Compactness of a sequence

Question

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 $\{x_n\}_{n\in \mathbb{N}}$.

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Note that:

This problem/question is under the framework of Measure of non-compactness 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.

Answer 1

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.