Continuity of $|.|$ in $W^{1,p}_0$

Question

please i dont understand this proof

We suppose that $u\rightarrow u$ on $W^{1,p}_0$ i dont understand why we must use the weak compactness and the uniform convexity of $W^{1,p}_0$ ?

Thank you

Answer 1

Assuming that $u_n\to u$ in $W_0^{1,p}(\Omega)$, it should be proven that $|u_n|\to |u|$ in $W_0^{1,p}(\Omega)$. Weak compactness is used to obtain a subsequence $\left\{u_{n_k}\right\}$ such that $|u_{n_k}|$ converges weakly to some $z$. You see that this is "almost" what we want. The statement that $|u_n|\to z=|u|$ for the whole sequence can be proven as a exercise, if you're not sure about it (try doing it by contradiction). Since you have only weak convergence, uniform convexity, by means of the second property in Wikipedia, solves this problem and gives you strong convergence.

If you're still not sure about all this, here's an alternative proof for this theorem:

Alternative Proof: Suppose, by contradiction, that the mapping $u\mapsto u^+$ is not continuous. Then we can find a sequence $(u_n)\subseteq W_0^{1,p}(\Omega)$ such that $u_n\to u$ but $u_n^+$ does not converge to $u^+$. By going to a subsequence if necessary, we can assume that there exists some $\varepsilon>0$ such that \begin{align*} \Vert u_n^+-u^+\Vert_{W_0^{1,p}(\Omega)}\geq \varepsilon\qquad\forall n.\tag{*} \end{align*}

Now, we can take a subsequence $v_k=u_{n(k)}$ such that $v_k\to u$ a.e. and $\nabla v_k\to\nabla u$ a.e.. By using the equation $$\nabla v^+(x)=\begin{cases}\nabla v(x)&\text{, if }v(x)>0\\0&\text{, otherwise}\end{cases}$$ which holds a.e. for $v\in W^{1,p}_0(\Omega)$, then its easy to see that $v_n^+\to v^+$ and $\nabla v_n^+\to\nabla v^+$ a.e. By the Dominated Convergence Theorem, it follows that $v_n^+\to v^+$ and $\nabla v_n^+\to\nabla v^+$ in $L^p(\Omega)$, which means that $v_n^+\to u^+$ in $W_0^{1,p}(\Omega)$. But this contradicts $(*)$ above.

Therefore, $u\mapsto u^+$ is continuous in $W_0^{1,p}(\Omega)$. Also, since $u^-=(-u)^+$ and $|u|=u^++u^-$ then $u\mapsto u^-$ and $u\mapsto |u|$ are continuous in $W_0^{1,p}(\Omega)$.QED

Just a quick observation: All this holds in $W^{1,p}(\Omega)$ as well.