Convolution of distributions is not associative

Question

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.

Answer 1

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 $0\ast R = 0$.

Answer 2

Although a much belated $\pm$-answer...

First, in general, it is very bad if what appears to be a reasonable operation is not associative. Thus, the classic non-associativity $$ 1 * (\delta' * H) \;=\; 1 * \delta \;=\; 1 \;\not=\; 0 \;=\; 0 * H \;=\; (1*\delta')*H $$ quoted in the question is rightfully disquieting.

Second, although we most often describe "convolution" formulaically, we might also ask what properties we want it to have... and some sort of associativity surely might be among them. I think it is worthwhile to observe that "convolution" can be characterized in a way that "makes it associative", but/and delegates the non-associativity in the iconic example to other issues.

That is, when a topological group $G$ acts continuously on a topological vector space $V$ , the space $C_c(G)$ of continuous, compactly supported functions $f$ on $G$ acts on $V$ by $$ f\cdot v \;=\; \int_G f(g)\,g\cdot v\;dg $$ with Haar measure. Note, in particular, the extreme asymmetry between the $f$ and the $v$. The space $V$ does not have to consist of functions on $G$. In this general situation, we can indirectly characterize "convolution" in $C_c(G)$ as the operation $C_c(G)\times C_c(G)\to C_c(G)$ such that $$ f\cdot(\varphi\cdot v) \;=\; (f*\varphi)\cdot v $$ for all $f,\varphi$ in $C_c(G)$ and for all $v$ in every $V$. (The integral exists and is unique and has good properties for a wide class of reasonable $V$...) Indeed, from this requirement we can deduce the usual formulaic description of "convolution".

In that context, in the iconic example, we can or should ask what's acting on what. Not just "what things can we define by formulas?". In particular, it turns out that compactly-supported distributions can naturally act on (all) distributions (potentially-misleadingly described by one form of "convolution") by $\varphi\cdot u=\varphi*u$). A prior step to see this non-formulaically is to understand how compactly-supported distributions naturally act on the space of smooth functions, for example.

That is, the "non-associativity" in the iconic example is "morally" due to its having no non-formulaic a-priori sense... or reason to be true... :)