If $f \in H(\Omega)$ holomorphic, then $|f(x) - f(y)| \geq (1/2)|f'(c)||x-y|$

Question

I was reading Rudin's proof on the Open mapping theorem. He declare $\Omega$ to be a region, which is connected and open.

Then he claims there is a neighbourhood of $c$ such that $|f(x) - f(y)| \geq (1/2)|f'(c)||x-y|.$ How is this obtained? Is it because being holomorphic yields the inequalities.

\begin{align} \left |\frac{ \phi(x) - \phi(y) }{x - y} - \phi'(c) \right | < (1/2)|\phi'(c)| &\iff -(1/2)|\phi'(c)| <\left| \frac{ \phi(x) - \phi(y) }{x - y} \right| - |\phi'(c)| < (1/2)|\phi'(c)| \\ &\iff (1/2)|\phi'(c)||x-y| < \left| \phi(x) - \phi(y) \right| \end{align}

I am not sure how to fix the second $\iff$ as it is only true in real analysis. Also I am not sure how to relax the last inequality to $\leq$

Answer 1

$|z_1-z_2| \geq |z_2|-|z_1|$ for all complex numbers $z_1,z_2$. Hence $|\frac {\phi(x)-\phi(y)} {x-y} -\phi'(c)| \geq |\phi'(c)|- |\frac {\phi(x)-\phi(y)} {x-y}|$ which gives $|\frac {\phi(x)-\phi(y)} {x-y}| \geq \frac 1 2 |\phi'(c)|$.