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Let $f, g\in C_0(\mathbb R^n)$ where $C_0(\mathbb R^n)$ is the set of all continuous functions on $\mathbb R^n$ with compact support. In this case $$(f*g)(x)=\int_{\mathbb R^n} f(x-y)g(y)\ dy,$$ is well defined.

How can I show $\textrm{supp}(f*g)\subseteq \textrm{supp}(f)+\textrm{supp}(g)$?

This should be easy but I can't prove it.

I tried to proceed by contradiction as follows: Let $x\in \textrm{supp}(f*g)$. If $x\not\in \textrm{supp}(f)+\textrm{supp}(g)$ then $(x-\textrm{supp}(f))\cap \textrm{supp}(g)=\phi$. This should give me a contradiction but I can't see it.

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If $f*g(x)\neq 0$ then $\int_{\Bbb R^n}f(x-y)g(y)dy\neq 0$, so there exists $y\in \Bbb R^n$ such that $f(x-y)g(y)\neq 0$, hence $g(y)\neq 0$ and $f(x-y)\neq 0$, take $z=x-y$ then $x=z+y$ with $f(z)\neq 0$ and $g(y)\neq 0$. Now we get $\{f*g\neq 0\}\subset \{f\neq 0\}+\{g\neq 0\}\subset \text{supp}(f)+\text{supp}(g)$, so $\text{supp}(f*g)\subset \text{supp}(f)+\text{supp}(g)$.

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  • $\begingroup$ perfect argument, thanks =) $\endgroup$ – PtF Aug 17 '14 at 22:44
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If $x\notin\operatorname{supp}f+\operatorname{supp}g$ then clearly $f*g(x)=0$, since the integrand in the definition is identically zero. Now note that $\operatorname{supp}f+\operatorname{supp}g$ is compact, being the sum of two compact sets.

You do need the compactness here: For a countexample with noncompact supports, let $f$, $g$ be nonnegative with supports $\{(x,y\}\colon x,y>0,\,xy>1\}$ and $\{(x,y\}\colon x<0,\,y>0,\,xy<-1\}$ respectively. The sum of the two supports is the open upper half plane, but the support of $f*g$ is the closed upper half plane.

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  • $\begingroup$ right, but I can't see where you used the compactness in your argument.. $\endgroup$ – PtF Aug 17 '14 at 22:38
  • $\begingroup$ The sum of the two supports is compact, hence closed. The set where $f*g$ is nonzero is contained in this set, hence so is the support of $f*g$. $\endgroup$ – Harald Hanche-Olsen Aug 17 '14 at 22:39
  • $\begingroup$ Now I understood, thanks =) $\endgroup$ – PtF Aug 17 '14 at 22:49

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