Sobolev spaces $\sin x\in W^{1,p}_0$ but $\sin x\notin W^{2,p}_0$ when $x\in (0,\pi)$

Question

Proof that $\sin x\in W^{1,p}_0$, but $\sin x\notin W^{2,p}_0$. I got the first question, but have no idea how to proof in general that the function $u(x)\notin W^{2,p}_0$? Please give me some ideas. Put $$\begin{equation*} u^{\delta}(x) = \begin{cases} \sin(x) - \sin(\delta), x\in(\delta,\pi-\delta)\\ 0,x\in(0,\delta)\cup(\pi-\delta,\pi) \end{cases} \end{equation*}$$ Then it's easy to show thad $\begin{equation*} D_xu^{\delta}(x) = \begin{cases} \cos(x) , x\in(\delta,\pi-\delta)\\ 0,x\in(0,\delta)\cup(\pi-\delta,\pi) \end{cases} \end{equation*}$ $u^{\delta}(x)\in W^{1,p}_0$ Beacuse $u^{\delta}(x)$ is finite function. End easly to see that $\lim_{\delta\to 0}||u^{\delta}-u,W^{1,2}(0,\pi)|| = 0$. Then $u(x) = \sin(x)\in W^{1,2}_0(0,\pi)$

Answer 1

Actually, $\sin$ is a smooth and bounded function, in fact you have $\sin \in W^{k,p}(0,\pi)$ for all $k\geq 0$ and $p\geq 1$. So what can make problems?

It is the boundary behaviour. Remember that $u \in W_0^{2,p}(0,\pi)$ also includes $u(0)=u(\pi)=u'(0)=u'(\pi)=0$. Now it is easy to check that $u'(0)=\cos(0)=1 \neq 0$.