I'm doing an exercise of generalized function, which claims that $$ \begin{aligned} C_c^\infty\times \mathscr{D}'&\rightarrow C^\infty\\ (\varphi,T)&\mapsto T*\varphi \end{aligned} $$ is a continuous map of $\varphi$ and $T.$ Remember that $\mathscr{D}=C_c^\infty,$ $T*\varphi(x)=\langle T_y,\varphi(x-y)\rangle.$
I solved the case of $\varphi.$ As we have a proposition of $\mathscr{D}'$ that: $$|\langle{T,\psi}\rangle|\le C_r\sup_{\begin{subarray}{c} |\alpha|\le m_r \end{subarray}}|\partial^\alpha \psi(x)|,\quad\forall \psi\in C_c^\infty(B_r).$$ If $\varphi_\nu\rightarrow 0,$ then we have $B_R\Subset \mathbb{R}^n$ such that $B_R\supset \operatorname{supp}\varphi_\nu.$ Now $\forall B_k\Subset \mathbb{R}^n,$ take $r\ge k+R,$ we can prove that $$\sup_{x\in B_k}|\langle T_y,\partial^\alpha\varphi_\nu(x-y) \rangle|\rightarrow 0,\quad\forall \alpha\in \mathbb{N}^n.$$ So $T*\varphi\rightarrow 0$ in $C^\infty$ sense.
How about the case with $T$? If $T_\nu\rightarrow 0,$ $T_\nu*\varphi\rightarrow 0$ pointwise. How to say the convergence is uniform on any given $K\Subset \mathbb{R}^n?$ It's enough to prove that it's equicontinuous. If we can prove that $|T_\nu*\varphi|\le M$ uniformly, then it's solved. Is it possible to prove that?