TITLE: Is there a way to reconstruct the convolution $(f * g)(x)$ of $f$ with a Gaussian $g$ from sampled values, $(f*g)(a), a \in A$?
QUESTION [6 upvotes]: Suppose that $f: \mathbb{R} \to \mathbb{C}$ is a function which has support in $[-1,1]$. Let $g = g_\sigma$ be a centered Gaussian with variance $\sigma^2$. Is there a way to reconstruct the convolution of $f$ with $g$ (on whole $\mathbb R$) if only sampled values of this convolution are given, i.e.
$$
\{ (f *g)(a) : a \in A \}
$$
where $A \subset \mathbb R$ is a set such as a uniform grid, $A=c\mathbb Z$ for some $c>0$? Can we make some strong assumptions on $f$ such that there exists an explicit reconstruction?

REPLY [2 votes]: I feel that a Fourier-based solution should be possible along these lines: Let $h(t)=\int f(x) \varphi (x-t) dx$ represent the convolution and $f(x)$ be sufficiently bounded. Using sampling rate $1$ for simplicity we have the exact result $$h(t) = \sum a(n) \frac{\sin \pi(t-n)}{\pi(t-n)}$$ where $$a(n) =\int h(t) \frac{\sin \pi(t-n)}{\pi(t-n)} dt.$$ If $h(t)$ was suitably band-limited one would actually have $a(n)=h(n)$. In fact, $h(n)$ are the only values given to us. The catch is that $h$ is not band limited although we do know that $f(x)$ is nicely bounded. Presumably if the sampling rate were increased (equivalent here to making $f(x)$ more bounded) the approximation involved in using $a(n)= h(n)$ would increase in accuracy. Since $h(t)$ is not band-limited, perfect reconstruction from the sampled $h(n)$ values may not be possible. On the other hand, the boundedness of $f$ and the fact that the Fourier transform of $\varphi$ is supported on all of $R$ gives some hope that the argument can be patched up to work. In particular, one could try to evaluate the integral for $a(n)$ by discretizing it given a sufficiently dense lattice for $h(t)$. As an aside, the moments solution that was proposed was extremely interesting. Can someone please try to complete the Fourier-based approach; it would be extremely worthwhile.