Can We Say "${y}_{n}=T{y}_{n-1}" in a Complete Metric Space?

[ozkan12]

Let $\left(X,d\right)$ be a complete metric space, and in this space we define iteration sequence by ${x}_{n}=T{x}_{n-1}={T}^{n}{x}_{0}$ ${x}_{0}$ is arbitrary point in $X$...Also, let $\left\{{y}_{n}\right\}$ be a arbitrary sequence in $X$ but $\left\{{y}_{n}\right\}$ is not iteration sequence...İn this case, Can we say that ${y}_{n}=T{y}_{n-1}$ ?

Thank you for your attention...:)

 

[Ackbach]

If $\{y_n\}$ is an arbitrary sequence, and $T$ is already given, there's definitely no way you could conclude that $y_n=Ty_{n-1}$. If $y_n=Ty_{n-1}$ were true (which it isn't), then $\{y_n\}$ would be an iteration sequence.

 

[math771]

No, we cannot say that ${y}_{n}=T{y}_{n-1}$ in this case. The definition of an iteration sequence specifically requires that ${x}_{n}=T{x}_{n-1}$, meaning that each term in the sequence is obtained by applying the transformation $T$ to the previous term. This may not be the case for the arbitrary sequence $\left\{{y}_{n}\right\}$, so we cannot make the assumption that ${y}_{n}=T{y}_{n-1}$.