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

[ozkan12]

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 ${y}_{n}\to {x}^{*}$

 

[Ackbach]

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?

 

[ozkan12]

Dear Ackbach,

İn an article, I see this case...I send link of article...İn article, in proof of theorem 3.3 last part, I see this...I hope you can explain...Best wishes...

http://www.fixedpointtheoryandapplications.com/content/pdf/1687-1812-2011-93.pdf

 

[Ackbach]

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
$$w_{\frac{\lambda\cdot m}{m}}(T^{m+1}x,y_{m+1}) \le
w_{\frac{\lambda(m-1)}{m}}(T^{m+1}x,Ty_m)+w_{\frac{\lambda}{m}}(Ty_m,Ty_{m+1}).$$

 

[ozkan12]

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 :)