Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $f$ is a holomorphic function on $\Omega^+$ (an open subset of the upper complex plane) that extends continuously to $I$ (a subset of $\mathbb{R}$). Let $\Omega^-$ be the reflection of $\Omega^+$ across the real axis. We can extend $f$ to be defined on $\Omega^+ \cup I \cup \Omega^-$ with $f(z) = \overline{f(\overline{z})}$ if $z \in \Omega^-$. And $f$ will be analytic on $I$. This is frequently proved using Morera's Theorem.

But is there a proof that $f$ has a derivative on $I$ based on showing that the derivative is bounded and hence must exist. First showing that the derivative is bounded around $c$. In other words $f'$ is bounded on the set $\{z | \operatorname{Im}(z)>0, |z-c| < \epsilon\}$. And then somehow showing that $f$ has a linear approximation at $c$, restricted to $\operatorname{Im}(z) \ge 0$).

share|cite|improve this question
Having linear approximation on half-planes is not enough. Consider the real-variable example: $f(x)=|x|$ has such approximations on $[0,\infty)$, and on $(-\infty,0]$. The derivative is bounded, too. – user53153 Jan 27 '13 at 2:18
Yes but here there is the constraint that $f$ is holomorphic on $\Omega^+$ which prevents such bad behavior on the real axis. – user782220 Jan 27 '13 at 3:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.