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?



Recovering a function from its Gaussian convolution
$\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.


