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Fall 2020 - Problem 8

Blaschke products

Let f ⁣:D→D\func{f}{\D}{\D} be holomorphic and satisfy f(12)=f(−12)=0f\p{\frac{1}{2}} = f\p{-\frac{1}{2}} = 0. Show that

∣f(0)∣≤14.\abs{f\p{0}} \leq \frac{1}{4}.
Solution.

For a∈Da \in \D, let φa ⁣:D→D\func{\phi_a}{\D}{\D} be the automorphism of the disk

φa(z)=z−a1−a‾z.\phi_a\p{z} = \frac{z - a}{1 - \conj{a}z}.

Then because ff has zeroes at 12\frac{1}{2} and −12-\frac{1}{2}, g(z)=f(z)φ1/2(z)φ−1/2(z)g\p{z} = \frac{f\p{z}}{\phi_{1/2}\p{z}\phi_{-1/2}\p{z}} is holomorphic. Hence, by the maximum principle, we see that ∣g(z)∣≤1\abs{g\p{z}} \leq 1 on D\D, and so

∣f(z)φ1/2(z)φ−1/2(z)∣≤1  ⟹  ∣f(0)∣≤∣φ1/2(0)∣∣φ−1/2(0)∣=14,\abs{\frac{f\p{z}}{\phi_{1/2}\p{z}\phi_{-1/2}\p{z}}} \leq 1 \implies \abs{f\p{0}} \leq \abs{\phi_{1/2}\p{0}}\abs{\phi_{-1/2}\p{0}} = \frac{1}{4},

which was what we wanted to show.