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.

Given a complex function $G(z)$, $z=x+iy, x,y\in \mathbb R$ which is analytic and bounded in the upper half-plane, i.e.,

$$|G(z)|\leq C, \forall z\in \mathbb C^{+}$$ does it imply that $G(x)$ is bounded on the real line (if we know that $G(x)$ has no poles on $\mathbb R$), i.e., $$|G(x)|\leq C_{o}, \forall x\in \mathbb R$$

EDIT: $\mathbb C^{+}=\{z\in \mathbb C, \text{Im}(z)>0\}$.

share|improve this question
    
Is $\mathbb{C}^+$ the set for which $\text{Re}(z) > 0$? –  Antonio Vargas Mar 21 '12 at 17:03
    
Sorry, I meant to type $\text{Im}(z) > 0$, of course. Ok. –  Antonio Vargas Mar 21 '12 at 17:19
    
Yeah, me too!! I fixed it above. Thanks –  Paul Mar 21 '12 at 17:27
    
Strange way to ask this. If your function extends continuously to the real line, then of course it is bounded there, by continuity. Now, there are "lacunary functions" such as $$ \sum z^{2^k} $$ which cannot be analytically continued past a natural boundary, in this case the unit circle. Also, in this case, doubtless no continuous extension to the closed unit disk is possible, by the maximum principle (no finite value at $1$ for a clear instance). However, you are asking about bounded in the interior, and it may be possible to extend continuously only. Not sure yet. –  Will Jagy Mar 21 '12 at 21:52

3 Answers 3

I borrowed some books, the most important being Boundary Behavior of Conformal Maps by Christian Pommerenke. Much of this subject goes back to Constantin Caratheodory.

The main thing I want to emphasize is that there is no guarantee that a bounded function on the open upper half plane extends continuously to the real line. The Pommerenke book concentrates on the unit disk, so here is some terminology for the theorem: he writes $\mathbb D$ for the open unit disk, then $\bar{\mathbb D}$ for the closed unit disk, finally $\mathbb T$ for the unit circle. We have Theorem 2.1 on page 20:

Let f map $\mathbb D$ conformally onto the bounded domain $G.$ Then the following four conditions are equivalent:

(i) $f$ has a continuous extension to $\bar{\mathbb D}$

(ii) $\partial G$ is a curve, that is $\partial G = \{\varphi(\zeta): \zeta \in \mathbb T \}$ with continuous $\varphi$

(iii) $\partial G$ is locally connected

(iv) $\mathbb C \backslash G$ is locally connected.

What this says to me is that there is no guarantee of a continuous extension to the boundary, but examples would be fairly difficult to describe, and it is possible that nobody has ever written one down. I would be very happy to learn of a reference where somebody gave a single example.

share|improve this answer

THE ANSWER IS NO.

Page numbers are from Boundary Behavior of Conformal Maps by Christian Pommerenke

The Pommerenke book concentrates on the unit disk, so here is some terminology for the theorem:

(page 1): he writes $\mathbb D$ for the open unit disk, then $\bar{\mathbb D}$ for the closed unit disk, finally $\mathbb T$ for the unit circle.

(page 30): A bounded simply connected domain $G$ in the complex plane comes equipped with a set called its prime ends, the set denoted $P(G).$

(page 32): For a prime end $p \in P(G),$ there is a set $I(p) \subseteq \mathbb C$ called the impression of $p.$ The impression of a prime end is a non-empty compact connected set.

(page 30): Theorem 2.15. Let f map $\mathbb D$ conformally onto the bounded simply connected domain $G,$ by the Riemann Mapping Theorem. There is an induced bijection $\hat{f}$ with $$ \hat{f} : \mathbb T \rightarrow P(G). $$

(page 35): Corollary 2.17(ii) with $\zeta \in \mathbb T,$ the mapping $f$ extends continuously to $\zeta,$ with limit $a,$ if and only if $$ I( \hat{f}(\zeta)) = \{a\}, $$ the singleton set consisting of the point $a.$

(page 35): Exercise 2.5.2. Let $$ G = \mathbb D \backslash \{ (1 - e^{-t})e^{it} : 0 \leq t < \infty \}. $$ Then $G$ has a prime end $p$ with $I(p) = \mathbb T.$ That is, $I(p)$ is not a single point.

Thus, there is some point $\zeta \in \mathbb T,$ with $\zeta = \hat{f}^{-1}(p), $ at which $f$ does not extend continuously. Put another way, $f(\zeta)$ cannot be defined in a way that makes $f$ continuous at $\zeta.$

To return to the upper half plane, let us call it $\mathbb U.$ Take a Möbius transformation $m$ that maps the real line to $\mathbb T$ minus one point of course, does NOT map $\infty$ to the point $\zeta$ where $f$ is discontinuous, and maps $\mathbb U$ conformally onto $\mathbb D.$ Then the function composition $$ f \circ m : \mathbb U \rightarrow G $$ and is thus bounded, but cannot be extended continuously at $m^{-1}(\zeta)$ on the real line.

share|improve this answer
    
I'm ashamed to say that I only skimmed this when it was posted and did not realize that it actually directly answered the OP's question in the negative. Nice work! –  Antonio Vargas Apr 1 '12 at 3:32
    
@Antonio, thanks. The 1992 Pommerenke book is very good. It just took me a few days to put it together. Evidently everything, perhaps with different examples, is in a 1913 paper by Caratheodory in German. I see, consecutive papers same issue, Mathematische Annalen volume 73, pages 305-320 and then 323-370. –  Will Jagy Apr 1 '12 at 4:37

The boundedness is not an issue, if it extent continuously, since your bound extent continuously.

Often function extent to the boundary under certain conditions in a certain measure theoretic sense, (Fatou's theorem and similar stuff for $H^p$ spaces)

More over Phragmen-Lindeloeff gives you a necessary condition, for the function to extent continuously.

Note that this is the important question, not the boundedness.

share|improve this answer
    
What do you mean by this? Are you saying that the above result is true and we can prove it by the above theorems! –  Paul Mar 21 '12 at 19:57
    
Phragmen-Lindelof does not apply, en.wikipedia.org/wiki/… –  Will Jagy Mar 21 '12 at 20:34
    
I was actually answering something else the other night... I hope I do now spell out, what I meant to say. –  plusepsilon.de Mar 22 '12 at 8:59

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.