Analysis Problem of the Day 93

Today’s problem appeared as Problem 8 on the UCLA Fall 2021 Analysis Qual:

Problem 8: Let $U,V,W$ be nonempty proper open simply connected subsets of $\mathbb{C},$ and $u \in U, v \in V$ are fixed.

a) Prove that for any compact $K \subseteq U,$ there exists a compact $L \subseteq V$ such that $f(K) \subseteq L$ whever $f: U \to V$ is holomorphic and $f(u)=v.$

b) Let $f_n: U \to V$ be holomorphic be a sequence of functions satisfying $f_n(u)=v$ for all $n$ and converging normally to $f: U \to V.$ Let $g:W \to U$ and $h: V \to W$ be conformal equivalences. Show that $f_n \circ g, h \circ f_n$ converge normally to $f \circ g, h \circ f,$ respectively.

Solution: a) Since all the sets are simply connected proper and open, by the Riemann mapping theorem they are conformally equivalent to the unit disk. Thus, without loss of generality, it suffices to prove the statement for $U=V=\mathbb{D}, u=v=0.$ But by the Schwarz lemma, any such map $f: \mathbb{D} \to \mathbb{D}$ with $f(0)=0$ satisfies $|f(z)| \leq |z|,$ so $f(K) \subset \overline{B}_{r_K}(0)$ for $r_K:= \sup_{k \in K} |k|.$ Conformally mapping $\overline{B}_{r_K}$ back to $V$ yields the desired compact set $L.$

b) Let $K$ be a compact set. By (a), there exists a compact $L$ such that $f_n(K), f(K) \subseteq L$ for all $n.$ Then, by uniform convergence on compact sets, for $|f_n-f|<\delta$ on $K$ for large enough $n$ and small enough $\delta>0$ with $f_n, f$ taking values in $L,$ we get by uniform continuity of $h$ on $L$ that $|h \circ f_n - h \circ f| < \epsilon$ for large enough $n,$ and since $K$ is arbitrary, this shows the normal convergence of $h \circ f_n$ to $h \circ f.$ Similarly, for any compact $K,$ $g(K)$ is compact since $g$ is continuous, so by uniform convergence of $f_n$ on $g(K)$ we get that $|f_n \circ g - f \circ g| < \epsilon$ for large enough $n,$ and since $K$ is arbitrary, this shows the normal convergence of $f_n \circ g$ to $f \circ g,$ and so we are done.

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