Is there a way to reconstruct the convolution $(f * g)(x)$ of $f$ with a Gaussian $g$ from sampled values, $(f*g)(a), a \in A$? 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?
 A: $\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}\newcommand\C{\mathbb C}$Suppose indeed that $A=c\mathbb Z$ for some $c>0$. By rescaling, the problem reduces to the following: for given real $a$ and $b$ such that $a<b$, recover the (say continuous) function $f\colon[a,b]\to\C$ given the numbers
$$\alpha_k:=\int_a^b dy\, f(y) e^{-(Ck-y)^2/2}$$
for some real $C>0$ and all $k\in\Z$. Note that
$$\alpha_k=e^{-(Ck)^2/2}\int_a^b dy\, f(y) e^{-y^2/2}e^{Cky}.$$
So, with the substitution $x=e^{Cy}$, the problem reduces to the following inverse moment problem: for given real $s$ and $t$ such that $s<t$, recover the continuous function $h\colon[s,t]\to\C$ given the power moments
$$\mu_k:=\int_s^t dx\, h(x) x^k$$
for all $k\in\N_0$.
Further, using the substitution $x=s+z$ and the binomial formula, we may assume that here $s=0$. Indeed, letting $H(z):=h(s+z)$ and $T:=t-s$, we have
$$\int_0^T dz\, H(z)(s+z)^k=\mu_k.$$
On the other hand,
$$z^k=((s+z)-s)^k=\sum_{q=0}^k\binom kq (-s)^{k-q} (s+z)^q.$$
Multiplying the latter identity by $H(z)$ and then integrating in $z$ from $0$ to $T$, we have
$$\int_0^T dz\, H(z)z^k=\nu_k:=\sum_{q=0}^k\binom kq (-s)^{k-q} \mu_q.$$
So, now it suffices to recover the function $H$ given its power moments $\nu_k$.
The resulting inverse moment problem can be solved by an explicit formula, as explained e.g. in the paragraph containing formula (2.2):
$$\sum_{0\le j\le ux}\sum_{k=j}^\infty\frac{(-1)^{k-j}u^k}{(k-j)!j!}\,\nu_k\underset{u\to\infty}\longrightarrow\int_0^x dz\,H(z)$$
for all $x\in[0,T]$, as desired. (This formula is stated in the linked paper only for nonnegative $h$. However, this inversion formula holds for complex-valued $h$ by linearity and in view of the decomposition of a real-valued function into its positive and negative parts.)
A: 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.
