Abstract Fourier series and Parseval’s identity

Jordan Bell

April 3, 2014

1 Orthonormal basis

Let $H$ be a separable complex Hilbert space.¹¹ 1 One talk do everything we are doing and obtain the same results for nonseparable Hilbert spaces, but one has to define what uncountable sums mean. This is done in John B. Conway, A Course in Functional Analysis, second ed., chapter I. If $e_{i}\in H$ , $i\geq 1$ , and $\left\langle e_{i},e_{j}\right\rangle=\delta_{i,j}$ , we say that the set $\{e_{i}\}$ is orthonormal. If $\textrm{span}\{e_{i}:i\geq 1\}$ is a dense subspace of $H$ , we say that $\{e_{i}:i\geq 1\}$ is an orthonormal basis for $H$ . We can write this in another way. If $S_{\alpha},\alpha\in I$ are subsets of $H$ , let $\bigvee_{\alpha\in I}S_{\alpha}$ be the closure of the span of $\bigcup_{\alpha\in I}S_{\alpha}$ . To say that $\{e_{i}\}$ is an orthonormal basis for $H$ is to say that $\{e_{i}\}$ is orthonormal and that $H=\bigvee_{i\geq 1}\{e_{i}\}$ .

2 Abstract Fourier series

If $a_{k}\in\mathbb{C}$ and the sequence $\sum_{k=1}^{n}a_{k}e_{k}$ converges in $H$ , we denote its limit by

\sum_{k=1}^{\infty}a_{k}e_{k}.

This is a definition of an infinite sum in $H$ . Since $H$ is complete, one usually shows that a sequence converges by showing that the sequence is Cauchy, and hence to show that $\sum_{k=1}^{n}a_{k}e_{k}$ converges it is equivalent to show that

\sum_{k=m+1}^{n}a_{k}e_{k}\to 0

as $m,n\to\infty$ . And showing this is equivalent to showing that

\left\langle\sum_{k=m+1}^{n}a_{k}e_{k},\sum_{k=m+1}^{n}a_{k}e_{k}\right\rangle\to 0

as $m,n\to\infty$ . This is equivalent to

\sum_{k=m+1}^{n}|a_{k}|^{2}\to 0

as $m,n\to\infty$ , and this is equivalent to the series

\sum_{k=1}^{\infty}|a_{k}|^{2}

converging. Thus, the series $\sum_{k=1}^{\infty}a_{k}e_{k}$ converges if and only if the series $\sum_{k=1}^{\infty}|a_{k}|^{2}$ converges.²² 2 Furthermore, using the triangle inequality rather than the orthonormality of the $e_{k}$ , one can check that if the series $\sum_{k=1}^{\infty}|a_{k}|$ converges then the series $\sum_{k=1}^{\infty}a_{k}e_{k}$ converges.

Let $\{e_{i}:i\geq 1\}$ be an orthonormal basis for $H$ ; it is a fact that one exists. Let $v\in H$ and define

s_{n}=\sum_{k=1}^{n}\left\langle v,e_{k}\right\rangle e_{k}.

If $1\leq i\leq n$ then

\left\langle v-s_{n},e_{i}\right\rangle=\left\langle v,e_{i}\right\rangle-\sum% _{k=1}^{n}\left\langle v,e_{k}\right\rangle\left\langle e_{k},e_{i}\right% \rangle=\left\langle v,e_{i}\right\rangle-\left\langle v,e_{i}\right\rangle=0,

hence

\left\langle v-s_{n},s_{n}\right\rangle=0.

It follows that

$\displaystyle\sum_{k=1}^{n}\|\left\langle v,e_{k}\right\rangle\|^{2}$	$\displaystyle=$	$\displaystyle\left\langle s_{n},s_{n}\right\rangle$
	$\displaystyle\leq$	$\displaystyle\left\langle s_{n},s_{n}\right\rangle+\left\langle v-s_{n},v-s_{n% }\right\rangle$
	$\displaystyle=$	$\displaystyle\left\langle v,v\right\rangle,$

where we used $\left\langle v-s_{n},s_{n}\right\rangle=0$ in the third line. Therefore the series $\sum_{k=1}^{\infty}|\left\langle v,e_{k}\right\rangle|^{2}$ converges, and so the sequence $s_{n}$ converges to some $v^{\prime}=\sum_{k=1}^{\infty}\left\langle v,e_{k}\right\rangle e_{k}\in H$ . Since $s_{n}$ converges to $v^{\prime}$ , in particular it converges weakly to $v^{\prime}$ , i.e., for any $w\in H$ ,

\lim_{n\to\infty}\left\langle s_{n},w\right\rangle=\left\langle v^{\prime},w% \right\rangle.

Therefore for any $j$ ,

\left\langle v-v^{\prime},e_{j}\right\rangle=\left\langle v,e_{j}\right\rangle% -\left\langle v^{\prime},e_{j}\right\rangle=\left\langle v,v_{j}\right\rangle-% \lim_{n\to\infty}\left\langle s_{n},e_{j}\right\rangle=\left\langle v,v_{j}% \right\rangle-\left\langle v,v_{j}\right\rangle=0;

this is because for $n\geq j$ we have $\left\langle v-s_{n},e_{j}\right\rangle=0$ and hence $\left\langle v,e_{j}\right\rangle=\left\langle s_{n},e_{j}\right\rangle$ . As $\left\langle v-v^{\prime},e_{j}\right\rangle=0$ for all $j$ , it follows that $v-v^{\prime}=0$ , i.e. $v=v^{\prime}$ . Hence,

v=\sum_{k=1}^{\infty}\left\langle v,e_{k}\right\rangle e_{k}.

We call this an abstract Fourier series for $v$ .³³ 3 If $H=L^{2}(\mathbb{T})$ , one checks that $e_{k}=e^{ik},k\in\mathbb{Z}$ , is an orthonormal basis for $H$ . Then, $\left\langle f,e_{k}\right\rangle=\frac{1}{2\pi}\int_{0}^{2\pi}f(t)e^{-ik}dt$ and $f$ is the limit in $H$ of $\sum_{k=0}^{n}\left\langle f,e_{k}\right\rangle e^{ik}$ . Thus in $H$ , $f=\sum_{k\in\mathbb{Z}}\left\langle f,e_{k}\right\rangle e^{ik}.$ It can be written as

v=\sum_{k=1}^{\infty}(e_{k}\otimes e_{k})v,

and thus can be written without $v$ as

\textrm{id}_{H}=\sum_{k=1}^{\infty}e_{k}\otimes e_{k};

$e_{k}\otimes e_{k}\in B(H)$ is a projection with rank 1, and the above series conveges in the strong operator topology on $B(H)$ . Writing the identity map in this way is called a resolution of the identity.

3 Parseval’s identity

On the one hand

\lim_{n\to\infty}\left\|s_{n}\right\|^{2}=\sum_{k=1}^{\infty}|\left\langle v,e% _{k}\right\rangle|^{2}.

On the other hand,

\lim_{n\to\infty}\left\|s_{n}\right\|^{2}=\left\|v\right\|^{2}.

Hence

\left\|v\right\|^{2}=\sum_{k=1}^{\infty}|\left\langle v,e_{k}\right\rangle|^{2},

which is Parseval’s identity.