# Commutativity of Convolution in higher dimensions

I have a basic question about how to show that convolution in dimension $n$ is commutative - or maybe it is rather a question about change of variables ..

So on $\mathbb{R}$ I know how to show commutativity:

We have the definition $$f \ast g := \int_{-\infty}^\infty f(u-t)g(t) \, dt$$

Now by changing the variable $t \mapsto s = u - t$ I get $$f \ast g := - \int_{\infty}^{-\infty} f(s)g(u - s) \, ds$$

And so the minus sign helps me to switch the boundary and get back to the original form.

Now, in more than one dimenstions the change of variables involves the absolute value of the determinant of the Jacobian, how do I switch the boundary in this case (i.e. reverse the order of the limits) ?

Thanks very much !

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In higher dimension, you make the change $s=u-t$, whose absolute value of the Jacobian is $1$, and $\mathbb R^n$ is mapped to $\mathbb R^n$. –  Davide Giraudo Feb 28 '12 at 19:12
@DavideGiraudo: but the assignment $s = u - t$ changes the sign of the limits, how do I revert this change ? Or is there no direction of integration in the higher dimensions ? Sorry if that question sounds too dumb I have very little knowledge of higher dimenstional calculus. –  harlekin Feb 28 '12 at 20:38
If you don't want to work with high dimensional integral, you can treat them as a succession of simple integrals, and you do the substitution $s_i:=u_i-t_i$ for each component (but you have to use Fubini's theorem). –  Davide Giraudo Feb 28 '12 at 21:39
ok, thanks for the help! –  harlekin Feb 28 '12 at 23:09

We can see, thanks to the formula of change of variables in $\mathbb R^n$, that the substitution $s=u-t$ maps $\mathbb R^n$ to $\mathbb R^n$, and the absolute value of the Jacobian is $1$.
If you don't want to use this formula, but just applying it to one-dimensional integrals, then write $u=(u_1,\ldots,u_n)$ and $t=(t_1,\ldots,t_n)$. Then $$f\star g(u)=\int_{\mathbb R^{n-1}}\int_{\mathbb R}f(u_1-t_1,\ldots,u_{n-1}-t_{n-1},u_n-t_n)g(t_1,\ldots,t_{n-1},t_n)dt_n\ldots d_1\ldots dt_{n-1}$$ and putting $s_n=u_n-t_n$ we get $$f\star g(u)=\int_{\mathbb R^{n-1}}\int_{\mathbb R}f(u_1-t_1,\ldots,u_{n-1}-t_{n-1},s_n)g(t_1,\ldots,t_{n-1},u_n-s_n)ds_n\ldots d_1\ldots dt_{n-1},$$ then switch the integrals and continue this process (write it by induction if you don't find this rigourous).