A condition for $\hat f$ to be integrable

Question

Let $f \in L^1 (\mathbb R^n)$. Suppose that $f$ is continuous at zero and that the fourier transform $\hat f$ of $f$ is non-negative. Does this imply that $\hat f \in L^1$ (and hence, by the inversion theorem, that $f$ is continuous)? If so, how could I go about proving it?

Answer 1

In A positive "Fourier transform" is integrable Zarrax gave a great answer to this question using mollifiers. He/she also provides a counterexample in the case $f$ is not continuous at $x=0$.

Answer 2

I don't think that $\hat f \in L^1$, but I think that you can get easily that $f$ is continuous.

If I remember right, $\hat{f}$ non-negative implies that $f$ is positive definite, and then Krein's inequality takes care of continuity...