Let $F(z)$ a complex analytic function defined on the whole interior of unit disk $\mathbb{D}$, such that $F(0)=0$, $F(1)=1$ (to be more precise, $\lim_{z\to1} F(z)=1$), which maps $\mathbb{D}$ into the left half-plane, that is, $\forall z\in\mathbb{D}\ \mathfrak{Re}(F(z))\leq1$.
I would like to have an estimate how close $F(\frac{1}{2})$ can get to $1$.
I have arrived at conjecture that $||F(\frac{1}{2})-1||\geq\frac{1}{3}$. (For real valued functions, that is equivalent to $F(\frac{1}{2})\leq\frac{2}{3}$.)
Further remarks. An example of a function with equality is $\frac{2z}{z+1}$. Another example where the inequality is strict, is $\log_2(z+1)$.
Any advice how to prove or disprove this or a reference to literature where this has already been solved would be appreciated. Borel–Carathéodory inequality seems to be related
(https://en.wikipedia.org/wiki/Borel%E2%80%93Carath%C3%A9odory_theorem),
but I am not sure if that could really help.
Best Answer
Note that $\Re (1-F)>0$ in the unit disc so using the disc self map $G=\frac{(1-F)-1}{(1-F)+1}=\frac{-F}{2-F}$ with $G(0)=0$, Schwarz lemma gives that $|G(1/2)|=|\frac{F}{2-F}(1/2)| \le 1/2$.
If $w=1-F(1/2)$, the above inequality reduces to $|\frac{w-1}{w+1}| \le 1/2$, which by squaring gives $3|w|^2+3 \le 10\ \Re w \le 10|w|$, so $|w| \in [1/3,3]$ and in particular $|1-F(1/2)| \ge 1/3$ with equality for $w=1/3, G(1/2)=-1/2, G(z)=-z, F(z)=\frac{2z}{z+1}$, which satisfies the constraint at $1$.