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I have the following question: Fix a point $a$ in a domain $U$. Let $F$ denote the family of functions that are analytic maps of $U$ into the open unit disk and map the point $a$ to zero. Prove that there exists a function $g\in F$ such that $|g'(a)|\geq |f'(a)|$ for each $f$ in $F$.

I think that I need to use Riemann mapping theorem but couldn't get the statement, any would be great.

user135582
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  • $U$ need not be simply connected, so the Riemann mapping theorem isn't (directly) applicable. Do you know Montel's theorem? – Daniel Fischer Aug 05 '15 at 15:29
  • Yes, but how we can use Montel's theorem? This family is normal since it is bounded so it is equicontinuous but how we get such function $g$? Thanks. – user135582 Aug 05 '15 at 15:48
  • In fact this is a major step in the usual proof of RMT. So it's just as well you can't use RMT here - if you did that we'd need a different proof of RMT... – David C. Ullrich Aug 05 '15 at 17:34

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Let $s = \sup \{ \lvert f'(a)\rvert : f \in F\}$, and pick a sequence $(f_n)$ in $F$ with

$$\lim_{n\to\infty} \lvert f_n'(a)\rvert = s.$$

Since $F$ is a normal family, the sequence $(f_n)$ has a subsequence that converges locally uniformly to a holomorphic function $g$ on $U$. Without loss of generality, suppose the entire sequence converges to $g$ locally uniformly. By the pointwise convergence, it follows that $\lvert g(z)\rvert \leqslant 1$ for all $z\in U$. Since $g(a) = \lim f_n(a) = 0$, it follows that $g$ is not a constant of modulus $1$, hence $\lvert g(z)\rvert < 1$ for all $z\in U$ and $g\in F$. By one of Weierstraß' theorems, we have

$$g'(a) = \lim_{n\to\infty} f_n'(a),$$

hence

$$\lvert g'(a)\rvert = s,$$

so a) $s < +\infty$, and b) the supremum is in fact a maximum.

Daniel Fischer
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