In our class notes we are asked to verify the following equality:
$E(t,x)\ast (g(x)\delta(t)) = t\int_{\omega\in S^2}{\frac{g(x-t\omega)}{4\pi}dS(\omega)}$ where $E(t,x)=\frac{1}{4\pi|x|}\delta(t-|x|)$ is the fundamental solution to the wave equation in $\mathbb{R}^3$.
My problem with this is that I am not sure how to deal with the convolution of these two distributions.
First, seeing as they both have compact support, then their convolution (as distributions) is well defined. Moreover, since $f(x)\delta_{a}(x)\ast g(x)\delta_{b}(x) = f(a)g(b)\delta_{a+b}(x)$, where $\delta_{a}(x)=\delta(x-a)$, then I would conclude that $E(t,x)\ast(g(x)\delta(t)) = \frac{g(x)}{4\pi|x|}\delta(t-|x|).$ However this is still only a distribution, and I cannot figure out how this would be equivalent to the given solution.
Second, if we appeal to the integral definition of the convolution, then we have $E(t,x)\ast(g(x)\delta(t)) = \frac{1}{4\pi}\int_{\mathbb{R}^3}\int_{-\infty}^{\infty}{\frac{1}{|y|}\delta(s-|y|)g(x-y)\delta(t-s)}dsdy$ which gives us an integral involving two delta distributions, again leaving me at a loss for how to proceed.
Any help in clarifying this equality would be greatly appreciated.