# weak derivative

• Nov 20th 2011, 05:31 PM
Veve
weak derivative
I have the function f(x)= x+1 for x greater or equal to 0 and -x-1 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.
• Nov 20th 2011, 07:11 PM
Jose27
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.