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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 $f(z)=\exp(i\psi(z))$

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  • $\begingroup$ Is that meant to say $\mathbb R \subseteq S \subseteq \mathbb C$? $\endgroup$ – joriki Dec 23 '11 at 3:35
  • $\begingroup$ Yes, I fixed it. $\endgroup$ – Eva Md Dec 23 '11 at 20:52
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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.

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  • $\begingroup$ To extend $f$ holomorphically into a neighborhood of the real line you can take $f(\bar{z}) = 1/\overline{f(z)}$ for $z$ in the lower half plane by the Schwarz reflection principle; then your argument gives the desired neighborhood of the real line. $\endgroup$ – t.b. Dec 24 '11 at 1:15

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