How can we proof the following argument:
Given an analytic function $f(z)$ ($z=x+iy, x,y\in \mathbb R$) in the closed upper half plane such that $|f(z)|<1$ for all $z$ in the open upper half plane, and $|f(x)|=1$ for all $x\in \mathbb R$, then:
there is an open simply connected set $S$ with $\mathbb R \subseteq S \subseteq \mathbb C$ such that $f(z)$ does not vanish on $S$, and there exists an analytic function $\psi:S \rightarrow \mathbb C$ such that $f(z)=\exp(i\psi(z))$
Best Answer
As far as I can tell this involves little to no complex analysis. Also, I'm assuming you want $S$ to be open in the upper half plane, not necessarily in all of $\mathbb{C}$.
To construct $S$, consider any $[a,b] \times [0, R] \subseteq \mathbb{H}$ (that is, $\{z: a \leqslant Re z \leqslant b, 0 \leqslant Im z \leqslant R \}$, for $R >> 0$, by compactness and hence uniform continuity there exists a subset $S_{a,b}$ of the form $[a,b] \times [0, \epsilon)$ on which $f$ is nonvanishing.
Now taking $[a,b]$ to be say $\ldots [-1,0], [0,1], [1,2], \ldots$, and patching your $S_{a,b}$ together nicely at the endpoints gives $S$.
$S$ is clearly simply connected (it deformation retracts onto the real line via "squishing").
Lastly, to construct $\psi$, we have $\Omega := S - \mathbb{R}$ is open and simply connected, $f$ is nonvanishing on $\Omega$, and hence $f = e^{g(z)}$ for $g$ holomorphic. Take $\psi := g/i$.
If this is unclear (or wrong!) lemme know.