I'm learning about distribution theory and I have a trouble proving that the convolution between a tempered distribution and a schwartz function is a tempered distribution. Let $S$ be the schwartz space and $S'$, its dual, the tempered distribution space.
Let $\varphi\in S$, $\psi\in S'$. The convolution between $\varphi$ and $\psi$ is defined as follows: $$(\psi*\varphi)(\phi) = \psi(\tilde{\varphi}*\phi) \text{ , for all } \phi\in S;$$ where $\tilde{\varphi}(x)=\varphi(-x)$. I want to prove that $\psi*\varphi\in S'$.
I have to see that $\psi*\varphi$ is well defined. As $\tilde{\varphi},\phi\in S\subset L^1$, $\tilde{\varphi}*\phi\in L^1$ and we can calculate its Fourier transform. In fact: $$\widehat{\tilde{\varphi}*\phi} = \widehat{\tilde{\varphi}}\widehat{\phi}.$$ Now, as $\widehat{\tilde{\varphi}},\widehat{\phi}\in S$, we have that the product $\widehat{\tilde{\varphi}}\widehat{\phi}\in S$. Finally, in consequence of the Fourier Inverse Formula, $\tilde{\varphi}*\phi\in S$. I think that proves that the functional is well-defined.
Now, I should prove that it is linear and continuous. The linearity is inmediate because of the linearity of the convolution (respect to $\phi$). My problem is proving the continuity. Let $(\phi_m)$ be a sequence in $S$ which converges to $0$ in $S$. Then $$\lim_{m\to\infty} \sup_{x\in\mathbb{R}^n} |x^{\beta}D^{\alpha}\phi_m(x)| = 0,$$ for all multi-index $\alpha, \beta$. I think I should prove that $(\tilde{\varphi}*\phi_m)$ also converges to $0$ and the continity will follow from the continuity of $\psi$. I'm trying to see if $x^{\beta}D^{\alpha}(\tilde{\varphi}*\phi_m)(x)$ is bounded, but I only have that $$|x^{\beta}D^{\alpha}(\tilde{\varphi}*\phi_m)(x)| = |x^{\beta}(\tilde{\varphi}*D^{\alpha}\phi_m)(x)|.$$ I'm trying to use the convergence of $(\phi_m)$ but I really don't know how to. Can someone please help with that? Thanks.
