first order weak derivative of function $ f(x) = |x| $

Question

let $f(x)= |x| $ how can I calculate the first order weak derivative of this function in $x=0$? Does anyone have an idea on how to calculate this?

Answer 1

$$f'(x)=-H(-x)+H(x)$$ where $H(x)=0$ when $-\infty\lt x\leq 0$ and $H(x)=1$ when $x\gt 0$

Answer 2

The direct answer is that since the weak derivative of a function is only defined almost everywhere, you can put whatever value of $f'(0)$ you want.

Always, remember that we say $g$ is the weak derivative of $f(x)=|x|$ if for any $\phi\in C_c^\infty(\mathbb R)$, we have $$ \int_{\mathbb R} f\,\phi'\,dx=-\int_{\mathbb R}g\,\phi\,dx \tag 1$$

Hence, define $$ f'(x)= \begin{cases} 1&x>0\\ -1&x<0\\ \alpha&x=0 \end{cases} $$ where $\alpha\in \mathbb R$ is an arbitrary constant, can you verify $(1)$?