Hi there.

Let $\displaystyle f(x) : = |x| , \ x \in (-1,1)$

Show that $\displaystyle f'(x) = sign x $ has no weak derivative.

What I tried yet is

Let $\displaystyle g \in C_0^\infty((-1,1))$

$\displaystyle \int_{-1}^1 (sign x) g' dx$

$\displaystyle = \int_{-1}^0 (-1) g' dx + \int_0^1 1 g' dx$

$\displaystyle = [-1 g]_{-1}^0 + [1 g ]^1_0 $

$\displaystyle = -g(0) - g(0) = -2g(0)$

What next?

Yours
Rapha