Bounded Function on Set S: Proving $|f(z)|\le1$ for All $z\in S$

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Discussion Overview

The discussion revolves around proving that a bounded function \( f \) defined on a specific set \( S \) in the complex plane satisfies the condition \( |f(z)| \le 1 \) for all \( z \in S \). The set \( S \) is defined by certain constraints on the real part and argument of \( z \), and the function is analytic and continuous on the closure of \( S \). Participants explore the implications of boundary conditions and the maximum modulus principle.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants express uncertainty about how to approach the problem and the relevance of the boundary conditions.
  • One participant suggests that the second inequality \( |f(z)| \le e^{\sqrt{x}} \) is unnecessary, arguing that the maximum modulus principle implies \( |f(z)| \le 1 \) based on the boundary condition alone.
  • Another participant counters this by providing a counterexample \( f(z) = e^{z^2} \), which is bounded on the boundary but unbounded in the interior, suggesting the need for the Phragmén-Lindelöf theorem.
  • A participant acknowledges a misunderstanding regarding the boundary conditions, noting that the inclusion of infinity in the boundary may affect the application of the maximum modulus principle.
  • There is a request for clarification on the solution, indicating confusion among participants about the problem's resolution.
  • One participant mentions the Phragmén-Lindelöf principle, suggesting that the problem may be a special case of this theorem.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the necessity of the second inequality or the correct application of the maximum modulus principle. Multiple competing views remain regarding the implications of the boundary conditions and the appropriate theorems to apply.

Contextual Notes

Participants express uncertainty about the implications of the boundary conditions, particularly regarding the inclusion of infinity in the boundary of \( S \). There are unresolved discussions about the application of the maximum modulus principle and the Phragmén-Lindelöf theorem.

Markov2
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Consider the set $S=\left\{ z\in \mathbb{C}:\text{Re}(z)>0,\text{ }\arg (z)\in \left( -\dfrac{\pi }{4},\dfrac{\pi }{4} \right) \right\},$ and a function $f\in H(S)\cap C(\overline S)$ so that for each $z\in\partial S$ is $|f(z)|\le1$ and for all $z=x+yi\in S$ is $|f(z)|\le e^{\sqrt x}.$ Prove that for all $z\in S$ is $|f(z)|\le1.$

Any ideas? Don't know how to start.
 
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インテグラルキラー;488 said:
Consider the set $S=\left\{ z\in \mathbb{C}:\text{Re}(z)>0,\text{ }\arg (z)\in \left( -\dfrac{\pi }{4},\dfrac{\pi }{4} \right) \right\},$ and a function $f\in H(S)\cap C(\overline S)$ so that for each $z\in\partial S$ is $|f(z)|\le1$ and for all $z=x+yi\in S$ is $|f(z)|\le e^{\sqrt x}.$ Prove that for all $z\in S$ is $|f(z)|\le1.$

Any ideas? Don't know how to start.
Well, what are some of your initial thoughts.
 
Don't know really, I wish I knew how to use the arg stuff and the closure stuff.
 
Markov said:
Consider the set $S=\left\{ z\in \mathbb{C}:\text{Re}(z)>0,\text{ }\arg (z)\in \left( -\dfrac{\pi }{4},\dfrac{\pi }{4} \right) \right\},$ and a function $f\in H(S)\cap C(\overline S)$ so that for each $z\in\partial S$ is $|f(z)|\le1$ and for all $z=x+yi\in S$ is $|f(z)|\le e^{\sqrt x}.$ Prove that for all $z\in S$ is $|f(z)|\le1.$

Any ideas? Don't know how to start.

The second inequality is entirely not necessary (unless you meant to ask something else). As the problem stands it is entirely trivial consequence of the maximum-modulos principle. If by approaching the boundary the modulos stays bounded by $1$ then the function everywhere inside the domain must also stay bounded by $1$.
 
ThePerfectHacker said:
The second inequality is entirely not necessary (unless you meant to ask something else). As the problem stands it is entirely trivial consequence of the maximum-modulos principle. If by approaching the boundary the modulos stays bounded by $1$ then the function everywhere inside the domain must also stay bounded by $1$.

This is not true: Take \(f(z)=e^{z^2}\) then \(f=1 on \partial S\) but it's unbounded. This is an application of the Phragmén-Lindelöf theorem, it shouldn't be too hard to find a good text with a proof.
 
Jose27 said:
This is not true: Take \(f(z)=e^{z^2}\) then \(f=1 on \partial S\) but it's unbounded. This is an application of the Phragmén-Lindelöf theorem, it shouldn't be too hard to find a good text with a proof.

I think my mistake is interesting enough to explain. This is what I did. As $\limsup_{z\to \zeta} | f(z) | \leq 1$ for all $\zeta \in \partial S$, I concluded that (by MM) that it follows that $|f(z)|\leq 1$. But I ignored an important case! Here, $\partial S$ includes $\infty$ also (if we view the boundary as belonging to the Riemann sphere topology). Thus, that condition is not true for all points on $\partial S$.
 
Okay, what's actually the solution? I'm confused now. :(
 
Markov said:
Okay, what's actually the solution? I'm confused now. :(

Hold on Markov. I will try to get back to you. I am really sorry because I have limited time and writing out these solutions takes up time.
It is not like I do not want to help you as much as I can it is sometimes hard to.
 
I understand, I'll wait you then!
 
  • #10
TPH, I really need help on this one, don't know how to solve it yet. :(
 
  • #11
One question: Do you know about the Phragmén-Lindelöf principle (look it up in Wikipedia, if you're not familiar with the name)? If so, this is just a special case since \(e^{\sqrt{x}}\leq e^{\sqrt{|z|}}\).
 

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