# Convolution and Dirac delta [closed]

I need to prove the following:

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

## closed as off-topic by Giuseppe Negro, AlexR, Dan, TMM, hardmathJan 16 '14 at 0:09

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• First thing, look up the definition of the convolution of a distribution with a test function. – Daniel Fischer Jan 15 '14 at 22:36

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 $\delta_0*\phi=\phi$.
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 $(\delta_0\ast\phi)(x) = \phi_x(0) = \phi(x)$.