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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 z ) P() F (X). Theorem 14. 14. , the same is true of A ( I + ,.\Ao E) == A - ,.\K E + ,.\E , showing that A + E E <1> . Thus , E E F(X ) , and the proof is complete. 0 1 4 . 3 '"" Related measures 0 such that I AT I m c i ! T I m , T E B( Z, X ), here does not depend on Z ( Theorem 14. 5 ). Thus, I AT I > c� ( T), T E B(Z, X) ( Theorem 14.28), and consequently, r(A) # 0 ( Corollary 14. 2 7 ). On the other hand, if A f/: + (X, Y) , then for each > 0 there is a K E K ( X , Y) such that I K I < E and o:(A-K) oo ( Theorem 14. 4 1 ). Let M N(A-K). Then I A I .JV! I 0. This implies that for anyE E > 0 there is an0 , K M m I I I N M such that I A I N I < E This implies that r(A) < . Hence , r(A) 0 and the proof is complete. Proof. If + (X, Y ). If A E (X, ==Y ), then there are operators Ao E B(Y, X ) and F E K (X) such Ao A I - F (Theorem 5. 4 ) . Thus � T == A0 AT - FT, consequently, 6. (T) == �(AoAT - FT) < �(Ao)�(AT) inf . We shall call it a + perturb�tion function if m(T, A) < 1 for A E + implies A - T E 4> + . Finally, we shall call it a a perturbation function if m(T, A) < 1 for A E + and o:(A - T) < o:(A) . Y In our discussions of perturbation functions , we shall assume that are Banach spaces. By Theorems 14.50 and 14. 5 1 , we see that II T il / J.L ( A ) is a + perturbation function , while II T il / J.Lo ( A) and II T II / r(A) are a per turbation functions . The following lemma shows that a + perturbation function is also a perturbation fu nction. perturbation function. a (Y, i (BA) 0 such that for every T in B{X, Y) satisfying I ITII < 1} , one has A + T E (X Y) with , 0 there is a K E K (X, Y) such that I l K II < E and o: (A - K) == oo . Thus there is an infinite dimensional subspace M such that the restriction of A to M is compact and has norm < E. -set, 129 _ ( X, Y ) , 122
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 - i0 u, v ) l 2 < M 2 b(e -i 0 u) b(v) .
B.
This implies -( 1 2 . 18) for u and v . The proof is complete.
D
1 2. 3.
269
Symmetric forms
Corollary 1 2 . 5 . If b( u, v) is symmetric b u t a ( u, v) is not and {1 2. 1 7) holds, then
( 12.22)
! a (u, v) l 2 < 4M 2 b(u)b(v) ,
u, v E D (a) n D(b) .
Proof. Set
( 1 2 . 23)
a 1 (u, v) ==
1
2
[ a(u, v) + a(v, u) ] ,
1 [ a( u , v) - a(v , u) ] . 2i Then a 1 and a 2 are symmetric bilinear forms, and a(u, v) == a 1 (u, v) + i a 2 (u, v) . ( 12 . 25) ( 1 2.24)
a 2 (u, v) =
( The forms a 1 and a2 are known as the real and imaginary parts of a , respectively. Note that , in general, they are not real valued. ) Now, by ( 1 2 . 1 7) ,
! aj ( u ) l
< Mb( u ) , j == 1 , 2 ,
u E D(a)
Hence, by ( 12.2 1 ) ,
n
D(b) .
which implies ( 12.22) .
D
Corollary 1 2 . 6 . If b(u, v) is a symmetric bzlinear form such that
( 1 2.26)
b(u)
>
u E D(b) ,
0,
then
( 1 2.27)
l b(u, v) l 2 < b(u) b(v) ,
u, v E D ( b) ,
and
(1 2.28)
b(u + v) 1 12 < b(u) 1 1 2
+
b(v) 1 1 2 ,
u, v E D ( b) .
Proof. By ( 1 2 . 26) , we have l b(u) l == b(u) . Setting a(u, v) == b(u, v) in Lemma 1 2.4, we get ( 1 2 . 27) . Inequality ( 12 . 28) follows from ( 1 2. 27) in the usual fashion. In fact ,
b( u + v ) == b(u) < b(v,)
+
This proves ( 1 2 . 28) .
+
b(u, v) + b(v , u)
+
b(v)
2b(u) 1 1 2 b(v) 1 1 2 + b(v) == [ b(u) 1 12 + b(v) 1 1 2 ] 2 . D
The following criteria for recognizing a symmetric bilinear form are some times useful. As in the case with most other staternents in this chapter, they are true only in a complex Hilbert space.
270
1 2.
BILINEAR FORMS
Theorem 1 2 . 7. The followzng statements are equivalent for a bilinear form:
{i) a ( u, v) is symmetric; {ii) SSm a( u)
==
0,
u
E D (a) ;
{iii) �e a(u, v) == R e a(v , u) ,
u, v E D(a) .
Proof. That (i) implies (ii) is trivial. To show that (ii) implies (iii) , note that a( i u + v) == a(u) + ia(u, v) - i a( v u) + a(v) . ,
Taking the imaginary parts of both sides and using (ii) , we get (iii) . To prove that (iii) implies (i) , observe that by (iii) , SSm a( u, v ) == SSm ( - i )a ( i u, v) == - �e a( iu, v ) ==
- Re a(v , iu) == - Re ( -i)a(v , u) == -SSm a( v u ) . ,
. This, together with (iii) , gives (i) , and the proof is complete.
0
1 2 . 4 . Closed forms
A bilinear form a(u, v) will be called closed if {un } C D (a) , Un --+ 0 as m, n --+ oo imply that u E D (a) and a(un - u) H, a(un - um ) as n --+ oo . The importance of closed bilinear forms may be seen from --+
u
in --+ 0
Theorem 1 2 . 8 . Let a( u , v) be a densely defined closed bilinear form with associated operator A . If W(a) is not the whole plane, a half-plane, a strip, or a line, then A is closed and 2
( 1 . 2 9)
a (A)
c
W(a) == W(A) .
In proving Theorern 1 2 . 8 , we shall make use of the following facts, some of which are of interest in their own right . They will be proved in Section 1 2 . 7. Theorem 1 2 . 9. The numerical range of a bilinear form is a convex set in the plane. Lemma 1 2 . 10. If W is a closed convex set zn the plane which is not the whole plane, a half-plane, a strip or. a line, then W is contained zn an angle of the form
( 1 2 . 30)
I arg (
z - zo) - Bo I < B <
7r
2
.
1 2. 4 .
Closed forms
271
a( u, W (a) . Fv D( a) ( 12.31 ) 1 Fv l 2 < C l a (v) l , v E D (a) , there are unique elements u E D(a) such that Fv = a(v, w ) , v E D (a), ( 12.32 ) and Fv = a(u, v), v E D (a) . ( 12.33 )
W(a)
Theorem 12 . 1 1 . Let �) be a closed bilinear form such that is not a half-plane, a strip, or a line, and such that 0 tt. Then for each linear functional on 3atisfying w,
Let us show how Theorems 1 2 .9 and 1 2 . 1 1 together with Lemma 12. 12 imply Theorem 1 2.8. Simple consequences of Theorem 12.9 and Lemma 12.8 are
W(a)
is not the whole plane, a half-plane, a strip, or Corollary 1 2 . 1 2 . If such that == 1 , > 0, ko is real, a line, then there are constants and
1, k, ko 111 k ( 12.34 ) l a (u) l < k[�e 1a(u) + ko l u l 2 ], u D(a) . Theorem 1 2 . 13. Let a(u, v) be a bilin ear form such that W (a) is not the whole plane, a half-plane, a strip, or a line. Then there is a symmetric bilinear form b( u, v ) with D(b) == .D(a) such that there is a constant C satisfying ( 12.35 ) c- 1 l a (u) l < b(u) < l a (u) l + C l u l 2 , u E D(a) . In particular, if {uk } D(a) , u E D(aj , uk u and a(uk - u) � 0, then a(uk , v) a(u, v), v E D(a) . ( 12.36 ) E
C
�
�
The proof of Corollary 12. 1 2 and Theorem 1 2 . 1 3 will be given at the end of this section. Let us now show how they can be used to give the proof of Theorem 12.8.
A { uk } D(A) Uk u, Auk l a (uj - uk ) l = I (Auj - Auk , Uj - uk ) l < I A uj - A u k I · l ui - u kl l � 0 as j, k Since a( u , v) is closed, this implies that u E D(a ) and that a(u k - u ) � 0. Thus, by Theorem 12. 13, a(uk , v) � a(u, v) as k v E D(a) . ( 1 2.37 ) Proof. To see that is closed, suppose that there is a sequence � � f in H. Then such that
� oo .
� oo ,
C
1 2. BILINEAR FORMS
272
Since
a(uk , v ) = (Auk , v ) , we have in the limit
v E D(a) ,
( f,v), v E D(a) , f. Thus , A is a closed operator. W(a) , let A be any scalar not in W (a) . Then
a(u, v) = showing that u E D (A) and Au = (12.38)
To show that cr(A) C by Theorem 1 2 . 3 (a) , A - .X is one-to'-one, and ( 12. 1 1 ) holds. In particular, ( 12 . 13) holds. Let
aA (u, v )
�
a(u, v ) - .X(u, v ) .
Then aA satisfies the hypotheses of Theorem 12. 1 1 . If f is any element of H, then (v, is a linear functional on D(aA ) D(a) , and
f)
=
l (v,
f ) l 2 < ll v ll 2 ll / ll 2 < C l aA (v ) l
by ( 12 . 13) . Hence, by Theorem 12. 1 1 , there is a u E H such that
aA (u, v ) = (J, v) , v E D (a) . This shows that u E D (A) and (A - .X)u = Since was any element of H, we see that R(A - .X) = H, and consequently, E p(A) . ( 1 2.39)
f. A f
-
Lastly, in order to prove W (a) = W (A) , we shall show that
W(a) c W(A) . The first inclusion is obvious, since u E D(A) implies (Au, u) = a(u) . To prove the second, we want to show that, for each u E D(a) , there is a sequence {u k } C D(A) such that a(uk) --+ a(u) . Let b(u, v) be a symmetric W(A)
( 12.40)
c
bilinear form satisfying ( 12 .35) . Then by Corollary 12.5,
v
1 l a( v ) - a (u) l < l a(v - u, ) l + ja(u, v - u) l � 2 C [b (v) 1 1 2 + b(u) 1 1 2 ] b( v - u) 12 . Thus, it suffices to show that for each u E D(a) , there is a sequence {u k } C D (A) such that b(uk -- u) --+ 0. Now consider D( a) as a vector space with scalar product b( u, v ) + ( u , v ) . This makes D (a) into a normed vector space X with norm [b(u) + l l u ll 2 ] 1 1 2 . (Actually, X is a Hilbert space, as we shall see later. ) We want to show that D (A) is dense in X. If it were not , then there would be an element w E X having a positive distance from D (A) . By . Theorem 2.9 there is a bounded linear functional F =F 0 on X which annihilates D(A) . Let be any scalar not in W(a) , and define aA as above. Then aA satisfies the hypotheses of
A
Theorem 12. 1 1 , . and by ( 12. 13) and ( 12 .35) ,
1 Fv l 2 < K [b(v) + ll v ll 2 ] < K' l aA (v) l ,
v E X.
273
12. 4. Closed forms
Thus , by Theorem 1 2. 1 1 , there is a w E X such that
Fv aA(v , ) v E X. Since F annihilates D(A) , we have aA(v, w ) == 0, v E D(A). ( 12.41
)
==
w ,
This, is equivalent to
((A - .X)v, ) 0, v E D (A). But we have just shown that R(A - .X) == H, so that there is a v E D (A ) such that (A - A)v == This shows that w == 0, which, by ( 12.41 ) , implies that F 0 , providing a contradiction. Hence, D (A) is dense in X , and the proof of Theorem 12.8 is complete. w
=
w.
=
D
Let us now give the simple proof of Corollary 12. 12.
W (a)
W(a). W(a) ( ) ( ) zo , Oo , ( 12.42 ) l �m{e - i90 [a (u) - zo] } l < tan (} �e {e - i90 [a (u) - zo] }. S et r == e - i Oo . Inequality ( 1 2 . 42 ) implies l �m ra(u) l < tan B [�e ra(u) + ko], ! l u l == 1 , u E D(a), where l + l c;}· m 'Fo l _ ko l �e ')'Zo tan (} This implies ( 12.34 ) with k == 1 + tan B. 0
Proof. By Theorem 12.9, By is 1t convex set. Hence, so is Lemma 12. 10, must satisfy 1 2 .30 for some 0. Now, 1 2 .30 is equivalent ta
=
We also give the proof of Theorem 12. 13.
k, ko such that ( 12.34 ) ra(u, v) [see ( 1 2 .23 ) ] ,
Proof. By Corollary 1 2 . 1 2 , there are constants [ , holds. Let be the real part of the bilinear form and set
b1 ( u, v)
( 12.43 )
b(u, v) == b 1 ( u, v) + ko ( u, v). Then by ( 12.34 ) , l a (u) l < kb(u), u E D (a). Moreover, by ( 12 .43 ) , b(u) == �e ra(u) + ko l u l 2 < l a (u) l + l ko l · l u l 2 . Thus ( 12.35 ) holds. To prove ( 1 2.36 ) , note that by Corollary 1 2 . 5 ,
274
1 2. BILINEAR FORMS
l a (uk , v) - a(u, v ) l 2
l a ( uk - u, v) l 2 < 4C2 b(uk - u)b(v) 4 C2 b( v )[ l a (uk - u) l C l uk - u l 2] � 0 as k �
<
oo .
+ This completes the proof. The proofs of Theorems in Section
12. 7.
0
12.9 and 12.11 and Lemma 12.10 will be given
1 2 . 5 . Closed extensions
A be a linear operator on a Hilbert space H. As we saw in Section true that a(A) W(A). In this section we shall concern 1ourselves 2 .1 , it maywithnotthebequestion of when A can be extended to an operator having this property. If D(A) An operator B is said to be an extension of an operator D(B) and Bx == Ax for x E D (A). We are going to prove Theorem 1 2 . 14. Let A be a dens ely defined l'lnear operator on H such that W(A) is not the whole plane, a half-plane, a strip, or a tine. Then A has a closed extension A such that a( A ) W(A) == W( A ). (12 . 44) The -proof of Theorem 12.14 will be based on the following two theorems. Theorem 1 2 . 15 . If A is a linear operator, then W(A) is a convex set in the plane. Theorem 1 2 . 1 6 . Let a(u, v) be a densely defined bilinear form such that W ( a) is not the whole plane, a half-plane, a str'lp, or a line. Suppose that { uk } D(a), Un � 0 , a(un - um ) � 0 imply a(un ) � 0. Then a( u, v) has a closed extension ii(u, v) such that D(a) is dense in D(a) and W(a) W(ii ) W(a). A few words of explanation might be needed for Theorem 12 .16. A bilinear form b(u, v') is called an extension of a bilinear form a(u, v) if D(a) v for u, v E D(a). A set U will be called dense in ) D(b) and b(u, v) , a(u, D(a) if for each E D (a) and each E > 0 there is a u E U such that a(w - u) < E and l w - u l < E . Theorem 1 2 .15 is an immediate consequence of Theorem 1 2.9. Theorem 12.16 will be proved at the end of this section. Now we shall show how they imply Theorem 12.14. Let
C
A
C
"'
c
C
c
c
c
w
1 2. 5.
Closed extensions
Proof. Let a( u, v ) be the bilinear form defined by
( 12.45) a(u, v) = (Au, v ) , u, v E D (A) , with D (a) = D (A) . Then W (a) = W(A) . By Theorem 12. 13, there is a symmetric bilinear form b( u, v ) satisfying ( 12.35) . Now suppose { un } c D(a) , Urn � 0 , and a (un - um ) � 0. Since a (un ) = a(un , U n - Um ) + (Aun , Um ) , we have, by Corollary 12.5 , (12.46) l a(un ) l < 2 C b(un ) 1 1 2 b(un - Um ) 1 /2 II Aun ll · ll um ll · Now by ( 12.35) , b(un - U m ) < l a(un - Um ) l + C ll un - um ll 2 � 0 as m, n � oo . Thus, there is a constant K such that ( 12.47) b(un ) < K 2 , n = 1 2 · · Now let c > 0 be given. Take N so large that 2 € , m, n > N. b(un - Um ) < 2 2 40 K
+
'
Thus
Letting m �
oo ,
l a(un ) l < E + II Aun ll · ll um ll ,
'
·
m, n > N.
we obtain
n > N.
l a(un ) l < E ,
This means that a( u n ) � 0 as n � oo . Thus, a( u, v) satisfies the hypotheses of Theo;rem 12. 16. Therefore, we conclude th�t a ( u, v ) has a closed extension a(u, v ) with D(a) dense in D(a) . Let A be the operator associated with a (u, v) . Then, by Theorem 12.8, A is closed, and "'
o- (A)
But, by Theorem 12. 16,
c
W (a) = W ( A ) .
W(a) = W(a) = W(A) .1
"'
All that remains is the minor detail of verifying that A is an extension of A. So, suppose u E D (A) . Then
( 12.48) a ( u, v ) = (Au, v) , v E D(a) = D (A) . Since a( u, v ) is an extension of a( u, v ) , ( 12.49) a(u, v ) = (Au, v ) , v E D(a) . Now, we claim that ( 12.49) holds for all v E D (a) . This follows from the fact that D (a) is dense in D(a) . Thus, if v E D (a) , there is a sequence { vn } C D(a) such that a(vn - v) � 0 and ll vn - v ii � 0. Now, W (a) is
276
1 2.
BILINEAR FORMS
not the whole plane, a half-plane, a strip, or a line. Thus, we may apply Theorem 1 2 . 1 3 to conclude that
a( U, Vn) � a( U , V) . Since
a( u, Vn) == (Au, Vn) ' we have in the limit that (12.49) holds. Thus , u E D( A ) and A u == A u. -
�
Hence, A i� an extension of A, and the pr'o of is complete.
D
Let us now give the proof of Theorem 12. 16. Proof. Define a(u, v ) as follows: u E D(a) if there is a sequence { un } c D ( a ) such that a ( un - u m ) � 0 and Un � u in H. If { vn } is such a sequence for v, then define
a( U , V) == nlim a ( Un , Vn) . ---+- oo
( 12.50)
This limit exists. To see this, note that
a ( un , Vn ) - a ( u m , Vm ) == a ( u n , Vn
- Vm ) + a ( 'lfn
-
Um , Vm ) ·
Now, by Theorem 12. 13, there is a symmetric bilinear form b( u, v ) satisfying Hence, by Corollary 12.5,
(12.35).
l a ( un , Vn ) - a ( um , Vm ) l < 2 C[b(un ) 1 1 2 b ( vn - Vm ) 1 1 2 + b(u n - Um ) 1 1 2 b( vm ) 11 2 ] .
( 12. 3 5 ) .
This converges to zero by Moreover, the limit in ( 12.50) is unique ( i.e. , it does not depend on the particular sequences chosen ) . To see this, let { u� } and { be other sequences for u and v, respectively. Set = UnI - Un , VnII == VnI - Vn . Then
v�}
u�
. b( UII - UII ) 1 / 2 _ I < b( UnI - Um ) 1 / 2 + b( Un - Um ) 1 / 2 � 0 as m, n � 00 , n m by Theorem 12. 13. Thus , a ( u� - u�) � 0, and similarly a(v� - v�) � 0. Since u� � 0, v� � 0 in H, we may conclude, by hypothesis, that a(u� ) � 0, a( v� ) � 0 as n � oo , which in1plies
b (u � ) � 0, b( v� ) � 0 as n �
oo .
Hence,
l a ( u� , v;t ) - a ( un , Vn ) l < l a ( u� , v� ) l + l a(u� , Vn) l
< 2 C [b (u� ) 1 1 2 b( v� ) 1 1 2 + b(u� ) 1 1 2 b( vn ) 1 1 2 ] � 0. From the way a( u , v) was defined, it is obvious th at
( 1 2.5 1)
W( a )
c
W(a)
c
W (a )
.
12.5. Closed extensions
277
To show that D (a) is dense in D(a) , note that if u E D (a) , there is a sequence {u n } C D(a) such that a(un - um ) � 0 , while Un � u in H and a(un ) � a(u) . In particular, for each n,
a(u n - Um )
�
a(un - u) as
m �
Now, let c > 0 be given, and ta,ke N so large that Letting
m �
oo ,
l a(un - u m ) l <
€,
m,
00 .
n > N.
we obtain
n > N,
l a(un - u) l < c, which shows that
a(un - u) � 0 as n � 00 . Consequently, D (a) is dense in D (a) . It remains only to show that a is closed. In order to do this , we note that W(a) is not one of the sets mentioned in Theorem 1 2 . 13 [see ( 12.5 1 )] . Thus, there is a symmetric bilinear form b( u, v) satisfying ( 1 2. 52)
"
( 1 2.53) Now, suppose {un} ( 12.53) ,
c
D (a) , a(un - Um )
�
0, and Un
--4
u in H. Then by
b ( Un - u m ) � 0 as n � oo . By the density of D (a) in D ( ) , for each n there is a Vn E D (a) such that
ii
"
l a(un - Vn) l < Thus, by {12.53) ,
1 2' n
1
. l l un - Vn ll < n
"b(Un - Vn) < c + 1 .
( 12.54)
n2
Since b is a symmetric form,
b ( vn - Vm) 1 /2 < b ( vn - Un ) 1/2 + b (un - Um ) 1 / 2 + b (um - Vm) 11 2 as
m,
n
---+-
oo .
Hence
a(Vn - Vm )
=
a( Vn - Vm)
�
0 as
m, n �
Since
ll vn - u ll < ll vn - un ll + ll un - u ll � 0 as n we see that u E D(a) and ( Vn - U ) � 0 as n 00
ii
by ( 12.52) . Thus ,
b ( vn - u)
-+
�
0 as n
� oo ,
00 .
--4 oo ,
--4
0
1 2. BILINEAR FORMS
278 which implies, by ( 1 2.54) , that
b (un � u)
�
0 as
n � oo ,
a(un - u)
�
0 as
n � 00 .
which, in turn, implies Hence, a is closed, and the proof is complete.
0
1 2 . 6 . Closable operators
In the preceding section we gave sufficient conditions for a linear operator
A on H to have a closed extension A satisfying a (A)
C
W(A) . Suppose we
are only interested in determining whether A has a closed extension. Then the condition can be weakened. In fact , we shall prove
A
Theorem 1 2 . 17. If is a densely defined linear operator on H such that is not the whole complex plane, then A has a closed extension.
W(A)
Before we give the proof of Theorem 1 2 . 1 7, let us discuss closed ex tensions in general. Let A be a linear operator from a normed vector space X to a normed vector space Y. It is called closable ( or preclosed) if { xk } C D (A) , X k � 0, A x k � y imply that y == 0. Clearly, every closed operator is closable. We also have Theorem 1 2 . 18. A linear operator has a closed extension if and only if it is closable.
We shall postpone the simple proof of Theorem 1 2 . 18 until the end of this section. By this theorem , we see that in order to prove Theorem 12. 1 7, it suffices to show that a densely defined operator on H is closable if W(A) is not the whole plane. To do this , we make use of
A
Lemma 1 2 . 1 9 . A convex set in the plane which is not the whole plane is contained in a half-plane.
From this lemma we have
W
Corollary 1 2 . 20 . If a ( u, v ) is a bilinear form such that ( a ) is not the whole plane, then there are constants [, ko with lrl == 1 such that
(12. 5 5)
�e [ra(u) + ko ll u ll 2] > 0 ,
u E D ( a) .
We shall also postpone the proof of Lemma 1 2 . 19 until the end of this section. Corollary 1 2 . 20 follows easily from the lemma. In fact , we know that (a ) is convex from Theorem 12.9. Hence, it must be contained in a half-plane by Lemma 12. 19. But every half-plane is of the form r z + ko] > 0 , r == 1 .
W
�e [
,I I
12. 6. Closable operators Thus,
279
�e [ra(u) + ko] > 0,
u E D (a) , l l u ll
=
1,
which implies ( 1 2 .55) . A consequence of Corollary 12.20 is Theorem 1 2 . 2 1 . Let a(u, v ) be a densely defined bilinear form such that W(a) is not the whole plane. Let A be the operator associated with a(u, v ) . If D(A) is dense in H, then A is closable. Proof. By Corollary 12.20, there are constants {, ko such that ( 1 2.55) holds. Set b( u, v ) = 1a ( u, v ) + ko ( u, v )
and
=
r A + ko . Then B is the operator associated with b( u, v ) . Moreover, A is closable if and only if B is. Hence, it suffices to show that B is closable. So suppose {un } C D(A) , Un � 0, Bun � f. Then for a > 0 and w E D (A) , we have b(un - o:w) b(un ) - o:b(un , w) - o:b(w, Un) + a 2 b(w) (Bun, Un) - a( Bun , w) - o: (Bw, Un) + a 2 b(w) --+ -a(f, w) + o:2 b(w) . B
Hence ,
�e [- (f, w) + o:b(w) ] > 0 ,
Letting o: � 0, we see that
�e (f, w)
( 12.56)
< 0,
a > 0 , w E D (A) . w E D (A) .
Since D(A) is dense in H, there is a sequence { vn } C D (A) such that Vn --+ f in H. Since (f, vn) < 0 , we have, in the limit, l l f ll 2 0 , which shows that 0 f = 0. Hence, A is closable, and the proof is complete.
�e
<
We can now give the proof of Theorem 12. 17. Proof. Set
a(u, v ) = (Au, v ) ,
u, v E D (A) . Then a(u, v ) is a bilinear form with D (a) = D (A) and W(a) = W(A) . Moreover, A is the operator associated with a( u, v ) . Thus a( u, v ) satisfies
all of the hypotheses of Theorem 1 2 . 2 1 . Thus, we may conclude that closable, and the proof is complete. We are now in a position to give the proof of Theorem 12. 18.
A is
D
12. BILINEAR FORMS
280
Proof. Suppose A has a closed extension A , and let {xn } be a sequence in D(A) such that X n � 0 in X while A x n � y in Y. Since A is an extension of A, X n E D(A) and Ax n � y . Since A is closed, we have AO == y , showing that y = 0. Hence, A is closable. Conversely, assume that A is closable. Define the operator A as follows: An element x E X is in D(A) if there is a sequence { x n } C D ( A) such that X n � x in X and A x n converges in Y to some element y . Define Ax to be y . This definition does not depend on the choice of the particular sequence { X n } . This follows from the fact that if {zn} C D (A) , Zn � x in X and A zn � w in Y, then X n - Zn � 0 �nd A( xn - Zn ) � y - w . Since A is closable, we see that y == w. Clearly, A is a linear extension of A , and we see that it is closed. To see this, suppose { x n } C D(A ) and Xn � x , A xn � y . Then for each n there is a sequence {wn k } C D (A ) such that Wnk � Xn and Awn k � A x n . In particular, one can find a Zn E D(A ) such that "
A
A
A
1
, ll zn - Xn ll < n
A
-
1
. II Azn - Ax n l l < n
Therefore, and
I I Azn - Y l l < II Azn - A xn ll + I I A xn - Y l l � 0. This shows that x E D( A ) and Ax = y . Hence, A is closed, and the proof is complete.
0
The operator A constructed in the proof of Theorem 12. 18 is called the closure of A. It is the "smallest" closed extension of A (we leave this as a simple exercise) . Now we give the proof of Lemma 1 2 . 19. Proof. Let W be a convex set in the plane, which is not the whole plane. Then W cannot be the whole plane either (this is left as an exercise) . Now consider the complex plane as a Hilbert space Z with scalar prod uct ( z 1 , z2 ) = z 1 z2 . Let zo be a point not in W . Then by Theorem 7. 1 7, there is a bounded linear functional f -=/= 0 on z such that
( 1 2.57)
�e f ( zo )
>
SR e f ( z) ,
z E W.
By Theorem 2. 1 , there is a complex number Hence, ( 12.57 ) becomes
( 12.58 )
f ( z ) = z i,
�e zo ')t > �e z')t,
r
#- 0 such that
z E Z. z E W.
1 2. 7.
28 1
Some proofs
We leave as an exercise the simple task of showing that the set of all complex z satisfying ( 12 . 58) is a half-plane. 0 We can also give a proof that does not make use of functional analysis. Proof. Let Q be a point not in W. If W is empty, there is nothing to prove. Otherwise, some ray from Q intersects W. The first point P of W encountered by the ray is a boundary point of W. Let V be the collection of all rays from P which intersect W at a point not equal to P, and let M be the set of all points on these rays. Clearly, M is a convex set . To prove this, note that if 81 and S2 are points in M, then the rays PS 1 and PS2 contain points T1 and T2 of W , respectively. Since W is convex, the segment T1 T2 is in W, and , hence, all rays between PS1 and PS2 are in lvf. This includes the points on the segment S1 S2 . Now, a convex set of points consisting of rays from P is merely an angle () with vertex at P ( the verification is left as an exercise) . Clearly, () n ; otherwise, one would have a line segment lying outside of M connecting points of !vf. If one extends either of the rays forming the sides of M, one obtains a half-plane free of W. This completes the proof. 0
<
1 2 . 7 . Some proofs
We now give the proofs that were deferred fron1 Section 1 2 .4. First , we give the proof of Theorem 12.9.
D ( a) satisfying ! l u l l v l == 1 . We < (} < 1 , we can find a w E D(a) ( 12.59) a(w).== ( 1 - B)a(u) + Ba(v). If a(u) a(v), we can take w == u and we are finished. Otherwise, there is a scalar r such that l r l == 1 and ( 1 2.60) SSm ra(u) == SSm ra (v) == d. Let o: , (3, be real numbers, and set (12.61) g ( o: , (3) == ra( o:e�'P u + (3v) - id l o:ei
==
Proof. Let and be two elements of want to show that for each (} satisfying 0 such that II �v I I 1 , and ==
_:
rp
+
Now pick
rp
so that
v,
SSm{ei'P [-ya(u, v) - (u, v)id] + e - i
282
1 2. BILINEAR FORMS
( This can always be done. ) Thus, from ( 12.60) and ( 12.61) , we see that g (a,
(3) is real. Now if
(12.62) then we have and
a( a e i 4?u) == a ((J v ) The first implies n 2 == {3 2 , while the second gives .
(3
n 2 a(u) == 2 a (v) . Since we are assuming a(u) -# a( v) , the only way ( 1 2.62) can hold is if n == {3 == 0. From this, we see that the function g (t , 1 - t) h (t) = 1 1 te i 4? u + (1 - t) v ll 2 is continuous and real valued in 0 t 1 . Moreover, h (O) == ra ( v ) - id, h ( 1 ) == ra (u) - id. Hence, there is a value t 1 satisfying 0 t 1 1 such that h ( t 1 ) == 0 [r a ( v) - id] + ( 1 - 0) [r a ( u) - i d] == r [Oa ( v ) + ( 1 0 ) a ( u)] - i d.
< < < <
-
Set
Then, ll w ll == 1 and
t 1 e i 4'u + ( 1 - t 1 )v w == ll t 1 e i0 u + (1 - t 1 )v ll 2 · h(t 1 ) == ra (w) - id.
This gives ( 12.59) , and the proof is complete.
D
We now give the proof of Lemma 12. 10. Proof. As we saw in the second proof of Lemma 12. 19, there is a half-plane V containing W such that the boundary line L of V contains a point P of W. The line L is called a support line for W. Suppose L C W. If W is not a half-plane or a line, then the interior of V contains a boundary point Q of W ( see the proof of Lemma 12. 19) . A support line £ 1 through Q cannot intersect L , for then we would have points of W on both sides of £ 1 ( namely, those on L) . Hence, £ 1 must be parallel to L . Moreover, every point in the strip between L and £ 1 is on a segment connecting Q and a point on L. This would mean that W is a strip, contrary to assumption. Hence, L must contain a point R not in W. Consequently, the ray emitted from R away from P must be free of W. For simplicity, assume that this ray is the positive real axis, R is the origin, P is the point - c , c > 0, and V is the upper half-plane. We can make the assertion that there is a 6 > 0 such that W is contained
1 2. 7.
283
Some proofs
in the angle 6 < () < 1r. This would give us exactly what we want. If it were not true, then for each E > 0, there would be a point z E W such that
0 < c �m z < E �e z . The segment connecting P and z must be in W. Moreover, the distance from this segment to the origin is less than d ==
c �m z E �e z · < E. < c + �e z c + �e z The quantity d is just the distance from the origin to the point where the segment intersects the imaginary axis. Use similar triangles . ) Since this is true for any E > 0, we see that there is a sequence of points of W converging to the o r igin, i.e. , to R. Since W is closed, we must have R W. But we 0 chose R not to be in W. This contradiction proves the lemma.
(
E
Theorem 12.11. Proof. Since . a (u, v) satisfies the hypothese of Theorem 12.13, we know that there is a symmetric bilinear form b( u, v ) such that D(b) == D(a) and ( 12. 3 5) holds . Moreover, since 0 tt. W(a), there is a 6 0 such that (12. 63) l a (u) l 6 > 0, u E D(a), ! l u l == 1. Hence, (12. 64) We now give the proof of
s
>
>
Thus , there are constants M1 , M2 , M3 such that
( 12. 65) Let X be the vector space D(a) equipped with the scalar product b(u, v ) . We can show that X is a Hilbert space. The only property that needs verification is completeness. Let { Un } be a Cauchy sequence in X; i. e . , suppose b(un - um ) � 0 as n � By (12. 65), { un } is a Cauchy sequence in H, and a( un - um) � 0 as � Since H is complete, there is an element u E H such that Un � u in H as Since ( u, v ) is a closed bilinear form, we know that u E D(a) and a(un -u ) � 0 as n � Thus b (un - u) � 0 by (12. 6 5), and X is complete. Now, for each w E X , ( 1 2. 6 6) Gv == a(v, w ), v E X, is a linear functional on X. It is bo u nded by (12.65) and Corollary 12.5. Hence, there is an element Sw E X such that (12. 67) Gv == b (v, Sw), v, E X m,
oo .
m, n
oo .
n � oo .
a
oo .
284
1 2. BILINEAR FORMS
( Theorem 2. 1 ) . As anyone can plainly see, S is a linear mapping of X into
itself. It is bounded since
b(Sw) 2 == a(Sw, w ) 2 < 4Mjb(Sw) b(w) , whence
( 12,68)
b(Sw) < 4Mjb(w) .
It is also one-to-one and has a closed range, since
Mf b(w) 2 < M? l a(w) l 2 == Mi l b(w, Sw) l 2 < Mi b( w )b(Sw) , by ( 12.65) , (12.66) and ( 1 2 .67) . This gives b(w) < M?b(Sw) /Mf . ( 12. 6 9) We claim that R(S) == X. To see this, let h be any element of X orthogonal to R( S) , i.e. , satisfying b(h, Sw) == 0, w E X. Then by ( 12.66) and ( 12.67) , a(h, w) == 0, w E X. In particular, this holds when w == h, showing that a(h) == 0, which implies b(h) 0 by ( 12.65) . Hence, h == 0. Now, let F be any linear functional in D (a) satisfying ( 12.3 1 ) . Then F is a bounded linear functional on X by ( 12.65) . Hence, there is an element f E X such that F v == b( v , f) , v E X. Since S is one-to-one and onto, there is a w E X such that Sw == f. Hence, Fv == b( v, f) == b(v , Sw) == a(v, w) , v E X, which proves ( 12.32) . The proof of ( 12.33) is almost identical and is omitted. =
0 1 2 . 8 . Some representation theorems
Theorem 12. 1 1 is a representation theorern similar to the Riesz representa tion theorem ( Theorem 2. 1 ) : There are some interesting consequences that follow from it. Theorem 1 2 . 2 2 . Let b( u, v ) be a closed, symmetric bilinear form on a Hilbert space H satisfying
(12.70) ll u l l 2 < Cb (u) , u E D( b) . Suppose that a(u, v ) is a bilinear form with D(b) C D(a) satisfying for m > 0 mb(u) < l a(u) l < Mb(u) , u E D (b) . ( 1 2 .71)
285
12. 9. Dissipative operators
Then for each linear function F on D (b) satisfying ( 12.72) 1 F v l 2 < Kb( v) , v E D(b) , there are elements w , u E D( b) such that ( 12.73 )
F v == a( v, w ) ,
v E D(b) ,
( 1 2.74 )
F v = a(u, v) ,
v E D(b) .
Proof. Let X be D(b) with scalar product b(u, v ) . Since b(u, v ) is a closed bilinear form, ( 12.70) and ( 12.71) imply that X is a Hilbert space (see the proof of Theorem 12. 1 1 ) . Now a(u, v) is a bilinear form on X and
< <
0 m l a(u) l < M, u E X, b(u) = 1 . ( 12.75 ) Let a be the restriction of a to X. Then by ( 1 2. 71 ) , a is a closed bilinear
form on X with D(a) == X. By ( 12.75) we see that W(a) is bounded and does not contain the point 0. Hence, a satisfies the hypotheses of Theorem D 12 . 1 1 . The conclusions follow immediately. An important case of Theorem 12.22
is
Corollary 12 . 2 3 . Let a(u, v ) be a bilinear form defined on the whole of H
such that ( 12.76 )
holds for positiv e m , M. Then for each bounded linear functional F on H, there are elements w, u E H such that
( 12. 77)
F v = a(v, w) ,
v E H,
( 12.78 )
F v == a(u, v) ,
v E H.
Proof. We merely take b( u, v ) to be the scalar product of H in Theorem 0
12.22.
Corollary 12.23 is sometimes called the Lax-Milgram lemma. 1 2 . 9 . Dissipative operators
In Section 12.5, we showed that if A is a linear operator on H such that
D(A) is dense in H and W(A) is not th� whole plane, a half-plane, a strip, or a line, then A has a closed extension A satisfying
( 12.79 )
a (A)
c
W (A) == W ( A ) .
286
1 2. BILINEAR FORMS
One of the questions we have deliberately deferred until now is: What hap pens when W (A) is one of these sets? We shall now discuss some of these cases. Further considerations are given in Section 1 2 . 10. Suppose W(A) is the whole plane. Then ( 12 . 79) is vacuously true for all extensions of A. On the other hand, the existence of a closed extension of A depends on whether A is closable ( Theorem 12. 18) . Hence, this case is not very interesting. In all other cases, including that of Section 1 2 . 5 , W ( A ) is contained in a half-plane ( Lemma 12. 19) . Thus , there ar� constants [ , k such that l r l == 1 and r (Au, u) - k ll u l l 2 < 0, u E D(A) .
�e
Set
B
( 12.80) Then D(B)
==
=
rA - k.
D (A) and
�e (Bu, u) < 0 ,
( 1 2.81 )
u E D(B) .
An operator B satisfying ( 1 2 . 8 1 ) is called dissipative. We are going to prove Theorem 1 2 . 24. Let B be a dissipative operator on H with D (B) dense in H . Then B has a closed dissipative extension B such that a ( B ) is contained in the hq/f-plane ,.\ < 0.
�e
In particular, if W(A) is a half-plane, then Theorem 12.24 gives a closed extension A of A satisfying ( 1 2 . 79) . In fact , all we ,.. need do is defin� B by ( 1 2 .80) for ,..appropriate [ , k. Then extend B to B by Theorem 1 2 . 24. The extension A defined by
,.. B + k A == r "'
--
clearly has all of the desired properties. Hence, T heorem 1 2 . 24 implies Theorem 1 2 . 25 . Let A be a densely defined lznear operator on H such that ,.. W (A) is a half-plane. Then A has a closed extenszon A satisfying (12. 79).
If W (A) is a line or a strip, then things are more complicated. We can use Theorem 1 2 . 24 to obtain a closed extension having one of the adjacent half-planes in its resolvent set as well. A study of the conditions under which this will be true is given in the next section. Let us now give the proof of Theorem 1 2 . 24.
1 2. 9.
Dissipative operators
Proof. By
(12.81),
287
�e ([I - B] u,
u
) > ll ul l 2 ,
showing that I - B is one-to-c.L1e. Hence, it has an inverse on B) . Define
(I - B) - 1 defined
R(I T == (I + B )(I - B) - 1 , where D ( T ) == R(I - B) . We claim that ( 1 2. 82) II Tu ll < ll u ll , u E D ( T ) . In fact , if (12. 83) then ( 1 2. 84) Tu == (I + B) v . Hence, 1 1 Tu ll 2 == l l v l l 2 + I I B v l l 2 + 2 �e (Bv, v) < l l v ll 2 + 11 Bv ll 2 - 2 �e (Bv , v) == II (I - B)v l l 2 == l l u l l 2 , by (12.81) and ( 12.83). Now, we can show that it is an easy matter to find an extension T of T in B(H) such that (12. 85)
[Remember that operators in B(H) are defined on all of H. ] To do this, first extend T to D ( T ) . This we do in the usual way. If u is an element in D (T) ,
()
then there is a sequence { un } of elements of D T converging to u in H. By { Tun } is a Cauchy sequence in H, and hence, has a limit z. Define Tu to be z . As usual, we check that this definition is independent of the
(1 2 . 82),
sequence chosen, and that T is an extension of T to D( T ) . Now, let u be any element of H. By the projection theorem (Theorem u == w y, where w E D T and y is orthogonal to D(T) . Define
2. 3 ),
()
,.. Tu == Tw.
Then
(12.85 ) holds. (12. 83) (12. 84) u == v - B v ,
Thus, T E B (H) , and Now by and
(12. 86) or ( 12. 87)
II Tu ll == II Tw l l < ll w ll < l l u ll ·
2 v == u + Tu,
Tu == v + Bv, 2B v == Tu - u.
+
288
1 2.
BILINEAR FORMS
The first equation in ( 1 2 87 ) together with (12.83) shows that I + T is one to-one and that its range is D(B) . Hence, we have .
B = (T - I) (I + T) - 1 . A candidate for the extension B is ( 1 2.88)
( 1 2.89) with D (B) = R(I + T ) . In order that ( 12.89) make sense, we must check that I + T is one-to-one. To see this, let u be an element of H such that ,..
( 1 2 . 90 )
(I + T)u == 0.
( 12.91)
l l g - v + nu ll 2 < ll v - nu ll 2 .
Let v be any element of H, and let n be a positive real number. Set g == (I + T ) v . Then, by ( 12. 85 ) , Expanding ( 12.91) out, we get
1 9 1 2 - 2 �e (g, v) + 2 n�e (g, u) < 0.
Divide by n and let n � oo. This gives ( 1 2.92 )
Since
�e (g, u) < 0,
g E R(I + T) .
R(I + T ) == D(B) , we see that R(I + T ) is dense in H. Hence, there is a sequence { gn } of elements in R(I + T ) such that 9n � u in H. Thus, ( 1 2.92 ) implies �e l l �t ll 2 < 0 , which shows that u = 0. Thus, the operator B given by ( 12.89 ) is well R(I + T )
:>
defined. It is clearly an extension of B. We claim that it is closed. To see this, suppose { un } is a sequence of elements in D(B) = R(I + T ) such that
Un � u, Bun � h in H as n � oo. Since, Un E R(I + T ) , there is a Wn E H such that Un == (I + T ) wn .
By ( 12.89 ) , Hence,
"'
"
2wn = Un - Bun � u - h as n � oo .
Since T E B(H) , this implies
2un == 2(1 + T )wn � (I + T ) (u - h) , 2Bun == 2 ( T - I) wn � ( T - I) (u - h) ,
289
12. 9. Dissipative operators
from which we conclude
2u
==
(I + T ) ( u ,- h ) , 2h
In particular, we see that u E R(I +
T)
==
==
�
('T - I) ( u - h) .
D (B) and
B u == ( T - I) (I + T ) - 1 u = ( i' - I) (u - h) ,..
==
h.
Hence, B is a closed operator. ,.. We must also show that B is dissipative. This follows easily, since
(Bu, u) where w
==
( [T - I] w, [I + T] w) (I + T ) - 1 u. Hence,
=
==
1 1 Tw ll 2 -- (w, Tw) + ( Tw, w) - ll w ll 2 ,
(12.93) by ( 12.85) . ,.. Finally, we ·must verify that ,.\ E p(B ) for we have
�e ,.\ > 0. Since B is dissipative, ,..
or ,..
�e ,.\ > 0. Thus , all we need show is
showing that B - ,.\ is one-to-one for that R(B - ,.\) == H for ,.\ > 0. Now,
�e
B - ,.\ == [ ( 1 - ,.\) T - (I + ,.\) ] (I + T ) - l . Thus, one can solve
( 12.94)
(B - ,.\ )u == f
if and only if one can solve
( 12.95)
[( 1 - ,.\ ) T .- ( 1 + ,.\)] w
=
f.
Note that ( 12.95) can be solved for all f E H when for ,.\ == 1 . If ,.\ -# 1 , all we need note is that
�e
�e ,.\ > 0. This is obvious
1 + ,.\ >1 1 - ,.\
,.\ > 0. Since I I T II < 1 , this shows that ( 1 + ,.\) / ( 1 - A) is in p( T ) for (Lemma 6.5) . Hence, ( 12.95) can be solved for all f E H. This completes the proof. 0
290
1 2. BILINEAR FORMS
1 2 . 10 . The case of a line or a strip Let us now outline briefly what one can do in case W (A) is a line -or a strip. Of course we can consider a line as a strip of thickness 0. So suppose that W(A) 'is a strip of thickness a - 1 , where a > 1 . As be fore, we can find an _ operator B of the form
B
( 12.96)
==
1A - k
such that W(B) is the strip 1 - a < �e z < 0. Thus , 1 - a < �e (Bu, u) < 0, u E D (B) , ll u l l ( 12.97) In particular, B is dissipative.
==
1.
B (a + B)(a - fJ ) - 1
B)
Now suppose B has a closed extension such that W( is the same strip and contains the two complementary ·half-planes. Set
p(B)
t == ( 12.98) Let u be any element of H, and set ( 1 2.99) v = (a - B ) - 1 u. Then
Tu == (a + B) v .
( 12. 100) Hence,
( 12. 101 ) showing that
( 1 2. 102) By ( 1 2.99) and ( 12. 100) , we have ( T + I)u = 2av , ( 12. 103)
( T - I)u
=
B
2 v.
Thus ( 12. 102) becomes
( 12. 104) or
1-a T II ( + l )u ll 2 a
< 1 1 Tu ll 2 - ll u ll 2 < 0,
(12. 105) In short , T is an extension of
-
T = (a + B) (a - B) - 1 , D(T) = R(a B) , vvhich satisfies ( 12. 105) and such that R( T ) = H. Conversely, if we can find such an extension T of T, then we can show that (12. 106)
( 12. 107)
1 2. 1 0. The case of a line or
a
291
strip
is a closed extension of B with W(B) = W(B ) :J o-(B) . In fact, we have, "' by the reasoning of the last section, that B is closed and dissipative, while (B ) contains the half-plane �e > 0. Moreover, ( 12. 105) and ( 12. 101) imply that W (B) is the strip 1 - a �e z < 0. However, we also want the half-plane �e 1 - a to be in We know that o: ( B - .X) = 0 and that R( B - .X) is closed for such points 12.3) . It suffices to show that the half-plane contains one point (Theorem ,.. in p(B) . Theil we can apply
p
A
<
A<
p(B) .
Theorem 1 2 .26. Let A be a closed linear operator on a Banach space X .
If ,.\ is a boundary point of not closed in X .
p(A), then either o: (A - .X) i= 0 or R(A - .X) zs
We shall postpone the proof of the theorem until the end of this section. To continue our argument , if the half-plane �e 1 - a contains a point of then the entire half-plane must be in (B) . Otherwise, it would contain a boundary point .X of B) . But this would imply, by Theorem 12.26, that either o:(B - .X) i= 0 or R(B - .A ) is not closed , contradicting the conclusion reached above. To complete the argument , let us show that the point ,.\ = - a is, indeed, in To see this , note that
p(B ),
A< p
p(
p (B).
( 12. 108) and since R ( T ) = R(a - B) = H , it follows that R (a + B ) = H This, coupled with the fact that o:(a + B) = 0, shows that - a is in So we must try to find an extension T of T satisfying ( 12. 105) and such that R( T ) = H . Let us consider first the case a = 1 (i.e. � the case of a line) . Now. ( 12 . 105) says
p(B).
I Tu l l u l , u E H =
( 12. 109)
.
.
A linear operator satisfying ( 12. 109) is called an isometry. If an isometry maps a normed vector space onto itself, it is called unitary. Thus, we want ,..
it to be an T to have-a unitary extension T. By continuity, we can extend "' isometry T of R(I - B) onto R(I + B) . Thus, to determine T, we need only · define it on R ( I + B) j_ . We can see that T would have to map R(I - B) j_ into R(I + B) j_ . This follows from the general property of isometries on Hilbert spaces:
(Tu, Tw) (u, w) =
(12. 1 10)
(see below) . Moreover, T must map onto R(I + B) j_ ; otherwise, we could not have R( ) = H . Thus , we have
T
292
1 2. BILINEAR FORI\IIS
B be a densely defined linear operator on H such that W(B) is the line �e ,.\ == 0. Then a necessary and sufficient condition that B have a closed extension B such that (12. 1 1 1) a (B ) W(B) == W(B) , is that there exist an isometry from R(I - B)j_ onto R(I + B)j_. In particular, Theorem 1 2 . 2 7. Let
A
c
this is true if they both have the same finite dimension or if they are both' separable and infinite dimensional.
R(I + B) j_ R(I - B)j_
and The last statement follows from the fact that have complete orthonormal sequences {
( 1 2. 1 12)
a
T
k == 1, 2 ,
·
·
·
.
If i= 1 (i.e . , in the case of a strip) , the situation is not so simple. It is necessary for T to map onto a closed subspace M such that
R(a - B) j_ H == R(a + B) tB M
and in such a way that ( 1 2. 1 05) holds. We shall j ust give a sufficient condi tion.
B be a densely defined linear operator on H such that W(B) a < �e < 0, a > 1 . If R(a - B) = R(a + B), then B has a closed extension B satisfying a(B) W (B) == W(B) . ( 12 . 1 13) Proof. On R(a - B)j_ == R(a + B)j_ we define T to be - I. Then T is isometric on this set. Thus , ( 12 . 1 04) [and hence, ( 1 2. 1 05)] holds for u E R(a - B) and for u E R(a - B)j_. For any u E H , u == u 1 + u 2 , where u 1 E R (a - B} and u2 E R(a - B)l_. Thus 1-a 1 - a i' 2 2 < I Tu 1 l 2 - l u 1 l 2 < 0 . I)u ( + I) u l = ( + . l T I I 1 a a But l 'tu1l l 2 - H�1 l 2 = I Tu 1 - u2 l 2 - 1 u 1 + u2 l 2 == 1 Tu l 2 - l u l 2 · Theorem 1 2 . 28. Let is the strip 1 -
z
""
c
Hence, ( 1 2. 1 04) holds, and the proof is complete. To prove ( 1 2 . 1 1 0) , expand both sides of This gives
I T(u + w) l 2 == l u + w l 2 · �e (Tu, Tw) == �e ( u , w ).
0
12. 1 0. The case of a lin e or
a
293
strip
Now, substitute iu in place of u. Thus, ( 1 2. 1 10 ) holds. In proving Theorem 12.26, we shall make use of Lemma 12 .29 . Let A be a closed linear operator on a Banach space X .
If ,.\ is a boundary point of p(A) and { ,.\ n } is a sequence of points �n p(A) converging to ,.\, then II (A - An ) - 1 1 1 � oo .
Proof. If the lemma were not true, there would be a sequence { ,.\ n } such that An � ,.\ as n � oo while
c
p(A)
( 12. 1 14 )
Since we have
<
1 2 1 I I (A - An ) - - (A - Am ) - 1 1 C I Am - An i � 0 as m, n � oo . Thus, (A - An ) - 1 converges to an operator E E B (X) as n � oo . Moreover, if x is any element in X, then Yn == (A - An ) - 1 x � Ex as n � oo . But
Ayn == (A - An ) Yn + AnYn � X + ,.\Ex as n � oo . Since A is closed, Ex is in D (A) and AEx == x + ,.\Ex, whence,
( 12. 1 15 )
(A - ,.\)Ex == x,
x E X.
Similarly, if x E D (A) , then (A - An ) - 1 (A - ,.\) X == X - ( ,.\ - An ) (A - An ) - 1 X � X as
n
� oo .
Hence,
( 12. 1 16 )
E(A - A )x == x,
x E D(A) .
This shows that ,.\ E p(A) , contrary to assumption. The proof is complete. 0 We can now easily give the proof of Theorem 12.26. Proof. If the theorem were not true, there would be a constant C such that
<
( 12. 1 17 ) l l x l l C I I (A - A )x l l , x E D (A) ( Theorem 3. 12 ) . Since A is a boundary point of p(A) , there is a sequence { An } of points in p(A) converging to ,.\ . Set Bn == (A - ,.\ n ) - 1 / I I (A - ,.\ n ) - 1 1 1 . .
294
12. BILINEAR FORMS
I Bn l = 1 . In particular, for each n there is an element Xn E X such ( 1 2. 1 18) xl n II 1 , I Bn xnl l > 1
Then that
2
=
·
Now, Hence
I (A - A)Bn l < I (A - An ) - 1 1 - 1 + ! An - .X I .
By Lemma 1 2 . 2 9 , this tends to 0 as n � oo. In particular, the norm of can be made less than 1 /3C for n sufficiently large. But by (A ( 1 2. 1 1 7) , we have
- .X ) Bn
1
2
< I Bnxn l < C I (A - A)Bn xn l
<
1
3 for large n. This contradiction shows that ( 12. 1 1 7) does not hold, and the proof is complete. 0 1 2 . 1 1 . Selfadjoint extensions
In Section 1 1 .3, we defined a selfadjoint operator as one that satisfies A* = A. In order for A* to be defined, D (A) must be dense. Moreover, A must be closed, since A* is. Note also that W(A) is a subset of the real axis, since (Au, u) = (u, A u ) = (A * u, u) = (Au , u) . Thus , W(A) is either the whole real axis, a half-axis, or an interval. We claim that
( 12. 1 1 9)
a (A)
c
W(A)
when A is selfadjoint. This follows from the fact that , by Theorem 12.3, o:(A - .X) = 0 and R(A - .X) is closed when rf:. W(A) . Moreover, if v is orthogonal to R(A - .X) , then
A
( v , (A - .X) u ) = 0 ,
u E D(A) ,
showing that v E D(A* ) = D (A) and ( [A - .X]v, u) = 0 ,
u E D(A) .
) A
A
Since D(A) is dense, we must have (A - .X v = 0. Now, if r/:. W (A) , then rf:. W (A) as well. To see this, note that if is not real, then neither of them is in W (A) . If is real, then .A = .X. By what we have just said, v = 0. This shows that R(A - .X) is dense as well as closed. Then E p(A - .X) , and the assertion is proved.
.X
A
A
295
12. 1 2. Problems
A
Now, suppose is a densely defined linear operator on H. Then is a subset of the real axis if and only if is symmetric, i.e. , if
W (A)
A (Au, v) == (u, Av), u, v E D (A). ( 12. 1 20) This follows immediately from Theorern 1 2 .7. If A is symmetric and ( 1 2. 1 1 9) holds, then we can show that A is selfadjoint . To see this , suppose that v E D (A*) . Let ,.\ be any nonreal point. Then ,.\ and .X are both in p(A), since neither of them is in W(A). Hence, there is a w E D(A) such that (A - .X)w == (A* - .X)v . Thus , ( v, (A - ,.\)u) == ([A * - .X]v , u) == ( [A - .X]w , u) == ( w , (A - ,.\ )u), u E D(A) , or (v - w, (A - ,.\)u) == 0, u E D(A). Since ,.\ E p(A) , this can happen only if v == w . Hence, we see that v E D ( A ) and Av == A*v . In particular , since every bounded operator satisfies ( 1 2. 1 19) '
( Theorem 12. 1 ) , we see that every bounded symmetric operator defined
everywhere is selfadjoint. Let us now pose and discuss the following question: Let A be a densely defined symmetric linear operator on a Hilbert space. Does it have a self adjoint extension? To examine the question, suppose first that is not the whole real axis . Then, by Theorem 12. 14 , has a closed extension satisfying
W(A) A A ( 1 2. 1 2 1 ) a ( A) W (A ) == W (A). In particular, A is symmetric and satisfies ( 1 2. 1 19) . Hence, A is selfadj oint On the other hand, if W(A) is the whole real axis , then A has a closed extension A satisfying ( 1 2 . 1 2 1 ) if and only if there is an isometry of R( i+ A)j_ onto R( i - A)j_ (just put B == iA i n Theorem 12.28) . Since, in this case, an extension violating ( 12 . 1 2 1 ) could not be selfadjoint , the condition is both necessary and sufficient for A to have a selfadjoint extension. The problem is, therefore, solved. A
c
1 2 . 1 2 . Problems
be a unitary operator on H. Show that for any subspace A L H one has A( Lj_) == A(L)j_.
( 1 ) Let
c
(2) Show that t he graph of the closure of an operator is the closure of the graph of the operator.
296
1 2. BILINEAR FORMS
(3) Show that a linear operator A is an isometry if and only if == I . Show that it is unitary if
A* A == AA*
( 4) Show that if
A* A == I.
A is normal, then . (Au, u) l I I A I == !lsup u ll = 1
is a densely defined linear operator on H. Show that A D(A*) is dense if and only if A is closed ( see Problem 1 1 .5) . (6) If A** exists, show that it is the smallest closed extension of A. (7) Show that if a(u, v) is a bilinear form such that W (a) consists of only the point 0, then a( u, v) 0 for all u, v E D( a). (8) Show that an operator A E B(H) is normal if and only if A* A AA*.
(5) Suppose
==
==
(9) Show that an unbounded linear functional cannot be closable. ( 10) Show that if either closable.
A or A' is densely defined , then the other is a( u,
( 1 1 ) Show that a bilinear form v ) is bounded if it is continuous in for each fixed v and continuous in v for each fixed \
u
u.
( 12 ) Show that in a complex Hilbert space, a symmetric bilinear form v) is determined by Is this true in a real Hilbert space?
a(u ). ( 13) Show that W # ffi. 2 implies W # IR 2 when W is convex. a(u,
(14) Show that the set of all z E IR 2 satisfying ( 12.58) forms a half-plane. ( 15) Show that a convex set of points consisting of rays from a point P is either an angle (} with vertex at P or the whole plane.
(16) Show that
r
and d can be chosen so that (i 2 . 60) holds.
Chapter
13
SELFADJ OINT OPERATORS 1 3 . 1 . Orthogonal projections
Let H be a complex Hilbert space, and let M be a closed subspace of H. Then, by the projection theorem (Theorem 2 . 3) , every element u E H can be written in the form
( 13. 1 )
u == u'
Set
+
u" ,
u' E M, u" E
Mj_ .
Eu == u' .
Then
E is a linear operator on H. It is bounded, since
(13.2) Since
Eu == u, u E M, I E I == 1 .
( 13 . 3) we have
( 13.4) We also have
( 13.5)
E
E 2 u == E ( Eu ) == Eu' == u' == Eu,
showing that is a projection. It is called the orthogonal projection onto M. Let us describe some other properties of
E.
(a) E is selfadjoint .
297
298
1 3. SELFADJOINT OPERATORS
Proof.
(u, Ev) == (Eu, Ev) == (Eu, v). D
R(E) == M. Proof. E M, then Eu == u. If u E R (E) , there is a v E H such that u D Ev == u . Thus, Eu == E2 v == Ev == u, by ( 13.5) . (c) u E M if and only if I Eu l == l u l · Proof. Since u == Eu + (I - E)u, we have (b)
If
( 1 3.6) If
I Eu l == l u l , then (I - E)u == 0, showing that u E M.
D
We also have Lemma 13 . 1 . If F is a bounded linear selfadjoint projection on H, then is closed and is the orthog onal projection onto R(
R( F)
F). F Proof. Set M == R(F). Then M is closed. This follows from the fact that if Fun then F 2 un Fv. Thus, v Fv, showing that v E M. Now, if u is any e!ement of H, then u Fu + (I - F)u. Moreover, Fu E M, while (I - F)u E Mj_ , since (Fv, (I - F)u) == ( [F - F2]v, u) 0, v E H. Thus F is the orthogonal projection onto M. D --4 v ,
==
--4
==
==
Suppose A E B (H) and M is a closed subspace of H. We say that M is invariant under A if A(M) C M. I f both M and Mj_ are invariant under then M is said to reduce A.
A,
Let M be a closed subspace of H, and let onto M. Then we have
E be the orthogonal projection
Lemma 1 3 . 2 . If A is in B(H) , then {i) M is invariant under if and only if {ii) M reduces A if and only if
A
AE EAE ; AE == EA. ==
299
1 3.2. Square roots of operators
A, AEu AE EAE u
then Proof. ( i ) If M is invariant under E M for all and E M, then Conversely, if Hence E M. ( ii ) Suppose M reduces We have
. == EAEu AEu EAEu == EAu .
==
u E H.== Hence,== Au AEu
Au A. EA == EAE + EA(I - E). Since M is invariant under A, EAE == AE. Moreover, for any u E H, (I E)u E Mj_ so that A(I - E)u E Mj_ by hypothesis. Hence, EA(I -E)u == 0, showing that the operator EA(I - E ) == 0. Hence, EA == EAE == AE. Conversely, if AE == EA, then EAE == AE 2 == AE , showing that M is invariant under A. Moreover, if u E M then Eu == 0. Thus EAu == AEu == 0 0. This shows that Au E M and the proof is complete. j_ ,
j_ ,
1 3 . 2 . Square roots of operators
Let us prove the following:
satisfyzng 0 < A < 1. Then there AB >be0anysuchrealthatnumber B2 == A. Proof. Suppose B exists. Set R 1 -A, S 1 - B. Then ( 1 - 5 ) 2 1 -- R, or 1 (13.7) S == 2 (R + S2 ). Conversely, if we can find a solution S of (13.7), then an easy calculation shows that B == 1 - S satisfies B 2 == A. So all we need do is solve ( 13. 7 ). To solve it, we use the method of iterations employed in Section 1.1. Set So == 0 and 1 (13. 8 ) Sn+ l == 2 (R + Sn2 ), n == 0 , 1 , · · · . If the sequence {Sn } has a limit S, then S is clearly a solution of (13.7). Now we claim that 0 < Sn < 1 , n == 0, 1 · . (13. 9 ) This is true for n == 0 , and if it is true for n, then 1 1 2 < Sn+ l 2 (R + Sn ) 2 (R + 1 ) < 1 , showing that it is also true for n + 1 . Hence , ( 13. 9 ) holds for all n by induction. Note also that Sn is a polynomial in R with nonnegative coefficients. Again, this is true for n == 0 , while ( 1 3. 8 ) shows that it is true for n + 1 whenever it is true for n.
Theorem 13.3. Let exists a real number
==
==
==
·
=
·
300
13. SELFADJOINT OPERATORS Next we claim that
> Sn , n 0 , 1 , This follows from the fact that 1 1 1 2 2 (R + Sn _ 1 ) == ( Sn + Sn- 1 ) ( Sn - Bn- 1 ) , Sn+ 1 - Sn == (R + Sn ) 2 2 2 which shows, in view of ( 13.9) , that ( 13. 10) holds for n == k whenever it holds for n == k - 1 . Since (13. 10) holds for n 0, it holds for all n by ( 13. 10)
==
Sn+ 1
·
·
·
.
-
==
induction. From ( 13.9) and ( 13. 10) we see that { Sn } is a nondecreasing sequence of real numbers bounded from above. By a theorem that should be well known to all of us , such a sequence has a limit . This is precisely what we wanted D to show. The proof is complete.
A2 n, A B2 2 C n B A.
A2 > B Ajn . M. C
n
Note that the restriction < 1 is not necessary. If 1 , we let be a positive integer satisfying < and we take == Then 0 1, and consequently, we has a positive square root If we take == == see that 0 and == Let us consider the following question: Given an operator does it have a square root? By a square root of an operator, we mean an operator such that == Let us describe a class of operators having square roots. An operator A E is called positive if
C>
B
B 2 A.
( 13. 1 1 )
B(H) (Au, u) > 0, u E H
i .e. , if W (A) is contained in the nonnegative real axis] . Actually, such an [operator should be called nonnegative. Still, no matter what we call it , we write A > 0 when A satisfies ( 13 . 1 1 ) . Since we are in a complex Hilbert space, we know that such operators are selfadjoint ( see Section 12. 1 1) . The expression A > B will be used to mean A - B > 0. An important observation is Lemma 13.4. If ( 13 . 1 2) M I < A < MI M > O , then ( 13. 13) I A I < M. ,
-
Proof. By ( 13. 12) ,
I (Au, u) l < M l u l 2 , u E H. Hence, by Lemma 12.4, I (Au , v) l < M l u l l v l , u, v E H. (13. 14)
·
' ' I
1 3. 2.
301
Square roots of operators
If we take v
==
Au, we obtain 1 Aul l 2 < M l u l I Au l , ·
which implies ( 13. 13) .
D
Returning to square roots, we shall prove the following:
If A A.
Theorem 1 3 . 5 . is a positive operator in B ( H ) , then there is a unique B > 0 such that B 2 == A. Moreover, B commutes with any E B ( H ) which commutes with
C
Proof. It suffices to consider the case
< A < I. To see this, note that the operator A / I A l always satisfies ( 13. 15) , and if we can find an operator G such that G 2 A / I A I , then B I A I 1 12 G satisfies B2 == A. Now, let us follow the proof of Theorem 13.3. Set R I - A. Then 0 < R < I. We wish to solve ( 1 3.7) . Set So == 0, and define Sn + l inductively by means of ( 13.8) . Then 0 < Sn < I, n 0 , 1, (13. 16) This is surely true for n == 0. Assume it true for n. Then 1 1 ( 13. 17) (Sn+ l u , u) == 2 (Ru , u) + 2 I Sn u l 2 , which shows immediately that Sn + l > 0. Moreover, by ( 13. 16) and Lemma 13.4, I Sn l t < 1 . Thus , ( 13. 1 7) implies Sn + l < I, showing that ( 13. 16) holds with n replaced by n + 1 . Consequently, ( 13. 1 6) holds for all n by induction. Next , we note that Sn is a polynomial in R with nonnegative coefficients. Again, this is true for n 0, and if it is true for n, then ( 13.8) shows that 0
(13. 15)
==
==
==
==
·
·
·
.
==
it is true for n + 1 . We can easily show that
Sn+ l - Sn 2 (Sn + Sn- d (Sn - Bn- 1 ) is a polynomial in R with nonnegative coefficients. This is true for n == 0, and (1 3. 18) shows that it is true for n == k if it is true for n == k - 1 . Now we can show that ( 13. 19) Rk > 0 , k 0, 1 , If k 2j, then (Rku, u) I RJ u l 2 > 0, while if == 2j + we have (Rk u, ) (RRJ u, Rj u) > 0. =
(13. 18)
1
==
==
==
k
1,
u
==
·
·
·
.
13. SELFADJOINT OPERATORS
302 From ( 13. 19) and the fact that each nonnegative coefficients, we see that Sn+l > Sn ,
( 13.20)
Sn+ l - Sn is a polynomial in R with 11 ==
0, 1 , To summarize , we see that the sequence { Sn } satisfies ( 13.21) 0 < Sn < Sn + l < I , n == 0, 1, ·
·
·
.
·
Sn
·
·
.
If the were real numbers and not operators, we could conclude that the sequence approaches a limit , and the first part of the theorem would be proved. The fact that we can reach a similar conclusion in the present case follows from Lemma 1 3 . 6 . If { Sn } is a sequence of operators in then there is an operator S in such that
B(H) Sn u � Su, u E H.
(13.22)
B(H) satisfying {13. 21),
Assume Lemma 13.6 for the moment, and let us continue the proof of Theorem 13.5. By ( 13.22) , we see that the operator S is a solution of (1 3.7) . Thus, == I - S is a square root of Now, let E be any operator Then it commutes with R == I that commutes with and since each must commute with each Sn . Then Sn is a polynomial in R,
B
A.
H) B( C A, A. C CSnu == Sn Cu , u E H. rraking the limit as � oo , we see that C commutes with S, and hence, with B == I - S. Now, suppose T > 0 is another square root of A. Then T commutes with A, since TA TT == T2 T == AT. Therefore, it commutes with B. Let u be any element of H , and set v == (B - T) u. Then ([B + T] v, v) ( [ B2 - T2]u, v) == ([A - A]u, v) == 0. But B 0 and T > 0. Hence, we must have (Bv, v) == (Tv, v) == 0. Since B > 0, there is an F E B( H ) such that F 2 == B. Consequently, 1 Fv l 2 == (F2 v, v) == (Bv, v) == 0, showing that Fv 0. But this implies Bv == F2 v 0. Similarly, we have Tv == 0. Thus , I (B - T)u l 2 == ( [B - T] 2u, u) == ( [B - T]v, u) == 0, n
=
2
==
>
==
=
/
1 3. 2. Square roots of operators
showing that
303
B u = Tu
B
for all u E H. Hence, is u:Gique. The proof of Theorem 13.5 will be complete once we have given the proof D of Lemma 13.6. Proof. Set
(
Then by 13.?.1 ) ,
( 13.23 )
Hence,
I Smn u l 4 = (Smn U , Smn u) 2 < (Smn U , u)(S�n u, SmnU) < [ ( Sn u,u) - (Sm u,u )] l u l 2 . This follows from Corollary 12.6 we take b(u) = (Smn U, u). Now, for each u E H the sequence { ( Sn u, u) } is nondecreasing and bounded from above. Hence, it is convergent . Thus, inequality ( 13.24 ) implies that {Sn u } is a Cauchy sequence in H. Consequently, the sequence converges to a limit which we denote by Su . Clearly, S s a linear operator on H. Moreover, (Su, u) = lim(Sn u, u), showing t � at 0 S < I. Thus , S E B(H), and the proof is complete. 0 ( 13.24 )
if
i
:S
A consequence of Theorem 13.5 is
A > 0, B > 0 and AB = BA, then BA > 0. Proof. By Theorem 13.5, A and B have square roots A 1 12 > 0 and B 1 1 2 > 0 which commute. Hence, (ABu, u) = I A I /2 B l f2 u l 2 > 0. eorollary 13.7. If
I
A fact that we shall need later is
M2
0
H,
E1
and Lemma 13.8. Let'M1 and be closed subspaces of and let be the orthog onal projections onto them, respectively. Then the following statements are equivalent:
E2
{a) {b) {c) {d)
E1 < E2 ; I E1 u l < I E2 u l , u E H; M1 M2 ; E2 E1 = E1 E2 = E1 . c
304
1 3. SELFADJOINT OPERATORS
2 == (E1 u, E1 u) == (Eru, u) == (E1 u, u) < E u I l 1 (E2 u, u) == I E2 u l 2 . (b) implies (c) . If u E M1 , then l u l == I E1 u l < I E2 u l < ! l u l [cf. Cc) of Section 13. 1] . Hence, ! l u l == I E2 u l , showing that u E M2 . where v E M1 then u == v + (c) implies (d) . If u is any element of and E M:f-. Thus, E1 u == v E M2 . Hence, E2 E1 u == E2 v == v == E1 u . Taking adjoints , we get E1 E2 == E1 . (d) implies (a) . E2 - E1 == E2 - E1 E2 = (I - E1 )E2 . Since E1 and E2 commute and I - E1 > 0, E2 > 0, we see by Corollary 13. 7 that (I - E1 )E2 > D 0. Hence, (a) holds, and the proof is complete. Pro of. (a) imp lies (b) .
H,
w
w,
1 3 . 3 . A decomposit ion of operators
Let us now show that a bounded, selfadjoint operator can be expressed as the aifference of two positive operators. Suppose A is a selfadjoint operator is positive. Hence, it has, a square root that in B (H) . Then the operator is positive and commutes with any operator commuting with A 2 (Theorem 13.5 ) . Denote this square root by I A I . Since any operator commuting with A also commutes with A 2 , we see that ! At--commutes .wJth any operator that commutes \vith A. Set 1 ( 13.25 ) A + = ( I A I + A) , A) .
A2
�
A- == -2 ( I A I -
These operators are selfadjoint and commute with any operator commuting with A . They satisfy + + A == A - A - , I A I == A + A - . ( 13.26 ) Moreover,
( 13.27 )
E be the orthogonal projection onto N(A+ ) (see Section 13. 1 ) . Thus, A + E == 0. ( 13.28 ) Let
Taking adjoints , we obtain
( 13.29 )
Now by ( 13.27 ) , R(A- )
( 13.30 )
and by adjoints
( 13.31 )
C
N(A + ) . Hence, EA - == A - ,
1 3. 3. A decomposition of operators
We see, therefore, that both do and [see ( 13.26)} . Note next that
IAI A
( 1 3.32) ( 13.33) ( 1 3.34)
305
A+ and A - commute with E. Consequently, so
EA == E(A + - A - ) == EA + - A - == -A - , E I A I == E(A+ + A - ) == A - , (I - E)A == A - EA == A + A - == A+ ,
and ( 13.35)
E, I - E I A I
Since are positive operators which commute, we see from and ( 1 3 . 33) and ( 13 .35) that
A+ >- 0 (Corollary 1 3 . 7) . Hence, by ( 13.26) , (1 3.37) I A I > A+ ' I AI > A- .
( 1 3.36)
'
Also, Therefore, ( 13.38) We now have the following:
B E B(H ) commutes with A, then 'lt commutes with E. Proof. As we mentioned above, B commutes with A + . Thus, BA + == A + B. This implies that N(A + ) is invariant under B (see Section 1 3 . 1 ) . Thus BE == EBE (Lemma 1 3 . 2) . Taking adjoints , we get EB == EBE , which 0 implies BE = EB. Lemma 1 3 . 9 . If
We also have
B ( H ) which commutes with A and B B > ± A . B > I A 1 . I A I 'lS the "smallest " operator having Proof. Since B - A 0 , I - E > 0 and they commute, we have (I - E)(B - A) > 0 (Corollary 1 3 . 7) . Thus by ( 1 3 . 34) , ( 1 3.39) ( I - E )B (I - E)A == A + .
Lemma 1 3 . 10 . Let be an operator 'ln satisfies Then Thus, these properties. 2:
>
306
1 3.
SELFADJOINT OPERATORS
Similarly, since B + A > 0 and E > 0, we have E(B + A) > O, which implies
(13 . 40)
by ( 13. 32) . Adding ( 13.39) and ( 1 3 .40) , we obtain B > A+ + A- = I A I , which proves the lemma.
D
In addition, we have
B( H) which commutes with A
Lemma 1 3 . 1 1 . Let B > 0 be an operator in and satisfies B > A. Then B > A + . Proof. By
(13. 3 9),
since BE > 0.
D
Also, we have Lemma 1 3 . 1 2 . Let B be a positive operator in A and satisfies B > - A . Then B > A - .
B(H) which commutes with
Proof. By ( 1 3.40 ) , EB > A - . But (I - E) B > 0. Hence, B > A -·.
0
1 3 . 4 . Spectral resolution
We saw in Chapter 6 that , in a Banach space X, we can define f(A) for any A E B ( X ) provided f(z) is a function analytic in a neighborhood of a(A) . In this section, we shall show that we can do better in the case of selfadjoint operators. To get an idea, let A be a compact , selfadj oint operator on Then by Theorem 1 1 . 3,
H.
(13. 4 1 )
where { 'Pk } is an orthonormal sequence of eigenvectors and the Ak are the corresponding eigenvalues of A. Now let p( t ) be a polynomial with real coefficients having no constant term
(13. 42)
p(t) =
m
L ak tk . 1
Then p(A) is compact and selfadjoint. I_Jet Then JL = p(A.) for so� A E a (A) (Theorem
JL
# 0 be a point in a(p(-(1.) ) . Now >.. # 0 (otherwise we
6. 8 ) .
1 3. 4. Spectral resolution
307
) p( O JL If <.p [p (A) - JL] <.p = L ak Ak <.p - JL<.p = L ak Ak <.p - JL<.p = [p(A) - JL] <.p = 0. Thus JL is an eigenvalue of p(A) and <.p is a corresponding eigenvector. This shows that would have = = 0) . Hence, it is an eigenvalue of A ( see Section 6. 1) . is a corresponding eigenvector, then
( 13.43)
p (t)
Now, the right hand side of ( 13.43) makes sense if is any function ,bounded on ( see Section 1 1 .2) . Therefore it seems plausible to define by means of ( 13.43) . Of course, for _such a definition to be useful, one would need certain relationships to hold. In particular, one would want We shall discuss this a bit later. to imply f A)g ) = f (t)g(t) = If is not compact , we cannot , in general, obtain an expansion in the form ( 13.41 ) . However, we can obtain something similar. In fact , we have
a(A)
p(A)
A
h ( t)
( (A h (A).
A be a selfadjoint operator in B (H) . Set = inf (Au , u), M = sup (Au, u). Then there is a family { E(A ) } of orthogonal proj ection operators on pending on a real parameter A and such that: Theorem 13. 13. Let m
llull =l
llull=l
H
de
E(A)u � E(Ao)u as Ao < A � Ao, u E H; E(A) = 0 for A < E(A ) = I for A > M; {3) (4) AE(A) = E(A ) A; if a < b > M and p(t ) is any polynomial, then {5) P(A) = 1b p( >..) dE(>.. ) . ( 1 3.44) This means the following. Let a = Ao < A 1 < · < An = b be any partition of [a, b], and let A� be any number satisfying A k - 1 < A� < A k· Then ( 13.45) L1 p (A� ) [E (Ak ) - E(Ak- 1)] � p(A) in B (H) as TJ = max ( A k - A k - 1 ) � 0. {2)
m,
m,
·
n
·
308
1 3. SELFADJOINT OPERATORS
A(A) A -.- A. Then ( 13.46) A(A 1 ) > A(A2 ) for A 1 < A2 . Let the operators I A (A) I , A + (A), A-(A) be defined as in the preceding section. Then ( 13.47) Proof. Set
=
by (13.38) , and hence
( 13. 4 8) (Lemma 13. 1 1 ) . Thus,
A(A) == I A (A) I A+ (A), A < (Theorem 13.5) . Similarly if A > M, then A(A) < 0, in which case , we have A(A) - I A (A) I == - A - (A), A > M. (13.50) In general, if A 1 < A 2 , we have, by ( 13.48) and Corollary 13.7, A+ (A2 ) [A+ (A 1 ) - A+ (A2 ) ] > 0. ( 13.49)
m
=
=
Hence,
( 13.5 1 ) This implies
N [A+ (AI) ] N [A+ (A2 )], A 1 < A2 . Let E(A) be the orthogonal projection onto N[A + (A) ] . Then ( 13.53) E(A I ) < E(A2 ), A 1 < A2 , by Lemma 13.8. Moreover, for A < we have ([A - A]u, u) > ( A)i l u l 2 , showing that N [A (A)] {0} , A < Thus, by ( 13.49) we have ( 13.54) E(A) 0, A < If A > !vf, we have A+(A) == 0 by ( 13.50) so that ( 13.55) E(A) I A > M. Set E(A1, A2 ) == E(A2 ) - E(A l ) · Then E(A 1 , A 2 ) > 0 for A 1 < A 2 . Therefore, ( 13.52)
c
rn ,
m -
==
m.
==
==
( 13.56)
m.
,
30 9
1 3.4. Spectral resolution
and
(13.57) ( Lemma 13.8) . Hence,
A(A2 )E(A 1 , A2 ) A(A2)E(A2 )E(At, A2 ) - A - (A2 )E(A l , A2 ) < 0 by (13.32) . Also, A(AI)E(A 1 , A2 ) == A(A 1 )[I - E(A 1 ))E(At, A2) == A+ (A 1 )E(AI, A2 ) > 0 ==
==
by ( 13.34) . Combinir1g these two inequalities , we obtain
(13.58)
a < m, b > M, and let a == Ao < A 1 < < An == b be any partition of [a , b] . Then by ( 1 3.58) , n n (13.59) L1 Ak - I[E (Ak ) - E( Ak - 1 ) ] < A L1 [E (Ak ) - E(Ak - 1 ) ] n == A < L A k [ E (A k ) - E(A k - 1 ) ] . 1 If A � is any point satisfying A k - l < A � < A k , then n n A - L A� [E (Ak ) - E(Ak - 1 ) ] < L (Ak - Ak - I ) [E ( .Ak ) - E(Ak - I ) ] < max (Ak - Ak - 1 )1 == rJl. Similarly, n A - L A� [E (.Ak ) - E(Ak - 1 ) ] > - TJI. ( 13.60) These inequalities imply, by Lemma 13.4, that n (13.61) I A - L A�[E (Ak ) - E(Ak - 1 ) ] 1 < Hence, n A == lim L1 .A. � [E ( Ak ) - E(Ak- 1 ) ] 1b AdE(A). Next , we can show that for each integer s > 0, Next , take
· · ·
1
1
1
TJ .
I
ry � o
( 13.62)
=
a
310
1 3. SELFADJOINT OPERATORS
This follows from the fact that for j < k , we have
E(Ak- I, Ak )E(Aj - I, Aj )
E(Ak )E(Aj ) - E(Ak )E(Aj - I) - E (Ak - I )E( >..j ) + E(Ak- I)E(Aj - I) = E( Aj ) - E(Aj - I) - E(Aj ) + E(Aj - I) = 0. But the left hand side of ( 13.62) approaches A 8 as 0, while the right hand side tends to TJ --4
Hence,
( 13.63 )
from which we conclude that ( 13.44 ) holds. We have proved ( 1 ) , ( 3 ) , ( 4 ) and ( 5) . It remains to prove ( 2 ) . Since
E(AI, A) > E(AI, J-L) , A > J-L, we see that E(AI, A) is nondecreasing in A. Hence, it approaches a limit G(AI) B( H) as A approaches AI from above (Lemma 13.6 ) . Statement (2) of the theorem is equivalent to the statement that G( >.. I ) = 0. To see that this is indeed the case, note that AIE ( AI, A) < AE(AI, A) < AE(>..I , A), A > AI, by ( 13.58 ) . Letting A AI, we obtain A l G(AI) < AG(AI) < AI G(AI), or 0 < A(AI)G ( AI) < 0, which implies, by Lemma 13.4, that A(>..I )G ( A l ) = 0. Hence A+(AI)G(AI) = (I - E(AI))A(AI)G(AI) = 0 by ( 13.34 ) . This shows that R[G (AI)) N[A+ (AI) ], E
--4
C
which implies
( 13.64)
But
E(AI)E(A l , A) = 0, AI < A
by ( 13.57 ) . This shows that
311
1 3. 5. Some consequences
This together with ( 13.64) implies 13. 13 is complete.
G(,.\ 1 )
==
0 , and the proof of Theorem
0
{ E(,.\)}
The family is called the resolution oj the identity corresponding to A . Theorem 13. 13 is a form of the spectral theorem. 1 3 . 5 . Some consequences
Let us see what consequences can be drawn from Theorem 13. 13.
1. If p
(,.\) > 0 in [a , b] , then p(A ) > 0.
Proof. This follows from the fact that n
L1 p (,.\k )[E (Ak ) - E(,.\k - 1 ) ] > 0. 0 2. We can define f (A ) for any real function f (,.\) continuous in [a , b] . Proof. We can do this as follows: If f (,.\) is continuous in [ a , b] , one can find a sequence {Pj (,.\)} of polynomials converging uniformly to f(,.\) on [ a , b] ( see Section 1 0 . 5 ) . Thus, for each E > 0, there is a number N such that ( 13.65) a < ,.\ < b. I Pj ( ,.\) - Pi ( ,.\) I < E, i , j > Thus, by our first remark, ( 13.66) - E l < Pi ( A ) - Pi ( A ) < c l, i,j > N , which implies , in view of Lemma 13.4, ( 13.67 ) i, j > N. I I Pj ( A ) - Pi ( A ) I I < Consequently, the sequence {pj (A) } converges in B (H) to an operator B. Moreover, it is easily verified that the limit B is independent of the particular choice of the polynomials 'Jlj (,.\) . We define f ( A ) to be the operator B . 0 3. f (A ) = I: f (,.\)dE(,.\). Proof. To see this, let {Pj (,.\) } be a sequence of polynomials converging uniformly to f (,.\) on [a , b] . Then for E > 0, ( 13.65 ) and ( 13.67 ) hold . Letting i oo, we have ( 13.68 ) IPj ( ,.\ ) f (,.\) 1 < E , j > N, 1V,
c'
---+
�
and
( 13.69 )
I I Pj ( A ) - / ( A ) I I < E ,
j > N.
312
1 3. SELFADJOINT OPERATORS
Now,
n
f (A) - L J (.X�) [E (.X k ) - E(Ak - 1 ) ] == { f (A) - pj (A)} 1
n
+ k==L[pj (A�) - f (.X�) ][E ( .Xk ) - E(.Xk - 1 )] == Q 1 + Q2 + Q3. 1 Let j be any fixed integer greater than N. Then - Ef < Q3 < Ef max (.Xk by ( 1 3.68) , while I Q 1 I < E by ( 13 .69 ) . Moreover, I Q 2 I � 0 as Ak - l ) � 0 (Theorem 13. 13) . Hence, we can take so small that 1 / (A) - L1 f (.X� )[E (.Xk ) - E(.Xk - 1 ) ] I I < 3c , 17
1] =
n
and the proof is complete.
0
f (.X) + g (.X) h(.X) in [a , b] implies f (A) + g(A) == h(A). 5. f (.X)g(.X) == h (.X) in [ a , b] implies f(A) g (A ) h (A) . Proof. If { Pi (.X)} converges uniformly to f (.X) in [a , b] and { Qj (.X)} converges uniformly to g( .X) there, then { pj (.X) + qj (.X)} converges uniformly to { f (.X) + } and {Pj (.X)qj (.X) } converges uniformly to { f (.X)g (.X)} there. Conse g(.X) quently, PJ (A) + qj (A) converges in B(H ) to f (A) + g (A), and 0 Pi (A) qj (A) converges in B(H) to f (A)g(A) . 6. f (A) commutes with any operator in B(H ) commuting with A. Proof. Any such operator commutes with each E(.X) (Lemma 13.9) . 0 4.
=
=
f (.X) > 0 for a < .X < b, then f (A) > 0. 8. l f (A) u l 2 = J: J (.X) 2 d(E(.X)u , u), where the right hand side is a Riemann-Stieltjes integral. 7.
If
313
1 3. 5. Some consequences
Proof. That the integral exists is obvious , since ing function of Now.
A. .
(E( A. )u, u) is a nondecreas
I / (A)u l 2 == lim( L f ( A.k ) [E ( A. k ) - E( A. k - I)]u, L f ( A.j) [ E ( A.j ) - E(Aj - I ) ]u) == lim L f ( A. k ) 2 ([ E ( A. k ) - E( A. k - l )] 'lt , u ) = 1b f (A) 2 d(E(A)u, ) w
by the statement following ( 13.62) .
9. l f (A) I < a<>.max. < l f ( A. ) I . b Proof. By Remark 8, l f (A) u l 2 < a<>.tnax.< f (A. ) 2 lim L ( [E ( A.k ) - E( A.k -l )]u, u) b 2 < c�f�b I J (A) I ) l u l 2 ·
0
/
0
a < o: < (3 < b and E ( o:) == E({3) , then f (A) = 1a f (A)dE(A) + hb f ( A)dE(A) . Proof. Just take o: and (3 as partition points in the definition of the inte 0 grals. 1 1 . If o: < A.o < and E(o: ) == E((3), then A.o E p(A) . Proof. Choose a < and b > so that a < o: < A. o < b� Set (3 ( A. - A.o) in [a , n ] and [(3 , b] , and define it in [o: , (3] in such a way gthat( A. ) it== is1 /continuous in [ a , b] . T-h en g (A) E B(H) and g (A)(A - Ao ) = 1b g(A)(A - A0 ) dE(A) == la dE(A) + f b dE(A ) a }(3 = 1 b dE(A) = I 10. If
j3
m
A�f
<
.
0
1 3, SELFADJOINT OPERATORS
314
Ao is a real point in p(A) , then there is an n < .Xo and a f3 > ,.\0 E(n) == E( /3). Proof. If the conclusion were not true, there would be sequences { nn } , { f3n } such that A o > On � .X o, Ao < f3n � .X o and E(n n ) # E ( f3n ) · Thus, there would be a Un E H such that l ttn l == 1 , E(nn )Un == 0, E( f3n )Un == Un (just take Un R [ E ( f3n ) ] n R [ E (nn ) ] j_ ) . This would mean n {f3 I (A - Ao)un l 2 = Jan (A - .Xo) 2 d(E ( ,.\)un , Un ) < (f3n - nn ) 2 l un l 2 � 0 as n � oo. But this would imply that .Xo E a(A), contrary to assumption . This com pletes the proof. 12.
If such that
E
D
1 3 . 6 . Ur1bounded selfadjoint operators
The spectral theorem (Theorem 1 3 . 13 ) is very useful for bounded selfadjoint operators. Is there a counterpart for unbounded selfadjoint operators? What would it say? To guess , let us propose a candidate for such a "theorem" using Theorem 13. 13 as a model. Firstly, we have to adjust the definitions of M and m to read m
=
inf ED( u A ) , II u ll = l
(Au, u),
M ==
sup u E D(A) , II u l l = l
(Au, u),
and we have to realize that we may have m == -oo or M = oo, or both. Secondly, ( 4) of that theorem cannot hold if is not the whole of H. Thirdly, the counterpart of ( 13.44) would be
D (A)
p(A) = 1: p (A)dE(A),
and we have to define such an integral. In dealing with a statement to replace ( 4) of Theorem 13. 13, it is con venient to use the following notation. We say that an operator is an extension of an operator S and write S C if S C and
T D ( ) D (T)
T
Su == Tu, u E D (S) . While we cannot expect statement ( 4) to hold in general, we would like a statement such as E(.X)A AE(.X) . ( 1 3 . 70 ) Even if we are willing to replace (4) with ( 13 . 70) , it is not clear how to find such a family {E(,.\) } . One attack is based on the following approach. c
13. 6. Unbounded selfadjoint operators
315
H {Hn } H. ( y ) == 0, "# n, E Hm , y E Hn , and each u E H there is a sequence { u 1 , u2 , } with Uj E Hj and n as j oo. U u j L j =l Note that the Uj are unique , for if uj E Hj and n as n --+ oo , j u u L j=l then n 0 as n --+ oo. (uj uj) L j= l Since Uj - uj U m - u� for j "# we have , f r each n l um - u�l l 2 == jL=l (uj - uj), Um - u� 0 as n --+ oo , showing that Um == u� . Assume that for each n we are given a bounded , self adjoint operator An on Hn . Then each A n has a spectral resolution {En (,.\)} of the identity on Hn by Theorem 1 3 . 13. Define ( 13.71 ) A Au == L u i i i= l
of closed subspaces of which are Suppose we can find a sequence pairwise orthogonal and span the entire space By this we mean x,
m
x
for
·
·
·
--+
--+
--+
--+
m,
j_
o
m,
--+
00
when
00
u == L Ui · i=l
( 1 3. 72)
Without further as s umptions, there is no reason to belie ve that the series (1 3.71 ) will converge in We define to be the set of those E that
H D (A) u such 2 < oo . (13. 73) A u L il l i I i=l Clearly, D(A) is a linear subspace of H. It is dense since it contains all elements of the form . u L i i= l H.
00
n
13. SELFADJOINT OPERATORS
3 16
A is symmetric . To show this , we note that if V E D (A) ' U == v u L L i i ' i=l i=l
Moreover, the operator
00
00
==
then
00
00
== (u, A Av). (A ) ) , (Au, v) L , v (u iv u i L i i i i i=l i=l ==
==
It is also selfadjoint. We show this by noting that if 00
00
U == u L f L fi E _ H i ' i=l i=l and ( u , Av) ( J , v), v E D(A), then Ai vi ) = L= ( h vi ) · , (u L i i=l il Take Vi == 0 for i #- m, Vm == v . Then (um , Am v) == ( fm , v), V E Hm . Since A m is selfadjoint on Hm , we see that A m u m == fm · Consequently, 2 == L l /ml l 2 < 00 , A u m L m I l m= l m=l ==
==
00
00
00
00
and
00
00
= A Au = L . = u f f i L i i i=l i=l Note that A reduces to A n on Hn . This follows from the fact that if Ut--== 0 then for i #Au == An Un . It is the only selfadjoint operator that does this. To see this, let A be selfadjoint on H satisfying A ui Ai ui , Ui E Hi . If u E D (A) (i.e. , it satisfies ( 1 3 . 72) and ( 1 3 . 73) ) , then n 2 n I Lm Ai uil l == L -ri Ai uil l 2 0 as Consequently, n n Au. Un L Ui --+ U, .ii Un = L A i ui i=l i=l n,
-
==
--+
m , n --+ oo .
m
==
�
1 3. 6.
Thus ,
Unbounded selfadjoint operators
3 _1 7
n ) , A v ( un , Av) = :L(u i i i i=ln = l:= l (Ai ui , vi ) i
v E D(A). In the limit , this gives (u, Av) = (Au, v ) , v E D (A). Since A is selfadjoint , we have u E D(A ) and A u = A u . Thus D(A) D(A) and Au = Au , u E D(A). But if u E D(A) , then (u, Av ) == (Au, v ), v E D(A ) , and consequently, (u, Av) = (Au , v),- v E D(A). This implie.s that u E D (A) and Au = A u . Hence, D (A) = D(A) and A == A. Next , suppose A is a selfadjoint operator on H, ana suppose that there is a sequence {Hn } of pairwise orthogonal closed subspaces that span the whole of H such that Hn D (A) for each n, and the restriction An of A to. Hn is a bounded, selfadjoint operator on Hn . Then each restriction An has a spectral family { En (,.\)} on Hn by Theorem 13. 13. Since En(,.\) is selfadj oint _ on Hn for each fixed ,.\ E IR, it follows, by what 'Ye have shown, that for each ,.\ there is a unique selfadjoint operator E ( ,.\) such that E(,.\) = n=:2:l En (,.\) un , when u satisfies ( 1 3 . 72) . From the fact that each En (,.\) is a proj ection and satisfies ( 1 )-(3) of Theorem 1 3 . 1 3 , it follows that E(,.\) has the sarne properties if we replace ( 3) by as ,.\ � { (3') E(A)u ____, : as ,.\ � for all
c
C
00
- oo �
oo ,
1 3. SELFADJOINT OPERATORS
318
E D(E(>..) ). To show this , let c > 0 be given. If u E D (E(>..) ) satisfies u (13.72) , take n so large that
for
00
n
and N so large that n
L I Ek (>.. ) uk l < c , >.. < - N. 1
Then we have
n -1
oo
1
n
I E ( >.)u l < I L Ek ( >.)uk l + I L Ek ( >.)uk l This proves the first statement in (3' ) ; the second is proved similarly. We also note that 1 3 .70 holds. If and satisfies ( 13.72) , then
) u E D(A) E(>.. ) Aui Ai Ei (A)ui , and L1 I Ai Ei (>..) uil l 2 L1 I Ei (>..) Ai uil l 2 < L I Ai uil l 2 < Thus, u E D (AE(A)) and ( �3.70) holds. The construction of the spectral family for an arbitrary selfadjoint oper ator A is based on the existence of a sequence {Hn } of pairwise orthogonal closed subspaces that span the whole of H such that lfn D (A) for each n, a�d the restriction An of A to Hn is a bounded, selfadjoint operator on Hn. We now demonstrate the existence of such a sequence. Let A be an arbitrary selfadjoint operator on H. We note that for each f E H, there is a unique x E D(A) such that (13.74) (x, y) + (Ax, Ay) ( J, y), y E D (A). see this , let [x , y] denote the left hand side of ( 13. 74 ) . It is scalar product on D (A) and converts it into a Hilbert space. Moreover, F ( y ) (y, f), y E D (A), is a bounded linear functional on this Hilbert space. Hence, there is a unique element x E D ( A) such that F (y) [ y , x] . This gives ( 13 . 74 ) . We write x B f and note tlrat B E B(H , D(A)), since [B f, B f] ( f, B f ) < 1 / I · I B /1 1 , and { I B/1 1 2 + I AB JI I 2 } � < 1 / 1 , f E H. (
00
=
=
00
00
oo .
1
C
=
a
1o
=
=
=
=
1 3. 6. Unbounded selfadjoint operators
319
B is symmetric on H, for x BJ, y == Bg , then [x , y] == [Bf, Bg] == ( /, Bg) == (B f , g). Also, if x == B j, then (Ax, Ay) == ( f - x, y), y E D ( A ) . Consequently, A x E D ( A) and A 2 x == f - x. Thus, (I + A 2 ) B == I. ( 13. 75) Conversely, if x E D ( A 2 ) and , (I + A 2 ) x == j , Then [x , y] == ( J, y), y E D(A). This means that x == B f . Hence, ( 13.76) Applying both sides to AB, we obtain B(I + A2 )AB c AB. But the left hand side equals BA ( I + A2)B == BA, in view of ( 13 . 75) . Thus , we have ( 13. 77) BA c AB . Take C == AB. Then BC == BAB c ABE == CB, and since BC is defined everywhere, we obtain BC == C B . Since B is symmetric and bounded ( and consequently selfadjoint ) and sat isfies if
Note also that
there is
a
==
0
< B < I, spectral family { F ( ,.\) } such that
B 11 >.dF(>.) . =
Since
B - 1 E B(H), we see that F{,.\) is continuous at F(,.\) ==
{ 0 , 0, I,
,.\ < ,.\ > 1 .
,.\
== 0 and satisfies
320
1 3.
SELFADJOINT OPERATORS
We obtain a pairwise orthogonal sequence of subspaces as follows. Consider the sequence of projections
Pn = F( 1/n) - F(1/(n + 1) ) , n = 1 , 2 , They are pairwise orthogonal and satisfy n Pk � I as n oo . L k= l We take Hn == R (Pn ) Then {Hn } is pairwise orthogonal sequence of closed subspaces that span H. Let ·
·
·
--4
a
·
1
1 .X '
Since
and
0,
Asn (.X) = 1 when
1
-- < .X < n+1 n' otherwise.
1 1 < -- < .X n+ 1 n .X = 0 elsewhere, we see that
Asn ( )
Bsn (B) == Sn (B)B == Pn .
Moreover, since C is bounded and commutes with B, it commutes with . Thus, F .X) , and
( Pn sn (B)
APn = ABsn ( B) = Csn ( B ) , and Pn A = Sn (B)BA Sn (B)AB = Sn (B)C. Hence, Hn D(A) for each n, and the restriction An == APn of A to Hn is bounded, defined everywhere and selfadjoint on Hn . Once we know that such a sequence {Hn } exists, we can construct the spectral family {E(.X) } above. Let { .Xk } , k = 0, ± 1 , ±2, , be a sequence of numbers such that A k - l < Ak and Ak � ± oo as k � ± oo . C
C
as
·
·
·
Set
and Lk =
1 p(A)dE(A), >.
k
A -1 k
where p ( t ) is a polynomial with real coefficients. Then L k is a bounded there. Thus, L = selfadjoint operator on Hk , and it coincide,s with
p(A)
k
1 3. 6. Unbounded selfadjoint operators
32 1
p(A k ) , where A k is the restriction of A to Hk . Let L denote the selfadjoint operator which reduces to L k on Hk . Thus, u E D ( L ) if u
00
==
Loo uk ,
== [E (.X k ) - E(A k - I )] u E Hk ,
Uk
-
an d -
oo
This means that
since .
Consequently,
Finally, we note that L == p(A) . For they are both selfadjoint operators whose restrictions to each Hk are bounded and selfadjoint, and they coincide on each Hk . Since· there can be only one selfadjoint operator with such properties, they must be equal. In summary, we have Theorem 13. 14. Let A be a selfadj oint operator on H. Then there is a family { E( .X) } of orthogonal projection operators on H satisfying {1) and
(2) of Theorem 13. 13 and
(3') ( 4 ')
(5')
E(.X)
{0 -t
as ,\ as ,\
I
E(.X) A p(A)
=
c
-t - oo
-t + oo
AE ( .X )
J: p(>..) dE(>.. )
13. SELFADJOINT OPERATORS
322 for any polynomial p( t ) . 1 3 . 7. Problems
A
( 1) Let be a closed, densely defined linear operator on and ( 1 < 1. -1 E
(2) Let
p(A* A)
1 + A * A) - 1 1
H. Show that
A be a densely defined , selfadjoint operator on H. Show that B (A - i)(A + i) - 1 ==
is unitary.
B B(H)
E be a unitary operator on H such that (3) Let one-to-one. Show that
B - I is
A == i(I + B)(I - B) - 1
is selfadjoint.
1
(4) In the notation of Lemma 13.2 . 6 , show that M if E1 E2 = 0.
..L
M2 if and only
1
(5) If E1 and E2 are orthogonal projections, show that E - E2 is an
1
orthogonal projection if and only if E > E2 .
A be a selfadjoint operator in B (H) , and let m and M be defined as in Theorem 13. 13. Show that both m and M are in ( A).
(6) Let a
(7) Let { E( .X ) } be a resolution of the identity corresponding to a self adjoint operator Show that for each u, v E H, the function (E(.X ) u , v ) is of bounded variation in the interval [m, M] .
A.
A
(8) Let be a normal operator in where S > 0 and U is unitary.
(9) If m > 0, show that A l/ 2
=
B(H). Show that A
1M ),_l/2 dE (A) .
=
SU
=
US,
13. 7. Problems ( 10) If
323
A is selfadjoint and J (>.. ) is continuous, show that 1 / (A) I = Amax 1 / (A) I . Eu( A )
( 1 1) Prove the second statement in ( 3 ) ( 12) If A is a positive selfadjoint operator, show that (A - I)(A + I) - 1 is a bounded selfadjoint operator satisfying I B I < 1 . '
.
Chapter 14
MEASURES OF OPERATORS
We have encountered many instances in which a number associated with an operator will reveal important information concerning the operator. Fqr instance, the norm of an operator will tell us much about it . The index of a Fredholm operator is very useful. We call any number associated with an operator a measure of that operator. In this chapter we shall discuss several measures associated with operators which provide much useful information with regard to them. Throughout this chapter, X, Z will denote infinite dimensional Banach spaces, and M, N, V, W will denote infinite dimensional, closed subspaces, unless otherwise designated.
Y,
14. 1 . A seminorm
A very useful seminorm is given by
I A I m d1m Minf0
.
I A if m <
It is easily checked that it is indeed a seminorm. ,Clearly, it satisfies We call dim M0 the codimension of the subspace M. We also have
I Af .
Lemma 14. 1. If
A
E
Y) and B B(Y, Z) , then f i B A f1 m f B I I A I m ·
B(X,
E
<
(14. 1) c
·
>
0 be given. Then there is a subspace M of X having finite codimension such that Proof. Let
I Ax l < �I I A I m
+
c ) ll x l i ,
x E M. 325
326
1 4. ME�4.SURES OF OPERATORS
Thus, Hence, Since
I BAx l < I B I · I Ax l < ( I A I m + E) l x l , ·I B A I m < I B I ( I A I m + E ) .
E was arbitrary, the lemma is pPOved .
We also have
X
E M.
0
I A I m = 0 if and only if A E K (X, Y) . Proof. Assume A E K (X, Y) , and let E > 0 be given. Since the image of l x l < 1 is relatively compact, it is totally bounded (Theorem 4. 17) . Hence, there are elements Y l , · · · , Yn E Y such that min I A x - Yk l < E , l x l < 1.: . ( 14.2) k Let y � · · , y� be functionals in Y' such that ( 1 4 .3) (see Theorem 2. 9) , and let M b e the set of all x E X that are annihilated by A' y � , · · · , A' y�, i.e. , M 0 [ A' y � , · · · , A' y�] . The subspace M has finite codimension in X by definition. Let x be any element in M satisfying < 1 , and let Yk be an element of Yl , · · · , Yn closest to Ax. Since x E M, lwex l have y� (Ax) A' y� (x) = 0. Consequently, (14. 4 ) I Ykl l = y� ( Yk ) = y� ( Yk - Ax) < I J Yk - Ax i l < E . Since H Ax l < I Ax - Ykl l + I Ykl l < 2E , we can conclude that I A I m < I A I M I < 2E . Since E was arbitrary, we see that I A I m = 0 . Conversely, assume that I A l i m = 0 . Then there is a subspace M of finite codimension such that ( 14.5) I Ax l < El l x l , x E M. Let P be a bounded projection onto M. Then I - P is an operator of finite rank on X. Thus , for x E X, I Ax l < I APx l + I A (I - P)x l < EI I Px l + I A I · I (I - P)x l < El l x l + ( I A I + E ) I (I - P) x l · Theorem
14.2.
,- ·.
=
=
2
14. 1 . A seminorm
3 7
-
Since I P is cornpact , there are elements XI , · · · , Xn of the set such that
l xl < 1
I ( / - P)(x - xk ) l < c/ (2 I A I + c), l x l < 1 . Now, let x be any element of X ·s uch that l x l < 1 , and let Xk be a member of XI, · · · , X n satisfying ( 14.6) . Then by ( 14.6) , I A (x - xk ) l < c l x - xk l + (2 I A I + c) I ( J - P)(x -xk ) l < c l x - xk l + c < 3c Hence, the elements Ax i , , Ax n form an c-net for the image of l x l < 1 . By Theorem 4. 1 7, this image is relatively compact . Consequently, A E D K(X, Y) . ( 14.6)
m1n
.
·
·
·
Corollary 14. 3 . For each A in B(X, Y),
(14. 7)
I A + Kl m = I AI m,
K E K(X, Y ) . -
.
Proof. Apply Theorem 14.2.
D
We also have
A is in q,+ (X, Y ) with i (A) < 0 if and only if there are an operator K E K (X, Y ) and a constant C such that (14.8) l x l C I (A - K)x l , x E D(A). Proof. If ( 14. 8 ) holds, then we know that R(A - K) is closed in Y and that A - K is one-to-one ( Theorem 3. 12) . Thus A - K E � + (X, Y ) with 0. Since f3 (A - K) > 0, we see that i (A - K) < 0. This implies a(thatA - K)�+= (X, A E Y ) and i (A) < 0 (Theorem 5. 24). Conversely, assume that A E � + (X, Y ) and i (A) < 0. Let , x n be a basis for N(A), where n = a( A). Let x� , � be functionals i'n such x that xj(xk ) = 8jk , 1 < j, k < n ( cf. Lemma 4. 14) . Since i (A) < 0, there is an n-dimensional subspace Yo Y such that R(A) Yo = {0} ( Lemma 5.3) . Let , Yn be a basis for Yo , and set n Kx = L xj(x) Yj · Since K is bounded and of finite rank, it is compact. The operator A - K is in � + (X, Y ) . For x E N(A - K ) , we have n Ax = L xj(x) Yi · Theorem 14.4. An operator �
XI , \" · · X'
··· ,
c
n
YI � · · ·
j= I
j= I
1 4. MEASURES OF OPERATORS
328
This can happen only if x E N (A) and xj (x) = 0 for 1 < j < n. But then X ==
fl.,
L o:k X k k= l
and xj (x) = O:j = 0, showing that x = 0. Thus, o: (A - K) = 0. Consequently, 0 inequality ( 14.8) holds (Theorem 3 . 12) . Next we have Theorem 14. 5 . An operator A in B(X, Y) is in + (X, Y) if and only if for each Banach space Z there is a constant C such that
( 14.9)
IITIIm
< C I I AT I I m ,
T E B ( Z, X ) .
Proof. If A E (X, Y) , let Ao be the operator satisfying Theorem 5 .4. Then F = AoA - I is in K(X) . Thus , for T E B ( Z, X) , we have
T = A0AT - FT. In view of Theorem 14 . 2 , th i s imp l ies I I T I I m = I I AoA T - F T I I m = II A o A T I I m
< II Ao ll
· I I AT I I m )
which is ( 14.9) . If A E + (X, Y) but not in (X, Y) , then (3(A) = oo . In particular, i (A ) < 0 . By Theorem 14.4, there is an operator K E K(X, Y) such that in equality ( 14.8) holds . Thus., if M is a subspace of Z with finite codimension the n II Tz l l < C I I ( A - K) Tz ll , z E Now for any > 0, there is such a subspace satisfying ,
IY1.
E
I I (A - K) T z l l < ( I I ( A - K) T II m + c ) l l z l l , Consequently, I I Tz l l '
This in1plies and since
E
< C ( I I (A - K) T I I m + c ) ll z ll ,
z E A1 .
z E l'v1 .
.
I I T I I m < C ( J I ( A - K ) T I I rn + c ) , was arbitrary, \Ve see that II T II rn < C I I (A - K ) T I I m = C I I AT I I rn
by Theore n1 14.2. This proves ( 14.9) . Conver sel y assume that for each Banach space Z, there is a constant C such that ( 1 4.9) holds . If A tt 4>+ (X, Y) , then there is a K E K(X, Y) such that n(A - K) = x ( Theore1n 9.43) . Take Z = N (A K) \V i th the nor1n of X, and let T be the i n1b cd d i n g of Z, in X. Then the operator T is ,
-
1 4. 2.
Perturbation classes
329
in B(Z, X) . Since Z is infinite dimensional, I I T I I m i= 0 . But (A - K )T = 0, and consequently, II AT I I m = I I (A - K ) T I I m = 0. This clearly violates ( 14 . 9 ) and the proof is complete. 0 ,
It should be noted that the constant C in ( 1 4 . 9 ) depends only on A and not on the space Z. Thus, the proof of Theorem 14.5 implies that if for each Z there is a constant C depending also on Z, such that ( 14.9) holds, then there is a constant C independent of Z for which it holds. 14. 2 . Perturbation classes
Let X be a complex Banach space, and let S be a subset of X. We denote by P(S) the set of elements of X that perturb S into itself, i.e. , P ( S ) is the set of those elements a E X such that a + s E S for all s E S. We shall assume throughout that S satisfies
( 14. 10)
aS C
S, n i= 0,
i.e. , o:s E S whenever s E S and n i= 0. First we have Lem ma 14.6. P(S) is a linear subspace of X. If, in addition, S is an open subset of X, then P( S) is closed. Proof. Suppose s E S, a, b E P(S) , a i= 0. Then aa + s = a( a + sf a) E S and (a + b) + s = a + (b + s) E S. Thus, P(S) is a subspace. Assume S open. Then for each s E S, there is a 6 > 0 such that ll q - s ll implies that q E S. If { xk } is a sequence of elements in P(S) converging to an element x E X , then for k sufficiently large, ll xk - x ll 6. Thus s + x - Xk E S. Since 0 Xk E P(S) , s + x E S. Thus, x E P(S) , and the lemma is proved.
<8
<
We also have Lemma 14.7. Let S, T be subsets of X which satisfy {14. 10). Assume that S is open, that S C T and that T does not contain any boundary points of S. Then P ( T) c P(S) . Proof. Suppose s E S, b E P(T) . Then
ab + s = a(b + s / n ) E T,
n -/= 0.
Since S is open, ab + s E S for I a I sufficiently small. It follows that ab + s E S for all scalars a. Otherwise for some ao the element ao b + s would be a boundary point of S which is in T. Thus b + s E S, and the proof is 0 complete.
14. MEASURES OF OPERATORS
330
G a x xE aE
Next , let B be a Banach algebra with identity e , and let be the set of regular elements in B. A subspace M of B is called a right ideal if E M for M when M, B. M, B. It is called a left ideal if If it is both a right and left ideal, it is called a two-sided ideal or rnerely an ideal. We have
ax E
xE aE
S, then P(S) is a left ideal. If SG S, then P(S) Proof. Suppose a G, b E P(S), s E Then ab + s == a(b + a- 1 s ) E S. Consequently, a b P(S). Since every element of B is the sum of two ele ments of G and P(S) is a subspace of B , the first statement follows. The second statement is proved in the same way. Lemma 14.8. If GS is a right ideal.
C
C
E
S.
E
D
In summary, we have Theorem 14.9. If
c
( 1 4. 1 1 ) then
S is an open subset of B which satisfies GS S, SG S,
P( S) is a closed, two-sided ideal.
The radical R of B is defined R ==
c
as
{b E B : e + ab E G for all a E G}.
We have Theorem 14. 10. P (G) == R.
1 G. a b E a E ab G. P(G) + a a- 1 + b) E1G. b E -1 b E a E G, a + b + a b) G. b E P(G), and + a- b E G. ,D Proof. Suppose Hence, e +bE Then and ( This means that R. Conversely, suppose R . If then e == a(e E Thus, Hence, the proof is complete.
==
Corollary 14. 1 1 . R is a closed, two-sided ideal.
E x E
An element B is called r'lght regular if it has a right 'lnverse, i.e. , B such that an element == e. It is called left regular if it has a left inverse, i.e. , an element B such that == e . If it is both right and left regular, then its right and left inverses coincide and it is regular in the sense defined above in Section 9. 1 . Let Gr denote the set of right regular elements of B , and let Gg denote the set of left regular elements. We have
y
Theorem 14. 1 2 .
xy yE
P(Gt ) == P(Gr) == R .
yx
1 4.2. Perturbation classes
331
G Gt Gt a E Gt, ak G ak a, bak - e G. bak - ba == b(ak - a) b 1 a. bak == e +1 b(ak - a)- 1 G (bak )ak -==1 b E ab == b- bab == b eb == e, a G b == a . a E Gt , P(Gt) P(G) bs. e + bs E G, b -1 s (e + bs) b(a + s) == e. a + s Gt , s E P(Gt) · Gr Gt. Let
==
1
From this we get
P( ) == P( t) == P(r) F( X ) . An interesting application can be given as follows. Let Z be any subset of the integers { 0, ± 1 , ± 2 , · · · , ±oo } , and let z be the collection of operators A E
==
U
U
==
,�
-
-
=:)
==
This proves ( 14. 1 2 ) in one direction. Conversely, since
A
332
1 4.
MEASURES OF OPERATORS
By hypothesis , there is an operator in A 1 E
c
for each ,.\. By the constancy of the index ( Theorem 5. 1 1 ) , A 1 ( I + ,.\A0E) .must be in
==
[I] .
This shows that I + ,.\Ao E E
Let X , Y be Banach spaces with dim X == oo , and let A be an operator in B ( X , Y) . We define r (A) == inf I I A I At I I ' Jvf
� ( A ) == sup in f I I A I N I I , M M NC
T (A) == sup inf I I A x l l , A1 llxxEll l\=1l v (A) ==
sup
d .I m
i\10 < oo
inf I I A x l l ,
x E 1\1
ll x ll = l
where !vi , N represent infinite dimensional, closed subspaces of X , and A I M denotes the restriction of A to !vi . We shall derive certain relationships among these quantities. Then we shall show how certain inequalities imply properties of A . Each of the quantities defined above has a counterpart on a subspace. For instance, we let f At ( A) == inf I I A I N I I , NCA1
r (A) == rx (A) ,
� A1 (A) == sup r N (A) , Nc A�J
� ( A ) == �x (A) .
Frorn the definitions we check easily that ( 1 4 . 13 )
f At ( A) == inf f N (A) == inf � N ( A) , NcA1 Nc M
( 14. 14)
� A t ( A) == sup � N ( A ) , Ne At
1 4 . 3.
333
Related measures
and
( 14. 15) We also have Theorem 14. 1 6 . f (A + B) < � (A) + r(B) . Proof. First we note that
( 14. 16) To see this , let E > 0 be given. By definition, there is an infinite dimensional subspace V of N such that I I A i v l l < r N (A) + E. Thus, I I (A + This implies
B) l v l < I I A i v l l + I B i v l < rN (A) + + I I B I N II · r N (A + B) < r N (A) + + l i B I N I I · c
Since this is true for every definition of � (A) , we have
E
c
> 0 , we obtain ( 1 4. 1 6) . By ( 14. 16) and the
r N (A + B) < � (A) + I I B I N I I · Thus
inf r N (A + B) < � (A) + inf I I B I N I I , N N from which the theorem follows .
0
Next , we have Theorem 14. 1 7. � (A + B ) < � (A) + � (B) . Proof. For any N, we have, by Theorem 14. 1 6 ,
f N (A + B) < � N (A) + f N (B) < � (A) + � (B ) . The theorem now follows from the definition.
0
Before we continue, we shall state a few lemmas which will be useful in the sequel. Lemma 14. 18. If M, N are subspaces of X such that dim M0 < oo and dim N == oo, then dim M n N == oo. Proof. We know that X == A1 EB L, where dim £ < oo (Lemma 5.3) . If dim M n N < oo , then M == M n N EB M1 and X == M1 EB N1 , where N1 == AI n N EB L and dim N1 oo. Moreover, M1 n N == {0} since every element in Af n N is in N1 . By Lemma 5 . 1 2 , dim N < dim N1 < oo , contrary to assumption. D
<
Lemma 14. 19. Let M be a subspace of X hav'lng 'lnfinzte cod'lmens'lon. Then there is an 'lnfinite dzmension,al subspace W of X such that W n !vi == { 0} .
334
1 4.
MEASURES OF OPERATORS
Proof. Let { x � } be a sequence of linearly independent functionals in M0 • Then there is a sequence { x k } of elements of X such that
xj ( x k ) == 6jk ,
j, k == 1 , 2 ,
·
·
·
( Lemma 5. 12) . The Xk are linearly independent . This follows from the fact that
implies xj
(� ) ak x k
= aj = O,
j = 1, 2
,
.
.
·
.
Let W be the set of linear combinations of the Xk · Then W n M dim W == oo.
If
f
==
{0} , and 0
Lemma 14.20. M and N are S'u bspaces o X having finite codimension in �' then M n N has finite codirnension in X. Proof. By Lemma 5.3 there are finite d-imensional subspaces M1 , N1 C X such that X == M EB lvf1 , X == lv EB N1 . If 1\I n N did not have finite codimension in M, then there would be an infinite dimensional subspace W c IV! such that W n (M n N) == {0} . But then W n N == {0} , and we see from Lemma 5. 12 that dim W < dim N1 < oo, providing a contradiction. Thus , M n N has finite codi1nension in M. Since 1\J has finite codimension in X, we see that M n N has finite codimension in X. D ..
T E B ( X, Y ) , we let Tl\1 denote the range of T I M · We Theorem 14. 2 1 . lfT E B ( X, Y) A E B(Y, Z) , then ( 14. 1 7) fi\tt (AT) < fAt (T)�Tl\1 (A) U'h en dim T A! If din1Tl\1 < oo , then rl'vt (AT) == 0. Proof. Let E > 0 be given. Then there is an N lvf such that I T I N I < r l\ f (T) Thus r N (AT) == VCinfN I (AT) I v l < VCN inf I A i r v l . I T h / 1 < I T INI I inf I A i r v l · VcN .A.ssurne that dim T N oo , and let b e any infinite din1ensional subspace of TN. Let U == N r- 1 W. Then U consists of those x E N such that Tx Thus. dim U == and TU W. Hence, inf I A hvl l ::::: rrN (A ) . inf I A i r v l For an operator have
a r�d
== 00 .
C
+ E.
==
E
�l,..
n
VC N
Ml
oo
==
==
lVCTN
335
1 4. 3. Related measures
Consequently,
( 14. 18) If dim TN < oo , then AT I N is a finite rank operator. In particular, it van ishes on an infinite dimensional subspace of N, and consequently, r (AT) 0. Now, by ( 14. 1 8) and the choice of N, we have fN (AT) < �TN (A)[f M (T) + E] < �TM (A) [f M (T) + E] . Since this is true for any 0, the result follows. 0 Corollary 14.22. � M (AT) < � M (T)�(A). Proof. For N M, we have N
==
c >
C
The corollary now follows from the definition.
0
�(AT) < �(A)�(T). Another quantity related to r and � is TM (A) Nsupc M xE Ninf, l x l = l I Ax l , T(A) Tx( A ). ( 14. 19) We have Theorem 14.24. T(A) < �(A). Proof. Suppose c > 0. Then for each N there is a V N such that II A I v i < rN (A) + E . Since Ex Ninf, l x l = l I Ax i l < I A I v i < �(A) + E, we have T(A) < �(A) + E. Since this is true for any E > 0, the result follows. 0 Corollary 14.23 .
==
==
C
We also have Lemma 14.25 .
r(A)T(T) < I AT I .
W such that I Tz l > [r (T) - c] l z l , z E W. If T(T) 0, the lemma is trivial. Otherwise, pick E < T(T). Now for z E W we have I ATz l < I AT I l i z I < I AT I I Tz i / [T (T) - E] . Set M TW. Then dim M and I Ax l < I AT I l x i /[T (T) - c] , x E M, Proof. By definition, for each
c
> 0 there is a
==
·
==
== oo
·
·
1 4, MEASURES OF OPERATORS
336 or
< II AT I I / [T (T) - E] . r (A) < I I AT I I / [ T ( T ) - c: ] .
I I A I M II This gives Letting
E � 0,
we obtain the desired result.
D
Also, we have Theorem 14.26.
( 14.20)
r (A)
=
inf
inf
Z TE B(Z, X )
I I AT I I . T ( T)
<
the right hand side of ( 14 . 20) . To show Proof. By Lemma 14.25, f (A) that it is equal, let c > 0 be given. Then there is an M c X such that IIAIMU
< r (A) + E.
We may assume that M is closed. Take Z == M, and let T be the embedding of M into X. Then == 1 , while AT == A I M · Thus , the right nand side D of ( 1 4.20) is f(A) + c. Since was arbitrary, the result follows.
<
T(T)
E
Corollary 14.27.
r (A) > inf
( 14. 2 1 )
-
inf
Z TE B ( Z, X )
I I AT II � ( T) .
Proof. Apply Theorems 14.24 and 14.26.
0
As before, we define We have Theorem 14.28. � (A )
E
< IIAIIm·
Proof. Let > 0 be given, and let V be a subspace having finite codimen sion such that I I A i v ll I I A I I m + c. Let M be given, and set N == M n V. Then dim N == oo and II A I N I I II A I I m € . Since N C M, we have
<
+
r A1 (A)
Since this is true for all !VI and
< II AIIm + E.
E > 0 , the result follows.
We also have Theorem 14.29. A E 4>+ (X, Y) if and only if f (A) # 0.
<
0
14.3. Related measures
337
A E
�
w
c
E
==
=
m
c
==
=
==
.
We now have
If �(B) < f(A), then A + B E + ( X, Y ) and i (A + B) i(A). Proof. By Theorem 14.16, f(A + >.. B ) > f(A) - >..r (B ) > 0 for 0 < >.. < 1 . Thus A + >.. B E + (X , Y) for such >.. ( Theorem 14. 2 9). The 0 rest follows from the constancy of the index ( Theorem 5.11) . Theorem 14.30.
=
It is clear from the definition that
v(A) < T(A).
(14. 2 2)
Another consequence is Lemma 14. 3 1 . A E
We also h ave
If 1 ( B) < v(A) , then A + B E + (X, Y) and i(A + B) i (A). Proof. If A + B f/: + ( X, Y) , then there are an infinite dirnensional subspace M and a K E K( X , Y) such that (A + B - K) I M 0 ( Theorem 9.43). Let E > 0 be given. Then there is subspace W having finite codimension such that I K i w l < E ( Theorem 14. 2). Let V be any subspace having finite codimension, and set N V n M n W. Then dim N oo , and I B x l I (A - K )x l > I A I - E, x E N, l x l 1 . Thus inf I Bx I > inf I Ax I - E. Theorem 14.32.
==
==
a
=
==
==
==
,
x E N , II x l l = l
x E N . II x l l = l
338
1 4. MEASURES OF OPERATORS
Since this is true for each subspace V having finite codimension, we have T( B) >
v(A) - E .
v(A). Thus A + B E + (X, Y). The same reasoning shows that A + ,.\B E + (X, Y ) for 0 < ,.\ < 1. Thus, �t he index is constant as before. 0 Corollary 14.33. If X == Y and T (A ) <. 1 , then I - A -E (X , Y) with i(I - A) == 0. Proof. Clearly v(I) == 1. Apply Theorem 14.32 and the fact that i(I) == Since this is true for each
E
> 0, we have T(B) >
[]
0.
We
also have
Theorem 14.34. Proof. Let
W
v(A) > r (A) .
E > 0 be given. Then there is an M such that I A I M i l < r(A) + c .
Let having finite codimension b e give·n , and set· dim N == oo , and
N
==
W n M. Then
I Ax l < sup = 1 I Ax l == I A I N I < r(A) + E . Since this is true for any subspace W having finite codimension, we have v ( A) < f(A) + c . Since E > 0 was arbitrary, we obtain the desired result . 0 In contrast 'vith Theorem 14.5 we have Theorem 14.35 . An operator A in B (X , Y ) is in
xE W, l l x l l == 1
If
tha t
and
x E N , I I x ll
339
1 4.4. Measures of noncompactness
14. 2 2 14. 2 3. A X Y (X, Y), i(A) ==such- oo.thatIn C K K(14. 4 )., ) l x l < C I (A - K)x l , (14.24) I Tz l < C I (A - K)Tz l , E Z. Let W be any infinite dimensional subspace of Z and - let c > 0 be given. Then there is a V W such that I [ (A - K)T] I v l < �[ (A - K)T] + E. Thus, by (14.24) and Corollary 14. 2 2, l i T ! vi i < C[� (AT) + c] . This means that fw (T) < C[� (AT) + c] . by Corollaries and If is not in then this case, there are an operator E and a constant x E X (Lemma This implies z
c
Since W was arbitrary, the result folows.
D
1 4 . 4 . Measures of noncompactness
X
X
Let be a Banach space. For a bounded subset f2 c we let q(f2) denote the infimum (greatest lower bound) of the set of numbers r such that n can be covered by a collection of open spheres of radius r. In particular, ( n ) == if and only if n is totally bounded, i.e. ' if and only if its closure is compact. It is for this reason that q(f2) is sometimes called the measure we let f2 w of noncompactness of n. For f2 and w bounded subsets of denote the set of all sums w 1/J , w E f2, 1/J E W . If f2 can be covered by n €-spheres and W can be covered by m ry-spheres, then n W can be covered by n m ry )-spheres. Thus ,
q
0
X,
+
+
+
(c + (14 . 25) q( n + w ) < q( n ) + q ( w ) . Next , let X , Y be Banach spaces, and let Sx denote the closed unit ball in X , i.e. , Sx { x E X : l x l < 1}. We define (14.26) I A I q q[A (Sx )] , A B(X , Y ). One checks easily that this is a seminorm on B(X, Y) and that I A I q == 0 if and only if A E K( X, Y). Thus we can consider (14.26 ) as a measure noncom p actness of A. We also have (14. 2 7) I AI q < I Al ' (14 . 28) I A + K l q == I A I q , E K( X, Y ). Moreover, if Z is Banach space and B E B ( Y Z), then (14. 29) =
E
=
of
a
K
,
340
1 4.
MEASURES OF OPERATORS
A close relationship between the m-seminorm and the q-seminorm is given by Theorem 14.36. For A in
(14. 30)
B{X, Y),
0 be given. Then we can find elements Yl , , Yn E Y such m in I A x - Yk l < I A i l q + t: , x E Sx . (14. 3 1) k Let y } , , y� be functionals in Y ' such that (14.32) I Y� I == 1, y� ( Yk ) == I Yk I , 1 < k < n , and let M be the set of all x E X that are annihilated by A' y� , , A' y�. This subspace M has finite codimension in X. Let x be any element in M n Sx , and let Yk be one of the elements Yl , which is closest to A x . , Yn Since E M, we have y� (Ax) == A'y� (x) == 0. Consequently, (14. 33) Thus, by ( 14.31) , (14. 34) Since. I Ax l < I Ax - Ykl l + I Yk l , this gives I Ax l < 2( 1 A I q + t: ), x E M Sx . From the definition we conclude that I A I m < 2( 1 A I q + t: ), and since t: was arbitrary, we obtain the right hand inequality in (14.30). To prove the other half, agaitl let t: > 0 be given . Then there is a subspace X having finite codimension such that (14. 3 5) I Ax I < ( I A I + E ) I I E Let P be a bounded proj ection onto Then I - P is an operator of finite rank on X. Thus , for x E X, I Axl l < I APx l + I A (I - P)x l < ( I A I nl + t: ) I Px l + I A I · I ( ! - P) x l < ( I A I nl + c ) l x l + (2 I A I + t: ) I ( - P ) x l , ( 14. 36)
Proof. Let E > that
·
·
·
·
·
·
·
·
·
·
x
n
!VI
c
n1
X
X
,
AI.
/
IYf.
J
·
·
341
1 4. 5. The quotient space
where we have used the fact that
I Px l < l x l + I ( I - P)x l and (14. 3 7) I AI m < I AI · Since I - P is compact , there are elements x 1 , · · · , x n E Sx such that ( 14.38 ) min I ( ! - P )( x - x k ) l < E/ (2 I A I + E ), x E Sx . k Now, let x be any element of Sx , and let Xk be a member of x 1 , · · · , Xn satisfying ( 14.38 ) . Then by ( 14.36 ) I A (x - Xk ) l < (I I A I m + c ) l x - xk l + (2 I A I + c ) I (I - P)(x - Xk ) l < 2( 1 A I m + E ) + E. Consequently, I A I q < 2 I A I m + 3E . Since E was arbitrary, we obtain the left hand inequality in (14. 3 0), and the proof is complete.
0
14 . 5 . The quotient space
Another measure of noncompactness , which is more widely used , is
( 14. 39)
I A I K == KEKinf( X,Y ) I A - K l · It is the norm of the quotient space B ( X, Y ) / K(X, Y) , which is complete ( cf. Section 3.5 ) . By (14.27) , (14.28), (14. 37) and (14.7), it follows that (14. 40) I AI q < I AI K, I AI m < I AI K· If the quotient space B ( X, Y )/ K ( X, Y) is complete with respect to the norm A I q orareI Aequivalent. I m , then it follows from the closed graph theorem that these three Inorms A Banach space X will be said to have the compact approximation propE X, erty with constant C if for each > and finite set of points x 1 , < and there is an operator K E K(X) such that
E 0
(14. 4 1)
I I - Kl C
· · · , Xn
We have Theorem 14.37. If Y has the compact approximation property with con
stant C, then
(14 . 4 2)
I A I K < C I A I q , A E B (X, Y) .
·
1 4. MEASURES OF OPERATORS
342
such Yl , K ,EYnKE(Y)Y such
Proof. Let E > 0 be given. Then there exist elements that ( 14.31 ) holds. By hypothesis , there exists an operator < and that
·
I I - Kl C
·
·
I Yk - Kykl l < E , 1 < k < n . Let x E S be given. Then there is a Yk among the Yl , , Yn such that ( 14.43)
x
·
·
·
( 14.44) Thus ,
1 ( 1 - K)Ax l < I ( ! - K)(Ax - Yk ) l + I ( ! - K) Ykl l < C I A x - Ykl l + E < C( I A I q + e) + E. Since E was arbitrary, we obtain ( 1 4.42) . The proof is complete. D Corollary 14.38. If Y has the compact approximation property, then the space B(X, Y)/ K( X, Y ) is complete with respect to the norms induced by I A I q and I A I m · 14. 6 . Strictly singular operators
S B(X,
Y) is called strictly singular if it does not have An operator E a bounded inverse on any infinite dimensional subspace of Important examples of strictly singular operators are the compact operators. We have
X.
Theorem 14.39. Compact operators are strictly singular.
X, Y), K( X. { xn }
K
Proof. Let be an operator in and suppose it has a bounded inverse B on a subspace M c is a bounded sequence in M, then If has a convergent subsequence ( cf. Section 4.3) . Hence , has a convergent subsequence in M, showing that M is finite dimensional D ( Theorem 4.6) .
{ Kxn }
Xn BKxn =
Before we proceed, we shall need a few simple lemmas.
E + (X, Y) if and only if it has a bounded inverse on A some subspace having finite codimension.
Lemma 14.40.
Proof. Lemma 9.40 and Theorem 9.41 .
A tt + (X, Y), a(A - K) I KI
0
Theorem 14.41 . For A in B(X, Y), if then for every E > 0 < E and such that there is a E = oo . ·rhus there is an infinite dimensional subspace M such that the restriction of to M
K K(X, Y )
is compact and has norm
< E.
A
1 4. 6.
343
Strictly singular operators
{ xk } { x� } 1 n 2 M { Xn , X n + 1 , · · · } Kx = k=Ln x�(x)Axk . As in the proof of Theorem 9.42 one verifies that K E K( X , Y) and I Kx l < k=Ln l x� l · I Axkl l · l x t l < k=Ln 2 - k l x l < 2 - n L 2 n - k l x l 2 - n L 2 -j l x l 2 1 - n l x l < c l x l · j =O k=n
Proof. By Theorem 9.42 there are sequences C X' such c X, that (9.47) holds. Let c > 0 be given, and take n so large that < E. Let be the closed subspace spanned by . Define 00
00
00
00
00
=
=
Now and by continuity
Kx = Ax, x E M.
0
This completes the proof. Corollary 14.42 . If S is strictly singular, then for each
an infinite dimensional subspace compact with norm < E.
E
> 0 there is
M such that the restriction of S to
M is
Proof. We merely note that S is not in + (X, Y) by Lemma 14.40 and apply Theorem 14.41 . 0 Corollary 14.43. Strictly singular operators are in F± (X, Y) . Proof. If the strictly singular operator S were not in F+ (X, Y) , then there would be an operator + (X, Y) such that - S) = oo ( Theorem 9.46) . Let S) . Then A I M = S I M · Now, A has a bounded inverse on some subspace X0 having finite codimension ( Lemma 14.40) . Moreover, is infinite dimensional ( Lemma ( 14. 18) ) . By Lemma for each E > 0 there is an infinite dimensional subspace c X0 n M such that is compact and has norm < € . Since == S N , the same is true of But this contradicts the fact that A + (X, Y) ( Theorem 14.4 1 ) . Similar reasoning applies to the set F_ (X, Y ) . 0
M = N(A -
AE
o: (A
Xo nM
AIN I E
N
14. 42,
SI N AI N ·
We continue with some important properties of strictly singular opera tors. Theorem 14.44. An operator S is strictly singular if and only if fM (S) for all M C X.
=
0
344
14.
MEASURES OF OPERATORS
S
Proof. If is strictly singular from X to Y, then it is strictly singular from M to Y for each infinite dimensional subspace M C X. Thus by Corollary 14.42, for each c > there is an infinite dimensional subspace N C M such that has norm < E. Thus, < . Since this is true for any > we have f M ( ) == Conversely, assume that S is not strictly singular. Then there is an infinite dimensional subspace M C X on which has a bounded such that inverse. Thus, there is a constant
0 SIN S 0.
rM (S) E
E 0,
S
Co l x l < Co i Sx l , x E M. 0 , then for 0 < CoE < 1 /2 , there is an infinite dimensional If r M (S) subspace N M such that S I N has norm < E. Consequently, xl l < Co i Sx l < CoE I x l < 21 1 x l , x E N, providing a contradiction. This completes the proof. 0 Corollary 14.45. T in B (X, Y) is strictly singular, if a'f!d only if � ( T ) 0. Theorem 14.46. T in B(X, Y) is strictly singular if and only if ==
C
==
fM (A + T)
(14.45)
==
rM (A) , A E B(X, Y) .
Proo.f. If T is strictly singular, then
for any A E B (X, Y) by Theorem 14. 16 and Corollary 14.45. For the same reasons, f M( A ) < f M ( A + · T) + � M (T) == f M ( A + T) .
for each M. Thus ,
(14.45) holds , then f M (T) = f M (O + T) == f M (O) == is strictly singular by Theorem 14.44.
(14.46)
� ( A + T) == � (A) , A E B(X, Y) .
Thus
(14.45)
holds. Conversely, if
0
T Corollary 14.47. T in B(X, Y) is strictly singular if and only if T
0
Proof. Clearly, ( 14.45) implies ( 14.46) . Thus, if is strictly singular, ( 14.46) holds by Theorem 14.46. On the other hand, if (14.46) holds, then == � ( + We now apply Corollary 14.45. = � (0) = D
�(T)
0 T)
0.
We also have Theorem 14.48. A is strictly singular if and only if 1(A)
=
0.
1 4 . 7.
345
Norm perturbations
Proof. If A is strictly singular, let E > 0 be given. Then every M contains an N such that II A I N I I < £ ( Corollary 14.4 2) . Thus
A xll II x E M, I I x l l = l
< E.
inf
This gives T( A ) < E . Since E was arbitrary, we have T(A) == 0, then inf II A xll == 0
T(A)
==
0. Conversely, if
x E M, l l x ll = l
for each M. This means that Hence, A is strictly singular.
A
cannot have a bounded inverse on any
M.
0
This should be contrasted with
B(X, Y ) is strictly singular zf and only T (A) , A E B(X , Y ) .
Theorem 14.49. The operator T E
zf
( 14.47)
T(A + T)
==
Proof. Suppose T is strictly singular, and let E > 0 be given. For A Y ) there is an M such that
B(X,
E
[T(A) - c ] l lxll , x E M. Moreover, there is an N C M such that II T I N II < E . Thus , II (A + T) xl l > [T ( A ) - 2 c ] l lxl l , x E N. Hence, T( A + T) > T(A) - 2 c . Since E was arbitrary, we have T(A + T) > T (A) . Since A was arbitrary and - T is strictly singular, T(A) T (A + T - T) > T ( A + T) . II A xl l
>
==
Thus, ( 14.47) holds . Conversely, if ( 1 4.47) holds , then
T(T)
==
T ( O + T)
==
T(O)
==
0.
Hence, T is strictly singular ( Theorem 14.48) . This completes the proof.
0
14.7. Norm perturbations
Suppose A is a + operator. Does there exist a quantity q(A) such that II T il < q (A) implies that A - T E + ? The answer is yes, as we shall see in this section. If A E Y ) , where Y are Banach spaces , define
X,
B(X,
( 1 4 . 48)
tto( A)
==
inf
o:( A - T) >a( A )
0,
II T i l ,
o:(A) o:(A)
< oo , ==
oo
346
1 4. MEASURES OF OPERATORS
and
(14. 49)
Jt(A)
We shall prove Theorem 14.50.
If
1,
=
inf
a(A -T)=oo
I TI .
is in B(X, Y) and
(14. 50) I T I < Jt(A), then A - T E + (X, Y) and (14. 5 1) i(A - T) i (A). If (14.52) l i T I Jto(A), then (14. 53) o: (A - T) < o: (A). Proof. Let > 0 be such that I T I + E < Jt(A). If A - T f/: + (X, Y) , then there is an operator K E K (X, Y) such that I K I < and o:(A - T -K ) oo ( Theorem 14. 4 1). Consequently, Jt(A) < l i T + K l < I T I + I K I < I T I + € < Jt(A), providing a contradiction. Thus A - T E + (X, Y) . The same is true for follows from the constancy of the A - OT for 0 < 1.IfThus , (14.51) index ( Theorem 5.11). (14. 52) holds, the same must be true of (14. 53). Otherwise, we would have to have J.Lo (A) < I T I · This completes the proof. 0 As we saw in Section 3.5 , R(A) is closed in Y if and only � f I Ax l (A) inf (14. 54) d (x , N(A)) is positive. There is a connection between this constant and those given by (14. 48) and (14. 49). In fact , we have Theorem 14.51. If o: (A) < oo , then (14.55) r (A) < Jto(A) < Jt(A). Corollary 14.52. If I T I < !(A), then A T E + (X, Y) and {14 . 51) and {14. 53) hold. It is surprising that the proof of Theorem 14.51 involves a lot of prepa =
<
c
c
() $
I
=
x� N(A)
-
ration. It is based upon the following lemma due to Borsuk.
=
347
14. 7. Norm perturbations
Lemma 14.53. Let M, N be subspaces of. a normed vector space X such that dim N < dim M. (In particular, dim N < oo.) lf G(x ) is a continuous
map of
R}
8BR n M == { x E M : ll x ll == into N, then there is an element xo E 8BR n M such that
G( - xo) == G(xo) .
The proof of Lemma 14.53 would take us too far afield , so we shall content ·o urselves with showing how it can be used to prove -Theorem 14. 5 1 . A not too obvious consequence of Lemma 14.53 is Lemma 14.54. Under the hypotheses of Lemma 14 . 53 there is an element u E M such that
! l u ll
( 14. 56)
== d( u, N) == 1 .
Once we know Lemma 14.54, we can easily give the proof of Theorem 14.5 1 . Proof. Suppose T E B ( X, Y) is such that Lemma 14.54 there is an element u E N ( A -
ll u ll Consequently,
o:(A - 7,) > o:(A) . T) such that
Then by
== d (u , N ( A )) == 1 .
Au ll == I I Tu l l < T . I I < (A) I r - l l u ll llull - I II for any su<;h T, we must have r (A)
Since this is true first inequality ( 14.55) . The second is obvious.
< J.Lo (A) . This is the 0
It remains to show how Lemma 14.53 implies Lemma 14.54. Proof. First , we note that we may assume that M is finite dimensional (just consider a subspace) . We can then assume that X is finite dimensional (same reason) . But we have no right to assume that X is strictly convex. This requires u, E X, < ! l u l l II llu unless u, v are linearly dependent. But we will assume it and pay the conse quences later. In this case, for each u E X there is an element w == G ( u) E N such that ll u - w ll == d(u , To see this , note that there is a sequence { w k } C such that ll u - wk ll � u , N). The sequence { W k } is bounded , and the finite dimensionality of N implies that there is a renamed subsequence converging to an element w E N. Thus, ll u - wk ll � ll u - w ll == d(u , N) . Next , we claim that
+ vi
+ vi , v N).
d(
(a)
w == G(u)
is unique;
N
348
1 4. MEASfJRES OF OPERATORS
G(u) is continuous; ,G ( - u) - G( u). Once ( a ) - ( c ) are known, we merely apply Lemma 14. 5 3 to conclude that there is an element uo E M such that l uo l 1 and G(uo) G( - uo) - G(uo). Consequently, G(uo) 0. But l uo - G(uo) l d(uo, N). Hence, u 0 satisfies the conclusion of Lemma 14. 5 4 . (b ) (c)
=
==
==
=
==
==
Let us now turn to the proof of ( a ) - ( c ) .
w be another point in N satisfying l u - w l l u - w l d(u, N ) , and let 0 < s < 1. Then l u - sw - ( 1 - s ) w l l s(u - w ) + ( 1 - s)(u - w) l < s l u - w l + ( 1 - s) l u - w l d(u, N) , since u -w and u - w are not linearly dependent. This violates the definition of d( u, N) and proves ( a) . To prove ( b ) , we first note that d ( u , N) is continuous in u. This follows from l u - v i < l u - uo l + l uo - v i · Thus, d(u, N) < l u - uo l + l uo - v i , v E N. This implies d( u , N) < l u - uo l + d( uo, N ), from which we obtain l d(u, N) - d(uo , N) I < l u - uo l · Hence, if u in X, then l uk - G(uk ) l d(uk , N) d(u, N) . This shows that the G(u k ) are bounded , and the finite dimensionality of N provides us with renamed subsequence converging to some element w E N. To prove ( a) , let
==
==
=
=
Uk �
=
Thus,
a
�
1 4. 7.
Norm perturbations
349
G(u). Uk u
showing that w = Moreover, the whole sequence must converge to w . Otherwise, there would be a subsequence converging to something else, violating ( a ) . Hence, � implies � This proves ( b ) .
G(uk ) G(u).
To prove ( c ) , we note that
d( - u·, N) inf I I - u - v i inf l u + v i d(u, N ). If G(u), h = G( - u), then l u - w l d(u, N) = d( - u, N) = I I - u - h l l u + h l , from which we conclude that h Now we have to pay the penalty for assuming that X is strictly convex. Let x 1 , · · · , X m be a basis for X. If u L1 nk Xk , y = L1 f3k x k , define (u,v)o L1 nkf3k · Then (u, v)o is a scalar product on For each integer n > 0, define ! l u l � l u l 2 + _!_n l u l 5 , where I I · !l o is the norm corresponding to ( · , · )o . We note that the norm l · l n is /strictly convex, since l u + v i � l u + v l 2 + _!_n [ l u l 5 + l v l 5 + 2(u, v)o] < ( l u l + l v l ) 2 + _!_ ( l u l o + l v l o ) 2 + � [ (u, v)o - l u l o l v l o ] n n < ( l u l n + l v l n ) 2 + � [(u, v) o - l u l o l u l o ] . n This shows that the only time we can have l u + � l i n = l u l n + l v l n is when i.e. , when u, v are linearly dependent . Thus, for each (u,thev)onorm= l u l o l v isl o ,strictly convex. By what we have already shown, there II · l i n is an element U n M such that l u n l n = dn ( u n , N ) = 1 , where dn is the distance as measured by the I I · l i n norm. Since l un l < l un l n 1 , there is a renamed subsequence such that Un u in X. Since l un - u l n l u - v i , v E N , 1 = dn (un , N) d(u, N), 1 l un l n l u l , =
=
vEN
vEN
=
w =
==
=
= -w.
=
m
m
m
=
X.
=
_
n
ES:
=
�
�
�
and the proof is complete.
=
�
D
350
1 4. MEASURES OF OPERATORS
14. 8 . Pert urbation functions
Let 1n (T, A) be a real valued function depending on two operators, Y) . Assume
B(X,
m(.XT, A)
( 1 4.57)
=
T, A
E
1 -X I m (T, A) .
We shall call m(T, A ) a
X,
Lemma 14.55. If m(T, A) is a + perturbation function and A E
m(T, A ) < 1 implies that
i ( A - T ) = i (A) .
( 14.58)
Proof. If m(T, A) < 1 , then ( 14.57) shows that A - OT E + for 0 < () < 1 . By the constancy of the index ( Theorem 5 . 1 1 ) , we see that ( 14.58) holds. 0
The injection modulus j( A ) of an operator A E
B(X, Y) is defined as
.X II x ll < II Ax l l , x E X } . If A is one-to-one , then j( A ) = r(A) . Otherwise, j (A ) = 0. ( 14.59)
j( A )
=
sup { .X :
� j
r (B)jv(T)
By Theorems 14.30 and 14.32, the quantities (T) r (A) and are + perturbation functions. We can improve on them by introducing
( 14.60)
. f p (T, A) = dim . In M0
. sup I n f < oo ..NcM xEN
II Tx ll II A X 1 1 .
\Ve have Theorem 14.56. The junction
sat isfies ( 14.61)
p(T, A) is a + perturbation function and
p(T, A )
<
� (T) /f (A) ,
and
( 1 4 . 6 2)
p(T, A ) < r (T) jv(A) .
1 4 . 8. Perturbation functions
35 1
p(T, < a < A Tx' I / I Ax' l < a. x' E I A-T
Proof. Suppose that 1 . Then there is a subspace M of finite A) codimension, on which is one-to-one, such that for �ach subspace C M there is a element N -nith We may assume that is not in + and , therefore, that its 1 . Assume also that restriction to M is not in + as well. Thus , for eaclr E > 0, there is a subspace C M such that
l x' l ==
N
N
I Ax - Tx l < c l x l ,- x E N, Hence, I Ax i l < I Tx l + c l x l , x E N. For x' chosen as above, we have I Ax' l < a l Ax' I + E . But !(A I M ) > and ! (A I M ) < I Ax' l < c/( 1 - a) , which is impossible if E is small enough. Thus, A - T E <1>+ , and therefore, p(T, A) is a perturbation function for <1> + . Given N, set Ax � infN I l ' c xE I Tx l which may be +oo , so that ( 14.63 ) c i Ax l < I Tx l , x E N. Then c i T I L I I < I B I L I for L N , and so cr(T I L ) < r (B I £ ). Then (T) . p(T, A) < dim infM 0
=
c
.
This completes the proof.
p(T,
0
Since A) is smaller than the perturbation function in ( 14.61 ) and ( 14.62 ) , it is better in the most obvious sense. For perturbation functions , smaller is better and smallest is best. The following theorem shows this. Theorem 14. 57. There is a smallest perturbation function. (i) : The smallest + perturbation function is
m 1 (T, A) == max { I -X I : .XA - T � + } ( ii) : The smallest perturbation function is ( 14.65 ) m2 (T, A) == max{ I -X I : .XA - T � } . ( 14.64 )
1 4. MEASURES OF OPERATORS
352
(iii) : The smallest perturbation function is ( 14.66) m3 (T, = max { ,.\ : either .AA - T f/: + or o:(,.\A - T) > o:(A) } . a
A)
l1
Proof. It is immediate from ( 1 4. 64) , ( 1 4.65) and ( 14.66) that m 1 , m 2 and m3 are perturbation functions. If m(T, is a + perturbation function and m(T, for some T and some A E <1> + , then by ( 14.64) there is < m 1 (T, a .A' satisfying m (T, < I -A ' I with ,.\'A + T not a + operator. But this is a contradiction because m(TI .A' , < 1 , so that A - (TI,.\') , and therefore .A' - T, is a + operator. Similar arguments show that m 2 and m3 are minimal. D
A)
A)
A)
A)
A
A)
The point of Theorem 14.57 is that there are minimal perturbation func tions. The formulas ( 14.64) , ( 1 4.65) and ( 14. 66) beg the question of whether, for given and T, the operator + T has the desired property. Therefore, while it is not possible to have, say, a better perturbation function than m 2 , it is certainly possible to have a formula for m 2 which is more informa tive than ( 14.65) . Recall that an operator A is Fredholm if and only if there is an operator such that = I + K1 on X and Ao == I + K2 on Y, where K1 , K2 are compact operators (Theorems 5.4) . Let [A ] denote the coset containing the operator in the quotient space B (X)I K(X) (cf. Section 3.5) . We have
A
A
Ao
AoA
A
A
be a Fredholm operator on X, and let Ao be as A above. Then, setting C = Ao T, the best Fredholm perturbation function m2 is given by
Theorem 14.58. Let
( 14.67)
m 2 (T, A) = lin1 ( T ( C n ) ) 1 / n
( 14. 68)
m 2 (T , A) == l im( � ( Cn ) ) 1 1n ,
( 14.69)
m 2 (T, A ) = ra ( [ C] ) .
,
Proof. Since Ao is also Fredholm, ,.\A - T E if and only if A0 (,.\A - T) = ,.\ - AoT + ,.\K 1 E if and only if ,.\ + AoT E <1>. Thus , ( 14.69) follows from the equation ra ( [ C] ) = sup { I ,.\ I : ,.\ C t/: } (Theorem 6. 13) . Since ra ( [C] ) = lim( T ( C n ) ) 1 1 n = lim ( � ( Cn ) ) 1 1 n , n oo -
�
(Theorem 6. 13) , we see that ( 14.67) and ( 1 4.68) hold as well.
D
A technical detail bars the extension of Theore1n 14.58 to 4> + perturba tions and m 1 . Namely, an operator A E + may not have a complemented range. One must consider , instead of + operators, the class f. of those
1 4. 8. Perturbation functions
353
operators in + which have complemented ranges and use the fact that such that belongs to
A AoA
Ao
m(T, A)
Ao
A m'(T, A) = m ( A T I)
p(T, A) T = Y = £2 , a = '2:: Qj ej by Ta = o:2 3 + o:4e5 + Let N be the closed subspace spanned by { e 2 , e4 , } . For M a subspace of finite codimension, M' = N M is infinite-dimensional, and j(T I M' ) = so that p(T, I ) = 1. However, (.X + T)(.X - T) = .X 2 since ITT2 I M=' I0, =and1, thus, m3 (T, I) = m2 ( T, I) m1 (T, I) = 0. Note that since � (T) T(T) 1 , the perturbation functions D.(T)/!( I ) and r(T)jv(I) are e
·
·
·
.
·
·
·
n
=
both 1 .
=
=
Next we have
A
Let H be a Hilbert space, and suppose that E B ( H ) , M is a subspace and > 0. Then for each c > 0 , there is an N C M such that + c. <
Lemma 14. 5 9 .
v(A I M ) I A I N I v(A I M )
M=
H. Then by definition, Proof. To simplify the notation, assume that for each c > 0 there is a subspace M' of finite codimension with > 0, and therefore E + , and there is a norm one element I E M' with +c < + c . Let N be_ the orthogonal complement of < the span of X I , let be the orthogonal complement of the span of I , and set M" n A- I (N ' ) n M' , which has finite codimension in M' . There is + c. Since a norm one element E M" with +c < < is orthogonal to X I and is orthogonal to I. E M" , and an Continuing in this way we obtain an orthonormal sequence orthogonal sequence with + c. On the closed span of < the we have
v(A I M' )
X A I Ax i i j(A I M' ) N' !(A) Ax =N x2 AAx2 l j(A I M" ) v(A) I Ax x2 x2 x2 {xn } { Axn } I Axn l v (A) { xn } , I A '2:: anxn l 2 '2:: l an i 2 1 Axn l 2 < (v(A) + e) l '2:: an Xn l 2 · =
0
We also have
14. MEASURES OF OPERATORS
354
� (T) and v(A) == r(A) . Consequently, the two perturbation functions �(T) jf (A) and T (T) jv(A)
Theorem 14.60. On a Hilbert space, T(T)
==
are equal.
Proof. To show that � (T) < T (A) , choose an M with r (T I M ) > 0. Then TI M E <1> + , and consequently, v (T I M ) > 0 ( Lemma 14. 3 1 . Given an E > 0 there is an N C M such that I I T I N I I < v(T I M ) + E < T(T I M ) + E < T(T) + E ( Lemma 14.59 . Hence, r (T I M ) < T(T) + E, and consequently, � (T) < T (T) . The reverse inequality follows from Theorem 14.24. It also follows from Lemma 14.59 that f (T) > v (T) , and the reverse inequality follows from Theorem 14.34. 0
)
)
1 4 . 9 . Factored pert urbat ion functions
Given a 4> + operator A and an operator T, the computation of I I T I I / r(A) involves the computation of the two useful numbers I I T I I and r(A) . After this computation, r(A) can, of course, be used in studying the pert'!lrbation of A by other operators , and I I T I I can be used in studying the perturbation of other + operators by T. In contrast , the computation of, say, p(T, A) must be completely redone if either T or A changes. This observation leads to the following definition. Definition. A perturbation function m is factored if it can be written in the form
( 14 . 70
)
m (T, A)
==
m 1 (T) jm 2 (A) .
We have Theorem 14.6 1 . Let the factored perturbatzon function m be given by (14 . 70}. For each numerator m 1 there is a best (i. e. , largest) denominator
d(A, m 1 ) for whzch m 1 (T) jd(A, m 1 ) zs a perturbatzon function of the sarne type as m (4>, + or a). For each denomznator m2 , there is a best (i. e. , smallest) numerator n(T, m2 ) for which n(T, m 2 ) /m 2 (A) is a perturbatzon
function of the same type as 1n .
Proof. Suppose that m is a + perturbation function. Set
( 14.71 )
{
inf m 1 (T) : A - T tf_ 4> + } . We claim that m 1 (T) j d(A) is the minimal 4>+ perturbation function with numerator m 1 ('T ) . First , if m 1 (T) jd(A) < 1 , then by the definition of d in 14. 71 , A + T is in 4> + . Second, suppose that m 1 (T) jd' (A) is a 4> + perturbation function. Let a 4> + operator A be given . For each E > 0 there is a T with d(T) + E > m 1 (T) and A + T tf_ 4>+ . This implies that
( )
d(A)
==
1 4. 9.
Factored perturbation functions
355
m 1 (T) I d' (A) > 1 , and consequently that d' (A ) < d(A) + c. Hence, for each A E <1>+ , we have d'(A) < d(A) . Note, in particular, that d(A) > � 2 (A) > 0. To define d(A) for a Fredholm perturbation function, replace + by iii ( 14. 71 ) . To define d(A) for a a perturbation function, set ( 14.72) d(A)
==
inf {m 1 (T)
:
either A - T f/: + or n(A - T) > n(A) } .
In either case, proceed as above to see that the quantity d so defined is maximal. Suppose again that m is a + perturbation function. To show that for each denominator m 2 there exists a best numerator, let N be the set of all those numerators n ' for which n ' (T) I m 2 (A) is a + perturbation function, and set ( 14. 73 )
n ( T)
==
inf { n' ( T ) : n ' E N} .
If n(T ) Im 2 (A) < 1 , then there is an n' E N such that n' (T) Im 2 (A) < 1 . Hence, A + T E + · That n(T) is the smallest member of N and that n(>.. T ) == 1 >-- l n(T) follow from ( 14. 73 ) Similar definitions and proofs establish the results for and a pertur0 bation functions. .
Note that one could alternatively define d (A) analogous to ( 14. 73 ) and n(T) analogous to ( 14.71 ) . Next , we have Lemma 14.62 . Suppose that m 1 (T) Im 2 (A) is a + perturbation function.
Let n(T) == n(T, m 2 ) be the best numerator for a given denominator m2 , and let d(A) == d (A m 1 ) be the best denominator for a given numerator m 1 , as described in Theorem 14. 61. Then: ,
(i) : n(>..T ) == l >.. l n(T) . (ii) : d (>.. A ) == 1 >-- l d(A) . (iii) : n(T + == n(T) for (iv) : d(A + == d(A) for
S) S)
S E F+ . S E F+ ·
Proof. Property ( i ) is immediate. To see that ( ii ) holds, note that
d (>.. A )
== == ==
�
inf { m1 (T) : >.. A + T t/: + } inf { m (>.. TI>.. ) : A + TI>.. t/: + }
! >.. l d(A) .
Define
n' (T)
==
inf { n (T +
S) : S E F+ } ·
356
1 4. MEASURES OF OPERATORS
S
If n ' (T) jm 2 (A) < 1 , then there is an E F+ such that A + T + S E + and so A + T E
S) : S F+ } ·
As above, m 1 ( T ) / d' (A) E + , and by the maximality of d, we have d' This completes the proof.
=
There is a corresponding theorem for perturbation functions with replaced by F in ( iii ) and ( iv ) .
d. 0
F+
This leads to Corollary 14.63. Given the numerator
II T II :
( i) : J.Lo (A) is the best denominator for a perturbation function; (ii) : J.L( A ) is the best denominator for a
Proof. Statements ( i ) and ( ii ) follow from Theorem 14.50 and Theorem 14.61. By Lemma 14.55, II Ti l /J.L( A is also a perturbation function, and since J.L(A) is the best demoninator for a + perturbation function, it is also the best denominator for a perturbation function. 0
)
We shall say that a + perturbation function m(T, A ) is A-exact if, given any operator A E
)
Lemma 14.64. lf m 1 ( T ) / m 2 ( A is an A-exact perturbation /unction, then the best denominator d ( A, m 1 ) for m 1 is m 2 (A) . Proof. See the proof of Theorem
14.61 .
0
A + perturbation function 1n which has m ( A , A) = 1 for each operator A E + is A-exact , since T = -A then satisfies m(T, A) = 1 and o: ( A + T ) = 00 .
1 4. 1 0.
Problems
357
14. 1 0 . Problems
( 1 ) Show that the expression ( 14.26) is a seminorm on B ( X, Y) . (2) Show that
I AI q
=
0 if and only if
A E K (X, Y) .
(3) Prove ( 1 4.28) . ( 4) Prove the second half of Lemma 14.8. (5) Prove: P ( Gr )
=
R.
(6) Prove ( 14. 13) , ( 14. 14) and ( 14. 15) . (7) If X is a normed vector space, V is a subspace of finite codimension, and W is an infinite dimensional subspace, show that dim V n W = 00 .
(8) Prove: If dim Mj <
oo ,
j
=
1, 2, then dim ( M1
(9) Prove ( 14.25) . ( 10) Show that
I A I q is a seminorm.
( 1 1 ) Prove ( 14.28) . ( 12) Prove ( 14.29) .
n
M2 ) 0 <
oo .
I •
,.,,..
•
'
'•
, ; .- ,o�
_,,.
' ' '
"'' ,.o
'
I
4.--,
Chapter
..
�
_,
......_
'
15
EXAMPLES AND AP P LI C ATIO N S
15 . 1 . A few remarks
Throughout these chapters , we have avoided involved applications. The primary reason for this is that we wanted to stress the beauty of the subject itself and to show that it deserves study in its own right . How successful we have been remains to be seen. However, lest we leave you with the wrong impression, we want to point out that functional analysis has many useful applications scattered throughout mathematics , science, and engineering. Indeed, a large part of its development came from these applications. Therefore, we think it appropriate to give some examples to illustrate some of the applications. In this connection, one must concede the fact that the more meaningful and useful the application, the more involved the details and technicalities. Since only a minimal background in mq,thematics has been assumed for this book, the choice of applications that need no further preparation is extremely limited. Moreover, our intentions at the moment are not to teach you other branches of mathematics , and certainly not to expound upon other branches of science. Faced with this dilemma, we have chosen a few applications that do not need much in the way of technical knowledge. It is, therefore, to be expected that these will not be the most interesting of the possible choices. Moreover, due to time and space limitations, it will not be possible to motivate the importance of those applications arising from other branches of science or the mathematical tools that are used in their connection. '
359
1 5. EXAMPLES AND APPLICATIONS
360
1 5 . 2 . A differential operator
Consider the Hilbert space L 2 = L 2 ( - oo, oo) of functions u(t) square inte grable on the whole real line ( see Section 1 .4) . Consider the operator
du(t) dt on L 2 , with D (A) consisting of those u E L 2 having continuous first deriva tives in £ 2 • At this point , one might question why we took £ 2 to be the space in which A is to act , rather than some other space that might seem more Au =
appropriate. The reason is that we want to consider an oversimplification of a problem which has considerable significance in physics. It will be more convenient for us to consider complex valued functions u(t) . Thus, L 2 will be a complex Hilbert space. It is easy to see that A is not a bounded operator on L 2 . In fact, let cp( t) ¢. 0 be any nonnegative continuously differentiable function which vanishes outside the interval l t l < 1 . For instance, we can take
cp(t)
(15.1)
=
{
a exp
(i-t) '
l t l < 1, ltl > 1.
0,
B y multiplying by a suitable constant [e.g. , the constant a in ( 15. 1 )] , we may assume that
1:
( 1 5.2) Set
'Pn
( 15.3)
=
ncp ( nt ) ,
=
1.
n = 1 , 2,
·
·
·
.
Then 'Pn E D (A) and ( 1 5.4) But
1: l iP (nt) l 2dt n 1: l iP ( ) l 2 dr 3 II A �Pn ll 2 n4 1: liP' ( nt) l 2 dt n II A�.pl l 2 •
II �Pn 11 2
=
n2
T
=
=
n l l �.p ll 2 .
=
=
Note that Acp i= 0; otherwise, cp would be a constant, and the only constant in £2 is 0. Hence , lf A cpn ii / I I 'Pn ll = n ii A cp l l / ll cp ll � oo as n � oo . This shows that A is not bounded. The next question we might ask about A is whether it is closed. Again, the answer is negative. To show this, we again make use of the functions ( 1 5.3) , where cp(t) is given by ( 1 5. 1 ) . For any u E L 2 , set ( 1 5.5)
Jn u (x)
=
1: IPn (X - t)u(t) dt.
1 5. 2.
361
A differential operator
Jnu E £2 . In fact , (15 . 6) Moreover, for any u E L 2 , we have (15.7) Assume (15. 6 ) and (15.7) for the moment , and let u(t) be the function in £ 2 given by 1 + t , - 1 < t < O, u(t) 1 - t, 0 < t < 1, 0 , l t l > 1. Clearly, u( t) is continuous in ( - oo, oo ) , but its derivative is not. Thus , u is not in D( A). However, Jnu is in D(A) for each n. This follows from the fact that the integral in (15. 5 ) really extends only from - 1 to 1, and u(t) is continuous. Thus, we can differentiate under the integral sign and obtain a continuous function. Since Jnu vanishes for l t l > 2 , the same is true for its derivative. This shows that Jnu E D (A) for each n. Now u' (t) may not be continuous , but it is in L 2 , since 1, - 1 < t < 0, u' (t) - 1 , 0 < < 1, 0, l t l > 1. We claim that
==
==
t
Moreover,
A Jnu(x) J�
==
==
-1
----+
==
----+
362
1 5. EXAMPLES AND APPLICATIONS
By ( 15.2) ,
1: 4'n(X - t) dt 1. ==
( 1 5.8) Thus,
1b I Jnu(x ) l 2 dx 1b 1: 4'n(x - t) ! u (t)i 2dt dx = 1: 1b 4'n(x - t ) l u (t) l 2dx dt. <
( The order of integration can be interchanged because the double integral is absolutely convergent . ) Si n ce
l b 4'n (x - t ) dx < /-00 4'n (X ) dx 1 , oo we have 1b I Jnu( x ) l 2 dx < 1: l u (t) l 2dt l u l 2 . Letting and b --+ we see that Jn u E £2 and that (15.6) holds. Turning to ( 1 5.7) , we note that it suffices to prove it for continuous reason is that such functions are dense in L2 ( in fact , so far uas EweL2.areTheconcerned, L2 is the completion of such functions with respect to the norm of L 2). Moreover, it follows from Lemma 10.2 that once we know that ( 1 5.6) holds for all u E L2, and ( 1 5.7) is proved for a set of u dense in L2, then ( 1 5 . 7) holds for all E £2. To prove ( 1 5. 7 ) for continuous u E L2, let E > 0 and u be given . Take R so large that 2 r u (t) dt . < l 1 16 jlti>R-1 Since u(t ) is uniformly continuous on bounded intervals , there is a 6 > 0 such that l u (s + t ) - u (t) 1 2 < 4�, l t l < R, l s i < Thus , E R 2 u(t) d t) J- R l u (s + - 1 t < 2 , l s i < 6. We may assume 6 < 1 . Then loo l u (s + t) - u(t) l 2dt < 2l00 ( l u (s t) l 2 + l u (t) 1 2 )dt a
-
-t
=
=
a --4 - oo
oo ,
u
€
8.
+
1 5. 3. Does
363
A have a closed extension ?
with a similar inequality holding for the interval ( - oo , -R) . Combining the inequalities , we have
(15. 9 ) Now, by
(15 . 10)
(15. 5 ),
1: i u (s + t) - u(t) ! 2 dt < E, l s i < 6. Jn u( x ) 1: IPn (s)u(x - s ) ds, =
(15. 8 ), we have (15.11) Jnu( x) - u(x) 1: IPn (s ) [u (x - s) - u(x)] ds . Applying Schwarz ' s inequality and integrating wit� .respect to x, we obtain (15.1 2 ) 1: I Jnu - u l 2 dx < 1: IPn (s) 1: i u ( x - s) - u(x) l 2 dxds . Now, by (15.1) and (15 . 3 ) ,
=
=
e
e,
0
and the proof is complete.
1 5 . 3 . Does A have a closed extension?
A L2. A 12.18, A } u A D(A) n L2, 0. UnD(A)0 Aun v n, (15.13) (Aun ,v) - (un ,v') . (Note that no boundary terms appear, because v vanishes outside some finite interval. ) Taking the limit as n oo, we get (15.14) ( J ,v) 0 for all such v. We maintain that this implies that f 0.
We have shown that the operator given in the preceding section is un bounded and not a closed operator in We now ask whether has a closed extension. By Theorem this is equivalent to asking if is clos able. For a change, this question has an affirmative answer. To prove that is closable, we must show that whenever { is a sequence of functions in such that --+ and --4 f in then f = To see this, let be any function in that vanishes for l t l large. Then, by integration by parts , we have, for each =
--+
=
=
364
1 5. EXAMPLES AND APPLICATIONS
R > 0, let !R be defined by t l < R, l f, { == fR 0, l t l > R.
We show this as follows: For each ( 15. 15) Since make
fE
£2 , the same is true of fR · Now, by ( 1 5 . 7) , for each II! - Jnf ll
( 15 . 16) by taking
n
<
E
> 0, we can
E
2
sufficiently large. Moreover, we can take
R so large that
( 15 . 1 7) In view of ( 15 .6) , this gives
Thus, ( 15 . 18) Now, for each n and R, the function JnfR vanishes for l t l large. More over, if we use the function cp given by ( 15. 1 ) , then Jn fR is infinitely - differ entiable (just differentiate repeatedly under the integral sign) . In particular, JnfR E D( A ) for each n and R so that
( J , JnfR) == Thus, by ( 15 . 18) , 11 / 11 2 ==
0.
( J , f - JnfR) < IIJII · II J - JnfR II
for all n and R. By ( 1 5. 18) this implies Since
e
ll f ll < E . was arbitrary, we must have f ==
0, and the proof is complete.
A function cp( t) which vanishes for l t l large is said to have
D
compact
support. Let C� denote the class of infinitely differentiable functions with
compact supports. We have just shown that C� is dense in
£2.
1 5 .4. The closure of A
We know that A has at least one closed extension. In particular, it has a closure A (see the proof of Theorem 12. 18) . By definition, D(A) consists of those u E £2 for which there is a sequence { un of functions in D (A) such that Un � u in and { Aun } converges in to some function f . Au is
£2
} £2
1 5. 4 .
365
The closure of A
defined to be f. Let us examine A a bit more closely. In particular, we want to determine p(A) . The problem of deciding which values of A are such that
(15.19)
(A - .X )u = f
£2
has a unique solution u E D(A) for each f E does not appear to be easy at all, since A is defined by a limiting procedure, and it is not easy to "put our hands on it ." To feel our way, assume that u and f are smooth functions satisfying In this case, A reduces to A, and becomes
(15.19)
(15.19). (15. 20)
u' - A u = f .
This is a differential equation that most of us have come across at one time or another. At any rate, if we multiply both sides of by e->..t and integrate between and x, we get
(15. 20)
0
(15. 2 1) Suppose A = a + general, we have
i(3. We must consider several cases depending on A . In
(15. 2 2) If o: >
0, we will have l u(x) l
� oo as x � oo
unless
(15. 23) Thus, the only way u can be in case,
(15. 24)
u(x) =
This function is , indeed, in
L2 for > 0 is when (15. 2 3) holds. In this o:
- 100 e>.(x - t) f (t) dt.
L2 since
by the Cauchy-Schwarz inequality. Since
100 X
ea.( x - t ) dt =
1- , Q:
366
1 5. EXAMPLES AND APPLICATIONS
00 1- oo l tt (x ) l 2dx < a1 1-00oo 1x00 ea(x - t) l f (t)i 2dtdx t (15.25) = � 1: 1 oo ea(x -t) dx i f (t) 1 2 dt = -\ 1 / 1 2 · Looking back, we note the following: If > 0, then for each continuous function f E L 2 , there is a unique continuously differentiable solution u E L2 of (1 5. 2 0) . This solution is given by (15. 2 4) and satisfies (1 5. 2 6) !lu l < I / I · Now, suppose f is any function in L 2 . Then, by the results of the pre ceding section, there is a sequence {fn } of functions in C(J converging to f in £2. For each fn , there is a solution Un D(A) of (A - A) Un = fn · ( 1 5.27) Moreover, by (15. 2 6), (15. 28) showing that {un } is a Cauchy sequence in L 2 . Since L 2 is complete, there is a u £2 such that Un . By definition, u E D ( A ) and (A - ;\ )u = f. u Since l 'ltn l < 1 /n l / a for each we see that u and f satisfy ( 15. 2 6) . This shows that when a > 0, equation (15.19) can be solved for each is true for a < 0. In this case, (15. 2 2) shows that l u (x) l f EasL2. The sameunless x u(O) = 1� e- At f (t)dt. (15. 29) Thus, the desired solution of (15 . 20) is u(x) = 1: eA(x-t) f(t) dt. (15. 30) Sirr1ple applications of Schwarz ' s inequality now show that u satisfies 15.31) Applying the limiting process as before, we see that (15.19) can be solved for all f E L 2 in this case as well. we have
0:
a
Q
E
as
E
�
m , n � oo ,
n,
�
oo
{
� - oo
1 5.4 . The closure of
367
A
(15. 20) (15.19), £2 , u
What about uniqueness? The fact that the solutions of are unique when they exist does not imply that the same is true for since solutions of are assumed continuously differentiable. To tackle this problem, suppose n =I= and that for some f E equation had two distinct solutions. Then their difference would be a solution of
(15. 20)
0
(1 5.19)
(A - .X)u 0. (15. 32) Since u E D LA) , there a sequence { un } of functions in D (A) such that while Aun � .Xu. Let v be any function in D(A), and assume that U(15.13) n u holds. Assuming this, we have in the limit , .X(u,v) ( ) Whence, (u, v' + ."Xv ) ::::; 0 for all v E D(A). Since �e( - �) �e .X 1= 0, we know, by the argument above, that for each E COO , there is a v E D (A) such that v' + �v =
is
�
= - u, v
g
'
.
=
=
-a
= g.
Hence,
(u , g)
=
0
OO . Since COO is dense in L2 ( cf. Section 15. 3), it follows that C u It remains to prove (15.1 3) for E D(A) . Since (15.13) holds whenever to show that for each E D(A), there is a sequence {
g
E
o.
v
v
�
k
� oo .
J: I ¢ (T ) I dT J: 1: i ¢ (x - t) l dx l w (t) 1 2 dt 2 (J: i ¢ (T) i dT) l w l 2 . In particular , if we take 1/;(t)
=
L2 ,
=
= u -
368
1 5.
EXAMPLES AND APPLICATIONS
v be any function in D(A). Let > 0 be given, and pick n so large l v - Jnv l < �' l v' - Jnv' l < � · Once n is fixed, we can take R so large that I Jn (VR - v) ll 2c , I AJn (VR - v) l < Since JnVR E COO and A Jn v = Jn v ' , (15. 3 3) is proved. We have seen that the half-planes �e .X > 0 and �e .X < 0 are in p(A) . What about the case = 0? A hint that things are not so nice comes from examining (15. 2 2). From this we see that u can be in L 2 only if (15. 2 3) and (15. 29) both hold. In this case, we must have 1: eif3t f (t) dt = 0. (15 .35) Thus, (15.35) is a necessary condition that (15. 2 0) have a solution. To make matters worse, if i t f3 e f ( t ) = - , t > a, t for some a > 0, then (15. 2 1) gives a x & x t e -�.f3 u(x) u(O) + L e-�{3. f (t) dt + 1a t = C + log a for x > a. This is clearly not in L 2 , even though f is continuous and in L 2 . However, the fact that we cannot solve (15. 2 0) for all continuous f E £ 2 when A = i {3 does not prove that i{3 is in a(A) . But this, indeed, is the case. To show it , let 7/J ( t ) be any function in COO such that 1 7/J I = 1. For\ c > 0 set (15.36) 7/Jc= (t ) = Jeeif3t 'l/J (ct) . E
Now, let that
€
<
2
.
a
=
0
X
Then
which shows that
But
2 dt = £ 2 1 1 ' 1 1 2 . 2 = £3 ' ( 1 t) 1 11 (A e i c: '1/1 ,B)'I/I '1/1 I 1: I 1 1 2 = c: 1: 1 '1/1 (c:t) 1 2 dt = I '1/1 11 2 1. 0, we see that 0, 1 7/Jc I = 1, I (A - i{3)7/Jc I 0 as 'l/le
Letting c
(15. 37)
�
which shows that i{3 cannot be in p(A) .
=
�
E �
369
1 5. 5. Another approach
We have therefore seen that
A E a (A) if and only if �e .A
=
0.
1 5 . 5 . Another approacn
We have just seen that a (A) consists of the imaginary axis. Most people would stop at this point since the original problem is solved. Some, however, may not be satisfied . Such people might be wondering what it was in the nature of the operator that caused its spectrum to be this particular set . In other words , is there a way of looking at the operator and telling what its spectrum is? In this section, we shall introduce a method that sheds much light on the matter. It employs the Fourier transform defined by ( 1 5.38)
Fu(x) v'2ii127r 1-00oo e -t.xt u(t) dt. =
Fu is bounded and continuous so long as u E £1 , i.e. , so long ( 1 5.39) 1: l u (t) l dt < However, we shall find it convenient to introduce a class S of functions having properties suitable for our purposes. A function u(t) is said to be in S if u is infinitely differentiable in ( - oo , oo ) and ti dk u(t) dtk is a bounded function for any pair of nonnegative integers j, Clearly, any function in COO is also in S, and since COO is dense in L 2 ( Section 15.3) , the same is true of S. We leave as a simple exercise the fact that u(t) / P(t) is in S whenever u E S and P is a polynomial that has no real roots. If u E S, then t k u(t) is in L 1 for each Hence, we may differentiate ( 1 5.38) under the integral sign as much as we like, obtaining ( 1 5.40) k 2 3 ... where k d k Dx dxk . Moreover, if u E S, so is u' . Thus, R Fu' v'27i Rlim-+oo 1- R e -ixt u'(t) dt. Integrating by parts , we obtain R 1-RR e- ixt u'(t) dt ix 1- R e- ixt u(t) dt + e- ixRu(R) - eixRu( - R). The function as
oo .
k.
k.
=
=
=
=
1
1'
'
'
'
370
1 5. EXAMPLES AND APPLICATIONS
u E S, we must have u( R) 0, u( - R) 0 as R Thus, (15. 4 1) Fu' ixFu . Repeated applications of (15. 4 1) give (15. 42) F[Dfu] ( ix) k Fu, k 1, 2, Combining (15. 4 0) and (15. 4 2), we get (15. 43) xj D�Fu ( -i)j +k F[Df (tku) ] , j, k 1, 2, . . . , which shows that Fu E when u E �
Since
�
� oo .
=
=
=
·
·
·
.
=
=
S.
S
We still need two important properties of the Fourier transform. The first is the inversion formula
1 00 1 (15. 44) u(') ..,fEr27r eiex Fu(x) dx, u E S, and the second is Parseval's identity (u, v) (Fu, Fv). (15.45) =
- oo
=
Let me assume these for the moment (they will be proved in the next section) and show how they can be used in our situation. Note that we can put in the following forms:
(15. 44) (15. 46) F[Fu( -x)](e) u(e), u E S. (15. 47) F[Fu(x)]( -e) u(�), u E S. These snow that for each function w ( x ) E S, there is a u(t) E S such that Fu. Now, suppose that f E £ 2 and that u E L 2 is a solution of (A - .X)u f (15. 48) By definition, there is a sequence { un } of functions in D( A ) such that Un u, (A - .X) un � f as n � oo in L 2 . Let be an arbitrary function in S. Then, by (15.13) and (15. 3 3), ( (A - .X)un , ) - (un , v' + X v ). Thus, in the limit't we get (J, ) - ( u, + .Xv). By (15. 4 5) and (15. 4 1), this becomes (Fj, Fv) - (Fu, F [y' + Xv]) - (Fu, [ix + X]Fv ) ([ix - .X]Fu, Fv). =
=
w=
=
�
v
v
v
=
=
=
=
=
v
'
.
1 5. 5.
371
Another approach S.
Moreover, we remarked above that for each This is true for all v E E there is a v E such that w = v. Hence,
S
w S,
F (Ff - [ix - A] Fu,w) 0 =
w E S. Since S is dense in £2 , this implies that Ff (15.49) Fu = A . Now, let us examine (15. 4 9). If n -# 0, then i x - ,.\ is a polynomial that does not vanish on the real axis. Hence, if f E S, the same is true of the right-hand side of (15. 49). Thus, we can solve (15. 49) for u, obtaining a function in S. Clearly, this function is a solution of (15.20). Moreover, I F/ I (15. 50) I Fu l < l n l ' which implies 11 f� ll , II Fu l which, in turn, implies, via (15. 4 5), (15. 5 1) ! l u l < � :�:I . The reasoning of the last section now shows that A E p(A ) . On the other hand, if then ix - A vanishes for x {3. If f E S, then 0, the only way we can have u E £ 2 is if F f ( /3 ) = 0. This is precisely (15. 3 5). Thus, (15. 4 9) cannot be solved for all /, and consequently, A E a(A for all
.
'lX -
<
o: =
=
). Now we can "see" why the imaginary axis forms the spectrum of A . If we write our original operator in the form
(15. 52)
A = P(Dt) ,
(15. 49)
then P(e) = e. The denominator appearing in the right-hand side of is P(i x ) - If this polynomial has no real roots , then E p(A) . Otherwise, it is in a (A) Thus, we can make the following conjecture: Let P( be any polynomial
A.
A �)
.
P
m
(� ) = L a k � k . 0
Consider the d_ifferential operator A given by
(15. 52):
372
1 5. EXAMPLES AND APPLICATIONS
i
(say, ix
is )- s
defined on a suitable class of functions n L2 S) . If A the closure of ,.\ ha no real roots. if the polynomial P( A, then ,.\ E p( A) if and We shall see, Section 15.8, that this is, indeed, the case. \Ve want to make a remark concerning the nature of the spectrum of = 0 A. It has no eigenvalues. For if f = 0, then ( 15 .49) implies that = 0. Since there and consequently, a sequence 1/Jc: therefore follows that the range of A i (3 not closed for all real {3. In particular, the spectrum of A coincides with its e ent al spectrum ee Section 7. 5) .
only
in
it
Fu is satisfying (15. 3 7), - is ss i (s
u
1 5 . 6 . The Fourier transform
We now give proofs of ( 1 5 .44) and ( 15.45) . To prove the former,
set
R GR (� ) vk21r J- R eiEx Fu(x) dx. We wish to show that G R � u i n £ 2 as R � oo. Now, R _2_ J GR(� ) 21r - R eiEx [!00 e - ix u (t) dt] dx R _!__ ! e i ( E-t) dx] dt . [ u(t) 21r J -R (We were able to interchange the order of integration because the double integral is absolutely convergent . ) Now, =
=
-
oo
=
-
t
oo
oo
( 1 5 . 53) Moreover, we know that
J
( 1 5 . 54)
oo
- oo
Hence,
Rd
sin s s
______
S
oo
=
1r .
( ) u t GR (� ) - u(� ) J t -- �u(�) sinR( t - �) dt (�) u( t cos R t ( 15.55) 7f� J: ( - ()! ( � = � ) dt . by integrating by parts. Set + u(�) u(t) ( ) ) ( )u u(�) u( ' t t t � !!_ ) ] ( 15. 56 ) El( t , .. dt [ t - � (t - �) 2 and ( t ) ( t - �) u' ( t) - u (t ) + u (�) . ( 1 5 . 57) = _!_ 1r
-
oo
=
C
==
=
g
=
373
1 5. 6. The Fourier transform
Expanding g ( t ) in a Taylor series with remainder, we get g (t) == g (�) (t - € )g'(€) + � ( t - � )�g" (�l), (15. 58) where �1 is between � and t . Since g'(t) == (t - �)u" (t) (15. 59) and g" (t) == ( t - � ) u"' ( t) + u" ( t ) , (15. 60) we have (15. 6 1) From (15.61) we see that H (t, � ) is continuous in t and � together . Moreover, since u E S, we see by (15. 5 9) and ( 15. 6 1) that (15. 62) l g (t ) l < C1 ( 1 + 1 � 1 ) (15. 63) l g ( t) l < C2l t - � 1 2 (1 + f � l ) . Substituting into ( 15. 5 5), we get I GR (€ ) - u (€ ) 1 < 1 1-00oo I H (t, € ) 1 dt 1 [ r I H (t , € ) 1 dt + r I H (t , €) 1 dt] < R ll t -f.l
1r R
1r
R
oo ,
J
we have R 1 R u (t) v ( ) dt - (Fu , Fv ) < 100 t
- oo
r
Jl t i>R /
ei txv ( t )
dt . I F u ( x ) l d x
374
1 5.
EXAMPLES AND APPLICATIONS
100 ! Fu(x) l dx }ltif >R l v (t) l dt This completes the proof. <
-
------t
oo
1 5 . 7. Multiplicat ion by
a
0
as R
------t
oo .
D
function
Let2 q(t) be a function defined on ( ) We can define an operator Q on2 L corresponding to q as follows: A function u E L2 is in D ( Q) if qu E £ . We then set Qu qu . We are interested in the following problem: We want to find conditions on q which will ensure that Q will be A-compact (see Section 7. 2) . One particular benefit we derive from such a situation is that it implies ( 15.64) (see Theorem 7. 28 ) . But O"e ( A ) a ( A) , ( 15.65) and a( A) is known to be the imaginary axis. It will therefore follow that ae (A + Q) consists of the imaginary axis. In order that Q be A-compact we must have D(Q) D( A) Moreover, if { un } is a sequence of functions in D(A) such that ( 1 5.66) this must imply that {Qun } has a convergent subsequence. This can be true only if ( 1 5.67) If ( 1 5.67) did not hold, there would be a sequence { un } of functions in D( A) In this case, Q could not be such that ( 1 5.66) holds, while l l qun l l A-compact. A weaker form of ( 1 5.67) is ll qu l l < cl ( l l u l l + I I Au l l ) , u D(A) . ( 1 5.68) By what we have just said, it is a necessary condition for Q to be A-compact. Let us see what ( 15.68) implies concerning the function q . Let <.p be a function in COO which satisfies - oo , oo
.
=
=
:J
.
� oo .
E
<.p(t) > 0,
<.p > 1, 7/Ja ( t)
- 00
0 < t < 1.
oo ,
For each real the function ) is in COO. Such a function is <.p( t certainly in D (A) . Hence, in order that ( 1 5.68) hold, we must have 1: I q,pa 1 2 dt < c1 ( II ,Pa II + II ,p� II ) , a,
=
-
a
1 5. 7.
Multiplication by
a
375
function
which gives Thus, we have shown that in order for (15. 68) to hold, we must have a+l 1 <M< a real. dt 2 (t) q i ( 15. 69) l a Next, I claim that for Q to be A-compact, one needs in addition to (15.69) the fact ti1at (15. 70) Suppose (15. 70) did not hold. Then there would be a sequence {ak} such that l ak l while a +l k 1 > > 0, k 1, 2 dt q (t) (15 . 7-l ) l 1 ak Let c.p (t) be defined above, and set ( 15. 72) k (t) (t - ak ), k 1, 2, Clearly, l vkl l + I Avkl l l c.p l + l cp' l C2 . . If Q were A-compact, there would be a subsequence ( again denoted by {vk} ) such that qvk converges in L2 to some function f. By (15. 7 1), 1: i qVk i 2 dt > 1:k + l i q (t)i 2 dt > which implies (15.73) On the other hand, for any finite interval I, we have vk (t) 0 on for k sufficiently large. This shows that j l f (t) i 2dt 0 for each bounded interval I. This clearly contradicts (15.73), showing that (15. 70) holds. Now let us show that (15. 69) and (15 . 70) are sufficient for Q to be A compact. Let us first prove (15. 68) from (15. 69). It clearly suffices to do so for u real valued. Let be any interval of length one and let x and x' be two points in it. For u E D (A) xwe have x 2u (x) - u2 (x') 1x' dt [u2 (t)]dt 2 1 u' (t)u(t) dt. oo ,
---+ oo
8
2, . . . .
=
as
v
= <.p
=
·
·
·
.
=
=
8,
=
=
q,
I
=
d
=
x
'
I
376
EXAMPLES AND APPLICATIONS
1 5.
Hence, ( 15 74) By the mean value theorem, we can pick x' so that u2 (x') 1 u2 dt. Hence, ( 15 . 75 ) Multiplying both sides by q2(x ) and integrating over I, we have ( 15 . 76) 1 q2 (x)u2 (x)dx < 1 q2 (x ) dx 1 [ (u') 2 + 2u2]dt < M 1 [(u') 2 + 2u2]dt by ( 1 5.69) . Summing over intervals of unit length, we obtain ( 15.77) 1: q2 (x )u2 ( x)dx < M 1: [(u') 2 + 2u2]dt, which implies ( 1 5.68) . Next consider the function qR [see ( 1 5 . 15) ] . Denote the corresponding operator by QR . We claim that, for R < the operator QR is A-compact.I, To see this, we note that for u E D(A), I a bounded interval and x , x' E we have u(x) - u(x') 1'x u'(t ) dt. x Thus, x ( 1 5 . 78) l u ( x ) - u(x' ) l 2 < l x - x' l 1x' l u'(t) l 2 dt < l x - x' l · 1 Au l 2 . Moreover, by taking x' to satisfy u(x') 1 u(t) dt which we can do by the theorem of the mean), we have, when I is of length (one, ( 15. 79) Combining ( 15.78) and ( 1 5 . 79) , one obtains ( 1 5 . 80) l u (x) l < l u l + I _Au l . Now, we claim that functions u E D(A) are continuous and satisfy ( 15.81 ) l u (x) l < ! l u l + I Au l , u E D ( A ) .
=
oo ,
=
=
,
377
1 5. 7. Multiplication by a function
l u (x ) - u(x') l 2 < l x - x' I · I Au l 2 , u E D (A) . In fact , if u E D (A), there is a sequence { Un } of functions in D ( A ) such that Un � u, Aun � Au in £2. By ( 1 5 . 80) , l un (x) - Um (x ) l < l un - Uml l + I Aun - Aum l , which shows that the sequence {un(x)} of continuous functions converges uniformly on any bounded interval. Since the limit must be u, we see that u is the uniform limit of continuous functions and consequently, is continuous. ( 1 5. 8 2 )
Moreover, by ( 15 .78) ,
l un(x) - un(x' ) l 2 < l x - x' l · I Aun l 2 for each Taking the limit as � we get ( 1 5 . 82) . To show that Q R is A-compact , suppose that { un } is a sequence of functions in D(A) satisfying ( 15 .66) . Then, by ( 1 5 $1) and ( 1 5 .82) , l un(x) I < C, ! un(x) - un (x') 1 2 < C2 l x - x' l · The first inequality says that the Un are uniformly bounded, while the second implies that they are equicontinuous ( i.e. , one modulus of continuity works n.
n
oo ,
for all of them) . We can now make use of the Arzela-Ascoli theorem, which says that on a finite interval, a uniformly bounded equicontinuous sequence has a uniformly convergent subsequence. Let us pick the interval to be and denote the subsequence again by Now,
[- R, R]
{ un } · J: l qR I 2 I un - Uml 2 dt J: l q l 2 l un - Um l 2 dt R < max l un(x ) - um (x) l 2 J i q i 2 dt lxi < R R 0 as < 2(R + 1) M max l un(x) - um (x) l 2 lx!
�
m, n �
:J
C
==
u
oo .
378
1 5. EXAMPLES AND APPLICATIONS
Since A is closed, we can make D ( A) into a Banach space by introducing the graph norm on it. Inequality (15. 84) states that Q is a bounded operator from D (A) to £2 with norm C3M or less. If we replace Q by Q - QR , we see that Q - QR is a bounded operator from D (A) to L2 w.ith l + 1 a a l + l 2 � q (t) dt. I Q - QRI I < c3 sup a j q (t) - qR(t) l 2 dt < c3 laisup l R> - 1 a Thus, I Q - QRI I 0 R A by (15.70). But we have shown that QR is a compact operator from D ( ) to L2 for each R < We now merely apply Theorem 4.11 to conclude that Q is a compact operator from D (A) to £2 • This completes the proof. Finally, we want to point out that if l + a l (1 5 .85) q (t)i 2 dt a real, < l a then (15.. 70) implies (15. 69). To see this, suppose (15. 70 ) and (15. 85) held, but (15 69) did not. Then there would be a sequence {ak } such that 1ak + l l q (t)l 2 dt ak (15. 7 0), the sequence {ak } is bounded. Hence, it has a subsequence ( also denoted by { ak } ) such that ak a , where a is a finite number. In particular, l ak - a l < 1 for sufficiently large. But for such 1a+2 l q (t) i 2dt > lak + l l q (t) i 2dt a- 1 ak which shows that (15. 8 5) is violated. Thus, we may conclude that Q is A-compact if and only if (15.70T and (15.85) hold. �
as
�
oo
oo .
0
oo ,
�
By
00 .
�
k
k,
--+ oo ,
1 5 . 8 . More general op erat ors
Consider the differential operator Bu '-- amAmu + am- l Am- l u + + aou, am -:/= 0, (15.86) where 0A Dt is the operator defined in Section 15. 2 . Thus, Ak dk / dtk and A 1. We can write B symbolically in the form B P (A), (15. 87) where P ( z) amzm + + ao. (15.88) ·
=
·
·
=
=
=
=
·
·
·
1 5. 8.
379
More general operators
u
We take D(B) to be the set of those E L 2 having continuous derivatives up to order m in L 2 . The argument used in Section 15.3 in connection with A shows that B is closable. Let B denote its closure. We are interested in determining a(B) . Let us employ the method of Section 15.5. Applying the Fourier trans form to the equation
.X)u J, we obtain, by the argument of that section, Ff ( 1 5.90) Fu P( i x ) - .X . Assume first that P( i x ) -# A for all real x. Then there exists a positive constant such that ( 1 5.91 ) I P (ix) - .X I > c, x real. (We leave this as an exercise.) In this case FI u l I FJ I ' (B -
(1 5.89)
=
=
c
<
c
and consequently,
!lul
(15.92) Now, if f is any function in Hence, there is a function of ( 1 5 .44) to both sides of
u
�
11 ! 1 1 . c
same is true of F f and of F f j[ P ( i x ) - .X]. S, Sthesatisfying ( 15.90) . Moreover, an application E
[P (ix) - .X]Fu Ff shows that u is a solution of ( 1 5.93) (B - A )u f. Thus, ( 1 5.93) can be solved when f E S. The solution is in S and satisfies ( 1 5.92) . Moreover, if f is any function in L 2 , there is a sequence of functions in S such that fn f in L2 . For each there is a solution Un E S of (B - .X)un fn · By ( 15.92) , - fm l /n 11 < U U 0 I n mI This shows that there is u E L 2 such that Un Since B is a closed u . operator, it follows that u E D(B) and satisfies ( 1 5.89) . All of this goes to show that P( { x) -# A for all real x implies that A E p( B) . =
==
n
�
==
-
c
a
�
as m , n � 00 . �
380
1 5. EXAMPLES AND APPLICATIONS
P(i/3) f (f3)
Next, suppose that there is a real f3 such that == ,.\ . In order for F u to be in £ 2 , we see by ( 15 .90) that we must have F == 0. Thus, ( 1 5.35) must hold. This shows that ,.\ E a( However, ,.\ is not an eigenvalue since == 0 implies u == 0 from ( 1 5.90) . Moreover, we can show that ,.\) is not closed. To this end, we use defined by ( 1 5.36 ) . To compute we make use of the function the formula
B). R(B -
f
P(Dt )'l/Jc , ( 1 5.94) P(Dt )(vw) L p( k) (Dt )vDf :, , where k d k P( ) (x ) dxP(x)k . We shall prove ( 15.94) in a moment. Meanwhile, let us use it. Thus , P(Dt )'I/Jc y'E f p(k\ Dt )eif3t Df 'lj; �t) y'Eeif3t f p(k) ( ij3)Df '1/J��t) . Since p ( O ) ( i/3) P( i/3 ) ,.\ , this gives [P ( Dt) - A]'I/Jc = y'Eeif3t L p(k) ( ij3)Df 'lj; �t ) . '1/Jc
=
m
.
0
==
=
0
=
0
==
==
m 1
Now, Hence,
D i k k /3 [I P (Dt ) - A]'I/Jc l < f l p( ) (i ) 1 e 1!1/J I Since l 'l/Jc l 1 , we see that R (B - ,.\) cannot be closed. To summarize, a (B) consists of those complex ,.\ which equal P(ix) for some real x. Moreover, ae(B) a ( B) . It remains to prove ( 15.94 ) . We know that P(Dt )(vw) is the sum of terms of the form aj k Djt vDtkw, Combining the coefficients of where the aj k are constants and j + k < Df w, we get P(Dt )(vw) L Tk (Dt )vDfw, �
1
==
==
m.
==
m
0
o
as c:
�
o.
38 1
1 5. 9. B-Compactness
where the
Tk ( z ) are polynomials.
Now take v ./
'
P (a + (3 )
( 1 5 . 95)
Taking the derivative of order shows that
==
==
e at
and w
==
ef3t . This gives
m
2: Tk( a )(3 k . 0
k with respect to (3 and then putting f3
==
This completes the proof.
0
0
We have shown that a ( A) consists of the imaginary axis, while a (B) consists of all scalars of the forn1 P ( i x) , x real. Thus, we have the following variation of the spectral mapping theorem
[
a P (A) ]
( 1 5 . 96)
==
P [ a ( A )] .
1 5 . 9 . B- Compactness
We now ask when Q is B-compact. The answer, which may seem a bit surprising, is that a necessary and sufficient condition that Q be B-compact is that ( 1 5 . 69) and ( 1 5 . 70) hold. This follows from the inequality
I Au l < C4 ( ll u ll + I I Bu l l ) ,
( 1 5.97)
u E D(B) ,
which we shall verify shortly. First , ( 1 5 . 97) implies that D( B) C D( A) . To see this , note that if u E D ( B ) , then there is a sequence { un } of functions in D(B) such that Un u, B un Bu in £ 2 . By ( 1 5.97) , Aun converges as well. Hence, u E D ( A ) and Aun A u. Applying ( 1 5.97) to Un and taking the limit , we have
---t
---t
---t
( 1 5.98)
< C4 ( 11 u ll + I I Bu ll ) ,
I I A u ll
u E D (B) .
Now ( 1 5.69) and ( 1 5 . 70) imply that Q is A-compact . Thus D (Q) :J D (A) D(B) . Moreover, if { un } is a sequ_e nce of functions in D (B) satisfying
:J
( 15.99) then, by ( 1 5 . 98) , the Un satisfy
< Cs ( l + C4 ) . Since Q is A-compact , it follows that { Qun } has a convergent subsequence. Thus, Q is B-compact . l
l un ll + II Aun ll
Conversely, if Q is B-compact , then we must have ( 1 5 . 1 00)
ll q u ll
< C6 ( ll u ll + I I Bu l l ) , ,
u E D ( B)
382
1 5. EXAMPLES AND APPLICATIONS
S ee (15.68)]. Making use of the functions 1/Ja ( t ) as in the derivation of [(15. 69), we have
1a+ l l q ( t ) i 2 dt < C6 ( I 'I/J I + I B'I/JI I ) c6c7 . This gives (15. 6 9). Similarly, if (15.70) did not hold , there would be a · sequence {a k } such that l a k l � oo and (15. 7 1) holds. Making use of the sequence { V k } defined by ( 15. 72), we prove a contradiction as in Section 15. 7. It remains to prove (15. 9 7). It is equivalent I Au l 2 < Cs ( l u l 2 + I Bu l 2 ) , Thus,
=
a
to
which is equivalent to
( 15.101)
(15. 4 2) and (15. 44). But (15.101) is a simple consequence of x2 < Cs (1 + I P (ix) l 2 ). (15.102) The verification of ( 15.102) for any polynomial of degree 1 or more is left as
i n view of
.
an exercise. Next , consider the operator
m- 1
Eu L qk (t )Ak u, where each of the functions q k ( t ) satisfies (15. 6 9) and (15. 70). We take D(E) D (B) . We claim that (15.104) I Eu l < Cg ( l u l + I Bu l ), u E D( B ) . This shows that E is B-closable. This means that 0, B 0 un implies Eun � 0. By using a limiting process, we obtain a unique extension E of E defined on D(B). Now we can show that E is ..8-compact . To see this, note that by (15. 6 8), (15.105) l qk Ak u l < Clo ( I Ak u l + I Ak+ l u l ), u D(B) . (15.103)
==
0
==
Un
�
E
Thus,
( 15.106)
m
I Eu l < ell L I Ak u l · 0
--+
1 5. 1 0. The adjoint of
If we can show that ( 15 . 107)
A
383
m Ak < ( L I u: l c12 l u l + I Bu l ), u E D(B) , 0
it will not only follow that ( 1 5 . 104) holds, but also that each term of E is ..8-closable and its extension to D ( B ) is ..8-compact . Moreover, E is the sum of the extension of its terms, so that E is B-compact as well. Thus, it ren1ains to prove ( 15. 1 07) . It is equivalent to -
-
1: ( 1 + x2 + . . . + x2m )! Fu(x) l 2 dx < c13 1: ( 1 + I P (ix) I 2 ) 1 Fu(x) l 2dx,
which follows from ( 1 5 . 108)
Again the verification of ( 15 . 1 08) is left as an exercise. Since E is B-compact , we can conclude that -
-
ae (B + E) ae ( B ) a ( B ) { P ( i x) : x real} . =
=
( 15. 109)
=
15. 10. The adjoint of A
£2, ,
Since D( A ) is dense in we know that A has a Hilbert space adjoint A* (See Section 7. 1 and Problem 4 of Chapter 1 1 . ) Let us try to determine A* . If and are in D ( A ) then
u v
(Au,v) -(u, Av) . To verify ( 1 5 . 1 10) , note that it holds for u E D ( A ) and v C(J by mere inte gration by parts [see ( 15 . 13)] . Next suppose u and v are in D(A). By ( 15 . 33) , there is a sequence {
=
E
---+
---+
=
---+
---+
---+
---+
=
we have ( 15 . 1 10) in the limit . From ( 15 . 1 10) we expect A * to be an extension of - A. In fact, we can show that (15. 1 1 1)
A * == -. A .
384
1 5. EXAMPLES AND APPLICATIONS
v
To see this, let be any function in D( .A * ) . Let ,\ =I= 0 be a real number. Then ± .A E p( .A ) . In particular, there is a E such that ( 15 . 1 1 2)
w D(A) - ( A + ..\ ) w = (A * - ..\ )v .
u E D(A) , we have ( ( A - ..\ )u, v) = (u, ( A * - ..\)v) = - (u , (A + ..\ )w) = ( ( A - ..\ )u, w) by ( 15 . 1 10) . Hence, (( A - ..\ ) u,v - w) = 0 for all u E D ( A ) . Since ,\ E p(A ) , R ( A - ..\) is the whole of £ 2 . Hence, v = w E D ( A) . By ( 1 5. 1 12) , we have A* v - Av . This proves ( 15 . 1 1 1 ) . By ( 1 5 . 1 1 1 ) , we have ( 15. 1 13 ) ( iA)* ( -i) ( - A) = iA , which shows that the operator iA is selfadjoint . Since a( .A ) is the imaginary axis , ( iA) consists of the real axis. Now, suppose q(t) is a real function satisfying ( 1 5.69) and ( 1 5 . 70) . Then Q is iA-compact and ( 1 5 . 1 1 4) O'e(iA + Q) = ae(iA) = a (i.A) consists of the real axis. All other points ,\ of the complex plane are in 4l i A + Q with iA + Q having index zerq. Thus , a point A that is not real can be in ( iA Q) only if it is an eigenvalue. But nonreal points ,\ cannot be eigenvalues of iA + Q since (iA u , u ) and (Q u, u ) are both real, the former because iA is selfadjoint (see Theorem 12. 7) and the latter because q is real Now, if
�
=
u
u
+
valued. Thus, if
( iA + Q - ..\)u = 0, then we have SSm ( (i A + Q - ..\ ) u , u ) = - c;:J m .A I u l 2 0. This shows that either ,\ is real or u 0. Thus � the operator iA + Q can have no nonreal eigenvalues. By what we have just said, this means that all nonreal points are in p (i .A + Q). Consequently, a(iA + Q) consists of the real axis. A repetition of the argument that proved ( 1 5. 1 1 1 ) now shows that iA + Q is selfadjoint . ==
==
15. 1 1 . An integral operator
Let us now consider the operator ( 15. 1 1 5)
V u(x ) = fox u (t) dt
385
1 5. 1 1 . An integral operator
defined on those continuous functions u(t} such that u and Vu are both in As before, it is easily ch·�cked that V is closable. In fact , we have for uE and v (x) E C0\
£2 .
D(V)
x 1 (Vu, v) = : [ fo u(t) dt] v(x) dx
t 00 ( 1 5. 1 16) = fo [1 v (x) dx] u (t) dt - 1� [ 1 v (x) dx] u(t ) dt. oo Let V denote the closure of V, and let us try to determine a ( V ) . If f and u are smooth functions, then the equation ( V - J-L)u f ( 1 5 . 1 1 7) is equivalent to u - ttu ' == f', J-Lu( O ) + f ( O ) 0. oo
==
==
If J-L =/=- 0 and we set A == 1/ J.-L , this becomes
( 1 5. 1 18)
U
u' - A == - Aj
1,
u( O ) + Aj(O) == 0.
We have already solved equations of the type ( 1 5 . 1 18) for a == �e A =I= 0 ( see Section 1 5 .4) . This will be the case when �e J-L =I= 0. Assuming a =I= 0 for the moment, we have by ( 1 5 . 2 1 ) and integration by parts
( 1 5 . 1 19) This gives
( 1 5. 1 20)
u (x) = - Aj ( x ) -
>.?
fo e>-(x - t) j (t)dt x
in view of the second member of ( 15. 1 18) . This gives a candidate for a solution of ( 15 . 1 18) . Is it in It will be only if
£2 ?
( 15 . 12 1 ) is. But
( 15 . 1 22) If x
a ---+
> 0 ( which is the case when oo unless
( 15. 1 23)
�e
11
> 0) , we will have jg(x) l
�
oo
as
386
1 5.
EXAMPLES AND APPLICATIONS
This immediately shows that we cannot solve ( 1 5. 1 18) for all assume that satisfies ( 1 5. 123) . Then ( 15 . 1 20) is equivalent to
f
f. Let us
x - Af ( x ) )..2 100 eA (x-t) f(t) dt. This is easily seen to be in £2. In fact , the reasoning in Section 15.4 gives Al : 1 :>.. 1 2 ) ( 1 5. 1 25) llull < IIIII · C 2 If < 0 (which is the case when �e < 0) , then ( 15. 1 22) shows us that unless as x l (x) I 1� e - At f (t) dt = 0. ( 1 5 . 1 26) +
u( ) =
( 1 5 . 124)
/-L
a
g
--+ - oo
--+ oo
On the other hand, if f satisfies ( 15. 126) , then ( 15 . 120) becomes
x
u( ) =
( 15 . 1 27)
-
Af ( x )
-
).. 2
£2
1� eA(x-t) f (t) dt,
which again is seen to be in and satisfy ( 15 . 125) . Moreover, if we apply the operator V to ( 15. 1 20) , we have Y Y 2 Vu( y ) = -AVf ( y ) - A
1 [[ eA(x-t) dx] f (t) dt y = - A lo e A (y -t) f( t) dt ,
( 1 5. 128)
which shows that ( 1 5 . 1 20) is a solution of ( 15 . 1 17) . If a > 0 and { 15 . 123) holds, then
Vu( y ) = )..
( 1 5 . 129)
ioo eA(y-t) f(t)dt,
L2. Similarly, if a < 0 and ( 15 . 126) holds, then V u( y ) = - :>.. 1� e A ( y - t) f(t) dt , ( 1 5 . 130) showing that V u E £ 2 in this case as well. Thus, for f E £ 2 and having a continuous first derivative in L 2 , we can solve ( 15. 1 1 7) for a =I= 0 provided f satisfies (15. 1 23) or ( 15 . 1 26) depending on whether a 0 or a < 0. Moreover, the solution u given ( 15. 120) is
which shows that Vu
E
>
continuously differentiable and satisfies ( 15. 125) . Now, suppose a > 0 and is any function in Define the function 1 ( by
h t)
f(t)
A et { ' h l (t) =
0,
t > 0,
t < 0.
by
£2 satisfying ( 15 . 123) .
1 5. 1 1 .
An integral operator
387
h 1 £2 .
Jn h lR ,
Then E ( If this is not obvious, then consider the functions which are in COO and converge tG See Section 15.3. ) Moreover, in ( 1 5 . 123) merely states that
h 1 £2 .
( 1 5 . 131)
{ fn } ) ( , h n 'l/J 1 J _ f 9n - n , ( 'lj;, h l ) ·
Now, we know that there is a sequence of functions in COO which converge to in =!= 0. Let Let 'lj; be any function in COO such that 'lj;,
f L2 .
( h1 )
_
Then
9n E COO and
( 1 5 . 1 32)
In addition But
I ( Jn , h 1 ) l l ( fn - J, h 1 ) l < l fn - f l · l h 1l l · These last two inequalities show that 9n in £ 2 . Since E COO and it f satisfies ( 1 5. 132) , we know that there is a solution of ( 1 5 . 1 33) (V - ..\ ) un 9n · Moreover, by ( 1 5. 125) , we see th at the Un converge in £ 2 to some function u . Thus, u E D(V) and it is a solution of ( 15. 1 1 7) . Thus, we have shown that for > 0 , ( 1 5. 1 17) has a solution for f E £ 2 if and only if ( 15. 123) [or ( 1 5 . 1 3 1 ) ] holds. Similarly, for < 0, equation ( 1 5 , 1 1 7) has a solution for f E £2 if and only if ( 15 . 126) holds. In particular, we see that R( V - J-L) is closed in L 2 for Re =!= 0. How about uniqueness? To get an idea, suppose u E D ( V) and v(x) is a continuous function in [0, oo] which vanishes for x large. Then ( 1 5 . 134) 100 VuVdx 100 [fox u (t) dt] v(x) dx 100 [100 v(x) dx] u (t) dt . Now, if u E D ( V), we can apply ( 15. 134) to a sequence { un } of functions in D(V) which converges to u in £2 and such that VUn Vu . This gives ( 15. 1 35) 100 ( Vu - J-LU)Vdx 100 [100 v(x) dx - J-LV (t )] u(t) dt. Now, if Vu - J-l U 0, we have ( 1 5. 1 36) 100 [100 v(x ) dx - J-LV (t )] u (t ) dt 0 =
g
�
=
a
a
11
=
=
�
=
=
=
388
1 5.
v.
EXAMPLES AND APPLICATIONS
'ljJ
for all such But we can show that for J-l -1= 0 and E CO there is a v ( x ) continuous in 0, oo ] , vanishing for x large and satisfying
[
( 15 . 1 37)
100 v ( x ) dx - J-LV (t) 'l/J(t) , t > 0. ==
In fact, the method used above gives
v ( x ) - �'l/J (x ) - :\2 100 e>. (t-x) 'ljJ(t) dt . Clearly, the function v given by ( 1 5. 138) is of the type mentioned. We leave the simple task of verifying that it is a solution of ( 1 5 . 137) as an exercise. Since CO is dense in £ 2 , it follows that u (t) 0 for t > 0. A similar argument holds for t < 0. For 0, it is even easier. We merely take v (�) -'l/J'( x ) . This clearly has the desired properties and is a solution of ( 1 5 . 1 37) . Thus, o:( V - J-L ) 0 for all Before we go further, we must take a look at D(V). In the case of A and B we know that D( A ) and D (B) are dense in £ 2 because they contain CO . In fact , a But a moment ' s reflection shows that this is not the case for function 'ljJ E CO is in D(V) if and only if 1� '1/J ( t) dt 100 '1/J (t) dt 0. ( 15 . 1 39) If ( 15. 139) holds, then V'ljJ (x) vanishes for l x l large, and hence, V?jJ E CO . On the other hand, if, say, 100 1;_,{ t) dt -1= 0, then V'ljJ( x ) # 0 as x showing that V'ljJ cannot be in \ L2 . It is, therefore, far from apparent that D(V) is dense in £ 2 . However, this indeed is the case. We shall prove it by showing that the set of functions 'ljJ E CO which satisfy ( 15. 1 39) is dense in £ 2 . To do so, it suffices to show for each c > 0 and each r.p E CO , there is a 'l/J E CO satis�-'y ing ( 15. 139) such that ( 1 5 . 140) l r.p - 'l/J II < c . So suppose c > 0 and r.p E CO are given. Let g(t) be any function in CQ which vanishes for t < 0 and such that g (t) > 0 and ( 15. 141 ) 1 00 g (t) dt 1 . Set h (t ) r.p (t ) - ECi g ( -Et) - EC29 (ct ) , ( 1 5 . 138)
==
==
J.-l ==
==
==
J-l ·
V.
=
c
=
=
� oo ,
� c
=
==
1 5. 1 1 .
An integral operator
where Cl
Clearly,
=
389
1:
h E COO . Moreover, 1: h (t)dt =
and similarly,
C l - CC I
1: g(-Et) dt == 0,
100 h (t) dt = 0.
h satisfies ( 1 5. 1 39) . Moreover, l h - cpl l 2 == c2 ci 1 o g2 ( -ct) dt + c2 c� lof oo 92 ( ct ) dt oo = c 2 (ci + c� ) 1 00 g 2 ( r ) dr . Thus, h can be made as close as desired to in the £ 2 norm by taking sufficiently small. This shows that D ( V) is dense in £ 2 . We now have the complete picture for �e J-l =f. 0. I� this case, ( J-l) == 0 and R( V - J-l) is closed in £ 2 . :tv1oreover , R(V - J-l) consists of those f E £ 2 which are orthogonal to the function h 1 . This means t h at the an r{ihilators of R( V - J-L) form a one-dimensional subspace. By (3. 1 2) , we see that V - J-l is a Fredholm operator with index equal to - 1 . It remains to consider the case �e J-l == 0. This means that A == i/3. From ( 15 . 1 22) we see that in order for f E COO to be in the range of V - J-l , it is necessary that Thus,
-
E
o:
"
.. -r -
E COO does satisfy ( 1 5. 1 42) , one checks easily that ( 1 5 . 1 20) is ( V - J-L ) == f . ( 1 5. 143) But the set of those f E COO satisfying ( 1 5 . 142) is dense in £ 2 . Thus, R(V-J-L) is dense in £ 2 for �e J-l == 0. However, R( V - J-l) cannot be the whole of L 2 . If it were, the fact that ( V - J-L) == 0 would imply that J-l E p(V) . But this would imply that E p(V) for close to J-l · However, we know that this is not the case.
However, if f a solution of
u
v
o:
v
Thus, a (V) consists of the whole complex plane. Ho�ever,
( - J-L) == - 1 , E
i V
u
390
1 5. EXAMPLES AND APPLICATIONS
1 5 . 1 2 . Problems
( 1 ) Verify that the function defined by ( 1 5 . 1 ) is infinitely differentiable in ( - oo , oo ) . (2) If P(t)· is a polynomial having no real roots, show that there is a positive constant c such that IP(t) l > c for real t. (3) Show that if u E S and P(t) is a polynomial having no real roots, then u P is in S.
j
( 4) Show that the operator Q of Section 15.7 is closed. (Hint: Use the fact that every sequence of functions converging in L 2 has a sub sequence that converges pointwise almost everywhere. Moreover, functions that agree almost everywhere are consiaered the same function in L 2 .] (5) Show that ( 15.68) holds if and only if the real and imaginary parts of q satisfy it for real u E
D (A) .
(6) For what values of a is the function q(t)
=
ltl a: A-compact?
(7) Prove ( 15. 108) for any polynomial P of degree details in the reasoning following ( 15. 1 07) .
m
> 1 . Supply the
(8) Find B* . Under what conditions is B selfadjoint? (9) If B is selfadjoint and q is a real valued function satisfying ( 15.69) and { 1 5 . 70) , is B + Q selfadjoint? Can you determine u(B + Q) ? {10) Show that the operator V of Section 15. 1 1 is closable. { 1 1) Prove { 1 5 . 125) . { 12) Show that the function given by { 15 . 1 38) is a solution of ( 15. 137) .
1 5. 1 2.
391
Problems
( 1 3) Prove that a solution of ( V - J..t ) u J.L =I= 0.
=
0 vanishes in ( - oo , 0] when
( 14) Using the fact that the set of those 'lj; E dense in L 2 , show that the set of those f is dense in £ 2 .
( 15. 1 39) is COOCOOsatisfying satisfying { 15 . 141) E
T V SLB"O
-
Appendix A
G lossary
Adjoint operator. If A maps X into Y and D (A) is dense in X, we say that y' E D(A' ) if there is an x ' E X ' such that ' (A. I ) x ' ( x ) == y ( Ax) , x E D (A) . Then we define A' y' to be x ' . a
( A) == dim N(A) .
Almost commuting operators . For any two operators A, B E B (X) , we shall write A '-.-/ B to mean that AB - BA E K (X) . The reason for this notation is that [A] [B] [B] [A] in this case. Such operators will be said to "almost cornmute." ==
Annihilators. For S a subset of a normed vector space X , a functional x ' E X ' is called an annihilator of S if x' (x) == O , x E S. The set of all annihilators of S is denoted by so . For any subset T of X', we call x E X an annihilator of T if x ' ( x) == 0, x ' E T . We denote the s�t of such annihilators of T by 0T .
Associated operator. With any densely defined bilinear form a ( u , v ) we can associate a linear operator A on H as follows : We say that u E D ( A) if u E D ( a ) and there is an f E H such that (A . 2 )
a ( u, v ) == ( f, v ) ,
u E D(a) . 393
A. Glossary
394
"'
The density of D(a) in H implies that f is unique. We define Au to be f. A is a linear operator on H; it is called the operator associated with the bilinear form a( u, v) . B = B [a, b) . The set of bounded functions on an interval [a, b] . The . norm IS II'PII = sup a<x
l cp ( x ) l .
,B(A) == dim N(A') . B {X, Y) . The set of bounded linear operators from X to Y. Banach algebra. A Banach space X having a "multiplication" satis fying:
(A.3)
If A, B
E X,
then AB
E
X,
(A.4)
II A BII
<
II A II . II BII ,
(A.5)
(aA + ,BB) C
=
a( A C)
C( a A + ,BB)
==
a(CA) + {3(CB) .
+
,B(BC) ;
and
(A.6)
Banach space. A complete normed vector space ( cf. p . ' 7) . Bessel's identity. n
(A.7 �
n
cp
�
""
n
II! - L: ai'Pi ll 2 == 11 ! 11 2 - 2 L ai ( f i ) + L a7 == 11! 11 2 - L a7 . 1
1
Bessel's inequality.
00
L a7
(A.8)
,
<
1
1
11 ! 11 2 •
1
Bilinear form. A bilinear form ( or sesquilinear functionaQ a( u, v) on a Hilbert space H is an assignment of a scalar to each pair of vectors u, v of a subspace D(a) of H in such a way that
(A.9) (A. lO)
w) == aa(u, w) ,Ba(v, w), a(u, av + ,Bw ) == a a(u, v) + i3 a(u, w),
a( au + ,Bv,
+
w E D(a) , u, v, w E D (a) .
u, v,
395
A . Glossary
The subspace D ( a ) is called the domain of the bilinear form. When we write a u) in place of a u, u .
(
( )
v ==
u
Bounded linear functional. A linear functional satisfying
I F (x) J < M l x l , x E X.
(A. l l )
X
Bounded linear operator. An operator A is called bounded if D(A) and there is a constant M such that
==
I Ax l < M l x l , x E X . The norm of such an operator is defined by (A. 1 3) IJ A I = ��� I JAxl x l i l . (A. 12)
[
Bounded variation. A function defined on an interval a , b] satisfying
n
L l g (ti ) - g (ti - 1 ) 1 < c
(A. 14) for any partition a
variation.
1
-
to < t 1 <
·
·
·
< tn
==
b is said to be of bounded
[ ]
[
C = C a,b . The set of functions continuous on a closed interval a , b] . The norm is == max (
I cp I
a< x < b
I cp x) I ·
Cauchy sequence. A sequence satisfying
l hn - hml l � 0 as m, n � oo.
10.
Cauchy-Schwarz inequalities. Inequalities ( 1 . 2 6 ) and ( 1 .28) on p.
Closable operator. Let A be a linear operator from a normed vector space to a nor med vector space Y. It is called closable ( or preclosed) if {xk } C D(A) , Xk � 0, Axk � y imply that y == 0. Every closed operator is closable.
X
)
Closed bilinear form. A bilinear form a( u , v will be called closed if un C D a , Un � u in H, a un - urn ) � 0 as m, n � oo imply that u D a and a un - u) � 0 as n � oo.
{ } () E () (
(
A. Glossary
396 Closed operator. An operator D(A) is a sequence satisfying
A is called closed if whenever { xn } Axn --+ y in Y,
c
Xn --+ X - in X , then x E D (A) and A x y . Closed set . A subset U of a normed vector space X is called closed if for every sequence { x n } of elements in U having a limit in the limit is (A. 1 5)
==
X,
actually· in U.
Commutative Banach algebra. A- Banach algebra B is called com mutative (or abelian) if
ab == ba,
(A . l6)
a, b E B.
Compact approximation property. A Banach space X has the com pact approximation property wzth constant C if for each c > 0 and finite set ·of points E X, there is an operator E K(X ) such that , < C and
II - Kl
(A. 1 7)
x 1 , Xn l x k - Kxkl l < E , ·
·
K
·
1 < k < n.
Compact operator. Let X, Y be normed vector spaces . A linear operator from X to Y is called compact (or completely continuous) if < C', the C X such that == X and for every sequence sequence has a subsequence which converges in Y. The set of all compact operators from X to Y is denoted by ( X Y) .
K D (K ) { KXn }
{ xn }
l xn l
K
,
Complete orthonormal sequence. An orthonormal sequence in a Hilbert space H is called complete if sums of the form
(A. 18)
s
=
{ 'Pi }
n L Oi 'Pi 1
are dense in H, i.e. , if for each f E H and form such that < E.
I ! - Sl
t
> 0, there is a sum
S of this
Completeness . Property ( 14) on p. 2 1 . The property that every Cauchy sequence converges to a liinit in the space. Dissipative operator. An operator B satisfying
(A. l9)
�e ( Bu, u ) < 0 ,
u E D(B) .
Dual space. The set of bounded linear functionals on a normed vector space X . It is denoted by X' .
397
A. Glossary
Essential Spectrum.
ae (A)
==
n
a(A + K) .
KEK(X)
It consists of those points of a(A) which cannot be removed from the spec trum by the addition to A of a compact operator. Euclidean n-dimensional real space, consisting of sequences of n real numbers
Rn .
( p. 9) A vector space
where addition is defined by
and multiplication by a scalar is defined by (
Extension of a bilinear form. A bilinear form b ( u, v ) is called an extension of a bilinear form a( u , v ) if D( a ) c D(b) and b ( u, v ) = a ( u , v ) for u, v ·E D(a) . -
Extension of an operator. An operator B is an extension of an operator A if D(A) C and Bx == Ax for x E D (A)
D (B)
.
Fredholm operator. Let X, Y be Banach spaces. Then the set
( 1 ) D(A) is dense in X,
( 2) .i4 is closed, (3) a (A) < oo,
(4) R(A) is closed in Y, (5) ,B(A ) < oo.
B(X)
Fredholm perturbation. An operator E E is called a Fredholm perturbation if A + E E ti> for all A E
398
A. Glossary
Functional. An assignn1ent F of a number to each element space and denoted by F(x) .
x of a vector
Hilbert space. A vector space which has a scalar product and is com plete with respect to the induced norm. Hilbert space adjoint. Let H1 and H2 be Hilbert spaces, and let A be �n operator in B(H1 , H2 ) . For fixed E H2 , the expression Fx == is a bounded linear functional on H1 . By the Riesz representation theorem == (Theorem 2. 1 ) , there is a E H1 such that for all x E H1 . Set z == Then is a linear operator from H2 to H1 satisfying
(Ax, y )
y
Fx (x, z ) A*y . A* (A.20) (Ax , y) (x , A* y) . A* is called the Hilbert space adjoint of A . z
==
Holder's inequality.
00
l xi PI zl q· Hyponormal operator. An operator A in B(H) is called hyponormal if (A. 22) I A* u l < I I Aull , u H, or, equivalently, if (A. 23) ( [AA* - A* A] u, u) < 0, u E H. L1
(A. 2 1 )
Xi Zi _< /
E
Of course, a normal operator is hyponormal. Ideal. Let B be a Banach algebra with identity e . A subspace M of B is called a right ideal if E M for E M, E B. It is called a left ideal if E M when x E M, E B. If it is both a right and left ideal, it is called a two-sided ideal or merely an zdeal.
xa a
ax
Index. i ( A)
==
x
a
a (A) - jj(A) .
A
Infinitesimal generator. A linear operator on X is said to be an infinitesimal generator of a semigroup {Et } if it is closed, is dense in X and consists of those E X for which (Etx)' exists , t > 0, and
x
( A.2 4 )
(EtX ) ' == AEtX,
XE
D(A).
Linear functional. A functional satisfying
(A.25)
D(A)
A . Glossary
for
a1 ,
399
0: 2 scalars.
Linear operator. subspace of X and b ) 0:1 , 0: 2 and all elements
An operator A is called linear if a ) D ( A ) is a A ( o: 1 X1 + o:2x2 ) == o: 1 Ax 1 + o:2 Ax 2- for all scalars x 1 , x2 E D ( A ) .
lp , where p is a real number satisfying 1 < p < oo. It is the set of all ) such that infinite sequences x = ( x 1 , , xj ·
·
·
,
·
·
·
l00 • A vector space consisting of infinite sequences of real numbers
( A.26 )
f
==
( 0:1 '
·
·
·
,
O:n ,
·
•
·
)
for which
Maximal element . An element x o is said to be maximal for a partially ordered set S if x o -< x implies x = x o. Maximal ideal. An ideal N in a Banach algebra B such that N =I= B , and every element of a E B can be written in the form a == a 1 + .Ae , where a1 E N and E is the unit element of B . Measure of noncompactness. Let X be a Banach space. For a bounded S ubset n c X we let q (O) denote the infimum ( greatest lower bound ) of the set of numbers r such that fl can be covered by a collection of open spheres of radius r. In particular, q(O) == 0 if and only if fl is totally bounded, i.e. , if and only if its closure is compact. It is for this reason that q(O) is sometimes called the measure of noncompactness of n .
A. Glossary
400
Measures of operators. Let X, Y be Banach spaces with dim X and let be an operator in B (X, Y) . We define
A
r(A) = infM I A I M I , Ll (A ) = supM NcinfM I A I Ni l , r (A) = supM inf U Ax l , z1 (A) sup OM < inf I Ax l , I A I m = . inf I A I M I , I A I q = q[A( Sx )] , inf I A I K = KEK(X,Y) I A - Kl , inf T I I J.to (A) = 0, J-L (A) = inf l i T I , x A . I mf T (A) = x¢N(A) d(x, N (lA) ) ,
=
oo,
xeM ll x ll = l
=
. dIm
oo
xeM ll x ll = l
dtm M0
'
a(A-T) >a(A)
o:(A) < oo ,
o:(A)
=
oo ,
a(A-T) =oo
AI M
de where M, N represent infinite dimensional, closed subspaces of X, notes the restriction of A to M, and Sx denotes the closed unit ball in X, . I.e. ' 1}. Sx = E X :
{x
l xl <
Multiplicative functional. A linear functional m on a commutative Banach algebra B is called multiplicative if m -:f. 0 and
( A 27) .
m(ab) = m(a)m(b) ,
a, b E B.
Norm. A real number assigned to elements of a vector space and having properties ( 1 1 ) - ( 13) on p. 7. Normal operator. An operator satisfying
(A. 28)
I A* !I I = I AJI I , f E H.
is called normal. Normed vector space. A set of objects satisfying statements ( 1 )-( 13) on pp. 6, 7.
401
A. Glossary
Numerical range. The numerical range W(A) of an operator A on a Hilbert space H is the set of all scalars A that equal (Au, u) for some u E D (A) satisfying l l u ll == 1 . The numerical range W(a ) of a bilinear form a ( u , v ) is the set of scalars A which equal a( u) for some u E D(a ) satisfying l l u l l == 1 . Operator. A mapping A which assigns to each element x of a set D(A) C X a unique element y E Y. The set D(A) is called its domain. The set R(A) of all y E Y for which there is an x E X such that Ax == y , is called its range. Orthogonal projection. Let H be a complex Hilbert space, and let M be a closed subspace of H. Then, by the projection theorem (Theorem 2.3) , every element u E H can be written in the form
( A .29)
u == u ' + u" ,
u ' E M, u" E M..l .
Set
Eu == u ' . It is called the
orthog onal projection onto M.
Parseval's equality.
00
1 !1 1 2 == L ( J, 'Pi ) 2 .
(A.30)
1
Partially ordered set . A set S is called partially ordered if for some pairs of elements x , y E S there is an ordering relation x -< y such that
( 1 ) X -< X ,
X
E 8,
(2) X -< y , y -< X
(3)
X
-<
y,
y
==?
X
==
y,
-< Z ==> X -< Z .
Positive operator. An operator A E B (H) is called positive if
(A .3 1 )
( Au , u) > 0,
uEH
[i.e. , if W(A) is contained in the nonnegative real axis] . Reflexive Banach space. A Banach space Y for which: the embedding of Y into Y " given by J E B(Y, Y " ) such that
(A. 32) is onto.
Jy(y' )
==
y' (y ) ,
y E Y, y' E Y '
40 2
A. Glossary
Relatively compact subset . A set V c X is called relatively compact if every sequence of elements of V has a convergent subsequence. The limit of this subsequence, however, need not be in V. Riesz operator. For a Banach space X , we call an operator E E B ( X ) a Riesz operator if E - A E q_) ( X ) for all scalars A =!= 0. We deno_t e the set of Riesz operators on X by R(X) . Saturated subspace. A-'closed subspace W E X' having the property that for each x ' E X' \ W there is an' x E OW such that x ' ( x ) =!= 0. Scalar product . A functional vector space satisfying
(f, g ) defined for pairs of elements of a
( af,g ) == a ( f,g ) ( f + g, h) == ( f, h) + (g, h) ( f, g ) == (g, f ) iv. ( f, f ) > 0 unless f == 0.
i. ii.
111 .
Schwarz's inequalities. Inequalities ( 1 . 2 6) and ( 1 . 28) on p. 10. Semigroup. A one-parameter family {Et } of operators in B(X) , t > 0, with the following properties:
( a) (b)
Es Et == Es + t , Eo == I.
s
> 0, t > 0,
Semi-Fredholm operators. q> + ( X, Y) denotes the set of all closed linear operators from X to Y, such that D(A) is dense in X, R(A) is closed in Y and ( A) < oo . q> _ ( X, Y ) denotes the set of all closed linear operators from X to Y, such that D(A) is dense in X, R ( A) is closed in Y and jj(A) < oo. Operators in either set are called semi-Fredholm.
a
Semi-Fredholm perturbations. F+ ( X ) denotes the set of all E E B(X) such that A + E E q> + VA E + . F_ (X) denotes the set of all E E B (X) such that A + E E _ \1 A E _ They are called semi-Fredholm perturbations. .
Seminormal operator. An operator A E B(H) is called seminormal if either A or A* is hyponormal.
403
A. Glossary
Separable. A normed vector space is called separable if it has a dense subset that is denumerable. Strictly singular operators. An operator S E B(X, Y) is called strictly singular if it does not have a bounded inverse on any infinite di mensional subspace of X. Strongly continuous semigroup. A one-parameter family {Et } , t > 0, of operators in B(X ) is called a semigroup if
(A. 33) It is called strongly continuous if (A.34)
Etx
is continuous in t > 0
for each
x E X.
x
Sublinear functional. A functional p ( ) on a vector space V is called sublinear if
p ( x + y) < p ( x ) + p(y), x, y E V, p( ax ) = ap(x), x E V, a > 0.
(A.35) (A.36)
Subspace. A subset U of a vector space V such that are scalars. U whenever are in U and a 1 ,
x 1 , x2
a2
a 1 x 1 + a2x2 is in
Supremum. The sup of any set of real numbers is the least upper bound, i.e. , the smallest number, which may be +oo, that is an upper bound for the set. (An important property of the real numbers is that every set of real numbers has a least upper bound. ) Symmetric bilinear form. A bilinear form a (u , v ) is said to be sym metric if
(A.37)
(
a v, u
Total subset. A subset of that annihilates it is 0.
) = a (u, v) .
X'
is called total if the only element of
X
Total variation. The total variation of a function g on an interval [a, b] is defined as n
V ( g) = sup :L l g (ti ) - g (ti- 1 ) 1 ,
1
where the supremum ( least upper bound) is taken over all partitions of [a, b] .
404
A . Glossary
Totally bounded subset. Let c > 0 be given. A set of points W c X is called an E-net for a set U c X if for every x E U there is a z E W such that ll x - z ll < E. A subset U c X is called totally bounded if for every E > 0 there is a finite set of points W c X which is an E-net for U.
S
x,
Totally ordered set . The set is called totally ordered if for each pair y of elements" of S, one has either x -< y or y -< x (or both) .
S
Upper bound. A subset T of a partially ordered set is said to have the element xo E as an upper bound if x -< xo for all x E T.
S
Vector space. A collection of objects which satisfies statements ( 1 )-= (9) , ( 1 5) on pp. 6, 7.
X
Weak convergence. A sequence { xk } of elements of a Banach space is said to converge weakly to an element x E X if
(A.38) for each
x x
'
E
X'.
'
( xk )
--+
x
'
( x ) as k
--+ oo
Weak* closed. A subset W of X' is called w eak* (pronounced "weak star" ) closed if x ' E W whenever it has the property that for each x E X, there is a sequence { x� } of members of W such that
(A.39)
x k ( x ) --+ x
'
(x ) .
Weak* convergence. A sequence { x �} of elements in weak * convergent to an element x ' E X' if
(A.40)
x � ( x ) --+ x
'
( x ) as n --+ oo,
x
E X.
X'
is said to be
-
Appendix
B
Major· Theorems We list here some of the important theorems that are proved in the text. For each we give the page number where it can be found.
X be a Banach space, and assume that K is an operator on X (i. e. , maps X into itself) such that Theorem 1 . 1 . (p. 7} Let
a) K(v + w) = Kv + Kw, b) K(-v) -Kv , c) II Kv ll < M ll v ll , d) 2:� II K n v ll < �
oo
for all v, w E
X. Then for each
(B . l )
u,
f
E =
X there is a unique f E X such that u + K f.
Theorem 1 .4. (p. 22} There is l2 and L2 such that if
( ao , a 1 , (B .2)
·
·
·
) corresponds to f, then n Jn II
L aj '/)j -
a one-to-one correspondence between
�
0
as n � oo ,
0
and (B. 3)
1 1 ! 11 2
00
= L a; , 0
OJ == ( f , cpJ ) ·
405
B. Major Theotems
406
) be a sequence of real numbers, be an orthonormal sequence in H. Then
Theorem 1 . 5 . (p. 23) Let (a 1 , a , 2
and let {
'Pn }
converges in H as
·
·
·
n L1 aiC,Oi
n �
Theorem 1 . 6 .
oo if, and only if, 00
L1 ar < oo . {p. 24} If { 'Pn } is complete, then for each f E H f L (f, cpi ) 'Pi 1 =
00
and
(B .4)
ll f ll 2 =
00
L1 ( f, C,Oi ) 2 ·
Theorem 2 . 1 . (Riesz Representation Theorem) (p. 29} For every bounded linear functional F on a Hilbert space H there is a unique element y E H such that
(B.5)
(
F x) = (x, y) for all x
E H.
Moreover, (B .6)
II Y II
=
sup x E H, x#O
I F (x) l II X I I ·
Theorem 2 .2. (p . 31) Let N be a closed subspace of a Hilbert space H , and let x be an element of H which is not in N. Set
d = inf ll x - z ll ·
(B.7)
zEN
Then there is an element z
E N such that ll x - z ll =
d.
(Projection Theorem) (p. 32) Let N be a closed subspace of a Hilbert space H. Then for each x E H , there are a v E N and a w orthogonal to N such that x = v + w . This decomposition is unique. Theorem 2 .3.
Theorem 2 . 5 . {Hahn-Banach Theorem) (p. 33} Let V be a vector space, and let p (x) be a sublinear functional on V. Let M be a subspace of V, and let f (x) be a linear functional on M satisfying ·
(B.8)
f (x) < p(x) ,
x
E M.
B. Major Theorems
407
Then there is a linear functional F(x) on the whole of V such that (B .9)
F(x) == f (x) ,
(B . 10)
F(x) < ( x ) ,
x E M,
p
x E V.
Theorem 2 . 6 . (p. 34) Let M be a subspace of a normed vector space
X, and suppose that f(x) is a bounded linear functional on M. Set
l f (x) l
11 !11 = x E M, x #O I I X I I . sup
Then there is a bounded linear functional F(x) on the whole of X such that (B. l l )
F(x) == f(x) ,
II F II
(B . 1 2)
=
x E M,
II I II ·
Theorem 2 . 7. (p. 36) Let X be a normed vector space and let xo =I= 0 be an element of X. Then there is a bounded linear functional F(x) on X
such that (B. 1�)
II F II
F ( xo ) == llxo l l ·
= 1,
Theorem 2. 9. (p. 37) Let M be a subspace of a normed vector space X, and suppose xo is an element of X satisfying
(B . 14)
d == d(xo , M) == inf l l xo - x l l x EM
Then there is a bounded linear functional F ( xo ) == d, and F ( x ) = 0 for x E M.
F on
> 0.
X such that
II F I J
Theorem 2 . 10. (p. 38) X' is a Banach space whether or not X is. Theorem 2 . 1 1 . (p. 4 1} l� == l q , wh e re q =
pj (p - 1 ) .
Theorem 2 . 1 2 . (p. 4 3) If f E l� , then there is a
f (x) =
and (B. 1 5)
00
L: Xi Zi , 1
ll f ll == ll z l l q ·
x E lp,
z
E l q such that
1,
408
B. Major Theorems
Theorem 2 . 14. (p. 5 0) For each bounded linear functional f on C [a, b] ·
there is a unique normalized function g of bounded variation such that f(x) =
(B . l6)
and
1b x(t) dfJ(t) ,
x E C [a, b] ,
V(g) = ll f ll ·
(B. l 7)
Conversely, every such normalized g gives a bounded linear functional on C [a, b] satisfying (B. 1 6} and (B. 1 7). Theorem 3 . 10.
(Closed Graph Theorem) (p.
Banach spaces, and A is a closed linear operator from then A E Y) .
X,
B(X,
Theorem 3 . 12. {p. 67) If
X
X
X,
62} If Y are to Y, with D(A) =
X, Y are Banach spaces, and A is a closed
linear operator from to Y, then R(A) is closed in Y if, and only if, there is a constant C such that (B . l8)
d(x, N (A) ) < C II Ax l l ,
x E D (A) .
X
Theorem 3.17. (Banach-Steinhaus Theorem) (p. 71) Let be a Banach space, and let Y be a normed vector space. Let W be any subset of
B( X, Y) such that for each x X, E
sup II Ax ll <
AEW
oo .
Then there is a finite constant M such that II A ll < M for all A E W. Theorem 3. 18.
(Open Mapping Theorem) (p. 71} Let A be a
X
closed operator from a Banach space to a Banach space Y such that R(A) = Y . If Q is any open subset of D(A) , then the image A(Q) of Q is 'open in Y. Theorem 4. 12.
X K.
be K XX and and
(p. 90) Let be a Banach space and let Set A == I an operator in Then, R(A) is closed in dim N(A) = dim N (A' ) is finite. In particular, either R(A) = N(A) = {0} , or R(A) and N (A) i: {0} .
K (X). #X
-
Theorem 4. 17. (p. 96} If a set U
C
X
is relatively compact, then it is totally bounded. If X is complete and U is totally bounded, then U zs relatively compact.
B. Major Theorems
409
Theorem 6.2. (p. 129) Under the hypothesis of Theorem 6. 1, a(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 r:J. then a (K - .\) == 0, R(K - .A) = X and K - A has an inverse in B(X) .
S,
Theorem 6.17. (p. 139} If f(z) is analytic in a neighborhood of a(A) ,
then
(B. l9)
i. e. ,
J.L
a
( f ( _4) )
E a (f(A) ) if and only if J-l
==
==
f ( a (A) ) ,
f (.A) for some A E a (A)
.
Theorem 6.26. (p. 148) Let V be a complex vector space, and let p be
a real valued functional on V such that (i) p (u + v ) < p(u) + p( v) , (ii) p(au )
==
u, v E V,
a complex , u E V.
l a lp(u) ,
Suppose that there is a linear subspace M of V and a linear (complex valued) functional f on M such that �e f (u) < p(u) ,
(B .20)
u E M.
Then there is a linear functional F on the whole of f (u) ,
(B.2 1 )
F(u)
(B.22 )
I F(u) l < p (u) ,
==
V
such that
u E M, u E V.
Theorem 7. 1 . (p. 1 57) If A E
such that
(a) N (Ao )
==
Yo ,
(b) R(Ao ) = Xo n D(A) , (c) A 0 A == I on Xo n D ( A ) , (d) AAo == I on R(A) . Moreover, there are operators F1 E B(X) , F2 E B ( Y ) such that (e) Ao A == I - F1 on D(A) ,
410
B. Major Theorems
(f) AAo
==
(g) R(F1 )
I - F2 on Y, ==
N(A) ,
(h) R(F2 ) = Yo,
N(F1 )
==
Xo ,
N(F2 ) = R(A) .
Theorem 7. 2 .
(p. 1 5 7) Let A be a densely defined closed linear operator from X to Y. Suppose there are operators A 1 , A2 E B(Y, X) , K1 E K(X) , K2 E K(Y) such that (B. 23)
and (B.24)
Then A
E
Theorem 7.3.
BA E
Z) and
(p. 1 5 7) If A E
(B.25)
==
then
i (A) + i (B) .
Theorem 7.8.
(p. 1 61} If A E
Z) ,
i (A + K)
==
is
in K(X, Y), then
i (A) .
(p. 1 61) For A E
(B . 27)
i(A + T)
=
i (A) ,
and (B .28)
a( A + T) < a(A) .
Theorem 7. 10.
(p. 1 62} If A E
Z).
Theorem 7. 15.
Z),
(p. 1 64) Let A be a densely defined closed linear operator from X to Y. If R(A) is closed in Y, then R(A' ) == N(A) 0 , and hence is closed in X' .
41-1
B. Major Theorems
Theorem 7. 1 6 . (p. 1 65) If A is a densely defined closed linear operator from X to Y and R(A') is closed in X' , then R(A) == 0N(A') , and hence, is closed in Y. Theorem 7. 1 7. (p. 1 66) If [! is a closed, convex subset of a normed
vector space X and xo E X is not in U, then there is an x' E X' such that �e x' (x o ) > �e x ' (x) ,
(B . 29)
x E U,
and �e x' (x o ) =I= �e x' (x 1 ) for some x 1 E U . Theorem 7. 19. (p. 1 69) If A is in B(X, Y), then A E
Y with D(A) dense in X. Then D (A') is total in Y' .
(p. 1 70) If A E
Theorem 7.23 . (p. 1 70) Let A be a closed linear operator from X to Y with D(A) dense in X. If A' E
i (A) == -i (A') .
Thegrem 7.27. (p. 1 72) )... tt ae (A) if and only if )...
0.
E
Theorem 7. 29. (p. 1 73} Let A be a closed linear operator from X to
Y with domain D(A) dense in X. Then A E + (X, Y) if and only if there
is a seminorm I I defined on D (A) , which is compact relative to the graph norm of A, such that ·
(B . 30) Theorem 7.35. (p. 1 78) Let X, Y,
Z be Banach spaces, and assume
that A is a densely defined, closed linear operator from X to Y such that R(A) is closed in Y and {J(A) < oo (i. e. , A E
Z.
(B . 3 1 )
(BA) ' == A' B ' .
Theorem 8. 1 . (p. 1 84) X is reflexive if and only if X' is.
412
B. Major Theorems
Theorem 8.2. (p. 1 84) Every closed subspace of a reflexive Banach
space is reflexive.
Theorem 8.3. (p. 1 85) A subspace W of X' is saturated if and only if
W == M0 for some subset M of X.
Theorem 8 .5. (p. 186) A subspace W of X' is saturated if and only if
. zt is weak* closed.
Theorem 8. 7. (p. 186) A Banach space X is reflexive if and only if
every closed subspace of X' is saturated.
Theorem 8.1 1 . (p. 188} If X' is separable, so is X. Theorem 8. 1 3 . (p. 1 89) If X is separable, then every bounded sequence
in X' -has a weak* con vergent subsequence.
Theorem 8 . 1 6 . (p. 1 91) If X is reflexive, then every bounded sequence
has a weakly convergent subsequence.
Theorem 8 . 1 8 . (p. 1 95) If X is a Banach space such that every total
subspace of X' is dense in X', then X is reflexive.
Theorem 9.5. {Spectral Mapping Theorem) (p. 203) If p (t) is a
polynomial, then (B . 32)
a
[p( ) ] a
==
p[a(a)] .
More generally, if f(z) is analytic in a neighborhood n of a ( a) , define (B .33 )
f (a)
=
� f ( z ) ( z e - a) - 1 dz , j 2 7r� Jaw
where w is an open set containing a ( a) such that w c n , and ow consists of a finite number of simple closed curves which do not inter-sect. Then (B.34)
a
[f (a)]
==
f [a (a)] .
Theorem 9.10. (p. 207) A complex number A is in a( a) if and only if there is a multiplicative linear functional m on B such that m ( a ) == A. Theorem 9. 1 1 . (p. 207) Every ideal H which is not the whole of B is
contained in a maximal ideal.
Theorem 9 . 16. (p. 21 0} An ideal N is maximal if and -only if the only ideal L satisfying B L ::) N is L == N.
#
41 3
B. Major Theorems
Theorem 9.26. (p. 214) If A E (X) , E E R(X ) an d A � E, then
A + E E (X) .
Theorem 9.29. {p. 214) The operator E in B(X) is in R (X) if and only if A + E E (X ) for a ll A E (X) such that A � E. Theorem 9.30. E1 + E2 E R(X) .
(p. 214) If E1 , E2 E R(X) and E1 � E2 , then
Theorem 9.39. (p. 21 6} F(X) is a closed two-sided ideal. oo
Theorem 9.43 . (p. 21 7} A in B(X) is in + if and only if a: ( A - K) < for all K E K( X ) .
Theorem 9 .44. (p. 21 8) A E if and only if o:(A - K) < jj(A - K) < oo for a ll K E K(X) . E E
oo
and
oo
Theorem 9.46. for all A E + .
(p. 218)
F+ (X) if and onlv if o:(A - E) <
oo
Theorem 9.47. for a ll A E .
(p. 21 8) E is in F(X) if and only if o:(A - E) <
Theorem 9.52. (p. 21 9) F+ (X) is a closed two-sided ideal. Theorem 9.54. (p. 220} A in B(X) is in _ if and only if ,B(A - K) <
oo
for all K in K(X).
oo
Theorem 9.56. for all A E _ .
oo
Theorem 9.57. (p. 221} E is in F(X) if and only if jj(A - E) < for a ll A E .
(p. 221} E E F_ (X) if and only if jj(A - E) <
Theorem 9.62. (p. 222) F_ (X) is a closed two-sided ideal. Theorem 10. 1 . (p. 230) 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 ( B . 35)
Then there is a family {Et } of operators in B(X) , t > 0, with the following propertzes:
414
B. Major Theorems
( a) (b)
Es Et = Es + t , s > 0 , t > 0, Eo = I, (C) II Et I I < e - a t , t > 0, (d) Etx is continuous in t > 0 for each x E X, Etx is differentiable in t > 0 for each x E D(A) , and (e) d x ( B . 36 ) = AEtx,
�;
Et ( A - A) - 1 == (A - A) - 1 Et ,
(f)
A > b, t > 0.
Theorem 10.3. (p. 23 5) Every strongly continuous, one-parameter
semigroup {Et } of operators in B (X) has an infinitesimal generator.
Theorem 1 1 .3. (p. 248) An operator is normal and compact if and
only if it is of the form (B .37 )
with { 'Pk } an orthonormal set and Ak
�
0 as k
� oo .
Lemma 1 1 .9. (p. 251) Every Jeparable Hilbert space has a complete
orthonormal sequence.
Theorem 1 1 . 1 9. (p. 261) Let A be _ a seminormal operator such that
a( A ) has no non-zero limit points. Then A is compact and normal.
Theorem 1 2 . 2 . (p. 266) If A is a closed, densely defined operator on H and A tt W(A) , then o:(A - A) == 0 and R(A - A) is closed in H . Theorem 1 2 . 3 . (p. 267) Let a(u, v ) be a densely defined bilinear form
with associated operator A . Then
{a) If A tt W (a) , then A - A is one-to-one and ( B.38)
l l u l l < C II (A -· A)u l l ,
u E D (A) .
{b) If A tt W (a) - and A is closed, then R(A - A) is closed in H . Theorem 1 2 .8. (p. 270) Let a(u, v ) be a densely defined closed bilinear
form with assoczated operator A . If W (a) is not the whole plane, a half plane, a strzp, or a line, then A is closed and ( B . 39 )
a (A)
c
W (a) == W ( A) .
415
B. Major Theorems
1.'heorem 12 .9. (p. 270} The numerical range of a bilinear form is a convex set in the plane. Theorem 1 2 . 14. (p. 274) Let A be a densely defined linear operator on H such that W(A) is not the whole plane, a half-plane, a strip, or a line. Then A has a closed extension A such that ,..
( B . 40 )
a
( A)
c
W (A)
=
W ( A) .
Theorem 1 2 . 1 7. (p. 278) If A is a densely defined linear operator on H such that W(A) is not the whole complex plane, then A has a closed extension. Theorem 1 2 . 18. (p. 278) A linear operator has a closed extension if and only if it is closable. Theorem 1 2 . 24. (p. 286) Let B be a dissipative operator on H with D(B) dense in H . Then B has a closed dissipative extension B such that a ( B ) is contained in the half-plane �e A < 0. ,..
Theorem 12.25. (p. 286) Let A be a densely defined linear operator on H. such that W (A) is a half-plane. Then A has a closed extension A satisfying ,..
a( A )
(B .41)
c
W (A) = W( A ) .
Theorem 1 2 . 28. (p. 292} Let B be a densely defined linear operator on H such that W(B) is the line �e A = 0. Then a necessary and sufficient condition that B have a closed extension B such that ,..
a( B )
(B.42)
c
W ( B ) = W (B)
,
is that there exist an isometry from R(I - B).l onto R(I + B) .l . In particular, this is true if they both have the same finite dimension or if they are both separable and infinite dimensional. Theorem 13.5. (p. ::101} If A is a positive operator in B (H) , then there is a unique B > 0 such that B 2 = A. Moreover, B commutes with any C E B (H) which commutes with A. -
Theorem 1 3 .4. (p. 307) Let A be a selfadjoint operator in B (H) . Set m
=
inf (Au, u) ,
ll u l l = l
M
==
sup (Au, 7.L ) .
l l ull = l
B. Major Theorems
416
Then there is a family { E(A) } of orthogonal projection operators on H de pending on a real parameter A and such that:
as
(2)
E(A)u � E(A o )u
(3)
E(A) = 0 for A <
(4)
A E(A)
(5)
if a <
== m,
A o < A � Ao ,
m,
u E H;
E(A) = I for A > M ;
E(A)A;
b > M and p(t) is any polynomial, then p( A )
(B.43)
=
1b p(A)dE(A) .
This means the following. Let a = Ao < A1 < < A n == b be any partition of [a, b] , and let A� be any number satisfying Ak - 1 < A� < Ak · Then n (B.44) L P (A � ) [ E(Ak) - E(Ak -1 ) ] � p(A) 1 ·
in B ( H ) as
17 =
·
·
max(Ak - Ak- 1 ) � 0.
Theorem 13. 14. (p. 321) Let A be a selfadjoint operator on H. Then
there is a family { E(A) } of orthogonal projection operators on H satisfying {1) and (2) of Theorem 13.4 and E(A) �
(3')
{
E(A ) A
( 4')
p( A )
(5')
for any polynomial p(t) .
=
as A � - oo as A � + oo
0 I c
A E(A)
1: p(A)dE(A)
Theorem 14. 16. (p. 333} f(A + B) < � (A) + f(B) . Theorem 14. 17. (p. 333) �(A + B) < �(A) + � (B) . Corollary 14.23. (p. 335) � (AT) < � ( A ) A (T) . Theorem 14.24. (p. 335) T (A) < � (A) .
417
B. Major Theorems
Theorem 14.26. (p. 336)
(B .45)
r(A) I
=
II ATII inf inf z TEB(Z,X) r (T) _
Theorem 14.28. (p. 336) � ( A ) < II A I I m · Theorem 14.29. (p. 336} A E
and i (A + B)
==
i (A) .
Theorem 14.32. {p. 337} If 1(B) < v(A) , then A + B E
(B.46)
I I A I I q /2 < I I A II m < 2 I I A I I q·
Theorem 14.37. (p. 34 1) If Y has the compact approximation property with constant C, then
(B.47)
II A II K < C II A I I q ,
A E B(X, Y) .
Theorem 14.39. (p. 342) Compact operators are strictly singular. Theorem 14.4 1 . (p. 342) For A in B(X, Y), if A fj_
S
Corollary 14.42 . (p. 343) If is strictly singular, then for each c > 0 there is an infinite dimensional subspace M such Jhat the restriction of S to M is compact with norm < c. Corollary 14.43. (p. 34 3) Strictly singular operators are in F± (X, Y) . Corollary 14.45 . (p. 344) T in B (X, Y) is strictly singular, if and only if �(T) == 0. Theorem 14.48. (p. 344) A is strictly singular if and only if 1(A) Theorem 14.5 1 . (p. 34 6) If o: (A) <
(B .48)
oo ,
then
!'(A) < J-Lo (A) < J.L(A) .
==
0.
4 18
B.
Major Theorems
Corollary 14.52 . (p. 34 6) If 1 1 T II < !'(A) , then A - T E
TVSLB"O
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Index
(J, g) , 10 II lim , 32 5 II ll q , 339 o ( A) , 101 ,B(A ) , 101 � ( A) , 332 � M ( A ) , 333 dim V, 80 r(A) , 332 rM ( A ) , 332 v ( A) , 332
·
·
·
·
•
·
B ( X, Y) , 56 B [a, b] , 1 6 C [a , b] , 6 F(X) , 2 1 5 F- (X) , 221 F+ (X) , 2 18 Ge , 330 Gr , 330 i ( A) , 101 j (A) , 350 Jx, 1 1 3 £2 ,
K (X, Y ) , 88 21 l oo , 1 2 l2 , 14 lp , 39 N (A) , 58
N BV[a , b] , 193 p(A) , 132 Pu (A) , 131 339 , 398 R(A) , 58 R(X) , 2 1 3 ra ( A) , 132 8° , 59 X \ M, 84 X' , 38 X/M, 68
q(O) ,
·
abelian Banach algebras, 206 adjoint operator, 57, 1 55 annihilators, 59 approximation by operators of finite rank, 252 axiom of choice, 207, 2 1 1
42&
424
Baire's category theorem, 61 Banach algebra, 20 1 Banach algebras, 201 Banach space, 7 Banach-Steinhaus theorem , 71 basis, 80, 8 1 Bessel's identity, 23 Bessel's inequality, 23 bounded functional, 29 bounded inverse theorem, 61 bounded linear functional , 29 bounded operator, 55 bounded set, 81 bounded variation, 46 Cauchy sequence, 7, 9 Cauchy-Schwarz inequality, 1 1 closed graph theorem, 62 closed operator, 62 closed range theorem , 70 closed set , 31 codimension , 325 commutative Banach algebras , 206 compact approximation property, 34 1 compact operators , 88 compact set , 8 1 compact support , 364 complement , 124 complementary subspaces, 124 complemented subspace, 1 24 complete orthonormal sequence, 23 completely continuous operators , 88 completeness, 7 complex Banach space, 133 complex Hahn- Banach theorem , 148 complexification, 147 conj ugate operator, 57 conj ugate space, 38 continuously embedded, 1 59 convex set , 166 coset , 67 differential equation, 225 differential equations, 1 dimension, 79 direct sum , 102 domain , 55 dual space, 29 , 38 eigenelement , 1 3 1 eigenvalue, 131 eigenvector, 131 equivalence relation, 67 equivalent norms, 80 essential spectrum , 171 Euclidean n-dimensional real space, 9 extension , 274, 3 1 4
Index
factor space, 68 factored perturbation function, 354 finite rank operator, 85 first category, 6 1 Fourier series, 1 7 Fredholm alternative, 90 Fredholm operators, 101 Fredholm perturbation , 2 1 5 functional, 29 geometric Hahn-Banach Theorem, 166 graph, 64 Hahn-Banach theorem, 33 Hilbert space adjoint , 247 Hilbert space, 1 1 Hilbert space , 243 Holder's inequality, 41 hyponormal operator, 257 ideal elements, 19 ideal , 207, 330 image set , 96 index, 101 infinite dimensional, 80 infinitesimal generator, 230, 235 injection modulus , 350 inner product , 1 1 integral operator, 253 invariant subspace, 298 inverse element , 202 inverse operator, 60 joint resolvent set , 208 j oint spectrum , 208 left ideal , 330 left inverse, 330 left regular elements, 330 linear operator , 55 linear space, 7 linearly independent vectors , 79 maximal element , 2 1 1 maximal ideal, 207 measure of an operator, 325 measure of noncompactness, 339 Minkowski functional , 167 Minkowski's inequality, 40 multiplicative functional , 206 norm , 7 norm of a functional, 32 norm of an operator, 56 normal operator, 246 , 248 normalized function of bounded variation , 49
425
Index
normed vector space, 7 nowhere dense, 6 1 null space, 58 open mapping theorem , 71 operational calculus, 134 operator , 3, 55 orthogonal projection, 297 orthogonal , 30 orthonormal sequence, 23 parallelogram law, 17 partially ordered sets, 2 1 1 partition, 45 perturbation classes, 329 pert urbation funct ion , 350 perturbation theory, 1 09 point spectrum, 131 positive operator , 300 projection theorem , 32 projection, 1 4 1 quotient space, 68 radical , 330 range, 58 real Banach space, 1 34 reduce, 298 reflexive Banach space, 1 70, 1 83 regular element, 202 relatively compact set , 400 relatively compact set , 95 resolution of the identity, 31 1 resolvent set, 13 1 , 171 resolvent , 202 Riemann-Stieltjes integral , 46 Riesz operator, 2 1 3 Riesz representation theorem , 29 Riesz theory of compact operators, 77 Riesz's lemma, 83 right ideal , 330 right inverse , 330 right regular elements, 330 saturated subspaces , 1 85 scalar product , 1 1 second category, 6 1 selfadjoint operator, 253 semi-Fred holm operators, 1 1 7 semi-Fredholm perturbation, 2 1 6 semigroup, 230 , 235 seminorm, 1 1 8, 1 50 seminormal, 257 separable spaces , 1 88 spectral mapping theorem , 1 33 spectral projectionb , 1 4 1 , 143 spect ral set , 1 4 1
spectral theorem, 3 1 1 spectral theory, 1 29 spectrum , 1 3 1 , 1 71 , 202 square integrable functions, 2 1 strictly convex , 34 7 strictly singular operator, 342 , 401 strong convergence , 1 90 strongly continuous, 235 sublinear functional , 133 subspace , 31 total subset , 1 69 total variation , 46 totally bounded , 95 totally ordered set , 2 1 1 transformation, 3 , 55 triangle inequality, 9 trivial Banach algebra, 204 two-sided ideal , 330 unbounded Fredholm operators , 1 55 unbounded operators, 1 55 unbounded semi-Fredholm operators , 1 73 uniform boundedness principle, 71 unit element , 20 1 upper bound , 2 1 1 variation of parameters, 1 vector space, 7 Volterra equation, 6 weak convergence , 1 90 weak* , 186 weakly compact , 198 Zorn's len1ma, 207, 2 1 1
