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.



  1. if $g$ is a Schwartz function, what is the Fourier transform of $x^\beta D^\alpha g$?
  2. if $g_n \to 0$ in $S$ and $g\in S$, where does $g\cdot\hat g_n$ converge?
  • 1
    $\begingroup$ Thank you very much! With your hints and using the continuity of the Fourier and Inverse Fourier Transform in S, I managed to prove it. $\endgroup$ Feb 8 '17 at 17:25
  • $\begingroup$ @JuanDavidSamboní you are welcome! $\endgroup$ Feb 8 '17 at 17:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.