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 $$ \frac{d}{dx}(f(x)*g(x)) = (\frac{d}{dx}f(x))*g(x)$$


  • 2
    $\begingroup$ Do you know how to differentiate under the integral sign? $\endgroup$ – J. M. is a poor mathematician Jul 31 '12 at 16:50
  • 1
    $\begingroup$ This can be helpful math.stackexchange.com/questions/12909/… $\endgroup$ – Norbert Jul 31 '12 at 16:52
  • 2
    $\begingroup$ 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

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.

  • 2
    $\begingroup$ 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

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.


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.

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


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.