1
$\begingroup$

Let $\Omega\subseteq\mathbb R^n$ be a bounded domain, $W_0^{1,2}(\Omega)$ be the Sobolev space and $C^{2,\alpha}$ be the Hölder space for some $\alpha\in (0,1]$. Suppose $(u_k)_{k\in\mathbb N}\subseteq C^{2,\alpha}(\Omega)$ converges to $u\in W_0^{1,2}(\Omega)$ with respect to the Sobolev norm and converges to $\tilde{u}\in C^0(\overline\Omega)$ with respect to the supremum norm. Why do we need to have $$u=\tilde{u}\;,$$ i.e. the representatives in $u$ are almost everywhere equal to $\tilde{u}$?

$\endgroup$
2
$\begingroup$

If you note $\|\cdot \|_\infty$ the sup ess norm, you have :

$$\| u - \tilde{u} \|_{\infty} \leq \| u - u_k\|_{\infty}+ \| u_k - \tilde{u} \|_{\infty}$$

Now let $\epsilon > 0$,

  • As $u_k \to \tilde{u}$ uniformly, there exist $k_1$ such that $\forall k > k_1, \ \| \tilde{u} - u_k\|_{\infty} < \frac{\epsilon}{2}$

  • As $u_k \to u$ in $W^{1,2}_0$, it converge in $L^2$, so there exist a subsequence $u_{i_n}$ that converge a.e to $u$. It means that there exists $i_{n_0} > k_1$ such that $\| u - u_{i_{n_0}} \|_{\infty} < \frac{\epsilon}{2}$

So, no matter what $\epsilon>0$ you take,

$$\| u - \tilde{u} \|_{\infty} \leq \| u - u_{i_{n_0}} \|_{\infty} + \| \tilde{u} - u_{i_{n_0}} \|_{\infty} \leq \frac{\epsilon}{2}+ \frac{\epsilon}{2} \leq \epsilon $$

This imply that $\| u - \tilde{u} \|_{\infty} = 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.