Convolution product $T*\varphi$ in the cas $T \in \mathcal{E}'$ and $\varphi \in \mathcal{E}$

Question

i have the follwing definition of convolution product. Let $T \in \mathcal{E}'(\mathbb{R}^n)$ and let $\varphi \in \mathcal{D}(\mathbb{R}^n)$. We have $$ T*\varphi(x)= <T,\varphi(x-y)>. $$ My question is: what's the formula of convolution product in the case: $T \in \mathcal{E}'(\mathbb{R}^n)$ and $\varphi \in \mathcal{E}(\mathbb{R}^n)$? There is an book where we can found the formula of convolution product $T * \varphi$ in the case $T \in \mathcal{E}'(\mathbb{R}^n)$ and $\varphi \in \mathcal{E}(\mathbb{R}^n)$?

Answer 1

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The specific question has been answered. Comments from the OP appear to indicate some confusion about the answer; I'm going to try to alleviate that by saying a bit about what this $\m E'$ thing is.

Officially: $\m E=C^\infty$. As such $\m E$ has a natural (metrizable) topology (where $f_n\to f$ if and only if $D^\alpha f_n\to D^\alpha f$ uniformly on compact sets, for every $\alpha$.) So $\m E$ is a toplogical vector space, and $\mathcal E'$ is precisely the dual.

Now, it's also said that $\m E'$ is the space of distributions with compact support. As is often the case, to make this literally true we need to read "is canonically isomorphic to" in place of "is". And to give a coherent explanation of why the two spaces are the same we need a different notation for the second space; let's write $\m D'_c$ for the space of distributions with compact support.

First, $\m E'\subset\m D'_c$: Suppose that $T\in\m E'$. Since the inclusion $\m D\subset\m E$ is continuous, the restriction of $T$ to $\m D$ is a distribution, and the fact that $T$ is continuous on $\m E$ turns out to imply that this distribution has compact support.

Having said that we can now give the literally true statement corresponding to the informal $\m E'=\m D'_c$: What's literally true is that $$\m D'_c=\{T|_{\mathcal D}:T\in\m E'\}.$$And that's the last time I'm going to put it that way; I'm going to continue to speak as though we're proving that $\m E'=\m D'_c$. That's the way people talk about this, may as well get used to it.

Conversely, $\m D'_c\subset \m E'$: Suppose that $T\in\m D'_c$. Fix a function $\psi\in\m D$ such that $\psi=1$ on a neighborhood of the support of $T$. Now if $\phi\in\m E$ then $\psi\phi\in\m D$, so we can define $\tilde T:\m E\to\Bbb C$by $$<\tilde T,\phi>=<T,\psi\phi>.$$Since $\phi\mapsto\psi\phi$ is a continuous mapping from $\m E$ to $\m D$ it follows that $\tilde T\in\m E'$.

(If I'm going to call the map $T\mapsto\tilde T$ canonical the definition of $\tilde T$ had better be independent of the choice of $\psi$. But this is clear: If $\psi_1$ and $\psi_2$ are both as above then $\psi_1\phi-\psi_2\phi$ vanishes on a neighborhood of the support of $T$, hence $<T,\psi_1\phi>=<t,\psi_2\phi>$.)

It's more or less clear that the maps $T\mapsto T|_{\m D}$ and $T\mapsto\tilde T$ are inverses of each other, and there's your canonical isomorphism.

(Again, for the sake of understanding the things you read about this, note that people are never this explicit: When they say an element of $\m E'$ is a distribution with compact support they are identifying $T$ and $T|_{\m D}$, and conversely when they say a distribution with compact support is an element of $\m E'$ they are identifying $T$ and $\tilde T$.)

Answer 2

An accessible introduction to convolution of distributions can be founded in the book Kesavan, S. Topics In Functional Analysis And Application, page 22. The definition is the same.