I'm trying to understand the proof of the following fact about tempered distributions. Let $U\in \mathscr{S}'$ be a tempered distribution, and let $\{\rho_\varepsilon\}$ be the family of standard Friedrichs mollifiers, then: $$\rho_\varepsilon\ast U\to U \ \ \text{in $\mathscr{S}'$}$$ The proof in my book proceeds as follows. Since the Fourier transform $\mathcal{F}:\mathscr{S}'\to \mathscr{S}'$ is an homeomorphism, we just need to prove that: $$\mathcal{F}(\rho_\varepsilon\ast U)\to \mathcal{F}U \ \ \text{in $\mathscr{S}'$}$$ Using the properties of the Fourier transform: $$\langle \mathcal{F}(\rho_\varepsilon\ast U),\psi \rangle=\langle \hat\rho_\varepsilon \hat{U},\psi\rangle=\langle \hat U,\hat \rho_\varepsilon \psi\rangle=\langle \hat U,\rho(\varepsilon\bullet)\psi\rangle$$ where $\rho(\varepsilon\bullet)$ is the function $x\mapsto \rho(\varepsilon x)$. Now my book concludes the proof by saying that it's obvious that $\rho(\varepsilon\bullet)\psi\to \psi$ in $\mathscr{S}$.
I tried to do prove this rigorously.
My attempt I have to prove that for any multi-indexes $\alpha,\beta$: $$||x^\alpha D^\beta (\rho(\varepsilon \bullet)\psi)||_\infty\to ||x^\alpha D^\beta \psi||_\infty$$ Using generalized Leibnitz formula: $$x^\alpha D^\beta (\rho(\varepsilon \bullet)\psi)|_x=x^\alpha \sum_{\gamma \leq \beta} \binom{|\beta|}{|\gamma|} \varepsilon^{|\gamma|}D^\gamma \rho|_{\varepsilon x} D^{\beta-\gamma} \psi|_x=x^\alpha D^\beta \psi|_x+ \sum_{\gamma \leq \beta,\gamma \neq 0} \binom{|\beta|}{|\gamma|} \varepsilon^{|\gamma|}x^\alpha D^\gamma \rho|_{\varepsilon x} D^{\beta-\gamma} \psi|_x $$
So I just need to bound uniformly (in $x$) the last sum with a term that tends to $0$ as $\varepsilon\to 0$. Since $\psi\in \mathscr{S}$, I can bound $ D^{\beta-\gamma} \psi|_x$ with the Schwartz seminorm $p_{|\beta|}(\psi)$ and since also $\rho\in \mathscr{S}$, I can bound $x^\alpha D^\gamma \rho|_{\varepsilon x}$ with $p_{\max(|\alpha|,|\beta|)}(\rho)$. Clearly $\binom{|\beta|}{|\gamma|}\leq |\beta|!$. So in the end:
$$\sum_{\gamma \leq \beta,\gamma \neq 0} \binom{|\beta|}{|\gamma|} \varepsilon^{|\gamma|}D^\gamma \rho|_{\varepsilon x} D^{\beta-\gamma} \psi|_x\leq \beta!p_{|\beta|}(\psi)p_{\max(|\alpha|,|\beta|)}(\rho)\sum_{\gamma\leq \beta,\gamma\neq 0} \varepsilon^{|\gamma|}\to 0 \ \ \ \text{ uniformly in $x$}$$
Is this correct?