Inclusion of infinite intersection 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 $$T\big(\bigcap_{n=0}^{+\infty}X_n\big)\subseteq \bigcap_{n=0}^{+\infty}X_n.$$

Edit: Unfortunately, the boundedness condition on $T$ created some confusion here. This question is under the framework of 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.
 A: No, take $E:=\mathbb{R}$, $x_0:=1$ and $T$ any continuous bounded function with $T(1)=-1$, $T(-1)=1$, $T(0)=2$.
A: While attempting to answer this question, I recalled that the term “bounded” can be pretty confusing in normed vector spaces if not clarified; in general they all take the form $$\|Tx\|\le L\|x\|+M\,,$$ where $L,M$ are some nonnegative real numbers and for all $x$ in the domain of the definition. In the context of affine operators, bounded mapping (which is also to mean continuous (affine) mapping), this is generally the understanding (with $M=\|T(0)\|$); in particular, for bounded linear operators, this becomes the familiar definition $\|Tx\|\le L\|x\|$, with $L$ being the operator norm; as explained Jochen’s comment above, your claim holds for bounded linear operators. A number of authors translate this definition naturally to bounded ‘nonlinear’ operators, with some requiring the generic definition above or specially to the case when $M=0$.
The case when $L=0$ in the above definition is usually the definition of bounded function (which can be confusing to the case for affine mappings unless the domain is restricted to the unit ball) and I suppose Pietro’s negative solution is for this definition. I will give a negative solution for the general definition above, though in my case $LM\ne 0$ and I don’t have a counterexample when $M=0$ is required (though I’m thinking you are rather interested in either the case $L=0$ or $M=0$).
Indeed let $(E,\|\cdot\|)\in\{(\ell^p,\|\cdot\|_p),(c_0,\|\cdot\|_\infty): 1<p<\infty\}$ with standard basis $\{e_n\}_n$. Now consider the mapping
$$T(\sum_na_ne_n):=(1-\sup|a_n|)e_1+ \sum_na_ne_{n+1}\,.$$
Writing $x:=\sum_na_ne_n $ and $y:=\sum_nb_ne_n $, then observe that
\begin{align}
\|Tx-Ty\|&=\left\|(\sup_n|b_n|-\sup_n|a_n|)e_1+ \sum_n(a_n-b_n) e_{n+1}\right\|\\
&\le\left|\sup_n|b_n|-\sup_n|a_n|\right|+\left\| \sum_n(a_n-b_n) e_{n+1}\right\|\\
&=|\|y\|_\infty-\|x\|_\infty|+\|x-y\|\\
&\le\|x-y\|_\infty+ \|x-y\|\\
&\le 2 \|x-y\|\,,
\end{align}
hence $T$ is Lipschitz; in particular, it is (uniformly) continuous; by the triangle inequality, we obtain $$\|Tx\|\le 2\|x\|+\|T(0)\|=2\|x\|+1\,.$$
Now, letting $x_n:=e_n$, we observe that $Te_n=e_{n+1}$.  We know that $e_n\rightharpoonup 0$, and by weak closure of closed convex sets (Mazur’s Lemma), it follows that that $0\in X_n$ for all $n$—that is, $0\in \bigcap_n X_n$; however, it is clear that $T(0)=e_1\notin X_2\supset\bigcap_n X_n$, which contradicts the claim.
