Let $E$ be a Banach space, $T:E\rightarrow E$ a continuous, norm-bounded, and **nonlinear** mapping., and $\{x\_n\}\_{n\in\mathbb N}$ such that $$x\_{n+1}=T(x\_n),\:\forall n\in \mathbb{N}:=\{0,1,\cdots\}.$$
Let $$X\_n=\overline{\text{Conv}}\{x\_n,x\_{n+1}\cdots\}.$$


Let $X\_{\infty}=\bigcap\_{n=0}^{+\infty}X\_n$. **We assume** that $X\_{\infty}$ is not empty, and compact.


So, I'm wondering, what we can or cannot say about the compactness of the sequence $\{x\_n\}\_{n\in \mathbb{N}}$.




---


**Note that:**


This problem/question is under the framework of [Measure of non-compactness](https://core.ac.uk/download/pdf/82269004.pdf) wich acts only on **norm-bounded** sets.


So, here **bounded** means that there exist $M>0$ such that: $$\left \| T(x) \right \|\leq M,\;\forall x\in E.$$


I don't think that this condition is useful in this question.



Compactness of a sequence
In general it may fail to be compact. Consider $E:=L\_2(\mathbb{R})$, and $x\_n:=\chi\_{[n,n+1]}$. Clearly, for any $n\in\mathbb{N}$, all functions in the set $\overline{\text{co}}\{x\_k:k\ge n\}$ have support in $[n,+\infty)$, and in the intersection we get a compact nonempty set, the singleton $X\_\infty=\{0\}$. However no subsequences of $x\_n$ converges strongly, since it already converges weakly to $0$, and $\|x\_n\|\_2=1$.


In general, in any first countable topological space, the compactness of a sequence $(x\_n)\_n$ is equivalent to $\bigcap\_{n\ge k}\overline{\{x\_n:n\ge k\}}$ being non-empty, since the latter is the set $\text{Lim}(x\_n)$ of the limits of all converging subsequences
. In a Banach space, this set may be any separable closed subset.


