Derivative of Analytic Function

We shall know well about Schwarz Lemma, which states that:

Analytic function $f(z): D\rightarrow D$ such that $f(0)=0$, then

$|f(z)|\le|z|$, and $|f'(0)|\le 1$.

We follow the scheme from the last post of Cartan Theorem, then we can get the estimate the derivative of $f(z)$.

Use the nested notation $f^{(k)}(z)=f\circ f\cdots\circ f(z)$, and if we want to estimate $|f''(0)|$, we shall write $f(z)$ as:

$f(z)=\alpha z+a_2 z^2+a_3 z^3+\dots$, where $\alpha=f'(0)$.

$f^{(k)}(z)=\alpha^k z+\displaystyle\alpha^{k-1}\frac{1-\alpha^k}{1-\alpha} a_2 z^2+\cdots$

While

$\displaystyle\frac{1}{2\pi}\int_0^{2\pi} f^{(k)}(e^{i\theta}) e^{-2i\theta}\mathrm{d}\theta\le 1$.

Which means

$\displaystyle|\alpha^{k-1}\frac{1-\alpha^k}{1-\alpha}||a_2|\le 1$ for all $k\in \mathbb{N}$.

However, this is just a thread for the estimation, we can easily see that $|a_2|\le 1$, as we put $k=1$.

The example is: $f(z)=e^{i\theta} z^2$. So, we can conclude that $|f''(0)|\le 2$.

And for the $k_{th}$ derivative $|\displaystyle\frac{\mathrm{d}^k}{\mathrm{d} z^k} f(z)|_{z=0}\le k$.