Change of variable in proof of convolution theorem?

Question

In the proofs of the convolution theorem I have seen, say here, there is a change of variable used in a nested integral. Specifically, within an outer integral with respect to x, there is an inner integral with respect to u, independent of x; and for which the substitution $w=u-x$ is made. Then $dw$ is supposedly equal to $du$. I don't quite understand why we don't have $dw=du-dx$, considering that u and x are independent. I imagine that someone may motivate this by saying something like 'within the second integral, x is a constant', and I can kind of get this by thinking about integrals as sums, and indeed in the inner summation we would perform it for each value of x (in the outer integral). However I am not convinced by this, and was wondering if anyone might have a better explanation, or some more rigorous motivation (although I'm not sure i'll understand something very rigorous!!!)

Answer 1

$$ h(u) = \int_{-\infty}^\infty f(x) g(u-x)\,dx. $$

In the expression $\displaystyle \int_{-\infty}^\infty \cdots\cdots\,dx,$ anything that remains fixed as $x$ goes from $-\infty$ to $+\infty$ is a “constant.” Thus $\dfrac d{dx} (u-x) = 0-1,\vphantom{\dfrac{\displaystyle\sum}{}}$ so if we set $y=u-x,$ then $dy = -dx.$

Answer 2

I'll prove it in the one-variable case. We have for $x\in\mathbb{R}$,

$$(f*g)(x)=\int_{-\infty}^\infty f(u)g(x-u)\,du.$$

Fixing $x\in\mathbb{R}$, define a function $w(u)=x-u$. Then $w'(u)=-1$, for all $u\in\mathbb{R}$. As $u:-\infty\to\infty$, we have $w:\infty\to-\infty$. Note that $w$ is strictly increasing and bijective. So $\frac{du}{dw}=\frac{1}{dw/du}=-1$. Hence by change of variables, we have

$$(f*g)(x)=\int_\infty^{-\infty}f(x-w)g(w)\frac{du}{dw}dw=-\int_\infty^{-\infty}f(x-w)g(w)dw=\int_{-\infty}^\infty f(x-w)g(w)dw=(g*f)(x)$$

The only key result here is the change of variables theorem. Check https://en.wikipedia.org/wiki/Integration_by_substitution