If $f*g(x)=0$ then is $f$ identically zero?

Question

Let $f$ be a function in $L^1(\mathbb R).$ The convolution of $f$ and $g$ is defined by $f * g(x) = \int_{\mathbb R} f(x-t)g(t)dt.$

Let $g(x) = e^{-x^2}.$ Suppose it is given that $f*g(x)=0$ for all $x,$ does it follow that $f$ is zero almost everywhere?

If $f \ge 0,$ then we know that $\int f = 0 \iff f = 0 $ a.e., so here it follows that $f=0$ a.e., but what about if $f$ is not a positive function?

Answer 1

An overkilled answer:

Yes, it is. As $f,g\in L^{1}(\mathbb{R}^{n})$, one has \begin{align*} \widehat{f\ast g}(\xi)=\widehat{f}(\xi)\widehat{g}(\xi)=\widehat{f}(\xi)e^{-|\xi|^{2}}. \end{align*} Therefore, $\widehat{f}(\xi)=0$. We know by inversion formula that $f(x)=(\widehat{f})^{\vee}(x)$ a.e. as long as both $f,\widehat{f}\in L^{1}(\mathbb{R}^{n})$, so $f(x)=0$ a.e. in this case.