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Super simple question about the integral of a convolution

I want to prove that

∫Rnf(x)∗g(x) dx=∫Rnf(x) dx⋅∫Rng(x) dx\displaystyle \int_{\mathbb{R}^{n}}f(x)\ast g(x)\,dx=\int_{\mathbb{R}^{n}}f(x)\,dx\cdot\int_{\mathbb{R}^{n}}g(x)\,dx

So, let's suppose that f,g∈S(Rn)f,g\in\mathcal{S}(\mathbb{R}^{n}). Then we can compute:

Unparseable latex formula:

\displaystyle \int_{\mathbb{R}^{n}}(f\ast g)(x)\,dx&=\int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}f(x-y)g(y)\,dy\right)\,dx



=∫Rn(∫Rnf(x−y) dx)g(y) dy\displaystyle = \int_{\mathbb{R}^{n}}\left(\int_{\mathbb{R}^{n}}f(x-y)\,dx\right)g(y)\,dy

Now, I guess I have to make a substitution here, but I don't know which one. Could anyone help?
u=x-y (to replace x)
Original post by DFranklin
u=x-y (to replace x)


Don't I have to deal with a g(y)↦g(x−u)g(y)\mapsto g(x-u); or can I treat the inner integral separately? I think you are right though; yy is still yy, after all.
Yes y is still y....
Original post by DFranklin
Yes y is still y....


Thanks again, Dr. Franklin!

Spoiler

Original post by RamocitoMorales
Don't I have to deal with a g(y)↦g(x−u)g(y)\mapsto g(x-u); or can I treat the inner integral separately? I think you are right though; yy is still yy, after all.


I don't understand your question here, but inside the inner integral, we treat y as constant, since in say R^2, we're integrating over a line of constant y. So z=x-y gives dz=dx
(edited 7 years ago)
Original post by atsruser
I don't understand your question here, but inside the inner integral, we treat y as constant, since in say R^2, we're integrating over a line of constant y. So z=x-y gives dz=dx


Yes, Dr. Franklin already said the same. I had already figured that x↦x−yx\mapsto x-y would have sufficed, but sought confirmation. Besides, I didn't include "super simple" in the title to deceive anyone. :wink:
Original post by RamocitoMorales
Yes, Dr. Franklin already said the same.

I didn't understand your question or his response, so I replied anyway.

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