# Derivative of sign function $\operatorname{sgn}(x)$ (in distribution sense).

In the book of Schilling and Partzsch : Brownian motion (in the part of the Tanaka formula), they say that the derivative of $$f(x)=\text{sgn}(x)$$ is given by $$f'(x)=\delta _0(x)$$ (in distribution sense). But I find $$f'(x)=2\delta _0(x)$$ and I don't see where is my mistake : so let $$\varphi$$ a test function.

$$\left=-\int_{\mathbb R}f\varphi '=\int_{-\infty }^0\varphi '-\int_0^\infty \varphi '=\varphi (0)+\varphi (0)=2\varphi (0)=\left<2\delta _0,\varphi \right>.$$

Did they do a mistake ?

• It should be $2\delta$, yeah. – user658409 Oct 20 '19 at 16:45

You're right. In terms of the Heaviside function, $$\operatorname{sgn}x=2H(x)-1$$.