Suppose $\phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$. Let $f$ be a function so that $\int |f| \phi dx<\infty$, namely, $f\in L^1(\phi)$.
Assume that $(f\ast \phi)(x)=\int f(x-y)\phi(y)dy=0$ for all $x\in \mathbb{R}$. How can one show that $f=0$ a.e.?
Note:
I have a heuristic argument: view $f$ as a tempered distribution. Then the Fourier transform of $f\ast\phi$ is $\phi\cdot \hat{f}$, because $\phi$ is the Fourier transform of itself. Since $\phi>0$, then $\hat{f}=0$. Then by the uniqueness of Fourier transform (of tempered distributions), $f=0$.
But how does one justify $\hat{f}=0$ when $\phi\cdot \hat{f}$ understood in the sense of a distribution?
edit New version.
This might be an overkill, but this is the only reasoning I can think of at the moment. The Wiener's theorem for $L^2$ says that the span of translations $\phi_y(x) = \phi(x + y)$ is dense in $L^2(\Bbb R)$ if and only if the real zeros of the Fourier transform of $\phi$ form a set of zero Lebesgue measure.
Clearly, the above condition holds for our choice of $\phi$, and for all $y\in\Bbb R$ the function $f$ is orthogonal to $\phi_y$ in the sense of $L^2(\Bbb R)$. As the latter functions form a dense set, we conclude that $f=0$ in the sense of $L^2$, and hence $f(x)=0$ a.e.
In addition, $L^2(\mathbb{R})$ is a dense subspace of $L^1(\phi)$ since simple functions with finite measure (two measures are equivalent) support are dense in both spaces, which completes the proof.