Intuituve affirmation of functions of $H^1_0(\Omega)$

Question

Consider $\Omega$ a open and bounded set.

Let $u \in H^{1}_0(\Omega)$ a continuous function. is true that $lim_{x \rightarrow y} u(x) = 0$ for $y \in \partial \Omega$ ? I dont know how to prove this.... Someone can give me a help to prove this affirmation ?

thanks in advance

Answer 1

As @Yiorgos pointed out, when $k>\frac{n}{2}$ (and the boundary is good) we have that $H_0^k(\Omega)\subset C(\overline{\Omega})$ and hence your statement is true.

On the other hand, for $k<\frac{n}{2}$ this might not be true and an counter example for your statement can be found here or here.

Answer 2

What you say is true (for sufficiently smooth boundary) for $u\in H^k_0(\Omega)$, with $\Omega$ open in $\mathbb R^n$, if $$ k>\frac{n}{2}, $$ according to the standard Sobolev imbedding.

In the case of $H^1_0(\Omega)$, the boundary trace of its elements are identified with elements of $H^{1/2}(\partial\Omega)$. But such traces are not defined pointwise, unless $n=1$.