I agree with your answer, but it could not be generalized to an infinite number of zeroes, a natural extension of the question. The Weierstrass factorization theorem allows to construct entire functions with prescribed zeroes and in some case these functions could be of exponential type, thus with compactly supported Fourier transforms, e.g $\sin z$.

The linear part of the equation $iu_t+\Delta u$ is easy to define for a distribution. However, taking $F(u)=u^2$ would require essentially that $u(t)$ belongs to an algebra, which is not the case for $H^1(\mathbb R^n)$ when $n\ge 2$: this means that I do not understand how the equation could make sense without some additional assumptions of regularity for $u$, e.g. $H^{\epsilon +n/2}(\mathbb R^n)$.

Nice. You mean that $f$ is scalar-valued and defined by $\sum_{n\ge 1}nx_n^n$. Your sum on $n$ does not converge, say if $x_1=1$.

@John Bentin : No, this is the same condition as the one for Young's inequality $L^p\ast L^q\subset L^r$ under $(\sharp)$.

$\hat \mu\hat f$ is a complex (valued) measure on the real line with finite total mass.

$\hat \mu\hat f$ should be a measure on the real line with finite total mass.

In that case, $\hat \mu=\mu$ by Poisson summation formula. It is probably possible to weaken my requirement to $$\hat \mu\hat f\text{ is a measure with a finite total mass}.$$ If $f=e^{-\vert x\vert}$, you find $\hat f\hat \mu$ with finite mass since $\sum_{n\ge 1}\frac{1}{n^2}<+\infty$.