Triangle Inequality and Convergence of ${y}_{n}$

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Discussion Overview

The discussion revolves around the convergence of a sequence ${y}_{n}$ in a metric space X, particularly in relation to the triangle inequality and its implications for convergence to a point ${x}^{*}$. Participants explore the conditions under which convergence can be asserted and the relationships between different sequences.

Discussion Character

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

Main Points Raised

  • One participant questions whether ${y}_{n}$ and ${y}_{m+1}$ refer to the same sequence, suggesting that ${y}_{n}$ is arbitrary and may not be directly related to ${y}_{m+1}$.
  • Another participant clarifies that the sequence ${y}_{n}$ is not arbitrary but satisfies specific conditions that allow it to approach a limit, referencing a particular inequality that governs the convergence behavior.
  • A participant cites an article and asks for clarification on how the limit $\lim_{{n}\to{\infty}} {y}_{n}={x}^{*}$ is established, presenting a mathematical expression that they believe supports their inquiry.
  • There is a reference to the triangle inequality being used in the proof of Theorem 3.3, with a specific mathematical expression provided to illustrate its application.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the sequence ${y}_{n}$ and its relationship to ${y}_{m+1}$. There is no consensus on whether the limit can be established based solely on the triangle inequality, and the discussion remains unresolved regarding the conditions necessary for convergence.

Contextual Notes

Participants highlight the need for clarity on the definitions and relationships between the sequences involved, as well as the assumptions underlying the convergence claims. The discussion reflects uncertainty about the implications of the triangle inequality in this context.

ozkan12
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Let ${y}_{n}$ be a arbitrary sequence in X metric space and ${y}_{m+1}$ convergent to ${x}^{*}$ in X...İn this case by using triangle inequality can we say that ${y}_{n}\to {x}^{*}$
 
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What's the relationship between $y_n$ and $y_{m+1}$? Is it the same sequence? On the one hand, it would seem not, because $y_n$ is arbitrary. On the other hand, the notation would indicate to me that they are the same sequence. Is it a subsequence?
 
Very helpful link, thank you!

The sequence $\{y_n\}$ is by no means arbitrary. It satisfies a very particular condition. It's not an iterative sequence, because $y_{n+1}\not\equiv T y_n$. However, the condition that the sequence satisfies ensures that $y_{n+1}$ is getting "closer and closer" - in the modular sense - to the quantity $Ty_n$ as $n\to\infty$. That's what
$$w_{\lambda}(y_{n+1},Ty_n)\le \varepsilon_n$$
guarantees.

To answer the question in the OP, there is a triangle inequality invoked in the proof of Theorem 3.3. It's used in the third line down from the word "Proof", where you have
$$w_{\frac{\lambda\cdot m}{m}}(T^{m+1}x,y_{m+1}) \le
w_{\frac{\lambda(m-1)}{m}}(T^{m+1}x,Ty_m)+w_{\frac{\lambda}{m}}(Ty_m,Ty_{m+1}).$$
 
Dear Ackbach

İn this case, How we say that $\lim_{{n}\to{\infty}} {y}_{n}={x}^{*}$ in this article...? I found something but I am not

sure...I wrote it..

${w}_{\lambda}\left({y}_{n},{x}^{*}\right)={w}_{\frac{\lambda}{2}}\left({y}_{n},{T}^{n}x\right)+{w}_{\frac{\lambda}{2}}\left({T}^{n}x,{x}^{*}\right)$
$\le\sum_{i=0}^{n-1}{k}^{n-1-i}{\varepsilon}_{i}+{w}_{\frac{\lambda}{2}}\left({T}^{n}x,{x}^{*}\right)$

By taking limit as $n\to\infty$ we get $\lim_{{n}\to{\infty}}{w}_{\lambda}\left({y}_{n},{x}^{*}\right)$...İs this true ?

Thank you for your attention :)
 

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