0

Open Mapping Theorem, with bounds

Suppose that $f \in H(V)$ and $|f''(z)| \leq B \,\, \forall z \in V$. Suppose that $z_0 \in V$ and $f'(z_0) \neq 0$. If $r > 0$ is small enough that $\overline{D(z_0, r)} \subset V$ and also $r < \frac{2|f'(z_0)|}{B}$ then $D(f(z_0), \delta) \subset f(D(z_0, r))$ for $\delta = r|f'(z_0)|-\frac{Br^2}{2}$.

In the proof I have that:

Since $|f''(z)| \leq B, \text{we must have} |f'(z)-f'(z_0)| \leq |z-z_0|B \,\,\forall z \in D(z_0, r)$.

How I get this result? Is something like mean value theorem over complex numbers or how?

MathLearner
  • 750
  • Yes, see for example https://math.stackexchange.com/a/208636/42969 (applied to $\phi = f'$). – Martin R Oct 19 '23 at 19:51
  • But if $V$ is not connected and don't exists a curve between $z$ and $z_0$? Exists another method ? – MathLearner Oct 19 '23 at 19:53
  • As I understand it, your question is about $D(z_0, r)$ which is convex, so that two points can be connected with a straight line. For general domains, such an estimate may not hold. – Martin R Oct 19 '23 at 19:55

0 Answers0