# 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.

• 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. – Jochen Wengenroth Oct 26 '20 at 15:40
• Can you explain why $T(X_n)\subseteq \overline{\text {Conv}}(T\{x_n, x_{n+1,...}\})$? – Motaka Oct 26 '20 at 15:50
• Continuity of $T$ yields $T(\overline{A})\subseteq \overline{T(A)}$ and linearity $T(\text{Conv}(B))\subseteq \text{Conv}(T(B))$. – Jochen Wengenroth Oct 26 '20 at 17:04
• Non linearity is assumed, unfortunately! – Motaka Oct 26 '20 at 19:18

No, take $$E:=\mathbb{R}$$, $$x_0:=1$$ and $$T$$ any continuous bounded function with $$T(1)=-1$$, $$T(-1)=1$$, $$T(0)=2$$.
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 $$(E,\|\cdot\|)\in\{(\ell^p,\|\cdot\|_p),(c_0,\|\cdot\|_\infty): 1 with standard basis $$\{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.
• I edited my post, for me the boundedness is to take $L=0$ in your answer. – Motaka Oct 27 '20 at 10:17