Is an $H_0^1$ function continuous to the boundary if it is continuous in the interior?

Question

Suppose $\Omega$ is a bounded domain in $\mathbb R^3$ with Lipchitz boundary $\partial\Omega$, and $u\in H_0^1(\Omega)\cap C(\Omega)$. Is $u$ continuous to the boundary i.e. do we have $u \in C( \overline{\Omega})$?

In other words, is is true that $H_0^1 (\Omega)\cap C(\Omega)\subset C(\overline \Omega)$?

Depending on the answer(s), I may have some follow-up questions (for what it's worth).

Thank you any and all in advance.

Edit: It seems the answer is no so I am adding follow-up questions: Can I get that $u$ is bounded and/or attains its maximum on $\overline\Omega$?

Answer 1

Not necessarily- let $\Omega = B_1 \cap \{x_3 > 0\}.$ Then $u(x) := (1-|x|^2)\frac{x_3}{|x|}$ is in $H^1_0(\Omega) \cap C^{\infty}(\Omega),$ but $u$ is discontinuous at the origin.

Answer 2

The answer to the follow-up question is negative too. For consider the half-ball $\Omega=\{x\,;\,x_3>0,\,|x|<1\}$. Choose a number $\alpha\in(1,\frac32)$, and a function $\phi\in C^\infty({\mathbb R}^3)$ such that $\phi(x)\equiv1$ for $|x|<\frac13$, while $\phi(x)\equiv0$ for $|x|>\frac23$. Then the function $u(x)=r^{-\alpha}x_3\phi(x)$ belongs to $H^1(\Omega)\cap C(\Omega)$. Its trace, being an element of $H^{1/2}(\partial\Omega)$, is a square-integrable function, hence can be determined by looking away from a negligible Lebesgue set. Thus we look at the trace away from the origin, where $u$ is continuous and vanishes at the boundary. Therefore the trace is $\equiv0$, that is $u\in H^1_0(\Omega)$. Yet, it is not a bounded function.