<Home>Qualifying Exams><Analysis>Fall 2012 - Problem 7

Fall 2012 - Problem 7

Schwarz lemma

Let ff be a function holomorphic in C\C and suppose that f(0)=0f\p{0} = 0, f(1)=1f\p{1} = 1, and f(D)⊆Df\p{\D} \subseteq \D, where D≔{z∈C∣∣z∣<1}\D \coloneqq \set{z \in \C \mid \abs{z} < 1}. Show that

  1. f′(1)∈Rf'\p{1} \in \R,
  2. f′(1)≥1f'\p{1} \geq 1.
Solution.

  1. Suppose f′(1)∉Rf'\p{1} \notin \R and let v∈Cv \in \C. By the chain rule,

    f′(1)v=lim⁡t→0+f(1+tv)−f(1)t=lim⁡t→0+f(1+tv)−1t.f'\p{1}v = \lim_{t\to0^+} \frac{f\p{1 + tv} - f\p{1}}{t} = \lim_{t\to0^+} \frac{f\p{1 + tv} - 1}{t}.

    If Re⁡v<0\Re{v} < 0 and tt is sufficiently small, then 1+tv∈D1 + tv \in \D. Since Im⁡f′(1)≠0\Im{f'\p{1}} \neq 0, we may pick vv such that Re⁡(f′(1)v)>0\Re\p{f'\p{1}v} > 0, e.g., Im⁡v=−Im⁡f′(1)\Im{v} = -\Im{f'\p{1}} and pick Re⁡v<0\Re{v} < 0 to be very small. Since 1+tv∈D1 + tv \in \D, we also have f(1+tv)∈Df\p{1 + tv} \in \D for all tt small, so in particular, Re⁡f(1+tv)−1<0\Re{f\p{1 + tv}} - 1 < 0. Thus, taking real parts,

    0<Re⁡(f′(1)v)=lim⁡t→0+Re⁡(f(1+tv)−1t)≤0,0 < \Re\p{f'\p{1}v} = \lim_{t\to0^+} \Re\p{\frac{f\p{1 + tv} - 1}{t}} \leq 0,

    a contradiction. Hence, f′(1)∈Rf'\p{1} \in \R to begin with.

  2. By the Schwarz lemma, ∣f(1−t)∣≤1−t\abs{f\p{1 - t}} \leq 1 - t for t∈(0,1)t \in \p{0, 1}. Thus, because ff is holomorphic in a neighborhood of 11 and f′(1)f'\p{1} is real,

    f′(1)=Re⁡f′(1)=lim⁡t→0+Re⁡(f(1−t)−1−t)=lim⁡t→0+Re⁡(1−f(1−t)t)≥lim⁡t→0+Re⁡(1−(1−t)t)=1.\begin{aligned} f'\p{1} = \Re{f'\p{1}} &= \lim_{t\to0^+} \Re\p{\frac{f\p{1 - t} - 1}{-t}} \\ &= \lim_{t\to0^+} \Re\p{\frac{1 - f\p{1 - t}}{t}} \\ &\geq \lim_{t\to0^+} \Re\p{\frac{1 - \p{1 - t}}{t}} \\ &= 1. \end{aligned}