Convergence of $\varphi_\epsilon * \psi$ in $\mathcal{D}(\mathbb{R}^n)$ Let $\varphi \in\mathcal{D}(\mathbb{R}^n)$ st $\int_{\mathbb{R}^n} \varphi(x)dx=1$ and $\varphi_\epsilon = \epsilon^{-n} \varphi(\frac{x}{\epsilon})$  for all $x \in \mathbb{R}^n, ~ \epsilon >0$. The claim is that  $\varphi_\epsilon * \psi$ converges to $\psi$ in $\mathcal{D}(\mathbb{R}^n)$ if $\epsilon \rightarrow 0$ for all $\psi \in \mathcal{D}(\mathbb{R}^n)$ ($*$ denotes the convolution).
I managed to prove that $\varphi_\epsilon * \psi  \in \mathcal{D}(\mathbb{R}^n)$  and that there's a $K \subseteq \mathbb{R}^n$ compact st $supp(\varphi_\epsilon * \psi ), supp(\psi) \subseteq K ~ \forall \epsilon >0$. Thus it remains to show
$$ \underset{x \in K}{sup} | \partial_x^{\alpha} (\varphi_\epsilon * \psi )(x) -\partial_x^{\alpha} \psi(x) | \underset{ \epsilon \rightarrow 0}{ \longrightarrow} 0 ~ \forall \alpha \in \mathbb{N}_0^n.$$
So far, I showed
\begin{align}  \underset{x \in K}{sup} | \partial_x^{\alpha} (\varphi_\epsilon * \psi )(x) -\partial_x^{\alpha} \psi(x) | &= \underset{x \in K}{sup} | \int_{\mathbb{R}^n} \partial_x^{\alpha} (\varphi_\epsilon(y) \psi(x-y)) dy -\partial_x^{\alpha} \psi(x) | \\ &= \underset{x \in K}{sup} | \int_{\mathbb{R}^n} \partial_x^{\alpha} (\varphi_\epsilon(y) \psi(x-y)) dy - \int_{\mathbb{R}^n}\partial_x^{\alpha} \psi(x) \cdot \varphi(y) dy| \\ &= \underset{x \in K}{sup} | \int_{\mathbb{R}^n} \partial_x^{\alpha} (\varphi_\epsilon(y) \psi(x-y))  -\partial_x^{\alpha} \psi(x)  \cdot \varphi(y) dy|. \end{align}
How do I proceed from here? I tried to use the definition of $\varphi_\epsilon$ but I'm having difficulties with the partial derivatives . In the lecture, we've shown the statement with the $sup...$ for $\alpha =0$, ie. without partial derivatives (maybe this could be helpful here).
Thank you!
 A: *

*Let $T$ such that $supp(\varphi ) \subset \{ |x| < T\}$, $\varphi = \varphi^+-\varphi^-$ both non-negative supported on $|x| < T$ and $C^+ = \int_{\mathbb{R}^n} \varphi^+(x)dx$.

*Note that since $\psi \in D(\mathbb{R}^n)$ we have $$|\psi(x+t)-\psi(x)| \le A |t|$$ where $A = \sup_i \sup_x |\partial^{x_i} \psi(x)|$

*Thus
$$| \varphi_\epsilon^+ \ast \psi(x)- C^+\psi(x)| =|\int_{\epsilon K} \epsilon^{-n}\varphi(t/\epsilon) (\psi(x-t)-\psi(x))dt| \\ \le  \int_{\epsilon K} \epsilon^{-n}\varphi^+(t/\epsilon)\epsilon T  A dt = \epsilon T A C^+$$

*Therefore
$$| \varphi_\epsilon \ast \psi(x)- \psi(x)| \le 
| \varphi_\epsilon^+ \ast \psi(x)- C^+\psi(x)|+| \varphi_\epsilon^- \ast \psi(x)- (C^+-1)\psi(x)| \le \epsilon T A (2C^+-1)$$


*

*And hence
$$|\partial^\alpha( \varphi_\epsilon \ast \psi(x))-\partial^\alpha \psi(x)|= | \varphi_\epsilon \ast \partial^\alpha \psi(x)- \partial^\alpha\psi(x)| \le \epsilon T A^\alpha (2C^+-1)$$
where $A^\alpha = \sup_i \sup_x |\partial^{x_i}\partial^\alpha \psi(x)|$

A: Using the change of variables $\frac{y}{\epsilon}=v\implies
    \epsilon^n\mathrm d v=\mathrm d y$, we have
    \begin{align*}
  \lim_{\epsilon \to 0} \int_{\mathbb R^n} \psi(x-y)
  \frac 1 \epsilon \Phi\left( \frac y \epsilon \right) \mathrm d y
  =
  \lim_{\epsilon \to 0} \int_{\mathbb R^n} \psi(x-v\epsilon)
   \Phi (v) \mathrm d v
\end{align*}
    By continuity of $\psi$, there is some
    $\delta>0$ with
    $$
||x-(x-v\epsilon)||<\delta \iff v\in B_{\delta/\epsilon}(0)\implies
\psi(x-v\epsilon)\sim\psi(x)
$$
    and
    \begin{align*}
  \lim_{\epsilon \to 0} \int_{\mathbb {R}^n} \psi(x-v\epsilon)
   \Phi (v) \mathrm d v&=
   \lim_{\epsilon \to 0} \int_{ B_\frac{\delta}{\epsilon}(0)}
   \psi(x-v\epsilon)
    \Phi (v) \mathrm d v+\lim_{\epsilon \to 0} \int_{\mathbb{R}^n\setminus B_{\delta/\epsilon}(0)} \psi(x-v\epsilon)
     \Phi (v) \mathrm d v
\end{align*}
    provided both limits exist. For the second limit, note that
    \begin{align*}
  &\lim_{\epsilon \to 0}|\int_{\mathbb{R}^n\setminus B_{\delta/\epsilon}(0)} \psi(x-v\epsilon)
   \Phi (v) \mathrm d v|\\
   &\leq \lim_{N\to \infty}M|\int_{\mathbb{R}^n\setminus [-N,N]^n}\Phi(v)\mathrm d v|\to 0\\
\end{align*}
    Since the tail of convergent integrals get arbitrarily small and $\psi$ is bounded by some $M$.
For the first limit, note that by our continuity estimate we have
\begin{align*}
  \int_{B_{\delta/\epsilon}}\psi(x-v\epsilon)\Phi(v)\mathrm dv\sim 
  \psi(x)\int_{B_{\delta/\epsilon}}\Phi(v)\mathrm dv  
\end{align*}
and thus in the limit we have
$$
\psi(x)\lim_{\epsilon\to 0}\int_{B_{\delta/\epsilon}}\Phi(v)\mathrm dv=\psi(x)
$$
since $\Phi$ integrates to 1.
