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I intend to prove the associativity of convolution but failed after several trials, i.e. $(f \ast g) \ast h = f \ast (g \ast h)$
where $(f \ast g)(t) = \int^{t}_{0}f(s)g(t-s)ds $

There are a number of proves considering $(f \ast g)(t) = \int^{\infty}_{-\infty}f(s)g(t-s)ds $. Those did not help since I have a different definition.

Would anyone be able to show the proof here?

Thanks.

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    $\begingroup$ Consider your functions $f(s), g(s), h(s)$ to take value $0$ if $s \notin [0, t]$. This makes the case with integration on $[0, t]$ equivalent to the case with integration on $(-\infty, \infty)$. $\endgroup$ Mar 3, 2017 at 18:13
  • $\begingroup$ I would like to insist on doing it by change of integration variables or other means, instead of changing the integration domain to $(-\infty, \infty)$. $\endgroup$
    – Simon
    Mar 3, 2017 at 18:15
  • $\begingroup$ Have you tried using Fubini's theorem (i.e. change the order of integration)? Usually this does the magic whenver repeated integral is involved. $\endgroup$
    – user160738
    Mar 3, 2017 at 18:31
  • $\begingroup$ I tried, but never succeed in arranging both sides in the same form. $\endgroup$
    – Simon
    Mar 3, 2017 at 20:51

1 Answer 1

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Using the following convolution of $f$ and $g$ \begin{align*} (f\star g)(t)=\int_0^tf(s)g(t-s)ds \end{align*}

we obtain \begin{align*} \color{blue}{((f\star g)\star h)(t)}&=\int_0^t(f\star g)(s)h(t-s)\,ds\\ &=\int_{s=0}^t\left(\int_{u=0}^sf(u)g(s-u)\,du\right)h(t-s)\,ds\\ &=\int\!\!\!\int_{0\leq u \leq s\leq t}f(u)g(s-u)h(t-s)\,du\,ds\\ &=\int_{u=0}^t\int_{s=u}^tf(u)g(s-u)h(t-s)\,ds\,du\\ &=\int_{u=0}^tf(u)\left(\int_{s=0}^{t-u}g(s)h(t-u-s)\,ds\right)\,du\\ &=\int_{u=0}^tf(u)(g\star h)(t-u)\,du\\ &\,\,\color{blue}{=(f\star (g\star h))(t)} \end{align*}

Note: This corresponds to the discrete case which could be somewhat easier to follow. \begin{align*} (f\star g)(n)=\sum_{k=0}^nf(k)g(n-k) \end{align*}

Here we get

\begin{align*} ((f\star g)\star h)(n)&=\sum_{k=0}^n(f\star g)(k)h(n-k)\\ &=\sum_{k=0}^n\left(\sum_{l=0}^kf(l)g(k-l)\right)h(n-k)\\ &=\sum_{0\leq l \leq k \leq n}f(l)g(k-l)h(n-k)\\ &=\sum_{l=0}^n\sum_{k=l}^nf(l)g(k-l)h(n-k)\\ &=\sum_{l=0}^nf(l)\left(\sum_{k=0}^{n-l}g(k)h(n-k-l)\right)\\ &=\sum_{l=0}^nf(l)(g\star h)(n-l)\\ &=(f\star(g\star h))(n) \end{align*}

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  • $\begingroup$ Thanks Markus. The solution is very neat and tidy. I think I worked out the wrong integral limit after exchanging the integration order. $\endgroup$
    – Simon
    Mar 4, 2017 at 6:35
  • $\begingroup$ @Simon: You're welcome! :-) $\endgroup$ Mar 4, 2017 at 7:36
  • $\begingroup$ @MarkusScheuer I started from ${(f\star (g\star h))(t)} $ and got a similar result. However, I cannot be sure that this is mathematically correct. Is it possible to do it that way? I looked at many proofs on the net and they always start from $ {((f\star g)\star h)(t)}$ not the other way around. $\endgroup$
    – Vesnog
    Jun 10, 2018 at 16:43
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    $\begingroup$ @Vesnog: It is just a matter of convenience from which side one might start. Recall that equality is an equivalence relation, which is in particular a symmetrical relation. You might also read the equality chain from my answer from the right-hand side to the left. $\endgroup$ Jun 10, 2018 at 17:39
  • $\begingroup$ Could you elaborate on why the exchange of the integral is justified here? $\endgroup$ Mar 31, 2020 at 8:18

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