Showing that a "convolution" operator is associative I am dealing with the following operator $*$ : $(A*B)(t) = \inf\limits_{\tau\in\mathbb{R}} (A(\tau) + B(t-\tau))$. I would like to show that it is associative, i.e : $((A*B)*C)(t) = (A*(B*C))(t)$ . 
I'm trying to follow common sense and simply use the definition the following way : $((A*B)*C)(t) = \inf\limits_{\tau\in\mathbb{R}} ((A*B)(\tau) + C(t-\tau))$, and then expand $(A*B)(\tau)$ the same way, but from here on I'm stuck in getting to the required equality. Does anyone have an idea of how this can be proved? perhaps I should approach this in a whole different way. Any help is appreciated. Thanks.   
 A: Your way will work. The main trick is making a substitution. In particular, the trick is to make a substitution of variables. You'll first get
\begin{align*}((A*B)*C)(t)&=\inf_{\tau_1\in \mathbb R}\inf_{\tau_2\in \mathbb R}(A(\tau_2)+B(\tau_2-\tau_1))+C(t-\tau_2)\\&=\inf_{\tau_1,\tau_2\in\mathbb R}A(\tau_2)+B(\tau_2-\tau_1)+C(t-\tau_2)\end{align*}
and, similarly,
$$(A*(B*C))(t)=\inf_{\tau_1',\tau_2'\in \mathbb R}A(\tau'_1)+B(\tau'_2)+C(t-\tau'_1-\tau'_2).$$
Then, you make the substitution $\tau_1'=\tau_2$ and $\tau_2'=\tau_2-\tau_1$, noting that the mapping $(\tau_1,\tau_2)\mapsto (\tau_1',\tau_2')$ is bijective, to yield that the second expression is equivalent to
$$(A*(B*C))(t)=\inf_{\tau_1,\tau_2\in \mathbb R}A(\tau_2)+B(\tau_2-\tau_1)+C(t-\tau_2).$$
Which is equivalent to the first expression, so you're done. 
A more symmetric way to do this would be to take
$$(A*B)(t)=\inf\{A(x_1)+B(x_2):x_1+x_2=t\}$$
which is equivalent to your definition, and then, by a similar method to above, conclude that, regardless of the grouping of terms, one has
$$(A*B*C)(t)=\inf\{A(x_1)+B(x_2)+C(x_3):x_1+x_2+x_3=t\}.$$
