Proof that if $f, g \in L^1(\mathbb{R}^n)$ with compact support, then the convolution $f*g$ has compact support I have this sketch of a proof, but I am unsure about it.
Problem: 
Let $f,g \in L^1(\mathbb{R}^n)$, where they have compact support. Show that $f * g (x) = \int f(x-y)g(y) \mathrm{d} y$ has compact support.
For $\int f(x-y)g(y) \mathrm{d} y \neq 0$, we must have that $\mathrm{supp}(fg)=\mathrm{supp}(f) \cap \mathrm{supp}(g) \neq \emptyset$. The intersection of two compact sets is compact, so $\mathrm{supp}(fg)$ is compact.
Is this enough? Something gives me the feeling that this is incomplete.
 A: Your idea is good but needs some finishing touches.
Indeed, for $f* g(x)$ to be non-zero, you have to be able to find some $y\in \mathbb{R}$ such that $f(x-y)\neq 0$ and $g(y)\neq 0$. Then, by definition, $x-y\in \textrm{supp}(f)$ and $y\in \textrm{supp}(g)$, implying that $x\in \textrm{supp}(f)+\textrm{supp}(g)$. Thus, $\textrm{supp}(f*g)\subseteq \overline{\textrm{supp}(f)+\textrm{supp}(g)}$, and we need simply argue that this latter set is bounded.
Indeed, if $y\in \textrm{supp}(f)$ and $z\in \textrm{supp}(g),$ then $||y+z||\leq ||y||+||z||,$ implying that $\textrm{supp}(f)+\textrm{supp}(g)$ is bounded when $\textrm{supp}(f)$ and $\textrm{supp}(g)$ are. Thus, $\textrm{supp}(f*g)$ is a bounded, closed set and hence, compact.
A: We can choose $r>0$ such that the supports of $f,g$ are both contained in $B(0,r).$ The support of $f(x-y)$ is then contained in $B(x,r).$ Therefore the support of $f(x-y)g(y)$ is contained in $B(x,r)\cap B(0,r).$ The last set is empty if $|x|>2r,$ which implies$f*g(x)=0$ if $|x|>2r.$ Thus the support of $f*g$ is contained in the compact set $\overline {B(0,2r)}.$
