Hi there.

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

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

What I tried yet is

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

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

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

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

= -g(0) - g(0) = -2g(0)

What next?

Yours
Rapha