Holomorphic self map on the open unit disc Does there exist a holomorphic function $f:\mathbb{D}\longrightarrow\mathbb{D}$ such that $f(1/2)=-1/2$ and $f'(1/4)=1$, where $\mathbb{D}=\{z:|z|<1\}$?
 A: No, there is no such function. Let $g \colon \mathbb{D}\to \mathbb{D}$ be holomorphic with $\left\lvert g'\left(\frac14\right)\right\rvert = 1$ and $g\left(\frac12\right) = -\frac12$. The Schwarz-Pick lemma
$$\frac{\lvert h'(z)\rvert}{1 - \lvert h(z)\rvert^2} \leqslant \frac{1}{1-\lvert z\rvert^2}$$
for all holomorphic $h \colon \mathbb{D}\to\mathbb{D}$, yields for $g\left(\frac14\right)$
$$\begin{align}
&&\frac{1}{1-\left\lvert g\left(\frac14\right)\right\rvert^2} &\leqslant \frac{1}{1 - \frac{1}{16}}\\
&\iff & 1 - \frac{1}{16} &\leqslant 1 - \left\lvert g\left(\frac14\right)\right\rvert^2\\
&\iff & \left\lvert g\left(\frac14\right)\right\rvert^2 &\leqslant \frac{1}{16}\\
&\iff & \left\lvert g\left(\frac14\right)\right\rvert &\leqslant \frac14.\tag{1}
\end{align}$$
On the other hand, a holomorphic map $\mathbb{D} \to \mathbb{D}$ does not increase the hyperbolic distance, so
$$\delta\left(-\frac12, g\left(\frac14\right)\right) \leqslant \delta \left(\frac12, \frac14\right).\tag{2}$$
The only possibility to simultaneously satisfy $(1)$ and $(2)$ is
$$g\left(\frac14\right) = - \frac14,\tag{3}$$
which implies equality in $(2)$ (and in $(1)$), whence $g$ must be an automorphism. But then $z \mapsto -g(z)$ is an automorphism with two fixed points in the disk, hence the identity, so we have $g(z) = -z$ for all $z$, whence $g'\left(\frac14\right) = -1 \neq 1$.
