Wiener's tauberian theorem for Hardy space For $a>0$ let us define $$H^2(-a,a)=\{f \ \mbox{is analytic in the strip $|\Im(z)|<a$}: \sup_{y\in [-a,a]}\int_{\mathbb{R}}|f(x+iy)|^2\,dx<\infty\}.$$ For $f\in H^2(-a,a)$, define $\|f\|=\sup_{y\in (-a,a)}\int_{\mathbb{R}}|f(x+iy)|^2\,dx<\infty$. We note that the function $e^{-z^2}\in H^2(-a,a)$ for any $a>0$. Can we have that
$\operatorname{span}(\{e^{-(z-b)^2}: \ b\in\mathbb{R}\})$ is dense in $H^2(-a,a)$ with respect to the norm $\|\cdot\|$?
 A: I'm not sure. (Edit: actually I believe that yes, those functions are dense. See below) Some comments on this $H^2(-a,a)$ thing that may be helpful:
First, it's not at all clear tat the norm you define is, or is equivalent to, a Hilbert-space norm. But in fact if $f\in H^2(-a,a)$ then the boundary values $f(x\pm ia)$ exist almost everywhere, and the norm above is equivalent to the $L^2$ norm of the boundary values.
Second, in fact $f\in H^2(-a,a)$ if and only if there exists $F\in L^2(\Bbb R)$ with $$\int e^{2a|\xi|}|F(\xi)|^2\,d\xi<\infty$$and $$F(z)=\int e^{iz\xi}F(\xi)\,d\xi;$$in fact $||f||$ is equivalent to $$\left(\int e^{2a|\xi|}|F(\xi)|^2\,d\xi\right)^{1/2}.$$So your question reduces to the question of whether the span of the functions $e^{ib\xi}e^{-\xi^2}$ is dense in $H=L^2\left(e^{2a|\xi|}d\xi\right)$.
I  think it's clear that this is so. Say $F_b(\xi)=e^{ib\xi}e^{-\xi^2}$. Say $F\in H$ and $F\perp F_b$ for every $b$. If you let  $G(\xi)=e^{2a|\xi|}e^{-\xi^2}F(\xi)$ then $G\in L^1(\Bbb R)$ and $\hat G=0$; hence $G=0$, so $F=0$.
About those preliminary results, the second in particular: I don't have a good reference. (i) Many years ago I heard Rudin say this had beenn proved by Bochner; I don't have a reference for that, to Rudin or to Bochner. (ii) I suspect this is a special case of results on "tube domains" in Stein&Weiss "Harmonic Analysis...".("Fourier Analysis..."?)
(iii) If you want to prove it yourself this might be  a way to get  started: Say $f_y(x)=f(x+iy)$.  Let $f_{\pm a}$ be a weak limit point of $f_y$ as $y\to\pm a$. Then a simple limiting argument shows that $$f(z)=\frac1{2\pi i}\left(\int\frac{f_{-a}(t)}{z-(t-ia)}\,dt-\int\frac{f_{a}(t)}{z-(t+ia)}\,dt\right).$$
Hint for showing the integral over the segments constituting the ends of the rectangles tends to $0$: If $0<\delta<a-|y|$ then 
$$\begin{align}|f(x+iy)|^2&=\frac1{\pi\delta^2}\left|\iint_{s^2+t^2<\delta^2}f^2(x+iy+s+it)\,dsdt\right|\\&\le\frac1{\pi\delta^2}\int_{-\delta}^\delta\int_{-\delta}^\delta|f(x+iy+s+it)|^2\,dsdt\end{align}$$
