On commutativity of convolution

[psie]

TL;DR Summary

My textbook claims the convolution is commutative. I think I have found a way to motivate this, but I don't quite understand the textbook's motivation.

I quote;

Throughout this section, we consider the measure space $(\mathbb R^d,\mathcal B(\mathbb R^d),\lambda)$ [$\lambda$ being Lebesgue measure]. Let $f$ and $g$ be two real measurable functions on $\mathbb R^d$. For $x\in\mathbb R^d$, the convolution $$f\ast g(x)=\int_{\mathbb R^d}f(x-y)g(y)\,\mathrm{d}y$$ is well defined provided $$\int_{\mathbb R^d}|f(x-y)g(y)|\,\mathrm{d}y<\infty.$$In that case, the fact that Lebesgue measure on $\mathbb R^d$ is invariant under translations and under the symmetry $y\to -y$ shows that $g\ast f(x)$ is also well defined and $g\ast f(x)=f\ast g(x)$.

I struggle with the last sentence and putting the pieces together. I know the abstract change of variables formula for $\nu$ on $(F,\mathcal D)$ being the pushforward measure of $\mu$ on $(E,\mathcal C)$ under $\varphi:E\to F$, i.e. $\nu(A)=\mu(\varphi^{-1}(A))$ for every $A\in\mathcal D$. Then for any $h:F\to\mathbb R$, $$\int_E h(\varphi(x))\,\mu(\mathrm{d}x)=\int_F h(y)\,\nu(\mathrm{d}y).$$I figured that $h(y)=f(x-y)g(y)$ and $\varphi(x)=x-y$, with $\mu=\lambda$ and $(E,\mathcal C)=(F,\mathcal D)=(\mathbb R^d,\mathcal B(\mathbb R^d))$. Since Lebesgue measure is invariant under translations, $\nu$ is just Lebesgue measure again. So I get indeed $f\ast g(x)=g\ast f(x)$. But what do they mean, or what's the relevance of, the invariance under the symmetry $y\to -y$? And how does one show Lebesgue measure is invariant under this symmetry?

 

[fresh_42]

I think your thoughts are way too complicated. Let $u=x-y.$
\begin{align*}
(f\ast g)(x)&=\int_{y\in \mathbb{R}^d} f(x-y)g(y)\,dy =- \int_{x-u\in \mathbb{R}^d} f(u)g(x-u) \,du \\[6pt]
&=\int_{u-x\in \mathbb{R}^d} f(u)g(x-u) \,du =\int_{u\in \mathbb{R}^d}g(x-u)f(u) \,du\\[6pt]
&=(g\ast f)(x)
\end{align*}

 

[psie]

\begin{align*}
(f\ast g)(x)&=\int_{y\in \mathbb{R}^d} f(x-y)g(y)\,dy =- \int_{x-u\in \mathbb{R}^d} f(u)g(x-u) \,du \\[6pt]
&=\int_{u-x\in \mathbb{R}^d} f(u)g(x-u) \,du =\int_{u\in \mathbb{R}^d}g(x-u)f(u) \,du\\[6pt]
&=(g\ast f)(x)
\end{align*}

Thank you.

What did you use in the second equality? In particular $$\int_{y\in \mathbb{R}^d} f(x-y)g(y)\,dy =- \int_{x-u\in \mathbb{R}^d} f(u)g(x-u) \,du.$$ Looks like a change of variables, but I'd expect one to take the absolute value of the determinant, if you look here (note, this is just one of many versions of the change of variables formula in higher dimensions). I don't see where the minus sign would come from, and how it disappears in the third equality.

 

[fresh_42]

Thank you.

What did you use in the second equality? In particular $$\int_{y\in \mathbb{R}^d} f(x-y)g(y)\,dy =- \int_{x-u\in \mathbb{R}^d} f(u)g(x-u) \,du.$$ Looks like a change of variables, but I'd expect one to take the absolute value of the determinant, if you look here (note, this is just one of many versions of the change of variables formula in higher dimensions). I don't see where the minus sign would come from, and how it disappears in the third equality.

It is a simple change of variables: $u=x-y$ so $\dfrac{du}{dy}=-1.$ I admit that it was a bit lazy. Let me see if I can rephrase it with $\varphi(y)=x-y$ and $d\varphi(y)=(-1)^d dy$ for the $d$-dimensional case.
\begin{align*}
(f\ast g)(x)&=\int_{y\in \mathbb{R}^d} f(x-y)g(y)\,dy \\[6pt]
&=(-1)^d \int_{x-\varphi(y)\in \mathbb{R}^d} f(\varphi(y))g(x-\varphi(y)) \,d\varphi(y) \\[6pt]
&=(-1)^d\int_{x-u\in \mathbb{R}^d}g(x-u)f(u) \,du\\[6pt]
&=(-1)^d\int_{-u\in \mathbb{R}^d}g(x-u)f(u) \,du\\[6pt]
&=\int_{u\in \mathbb{R}^d}g(x-u)f(u) \,du\\[6pt]
&=(g\ast f)(x)
\end{align*}
The transformation matrix is $\varphi = -I$ with $\det \varphi=(-1)^d.$

The crucial points are the last two equations:

1. $x+y$ over $\mathbb{R}^d$ covers the same integration range as $y$ does over $\mathbb{R}^d.$ A shift translation by a finite fixed vector does not change the integration of infinite integration limits.

2. Orientation does change the integral, namely by a factor $\pm 1.$ However, it changes the orientation of the differential form as well as the orientation of the integration range. Both cancel each other so that we get commutativity instead of anti-commutativity. I didn't use the transformation theorem for this, only the fact that differential forms build a Grassmann algebra, i.e. that they are oriented. I was lazy once more since I should have written $dy=dx_1\wedge \ldots \wedge dx_d$ and so on.