Bound on the derivative of a holomorphic function from right half plane to unit disc Let $D=\{z\in\mathbb{C}:|z|<1\}$ and let $V=\{z\in\mathbb{C}:\Re(z)>0\}$.
Let $f:V\to D$ be a holomorphic function. Prove that
$$\forall z\in V:|f'(z)|\leq\frac{1-|f(z)|^2}{2\Re(z)}.$$
By taking $B_z=\{\xi\in V:|\xi-z|<\Re(z)\}$ and using Cauchy's formula, I only managed to obtain
$$|f'(z)|\leq\frac{\sup\limits_{\partial B_z}|f|}{\Re(z)}\leq\frac{1}{\Re(z)}.$$
Edit:
It seems to be very much related to Schwarz lemma. Maybe it is possible to use some mapping from $D$ to $V$, say $\phi:D\to V$ and then look at$f\circ \phi:D\to D$.
 A: As one readily verifies,
$$\psi \colon z \mapsto \frac{z-1}{z+1}$$
maps $V$ biholomorphically to $D$. With
$$\phi = \psi^{-1} \colon w \mapsto \frac{1+w}{1-w}\,,$$
the differential version of the Schwarz-Pick lemma applied to $g = f\circ \phi \colon D \to D$ tells us that
$$\frac{\lvert g'(w)\rvert}{1 - \lvert g(w)\rvert^2} \leqslant \frac{1}{1 - \lvert w\rvert^2}\tag{1}$$
for all $w \in D$. Writing $z = \phi(w)$ (and consequently $w = \psi(z)$), we have $g(w) = f(z)$ and $g'(w) = f'(z)\cdot \phi'(w)$. Since $\phi'(w) = \frac{2}{(1-w)^2}$, plugging into $(1)$ and rearranging yields
\begin{align}
\lvert f'(z)\rvert &\leqslant \frac{1 - \lvert f(z)\rvert^2}{(1 - \lvert w\rvert^2)\cdot \lvert\phi'(w)\rvert} \\
&= \bigl(1 - \lvert f(z)\rvert^2\bigr)\cdot \frac{\lvert 1-w\rvert^2}{2(1 - \lvert w\rvert^2)} \\
&= \frac{1 - \lvert f(z)\rvert^2}{2}\cdot \frac{\lvert 1 - \psi(z)\rvert^2}{1 - \lvert \psi(z)\rvert^2} \\
&= \frac{1 - \lvert f(z)\rvert^2}{2}\cdot \frac{\bigl\lvert\frac{2}{z+1}\bigr\rvert^2}{1 - \bigl\lvert \frac{z-1}{z+1}\bigr\rvert^2} \\
&= \frac{1 - \lvert f(z)\rvert^2}{2}\cdot \frac{4}{\lvert z+1\rvert^2 - \lvert z-1\rvert^2} \\
&= \frac{1 - \lvert f(z)\rvert^2}{2}\cdot \frac{4}{4 \Re (z)} \\
&= \frac{1 - \lvert f(z)\rvert^2}{2\Re (z)}\,.
\end{align}
