I'm studying differential equations using MIT's online materials. In the notes on Laplace transforms it is asserted without proof that convolution is commutative, distributive over addition, and associative. The first two are easy to show, but I'm struggling with the last.
We have \begin{align} f * (g * h) & = \int_0^t f(\tau)(g * h)(t - \tau)\,d\tau\\ & = \int_0^tf(\tau)\left[\int_0^{t - \tau}g(u)h(t - \tau - u)\,du\right]\,d\tau\\ & = \int_0^t\int_0^{t - \tau}f(\tau)g(u)h(t - (\tau + u))\,du\,d\tau\\ \end{align} Let $v = \tau + u$. Then $du = dv$. So we have $$ \int_0^t\int_\tau^t f(\tau)g(v - \tau)h(t - v)\,dv\,d\tau $$
At this point, I suspect we are to apply Fubini's theorem and interchange the integrals, but I am troubled by the presence of the $\tau$ in the limits of integration in the inner integral.
Note that I am aware of this question, as well (from googling) that the general definition of convolution is
$$ \int_{-\infty}^{\infty}f(t)g(t-\tau)\,d\tau $$ which makes my problem go away. Still, I would like to understand how the proof works given the definition presented in the course.
$$ \int_0^t\int_\tau^t f(\tau)g(v - \tau)h(t - v)\,dv\,d\tau = \int_0^t \int_0^v f(\tau)g(v - \tau)h(t - v) \, d\tau \, dv $$
The above is what an application of Fubini's theorem should give you since in both cases the bounds say that the region over which you're integrating is $$ \{ (\tau,v) : 0 \le \tau \le v \le t \}. \tag 1 $$ On the left side, the outer integral says $0\le \tau\le t$ and the inner one says that for every fixed value of $\tau$, the other variable $v$ goes from $\tau$ up to $t$.
On the right side, the outer integral says $0\le v\le t$ and the inner one says that for every fixed value of $v$, the other variable $\tau$ goes from $0$ up to $v$.
In both cases, you can write $\displaystyle\int_{-\infty}^\infty$, and the function getting integrated will be $0$ everywhere outside of the region $(1)$ since the functions have the value $0$ at negative arguments.