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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.

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    $\begingroup$ This just follows from $T(X_n)\subseteq \overline{\text{Conv}} T(\{x_n,x_{n+1},\ldots\}) \subseteq X_{n+1}$ -- not really research level. $\endgroup$ – Jochen Wengenroth Oct 26 '20 at 15:40
  • $\begingroup$ Can you explain why $T(X_n)\subseteq \overline{\text {Conv}}(T\{x_n, x_{n+1,...}\})$? $\endgroup$ – Motaka Oct 26 '20 at 15:50
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    $\begingroup$ Continuity of $T$ yields $T(\overline{A})\subseteq \overline{T(A)}$ and linearity $T(\text{Conv}(B))\subseteq \text{Conv}(T(B))$. $\endgroup$ – Jochen Wengenroth Oct 26 '20 at 17:04
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    $\begingroup$ Non linearity is assumed, unfortunately! $\endgroup$ – Motaka Oct 26 '20 at 19:18
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No, take $E:=\mathbb{R}$, $x_0:=1$ and $T$ any continuous bounded function with $T(1)=-1$, $T(-1)=1$, $T(0)=2$.

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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.

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  • $\begingroup$ I edited my post, for me the boundedness is to take $L=0$ in your answer. $\endgroup$ – Motaka Oct 27 '20 at 10:17
  • $\begingroup$ Ah, I supposed just about that. Then I guess Pietro’s answer does it. $\endgroup$ – Jack L. Oct 27 '20 at 10:25

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