Application of Schwarz's lemma to $f(0)=1/2$

Question

I am trying to find the maximum of $|f'(0)|$ given $f : D(0,1)\rightarrow D(0,1)$ is analytic function such that $f(0)=1/2$. So if $T_{\alpha}=\dfrac{z-\alpha}{1-\overline{\alpha}z}$, I considered $g(z)=T_{1/2}\circ f(z)$. Then $g(0)=0$ and $|g(z)|\leq 1$ for all $z$ in unit disk. With this I got $|f(z)|\leq |T_{-1/2}(z)|$. But then if I use directly $g'$ and calculate the derivative, I got $|f'(0)|\leq \frac{9}{8}$ but if I use previous inequality of $|f(z)|$, I got the upper bound for $|f'(0)|$ is $\frac{3}{2}$. Which one is correct (is $\frac{9}{8}$ actually the sharpest bound)? If $\frac{9}8$ is the sharpest one, what is example of $f$ that achieves the bound?

Answer 1

You are right that $g(z)=T_{1/2}\circ f(z)$ satisfies the conditions of the Schwarz lemma, so that $|g'(0)| \le 1$. Since $$ T_{1/2}'(z) = \frac{3/4}{(1-z/2)^2} $$ and $$ g'(0) = T_{1/2}'(f(0)) f'(0) = T_{1/2}'\left(\frac 12\right) f'(0) = \frac 43 f'(0) $$ it follows that $$|f'(0)| = \frac 34 |g'(0)| \le \frac 34 \, . $$ This bound is sharp, as can be seen by choosing $f = T_{1/2}^{-1} = T_{-1/2}$.

One can also use the Schwarz-Pick theorem: $$ {\frac {\left|f'(z)\right|}{1-\left|f(z)\right|^{2}}}\leq {\frac {1}{1-\left|z\right|^{2}}}. $$ with $z = 0$, $f(0) = 1/2$.