Analysis Problem of the Day 44

Today’s problem appeared as Problem 7 on the UCLA Spring 2023 Analysis Qual:

Problem 7: Let $f: \mathbb{D} \to \mathbb{D}$ be holomorphic and such that $\sup_{\mathbb{D}} |f(z)| \leq r<1.$

a) Show that $f$ has a fixed point in $a \in \mathbb{D}.$

b) If $f^{[0]}=f,f^{[n+1]}=f \circ f^{[n]},$ show that $f^{[n]} \to a$ normally on $\mathbb{D}.$

Solution: a) By Rouche’s theorem, $|f(z)|< |z|$ on $|z|=1$ implies $z$ and $z-f(z)$ have the same number of zeros in $\mathbb{D},$ namely, $1.$ Thus, there exists a unique fixed point $a$ of $f$ inside $\mathbb{D}.$

b) Let $B_a = \frac{z-a}{1-\overline{a} z}$ be a Blaschke factor, i.e. an automorphism of the unit disc, mapping $a \to 0,$ and consider the map $g: \mathbb{D} \to \mathbb{D}$ given by $g = B_a \circ f \circ B_a^{-1}.$ Notice that since $|z| \leq r$ is compact in $\mathbb{D}$ and $B_a$ satisfies $|B_a(z)| \to 1$ as $|z| \to 1,$ one has that $|B_a(f(z))| \leq r' < 1$ for $|z| \leq 1$ since $|f(z)| \leq r.$ Thus, $|g(z)| \leq r',$ so $\frac{g}{r'}$ satisfies the conditions of the Schwarz lemma, i.e. $|g(z)| \leq |r'z|.$ Iterating this, one obtains $|g^{[n]}(z)| \leq (r')^{n+1}|z| \to 0$ normally as $n \to \infty$ since $r'<1.$ However, $g^{[n]}(z)=B_a \circ f^{[n]} \circ B_a^{-1} \to 0$ normally, so by continuity of Blaschke factors, one has $f^{[n]}(z) = (f^{[n]} \circ B_a^{-1})(B_a(z))\to a$ normally on compact subsets of $\mathbb{D}.$

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