I'm trying to brush up on some complex analysis for a qualifying exam. I'm quickly realizing that I have never done many applications of Schwarz. In particular I don't have much experience with those types of questions that ask to find a bound for $|f(z_0)|$ for some specified $z_0$. So my questions are

(1) Can you suggest any texts, preferably accessible online, that gives examples of such problems with good explanations?

(2) So far, in reviewing past quals, I have only come upon one of these questions, so I have only one example to illustrate the type of problem. I think I was able to solve it, but I was mostly fumbling through it, are there general ways to approach it that would be better?

The question was, suppose that $f(z)$ is a holomorphic on $|z| <1$ and satisfies $|f(z)| < 1$ and $f(1/2)$ = 0. Use Schwarz Lemma to show that $|f(4/5)| \le 1/2$.

I know that in general $\frac{a-z}{1-\bar{a}z}$ maps $a$ to $0$ and maps the unit disk to the unit disk. So $g(z) = f(\frac{1/2-z}{1-\bar{1/2}z})$ has $g(0)=0$. And for all $z$ s.t. $|z|<1$, since that above function maps the unit disk to the unit disk $\frac{1/2-z}{1-\bar{1/2}z} \in$ the unit disk. Thus $|g(z)| = |f(\frac{1/2-z}{1-\bar{1/2}z})| \le 1$. So Schwarz lemma applies to $g$ and we can conclude $|g(z)| \le |z|$. But $z = 1/2 \rightarrow \frac{1/2-z}{1-\bar{1/2}z}=4/5$ so the desired bound follows.

Can you suggest better approaches? Thanks!

  • $\begingroup$ You have a sign error, the Möbius transformation is $T \colon z \mapsto \frac{z+1/2}{1+z/2}$ to have $g = f\circ T$ with $g(0) = 0$. But apart from that, that is the correct approach for this problem. $\endgroup$ – Daniel Fischer Aug 10 '13 at 22:04
  • $\begingroup$ @DanielFischer Oops, thanks. $\endgroup$ – Fractal20 Aug 11 '13 at 0:39
  • $\begingroup$ Wait I don't think that is right either, shouldn't it be $\frac{1/2-z}{1-z/2}$? $\endgroup$ – Fractal20 Aug 11 '13 at 0:47
  • $\begingroup$ Wouldn't anything of the form of a Mobius transformation work, as long plugging in 1/2 gave zero and the ad-bc wasn't zero? $\endgroup$ – Fractal20 Aug 11 '13 at 0:59
  • 1
    $\begingroup$ You have $f(\frac12) = 0$, and you want $g(0) = 0$, so you need $T(0) = \frac12$ if you write $g(z) = f(T(z))$. Any $T \in \operatorname{Aut}(\mathbb{D})$ with $T(0) = \frac12$ works, so $\frac{1/2-z}{1-z/2}$ would work too. I chose the one I gave since you had it in "normal form" $\frac{z-w}{1-\overline{w}z}$ originally, and I kept "normal form". Note: with the one you have now, you need $g(-\frac12)$. $\endgroup$ – Daniel Fischer Aug 11 '13 at 12:09

Instead of introducing auxiliary Möbius transformations every time, one could prove the general Schwarz-Pick lemma $$\left|\frac{f(a)-f(b)}{1-f(a)\overline{f(b)}}\right|\le \left|\frac{a-b}{1-a\overline{b}}\right|,\quad a,b\in\mathbb D \tag1$$ valid for any holomorphic map $f:\mathbb D\to\mathbb D$. The proof is, of course, an application of Schwarz lemma to the composition of $f$ with Möbius transformations.

You can think of the quantity $d(a,b)=\left|\frac{a-b}{1-a\overline{b}}\right|$ as a kind of distance adapted to the geometry of the unit disk. Then (1) simply says that this distance does not increase under $f$. In your case, $$d(f(4/5),f(1/2))\le d(4/5,1/2)=1/2$$ and since $f(1/2)=0$, the left side simplifies to $d(f(4/5),0)=|f(4/5)|$.

If, instead of $f(1/2)=0$, you were told that $f(1/2)=1/3$, extra work would be needed to decipher the inequality $$\left|\frac{f(4/5)-1/3}{1-f(4/5)/3}\right|\le \frac12$$ Geometric interpretation helps here: the inequality describes a non-Euclidean disk with non-Euclidean center $1/3$. Its farthest point from the origin $0$ will be found on the radius through $1/3$. This reduces the task to solving the equation $$\frac{x-1/3}{1-x/3}=\frac12$$ for real $x$. The solution is $5/7$. Thus, $|f(4/5)|\le 5/7$ in this version of the problem.

The quantity $\rho(a,b)=\tanh^{-1}d(a,b)$ is known as the Poincaré metric on the unit disk, and has a natural generalization to other domains in $\mathbb C$. The Schwarz-Pick lemma generalizes accordingly: a holomorphic map $f:\Omega_1\to\Omega_2$ is a contraction with respect to the hyperbolic metrics of $\Omega_1$ and $\Omega_2$.

The hyperbolic metric and geometric function theory by Beardon and Minda is an introductory article which explains all this and more; probably more than you need to know for a qualifying exam.

  • $\begingroup$ There is also the excellent Complex Analysis: A Geometric Viewpoint by Steven G. Krantz. $\endgroup$ – lhf Aug 12 '13 at 0:25
  • $\begingroup$ @user89958 Thanks that is great. I've never studied any non-Euclidean geometry, so I'm having difficulty having any intuition about the second part of your comment. Suppose $f(1/2)=1+i$ then would the farthest point from the origin be found on the radius through $1+i$ and thus we would solve $\frac{x(1+i) - 1/3}{1-(x/3)*(1+i)} = 1/2?$. I glanced through that link, but I'm afraid it seems to deep for me to delve into with the amount of time I have. So my apologies, since I presume the answers are all in there. $\endgroup$ – Fractal20 Aug 12 '13 at 15:35

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