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

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The discussion revolves around proving that for an analytic function f defined on the unit disk with f(1)=0, the inequality |f(z)/z| ≤ |2(z-1)/(4-z)| holds for all z in the disk. Participants express confusion about how to utilize the condition f(1)=0 and suggest defining new functions like g(z)=(z-1)f(z) or g(z)=zf(z) to approach the problem. However, one contributor points out a flaw in the original statement by providing a counterexample with f(z)=a(z-1), indicating that the inequality could lead to incorrect conclusions about f(0). The conversation highlights the challenges in applying the modulus condition effectively within the constraints of analytic functions. Overall, the discussion emphasizes the need for careful consideration of the implications of the given conditions in complex analysis.
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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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