# Find $\sup\{|f'(i)|: f \in \mathcal{F} \}$ where $\mathcal{F}$ is a normal family.

I've been working on the following problem:

Let $D = \{ z \in \mathbb{C}: \operatorname{Im} z > 0 \}$ and $\mathcal{F} = \{ f : f \text{ analytic on }$D$, |f(z)| < 1 \text{ for } z \in D\}$. Find $\sup\{|f'(i)|: f \in \mathcal{F} \}$. Is the supremum attained?

I know that by the second theorem of Montel, $\mathcal{F}$ is a normal family since every function omits the same two points (in fact, every function omits all points with in the complex plane with $|z| \geq 1$).

By Cauchy's inequalities $$|f'(i)| < \frac{1}{r} \quad \text{ for any } r \in (0,1).$$ Taking the limit as $r \to 1$, we have $$|f'(i) | \leq 1.$$

I don't know how to proceed from here to prove that $1$ is actually the supremum, nor how to determine whether it is attained. The little material I've seen on problems of this type doesn't deal with strict inequalities (i.e. they address families with $|f(z)| \leq 1$ rather than $|f(z)| < 1$), and the strict inequality seems to make it more complicated.

Any advice would be appreciated. Thanks.

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The supremum is attained by a conformal map onto the unit disk sending $i$ to 0. (In particular, it is not 1). Idea: consider the composition $f\circ g$ where $g$ maps the unit disk onto the halfplane. – user31373 Jul 7 '12 at 19:45
I see; thanks. I can compose each $f$ with two well-chosen conformal maps and then apply the Schwarz lemma. – ec92 Jul 7 '12 at 23:17

Define a function $g$ by $$g(z) = i\frac{1 +z}{1-z} \quad \Rightarrow \quad g'(z) = \frac{2i}{(1-z)^2}.$$ This is the inverse of the Cayley transform, and maps the unit disk onto the upper half plane. Next, for each $f \in \mathcal{F}$, define a map $$h(z) = \frac{z-f(i)}{1-\overline{f(i)}z} \quad \Rightarrow \quad h'(z) = \frac{1-|f(i)|^2}{(1-\overline{f(i)}z)^2}.$$ This is a conformal map of the unit disk onto itself. Now consider the composition $h \circ f \circ g$, which maps the unit disk onto itself and fixes zero. By the chain rule and Schwarz' lemma, $$|(h \circ f \circ g)'(0) |= |h'(f(g(0)))| \cdot |f'(g(0))| \cdot |g'(0)| = |h'(f(i))| \cdot |f'(i)| \cdot 2 = \frac{2|f'(i)|}{1-|f(i)|^2} \leq 1.$$
Therefore for every $f$, $$|f'(i)| \leq \frac{1-|f(i)|^2}{2} \leq \frac{1}{2}.$$
In particular, for $f \in \mathcal{F}$ given by $\displaystyle f(z) = \frac{z-i}{z+i}$, we have $\displaystyle f'(i) = \frac{2i}{(z+i)^2} \bigg|_{z=i} = \frac{1}{2i}$. Thus we conclude that $1/2$ is the required supremum, not merely an upper bound, and that it is attained.