I am trying to show that a harmonic function has constant sign on a open connected domain if it satisfies certain conditions on the boundary:
Suppose $U \subset \mathbb{C}$ is the top right quadrant ( both $x=Re(z)$ and $y=Im(z) >0$ ). Note this is open and connected.
Suppose further that, $h:U \to \mathbb{R}$ is harmonic on $U$ and extends continuously to the boundary:
$lim_{x\to 0}h(x+iy)=0$, $\forall 0<y<\infty$
$lim_{y\to 0}h(x+iy)=0$, $\forall 0<x<\infty$
$lim_{x\to \infty}h(x+iy)=0$, $\forall 0<y<\infty$
$lim_{y\to \infty}h(x+iy)=+\infty$, $\forall 0<x<\infty$
Can we therefore conclude, by the maximum principle, that $0<h(x+iy)<\infty$ for all $z=x+iy \in U$?
Moreover, it seems like it can be generalised to any simply connected domain $U \subset \mathbb{C}$ which is not the whole of $\mathbb{C}$! ( The domain U i gave above was in particular simply connected )
By the Riemann mapping theorem, there exists a biholomorphic ( holomorphic bijection whose inverse is also holomorphic ) map $\phi :U \rightarrow \mathbb{D}$.
Now, since $h:U\rightarrow \mathbb{C}$ is harmonic, it follows that the composition $\phi^{-1} \circ h : \mathbb{D} \rightarrow \mathbb{R} $ is also harmonic.
By the maximum principle for harmonic functions, $\phi^{-1} \circ h$ attains its extreme values on its boundary.
Caratheodory proved that the Riemann map above extends to a bijection between the boundary of $U$ and the unit circle. In our case this says that the extreme values of $\phi^{-1} \circ h$ on the unit circle correspond to the extreme values of h on the boundary of U.