I tried to answer a question from a user, as in the following link
Properties of the solution of the schrodinger equation
As in the comments I am aware that my answer is not satisfactory, but intuitively it seems so.
In other words it is well known that if one of the two functions $f$ or $g$ has compact support, and we assume that it is well-defined convolution of them, then we have that
$$\mathrm{supp}(f \ast g) \subset \mathrm{supp}(f)+\mathrm{supp}(g)$$
My question is a reversal, that is, if $f$ has not compact support and $g$ has compact support, we may conclude that the convolution does not have compact support?
Answer
This is not true. It is even possible that the convolution is identically zero. Consider $f(x) \equiv 1$, which clearly does not have compact support, and let $g(x)$ be any odd function with compact support. Then
$$(f*g)(x) = \int_{-\infty}^\infty g(y) \, dy = 0.$$
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