I am working on a rather brief introduction to distributions and I am quite stuck on the section on convolution. I have two questions in particular:
The convolution of a distribution $u$ and a compactly supported smooth function $v$ is defined as $u(v_x)$, where $v_x(y) = v(x-y)$. It is clear that this is a function, but, to me at least, it is not clear why it is smooth. I have seen one proof which uses Taylor's theorem with remainder, but is there another way to show this?
Secondly, is $u$ is a distribution and $v, \phi$ are compactly supported smooth functions, then why do we have $\int u*v(x)\phi(x)dx = u(\tilde{v}*\phi), $ where $\tilde{v}(x)=v(-x)$?
I am obviously new to this, so any good references would be appreciated.
For a test function $\varphi \in D(\mathbb{R})$ and a distribution $ T \in D'(\mathbb{R})$ let $$ \phi(x) = T \ast \varphi(x)= \int_{-\infty}^\infty T(y) \varphi(x-y)dy$$ Claim : for every $k$, $\phi^{(k)}(x) = \int_{-\infty}^\infty T(y) \varphi^{(k)}(x-y)dy$ is well-defined. Thus $\phi \in C^\infty$
$\qquad $ (note $\varphi^{(k)}(x+\epsilon-.)-\varphi^{(k)}(x-.) \to 0$ in the test function topology)
Therefore if $T$ is a compactly supported distribution then $T \ast \varphi$ is a test function and we can define $S \ast T \ast \varphi = S \ast (T \ast \varphi)$ for any distribution $S$. Moroever if $S$ is compactly supported then the convolution is commutative $S \ast T = T \ast S$.
Your last point follows from $\int_{-\infty}^\infty U(y) \varphi(y)dy = (U \ast \varphi(-.))(0)$