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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 $u(x) = \sin(x)\in W^{1,2}_0(0,\pi)$

Sneach hcaens
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  • What is the domain? Something like $[0,2\pi]$? Then probably because $u'$ is not zero on the boundary, note the difference between $W_0^{2,p}$ and $W_0^{1,p} \cap W^{2,p}$. – Cahn Mar 04 '20 at 08:16
  • @Fritz, i am sorry, i forgot about domain. It's $(0,\pi)$ – Sneach hcaens Mar 04 '20 at 15:50
  • Have you tried to use the definition of $W^{k,p}_0$? Which $p$ are you interested in? When you try to repeat your proof for $\sin x \in W^{1,p}_0$ for the space $W^{2,p}_0$, what happens, where does it go wrong? – supinf Mar 04 '20 at 16:08
  • I edited, how i proof about@supinf – Sneach hcaens Mar 04 '20 at 16:48

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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 $u'(0)=\cos(0)=1 \neq 0$.

Cahn
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  • Why if $u(x)\in W^{2,p}(a,b)_0$ then $u(a)=u(b)=u'(a)=u'(b)=0$? I know that function $u(x)$ can be approximated by $u_m(x)\in C^{\infty}_0$, it means $u_m(a)=u_m(b)=u'_m(a)=u'(b)_m=0$. But $u_m$ approximates according to the norm of space $W^{2,p}_0$ and from the integral convergence does not follow pointwise? – Sneach hcaens Mar 04 '20 at 19:25
  • Do you know how to prove that $H_0^1=\overline{C_c^\infty}$ can equivalently be written as $H_0^1(\Omega)={u \in H^1(\Omega): u|_{\partial \Omega}=0}$? It is essentially the same proof. You should be able to find it e.g. in Evans. – Cahn Mar 05 '20 at 20:20
  • No, i don't know, what is the name of book?@Fritz And is it right only for demention of space n = 1? – Sneach hcaens Mar 05 '20 at 20:31
  • I am talking about the book "Partial Differential Equations" from Evans, see p.275 Theorem 2 in Section 5.5. This is true for general dimensions $n$, the boundary of the domains just has to fulfill some regularity (Lipschitz for $W_0^{1,p}$ and $C^{1,1}$-boundary for $W_0^{2,p}$ as far as I remember). – Cahn Mar 05 '20 at 20:47
  • Thank you, mr.Fritz. – Sneach hcaens Mar 05 '20 at 21:30