Why does convolution of delta function commute (test distribution perspective)?

Question

If I understand correctly, for test functions $f$ we define

$$ \langle\delta, f\rangle = f(0)$$

and convolution is defined as

$$ \langle g * T, f\rangle = \langle T, g^- * f\rangle,$$

where $f$ and $g$ are test functions.

From this it follows that

$$ \langle g * \delta, f\rangle = \langle\delta, g^- * f\rangle = (g^- * f)(0) = \int g^-(-y) f(y) dy = \int g(y) f(y) dy = \langle g, f\rangle$$

so that $g * \delta = g$.

I don't see how any of this would imply that $ \delta * g = g$ as well. In fact, I don't even see how this expression makes sense, since $\langle \delta * g, f\rangle = \langle g, \delta * f\rangle $. Since $\delta$ is only defined as a distribution, I don't think $\delta * f$ has any meaning, let alone is a test function. Can someone clear this up for me?

Answer 1

To your question: Your definition of convolution is a map

$$ *:(\mbox{test functions})\times (\mbox{distributions})\to \mbox{distributions}. $$

So you're right, "$\delta * g$" does not make sense, as $\delta$ is not a test function.

Answer 2

The canonical definition of convolution of two arbitrary distributions with well associated supports is the following.

Let $S$ and $T$ be two distributions, respectively, with supports $A$ and $B$, and let $\varphi$ is a test function with support $K$. You define $$ \langle S*T,\varphi\rangle=\langle S_x\otimes T_y,\alpha(x)\varphi(x+y)\rangle, $$ where $\alpha$ is any test function that is equal to $1$ in a neighborhood of $A\cap(K-B)$. Then $S*T=T*S$ for any distributions with well associated supports.

Recall that two closed sets $A$ and $B$ are said to be well associated if $A\cap (K-B)$ is compact for any compact set $K$.