# Showing that a complex-valued function is constant

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?

• Liouville's theorem is for entire functions. – José Carlos Santos Mar 26 at 15:51
• 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. $f(x + yi) = y$. – Trevor Gunn Mar 26 at 15:53
• 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? – Aidan Mar 26 at 16:08
• 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 – Aidan Mar 26 at 16:19

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, $$[0,1]$$.

• 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. – JOJO Mar 26 at 15:57
• Yes, and your function is not defined on all of $\mathbb{C}$. – cmk Mar 26 at 16:03

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.

• The identity theorem (and holomorphicity in general) is defined on open sets, and $H\cup \partial H$ is not open. – cmk Mar 26 at 17:23
• You are right. Indeed $f$ is analytic on some open $X$ containing $H\cup\partial H$. – Nitin Uniyal Mar 27 at 1:45
• $f$ isn't defined outside of $H\cup \partial H$, though. You need an extension procedure, like Schwarz reflection. – cmk Mar 27 at 13:05
• $f$ is analytic on a closed disc $D\implies f$ is analytic on an open disc $D'$ containing $D$. – Nitin Uniyal Mar 27 at 13:39
• $f$ 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. – cmk Mar 27 at 13:54