Convolution and Dirac delta

Question

I need to prove the following:

$\delta_0 * \phi = \phi$, where $\phi$ is a test function.

Thank you for your help.

Answer 1

By definition $\delta_0[\psi]=\psi(0)$ for any test-function $\psi$. Now convolution of a distribution with a test function is defined as $$ (\delta_0 * \phi)(x) = \delta_0[\phi(x-\cdot)] = \phi(x-0)=\phi(x),$$ thus $\delta_0*\phi=\phi$.

Answer 2

The definition of the Dirac delta distribution is such that $\langle \delta_0,\phi\rangle = \phi(0)$. Thus the convolution (when extended to distributions) is given by $(\delta_0\ast\phi)(x) = \langle \delta_0,\phi_x\rangle$, where $\phi_x(t) = \phi(x-t)$. Then we have $(\delta_0\ast\phi)(x) = \phi_x(0) = \phi(x)$.