Take the 2-minute tour ×
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.

Let $f$ be an analytic function in $\Omega = \{z : \mathcal R z > 0\}$.
Assume $f$ is continuous on $\bar \Omega$ and $|f(z)| \le 1$, when $z \in i\Bbb R$.
Is $|f(z)| \le 1$ for all $z \in \Omega$?

I think the answer is yes, but I don't see how to prove it.
As $f$ is only continuous on $\bar \Omega$, the maximum modulus principle doesn't apply.

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

The problem is that $\Omega$ is unbounded, otherwise the maximum modulus principle would indeed be useful.

A counterexample is $f(z) = \exp(-z^4)$. If $z = iy$, then $f(z) = \exp(-y^4)$, so $|f(z)| \le 1$.

(Note that for this example, it's even the case that $|f(iy)| \to 0$ as $y \to \pm\infty$.)

share|improve this answer
add comment

$$f(z)=e^z$$

Then $|f| = 1$ on $i \mathbb R$.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.