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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?

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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.

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    Thanks, although $f$ may not be integrable with respect to the Lebesgue measure, but only integrable with respect to the Gaussian measure. – 2017-02-07
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    My bad, I misread the conditions on $f$. Let me find a workaround=) – 2017-02-07
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    Thanks for your reply. I now have an even more fundamental question: if we only assume $f\in L^1(\phi)$, the function $f$ may not be a tempered distribution (regarding to the Schwartz Class)? – 2017-02-07
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    @Uchiha Of course. take, for example, $f(x)=e^x$. It is integrable with respect to $\phi$, yet it is a classic example of a non-tempered distribution. – 2017-02-07
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    Then in this case, the argument using tempered distribution does not completely solve the problem, right? – 2017-02-08
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    Thanks for the update. If $f\in L^2(\mathbb{R})$ this is a good answer. But still one can allow $f\notin L^2(\mathbb{R})$. – 2017-02-08
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    I realize that $L^2(\mathbb{R})$ should be a dense subspace in $L^1(\phi)$. Then it solves the problem. – 2017-02-08