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Super simple question about the integral of a convolution Watch

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    I want to prove that

    \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\in\mathcal{S}(\mathbb{R}^{n}  ). Then we can compute:

    \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

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

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    Although I wonder if you could perhaps post in my thread about the Abel transform...
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    (Original post by RamocitoMorales)
    Don't I have to deal with a g(y)\mapsto g(x-u); or can I treat the inner integral separately? I think you are right though; y is still y, 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
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    (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\mapsto x-y would have sufficed, but sought confirmation. Besides, I didn't include "super simple" in the title to deceive anyone.
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    (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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