Let $\vec{v}\in R^n$ be a probability distribution over a domain of size $n$. I.e., $\vec{v}$ has nonnegative entries that sum up to one.
What can we say about the zeros of $DFT(\vec{v})$, the discrete Fourier transform of $\vec{v}$?
Clearly, for the uniform distribution $\vec{v}=(1/n,...,1/n)$, $DFT(\vec{v})=(1,0,...,0)$. What I am trying to prove or disprove is that for a prime $n$, this is the only case in which $DFT(\vec{v})$ has zeros.
In other words, if $\vec{v}$ is a non-uniform distribution over a domain of size $n$ which is a prime, then $DFT(\vec{v})$ has no zeros.
Counterexample: Take $n=5$; define $$v_j=\frac1{10}\left(2+2\cos(2\pi j/5)\right)=\frac1{10}\left(2+e^{2\pi ij/5}+e^{2\pi i4j/5}\right)\quad(j=0,1,\dots,4).$$The first form makes it clear that $v_j\ge0$, while the second makes the DFT clear.