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Fall 2009 - Problem 11

Blaschke products, Schwarz lemma

Let $f$ be an analytic function in the open unit disc $D = \set{z \mid \abs{z} < 1}$ that obeys $\abs{f\p{z}} \leq 1$ for all $z \in D$ . Let $g$ be the striction of $f$ to the real interval $\p{0, 1}$ and assume

\lim_{r\to1} g\p{r} = 1 \quad\text{and}\quad \lim_{r\to1} g'\p{r} = 0.

Prove that $f$ is constant.

Solution.

If $\abs{f\p{z}} = 1$ anywhere in $D$ , then $f$ is constant by the maximum principle. Now suppose that $\abs{f\p{z}} < 1$ for all $z \in D$ .

If we can apply the Schwarz lemma, then we know that $f$ grows at most like $\abs{z}$ , which "barely" approaches $1$ as $r \to 1$ . But the extra decay from $\abs{f'\p{z}}$ tending to $0$ will cause $f$ to grow too slowly to reach $1$ , which will be our contradiction.

In order to apply the Schwarz lemma, we need to perform a change of variables first. Consider the Blaschke product $\func{\phi_a}{D}{D}$ where $a \in D$ , given by

\phi_a\p{z} = \frac{z - a}{1 - \conj{a}z},

which is a biholomorphic function. By a quick calculation, $\phi_a$ has derivative

\phi_a'\p{z} = \frac{1 - \conj{a}z + \p{z - a}\conj{a}}{\p{1 - \conj{a}z}^2} = \frac{1 - \abs{a}^2}{\p{1 - \conj{a}z}^2}.

By assumption, $\abs{f\p{0}} < 1$ , so $\phi_{f\p{0}}$ is well-defined. Moreover,

\lim_{r\in\R,r\to1} \p{\phi_{f\p{0}} \circ f}\p{z} = \lim_{r\in\R,r\to1} \frac{f\p{z} - f\p{0}}{1 - \conj{f\p{0}}f\p{z}} = \frac{1 - f\p{0}}{1 - \conj{f\p{0}}}.

Hence, we may define $h = e^{i\theta}\phi_{f\p{0}} \circ f$ , where $\theta$ is chosen so that $\lim_{r\in\R,r\to1} h\p{r} = e^{i\theta}\phi_{f\p{0}}$ . Observe that $h\p{0} = 0$ and that

\begin{aligned} \lim_{r\in\R,r\to1} \abs{h'\p{r}} &= \lim_{r\in\R,r\to1} \frac{1 - \abs{f\p{0}}^2}{\abs{1 - \conj{f\p{0}}f\p{r}}^2} \abs{f'\p{r}} \\ &= \frac{1 - \abs{f\p{0}}^2}{\abs{1 - \conj{f\p{0}}}^2} \lim_{r\in\R,r\to1} \abs{f'\p{r}} \\ &= 0. \end{aligned}

Since $\phi_{f\p{0}}$ has image in the disk, we still have $\func{h}{D}{D}$ and so we may apply the Schwarz lemma to get $\abs{h\p{z}} \leq \abs{z}$ .

We now restrict ourselves to $r \in \p{0, 1}$ . Let $\epsilon > 0$ . There then exists $0 < \delta < 1$ such that if $r > 1 - \delta$ , then $\abs{h'\p{z}} < \epsilon$ . For these values of $r$ , we have

\begin{aligned} \abs{h\p{r}} &= \abs{\int_0^r h'\p{t} \,\diff{t}} \\ &\leq \abs{\int_0^{1-\delta} h'\p{t} \,\diff{t}} + \abs{\int_{1-\delta}^r h'\p{t} \,\diff{t}} \\ &\leq \abs{h\p{1 - \delta} - h\p{0}} + \int_{1-\delta}^r \abs{h'\p{t}} \,\diff{t} \\ &\leq \abs{h\p{1 - \delta}} + \int_{1-\delta}^r \epsilon \,\diff{t} \\ &= 1 - \delta + \epsilon\p{r - 1 + \delta}. \end{aligned}

Letting $r \to 1$ , we have

1 \leq 1 - \delta\p{1 - \epsilon}.

In particular, if $\epsilon = \frac{1}{2}$ , then $1 \leq 1 - \frac{\delta}{2} < 1$ , which is impossible. Hence, $f$ must have been constant to begin with.