Set-up

Let $w > 0$ be the (two-sided) bandwidth. The variable $\xi$ below will denote frequency, and the variable $t$ will denote time. (Both $w$ and $\xi$ are measured in cycles per unit time.)

Consider the space of functions: $$ \Hilb' = \{ g \in \Le^2(\real) \colon g(\xi) = 0 \text{ for almost all } \abs{\xi} > w/2 \} $$ which is clearly seen to be a complex Hilbert space with the usual inner product for $\Le^2(\real)$ .

Let $\FT$ denote the Fourier transform on $\Le^2(\real)$ , which is a unitary transform by Plancherel's theorem. So, $$ \Hilb = \{ f \in \Le^2(\real) \colon (\FT f)(\xi) = 0 \text{ for almost all } \abs{\xi} > w/2 \} = \FT^{-1} \Hilb' $$ is also a Hilbert space.

Computation of orthonormal basis

One orthonormal basis for $\Hilb'$ consists of the usual Fourier basis functions on the interval $[-w/2, w/2]$ , extended to be zero on $\real \setminus [-w/2, w/2]$ :

Expansion by orthonormal basis

Given $f \in \Hilb$ , let $g = \FT f \in \Hilb'$ . We can expand $g$ in a Fourier series with respect to the basis $\{ \phi_n \}$ : $$ g(\xi) = \sum_{n \in \intset} \langle g, \phi_n \rangle \, \phi_n(\xi)\,, $$ with the infinite sum converging in $\Le^2(\real)$ . Taking $\FT^{-1}$ of both sides, we obtain: $$ f(t) = (\FT^{-1} g)(t) = \sum_{n \in \intset} \langle g, \phi_n \rangle \, (\FT^{-1} \phi_n)(t) = \sqrt{w} \sum_{n \in \intset} \langle g, \phi_n \rangle \, \sinc(wt-n)\,. $$ Moreover, $$ \langle g, \phi_n \rangle = \frac{1}{\sqrt{w}} \int_{-\infty}^\infty g(\xi) \, e^{2\pi i n \xi/w} \, d\xi = \frac{1}{\sqrt{w}} (\FT^{-1} g)\Bigl( \frac{n}{w} \Bigr) = \frac{1}{\sqrt{w}} f\Bigl( \frac{n}{w} \Bigr)\,. $$ (Since $g$ is also in $\Le^1$ , its inverse Fourier transform $\FT^{-1}g$ is a continuous function. Provided that we modify $f$ on a set of measure zero, we can assume that $f = \FT^{-1} (\FT f) = \FT^{-1} g$ is continuous. So it is legal to talk about the pointwise values $f(n/w)$ .)

Result

Hence, we arrive at the representation: $$ f(t) = \sum_{n \in \intset} f\Bigl( \frac{n}{w} \Bigr) \, \sinc(wt - n)\,, $$ thereby reconstructing any $f \in \Hilb$ -- a square-integrable band-limited function -- from its samples at every time period of length $1/w$ .

Uniform and absolute convergence of series

The infinite series for $f$ converges in $\Le^2$ by construction, but in fact it also converges uniformly and absolutely. To see this, first note that by the Cauchy-Schwarz inequality, $$ \sum_{n \in \intset} \abs{ f(\tfrac{n}{w}) } \, \abs{\sinc (wt-n)} \leq \Bigl( \sum_{n \in \intset} \abs{f(\tfrac{n}{w})}^2 \Bigr)^{1/2} \Bigl( \sum_{n \in \intset} \sinc^2(wt -n) \Bigr)^{1/2}\,. $$ The series $\sum_n \abs{f(n/w)}^2$ converges by Parseval's theorem ($w^{-1/2} f(n/w)$ are the Fourier coefficients of $g$ ). Also, the series $\sum_n \sinc^2(wt-n)$ is uniformly bounded for all $t \in \real$ . To prove this, it suffices to restrict to $t$ bounded inside $[0, 1/w]$ as the function $t \mapsto \sum_n \sinc^2(wt-n)$ is $1/w$ -periodic; and then it becomes an easy estimate using the fact that $\sum_n n^{-2} < \infty$ . It follows that the series $\sum_n \abs{f(n/w)} \abs{\sinc(wt-n)}$ is uniformly bounded for all $t$ , and its tail tends to zero uniformly in $t$ .

Illustrations

Figure 1: Reconstructing a Bessel function (of the first kind), using a $\sinc$ series with . Theoretically exact bandwidth is $w = 1/\pi \approx 0.32$ .

Figure 2: Effect of under-sampling and over-sampling beyond the actual bandwidth of the function

Figure 3: The basis functions $\sinc(t)$ , $\tfrac12 \sinc(t\pm 1)$ , $\tfrac14 \sinc(t \pm 2)$

• Python program to produce the three figures