Can Analytic Functions in the Unit Disk Meet Specific Modulus Conditions?

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Discussion Overview

The discussion revolves around the properties of analytic functions defined on the unit disk, specifically focusing on a condition involving the modulus of a function related to its value at a specific point. Participants are attempting to prove an inequality involving an analytic function that vanishes at a given point.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • Participants are tasked with proving that for an analytic function \(f\) mapping the disk \(B(0,2)\) to itself and satisfying \(f(1)=0\), the inequality \(\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|\) holds for all \(z\in B(0,2)\.
  • Some participants express confusion about how to utilize the condition \(f(1)=0\) effectively in their proofs.
  • One participant suggests defining a new function based on the initial condition but struggles to formulate a clear approach.
  • Another participant proposes potential definitions for a function, such as \(g(z)=(z-1)f(z)\) or \(g(z)=zf(z)\), but also finds it challenging to progress from there.
  • A later reply challenges the validity of the original inequality by providing a counterexample with the function \(f(z)=a(z-1)\), indicating that the inequality may not hold universally as it implies \(f(0)=0\), which is not necessarily true.

Areas of Agreement / Disagreement

Participants generally express uncertainty and confusion about the problem, with no consensus on a clear method to prove the inequality or on the implications of the condition \(f(1)=0\). There are competing views on the validity of the proposed inequality.

Contextual Notes

Participants note limitations in their understanding of how to apply the initial condition and the implications of the inequality, suggesting that further exploration of definitions and properties of analytic functions is needed.

Markov2
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Let $f:B(0,2)\to B(0,2)$ be an analytic function so that $f(1)=0.$ Prove that for all $z\in B(0,2)$ is $\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|.$

What's the trick here? I don't see it.
 
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インテグラルキラー;490 said:
Let $f:B(0,2)\to B(0,2)$ be an analytic function so that $f(1)=0.$ Prove that for all $z\in B(0,2)$ is $\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|.$

What's the trick here? I don't see it.

You have to have some starting ideas, even failed ones. Give us some of them.
 
I don't know how to use the data $f(1)=0,$ I thought on defining a function, but, I don't get a thing.
 
I can't solve it yet, how to start?
 
Does anyone have any idea? I think a function needs to be defined by using the initial condition, but I can't think of it.
 
Perhaps on taking $g(z)=(z-1)f(z)$ or $g(z)=zf(z),$ but I still can't get it.
 
Markov said:
Let $f:B(0,2)\to B(0,2)$ be an analytic function so that $f(1)=0.$ Prove that for all $z\in B(0,2)$ is $\left| \dfrac{f(z)}{z} \right|\le \left| \dfrac{2(z-1)}{4-z} \right|.$

What's the trick here? I don't see it.

This is false as written: Take \(f(z)=a(z-1)\) for \(a\in \mathbb{R}\) small enough. This is because your inequality implies that \(f(0)=0\) which need not happen.
 

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