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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.

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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 $0\ast R = 0$.

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  • $\begingroup$ Very clear explanation, TZakrevskiy! Thanks a lot! $\endgroup$ – Mark_Hoffman Nov 15 '13 at 9:51
  • $\begingroup$ @Mark_Hoffman : you are welcome! $\endgroup$ – TZakrevskiy Nov 15 '13 at 17:43
  • $\begingroup$ I couldn't prove these properties, can you please suggest some books to prove them on my own? $\endgroup$ – Serkan Yaray Jan 26 '16 at 21:12
  • $\begingroup$ @SerkanYaray It would be better to ask it as a separate question, because my knowledge of english literature on distributions is quite limited. I can recommend you a book in Russian or in French, though. $\endgroup$ – TZakrevskiy Jan 27 '16 at 8:53

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