Integral of a function times a Fourier transform is zero This comes from Hall's Quantum Theory for Mathematicians, Lemma 9.33. I'm having trouble with one of the arguments in the proof. I believe it boils down to the following:
Let $f\colon \mathbb{R}^n \to \mathbb{R}$ such that $\int f(x) \hat g(x) \,\mathrm{d}x = 0$ for all smooth $g$ with compact support, where $\hat g$ denotes the Fourier transform of $g$. I then want to show that $f = 0$. In the proof $f$ is a difference of an $L^2$-function and an $L^2$-function times a coordinate function.
Hall cites the Stone-Weierstrass theorem and the theorem about density of continuous functions with compact support in $L^p$, but he's not super clear. (He also starts talking about the functions as if they are defined on $\mathbb{R}$, so I don't know what is going on.)
Also, do feel free to change the title to something more descriptive, I wasn't sure how to title my question.


 A: Since $g\mapsto \hat g:L^2(\mathbb R^n)\to L^2(\mathbb R^n)$ is an isometric isomorphism, we can choose an approximate identity $(\hat \phi_n).$ Now $\phi_n$ is smooth for each integer $n$ and  there are compactly supported smooth functions $(\psi_{jn})$ such that $\|\psi_{jn}-\phi_n\|\to 0$ and so $\|\widehat{\psi_{jn}-\phi_n}\|=\|\hat\psi_{jn}-\hat\phi_n\|\to 0$ as well. Then,
$\tag1 \|\hat\psi_{jn}*f-f\|\le \|\hat \psi_{jn}*f-\hat \phi_n*f\|+\|\hat \phi_n*f-f\|.$
We have, by Cauchy-Schwarz,
$\|\hat \psi_{jn}*f-\hat \phi_n*f\|\le \|\hat \psi_{jn}-\hat \phi_n\|\cdot \|f\|$ and so $\|\hat \psi_{jn}*f-\hat \phi_n*f\|<\epsilon$ if $j$ is large enough. On the other hand, $\|\hat \phi_n*f-f\|<\epsilon$ if $n$ is large enough, because $(\hat\phi_n)$ is an approximate identity.
But $\hat \psi_{jn}*f(x)=\int \hat \psi_{jn}(x-y)f(y)dy=0$ by hypothesis and so $(1)$ becomes
$\tag2 \|f\|\le \|\hat \psi_{jn}*f-\hat \phi_n*f\|+\|\hat \phi_n*f-f\|<2\epsilon.$
It follows that $f=0$ almost everywhere.
