# prove that if $f$ is not surjective, then there exist $g\in M$ s.t. $|g'(z_0)|\gt |f'(z_0)|$

$N$ is aconnected, simply connected, nonempty, proper open subset of $\Bbb C$

Let $z_0\in N$ is assumed to be fixed.

Let M = the set of all analytic injective functions $f: N \to D$ for unit disk $D$.

And $f(z_0)=0$

Let $f\in M$

I want to prove that if $f$ is not surjective, then there exist $g\in M$ s.t. $|g'(z_0)|\gt |f'(z_0)|$

This question seems so difficult to me. Hopefully help me. Thank you.

Suppose $w_0 \in \mathbb{D}\setminus f(N)$. Then

$$h(z) = \frac{f(z)-w_0}{1-\overline{w_0}f(z)}$$

gives an injective holomorphic function $h\colon N \to \mathbb{D}$ that doesn't attain the value $0$. Since $h$ is injective, it is a homeomorphism between $N$ and $h(N)$, which therefore is simply connected. Hence there exists a holomorphic branch of the square root on $h(N)$, let's call it $r$. Now consider the function

$$k(z) = \frac{r(h(z)) - r(-w_0)}{1-\overline{r(-w_0)}r(h(z))}.$$

Write it as $k = S \circ r \circ T \circ f$, with the two automorphisms

$$T\colon w \mapsto \frac{w-w_0}{1-\overline{w_0}w};\quad S\colon w \mapsto \frac{w-r(-w_0)}{1-\overline{r(-w_0)}w}$$

of the unit disk. Since $r$ maps $h(N)$ into the unit disk, we have $k(N) \subset \mathbb{D}$, and $S\circ r\circ T\colon f(N) \to \mathbb{D}$. Also, $k(z_0) = S(r(h(z_0))) = S(r(-w_0)) = 0$, so we must see

$$\lvert (S\circ r\circ T)'(0)\rvert > 1.$$

We have $T'(0) = \dfrac{1-\lvert w_0\rvert^2}{(1-\overline{w_0}\cdot0)^2} = 1 -\lvert w_0\rvert^2$, further $r'(-w_0) = \dfrac{1}{2r(-w_0)}$, and $S'(r(-w_0)) = \dfrac{1-\lvert r(-w_0)\rvert^2}{(1-\overline{r(-w_0)}r(-w_0))^2} = \dfrac{1}{1-\lvert r(-w_0)\rvert^2}$. Multiplying that yields

$$\lvert (S\circ r\circ T)'(0) \rvert = \frac{1-\lvert w_0\rvert^2}{2\sqrt{\lvert w_0\rvert}(1-\lvert w_0\rvert)} = \frac{1+\lvert w_0\rvert}{2\sqrt{\lvert w_0\rvert}} > 1.$$

• Is this completely my question's answer? Right? Commented Nov 24, 2013 at 21:26
• Thank you, by the way:) I am studying your answer step by step right now:) Commented Nov 24, 2013 at 21:28
• Yes. $k \in M$, and $\lvert k'(z_0)\rvert = \lvert (S\circ r\circ T)'(0)\rvert \cdot \lvert f'(z_0)\rvert > \lvert f'(z_0)\rvert$. Commented Nov 24, 2013 at 21:29

This is a consequence of P. Koebe's proof of the Riemann mapping theorem. During the proof, it is proved that the function $f$ that maps $N$ biholomorphically onto the unit disk and satisfies $$f(z_0)=0,f'(z_0)>0$$is unique, and also maximizes $|f'(z_0)|$.

Now, if the function $f$ in your problem maximizes the (absolute value of the) derivative at $z_0$, a simple rotation will produce $\tilde{f}$ which satisfies the normalisation conditions above, and will have to coincide with the established biholomorphism. But this will result in the surjectivity of $f$!

The proof of Riemann's mapping theorem can be found in Ahlfors' Complex Analysis text, chapter 6.