In some class notes I have found the following statement:

Let $f(x)$ be a continuous funtion, $\delta(x)$ the Dirac delta function and $\ast$ the convolution operation given by $(f \ast g)(x) = \int_{-\infty}^{\infty} f(\tau) g(x-\tau) d\tau$, then:

$f(x) \ast \frac{d^2}{dx^2} \delta(x) = \frac{d}{dx} f(x)$

Is that true? Is some miscopied note? I could not find a proof.


1 Answer 1


$f \ast g^{(n)} = f^{(n)} \ast g$. The proof is integration by parts + definition of the distributional derivative. With $g = \delta$ you get $f \ast \delta^{(n)} = f^{(n)}$. In other words the derivative is a convolution operator which commutes with other convolution operators.

  • $\begingroup$ Yes. I know about $f' \ast g = f \ast g'$. My main doubt here is the decreasing in the derivative order. $\endgroup$
    – Lin
    Commented Aug 4, 2019 at 3:50
  • $\begingroup$ Then read my answer again $\endgroup$
    – reuns
    Commented Aug 4, 2019 at 3:56
  • $\begingroup$ @Lin I think "try writing out the convolution on your LHS and then using integration by parts to evaluate; you should see that a whole lot of things evaluate to zero" may help you, as well. $\endgroup$
    – nitsua60
    Commented Aug 4, 2019 at 4:04

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