
weak derivative
I have the function f(x)= x+1 for x greater or equal to 0 and x1 for x less than 0 ,
I have to prove that f has no weak derivative.
I computed <Df,g>= 2g(0)+\int_0^\infty g(x)dx  \int_{\infty}^0 g(x)dx , where g is a test function. But how do I prove that this is of the form \int_{infty}^\infty h(x)g(x)dx and that Df is not in L^1_{loc}(R)?
Thanks.

Re: weak derivative
Notice that $\displaystyle f$ has classical derivatives on $\displaystyle (\infty,0)$ and on $\displaystyle (0,\infty)$, so any weak derivative of $\displaystyle f$ must coincide, on these intervals, with the strong derivative, so the only possible choice of derivative is $\displaystyle g(x)=1$ if $\displaystyle x>0$ and $\displaystyle g(x)=1$ for $\displaystyle x<0$. Now take $\displaystyle h \in C_c^{\infty}(\mathbb{R})$, we calculate
$\displaystyle \int_{\mathbb{R}} fh'= 2\left( xh(x)_{0}^{\infty}\int_0^{\infty} h(x)dx\right)  2h(0)$
$\displaystyle \int_{\mathbb{R}} gh= 2\int_0^{\infty} h(x)dx$
By the definition of weak derivative, the sum of these two quantities must be zero, so we arrive at $\displaystyle h(0)=0$. Since $\displaystyle h$ was arbitrary we get our contradiction. Notice that the contradiction works only because we allow the support of $\displaystyle h$ to be outside the sets where we know the function is differentiable.