Is it possible to have an even probability measure $\mu$ (that is $\mu(A)=\mu(A)$ for any set $A\subset \mathbb{R}^d$) supported on the unit sphere $S^{d1}$ such that its Fourier Transform $$ \widehat{\mu}(\xi) = \int_{S^{d1}} e^{2\pi i x\cdot \xi} d\mu(x) $$ is nonnegative, that is $\widehat{\mu}(\xi)\geq 0$. If not, then how small $\left\inf_{\xi \in \mathbb{R}^d} \{\widehat{\mu}(\xi)\}\right$ can be?

4$\begingroup$ Average over rotations. $\endgroup$ – fedja Dec 21 '17 at 1:01

4$\begingroup$ Hint: If $\mu$ vanishes at 0, what does that imply about $\hat\mu$? $\endgroup$ – Michael Renardy Dec 21 '17 at 1:32

1$\begingroup$ If we average on rotations, we get a measure constant on the sphere, this give us a Bessel function which oscillates a lot and is negative at some points. Your idea actually suggests the question if a uniform measure is indeed the best possible, having the smallest $\inf$. Which I already considered. Also, a priori $\mu$ is supported in the sphere, so "at zero" it does vanishes. $\endgroup$ – Felipe Ferreira Dec 22 '17 at 4:56
Not to leave this question "unanswered", while the answer (to the qustion in the title) is obvious: because $\mu$ vanishes on a neighborhood of $0$, $\int\hat\mu=0$, so that $\hat\mu$ (real, for even $\mu$) has negative and positive values. Since it is also continuous, it vanishes somewhere.
What seems not obvious is: how small can $\inf \hat\mu(\xi)$ be?

1$\begingroup$ Perhaps it is worth emphasizing that the Fourier transform of a compactly supported (regular Borel) measure is indeed a continuous function, so that it makes sense to talk about pointwise values. Thus, approximating $\widehat\mu$ by truncations does give the correct limiting value for $\int \widehat \mu$. $\endgroup$ – paul garrett Dec 23 '17 at 21:12

3$\begingroup$ As hinted in the comment by fedja on the original question, an averaging argument shows that $\inf \hat\mu(\xi)$ is minimized when $\mu$ is the uniform distribution on the sphere, so it comes down to finding the minimum value taken by the relevant Bessel function. $\endgroup$ – Noam D. Elkies Dec 24 '17 at 4:45

1$\begingroup$ This answers in part the question, I would say. But, for instance, if $\mu$ is a finite sum of deltas on the sphere then $\hat{\mu}$ is a sum of cosines, which is not integrable. This can be fixed by an approximation of the identity argument, showing that $\int \hat{\mu} \hat{\phi}$ is as close to zero as one wishes by choosing $\phi$ close to a delta at zero with $\hat{\phi}\geq 0$ and close to $1$. But, I still do not know how to show that the argument over rotations would imply that the the constant measure (producing a bessel) is the one with the smallest $\inf \hat{\mu(\xi)}$. $\endgroup$ – Felipe Ferreira Dec 24 '17 at 20:35