Suppose \(f:\DD\to\DD\) is analytic, has a single zero of order \(k\) at \(z=0\), and satsifies \(\lim_{\abs z \to 1} \abs{f(z)} = 1\). Give with proof a formula for \(f(z)\).
Solution
Goal: Suppose \(f: \DD \to \DD\) is analytic, has a single zero of order \(k\) at \(z = 0\), and satisfies \(\lim_{|z| \to 1}|f(z)| = 1\). Give, with proof, a formula for \(f(z)\).
1. \(f\) has only finitely many zeros in \(\DD\).
by the hypothesis, there is \(r < 1\) with \(|f(z)| > \tfrac12\) for \(r < |z| < 1\); hence all zeros lie in the compact set \(\{|z| \le r\}\) and, being isolated (if \(f \not\equiv 0\); \(f \equiv 0\) is excluded since \(|f| \to 1\)), are finite.
Let \(a_1, \dots, a_m\) be the zeros other than \(0\) (with multiplicity).
2. Form the finite Blaschke product \(B(z) = z^k \prod_{j=1}^m \frac{z - a_j}{1 - \bar a_j z}\).
\(|a_j| < 1\), so each factor is a disk automorphism; \(B\) is holomorphic on \(\DD\) with zeros exactly those of \(f\) (same multiplicities) and \(|B(z)| = 1\) for \(|z| = 1\) (each factor has modulus 1 on the circle).
3. \(h := f/B\) is holomorphic and zero-free on \(\DD\).
\(B\) and \(f\) have identical zeros with identical multiplicities (<1>1, <1>2), so the singularities of \(f/B\) at the zeros are removable and \(h \ne 0\).
4. \(|h(z)| \to 1\) as \(|z| \to 1\).
\(|h(z)| = |f(z)|/|B(z)|\); by 2, \(|B(z)| \to 1\) as \(|z| \to 1\), and by hypothesis \(|f(z)| \to 1\).
5. \(|h| \equiv 1\) on \(\DD\).
5.1. \(\log|h|\) is harmonic on \(\DD\) with boundary limit \(0\).
\(h\) is zero-free (<1>3), so \(\log|h| = \Re\log h\) is harmonic; 4 gives the boundary limit.
5.2. \(\log|h| \equiv 0\).
5.3. \(h \equiv \alpha\) with \(|\alpha| = 1\).
\(|h| \equiv 1\) by <2>1–<2>2; a holomorphic function of constant modulus 1 is constant with modulus 1.
6. \(f(z) = \alpha z^k \prod_{j=1}^m \frac{z - a_j}{1 - \bar a_j z}\) with \(|\alpha| = 1\) and \(0 < |a_j| < 1\); conversely every such function satisfies the hypotheses.
\(f = Bh\) with \(h = \alpha\) by <1>3–<1>5. Conversely, for any such product, \(f: \DD \to \DD\), \(f\) has a single zero of order \(k\) at \(0\) (and the other listed zeros), and \(|f(z)| \to 1\) as \(|z| \to 1\).
7. Q.E.D.
<1>1–<1>6 give the formula with proof.