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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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    @GiuseppeNegro : Maybe I was hasty; I was just assuming everything was well-behaved except in the respects mentioned.2013-10-08

3 Answers 3

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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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    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).2012-07-31
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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.

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