# 1.13 Infinite-dimensional hilbert spaces

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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 $\left(X,R,+,·\right)$ be a vector space. An infinite sequence of orthonormal vectors $\left\{{e}_{1},{e}_{2},...\right\}\subseteq X$ is said to be a complete orthonormal sequence (CONS) in $X$ if for every $x\in X$ there exists a sequence ${\alpha }_{1},{\alpha }_{2},...\in R$ such that $x={\sum }_{i}{\alpha }_{i}{e}_{i}$ .

For the sake of concreteness, an infinite sum is defined as $x={\sum }_{i=1}^{\infty }{\alpha }_{i}{e}_{i}={lim}_{n\to \infty }{\sum }_{i=1}^{n}{\alpha }_{i}{e}_{i}$ . It is easy to see that ${\alpha }_{i}=〈x,,,{e}_{i}〉$ .

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, $R=\mathbb{C}$ , a CONS is given by ${e}_{k}\left(t\right)=\frac{1}{\sqrt{2\pi }}{e}^{jkt}$ for $k=0,±1,±2,...$ . These vectors are orthonormal:

$〈{e}_{k},,,{e}_{l}〉={\int }_{0}^{2\pi }\frac{1}{2\pi }{e}^{jkt}{e}^{-jlt}dt={\int }_{0}^{2\pi }\frac{1}{2\pi }{e}^{j\left(k-l\right)t}dt=\left\{\begin{array}{cc}1\hfill & k=l\hfill \\ 0\hfill & k\ne l\hfill \end{array}\right)$

The coefficients are given by

${c}_{k}=〈x,,,{e}_{k}〉={\int }_{0}^{2\pi }x\left(t\right){e}^{-jkt}dt,$

and we obtain $x={\sum }_{k}{c}_{k}{e}_{k}$ . This is the sequence behind the Fourier series representation.

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

${e}_{k}\left(t\right)=\frac{1}{\sqrt{2B}}\mathrm{sinc}\left(2,B,\left(t,-,\frac{k}{2B}\right)\right),$

where $\mathrm{sinc}\left(t\right)=\left(sin\left(\pi t\right)\right)/\left(\pi t\right)$ . It is possible to show that the functions are orthogonal to each other, i.e.,

$〈{e}_{k},,,{e}_{l}〉={\delta }_{k,l}\left\{\begin{array}{cc}1\hfill & k=l\hfill \\ 0\hfill & k\ne l\hfill \end{array}\right).$

If $x$ is bandlimited, then it follows that $x\left(t\right)={\sum }_{k}{c}_{k}{e}_{k}\left(t\right)$ , with ${c}_{k}=〈x,,,{e}_{k}〉=x\left(k/\left(2B\right)\right)$ . 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 $x\in X$ :

${\parallel x\parallel }^{2}=\sum _{i}|〈x,,,{e}_{i}〉{|}^{2}=\sum _{i}{|{c}_{i}|}^{2}.$

The sequence $\left\{{e}_{i}\right\}$ is CONS if and only if

$x=\sum _{i=1}^{\infty }〈x,,,{e}_{i}〉{e}_{i}=\underset{n\to \infty }{lim}\sum _{i=1}^{n}〈x,,,{e}_{i}〉{e}_{i}=\underset{n\to \infty }{lim}{x}_{n},$

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

${\parallel x\parallel }^{2}=\underset{n\to \infty }{lim}\parallel x-{x}_{n}{\parallel }^{2}+\underset{n\to \infty }{lim}\parallel {x}_{n}{\parallel }^{2}=0+\underset{n\to \infty }{lim}\sum _{i=1}^{n}|〈x,,,{e}_{i}〉{|}^{2}=\sum _{i=1}^{\infty }{|〈x,,,{e}_{i}〉|}^{2}.$

Theorem 2 Let $X$ be a Hilbert space with a CONS $\left\{{e}_{1},{e}_{2},...\right\}$ . Then for any $x,y\in X$ , Parseval's relation holds: $〈x,,,y〉={\sum }_{i}〈x,,,{e}_{i}〉\overline{〈y,,,{e}_{i}〉}$ .

Using the CONS, we can write the partial sums ${x}_{n}={\sum }_{i=1}^{n}〈x,,,{e}_{i}〉{e}_{i}$ and ${y}_{n}={\sum }_{i=1}^{n}〈y,,,{e}_{i}〉{e}_{i}$ . We then have

$\begin{array}{cc}\hfill |〈{x}_{n},,,{y}_{n}〉-〈x,,,y〉|& =|〈{x}_{n},,,y〉-〈{x}_{n},,,y〉+〈{x}_{n},,,{y}_{n}〉-〈x,,,y〉|\hfill \\ & =|〈{x}_{n},,,{y}_{n},-,y〉+〈{x}_{n},-,x,,,y〉|\hfill \\ & \le |〈{x}_{n},,,{y}_{n},-,y〉|+|〈{x}_{n},-,x,,,y〉|\hfill \\ & \le \parallel {x}_{n}\parallel \parallel {y}_{n}-y\parallel +\parallel {x}_{n}-x\parallel \parallel {y}_{n}\parallel \hfill \end{array}$

Letting $n\to \infty$ we have that the upper bound goes to zero, and therefore as $n\to \infty$ , we have $〈{x}_{n},,,{y}_{n}〉\to 〈x,,,y〉$ . Therefore,

$\begin{array}{cc}\hfill 〈x,,,y〉& =\underset{n\to \infty }{lim}〈{x}_{n},,,{y}_{n}〉=\underset{n\to \infty }{lim}\sum _{i=1}^{n}\sum _{j=1}^{n}〈〈x,,,{e}_{i}〉,{e}_{i},,,〈y,,,{e}_{j}〉,{e}_{j}〉\hfill \\ & =\underset{n\to \infty }{lim}\sum _{i=1}^{n}\sum _{j=1}^{n}〈x,,,{e}_{i}〉\overline{〈y,,,{e}_{j}〉}〈{e}_{i},,,{e}_{j}〉=\underset{n\to \infty }{lim}\sum _{i=1}^{n}〈x,,,{e}_{i}〉\overline{〈y,,,{e}_{i}〉}\hfill \\ & =\sum _{i=1}^{\infty }〈x,,,{e}_{i}〉\overline{〈y,,,{e}_{i}〉}.\hfill \end{array}$

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