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?


1 Answer 1


$\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.

  • $\begingroup$ Thank you for a neat answer. $\endgroup$
    – user477138
    Commented Feb 15, 2022 at 18:52

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