I can't seem to see why this should be true. By Riemann Mapping Theorem we know $\mathbb{H}$ (upper half plane) is conformally equivalent to $\Delta$ (open unit disk), so what's the problem with sending the line Im($z$) = 1 to the real axis? I thought of using Open Mapping Theorem or Max Mod to get a contradiction but didn't get far with either of those. Any thoughts?

EDIT: Let $g(z) = i+z$ and let $F(z) = (f\circ g)(z) = f(i+z)$. Then $F$ is holomorphic on $\mathbb{H}$ and continuous and real valued on $\mathbb{R}$, so by the reflection principle $F$ extends to an entire function $h_1(z)$ that satisfies $h_1(z) = \overline{F(\overline{z})} = \overline{f(i+\overline{z})}$ and agrees with $F$ on $\mathbb{H}\cup\mathbb{R}$. But then $h_1$ is bounded since $f$ is, and thus $h_1$ is constand, whereby $f$ is also constant.

  • $\begingroup$ Have you considered Liouville's Theorem? $\endgroup$ – PhysMath Aug 10 '19 at 3:01

For $\Im(z) < 0$ let $f(i+z) = \overline{f(i+\overline{z})}$. Then $f$ is entire and bounded.

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  • $\begingroup$ hmm ? OP's function $f(i+x)$ is real for $x$ real so it satisfies $f(i+z) = \overline{f(i+\overline{z})}$ for $\Im(z) \in (-1,1)$ $\endgroup$ – reuns Aug 10 '19 at 5:53
  • $\begingroup$ Maybe clearer, the reflection principle extends holomorphically the original function restricted to $\Im z \ge 1$ to the whole plane and by the identity principle, the extension coincides with the original function on the strip $0 < \Im z < 1$ so Liouville applies etc $\endgroup$ – Conrad Aug 10 '19 at 14:54
  • $\begingroup$ @Conrad, thanks for the response. I'm having trouble understanding the last bit. For the identity principle to apply, don't both functions have to be defined on the same region to begin with? So if $f$ is the original function and $g$ is it's extension to $\mathbb{C}$, if I understand your answer correctly, then $f$ and $g$ agree on all of $\mathbb{C}$ since they agree on the strip? But $f$ is not defined on all of $\mathbb{C}$ to begin with. $\endgroup$ – aleph0 Aug 12 '19 at 23:40
  • $\begingroup$ Thanks for the replies, @reuns and I think I get it now. Care to check my edit? $\endgroup$ – aleph0 Aug 13 '19 at 1:19
  • $\begingroup$ $f$ is originally defined on the upper plane; however the condition on the line $y=1$ allows an extension of $f$ restricted to the domain $y \ge 1$ to the whole plane, while the identity principle ensures this extension coincides with $f$ on the upper plane (or if you want in the strip $0 < y <1$) where $f$ was originally defined $\endgroup$ – Conrad Aug 13 '19 at 1:25

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