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Assume that $f(x),g(x)$ are positive and are in $L^1$. Moreover, they are differentiable and their derivative is integrable. Let $h(x)=f(x)*g(x)$, the convolution of $f$ and $g$. Does the derivative of $h(x)$ exist? If yes, how can we prove that $$ \frac{d}{dx}(f(x)*g(x)) = (\frac{d}{dx}f(x))*g(x)$$

Thanks

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    Do you know how to differentiate under the integral sign?2012-07-31
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    This can be helpful http://math.stackexchange.com/questions/12909/will-moving-differentiation-from-inside-to-outside-an-integral-change-the-resu2012-07-31
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    If I'm not mistaken, if _either_ $f$ or $g$ is differentiable, then $f*g$ is differentiable. If $f$ is differentiable, then $(f*g)'=f'*g$. If they're both differentiable then $(f*g)'=f'*g=f*g'$.2012-07-31
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    @MichaelHardy: Sorry to revive this old comment, I arrived at it following some links to recent questions. I think that what you say is not true if $'$ stands for a classical derivative. To obtain such a result you need the dominated convergence theorem, which is available only with some additional assumption (for example, $f'\in L^\infty$ will do). However, the result is certainly true if $'$ stands for derivative in some other sense, such as a Fourier multiplier or something like that. Do you agree?2013-10-08
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    @GiuseppeNegro : Maybe I was hasty; I was just assuming everything was well-behaved except in the respects mentioned.2013-10-08

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