Boundary Trace in Half Space Let $\Omega$ denote the upper half-plane in $\mathbb{R}^2$, i.e.
$$
\Omega = \left\{ (x_1,x_2) : x_2 > 0\right\}. 
$$
Is it possible to find a function in $H^1_0(\Omega)$ whose boundary trace vanishes in $L^2(\partial \Omega)$ but not pointwise?
I saw this question in a set of notes and, honestly, I'm not even sure how to understand the question itself. Functions in $H^1(\Omega)$ are actually equivalence classes of functions equal almost everywhere. So how does one even go about discussing pointwise convergence of the trace?
Edit
It turns out that one can find a function $u \in H_0^1(\Omega)\cap C(\Omega)$ whose boundary trace vanishes in $L^2(\partial \Omega)$, with the property that
$$
\lim_{\substack{x \to x_0\\ x \in \Omega}} u(x) \neq 0
$$
for at least one point $x_0 \in \partial\Omega$. Still, I do not see how to construct such a function $u$.
 A: Yes, it is a weird question. The trace of a function $u\in H^1(\Omega)$ is a function $v$ in $L^1(\partial \Omega)$ and as such $v$ is not a function but an equivalence class of functions. Thus if the $L^2(\partial\Omega)$ norm of the trace $v$ of $u$ is zero, this means that $v=[0]$ so the trace of $u$ is zero. 
Even more, $H^1_0(\Omega)$ is the closure of $C^\infty_c(\Omega)$ so all functions in $H^1_0(\Omega)$ have trace zero as soon as the boundary is regular enough. In your example is a line so as nice as it gets.
As for talking about the pointwise value of the trace, you could still choose a good representative in the equivalence class of $v$. Usually, this is done by using Lebesgue points with respect to the surface measure on $\partial \Omega$.
Edit
Concerning your edit, I don't think such a function exists. Assume that 
$$\lim_{\substack{(x,y) \to (x_0,0)\\ (x,y) \in \Omega}} u(x,y) =\ell>0.$$
Then using the definition of limit you can find $r>0$ such that $u(x,y)\ge\frac\ell2$ for all $(x,y)\in\Omega$ with $\Vert (x,y)-(x_0,0)\Vert<r$ (and this remains true a.e. if you change representative). But this imply that the trace $v(x)\ge \frac\ell2$ for all $x\in [x_0-\frac{r}4,x_0+\frac{r}4]$ (for a proof of this fact, just follow the proof of trace positive but with $0$ replaced by $\ell/2$)
