Weak derivative of $u(x)=|x|$ not belong in $W^{1,p}(-1,1)$

Question

Consider the functión $u \in W^{1,p}(-1,1) $, defined by $u(x)=|x|$, we know its weak derivative is $$g(x)=\left \{ \begin{matrix} 1 & \text{if }x\in(0,1) \\ -1 & \text{if } x \in (-1,0) \end{matrix} \right..$$ By intregation by parts is straightforward verify this. But I want to prove that $g \notin W^{1,p}(-1,1)$. If we suposse that $g$ has a weak derivative then exist a $h \in L^{p}(-1,1)$, which satisfies $$\varphi(1)-\varphi(0)+\varphi(-1)-\varphi(0)=\int_{-1}^{1} h(t)\varphi(t)dt$$for any $\varphi \in C_{c}^{1}(-1,1)$. I tried to get a contradiction evaluating by concrete test functions like $\varphi(t)=t$ or $\varphi(t)=1$ but I don't get any interesting. Which kind functions would help me? or there is another aproach?

Answer 1

For any $\varphi \in C_c^1((-1,1))$ satisfying $\varphi(0) = 0$, we have $\int_{-1}^1 h(t) \varphi(t)\,dt = 0$. However, the set of all such $\varphi$ is dense in $L^q((-1,1))$. Hence we must have $h=0$ which is absurd.

(Or, find a sequence of such $\varphi_n$ which converges a.e. and boundedly to $\operatorname{sgn} h$, and use dominated convergence to conclude $\int_{-1}^1 |h(t)|\,dt = 0$.)

Answer 2

Any function in $W^{1,p}(-1,1)$ is absolutely continuous (to be precise, any $f\in W^{1,p}(-1,1)$ has an absolutely continuous representative).

Your function $g$ is clearly not even continuous, hence it cannot belong to $W^{1,p}(-1,1)$.