Consider $\Omega \subset R^n$ an open, bounded and convex set. Then your boundary is Lipschtz.
Then we can define the trace operator T. Consider $K \subset \partial \Omega$ a compact set with non empty interior and $g : \partial \Omega \rightarrow R$ defined by $g(x) = 1 $ for $x \in K$ and $g(x)=0$ for $x \in \partial \Omega \setminus K .$
Consider the following problem:
Find $u \in H^{1}(\Omega)$ a weak solution for the problem (lets call problem 1)
$$ \left\{ \begin{array}{ccccccc} \Delta u = 0, \ in \ \Omega \\ u = g \ in \ \partial \Omega \ (\ in \ trace \ sense\ ) \\ \end{array} \right. $$
This problem has a unique solution in the Sobolev space $H^{1}(\Omega)$.
Now if we use the classic Perron method (the method works by the convexity of $\Omega$) we can construct a function $u: \Omega \rightarrow R$ satisfying:
i) $\Delta u = 0 \ on \ \Omega$ and $u \in C^{2}(\Omega)$
ii)$\displaystyle\lim_{y \rightarrow x} u(y) = g(x)$ if $g$ is continuous in $x \in \partial \Omega$
My question is : the function given by the Perron method and the unique solution for the problem (1) are the same ?
I am studying an article an appears the author use this. I searched a lot in the literature but I did not find anything... Please, someone can answer the question or say to me a reference?
thanks in advance.
Unfortunately the statement that u is in H^1 is not correct. The boundary data here is not in H^(1/2)(bdy) as it is a discontinuous step function. Similarly the Perron construction doesn't work as the boundary data is not continuous. Nevertheless there is a unique solution of this problem in certain H^s spaces with s < 1. See the paper by Auchmuty on Reproducing Kernels for spaces of Real Harmonic functions in SIAM J Math Anal, 41, 1994-2001.