Let $$H^1(-1,1)=W^{1,2}(-1,1)=\left\{u\in L^2(-1,1)\mid \exists g\in L^2(-1,1): \forall \varphi\in \mathcal D(-1,1), \int u\varphi'=-\int g\varphi\right\}.$$
I just started Sobolev space, so how can I show that $|x|\in H^1(-1,1)$ ?
What I tried is to set $$g(x)=\begin{cases}-1&x\in (-1,0)\\ 0&x=0\\ 1&x\in (0,1)\end{cases}.$$ Then, $$\int |x|\varphi'(x)dx=-\int_{-1}^0x\varphi'(x)dx+\int_0^1 x\varphi'(x)dx$$ $$\underset{IPP}{=}-\varphi(-1)+\int_{-1}^0 \varphi(x)dx+\varphi(1)-\int_0^1 \varphi(x)dx=\varphi(1)-\varphi(-1)+\int_{-1}^1 g(x)\varphi(x)dx.$$ Therefore, I'm very close of the solution, but I still have $\varphi(1)-\varphi(-1)$ that disturbes me.