Since \(D\) is nonempty, we can pick \(z_0\in D\). Suppose there exists \(z_1\in D\) with \(u(z_1)\neq u(z_0)\). Since \(D\) is a domain, we can find a continuous (in fact, piecewise linear) arc \(\gamma :[0,1]\to D\) such that \(\gamma (0) = z_0\) and \(\gamma (1) = z_1\). Set \(h(t) = u(\gamma (t))\) for \(t\in [0,1]\). Then \(h:[0,1]\to \mathbf{R}\) is continuous, so, by the Intermediate Value Theorem, \(h\) takes each of the infinitely many values lying between \(u(z_0)\) and \(u(z_1)\) on the real line. But this contradicts the assumption that \(u\) takes only finitely many values.

Suppose \(0{\lt}r{\lt}a\). Then we have \(f(z_0) = (1/2\pi )\int _0^{2\pi } f(z_0 + re^{i\theta })\, d\theta \). Therefore, since \(|f(z_0)|\) is the maximum value of \(|f|\) on \(\Omega \),

It follows then that \(|f(z_0)| = (1/2\pi ) \int _0^{2\pi } |f(z_0 + r e^{i\theta })|\, d\theta \). So,

But then, since \(|f(z_0)| \geq |f(z_0 + re^{i\theta })|\) for all \(\theta \), it follows that \(|f(z_0)| = |f(z_0 + re^{i\theta })|\) for all \(\theta \). So, we get that \(|f(z_0)| = |f(z)|\) for all \(z\in \Delta (z_0,a)\). In other words, the function \(|f|\) is constant on \(\Delta (z_0, a)\). But then this implies that \(f\) is constant on \(\Delta (z_0,a)\).

Suppose \(u\) is locally constant at \(z_0\in \Omega \). To show that \(u\) is continuous at \(z_0\), we need to show that, for every \(\epsilon {\gt} 0\), there exists a \(\delta {\gt} 0\), such that \(|z - z_0| {\lt} \delta \Rightarrow |u(z) - u(z_0)| {\lt} \epsilon \). But, since \(u\) is locally constant, we can find a \(\delta {\gt} 0\) such that \(|z - z_0| {\lt}\delta \Rightarrow u(z) = u(z_0)\). So the required inequality is obvious no matter what positive real number \(\epsilon \) we pick.

The last statement of the proposition follows directly from the first.

Define a function \(u:D\to \mathbb {R}\) by setting

I claim that \(u\) is locally constant.

To see this, pick \(z\in D\). If \(u(z) = 0\), then \(|f(z)|\) is the maximum value of \(|f|\) on \(D\). And, since \(D\) is open, we can find \(\delta {\gt} 0\) such that \(\Delta (z, \delta )\subseteq D\). But then, by Lemma 2.2, \(f(w) = f(z)\) for all \(w\in \Delta (z,\delta )\). So \(u(w) = u(z)\) for all \(w\in \Delta (z,\delta )\).

On the other hand, if \(u(z) = 1\), then, since \(f\) is continuous at \(z\), we can find \(\delta {\gt}0\) such that \(|w-z|{\lt}\delta \Rightarrow |f(w) - f(z)| {\lt} |f(z) - f(z_0)|\). Then \(|w-z|{\lt}\delta \Rightarrow f(w)\neq f(z_0)\Rightarrow u(w) = 1 - u(z)\).

Now, since \(u\) is locally constant on \(D\), it is continuous. And, since \(D\) is a domain and \(u\) is continuous, \(u\) must be constant by Proposition 2.1. So, as \(u(z_0) = 0\), \(u(z) = 0\) for all \(z\in D\). Thus \(f(z)\) is constant on \(D\).

Suppose \(u(z_0)\) is the maximum value of \(u\) on \(D\). Then, since \(e^{f(z)} = e^{u(z)}(\cos v(z) + i\sin v(z))\), \(e^{u(z_0)}\) is the maximum value of \(|e^{f(z)}\) on \(D\). Since \(f\in A(D)\), \(e^f\in A(D)\) as well. So, by the Maximum Modulus Principle applied to \(e^f\), \(e^f\) is constant on \(D\). Therefore, the function \(g(z) = f(z) - f(z_0)\) is an analytic function on \(D\) taking values in the set \(2\pi i\mathbb {Z}\). So \(h(z) = g(z)/2\pi i\), is an analytic function on \(D\) taking values in \(\mathbb {Z}\). In particular, \(h(z)\in \mathbb {R}\) for all \(z\in D\). So it follows that \(h\) is constant.