# Convolution of distributions is not associative

I need some help with this exercise:

It proposes to show that convolution of distributions is not associative:

If $T=T_1$ (distribution given by f=1), $S=\delta'$, and $R=T_H$ (we denote as $H$ the Heaviside function, in $\mathbb{R}$), then:

$$T\ast(S\ast R)\neq(T\ast S)\ast R$$

I'm not very familiarized with convolutions of distributions, so any help will be very useful. Thanks in advance.

We have a couple of useful properties of convolution. First, $\delta\ast Z=Z$ for any distribution $Z$. Second, $(Z_1\ast Z_2)' = Z_1'\ast Z_2=Z_1\ast Z_2'$ for any distributions $Z_i$ for which the convolution is defined.
Therefore, $S\ast R = \delta'\ast T_H = (\delta\ast T_H)'= T_H'= \delta$, then $T\ast \delta = T$.
On the other hand, $T\ast S = T_1\ast \delta'= (T_1\ast \delta)' = T_1 ' = 0$, and obviously $0\ast R = 0$.