Convolution is associative in $L^1(-\infty$,$\infty)$

Question

Let $f, g, h \in L^1(-\infty$,$\infty)$. Prove that convolution operation is associative. How to apply Tonelli and Fubini's Theorem for the proof?

Answer 1

Claim: If $f, g \in L^1$ then so is $f\ast g$.

Proof:

$$ \begin{split} &\int_{\mathbb{R}} |(f \ast g)(x)| dx \\\\ &= \int_{\mathbb{R}} \left| \int_{\mathbb{R}} f(x-y) g(y) dy \right| dx \\\\ &\le \int_{\mathbb{R}} \int_{\mathbb{R}} |f(x-y)| |g(y)| dy dx \\\\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} |f(x-y)| |g(y)| dx dy \\\\ &= \int_{\mathbb{R}} \left(\int_{\mathbb{R}} |f(x-y)| dx \right) |g(y)| dy \\\\ &= \int_{\mathbb{R}} ||f||_1 |g(y)| dy \\\\ &= ||f||_1 ||g||_1 < \infty, \end{split} $$ where the second equality is the nonnegative version of Fubini's theorem. From this it follows that $(f \ast g) \ast h \in L^1$ if $f,g,h \in L^1$.

Note then that

$$ \begin{split} &((f \ast g) \ast h)(x) \\\\ &= \int_{\mathbb{R}} (f \ast g)(x-y) h(y) dy \\\\ &= \int_{\mathbb{R}} \left( \int_{\mathbb{R}} f(x-y -t) g(t) dt \right) h(y) dy \\\\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} f(x-y-t) g(t) h(y) dt dy\\\\ &= \int_{\mathbb{R}} \int_{\mathbb{R}} f(x-y-t) g(t) h(y) dy dt \end{split} $$ Consider the transformation $T : \mathbb{R}^2 \to \mathbb{R}^2$ given by

$T(y,t) = (y, t-y)$ which has matrix representation

$$ \begin{bmatrix} 1 & 1 \\\\ 0 & -1 \end{bmatrix} $$

hence $det(T) = -1$ This transformation takes

$x -y -t \to x - y - (t-y) = x-t$, so our integral becomes

$$ \begin{split} &\int_{\mathbb{R}} \int_{\mathbb{R}} f(x-t) g(t-y) h(y) dy dt\\\\ &= \int_{\mathbb{R}} f(x-t) \left( \int_{\mathbb{R}} g(t-y)h(y) dy \right) dt \\\\ &= \int_{\mathbb{R}} f(x-t) (g \ast h)(t) dt \\\\ &= (f \ast (g \ast h))(x) \end{split} $$