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Given that the convolution $f*g$ exists (in $\mathbb{R}^{n}$), how would one show that for example $g*f$ or $f(-x)*g(-x)$ exists?

Would it be enough to say that $f$ and $g$ are both locally integrable? Or that both are continuous and one of them has compact support? If that's the case, does it not automatically follow from the fact that $f*g$ exists?

Drn004
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1 Answers1

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You can do the first of these through change of variables:

with $y=x-z$, $$(f*g)(x)=\int_{\mathbb{R}^n}f(y)g(x-y)dy=(-1)^n\int_{\mathbb{R}^n}f(x-z)g(z)(-1)^ndz=(g*f)(x).$$

The second one isn't defined because you can't take the convolution of two numbers. Do you mean $f_2*g_2$ where $f_2(x)=f(-x)$ and $g_2(x)=g(-x)$? In that case, with $y=-z$, $$(f_2*g_2)(x)=\int_{\mathbb{R}^n}f_2(y)g_2(x-y)dy=(-1)^n\int_{\mathbb{R}^n}f(z)g(-x-z)(-1)^ndz=(f*g)(-x).$$

The $(-1)^n$ factors come from, first, the substitution and, second, swapping the bounds of the integrals.

You don't need any more assumptions than $f*g$ exists.

mrob
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  • I do know how to prove their equality, I was specifically asking how I should go about showing their existence beforehand. Regarding the second one, that was exactly how my professor put it, and I need to show that $f(-x)g(-x)=(fg)(x)$ as well. I'm quite confused about the $(-1)^n$ factor as well. When substituting, don't we take the absolute value of the Jacobian? – Drn004 Nov 16 '16 at 23:19
  • You know the convolution exists, and they're equal. How did you show $f*g$ exists? You can show the others exist in the same way. – mrob Nov 17 '16 at 04:37
  • Also, I don't know what $f(-x)*g(-x)$ means. You can't take the convolution of two numbers. For the $(-1)^n$ factors, write the integral as $n$ integrals. The substitution gives $(-1)^n$ and then the swapped bounds gives another $(-1)^n$. – mrob Nov 17 '16 at 04:40