Finding convolution of two distributions.

Question

I am trying to show $\delta'*H =\delta $ and $1*\delta' = 0$,where $H$ denotes heavyside function. I tried this but after some simplication I am not getting anything interesting.

Answer 1

For a test function $\phi$, let $\psi(x) = \langle H,\tau_{-x}\phi \rangle$.

$\langle \delta ' * H, \phi \rangle =^{\text{def}} \langle \delta',\psi \rangle = -\psi'(0)$.

We know that $\psi(x) = \int_0^\infty \phi(y-x)dy$

Hence $\psi'(x) = -\int_0^\infty \phi'(y-x)dy = -\phi(-x)$. This uses the fact that $\phi$ is compactly supported.

Thus $\psi'(0) = -\phi(0)$ i.e. $\delta' * H = \delta$.

Answer 2

First try to show that $u * v' = u' * v.$ Then both will be trivial.