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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 \begin{equation} f \ast g := \int_{-\infty}^\infty f(u-t)g(t) \, dt \end{equation}

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

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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    ok, thanks for the help!2012-02-28

1 Answers 1

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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).