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

Question

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.

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

Did they do a mistake ?

Answer 1

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