Proof of associativity of convolution

Question

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.

Answer 1

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*}