# How to observe such an inequality?

Yesterday somebody post a problem on my college bbs:

Let $f(z)=1+\sum\limits_{n=1}^{+\infty}a_{n}z^{n}$ be a analytic function with positive real part on the unit disc $D=\{z:|z|<1\}$, please show that $|a_{1}^{2}-a_{2}|\leqslant2$ and $|a_{1}^{3}-2a_{1}a_{2}+a_{3}|\leqslant2$.

An equivalent form is that: For any analyitic function $g(z)=\sum\limits_{n=0}^{+\infty}b_{n}z^{n}$ mapping the unit disc into itself, it holds that $|b_{0}^{2}+b_{1}|\leqslant1$ and $|b_{0}^{3}-2b_{0}b_{1}+b_{2}|\leqslant1$.

I've got a solution, but I believe that there is a more enlightening one explaining why we pay attention to the polynomial $b_{0}^{3}-2b_{0}b_{1}+b_{2}$. (Notice that the equality holds in so many cases!)

EDIT(My solution): We work with the equivalent form, and firstly note that WLOG we can suppose $b_{0}$ to be real and nonnegative. When $g$ is noncostant, $h(z)=\frac{g(z)-b_{0}}{1-\bar b_{0}g(z)}$ satisfy $h(0)=0$ and $|h(z)|\leqslant1$, and then the Schwarz lemma gives that $|h'(0)|=\frac{|b_{1}|}{1-|b_{0}|^{2}}\leqslant1$, thus $|b_{0}|^{2}+|b_{1}|\leqslant1$ (which also holds for $g(z)=b_{0}$). By Schwarz lemma, $h(z)/z=\sum\limits_{n=0}^{+\infty}c_{n}z^{n}$ satisfied the same condition with $g$, so doing the same for it gives $|c_{0}|^{2}+|c_{1}|\leqslant1$. Finally combining the identity $b_{0}^{3}-2b_{0}b_{1}+b_{2}=b_{0}^{3}+2(b_{0}^{3}-b_{0})c_{0}+(b_{0}^{3}-b_{0})c_{0}^{2}+(1-b_{0}^{2})c_{1}$ and the inquality $|b_{0}^{3}+2(b_{0}^{3}-b_{0})z+(b_{0}^{3}-b_{0})z^{2}|\leqslant |z|^{2}(1-b_{0}^{2})+b_{0}^{2}$ (for $0\leqslant b_{0}\leqslant1$) gives the second inequality.

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Please explain your own solution. – Mårten W Feb 7 '13 at 11:36
All right, Marten. I shall edit my post. – Jun Su Feb 7 '13 at 12:01