For an exercise I have to show the following:
Let $u_j \to u$ in $\mathcal{D'(\mathbb{R}^n)}$ and let $\phi_j \to \phi$ in $C^{\infty}_0(\mathbb{R}^n)$. Show that $$ \lim_{j\to \infty} u_j * \phi_j = u*\phi $$ in $C^{\infty}(\mathbb{R}^n)$.
What I have tried so far
This means that for all compact sets $K \subset \mathbb{R}^n$, and all orders of differentiations $\alpha \in \mathbb{Z}_{\geq 0}$, it must hold that \begin{equation} \partial^{\alpha}(u_j*\phi_j) \to \partial^{\alpha}(u*\phi) \tag{1} \end{equation} uniformly on $K$. I can use that $\partial^{\alpha}(u_j*\phi_j) = u_j*(\partial^{\alpha} \phi_j)$, and since $\partial^{\alpha} \phi_j \in C^{\infty}_0(\mathbb{R}^n)$, we only have to show (1) for $\alpha = 0$, because the other values follow by induction.
Let $K \subset \mathbb{R}$. Then we know that \begin{equation} |(u_j * \phi_j)(x) - (u*\phi)(x)| \leq |(u_j*\phi_j)(x) - (u_j*\phi)(x)| + |(u_j*\phi)(x) - (u*\phi)(x)| \end{equation} I will now show that both terms go to zero. We use the translation operator $T_x$ ($T_x \phi(y) = \phi(y-x)$) and the reflection operator $S$ ($S\phi(y) = \phi(-y)$). Now the first term is given by \begin{equation} (u_j*\phi_j)(x) - (u_j*\phi)(x) = u_j(T_x \circ S(\phi_j-\phi)) \end{equation} Because of the convergence of $\phi_j$ we know that there exists a compact set $A$ such that $\text{supp }\phi_j \subset A$ and $||\phi_j-\phi ||_{C^k} \to 0$ for all $k$. Hence we find that \begin{equation} \text{supp }(T_x \circ S(\phi_j-\phi)) = x - \text{supp }(\phi_j-\phi) \subset K+(-A) := B \end{equation} Since $B$ is the sum of two compact sets, it itself is compact. We can now use the uniform boundedness theorem to find that \begin{equation} |u_j (T_x \circ S(\phi_j -\phi))| \leq c || T_x \circ S(\phi_j -\phi)||_{C^k} = c ||\phi_j -\phi||_{C^k} \to 0 \end{equation}
We now look at the second term. Let $v_j = u_j-u \to 0$ in $\mathcal{D}'(\mathbb{R}^n)$. Now by the same arguments as above we find by the uniform boundedness theorem \begin{equation} |v_j(T_x \circ S\phi)| \leq c || \phi||_{C^k} \leq M \end{equation} Now for $x,y \in K$ we find that \begin{equation} |v_j*\phi(x) - v_j*\phi(y)| \leq |x-y| \sup_{z \in K} \partial (v_j*\phi)(z) \leq |x-y|M \end{equation} So that for all $\epsilon > 0$ we can pick $\delta = \frac{\epsilon}{M}$ and $|x-y| < \delta$ implies $|v_j*\phi(x) - v_j*\phi(y)|<\epsilon$. We now invoke the Arzelà-Ascoli theorem to find a uniformly convergent subsequence $v_{j_k}*\phi \to 0$. Since $v_j \to 0$ it must be that $v_j*\phi \to 0$ uniformly on $K$.
My question
I feel the first part of the proof is good and complete, but I am not so sure about the second part. Could you please point out if there are any mistakes, or if there are some missing arguments?