I have a problem:
Let $K$ be a nonempty, closed, bounded and convex subset of reflective Banach space $X$. Suppose that $$T:K \to K$$ is nonexpansive.
Prove that $\exists\ C$ such that $$T: C \to C$$ where $C=\overline{\text{conv}}T(C)$ and if $C \ne \left \{ \text{one point} \right \}$ then $C$ is a diametral set.
It means that: $\forall x \in C: \ \sup_{y \in C}\left \| x-y \right \|=\text{diam}\ C$.
=====================================
I have trouble when I try to find a solution to the problem.
I'll write here:
Consider the family $\mathcal{F}$ of all nonempty, closed and convex (thus weakly compact) subset of $K$ which are $T$ - invariant, and order this family by set inclusion: For $K_1, \ K_2 \in \mathcal{F},\ K_1 \le K_2$ provided $K_2 \subset K_1$. By Zorn's Lemma, $\mathcal{f}$ has a minimal $C$.
We'll show that $C=\overline{\text{conv}}T(C)$:
We consider $D=\{z \in C: C \subset \overline{B}(z,r)\} \ne \emptyset$.
Take $z \in D$, because $T$ is nonexpansive, we have: $$T(C) \subset B(Tz,r)$$ Hence, $\overline{\text{conv}}T(C) \subset B(Tz,r)$.
Obviously $\overline{\text{conv}}T(C)$ is closed, convex in $K$ then weakly compact. We have $$\overline{\text{conv}}T(C) \subset \overline{\text{conv}}C=C$$ then $$T(\overline{\text{conv}}T(C))\subset T(C) \subset \overline{\text{conv}}T(C)$$
Therefore, $\overline{\text{conv}}T(C) \in \mathcal{F}$.
Because $\overline{\text{conv}}T(C) \subset C$ and $C$ is minimal then $C=\overline{\text{conv}}T(C)$.
How to prove that "if $C \ne \left \{ \text{one point} \right \}$ then $C$ is a diametral set"??? Any help will be appreciated. Thanks!
$\operatorname{conv} A$ is the convex hull of $A$, and $\overline{\operatorname{conv}} A$ denotes the closed convex hull of $A$, that is, $\overline{\operatorname{conv} A}$. So the requirement is that $C$ be equal to the closure of the convex hull of its $T$-image.
Because in a reflexive space, every closed convex and bounded subset is weakly compact (compact in the weak topology).
The weak compactness guarantees the existence of a minimal nonempty closed convex $T$-invariant subset, that is, the existence of a nonempty closed convex $C\subset K$ with $T(C)\subset C$ such that no proper nonempty subset of $C$ is closed, convex, and $T$-invariant.
The minimality then guarantees $C = \overline{\operatorname{conv} T(C)}$, since for a nonempty closed convex $T$-invariant $C\subset K$ we have $\overline{\operatorname{conv} T(C)} \subset C$, and since then $T(\overline{\operatorname{conv} T(C)}) \subset T(C) \subset \overline{\operatorname{conv} T(C)}$, the subset $\overline{\operatorname{conv} T(C)}$ of $C$ is also nonempty, closed, convex and $T$-invariant. Minimality says it's $C$.
Such a minimal $C$ either consists of a single fixed point of $T$, or $T$ is fixed-point free on $C$.
In the latter case, $C$ is a diametral set,
$$\sup_{y\in C} \lVert x-y\rVert = \operatorname{diam} C$$
for all $x\in C$ (filling in the details of the proof sketch in Kazimierz Goebel and Brailey Sims, More on minimal invariant sets for nonexpansive mappings):
Suppose there is an $x \in C$ with $\sup\limits_{y\in C} \lVert x-y\rVert = r < \operatorname{diam} C$. Let
$$C_r = \left\{ x\in C : \sup_{y\in C}\lVert x-y\rVert \leqslant r\right\}.$$
Then $C_r \subsetneq C$, since there exist $a,b\in C$ with $\lVert a-b\rVert > r$. $C_r$ is closed, since if $\sup\limits_{y\in C} \lVert x-y\rVert = r+\delta$ with $\delta > 0$, we have $\sup\limits_{y\in C}\lVert z-y\rVert \geqslant r + \delta/2$ for all $z \in C\cap B_{\delta/2}(x)$, and $C_r$ is convex, since for $x_1,x_2\in C_r$ and $\lambda\in [0,1]$ we have
$$\lVert (1-\lambda)x_1 + \lambda x_2 - y\rVert \leqslant (1-\lambda)\lVert x_1-y\rVert + \lambda \lVert x_2-y\rVert \leqslant (1-\lambda)r+\lambda r = r$$
for all $y\in C$.
Also, $C_r$ is $T$-invariant: Let $x\in C_r$ and $y \in \operatorname{conv} T(C)$. Then there are $\lambda_1,\dotsc,\lambda_n \in [0,1]$ with $\sum\limits_{i=1}^n \lambda_i = 1$ and $y_1,\dotsc,y_n\in C$ with $y = \sum\limits_{i=1}^n \lambda_i T(y_i)$, and hence
$$\lVert T(x)-y\rVert = \left\lVert T(x)-\sum_{i=1}^n \lambda_i T(y_i)\right\rVert \leqslant \sum_{i=1}^n \lVert T(x) - T(y_i)\rVert \leqslant \sum_{i=1}^n \lVert x-y_i\rVert \leqslant r.$$
Since $\operatorname{conv} T(C)$ is dense in $C$, it follows that $\lVert T(x)-y\rVert \leqslant r$ for all $y\in C$, whence $T(x)\in C_r$.
But that contradicts the minimality of $C$ among the closed convex $T$-invariant subsets. Hence $C$ is diametral if it contains more than one point.