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Given differentiable functions $f,g$, one can make the following statement about the derivatives of their convolution:

$(f \star g)' = f' \star g = f \star g'$

Suppose I pick $g$ as a non differentiable function such as $g(x) = |x|$, does this property still hold? (plotting $(|x| \star |x|)'$ and $(|x|' \star |x|)$ in Matlab shows different functions)

If the above property is true then by definition of convolution $f \star g' (x) = \int f(y) g'(x-y) dy$

So when can we say the convolution is not differentiable whenever $g'(x-y)$ is not differentiable?

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  • $\begingroup$ How are you defining $(|x| \star |x|)'$? Doesn't that diverge? $\endgroup$
    – joriki
    May 3, 2011 at 16:29
  • $\begingroup$ True, I assumed $g(x) = |x|$ is defined between $[-5,5]$ just for the purposes of visualization. We can consider functions with compact support if needed. $\endgroup$
    – snel
    May 3, 2011 at 16:34
  • $\begingroup$ Your identity comes about because you can integrate the convolution by parts to transfer the derivative from one function to the other, so it holds iff the boundary terms are zero. $\endgroup$
    – joriki
    May 3, 2011 at 16:39
  • $\begingroup$ I'm not sure I fully understand. Are you saying $f \star g'(x) $ is not differentiable when $g'$ is not differentiable iff $(f \star g)'$ goes to zero at the boundary? $\endgroup$
    – snel
    May 3, 2011 at 16:53
  • $\begingroup$ No, I'm saying that the identity requires the boundary terms to vanish, so the fact that it doesn't hold for $|x|$ need not have anything to do with differentiability; one wouldn't expect it to hold for that other reason. $\endgroup$
    – joriki
    May 3, 2011 at 17:16

1 Answer 1

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If also the weak derivatives exist then the derivative is understood in the weak sense.

$f'$ is the unique weak derivative of $f$ is for all $C_c^\infty(\mathbb R)$ functions $\phi$ we have

$$\int f(x) \phi'(x) \, dx = - \int f'(x) \phi(x) \, dx$$

So plug in and use Fubini.

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