Let $E$ be a Banach space, $T:E\rightarrow E$ a continuous bounded **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 $$X\_n=\overline{\text{Conv}}\{x\_n,x\_{n+1}\cdots\}.$$


I want to prove that $$T\big(\bigcap\_{n=0}^{+\infty}X\_n\big)\subseteq \bigcap\_{n=0}^{+\infty}X\_n.$$




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**Edit:** Unfortunately, the boundedness condition on $T$ created some confusion here. This question is under the framework of [Measure of non-compactness](https://en.wikipedia.org/wiki/Measure_of_non-compactness) wich is acting 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 to prove this inclusion, this is why I had to NOT mention it in this question.



Inclusion of infinite intersection
No, take $E:=\mathbb{R}$, $x\_0:=1$ and $T$ any continuous bounded function with $T(1)=-1$, $T(-1)=1$, $T(0)=2$.


