Boundary value of Sobolev space

Question

Let $D$ be a regular domain in $\mathbb R^2$. Suppose that $u \in H_0^1(D) \cap C(D)$. Does this imply $u \in C(\overline D)$ and $u|_{\partial D} = 0$?

Answer 1

The answer is no. Namely, for every nonempty open set $\Omega\subset\mathbb R\times\mathbb R$ there is a continuous function $u$ in $W^{1,2}_{\boldsymbol.}(\Omega)$ that has no continuous extension defined on the closure of $\Omega\,$. The (counter)example is constructed as follows.

Let $\langle\,x_i:i\in\mathbb N_0\,\rangle$ be any injective sequence in $\Omega$ with at least one accumulation point, and such that all accumulation points are on the boundary of $\Omega\,$. Then one can choose a stictly positive sequence $\langle\,R_i:i\in\mathbb N_0\,\rangle$ in $\ell^{\,2}(\mathbb N_0)$ such that $\langle\,{\rm B\,}(x_i,R_i):i\in\mathbb N_0\,\rangle$ is a sequence of disjoint balls in $\Omega\,$. For $s\le 0$ letting $\ln\kern.3mm s=-\infty=-\ln\kern.3mm(+\infty)$ and defining $u_i:\Omega\to[\,0\,,1\,]$ by $$ x\mapsto\inf\,\{\,\sup\,\{\,R_i\ln\kern.5mm(1- \ln\frac{|\,x-x_i\,|}{R_i}\Bigg)\,,0\,\}\,,1\,\} $$ then the function $u=\sum_{\,i\,\in\,\mathbb N_0\,}u_i$ has the claimed property.

If there are points that you do not understand, leave a comment specifying the unclear parts.

Answer 2

This is not true. Take a sequence of points $(x_n) \subset D$ which converge to a point on the boundary (or even worse: whose accumulation points are $\partial D$).

Then, for every point $x_n$ we can find $\varepsilon_n > 0$ and $\varphi_n \in C_c^\infty(D)$ with $\varphi_n \ge 1$ on the ball $B_{\varepsilon_n}(x_n)$, $\|\varphi_n\|_{H_0^1(D)} \le 2^{-n}$ such that $\varphi_n$ is supported on $B_{2\varepsilon_n}(x_n)$. Then, consider $$ u := \sum_{n \in \mathbb N} \varphi_n. $$

It is easy to check that $u$ has the desired properties.