<Home>Qualifying Exams><Analysis>Spring 2010 - Problem 12

Spring 2010 - Problem 12

Montel's theorem, normal families, Urysohn subsequence principle

Let $F$ be a function from $\D$ to $\D$ such that whenever $z_1, z_2, z_3$ are distinct points of $\D$ there exists an analytic function $f_{z_1,z_2,z_3}$ from $\D$ to $\D$ such that $F\p{z_j} = f_{z_1,z_2,z_3}\p{z_j}$ . Prove that $F$ is analytic at every point of $\D$ .

Hint: Fix $z \in \D$ and let $\D \ni z_n \to z$ , $z_n \neq z$ . Show that the sequence

\frac{F\p{z_n} - F\p{z}}{z_n - z}

is bounded and prove that every two of its convergent subsequences have the same limit.

Solution.

Let $0 < R < 1$ , and let $K = \cl{B\p{0, R}}$ , which is a convex compact subset of $\D$ . We will show that

\sup_{z,w \in K; z \neq w} \abs{\frac{F\p{z} - F\p{w}}{z - w}} < \infty.

Observe that $\set{f_{z_1,z_2,z_3} \mid z_1, z_2, z_3 \in \D}$ is uniformly bounded by $1$ , since their image lies in the unit disk. Thus, $\set{f_{z_1,z_2,z_3}' \mid z_1, z_2, z_3 \in \D}$ is also uniformly bounded: by Cauchy's integral formula,

\abs{f_{z_1,z_2,z_3}'\p{z}} \leq \frac{1}{2\pi} \int_{\partial K} \abs{\frac{f_{z_1,z_2,z_3}\p{\zeta}}{\p{\zeta - z}^2}} \,\diff{s} \leq \sup_{z_1,z_2,z_3\in\D}\,\frac{\norm{f_{z_1,z_2,z_3}}_{L^\infty\p{\D}}}{R} \leq \frac{1}{R}.

Thus, for distinct $z, w \in K$ ,

\begin{aligned} \abs{F\p{z} - F\p{w}} &= \abs{f_{z,w,z_3}\p{z} - f\p{w}_{z,w,z_3}} \\ &\leq \int_z^w \abs{f_{z,w,z_3}'\p{\zeta}} \,\diff{s} \\ &\leq \frac{\abs{z - w}}{1}{R} \\ \implies \abs{\frac{F\p{z} - F\p{w}}{z - w}} & \leq \frac{1}{R}, \end{aligned}

where we integrated along the segment from $z$ to $w$ .

Fix $z \in \D$ . Since $\abs{z} < 1$ , we may pick $\abs{z} < R < 1$ , which means that $\frac{F\p{\zeta} - F\p{z}}{\zeta - z}$ is bounded above by $\frac{1}{R}$ . We will show that if $\set{z_n}_n, \set{w_n}_n \subseteq \D \setminus \set{z}$ are sequences which converge to $z$ and such that both

\set{\frac{F\p{z_n} - F\p{z}}{z_n - z}}_n \quad\text{and}\quad \set{\frac{F\p{w_n} - F\p{z}}{w_n - z}}_n

converge, then they converge to the same limit. For $n \geq 1$ , let $g_n = f_{z_n,w_n,z}$ and observe that

\tag{1} \frac{F\p{z_n} - F\p{z}}{z_n - z} = \frac{g_n\p{z_n} - g_n\p{z}}{z_n - z} \quad\text{and}\quad \frac{F\p{w_n} - F\p{z}}{w_n - z} = \frac{g_n\p{w_n} - g_n\p{z}}{w_n - z}.

Notice that $\set{g_n}_n$ is uniformly bounded by $1$ , so it admits a locally uniformly convergent subsequence $\set{g_{n_k}}_k$ , which converges to some $g$ . Let $\epsilon > 0$ . By locally uniform convergence, $g$ is also holomorphic, so there exists $\delta > 0$ such that if $\abs{\zeta - z} < \delta$ , then

\tag{2} \abs{\frac{g\p{\zeta} - g\p{z}}{\zeta - z} - g'\p{z}} < \epsilon.

Similarly, by uniform convergence on $K$ , there exists $N \in \N$ such that if $k \geq N$ , then

\tag{3} \abs{\frac{g_{n_k}\p{\zeta} - g_{n_k}\p{\eta}}{\zeta - \eta} - \frac{g\p{\zeta} - g\p{\eta}}{\zeta - \eta}} < \epsilon

for any distinct $\zeta, \eta \in K$ . Thus, for $k$ large enough, $\abs{z_{n_k} - z}, \abs{w_{n_k} - z} < \delta$ and we get

\begin{aligned} \abs{\frac{F\p{z_{n_k}} - F\p{z}}{z_{n_k} - z} - \frac{F\p{w_{n_k}} - F\p{z}}{w_{n_k} - z}} &= \abs{\frac{g_{n_k}\p{z_{n_k}} - g_{n_k}\p{z}}{z_{n_k} - z} - \frac{g_{n_k}\p{w_{n_k}} - g_{n_k}\p{z}}{w_{n_k} - z}} && (\text{by (1)}) \\ &\leq \abs{\frac{g_{n_k}\p{z_{n_k}} - g_{n_k}\p{z}}{z_{n_k} - z} - \frac{g\p{z_{n_k}} - g\p{z}}{z_{n_k} - z}} + \abs{\frac{g_{n_k}\p{w_{n_k}} - g_{n_k}\p{z}}{w_{n_k} - z} - \frac{g\p{w_{n_k}} - g\p{z}}{w_{n_k} - z}} + \abs{\frac{g\p{z_{n_k}} - g\p{z}}{z_{n_k} - z} - \frac{g\p{w_{n_k}} - g\p{z}}{w_{n_k} - z}} && (\text{triangle inequality}) \\ &\leq 2\epsilon + \abs{\frac{g\p{z_{n_k}} - g\p{z}}{z_{n_k} - z} - g'\p{z}} + \abs{\frac{g\p{w_{n_k}} - g\p{z}}{w_{n_k} - z} - g'\p{z}} && (\text{by (3) and the triangle inequality}) \\ &\leq 4\epsilon. && (\text{by (2)}) \end{aligned}

Thus, because $\set{\frac{F\p{z_n} - F\p{z}}{z_n - z}}_n$ and $\set{\frac{F\p{w_n} - F\p{z}}{w_n - z}}_n$ were both convergent and we found subsequences which converge to the same limit, it follows that both sequences converge to the same limit as well. Call this limit $F'\p{z}$ .

Now let $\set{z_n}_n \subseteq \D \setminus \set{z}$ be any sequence which converges to $z$ . As before, $\set{\frac{F\p{z_n} - F\p{z}}{z_n - z}}_n$ is bounded above. Let $\set{z_{n_k}}_k$ be any subsequence, and by boundedness, it admits a further subsequence which converges. By our previous argument, this subsequence must be equal to $F'\p{z}$ . We have shown that any subsequence of $\set{\frac{F\p{z_n} - F\p{z}}{z_n - z}}_n$ has a further subsequence which converges to $F'\p{z}$ , and so

\lim_{n\to\infty} \frac{F\p{z_n} - F\p{z}}{z_n - z} = F'\p{z}.

Thus, as our choice of $\set{z_n}_n$ was arbitrary, it follows that

\lim_{w \to z; w \neq z} \frac{F\p{w} - F\p{z}}{w - z} = F'\p{z},

so $F$ is analytic at $z$ . Finally, our choice of $z \in \D$ was also arbitrary, so $F$ is analytic on all of $\D$ , which completes the proof.