# Definition of convolution for $u \in \mathcal{E}'(\mathbb{R}^n)$ and $\varphi \in C^{\infty}(\mathbb{R}^n).$

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 $$\varphi \in C^{\infty}(\mathbb{R}^n)$$?

• 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 $\mathcal{E}'\times C^\infty$? Oct 19, 2021 at 21:08
• How to do such an extension? @md2perpe Oct 19, 2021 at 22:19

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 $$\langle u, \varphi \rangle := \langle u, \rho\varphi \rangle.$$