What does triple convolution actually look like? I have to prove associativity of the convolution of three functions. I'm having trouble picturing how the variables will look. For periodic functions $f$, $g$, and $h$, I have the definition
$$
(f * g) (x) = \int_\Bbb{T} f(y) g(x - y) \, dy.
$$
What exactly will $f * (g * h)$ look like?  Will a third variable be introduced?  Thanks!
 A: Break it down into steps, and remember that all variables introduced other than $x$ are dummy variables (they only exist within the integrals):
$$
\begin{align}
\big( f * (g * h) \big)(x) &= \int_{\Bbb{T}} f(y) \cdot (g * h)(x - y) \, dy \\
&= \int_{\Bbb{T}} f(y) \cdot \left( \int_{\Bbb{T}} g(z) \cdot h(x - y - z) \, dz \right) \, dy \\
&= \iint_{\Bbb{T} \times \Bbb{T}} f(y) \cdot g(z) \cdot h(x - y - z) \, dz \, dy
\end{align}
$$
and similarly,
$$
\begin{align}
\big( (f * g) * h \big)(x) &= \int_{\Bbb{T}} (f * g)(u) \cdot h(x - u) \, du \\
&= \int_{\Bbb{T}} \left( \int_{\Bbb{T}} f(v) \cdot g(u - v) \, dv \right) \cdot h(x - u) \, du \\
&= \iint_{\Bbb{T} \times \Bbb{T}} f(v) \cdot g(u - v) \cdot h(x - u) \, dv \, du.
\end{align}
$$
You can check that these two double integrals are equivalent under the change of variables
$$
\left\{
\begin{align}
y &= v \\
z &= u - v.
\end{align}
\right.
$$
A: The natural generalisation of that formula is $(f*g*h)(z)=\int_T\int_Tf(x)g(y)h(z-x-y)\mathrm dx\mathrm d y$, which shows that convolution is associative (you can easily equate this formula with those obtained from repeated two-argument convolution using either of the two possible groupings). The main point is that the sum of it arguments to $f,g,h$ is always the argument to $f*g*h$. You can compare this with the way monomials in (formal) power series are combined under multiplication, which is in fact a discrete form of a convolution product (viewing a power series as a map from powers to the corresponding coefficients).
