Convolution and associatuvity

Question

we have the following proposition: if $u, v$ and $w$ are distributions with convolutifs supports, then $$ u*(v*w)= (u*v)*v $$ where $*$ designate convolution.

As example, it puposed to compare between $1*(\delta' * H)$ and $(1*\delta')*H$, where $H$ is Heaviside, $\delta$ is Dirac.

My question is: why the products 1*(\delta' * H)$ and $(1*\delta')*H$ are well defined and why we can calculate them?

Kin regards

Answer 1

The convolution $u*v$ of two distributions $u$ and $v$ exist if (but not only if) at least one of $u$ and $v$ has compact support.

In the convolutions given, $\delta'$ has compact support, while $1$ and $H$ do not. Therefore $\delta'*H$ exists. It equals $(\delta*H)' = H' = \delta$ which also has compact support, so $1*(\delta'*H)$ is defined. This equals $1*\delta = 1$. Thus, $1*(\delta'*H) = 1$. On the other hand, $1*\delta' = (1*\delta)' = 1' = 0,$ so $(1*\delta')*H = 0*H = 0.$ Thus, $1*(\delta'*H)$ and $(1*\delta')*H$ both exist, but they do not coincide.

Answer 2

As you may know $\delta(x) \ast H(x) $ is an integral $$ (\delta \ast H)(x) = \int \delta(x-y) H(y) \mathrm{d}y = H(x) $$ More explicitly $H(y)$ is the value of $H$ at $x = y$ and the area of the delta function is assumed to be one, thus $ \int \delta(x) dx = 1 $