$\{e_{\lambda }{\}}_{\lambda \in \mathrm{\Lambda }}$ in a Hilbert space is maximal if $u\in H$ and $⟨u,e_{\lambda }⟩=0$ for all $\lambda \in \mathrm{\Lambda }$ implies $u=0$.

Let $H$ be a separable Hilbert space. Then $H$ has a countable maximal orthonormal set.

Let $H$ be a Hilbert space. An orthonormal basis of $H$ is a countable maximal orthonormal subset $\{e_n\}$ of $H$.

If $\{e_n\}$ is an orthonormal basis of a hilbert space $H$, then for all $u\in H$ we have

ie like in finite-dimensional linear algebra, we can write any element as a linear combination of the basis elements. But in this space, there might be an infinite number of such elements.

First, we prove ${\{\sum _{n=1}^m⟨u,e_n⟩e_n\}}_m$ is Cauchy. Let $ϵ>0$. Then by Bessel's inequality, we have

Thus, for all $M\in \mathbb{N}$ such that for all $N\ge M$ we have $\sum _{m=N+1}^{\mathrm{\infty }}|⟨u,e_n⟩{|}^2<ϵ$

Then for all $m>\ell \ge M$ we compute

thus the sequence is indeed Cauchy. Since $H$ is complete, there exists some $\overline{u}\in H$ where

By continuity of inner product, we know that for all $\ell \in \mathbb{N}$

Thus $⟨u-\overline{u},e_{\ell }⟩=0$ for all $\ell$ if and only if $u-\overline{u}=0$.

If a Hilbert space $H$ has an orthonormal basis, then $H$ is separable.

Suppose $\{e_n\}$ is an orthonormal basis for $H$. Then

is a countable. This is because elements in each component (index by $m$) have a bijection with ${\mathbb{Q}}^{2m}$, which is countable. Then since the $m$ are countable, the union is countable and therefore yields a countable set.

So by all elements of hilbert spaces with orthonormal bases can be written as sums of the basis elements, $S$ is dense in $H$

• (since every element $u$ can be expanded in the series described above, and so the partial sums converge to $u$)
  Thus for any $ϵ>0$, we have take a sufficiently long partial sum of length $L$ to get within $\frac{ϵ}{2}$ of $u$ and approximate each coefficient with a rational number.

So the sums will be in one of the parts of $S$ above, ie $S$ is a countable dense subset of $H$, ie $H$ is separable.

Let $H$ be a Hilbert space and let $\{e_n\}$ be a countable orthonormal basis of $H$. Then for all $u\in H$,

We know that

from all elements of hilbert spaces with orthonormal bases can be written as sums of the basis elements. If the sum over $n$ is a finite sum, then we just expand the inner product $||u|{|}^2=⟨u,u⟩$. Otherwise, by continuity of inner product, we have

If $H$ is an infinite-dimensional separable Hilbert space, then $H$ is isometrically isomorphic to ${\ell }^2$.

ie, there exists a bijective bounded linear operator $T:H\to {\ell }^2$ such that for all $u,v\in H$ we have

Since $H$ is separable, it has an orthonormal basis (Hilbert spaces are separable if and only if they have an orthonormal basis). And we can write the Fourier-Bessel series for each element $u\in H$:

By Parseval's identity. So we can define our map $T$ as

ie, $Tu$ is the sequence of coefficients in the expansion. And this sequence is in ${\ell }^2$ (ell-2).

• $T$ is linear by construction,
• It is surjective because every sum $\sum _{n=1}^{\mathrm{\infty }}{c}_n{e}_n$ is Cauchy in $H$
• It is one-to-one because every $u$ is expanded in this way, so any two expansions that are the same will be have the same infinite sum.

The subset of functions ${\left\{\frac{e^{inx}}{\sqrt{2\pi }}\right\}}_{n\in \mathbb{Z}}$ is an orthonormal subset of $L^2([-\pi ,\pi ])$

if we don't like working with complex exponentials we can use

And work out everything we need.

Note that

Since the exponential is periodic in $2\pi$. Normalizing by $2\pi$ yields

Giving us our orthonormal set.

For a function $f\in L^2([-\pi ,\pi ])$, the Fourier coefficient $\hat{f}(n)$ of $f$ is given by

And the $N$th partial Fourier sum is

this follows from all elements of hilbert spaces with orthonormal bases can be written as sums of the basis elements and Hilbert spaces are bijective with ell-2

The Fourier series of $f$ is the formal series

Does convergence (in $L^2$) $\sum _{n=1}^{\mathrm{\infty }}\hat{f}{e}^{inx}\to f$ hold for all $f\in L^2([-\pi ,\pi ])$? ie, does

• ie, is $\left\{\frac{e^{inx}}{\sqrt{2\pi }}\right\}$ a maximal orthonormal set in $L^2([-\pi ,\pi ])$?
• ie, does $\hat{f}(n)=0\mathrm{\forall }n\in \mathbb{Z}⟹f=0$ ?

It turns out, yes, but we will need to build some more framework to show this.

For all $f\in L^2([-\pi ,\pi ])$ and all $N\in \mathbb{N}\cup \{0\}$,

The function $D_N$ is continuous and also smooth and is called the Dirichlet kernel.

For any $f\in L^2([-\pi ,\pi ])$, we know that

Where $(\ast )$ is by linearity of the integral

Which gives us the desired result

Let $f\in L^2([-\pi ,\pi ])$. The $N$th Cesaro-Fourier mean of $f$ is

Where $S_{k}f(x)$ is the $k$th partial Fourier sum.

• If the Fourier coefficients are all zero, then the function is zero
• Want to show that if the partial Fourier sums converge, then it must be a sum of zeros. But doing this with the $S_N$ directly is the original question.
• The "averaged" version (the Cesaro-Fourier mean) is easier to work with, so we will try and show convergence here instead

We know from real analysis that the Cesaro means of a sequence behave better but do not lose any information.