Convolution and second derivatives of Dirac Delta function

Question

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.

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.