# Harmonic functions in the half space are unique $\iff$ dirichlet problem solutions are unique?

Let $$\Omega$$ be an open connected subset of $$\mathbb{R}^N$$ satisfying the following property:

$$x=(x_1,\cdots,x_N)\in \Omega\implies (x_1,\cdots,-x_N)\in\Omega$$

Let $$u$$ be an harmonic function in

$$\Omega_+=\{x=(x_1,\cdots,x_N)\in\Omega: x_N>0\}$$

continuous in $$\Omega_{-} = \{x=(x_1,\cdots,x_N)\in\Omega:x_N\ge 0\}$$ and null when $$x_N=0$$. Show that there exists only one harmonic function $$\overline{u}$$ in $$\Omega$$ which coincides with $$u$$ in $$\Omega_+$$

Doesn't this follow from the unicity of the dirichlet problem? We have a function $$u$$ harmonic in the superior half plane and continuous in the superior + boundary. Suppose that there are two harmonic functions $$u_1,u_2$$ with such properties. Then $$u_1-u_2$$ is also harmonic. By the maximum principle, its max is attained in the border, but $$u_1-u_2=0$$ in the border. We can say the same for the minimum value. That is, the minimum is on the boundary and is $$0$$. Therefore $$u_1-u_2=0$$ everywhere, that is, $$u_1=u_2$$

The problem is that the hypothesis that $$x=(x_1,\cdots,x_N)\in \Omega\implies (x_1,\cdots,-x_N)\in\Omega$$ is useless. Maybe I'm missing something?

You are actually making two mistakes: First,the boundary of $$\Omega_+$$ need not be contained in the hyperplane $$x_N=0.$$ There's all the stuff above as well. For example, let $$N=2$$ and let $$\Omega$$ be the open unit disc. Then $$\partial \Omega_+$$ equals $$\{e^{it}: t\in [0,\pi]\} \cup [-1,1].$$ Secondly, even if all of $$\partial \Omega_+$$ lies in that hyperplane, as is true for $$\Omega = \{x_N>0\},$$ the unicity of the Dirichlet problem can fail. For example $$u(x)\equiv 0$$ and $$u(x)=x_N$$ both equal $$0$$ on $$\partial \Omega_+$$ in this case.
The key is boundedness: The solution to the Dirichlet problem for all bounded $$\Omega$$ is unique if it exists.