Convolution of the delta distribution, basic property

Question

I am trying to prove that for all $\phi \in S(\mathbb{R}), \phi*\delta = \phi$. I tried with a direct calculation of $ \phi*\delta(\psi)$. At the very end, I got (expanding convolution and exchange the order of the integral, where $\tilde{\phi}(y-x) = \phi(x-y)$ $$ \int \delta(y)(\tilde{\phi} * \psi)(y) dy $$ But I am not sure if it's legitimate to write the above integral as $$ T(\tilde{\phi} * \psi) = T_{\phi}(\psi) $$ Or is there another way to show this property?

Answer 1

You are nearly finished. Note that by definition of the delta distribution it holds that $$\langle \phi*\delta,\psi\rangle = \int (\phi*\delta)(x) \psi(x) dx = \int \int \underbrace{\phi(x-y)}_{\tilde{\phi}(y-x):=} \delta(y) \psi(x) dy dx = \int \delta(y)(\tilde{\phi}*\psi)(y)dy =\langle \delta, \tilde{\phi}*\psi\rangle = (\tilde{\phi}*\psi)(0) = \int \phi(x) \psi(x) dx = \langle \phi,\psi\rangle,$$ where the second to last step follows since inserting 0 just gives you this integral. Note that I use $\langle .,.\rangle$ as a notation for the integral. Since this equality holds for all test functions $\psi$, you get $\phi*\delta=\phi$.