Suppose $\{f_i\}$ is a sequence of locally integrable functions in $\mathbb R^n$, such that $$\lim_{i\rightarrow \infty}(f_i*\phi)(x)$$ exists, for each $\phi\in D(\mathbb R^n)$ and each $x\in \mathbb R^n$. Prove that then $\{D^\alpha(f_i*\phi)\}$ converges uniformly on compacts sets, for each multi-index $\alpha$.
My approach: Since each $\{f_i\}$ is given to be locally integrable hence one can treat them as a distribution and since the limits exists so one can think that the limit is again given by some distribution say $\Lambda$, but the think i am stuck in proving the convergence is uniform on compact sets. Any hint or suggestion on how to go further from here.
Thanks
Fix $\phi \in \mathscr{D}(\mathbb{R}^n)$, and a (nonempty) compact $K \subset \mathbb{R}^n$. We want to show that $(f_i \ast \phi)$ converges uniformly on $K$. Let $L = \operatorname{supp} \phi$ - we can assume $L \neq \varnothing$ since the assertion is trivial for $\phi = 0$ - and define $M = K - L$ (Minkowski sum). The map $\eta \colon x \mapsto \tau_x\check{\phi}$ (where $\tau_x\check{\phi}(t) = \phi(x-t)$) from $K$ to $\mathscr{D}_M = \{ \psi \in \mathscr{D}(\mathbb{R}^n) : \operatorname{supp} \psi \subset M\}$ is continuous, so $\eta(K)$ is a compact subset of $\mathscr{D}_M$. By the Banach-Steinhaus theorem, the assumption implies that $T_{f_i}\lvert_{\mathscr{D}_M}$ is an equicontinuous family, since $\mathscr{D}_M$ is a Fréchet space.(1)
For an equicontinuous family, pointwise convergence and uniform convergence on compact sets are equivalent, hence it follows that $\bigl(T_{f_i}\lvert_{\mathscr{D}_M}\bigr)_{i \in \mathbb{N}}$ converges uniformly on $\eta(K)$. But
$$T_{f_i}\lvert_{\mathscr{D}_M}(\eta(x)) = T_{f_i}\bigl(\tau_x\check{\phi}\bigr) = (f_i \ast\phi)(x),$$
so this is just the uniform convergence of $(f_i \ast \phi)$ on $K$.
Since $D^{\alpha}(f_i \ast \phi) = f_i \ast D^{\alpha}\phi$, the remaining part follows.
(1) It is not actually necessary to consider $\mathscr{D}_M$, we could work with $\mathscr{D}(\mathbb{R}^n)$, since that is a barrelled space, and the Banach-Steinhaus theorem is naturally a theorem about barrelled spaces. However, Rudin doesn't treat barrelled spaces - he mentions them in passing - and therefore it's probably safer to make the detour through $\mathscr{D}_M$.