Convolution and second derivatives of Dirac Delta function

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.

$$f \ast g^{(n)} = f^{(n)} \ast g$$. 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.
• Yes. I know about $f' \ast g = f \ast g'$. My main doubt here is the decreasing in the derivative order. – Lin Aug 4 at 3:50