Analysis Problem of the Day 50

Today’s problem appeared as Problem 12 on the UCLA Fall 2021 Analysis Qual:

Problem 12: Let $f$ be holomorphic on $B:=\{|z|<2\}$ and satisfy $|f(z)|<1$ for all $z \in B.$ Additionally assume that $f(1)=f(-1)=f(i)=f(-i)=0.$

a) Show that $|f(0)| \leq \frac{1}{16}.$

b) Find an $f: B \to \mathbb{D}$ for which $|f(0)|=\frac{1}{16}.$

Solution: a) Notice that if one rescales $f$ by 2, one obtains a function $g: \mathbb{D} \to \mathbb{D},g(z) = f(2z),$ with $g(\frac12)=g(-\frac12)=g(\frac{i}{2})=g(-\frac{i}{2})=0.$ In particular, one may divide out by the corresponding Blaschke products $B_{\pm \frac12}, B_{\pm \frac{i}{2}},$ which have magnitude 1 on the boundary, to obtain a holomorphic function $\widetilde{g}: \mathbb{D} \to \mathbb{D}$ (since by the maximum modulus principle, $|\widetilde{g}(z)| \leq \limsup_{|z| \to 1} |\widetilde{g}(z)| \leq 1$ on $\mathbb{D}$ ). In particular, $|f(0)|=|\frac{1}{16} \widetilde{g}(0)| \leq \frac{1}{16}.$

b) Note that it suffices to take $\widetilde{g} \equiv 1,$ so

$f(z) = B_{\frac12}(\frac{z}{2}) B_{-\frac12}(\frac{z}{2}) B_{\frac{i}{2}}(\frac{z}{2}) B_{-\frac{i}{2}}(\frac{z}{2})= \frac{(\frac{z}{2})^4-\frac{1}{16}}{1+\frac{\bar{z}^4}{256}}=\frac{16 z^4-16}{256+\bar{z}^4}.$

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