Let $f$ be holomorphic in a domain $G$ not convex, satisfying $|f'(z)| < 1$. Does the claim $|f(z_1) - f(z_2)| < |z_1 - z_2|$ necessarily hold?

Question

Let $f$ be holomorphic in a domain $G$, satisfying $|f'(z)| < 1 \ \forall z \in G$.

Does the claim $|f(z_1) - f(z_2)| < |z_1 - z_2|$ necessarily hold for a domain that is not convex?

My intuition tells me that this inequality might not always hold, but I am having trouble finding a counter-example, I thought about a region of a "track" taking a circle, now taking its center and making a smaller circle, now I will delete this circle and I will delete the lower half of the bigger circle now I will get this track and I will delete its border in order to make G a domain ( a simply connected and open set). and I want my $f(z)$ to be a function such that $|f(z_1) - f(z_2)|$ will give me the length between $z_1$ and $z_2$ along the arc of the track they are on, and I could take $2$ points that will be in opposite sides of the track and it might work, although I am not sure what is $f(z)$ exactly and what will be its derivative.

I would appreciate any help!

Answer 1

If the domain is not convex than such an estimate does not necessarily hold. As an example, let $$ G = \{ z : |z| > 1 \} \setminus (-\infty, -1] $$ be the exterior of the unit disk without the negative real axis, and $$ f(z) = \log(z) = \ln(|z|) + i \arg(z) \ , \, -\pi < \arg(z) < \pi \, , $$ be the principal branch of the logarithm on $G$. Then $|f'(z)| = 1/|z| < 1$ in $G$.

For $$ z_1 = 2 e^{i (\pi - \epsilon)} \, , \, z_2 = 2 e^{i (-\pi + \epsilon)} $$ with $0 < \epsilon < \pi$ is $$ |f(z_1) - f(z_2)| = 2 \pi - 2\epsilon \, , $$ but $$ |z_1 - z_2| = 2 |\sin(\pi - \epsilon) - \sin(\epsilon-\pi)| = 4 |\sin(\epsilon)| < 4 \epsilon \, , $$ so $$ \frac{|f(z_1) - f(z_2)|}{|z_1 - z_2|} $$ can be arbitrarily large.

Remark: If $f$ is holomorphic (or just continuous) in a domain $G \subset \Bbb C$ and satisfies a Lipschitz condition then $f$ can be continuously extended to the boundary of $G$ (see for example here). Therefore is suffices to find a function for which such an extension is not possible, like the complex logarithm.