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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
1 such that N(A' ) == N(A' m ) for k > m . If both r (A) < oo and r' (A) < oo , let j == 1nax [m , n ] . Then o:(A k ) == o:(AJ ) , (3 (A k ) == (3 (Ai ) for k > j . Hence, i (A k ) == n(A k )  ,B(A k ) == o:(AJ )  j](AJ ) == i (AJ ) for k > j . But i (A k ) == k i (A) . Hence, (k  j ) i (A) == 0 for all k > j , showing that we must have i (A) == 0. But if this is the case, we must also have m == n . To recapitulate, if A E (X ) , r (A) < oo and r' (A) < oo , then i (A) == 0, and there is an n > 1 such that ( 5 . 29) Moreover, we claim that
(5.30) This follows from the fact that if x is in this intersection, we have, on one hand , that An x == 0 and , on the other, that there is an x 1 E X such that x == A n x 1 . Thus , A 2 n x 1 == 0 , or X I E N(A 2 n ) == N (A n ) . Hence, , x 8 be a basis for N (A n ) . Then there are x == An x 1 == 0. Next , let x 1 , bounded linear functionals x � , · · · , x � , which annihilate R(An ) and satisfy ·
(5 .31 )
·
·
x� ( xk ) == §Jk ,
1 < j, k
.. ) # 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,
n1
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 . z1
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. ,
JL
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. , JL 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,
( JL  A)  1 x + (JL  A) ( JL  A)  1 u.
SuDbvttuting for u, we get ( 6.27 ) . Note that it follows from ( 6.27 ) that (A  A)  1 and (JL  A)  1 commute. Substituting for (A  A)  1 in the right hand side of ( 6. 27) , we get
(A  A)  1
==
(JL  A)  1 + ( JL  A ) ( JL  A)  2 + ( JL  A ) 2 ( A  A)  1 (JL  A)  2 .
Continuing in this way, we get k
(A  A)  1 == l: (JL  A ) nl (JL  A)  n + (JL  A )n (A  A )  1 (JL  A)  n . 1
Since
II ( JL  A )n ( A  A)  1 ( JL  A)  n i l < I JL  A I n · II ( A  A)  1 1 1 · II (JL  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 finitedimensional. 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 finitedimensional. 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 ndimensional 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 onetoone 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 finitedzmensional, 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 finitedimensional, 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,
sup I I = IIAll · x# o l l (x , O ) I I
Hence,
II A II = I I A II ·
If ,.\ is real, then "'
(A  ,.\) (x , y ) = ( (A  ,.\)x, (A  ,.\)y ) . This shows that ,.\ E p ( A ) if and only if ,.\ E p( A) . Similarly, if p ( t) is a
polynomial with real coefficients , then
p( A ) (x , y ) = (p(A)x, p (A)y ) , showing that p ( A ) has an inverse in B(Z) if and only if p ( A ) has an inverse in B (X ) . Hence, we have Theorem 6 . 25 . Equation {6. 1 2} has a unique solution for each y in X if arLd only if p (,.\) # 0 for all ,.\ E a ( A ) .
In the example given at the end of Section 6. 1 , the operator A has eigenvalues i and  i . Hence,  1 is in the spectrum of A 2 and also in that of A 2 . Thus , the equation
(A 2 + 1 )x = y cannot be solved uniquely for all y.
6 . 6 . Tne complex HahnBanach theorem
In case you have not already figured it out, the only theorem that we proved in the preceding chapters that needs modification for complex vector spaces is the HahnBanach theorem. We now give a form that is true for a complex Banach space. Theorem 6. 26. 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) == l a l p(u) ,
u, v E V,
a complex, u E V.
Suppose that there is a linear subspace M of V and a linear (complex valued) functional f on M such that (6.42)
SR e f ( u) < p( u) ,
uE
J\;f .
6. 6. The complex HahnBanach theorem
1 49 w
Then there is a linear functional F on the
h ole of V such that
(6.43)
F(u) == f (u ) ,
u E M,
(6.44)
I F (u) l < p(u) ,
u E V.
Proof. Let us try to reduce the "complex' ' case to the "real" case. To be sure, we can consider V as a real vector space by allowing multiplication by real scalars only. If we do this , M becomes a subspace of a real vector space V. Next, we can define the real valued functional
u E M.
!1 (u) == SRe f (u) , Then by (6.42) ,
!1 (u) < p(u) , u E M. We can now apply the "real" HahnBanach theorem ( Theorem 2 . 5) to con clude that there is a real functional F1 ( u) on V such that F1 (u) == f1 (u) ,
u E M,
F1 ( u) < p(u) , u E V. Now, this is all very well and good, but where does it get us? We wanted to extend the whole of J , not j ust its real part . The trick that now saves us is that there is an intimate connection between the real and imaginary part of a linear functional on a complex vector space. In fact ,
! (iu) == SRe f (iu) == SRe i f ( ) == �m f (u) . 1
u
Hence, f (u) == /1 (u)  i/1 (iu) . This suggests a candidate for F(u) . Set F(u) == F1 (u)  iF1 (iu) ,
u E V.
F( u) is clearly linear if real scalars are used . To see that it is linear iii the "complex" sense, we note that F ( iu) == F1 (iu)  iF1 (  u) == i [F1 (u)  i F1 (iu)] == iF(u) . Second, we note that F(u) == f (u) for u E must show that (6.44) holds . Observe that (6.45)
p (u) > 0,
M.
To complete the proof, we
u E V.
In fact , by (ii) we see that p(O ) == 0, while by (i) , we see that p(O) < p(u) + p( u) == 2p(u) . Hence, (6.44) holds whenever F(u) == 0. If F(u) # 0, we write it in polar form F (u) == I F(u) l e i 0 . Then 0 I F (u) l == e  iO F (u) == F( e  i 0 u) == F1 (e  i 0 u) < p(e  i u) == p(u) . Th1s completes the proof.
0
6. SPECTRAL THEORY
150
A functional satisfying (i) and (ii) of Theor�m 6 . 1 is called a seminorm. As a corollary to Theorem 6.26 we have Corollary 6.2 7. Let M be a subspace of a complex normed vector space X. If f is a bounded linear functional on M, then there is a bounded linear functional F on X such that
F(x)
=
f (x) ,
I I F II =
x
1 !1 ·
E M,
This corollary follows from Theorem 6.26 as in the real case. In all of our future work, if we do not specify :r;eal or complex vector spaces, normed vector spaces , etc. , it will mean that our statements hold for both. 6 . 7 . A geometric lemma
In Section 6.3, we made use of the following fact: Lemma 6 . 28. Let n be an open set in IR 2 , and let K be a bounded closed set in n . Then there exists a b.ounded open set w such th at {1) w
:J
K,
{2) w
c
n,
{3) ow consists of a finite number of simple polygonal closed curves which
do not intersect. Proof. By considering the intersection of n with a sufficiently large disk, we may assume that n is bounded. Let 6 be the distance from K to an . Since both of these sets are compact and do not intersect , we must have 6 > 0. Cover IR2 with a honeycomb of regular closed hexagons each having edges of length less than 6/4. Thus , the diameter of each hexagon is less than 6 /2 Let R be the collection of those hexagons contained in n , and let W be the union of all hexagons in R. Let w be the interior of W. Then w :J K. If X E K, then its distance to an is > 6 . Thus, the hexagon containing X and all adjacent hexagons are in n. This implies that x E w. Next we note that w c n. For if X E w , then X is in some hexagon contained in n . Thus, it remains only to show that ow satisfies (3) . Clearly, ow consists of a finite number of sides of hexagons. Thus , we can prove (3) by showing .
6.8. Prob.lems
151
(a) that aw never intersects itself and (b) that no point of aw is an end point of ow. Now, every point x E 8w is either on an edge of some hexagon or at a vertex. If it is on an edge, then the whole edge is in ow, in which case, all points on one side of the edge are in w and all points on the other side are not in w. If x is a vertex, then there are three hexagons meeting at x . Either one or two of these hexagons are in n. In either case, exactly two of the three edges meeting at x belong to ow. Thus (a) and (b) are true, and the proof is �omplete. 0 6 . 8 . Problems
( 1 ) Show that if X is infinite dimensional and K E K(X) , then 0 E
a( K) .
(2) Let A and B be operators in B (H) which commute ( i.e. , AB BA) . Show that
ru (AB) < ru (A)ru (B) ,
ru (A + B) < ru (A) + ru (B) .
(3) Suppose A E B(X) , and f (z) is an analytic function on set n containing l z l < ru (A) . If
l f(z) l < M,
==
an
open
l z l < ru (A) ,
show that
ru [f (A)] < M. ( 4) Let A be an operator in X and p( t) a polynomial. Show that if A is not empty, then p(A ) is a closed operator. (5) Suppose that ,.\0 is an isolated point of a(A) , A E B(X ) , and f(z) is analytic in a neighborhood of Ao . If f(A) == 0, show that f(z) chas a zero at .Ao . (6) Let ,.\0 be an isolated point of a(A) , A E B(X) , with Ao E A . Let x ' be any functional in X'. Show that f(z) x' [(z  A)  1 ] has a pole at z == Ao . =
(7) Show that a projection P E B(X) is compact if and only if it is of finite rank.
6. SPECTRAL THEORY
152 (8) If A E B(X ) , show that
z(z  A)  1 � I as l z l
� oo .
(9) Let A be an operator in B(X ) , and suppose a (A) is contained in the halfplane � e z > 6 > 0. Let r be a simple closed curve in �e z > 6 containing a ( A) in its interior. Consider the operator
T=
J z 1 1 2 (z  A)  1 dz . � 2 7r'l !r
Show that T is well defined and that T2 == A . What can you say about a (T) ?
(10) Suppose A, B E B(X) with 0 E p(A ) and II A  B ll < 1 / I I A 1 11 . Show that 0 E p(B) and A1 11 II 1 I I B 1 1 < 1 I A 1  I  II · II A  B l l " ( 1 1) Show that every orthonormal sequence in a separable Hilbert space can be made part of a complete orthonormal sequence.
( 1 2) Let M be a closed subspace of a Banach space X . Show that there is a projection P E B(X) such that R(P) == M if and only if there is a closed subspace N E X such that X == M EB N . ( 13) Let P, Q be bounded projections on a Banach space X such that l i P  Q l l < 1 . Show that there is an operator A E B(R(P) , R( Q ) ) such that A 1 E B (R(Q) , R(P) ) . ( 14) If p (x) is a seminorm, show that l p (x 1 )  p ( x2 ) l < p ( x 1  x2 ) . ( 1 5) If p( x) is a seminorm on a vector space V, let Q c == { x E V : p( x ) < c} , Show that (a)
0 E Qc
c
>
0.
1 53
6.8. Problems and (b)
l n l < 1 implies a x E Qc for x E Q c .
( 16) Under the same hypothesis, show that for each x E V, there is an n > 0 such that a x E Qc . ( 1 7) Under the same hypothesis, show that
p (x)
=
inf{o: > 0, o:  1 x E Q 1 } .
Chapter
7
UNB O UND ED _/
O P ERAT O RS
7 1 .
.
Unbounded Fredholm operators
In many applications, one runs into unbounded operators instead of bounded ones. This is particularly true in the case of differential equations. For instance, if we consider the operator djdt on C [O, 1) , it is a closed operator with domain consisting of continuously differentiable functions. It is clearly unbounded. In fact , the sequence X n (t) == t n satisfies l l x n ll == 1 , ll dx n /dt ll == n � oo as n � oo . It would , therefore, be useful if some of the results that we have proved for bounded operators would also hold for unbounded ones. We shall see in this chapter that , indeed, many of them do. Unless otherwise specified, X, Y, Z, and W will denote Banach spaces in this chapter. Let us begin by trying to enlarge the set
0 such that rd(y', N(A')) < I A ' y' l , y ' E D (A). (7.19) Let S be the set of those x E D(A) such that l x l 1. We shall show that r, then y E A(S) . (i) If y E 0N(A') and I Y I r, the n y E A(S) for ( ii) If y E A(S) for all y E 0N(A') such that I Y I all such y . =
�e x' (x) , x E U, and �e x' (xo) i= �e x' (x 1 ) for s om e x 1 E U.
Before proving Theorem 7. 17, let us show how it implies (i) . We first
note that A(S) and, hence, A(S) are convex. Thus if y tJ. A(S) , then by Theorem 7. 17, there is a nonvanishing functional y ' E Y' such that
�e y' ( y ) > �e y ' (Ax) , x E S. If x E S, set y ' (Ax) = I Y ' (Ax) l e i0 • Then xe  i9 E S, and hence, SRe y' ( y ) > SR e y' ( e iB Ax) = I Y ' (Ax) l , x E S.

Thus, y ' ( y ) i= 0 and
I y' (Ax) I < I y' { y) I · II x I I ,
x E D (A) .
I A' y' ( X ) I < I y' ( y ) I . II X II '
X E D (A) '
This shows that y ' E D (A') and or
(7.23)
II A' y ' II
r. This is equivalent to (i) . (7.24)
It remains to prove Theorem 7. 17. To do this , we introduce a few con cepts. Let be a convex subset of a normed vector space X. Assume that 0 is an interior point of U, i.e. , there is an c > 0 such that all x satisfying ll x ll < E are in For each x E X, set
U
U.
p( x)
=
inf
a>O, ax EU
a 1 .
U.
U
Since ax E for a sufficiently This is called the Minkowski functional of small, p(x) is always finite. Other properties are given by Lemma 7. 18. p (x) has the following properties:
(a) p(x + y) < p(x) {b) p(ax)
=
ap(x) ,
+
p(y) ,
x, y E X;
x E X, a > 0;
(c) p( x ) < 1 implies that x is in U; {d) p(x) < 1 for all x in U. Proof. (a) Suppose o:x E . IS convex,
U and {3y E U , where o: > 0 and f3 > 0. Since U
o:  1 o:x + 13  1 f3y o:  1 + {3  1
is in U. Hence,
p( x + y) < a  1 + 13 1 . Since this is true for all o: > 0 such that o:x E and (3 > 0 such that (a) follows. (3y E To prove (b) we first note that p(O) = 0. If a > 0, then p(o:x) = inf /3  1 = o: inf r  1 = ap(x) .
U
U,
/3>0, af3xEU
r>O, rxEU
Concerning (c) , assume that p ( x ) < 1 . Then there is an a > 1 such that Moreover, if ax E U. Since is convex and 0 E U, we see that x E 0 1 x E U, and hence, p(x) < 1 . This proves (d) . xE
U,
U
U.
168
7.
UNBOUNDED OPERATORS
Now we can give the proof of Theorem 7. 1 7. Proof. Since U is closed and xo t/:. U , there is an 'f/ > 0 such that l l xxo l l < rJ implies that x tt. U . Let uo be a point of U , and let V be the set of all sums of the form v == u + y  uo ,
I YI I YI
where u E U and < ry . Clearly, V is a convex set . Moreover, it contains all y such that < ''7 · Hence, 0 is an interior point, showing that the Minkowski functional p( x ) for V is defined. Assume for the moment that X is a real vector space, and set w o == xo  uo . For all vectors of the form o:wo , define the linear functional
f ( o: wo) For o: > 0, we have .f (o:wo) have
==
f(o:wo)
==
o:p ( wo ) .
p(o: wo ) by ( b ) of Lemma 7. 18. For o: < 0, we
==
o:p( wo) < 0 < p(o:wo) .
Hence,
f ( o:wo) < p( o:wo) , o: real. Properties (a ) and ( b ) of Lemma 7. 18 show that p( x ) is a sublinear func tional. We can now apply the HahnBanach theorem ( Theorem 2.5) to conclude that there is a linear functional F ( x ) on X such that F( o: wo)
==
o:p( wo ) ,
o: real,
and
F ( x ) < p(x) ,
x E X. The functional F(x) is bounded. For if x is any element of X , then y == < 1] , and hence, is in V. Thus, p(y) < 1 by Lem1na ryx/ 2 ll x ll sat isfies 7. 18. Henc e, F(y) < 1 , or F( x ) < 2 ry  1 l l x ll · Now, wo tt. V. Hence,
I YI
(7.25)
F( xo)  F (uo)
==
p(wo)
>
1.
On the other hand , if u E U, then u  uo E V , and , hence,
F(u)  F(uo) < p(u  uo ) < 1 . Thus ,
F( xo )
>
F(u) ,
==
F( x )  iF(i x )
u E U.
This proves (7. 22) for the case when X is a real vector space. The complex case is easily resolved. In fact , we first treat X as a real space and find F(x) as above. We then set
G( x )
Total subsets
7. 4.
169
and verify, as in the proof of Theorem 6.26, that G(x) is a complex, bounded linear functional on X. Since, �e G (x) = F(x) , (7.22 ) is proved. The last statement follows from (7.25 ) . 0 As a consequence of Theorem 7. 16 � we have Theorem 7. 19. If A is in B{X, Y), then A E (X, Y) if and only if A' E
(Y', X') .
Proof. That A E ( X, Y) implies that A' E (Y', X') is Theorem 5 . 15. If A' E (Y' , X') , then ,B(A) = o:(A') < oo and o: (A) < o:(A'') = ,B(A') < oo by ( 5.24 ) . We now use Theorem 7. 16 to conclude that R(A) is closed, and 0 the proof is complete.
What can be said if A is not in B(X, Y) ? We shall discuss this in the next section. 7.4.
Total subsets
Suppose A E (X, Y) . What can be said about A'? Of course, o:(A') = ,B(A) < oo . If A'' exists , then the proof that ,B(A') = a(A) is the same in the unbounded case as in the bounded case ( see Section 5.4 ) . As we just noted in the last section, adjoint operators are always closed. Moreover, R(A') is closed by Theorem 7. 15. Thus, the only thing needed for A' to be in (Y', X') is the density of D (A') in Y' . For this to be true, the only element ...of y E Y which can annihilate D (A') is the zero element of Y. A subset of Y' having this property is called total. We have the following. Theorem 7.20 . Let A be a closed linear operator from X to Y with D(A)
dense in X. Then D (A') is total in Y' .
We shall give the simple proof of Theorem 7.20 at the end of the section. We should note that there may be total subsets of Y' that are not dense in Y' . The reason for this is as follows : We know that a subset W C Y' is dense in Y ' if and only if the only element y" E Y" which annihilates W is the zero element of Y " . Now you may recall ( see Section 5.4 ) that we have shown that there is a mapping J E B(Y, Y") such that "
(7.26 )
y E Y, y ' E Y' . According to the terminology of Section 7. 1 , this gives a continuous embed ding of Y into Y". Now from (7.26 ) we see that W C V' is total if and only if the only element of R( J) which annihilates W is 0. If R( J) is not the whole of Y ", it is conceivable that there is a y" =I= 0 which is not in R( J� which annihilates W . We shall show that this, indeed , can happen. Jy(y ')
==
y ' (y) ,
7. UNBOUNDED OPERATORS
1 70
R( J)
Of course, this situation cannot occur if that Y is reflexive. Thus, we have
=
ln this
Y".
case ,
we say
Lemma 7.2 1 . If Y is a reflexive Banach space, then a subset W of
Y'
total if and only if it is dense in
Y' is
Combining Theorem 7. 20 and Lemma 7.2 1 , we have Theorem 7.22. If A and i (A') =  i ( A )
(X , Y) and Y is reflexive, t(l,en
E
A1
E
(Y' , X' )
.
The converse is much easier. In fact , we have Theorem 7.23. Let A be a closed linear operator from X to Y wi th D(A) dense in X. If A' E X') , then A E ( X , Y) w.iJ,h i A ) = i(A') .
(Y' ,
Proof. Clearly, ,B(A)
=
(
oo ,
a (A' )
0 such that JL E
0 such that A + B E
+ (X, Y) and (7. 36)
o:(A + B) < o:(A)
for each linear operator B from X to Y with D(B) ::) D (A) satisfying I I B x l l < TJ ( I I x l l + I I Ax ll ) ,
(7.37)
x E D (A) .
Proof. By (7.35) ,
ll x ll + I I A x l l < (C + 1 ) II (A + B) x ll + I I Px ll + (C + 1 ) 1 1 B x ll < (C + l ) I I (A + B) x ll + I I P x l l + (C + 1 ) ry ( l l x l l + I I A x ll ) for x E D (A) . Take (7.38)
'rJ
==
1/2 (C + 1 ) . Then
ll x ll + I I A x l l < 2 (C + 1 ) I I (A + B) x l l + 2 I I Px ll ,
x E D (A) ,
which shows that A + B E + (X, Y) by Theorem 7.29. Moreover, by (7.38) , (7.39)
l l x l l < 2 (C + 1 ) I I (A + B) x l l ,
which shows that N(A + B) n X0 5. 12.
==
X E Xo n D (A) ,
{0} . Thus, (7. 36) follows from Lemma 0
Theorem 7.32 . If A E + (X, Y) , B E + (Y, Z) and D{BA) is dense in X, then BA E + (X, Z) .
176
7. UNBOUNDED OPERATORS
BA
Proof. That is a closed operator follows as in the proof of Theorem 7.3 (in fact, there we used only the facts that is closed, and < oo , that and are closed) . To prove the rest , we note that
(B) R(B) o: A B l x l < C1 I Ax l + l x l 1 , x E D (A), and I Y I < C2 I By J I + I Y I 2 , y E D (B), where I · h and l · l 2 are seminorms that are defined on D(A) and D(B), re spectively, and compact relative to the graph norms of A and B, respectively (Theorem 7.29) . Thus , (7.40) I Ax l < C2 I BAx l + I Ax l 2 , x D(BA), and hence, (7.41) l x l < C1 C2 I BAx l + C1 I Ax l 2 + l x h , x E D(BA). We note that ( 7.42) I Ax l < C3 ( l x l + I BAx l ), x E D (BA). Assuming this for the moment , we set l x l 3 C1 I Ax l 2 + l x ll· . Then I · 1 3 is a seminorm defined on D (BA) and is compact relative to the graph norm of BA . For if { x n } is a sequence in DfBA) such that l xn l + I BAxn l < c4, then I Ax n l < c3c4 by (7.42) , and hence there is a subsequence (assumed to be the whole sequence) such that l xn  Xm h 0 as m, n Of this sequence, there is a subsequence (need say it?) such that I Axn  Axm l 2 � 0 as m, n � This now gives the desired result by Theorem 7.29. Thus, it remains to prove (7.42) . Since BA is a closed operator, D(BA) can be made into a Banach space U by equipping it with the graph norm. Now, the restriction of A to D (BA) is a closed linear operator from U to Y which is defined on the whole of U. Hence, A E B(U, Y) by the closed E
=
�
� oo .
I
oo.
graph theorem (Theorem 3 . 10) . This is precisely the statement of (7.42) . D The proof is complete.
D (BA)
To see that we cannot conclude that is dense in X as we did be in Theorem 7.3, we use the example given after Theorem 7. 13. Let defined as it was there, and let w(t) be any function in that is not in (take a function which does not have a continuous derivative) . Let
D(B)
C
B
7. 7.
177
The adjoint of a product of operators
by w, and let A be the C ,generated E N. Then A E + ( N, C) and C Ax = x D (BA) {0} .
N be the onedimensional subspace of operator from N to defined by = But E
B � (C) .
x
The following is useful.
tt. + if and only if there is a bounded sequence { u k} A D(A) having no convergent subsequence such that {Auk } converges. Proof. Suppose A tt. + . If o:(A) then there is a bounded sequence in N(A) having no convergent subsequence. If o: (A) then R(A) is not closed. Let P be a bounded projection onto N (A) . Then there is a sequence {wk } D (A) such that I ( !  P)wkl l l ! A wk l Put k. uk = 1 (I(1  P)w P)wkl l Then l u kl l = 1 and I A u kl l 0 . If { uk } had a convergent subsequence, the limit u would be in N(A) while Pu 0. Consequently, we would have u = 0. But this is impossible since l ukl l 1 . Hence, there is no convergent subsequence. On the other hand , if A E + and such a sequence existed, the limit f of A u k would be in R(A). Thus, there would be a u E D(A) such that A u f. If we take Vk u k  u, then { vk } is bounded, has no convergent subsequence, and Avk 0 . Set gk (I  P)vk . Then { gk } has no convergent subsequence ( otherwise, { v k } would have one, since P is a compact operator) . Thus, there is a co > 0 such that l gkl l > co for k sufficiently large. Hence, l 9kl l > co I Agkl l I Avkl l D This contradicts the fact that R(A) is closed. Corollary 7.34. A tt. �+ if and o nly if there is a bounded sequence { u k } D(A) having no convergent subsequence such that Auk 0 . C
Lemma 7.33.
= oo,
< loo ,
__
C
+ oo .
+
==
==
==
==
==
+
t
00
.
+
c
Other statements of Section 5 . 6 hold for unbounded operators as well. The proofs are the same. 7. 7 . The adjoint of a product of operators
A, B are bounded operators defined everywhere, it is easily checked that ( 7.43) (BA) ' A' B'
If
==
1 78
7.
UNBOUNDED OPERATORS
D(AB) D(BA) E D(A' B' ) � A' B ' We can prove this by noting that if E D(A' B'), then (7.45) ( BAx) B' (Ax) A' B' (x), x E D (BA). A, B (B (B D(A' B') D [(BA)'] (BA) ' (7.44)
However, if are only densely defined , need not be dense, and consequently, A ) ' need not exist . In fact , we gave an example in Sec tion 7.6 in which D A ) == {0} . If is dense, then it follows that and c
z
=
'
z
'
z
' z ,
1=
z
'
==
z
'
'
z
'
Consequently, z ' E D [(BA)'] and ( 7.44) holds. If, in additJon, is bounded and defined everywhere, then ( 7.43) will hold. (We leave this as an exercise.)
B
Now we shall show that (7.43) can hold even when neither operator is bounded. We shall prove Theorem 7.35 . Let X, Y, Z be Banach spaces, and assume that A is a densely defined, closed linear operator from X to Y such that is closed in Y and {3(A) < oo (i. e. , A E _ (X, Y)). Let be a densely defined linear operator from Y to Z . Then exists and (7.43) holds.
R(A)
B
(BA)'
That D(BA) is dense in X follows from
B satisfy the hypotheses of Theorem 7. 35 and x is any element in ·n (A) , then there is a sequence { xk } D(BA) such that in X . Consequently, D(BA) is dense in X. A x k Ax in Y and Xk Proof. Since D(B) is dense in Y, we see by Lemma 7.4 that Y R( A ) EB Yo , (7.46) where Yo D(B). Let x be any element in D(A). Then there is a sequence { Yk } D(B) such that Yk Ax in Y. By (7.46) , we can write Yk Y 1 k + Y2 k , where Yl k E R(A) and Y2 k E Yo . Let R(A), E y, y { Py 0, y E Yo . In view of Lemma 5.2, P E B( Y) . Since Yk , Y2 k are both in D(B), the same is true of Yl k · Thus, Y l k E R(A)nD(B). Consequently, there is an X k E D (BA) such that Ax k Y l k · Hence, Ax k Y l k Pyk . Since N(A) PAx Ax is closed in X, we note that X/ N(A) is a Banach space. Let [ x ] denote the coset containing x . Since R(A) is closed in Y, there is a constant C such that l [x] l < C I Ax l , x E D(A). (7.47) Lemma 7.36. If A,
C
4 x
4
==
C
C
4
==
==
==
==
==
+
==
7. 8.
179
Problems
D(�) { xk } x D(BA) Axk Ax } N [xk] [x] X/N(A). { xok (A) X Xok x k X. Xk  Xok D(BA) A(xk  Xok ) == Axk Ax ,
there is a sequence C By what we have shown, for each E � such that in Y. By (7.47) , + in This means that there is a sequence C such that 4 in and Since + the proof is E complete. 0 We can now give the proof of Theorem 7.35.
D (A'B') D[(BA) '] and that (7.44) D[ (B A) '] D (A' B' ). Suppose z' E D[ (BA)'] , D (B) nR(A). Then there is an x E D(BA) such Ax == xo N (A), z'(By) == z'[BA(x  xo) ] == (BA)' z'(x  xo)· Consequently, l z'(By) l < I (BA)'z' l inf l x  xo l == I (BA)'z' l · l [x]l l < CI I (BA)' z' J I · I Y I
Proof. We have already proved that holds. We must show that and let y be any element in that y . If E we have
C
C
xo EN(A)
by ( 7.47 ) . On the other hand, if y E Yo , then
I z' (BY) I < Ct i l z' I · I Y I , since B is bounded on Yo . By (7.46 ) , every E D(B) is the sum of an element in D ( B) n R (A) and an element in Yo . Thus, l z' (By) l < C2I I Y I , y E D (B). This shows that z ' E D(B'). To see that B ' z ' E D (A' ), note that B' z' (Ax ) == z' (BAx) == (BA) ' z' ( x), x E D (BA). If x is any element in D (A), then there is a sequence { x k } D (BA) such that Ax k Ax in Y and Xk x in X. Thus, B' z' (Axk ) (BA) ' z' ( xk ), k 1 , · Taking the limit as k we see that B' z' (Ax) (BA) ' z' (x), x E D (A) . Hence, B'z ' E D (A') , and ( 7.43 ) holds. This completes the proof. D y
+
C
+
==
+ oo ,
=
2,
·
·
.
=
7 .8. Problems
X Y with D (A) 'ljJ dense in X . A' B'. B B( Y, (2) Let A be a closed linear operator on X such that (A  .x)  1 is compact for some A E p(A). Show that ae(A) is empty. ( 1 ) Let A be a linear operator from to If is in Z) , show that (BA)' ==
7. UNBO UNDED OPERATORS
180
p (ae(A))on X .ae(p(A)) for any polynomial p(t) and any A �4) Let A be a closed operator on X such that 0 E p(A). Show that ,.\ =I= 0 in a (A) if and only if 1 /A is in a (A  1 ) . (5) Show that if A E + (X, Y ) and is a closed subspace of X , then A(M) is closed in Y. (6) Show that if U is a closed convex set and p(x) is its Minkowski functional, then u E if and only if p( u) < 1 . (7) Let U be a convex set , and let u be an interior point of U and y a boundary point of U. If 0 < < 1 , show that ( 1  O)u + Oy is an interior point of U . (3) Show that linear operator
C
is
M
U
()
(8) Examine the proof of Theorem 7. 3 to determine the exact hypothe ses that were used in proving each of the statements (a) , (b) , and (c) . (9) In Lemma 7 . 4, show that
D
n
R is dense in R.
( 10) Show that y' (y) =I= 0 in the proof of (i) in Theorem 7. 16. ( 1 1 ) Let A be the operator on l2 defined by A(x i , X2 , . . . ' Xn , . . . ) = (xi, 2x2 , . . . 'nxn , . . . ) , where D(A) consists of those elements (x 1 , ) such that L l nxn l 2 < oo . (a) Is closed? (b) Does A' exist? (c) What are a(A), ae(A)? Y be Banach spaces, and let A be a linear operator from X (12) toLetYX,such that D (A) == X . Show that A E B(X, Y) if and only if D(A') is total in Y'. ( 1 3) Prove: Every operator in B(Y' , X ') is the adj oint of an operator in B(X, Y ) if and only if Y is reflexive. 00
·
·
·
1
A
A,
7. 8.
Problems
181
X', and X is reflexive, show that W ( 0W ) 0 • ( 1 5) Let A he a onetoone, linear operator from X to Y with D (A) dense in X and R (A ) dense in Y. Show that {A')  1 exists and equals (A  1 )' . ( 16) Under the same hypotheses, show that A  1 is bounded if and only if (A')  1 is. ( 1 7) Let X , Y be normed vector spaces, and let A be a linear operator from X to Y such that D(A) is dense in X and N(A) is closed. Assume that ! (A) > 0, where l . (A) == xE1B(. fA) d(x,I ANx(A)) Show that ! (A') == !(A) and that R(A') is closed: (18) If p(x) is a seminorm on a vector space V, and xo E V, show that there is a functional F on V satisfying F(xo) == p(xo) and I F (x) l < p(x), x E V. ( 19) For X a normed vector space and c > 0 , show that for every xo E X satisfying l x o l > c there is an x ' E X' such that x'(xo) > 1 and l x' (x) l < 1 , l x l < c. � E X ' and (/31 , f3n ) E (20) Suppose X is a Banach space, x � , , x , IRn . Show that for each c > 0 one can find an element x o E X satisfying l x o l < M + c and x�(xo) == f3k , 1 < k < n, (14) If W is a subspace of
==
1
·
if and only if
n
·
·
n
I 2:1 : ak ,Bkl < M i l 2:1 : ak x/J holds for every choice of the o: k . ( 2 1 ) Show that the closure of a convex set is convex.
·
·
·
Chapter
8
REFLEXIVE B ANAC H S PAC E S
8 . 1 . Properties of reflexive spaces
We touched briefly on reflexive spaces in Section 7.4 when we discussed total subsets. Recall that we showed, in Section 5.4, that corresponding to each element in a Banach space there is an element in such that
X Jx X" Jx (x' ) x' (x), x' E X' ,
x
==
(8. 1) and
I Jx l == l x l , x E X. Then we called X reflexive if R( J ) X" , i.e. , if for every x " E X" there is an E X such that Jx x" . One advantage of reflexivity was already noticed, xnamely, the fact that total subsets are dense. This allowed us to conclude that if Y is reflexive and A E ( X, Y ) , then A' E ( Y ',X') ( Theorem 7. 22) . (8.2)
==
==
In this chapter, we shall see that there are other advantages as well. Let us first mention some reflexive Banach spaces. Every Hilbert space is reflexive, since, by the Riesz representation theorem ( Theorem 2 . 1 ) , every z ) , where z Hence, can be made into is of the form a Hilbert space. Thus, elements of are of the same form. If p > 1 , we proved that l� can be identified with lq , where 1/p + 1 /q == 1 ( Theorem 2 . 1 1 ) . The same theorem shows that l � cttn be identified with lp . Hence, l� == lp, showing that lp is reflexive for p > 1 . We shall show that l 1 is not reflexive. We now discuss some properties of reflexive spaces. Assume that is a Banach space unless otherwise specified.
x' E X'
(x,
X"
E X.
X'
X
183
8. REFLEXIVE BANACH SPACES
184
Theorem 8. 1 . X is reflexive if and only zf
X' zs.
X is reflexive, and let J1 be the mapping of X' into X"' J1 x ' ( x ") x " (x ') , x " E X " . (8.3) Let x"' be any element of X"'. Then by (8.2) , x'" (Jx) is a bounded linear functional on X. Hence, there is an x' E X' such that E X. (x) , x "' ( Jx ) But E X. x' ( x ) Jx ( x' ) , Hence, E X. (8.4) x "' ( Jx ) Jx ( x ' ) , Now, since X is reflexive, we know that for any x" E X" there is an x E X such that J x x" . Substituting in (8.4) , we have x "' ( x ") x " ( x ' ) , x " E X " . If we now compare this with (8.3) , we see that .J1 x ' x"'. Since x"' was any element of X"', we see that X' is reflexive. Now assume that X' is reflexive. We note that R( J) is a closed subspace of X". For if J x n x" in X" , then { x n } is a Cauchy sequence in X by (8 2 ) . Since X is complete, there is an x E X such that x in X. Hence, Jx in X", showing that x" Jx . Now if R(J) is not the \vhole of Jxn of X" not in R(J) . Then , by Theorem 2 . 9 , there X", let x" be any element is an x"' E R(J)0 such that x" ' ( x" ) # 0. Since X' is reflexive, there is an x' E X' such that J1 x' x"'. Hence, (8.5) Jx(x ' ) J1 x' (J x ) x "' (J x ) 0, x E X , Proof. Suppose given by
==
== x '
X
X
==
X
==
==
==
==
�
Xn �
.
==
�
==
==
==
==
and
(8.6)
x " ( x' )
==
J1 x ' ( x ")
==
x "' ( x") # 0.
By (8.5) , we see that
x' ( x )
==
Jx ( x ' )
==
0,
x
E X,
showing that x ' == 0. But this contradicts (8.6) . Hence, R(J) the proof is complete .
==
X" , and 0
We also have Theorem 8 . 2 . Every closed subspace of a refiexzve Banach space is refiex
zve.
8. 2.
Saturated subspaces
185
Z Z
X. X',
Z
Proof. Let be a closed subspace of a reflexive space Then is a the restriction Banach space. Let z" be any element of For any x ' E x� of x' to is an element of Thus, z" (x� ) is defined, and
Z" .
Z'.
l z" ( x �) l < ll z" ll · ll x� ll < ll z" l l · l l x' ll · Hence, there is an x" such that
E X"
x" ( x')
=
z" ( x � ) .
X is reflexive, there is an x E X such that x" Jx . Hence, (8. 7) z" ( x� ) x' ( x) , x' E X'. If we can show that x E Z , it will follow that ( x�) x� ( x) , x' E X' . Since, for every z' E Z', there is an x' E X ' such that x � z' (the Hahn Banach theorem) , we have ( z ' ) z ' ( x) , z ' E Z' , showing that Z is reflexive. Thus, it remains to show that x E Z. If it were not, there would be an x ' E such that x' ( x ) =1. 0. But this contradicts (8. 7) , which says that every x' E annihilates x as well. This completes Since
==
=
z
"
=
==
z
"
==
zo
zo
the proof.
0
8 . 2 . Saturated subspaces
X,
EX
then for each x \ M, there is an If M is a closed subspace of x' M0 such that x '(x) =I= 0 (Theorem 2.9) . Now suppose W is a closed subspace of Does W have the property that for each x' X ' \ W there W such that x ' (x) =I= 0? Subspaces of is an x having this property are called saturated. If is reflexive, then any closed subspace W of is saturated, because there is an x" W0 such that x" ( x') =I= 0. We then take x = J  1 x" . In this section, we shall see that this property characterizes reflexive spaces. We now investigate some properties of saturated subspaceso
E
E
X'.
E
X'
X
X'
E
Theorem 8.3. A subspace W of some subset M of X.
X' is saturated if and only if W
==
M0 for
Proof. If W is saturated, set M == W. Then clearly, W c M0 Now suppose x' tt. W . Then there is an x E W == M such that x' (x) =I= Oo Therefore, x ' is not in lV/0 • This shows that W == M0 Conversely, assume that W == M0 for some set M C If x' tt. W, then there is an x M such 0 that x ' ( x ) "# 0. But M C 0(M0 ) == Wo Hence, W is saturated. 0
X.
Corollary 8.4. W is saturated if and only if W
0
==
(W)0 •
E
186
8.
REFLEXIVE BANACH SPACES
X'
A subset W of is called weak* (pronounced "weak star" ) closed if there is a sequence W whenever it has the property that for each of members of W such that
x' �}E {x
x E X,
x�(x) x' (x). ' implies (8.8) . As we shall see, it A weak* closed set is closed, since x� x is possible for a closed subset of X' not to be weak* closed. In particular, we have Theorem 8 . 5 . A subspace of X ' is saturated if and only if it is weak* closed. Proof. Suppose W is saturated , and let x' E X' be such that for each X , there is a {x� } W satisfying (8.8) . If x' tf:. W, then there is anW x EE W xbecause such thatlimx'(x) # 0. 0.But then (8.8) cannot hold for any { x� } x' ( x ) x� ( x) Now assume that W is weak* closed. If W X' , there is nothing to prove. Otherwise, let x� be son1e element in X ' \ W. Thus , tlrere is an element x o E X such that no sequence {x � } W satisfies (8.8) for x x o . Hence, inf l x ' (xo)  x�(xo ) I 0. (8.9) Since 0 E W, we see that �;� (xo) # 0. Moreover, we claim that (8.9) implies that x'(xo) 0 for all x' E W, i .e. , xo E W. For if x' E W and x'(x o) # 0, set o: x� ( x o) / x ' ( xo) and Then E and ( xo) ::::: o: x' ( x o) '. o:x xo), contradicting (8.9) . This shows that W is saturated, and the proof0 xis�(complete. �
(8.8)
+
W
c
c
==
==
==
==
c
>
x' EW
==
==
y
==
'
y
'
H'
y
'
==
We also have
X' 2s always saturated. Proof. Suppose that x ; , x� form a basis for W X' and assume that x� t/:. W. Then the functionals x� , x� , , x� are linearly independent . By Lemma 4. 14, there are elements EX such that (8 . 1 0) In particular, x�(x o ) 1 while x�(xo) 0 for 1 < j < Hence, x0 E 0W,
Theorem 8.6. A jinitedimens2onal subspace of · ·
·
,
· · ·
c
Xo , X I , · · · , X n
==
showing that W is saturated.
==
n.
D
Theorem 8. 7. A Banach space X is reflexive if and only 2/ every closed
subspace of
X' 2s saturated. I
8. 2. Saturated subspaces
187
In proving Theorem 8.7, we shall make use of the simple Lemma 8.8. Let X be a normed vector space, and let x' be an element of Let M be the set of those x in X such that x' (x) 0 (i. e. , M 0[x']).
X' .
= = Let y be any element not in M, and let be the onedimensional subspace of X spanned by y {i. e. , [yj). Then X = N EB M. Proof. Clearly, N M = {0}. Moreover, for any x E X , set x ' (x) (8. 1 1 ) x = x' ( y ) 0 , showing that E M. Since x z + the proof is Then x' ( z ) N
N
n
z
y.
o:y ,
z
complete.
D
We can now give the proof of Theorem 8.7. Proof. We have already given the simple argument that all closed suhspaces of X' are saturated if is reflexive. Now assume that all closed subspaces of X' are saturated. Let x� be any element of and let W be the set of all x' E which annihilate x� . Clearly, W is a closed subspace of Let By hypothesis it is saturated. Since x� W is not the whole of x� be any element in W. Then there is an element X t W such that x� (xt ) i= Moreover, by Lemma 8.8, every x' can be written in the form
X
X'
where
w
'
X" ,
i= 0 ,
X' \
0.
(8. 12)
i= 0
E W. Set
E X'
E
X'.
X'.
x' = ax� + w' ,
= x� (x t )x� (x')  x� (x� )x' (x t ) , x' E X'. Then g E X " , and g(x') == 0 for x' E W and for x' = x� . Hence, g (x') 0 by (8. 12) . This shows that x� (x ' ) = x' (f3xt ) , x' E X' , where {3 = x� (x� ) /x� ( x t ) · Hence, x� = J(f3x t ) · Since x� was arbitrary, it follows that X is reflexive. g (x')
=
D
We also have Corollary 8.9. A finitedimensional Banach space X is reflexive.
In proving Corollary 8.9, we shall make use of Lemma 8 . 10. If dim
X=n
0, there is an X k satisfying < c . We investigate some properties of separable spaces. First we have
{ xk }
l x  xkl l
x
Theorem 8.1 1 . If X' is separable, so is X.
{ x�}
Proof. Let be a dense set in X' . For each n, there is an Xn E X such that = 1 and
l xn l
8. 3.
Separable spaces
189
of the set of linear combinations of [ { xn }],betheanyclosure element of not in M. Then there is an . X xo n x� l x� l 1 and xti(xo) "# 0 (Theorem 2. 9) . In particular, x�(xn ) 0 , n 1 , 2 , · Thus, l x�l l / 2 < l x� (xn ) l l x� (xn )  x� ( xn ) l < l x �  x� l · l x n l H x�  x� l · Hence, 1 l x � l < l x �  x � l + l x � l < 3 l x �  x� l , showing that none of the x� can come closer than a distance 1 / 3 from x0. This contradicts the fact that { x�} is dense in X'. Thus we must have M X . But M is separable. To see this, _we note that all linear combinations of the Xn with rational coefficients form a denumerable set. This set is dense in M. Hence, M is separable, and the proof is complete. We shall see later that it is possible for X to be separable without X' being
by (2.28) . Let M = the If M # X, let E M0 such that
==
X
=
==
=
·
·
.
==
=
=
0
so. However, this cannot happen if X is reflexive.
Corollary 8. 12. If X is reflexive and separable, then so is X' .
{xk }
X, x" x x". I Jxk  x" l
be a sequence which is dense in be any Proof. Let and let element of X " . Then there is an E X such that J = Moreover, for any c > 0, there is an such that < € . Thus, < c by 8. . This means that X " is separable. We now apply Theorem 8.11 to 0 conclude that X' is separable.
Xk
( 2)
x
l xk  x l
�} of elements in X' is said to be weak * convergent to an { x element x ' E X' if (8. 14) x� ( x ) x' ( x) as n x E X. weak* convergent sequence is always bounded. This follows directly from the BanachSteinhaus theorem (Theorem 3.17). There is a partial A sequence
�
� oo ,
A
converse for separable spaces .
Theorem 8 . 1 3 . If X is separable, th�n every bounded sequence in X' has a
weak * convergent subsequence.
{ x�}
{ xk }
be a sequence be a bounded sequence in X' , and;let Proof. Let dense in Now is a bounded sequence of scalars , and hence, it contains a convergent subsequence. Thus , there is a subsequence of converges as � oo . Likewise, there is a subse such that quence such that of converges. InductiVely, there is a = subsequence converges. Set such that of
X.
x�(x 1 ) n1 � 1 (x 1 ) {x�} x �2 (x2 ) { x�2} { x� 1 } x { x�k } { x�k  l} x� k ( x k )
{ x�1}
Zn x�n.
190
8. REFLEXIVE BANACH SPACES
{ z�} is a subsequence of { x�} , and z�(xk ) converges for each Xk . Let E > 0 be given. There is an X k such that l x x kl l < c /3C , where x EisXsuchandthat C (8. 15) I x� I < C, 1 , 2 , Thus, l z� (x)  z:n ( x) l < l z� (x)  z� (xk ) l + l z� ( xk )  z:n (xk ) � + l z:n ( xk)  z:n (x) l < 2e/3 + l z� (x k )  z� (xk ) l . Now take m and so large that the last term is less than c /3. This shows that Zn ( x) is a convergent sequence for each x E X. Set F(x) lim z� ( x). Clearly, F is a bounded linear functional on X. This proves the theorem. D Then
· · · .
n =
n
==
n + oo
We also have Theorem 8. 14. Every subspace of a separable space is separable.
X,
{ xk }
Proof. Let M be a subspace of a separable space be a dense and let sequence in For each pair of integers j, k, we pick an element M, if there is one, such that
X.
l xjk  Xk I < 1 /j.
{ xj k } { xk }
Xj k E
of elements of .l'vf is If there is not any, we forget about it . The set denume�able. We claim that it is dense in M. To see this , let E M and E > 0 be given. Take j so large that 2 < jE. Since is dense in there is a k such that < 1/j. Since for this choice of j and k. M, this shows that there is an Hence, < < 2 /j < E. 0 This completes the proof.
xX ,
l x  xk l
xE
xj k l x  Xjk l l x  Xk l + l x k  Xjk "
8 . 4. Weak convergence
{ xk } Q,f elements of a Banach space X is said to converge weakly x E X if x' (xk ) x ' (x) as k E X'. We shall investigate how this convergence compares with
A sequence to an element (8. 16)
�
� oo
for each �' convergence in norm ( sometimes called strong convergence for contrast ) '. Clearly, a sequence converging in norm also converges weakly. We have Lemma 8 . 1 5 . A weakly convergent sequence is necess{Lrily bounded.
191
8. 4. Weak convergence
{ Jxk } of elements of X". For each x' we ' J ) (x x < I l k k Then by the BanachSteinhaus theorem (Theorem 3 . 1 7) , there is a constant D C such that I Jx kl l < C. Thus, l x kl l < C, and the proof is compl ete. Proof. Consider the sequence have sup
oo .
We also have Theorem 8 . 1 6 . If X is reflexive, then every_ bounded sequence has a weakly
convergent subsequence.
X
{ Xn }
Proof. Suppose is reflexive, and let be a bounded sequence in X. Set M == As the closure of the set of linear combinations of the we observed before, M is separable (see the proof of Theorem 8. 1 1 ) . Since it is a closed subspace of a reflexive space, it is reflexive (Theorem 8.2) . Thus , M ' is ·separable (Corollary 8. 12) . Now is a bounded sequence in M" . Hence, by Theorem 8. 13, it has a subsequence (also denoted by suGh that converges for each E M'. This is the same as saying that converges for each E M' . Now let be any element of X' . Then the restriction to M is in lvl' . Thus , of converges. This means that converges weakly, and the proof is complete. 0
[{ Xn }] ,
Xn .
{Jxn }
{ JXn ( x')} x'( xn ) x' x� }x' { Xn
x'
x'
{Jxn })
x'(xn ) x�(xn ) ==
We now realize that weak convergence cannot be equivalent to strong convergence in a reflexive, infinite dimensional space. The reason is that if == 1 has a weakly convergent X is reflexive, every sequence satisfying subsequence (Theorem 8. 16) . If this subsequence converged strongly, then it would follow that X is finitedimensional (Theorem 4.6) . On the other hand, we have
l xn l
Theorem 8 . 1 7. If X is finitedimensional, then a sequence converges weakly
if and only if it conve1yes in norm.
{ xk } X'
. x x� , , x �
Proof. Let be a sequence that is weakly convergent to Since X is be a basis for X' . finitedimensional, so is X' (Lemma 8. 10) . Let Then every x' E can be put in the form ·
·
·
n x' == L O:i X� . J
Since all norms are equivalent on X' (1.,heorem 4. 2) , we can take
n I I x' l == L I o: I · 1 j
192
8. REFLEXIVE BANACH SPACES
E > 0 , there is an N such that l xj ( x k  x ) l < 1 < j < for all k > N. Then for any x ' E X', l x' (xk  x ) l < 2:: l ai l · l xj ( xk  x ) l < c l� x' l l Now for each
E,
n
n
1
for such k. In view of (2.28) , this gives
{ }
>
k
l l xk  x l l < E ,
N,
\Vhich tneans that x k converges to x in norm.
D
8 . 5 . Examples LPt
llS
now apply the concepts of the preceding sections to some of the spaces we have en
00
(8. 24) Taking
2: !x)n) I >
j=l
e,
n
= 1 , 2, · · ·
.
Zj = §j k , j = 1 , 2, · · · , we have by (8. 23) , k = 1 , 2, · xkn) 0 as �
n � oo ,
such that
·
·
.
19 5
8. 5. Examples
mo no == 0, and inductively define the sequences {mk } , {nk } as mk  1 and nk_ 1 are given, let nk be the smallest integer n > nk 1 ffik 1 nk ) L l x) l < e / 5 , (8. 25) j= 1 and let 'n1: k be the smallest integer m > m k  1 such that 00 (8.26) L l xY'tk ) l < e /5. J = mk Now, let z == ( z 1 , ) be the vector in l 00 defined by zj == sgn xj(nk ) , mk 1 < j < mk , k == 1 , 2 where sgn o: is the signum function defined to be o:/ l o: l for o: # 0 and 0 for o: == 0 . Thus by (8.25) and (8.26) , 1 00 00 ffi k n n n n ) k ) l < 4e/5. k k k ) ) ) 2 ) 2 ) ) < x x ) ( x x zj L L L l l l l l j =mk j=1 j=1
Set == follows. If such that
z2 ,
·
·
·
,
·
..
·
,
+
By (8.24) , this gives
nk ) l > e/5, ) Z x j j=1 00
IL
k = 1 , 2,
·
·
·
.
This contradicts (8.23) , and the proof is complete. (8) The last paragraph provides another proof that l1 is not reflexive. If it were, every bounded sequence would have a weakly convergent subse quence (Theorem . 8. 16) . But weak and strong convergence are equivalent in l 1 . Hence, this subsequence converges in norm. This would show that the surface of the unit sphere is compact, and, consequently, that is finite dimensional (Theorem 4.6) . Since we know otherwise, it follows that l1 is not reflexive.
l1
(9) If X is a Banach space that is not reflexive, X' has closed subspaces that are not saturated (Theorem 8. 7) and , hence, not weak* closed (Theorem 8.5) . It also has total subspaces which are not dense. This follows from Theorem 8 . 1 8 . If X is a Banach space such that every total subspace of X' is dense in X' , then X is refiexi�e.
8. REFLEXIVE BANACH SPACES
196
X
E X" X'.
Proof. If were not reflexive, tltere would be an x � which is not in which annihilate x� . Since J) . Let W be the set of those x ' x � # 0, W is a subspace of that is not dense in We claim that it is total . in If we can substantiate this claim, it would follow that W violates the hypothesis of the theorem, providing a contradiction. To prove that W is total, we must show that for each x # 0 E there is an x ' E W such that x ' ( x ) # 0. Let x � be such that x�( x �) =I= 0, and suppose we are given an x # 0 in If x� x ) == 0, let x' be any element of such that x ' ( x ) # 0. By Lemma 8.8,
R(
X'.
E X'
X'
X,
E X' X. (
X'
x
(8. 27)
where x � E W. Thus x � ( x ) == there is a 13 # 0 such that
I
x
==
'
I
nx 0
I
+ x1 ,
( x ) =I= 0, and we are through. Otherwise,
(8.28 ) (just take (3 == x� ( x� ) / x�( x ) ) . Since x� is not in R(J) , there is an such that x
(8.29)
'
((3 x ) # x �
(
x
'
)
(otherwise we would have x � == J ( (J x ) ) Decomposing we see by (8. 28) and (8.29) that .
x
x
'
x
'
E
X'
in the form (8. 27) ,
� ((Jx ) # x � ( x � ) .
But, x�( x� ) == 0, since x � E W. Hence, x � ( x ) =I= 0 . The proof is complete.
x
� (j1 x ) # 0, which means that D
8 . 6 . Completing a normed vector space
In Section 1 .4, we mentioned that one could always complete a normed vector space i.e. , find a Banach space Y containing such that
X,
X (a) the norm of Y coincides with the norm of X on X. (b) X is dense i n Y.
We now give a proof of this fact.
X
X
Proof. Let be any normed vector space. Consider the mapping J of into defined by (8 . 1 ) . By (8. 2) , R(J) is a normed vector space. Now is complete (Theorem 2. 10) , and hence, the closure Y of R(J) in is a Banach space. Hence, Y is a Banach space containing R(J) and satisfies
X"
X"
X" /
8. 7. Problems
197
( a) and ( b ) with respect to it . Finally, we note that we can identify X with D R(J) by means of (8 . 1 ) and (8.2) . This completes the proof.
8. 7. Problems
( 1 ) If .L�;J is a closed subspace of a separable normed vector space X , show that X/ M is also separable.
{ xn } is a bounded sequence in a normed vector space X x' (xn ) x' (x)
(2) Suppose satisfying
as
�
n � oo
x'
in a set M C X' such that M is dense in X' . Show that for all converges weakly to
x. ( 3) Show that the sequence xj ( O: l j , ) converges weakly in lp , 1 converges if and only if it is bounded and for each O: j n as Xn
==
p < oo ,
·
·
·
0 and a nondenumerable subset W of X such that > 6 for all y E W, y.
l x  Yl
x,
x i=
( 1 6) Let X be a separable Banach space. Show that there exists a map A E B (l 1 , X) which is onto and such that A' maps X' isometrically into l 00 •
x�
converges strongly if and only ( 1 7) If X is a Banach space, show that if converges uniformly for each E X such that = 1.
x�(x)
x
l xl
8. 7. Problems
199
(18) For X �a Banach space, show that Xk � x strongly and x� the weak* sense implies x� (xk) � x' (x) .
�
x' in
( 19) Show that one can come to the same conclusion asin Problem 18 if Xk � x weakly and x� 4 x' strongly.
(2 0 )
If X, Y are Banach spaces and K E B (X, Y ) is such that K' E K ( Y', X') , show that K E K ( X, Y ) .
( 21 ) Let X, Y be Banach spaces, with  Y reflexive and X separable. If Ak is a bounded sequence in B (X, Y ) , show that it has a renamed subsequence such that Akx converges weakly for each x E X .
(22) (23)
If X, Y are Banach spaces and K E B (X, Y ) , show that K E K (X, Y ) if and only if KX n � K x strongly in Y wh·e never Xn � x weakly in X . If
z ( t) is given by z(t) =
0, 1, 0,
a < t < r, r < t < s, s < t < b,
for a < r < s < b, show that the total variation of z (t) in [a, b] is
2.
Chapter
9
B ANAC H AL GEB RA S
9 . 1 . Introduction
B(X,
Y) is a Banach If X and Y are Banach spaces, then we know that space (Theorem 3.2) . Moreover, if Y = X, then elements of B (X) can be "multiplied;" i.e. , if A, B are in B (X) , then E ' B(X) and
AB (9. 1.) I AB I < I A I . I B I , SinCe I ABx l < I A I I Bx l < I A I · I B I · l x l · Moreover, we have trivially (9.2) ( aA + ,BB)C = a (AC) (3(BC), and (9.3) C ( aA + ,BB) a (CA) + (3(CB). .
·
+
==
A Banach space having a "multiplication" satisfying (9. 1 )(9.3) is called a Banach algebra. In this chapter, we shall study some of the properties of such algebras. Among other things, we shall see that we can obtain a simple proof of that part of Theorem 6.23 not yet done. Another property of B(X) is that it has an element I such that
(9.4)
B
AI = IA == A
for all A E (X ) . Such an element is called a unit element. Clearly, it is unique. Unless otherwise specified, whenever we speak of a Banach algebra, we shall assume that it has a unit element . This is no great restriction, for
2 01
9. BANACH ALGEBRAS
202
B
we can always add a unit element to a Banach algebra. In fact , if is a Banach algebra without a unit element , consider the set C of pairs ( a, o: ) , where a E and o: is a scalar. Define
B
( a, o: )
+
(b, {3)
==
( a + b, o: + {3)
(3 ( a, o: ) == ((3 a, (3o: ) (a, o: ) ( b, {3) == ( ab + o: b + {3a , o: {3 ) ,
and
I (a, o:) I == I a ll + l o: l .
We leave it as a simple exercise to show that C is a Banach algebra with unit element (0, 1 ) . An element a of a Banach algebra is called regular if there is an element a  1 E such that
B
B
( 9.5)
B.
where e is the unit element of The element a  1 is unique and is called the inverse of a. The resolvent p( a) of an element a E is the set of those scalars ,.\ such that a  ,.\e is regular. The spectrum a (a) of a is the set of all scalars ,.\ not in p(a) . are true for an arbitrary Banach algebra, Some of the theorems for and the proofs carry over. We list some of these here. In them, we assume that is a complex Banach algebra.
B
B(X)
B
Theorem 9 . 1 . If 00
L l l an ll
(9.6)
< oo ,
0
then e a is a regular element, and (9.7)
( Theorem 1 . 1 . )
I
Corollary 9.2. If a ll < 1 , then e  a is regular, and {9. 7) holds. Theorem 9.3. For any a in B,
(9.8)
r (a) u
==
n
lim ll an ll 1 / n + oo
9. 1 . Introduction
203
exists, and (9.9)
r
a
(a)
==
max
,\ E u ( a)
I ,.\ I ·
l z l > r (a) , then ze a is regular, and (9. 10) ( ze  a)  1 L z  n an  1 . 1
If
u
00
==
( Theorems 6. 13 and 6 . 14 . )
B(X)
p(A)
Our proof that is an open set for A E depended on properties of Fredholm operators ( see the proof of Theorem 6.3) . Let us give another proof that works in Banach algebras.
11 < a c E > l x  a l < E . l xa  1  e l <  1 xa 1 l a zx x z yx a  1 zxa  1 a a  1 ea z e. y a  1 z . xy x y Theorem 9.4. If ,.\ , JL are in p(a), then (9.11) (,.\e  a )  1  (f.le  a)  1 (JL  ,.\ )( ,.\e  a)  1 (Jte  a)  1 If 1 ,.\  JL I · I (Jte  a)  1 1 < 1 , then (,.\e  a)  1 L (JL  ,.\) n  1 (Jte  a )  n . (9 . 1 2 ) 1 ( Theorem 6 . 1 8. ) Theor e m 9.5. (Spectral mapping theorem) If p (t) is a polynomial, then a(p(a )] p[a (a)]. (9 . 13 ) More generally, if f(z) is analytic in a neighborhood f2 of a ( a), define 1 f ( z )( z e  a )  1 dz, (9. 14) f (a) � 2 7r z Jaw where is an open set containing a( a) such that consists of n, and a finite number of simple closed curves which do not intersect. Then a[! (a)] f [a (a) ] . (9. 15) a
Proof. Let be a regular element , and suppose 0 is such that 1 . Let be any element satisfying 1 , showing Then that == is regular ( Theorem 9. 1 ) . Let be its inverse. Then xa  1 == Set == Then == e and == == == e. D This shows that has as an inverse.
==
00
==
==
=
w
w
==
c
ow
204
9. BANACH ALGEBRAS (Theorem 6. 1 7. )
A Banach algebra B is called trivial if just the element 0 . We have
e
==
0 . In this case, B consists of
a( a) is not empty. Proof. Suppose a # 0 and p(a) is the whole complex plane. Let a' # 0 be any element of B' (the dual space of B considered as a Banach space) . Since the series in ( 9 . 12) converges in norm for I ,.\  J.L I · I I ( J.L e  a )  1 1 < 1 , we have a' [(.Xe  a)  1 ] L (J.L  ,.\) n  1 a' [( J.Le  a)  n ] . 1 1 This shows that f ( ) [ ( . a)  ] is an entire function of Now, by (9. 10) , Theorem 9.6. If B is nontrivial, then for each a in B,
00
==
z
==
a
'
ze
z.
00
(9. 16) If
1
l z l > k l a l , k > 1 , then ) ! < � l a' l · l a l n  l ( � I1 n n a k l l 1 k l a l · This shows that f ( ) 0 In particular, f ( ) is bounded in the z l l whole complex plane. By Liouville ' s theorem , f ( ) is constant , and since 0 as we must have f ( ) 0. Since this is true for any f( ) a' E B ' , we see that ( z e  a )  1 0. [See (2.28) .] But this is impossible since it would imply e ( e  a )(ze  a )  1 0 . Thus, a(a ) cannot be empty. Since a(O) {0} when B is nontrivial, the (9. 17)
_
z
z
�
�
as
� oo .
z � oo ,
z
==
z
z
==
z
==
==
proof is complete .
D
Thus, we have proved Corollary 9. 7. If X is a complex Banach space, and A is in 'B(X), then
a(A) is not empty.
9.2. An example
205
9 . 2 . An example
Let X be a complex Banach space, and consider the Banach algebra B (X ) . We know that the subspace K(X) of compact operators on X is closed in B(X) (Theorem 4. 1 1 ) . Let C be . the factor space B (X )/ K (X) ( see Section 3.5) . Let [A] denote the coset of C containing A. If we define (9. 18)
[A] [B]
==
[AB] ,
then it is easily checked that C is a complex Banach algebra with unit element [I] . In this framework , we have Theorem 9.8. p( [A] ) == A . Proof. By considering A + ,.\ in place of A , it suffices to prove that A E (X) if and only if [A] is a regular element of C. If [A} is a regular element of C, then there is an Ao E B (X ) such that
( 9. 1 9)
[Ao] [A] == [A] [Ao]
==
[I] ,
or (9. 20)
[AoA] == [AAo]
==
[I] .
Thus, (9. 2 1 )
AoA == I  K1 ,
AAo == I  K2 ,
where K1 , K2 E K (X) . By Theorem 5.5, we see that A E ( X ) . Conversely, if A E (X ) , then there are Ao E B (X) and Ki E K (X) such that (9 . 2 1 ) holds (Theorem 5.4) . This leads t o (9.20) and (9 . 19) , which shows that [A] 0 is a regular element. This completes the proof. We can now prove Theorem 9.9. If there is an operator A 'ln B(X) such that A consists of the whole complex plane, then X is jin'lte dimensional. Proof. If
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
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, eat , 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
..  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 oneparameter 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, JL 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
.  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 oneparameter 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 oneparameter, 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, oneparam·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 BanachSteinhaus 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 (A2  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 ) (AA)x (AA ) 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, oneparameter 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
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, N. Thus
k 1 X�Ct I An u  Au l 2 = nI:+ l I .Xkl 2 l (u, N, which means that An in norm as Hence, we see that A is A compact by Theorem 4. 1 1 . That is normal follows from 1 1 . 24 . This 00
+
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, 0 such that
IlK  Fl > 6 for all operators F of finite rank. Now n Fn u = Pn Ku = L (Ku, 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 onetoone. 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 halfplane, 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 halfplane, 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
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 halfplane, 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 oneto'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 �
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 halfplane, 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
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 halfplane.
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 halfplane by Lemma 12. 19. But every halfplane 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
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 ! Atcommutes .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 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, JL) , A > JL, 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 RiemannStieltjes 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] . Th 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 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 onetoone. 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 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 ndimensional 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.
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)
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.
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
0 , the result follows.
We also have Theorem 14.29. A E 4>+ (X, Y) if and only if f (A) # 0.
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:(AK) oo ( Theorem 14. 4 1 ). Let M N(AK). 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
�
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
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 ) 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
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 =
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 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
+ . Finally, we shall call it a a perturbation function if m(T, A) < 1 for A E
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 + , 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 + , 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 o: ( A ) We have .
)
Lemma 14.64. lf m 1 ( T ) / m 2 ( A is an Aexact 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 Aexact , 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
1/6, then (15. 9) will hold for s in thisinterval. This will give I ! Jnu  u l 2 < 1: IPn(s) ds so that by
=
=
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
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 CauchySchwarz 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 100oo 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(xt) 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 { 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
= c, x real. (We leave this as an exercise.) In this case FI u l I FJ I ' (B 
(1 5.89)
=
=
c
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  JL) 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  JLU)Vdx 100 [100 v(x) dx  JLV (t )] u(t) dt. Now, if Vu  Jl U 0, we have ( 1 5. 1 36) 100 [100 v(x ) dx  JLV (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 Jl 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  JLV (t) 'l/J(t) , t > 0. ==
In fact, the method used above gives
v ( x )  �'l/J (x )  :\2 100 e>. (tx) '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  JL ) 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 ==
==
==
Jl ·
V.
=
c
=
=
� oo ,
� c
=
==
1 5. 1 1 .
An integral operator
where Cl
Clearly,
=
389
1:
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 oneparameter 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,
SemiFredholm 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 semiFredholm.
a
SemiFredholm 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 semiFredholm 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 oneparameter 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 Enet 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 Enet 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 onetoone 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 . {HahnBanach 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. (BanachSteinhaus 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
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, oneparameter
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 nonzero 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 onetoone 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 halfplane, 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 halfplane �e A < 0. ,..
Theorem 12.25. (p. 286) Let A be a densely defined linear operator on H. such that W (A) is a halfplane. 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
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