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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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    $\begingroup$ Do you know how to differentiate under the integral sign? $\endgroup$ – J. M. is a poor mathematician Jul 31 '12 at 16:50
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    $\begingroup$ This can be helpful math.stackexchange.com/questions/12909/… $\endgroup$ – Norbert Jul 31 '12 at 16:52
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    $\begingroup$ 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 $(f*g)'=f'*g=f*g'$. $\endgroup$ – Michael Hardy Jul 31 '12 at 17:44
  • $\begingroup$ @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? $\endgroup$ – Giuseppe Negro Oct 8 '13 at 17:06
  • $\begingroup$ @GiuseppeNegro : Maybe I was hasty; I was just assuming everything was well-behaved except in the respects mentioned. $\endgroup$ – Michael Hardy Oct 8 '13 at 18:11
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Using this thread, and the fact that if $f_1$ and $f_2$ are two integrable functions, $\mathcal F(f\star g)=\mathcal F(f)\cdot\mathcal F(g)$, we have $$\mathcal F\left(\frac d{dx}(f\star g)\right)(x)=ix\mathcal F\left((f\star g)\right)(x)=ix \mathcal F(f)(x)\cdot \mathcal F(g)(x),$$ and $$\mathcal F\left(\left(\frac d{dx}f\right)\star g\right)(x)=\left(\mathcal F\left(\frac d{dx}f\right)\right)\cdot\left(\mathcal F(g)(x)\right)=ix \mathcal F(f)(x)\cdot \mathcal F(g)(x).$$ We conclude by uniqueness of Fourier transform.

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    $\begingroup$ How you can take Fourier transform while we don't know it has Fourier transform or not? In other words, we don't know $\frac{d}{dx} (f*g)$ is in $L^1$? For the second Fourier transform, it is correct since we know that $f'*g$ is in $L^1$. $\endgroup$ – rfvahid Jul 31 '12 at 17:09
  • $\begingroup$ Indeed, it deserves more details. I think an approximation argument can work (approximate in $L^1$ $f$ and $g$ by $C^1$ functions with compact support). $\endgroup$ – Davide Giraudo Jul 31 '12 at 17:20
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Note that, if $ f\in L_1(R)$ then it is Fourier transformable. Since,

$$ \left|\int_{-\infty}^{\infty} f(x) e^{-ixw}\right| \leq \int_{-\infty}^{\infty} |f(x)| < \infty$$.

To prove that the convolution of two $L_{1}(R)$ functions is again an $L_{1}(R)$ function, let

$$ h(x) = \int f(t) g(x-t) dt $$

$$ \int |h(x)|dx \leq \int\int |f(t)| |g(x-t)| dt dx = \int |f(t)|\int |g(x-t)|dxdt = \int |f(t)| ||g||_1 dt = ||f||_1 ||g||_1 \Rightarrow h \in L_1(R)\,.$$

The change of the order of integration is justified by Fubini's theorem. So, you can use the Fourier technique as in Davide's answer.

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Definition: $$h(x)=f*g(x)=\int f(x-t)g(t)dt$$

Let's calculate derivative:

$$\frac {dh}{dx}=\underset{dx\rightarrow0}{\lim} \frac {(\int f(x+dx-t)g(t)dt-\int f(x-t)g(t)dt)}{dx}=\underset{dx\rightarrow0}{\lim}(\int \frac{(f(x+dx-t)-f(x-t))}{dx}g(t)dt$$

If we assume that there exists some integrable function $q(t)$, such that for $t$ almost everywhere $$ \left| \frac{(f(x+dx-t)-f(x-t))}{dx} \right| < q(t) $$ then by the Lebesgue dominated convergence theorem we can push the limit inside integral.

$$\frac {dh}{dx}=\frac{d}{dx}(f*g(x))=\int f'(x-t)g(t)dt=f'*g$$

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