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?
2 Answers
$\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\N{\mathbb N}\newcommand\C{\mathbb C}$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 $$\alpha_k:=\int_a^b dy\, f(y) e^{(Cky)^2/2}$$ 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 $T:=ts$, we have $$\int_0^T dz\, H(z)(s+z)^k=\mu_k.$$ On the other hand, $$z^k=((s+z)s)^k=\sum_{q=0}^k\binom kq (s)^{kq} (s+z)^q.$$ Multiplying the latter identity by $H(z)$ and then integrating in $z$ from $0$ to $T$, we have $$\int_0^T dz\, H(z)z^k=\nu_k:=\sum_{q=0}^k\binom kq (s)^{kq} \mu_q.$$ 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): $$\sum_{0\le j\le ux}\sum_{k=j}^\infty\frac{(1)^{kj}u^k}{(kj)!j!}\,\nu_k\underset{u\to\infty}\longrightarrow\int_0^x dz\,H(z)$$ 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 complexvalued $h$ by linearity and in view of the decomposition of a realvalued function into its positive and negative parts.)

$\begingroup$ 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? $\endgroup$– J. SwailDec 15, 2020 at 16:59

$\begingroup$ @J.Swail : You don't actually have to use induction here. Or, one may say, this induction can be explicitly resolved. I have now added the corresponding details. $\endgroup$ Dec 15, 2020 at 18:50
I feel that a Fourierbased solution should be possible along these lines: Let $h(t)=\int f(x) \varphi (xt) dx$ represent the convolution and $f(x)$ be sufficiently bounded. Using sampling rate $1$ for simplicity we have the exact result $$h(t) = \sum a(n) \frac{\sin \pi(tn)}{\pi(tn)}$$ where $$a(n) =\int h(t) \frac{\sin \pi(tn)}{\pi(tn)} dt.$$ If $h(t)$ was suitably bandlimited 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 bandlimited, 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 Fourierbased approach; it would be extremely worthwhile.