Consider a smooth, bounded and convex domain $K$ in $R^n$ such that $K\subset \{ x_1 = 0 \}$ and $\Omega $ a bounded convex domain (not necessarily smooth) such that $\partial \Omega \supset K$.
Exists $u: \overline{\Omega} \rightarrow R$ where $$ \left\{ \begin{array}{ccccccc} \Delta u = 0, \ in \ \Omega (\ in \ the \ weak \ sense\ ) \\ \hspace{-1.5cm}u = 1 \ in \ K \ (\ pointwise \ ) \\ \hspace{-0.7cm}u = 0 \ on \ \partial \Omega - K (pointwise) \\ \end{array} \right. $$
?
I am making this question because in this article (in page 2) http://hal.inria.fr/docs/00/12/87/60/PDF/fbpLaplacian.pdf , the author says the answer is positive when $\Omega$ is smooth, but i believe that he did a typo (in relation of the smoothness). I dont know if the answer for my question is positive. I believe the answer is positive because of the variational methods. But i dont know much about variational methods.
Someone can give me a help with my question or say to me a reference ?
My english is terrible sorry..
A solution of $\Delta u=0$ in the weak sense is also a solution in the classical sense (Weyl's lemma).
What you wrote in the question is not what they write in the paper. In the paper, $u$ is not defined on $\overline{\Omega}$, it is defined on $\Omega$. The Dirichlet problem with discontinuous boundary condition is not fulfilled pointwise: it is not true that $u\to 1$ as $x\to x_0\in K$, for every $x_0\in K$. What they mean is that $u_\Omega$ is the harmonic measure of $K$. They prefer not to discuss its properties or existence, because they assume the reader knows what it is. The discontinuous boundary conditions can be understood in the $H^1_0$ sense, or in the sense of Perron. Some references: