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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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    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.2011-05-03

1 Answers 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.