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Let $H$ be the upper half plane and let $f : H → \mathbb C$ be holomorphic on $H$ and continuous on $\bar{H}$. Suppose that $f$ is constant on the real line, that is, there is $c ∈ \mathbb C$ such that $f(x) = c$ for all $x ∈ \mathbb R$. Show that $f$ is constant.

My attempt:

I tried proving this using Liouville's theorem.

Since $f$ is continuous on $\bar{H}$, then it is continuous at each $p \in \bar{H}$, so for all $\epsilon > 0$, there exists $\delta > 0$ s.t if $|z-p|< \delta$, then $|f(z) - f(p)|< \epsilon$.

Choose $\epsilon = 1$

Now we can rewrite $|f(z)|$ as: $|f(z) - f(p) + f(p)| ≤ |f(z) - f(p)| + |f(p)| ≤ 1 + |f(p)|$

Let $M := 1 + |f(p)|$

Hence $|f(z)| ≤ M$ and hence $f$ is bounded on $\bar{H}$

But since $p \in \bar{H}$, it can also lie on the real line, so $f(p) = c$ for some $c \in \mathbb C$ and thus $|f(p)| = |c|$ and hence $f$ is bounded on $H$.

Therefore, by Liouville's theorem, if a function is holomorphic and bounded, it must be constant.

Hence $f$ is constant.

Is my attempt correct? Any other answers?

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  • $\begingroup$ Liouville's theorem is for entire functions. $\endgroup$ – José Carlos Santos Mar 26 at 15:51
  • $\begingroup$ Are you saying that every continuous function $f : \bar H \to \mathbb{C}$ is bounded? Because you know that isn't true right? E.g. $f(x + yi) = y$. $\endgroup$ – Trevor Gunn Mar 26 at 15:53
  • $\begingroup$ The issue with your bound is that it only holds for $|z-p|<\delta$, so it doesn't provide a global bound. This problem seems like an identity theorem problem. Do you know anything more about this function? Can it be extended holomorphically beyond H? $\endgroup$ – Aidan Mar 26 at 16:08
  • $\begingroup$ Alternatively, consider the map from H to the unit disc given by the Riemann mapping theorem. If this map take the real line to the boundary of the disc, you can consider f as a holomorphic function on the disc, and so f must be constant by the Cauchy integral formula. That might be an approach $\endgroup$ – Aidan Mar 26 at 16:19
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You can't apply Liouville, since your function is not entire. Instead, you can use the Schwarz reflection principle.

WLOG, let $f=0$ on $\mathbb{R}.$ Let $\Omega ^+$ be the upper half plane, $\Omega ^-$ be the lower half plane, and $L=\mathbb{R}.$ Let $\Omega=\Omega ^+\cup L\cup \Omega ^-,$ which is open and connected.

By assumption, $f$ is holomorphic on $\Omega ^{+}$, continuous on $\Omega ^{+}\cup L,$ and real-valued on $L$. By the Schwarz reflection principle, the function $g:\Omega\rightarrow\mathbb{C}$ given by $$g(z)=\begin{cases} f(z)\textrm{ if } z\in\Omega ^{+}\cup L\\ \overline{f(\bar{z})}\textrm{ if }z\in\Omega ^-\end{cases}$$ is holomorphic. However, $g=f=0$ on $L$, and since $L$ contains a limit point, the uniqueness of analytic continuation guarantees that $g=0$ on $\Omega,$ and hence $f=0$ on $\Omega^+\cup L.$ That is, $f=0$ on $\mathcal{H}\cup \mathbb{R}=\overline{\mathcal{H}}.$

Since $f$ is continuous on $\overline{\cal{H}}$, we could have just assumed, for example, that $f=0$ on, say, $[0,1]$.

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  • $\begingroup$ Here's what I got from Wikipedia : That is, every holomorphic function $f$ for which there exists a positive number $M$ such that $|f(z)| ≤ M$ for all $z$ in $\mathbb C$, is constant. $\endgroup$ – JOJO Mar 26 at 15:57
  • $\begingroup$ Yes, and your function is not defined on all of $\mathbb{C}$. $\endgroup$ – cmk Mar 26 at 16:03
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Since $f$ is analytic on $H$ and continuous on its boundary hence $f$ is analytic on $H \cup\partial H$. Now the analytic function $f=c$ on the connected subset $\mathbb R$ of $H \cup\partial H$ is sufficient to conclude that $f=c$ on whole domain by Identity theorem.

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  • $\begingroup$ The identity theorem (and holomorphicity in general) is defined on open sets, and $H\cup \partial H$ is not open. $\endgroup$ – cmk Mar 26 at 17:23
  • $\begingroup$ You are right. Indeed $f$ is analytic on some open $X$ containing $H\cup\partial H$. $\endgroup$ – Nitin Uniyal Mar 27 at 1:45
  • $\begingroup$ $f$ isn't defined outside of $H\cup \partial H$, though. You need an extension procedure, like Schwarz reflection. $\endgroup$ – cmk Mar 27 at 13:05
  • $\begingroup$ $f$ is analytic on a closed disc $D\implies f$ is analytic on an open disc $D'$ containing $D$. $\endgroup$ – Nitin Uniyal Mar 27 at 13:39
  • $\begingroup$ $f$ is assumed to be analytic on an open set and continuous on the closure, that's it. If one wants to define analytic on a closed set as meaning that it's analytic on an open set containing that closed set, then that's fine, but a priori, $f$ is not even defined outside of $\bar{H}$. What you're saying is circular. You can extend, but you need something like Schwarz reflection. $\endgroup$ – cmk Mar 27 at 13:54

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