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.