Shannon sampling for band-limited $L^2$ functions

By [F1], $G_h(\theta )=h^{-1\mathrm{/}2}\widehat{f}(-\theta \mathrm{/}h)$ on the fundamental interval and $\widehat{G_h}(k)=h^{1\mathrm{/}2}f(hk)$; the expansion clause of [F2] gives $G_h={\sum }_k\widehat{G_h}(k)e_k$ as the $L^2(\mathbb{T})$ limit of the partial sums ${\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}{h}^{1\mathrm{/}2}f(hk)e_k$. Substituting $\theta =-h\xi$, which maps $[-1\mathrm{/}2,1\mathrm{/}2]$ diffeomorphically onto $B$ with $\mathrm{\mid }d\theta \mathrm{/}d\xi \mathrm{\mid }=h$ and converts the torus integral into $h{\int }_B$ by [F4], turns this into convergence in $L^2(B)$ of ${\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}hf(hk)e^{-2\pi ihk\xi }$ to $\widehat{f}(\xi )$; since $\widehat{f}$ vanishes off $B$, this is an identity $\widehat{f}={\sum }_k{c}_k{\phi }_k$ in $L^2(\mathbb{R})$ with $c_k:=hf(hk)$ and ${\phi }_k(\xi ):=e^{-2\pi ihk\xi }{\mathbf{1}}_B(\xi )$.

Each ${\phi }_k$ lies in $L^1\cap L^2$ and its integral transform is computed from the primitive [F5]: with $u:=x+hk$, ${\int }_B{e}^{-2\pi iu\xi }d\xi =(e^{-\pi iu\mathrm{/}h}-e^{\pi iu\mathrm{/}h})\mathrm{/}(-2\pi iu)=\sin (\pi u\mathrm{/}h)\mathrm{/}(\pi u)=h^{-1}\mathrm{sinc}(u\mathrm{/}h)$ for $u\mathrm{\ne }0$ (using oddness of sine from [F5]), while for $u=0$ the integral is $h^{-1}=h^{-1}\mathrm{sinc}(0)$. Hence $\mathcal{F}{\phi }_k(x)=h^{-1}\mathrm{sinc}((x+hk)\mathrm{/}h)$; by the agreement clause of [F3] this bounded continuous function represents ${\mathcal{F}}_2{\phi }_k$, so the inverse transform ${\mathcal{F}}_2^{-1}{\phi }_k=R{\mathcal{F}}_2{\phi }_k$ is represented by $x\mapsto h^{-1}\mathrm{sinc}(-x\mathrm{/}h+k)=h^{-1}\mathrm{sinc}(x\mathrm{/}h-k)$, where evenness of $\mathrm{sinc}$ was used [F3, F5].

Since ${\mathcal{F}}_2^{-1}$ is continuous complex-linear [F3], it may be applied termwise to the $L^2$-convergent series of step 1.1: $f={\mathcal{F}}_2^{-1}\widehat{f}={\sum }_k{c}_k{\mathcal{F}}_2^{-1}{\phi }_k={\sum }_{k}f(hk)\mathrm{sinc}(\cdot \mathrm{/}h-k)$ in $L^2(\mathbb{R})$, which is the asserted $L^2$ identity.

Assume now ${\sum }_k\mathrm{\mid }f(hk)\mathrm{\mid }<\mathrm{\infty }$, and consider the series of step 3.1. For every $x$, $\mathrm{\mid }f(hk)\mathrm{sinc}(x\mathrm{/}h-k)\mathrm{\mid }\le \mathrm{\mid }f(hk)\mathrm{\mid }$ by $\mathrm{\mid }\mathrm{sinc}\mathrm{\mid }\le 1$ [F5], so the series $S(x):={\sum }_{k}f(hk)\mathrm{sinc}(x\mathrm{/}h-k)$ converges absolutely and uniformly on all of $\mathbb{R}$ by the Weierstrass majorant ${\sum }_k\mathrm{\mid }f(hk)\mathrm{\mid }$; each term is continuous [F5], so the real and imaginary parts of $S$ are continuous by [F6], and $S$ is continuous.

The partial sums converge pointwise everywhere to $S$ by step 4.1 and in $L^2(\mathbb{R})$ to the class $f$ by step 3.1; by [F7] a subsequence converges to $f$ almost everywhere, so $S=f$ almost everywhere. Both $S$ and the continuous representative $f$ are continuous [F6], and a continuous function vanishing almost everywhere vanishes identically: if $(S-f)(x_0)\mathrm{\ne }0$, continuity would make $S-f$ nonzero on a ball about $x_0$, which contains a nondegenerate box of positive measure [F7] on which $S-f$ is nonzero, contradicting almost-everywhere equality. Hence $f(x)=S(x)$ for every $x$, which is the pointwise identity under the extra hypothesis. Without that hypothesis only step 3.1, an $L^2$ statement, is asserted. Countable Choice enters only through the integration, Fourier and Riesz–Fischer suppliers quoted above.