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.

About YiMin

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