proof of sampling theorem
0.1 Setup
Let $w>0$ be the (twosided) bandwidth. The variable $\xi $ below will denote frequency, and the variable $t$ will denote time. (Both $w$ and $\xi $ are measured in .)
Consider the space of functions:
$${\mathscr{H}}^{\prime}=\{g\in {\mathbf{L}}^{2}(\mathbb{R}):g(\xi )=0\text{for almost all}\xi w/2\}$$ 
which is clearly seen to be a complex Hilbert space^{} with the usual inner product^{} for ${\mathbf{L}}^{2}(\mathbb{R})$.
Let $\mathcal{F}$ denote the Fourier transform^{} on ${\mathbf{L}}^{2}(\mathbb{R})$, which is a unitary^{} transform by Plancherel’s theorem. So,
$$\mathscr{H}=\{f\in {\mathbf{L}}^{2}(\mathbb{R}):(\mathcal{F}f)(\xi )=0\text{for almost all}\xi w/2\}={\mathcal{F}}^{1}{\mathscr{H}}^{\prime}$$ 
is also a Hilbert space.
0.2 Computation of orthonormal basis
One orthonormal basis^{} for ${\mathscr{H}}^{\prime}$ consists of the usual Fourier functions^{} on the interval $[w/2,w/2]$, extended to be zero on $\mathbb{R}\setminus [w/2,w/2]$:
$${\varphi}_{n}(\xi )=\{\begin{array}{cc}\frac{1}{\sqrt{w}}{e}^{2\pi in\xi /w},\hfill & \xi \le w/2\hfill \\ 0,\hfill & \xi >w/2,\hfill \end{array}\mathit{\hspace{1em}}n\in \mathbb{Z}.$$ 
Mapping these by ${\mathcal{F}}^{1}$ produces an orthonormal basis for $\mathscr{H}$:
$({\mathcal{F}}^{1}{\varphi}_{n})(t)$  $={\displaystyle \frac{1}{\sqrt{w}}}{\displaystyle {\int}_{w/2}^{w/2}}{e}^{2\pi in\xi /w}{e}^{2\pi i\xi t}\mathit{d}\xi $  
$={\displaystyle \frac{1}{\sqrt{w}}}{\displaystyle {\int}_{w/2}^{w/2}}{e}^{2\pi i\xi (tn/w)}\mathit{d}\xi $  
$={\displaystyle \frac{1}{\sqrt{w}}}\left(w\mathrm{sinc}(w(tn/w))\right)=\sqrt{w}\mathrm{sinc}(wtn),$ 
where we have used the fact that the Fourier transform of $t\mapsto w\mathrm{sinc}(wt)$ (normalized sinc function^{}) is the rectangular box function of bandwidth $w$, and vice versa.
0.3 Expansion by orthonormal basis
Given $f\in \mathscr{H}$, let $g=\mathcal{F}f\in {\mathscr{H}}^{\prime}$. We can expand $g$ in a Fourier series with respect to the $\{{\varphi}_{n}\}$:
$$g(\xi )=\sum _{n\in \mathbb{Z}}\u27e8g,{\varphi}_{n}\u27e9{\varphi}_{n}(\xi ),$$ 
with the infinite sum converging in ${\mathbf{L}}^{2}(\mathbb{R})$. Taking ${\mathcal{F}}^{1}$ of both sides, we obtain:
$$f(t)=({\mathcal{F}}^{1}g)(t)=\sum _{n\in \mathbb{Z}}\u27e8g,{\varphi}_{n}\u27e9({\mathcal{F}}^{1}{\varphi}_{n})(t)=\sqrt{w}\sum _{n\in \mathbb{Z}}\u27e8g,{\varphi}_{n}\u27e9\mathrm{sinc}(wtn).$$ 
Moreover,
$$\u27e8g,{\varphi}_{n}\u27e9=\frac{1}{\sqrt{w}}{\int}_{\mathrm{\infty}}^{\mathrm{\infty}}g(\xi ){e}^{2\pi in\xi /w}\mathit{d}\xi =\frac{1}{\sqrt{w}}({\mathcal{F}}^{1}g)\left(\frac{n}{w}\right)=\frac{1}{\sqrt{w}}f\left(\frac{n}{w}\right).$$ 
(Since $g$ is also in ${\mathbf{L}}^{1}$, its inverse Fourier transform ${\mathcal{F}}^{1}g$ is a continuous function^{}. Provided that we modify $f$ on a set of measure zero, we can assume that $f={\mathcal{F}}^{1}(\mathcal{F}f)={\mathcal{F}}^{1}g$ is continuous. So it is legal to talk about the pointwise values $f(n/w)$.)
0.4 Result
Hence, we arrive at the representation:
$$f(t)=\sum _{n\in \mathbb{Z}}f\left(\frac{n}{w}\right)\mathrm{sinc}(wtn),$$ 
thereby reconstructing any $f\in \mathscr{H}$ — a squareintegrable bandlimited function — from its samples at every time period of length $1/w$.
0.5 Uniform and absolute convergence of series
The infinite series for $f$ converges in ${\mathbf{L}}^{2}$ by construction, but in fact it also converges uniformly and absolutely. To see this, first note that by the CauchySchwarz inequality,
$$\sum _{n\in \mathbb{Z}}f(\frac{n}{w})\mathrm{sinc}(wtn)\le {\left(\sum _{n\in \mathbb{Z}}{f(\frac{n}{w})}^{2}\right)}^{1/2}{\left(\sum _{n\in \mathbb{Z}}{\mathrm{sinc}}^{2}(wtn)\right)}^{1/2}.$$ 
The series ${\sum}_{n}{f(n/w)}^{2}$ converges by Parseval’s theorem (${w}^{1/2}f(n/w)$ are the Fourier coefficients of $g$). Also, the series ${\sum}_{n}{\mathrm{sinc}}^{2}(wtn)$ is uniformly bounded for all $t\in \mathbb{R}$. To prove this, it suffices to restrict to $t$ bounded inside $[0,1/w]$ as the function $t\mapsto {\sum}_{n}{\mathrm{sinc}}^{2}(wtn)$ is $1/w$periodic; and then it becomes an easy estimate using the fact that $$. It follows that the series ${\sum}_{n}f(n/w)\mathrm{sinc}(wtn)$ is uniformly bounded for all $t$, and its tail tends to zero uniformly in $t$.
0.6 Illustrations

•
http://aux.planetmath.org/files/objects/8650/sample.pyPython program to produce the three figures
Title  proof of sampling theorem 

Canonical name  ProofOfSamplingTheorem 
Date of creation  20130322 16:28:57 
Last modified on  20130322 16:28:57 
Owner  stevecheng (10074) 
Last modified by  stevecheng (10074) 
Numerical id  13 
Author  stevecheng (10074) 
Entry type  Proof 
Classification  msc 42A38 
Classification  msc 94A20 
Related topic  PlancherelsTheorem 