$f: \mathbb{H} \to \mathbb{C}$ a bounded holomorphic function with $\lim\limits_{\operatorname{Im}(z) \to 0, z \in \mathbb{H}} \mathrm{Re}(f(z))= 0$.

Question

$f: \mathbb{H} \to \mathbb{C}$ a bounded holomorphic function with $\lim\limits_{\operatorname{Im}(z) \to 0, z \in \mathbb{H}} \operatorname{Re}(f(z))= 0$. How does it follow that $f$ is constant?

$\mathbb{H}$ is the upper half plane, i. e. $\{ z \in \mathbb{C} | \operatorname{Im} z > 0\}$

I'm tempted to use the Schwarz reflection principle, but I'm not sure whether I'm applying it correctly.

Since $\operatorname{Re} (f(z)) \equiv 0$ near the real axis, it follows for continuity reasons, that $\operatorname{Re} f(z) \equiv 0$ on the real axis. Hence by the Cauchy-Riemann equations I find that $\operatorname{Im} (f(z)) \equiv c$, where $c$ is a constant. If I could deduce that $c=0$, $f$ would only take real values on the real axis, I could use the Schwarz reflection principle to extend $f$ to a bounded holomorphic function on $\mathbb{C}$, which is, by Liouville, constant.

Is my approach more or less correct, or am I completely on the wrong track?

Answer 1

Hence by the Cauchy-Riemann equations I find that $Im(f(z))\equiv c$, where $c$ is a constant.

That is not correct. Consider $f(z) = iz$, then $\operatorname{Re} f(z) = 0$ for $\operatorname{Im} z = 0$, but $\operatorname{Im} f(z) = \operatorname{Re} z$ is not constant on $\mathbb{R}$.

The idea to use the reflection principle is good, however. But you need to apply it a little differently.

Consider $g(z) = - i\cdot f(z)$. That is a (bounded) holomorphic function on $\mathbb{H}$ with $\lim\limits_{\operatorname{Im} z \to 0} \operatorname{Im} g(z) = 0$, which is exactly the standard formulation of the reflection principle. Hence $$G(z) = \begin{cases}\quad g(z) &, z \in\mathbb{H}\\ \quad\overline{g(\overline{z})} &, \overline{z} \in\mathbb{H}\\ \lim\limits_{\varepsilon\searrow 0} g(z+i\varepsilon) &, z\in\mathbb{R} \end{cases}$$ is an entire holomorphic function, and bounded, hence constant. Thus $g$ is constant, so $f$ is constant.

The roles of the real and imaginary part of a holomorphic function in the reflection principle are pretty much symmetric, in fact, at its heart, the reflection principle is a theorem about harmonic functions.

If $\Omega \subset \mathbb{C}$ is a domain that is symmetric with repsect to the real axis (that means $z\in\Omega \iff \overline{z}\in\Omega$), and $u \colon \Omega\cap\mathbb{H}\to \mathbb{R}$ is harmonic with $\lim\limits_{\operatorname{Im} z \to 0} u(z) = 0$, then

$$U(z) = \begin{cases} u(z) &, z\in\mathbb{H}\\ -u(\overline{z}) & \overline{z}\in\mathbb{H} \\ 0 &, z\in\mathbb{R} \end{cases}$$

is a harmonic function on $\Omega$.

That is the core of the reflection principle, from which the reflection principle for holomorphic functions is derived by using the fact that locally (not necessarily globally), each real-valued harmonic function has a harmonic conjugate [that means, a harmonic $v$ such that $z\mapsto u(z) + i\cdot v(z)$ is holomorphic, which entails that locally, every real-valued harmonic function is the real part of a holomorphic function $f$ (and also the imaginary part of the holomorphic function $i\cdot f$)].