Fix a "test" function $f(x)=x\exp(-x^2)$, which is nonzero except $x=0$. Suppose that $g$ is a function with some necessary regularity. Consider the convolution.
$$
(f\ast g )(x)=\int\_{-\infty}^{+\infty} f(y)g(x-y)dy.
$$
Assume that $(f\ast g) (x)=0$ on an open interval $x\in (a,b)$. Does this imply that $g(x)$ takes a constant value almost everywhere on $\mathbb{R}$? (The converse is true due to the symmetry of $f(x)$.) 



Convolution vanishes on an interval
Yes, because $f\*g$ is indeed *real-analytic* (provided $g$ doesn't grow too fast at $\infty$), so that, being $0$ on an interval, it is $0$ everywhere. Then the Fourier transform of $f\*g$, which is a Gaussian times the Fourier transform of $g'$, is $0$, and as the Gaussian is nonzero everywhere this implies $g'=0$ as a distribution.


