Let $\phi$ be a real compactly supported smooth function on $\mathbb R$ with total integral zero. Define $\phi_t=\frac{1}{t} \phi(\frac{x}{t})$. I also suspect that they must be even, but the notes I am working from are imprecise.
I want to show that $$f(x) \left(\int_0^\infty |\hat\phi(t)|^2 \frac{dt}{t}\right)=\int_0^\infty (\phi_t\ast\phi_t\ast f)(x) \frac{dt}{t}.$$
This can be thought of as a reproducing formula for $f$ because the integral on the left side is a constant (with respect to $x$).
I am attempting to prove this but coming up short. I began by taking the Fourier transform of the right hand side and recalling that it takes convolutions into pointwise multiplication. We get
$$f(\xi) \int_0^\infty \hat\phi_t(\xi)^2 \frac{dt}{t}.$$
We can rewrite the term being integrated because the Fourier transform of ${\frac{1}{t}\phi(\frac{x}{t})}$ is $\hat\phi(t\xi)$.
$$f(\xi) \int_0^\infty \hat\phi(t\xi)^2 \frac{dt}{t}.$$
I do not see how to continue. In particular, I do not see how to make use of the fact that the total integral of $\phi$ is 0.
EDIT: A proof has been given in the comments, but I still don't know why the integral of $\phi$ must be zero! This condition appears in other statements of this formula I have found, not just this simplified one, but it does not seem necessary. I still welcome a definitive answer on this matter.