Consider a smooth ,convex and bounded domain $K \subset \{ x_1 = 0 \} \subset R^n$ . consider a smooth open set $\Omega \subset R^{n}_+ = \{ x = (x_1,..,x_n)\in R^n ; x_1 > 0\} $ with $K \subset \partial \Omega$.
Consider $u \in H^{1}(\Omega)$ the weak solution for the problem
$$\left\{ \begin{array}{11} \Delta u &=& 0 \ \mbox{i}n \ \Omega \\ u &=& 1 \ \mbox{i}n \ K \ \\ u &=& 0 \ \mbox{o}n \ \partial \Omega\setminus K \\ \end{array} \right. $$
where the boundary conditions is in the trace sense. By the maximum principle i know that $0 \leq u \leq 1$.
Fix $0 < l < 1$. The set $\{ x \in \Omega ; u(x) = l\}$ is always non empty ? Drawing a picture its easy to see this. but i dont know how to prove this. someone can give me a help to prove or disprove this ? by regularity theory $u$ is continuous.maybe this helps.. Thanks in advance!
Take a look in Brezis book page 303 (Estimates near the boundary). First, fix a point $p\in \partial\Omega\setminus\overline{K}$. Let $\theta_i,U_i$ be as in the book, consider the function $v=\theta_i u$ and assume without loss of generality that $\partial K$ is not contained in $\overline{U_i}$.
As the trace of $u$ near $p$ is zero, you have that $v\in H_0^1(\Omega\cap U_i)$. Now you can follow Brezis proof to conclude that $v\in H^m(\Omega)$ for all $m$, which implies that $u$ is continuous near the point $p$, hence, $u(x)=0$ in a neighbourhood of $p$ in $\partial\Omega$.
Now let $q\in K$. In the same way as before, assume that $\partial K$ is not contained in $\overline{U_i}$. The trace of the function $u-1$ is zero near $q$, hence, you can consider the function $\theta_i(u-1)$ and apply the same argument as before to conclude that $u-1$ is continuous near the point $q$, therefore, $u(x)=1$ in na neighbourhood of $q$ in $\partial \Omega$.
Once $u\in C(\Omega)$ we conclude that your set is not empty.