Convolution of distributions, compact support

Question

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.

Answer 1

$\DeclareMathOperator{\supp}{supp}$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.