# Recovering a function from its Gaussian convolution

Let $$\varphi(x)=\frac{1}{\sqrt{2\pi}}\exp(-x^2/2)$$ be the Gaussian density and $$f:\mathbb{R}\to\mathbb{R}$$ another measurable function.

Under what conditions can $$f$$ be recovered from its convolution with $$\varphi$$? In other words, under what conditions does $$f\ast\varphi=0$$ imply that $$f$$ is zero a.e?

If $$f\in L^1(\mathbb{R})$$, then it has a Fourier transform and the statement follows since $$\varphi$$ has a Fourier inverse. What about other conditions on $$f$$? For example, what if it is bounded by a polynomial? Or a subexponential function?

$$\newcommand{\vpi}{\varphi}\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}\newcommand{\ep}{\varepsilon}$$The minimal condition $$\begin{equation*} |f|*\vpi<\infty \tag{1}\label{1} \end{equation*}$$ is already enough for the recovery of $$f$$.
Indeed, since $$\vpi(x-u)=\vpi(u)e^{xu}e^{-x^2/2}$$, condition \eqref{1} can be rewritten as $$\int_\R|f(u)|\vpi(u)e^{xu}\,du<\infty$$ for all real $$x$$ or, equivalently, as
$$\int_\R|f(u)\vpi(u)e^{zu}|\,du<\infty$$ for all complex $$z$$. Letting now $$\begin{equation*} g(z):=\int_\R f(u) \vpi(u) e^{zu}\,du, \end{equation*}$$ we have an entire function $$g\colon\C\to\C$$ such that $$\begin{equation*} g(x)=e^{x^2/2}\int_\R f(u) \vpi(x-u)\,du=e^{x^2/2}(f*\vpi)(x)=0 \end{equation*}$$ for all real $$x$$.
So, $$g=0$$. In particular, $$\begin{equation*} 0=g(it)=\int_\R f(u) \vpi(u) e^{itu}\,du \end{equation*}$$ for all real $$t$$ -- that is, the Fourier transform of the integrable function $$f\vpi$$ is $$0$$. It follows that $$f\vpi=0$$ almost everywhere (a.e.) and thus $$f=0$$ a.e., as desired.