Suppose \(f\) is analytic in an open set containing the unit disc \(\mathbb D\) and \(|f(z)| =1\) when \(|z|\)=1. Show that either \(f(z) = e^{i \theta}\) for some \(\theta \in \mathbb R\) or there are finite number of \(z_k \in \mathbb D\), \(k \leq n\) and \(\theta \in \mathbb R\) such that \[ \displaystyle f(z) = e^{i\theta} \prod_{k=1}^n \frac{z-z_k}{1 - \bar{z}_k z } \, . .\]
Also cf. Stein et al, 1.4.7, 3.8.17
Solution
1. Finiteness of zeros in \(\mathbb{D}\):
1.1. Since \(f\) is analytic on an open set containing the compact closure \(\overline{\mathbb{D}}\), and \(|f(z)| = 1\) on the boundary \(\partial\mathbb{D} = \{|z| = 1\}\), \(f\) cannot be identically zero and has no zeros on \(\partial\mathbb{D}\).
\(|f(z)| = 1 > 0\) on \(\partial\mathbb{D}\).
1.2. By the Identity Theorem, the zeros of \(f\) in \(\overline{\mathbb{D}}\) are isolated.Since \(\overline{\mathbb{D}}\) is compact, \(f\) has only finitely many zeros \(z_1, z_2, \ldots, z_n \in \mathbb{D}\) (counted with multiplicity).(If \(f\) has no zeros in \(\mathbb{D}\), we take \(n = 0\) and the empty product to be \(1\)).
zeros of non-constant holomorphic functions in a compact domain are finite.
2. Properties of the Blaschke product:
2.1. For each \(k \in \{1, \dots, n\}\), define the Blaschke factor:\[ B_k(z) = \frac{z - z_k}{1 - \overline{z_k} z}. \]For \(|z| = 1\), write \(z = e^{i\phi}\):\[ |B_k(e^{i\phi})| = \left| \frac{e^{i\phi} - z_k}{1 - \overline{z_k} e^{i\phi}} \right| = \left| e^{i\phi} \frac{1 - z_k e^{-i\phi}}{1 - \overline{z_k} e^{i\phi}} \right| = 1 \cdot \frac{|1 - z_k e^{-i\phi}|}{|\overline{1 - z_k e^{-i\phi}}|} = 1. \]
\(|w| = |\overline{w}|\) for all \(w \in \mathbb{C}\).
2.2. Define the finite Blaschke product \(B(z) = \prod_{k=1}^n B_k(z)\).\(B(z)\) is holomorphic on a neighborhood of \(\overline{\mathbb{D}}\), satisfies \(|B(z)| = 1\) for all \(|z| = 1\), and has exactly the zeros \(z_1, \dots, z_n\) in \(\mathbb{D}\) with the same multiplicities as \(f\).
product of Blaschke factors.
3. Quotient function and Maximum Modulus Principle:
3.1. Define \(g(z) = \frac{f(z)}{B(z)}\).Since the zeros of \(B(z)\) cancel with the zeros of \(f(z)\) in \(\mathbb{D}\), all singularities of \(g\) in \(\mathbb{D}\) are removable.Thus \(g\) extends to a holomorphic function on a neighborhood of \(\overline{\mathbb{D}}\) with \(g(z) \neq 0\) for all \(z \in \overline{\mathbb{D}}\).
Riemann Removable Singularity Theorem.
3.2. On the boundary circle \(|z| = 1\):\[ |g(z)| = \frac{|f(z)|}{|B(z)|} = \frac{1}{1} = 1. \]
\(|f| = 1\) and \(|B| = 1\) on \(\partial\mathbb{D}\).
3.3. By the Maximum Modulus Principle applied to \(g(z)\):\[ |g(z)| \le 1 \quad \text{for all } z \in \mathbb{D}. \]
Maximum Modulus Principle.
3.4. Since \(g(z)\) has no zeros in \(\overline{\mathbb{D}}\), the function \(\frac{1}{g(z)}\) is holomorphic on \(\overline{\mathbb{D}}\) and satisfies \(\left|\frac{1}{g(z)}\right| = 1\) on \(\partial\mathbb{D}\).Applying the Maximum Modulus Principle to \(\frac{1}{g(z)}\) yields:\[ \left| \frac{1}{g(z)} \right| \le 1 \implies |g(z)| \ge 1 \quad \text{for all } z \in \mathbb{D}. \]
Maximum Modulus Principle for reciprocal.
3.5. Thus \(|g(z)| = 1\) for all \(z \in \mathbb{D}\).A holomorphic function of constant modulus on a connected domain is constant:\[ g(z) = c = e^{i\theta} \quad \text{for some } \theta \in \mathbb{R}. \]
Cauchy–Riemann equations for constant-modulus holomorphic functions.
4. Conclusion:\(f(z) = e^{i\theta} B(z) = e^{i\theta} \prod_{k=1}^n \frac{z - z_k}{1 - \overline{z_k} z}\). Q.E.D.
2 and 3.