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,en)1 2 ~p~(<j>)
2 nb)>o b -(b 21t + ~-~~cos 3 1t ....... 2 .
so
n
n=l
~ cos nt . .f b . . . 1 S mce ......, -2 -ls um orm y convergent on [-1t,1t] y We1erstrass' M-test, 1t follows that
n
n=l
t2 1t 2 ~ cos nt) . . t- ( 21t + 3--2IS continUOUS on [-1t,1t], (see AMS §3).
n n=l ....
n
But also from the uniform convergence of
I n=l
J (f_ + ~ _ ~ f. {t _
-n
21t
3
1t n=l
cos2 nt it follows, (see AMS §7), that n
~
~ f.
cos nt)} dt=f-(_f_ + t _ sin nt) n2 2 61t 3 1t n=l n3 2 I 2 2 2 = 1t - 3 1t - 3 1t = 0 ·
]n:-n:
Since the integrand is continuous and positive we conclude that b=(b 2 21t
+~-~f. 3
1t n=l
cos nb)· n2
'
that is,
0 choose mEN such that II xm II 11m< v + £.
For any n E N write n = Pnm + qn where Pn and qn are integers and Pn ~ 0 and
0 ~ qn
~
m- 1. Then by the submultiplicative property, for any n EN
v ~II xn 11 11n =II xmpn xqn 11 11n ~II Xm IIPn/n II X llq"/n
q Since nn --7 0 as n --7 Therefore
oo
< (v + £) mpn/n II x llq 010 • mp we have ~ --7 I as n --7 oo.
lim II x" 11 11" exists and is equal to v.
D
n~~
4.11.13 Remark. Given a unital Banach algebra (A, 11·11) and an element a E A, Theorem 4.11.1 0 shows that e- Ax is regular for scalar A if lim sup II ('Ax)n 11 11" < I; that is, for all n~
scalars A such that I A I< I I lim sup II x" 11 11". n~
For that range of scalars A, 00
(e-'Axr' = e
+I
A"x".
n=l
This series is called the Neumann series for x.
D
We now apply our theory to the solution of an integral equation. 4.11.14 Example. Volterra integral equations Consider the integral equation of the form X
f(x) = g(x) + A where
Jk(x,t) f(t) dt
g is a given continuous function on [a,b], f is the required solution function on [a,b ], k the kernel of the equation is a given continuous function on the triangular region { (x,t) : a~ t ~ x, a~ x ~ b},
and A is a scalar.
84
§4. Duals and operator algebras
The equation has a unique solution function f
E
'C[a,b] for each given g
E
'C[a,b] and
parameter A..
Proof. The Volterra operator K on t: [a,b] defined by X
(K(f))(x) =
Jk(x,t) f(t)
is a continuous linear operator on ('C[a,b], 11·11 00 ) , (see AMS §7). In terms of the Volterra operator, the integral equation can be written (I- A.K)(f) = g. From Remark 4.11.13 we see that I - A.K is regular for all I A. I < I I lim sup II K 0 11 11" n-;oo
and the solution function will be given by f =(I +
~~ A.Kn )cg);
that is, we have a series solution for the Volterra integral equation. In fact, we show that lim sup II K 0 II lin = 0 n-;oo
and so we have such a solution for all scalars A.. Now
IK(f)(x) I ~ (x-a) sup{ I k(x,t) II f(t) I : a~ t :S: x} :S: M II f lloo (x-a) where M =sup{ I k(x,t) I: a~ t ~ x, a~ x ~ b}. We prove by induction that
I K 0 (f)( X) I ~ M 0 II f lloo
(x-~)n n.
for all a :S: x
~
b:
The previous statement is case n = I. Suppose that the statement is true for some n EN. Then I Knt 1(f)(x) I =I K(K"f)(x) I =I
Jk(x,t) (K"f)(t) dt I a
~M
nt]
X
1: f lloo
n.
J(t-a)" dt
= Mnt-111 f II
nt-] (x-a) oo (n+l)!
which concludes the induction. Using this fact we obtain for all n EN II K 0 f II= max {I (K 0 f)(x) I : a~ x :S: b} ::;; M 0 II f lloo so
II Kn II = sup {II Kn II : II f II
~
I } ::;; Mn
(b-~)n . n.
(b-~)n n.
Spaces of continuous linear mappings
Since
(~) 11n ---7 0 n.
85
as n ---7 oo we conclude that lim sup II Kn 11 11n = 0. n-->oo
It follows that, for every g E 'C[a,b] and parameter 'A there exists a unique solution of the
Volterra integral equation given by the Neumann series
0 4.11.15 Remark. In AMS §5 we established the existence of a unique solution using Banach's Fixed Point Theorem and that theorem provided a similar way of finding a solution by iteration. 0 Corollary 4.11.11 tells us that in any unital Banach algebra, B(e; I) is an open ball of regular elements. As we might expect this implies a topological property for the set of regular elements. 4.11.16 Theorem. In any unital Banach algebra (A, 11·11), the set of regular elements is
open.
Proof. Given a regular element x E A andy E A, II e-x-Iy II= II x-I(x-y) II::;; II x-I 1111 x-y II. So from Corollary 4.11.11 we have that x-Iy is regular when II x-y II< 1111 x-I II. But in this casey= xx-Iy is itself regular being as it is, the product of regular elements x and x-Iy. Therefore, {yEA:IIx-yll< 1/llx-111} is a set of regular elements and this implies that the set of regular elements is open.
0
4.11.17 Corollary. In any unital Banach algebra the set of singular elements is closed. Another topological property relating to the regular elements is as follows. 4.11.18 Theorem. On a unital normed algebra (A, 11·11), the mapping x
H
x-1 is a
homeomorphism of the set of regular elements onto itself.
Proof. Since the mapping x
H
x-I is its own inverse mapping it is sufficient to prove
continuity. For regular elements x,y E A, II x-I- y-1 II= II x-I(x-y) y-1 II::;; II x-I 1111 y-1 1111 x-y II. II y-I II ::;; II x-I - y-I II+ II x-I II::;; II x-I 1111 y-I 1111 x-y II+ II x-I II. But So for all regular yEA where II x-y II< I/211 x-I II we have II y-111 < 2 II x-I II and so II x-I- y-1 II< 2 II x-I 11 2 II x-y II, which implies that the mapping x H x-I is continuous at x.
0
86
§4. Duals and operator algebras
4.12 EXERCISES I.
For the following continuous linear mappings T determine II T II. Given k E I 1,2, •.• , n}, the projection Pk: (IR n, 11·11 2) --7IR where (i) (ii) (iii)
2.
Pk0"1, A2, · · · , An) = Ak· T: (IR 3, 11·11 2) --7 (IR 11-11 2) where T (AI,~. A3) = (A3, A1, ~). Do: ('C 1[0,1], 11·11') --t!R where II f II'= II f lloo +II f' lloo and D 0(f) = f'(O).
J.
(iv)
the shiftS: (.t 1, 11·11 1) --7 (.t 1, 11-11 2) where S(IAI, A2o ···,An,··.})= 10, (AI, A2o · .. ,An··.}.
(i)
Given a sequence I a. 1, a.2, ... , a.n, ... } of scalars prove that la.1A 1, a. 2A2o ... , a.n An, ... }E .t 2 for all IA 1, ~ •... ,An, ... }E .t 2 if and only if I a. 1, a.2, ... , Un, ... } is bounded.
(ii)
A diagonal operator Ton (.t 2, 11-11 2) is defined by a bounded sequence of scalars I a. 1, a. 2, ... , a.n , ... } where for X= IA1, A2, ···,An,·· .}E .t2, Tx = la.1A1, U2A2, · · · • UnAn, · · .}.
(a) Prove that Tis continuous and II T II= sup {I Un I : n EN} . (b) Prove that T is a topological isomorphism onto if and only if (iii)
inf { I a.n I : n E N } > 0. Deduce that (m, IHioo) is isometrically isomorphic to a linear subspace of (~(.t 2 ),
3.
Given y
11·11) and that (~(.t 2 ), 11·11) is not separable.
={J.11,J.12, ... , lln• ... }
by fy(x) =
lA.n lin
E
.t 1 consider the functional fy defined on .t 1
for all x ={AI,~ •... , An, ... }
E
.tl.
(i)
Prove that fy is linear and continuous on .t 1 with respect to both the 11-11 2 and 11·11 1 norms.
(ii)
Determine II fy II in both (.t 1, 11·11 2) and (.t 1, 11·11 1) and find whether fy attains its norm on the unit sphere in each case.
4.
Consider the functional F defined on the real linear space 'C[-7t,7t] by 1t
F(f) =
Jf(t) sin t dt. -1t
(i)
Prove that F is linear and determine whether F is continuous on 'C [-7t,7t] with respect to norms ll·ll2, ll·lloo and 11·11 1.
(ii)
Wherever possible for 'C[-7t,7t] with norms 11·11 2, ll·lloo and 11·11 1 determine II F II in the corresponding dual space, and find whether F attains its norm on the unit sphere of the space.
Spaces of continuous linear mappings
5.
(i)
87
Consider a continuous linear mapping T of a Banach space (X, 11·11) into a normed linear space (Y, 11·11') where T(X) is dense in (Y, 11·11') and T-1 exists and is continuous on (T(X), 11·11'). Prove that T maps X onto Y and (Y, 11·11') is complete.
(ii)
Consider Banach spaces (X, 11·11) and (Y, 11·11') and a continuous linear mapping T from a dense linear subspace of (X, 11·11) onto a dense linear subspace of (Y, 11·11'). Given T, the unique continuous linear extension of T mapping (X, 11·11) into (Y, 11·11'), prove that (a) ifT is an isometric isomorphism then Tis an isometric isomorphism onto (Y, 11·11'), (b) if T is a topological isomorphism then T is a topological isomorphism onto (Y, 11·11').
(iii)
Deduce that (a) isometrically isomorphic normed linear spaces have isometrically isomorphic completions, (b) topologically isomorphic normed linear spaces have topologically isomorphic completions.
6.
A real trigonometric polynomial t on IR of degree n is of the form n
L
(ak sin k8 + bk cos k8) where ak, bk E IR fork E {0, 1, ... , n}. k=O Consider the real linear space 'J' n of trigonometric polynomials of degree less
t(S) =
than or equal to n and the linear operator D on 'J' n defined by D(t) = t'(S). (i)
Prove that Dis continuous on ('J' n• 11·11) for any norm 11·11 on 'J' n·
(ii)
For norm 11-11 2 on 'J' n where 11·11 2 is generated by the inner product 1t
(t, s) =
J ts(S)
de
-1t
(iii)
prove that II D II = n. For norm ll·lloo on 'J' n where II t lloo =max {I t(S) I : e prove that II D II
E
[-1t,1t]}
=n.
(This result was first proved by S. Bernstein about 1900. For an elementary proof see R.P. Boas, Math.Mag. 42 (1969), 165-174.) 7.
Consider ("C [0, 1], IHioo) and the set of continuous linear functionals {Px : x E [0, 1] where Px(f) = f(x) for all f E "C [0, 11}. (i) Prove that this set is contained in the unit sphere of ("C[O, 1], IHioo)*. (ii)
Show that ("C [0, 1], ll·lloo)* is not rotund.
88 8.
§4. Duals and operator algebras
(i)
Consider a normed linear space (X, 11·11) with a Schauder basis {en}. Prove that {en} is a monotone basis if and only if II T n II :::; 1 for all n EN where Tn is the finite rank operator defined for each =
x
=I
n
"-nen by T n(x) =
n=l
(ii)
I
A.kek .
k=l
Prove that if a Hilbert space has a countable orthonormal basis then it has a monotone Schauder basis.
9.
(i)
A linear space X has norms II· II and 11·11' where for some M > 0,
(ii)
Deduce that if 11·11 and 11·11' are equivalent norms for X then
II x II :::; M II x II' for all x EX. Prov.e that (X, II· II)* s;;; (X, 11·11')*. (X, 11·11)* = (X, 11·11')* and 11·11* and 11·11'*, norms on X* induced by 11·11 and 11·11', are equivalent norms on X*. (iii)
For t: [-1t,1t). prove that ('C:[-1t,1t), 11·111)* C ('C:[-1t,1t), 11·112)* C ('C:[-1t,1t), 11·11 00 )*.
*
10.
(i)
*
.
Given a linear space X with algebraic dual x#, prove that each x EX generates a linear functional "x on X# defined by "x(f) = f(x) for all f EX#.
(ii)
Hence, or otherwise prove that a linear space X is finite dimensional if
(iii)
and only if its algebraic dual x# is finite dimensional. A normed linear space (X, 11·11) has a finite dimensional dual space X*. Prove that X is finite dimensional.
11.
Prove that (.tJ, 11-iloo)* is isometrically isomorphic (co, 11·11 00 )* and that
(~((.t 1 , 11·11 (c 0 , ll·lloo)), 11·11) is isometrically isomorphic to (~(co, ll·lloo), 11·11). 00 ),
12.
(i)
Given a closed hyperplane M
={x EX: f(x) =A.} in a real normed
linear space (X, 11·11), and a point x0 EX, prove that I A. I
(a)
d(O, M)
= iTTTI ,
(b) d(x 0 , M)
I A.-f(x 0 ) I and II f II
(c) there exists a y0 EM such that II x0-y 0 II= d(x 0 , M) if and only if there (ii)
exists an x EX such that f(x) = II f II II x II. Given closed hyperplanes M 1 {x EX : f(x) = A. 1} and
=
M 2 = {x EX : f(x) = A.2} in (X, 11·11), prove that d(MI,M2) =
I A. 1-A. 2 I II f II
89
Spaces of continuous linear mappings
13.
Given a real normed linear space (X, 11·11), prove that a linear functional
(i)
f is continuous if and only if for any scalar A., the half-space { x EX : f(x) ~ A.} is closed. (ii)
Given an open ball B(x 0 ; r) contained in a closed half-space H
= {x EX: f(x)
~A.}
prove that
inf f(B(x 0 ; r))-A. f(x 0 )-'A d(B(xo; r), H) = II f II = -1-1f-1-1 - r.
14.
Given a nonempty subset A of a real normed linear space (X, 11·11), a hyperplane M
= { x EX : f(x) = A.} is called a hyperplane of support for A if A is contained in
one or other of the half-spaces { x EX : f(x) ~ A.} or {x EX : f(x) ~ A.} and there exists at least one x0 E A such that f(x 0) = A.. (i)
Given a compact set Kin a normed linear space (X, 11·11) and a closed hyperplane M
={x EX: f(x) = 0}, prove that there exists an x0 E K such
that d(x 0 , M) = d(K, M) and that
M 0 = {x EX: f(x) = f(x 0 )} is a hyperplane of support forK at x0 . (ii)
Consider the closed unit ball B[O; 1] in (X, 11·11). Prove that M 1 = {x EX: f(x) = f(x 1)} is a hyperplane of support forB[O; 1] at x 1 EB[O; 1], if and only if I f(x 1) I= II f 1111 x 1 II.
15.
Consider the functional f defined on (c0 , ll·lloo) for x =
by
f(x)
= {A. 1,A. 2, ...
, An, ... }
')...
= I. ~ k=l 2
(i)
Prove that f is linear and continuous and II f II = I, but show that f does not
(ii)
Show that for any x0 10M= {xE X: f(x) = 0} there is no closest point to x0
(iii)
Show that the closed unit ball has no hyperplane of support of the form
attain its norm on the unit sphere. inM. {x EX: f(x) =A.).
16.
Given a normed linear space (X, 11·11) and elements x EX and f EX*, an operator x ® f is defined on X by x ® f(y) = f(y) x (i)
for all y EX.
Prove that x ® f is a continuous linear operator on X and II X® f II= II X 1111 f II.
(ii)
Prove that every continuous linear operator T on X with n-dimensional range T(X) has the form T = x 1 ® f 1 + x2 ® f 2 + ... + xn ® fn where x 1,x 2 , ... , Xn EX and f 1,f2 , ... , fn EX*.
90
§4. Duals and operator algebras
(iii)
Prove that X is finite dimensional if and only if every linear operator on X is continuous.
17.
Given a normed linear space (X, 11·11) with proper closed linear subspace M, consider the quotient space XIM with norm 11·11': XIM ~ R defined by
II [x]ll' = inf {II x+m II: m EM}. (i)
(a) Prove that the quotient mapping 1t:
X~
XIM defined by
1t(x) = x + M is a continuous linear mapping and ll1t II = I. (b) Prove that 1t attains its norm on the closed unit ball of (X, II· II) if and only if for each x E X there exists an m E M such that
II x-m II = d(x, M) . (ii)
The linear subspace M is said to have codimension n in the linear space X if there exists a linearly independent set {e 1,e 2, ... , en} in X\ M such that each x EX can be represented in the form x
= A. 1e 1 + A.2e 2 + ... + Anen + z
for scalars {A. I> ~ •••. , An} and z EM. Prove that if M has codimension n in X then XIM has dimension n. (iii)
A linear functional f on X has M s;;; kerf. (a) Define a linear functional F on XIM by F(x+M) = f(x). Prove that F is continuous on (XIM, 11·11) if and only iff is continuous on (X, 11·11) and II F II = II f II. (b) Prove that if M has finite codimension in X then f is continuous on (X, 11·11).
18.
Given a normed linear space (X, 11·11), prove that 13(X) is a commutative algebra if and only if X is one dimensional. (Hint: If X has more than one dimension use Exercise 16 to construct noncommutative operators with one dimensional range.)
19.
Given a unital normed algebra (A, 11·11) consider the left regular representation map ·a H Ta of A into 13(A) defined by Ta(x) = ax
for all x EA.
Prove that (A, 11·11) is isometrically algebra isomorphic to a subalgebra of (13(A), 11·11) under left regular representation.
Spaces of continuous linear mappings
20.
Consider a unital Banach algebra (A, 11·11). (i)
Prove that if II x II < I then II (e-x)
(ii)
-1
-(e-x)
II~
2
II X 11 I=iTXTf .
Prove that if x E A is regular andy
E
A is such that II yx-1 II < I then
x-y is regular and (x-yr 1 = x- 1 +
f x-\yx- 1)k. k=l
21.
22.
Given a unital Banach algebra A, prove that (i)
if x and xy are regular then y is also regular,
(ii)
if xy and yx are regular then both x and y are regular,
(iii)
if xy
=e and yx =z * e then z2 =z where z * 0, e.
Solve the Volterra integral equations X
(i)
f(x)
= I + A. Jex-t f(t) dt
for 0
~
x
~I,
for 0
~
x ~I.
0 X
(ii)
f(x) = I + A.
J(x-t) f(t) dt 0
91
92
Spaces of continuous linear mappings
§5. THE SHAPE OF THE DUAL
To gain a firmer grasp of the concept of the dual of a normed linear space we need to examine the particular shape of the dual space for several example normed linear spaces.
5.1 Finite dimensional normed linear spaces Given a n-dimensionallinear space Xn over (C (or IR) with basis {e 1, e 2,
... ,
en} we recall from Theorem 4.10.5 that the algebraic dual Xn #is also a linear
space with basis {f1, f 2,
... ,
fnl dual to {e 1, e 2,
•.. ,
enl where
k}
whenj = j ;t k n
Furthermore, since f =
L
f(ek) fk,
k=l
x:
0
is isomorphic to u:n (or IRn) under the mapping
fH (f(e 1), f(e 2),
.•. ,
f(en)).
and every linear functional f on Xn is of the form n
f(x) =
L
n
L
"-k f(ek)
where x =
"-kiik
for some (a. 1, a. 2 ,
A.kek. k=l Being an isomorphism onto [n implies that every linear functional f on Xn has the form k=l n
f(x)
=L
... ,
a.n)
E
u:n
k=l n
where x
=L
A.kek.
k=l
X~ =
x:.
Now given any norm 11·11 on Xn, we have from Theorem 2.1.12 that However, the actual norm on X~ does depend on the norm given on Xn. To
obtain some idea of the form of the actual dual (Xn, 11·11)* we determine the corresponding norm 11·11 on u:n (or IR n) so that (
f(tk) (FCtk,n)- FCtk-l•n)) ·
k=2
But since for each k
E {
1, 2, ... , m}, 0
~
F(<j> 1k,n)- a(tk) <Elm, we have
m
I F(f0 ) -
L
f(tk) (a(tk)- a(tk_ 1)) I< 2 E II f II
k=l
using the fact that a(t0) = 0. m
But
L
f(tk) (a(tk)- a(tk_ 1)) is a Riemann-Stieltjes sum S(P, f, a) for f with respect to a k=l where Pis a partition which is a refinement of Pe. Therefore,
§5. The shape of the dual
107
I F(f)- S(P, f, a) I :-:; I F(f)- F(fo) I + I F(fo)- S(P, f, a) I < (II F II + 2 II f II) e and we conclude that b
F(f) =
Jf(t) da(t) .
D
We now extend this result to the representation of continuous linear functionals on the space. To do so we introduce the following ideas.
5.4.7 Definitions. Given a real function a on [a,b] and a partition P of [a,b], a= t0 < t 1 < ... < t 0 = b we write n
V(P, a)=
I.
I a(tk)- a(tk_ 1) I
k=l
The function a is said to be of bounded variation if the set of real numbers {V(P, a): P any partition of [a,b]} is bounded. If a is of bounded variation V(a) = sup{V(P, a): P any partition of [a,bl} is called the total variation of a on [a,b].
5.4.8 Remark. Clearly any bounded monotone function on [a,b] is of bounded variation. But further, functions of bounded variation have the following characterisation by monotone functions. A function a on [a,b] is of bounded variation if and only if a can be expressed as the difference of two monotone increasing functions; (see T.M. Apostol, Mathematical Analysis, p. 168). D We express our general representation theorem in terms of functions of bounded variation.
5.4.9. The Riesz Representation Theorem for ('C [a,b], 11·11 00 ) . For any continuous linear functional F on a real Banach space ('C[a.b], 11·11 00 ) , there exists a function a of bounded variation on [a,b] such that b
F(f) = and
Jf(t) da(t)
V(a) =II F II.
for all f E 'C[a,b],
108
Spaces of continuous linear mappings
Proof. From Lemma 5.4.5 there exist positive linear functionals F+ and F- on t; [a,b] such that F = F+ - F -0 By Theorem 5.406 there exist bounded monotone increasing functions a+ and a- on [a,b] such that b
F+(f) =
b
Jf(t) da+(t)
and F -(f)=
Jf(t) da-(t)
for all f
E
t;[a,b].
b
Then
Jf(t) d(a+- a-)
F(f) = P(f)- F -(f)=
by Remark 5.4o2(iii)(b)o But from Remark 5.408, a= a+- a- is of bounded variation and we have b
J f(t) da(t)
F(f) =
for all f E t; [a,b]o
a
Since F has this representation, given f [a,b], a= t0 < t 1 < o o o< tn
E
t; [a,b] and £ > 0 there exists a partition P of
=b such that n
I F(f)I F(f) I 0 consider a partition
P of [a,b], a= t0 < t 1 < 0 0 0 < tm = b such that m
I V(a)-
I.
I a(tk)- a(tk_ 1) I
E
{I, 2, 0 0 0, m}o
Consider the continuous function f0* on [a, b] defined by
*
fo(x) = Et,,n(x) +
m
I. Ek(tk,n- tk-I•n)(x)
k=2 where if = -1
*
Now F(f0 ) = e 1F(t 1,n) +
if m
I. Ek(F(tk,n)- F(tk-l•n))o
k=2
§5. The shape of the dual
So we have
*
I F(f0)-
*
I F(f0 ) -
then
m
L £k
109
(a(tk)- a(tk_ 1)) I< 2£ using the fact that a(t0) = 0,
k=l
m
L I (a(tk)- a(tk_ 1)) I< 2£ k=l
and we deduce that I F(f0*)- V(a) I< 3£. Therefore
V(a)-3£ [a,b], 11·11=)* and deduce that ('f'[a,b], ll·lloo)* is a Hilbert space. (Hint: Use Weierstrass' Approximation Theorem, AMS §9.)
9.
(i)
Given a bounded monotone increasing function a on [a,b] prove that the function F on (t'>[a,b ], 11·11 00 ) defined by b
F(f) =
Jf(t) da(t) a
is a positive linear functional and II F II= V(a). (ii)
Given a function a of bounded variation on [a,b] prove that the function F on (t: [a,b], ll·lloo) defined by b
F(f) =
Jf(t) da(t) a
is a continuous linear functional and II F II= V(a).
112
Spaces of continuous linear mappings
(iii)
Given a continuous function a on [a,b], prove that the function F on ('C [a,b], ll·lloo) defined by b
Jf(t) a(t) dt
F(f) =
a
b
is a continuous linear functional and II F II
= JI a(t) I dt. a
(iv)
For the linear functionals F 1 and F2 on ('C[a,b], ll·lloo) where I
F1 =
I
Jt2 f(t) dt and F2 = J(I -2t) f(t) dt 0
0
prove that both F 1 and F 2 are continuous and determine II F 1 II and II F 2 11.
III.
THE EXISTENCE FUNCTIONALS
OF
CONTINUOUS
LINEAR
Given any linear space X, it follows from the existence of a Hamel basis for X and the fact that any linear functional is determined by its values on the Hamel basis, that the algebraic dual X# is generally a "substantial" space. We know, from Remark 4.10.2, that for an infinite dimensional normed linear space (X, 11·11), the dual X* is a proper linear subspace of X#. For the development of a theory of normed linear spaces in general, quite apart from particular examples or classes of examples, it is important to know that given any normed linear space (X, 11·11), its dual X* is also "substantial enough" and by this we mean that we have a dual which generalises sufficiently the properties we are accustomed to associate with the dual of a Euclidean space or indeed, with the duals of the familiar example spaces. We now use the Axiom of Choice in the form of Zorn's Lemma, (see Appendix A. I), to prove the Hahn-Banach Theorem, an existence theorem which is crucial for the development of our general theory. The theorem assures us that for any nontrivial normed linear space there is always an adequate supply of continuous linear functionalii. The immediate application of this result is in the study of the structure of the second dual X** of a normed linear space (X, II· II) and of the relation between the space X and its duals X* and X**. §6. THE HAHN-BANACH THEOREM There are several forms of the Hahn-Banach Theorem, some more general than others. The form sufficient for our purpose asserts the existence of norm preserving extensions of continuous linear functionals from a linear subspace of a normed linear space to the whole space. It is an interesting exercise to generalise this form, (see Exercise 6.13.5), but we will make no use of the generalisation. In fact we mainly use our restricted form of the theorem through its corollaries. The proof of the Hahn-Banach Theorem is an application of Zorn's Lemma but we isolate the computational first step of that proof in the following lemma.
114
The existence of continuous linear functionals
6.1 Lemma. Given a normed linear space (X, 11·11), consider a continuous linear
functional f defined on a proper linear subspace M of X. Given x0 EX\ M, f can be extended to a continuous linear functional f on M0 = sp {x0 ,M} such that II f0 II on M0 is equal to II f II on M.
Proof. We may suppose that II f II = I.
Case 1: X a real linear space Since M is of codimension I in M0 , any x E M0 can be represented uniquely in the form where A. E lR andy EM .. x = A.x 0 + y For fo to be a linear extension off on M we must have fo(Y) = f(y)
for ally EM
f0 (x) = Af0(x0 ) + fo(Y)
and
= Af0 (x 0) + f(y)
and we are free to choose a value for f0(x 0). We show that a value for f0(x 0) can be chosen such that f0 is continuous on M0 and II f0 II= I. Now for f0 continuous on M0 , II f0 II = sup {I f(x) I : II x II ::; I, x E M0 }
~sup{ I f(y) I: II y II::; I, y EM}= II fil =I. But fo is continuous and II f0 II = I if I f0(x) I::; II x II
for all x E M0 .
Now this will happen if the value f0 (x0) satisfies - II A.x 0+y II ::; A.f0(x 0) + f(y)::; II A.x 0+y II; that is, - f (y/A-)- II x0 + y/A. II::; f 0(x0 )::;- f (y/A-) +II x0 + y!A. II
when A."# 0.
(*)
(Notice that when A.< 0, we have I I - f (y/A.) +- II h 0 + y II ::; f0(x 0)::;- f (y/A.) --II A.x 0 + y II A, A, - f (y/A-) -II - (x 0 + y!A.) II ::; f0 (x 0)
::; -
f (y/A.) +II - (x 0 + y/A.) II .)
Now for any tw0 elements y 1, y2 EM, we have But Therefore
f(y 2)- f(yJ) = f(YrYI)::; II YrYI II. II YrYI II::; II x0+y 2 II+ II x0+y 1 II. -f(y 1) -II x0+y 1 II::;- f(y2) +II xo+Y2II.
So
sup{-f(y)-11 x0+y II: y EM} and inf{- f(y) +II x0+y II : y EM}
and
both exist
sup{- f(y) -II x0+y II: y EM} ::; inf{- f(y) +II x0+y II : y EM} .
So it is possible to choose a value for f0 (x 0) between these two and then that value will satisfy inequality (*) which gives II f0 II = I as required.
115
§6. The Hahn-Banach Theorem
Case 2: X a complex linear space We have seen from Theorems 4.10.15 and 4.10.16 that a complex normed linear space (X, 11·11) can be regarded as a real normed linear space (XJR, 11·11) and there is a one-to-one norm preserving correspondence f H fiR of X* onto (XJR)* where f and fiR are related by f(x) =fiR (x)- i fiR (ix)
for x EX.
Now M 0 "' sp { x 0 , M} as a real normed linear space is M 0 JR = sp { x0 , ix 0 , M lR } . From Case I applied twice fiR on MJR can be extended to a continuous linear functional f0 JR on M 0 1R such that II fOJR II = I. But then f0 given by for all x EM 0 extends f on M as a continuous linear functional f0 on M 0 such that II f0 II = I.
0
The Hahn-Banach Theorem uses Zorn's Lemma to carry extensions, like those achieved in Lemma 6.1 for one extra dimension, to the whole space.
6.2 The Hahn-Banach Theorem. For a normed linear space (X, 11·11), consider a continuous linear functional f defined on a proper linear subspace M of X. Then f can be extended as a continuous linear functional fo on X such that II f0 II on X is equal to II f II on M.
Proof. Consider the set ~ of all possible extensions of f as norm preserving continuous linear functionals on linear subspaces containing M, with partial order relation :::; on ~: f 1 :::_ f2 if dom f 1 ~ dom f2 and f 2 is an extension of f 1. From Lemma 6.1, we see that ~ is nonempty. Suppose that lfa} is a totally ordered subset of~. We show that {fa} has an upper bound in ~. For x E U dom fw there exists an a 0 such that x
a
E
dom fa 0 , so we define a
functional fl on U dom fa by
a
f 1(x) =fa (x), 0
and since {fa} is a totally ordered set, f 1 is well defined. Now since {fa} is a totally ordered subset, dom f 1 = U dom fa is a linear subspace of X
a
containing M. But also from the definition we see that f 1 is a linear extension of fa for all a. As such an extension, sup {I f(x) I : II x II : II f II which contradicts the defining 0
0
property for elements of~ that II fa II = II f II. So we conclude that II f 1 II = II f II and then 0
f 1 E~.
116
The existence of continuous linear functionals
Since fa::; f 1 for all a, we have that f 1 is an upper bound for {fa} . It now follows from Zorn's Lemma that
~
has a maximal element f0 .
Now dom f0 = X, for otherwise by Lemma 6.1, for x0 EX\ dom f0 we could extend f0 as a norm preserving continuous linear functional on sp { x0, dom f0 } but then fo would not D be maximal in ~. The Hahn-Banach Theorem has the following important corollary which guarantees the existence of certain continuous linear functionals which attain their norm on the closed unit ball, or considered geometrically guarantees the existence of tangent hyperplanes to the closed unit ball. It is this first corollary rather than the theorem itself which we will use in the subsequent development of our analysis. Now the norm of a continuous linear functional f on a normed linear space (X, 11·11) is given by II f II =sup {I f(x) I : II x II::; I}. Given a continuous linear functional f there does not necessarily exist an x EX, II x II = I such that f(x) = II f II. However, the corollary tells us that given an x EX, II x II = I there does exist a continuous linear functional f such that f(x) = II f II. 6.3 Corollary. Given a normed linear space (X, 11·11), for each non-zero x0 EX, there exists a non-zero continuous linear function f0 on X such that f0(x 0) = II f0 1111 x0 II.
Proof. Consider the one dimensional linear subspace M"' sp{x 0 }. Define the functional f onMby f(A.x 0 ) = A. II x0 II
for scalar A..
Then f is a continuous linear functional on M such that f(x 0) =II x0 II and II f II = I. By the Hahn-Banach Theorem 6.2, f can be extended as a continuous linear functional f0 on X such that II f0 II = I. Then f0 on X satisfies f0(x 0) =II f0 1111 x0 II.
D
Let us explore the geometrical interpretation of this result. 6.4 Remark. Corollary 6.3 implies that given x0 EX, II x0 II = I there exists a continuous linear functional f0 on X, II f0 II = I such that f0 (x 0) = I. Geometrically this says that for each x0 EX on the boundary of the closed unit ball B [0; I] there exists a closed tangent hyperplane Mr0 "' { x EX: f0(x) = I} to B[O; I] at x0 where II f0 II= I. Clearly, I
d(O, Mr0) = ffTol1 = I = II x0 II and x0 E Mro· This is the sort of geometrical property we assume quite naturally in Euclidean space.
D
We now show how Corollary 6.3 implies that X* is "substantial" in another sense.
§6. The Hahn-Banach Theorem
117
6.5 Definitions. (i) Given a linear space X, a linear subspace Y of xlf is said to be total on X if for for x EX, f(x) = 0 for all fEY implies that x = 0; or contrapositively, for x "# 0 there exists an fEY such that f(x)
"#
0.
(ii) Given a linear space X, a linear subspace Y of X# is said to separate the points of X if for x,y EX, x "# y there exists an fEY such that f(x) "# f(y).
Clearly, the fact that Y is a linear subspace of X# implies that these properties are equivalent.
6.6 Remark. Given a normed linear space (X, 11·11), Corollary 6.3 implies that the dual X* is total on X or equivalently separates the points of X.
D
The second corollary to the Hahn Banach Theorem implies that in general there are sufficient continuous linear functionals on a normed linear space, not only to separate points, but also to separate points from proper closed linear subspaces. 6.7 Corollary. Given a normed linear space (X, 11·11), for any proper closed linear subspace M and xo EX \ M, there exists a continuous linear functional fo on X such that I fo(M) = 0, f0(x 0) = I and II fo II= d(xo, M)
Proof. Consider M0 = sp{x0 , M). Now any x E M0 has the form x = A.x 0 + y where A. is a scalar and y EM. Define a linear functional f on M0 by f(x) = A.. Then clearly f(M) = 0 and f(x 0) = 1. Since kerf= M which is closed, then f is continuous on M 0. But also, II f II = d(0,1Mr) . However, d(O, Mr)
=inf{ II x II : f(x) = I = f(xo)} = inf{ II x II : f(x-x0) = 0} = inf{ II x0-y II : f(y) = 0} =d(x0, M).
I Sollfll=d(xo, M).
By the Hahn-Banach Theorem there exists a continuous linear functional f 0 on X an extension of f on M0 such that 1
llf0 11=11fll=d(xo, M).
D
6.8 Remark. Corollary 6.7 implies that given a set A in a normed linear space (X, 11·11), a point x0 E spA if and only if for every continuous linear functional f on X where f(A) = 0 we have f(x 0) = 0. If for every continuous linear functional f on X where f(A) = 0 we have f(x 0) = 0 then d(x 0 , spA) = 0, for otherwise by Corollary 6.7 there exists a continuous linear functional f0 on X such that f0(spA) = 0 and f0(x 0) = 1. The converse is immediate. D
118
The existence of continuous linear functionals The Hahn-Banach Theorem can be used to reveal a great deal about the relation
between a normed linear space and its dual. To illustrate this we consider separability of the spaces. The separability of a normed linear space does not necessarily imply the separability of its dual. For example the Banach space (1 1, 11·11 1) is separable but its dual is isometrically isomorphic to (m, 11·11 00 ) which is not separable (see Example 1.25.3 and Exercise 1.26.16). However, Corollary 6.7 enables us to establish the following relation. 6.9 Theorem. A normed linear space (X, 11·11) is separable if its dual (X*, 11·11) is
separable.
Proof. Consider a countable set {fn EX*: n EN} dense in (X*, 11·11). For each n EN there exists an Xn EX, II Xn II ~ I such that I f 0 (xn) I ;::>:~II fn II. Consider M = sp { x 0 : n E N }. Suppose that M *X. Then for x0 EX\ M we have from Corollary 6.7 that there exists an f0 EX* such that f0(M) = 0 and f0(x0) 0. But then
*
I
2 II fn II II f0 II
and
~I
~II
fn(Xn) I =(fn- f0)(xn) I ~II f 0 - f0 II
fn- f0 II + II f 0 II
~
3 II f 0 - f0 II for all n EN.
But this contradicts the density of {fn EX* : n EN} in (X*, 11·11). So we conclude that X= sp { xn: n EN}; that is, (X, 11·11) is separable.
D
The Hahn-Banach Theorem is useful in determining the form of the dual for subspaces and quotient spaces of a normed linear space. 6.10 Definition. For a linear subspace M of a normed linear space (X, II· II) the
annihilator of M is the subset Mj_ = {f EX* : f(x)
=0
for all x EM}.
It is evident that Mj_ is always a closed linear subspace of (X*, II· II); (see Exercise
6.!3.8(i)). 6.11 Theorem. Consider a linear subspace M of a normed linear space (X, 11·11). (i) M* is isometrically isomorphic to X*!Mj_.
(ii) /fM is a closed linear subspace, then (XIM)* is isometrically isomorphic to Mj_.
Proof. (i)
Given f EM* we have from the Hahn-Banach Theorem 6.2 that there exists an
extension f0 EX* and II f0 II
= II f II.
Consider the mapping T: M*
--7
X*/Mj_ defined by T(f)
= fo + Mj_.
§6. The Hahn-Banach Theorem
119
If fb EX* is another extension off then f0 - fb E M_j_ and so this mapping is well defined. Clearly T is linear. Since each element in fo + M_j_ is an extension off,
II f II ::;; inf {II f0+g II : gEM_!_} = II f 0+M_!_ II = II T(f) II. But also since f 0 is a norm preserving extension off,
II f II = II f 0 II ~ inf {II f 0+g II : g EM_!_} = II f 0+M_!_ II = II T(f) II. So T is an isometric isomorphism. But since the restriction of any continuous linear functional on X is a continuous linear functional on M we conclude that T is onto. (ii)
Consider the quotient mapping rc:
X~
X/M defined by rc(x)
= x+M and the
mapping T: (XIM)* ~X* defined by T(h)
= h o rc.
Clearly T is linear. But also
I T(h)(x) I = I h(x+M) I::;; II h 1111 x+M 11·::;; II h 1111 so
II T(h) II::;; II h II
X
II
for all
X
EX
for all h E (XIM)* .
Ifx EM then T(h)(x) = h(x+M) = h(M) = 0 since M is the zero of X/M. Therefore T maps (XIM)* into M_j_. Further for f E M_j_ we can define unambiguously an hE (XIM)* by h(x+M)
= f(x).
Clearly, such an h is a linear functional on XIM and
I h(x+M) I = I f(x) I = I f(x+m) I ::;; II f 1111 x+m II =II f 1111 x+M II
for all mE M
for all x EX .
So h is continuous on XIM and II h II ::;; II f II. But then T maps (X!M)* onto M_j_. Moreover, T is one-to-one because if T(h) = 0 then h(x+M) = 0 for all x EM and so h = 0. So the mapping T can be expressed as T(h)(x) and then
=h(x+M) =f(x)
II h II::;; II f II= II T(h) II
for all x EX
for all hE (X!M)*.
We conclude that
II T(h) II = II h II for all h E (X!M)* soT is an isometric isomorphism of (XIM)* onto M_j_.
D
6.12 Remark. The proofs of Theorem 6.11 can be given using the techniques of conjugate mappings presented in Chapter 5, Section 12. D
120
The existence of continuous linear functionals
6.13 EXERCISES I. A normed linear space (X, 11·11) is said to be smooth if for each x EX, II x II = I, there exists only one continuous linear functional f on X where II f II = I and f(x) = I. (i) Show that (a) (IR 3 , 11·11 1) is not smooth (b) (c 0 , ll·lloo) is not smooth.
and
(ii) Prove that Hilbert space 2.
(i)
i~
smooth.
Prove that a normed linear space (X, 11·11) is rotund if and only if for every f EX* where II f II = f(x) = f(y) = I for x,y EX where II x II =II y II = I we have x =y.
(ii) Prove that (a) if X* is smooth then X is rotund
(b) if X* is rotund then X is smooth.
and 3.
Prove that a normed linear space (X, II· II) is smooth at x EX, II x II l.f
i!ffi.
II x+Ay II- II x II
A.---70
A
.
f
eXIStS Or a
II y
E
= I if and only
X,
and if it is smooth at x then this limit is f(y) where f EX*, II f II= I and f(x) =II x II. 4.
(i)
Given a normed linear space (X, 11·11), prove that for any x EX, II x II
= sup {I f(x) I : f EX*, II f II ~ I } .
(ii) A linear subspace Y of X* is said to be nanning if
II x II
= sup {I f(x) I : fEY, II f II ~ I } .
Prove that such a subspace Y is total on X. (iii) Prove that if a linear subspace Y of X* is dense in X* then Y is norming. But give an example of a linea~ subspace Y of X* which is norming but not dense in X*. 5.
A convex functional <jl on a linear space X is a real functional on X defined by
<jl((l-A)x+Ay) ~ (1-A) <jl(x) + A<jl(y) for all x,y EX and 0 ~ A~ I. Prove the more general form of the Hahn-Banach Theorem. Consider a convex functional <jl on a linear space X, a proper linear subspace M of X and a linear functional f on M such that
Re f(x)
~
(x)
for all x EM.
Then there exists a linear functional f 0 on X an extension off on M, such that
Re f0(x) 6.
~
<jl(x)
for all x EX.
Consider a normed linear space (X, 11·11), a proper closed linear subspace M of X and x0 EX \M.
§6. The Hahn-Banach Theorem (i)
121
Prove that there exists a continuous linear functional f on X such that II f II= I, f(M) = 0 and f(x 0) = d(x 0 , M).
(ii) Prove that there exists a closed hyperplane M0 containing M such that
d(x 0, M 0) = d(x 0, M). 7.
In the Banach space (m, ll·lloo), consider m 0 the smallest closed linear subspace of (m, 11·11 00 ) containing all sequences of the form
P-1, where x (i)
= {1.. 1,
Az-AJ, A3-Az, · · · , An-An-I, · · · l
Az, A3, ... , An, ... } Em.
Show that e = {I, I, I, ... } e m 0 , and prove that there exists a continuous linear functional f* on (m, ll·lloo) such that II f* II= I, f*(e) =I and f*(m 0) = 0.
(ii) The Banach limit of a bounded sequence x
={A1, A.z, ... , An, ... } is
defined by LIM An= f*(x). Prove that
8.
(i)
(a)
LIM An= LIM An+ I·
(b)
LIM An ~ 0 if An ~ 0 for all n E M.
(c)
LIM (aAn+~fln) =a LIM An+~ LIM fln for scalars a,~ andy= {f.L 1, flz, ... , fln, ... } Em.
An ~ lim sup An A.o =lim An·
(d)
lim inf An ~ LIM
(e)
If x E c then LIM
if An E lR for all n E M.
Given a nonempty subset M of a normed linear space (X, 11·11), prove that Mj_ is a closed linear subspace of (X*, 11·11).
(ii) Given a nonempty subset N of (X*, 11·11) we define the set Nj_ = {x EX: f(x) = 0 for all fEN}. Prove that Nj_ is a closed linear subspace of (X, 11·11). (iii) Prove that (Mj_)j_ = M if and only if M is a closed linear subspace of
(X, 11·11). (iv) Prove that N,;;;;; (N j_)j_ but by considering the linear space c0 as a closed linear subspace of (m, ll·lloo) show that j_ (co j_) "' co. (v) Prove that if N is a finite dimensional linear subspace of (X*, 11·11) then N = (N j_)j_. 9.
Given a linear subspace M of a normed linear space (X, II· II), prove that for any fEX*,
d(f,Mj_)=llfiMII
and there exists an f0 E Mj_ such that II f-f0 II
= d(f, M_j_).
122
§7. The natural embedding and reflexivity
§7. THE NATURAL EMBEDDING AND REFLEXIVITY
It is Corollary 6.3 to the Hahn-Banach Theorem which enables us to develop a significant theory of dual spaces and to define the important class of reflexive spaces. 7.1 Definitions. Given a normed linear space (X, 11·11) and its dual X* with norm 11·11 defined by II f II = sup {I f(x) I : II x II ~ I}. the dual space (X*, 11·11) has its own dual (X*)* usually written X** and called the second
dual space (or second conjugate space) of (X, 11·11). We will usually denote elements of X
by x,y,z, elements of X* by f,g,h and elements of X** by F,G,H. The norm of X** is of course defined by II F II = sup {I F(f) I : II f II ~ I}.
The definition of the second dual (X**, 11·11) prompts us to enquire into its relation to the original space (X, 11·11) from which it is generated. For elements x EX and f EX*, we can consider f fixed and x varying over X as we do when we think off as a functional on X. But alternatively, we can consider x fixed and f varying over X*, and when we do this we have x acting as a functional on X*. 7.2 Theorem. Given a normed linear space (X, 11·11) and x EX, the functional~ defined by
~(f)= f(x)
for all f EX*
is a continuous linear functional on X* and~ as an element of(X**, 11·11) satisfies
11~11 =llxll. Proof. Clearly, ~is linear: ~ (f+g) = (f+g)(x) = f(x) + g(x) =~(f)+~ (g) for f, g EX* and
~(at)= uf(x) = u~ (f)
for scalar a and f EX*,
using the fact that in X* addition and multiplication by a scalar are defined pointwise. But also~ is continuous since I ~ (f) I = I f(x) I ~ II x 1111 f II
for all f EX*.
As a continuous linear functional on X* we have that~ EX**. But II~ II = sup {I f(x) I : II f II ~ I} ~II x II. However, by Corollary 6.3 applied to (X, 11·11), for x EX there exists an f EX* such that II f II = I and f(x) = II x II, so that II QII =sup {I f(x) I : II f II ~ I} ~II xll.
Therefore, II ~ II = II x II.
0
The existence of continuous linear functionals
123
Given a nonned linear space (X, II· II) and x eX the identification given in Theorem 7.2 of
Q
as an element of X** suggests that we investigate the mapping
x f--t Q of X into X**. 7.3 Theorem. Given a normed linear space (X, 11·11), the mapping x f--t Qinduced by the
definition of
Qas A
x(f) = f(x)
for all f eX*
is an isometric isomorphism ofX into X**.
Proof. The mapping x f--t Qis linear: 1\ A A AA x+y(f) = f(x+y) = f(x) + f(y) = x(f) + y(f) = (x+y)(f) for all f eX* 1\ A A x+y = x +y. so 1\ A ax(f) = f( ax) = af(x) = ax(f) for all scalar a and f eX* Also 1\ A ax= ax so using the fact that f is linear. From Theorem 7.2 we have that II QII = II x II for all x eX so we conclude that the mapping x f--t Qis an isometric isomorphism.
0
This identification of the original space (X, 11·11) as part of the second dual (X**, 11·11) suggests that we give a name to this mapping. f--t Q of X into X** induced by the definition of Qas Q(f) = f(x) for all f eX*, is called the natural
7.4 Definition. Given a normed linear space (X, 11·11), the mapping x A
embedding of X into X**. We denote by X the image of X in X** under the natural embedding.
Figure 11. The natural embedding x f--t
Qof X into X**.
Now it will be important to determine whether a normed linear space is isometrically isomorphic to its second dual under the natural embedding.
124
§7. The natural embedding and reflexivity
7.5 Definition. A normed linear space (X, 11·11) where the natural embedding x H ~ maps X onto X** is said to be reflexive. 7.6 Remark. Since a dual space is always complete we deduce that a normed linear space (X, 11·11) is isometrically isomorphic to its second dual (X**, 11·11) only if (X, 11·11) is a Banach space. So completeness is a necessary condition for a normed linear space to be 0 reflexive. But it is not a sufficient condition as is shown in the examples below. 7.7 Examples of reflexive spaces We now give some examples of classes of normed linear spaces which are reflexive. 7. 7 .I Finite dimensional normed linear spaces. A finite dimensional linear space is algebraically reflexive. Given an n-dimensionallinear space Xn we have from Theorem
4.1 0.5 that the algebraic dual X~ is n-dimensional and so the second algebraic dual X~# is also n-dimensional. Now the mapping x H ~of Xn into X~ induced by the definition of~ as
~(f)= f(x)
for all f EX~
is linear. But again from Theorem 4.1 0.5 we see that X~ is total on X 0 and this implies A
that if ~(f)= 0 for all f EX~ then x = 0, so the mapping x H ~is one-to-one. But then Xn, A
the image under this mapping, is n-dimensional so X0 = X~#· For an n-dimensional normed linear space (Xn, 11·11) we have
x:
=
X~ so it is clear that (Xn, II· II) is reflexive. 0
7.7.2 Hilbert space. From the Riesz Representation Theorem 5.2.1 we have that for any given continuous linear functional f on a Hilbert space H there exists a unique z E H such that f is of the form f(x) = (x, z)
for all x E H
and we denote this functional by fz. Now the dual space H* is itself a Hilbert space with inner product defined by (fx, fz) = (z, x) for all x,z E H; (see Exercise 5.5.7). Applying the Riesz Representation Theorem 5.2.1 to H* we have for any given continuous linear functional F on H* there exists a unique fx E H* such that F is of the form F(fz) = (fz. fx) = (x, z)
for all fz E H* for all z EH.
But
~(fz) = fz(x) = (x, z)
so
F(f) = ~(f)
that is,
F =~
and we conclude that
H = H**.
for all z E H
for all f E H* ;
A
0
The existence of continuous linear functionals
7.7.3 The (.lp, ll·llp) spaces where I < p
... } e lq and {1.. 1, 1..2 , ••• , A.k, ... } e lp.
So { F(f 1), F(f2), ••• , F(fk), ... } e lp. But then A
F(f) = x(f) for all f e l F= Q A ** and we conclude that lp = l P .
**
P
where x = { F(f1), F(f2 ),
.•.
F(fk), ... } e lp;
that is,
O
7.8 Remark. It should be noted that it is not sufficient for reflexivity that a Banach space (X, 11·11) be isometrically isomorphic to (X**, 11·11). R.C. James, Proc. Nat. Acad. Sci.
USA 37 ( 1951 ), 174.--177, has given an example of a nonreflexive Banach space with just this property. For reflexivity we must have (X, 11·11) isometrically isomorphic to (X**, 11·11) under the natural embedding. That is why in Examples 7.7.2 and 7.7.3, although it is obvious from Example 5.3.3 that (.lp, ll·llp) is isometrically isomorphic to (lp, ll·llp)** and from Exercise 5.5.7 that a Hilbert space H is isometrically isomorphic to H**, yet for reflexivity we need to establish the isometric isomorphism under the natural embedding and this requires a little more careful computation. 0
126
§7. The natural embedding and reflexivity
7.9 Techniques to prove nonreflexivity To prove nonreflexivity directly we have to show that the natural embedding is into but not onto. Such computation usually involves a knowledge of the form of the dual and the form of the second dual of the space and in other than a few cases this can be quite a complication.
So the nonreflexivity of a normed linear space is usually established by
indirect argument. It is clear that if a normed linear space is not complete then it is not reflexive. But we can sometimes quickly determine non-reflexivity by an appeal to separability. 7.9.1 Theorem. A separable Banach space (X, II· II) with a nonseparable dual
(X*, 11·11) is not reflexive. Proof. It follows from Theorem 6.9 that the second dual (X**, 11·11) cannot be separable. But then (X, 11·11) and (X**, 11·11) cannot be isometrically isomorphic, (see Exercise 1.26.15). D 7.9.2 Example. The Banach space (c 0 , ll·lloo) is separable, (see Example 1.25.2(ii)). But from Example 5.3.1, (c 0, 11·11 00 )* is isometrically isomorphic to (1 1, 11·11 1) and in Exercise 5.5.3 we prove that (1 1, 11·11 1)* is isometrically isomorphic to (m, ll·lloo). So (c 0, 11·11)** is isometrically isomorphic to (m, 11·11 00 ). But (m, ll·lloo) is not separable, (see Example 1.25.3). Therefore we can conclude that (c0 , 11·11 00 ) is not reflexive.
D
A more generally applicable indirect method is a consequence of the following important property of reflexive spaces. 7.9.3 Theorem. On a reflexive normed linear space (X, 11·11) every continuous linear
functional f attains its norm on the closed unit ball of(X, II· II).
Proof. Applying Corollary 6.3 to (X*, 11·11) we have that for any given f EX* there exists a continuous linear functional F on X* such that F(f) = II f II II F II. But we are given that II
X= X**, so F =Q Then
for some x EX.
f(x) = Q(f) =II f 1111 x II.
D
7.9.4 Remark. It follows from Theorem 7.9.3 that if a Banach space has a continuous linear functional which does not attain its norm on the closed unit ball then the space is not reflexive. This provides a technique for proving nonreflexivity which does not even D necessitate a knowledge of the form of the dual space.
127
The existence of continuous linear functionals
7.9.5 Example. Consider the Banach space ('C[-1t,1t], 11·11
00 ) .
We exhibit a continuous
linear functional on the space which does not attain its norm on the closed unit ball, (see Exercise 4. 12.4 ). Consider the linear functional F defined by 1t
F(f)
Jf(t) sin t dt .
=
-1t
Now
_[1 sin t I dt}fll~
IF(f)l$;(
forallfE'C[-1t,1t].
1t
JI sin t I dt = 4. If we choose f
So F is continuous and II F II$;
0
on [-1t,1t] defined by
-1t
$; t < 0 }· Q$;t$;1t
f 0 (t) = -1 I
-7t
1t
Then
F(f0)
= JI sin t I dt = 4 -1t
and since II fo lloo = I we would have II F II = 4 and F attain its norm at f0 . But fo,; 'C[-1t,1t]. However, we can modify f0 to show that II F II= 4.
-1t
Figure 12. A modification of f0 to give a sequence { fn} in 'C [-1t,1t] so that F(fn)
--7
4 as n
--7
oo.
We define a sequence {fn} in 'C[-1t,1t] by I
$;t 0 there exists an x0 E B where f(x 0) > 0 such that k II x0-u II :5 f(x 0)- f(u) and f(y)- f(x 0) < k II y-x0 II for all y E B, y -T- x0 . Proof. We define a sequence {x0 } in B inductively as follows. Choose x 1 = u. For each n EN consider the set
={y E B : f(y)- f(xn) > K II y-xn II}. If sn = 0, write xn+l = xn and if sn "* 0, choose xn+l E sn such that sn
I
f(Xn+l) ~ 2 (f(x 0 ) +Sup{ f(x) : X E S0 } ). For the sequence {x0 } so defined k II X0 -x 0 _J II < f(x 0 ) - f(x 0 _J) so
for all n EN for all m > n.
(i)
(ii)
134
The existence of continuous linear functionals
Now the sequence { f(xn)} is increasing and is bounded above, so it is convergent. But then this implies that the sequence {xn} is Cauchy. Since (X, 11·11) is complete, the sequence {xn} is convergent to some x 0 EB. Since f is continuous at x0, inequality (ii) implies that k II xo-xn II ~ f(xo)- f(xn) and in particular
k II x0-u II
~
for all n E :f::I
f(x 0)- f(u).
Suppose there exists avE B, v oF- x0 such that f(v) Then
~
f(x 0) + K II v-x 0 II.
f(v) > f(x 0) = lim f(xn) n---;~
since f is continuous at x0 . But then v E Sn for all n E :f::I which implies that 2 f(xn+t)- f(xn) ~ sup{f(x) : x E Sn} ~ f(v) and so
f(x 0 )
~
f(v).
But this is a contradiction so f(y)- f(x 0) < k II y-x 0 II for ally E B, y
oF-
x0 .
0
8.5 Theorem. For a real Banach space (X, 11·11), given f EX*, II f II = I and E > 0 there exists
a g EX*, II g II = I and x 0 EX, II x 0 II = I such that g(x 0) = I and I g(x) I ~ E
Proof.
for x E kerf and II x II
~
I.
We need to find agE X*, II g II= I and x0 E K, II x0 II= I such that g(x 0 ) = I
and I g(z) I~ I for all z E T
= { x EX : x E kerf and II x II ~ _!_}. E
Consider K
= co {B,
From Lemma 8.3, K is an equivalent norm ball which contains B. If x0 E B
n bdy K, from Corollary 6.3 we see that there exists agE X*, such that g(x 0) = sup {I g(x) I : x E K} ~ {I g(x) I : x E B}
= II g II ~ g(x0).
Figure 14. K is to one side of the hyperplane {x EX : g(x)
= I}.
T}.
§8. Subreflexivity So to establish the theorem we need to show that B
n bdy K
135
* 0.
Consider u E B but u il bdy K. We may consider f(u) > 0. There exists an a > I such that au E K. Then there exists 0 < 'A < I such that au= 'Ax+ (1-'A)z So
for some x EB and z ET.
x-u = (a-l)u + (1-'A)(x-z) II x-u II :(a-A.) f(u);:: II x-u II f(u; . 1+-
(i)
E
Figure 15. u E B but u il bdy K and au E K for a> I. . f(u) h . In Lemma 8.4, choosmg k = - 1 we ave that there ex1sts an x0 E B where f(x 0) > 0,
1+E
such that
f(u; II x0-u II : 0 such that oB' ~ T(B). For y E oB' we have -0 y E T(B) and so there exists an x 1 E B such that y 1 = Tx 1 and II y-y 1 II < Now
z·
~ B' ~ T(kB) so there exists an x2 Ek B such that y2 = Tx 2 and II (y-y 1)- y 2 II 0 such that x + rB
~G.
Now Haire's Theorem 9.11 gives us that (Y, 11·11') is second category so from Lemmas 10.5 and 10.6 we deduce that there exists a o > 0 such that oB' s;;;; r T(B). Now y + oB' s;;;; y+rT(B) = T(x+rB) s;;;; T(G) soy E int T(G). This implies that T(G) is open in (Y, 11·11').
D
10.8 Remark. We note that Theorem 10.7 could be generalised to have range space (Y, 11·11') a normed linear space of second category. However, we have given the most
common form of the statement of the theorem and it is generally in this context where it has application. D
!56
The fundamental mapping theorems
Most applications of the Open Mapping Theorem are derived from the following corollary which gives a simple criterion for a continuous linear mapping to be a homeomorphism. 10.9 Corollary. A continuous linear one-to-one mapping T of a Banach space (X, 11·11)
onto a Banach space (Y, 11·11') is a topological isomorphism.
Proof. As Tis one-to-one and onto its inverse T-1 exists and is a linear mapping from (Y, II· II') onto (X, 11·11). From the Open Mapping Theorem 10.7, Tis an open mapping so it is also a homeomorphism. 0 Corollary 10.9 has a particularly useful application in determining whether norms are equivalent. 10.10 Corollary. For a linear space X, ifll·ll and 11·11' are norms such that both (X, 11·11) and (X, II· II') are complete and there exists a K > 0 such that
II x II $; K II x II' for all x EX then 11·11 and 11·11' are equivalent norms for X.
Proof. Consider the identity mapping id: (X, 11·11')
-7 (X,
11·11). The inequality
II x II $; K II x II' for all x E X implies that id is continuous. But id is linear one-to-one and onto so by Corollary 10.9, it is a topological isomorphism which implies that 11·11 and 11·11' are equivalent norms for X. 0 The second mapping theorem which is in many cases easier to apply than the Open Mapping Theorem is the Closed Graph Theorem and its proof can be derived directly from the Open Mapping Theorem. 10.11 Definitions. Given metric spaces (X, d) and (Y, d') the product metric dn for
X x Y is defined by dn((x,y), (x',y')) = max { d(x, x'), d'(y, y')} and we call (X x Y, dn) the product space of (X, d) and (Y, d'). It is easy to see that a sequence {(Xn.Yn)} is convergent to (x,y) in (X x Y, dn) if and only if (xnl is convergent to x in (X, d) and IYnl is convergent toy in (Y, d'). Given normed linear spaces (X, 11·11) and (Y, 11·11') over the same scalar field, X x Y is a linear space, the product norm 11-iln for X x Y is defined by II (x,y) lin= max {II x II, II y II'}
§ 10. The Open Mapping Theorem
157
and we call (X x Y, ll·llrr) the product space of (X, 11·11) and (Y, 11·11'). The product norm generates the product metric. Given a mapping T of a set X into a set Y the graph ofT is the subset GT of X x Y defined by GT
={(x,y) : y = Tx, x EX}.
When X andY are linear spaces over the same scalar field then Tis linear if and only if GT is a linear subspace of X x Y. We say that a mapping T of a metric space (X, d) into a metric space (Y, d') has a closed graph if GT is closed in (X x Y, drr). For mappings between metric spaces there is a close relation between continuity and having closed graph. We will explore this relation but first we give a more convenient way of expressing the fact that a mapping has a closed graph. 10.12 Lemma. A mapping T of a metric space (X, d) into a metric space (Y, d') has a closed graph if and only if for every sequence {xnl in X where {xnl is convergent toxin (X, d) and {Txnl is convergent toy in (Y, d'), we have y = Tx.
Proof. Suppose that GT is closed in (X x Y, drr) and that {Xn} is convergent to x in (X, d) and {Txn} is convergent toy in (Y, d'). Then { (xn,TXn)} is convergent to (x,y) in (X x Y, drr) . But since GT is closed, (x,y)
E
GT and therefore y = Tx.
Conversely, if (x,y) is a cluster point of GT in (X x Y, drr) then there exists a sequence { (xn,TXn)} in GT which is convergent to (x,y) in (X x Y, drr)· Then {Xn} is convergent to x in (X, d) and {Txnl is convergent toy in (Y, d').
D
From Lemma I 0.12 it is clear that any continuous mapping between metric spaces always has a closed graph. But further, it is not difficult to prove that a mapping T from a metric space (X, d) into a compact metric space (Y, d') with closed graph is continuous, (see AMS, §8). It is reasonable to ask whether there is a simple criterion for a linear mapping between normed linear spaces with a closed graph, to be continuous. We should note that not every linear mapping between normed linear spaces with closed graph is necessarily continuous. 10.13 Example. The differential operator D from ("CI[O,l], ll·lloo) into ("C[O,I], ll·lloo) defined by D(f) = f' has a closed graph; (this is the classical uniform convergence theorem for differentiation, see AMS §8). However, D is not continuous, We note that ("C [0, I], ll·lloo) is complete but ("C 1[0, I], ll·lloo) is not complete, (see AMS, §4).
D
!58
The fundamental mapping theorems The Closed Graph Theorem provides a criterion for the continuity of linear mappings
between normed linear spaces. Its proof is an elegant application of the Open Mapping Theorem. 10.14 The Closed Graph Theorem. A linear mapping T of a Banach space (X, 11·11) into a Banach space (Y, II· II') with a closed
graph, is continuous. Proof. Consider X renormed with norm 11·11 1 defined by II x 11 1 =II x II+ II Tx II'. Then
II Tx
II'~
II x II+ II Tx II'= II x 11 1 for all x EX soT is a continuous linear mapping from (X, 11·11 1) into (Y, 11·11').
We show that 11·11 1 and II· II are equivalent norms for X. llxll~llxii+IITxll'=llxll 1
Now
forallxEX
so if we show that (X, 11·11 1) is complete then from Corollary I 0.10 to the Open Mapping Theorem we will have that 11·11 and 11·11' are equivalent. We show that (X, 11·11 1) is complete. Consider a Cauchy sequence {xnl in (X, 11·11 1). Then since II Xm-Xn 11, =II Xm-Xn II+ II Txm-TXn II' we have that {xnl is a Cauchy sequence in (X, 11·11) and {Txnl is a Cauchy sequence in (Y, 11·11'). As both (X, 11·11) and (Y, 11·11') are complete there exist an x EX such that {xnl is convergent x in (X, 11·11) and ayE Y such that {Txnl is convergent toy in (Y, 11·11'). ButT has closed graph so y = Tx. Then II Xn-X 11 1 =II Xn-X II+ II Txn-Tx II' so {xnl is convergent toxin (X, 11·11 1), and we conclude that (X, 11·11 1) is complete.
10.15 EXERCISES I.
Consider an open mapping T of a metric space (X, d) into a metric space (Y, d'). (i)
Prove that T maps compact sets in (X, d) to compact sets in (Y, d').
(ii) Show that T does not necessarily map closed sets in (X, d) to closed sets in (Y, d').
2.
(i)
Consider the Banach space ('C [0, 1], ll·lloo) and 'C [0, I] with norm 11·11 1 and norm 11·11 2. Deduce from the Open Mapping Theorem that ('C[O,l], 11·11 1) and ('C[O,l], 11·11 2 ) are not complete.
(ii) A mapping T on 'C [0, I] is defined by t
T f(t) =
Jf(s) 0
ds.
0
159
§10. The Open Mapping Theorem
(a)
Prove that Tis a continuous one-to-one mapping of (t: [0, 1], ll·lloo) onto the linear subspace t:6 [0, I]
(b)
Show that
T- 1 is
= { f e t: 1[0, I]
: f(O) = 0}.
not continuous and explain why this does not contradict
the Open Mapping Theorem. 3.
M and N are closed linear subspaces of a Banach space (X, 11·11) such that X= M EB N. Prove that the norm 11·11' on X defined for z eX where z = x + y, x eM and y eN by
II z II' =II x II + II y II is an equivalent norm for X. 4.
Consider the finite dimensional linear space Xm over lR with basis {e 1, e2 , ... , em} and any norm on Xm. Assuming that (Xm, 11·11) is second category and using the Open Mapping Theorem with the mapping X f-7
(A('
Az'
0
°
0
'
Am)
of Xm into lR m where x = A1e 1 + A2e 2 + ... + A.mem, prove that (Xm, 11·11) is topologically isomorphic to (IRm, 11·11 2). 5.
(i)
Consider a linear mapping T of a normed linear space (X, 11·11) into a normed linear space (Y, 11·11'). (a) Prove that if T has closed graph then T has closed kernel. (b) But show that a linear mapping with a closed kernel does not necessarily have a closed graph; (see AMS §7).
(ii) (a) Prove that a linear functional on a normed linear space is continuous if it
has a closed graph. (b) Show that a linear mapping of a normed linear space into a finite dimensional normed linear space with closed graph is not necessarily continuous. 6.
For the mapping T of (t: 1[0,1], ll·lloo) into (t:[O,I], ll·lloo) defined by T(f)
=f' + f
prove that (i)
T has closed graph,
(ii) T is not continuous, (iii) (t: 1[0, I], ll·lloo) is not complete.
160
7.
The fundamental mapping theorems
A linear space X is a Banach space with respect to both norms 11·11 and II· II' and has the property that if a sequence {Xn} in X is convergent with respect to both norms 11·11 and II· II' then the limit point is unique. Prove that norms 11·11 and 11·11' are equivalent.
8.
(i)
For a linear mapping T of a Banach space (X, 11·11) into a Banach space (X, 11·11'), prove that iff o T EX* for all fEY* then T is continuous.
(ii) For linear operators T and S on a Hilbert space H we have (Tx, y)
= (x, Sy) for all x,y E H.
Prove that both T and S are continuous. (iii) For a linear operator T on a Hilbert space H, prove that the functional n0, so I f0 (Xo) IS: no II x0 lloo for all n EN;
D
164
The fundamental mapping theorems
that is, {fn: n EN} is pointwise bounded. Nevertheless, II fn II = n for every n E N, so {fn: n EN} is not uniformly bounded. But notice that the normed linear space
(E 0 , 11·11
00 )
D
is not complete.
On the other hand, for a set of continuous linear mappings on a Banach space, pointwise boundedness does imply uniform boundedness. This is the content of the Uniform Roundedness Theorem. This theorem, like the Open Mapping Theorem uses Baire category arguments and in particular Haire's Theorem 9.11 that a complete normed linear space is second category. 11.4 The Uniform Roundedness Theorem.
If a set 'J' of continuous linear mappings from a Banach space (X, 11·11) into a normed linear space (Y, 11·11') is pointwise bounded then it is uniformly bounded.
Proof. For each n EN, write Fn = { x EX : II Tx II' : 0 such that
II T n II ~ M
for all n e N.
II T 0 (x) II' ~ M II x II
Therefore, But for each x eX,
II Tx II'
~
for all n e N and x eX.
II Tx-Tn(x) II'+ II Tn(x) II' ,
so as {Tn} is pointwise convergent to T we conclude that
II Tx II'
~
M II x II
for all x eX;
that is, T is continuous.
0
However, the Banach-Steinhaus Theorem does not imply that pointwise convergence of a sequence of continuous linear mappings on a Banach space is any stronger than pointwise convergence; it does not provide a generalisation of Dini's Theorem. 11.10 Example. Consider the Banach space (c0 , 11·11 00 ) and the sequence of continuous
linear functionals {fnl on c 0 where for x =
p. 1, ~ •.•• , A.n •... },
fn(X) =An. Now {fn} is pointwise convergent to the zero functional. From the Uniform Boundedness Theorem 11.4, the set {fn : n e N} is uniformly bounded. But for en= {0, ... , 0, 1, 0, ... }, nth place
II fn II ~ f 0 (en) = 1 so {fnl is not convergent to the zero functional under the dual norm on (c0 , 11-iloo).
0
Another important application of the Uniform Boundedness Theorem is in characterising the boundedness of a set in a normed linear space. 11.11 Definitions. A nonempty set A in a normed linear space (X, 11·11) is said to be weakly bounded if f(A) is a bounded set of scalars for each f eX*. 1\
A nonempty set B in the dual (X*, 11·11) is said to be weak * bounded if x(B) is a bounded set of scalars for each x e X. There are occasions when we deduce the boundedness of a set from its weak boundedness or weak* boundedness, which is sometimes easier to test. 11.12. Theorem. (i) A nonempty set A in a normed linear space (X, 11·11) is bounded if and only if it is weakly
bounded. (ii) lf(X, 11·11) is complete, a nonempty set Bin the dual (X*, 11·11) is bounded it is weak
* bounded.
if and only if
167
§II. The Uniform Boundedness Theorem
Proof. For any x EX and f EX*, I f(x) I
~
II f 1111 x II.
So if A is bounded in (X, 11·11) then it is weakly bounded and if B is bounded in (X*, 11·11) then it is weak * bounded. Conversely, A
(i)
f(A)
= {f(x) : x E A} = {x(f) : x E A} is bounded for each f EX*; that is, the set
{ ~ : x E A} is pointwise bounded on (X, 11·11). But (X*,II·II) is always complete so from the Uniform Roundedness Theorem, the set A
{ x: x E A} is uniformly bounded on (X*, 11·11), which implies that A is bounded in (X, 11·11). A
(ii)
x(B)
={f(x) : x E B} is bounded for each x EX; that is the set {f: fEB} is pointwise
bounded on (X, 11·11). But here we assume that (X, 11·11) is complete so from the Uniform Roundedness Theorem the set {f: fEB} is uniformly bounded on (X, 11·11), which implies that B is bounded in
0
(X*, 11·11). 11.13
Remark. In Example 11.3 we have a set in the dual of an incomplete normed linear
space which is weak * bounded but not bounded. So the completeness condition in Theorem 0 I 1.12(ii) is significant. Another application of the Uniform Roundedness Theorem concerns bilinear mappings. 11.14 Definitions. Given linear spaces X, Y and Z over the same scalar field, a mapping
p of X x Y into Z is said to be a bilinear mapping if for given x EX, the associated mapping p,: Y ~ Z defined by p,(y) = P(x,y) and for given y E Y, the associated mapping Py: Y ~ Z defined by Py(Y) = P(x,y) are both linear. If (X, 11·11), (Y, II· II') and (Z, II· II") are normed linear spaces and for given x EX, Px is
continuous and for given y E Y, Py is continuous then p is said to be separately continuous. If p is continuous on X x Y with the product norm then p is said to be jointly continuous. If a bilinear mapping is jointly continuous then clearly it is separately continuous. The Uniform Roundedness Theorem enables us to establish a converse result. ll.IS Theorem. Consider normed linear spaces (X, 11·11), (Y, 11·11') and (Z, 11·11") and a
bilinear mapping then
p of X x Y into Z which is separately continuous.
p is jointly continuous.
If (X, 11·11) is complete
168
The fundamental mapping theorems
Proof. For each y E Y, II y II = I consider the associated continuous linear mapping
~Y
of
(X, 11·11) into (Z, 11·11"). Then there exists a Ky > 0 such that II ~y(x) II" :;::; Ky II x II But then the set { ~Y : y E Y, II y II
for all x EX.
= I } is pointwise bounded on (X, 11·11).
Since (X, 11·11) is
complete, by the Uniform Boundedness Theorem the set is uniformly bounded on (X, 11·11); that is, there exists a K > 0 such that II ~y(x) II" :;::; Kll x II for all x EX and y E Y, II y II= I. But since ~is bilinear this implies that II ~(x,y) II" :;::; Kll x 1111 y II for all x EX and y E Y
0
and this inequality gives us that~ is jointly continuous. We now present an interesting application which exhibits the power of the Uniform Boundedness Theorem. Although we have seen, (from Example 3.16) that every f E 'C [-7t,7t] has mean square representation by its Fourier series, it was shown by duBois Reymond in 1876 that the Fourier series for such a function may actually fail to converge pointwise to the function. He actually constructed a continuous function whose Fourier series is divergent at 0. The proof given here is an existence proof and it does also give us
more information about the pointwise convergence of a Fourier series, (see Exercise 11.17 .5). 11.16 The du Bois Reymond Theorem.
There exists an f
E
'C[-1t,1t] whose Fourier series is divergent at 0.
Proof. Given f E 'C [-7t,7t] we write
a
n
°
I,
Sn (t; f)= 2 +
(ak cos kt
+ ~k sin kt)
k=l
the nth partial sum of the Fourier series for f, where a 0 , ak and ~k are the Fourier coefficients for f defined with respect to the orthonormal set I I I . {{21[ '{1[ cos nt , {1t sm nt : n E
w}
N
.
We show that { Sn(O; f)} is divergent for some f E 'C[-1t,1t]. Suppose that { Sn(O; f)} converges for every f E 'C[-1t,1t]. Now for any f E 'C[-1t,1t].
sn (0; f)=
a
20 +
n
I,
ak.
k=l
From the formula for the Fourier coefficients we have 1t
Sn (0 ; f)=~
f f(t) Dn(t) dt
-lt
I
where
Dn(t) =
n
2 + I,
cos kt.
k=l
But
sin (n+t)t . t fortE(-7t,7t)\{O}. 2 s1n 2
I69
§II. The Uniform Boundedness Theorem
Notice that cjl 0 (f) = S0 (0; f) is a linear functional on 'C[-1t,1t] and 1t
I cjl 0 (f)
I:;::;~
II flloo
f I D (t) I dt. 0
-lt
So 0. For x E H, II x II
= I, II (T-'AI) x II ~ I ((T-'AI)x, x) I =I (Tx, x)- A I ~ d
(*)
II (T-'AI)x II ~ d II x II for all x E H, so which implies that T-'AI is a topological isomorphism.
Now (T-'AI)(H) is a closed linear subspace of H. Suppose that (T-'AI)(H) is a proper subspace of H. Then by Corollary 2.2.21 there exists a z E H, II z II = I such that z is orthogonal to (T-'AI)(H), so (( T-'AI)z, z) = 0 which contradicts(*). So T-'AI is a topological isomorphism of H onto H.
D
Using the characterisation for self-adjoint operators given in Theorem 13.10.8 we can make the following deduction. 13.10.10 Corollary. Given a self-adjoint operator Ton a complex Hilbert space H,for any complex number A;; a+ i~ where~* 0, the continuous linear operator T-'AI has a continuous inverse on H. Theorem 13.10.8 also enables us to define other sets of self-adjoint operators using the order relation of the real numbers. 13.10.11 Definition. A self-adjoint operator T on a Hilbert space H is said to be a positive operator if W(T) ~ 0; that is, Tis a positive operator if A~ 0 for all A E W(T). 13.10.12 Remarks. (i)
The set of positive operators is always nontrivial; it contains the zero and identity
operators and for any continuous linear operator T on H we have that T*T and TT* are
§ 13. Adjoint operators
189
positive operators since (T*Tx, x) = (x, T*Tx) = (Tx, Tx) =II Tx 112:?: 0 for all x E H. (ii)
The set of positive operators is a positive cone in B (H); that is, for a positive
operator T and a :?: 0 we have that aT is also a positive operator and for positive operators T 1 and T 2 we have that T 1+T 2 is also a positive operator. This positive cone is also closed in B (H): For a sequence {T n} of positive operators convergent to a selfadjoint operator T in (B(H), 11·11) we have I (T 0 x, x)- (Tx, x)
I~
II T 0 -T 1111 x 112 for all x EH
so (Tx,x):?: 0 for all x E H. (iii) The positive cone of positive operators induces a partial order relation on the set of self-adjoint operators. For self-adjoint operators T 1 and T 2 we say that T 1 ~T 2 if T 2-T 1 :?: 0. From Lemma 13.10.6(i) it follows that if T 1 ~ T2 and T 2 ~ T 1 then T 1 = T 2.
0
As in Corollary 13.10.10 we have the following property for positive operators. 13.10.13 Corollary. Given a positive operator Ton a complex Hilbert space H, for any complex number A= a+ ip where a< 0 or p 0, the continuous linear operator
*
T-'AI has a continuous inverse on H. The following particular case is important. 13.10.14 Corollary. For any continuous linear operator Ton a Hilbert space H, the
continuous linear operators I+T*T and I+TT* are topological isomorphisms ofH onto H. 13.10.15 Remark. In the attempt to establish the Gelfand-Naimark Representation Theorem forB* algebras mentioned in Remarks 13.9(iii), a key step was to use the B* algebra structure to prove that e+x*x and e+xx* are regular for every element x of a unital B* algebra. We notice that the proof of Corollary 13.10.14 uses the underlying Hilbert space structure on which the operators act. The Gelfand-Naimark Theorem had to achieve this result without the assumption of any such underlying structure.
0
13.11 Normal and unitary operators For a complex Hilbert space, further structural properties of B(H) are revealed by drawing attention to other special sets of continuous linear operators on H. 13.11.1 Definitions. A continuous linear operator Ton a Hilbert space H is said to be a normal operator if T*T = TT*, and is said to be a unitary operator if T*T = TT* = I. We will first explore the properties of normal operators.
190
Types of continuous linear mappings
13.11.2 Remarks. (i)
Clearly self-adjoint operators are normal so the set of normal operators contains the
closed real linear subspace of self-adjoint operators in n(H). But also given any normal operator Ton Hand any scalar A., then A.T is also normal. Further, the set of normal operators is closed in n (H): For a sequence {T n} of normal operators convergent to a continuous linear operator T in (n(H), 11·11) we have II T*T-TT* II :::; II T*T-T*T0 II + II T*T0 -T 0 *T0 II + II T 0 *T0 - TnTn* II +II TnT n*-TnT* II+ II TnT*-IT* II :::; 2 II T* 1111 T-T n II + 2 II (T-T n)* 1111 T n II and so T*T = TT*.
(ii)
Given a complex finite dimensional inner product space Hn with an orthonormal
basis and a linear operator Ton Hn we saw in Example 13.6 that the matrix representation ofT* is the conjugate transpose of the matrix representation ofT. The linear operator T 0 commutes with its adjoint T* if and only if the matrix representations do the same.
Normal operators have the following characterisation. 13.11.3 Theorem. A continuous linear operator T on a complex Hilbert space H is
normal if and only if II Tx II = II T*x II for all x E H. Proof. Now II Tx 112 = (Tx, Tx) = (x, T*Tx) and II T*x 112 = (T*x, T*x) = (x, TT*x) so II Tx II = II T*x II for all x E H if and only if (x, (T*T-TT*)x) = 0 for all x E H; that is, by Lemma 13.10.6(i) if and only if T*T = TT*.
0
The limit developed in Proposition 4.11.12 is very significant for normal operators. 13.11.4 Theorem. A normal operator Ton a complex Hilbert space H has the
properties (i) (ii)
II T211 =II T 11 2 and lim IIT0 11 11"=11TII. n-7~
Proof. (i)
From Theorem 13.11.3 we have that II T2x II = II T*Tx II
so
for all x EH
II T2 II = II T*T II.
But from Theorem 13.7(vi) we have that II T*T II= II T 11 2. Therefore (ii)
II T211 =II T 112.
Since powers ofT are also normal operators we have that II T 2" II = II T 11 2" for all n E N .
§ 13. Adjoint operators
191
We have from Proposition 4.11.12 that lim II T 0 11 11" exists. So lim II T 0 11 11" = lim II T 2" 11 112" =II T II.
D
We now discuss the properties of unitary operators. 13.11.5 Remark. (i)
The algebraic formulation of the definition tells us that T is unitary if and only if T
has a continuous inverse and T -I = T*. (ii)
Such operators generalise rotation operators in Euclidean space. For example, in (IR 2 , 11·11 2) the rotation operator Te which rotates by an angle e about the origin with matrix representation, Te= [
cos e
-sin e ]
sin e *
T
has
e
=[
cos e
cos e
sine ]
-sine
cos e
and clearly T8* = T8-1 . Equivalently, in ( 0 such that II Xn II :-::; r x x xn for all n EN. Then II __n II:-::; I for all n EN. If {T( __n)} has a subsequence {T(__l 0 such that II f lloo 0 there exists a 8 > 0 such that
Ik(x,t)- k(x 2,t) I < £
for all t E [a,b] and x 1,xz E [a,b] and I x 1-x 2 1 < o.
Therefore, for any fEB b
I (Kf)(x 1) - (Kf)(x 2) I 0 there exists avE N such that I an I v. Then II(T-Fn)xll 2 =
~
I.
2
2
lakiiA.kl <e
k=n+l IIT-Fnllv;
So
,------:::---~
I.
k=n+l
2
IA.kl.
that is, Tis the limit of continuous finite rank operators Fn which are compact. By Theorem 15.14(ii) we conclude that Tis compact.
D
The following structural property of 'K(X) is of interest. 15.16 Theorem. Given a compact operator T and a continuous linear operatorS on a normed linear space (X, 11·11), then STand TS are both compact operators on (X, 11·11).
Proof. Consider a bounded sequence {x0 } in (X, 11·11). Then since Sis continuous the sequence {Sx 0
}
is also bounded. Since T is compact the sequence { T(Sxn)} has a
convergent subsequence; that is, TS is a compact operator. But also the sequence { Txn} has a convergent subsequence { Tx 0 k} and since S is continuous the sequence { S(Txnk)} is also convergent; that is, ST is a compact operator. D
§ 15. Compact operators
211
This result has a consequence for the continuity of inverses.
!5.17 Corollary. Given a compact operator Ton an infinite dimensional normed linear space (X, 11·11), if T -I exists on (X, 11·11) then T -I is not continuous.
Proof. If T- 1 is continuous on X then by Theorem 15.16 we have that I= T- 1T is also compact. But from Remark 15.5 we see that I is not compact so T- 1 is not continuous. 0 We now show how the compactness property is inherited by conjugate mappings.
15.18 Schauder's Theorem. A compact mapping T of a normed linear space (X, II· II) into a normed linear space (Y, II· II') has compact conjugate mapping T'of (Y*, 11·11') into (X*, 11·11).
Proof. Since T is compact, for the closed unit ball B in (X, 11·11), T(B) is compact in (Y, 11·11'). Consider {f0 } a sequence in the closed unit ball of (Y*,II·II'). In ('C(T(B), ll·lloo) the set A= {f 0 l_: n EN} is bounded and since T(B)
I f0 (y)- f 0 (y 1) I 0 the set
{a:
I (Tew ea) I ~f.} is finite.
(iii) There exists a sequence {F0 } of continuous finite rank operators on H such that
II T-F0 ll-7 0 as n
--7
oo.
212
Types of continuous linear mappings
Proof. (i) =} (ii) ForT compact suppose that (ii) does not hold. Then for some orthonormal set {ea} and some r > 0, the set
{a : I (Tea, ea) I ~ r}
is infinite. So it has a countably
infinite subset which gives rise to an orthonormal sequence {en} such that I (Ten, en) I ~ r for all n E M . Since Tis compact there is a subsequence {enk} such that {Tenk} converges to an x E H.
Discarding a finite number of terms of this sequence we may suppose that
II Te 0 k- x II < ~
for all k E N.
I (Tenk' enk)- (x, e0 k) I= I (Te 0 k- x, e0 k) I~ II Te 0 k- x II < ~
Then
for all kEN.
and so
But this contradicts Bessel's inequality, Theorem 3.13. (ii) =}(iii) Given n EN, consider the family I
I (Teao ea) I > ~ for all
::r of all orthonormal sets {eal in H where
a.
::r partially ordered by set inclusion. Now the union of any totally ordered subfamily of orthonormal sets from ::r is itself a member of ::r and is an upper bound for this subfamily. It follows by Zorn's Lemma that ::r has a maximal member {e~}. As a member of ::r, {e~} is finite.
By (ii) each such orthonormal set is finite. Consider
Consider the finite dimensional linear subspace M = sp {e~}. Then
I (Tx, x) I
For any f EA* we have foR(A)=
_l
(f(e)
+
A But f o R is analytic on p(x)
;;;?. { AE
L
n=1
f(x"))· A."
v(x)} so this is the Laurent series expansion
D
for f o R on { AE v(x)}.
Lemma 16.6 enables us to establish the following spectral properties. 16.7 Theorem. Given an element x in a complex unital normed algebra (A, 11·11), (i)
cr(x)
* 0 and
(ii) when A is complete v(x)= lim llx"lllin =inf{llx"ll 110 :nEN}. n~~
Proof. (i)
Suppose that cr(x) = 0, then p(x) = (C and for any f E A*, f o R is an entire function.
But also since f E A*, I f o R(A.) I ~II f 1111 R(A.) II so from Lemma 16.6(i), f o R(A.)
~
0 as A.
~ oo
By Liouville's Theorem we deduce that f o R(A.) = 0 for all A. E
Now for v(x) > 0 there exists a A E a: such that I A I < v(x) where x- Ae is singular. If also lim sup II xn II lin< I A I n-+x>
then by Theorem 4.11.1 0, x - Ae is regular. So we conclude that v(x) = lim sup II xn 11 11n. n-+x>
From Proposition 4.11.12 we have v(x)= lim 11xn11 11n =inf{llxnll 11n:nEN}.
D
n~~
16.8 Remark. We note that we invoked the completeness condition when we used series arguments like those of Theorem 4.11.1 0 and Corollary 4.11.11. Completeness guarantees the boundedness of the spectrum and a meaningful definition of spectral radius. D We now present a useful algebraic property of the spectrum. 16.9 The Spectral Mapping Theorem for polynomials. Consider a complex unital normed algebra (A, II· II) and a polynomial p with complex
coefficients. For every x E A, cr(p(x)) = p(cr(x)) = {p(A): A E<J(x) }.
Proof. Clearly p(x) EA. For any A E a: consider the polynomial p(t) - A. Now if p is of degree n then by the Fundamental Theorem of Algebra, p(t)- A has n roots, A1, "-2,
... , An
and we can write n
for some a E a:. i=l If A E cr(p(x)) then there exists some i0 E { 1, 2, ... , n} such that x- Ai 0e is singular, that p(x)- Ae =a
IT (x-Aie)
is Aio E cr(x). Then A= p(Ai 0 ) E p(cr(x)) and so cr(p(x)) (;;;; p(cr(x)). Conversely, if A E p(cr(x)) then there exists i0 E { 1, 2, ... , n} such that Aio E cr(x) and A= p(Ai 0 ).
However, if Ae cr(p(x)) then every factor x- Aie, of p(x)- Ae is
regular so Ai ecr(x) for all i E { 1, 2, ... , n}. So we conclude that p(cr(x)) (;;;; cr(p(x)).
D
221
§ 16. The spectrum
!6.1 0 EXERCISES
I.
An element z of a complex unital Banach algebra (A, 11·11) is called a topological divisor of zero if there exists a sequence {z0 } in A where II z0 II = 1 for all n EN
and either z0 z --7 0 or z:z, (i)
--7
0 as n --7 ""· Prove that
every divisor of zero is a topological divisor of zero,
(ii) the set of topological divisors of zero is closed in A, (iii) every topological divisor of zero is a singular element, (iv) the boundary of the set of singular elements is contained in the set of topological divisors of zero. 2.
Consider a complex unital Banach algebra A and a complex unital Banach subalgebra A'. Prove that for any element x E A', (i)
<J A'(x) ~
crA(x) and
(ii) acr A(x) ~ acr A'(x). 3.
Consider a complex unital Banach algebra (A, 11·11). Prove that each of the following conditions implies that (A, 11·11) is isometrically isomorphic to (i)
a:.
A is a division algebra; that is, every nonzero element of A has an inverse.
(ii) Zero is the only topological divisor of zero. (iii) There exists an M > 0 such that II xy II ~ M II x 1111 y II for all x, y EA. (iv) For every regular element x E A, II x- 1 II=~ . 4.
Consider a complex unital normed algebra A. Prove that (i) if x E A is regular then x- 1 - e/A = (x- 1/A)(Ae-x) and deduce that cr(x -1) = { IIA : AE cr(x)}, (ii) if we are given that e- xy is regular for x, y E A then e- yx is regular where (e-yxr 1 = e + y(e-xyr 1 x and deduce that for any x, yEA cr(xy) \ {0} = cr(yx) \ {0}.
5.
Consider the complex unital commutative Banach algebra (t:[O, 1], II· II"") and f E t; [0, I]. Prove that (i)
f is either regular or is a topological divisor of zero,
(ii) A E cr(f) if and only if XE crcf), (iii) f = 6.
f
if and only if cr(f) ~ IR.
Consider a complex unital Banach algebra A. Prove that for any given x E A the spectral radius v(x) has the following properties (i)
v('Ax) = I A I v(x),
(ii) v(xy) = v(yx), (iii) if xy = yx then v(xy)
~
v(x) v(y) and v(x+y)
~
v(x) + v(y).
222
Spectral theory
§17. THE SPECTRUM OF A CONTINUOUS LINEAR OPERATOR
We now consider the particular case of the algebra of continuous linear operators on a normed linear space. Although Section 16 gives us general structural properties of these as elements of the unital normed algebra of operators, here it is important to determine whether a particular continuous linear operator has an inverse on the space and whether the inverse is also continuous. 17. 1 The spectrum in linear spaces We begin with purely algebraic considerations in ;r; (X), the algebra of linear operators on a linear space X. It is clear that a linear operator T on a linear space X has an inverse T -I on X if and only if T is one-to-one and onto. So we have the following definitions. 17.1.1 Definitions. Given a linear operator Ton a linear space X, an eigenvalue ofT is a scalar A where there exists an xi' 0 such that Tx =Ax. Given an eigenvalue A ofT, the
elements x EX such that Tx = Ax are called eigenvectors ofT and the linear space { x EX : Tx = Ax} is called the eigenspace associated with the eigenvalue A. The set of eigenvalues ofT is called the point spectrum ofT and is denoted by Pcr(T). Clearly A E Pcr(T) if and only if T- AI is not one-to-one. There is a fundamental relation between the eigenvalues and their associated eigenspaces. 17.1.2 Theorem. Given a linear operator Ton a linear space X with distinct eigenvalues At, A2, ... , A0 , any set { x 1, x2, ... , x0 } of corresponding eigenvectors is linearly
independent. Proof. We use proof by induction. Now {xtl is linearly independent. Suppose that {x 1, x2, ... , xk} is linearly independent and consider a 1x 1 + a 2x2 + ... + akxk + ak+lxk+l = 0. Then T(a 1x 1 + a 2x 2 + ... + ak+ 1xk+l) = a 1Tx 1 + a 2Tx 2 + ... + ak+ 1Txk+l = atAtxl + a2A2x2 + ... + ak+tAk+lxk+l = 0.
Then al(At-Ak+l)xl + aiA2-Ak+l)x2 + ... + ak(Ak-Ak+l)xk = 0 and since { x 1, x2, ... , xk} is linearly independent at(At-Ak+l) = a2(ArAk+l) = · · · = ak(Ak-Ak+l) = 0. Since the eigenvalues are distinct, we have a 1 = a 2 = ... = ak = 0. Since xk+l i' 0 we conclude that a 1 = a 2 = ... = ak and so { x 1, x2, ... , xk+l} is linearly independent. The following is an elementary consequence.
= ak+l = 0 D
223
§ 17. The spectrum of a continuous linear operator
17.1.3 Corollary. Given a linear operator Ton a linear space X, for eigenspaces M 1 and M 2 associated with eigenvalues A. 1 and A.z, M 1 n M 2 = {01. So we have some insight into the structure of the point spectrum of a linear operatQr on a finite dimensional linear space. 17.1.4 Corollary. For a linear operator Ton an n-dimensionallinear space Xn, the point spectrum Pcr(T) contains no more than n elements. Proof. Xn contains the direct sum of its eigenspaces so T cannot have more than n eigenvalues. D For finite dimensional linear spaces, the failure of a linear operator to have an inverse can be reduced to its failing to be one-to-one. 17 .1.5 Theorem. A linear operator T on an n-dimensionallinear space Xn is one-to-one
if and only if it is onto. Proof. Consider { e 1, e2 , ••. , en} a basis for Xn. Suppose that T is one-to-one and consider the set { Te 1, Te 2 ,
I
••• ,
I
Ten}. If for scalars
akek=Osinc~
f
akek)=Oso k=l k=l Tis one-to-one. But { e 1, e 2 , •.. , en} is a basis for Xn so a 1 = a 2 = ... =an = 0. Then {Te 1, Te2 , ... , Ten} is linearly independent and so is a basis for Xn. We conclude {a 1,a2 ,
•••
,an} we have
akTek=OthenT(
k=l
that T is onto. Conversely, if T is onto then for each ek E Xn where k
E { I,
2, ... , n I there exists n
an xk
E
Xn such that ek = Txk. Then for such a set { x 1, x2 ,
..• ,
xn}, if
I. akxk = 0 then k=l
n
n
I. ak Txk = 0 and I. akek = 0 so a 1 = a 2 = ... =an = 0 since { e 1, e2, ••• , en} is k=l k=l basis for Xn. Then {x 1, x2 , •.. , Xn} is linearly independent and so is a basis for Xn·
a
n
IfTx = 0 then x =
I. ~k xk for scalars ~k where k E {I, 2, ... , nl. k=l
n
0=
So
n
I. ~k Txk = I. ~kek
k=l k=l and then ~ 1 = ~ 2 = ... = ~n = 0 since { e 1, e2 , . . . , en} is a basis for Xn. Therefore D x = 0 and we conclude that T is one-to-one. 17.2 The spectrum in normed linear spaces Given a continuous linear operator Ton a normed linear space (X, 11·11), Tis a regular element of ~(X) if and only if T -I exists and is continuous on X. When X is finite dimensional we have seen in Theorem 17 .1.5 that T -I exists on X if and only if T is
224
Spectral theory
one-to-one and the continuity ofT and T -I is automatic by Corollary 2.1.1 0. When X is infinite dimensional the situation is a little more complicated. In general, a continuous linear operator Ton a Banach space (X, 11·11) is not invertible in (i)
~(X)
if and only if
T is not one-to-one, or
(ii) T is one-to-one but T(X) is not dense in (X, 11·11), or (iii) Tis one-to-one and T(X) is dense in (X, 11·11) but T- 1 is not continuous on T(X). So we are led to the following decomposition of the spectrum ofT. 17 .2.1 Definitions. The spectrum of a continuous linear operator T on a complex Banach space (X, 11·11) ,
cr(T)
={A
E
a: : AI- Tis singular in ~(X)}
can be separated into three disjoint component sets: the point spectrum consists of the eigenvalues ofT Pcr(T) {A E a: : AI - T is not one-to-one},
=
the residual spectrum is the set Rcr(T) = {A E
a: : AI - T is one-to-one but (AI-T)(X) is not dense}
and the continuous spectrum is the set Ccr(T) = {A E a: : AI - T is one-to-one and (AI-T)(X) is dense but (AI-Tr 1 is not continuous on (AI-T)(X)}. So
cr(T) = Pcr(T) u Rcr(T) u Ccr(T).
17.2.2 Remarks.
(i) Theorem 17 .1.5 shows that for a continuous linear operator T on a complex finite dimensional normed linear space, cr(T) = Pcr(T). (ii) We recall from Corollary 1.24.10 that for a continuous linear operator Ton a Banach space (X, 11·11), if AI- T is one-to-one and (AI- T)(X) is dense and (AI- Tr 1 is continuous on (AI- T)(X) then AI- Tis onto and so AI- Tis regular in ~(X). 0
I 7. 3 The spectrum and numerical range It is important to be able to determine the spectrum of a continuous linear operator T on a complex Banach space (X, 11·11), but it is not always a simple set to specify. From Section 16 we saw that cr(T) is a compact set contained in the closed disc of radius II T II. In Hilbert space H the closure of the numerical range ofT is a smaller set contained in the closed disc of radius II T II which also contains the spectrum. In Theorem 13.10.9 we showed that for any A f1' W (T), T - AI is regular in
~(H).
This result has the following
expression in terms of the spectrum ofT. 17.3.1 Theorem. For a continuous linear operator Ton a complex Hilbert space,
cr(T) !:;;; W (T).
§ 17. The spectrum of a continuous linear operator
225
Theorem 13.10.8 then has an immediate implication for the spectrum of self-adjoint operators. 17.3.2 Corollary. For a self-adjoint operator Ton a complex Hilbert space, cr(T) ~[-II T II, II T II]. For the spectrum of positive operators we have the following result. 17.3.3 Corollary. For a positive operator Ton a complex Hilbert space, cr(T) ~ [0, II T II]. For the spectrum of normal operators, Theorems 13.11.4 and 16.2 have the following implication. 17 .3.4 Corollary. For a normal operator Ton a complex Hilbert space, v(T) = w(T) = II T II
and there exists a A E cr(T) such that I A I = II T II. For the spectrum of unitary operators, numerical range considerations do not give us further insight. Theorem 13.11.9 has the following expression in terms of the spectrum. 17.3.5 Corollary. For a unitary operator Ton a complex Hilbert space cr(T) ~ { A E ([ : I A I = I }. In the 1960s there was a successful generalisation of the concept of numerical range of a continuous linear operator on a normed linear space. This development was largely due to F.F. Bonsall and an exposition of the basic theory can be found in F.F. Bonsall and J. Duncan, Numerical Ranges I and II, Cambridge University Press (1971) and (1973).
17.3.6 Definition. Given a complex normed linear space (X, 11·11), for each x EX, II x II = I, consider the set D(x) = { f EX* : II f II = I, f(x) = I}. Corollary 6.3 to the HahnBanach Theorem guarantees that for each x EX, II x II = 1, we have that D(x) is nonempty. For a continuous linear operator T on X we call the set of complex numbers V(T) = {f(Tx): x EX, II x II= I, f ED(x)} the numerical range ofT. Clearly, when X is an inner product space, V(T) = W(T). The result which gave the stamp of success to this work is the generalisation of Theorem 17 .3. I. 17.3.7. Theorem. For a continuous linear operator Ton a complex Banach space
(X, 11·11), cr(T)
~
V(T).
226
Spectral theory
Proof. For A f1' V(T) we write d
=d('A, V(T)) > 0.
For x EX, II x II = I and f E D(x),
II (T-'AI)x II~ I f((T-'AI)x) I~ d
II (T-'AI)x II ~ d II x II
so
for all x E X
which implies that T -'AI is a topological isomorphism. Now (T-'AI)(X) is a closed linear subspace X. Suppose that (T-'AI)(X) is a proper subspace. Then by Corollary 8.7 there exists an x0 f1' (T-'AI)(X), II x0 II= 1 and an f0 E D(x 0) such that d I f0(x) I~ - - for all x E (T-'AI)(X) and II x II~ I. 211T-'AIII d But then 2 ~ I f0( (T-'AI)x 0 ) I = I f0(Tx 0) - 'A I ~ d('A, V (T)) ~ d, a contradiction. So we conclude that T -AI is onto and A f1' cr(T).
D
17.4 The spectrum of a conjugate operator In Theorem 12.11 we studied the relation between the regularity of a continuous linear operator on a normed linear space and its conjugate on the dual space. This result has the following implications for the spectrum. 17 .4.1 Theorem. For a continuous linear operator Ton a complex normed linear space (X, 11·11) with conjugate operator T' on (X*, 11·11), (i)
cr(T' ) \;;; cr(T) and
(ii)
if (X, 11·11) is complete then cr(T') = cr(T). We are particularly interested in the relation on Hilbert space where for A scalar, T-'AI
is regular if and only if T* -'AI is regular. 17.4.2 Corollary. For a continuous linear operator Ton a complex Hilbert space H with
adjoint T* on H,
cr(T*) = {I: 'A E cr(T)}.
The point spectrum of a normal operator on Hilbert space has particularly satisfying properties which suggest a decomposition of the space. 17.4.3 Theorem. Consider a normal operator Ton a complex Hilbert space H.
-
-
if and only if 'A is an eigenvalue ofT*, and A and A have the
(i)
A is an eigenvalue ofT
(ii)
If A and 11 are distinct eigenvalues forT then the corresponding eigenspaces Mt.. and
same eigenspace: M11 are orthogonal.
Proof. (i)
Since T -'AI is also normal, by Theorem 13.11.3 we have that II (T-'AI)x II = II (T*-AI)x II
and our result follows.
for all x E H,
§17. The spectrum of a continuous linear operator
(ii)
227
For x E MA. andy EMil we have A.(y, x) = (y, A.x) = (y, T*x)
by (i)
= (Ty, x) = (!J.y, x) = !J.(y, x). Since A."* 1J. then (y, x) = 0.
D
17.5 EXERCISES
1.
(i)
Given a continuous linear operator Ton a complex Banach space with conjugate operator T', prove that (a)
(ii)
R cr(T) k: Pcr(T' ) ,
u Rcr(T'),
(b)
Pcr(T) k: Pcr(T')
(c)
Ccr(T') k: Ccr(T).
Given a continuous linear operator T on a complex Hilbert space with adjoint T*, prove that
2.
(i)
(a)
A. E Rcr(T) if and only if A. E Pcr(T*),
(b)
if A E Pcr(T) then A E Pcr(T*)
(c)
A. E Ccr(T*) if and only if A. E Ccr(T).
u Rcr(T*),
Consider the right shift operator SIR on (1 2 , 11·11 2) defined for X=
{A. 1, A. 2, ... , An, ... } by SIR (x) = { 0, A.1, A.2, ... , A.n, ... } .
Prove that
(ii)
(a)
Pcr(SIR) = 0,
(b)
Rcr(SIR)={A.EU::IA.I ... , an, ... } where for X=
{A. 1 , A. 2, ... , An, ... } E 1 2 , Tx = { a 1 A. 1, a 2 A. 2 ,
(i)
... ,
Prove that (a)
Pcr(T) = {an : n EN},
(b)
cr(T) = {an : n E N } .
anA.n, ... } .
228
Spectral theory
(ii)
Show that every nonempty compact subset of the complex plane is the spectrum of some diagonal operator; (See Exercise 4.12.2.)
4.
Consider the complex Banach space ('C[O,l], 11·11 00 ). (i)
For the integral operator I defined by X
Jf(t) dt
I(f)(x) =
0
prove that
(ii)
(a)
Jl. ~l
ll I' lUi•
"11111111illlllllll
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