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Let $f : D \rightarrow D$ be a holomorphic function such that $f(0) = 0$ where $D = \{z \in \mathbb{C} : |z| < 1\}.$

By Cauchy inequlity, $$|f''(0)| \leq \frac{2!\sup_{z \in \gamma_r} |f(z)|}{r^n}$$ for any $r < 1.$

By Schwarz Lemma, $|f(z)| \leq |z|$ for all $z \in D.$ So $$|f''(0)| \leq \frac{2!\sup_{z \in \gamma_r} |f(z)|}{r^n} \leq \frac{2\sup_{z \in \gamma_r} |z|}{r^2} = \frac{2}{r}$$ for any $r < 1$, $\gamma_r$ is a circle with radius $r$.

Let $r \rightarrow 1^-$ yields that $|f''(0)| \leq 2$ which is the smallest ($2/r > 2$ for any $r < 1$.)

I am not sure if this is a sharp bound for function $f$. I try to construct function $f$ by using function of the form $$f_a(z) = \frac{z-a}{1-\overline{a}z}$$ but still struggling.

Precisely, I would like to find $L > 0$ such that

1) $L$ bounds $|f''(0)|$ for any holomorphic functions $f : D \rightarrow D$ with $f(0) = 0$

2) there exists a holomorphic function $g : D \rightarrow D$ with $g(0) = 0$ satisfying $|g''(0)| = L.$

Any suggestions how to solve it ?

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    $\begingroup$ $f(z) = z^2$ shows that your bound $|f''(0)| \leq 2$ is sharp. $\endgroup$ – Martin R May 26 '18 at 23:05
  • $\begingroup$ And indeed, $z^n$ saturates Cauchy's inequality for order $n$. $\endgroup$ – Chappers May 26 '18 at 23:07
  • $\begingroup$ Oh ... ok silly me ... Thank you very much $\endgroup$ – Both Htob May 26 '18 at 23:07

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