# Proof that convolution is associative using Fubini's theorem.

The question is to prove that $$((f *g)*h) (x) = (f*(g*h)) (x)\\$$

So far, I have the following:

\begin{align*} ((f \ast g) \ast h) (x) &= \int^x_0 (f \ast g)(t) h (x-t) dt\\ &=\int^x_0 \left[\int^t_0 f(h)g(t-h) dh\right] h(x-t) dt\\ &= \int^t_0 \int^x_0 g(t-h) h(x-t) f(h) dt \quad dh\\ &= \int^t_0 \int^{\alpha = x - h}_{\alpha = -h} g(\alpha) h(x - h - \alpha) f(h) d\alpha \quad dh, \quad \text{where } \alpha = t - h \Rightarrow d\alpha = dt\\ \end{align*}

I'm stuck here because somehow, I need: $$\int^{\alpha = x - h}_{\alpha = 0} \text{instead of} \int^{\alpha = x - h}_{\alpha = -h}$$

I'm aware that there are different definitions of convolution that go from $$(-\infty, +\infty)$$ that might simply this, but I'm wondering how to do for the above bounds.

Is there something I did incorrectly? It's been a while since I've learned Fubini's theorem, so I may be missing up the bounds somehow. Thank you!

• When you interchange the integrals, the limits change. This is because the double integral is happening over a triangular region. See math.stackexchange.com/questions/2170534/… Oct 24 '20 at 3:58
• You can also integrate over $\mathbb{R}$ using characteristic functions and observing that $$\chi_{[0, x]}(t)\chi_{[0, t]}(h) = \chi_{[0, x]}(h)\chi_{[h, x]}(t) = \chi_{[0, x]}(h)\chi_{[0, x-h]}(t-h).$$ Oct 24 '20 at 4:07

We obtain \begin{align*} \color{blue}{((f\ast g)\ast h)(x)}&=\int_0^x(f\ast g)(t)h(x-t)\,dt\\ &=\int_{t=0}^x\left(\int_{u=0}^tf(u)g(t-u)\,du\right)h(x-t)\,dt\tag{1}\\ &=\int\!\!\!\int_{0\leq u\leq t\leq x}f(u)g(t-u)h(x-t)\,du\,dt\tag{2}\\ &=\int_{u=0}^x\int_{t=u}^xf(u)g(t-u)h(x-t)\,dt\,du\tag{3}\\ &=\int_{u=0}^x\int_{t=0}^{x-u}f(u)g(t)h(x-t-u)\,dt\,du\tag{4}\\ &=\int_{u=0}^xf(u)\left(\int_{t=0}^{x-u}g(t)h(x-u-t)\,dt\right)\,du\\ &=\int_{u=0}^xf(u)(g\ast h)(x-u)\,du\\ &\,\,\color{blue}{=(f\ast (g\ast h))(x)} \end{align*}
• In (1) we introduce a new integration variable $$u$$. It is crucial to use different names for different objects to not mix up function names ($$h$$) with variable names.
• In (4) we substitute $$t\to t+u$$ to start the region of integration with $$t=0$$. In this substitution $$u$$ is to be seen as constant, so that we have again $$dt$$.