- Email: [email protected]
Chapter VI Hilbert Spaces
CHAPTER VI
HILBERT SPACES
Hilbert spaces constitute at present the most important examples of Banach spaces, not only because they are the most natural and closest generalization, in the realm of “ infinite dimensions,” of our classical euclidean geometry, but chiefly for the fact that they have been, up to now, the most useful spaces in the applications to functional analysis. With the exception of (6.3.1),all the results easily follow from the definitions and from the fundamental Cauchy-Schwarz inequality (6.2.4).
1. H E R M I T I A N FORMS
For any real or complex number A, we write X for its complex conjugate (equal to A if A is real). A hermitiun form on a real (resp. complex) vector space E is a mapping f of E x E into R (resp. C) which has the following properties : (1) (11)
(W (W (V)
f(x + x‘,Y ) =f(x, Y ) + f ( x ’ , Y ) ,
+ Y’) = A x ,Y ) +m, Y’), f ( kr) = Af(x,Y ) ,
f(x, Y
f ( x , AY) = m x ,Y>, f ( Y >4
=f(% Y ) .
(Observe that (11) and (IV) follow from the other identities; (V) implies that f(x, x) is real.) When E is a real vector space, conditions (I) to (IV) express that f is bilinear and (V) boils down to f ( y , x) = f ( x , y), which 115
116
VI
HILBERT SPACES
expresses t h a t f i s symmetric. For any finite systems ( x i ) ,( y j ) , (ai), (Bj), we have (6.1 .I)
by induction on the number of elements of these systems. From (6.1.1) it follows that if E is finite dimensional and ( a i ) is a basis of E, f is entirely determined by its values a i j = f ( a i , a j ) , which are such that (by V)) aJ,l. = a I,J.’
(6.1.2)
Indeed we have then, for x = (6.1.3)
1 S i a i ,y = i
S ( X .,Y>
viai I
C aij ti
i, i
iij
.
Conversely, for any system ( a i j ) of real (resp. complex) numbers satisfying (6.1.2), the right hand side of (6.1.3) defines on the real (resp. complex) finite dimensional vector space E a hermitian form. Example (6.1.4) Let D be a relatively compact open set in R2, and let E be the real (resp. complex) vector space of all real-valued (resp. complex-valued) bounded continuous functions in D, which have bounded continuous first derivatives in D. Then the mapping
(where a, h, c are continuous, bounded and real-valued in D) is a hermitian form on E. A pair of vectors x, y of a vector space E is orthogonal with respect to a hermitian form f on E if f ( x , y ) = 0 (it follows from (V) that the relation is symmetric in x, y ) ; a vector x which is orthogonal to itself (i.e.f(x, x) = 0) is isotropic with respect to$ For any subset M of E, the set of vectors y which are orthogonal to all vectors x E M is a vector subspace of E, which is said to be orthogorial to M (with respect t o f ) . It may happen that there exists a vector a # 0 which is orthogonal to the whole space E, in which case we say the formfis degenerate. On a finite dimensional space E, nondegenerate hermitian forms f defined by (6.1.3) are those for which the matrix ( a i j ) is invertible.
2
POSITIVE HERMITIAN FORMS
117
PROBLEMS
1. (a) Let f be a hermitian form on a vector space E. Show that if E is a real vector space, then 4f(x, Y> = f ( x
+ Y , x +v) -f(x
and if E is a complex vector space
4fh, Y ) =f(x
+ y. x + y ) - f ( x
-Y ,x
-y)
-Y, x
-Y)
+ i f ( x + iy, x + iy) - if(x - iy, x
- iy).
(b) Deduce from (a) that if f ( x , x ) = 0 for every vector in a subspace M of E, then f ( x , y ) = 0 for any pair of vectors x , y of M. (c) Give a proof of (b) without using the identities proved in (a). (Write that f(x hy, x +Xu) = O for any h.) 2. Let E be a complex vector space. Show that iffis a mapping of E x E into C satisfying conditions (I), (II), (III), (IV), and such that f ( x , x ) E R for every x E E, then f is a hermitian form on E.
+
2. POSITIVE H E R M I T I A N FORMS
We say a hermitian form f on a vector space E is positive iff(x, x) 2 0 for any x E E. For instance, the form q defined in example (6.1.4) is positive if a, b, c are 2 0 in D. (6.2.1)
then
(Cauchy-Schwarz inequality)
for any pair of vectors
X,
I f f is a positive hermitian form,
If(x, Y>12 < f ( x , X)f(Y, Y ) y in E.
Write a =f ( x , x), b = f ( x , y ) , c = f ( y , y ) and recall a and c are real and 2 0. Suppose first c # 0 and write that f ( x I y , x + l y ) 2 0 for any scalar 1, which gives a + b2 + bA + c3J 0 ; substituting I = -b/c yields the inequality. A similar argument applies when c = 0, a # 0 ; finally if a = c = 0, the substitution I = - h yields -266 2 0, i.e. b = 0.
+
(6.2.2) In order that a positii!e hermitian form f on E be nondegenerate a necessary and suficient condition is tliat there exist no isotropic vector for f otlier than 0 , i.e. that f ( x , x) > 0 for any x # 0 in E.
Indeed, f ( x , x) all y E E.
=0
implies, by Cauchy-Schwarz, that f (x, y ) = 0 for
118
(6.2.3)
VI
HILBERT SPACES
(Minkowski's inequality) I f f is a positi1.e hermitian form, tlien (f(x
+y, x +
d ( A x , x))''2 + ( f b ,
for any pair of vectors x, y in E.
+
As f ( x y , x equivalent to
+ y ) = f ( x , x) + f ( x , y ) + A x , y ) + f ( y , y ) , the inequality is
2,gf(x,Y)= f ( x , Y >+f(x,P,)d 2 ( f ( x , x)f(Y,Y))'12 which follows from Cauchy-Schwarz. The function x -+ ( f ( x , x))'12 therefore satisfies the conditions (I), (Ill), and (IV)of (Section 5.1); by (6.2.2), condition (11) of Section 5.1 is equivalent to the fact that the form f is nondegenerate. Therefore, when f is a nondegenerate positive hermitian form (also called a positive dejnite form), ( f ( x , x))'I2 is a norm on E. A preliilbert space is a vector space E with a given nondegenerate positive hermitian form on E; when no confusion arises, that form is written (x I y ) and its value is called the scalar product of x and y ; we always consider a prehilbert space E as a normed space, with the norm llxll = (x I x)''', and of course, such a space is always considered as a metric space for the corresponding distance Ijx - y ( / . With these notations, the Cauchy-Schwarz inequality is written
I(xlu)l d IIxIl
(6.2.4)
IIYII
and this proves, by (5.5.1), that for a real prehilbert space E, (x, y ) -+ (x 1 y ) is a continuous bilinear form on E x E (the argument of (5.5.1) can also be applied when E is a complex prehilbert space and proves again the continuity of (x, y ) (x I y ) , although this is not a bilinear form any more). We also have, as a particular case of (6.1 .I): --f
(6.2.5) (Pythagoras' theorem) onal vectors, IIX
In a prehilbert space E, i f x , y are orthog-
+ YIl2 = l/xll2+ llY/I2.
A n isomorphism of a prehilbert space E onto a prehilbert space E' is a linear bijection of E onto E' such that ( f ( x )If(y)) = (x Iy) for any pair of vectors x, y of E. It is clear that an isomorphism is a linear isonzetry of E onto E'. Let E be a prehilbert space; then, on any vector subspace F of E, the restriction of the scalar product is a positive nondegenerate hermitian form ; unless the contrary is stated, it is always that restriction which is meant when F is considered as a prehilbert space.
3 ORTHOGONAL PROJECTION ON A COMPLETE SUBSPACE
119
A Hilbert space is a prehilbert space which is coniplete. Any finite dimensional prehilbert space is a Hilbert space by (5.9.1); other examples of Hilbert spaces will be constructed in Section 6.4. If in example (6.1.4) we take a > 0, b 2 0 , c 3 0, it can be shown that the prehilbert space thus defined is not complete.
PROBLEMS
1. Prove the last statement in the case a = 1 , b = c = 0 (see Section 5.1, Problem 1). 2. (a) Let E be a real nornied space such that, for any two points ,Y, y of E, 1I.y
+ Yl12 + Ilx - Y1IZ
= 2(11x1I2
+ llvIl2).
Show that f ( x , y) = IIx i-V I -~ Ilxl/* ~ - /IyIl2 is a positive nondegenerate hermitian form on E. (In order to prove that f ( h x , y ) = h f ( x , y ) , use Problem I of Section 5.5.) (b) Let E be a complex nornied space, Eo the underlying real vector space. Suppose there exists on Eo a symmetric bilinear form f ( x ,y) such that f ( x , x) = l1x/l2for every x E E o . Show that there exists a herniitian form g(x, y ) on E such that f ( x , y ) = .jR(g(x, y ) ) , hence g(x, x) = 1/x1l2for x E E. (Using the first formula of Problem 1 of Section 6.1, prove that f ( i x , y ) = - f ( x , i-v).) (c) Let E be a complex normed space such that Ilx - Y1I2
+ Ilx + Y / I 2 < 2(11x1I2 + llvll’)
for any pair of points x , y of E. Show that there exists a nondegenerate positive hermitian fornif(x,y) on E such thatf(x, x) = XI/^. (Use (a) and (b)). 3. Letfbe a positive nondegenerate hermitian form. In order that both sides of (6.2.1) be equal, it is necessary and sufficient that x and y be linearly dependent. In order that both sides of (6.2.3) be equal, it is necessary and sufficient that x and ,y be linearly dependent, and, if both are # 0, that y = Ax, with real and > 0. 4. Let a, b, c, d be four points in a prehilbert space E. Show that
/(a- c ( / . / / b- d ( (< /la - bll . JJc- dll
5.
+ 116
- cii
. / / a- dll.
(Reduce the problem to the case a = 0, and consider in E the transformation x+x/llxl~’, defined for x # 0.)When are both sides of the inequality equal? be a finite sequence of points in a prehilbert space E. Show that Let ( x i ) ,
If /lxi - x,Jl > 2 for i # j , show that a ball containing all the x i has a radius >(2(n
-
1)jn)’”.
3. O R T H O G O N A L PROJECTION O N A COMPLETE SUBSPACE
(6.3.1) Let E be a prehilbert space, F a complete tvctor subspace of E (i.e. a Hilbert space). For any x E E, there is one and only one point y = P F ( x )E F
120
VI
HILBERT SPACES
such that I/x - J~II = d(x, F). Tlze point y = P,(x) is also the only point z E F such that x - z is orthogonal to F. The mapping x P,(x) of E onto F is linear, continuous, and of norm 1 if F # { O } ; its kernel F’ = PF1(0) is the subspace orthogonal to F, and E is the topological direct sum (see Section (5.4)) of F and F‘. Finally, F is the subspace orthogonal to F‘. --f
Let CI = d(x, F); by definition, there exists a sequence (y,) of points of F such that lim 1I.x - ynl/ = a ; we prove (y,,) is a Cauchy sequence. fl-+
co
Indeed, for any two points u, u of E, it follows from (6.1.1) that (6.3.1.1)
hence
112.4
llym
+
+ -
U(l2
1/24
PJ12= 2(llu112
+ 11z~112>
- yn/I2= 2(llx - ~ , ~ ,+l /IIx ~ - ~ , , l l ~-) 4llx - ! d Y m
+ y,)112.
But
+ y,,) E F, hence jlx - )(ym + y,,)lI2 > a 2 ;therefore, if no is such that for n 2 n o , ( / x- y , ( / 2< g2 + E , we have, form > no and n 3 n o , I/ym- Y,;;? < 4e, +(yJ,
which proves our contention. As F is complete, the sequence (y,) tends to a limit y E F, for which IJx- yll = d(x, F). Suppose y’ E F is also such that /Ix - y’ll = d(x, F); using again (6.3.11), we obtain / ( y- y’(I2= 4a2- 4 /Jx- ) ( y y’)1I2, and as j ( y y ’ ) E F, this implies (ly - y’1I2 < 0,i.e. y’ = y. Let now z # 0 be any point of F, and write that (lx - ( y Az)/l* > a2 for any real scalar A # 0; this, by (6.1 .I), gives
+
+
21B(x - y 1 z )
+
+ R 2 / ( 2 / ( 2> 0
and this would yield a contradiction if we had 9 ( x - y 1 z ) # 0, by a suitable choice of 1. Hence B(x - y 1 z ) = 0, and replacing z by iz (if E is a complex prehilbert space) shows that #(x - y I z ) = 0, hence (x - y I z ) = 0 in every case; in other words x - y is orthogonal to F. Let y’ E F be such that x -y’ is orthogonal to F; then, for any z # 0 in F, we have JIx- (y’ z)1I2 = /Jx- y‘/12 j/zIlz by Pythagoras’ theorem, and this proves that y’ = y by the previous characterization of y . This last characterization of y = P,(x) proves that P , is linear, for if x - y and x’- y’ are orthogonal to F, then Ax - Ay is orthogonal to F and so is (x + x’)- ( y + y’) = (x - y ) (x’ - y ’ ) ; as y y’ E F and Ay E F, this shows that y y’ = P F ( x x’)and Ay = P , ( ~ x ) .By Pythagoras’ theorem, we have
+
+
+
(6.3.1.2)
+
llX/12
=
+
IIpF(x)112
+
+ (Ix
- pF(x)l/2
and this proves that IlPF(x)l1< I/x/J,hence (5.5.1) P, is continuous and has norm d 1 ; but as P F ( x )= x for x E F, we have /lP,I/ = 1 if F is not reduced to 0. The definition of P, implies that F’ = PF’(0) consists of the vectors x orthogonal to F; as x = P,(x) + (X - PF(,y)) and x - P F ( x )E F’ for any x E E, we have E = F F’; moreover, if x E F n F‘, x is isotropic, hence
+
3 ORTHOGONAL PROJECTION ON A COMPLETE SUBSPACE
x
= 0, and
121
this shows that the sum F + F' is direct. Furthermore, the mapping
x -iP F ( x )being continuous, E is the topological direct sum of F and F' (5.4.2). Finally, if x E E is orthogonal to F', we have in particular (x I x - P F ( x ) )= 0 ; but we also have (PF(x) I x - PF(x))= 0, hence /Ix - PF(x)(I2= 0, i.e. x = PF(x) E F. Q.E.D. The linear mapping PF is called the orthogonal projection of E onto F, and its kernel F' the orthogonal supplement of F in E. Theorem (6.3.1) can be applied to any closed subspace F of a Hilbert space E (by (3.14.5)), or to any Jinite dimensional subspace F of a prehilbert space, by (5.9.1).
(6.3.2) Let E be a prehilbert space; then, for any a E E, x -i (x I a) is a continuous linear form of norm Ilall. Conversely, if E is a Hilbert space, for any continuous linear form u on E, there is a unique vector a E E such that u(x> = (x I a)for any x E E. ByCauchy-Schwarz,I(x I a)l < llall * IIxII, whichshows(by5.5.1))~+ (x la) is continuous and has a norm < llalj; on the other hand, if a # 0, then for x, = a / ~ ~ we a ~have ~ , (x, I a) = llall; as llxoII = 1, this shows the norm of x - i ( x l a ) is at least Ilall. Suppose now E is a Hilbert space; the existence of the vector a (=O) being obvious if u = 0, we can suppose u # 0. Then H = u-'(O) is a closed hyperplane in E; the orthogonal supplement H' of H is a one-dimensional subspace; let b # 0 be a point of H'. Then we have ( x I b) = 0 for any x E H. But any two equations of a hyperplane are proportional, hence there is a scalar L such that u(x) = L(x 16) = (xI a) with a = Xb (see Appendix) for all x E E. The uniqueness ofa follows from the fact that the form (x I y ) is nondegenerate.
PROBLEMS 1. Let B be the closed ball of center 0 and radius 1 in a prehilbert space E. Show that for each point x of the sphere of center 0 and radius I , there exists a unique hyperplane of support of B (Section 5.8, Problem 3) containing x . 2. Let E be a prehilbert space, A a compact subset of E, 6 its diameter. Show that there exist two points a , b of A such that /la- bll = 8 and that there are two parallel hyperplanesof support of A (Section 5.8, Problem 3)containingaand b respectively, and such that their distance is equal to 6. (Consider the ball of center a and radius 8 and apply the result of Problem 1 .)
3.
Let E be a Hilbert space, F a dense linear subspace of E, distinct from E (Section 5.9, Problem 2). Show that there exists in the prehilbert space F a closed hyperplane H such that there is no vector # 0 in F which is orthogonal to H.
VI
122
4.
HlLBERT SPACES
Let X be a set, E a vector subspace of Cx, on which is given a structure of complex Hilbert space. A mapping ( x , y ) K(x, y ) of X x X into C is called a reproducing kernel for E if it satisfies the two following conditions: ( I ) for every y E X, the function K(. , y ) : x --f K(x, y ) belongs to E; (2) for any function f~ E, and any y E X, f(y) = (fI K ( . , A). (a) Show that K is a mapping of positive fype of X x X into C, i.e., for any integer of points of X , the mapping n 3 1 and any finite sequence ( x i ) , --f
+x W x i , xi)&
((Ad, (pi))
I. J
pi
of C2" into C is a positiue hermitian form. This in particular implies K(x, x ) 3 0 for every x E X, K(y, x ) = K(x, y ) and IK(x, y)12 < K(x, x ) K ( y ,y ) for x, y in X. Show that f0rj.E E, one has lf(y)I < llfll . (K(y, y))"' for y E X. (b) Show that if (f.)is a sequence of functions of E which converges (for the Hilbert space structure) to f~ E, then, for every x E X , the sequence (f.(x)) converges to f ( x ) in C; the convergence is uniform in any subset of X where the function x + K(x, x ) is bounded. n a finite sequence of points of X, (al)lslsn a sequence of n com(c) Let ( x i ) l i 1 Bbe plex numbers. Suppose det(K(x, , x i ) ) # 0, so that the system of linear equations J= 1
cj K(xi, x J )= al (1
< i < n) has a unique solution (ci). Show that among the func-
tions f~ E such that f(x1) = a , for 1 f i S n, the function fo
n
= cj J= I
K(. , x i ) has the
smallest norm. In particular, among all functions f~ E such that f ( x ) = 1 for a point E X where K(x, x ) # 0, the function K(. , x ) / K ( x ,x ) has the smallest norm. (d) If X is a topological space and if all the functionsf€ E are continu0u.s in X, then the functions K(.,x) (where x takes all values in X, or in a dense subset of X) form a total subset of E (show that there is no element h # 0 in E which is orthogonal to all the elements K(. , x ) ) . In particular, if X is a separable metric space, E is a separable Hilbert space. 5. (a) The notations being those of Problem 4, in order that there exist a reproducing kernel for E, a necessary and sufficient condition is that for every x E X, the linear form f + f ( x ) be continuous in E. The reproducing kernel is then unique. (b) Deduce from (a) that if there exists a reproducing kernel for E, there also exists a reproducing kernel for every closed vector subspace El of E. If K, is the reproducing kernel for E l , show that for every function f~ E, the function y --f (f1 K,(. ,y ) ) is the orthogonal projection o f f i n El. If E, is the orthogonal supplement of El and K2 the reproducing kernel for E, , then K I $- K 1 is the reproducing kernel for E. 6. Let X be a set, E a vector subspace of Cx, on which is given a structure of complex prehilbert space. In order that there should exist a Hilbert space c Cxcontaining E, such that the scalar product on E is the restriction of the scalar product on B, and that there exists a reproducing kernel for B, it is necessary and sufficient that E satisfy the two following conditions: ( I ) for every x f X, the linear form f + f ( x ) is continuous in E; (2) for any Cauchy sequence (f,)in E such that, for every x E X, lim f.(x) = 0, x
one has lim llfnll 20'"
= 0.
n-m
(To prove the conditions are sufficient, consider the subspace
f?
of Cx whose elements are the functionsffor which there exists a Cauchy sequence (fn) in E such that limfn(x) = f ( x ) for every x E X. Show that, for all Cauchy sequences (f.) n-r m
having that property, the number lim llfnll is the same, and if l i f l l is that number, this n-rm
defines on f a structure of normed space which is deduced from a structure of prehilbert
4 HILBERT SUM OF HILBERT SPACES
123
space which induces on E the given prehilbert structure; finally show that E is dense in 8 and that 8 is complete, hence a Hilbert space, and apply Problem 5(a) to 8.) 7. Let X be a set, f a mapping of X into a prehilbert space H; show that the mapping ( x , y ) + ( f ( x )1 f(y)) of X x X into C is of positive type (Problem 4(a)). 8 . Let X be a set, K a mapping of positive type of X x X into C (Problem 4(a)). (a) Let E be the set of mappings u : X -+ C such that there exists a real number a > 0 having the property that the mapping ( x , Y ) +aK(x, Y ) - 4x)UO
is of positive type; let m(u)be the smallest of all real numbersa 3 Ohavingthat property. Show that m(u) is also the smallest number c such that, for every finite sequence ( x i ) of elements of X, the inequality
holds for all complex numbers h, , p (use the Cauchy-Schwarz inequality). For every E X, show that Iu(x)12 < K(x, x)m(u). (b) Show that E is a vector subspace of Cx, that ( m ( ~ ) ) is ” ~a norm on E and that m(u u ) 4- m(u - u ) 2 2(m(u) m(u)).Conclude that there is a nondegenerate positive hermitian form g(u, u) on E x E such that g(u, u ) = m(u), and that for this form E is a Hilbert space; one writes g(u, u ) = (u I u). (Use Problem 2(c) of Section 6.2 to prove the existence of g ; to show that E is complete, use the last inequality proved in (a)). (c) For every x E X, show that the function K ( . , x ) belongs to E and that (K(. , x) I K ( . ,y ) ) = K(x, y ) for all ( x , y ) E X x X (use Cauchy-Schwarz inequality). Prove that if X is a topological space and if K is continuous in X x X, the mapping x + K(. , x) of X into E is continuous. (d) Deduce from (a) that the Hilbert space E defined in (b) has a reproducing kernel, and if F is the closed vector subspace of E generated by the functions K(. ,x), the reproducing kernel for F (Problem 5(b)) is equal to K. 9. Let E be a prehilbert space, N a finite dimensional vector subspace of E, M a vector subspace of E having infinite dimension, or finite dimension >dim(N). Show that there exists in M a point x # 0 such that llxll= d(x, N). (Consider the intersection of M and of the orthogonal supplement of N.) x
+
+
4. H I L B E R T SUM O F HILBERT SPACES
Let (En) be a sequence of Hilbert spaces; on each of the En, we write the scalar product as ( x , ) y n ) .Let E be the set of all sequences x = (xl, x 2 , . . . , x,, . . .) such that x, E En for each n , and the series ( I I x , ~ ~ ~ ) is convergent. We first define on E a structure of vector space: it is clear that if x = (x,) E E, then the sequence (Axl,.. . ,Ax,,,. . .) also belong to E. On the other hand, if y = (y,) is a second sequence belonging to E, we observe that IIx, ynl12< 2(/1~,11~ liyn112) by (6.3.1.1), hence the series (llx, + y,I12) is convergent by (5.3.1), and therefore the sequence (xl + y , , ..., x, + y n r ...) belongs to E. We define x + y = (x,+y,,),
+
+
124
V
HILBERT SPACES
Ax = (Ax,), and the verification of the axioms of vector spaces is trivial. On the other hand, from the Cauchy-Schwarz inequality, we have
IIxnII *
I(xn IYn)I G
IIYnIt G
HIIXnII’
+ IIynII’).
Therefore, if x = (x,) and y = (y,) are in E, the series (of real or complex numbers) ((x, I y,)) is absolutely convergent. We define, for x = (x,) and
y = (y,) in E, the number (x 1 y ) =
00
,= 1
(x, 1 y,); it is immediately verified
that the mapping (x, y ) -.+ (x I y ) is a Hermitian form on E. Moreover we have (x Ix) =
1 llxn\1’, hence ( x l y ) is a positive nondegenerate
n= 1
hermitian
form and defines on E a structure of prehilbert space. We finally prove E is in fact a Hilbert space, in other words it is complete. Indeed, let (xc”’)), where x(”’)=(x$)), be a Cauchy sequence in E: this means that for any E > 0 there is an m, such that for p 2 m, and q >, m, , we have m
(6.4.1)
For each fixed n, this implies first
IIxp) - xP)1l2< E ,
( x ~ ~ ) ) ~ = ~is , a~ ,Cauchy ... sequence in
hence the sequence
En,and therefore converges to a
limit y , . From (6.4.1) we deduce that for any given N N
C IIxn(P) - xn(4)II2 < &
n= I
as soon as p and q are 2 m o , hence, from the continuity of the norm, we deduce that
N
n= 1
1 1 ~ : ~ ) - y,II’
integers N, we have
m “.= .
for p 2 m,, and as this is true for all
IIxp)- ynllZG E .
I.
This proves first that the
sequence (xp’ - y,) belongs to E, hence y = (y,) also belongs to E, and - y11’ < E for p 2 m,, which ends the proof by showing we have (Idp) that the sequence (x(”’)) converges to y in E. We say that the Hilbert space E thus defined is the Hilbert sum of the sequence of Hilbert spaces (En). We observe that we can map each of the En into E by associating to each x, E En the sequence j,(x,) E E equal to (0, . . . , 0 , x, 0,. . .) (all terms 0 except the nth equal to x,); it is readily verified that j , is an isomorphism of En onto a (necessarily closed) subspace EA of E; j , is called the natural injection of En into E. From the definition of the scalar product in E, it follows that for m # n, any vector in EL is orthogonal to any vector in EL; furthermore, from the definition of the norm in E, it follows that for any x = (x,) E E, the series (j,(x,)) is convergent
4
in E, and x =
m n= 1
HILBERT SUM OF HILBERT SPACES
125
j,,(x,) (observe that the series (j,(x,)) is not absolutely con-
vergent in general). This proves that the (algebraic) sum of the subspaces EA of E (which is obviously direct) is dense in E, in other words that the smallest closed vector subspace containing all the EA is E itself. Conversely: (6.4.2) Let F be a Hilbert space, (F,) a sequence of closed subspaces such that: (1) for m # n, any vector of F , is orthogonal to any vector of F,; (2) the algebraic sum H of the subspaces F , is dense in F . Then, i f E is the Hilbert sum of the F,, there is a unique isomorphism of F onto E which on each F, coincides with the natural injection j , of F, into E.
Let F; = j,,(F,,), and let h, be the mapping of FA onto F,, inverse to j , . Let G be the algebraic sum of the FA in E; that sum being direct, we can define a linear mapping h of G into F by the condition that it coincides with h, on each FA. I claim that h is an isomorphism of G onto the prehilbert space H (which, incidentally, will prove that the (algebraic) sum of the F, is direct in F ) ; from the definition of the scalar product in E, we have to check that
for Xk E Fk,yk E F k ; but by assumption (x, I yk) = 0 if h # k, and the result follows from the fact that each j k is an isomorphism. There is now a unique continuous extension T; of h which is a linear mapping of = E into = F, by (5.5.4); the principle of extension of identities (3.1 5.2) and the continuity of the scalar product show that h is an isomorphism of E onto a subspace of F, which, being complete and dense, must be F itself; the inverse of h satisfies the conditions of (6.4.2). Its uniqueness follows from the fact that it is completely determined in G and continuous in E (3.15.2). Under the condition of (6.4.2), the Hilbert space F is often identified with the Hilbert sum of its subspaces F , .
e
Remark (6.4.3)
We can also prove (6.4.2) by establishing first that the sum of the F, n
is direct; indeed, if E x i = 0 with x i € Fi (1 < i < n) we also have i = l.
= 0 for any j
< n, and as (xjIxi) = 0 for i # j , this boils down
126
VI HILBERT SPACES
to llxil12 = 0, hence xi = 0 for 1 < j < n. Then we define the inverse mapping g of h by the condition that it coincides with j,, on each F,,: we at once verify, as above, that g is an isomorphism of H onto G, and then (5.5.4) is applied in the same way. We observe that this argument still applies when F is a prehilbert space and the F, are complete subspaces of F; it proves the existence of an isomorphism of F onto a dense subspace of the Hilbert sum E of the F,,, which coincides withj,, on each F, .
PROBLEM
Let X be a set, E l , Ez two vector subspaces of Cx, possessing reproducing kernels K,, K2 (Section 6.3, Problem 4). Let E = El E2 c Cx,and let F be the Hilbert sum of the Hilbert spaces El, Ez;the kernel of the surjective mapping u : (fl, fz)-fl +fz of F into E is the subspace N of F consisting of the pairs (f,-f) where YE El n EZ. Show that N is closed in F; if H is the orthogonal supplement of N in F, the restriction u of u to H is a bijection of H onto E; transporting by u the Hilbert structure of H, E is defined as a Hilbert space. Show that for that Hilbert space structure, E has a reproducing kernel equal to K1 Kz; the norm llfll in E is equal to inf(llflII: llfzll~)l~zwhen (fl,f2) takes all values in F such thatf=fl +fz(llfl Illand llf~ [I2 being the norms in El and E2 ,respectively).
+
+
+
5. ORTHONORMAL SYSTEMS
If (with the notations of Section 6.4) we take for each E,, a one-dimensional space (identified to the field of scalars with the scalar product (5 I q) = (fj), the Hilbert sum yields an example of an infinite dimensional Hilbert space E, which is usually written l2 (with index R or C to indicate if necessary what the scalars are); the space I: (resp. 1;) is therefore the space of all sequences x = (5,) of real (resp. complex) numbers, such that convergent, with the scalar product ( x I y ) =
c m
n=l
{, ij,
gi
l{,,I2 is
n= 1
.
In 12, let en be the sequence having all its terms equal to 0 except the nth term equal to 1; we then have (em1 en)= 0 for m # n, and IJe,,JI= 1 for each n, and we have seen in Section 6.4 that for every x = (t,) in f 2 , we can write x =
c <,en, the series being convergent in 12. We observe that this m
n= 1
shows the sequence (en)is total in 12, hence (5.10.1) 1’ is separable. Let us now consider an arbitrary prehilbert space F; we say that a (finite or infinite) sequence (a,) in F is an orthogonal system if (a, I a,) = 0 for m # n and a,, # 0 for every n ; we say that (a,) is an orthonormal system
5 ORTHONORMAL SYSTEMS
127
if in addition lla,,II = 1 for each n. From any orthogonal system (a,,) we deduce at once an orthonormal system by “normalizing” (a,,), i.e. considering the sequence of the b,, = a,,/\lanl\.We have just seen an example of an orthonormal system in 1’; another fundamental example is the following: (6.5.1) Let I be the interval [- 1, 11 of R, and let F = VJI) be the vector space of all continuous complex-valued functions defined on I. We define on F a scalar product by
J -1
(the fact that this is a nondegenerate positive hermitian form is readily verified). For each positive or negative integer n, let
q,,(t) = ennit. It is readily verified that (q,,/,/2)is an orthonormal system in F, called the trigonometric system. Let now (a,) be an arbitrary orthonormal system in a Hilbert space F; for each x E F, we say that the number c,,(x) = ( x I a,,) is the nth coefficient (ornth coordinate) ofx withrespect to the system (a,,) (“nth Fourier coefficient ” of x for the system (6.5.1)). (6.5.2) In a Hilbert space F, let (a,,) be an orthonormal system, V the closed subspace of F generated by the a,, . Then,for any x E F; 1.
the series
m
1 I(x I a,,)1’
n= 1
is convergent, and we have
and
2.
the series of general term (x I a,)a,, is convergent in F and we have m
1 (x I an)% =
n= 1
P”(X).
Conversely, let (A,,) be a sequence of scalars such that
m
1 1A,,I2 is convergent.
n=1
Then, there exists a unique vector y E V such that ( y I a,,) = A,, for every n ; any other vector x E F such that (X I a,,) = Anfor every n is such that x = y + z , with z orthogonal to V, and conversley.
128
VI HILBERT SPACES
For any x E F, we can write x = Pv(x) + 2 , with z orthogonal t o V (6.3.1) and we have therefore (x I a,) = (P,(x) I a,). To prove the theorem, we may therefore assume V = F; but then, the one-dimensional subspaces F, generated by the vectors a, satisfy the assumptions of (6.4.2), and the results are mere restatements of (6.4.2) for that particular case (taking into account the definition of a Hilbert sum). The most interesting case is that in which V = F, i.e., the orthogonal system (a,) is total. It is then called a Hilbert basis for F; (c,) is such a basis for 12. It will be proved in (7.4.3)that the trigonometric system (6.5.1) is total. For a Hilbert space F and a total orthonormal system (a,), we can replace everywhere P, by the identity in (6.5.2); the relations n
n=
I
are then called Parseral’s identities. It follows at once from (6.5.2) that these identities represent not only necessary but su#cient conditions for (a,) to be a total system in a Hilbert space. (6.5.3) It1 a Hilbert space F, a necessary arid sufJicient condition for an orthogonal system (a,) to be total is that the relations (x I a,) = 0 for every n imply .x = 0. Indeed, by (6.5.2) this means that the relation Pv(x) = 0 implies x = 0, and this is equivalent to the relation V = F, since P,(x - P v ( x ) )= 0.
Remark (6.5.4) Suppose E is a prehilbert space and the orthonormal system (a,) in E is total. Then the results ( I ) and (2) of (6.5.2) are still valid, with Pv(x) replaced by x; this follows by the same argument as in (6.5.2), using the Remark (6.4.3).
PROBLEMS
1. Let E be a Hilbert space with a Hilbert basis (e,,),,zl.Let A be the subset of E
consisting of the linear combinations x (n arbitrary).
n
= k= 1
k= I
6 ORTHONORMALIZATION (a> Show that the closure where An > 0, the series
-
A is the set of all the sums of the series
C A,, is convergent I
"=I
129
"= 1
and has a sum equal to I .
(b) Prove that the diameter of A is equal to d2 but that there is no pair of points A such that /la- b/l = 42 (compare to Section 6.3, Problem 2). Let E be a Hilbert space with a Hilbert basis Let a, = eZn and
a, h of
b, = ezn
1 ++ 1 e z n f lfor every n > 0; letA (resp. B) be the closed vector subspace of E I1
generated by the a. (resp. b"). Show that: (a) A n B = { O } , hence the sum A B is direct (algebraically). (b) The direct sum A B is not a topological direct sum (consider in that subspace the sequence of points h, - a, and apply (5.4.2.)). (c) The subspace A B of E is dense but not closed in E (show that the point m
"=O
+
+ +
(b, - an)does not belong to A
+ B).
Show that the Banach space 9 ( P ; 1 2 ) can be identified with the space of double
sequences U = (a,,,,) such that: (I) the series ( 2 ) sup n
m
nt=O
m
nr= 0
is convergent for every n:
la,,,nlzis finite. The norm is then equal to IlUIj = sup
"
method as in Section 5.7, Problem 2(b)). (a) Let u be a continuous linear mapping of show that the series
c m
fl=O
and that their sums are on E and use (6.3.2).)
/cz,,,12 and
<
c m
P
/a,,,,Jz are
into itself, and let u(e,,) =
(Observe that x
--z
x
m=0
la,nnlz <1
for all values of m and
11,
m
1 a,m,e,;
lll=O
( ~ ( xI e,") ) is a continuous linear form
(b) Give an example of a double sequence (a,",J such that m
(same
convergent for all values of m and n,
Wt=O
lIi(l/'.
imYo)
112
(a,,,('
m
"=O
<
(anln/* 1
and
but such that there is no continuous linear
mapping it of P into itself satisfying the relations (u(e,) I e,,,)= a,,,, for all pairs (m, n). (If V is a subspace of I z generated by the vectors en with n E H, where H is a set of p integers, show that there is a linear mapping up of V into itself such that (//,,(en)I e",)= I/t'p for all indices m, 11 of the set H, but l ! i t p l l > v'p.)
6. ORTH 0 NORMALl ZATl O N
(6.6.1) Let E be a separable preliilbert space, (6,) a total sequence of linearly independent cectors in E (see (5.10.1)), and let V, be the n-dimensional subspace of E generated by b,, . . . , 6,. Then if we define c, = b, - Pv,-,(b,), (c,) is a total orthogonal system, such that cl, . . . , c, generate V, for each n.
We use induction on n, assuming that cl, . . . , c , - ~ is an orthogonal system generating V n - l ; then, by definition of Pv,-l (6.3.1),c, is orthogonal
130
VI
HILBERT SPACES
to V,,-l, which proves that (ciIc j ) = 0 for 1 Q i < j < n ; moreover, as b, # V,-l by assumption, c, # 0, hence e l , . , . , c, is an orthogonal system; moreover, b,, - c,, E V,,...l, hence c l , . , . , c, generate the same subspace as the union of V,-l and {b,,},i.e. V , . This completes the proof.
If we normalize the system (c,,), by putting a, = c,,/~lc,,Il, the system (a,) is said to be deduced from (h,,) by the orthonormalization process. For instance, in the space F = %'Jl) considered in (6.5.1), the sequence (t") is total (as will be proved in (7.4.1)) and obviously consists of linearly independent vectors. If we denote by (Q,) the orthonormal system deduced from (t") by orthonormalization, it is clear that Q,,(t) = a,, t" + * . * , polynomial of degree n (with a, # 0) with real coefficients; the Q, are (up to a constant factor) the Legendre polynomials (see Section 8.14, Problem 1). (6.6.2) Any separable prehilbert space (resp. Hilbert space) is isomorphic to a dense subspace of I' (resp. to 1'). As there exists in a separable prehilbert space a total orthonormal system by (6.6.1), the result follows at once from (6.5.2).
PROBLEMS
1. Let E be a separable noncomplete prehilbert space. Show that there exists in E an orthonormal system which is not total, but which is not properly contained in any orthonormal system (imbed E as a dense subspace of a Hilbert space, and use problem 3 of Section 6.3). 2. Let E be a n infinite dimensional separable Hilbert space, V a closed vector subspace of E. Show that if V is infinite dimensional, there exists an isometry of E onto V (write E as the direct sum of V and its orthogonal supplement V', and take Hilbert bases in V and V'. 3. Let (xi)l<, C n be a finite sequence of points in a prehilbert space E. The Gram determinant of that sequence is the determinant G ( x l ,xz , . . . ,x.) = det((xi 1 x i ) ) . (a) Show that G ( x , ,. . . , x.) 2 0 and that G(x1,. . , x,) = 0 if and only if the x I are linearly dependent. (Consider a Hilbert basis of the sub space generated by the xi, and express the x1 as linear combinations of that basis.) (b) Suppose the xi are linearly independent, and let V be the n-dimensional subspace which they generate. Show that the distance of a point x to V is equal to
.
(G(x,X I , * 4.
I . ,
xn)/G(xi,. . . ,x,))"~
(find the projection of x on V, writing it as a linear combination of the xi). Let M be a compact subset of a Hilbert space E; if El is the smallest closed vector subspace of E containing M, show that El is separable.
6
ORTHONORMALIZATION
131
5. Let E c Cx a separable Hilbert space having a reproducing kernel (Section 6.3, Problem 4). (a) Let (At) be a Hilbert basis for E. Show that for every (x, y ) E X x X one has K(x, y ) = c f . ( x ) f , o , where the series converges in C. For any function g E E, if c, = (g
Ifn),
I!
one has, for every x E X, g(x)
=c n
cnfn(x), where
the series converges in C;
in addition, that series is uniformly convergent in every subset of X where K(x, x ) is bounded. (b) Conversely, let (fn) be a sequence of complex functions defined in a set X, such Ifn(x)12 < ca. For every sequence (c,) E I;, the series that for every x E X,
+
cnf.(x) is then convergent in
C for every x
E
X. The functions which are the sums of
such series form a vector subspace E c Cx,on which one defines a structure of separable Hilbert space by taking as scalar product, for u c,f,, u d. f,, the number
=c
(u I u ) =
n
c. n
=c n
-
d, . This space has a reproducing kernel K(x, y ) =cfn(x)J,(y). "
HILBERT SPACES
Hilbert spaces constitute at present the most important examples of Banach spaces, not only because they are the most natural and closest generalization, in the realm of “ infinite dimensions,” of our classical euclidean geometry, but chiefly for the fact that they have been, up to now, the most useful spaces in the applications to functional analysis. With the exception of (6.3.1),all the results easily follow from the definitions and from the fundamental Cauchy-Schwarz inequality (6.2.4).
1. H E R M I T I A N FORMS
For any real or complex number A, we write X for its complex conjugate (equal to A if A is real). A hermitiun form on a real (resp. complex) vector space E is a mapping f of E x E into R (resp. C) which has the following properties : (1) (11)
(W (W (V)
f(x + x‘,Y ) =f(x, Y ) + f ( x ’ , Y ) ,
+ Y’) = A x ,Y ) +m, Y’), f ( kr) = Af(x,Y ) ,
f(x, Y
f ( x , AY) = m x ,Y>, f ( Y >4
=f(% Y ) .
(Observe that (11) and (IV) follow from the other identities; (V) implies that f(x, x) is real.) When E is a real vector space, conditions (I) to (IV) express that f is bilinear and (V) boils down to f ( y , x) = f ( x , y), which 115
116
VI
HILBERT SPACES
expresses t h a t f i s symmetric. For any finite systems ( x i ) ,( y j ) , (ai), (Bj), we have (6.1 .I)
by induction on the number of elements of these systems. From (6.1.1) it follows that if E is finite dimensional and ( a i ) is a basis of E, f is entirely determined by its values a i j = f ( a i , a j ) , which are such that (by V)) aJ,l. = a I,J.’
(6.1.2)
Indeed we have then, for x = (6.1.3)
1 S i a i ,y = i
S ( X .,Y>
viai I
C aij ti
i, i
iij
.
Conversely, for any system ( a i j ) of real (resp. complex) numbers satisfying (6.1.2), the right hand side of (6.1.3) defines on the real (resp. complex) finite dimensional vector space E a hermitian form. Example (6.1.4) Let D be a relatively compact open set in R2, and let E be the real (resp. complex) vector space of all real-valued (resp. complex-valued) bounded continuous functions in D, which have bounded continuous first derivatives in D. Then the mapping
(where a, h, c are continuous, bounded and real-valued in D) is a hermitian form on E. A pair of vectors x, y of a vector space E is orthogonal with respect to a hermitian form f on E if f ( x , y ) = 0 (it follows from (V) that the relation is symmetric in x, y ) ; a vector x which is orthogonal to itself (i.e.f(x, x) = 0) is isotropic with respect to$ For any subset M of E, the set of vectors y which are orthogonal to all vectors x E M is a vector subspace of E, which is said to be orthogorial to M (with respect t o f ) . It may happen that there exists a vector a # 0 which is orthogonal to the whole space E, in which case we say the formfis degenerate. On a finite dimensional space E, nondegenerate hermitian forms f defined by (6.1.3) are those for which the matrix ( a i j ) is invertible.
2
POSITIVE HERMITIAN FORMS
117
PROBLEMS
1. (a) Let f be a hermitian form on a vector space E. Show that if E is a real vector space, then 4f(x, Y> = f ( x
+ Y , x +v) -f(x
and if E is a complex vector space
4fh, Y ) =f(x
+ y. x + y ) - f ( x
-Y ,x
-y)
-Y, x
-Y)
+ i f ( x + iy, x + iy) - if(x - iy, x
- iy).
(b) Deduce from (a) that if f ( x , x ) = 0 for every vector in a subspace M of E, then f ( x , y ) = 0 for any pair of vectors x , y of M. (c) Give a proof of (b) without using the identities proved in (a). (Write that f(x hy, x +Xu) = O for any h.) 2. Let E be a complex vector space. Show that iffis a mapping of E x E into C satisfying conditions (I), (II), (III), (IV), and such that f ( x , x ) E R for every x E E, then f is a hermitian form on E.
+
2. POSITIVE H E R M I T I A N FORMS
We say a hermitian form f on a vector space E is positive iff(x, x) 2 0 for any x E E. For instance, the form q defined in example (6.1.4) is positive if a, b, c are 2 0 in D. (6.2.1)
then
(Cauchy-Schwarz inequality)
for any pair of vectors
X,
I f f is a positive hermitian form,
If(x, Y>12 < f ( x , X)f(Y, Y ) y in E.
Write a =f ( x , x), b = f ( x , y ) , c = f ( y , y ) and recall a and c are real and 2 0. Suppose first c # 0 and write that f ( x I y , x + l y ) 2 0 for any scalar 1, which gives a + b2 + bA + c3J 0 ; substituting I = -b/c yields the inequality. A similar argument applies when c = 0, a # 0 ; finally if a = c = 0, the substitution I = - h yields -266 2 0, i.e. b = 0.
+
(6.2.2) In order that a positii!e hermitian form f on E be nondegenerate a necessary and suficient condition is tliat there exist no isotropic vector for f otlier than 0 , i.e. that f ( x , x) > 0 for any x # 0 in E.
Indeed, f ( x , x) all y E E.
=0
implies, by Cauchy-Schwarz, that f (x, y ) = 0 for
118
(6.2.3)
VI
HILBERT SPACES
(Minkowski's inequality) I f f is a positi1.e hermitian form, tlien (f(x
+y, x +
d ( A x , x))''2 + ( f b ,
for any pair of vectors x, y in E.
+
As f ( x y , x equivalent to
+ y ) = f ( x , x) + f ( x , y ) + A x , y ) + f ( y , y ) , the inequality is
2,gf(x,Y)= f ( x , Y >+f(x,P,)d 2 ( f ( x , x)f(Y,Y))'12 which follows from Cauchy-Schwarz. The function x -+ ( f ( x , x))'12 therefore satisfies the conditions (I), (Ill), and (IV)of (Section 5.1); by (6.2.2), condition (11) of Section 5.1 is equivalent to the fact that the form f is nondegenerate. Therefore, when f is a nondegenerate positive hermitian form (also called a positive dejnite form), ( f ( x , x))'I2 is a norm on E. A preliilbert space is a vector space E with a given nondegenerate positive hermitian form on E; when no confusion arises, that form is written (x I y ) and its value is called the scalar product of x and y ; we always consider a prehilbert space E as a normed space, with the norm llxll = (x I x)''', and of course, such a space is always considered as a metric space for the corresponding distance Ijx - y ( / . With these notations, the Cauchy-Schwarz inequality is written
I(xlu)l d IIxIl
(6.2.4)
IIYII
and this proves, by (5.5.1), that for a real prehilbert space E, (x, y ) -+ (x 1 y ) is a continuous bilinear form on E x E (the argument of (5.5.1) can also be applied when E is a complex prehilbert space and proves again the continuity of (x, y ) (x I y ) , although this is not a bilinear form any more). We also have, as a particular case of (6.1 .I): --f
(6.2.5) (Pythagoras' theorem) onal vectors, IIX
In a prehilbert space E, i f x , y are orthog-
+ YIl2 = l/xll2+ llY/I2.
A n isomorphism of a prehilbert space E onto a prehilbert space E' is a linear bijection of E onto E' such that ( f ( x )If(y)) = (x Iy) for any pair of vectors x, y of E. It is clear that an isomorphism is a linear isonzetry of E onto E'. Let E be a prehilbert space; then, on any vector subspace F of E, the restriction of the scalar product is a positive nondegenerate hermitian form ; unless the contrary is stated, it is always that restriction which is meant when F is considered as a prehilbert space.
3 ORTHOGONAL PROJECTION ON A COMPLETE SUBSPACE
119
A Hilbert space is a prehilbert space which is coniplete. Any finite dimensional prehilbert space is a Hilbert space by (5.9.1); other examples of Hilbert spaces will be constructed in Section 6.4. If in example (6.1.4) we take a > 0, b 2 0 , c 3 0, it can be shown that the prehilbert space thus defined is not complete.
PROBLEMS
1. Prove the last statement in the case a = 1 , b = c = 0 (see Section 5.1, Problem 1). 2. (a) Let E be a real nornied space such that, for any two points ,Y, y of E, 1I.y
+ Yl12 + Ilx - Y1IZ
= 2(11x1I2
+ llvIl2).
Show that f ( x , y) = IIx i-V I -~ Ilxl/* ~ - /IyIl2 is a positive nondegenerate hermitian form on E. (In order to prove that f ( h x , y ) = h f ( x , y ) , use Problem I of Section 5.5.) (b) Let E be a complex nornied space, Eo the underlying real vector space. Suppose there exists on Eo a symmetric bilinear form f ( x ,y) such that f ( x , x) = l1x/l2for every x E E o . Show that there exists a herniitian form g(x, y ) on E such that f ( x , y ) = .jR(g(x, y ) ) , hence g(x, x) = 1/x1l2for x E E. (Using the first formula of Problem 1 of Section 6.1, prove that f ( i x , y ) = - f ( x , i-v).) (c) Let E be a complex normed space such that Ilx - Y1I2
+ Ilx + Y / I 2 < 2(11x1I2 + llvll’)
for any pair of points x , y of E. Show that there exists a nondegenerate positive hermitian fornif(x,y) on E such thatf(x, x) = XI/^. (Use (a) and (b)). 3. Letfbe a positive nondegenerate hermitian form. In order that both sides of (6.2.1) be equal, it is necessary and sufficient that x and y be linearly dependent. In order that both sides of (6.2.3) be equal, it is necessary and sufficient that x and ,y be linearly dependent, and, if both are # 0, that y = Ax, with real and > 0. 4. Let a, b, c, d be four points in a prehilbert space E. Show that
/(a- c ( / . / / b- d ( (< /la - bll . JJc- dll
5.
+ 116
- cii
. / / a- dll.
(Reduce the problem to the case a = 0, and consider in E the transformation x+x/llxl~’, defined for x # 0.)When are both sides of the inequality equal? be a finite sequence of points in a prehilbert space E. Show that Let ( x i ) ,
If /lxi - x,Jl > 2 for i # j , show that a ball containing all the x i has a radius >(2(n
-
1)jn)’”.
3. O R T H O G O N A L PROJECTION O N A COMPLETE SUBSPACE
(6.3.1) Let E be a prehilbert space, F a complete tvctor subspace of E (i.e. a Hilbert space). For any x E E, there is one and only one point y = P F ( x )E F
120
VI
HILBERT SPACES
such that I/x - J~II = d(x, F). Tlze point y = P,(x) is also the only point z E F such that x - z is orthogonal to F. The mapping x P,(x) of E onto F is linear, continuous, and of norm 1 if F # { O } ; its kernel F’ = PF1(0) is the subspace orthogonal to F, and E is the topological direct sum (see Section (5.4)) of F and F‘. Finally, F is the subspace orthogonal to F‘. --f
Let CI = d(x, F); by definition, there exists a sequence (y,) of points of F such that lim 1I.x - ynl/ = a ; we prove (y,,) is a Cauchy sequence. fl-+
co
Indeed, for any two points u, u of E, it follows from (6.1.1) that (6.3.1.1)
hence
112.4
llym
+
+ -
U(l2
1/24
PJ12= 2(llu112
+ 11z~112>
- yn/I2= 2(llx - ~ , ~ ,+l /IIx ~ - ~ , , l l ~-) 4llx - ! d Y m
+ y,)112.
But
+ y,,) E F, hence jlx - )(ym + y,,)lI2 > a 2 ;therefore, if no is such that for n 2 n o , ( / x- y , ( / 2< g2 + E , we have, form > no and n 3 n o , I/ym- Y,;;? < 4e, +(yJ,
which proves our contention. As F is complete, the sequence (y,) tends to a limit y E F, for which IJx- yll = d(x, F). Suppose y’ E F is also such that /Ix - y’ll = d(x, F); using again (6.3.11), we obtain / ( y- y’(I2= 4a2- 4 /Jx- ) ( y y’)1I2, and as j ( y y ’ ) E F, this implies (ly - y’1I2 < 0,i.e. y’ = y. Let now z # 0 be any point of F, and write that (lx - ( y Az)/l* > a2 for any real scalar A # 0; this, by (6.1 .I), gives
+
+
21B(x - y 1 z )
+
+ R 2 / ( 2 / ( 2> 0
and this would yield a contradiction if we had 9 ( x - y 1 z ) # 0, by a suitable choice of 1. Hence B(x - y 1 z ) = 0, and replacing z by iz (if E is a complex prehilbert space) shows that #(x - y I z ) = 0, hence (x - y I z ) = 0 in every case; in other words x - y is orthogonal to F. Let y’ E F be such that x -y’ is orthogonal to F; then, for any z # 0 in F, we have JIx- (y’ z)1I2 = /Jx- y‘/12 j/zIlz by Pythagoras’ theorem, and this proves that y’ = y by the previous characterization of y . This last characterization of y = P,(x) proves that P , is linear, for if x - y and x’- y’ are orthogonal to F, then Ax - Ay is orthogonal to F and so is (x + x’)- ( y + y’) = (x - y ) (x’ - y ’ ) ; as y y’ E F and Ay E F, this shows that y y’ = P F ( x x’)and Ay = P , ( ~ x ) .By Pythagoras’ theorem, we have
+
+
+
(6.3.1.2)
+
llX/12
=
+
IIpF(x)112
+
+ (Ix
- pF(x)l/2
and this proves that IlPF(x)l1< I/x/J,hence (5.5.1) P, is continuous and has norm d 1 ; but as P F ( x )= x for x E F, we have /lP,I/ = 1 if F is not reduced to 0. The definition of P, implies that F’ = PF’(0) consists of the vectors x orthogonal to F; as x = P,(x) + (X - PF(,y)) and x - P F ( x )E F’ for any x E E, we have E = F F’; moreover, if x E F n F‘, x is isotropic, hence
+
3 ORTHOGONAL PROJECTION ON A COMPLETE SUBSPACE
x
= 0, and
121
this shows that the sum F + F' is direct. Furthermore, the mapping
x -iP F ( x )being continuous, E is the topological direct sum of F and F' (5.4.2). Finally, if x E E is orthogonal to F', we have in particular (x I x - P F ( x ) )= 0 ; but we also have (PF(x) I x - PF(x))= 0, hence /Ix - PF(x)(I2= 0, i.e. x = PF(x) E F. Q.E.D. The linear mapping PF is called the orthogonal projection of E onto F, and its kernel F' the orthogonal supplement of F in E. Theorem (6.3.1) can be applied to any closed subspace F of a Hilbert space E (by (3.14.5)), or to any Jinite dimensional subspace F of a prehilbert space, by (5.9.1).
(6.3.2) Let E be a prehilbert space; then, for any a E E, x -i (x I a) is a continuous linear form of norm Ilall. Conversely, if E is a Hilbert space, for any continuous linear form u on E, there is a unique vector a E E such that u(x> = (x I a)for any x E E. ByCauchy-Schwarz,I(x I a)l < llall * IIxII, whichshows(by5.5.1))~+ (x la) is continuous and has a norm < llalj; on the other hand, if a # 0, then for x, = a / ~ ~ we a ~have ~ , (x, I a) = llall; as llxoII = 1, this shows the norm of x - i ( x l a ) is at least Ilall. Suppose now E is a Hilbert space; the existence of the vector a (=O) being obvious if u = 0, we can suppose u # 0. Then H = u-'(O) is a closed hyperplane in E; the orthogonal supplement H' of H is a one-dimensional subspace; let b # 0 be a point of H'. Then we have ( x I b) = 0 for any x E H. But any two equations of a hyperplane are proportional, hence there is a scalar L such that u(x) = L(x 16) = (xI a) with a = Xb (see Appendix) for all x E E. The uniqueness ofa follows from the fact that the form (x I y ) is nondegenerate.
PROBLEMS 1. Let B be the closed ball of center 0 and radius 1 in a prehilbert space E. Show that for each point x of the sphere of center 0 and radius I , there exists a unique hyperplane of support of B (Section 5.8, Problem 3) containing x . 2. Let E be a prehilbert space, A a compact subset of E, 6 its diameter. Show that there exist two points a , b of A such that /la- bll = 8 and that there are two parallel hyperplanesof support of A (Section 5.8, Problem 3)containingaand b respectively, and such that their distance is equal to 6. (Consider the ball of center a and radius 8 and apply the result of Problem 1 .)
3.
Let E be a Hilbert space, F a dense linear subspace of E, distinct from E (Section 5.9, Problem 2). Show that there exists in the prehilbert space F a closed hyperplane H such that there is no vector # 0 in F which is orthogonal to H.
VI
122
4.
HlLBERT SPACES
Let X be a set, E a vector subspace of Cx, on which is given a structure of complex Hilbert space. A mapping ( x , y ) K(x, y ) of X x X into C is called a reproducing kernel for E if it satisfies the two following conditions: ( I ) for every y E X, the function K(. , y ) : x --f K(x, y ) belongs to E; (2) for any function f~ E, and any y E X, f(y) = (fI K ( . , A). (a) Show that K is a mapping of positive fype of X x X into C, i.e., for any integer of points of X , the mapping n 3 1 and any finite sequence ( x i ) , --f
+x W x i , xi)&
((Ad, (pi))
I. J
pi
of C2" into C is a positiue hermitian form. This in particular implies K(x, x ) 3 0 for every x E X, K(y, x ) = K(x, y ) and IK(x, y)12 < K(x, x ) K ( y ,y ) for x, y in X. Show that f0rj.E E, one has lf(y)I < llfll . (K(y, y))"' for y E X. (b) Show that if (f.)is a sequence of functions of E which converges (for the Hilbert space structure) to f~ E, then, for every x E X , the sequence (f.(x)) converges to f ( x ) in C; the convergence is uniform in any subset of X where the function x + K(x, x ) is bounded. n a finite sequence of points of X, (al)lslsn a sequence of n com(c) Let ( x i ) l i 1 Bbe plex numbers. Suppose det(K(x, , x i ) ) # 0, so that the system of linear equations J= 1
cj K(xi, x J )= al (1
< i < n) has a unique solution (ci). Show that among the func-
tions f~ E such that f(x1) = a , for 1 f i S n, the function fo
n
= cj J= I
K(. , x i ) has the
smallest norm. In particular, among all functions f~ E such that f ( x ) = 1 for a point E X where K(x, x ) # 0, the function K(. , x ) / K ( x ,x ) has the smallest norm. (d) If X is a topological space and if all the functionsf€ E are continu0u.s in X, then the functions K(.,x) (where x takes all values in X, or in a dense subset of X) form a total subset of E (show that there is no element h # 0 in E which is orthogonal to all the elements K(. , x ) ) . In particular, if X is a separable metric space, E is a separable Hilbert space. 5. (a) The notations being those of Problem 4, in order that there exist a reproducing kernel for E, a necessary and sufficient condition is that for every x E X, the linear form f + f ( x ) be continuous in E. The reproducing kernel is then unique. (b) Deduce from (a) that if there exists a reproducing kernel for E, there also exists a reproducing kernel for every closed vector subspace El of E. If K, is the reproducing kernel for E l , show that for every function f~ E, the function y --f (f1 K,(. ,y ) ) is the orthogonal projection o f f i n El. If E, is the orthogonal supplement of El and K2 the reproducing kernel for E, , then K I $- K 1 is the reproducing kernel for E. 6. Let X be a set, E a vector subspace of Cx, on which is given a structure of complex prehilbert space. In order that there should exist a Hilbert space c Cxcontaining E, such that the scalar product on E is the restriction of the scalar product on B, and that there exists a reproducing kernel for B, it is necessary and sufficient that E satisfy the two following conditions: ( I ) for every x f X, the linear form f + f ( x ) is continuous in E; (2) for any Cauchy sequence (f,)in E such that, for every x E X, lim f.(x) = 0, x
one has lim llfnll 20'"
= 0.
n-m
(To prove the conditions are sufficient, consider the subspace
f?
of Cx whose elements are the functionsffor which there exists a Cauchy sequence (fn) in E such that limfn(x) = f ( x ) for every x E X. Show that, for all Cauchy sequences (f.) n-r m
having that property, the number lim llfnll is the same, and if l i f l l is that number, this n-rm
defines on f a structure of normed space which is deduced from a structure of prehilbert
4 HILBERT SUM OF HILBERT SPACES
123
space which induces on E the given prehilbert structure; finally show that E is dense in 8 and that 8 is complete, hence a Hilbert space, and apply Problem 5(a) to 8.) 7. Let X be a set, f a mapping of X into a prehilbert space H; show that the mapping ( x , y ) + ( f ( x )1 f(y)) of X x X into C is of positive type (Problem 4(a)). 8 . Let X be a set, K a mapping of positive type of X x X into C (Problem 4(a)). (a) Let E be the set of mappings u : X -+ C such that there exists a real number a > 0 having the property that the mapping ( x , Y ) +aK(x, Y ) - 4x)UO
is of positive type; let m(u)be the smallest of all real numbersa 3 Ohavingthat property. Show that m(u) is also the smallest number c such that, for every finite sequence ( x i ) of elements of X, the inequality
holds for all complex numbers h, , p (use the Cauchy-Schwarz inequality). For every E X, show that Iu(x)12 < K(x, x)m(u). (b) Show that E is a vector subspace of Cx, that ( m ( ~ ) ) is ” ~a norm on E and that m(u u ) 4- m(u - u ) 2 2(m(u) m(u)).Conclude that there is a nondegenerate positive hermitian form g(u, u) on E x E such that g(u, u ) = m(u), and that for this form E is a Hilbert space; one writes g(u, u ) = (u I u). (Use Problem 2(c) of Section 6.2 to prove the existence of g ; to show that E is complete, use the last inequality proved in (a)). (c) For every x E X, show that the function K ( . , x ) belongs to E and that (K(. , x) I K ( . ,y ) ) = K(x, y ) for all ( x , y ) E X x X (use Cauchy-Schwarz inequality). Prove that if X is a topological space and if K is continuous in X x X, the mapping x + K(. , x) of X into E is continuous. (d) Deduce from (a) that the Hilbert space E defined in (b) has a reproducing kernel, and if F is the closed vector subspace of E generated by the functions K(. ,x), the reproducing kernel for F (Problem 5(b)) is equal to K. 9. Let E be a prehilbert space, N a finite dimensional vector subspace of E, M a vector subspace of E having infinite dimension, or finite dimension >dim(N). Show that there exists in M a point x # 0 such that llxll= d(x, N). (Consider the intersection of M and of the orthogonal supplement of N.) x
+
+
4. H I L B E R T SUM O F HILBERT SPACES
Let (En) be a sequence of Hilbert spaces; on each of the En, we write the scalar product as ( x , ) y n ) .Let E be the set of all sequences x = (xl, x 2 , . . . , x,, . . .) such that x, E En for each n , and the series ( I I x , ~ ~ ~ ) is convergent. We first define on E a structure of vector space: it is clear that if x = (x,) E E, then the sequence (Axl,.. . ,Ax,,,. . .) also belong to E. On the other hand, if y = (y,) is a second sequence belonging to E, we observe that IIx, ynl12< 2(/1~,11~ liyn112) by (6.3.1.1), hence the series (llx, + y,I12) is convergent by (5.3.1), and therefore the sequence (xl + y , , ..., x, + y n r ...) belongs to E. We define x + y = (x,+y,,),
+
+
124
V
HILBERT SPACES
Ax = (Ax,), and the verification of the axioms of vector spaces is trivial. On the other hand, from the Cauchy-Schwarz inequality, we have
IIxnII *
I(xn IYn)I G
IIYnIt G
HIIXnII’
+ IIynII’).
Therefore, if x = (x,) and y = (y,) are in E, the series (of real or complex numbers) ((x, I y,)) is absolutely convergent. We define, for x = (x,) and
y = (y,) in E, the number (x 1 y ) =
00
,= 1
(x, 1 y,); it is immediately verified
that the mapping (x, y ) -.+ (x I y ) is a Hermitian form on E. Moreover we have (x Ix) =
1 llxn\1’, hence ( x l y ) is a positive nondegenerate
n= 1
hermitian
form and defines on E a structure of prehilbert space. We finally prove E is in fact a Hilbert space, in other words it is complete. Indeed, let (xc”’)), where x(”’)=(x$)), be a Cauchy sequence in E: this means that for any E > 0 there is an m, such that for p 2 m, and q >, m, , we have m
(6.4.1)
For each fixed n, this implies first
IIxp) - xP)1l2< E ,
( x ~ ~ ) ) ~ = ~is , a~ ,Cauchy ... sequence in
hence the sequence
En,and therefore converges to a
limit y , . From (6.4.1) we deduce that for any given N N
C IIxn(P) - xn(4)II2 < &
n= I
as soon as p and q are 2 m o , hence, from the continuity of the norm, we deduce that
N
n= 1
1 1 ~ : ~ ) - y,II’
integers N, we have
m “.= .
for p 2 m,, and as this is true for all
IIxp)- ynllZG E .
I.
This proves first that the
sequence (xp’ - y,) belongs to E, hence y = (y,) also belongs to E, and - y11’ < E for p 2 m,, which ends the proof by showing we have (Idp) that the sequence (x(”’)) converges to y in E. We say that the Hilbert space E thus defined is the Hilbert sum of the sequence of Hilbert spaces (En). We observe that we can map each of the En into E by associating to each x, E En the sequence j,(x,) E E equal to (0, . . . , 0 , x, 0,. . .) (all terms 0 except the nth equal to x,); it is readily verified that j , is an isomorphism of En onto a (necessarily closed) subspace EA of E; j , is called the natural injection of En into E. From the definition of the scalar product in E, it follows that for m # n, any vector in EL is orthogonal to any vector in EL; furthermore, from the definition of the norm in E, it follows that for any x = (x,) E E, the series (j,(x,)) is convergent
4
in E, and x =
m n= 1
HILBERT SUM OF HILBERT SPACES
125
j,,(x,) (observe that the series (j,(x,)) is not absolutely con-
vergent in general). This proves that the (algebraic) sum of the subspaces EA of E (which is obviously direct) is dense in E, in other words that the smallest closed vector subspace containing all the EA is E itself. Conversely: (6.4.2) Let F be a Hilbert space, (F,) a sequence of closed subspaces such that: (1) for m # n, any vector of F , is orthogonal to any vector of F,; (2) the algebraic sum H of the subspaces F , is dense in F . Then, i f E is the Hilbert sum of the F,, there is a unique isomorphism of F onto E which on each F, coincides with the natural injection j , of F, into E.
Let F; = j,,(F,,), and let h, be the mapping of FA onto F,, inverse to j , . Let G be the algebraic sum of the FA in E; that sum being direct, we can define a linear mapping h of G into F by the condition that it coincides with h, on each FA. I claim that h is an isomorphism of G onto the prehilbert space H (which, incidentally, will prove that the (algebraic) sum of the F, is direct in F ) ; from the definition of the scalar product in E, we have to check that
for Xk E Fk,yk E F k ; but by assumption (x, I yk) = 0 if h # k, and the result follows from the fact that each j k is an isomorphism. There is now a unique continuous extension T; of h which is a linear mapping of = E into = F, by (5.5.4); the principle of extension of identities (3.1 5.2) and the continuity of the scalar product show that h is an isomorphism of E onto a subspace of F, which, being complete and dense, must be F itself; the inverse of h satisfies the conditions of (6.4.2). Its uniqueness follows from the fact that it is completely determined in G and continuous in E (3.15.2). Under the condition of (6.4.2), the Hilbert space F is often identified with the Hilbert sum of its subspaces F , .
e
Remark (6.4.3)
We can also prove (6.4.2) by establishing first that the sum of the F, n
is direct; indeed, if E x i = 0 with x i € Fi (1 < i < n) we also have i = l.
= 0 for any j
< n, and as (xjIxi) = 0 for i # j , this boils down
126
VI HILBERT SPACES
to llxil12 = 0, hence xi = 0 for 1 < j < n. Then we define the inverse mapping g of h by the condition that it coincides with j,, on each F,,: we at once verify, as above, that g is an isomorphism of H onto G, and then (5.5.4) is applied in the same way. We observe that this argument still applies when F is a prehilbert space and the F, are complete subspaces of F; it proves the existence of an isomorphism of F onto a dense subspace of the Hilbert sum E of the F,,, which coincides withj,, on each F, .
PROBLEM
Let X be a set, E l , Ez two vector subspaces of Cx, possessing reproducing kernels K,, K2 (Section 6.3, Problem 4). Let E = El E2 c Cx,and let F be the Hilbert sum of the Hilbert spaces El, Ez;the kernel of the surjective mapping u : (fl, fz)-fl +fz of F into E is the subspace N of F consisting of the pairs (f,-f) where YE El n EZ. Show that N is closed in F; if H is the orthogonal supplement of N in F, the restriction u of u to H is a bijection of H onto E; transporting by u the Hilbert structure of H, E is defined as a Hilbert space. Show that for that Hilbert space structure, E has a reproducing kernel equal to K1 Kz; the norm llfll in E is equal to inf(llflII: llfzll~)l~zwhen (fl,f2) takes all values in F such thatf=fl +fz(llfl Illand llf~ [I2 being the norms in El and E2 ,respectively).
+
+
+
5. ORTHONORMAL SYSTEMS
If (with the notations of Section 6.4) we take for each E,, a one-dimensional space (identified to the field of scalars with the scalar product (5 I q) = (fj), the Hilbert sum yields an example of an infinite dimensional Hilbert space E, which is usually written l2 (with index R or C to indicate if necessary what the scalars are); the space I: (resp. 1;) is therefore the space of all sequences x = (5,) of real (resp. complex) numbers, such that convergent, with the scalar product ( x I y ) =
c m
n=l
{, ij,
gi
l{,,I2 is
n= 1
.
In 12, let en be the sequence having all its terms equal to 0 except the nth term equal to 1; we then have (em1 en)= 0 for m # n, and IJe,,JI= 1 for each n, and we have seen in Section 6.4 that for every x = (t,) in f 2 , we can write x =
c <,en, the series being convergent in 12. We observe that this m
n= 1
shows the sequence (en)is total in 12, hence (5.10.1) 1’ is separable. Let us now consider an arbitrary prehilbert space F; we say that a (finite or infinite) sequence (a,) in F is an orthogonal system if (a, I a,) = 0 for m # n and a,, # 0 for every n ; we say that (a,) is an orthonormal system
5 ORTHONORMAL SYSTEMS
127
if in addition lla,,II = 1 for each n. From any orthogonal system (a,,) we deduce at once an orthonormal system by “normalizing” (a,,), i.e. considering the sequence of the b,, = a,,/\lanl\.We have just seen an example of an orthonormal system in 1’; another fundamental example is the following: (6.5.1) Let I be the interval [- 1, 11 of R, and let F = VJI) be the vector space of all continuous complex-valued functions defined on I. We define on F a scalar product by
J -1
(the fact that this is a nondegenerate positive hermitian form is readily verified). For each positive or negative integer n, let
q,,(t) = ennit. It is readily verified that (q,,/,/2)is an orthonormal system in F, called the trigonometric system. Let now (a,) be an arbitrary orthonormal system in a Hilbert space F; for each x E F, we say that the number c,,(x) = ( x I a,,) is the nth coefficient (ornth coordinate) ofx withrespect to the system (a,,) (“nth Fourier coefficient ” of x for the system (6.5.1)). (6.5.2) In a Hilbert space F, let (a,,) be an orthonormal system, V the closed subspace of F generated by the a,, . Then,for any x E F; 1.
the series
m
1 I(x I a,,)1’
n= 1
is convergent, and we have
and
2.
the series of general term (x I a,)a,, is convergent in F and we have m
1 (x I an)% =
n= 1
P”(X).
Conversely, let (A,,) be a sequence of scalars such that
m
1 1A,,I2 is convergent.
n=1
Then, there exists a unique vector y E V such that ( y I a,,) = A,, for every n ; any other vector x E F such that (X I a,,) = Anfor every n is such that x = y + z , with z orthogonal to V, and conversley.
128
VI HILBERT SPACES
For any x E F, we can write x = Pv(x) + 2 , with z orthogonal t o V (6.3.1) and we have therefore (x I a,) = (P,(x) I a,). To prove the theorem, we may therefore assume V = F; but then, the one-dimensional subspaces F, generated by the vectors a, satisfy the assumptions of (6.4.2), and the results are mere restatements of (6.4.2) for that particular case (taking into account the definition of a Hilbert sum). The most interesting case is that in which V = F, i.e., the orthogonal system (a,) is total. It is then called a Hilbert basis for F; (c,) is such a basis for 12. It will be proved in (7.4.3)that the trigonometric system (6.5.1) is total. For a Hilbert space F and a total orthonormal system (a,), we can replace everywhere P, by the identity in (6.5.2); the relations n
n=
I
are then called Parseral’s identities. It follows at once from (6.5.2) that these identities represent not only necessary but su#cient conditions for (a,) to be a total system in a Hilbert space. (6.5.3) It1 a Hilbert space F, a necessary arid sufJicient condition for an orthogonal system (a,) to be total is that the relations (x I a,) = 0 for every n imply .x = 0. Indeed, by (6.5.2) this means that the relation Pv(x) = 0 implies x = 0, and this is equivalent to the relation V = F, since P,(x - P v ( x ) )= 0.
Remark (6.5.4) Suppose E is a prehilbert space and the orthonormal system (a,) in E is total. Then the results ( I ) and (2) of (6.5.2) are still valid, with Pv(x) replaced by x; this follows by the same argument as in (6.5.2), using the Remark (6.4.3).
PROBLEMS
1. Let E be a Hilbert space with a Hilbert basis (e,,),,zl.Let A be the subset of E
consisting of the linear combinations x (n arbitrary).
n
= k= 1
k= I
6 ORTHONORMALIZATION (a> Show that the closure where An > 0, the series
-
A is the set of all the sums of the series
C A,, is convergent I
"=I
129
"= 1
and has a sum equal to I .
(b) Prove that the diameter of A is equal to d2 but that there is no pair of points A such that /la- b/l = 42 (compare to Section 6.3, Problem 2). Let E be a Hilbert space with a Hilbert basis Let a, = eZn and
a, h of
b, = ezn
1 ++ 1 e z n f lfor every n > 0; letA (resp. B) be the closed vector subspace of E I1
generated by the a. (resp. b"). Show that: (a) A n B = { O } , hence the sum A B is direct (algebraically). (b) The direct sum A B is not a topological direct sum (consider in that subspace the sequence of points h, - a, and apply (5.4.2.)). (c) The subspace A B of E is dense but not closed in E (show that the point m
"=O
+
+ +
(b, - an)does not belong to A
+ B).
Show that the Banach space 9 ( P ; 1 2 ) can be identified with the space of double
sequences U = (a,,,,) such that: (I) the series ( 2 ) sup n
m
nt=O
m
nr= 0
is convergent for every n:
la,,,nlzis finite. The norm is then equal to IlUIj = sup
"
method as in Section 5.7, Problem 2(b)). (a) Let u be a continuous linear mapping of show that the series
c m
fl=O
and that their sums are on E and use (6.3.2).)
/cz,,,12 and
<
c m
P
/a,,,,Jz are
into itself, and let u(e,,) =
(Observe that x
--z
x
m=0
la,nnlz <1
for all values of m and
11,
m
1 a,m,e,;
lll=O
( ~ ( xI e,") ) is a continuous linear form
(b) Give an example of a double sequence (a,",J such that m
(same
convergent for all values of m and n,
Wt=O
lIi(l/'.
imYo)
112
(a,,,('
m
"=O
<
(anln/* 1
and
but such that there is no continuous linear
mapping it of P into itself satisfying the relations (u(e,) I e,,,)= a,,,, for all pairs (m, n). (If V is a subspace of I z generated by the vectors en with n E H, where H is a set of p integers, show that there is a linear mapping up of V into itself such that (//,,(en)I e",)= I/t'p for all indices m, 11 of the set H, but l ! i t p l l > v'p.)
6. ORTH 0 NORMALl ZATl O N
(6.6.1) Let E be a separable preliilbert space, (6,) a total sequence of linearly independent cectors in E (see (5.10.1)), and let V, be the n-dimensional subspace of E generated by b,, . . . , 6,. Then if we define c, = b, - Pv,-,(b,), (c,) is a total orthogonal system, such that cl, . . . , c, generate V, for each n.
We use induction on n, assuming that cl, . . . , c , - ~ is an orthogonal system generating V n - l ; then, by definition of Pv,-l (6.3.1),c, is orthogonal
130
VI
HILBERT SPACES
to V,,-l, which proves that (ciIc j ) = 0 for 1 Q i < j < n ; moreover, as b, # V,-l by assumption, c, # 0, hence e l , . , . , c, is an orthogonal system; moreover, b,, - c,, E V,,...l, hence c l , . , . , c, generate the same subspace as the union of V,-l and {b,,},i.e. V , . This completes the proof.
If we normalize the system (c,,), by putting a, = c,,/~lc,,Il, the system (a,) is said to be deduced from (h,,) by the orthonormalization process. For instance, in the space F = %'Jl) considered in (6.5.1), the sequence (t") is total (as will be proved in (7.4.1)) and obviously consists of linearly independent vectors. If we denote by (Q,) the orthonormal system deduced from (t") by orthonormalization, it is clear that Q,,(t) = a,, t" + * . * , polynomial of degree n (with a, # 0) with real coefficients; the Q, are (up to a constant factor) the Legendre polynomials (see Section 8.14, Problem 1). (6.6.2) Any separable prehilbert space (resp. Hilbert space) is isomorphic to a dense subspace of I' (resp. to 1'). As there exists in a separable prehilbert space a total orthonormal system by (6.6.1), the result follows at once from (6.5.2).
PROBLEMS
1. Let E be a separable noncomplete prehilbert space. Show that there exists in E an orthonormal system which is not total, but which is not properly contained in any orthonormal system (imbed E as a dense subspace of a Hilbert space, and use problem 3 of Section 6.3). 2. Let E be a n infinite dimensional separable Hilbert space, V a closed vector subspace of E. Show that if V is infinite dimensional, there exists an isometry of E onto V (write E as the direct sum of V and its orthogonal supplement V', and take Hilbert bases in V and V'. 3. Let (xi)l<, C n be a finite sequence of points in a prehilbert space E. The Gram determinant of that sequence is the determinant G ( x l ,xz , . . . ,x.) = det((xi 1 x i ) ) . (a) Show that G ( x , ,. . . , x.) 2 0 and that G(x1,. . , x,) = 0 if and only if the x I are linearly dependent. (Consider a Hilbert basis of the sub space generated by the xi, and express the x1 as linear combinations of that basis.) (b) Suppose the xi are linearly independent, and let V be the n-dimensional subspace which they generate. Show that the distance of a point x to V is equal to
.
(G(x,X I , * 4.
I . ,
xn)/G(xi,. . . ,x,))"~
(find the projection of x on V, writing it as a linear combination of the xi). Let M be a compact subset of a Hilbert space E; if El is the smallest closed vector subspace of E containing M, show that El is separable.
6
ORTHONORMALIZATION
131
5. Let E c Cx a separable Hilbert space having a reproducing kernel (Section 6.3, Problem 4). (a) Let (At) be a Hilbert basis for E. Show that for every (x, y ) E X x X one has K(x, y ) = c f . ( x ) f , o , where the series converges in C. For any function g E E, if c, = (g
Ifn),
I!
one has, for every x E X, g(x)
=c n
cnfn(x), where
the series converges in C;
in addition, that series is uniformly convergent in every subset of X where K(x, x ) is bounded. (b) Conversely, let (fn) be a sequence of complex functions defined in a set X, such Ifn(x)12 < ca. For every sequence (c,) E I;, the series that for every x E X,
+
cnf.(x) is then convergent in
C for every x
E
X. The functions which are the sums of
such series form a vector subspace E c Cx,on which one defines a structure of separable Hilbert space by taking as scalar product, for u c,f,, u d. f,, the number
=c
(u I u ) =
n
c. n
=c n
-
d, . This space has a reproducing kernel K(x, y ) =cfn(x)J,(y). "