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Describes the extension of Hilbert spaces to infinite dimensions, including complete orthonormal sequences and Parseval's relation.

While up to now bases have been linked to finite-dimensional spaces and subspaces, it is possible to extend the concept to infinite-dimensional spaces.

Definition 1 Let ( X , R , + , · ) be a vector space. An infinite sequence of orthonormal vectors { e 1 , e 2 , ... } X is said to be a complete orthonormal sequence (CONS) in X if for every x X there exists a sequence α 1 , α 2 , . . . R such that x = i α i e i .

For the sake of concreteness, an infinite sum is defined as x = i = 1 α i e i = lim n i = 1 n α i e i . It is easy to see that α i = x , e i .

Lemma 1 An orthonormal sequence is complete if and only if the only vector in X orthogonal to each of the e i 's is the zero vector.

Example 1 For the space X of finite-energy complex-valued functions, R = C , a CONS is given by e k ( t ) = 1 2 π e j k t for k = 0 , ± 1 , ± 2 , . . . . These vectors are orthonormal:

e k , e l = 0 2 π 1 2 π e j k t e - j l t d t = 0 2 π 1 2 π e j ( k - l ) t d t = 1 k = l 0 k l

The coefficients are given by

c k = x , e k = 0 2 π x ( t ) e - j k t d t ,

and we obtain x = k c k e k . This is the sequence behind the Fourier series representation.

Example 2 Let X be the space of bandlimited functions x ( t ) (i.e., the set of functions with Fourier transform X ( f ) such that | X ( f ) | = 0 for all f [ - B , B ] ). A CONS for this space is given by

e k ( t ) = 1 2 B sinc 2 B t - k 2 B ,

where sinc ( t ) = ( sin ( π t ) ) / ( π t ) . It is possible to show that the functions are orthogonal to each other, i.e.,

e k , e l = δ k , l 1 k = l 0 k l .

If x is bandlimited, then it follows that x ( t ) = k c k e k ( t ) , with c k = x , e k = x ( k / ( 2 B ) ) . The preservation of the norm in the coefficients can also be extended from ONBs to CONS.

Theorem 1 (Completeness Relation) An orthonormal sequence e 1 , e 2 , . . . is complete for X if and only if the completeness relation holds for all x X :

x 2 = i | x , e i | 2 = i | c i | 2 .

The sequence { e i } is CONS if and only if

x = i = 1 x , e i e i = lim n i = 1 n x , e i e i = lim n x n ,

where x n = i = 1 n x , e i e i is the partial sum. We then have x 2 = x - x n 2 + x n 2 as these two components are orthogonal to each other. Applying a limit on both sides for n , we have

x 2 = lim n x - x n 2 + lim n x n 2 = 0 + lim n i = 1 n | x , e i | 2 = i = 1 | x , e i | 2 .

Theorem 2 Let X be a Hilbert space with a CONS { e 1 , e 2 , . . . } . Then for any x , y X , Parseval's relation holds: x , y = i x , e i y , e i ¯ .

Using the CONS, we can write the partial sums x n = i = 1 n x , e i e i and y n = i = 1 n y , e i e i . We then have

| x n , y n - x , y | = | x n , y - x n , y + x n , y n - x , y | = | x n , y n - y + x n - x , y | | x n , y n - y | + | x n - x , y | x n y n - y + x n - x y n

Letting n we have that the upper bound goes to zero, and therefore as n , we have x n , y n x , y . Therefore,

x , y = lim n x n , y n = lim n i = 1 n j = 1 n x , e i e i , y , e j e j = lim n i = 1 n j = 1 n x , e i y , e j ¯ e i , e j = lim n i = 1 n x , e i y , e i ¯ = i = 1 x , e i y , e i ¯ .

Questions & Answers

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Source:  OpenStax, Signal theory. OpenStax CNX. Oct 18, 2013 Download for free at http://legacy.cnx.org/content/col11542/1.3
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