Convolution of distributions is commutative

Question

Let $u,v\in \mathcal{E}'(\mathbb{R}^n)$ be two compactly supported distributions. Define $u*v$ to be the distribution $u*v(\phi) = u(Rv*\phi)$, where $v*\phi =v(\tau_xR\phi)$ for $\phi\in C^\infty$, and $\tau_x\phi(t) = \phi(t-x)$, $R\phi(t) = \phi(-t)$. The book I'm reading claims that this convolution is commutative, that is, $u*v = v*u$. Is this true? I get that $u*v(\phi) = u(v(\tau_{-x}\phi))$, while $v*u(\phi) = v(u(\tau_{-x}(\phi))$, how are these equal?

Answer 1

Apart from citing sources for the proof... there might be some interest in setting a context in which the commutativity of the group $\mathbb R$ makes the conclusion obvious.

And, although I myself was late to understand that this could be an issue, there is a question of how compactly-supported distributions on a Lie group $G$ might act on an arbitrary (continuous...) representation of $G$.

Only relatively recently I came to understand that compactly-supported distributions do not naturally act on all repns. They do naturally act on smooth functions on $G$, and that's really the extent.

And... how to say what "convolution" is? It should be an operation on functions $f,F$ such that in the integrated action of functions (on a group) on a repn of the group, $(f*F)\cdot v=f\cdot (F\cdot v)$.

(Many older sources do not talk about repns or group actions... but only talk about "convolution" as specific integrals... because this is obviously more elementary.)