4
$\begingroup$

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?

Improve this question
$\endgroup$
5
  • $\begingroup$ How are you defining $(|x| \star |x|)'$? Doesn't that diverge? $\endgroup$ – joriki May 3 '11 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 '11 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 '11 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 '11 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 '11 at 17:16
1
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.