1
$\begingroup$

For $u \in \mathcal{D}'(\mathbb{R}^n)$ and $\varphi \in C^{\infty}_0(\mathbb{R}^n),$ a convolution is given by $$ (u* \varphi)(x) = \langle u, \varphi (x- \cdot)\rangle$$ and $u* \varphi \in C^{\infty}(\mathbb{R}^n).$

How can i know that the above definition is good for $u \in \mathcal{E}'(\mathbb{R}^n)$ and $\varphi \in C^{\infty}(\mathbb{R}^n)$?

$\endgroup$
2
  • $\begingroup$ Also in that case, the right hand side of the definition is well-defined. Do you know how $\langle u, \varphi \rangle$ extends from $\mathcal{D}'\times C^\infty_c$ to $\mathcal{E}'\times C^\infty$? $\endgroup$
    – md2perpe
    Oct 19, 2021 at 21:08
  • $\begingroup$ How to do such an extension? @md2perpe $\endgroup$ Oct 19, 2021 at 22:19

1 Answer 1

1
$\begingroup$

Let $u\in\mathcal{D}'$ have compact support and $\varphi\in C^\infty$. Take $\rho\in C^\infty_c$ such that $\rho\equiv 1$ on a neighborhood of $\operatorname{supp} u.$ Then $\rho\varphi\in C^\infty_c$ so $\langle u, \rho\varphi \rangle$ is defined. Its value is also independent of choice of $\rho$ since different choices will only differ outside of the support of $u$. We can therefore extend $\langle \cdot, \cdot \rangle$ to $\mathcal{E}'\times C^\infty$ by setting $$\langle u, \varphi \rangle := \langle u, \rho\varphi \rangle.$$

The extension of the convolution follows from this extension.

$\endgroup$

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.