Assume you are given a non-negative, continuous, radial function $f\in L^q(\mathbb{R}^3)$ (for any $q\geq 1$).


Are there any conditions which would guarantee that the Fourier transform of $f$, that is $\hat{f}(p)$, is also non-negative? 



Conditions for positivity of Fourier transform
the three-dimensional Fourier transform $F(\vec{p})$, of the radial function $f(r)$ has Fourier transform


$$F(\vec{p})=\int\_0^\infty dr\int\_0^\pi d\theta \int\_0^{2\pi}d\phi\;e^{ ipr\cos\theta}f(r) r^2\sin\theta$$
$$\qquad=\frac{4\pi}{p}\int\_0^\infty rf(r)\sin(pr)\,dr,\;\;{\rm with}\;\;p=|\vec{p}|.$$


so you're asking when the Fourier-sine-transform $S(p)$ of $rf(r)$ will be (pointwise) positive for $p>0$. A sufficient condition is that $rf(r)$ is a decreasing function of $r>0$, see [On positivity of Fourier transforms.](http://journals.cambridge.org/download.php?file=%2FBAZ%2FBAZ74_01%2FS0004972700047511a.pdf&code=c132cf865945b7fc535dff80e4f96db5)


For a related question, see [this MO post.](http://mathoverflow.net/questions/193849/a-calculus-question-related-to-the-nonnegative-definite-functions/193866#193866)


For the connection to [Bochner's](http://en.wikipedia.org/wiki/Bochner_theorem) theorem: the OP's question amounts to finding a function that is both positive and positive-definite, since positive functions have positive-definite Fourier transforms.


