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Let $D$ be a bounded region (open & connected subset of the complex plane) containing $0$ and let $f : D \rightarrow D$ be a holomorphic function such that $f(0) = 0$.

I was told that there exists a "Schwarz's lemma"-like result (apparently using something called Cartan's iteration trick but I don't know what is that) stating that $|f'(0)| \leq 1$ and if $f'(0) = 1$ then $f(z) = z$ on $D$.

Any idea where to start in order to prove this result ?

The "if $f'(0) = 1$ then $f(z) = z$ on $D$" thing was shown here : Holomorphic function $\varphi$ with fixed point $z_0$ such that $\varphi'(z_o)=1$ is linear? so it seems naturel to expect the first bit to be true as well.

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M.G
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1 Answers1

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If $D$ is simply connected this follows from the Riemann mapping theorem.

If not, let's see. There exists $r>0$ such that $B(0,r)\subset D$. And there exists $M$ so $|z|\le M$ for all $z\in D$, hence $|f(z)|\le M$. So we have $|f(z)|\le M$ in $B(0,r)$, hence Cauchy's estimates show there exists $c$ such that $$|f'(0)|\le c$$for every holomorphic $f:D\to D$.

Now let $g=f\circ f$. Then $g'(0)=f'(f(0))f'(0)=f'(0)^2$. So in fact we have $$|f'(0)|\le c^{1/2},$$since $|g'(0)|\le c$. The same trick shows that $$|f'(0)|\le c^{1/n}$$for every $n$, hence $|f'(0)|\le 1$.

David C. Ullrich
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