$f$, $g$ have compact support implies their convolution does?

Question

If $f$ and $g$ have compact support, then does it follow that $f * g$ has compact support?

Answer 1

If $f$ has support on $[a,b]$ and $g$ has support on $[c,d]$ then the convolution also has compact support, on the interval [a+c, b+d].

To see this, note that $f(t) = 0$ when $t < a$ and $g(x-t) = 0$ when $t > x - c$. Thus, the product $f(t) g(x-t) = 0$ for any $t$ when $x < a+c$. The argument for the upper bound is identical.

Answer 2

Note: General rule in $\mathbb{R}^n$ \begin{align} \operatorname{supp}[f\ast g] \subset \overline{\operatorname{supp}[f]+\operatorname{supp}[g]} \end{align}