# Convolution of distributions, compact support

We denote by $$E'$$ the dual of $$C^\infty(\mathbb R^n)$$( i.e. the set of distribution with compact support) and with $$D'$$ the dual of $$C^\infty(\mathbb R^n)$$ with compact support. Considering $$u\in D'$$, $$v\in E'$$, $$\varphi \in C^\infty_c(\mathbb R^n)$$, we define: $$\langle u*v,\varphi \rangle := \langle u \otimes v, \tilde \varphi\rangle= \langle u(x), \langle v(y),\varphi(x+y)\rangle\rangle.$$ In order to be consistent with the definition of $$D'$$, I have to verify that $$x\to \langle v(y),\varphi(x+y)\rangle$$ is a function in $$C^\infty(\mathbb R^n)$$ with compact support, but I don't know how to show it.

• It is immediate that $\langle v(y),\varphi(x+y)\rangle$ is continuous, smooth, and with compact support. Dec 6, 2022 at 20:32
• Could you show me why that function has compact support? Dec 6, 2022 at 21:42
• You said that $v$ is a distribution with compact support and $\varphi$ is $C^\infty_c$, it is completely obvious that their convolution vanishes for $|x|$ large enough. Dec 6, 2022 at 22:18

$$\DeclareMathOperator{\supp}{supp}$$By considering how to differentiate a distribution respect to a parameter, the smoothness is not difficult to prove, so I'll focus on proving the compactness of the support of $$x\mapsto \langle v(y),\varphi(x+y)\rangle$$. Let $$V=\supp v\Subset\Bbb R^n$$ (in the sense of distributions) and $$W=\supp\varphi\Subset\Bbb R^n$$ and consider their Minkowski sum $$V+ W=\left\{u+w : u\in V \wedge w\in W\right\}.$$ Note that $$V+W\Subset\Bbb R^n$$ i.e. it has a compact closure in $$\Bbb R^n$$.
Now, if $$x$$ is chosen in such a way that $$x+y\notin V+ W,\; \forall y\in V$$ then the compact set $$W_x=\supp \varphi(x+y) = \supp \varphi\circ\mathscr{t}_x$$ supporting the test function $$\varphi(x+y) = \varphi\circ\mathscr{t}_x$$ ($$\mathscr{t}_x$$ is a translation in $$\Bbb R^n$$) surely does not share any point with $$V \implies \langle v(y),\varphi(x+y)\rangle=0$$. This implies that $$\supp \langle v(y),\varphi(x+y)\rangle\subset V+W$$ and, being it closed by definition, it is compact.