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?

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.)
• That looks very nice, thank you! Just one (maybe trivial) question: If we set $s=0$ and use the binomial formula, the term $x^k$ turns into a polynomial in $x$ and $\mu_k$ turns into a sum. Do we then have to start with $k=0$ and obtain the $k-$th moment by induction? Dec 15, 2020 at 16:59
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.