Let $f: \mathbb D \to \mathbb D$ be analytic or holomorphic with $f(0)=\frac{1}{2}$ and $f(\frac{1}{2}) = 0$ where $\mathbb {D} = \{ z: |z| \leq 1\}$. Then find $|f^{'}(0)|$ and $|f^{'}(\frac{1}{2})|$.
I tried Cauchys Inequality Theorem $ |f^{(n)}(z)| \le \frac{n!M}{r^{n}}$.
Since n=1 the inequality becomes $ |f^{'}(z)| \le \frac{M}{r}$. The value of r=1 since ($\mathbb {D} = \{ z: |z| \leq 1\}$)and hence i arrive at $ |f^{'}(z)| \le {M}$.Now how to find the value of $M$.Please Do reply.
On
Consider the Blaschke factor $$g(z)=\frac{\frac{1}{2}-z}{1-\frac{1}{2}z}=\frac{1-2z}{2-z}.$$
Then $g$ is holomorphic on $\mathbb D$, and maps $\mathbb D$ to $\mathbb D$. Moreover, $g(0)=\frac{1}{2}$ and $g\left(\frac{1}{2}\right)=0$. Then, $f\circ g$ maps $\mathbb D$ to $\mathbb D$ and fixes $0$ and $\frac{1}{2}$, therefore, using the Schwartz lemma, we see that $(f\circ g)(z)=cz$ for some $c\in\mathbb C$ with $|c|=1$. Since $f\circ g$ fixes $\frac{1}{2}$, we get that $c=1$, therefore $$f\left(\frac{1-2z}{2-z}\right)=z.$$
We now differentiate: $$1=f'\left(\frac{1-2z}{2-z}\right)\left(\frac{1-2z}{2-z}\right)'=-f'\left(\frac{1-2z}{2-z}\right)\frac{3}{(2-z)^2},$$
therefore, by setting $z=0$ we get that $$1=-f'\left(\frac{1}{2}\right)\frac{3}{4}\Rightarrow f'\left(\frac{1}{2}\right)=-\frac{4}{3},$$ and similarly, $f'(0)=-\frac{3}{4}$.
Schwarz–Pick theorem: If $f:\mathbb D\to \mathbb D$ is analytic and: $f(z_1)=w_1\,$, $f(z_2)=w_2$, then: $$\left |\frac{w_1-w_2}{1-w_1\overline w_2}\right|\le \left |\frac{z_1-z_2}{1-z_1\overline z_2}\right|$$ And: $$|f'(z_j)|\le \frac{1-|w_j|^2}{1-|z_j|^2}\,\,\,\,,\, j=1,2$$ If equality obtains in the first expression for some $z_1\neq z_2$ or if equality obtains in the second expression, then F must be a conformal self-map of the disc.
Because of $f(0)=\frac12$ and $f(\frac12)=0\,$, so in the first expression we have equality. So: $$|f'(0)|=\frac{1-|\frac12|^2}{1-0}=\frac34$$ $$|f'(\frac12)|=\frac{1-0}{1-|\frac12|^2}=\frac43$$