , set I
g (z)
==
{�
[ (z) - J.-L] / ( z - A) , f (A) ,
z # A, Z - A.
Then g ( z ) is analytic in a neighborhood of a ( A) and g ( z ) ( z - A) == f ( z ) - f.-L· Hence , g ( A) ( A - A) == (A - A ) g (A ) == f ( A) - f.-L· If f.-L were in p ( f ( A) ) , then we would have h ( A) (A - A) == (A - A) h ( A ) == I , where
==
g (A) [f ( A) - J.-L] - 1 . This would mean that A E p(A) , contrary to assumption. Thus f.-L E a(f(A) ) , D and the proof is complete. h (A)
It remains to prove Theorem 6 . 10 and Lemma 6 . 1 5 . For both of these, we shall employ Theorem 6 . 18. If A , f.-L a re 'ln p(A) , then (A - A) - 1 - (J.-L - A) - 1 == (J.-L - A) (A - A) - 1 (f.-L - A) - 1 . ( 6 . 27) Moreo ver, 'lj A - f.-L - A) - 1 < 1 , then
I
I
·
I (Jl
I00
(6.28) 1 and t he series con?)erges zn B{X) .
140
6. SPECTRAL THEORY
Proof. Let x be an arbitrary element o f X , and set u (A - A)u == x and ( J.t - A)u == x + ( J.t - A)u. Hence,
u ==
== (A - A) - 1 x .
Thus,
( J-L - A) - 1 x + (J-L - A) ( J-L - A) - 1 u.
SuDbvttuting for u, we get ( 6.27 ) . Note that it follows from ( 6.27 ) that (A - A) - 1 and (J-L - A) - 1 commute. Substituting for (A - A) - 1 in the right hand side of ( 6. 27) , we get
(A - A) - 1
==
(J-L - A) - 1 + ( J-L - A ) ( J-L - A) - 2 + ( J-L - A ) 2 ( A - A) - 1 (J-L - A) - 2 .
Continuing in this way, we get k
(A - A) - 1 == l: (J-L - A ) n-l (J-L - A) - n + (J-L - A )n (A - A ) - 1 (J-L - A) - n . 1
Since
II ( J-L - A )n ( A - A) - 1 ( J-L - A) - n i l < I J-L - A I n · II ( A - A) - 1 1 1 · II (J-L - A) -l l l n 0 -----*
we see that ( 6 . 28) follows.
as n
---t oo ,
D
The analyticity claimed in Theorem 6 . 10 is precisely the expression ( 6.28 ) of Theorem 6. 18. We can now give the proof of Lemma 6. 1 5 . Proof. Suppose h(z) == f (z)g(z) , where both f (z) and g (z) are analytic in an open set n =:) a (A) . Let WI and w2 be two open sets \ such that a(A) C W I , W I C w2 , W2 C f2 , and whose boundaries consist of a finite number of simple closed curves. Then
1 f (z) (z - A) - 1 g(A)dz 7r !awl == - � 1 f (z) (z - A) - 1 1 g (() (( - A) - 1 d(dz 47r law 1 Jaw2 1 - (( A) - 1 A) (z 1 f (z) 1 g(() d(dz == - � (-z 4 7r faw1 Jaw2 g( ( ) d( = - � 1 f (z) (z - A) - 1 1 dz 4 7r faw1 faw2 ( - z � 1 g (() (( A) - 1 1 f (z) dz d( , 47r Jaw2 faw1 z - (
f(A) g (A) =
_
� 2 z
_
6. 4. Spectral projections
141
by ( 6. 27) . Since ow 1 is in the interior of w2 ,
1 g(() d( = 21rig(z) Jaw2 ( - z
for z on aw l ' while for ( on ow 2 . Hence,
f(A)g(A) =
1 f (z) dz = 0 !awl Z - (
1 � 2 z
This completes the proof.
7r !awl
f (z)g (z) (z - A) - 1 dz == h(A) . 0
6 .4 . Spectral proj ect ions
In the previous section, we saw that we can define f (A) whenever A E B(X) and f ( z) is analytic i n some open set n containing a (A) . If a (A) is not connected, then n need not be connected, and f(z) need not be the same on different components of n . This leads to some very interesting consequences . For example , suppose a1 and a2 are twQ._subsets of a(A) such that a( A) == Cf l U Cf2 , and there are open sets f2 1 ::J C! I and f2 2 :) 0"2 s.uch that f2 1 and f2 2 do not intersect . Such sets are called spectral sets of A'. We can then take f(z) to be one function on 0 1 and another on n� , and j (A) is perfectly well defined . In particular, we can take f ( z) == 1 on !1 1 and f ( z) == 0 on f2 2 . We set P == f (A) . Thus, if w is an open set containing a1 such that w c 0 1 and such that ow consists of a finite number of simple closed curves , then
(6.29)
P=
1 (z - A) - 1 d z. 7r Jaw
� 2 z
CleaTly, P2 == P (Lemma 6. 1 5 ) . Any operator having this property is called a proJectzon. For any projection P, x E R(P) if and only if x == Px . For if x == Pz , then Px == P 2 z == Pz == x . Thus, N(P) n R(P) == { 0 } . Moreover,
X == R(P) EB N(P) , for if x E X , then Px E R(P) , and (I - P) x E N(P) , since P(I - P)x == (P - P 2 )x == 0. In our case, P has the additional property of being in B(X) . Thus, N (P) and R (P) are both closed subspaces of X. Now, A maps N (P) and R(P) into themselves . For if P x == 0, then P Ax == APx == 0. Similarly, if Px == x, then PAx == AP x == Ax. Let A 1 be the restriction of A to R (P) and A 2 its restriction to N (P) . Thus, we can consider \A split into the "sum" of A 1 and A 2 . Moreover, if we consider A 1 as an operator on R( P) and A 2 as an operator on N ( P} , we have Theorem 6 . 1 9 . a (Ai ) == ai , i == 1 , 2 . (6.30)
6. SPECTRAL THEORY
142
Proof. Let JL be any point not in a 1 . Set g(z) f (z) / (JL - z) , where f (z) is identically one on a 1 and vanishes on a2 . Then g(z) is analytic in a neighborhood of a (A) and fg P and g . Hence, ( JL - A)g(A) g(A) . Thus , g(A) maps R(P) into itself, and its restriction to Pg (A) R( P) is the inverse of JL - A 1 . Hence, JL E p( A 1 ) . Since I - P is also a projection and R(I - P) N(P) , N(I - P) R(P) , we see, by the sa1ne reasoning, that if JL is not in a 2 , then it is in p(A 2 ) . Hence, a (A i ) C O"i for i 1 , 2. Now, if JL E p(A 1 ) n p(A 2 ) , then it is in p(A) . In fact , we have ==
==
==
==
=
(JL - A) - 1
=
==
==
(JL - A 1 ) - 1 P +
(f.-l - A2 ) - 1 (I - P) .
Hence, the points of a 1 must be in a (A 1 ) and those of a 2 must be in a(A 2 ) . D This completes the proof. Next, suppose that a 1 consists of j ust one isolated point A 1 . Then a(A 1 ) consists of precisely the point A 1 . Hence, ra (AI - A 1 ) 0 , i .e. , ==
(6 . 3 1 )
( Theorem 6. 13) . In particular, (6.32) for all x E R(P) . Conversely, we note that every x E X which satisfies (6. 32) is in R(P) . Suppose (6.32) holds. Then n 0, II (A - A 1 ) n (I - P)x ll l l �
since I - P com1nutes with A - A 1 and is bounded. Set Then But A 1 E p(A 2 ) , and hence,
(I - P)x
or
11
I
==
(A 2 -
J
A l ) - nzn .
I
I I (I - P) X I n < (A 2 - A 1 ) - l I I II Zn 1 I n Hence, (I - P) x 0 � showi ng that x E R(P) . ·
�
0·
==
\iVe call P the spectral projectzon associated with a1 . As an application, we have
143
6.4. Spectral projections
Theorem 6.20. Suppose that A is in B{X) and that AI is an isolated point of a (A) {i. e. , the set consisting of the point AI is a spectral set of A). If R(A - AI ) is closed and r (A - AI ) < oo {see Section 5. 5), then AI E A . Proof. We must show that j3(A - AI ) < associated with the point A I , and set
oo .
Let P be the spectral projection
00
No == UI N [(A - AI ) n ] . By hypothesis, No is finite-dimensional. We must show that this implies that No == R(P) . Once this is known, the theorem follows easily. For, by ( 6.3 0 )
(6.33)
'
X == No EB N(P) ,
and if A 2 denotes the restriction of A to N( P) , then AI E p(A2) ( Theorem 6. 19) . In particular , this means that R(A 2 - A I ) == N(P) , and since R(A AI ) :J R( A2 - AI ) , we see that N [ (A - AI )'] c N ( P) 0 • This latter set is finite-dimensional. For if x � , , x � are linearly independent elements of , Xn E X such that N ( P ) 0 , then there are elements X I , ·
·
·
·
·
·
xj ( xk ) == §j k , 1 < j, k < n ( Lemma 4. 14) . Clearly, the Xk are linearly independent , and if M is the n-dimensional subspace spanned by them, then M n N (P) == {0} ( see the proof of Lemma 5.3) . Thus n < dim No ( Lemma 5 . 12) , and consequently, ,6(A - AI ) < 00 . To prove that N0 == R(P) , we first note that N0 c R( P) , since x E R( P ) if and only i f it satisfies (6.32) . Now, suppose No is not all of R( P) Set
Xo
=
R(P) /No and define the operator B on Xo by means of
B [x]
==
[(A - AI )x] ,
.
x E R( P ) ,
where [x] is any coset in X0 ( see Section 3.5) . Clearly, B is one-to-one on Xo , and B n [X] == [ (A - A I ) n X] , n == 1 , 2 , · · . If we can show that the range of B is closed in X0 , it will follow that there is a constant c > 0 such that ·
I I [x] I I < c ii B [x] II ,
x E R(P)
( Theorem 3. 12) . Consequently, we shall have I t[x} II < en II [ (A - A I ) n X] II < en II (A - A I ) n X II , n == 1 , 2, · · · . Now, if x is an element in R( P) which is not in No , then II [ x] I I # 0. Hence, lim inf I I (A - AI ) n x ll l /n > 1 / c > 0, which contradicts (6.32) . This shows that there is no such x .
144
6. SPECTRAL THEORY
Thus , it remains only to show that R(B) is closed in Xo . To this end , we employ a simple lemma.
Lemma 6 . 2 1 . If X1 and X2 are subspaces of a normed vector space X, let X I + X2 denote the set of all sums of the form X I + X 2 , where Xi E Xi , i == 1 , 2 . If XI is closed and X2 is finite-dzmensional, then X I + X2 is a closed subspace of X.
Returning to the proof of Theorem 6.20, let AI be the restriction of A to R(P) . Then R(AI - )q ) == R( A - )q ) n R(P) . For, if y E R(A - A 1 ) n R(P) , then y == (A - Al ) x for some x E X, and Py == y. Thus, ( A - AI )P x == P(A - A l )x == Py == y, showing that y E R(AI - AI ) · In particular, vve see from this that R(AI - A I ) is closed in R(P) . Thus , by Lemma 6.21 , the same is true of R(AI - AI ) + No . Now all we need is the simple observation that R(B) == [R(AI - AI ) + No ] /N0 . This latter set is closed ( i.e. , a Banach space ) by Theorem 3. 1 3 . Thus, the proof of Theorem 6.20 will be complete D once we have given the simple proof of Lemma 6 . 2 1 . Proof. Set X3 == XI n X2 Since X3 is finite-dimensional, there is a closed subspace X4 of X1 such that X I == X3 EBX4 ( Lemma 5. 1 ) . Clearly, X I + X2 == D X4 EB X2 , and the latter is closed by Lemma 5 . 2. .
As a consequence of Theorem 6.20 we have Corollary 6.22. Under the hypotheses of Theorem 6. 20, there are operators E in B(X) and K in K(X) and an integer m > 0 such that
(6.34)
(A - AI ) m E == E(A - AI ) m == I - K.
Proof. By Theorem 6 20,
( X , Y) with N(A) and R (A) unchanged. Proof. Let P be the operator embedding W into X. Let Q be the linear operator from X to W with D (Q) == R(P) defined by Q x == w when x == Pw. Since P is one-to-one, Q is well defined. Moreover, one checks easily that Q E ( X W) . Hence, by Theorem 7. 3, AQ E ( X Y) . But in the new D terminology, AQ is the same as A . This completes the proof. ,
,
We now give the proof of Lemma 7.4. Proof. By Lemma 5.3, there is an n-dimensional subspace V of X such that X == R tfJ V. Let XI , · · · , Xn be a basis for V. Then by Lemma 5.2,
(7. 12)
ll xo l l +
n
n
1
1
L l nk l < C l l xo + L nk xk ll ,
xo E R, nk scalars,
where we �ave made use of the fact that all norms are equivalent on sub spaces that are finite-dimensional (Theorem 4.2) . Now by the density of D, there is for each k an element E D such that
Xk
Thus,
ll xo ll + or
(7. 13) The
2 ll xo ll +
n
n
1
1
n
L i aki < C ll xo + L ak:hi i + 2 L i aki ,
n
n
1
1
L i aki < 2 C II xo + L ak:hi i ,
Xk are linearly independent, for if
1
1
xo E R,
ak scalars.
161
7. 2. Further properties
then each o: k == 0 by (7. 13) . Let W be the n-dimensional subspace spanned by the x k . Then W n R = {0} . For if x E R and
then, by (7. 1 3) ,
/
n
ll x l l + L l o:k l < 2 C II x - x ll = 0. 1 We claim that X == R EB W. To see this, note that if x is an element of X not in R EB W, let W1 be the subspace spanned by x and the x k . Then W1 n R = {0} , and by Lemma 5. 1 2 , dim W1 < dim V = n, which pr-ovides a contradiction. This completes the proof.
0
7 . 2 . Further prop erties
We now prove the counterparts of Theorems 5. 10 and 5. 1 1 . First we have Theorem 7. 8. If A E (X, Y) and K is in K(X, Y), then A + K E (X, Y)
and
i (A + K) = i (A) .
(7. 14)
Proof. The fact that A + K E ( X, Y) follows from Theorems 7. 1 and 7.2 as before. To prove (7. 14) , we need a trick. Since A is closed , one can make . D (A) into a Banach space W by equipping it with the graph norm,
ll x ii D ( A ) = ll x ll + l f A x ll · Moreover, W is continuously embedded in X and is dense in X . Hence, A E ( W, Y) ( Lemma 7.7) . In addition, an operator A0 , given by Theorem 7. 1 , is in
(D(BA) , V) ( Theorem 7.3 ) . Moreover, we know that A E B (D (A) , Y) and F3 E B (Y, N(B) ) . Consequently, F3 A E B (D (A) , N(B) ) , and it is compact on D (BA) . Thus , the restriction of A to D ( BA) is in 0 (D (BA) , V) . By Lemma 7.7, it is in
.
< <
<
<
Let F+ (X) denote the set of all E E B (X) such that A + E E
+ . It is called the set of semi-Fredholm perturbations. We have Corollary 9.45 . If E1 , E2 E F+ (X) , t hen E1 + E2 E F+ (X) . Proof. Just use the definition.
D
We also have Theorem 9.46. E E F+ (X) if and only if o:(A - E)
<
oo
for all A E <1>+ .
Proof. If E E F+ (X) and A E <1> + , then A - E E + by definition. Hence, o:(A - E) oo . If A E + and A -. E fl. <1> + , then there is a K E K(X) such that o:(A - E - K) == oo ( Theorem 9.43) . Set B == A - K. Then B E + while o:(B - E) == oo . Thus , E fl. F+ (X) . This proves the theorem. D
<
Theorem 9.47. E is in F(X) if and only if o:(A - E)
<
oo
for all A E <1> .
Proof. If E E F(X) and A E <1> , then A - E E
. Let A be o:(A - E) oo . Conversely, suppose o:(A - E) any particular operator in
for each K E K(X) and ,.\ # 0. Hence, o:(A - ,.\E - K ) oo for all ,.\ and for all K E K(X) . By Theorem 9.43 , A - ,.\E E
, the result follows. Corollary 9.58. F_ ( X)
C
F (X ) .
Lemma 9.59. If Ek E F_ ( X ) and Ek
--4
E, then E E F_ ( X) .
Proof. Use the same reasoning as in the proof of Corollary 9.38. Use The 0 orem 5 . 23 in place of Theorem 5. 1 1 .
We also have Lemma 9.60. If E E F_ (X ) , then, AE an d EA are in F_ ( X ) for all A <1>.
E
9. BANACH AL GEBRAS
222
Proof. If A E
_ , then E + AoC E _ . Thus, A(E + Ao C ) E _ together with AE + C . This means that AE E F_ (X ) . A similar argument works for EA. 0 Lemma 9.61 . If E E F_ (X) , then BE and EB are in F_ (X) for all B E
B (X) .
0
Proof. See the proof of Corollary 9.35. Theorem 9.62. F_ (X) is a closed two-sided ideal. Proof. See the proofs of Lemmas 9.49 and 9.5 1 .
D
9 . 9 . Remarks
We note that R(X) is not an ideal, and we see from T]leorem 9.39 that }, ( X) is the largest ideal contained in R(X) . Moreover, operators in R(X) are characterized by the fact that each of them behaves like a Fredholm perturbation with respect to Fredholm operators which almost commute with it (Theorem 9.29) . Theorem 9.43 says that an operator in + cannot coincide with a com pact operator on any infinite dimensional subspace, and that this property characterizes these operators . Theorem 9 .46 says that an operator is in F+ (X) if and only if it does not coincide with a + operator on any infinite dimensional subspace. Theorem 9.47 makes a similar statement for the set F(X) . Theorem 9.54 states that A E _ if and only if A' does not coin cide with the adjoint of any compact operator on an infinite dimensional subspace of X' . 9 . 1 0 . Problems
( 1 ) Prove
(2) Let {a n } be a sequence of elements of a Banach algebra such that an 1 exists for each n and an � a. Show that either (i) a - 1 exists or (ii) there is a sequence {bn } such that ll bn ll 1 and abn � 0. =
9. 1 0. Problems
223
(3) If (AI , , An ) E show that ·
·
·
a
(a1 ,
·
·
·
, an ) and P( ZI ,
·
·
·
, Zn ) is a polyno1nial,
( 4) Let B be a Banach algebra with unit . Show that for each ao E B and each � > 0, there is a 6 > 0 such that a - ao < 8 implies that a( a) lies in the set of all A satisfying a(ao)) < £ .
I d(.X,
I
(5) If B is a Banach algebra without a unit element , consider the set C of pairs ( a, o: ) , where a E B and o: is a scalar. Define
(a, a ) + ( b, /3 ) == (a + b, a + /3 ) (3(a, a ) == (/3a, f3a ) (a, a) (b, /3 ) = (ab + ab + {3a, a/3 ) , t
and '
I (a, n) I I a ll + I a I . =
Show that C is a Banach algebra with unit element (0, 1) . (6) Prove Lemma 9 . 5 l . and Corollary 9.52. ( 7) Prove Lemma 9.61 and Corollary 9.62.
Chapter
10
SEMI GROUPS
10. 1 . A differential equation
Suppose A and uo are given numbers, and we want to solve the equation ( 10. 1 )
du = Au, dt
t > 0,
with the initial condition ( 10.2)
u (O) = uo
for a function u( t) in t > 0. The answer is ( 10.3)
u( t) = u0etA .
Now we want to put the problem in a more general framework. Let X be a Banach space. We can define functions of a real variable t with values in X. This means that to each value t in the "domain" of the function u, we assign an element u(t) E X. We say that u (t) is continuous at to if for each e € > 0 there is a 6 > 0 such that I t - t o l < 6 implies that l l u(t) - u(to) ll < E. We say that u(t) is differentiable at to if there is an element v E X such that for each E > 0, there is a 6 > 0 such that I t - to I < 6 implies
u(t) - u(to) - v < E. t - to
----
In such a case, we denote v by u' (t 0 ) or du(t o ) /dt. Of course, a function differentiable at t o is continuous there. To complete the picture, let A be any operator in B (X) , and let uo be any element of X. Then the equations ( 10.4)
u' (t) = Au,
t > 0, 225
226
1 0.
SEMIGROUPS
and u ( O ) = uo
( 1 0 .5)
make sense. So we have created a differential problem in a Banach space. Does it have a solution? Our first instinct is to look at ( 10.3) . However, it does not seem to make much sense within the present framework. First , is an operator and, hence, must operate on something. Second, what interpretation can we give to e t A ? Let us tackle the second point first. When is a number,
A A
00
'"' -k1' tk Ak ' and the series converges for each finite A. If the convergence is in norm, then the series in ( 10.6) makes sense for an operator A E B (X) . And, indeed, it etA =
( 1 0.6)
� 0
•
does converge in norm. This follows immediately from the fact that the . series
converges, and hence,
n 1kk n 1 k k L k ! t A < L k ! l t l i A I � 0 as m
m
m, n
�
oo .
Thus we can define e t A by means of ( 1 0.6) . With this definition, e tA is in B(X) , and e t A < e l t i iiA II ( 10. 7) ' . s1nce
I I
n kk L ;, t A < L ;, l t l k i A I k = e l t i iiA II . oo
0
.
0
.
Once we have defined et A , we can resolve the other difficulty by replacing ( 10.3) by u ( = e t A uo . ( 10.8)
t)
Now, the question is whether or not ( 10.8) is a solution of ( 10.4) and ( 10.5) . In order to answer this question, we must try to differentiate ( 10.8) . Thus u ( + h) - u ( ) ( 10.9)
t
h
t
"Wait a minute!" you exclaim. "The property e B+C = e 8 e0 ( 10. 10)
A differential equation
1 0. 1 .
227
is well known for numbers, but far from obvious for operators." True, but let us assume for the moment that ( 10. 10) does apply to operators in == that We shall prove this at the end ( i.e. , that satisfy of the section. = Since we suspect that let us examine the expression e hA _ J _A ==
commute
B(X)
BC CB). u'(t) Au(t), ] u. u(t + h) - u (t) _ Au [ h h e hA - I == " _!_ h k - 1 A k h 1 k '.
Now,
00
'
�
and hence, e hA I
_ - A < �00 1! l h l k - l i A I k = el h i ii A II _ - I A I � 0 as h � 0. k h l hl This shows that u'( t ) exists for all t and equals Au(t). Clearly, ( 10 . 5) holds. 1
Hence, ( 10.8) is indeed a solution of ( 1 0.4) and ( 10.5) . Now, let us give the proof of ( 10. 10) for operators in Proof. Since
B(X) that commute.
B, C commute, k ( B + C) k = m"= m .' (kk_.' m ) ,' Bm ck- m . �
0
Thus,
n m m m k = B B . L �! (B + C) k = L c L c L � ! ! ! � n m (k m m) m+n k=O m=O N
N
k
0
Hence,
( 10. 1 1 )
Moreover, the norm of this expression is not greater than m,n�N m+ n > N
N
k B ) ( C + L I � I I I k=BO !ii CII B II + IICII ii U
�� e
e
_
eii
== 0
228
1 0.
as N --+
oo .
SEMIGRO UPS
But the left hand side of ( 10. 1 1 ) converges in norm to
eBeC
_
e B+C
'
and, hence, ( 10. 10 ) holds. This completes the proof.
D
1 0 . 2 . Uniqueness
Now that we have solved ( 10.4) and ( 10.5 ) , we may question whether the solution is unique. The answer is yes, but we must reason carefully. Suppose there were two solutions. Then their difference would satisfy ( 10. 12)
u' (t) == Au,
t > 0,
u(O) == 0.
( 10. 13) In particular, we would have ( 10. 14)
e - tA (u' - Au) == 0.
Now, if A were a number, this would reduce to
( 10. 15 )
(e - t A u) ' == 0 ,
t > 0.
However, in the present case , care must be exercised. Set v(t) == e - t A u. Then
e - ( t + h ) A u(t + h ) - e - t A u(t) h - hA t) h e t u( ) + u( + h A ( ) t + v (t) , eh h and this converges to the left hand side of ( 10. 14 ) as h --+ 0, since e - hA - I --+ h in norm as h --+ 0 . Once we have ( 1 0. 15 ) , we expect that it implies, in view of ( 10. 1 3 ) , that v( t + h ) - v(t) h
] [
[
---
-
I]
A
( 10. 1 6 )
To see this, let f be any element of X', and set ( 10. 1 7 )
F(t) = f (e - t A u) .
One easily checks that F(t) is differentiable in t > 0 , continuous in t > 0 and satisfies F' ( t) = 0 , t > 0 ,
and
F(O) == 0 .
1 0. 3.
229
Unbounded operators
Since F is a scalar function, this implies th.at F vanishes identically. bince this is true for any f E X' , we see that ( 1 0. 16) does indeed hold. To complete the proof, we merely note that
u(t) = etA ( e tAu) -
=
0,
t > 0.
to < t < t 1 with values in X, we can 1t1 u( s) ds = lim Ln u (s� )(si - si 1 ), n�o 1 t where to = so < s 1 < · · · < S n t 1 , = max l si- - Si - l l , and s� is any number satisfying S i - 1 < s� < S i · When u(t) is continuous , the existence of the integral is proved in the same way as in the scalar case. We also have the estimate tl tl i. i. d d ( 10. 18) ( s) s ( s) s. < u u I I to to The existence of the integral on the right follows from the continuity of l u (s) l , which in turn follows frorn ( 10. 19) l l u (s) l - l u (t) l t I < l u (s) - u (t) l · Moreover, the function U (t) = iftto u(s)ds is differentiable in t o < t < t 1 , and U '(t) = u(t). Again, the proof is the same as in the scalar case. In particular, if u ' ( t) is continuous in to < t < t 1 , then ( 10.20) u(t) - u (to) = irto u' (s)ds, to < t < t l . This follows from the fact that both sides of ( 10.20) are equal at t 0 , and their derivatives are equal for all t in the interval. u(t)
If is a continuous function in define the Rie1nann integral 0
=
1 0 . 3 . Unbounded operators
rJ
A
In Section 10. 1 , we solved ( 10.4) and ( 10.5) for the case E B (X) . Suppose, as is more usual in applications, is not bounded, but rather a closed linear operator on X with domain D(A) dense in X. Can we solve ( 10.4) , ( 10.5)? First , we must examine ( 10.4) a bit more closely. Since D (A) need not be the whole of X, we must require to be in D(A) for each > 0 in order that it be a solution of ( 1 0.4) . With this interpretation, ( 10.4) makes sense.
A
u(t)
t
-
230
1 0. SEMIGRO UPS
Many sets of sufficient conditions are known for ( 10.4) , ( 10.5) to have- a solution. We shall consider one set . We have Theorem 1 0 . 1 . Let A be a closed linear operator with dense domain D(A) on X having the interval [b, oo ) in its resolvent set p(A) , where b 0 , and
>
such that there is a constant a satisfying ( 10.21)
Then there is a family {Et } of operators in properties:
( a) (b) (C) (d) (e)
B ( X ) , t > 0 , with the following
> > > > >
Es Et = Es + t , s 0 , t 0, Eo = I, e-at , t 0, Et Etx is continuous in t 0 for each x E X, Etx is differentiable in t 0 for each x E D(A) , and
I I<
dEt x AEt X, ( 10. 22) dt -Et (>.. - A) - 1 = (>.. - A) - 1 Et , >.. b, t (f )
> > 0.
Before proving the theorem , we show how it gives the solution to our problem provided uo E D (A) . In fact , ( 10.23)
u (t) = Etuo ,
t > 0,
is a solution of ( 10.4) and ( 1 0.5) . To see this , note that Et maps D (A) into itself. The reas on for this is that if v E D ( A ) , then by (f) , Etv = Et (b - A) - 1 ( b - A)v = b A) - 1 Et (b - A)v,
( -
which is clearly an element of D (A) . Thus, if uo E D (A) , so is u( t ) for each 0. By ( b ) , we see that ( 10.5) is satisfied, while ( 10.22) implies ( 10.4) .
t>
A one-parameter family {Et } of operators satisfying ( a ) and ( b ) is called a semigroup. The operator A is called its infinitesimal generator. In proving Theorem 10. 1 we shall make use of Lemma 10.2. Let D be a dense set in X , and let
operators in B (X) satisfying
{BA} be a family of
( 10.24)
If BAx converges as >.. that ( 10.25)
� oo
for each x E D, then there is a
IBI <M
B in B ( X ) such
231
1 0. 3. Unbounded operators
and ( 10.26)
>x
B>.. x Bx --4
as
A
x
--4 oo ,
E
X.
x be any element of X. Then we can l x - XI I < 3�
0 be given, and let Proof. Let c find an element E D such that ( 10. 27)
·
Thus,
I B>.. x - BJLx l < I B>.. ( x- x) I + I B>.. x - Bp,x i + I Bp, (x - x) l < �; + I B.x X - Bp,x l · We now take A, so large that I B>.. X - BILXI I < �. Thus Bx lim B>.. X exists for each x E X. Clearly, B is a linear operator. Moreover, H Bx l = lim I B>..x l < M l x J I , JL
,
and the proof is complete.
D
We are now ready for the proof of Theorem 10. 1 . Proof. Assume first that a
( 1 0 .28)
> 0. Set
We are going to show that
A>.. E B(X ), A > b, ( atA ) t > 0 , A > b, (2) l e t A>- 1 < exp a+A ' (3) x E D(A) , (4) X E X, t > 0. The last assertion states that for each t > 0 and each x E X , e tA >. x converges to a limit in X as A � We define this limit to be Et x. Clearly, Et is a linear operator. It is in B( X ) by (2) and Lemma 10.2. Moreover, I Et x l < l etA>.x l + I (Et - etA>.)x l < exp ( ��� ) l x l + I (Et - etA>.)x l · Letting A we get (c) . We now show ttiat (a) - (f) follow from ( 1 ) (4) . We have just seen that (2) implies (c) . To obtain (a) , note that II Es+ t X - .fis Et x ll < I (Es + t e ( s+ t ) A >. )x l
(1)
oo .
� oo ,
232
1 0. SEMIGROUPS
+ l esA>. (etA>. - Et )x l + l (esA>. - Es )Et x l · By (2) and (4) , the right hand side tends to 0 ,.\ � To prove (d) , note that etA>. x - etoA>. x ifto\ esA>. x ) 'ds itto esA>. A.xxds by ( 10. 20) . Hence, t t l etA>.x - etoA>.x l < irto l eSA>.A.xx l ds < irto I A.xx l ds (t - to) I AAx l · Thus , we have by (3) and (4) Assume that x E D(A), and let ,.\ � I Et x - Eto x l < (t - to) I Ax l · oo .
as
=
=
=
oo .
This shows that ( 10.29 )
Et x � Et0 X t � to, x E D(A). as
By (c) and Lemma 10. 2 , this implies (d) . Moreover, once we know that is continuous in we can take the limits in the integrals to obtain
t,
Et x
t Et x - Et0 X it{o Es Axds, x· E D(A). ( 10.30 ) In partic ular, ( 10.30) implies that Et x is differen\iable, and d Et x - Et AX, X E D (A) . ( 10. 3 1 ) dt =
However, it follows from (f) that
AEt X Et AX, X E D(A). In fact , we have AEt x bEt x - (b - A)Et (b - A) - 1 (b - A)x bEt x - Et (b - A)x Et Ax. =
( 1 0. 32 )
==
=
=
This gives (e) . (Yes, I know that I have not proved (f) yet. I just wanted to see if you would notice. As you will see, its proof will not involve (e) .) To prove (f) , note that
L �! tk A� (A - A) - 1 x 1 N
=
for each N, and hence,
( 10.33 ) Taking the limit as ,.\ �
oo ,
we get (f) .
A) - 1 L �! tk A� x , 1 N
(,.\ -
1 0. 3.
Unbounded operators
233
It only remains to prove ( 1 ) - (4) . Statement ( 1 ) follows from the fact that ( 10.34) and hence , ( 10.3 5 ) Thus. ( 10.36) so that ( 10.37)
el tA>. I :::; -t>. exp ( at�2,\) = exp (::�) < 1 , e
which is (2) . Now by ( 10.34) ,
I A( .X - -A) - 1 1 < 1 + a � A < 2, while A x I 1 (10.39) < (.X I A - A) x l - (a + .X)l -+ 0 as A -+ oo , x E D(A). Therefore, it follows from Lemma 10.2 that (10.40) A(.X - A) - 1 x � 0 as .X � oo , x E X. In view of ( 10.34) , we get .X(.X - A) - 1 x � x as .X � oo , x E X. ( 1 0 .4 1 ) Consequently, .X( ,\ - A) - 1 Ax -+ Ax as .X � oo , x E D(A). (10.38)
This gives (3) . To obtain (4) , let A, J-L be any two positive numbers, and set
Vs == exp [stAA + ( 1 - s)tA1J . If v(s) = V8 x, then v' ( s ) = t (A A - A ) v ( s) . Consequently, 1 v( 1) - v(O) = t(AA - Ap.) 1 v(s) ds by ( 1 0.20) . This means 1 ( )x = t 1 V8 (A>. - Ap. )xds . 11
et A ,\ - et A �'
Now
234
1 0. SEMIGROUPS
so that
IJ V., If
<
( 1 - s)tJL] exp [ ( :�: ) c 1 ::);JL2 ) ] exp [ (:�a: ) c1:;);aJL)] exp ( - sU -
+
< L
-
Hence,
f l (e tA>. - e tA�-< ) x ll
< t 11 ds i ( A>. - AJL ) x ! l , x E X.
x E D (A) , then this implies that ( etA>. - etA�L )x � 0 .A, 11 � oo in view of (3) . Thus e t A>. x approaches a limit .X � oo for each x E D(A). Denote this limit by Et x, and apply Lemma 10 . 2 . This completes the proof when a > 0. Now suppose that a < 0 . Set B == A + a - 1 . Then p ( B) contains an interval of the form (b 1 , oo) for some b 1 0 , and Now if
as
-as
2:
By what has already been proved, there is a family B (X) sati sfying ( a ) , ( b ) , ( d ) , ( f ) ,
I and
{ Et } of operators in
El t 11 < e- t ,
dEtx _- BEt X, E D (A), t > 0 . __dt_ t - et- at t , t > 0. X
Set
F,
_
E
Then and
dFt x = (1 - a)Ft x + et -at BEtX == ( B + 1 a)Ft x == AFt x . dt Thus , the family {Ft } satisfies all of the requirements of the theorem. This completes the proof. D -
1 0. 4 .
235
The infinitesimal generator
10 .4. The infinitesimal generator
As noted before, a one-parameter family is called a semigroup if
{Et } , t > 0, of operators in B (.X)
( 10.42) It is called strongly continuous if
Et x is continuous in t > 0 for each _ x E X. A linear operator A in X is said to be an infinitesimal generator of { Et } if it is closed, D(A) is dense in X and consists of those x E X for which (Et x) ' exists, t > 0 , and ( 1 0 .44) (Etx)' AEt x, x E D (A). Clearly, an infinitesimal generator is unique. Theorem 10. 1 states that every closed linear operator A satisfying ( 10. 2 1 ) is an infinitesimal generator of a one-parameter, strongly continuous semigroup {Et } satisfying (c) and (f) . ( 10.43)
=
We now consider the opposite situation. Suppose we are given a o �e parameter, strongly continuous semigroup. Does it have an infinitesimal generator? The answer is given by Theorem 10.3. Every strongly continuous, one-param·e ter semigroup of operators in B (X ) has an infinitesimal generator.
1 Xs s 0 Etxdt, X Then W is dense in X, since Xs Eox x s 0. Set ( 10.45) Ah Eh - I , h > 0.
Proof. Let W be the set of all elements of X of the form 1 8 = E X , s > 0. I
-
---+-
=
Then ( 10.46)
=
as
----+-
h
AhXs (Eh - I)shJ; Et xdt J;+h Et xdt - J; Etxdt sh fss+h Et xdt - J0h Etxdt sh =
_
_
{Et}
1 0. SEMIGROUPS
236
T hus,
( 10 . 4 7) T his shows that the set
D of those x E X forwhich Ahx converges E
as
h�0
Ax to be the limit of Ahx as h � 0 . Clearly, A is a linear operator, and D(A) = D is dense in X. It remains to show that A is closed and that ( 10. 44 ) holds. The latter i s si mple , since Et +h X - EtX ( 10. 48) = EtA h x = A h Etx. contains
W and
hence, is dense in
X. f:or x
D,
we define
h
shows that Etx E D (A) for x E D(A) and, hence, (Etx)' = EtAX = AEtx. To prove that A is closed, we make use of the facts that for each s > 0 we have T his
( 10. 4 9)
Ms
=
sup O� t < s
I Etf l < oo
and
( 10.50) Assuming these
A8z = Azs for
the moment ,
=
(Az)8,
we note
z
E
D (A) .
that
( 10.5 1) Now suppose {x ( n ) } is a sequence of elements in D(A) s·uch that x ( n ) � x, Ax ( n ) � y in X as n � oo . ( 10. 52) -
By (10.5 1 ) , this i mplies (Ax (n ) ) s
Ys as n � oo . But by ( 10.50) , (Ax (n ) ) s = A8X (n ) , �
showing that ( 10.53) But Hence ( 10.54) from which it follows that A8x --4 y as s � 0. T hus x E D (A) and Ax = y by the definition of A. This shows that A is a closed operator. Therefore, it remains only to prove ( 10.49) and ( 10.50) . If ( 10.49) were not true, there would be a sequence t k � Jo , 0 < t k < s, such that oo . By ( 10.43) ,
I Etk l i --t
( 10.55)
Etk X
�
Et0 X as k � oo ,
x E X.
1 0. 4.
237
The infinitesimal generator
In particular, sup II Et k x ll < oo ,
( 10. 56)
k
x E X.
But this contradicts the Banach-Steinhaus theorem ( Theorem 3. 17) , which says that ( 10. 56) implies that there is a constant M < oo such that ( 10.57) Thus ( 10.49) holds. To prove ( 1 0. 50) , note that by ( 10.46) ,
A sZh == A h Zs == (A h z) s ·
( 10.58)
Letting h --+ 0, we obtain ( 10.50) in view of ( 10. 5 1 ) . This completes the proof of Theorem 10.3. 0 In Theorem 10. 1 the assumptions on A seem rather special. Are they nec essary? The next theorem shows that they are. Theorem 10.4. If the family { Et } satisfies (a) -
generator A satisfies {1 0. 21).
(d), then its infinitesimal
Proof. We first note that for any s > 0 and ,.\ > 0 , we have ,.\ E p(A 8 ) and s 1 (1 0.59) < II (,.\ - A s ) - 1 II < ,.\ ' 1 + As _ e - a s where A s is defined by ( 10.45) . In fact,
,.\ _ A s =
s
(As - Es + I)
==
(,.\s
+ 1 ) [1 -s( Es /,.\s + 1 )] .
Since I I Es l l < 1 by ( c ) , this operator is invertible for ,.\ > 0, by Theorem 1 . 1 , and a simple calculation gives ( 10.59) . In particular, we have
s e - a s ) ll x ll A + ( 10.60) x E X. 1 1 ( ,.\ - A s ) x ll > ' s 0, obtaining by ! ' Hospital ' s rule If x E D (A) , we can take the limit as s (1
--+
I I (,.\ - A) x l l > (,.\ + a) ll x l l , x E D ( A ) . This shows that ,.\ - A has an inverse for ,.\ > 0 and that its range is closed. If we can show that its range is also dense in X , it will follow that it is the whole of X , and ( 10 . 2 1 ) will follow from ( 10.61 ) . Thus it remains to show that R(,.\ - A) is dense in X. To do this, we shall show that it contains D (A) . (10. 6 1 )
In other words, we want to show that we can solve (10.62)
for x E D(A) provided (10.63)
y
(,.\ - A) x == y
E D( A ) . Set
238
1 0. SEMIGROUPS
x(s) E D (A) . This follows from the fact that (A - As ) - 1 Ah y Ah (A - As ) - 1 y . (10. 64) Since A h y converges as h --+ 0 , then the same is true of A h (A - A s )- 1 y . In view of (10. 5 9). , we see by (10. 6 3) that x ( s ) converges in X as s --+ 0. In fact , we have I [ (A - As ) - 1 - (A - At ) - 1 ] YI I I (A-2 - As ) - 1 (A - At ) - 1 (At - As ) YI I < A 1 (A t - A s ) YI I --+ as s , t Moreover, (A A) y l i ---+ 0 as S ---+ 0. I s � s 1 _ ( < A) (A ) (A-A)x (A-A ) YI I I � s s - y l i I Thus (A - A)x( s ) as x --+ 0. Let x be the limit of x( s ) . Then Ax( s ) y AX - y. Since A is a closed operator, x E D( A) and Ax AX - y . Thus, x is a solution of (10. 6 2). This shows that R(A - A) D(A), and the proof is 0 We claim that
= I
o
--+
o.
=
--+
=:)
complete.
=
--+
10 . 5 . An approximation theorem
{et8}_,
We shall show how to approximate a semigroup by a family of the form where B E B(X) . We have
{ Et }
Theorem 10.5. Let be a strongly continuous, one-parameter semi group, and let be defined by {1 0. 4 5). Then
Ah
(10. 65)
Moreover, for each interval.
x E X, the convergence is uniform for t in a bounded
Proof. We shall verify that
Ah
Et
(i) : commutes with for each t > 0 , (ii) : for each s > there is an Ns such that
0, ( 10. 66) l etAh I < N8 , 0 < t < s, 0 < h < 1, (iii) : A h x --+ Ax as h 0 , x E D(A), where A is the infiliitesimal generator of { Et } · Assuming these for the moment, let us see how they imply (10. 6 5). Let n --+
be
a
positive integer, and set
T =
tjn. Then
1 0. 5.
239
An approximation theorem
t s, l (etAh - Et )x l
Hence, if <
l (erAh - Er ) x l [� Ns Ms] = nNs Msl l (erAh - Er )x l < (s/r )Ns Msl l ( e r A h - Er )x l , where Ms is given by ( 10. 4 9) and N8 is given by ( 10. 6 6). Now if x E D (A) , we have (erAh - E____;,r )x = (erAh - I)x + ( I - Er )X --+ ( Ah - A) x as T --+ 0. T T T Hence, (10.67) (etAh - Et )x 0 as h 0 , x E D (A) . Moreover, if t < s and x is fixed, this convergence is uniform. To see this , first take ( Ah - A ) x as small as you like, and then let T --+ 0. Since l etAh - Et J I < Ns + Ms , (10.67) follows from Lemma 10. 2 . We leave it as an exercise to show tl!at the convergence is uniform in t on bounded sets for each x E X. It remains to prove (i) - (iii) . Now, (i) follows from (a) , and (iii) is j ust (10. 22). To prove (ii) , note that (10. 68) I Et l < Mt+l ' t > 0, where M = M1 is Ms for s = 1. This can be proved by noting that if j is the largest integer less than or equal to t, then I Et l < I E{I I · I Et -j l < Mj + l . Now, 1 a l e Eh l < L k , l a l k i 1 Ehk l < L k\ [ a l k Mhk+ l < Me i a i Mh Hence, in view of (10. 4 5), tl e Ah I < M. exp [ t(Mhh - 1) ] . Since M > 1 , we have (Mh 1) <- max dMh = log M, h dh and hence, (10. 69) This gives (10 . 66), and the proof is complete.. 0 _ _ _
<
_
�
----+
00
0
00
.
0
---
O
lV!
1 0. SEMIGROUPS
240
As an application, let us prove the Weierstrass approximation theorem, which states that every function continuous in the interval 0 < < 1 can be uniformly approximated as closely as desired by a polynomial. By defining
t
x(t) == x(1), t > 1 , we can consider every function iri C == C [O, 1] to be contained in the Banach space of functions bounded and 1Uniformly continuous in the interval 0 < t < with norm l x l == O <supt
+
the semigroup is strongly continuous. By Theorem 10.5, 00 1 '" "" k ' = lim + s ) lim
= h�o e8Ah x h�o k=O sk A�x (t). Thus, for any c > 0 , there are an h > 0 and an N such that 1 k k � + s) x(t k= k! s Ah x(t) < 0 < s, t < 1 . O Setting t 0, we obtain precisely the desired result. x(t
=
L.....t
.
c,
L.....t
1 0 . 6 . Problems
of operators . Suppose ( 1 ) Let { Et } be a strongly continuous semigroup for some fixed t 1 > 0. Show that Et is cornpact for Eallt 1 t is compact > t1 . (2) Show that dkk Et X = Et Ak x, X E D(Ak ), t > 0, dt where { Et } is a strongly continuous semigroup and A is its infini tesimal generator. (3) Show that the set of functions bounded and uniformly continuous in the interval 0 < < oo with norm
t
l x l == O <supt< oo l x (t) l
forms a Banach space.
1 0. 6.
241
Problems
(4) Show that ( 10. 1 7) is differentiable in t > 0, continuous in t > 0 and satisfies ( 0) == 0. ( t) == ) , t > 0 ,
F'
F
(5) Show that the Riemann integral
rtl
ito
u(s) ds
exist.s for u(t) continuous in [to , t 1 ] with values in a Banach space X. (6) Prove the estimate ( 10. 18) . (7) Under the same hypotheses, show that the function U(t)
=
{t U(s) ds ito
is differentiable in the interval and satisfies U' ( t) :.... u( t) . (8) Prove ( 10.20) for u(t) satisfying the same hypotheses. (9) Show that if the convergence in ( 1 0. 67) is uniform in t on bounded sets for each x E D(A) , then it is uniform for e�h x E X.
Et }
( 10) Prove the following. Let { be a strongly continuous, one parameter semigroup of operators on X with infinitesimal genera tor A. Let { Gr } be a family of bounded operators depending on a parameter r , 0 < r < ro such that
GrEt Et Gr , sup l e t Gr l and Gr u � Au as Then etGr u Et u as
t > 0, 0 <
==
O< t
�
<
Mr , r �
r �
0,
0<
0, x
r
r
< ro , <
ro ,
u E D (A) . E X, 0 < t < T.
Moreover, the convergence is uniform.
� �-
-'
- -�-
' "'·
...
• �-· .
• •
•
Chapter
•J
,.
""
11
HILBERT SPACE
1 1 . 1 . When is a Banach space a Hilbert space?
Since a Hilbert space has a scalar product , one should expect that many more things are true in a Hilbert space than are true in a Banach space that is not a Hilbert space. This supposition is, indeed, correct, and we shall describe some of these properties in the next few chapters. There will be many occasions when it will be preferable to work in a complex Hilbert space. This is a complex Banach space having a scalar product ( ) which is complex valued and satisfies ·
,
·
(i) : ( o: u, v ) = o: ( u, v ) , ( ii) : ( u + v ' w )
=
o: complex,
(u ' w) + (v ' w ) '
(iii) : ( v' u)
=
(u ' v ) '
(iv) : (u, u)
=
llull2.
Before studying Hilbert spaces , we should learn how to recognize them. It may seem simple. We first need to see if it has a scalar product. However, this is not as simple as it may first appear. If we are given a Banach space and not given a scalar product for it , this does not necessarily mean that one does not exist. Perhaps we could find one if we searched hard enough, and of course, as soon as we find one, the Banach space becomes a Hilbert space. The problem reduces to the following: Given a Banach space X , does there exist a scalar product on X which will convert it into a Hilbert space with the same norm? We saw way back in - Section 1.3 that a necessary 243
244
1 1 . HILBERT SPACE
condition is that the parallelogram law (11.1)
l x + Yl l 2 + l x - Yl l 2 = 2 l x l 2 + 2 I Y I 2 , x, Y E X
should hold. We shall see that this is also sufficient . We have Theorem 1 1 . 1 . A Banach space X can be converted into a Hilbert space with the same norm if and only if (1 1 . 1) holds. Proof. The simple proof of the "only if' part was given in Section 1 .3. To prove the "if' part , assume that ( 1 1 . 1 ) holds. We must find a scalar product for X. First, assume that X is a real Banach space. If a scalar product existed, it would satisfy
( 1 1 .2)
l x + Yl l 2 = l x l 2 + 2(x, Y) + I Y I 2
·
This suggests that we define the scalar product to be
( 1 1 .3)
This does not look like a scalar product , but things are not always what they seem. If we take this as the definition, we get immediately that
(x,y) = (y,x), ( 1 1 .5) ( x, x) = l x l 2 ' ( 1 1 .6) (xn , y) --+ (x, y) if Xn X in ( 1 1 . 7) (O ,y) = 0. Moreover, 2(x, y) + 2( z , y) = l x + Yl l 2 - l x l 2 + l i z + Y f l 2 - l z l 2 - 2 I Y I 2 · By the parallelogram law, 2 l x + Yl l 2 + 2 l z + Yl l 2 = l x + z + 2y l 2 + l x - z l 2 ( 1 1 .4)
--+
X,
and Hence,
(x, y) + ( z , y) = [ l x + z + 2y l 2 - l x + z l 2 - 4 I Y I 2]/4 = i � (x + z) + Y I 2 - l � (x + z ) l 2 - I Y I 2 = 2 ( � (x + z) , y) . Now, if we take z = 0 in ( 1 1 . 8 ), we see by ( 1 1 . 7) that ( 1 1 .9) (x,y) = 2 (� ,y) . ( 1 1 .8)
Applying this to the right hand side of ( 1 1 . 8) , we obtaj!l ( 1 1 . 10)
(x, y) + (z, y) = (x + z, y).
1 1 . 1 . When is a Banach space a Hilbert space?
245
It only remains to prove that
( y) ( ox , y), y E X for any E IR. By repeated applications of ( 1 1 . 10) , we see that n ( x , y) ( n x , y) , x , y E X for any positive integer n , and hence, if m, n are positive integers , ( : ) (x,y) = ( : ) (m (�) ,y) ( : ) m ( : ,y) n y y) ' ( � ' ) c: Thus , ( 1 1 . 1 1 ) holds for a a positive rational number. Moreover, by ( 1 1 . 10) (x,y) ( -x, y) ( x - x,y) 0. Hence, ( y) - ( x , y) . This shows that ( 1 1 . 1 1 ) holds for all rational numbers a. Now if a is any real number, there is a sequence { a n } of rational numbers converging to a. Hence , by ( 1 1 .6) , a ( x , y) lim an ( x , y) lim ( an x , y) ( a x , y ) . a x,
(11.11)
==
x,
n
==
=
=
=
'
+
==
-x,
==
==
==
==
==
This completes the proof for a real Banach space. If X is a complex Banach space, then any prospective scalar product must satisfy Hence , we must have ( 1 1 . 12)
y) ( y) SSm ( x , y ) �e ( - i )( y) - �e ( i x , y) . Thus, it seems reasonable to define ( x , y) by ( x , y) Re ( x, y) - i �e ( ix , y) , ( 1 1 . 13) where R e ( x , y) is given by ( 1 1 . 1 2) . We then have ( ix , y) Re (ix , y) - i �e ( - x , y) i [Re ( x , y) - i �e (ix , y) ] i ( x , y) . This shows that ( 1 1 . 1 1 ) holds with respect to ( x , y) for all complex a. Now by ( 1 1 . 1 2) ' � e ( i x , i y) �e (x, y).
The arguments given above show that �e ( x , has all of the properties of a real scalar product . Moreover, if x , is a complex scalar product , then ==
x,
==
==
==
==
==
==
246
1 1 . HILBERT SPACE
Hence,
(y, x) = �e (y, x) - i �e (iy, x ) = �e (x, y) - i �e ( y i x ) = �e ( x , y) + i �e ( ix, y) -
,
= (x, y) .
l xl 2.
Thus (x, x) is real and must, therefore, equal �e(x, x) , which equals The other properties of ( x , y) follow from those of �e ( x, y) , and the proof is complete. 0 1 1 . 2 . Normal operators
{ 4'n } be an orthonormal sequence (finite or infinite) in a Hilbert space {>..k } be sequence (of the same length) of scalars satisfying 1 -Ak l < c. Then for each element f E H, the series L Ak ( f, 4'k ) 4'k
Let H. Let
a
converges in H (Theorem 1 .5) . Define the operator A on H by ( 1 1 . 14) Clearly, A is a linear operator. It is also bounded, since ( 1 1 . 15) by Bessel ' s inequality (cf. ( 1 .60) ) . For convenience, let us assume that each =I= 0 (just remove those that vanish) . In this corresponding to the case, N(A) consists of precisely those E H which are orthogonal to all of Clearly, such are in N(A) . Conversely, if E N (A) , then the
Ak
4'k
4'k ·
f
Ak f
f
4'k ) = .Ak ( f, 4'k ) · Hence, (/, cpk ) = 0 for each k. Moreover, each A k is an eigenvalue of A with eigenvector. This follows immediately from ( 1 1 . 14) . 4'kSincetheu(Acorresponding ) is closed, it also contains the limit points of the A k · Next , we shall see that if A =I= 0 is not a limit point of the A k , then A E p(A ) . To show this, we solve (.A - A ) u f ( 1 1 . 16) for any f E H. Any solution of ( 1 1 . 16) satisfies ( 1 1 . 17) .Au = f + Au = f + L Ak (u, 4'k ) 4'k · 0 = (A J,
=
Hence,
1 1 . 2. Normal operators
247
or
( 1 1 . 18) Substituting back in ( 1 1 . 1 7) , we obtain
Au = 1 + L Ak V,_cp��'Pk . Since A is not a limit point of the A k , there is a 6 > 0 such that I A - Ak I > 6' k = 1 ' 2 ' . . . . ( 1 1 . 19)
Hence, the series in ( 1 1 . 19) converges for each f E H. It is an easy exercise to verify that ( 1 1 . 19) is indeed a solution of ( 1 1 . 16) . To see that (.X - A) - 1 is bounded, note that
I A I · ! l u l < l f l + C l f l /6 (cf. ( 1 1 . 1 5) ) . Thus , we have proved Lemma 1 1 . 2 . If the operator A is given by {1 1 . 14), then a(A) consists of the points A k , their limit points and possibly 0. N(A) consists of those u which are orthogonal to all of the 'Pk · For ,.\ E p(A) , the solution of {1 1 . 1 6) is given by (1 1 . 1 9) . ( 1 1 . 20)
.We see from all this that the operator ( 1 1 . 14) has many useful proper ties. Therefore, it would be desirable to determine· conditions under which operators are guaranteed to be of that form. For this purpose, we note an other property of It is expressed in terms of the Hilbert space adjoint of A. Let H1 and H2 be !filbert spaces, and let be an operator in B (H1 , H2 ) . For fixed E H2 , the expression Fx = ( A x , is a bounded linear functional on H1 . By the Riesz representation theorem (Theorem 2. 1 ) , there is a z E H1 such that F x == ( x , z ) for all x E H1 . Set z is a linear Then operator from H2 to H1 satisfying ( Ax , (1 1.21) (x, A ) .
A.
A y)
y
= A* y .
y) =
A.
A*
*y
A* is called the Hilbert space adjoint of Note the difference between A* and the operator defined in Section 3. 1 . As in the case of the operator A' , we see that A * is bounded and
A'
I A* I = I A I . The proof is left as an exercise. Returning to the operator A, we remove the assumption that each A k -:/= 0 and note that (Au,v) = L ,.\k (u,
1 1 . HILBERT SPACE
248
showing that ( 1 1 .23)
( If H is a complex Hilbert space, then the complex conj ugates A k of the A k are required. If H is a real Hilbert space, then the A k are real, and it does not matter. ) Now, by Lemma 1 1 .2 , we see that each Ak is an eigenvalue of with 'Pk a corresponding eigenvector. Note also that
A*
( 1 1 . 24)
showing that ( 1 1 .25)
I A* ! I == I Af l l ,
f E H.
An operator satisfying ( 1 1 . 25) is called normal. An important characteriza tion is given by Theorem 1 1 .3. An operator is normal and compact if and only if it is of oo . 0 the form {1 1 . 14) with { 'Pk } an orthonormal set and A k
� as k �
In proving Theorem 1 1 . 3 , we shall make use of the following properties of normal operators: Lemma 1 1 .4. If A is normal, then
I (A* - � )u ll == I (A - A.)u ll , u E H. Corollary 1 1 . 5 . If A is normal and A<.p == A.rp , then A * <.p A.rp. Lemma 1 1 .6. If A is normal and compact, then it has an eigenvalue >.. such that I -X I == I A I · ( 1 1 .26)
=
Let us assume these for the moment and give the proof of Theorem 1 1 .3.
A <.po I A-o l I A I A* (A v , <.po ) == ( v, A* <.po ) == A.o \ v, <.po ) == 0 by Corollary 1 1 .5. A similar argument holds for A * . Let A of A to H . For u, v E H , we have
Proof. Let be a normal compact operator on H. If A == 0, then the theorem is trivially true. Otherwise , by Lemma 1 1 .6 , there is an eigenvalue be a corresponding eigenvector with Ao such that == # 0. Let norm one, and let H 1 be the subspace of all elements of H orthogonal to c.p0 . We note that A and map H1 into itself. In fact , if v E HI , then I
be the restriction
1 (u, Aiv) == (A 1 u, v) == (Au, v) == (u, A* v ) , showing that A i == A * on H1 . This implies_ th_at AI is normal as well compact on the Hilbert space H1 . Now, if AI 0, then Au == A.o ( u, u E H, 1
<.po )<.po ,
==
as
1 1 . 2.
Normal operators
249
and the proof is complete. Otherwise, by Lemma 1 1 . 6 there such that
is
a pair
,.\ 1 ,
I A ll = I A ll l ,
Moreover, by Theorem 6 . 2 ,
,.\k � 0 k � oo . Now, let u be any element of H. Set Un = u - I: (u,
(1 1 . 27)
n
0
Then
(un ,
0<
k<
n.
( 1 1 . 28)
n n l un l 2 = l u l 2 - 2 L l (u,
0
0
n
00
0
and the theorem is proved in one direction . To prove it in the other direction, � 0. Set suppose is of the form ( 1 1 . 14) with
A
,.\k n An u = 2: Ak (u, 'Pk )
n
= 1, 2,
··· .
250
11.
HILBERT SPACE
An
is a linear operator of finite rank for each n . Now, for any c > 0, Then there is an integer N such that < E for > N. Thus
k 1 -X�Ct I An u - Au l 2 = nI:+ l I .Xkl 2 l (u,
---+-
completes the proof.
n --4 oo .
A
(
)
0
It remains to prove Lemmas 1 1 .4 and 1 1 .6. The former is very simple. In fact, one has 2 �e = == 2 �e
I (A* - � )u l 2
-=
I A*u l 2 - ( �u, A*u) + I -A I 2 I u2 l 2 I Au l 2 - (Au, .Xu) + I -A I 2 I u l I (A - .X)u Wl .
Corollary 1 1 .5 follows immediately. In proving Lemma 1 1 .6 we shall make use of
A is normal, then ( 1 1 .30 ) ra (A) = I A I · Once this is known, it follows from Theorem 6. 13 that there is a A E such that I -X I = I A I · If A == 0 , then clearly, 0 is an eigenvalue of A. o-(A) Otherwise, A i= 0, and it must be an eigenvalue by Theorem 6.2. Lemma 1 1 . 7. If
Therefore, it remains to give the proof of Lemma 1 1 . 7. Proof. Let us show that
( 1 1 .3 1 )
when
A
···
=1 2 is normal. For this purpose we note that n
'
'
'
1 Ak u l 2 = (Ak u ,kAk u )k = . ( A* A u, A - 1 u) < I A* A k u l · I A k - l u l k = I Ak + l u l · I A - l u l , which implies Ak l 2 < I Ak+ l i · I Ak - l l , k = 1 , 2 , . . . . ( 1 1 .32 ) I Since I A n l < I A I n for any operator , it suffices to prove ( 1 1 .33) I A I n < I An I , = 1 , 2 , · · · . n
1 1 .2. Normal operators
251
Let k be any positive integer, and assume that ( 1 1 . 33) holds for all Then by ( 1 1 .32) we have
n
< k.
I A I 2k < I Ak + l l . I A I k - l '
which gives ( 1 1 .33) for n == k + 1 . Since ( 1 1 .3 1 ) holds for n = 1 , the induction argument shows that it holds for all n . Once ( 1 1 .3 1 ) is known, we have 1 ( 1 1 . 34) ro11 = lim ==
(A)
n-+oo
I An I n I A I ,
0
and the proof is complete. We also have
A{ 4'k }
Corollary 1 1 .8. If is a normal compact operator, then there is an or thonormal sequence of eigenvectors of such that every element in H can be written in the form
( 1 1 .35)
u
A
N (A).
N (A) is separable, then A has a complete_ Proof. By Theorem 1 1 .3 , has an orthonormal set { 4'k } of eigenvectors such that ( 1 1 . 14) holds. If u is any element of H , set h = u - L (u, cpk ) 4'k · Then, by ( 1 1 . 14) , A h == 0, showing that h E N (A ) . This proves ( 1 1 . 35) . If N (A) is separable, then we shall see that it has a complete orthonormal set {1/Ji } . Once we know this , we have u = L (h, 1/;j ) 1/lj + L (u,
_,
·
·
·
·
·
·
Lemma 1 1 .9. Every separable Hilbert space has a complete orthonorrnal sequence.
{ Xn }
Proof. Let H be a separable Hilbert space, a�d let be a dense sequence in H. Remove from this sequence any element which is a linear combination of the preceding Let Nn be the subspace spanned by x1 , · · · X n , and let Such an be an element of having norm one and orthogonal to
'Pn
Xj ·
Nn
, Nn- 1·
1 1 . HILBERT SPACE
252
element exists by Corollary 2 .4. Clearly, the sequence {
xn }
00
U Nn , 1
and each element in W is a linear combination of a finite number of the
H.
1 1 . 3 . Approximation by operators of finite rank
K
in a Hilbert In Section 4.3 we claimed that , for each compact operator space, one can find a sequence of operators of finite rank converging to in norm. We are now in a position to prove this assertion. First , assume that is separable. Then, by Lemma 1 1 .9 , we know that has a complete orthonormal sequence {
H
( 1 1 .36)
K
H
Pn
n
Pnu == l: (u ,
Then by Theorem 1 .6 , ( 1 1 .37) and ( 1 1 .38)
l j ( J - Pn ) l < 1 , == 1 ' 2 ' · · · ' Pn u � u as � u E H. n
oo ,
n
Now, if the assertion were false, there would be an operator a number 6 > 0 such that
IlK - Fl > 6 for all operators F of finite rank. Now n Fn u = Pn Ku = L (Ku,
K E K(H) and
( 1 1 .39)
1
is an operator of finite rank. Hence, ( 1 1 .39) implies that for each a satisfying
n
there is
Un E H ( 1 1 .40 ) Since the sequence { un } is bounded, it has a subsequence { Vj } such that Kvj converges to some element w E H. But I (K - Fn )un l < I ( ! - Pn )(Kun - w) l + I (/ - Pn )w l < I K un - w l + I ( ! - Pn )w l ·
253
1 1 .4. Integral operators
(11.38) I ( / - Pn )w l < 6/4, and, for an infinite number of indices , I Kun - w l < 6j4. This contradicts (11. 4 0) and proves the assertion in the case when H is sep�rable. Now let H be any Hilbert space, and let K be any operator in K(H) . Set A == K * K. Then (11.4 1) A* == A. Operators satisfying ( 11. 4 1) are called selfa djoint. In particular, A is normal. By Theorem 11. 3 , A is of the form (11.14) for some orthonormal sequence { 'Pk }. Let Ho be the subspace of all linear combinations of elements of the form Kn k , = 0 , 1 , 2 , k == 1 , 2, Then H is a separable closed subspace of H , and K maps o into itself. By what we have already proved, there is a sequence {Fn } of finite - rank operators on H o converging in norm to the restriction K of K to H Now every element u E H can be written in the form u = v w, where v E H o , and w is orthogonal to Ho . Observe that K w == 0. To see this, note that w is orthogonal to each 'P k , and hence, is in N(A) ( Lemma 11.2). Consequently, 0 = ( Aw,w ) = (K*Kw,w) = 1 Kw l 2 , showing that K w = 0. Thus , if we define == 1 ' 2 ' ' then Gn is of finite rank, and I G n - K l � 0 as � This completes the proof. Now, for
n
sufficiently large, we have by
<.p
n
·
·
·
;
·
·
.
·
_lf
0
"'
+
n
·
·
O
·
·
n
oo .
1 1 . 4. Integral op erators
Let us now describe an application of the theorems of Section integral operators of the form
11. 2 to some
Ku(x) = 1b K(x, y)u(y)dy,
- oo
oo
oo .
1 1 . HILBERT SPACE
254
K
given by {1 1 .42) is a compact operator on Lemma 1 1 . 10. The operator if K(x, y) satisfies {1 1 . 43).
L2(a, b)
We shall save the proof of Lemma 1 1 . 10 until the end of this section. Let us see what conclusions can be drawn from it. By Theorem 1 1 . 5 , we have Theorem 1 1 . 1 1 . If
K(x, y)K(x, z ) = K( z , x) K (y, x), a � x, y , z < b, then there exists an orthonormal sequence { 'P k } (finite or infinite) of func tions in L 2 (a , b) such that
( 1 1 .44)
( 1 1 .45 )
L2 (a, b) . Moreover, 2 dx dy = L 1 Akl 2 , m = 1 , 2, ) ( (x) cp Y k I >.kcpk lb lb I K (x, y) - L k> m k<m In particular, the sequence { 'P k } is infinite unless K(f, y) is of the form ( 1 1 .47 ) K(x, y) = L Ak 'Pk (x) cpk (y) .
where the series converges in ( 1 1 .4 6 )
0
a
a
0
0
0
N 1
If
( 1 1 .48 ) then the convergence in {1 1 .4 5) is uniform.
( ) K*v(y) = j K(x , y)v(x)dx o
Proof. In view of Lemma 1 1 . 10 , we can prove 1 1 . 45 by verifying _that the operator ( 1 1 .42 ) is normal and applying Theorem 1 1 . 3. This is simple, since
( 1 1 .49 ) Hence,
while
KK*u(z ) = j K(z,x)K*u (x)dx = jj K(z,x )K(y, x)u(y)dx dy, K* Ku(z ) = j K(x , z )Ku(x)dx = jj K(x , z )K(x, y)u(y)dxdy o
By ( 1 1 .44 ) , we see that
( 1 1 . 50)
KK * == K* K
'
255
1 1 .4. Integral operators which implies that K is normal� since
I
I K* u ll 2
Thus,
=
(K
K * u , u)
= ( K* Ku , u) = II KuiJ 2 .
( 11.45) holds. To prove ( 11. 46) , note that for u, L2 (a, b) , we have II [K(x, y) - I: Ak<;?k (x)
= (Ku , v) - I: Ak ( u , <,Ok) (cpk, v ) = 0
( 11.45) . Since this is true for any u , v E L2 (a, b) , it follows that (1 1.51) where the series converges in L2 ( Q ) . This gives (11 ..46) . The last statement
by
of the_theorem follows from the inequality 1Ku ( x) l 2 =
Hence
I K(x , y)u (y)dy 2 < I I K (x , y) i 2dy I iu(y) f2dy
( 11.52 )
I K u (x ) f < M llull ,
< M2 ll u l l 2 •
a < x < b.
uk � u in £2 (a, b) , then ( 11.53) I Kuk (x) - Ku(x) l < M l l uk - u ll � 0, showing that Ku (x) converges uniformly to K u( x ) . This completes the proo[
In particular, if
k
0
It remains to prove Lemma easy consequence of
11.10. We do this by showing that it is an
{ 'Pk }
is a complete orthonormal sequence in a Hilbert Lemma 1 1 .12. If space H, and K is an operator in B (H) satisfying
( 11. 54) then K is in
K( H).
11.1 2 implies Lemma 11.10. Proof. As we saw in Section 1.4, L 2 (a, b) has a complete orthonormal se quence {
finite values, a slight change of variables is needed. Otherwise, we can use
1 1 . HILBERT SPACE
256
the fact that ( a, b) is the denumerable union of bounded intervals. The details are left as an exercise.] Thus, we have by Theorem 1 .4 1 , ( 1 1 .55)
j, k
2 = d ( ( ) x ( y ) x, x)d K Y Y j j
D
1 1 . 10.
Now, we give the simple proof of Lemma 1 1 . 12. Proof. First note that 00
Ku = L1 (u, 'Pk )Kcpk , u E H. In fact , since n L1 ( u ,
1
00
00
00
Hence, ( 1 1 .58)
00
I l K - Knl l 2 < nL I Kcpkl l 2 --4 0 +l
as
n
--4 oo .
257
1 1 . 5. Hyponormal operators
This shows that K is the limit in norm of a sequence of operators of finite 0 rank. Hence, K is compact, and the proof is complete. 1 1 . 5 . Hyponormal operators
A in B(H) is called hyponormal if ( 1 1 .59) H A*u l < I Au l , u E H, or, equivalently, if ([AA* - A* A]u, u) < 0, u E H. ( 1 1 .60) Of course, a normal operator is hyponormal. An operator A E B(H) is called semi normal if either A or A * is hyponormal. In analogy with Lemma 1 1 .7, we have Theorem 1 1 . 13. If A is seminormal, then ( 1 1. 61) ra (A) I A I · Proof. First, assume that A is hyponormal. Then ( 1 1 .31) holds. The proof is the same as that for normal operators with the exception that in this ca.Se we have I A* Ak u l · I Ak-l u l < � I Ak+ I u l · I Ak-l u l in the proof of ( 1 1 .32) . Thus ( 1 1 .34) holds. If A * is hypo normal, then the result just proved gives ( 1 1 .62) ra (A*) I A* I · But in general we have ( 1 1 .63) (A*) n (An ) * , 1 , 2, · · · . An operator
==
==
n ==
==
(This follows immediately from the definition. ) Thus, by ( 1 1 .22) and ( 1 1 .63) , we have showing that
( 1 1 .64)
ra (A*)
A
ru ( ) . We now see that ( 1 1 .61) follows from ( 1 1 .22) , ( 1 1 .62) and ( 1 1 .64) . This ==
0
completes the proof. In Section 7.5 we defined the essential spectrum of an operator
o-e ( A) n ( A + K) . It was shown that .A fj. O"e(A) if and only if A E ( 1 1 .65)
A to be
a
==
KEK(H)
A
i (A
- .X) == 0 and (Theorem 7.27) . Let us show that we can be more specific in the case of seminormal operators.
1 1 . HILBERT SPACE
258
A
a(A) ae (A)
Theorem 1 1 . 14. If is a seminormal operator, then A E \ if and only if A is an isolated eigenvalue with < oo (see Section 5. 5).
r(A - .X)
Before proving Theorem 1 1 . 14 , we note that some brief comments are in order. Let M be a subset of . a Hilbert space H. By M.l we denote the set of those elements of H that1 are orthogonal to M. There is a connection between this set and the set of annihilators of M ( see Section 3.3) . In if and only if the element given by the Riesz fact, a functional is in representation theorem ( Theorem 2 . 1 ) is in M .l . In particular, if either of these sets is finite dimensional, then dim M 0 = dim M .l . ( 1 1 .66)
0 M M0
Now we know that ( 1 1 .67)
( see (3. 12) ) . Similarly, ( 1 1 .68)
Hence,
( 1 1 .69)
N(A' ) = R(A) 0 N(A*) = R(A) .i . ,B (A) = dim N(A*).
Another fact we shall need is Lemma 1 1 . 1 5 . If
.X.
Proof.
A is hyponormal, then so is B = A - A for any complex
I B *u l 2 = ( [AA* - .XA* - .XA + I .X I 2 ]u, u) < ( [A * A - .XA* - .XA - I .X I 2 ]u , u) = 1 Bu l 2 ·
This gives the desired inequality.
D
We shall also make use of
A
Lemma 1 1 . 1 6. If is hyponormal and maps a closed subspace M into itself, then the restriction of to M is hyponormal.
A
A1 be the restriction of A to M. Then for u, v E M, we have (u, A*v) = (Au, v) = (A 1 u, v) = (u, Ai v). ( 1 1 . 70) In particular, I Ai u l 2 = (Ai u, Ai u) = (Ai u, A*u).
Proof. Let
Hence,
1 1 . 5.
259
Hyponormal operators
0
Thus, A 1 is hyponormal. Now we are ready for the proof of Theorem 1 1 . 14.
Proof. Suppose A E a (A) is not in ae (A) . Set B = A - A. Then B E iJJ ( H) , and i (B) = 0. By Lemma 1 1 . 15, B is also seminormal. If B is hyponormal, then
N(B)
c
N(B * ) .
Since i (B) = 0, the dimensions of these subspaces are finite and equal [see ( 1 1 .69)] . Hence,
N (B * ) = N (B) .
( 1 1 .71) If B* is hyponormal, then
N(B * )
c
N (B) [here we have made use of the fact that A * * = A for A E B ( H ) ] . Again, since they have the same finite dimension, ( 1 1 .71) holds in this case as well. By ( 1 1 .68) , ( 1 1 . 71) and the fact that R(B) is closed, we have ( 1 1 .72) H = N (B) E9 R(B) . We see from this that B is one-to-one on R(B) and maps R(B) into itself. Thus, if B 2 v = 0, we must have B v = 0. Hence, N(B 2 ) = N(B) , and consequently, from which we see that
k= 1 2 ... '
'
'
r (B) = n(B) < oo . Since i (B) = 0, we also have r ' (B) < oo . It now follows from Theorem 6.23 that 0 is an isolated point of a(B) . Hence, A is an isolated point of a ( A) with r(A - .X) < oo. Since A E A with i (A - .X) = 0, A must be an eigenvalue of A. This proves the theorem in one direction. The proof of the rest of the theorem follows easily from Lemma 1 1 . 1 7. If B is hyponormal with 0 an isolated point of a (B) and either n (B) or {3(B) is finite, then B E cJ.> (H) and i (B) = 0.
We shall postpone the proof of Lemma 1 1 . 1 7 a moment and suppose that A is seminormal with A an isolated eigenvalue such that r(A - .X) < oo . Set B = A A. If A is hyponormal, so is B. Moreover, 0 is an isolated point of a(B) with n (B) < oo . Hence, by Lemma fl . l7 , we see that B E cJ.> (H) with i (B) = 0. This is precisely what we want . If A * is hyponormal, so is B* . Moreov;er, 0 is an isolated point of a(B* ) , and {3(B* ) = a(B) is -
1 1 . HILBERT SPACE
260
finite. Another application of Lemma 1 1 . 17 gives the desired result . Hence, Theorem 1 1 . 14 will be proved once we have given the proof of Lemma 1 1 . 17. D We now give the proof of Lemma 1 1 . 17. Proof. Set
1 1 ( z - B) - l dz, p = 2 7r 'l Jlzl=c where c > 0 is so small that the points 0 < l z l < c are in p(B) . Then by (6.30) , we have .
H = R(P) EB N(P) ,
( 1 1 . 73)
and B maps both of these closed subspaces into themselves. Let B 1 and B2 denote the restrictions of B to R(P) and N(P) , respectively. Then a(B 1 ) consists precisely of the point 0 ( Theorem 6. 19) . Hence, ru (Bl ) = 0 ( Theorem 6. 12) . But B 1 is hyponormal ( Lemma 1 1 . 16) , so that ru (Bl ) = II B 1 II ( Theorem 1 1 . 13) . Hence, B 1 = 0 , showing that R(P) C N(B) . But, in general, we have N(B) C R(P) ( see (6. 32) ) . Hence,
R(P) = N(B) .
( 1 1 . 74)
Next , note that 0 E a(B2 ) ( Theorem 6. 19) . In particular, this gives N(P) = R(B2 ) c R(B) . ( 1 1 . 75) Thus, we have ( 1 1 . 76)
R(P) = N(B)
c
N(B * ) = R(B) j_
c
N(P) j_ .
This, together with ( 1 1 . 73) , shows that
R(P) = N(P) .i . For if u E N(P) .l , then u = u 1 + u2 , where u 1 E R(P) , u 2 E N(P) . Now u 1 E N(P) .l by ( 1 1 . 76) , showing that u2 E N(P).l as well. This can happen only if u2 = 0, and this implies that u = u 1 E R(P) . By ( 1 1 . 76) and ( 1 1. 77) , ( 1 1 . 77)
we have
N(B) = N(B * ) ,
( 1 1 . 78) and
R(B)
c
N (B * ) .i = R(P) .i = N(P) .
This, together with ( 1 1 . 75) , gives
R(B) = N(P) . From ( 1 1 .79) we see that R(B) is closed, and from ( 1 1 . 78) we see that o:(B) · = {3(B) . Since one of them is assumed finite, we see that B E 'P (H) D and i (B) = 0. This completes the proof. ( 1 1 . 79)
1 1 . 5. Hyponormal operators
26 1
There is a simple consequence of Lemma 1 1 . 17.
a(A), A . A ,.\ Proof. Set B == A - ,.\. If A is hyponormal, so is B ( Lemma 1 1 . 15) . If n(B) == 0, then by Lemma 1 1 . 17, we have B E (H) and i ( B ) == 0. But then R(B) == H , showing that ,.\ E p (A) . Thus, we must have n(B) > 0, which is what we want to show. If A* is hyponormal, so is B* . If ( B) == 0, then (3 ( B * ) == 0, and by Lemma 1 1 . 17, B* E
a
We also have the following:
A A {
a(A)
Theorem 1 1 . 1 9 . Let be a seminormal operator such that has no non- z ero lzmit points. Then �s compact and normal. Thus, it is of the form {1 1 . 14) with the orthonormal and � 0.
,.\k Proof. We follow the proof of Theorem 1 1 . 3 . If A == 0, the theorem is trivially true. Thus, we n1ay assume that A #- 0. First , assume that A is hyponormal. By Theorem 1 1 . 1 3, there is a point A o E a(A) such that 0 ( see Theorem 6 . 13) . By hypothesis, A o is not a lirnit ,.\o l ==ofI aA(A). I #- Hence, 1point it is an eigenvalue ( Corollary 1 1 . 18 ) . Let <po be an eigenvector with norm one, and let HI be the subspace of all the elements of H orthogonal to <.po . We see that A maps H1 into itself. This follows from the fact that A - Ao is hyponormal, showing that <po E N (A* - �o). Hence, if v E H1 , then (Av , <po) == ( v , A * <po) == Ao ( v , <po) == 0 , and consequently, Av E HI . Thus, the restriction A I of A to H 1 is hyponor mal ( Lemma 1 1 . 1 6) . If A 1 == 0, we are finished. Otherwise, we repeat the process to find a pair A I , 'P I such that I AII == I A II I , 'P I E HI , I 'P I I == 1 , A l 'P I == A l 'P l· We let H2 be a subspace orthogonal to both
262
1 1 . HILBERT SPACE
Ak
that A is of the form ( 1 1 . 14 ) with --4 0. Thus, A is normal and compact ( Theorem 1 1 . 3 ) . If A* is hyponormal, then we apply the same reasoning to A* , making use of the fact that 0 is the only possible limit point of a (A* ) . Thus, A* is of the form- 1 1 1 . 14 ) with --4 0. By ( 1 1 .23 ) the same is true of, A = A** . This completes the proof. 0
Ak
Corollary 1 1 .20. If A is seminormal and compact, then it is normal. 1 1 . 6 . Problems
( 1 ) Suppose A is a seminormal operator in B(H) such that a( A ) con sists of a finite number of points. Show that A is of finite rank. ( 2 ) Show that if A is hyponormal and a ( A) has at most a finite number of limit points, then A is normal. ( 3 ) If the operator K is defined by ( 1 1 .42 ) , show that M
=
1b 1 b I K (x, y) i 2 dx dy < 1
implies that I - K has an inverse in B(H) . Do this by showing that II K II 2 < M .
( 4 ) Define A* for unbounded operators. Show that if A is closed and densely defined, then A** exists and equals A. (5) An operator A on H is said to have a singular sequence if there is a sequence of elements in D (A) converging weakly to 0 such that ll u n ll = 1 , Aun --4 0. If A is closed, show that it has a singular sequence if and only if either R(A) is not closed or o:(A) = oo . Thus A has a singular sequence if and only if it is not in + (H) .
{ u.n}
( 6 ) Show that if A is normal, then A - >.. has a singular sequence if and only if >.. E O"e (A) . (7) If the operator A is -given by ( 1 1 . 14 ) , show that 0 E p(A) if and is complete. only if the sequence
{ 'Pk }
(8) Prove ( 1 1 .22 ) .
1 1 . 6. Problems
263
(9) Prove that L 2 ( - oo, oo) is separable. ( 10) Show that if A is seminormal and An = 0 for some positive integer n , then A = 0.
( 1 1 ) Prove the projection theorem and the Riesz representation theorem for complex Hilbert spaces. ( 12) Show that A is normal if and only if AA* = A* A. ( 1 3) Show that if K ( x , y ) = K (y, x ) , then the operator ( 1 1 .42) is self adjoint. ( 14) Show that the operator A given by ( 1 1 . 14) has ,.\ in its resolvent set if ,.\ is not a limit point of the Ak .
Chapter
12
BILINEAR FORM S
12 . 1 . The numerical range
Let A be a bounded linear· operator on a Hilbert space H. If A E a (A) , then either N (A - A) # { 0 } or N (A * - � ) # { 0 } or R(A) is not closed in H [otherwise, the bounded inverse theorem would imply that A - A has an inverse in B( H)] . Thus, we have a u E H satisfying either ( 1 2. 1 )
l l u l l == 1 , Au == AU,
or ( 12 . 2)
! l u ll == 1 , A* u == AU,
or we have a sequence { uk } of elements of H satisfying ( 1 2.3)
l l u k ll == 1, ( A - A)u k � 0
as k � oo .
In the case of ( 1 2. 1 ) or ( 12 . 2 ) , we have ( 1 2.4)
(A u, u) == A, ! l u l l == 1 ,
while if ( 12 .3) holds, we have ( 1 2 . 5) To put this in a concise form, let W (A) be the set of all scalars A that equal (Au, u) for some u E H satisfying I I u I I == 1 . The set Hl (A) is called the numerzcal range of A . Now ( 1 2.4) says that A E H1 (A) . while ( 1 2 . 5 ) says that A E Hl ( A) . Hence, we have Theorem 1 2 . 1 . If 11 is in B (H) , t hen a (A)
C
ll � ( A ) . 265
266
1 2. BILINEAR FORMS
If we try to extend the concept to unbounded operators, we must proceed cautiously. First , we can only define for Hence, ,.\ if there is a such that == and == ,.\ . Second, we must define for an unbounded operator As in the case of A' , this can be done in a unique way only if is dense in H (see Section In this case, we say that if, there is an f H such that
u E D ( A). E W(A) (Au, u) E D (A) u l A.1 (Au, u) u ! l A* D (A) 7.1 ) . E v E D (A*) (12. 6 ) ( u, f) == (Au, v), u E D (A) . We then define A*v to be f. Let us assume once and for all that D (A) is dense in H. Then A* exists . Now for an unbounded closed operator we also have that ,.\ E a(A) implies that either N(A - ,.\) # {0} or N(A* - �) # {0} or that R(A) is not closed in The first possibility implies that there is a u E D (A) satisfying while the third implies the existence of a sequence { u k } of elements (12.4), in D(A) satisfying (12.5 ) . However, the second need not imply the existence of a u E D(A) satisfying (12. 4 ). Thus, all we can conclude at the moment is Theorem 1 2 . 2 . If A is a closed, densely defined operator on H and ,.\ ¢:. W(A), then a(A - ,.\) == 0 and R(A - ,.\) is closed in H . In particular, it may very well happen that a(A) is not contained in W(A). As we see from Theorem 12.2, the fault would be in R(A - ,.\) not being large enough. The question thus arises as to whether we can enlarge D(A) in such a way to make a(A) W(A). We are going to see in Sec tions 12.5 , 12. 9 , and 12.10 that this indeed can be done under very general If.
--
c
circumstances . In the considerations which follow it would be a nuisance to consider real Hilbert spaces. So as usual, we take the easy way out and assume throughout this chapter that we are dealing with a complex Hilbert space . 1 2 . 2 . The associated op erator
a( u, v)
Let H be a Hilbert space. A- bilinear form (or sesquzlinear functional) of a subspace on H is an assignment of a scalar to each pair of vectors of H in such a way that
,v u D(a) (12.7) a( au + (3v, w) == aa(u, w ) + (3a (v , w), u, v, w E D(a), (12. 8 ) a(u, av + (3w) == O:a(u, v) + (3a(u, w), u, v, w E D(a). The subspace D (a) is called the domain of the bilinear form. Examples of bilinear forms are the scalar product of H and an expression of the form where A is a linear operator on H. The domain of the scalar product (Au, v), is the whole of H, while that of ( A u v) is D(A). When v == u we shall write a(u) in place of a (u, u). ,
267
The associated operator
1 2. 2.
(a)
As in the case of an operator, we can define the numerical range W of a bilinear form for some as the set of scalars which equal uE satisfying == 1 . we can associate a linear With any densely defined bilinear form and there is operator A on H as follows: We say that E if E a constant C such that
D (a)
(1 2.9)
a( u) ,.\ a(u , v) u D(A) u D(a) l a ( u, v) l < C l v l , v E D ( a).
a( u, v) !lul
Now by the Hahn-Banach theorem (Theorem 2 . 5 ) and the Riesz representa tion theorem (Theorem 2. 1 ) , inequality ( 1 2.9) implies that there is an f E H such that
a(u, v) == ( J, v ) , u E D(a) (you will have to take conj ugates to apply them) . Moreover, the density of D(a) H implies that f is unique. Define Au to be f. Clearly, A is a linear operator on H, and we shall call it the operator associated with the bilinear form a( u, v) . We shall prove the following: Theorem 1 2 . 3 . Let a( v) be a densely defined bilinear form with associ ated operator A . Then ( 1 2. 10)
in
u,
,.\ tt W (a), then A - ,.\ is one-to- one and ( 12. 1 1 ) ! l u l < C I I (A - ,.\) u l , u E D(A). {b) If ,.\ t/: W (a) and A is closed, then R( A - ,.\ ) is closed in H . Proof. (a) Since ,.\ f/: W (a), there is a 6 > 0 such that ( 12. 12) l a (u) - ,.\ 1 > 6, ! l u l == 1, u E D (a) . {a) If
Thus
( 1 2 . 13)
u E D(A) and ( A - ,.\)u == f, then a(u, v) - ,.\(u, v) == (f, v), v E D(a) . ( 12. 14) In particular, a ( u) - ,.\ l u l 2 == (f, u ) ,
Now if
and
( 1 2 . 15) Combining this with ( 1 2 . 13) , we obtain
! l u l < 11 � 1 1 ,
268
1 2.
BILINEAR FORMS
whi_ch is clearly ( 12. 1 1 ) . Now, ( 1 2. 1 1 ) implies that A - ,.\ is one-to-one. Thus, (a) is proved. D (b) Apply Theorem 3. 1 2 . 1 2 . 3 . Symmetric forms
A bilinear form a ( u, v) is said to be symmetric If ( 1 2. 16)
a (v , u) == a(u, v) .
An important property of symmetric forms is given by Lemma 1 2 .4. Let a(u, v) and b(u, v) be symmetric bilinear forms satisfying
( 1 2 . 1 7)
l a (u) l < Mb(u) ,
u E D (a)
n
D ( b) .
Then ( 12. 18)
l a (u, v) l 2 < M 2 b(u) b(v ) ,
u, v E D (a) n D (b) .
Proof. Assume first that a( u, v) is real. - Then
( 12. 19)
a(u ± v) == a(u) ± 2a(u, v) + a(v) ,
and hence, 4a( u , v) = a (u + v) - a(u - v) . Thus 4 l a(u, v) l < M[b(u + v) ·+ b(u - v)] == 2M [b(u) + b(v)] by ( 1 2 . 17) and ( 1 2. 19) . Replacing u by o:u and v by vjo:, o: E JR, we get ( 12 . 20)
[
2 l a(u, v) l < M a2 b(u) +
b��) ]
If b(u) == 0, we let a � oo , showing that a (u , v ) == 0. In this case, ( 1 2 . 18) holds trivially. If b( v) == 0, we let o: � 0. In this case as well, a( u, v) == 0, and ( 1 2 . 18) holds. If neither vanishes, set 4
b(v)
0: == b( u) .
This gives (12.21) which is just ( 1 2. 18) . If a( u, v ) is not real, then a( u, v ) == e i 0 I a ( u, v) I for some Hence a( e - i0 u, v) is real. Applying ( 1 2 . 1 8) to this case, we have l a ( e