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.