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Let $f,g : \mathbb R \rightarrow \mathbb R$ continuous and compactly supported. I want to show that $f*g$ is continuous and compactly supported. I am 100% sure how to do it.

I began as follows:

\begin{align*} |(f*g)(x)-(f*g)(x')| \leq \int_{\mathbb R} |f(y)||g(x-y)-g(x'-y)| dy \\ \leq M \int_{\mathbb R} |g(x-y)-g(x'-y)| dy \end{align*} where $f$ is bounded by $M$ on $\mathbb R$. What must I do now ?

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    Instead of $\int_{\mathbb R}$, write $\int_I$ where $I$ is some compact interval independant of $x$ and $x'$. (The point $x$ is fixed and you may assume that $x'\in [x-1,x+1]$ to check continuity at $x$). Then use the uniform continuity of $g$. – Etienne Jul 18 '13 at 10:17
  • Oh yes. I forgot the uniform continuity. Thanks. But how can the support of $g(x-y)$ as a function of $y$ be independent of $x$ ? –  Jul 18 '13 at 10:20
  • It's not independent of $x$, but for all $x$, it is compact. The support of $y \mapsto g(x-y)$ is $x - \operatorname{supp} g$. – Daniel Fischer Jul 18 '13 at 10:24
  • Yes. I thus will show continuity in every point of $\mathbb R$ by using the uniform continuity of $g$ and a fixed compact support for this point. –  Jul 18 '13 at 10:26
  • If $F = supp(f)$ and $G = supp(g)$ then I found that $supp(fg) = F+G$. I know that $F+G$ is bounded so assume $F+G \subseteq B$ where $B$ is a compact set. Then $fg$ is supported on $B$ ? –  Jul 18 '13 at 10:48
  • @Daniel. Of course, you are right. It is independent of $x$ ... once $x$ is fixed! – Etienne Jul 18 '13 at 11:26
  • ??? It depends continuously on $x$, and that is all you need. – Daniel Fischer Jul 18 '13 at 11:29
  • That was just a joke. Apparently, not that funny. – Etienne Jul 19 '13 at 13:26
  • @Epsilon It seems ${\rm supp}(f*g) \subset {\rm supp}(f) + {\rm supp}(g)$; they are not equal? – swoopin_swallow Nov 02 '16 at 17:16

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