Prove that $\exists\ C$ such that $T: C \to C$ with $C=\overline{\text{conv}}T(C)$

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: $\forall x \in C: \ \sup_{y \in C}\left \| x-y \right \|=\text{diam}\ C$.

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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!

• If $C = \varnothing$ is allowed, we can always choose $C=\varnothing$. If that is not allowed, look at $X = \mathbb{R}$, $K = [1,\infty)$, and $T(x) = x+1$. – Daniel Fischer Feb 12 '14 at 16:15
• Thanks Daniel Fischer! Yes, if we take $T(x) = x+1$ then $\left \| Tx-Ty \right \|=\left \| x-y \right \|$. Hence, we have $$T: K = [1,\infty) \to [1,\infty)$$ is nonexpansive. And if $C=\emptyset$ then $$\forall x \in C: \ \sup_{y \in C}\left \| x-y \right \|=\text{diam}\ C=0$$. I'm correct since your hint? – kimtahe6 Feb 12 '14 at 16:43
• I'm sorry when I have a question If $C \ne \emptyset$ then How to show that...? – kimtahe6 Feb 12 '14 at 16:48
• The hint is that in the example $C = \varnothing$ is the only set satisfying the conditions, so if that isn't allowed, as stated, the proposition is false. Is there perhaps an assumption that $K$ is bounded? – Daniel Fischer Feb 12 '14 at 16:50
• Yes, I think so, maybe! And can you show that? Thanks! – kimtahe6 Feb 12 '14 at 16:55

But I don't understand $C=\overline{\operatorname{conv}}T(C)$?

$\operatorname{conv} A$ 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.

and why we need $X$ be a reflective Banach space?

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.

• Yes, you're right. Now I remember that If $X$ be a reflective Banach space then we have $K$ be a weakly compact. I thought about my problem and I try to use: 1/ Theorem: Suppose $K$ is a nonempty, weakly compact, convex subset of a Banach space. Then for any mapping $$T: K \to K$$ there exists a closed convex subset of $K$ which is minimal $T$ - invariant. 2/ Lemma: If $K$ is a noempty, closed, convex and minimal $T$ - invariant set, then $$K=\overline{\text{conv}}T(K)$$ Do you think so? Daniel Fischer – kimtahe6 Feb 13 '14 at 4:05
• "I haven't yet figured out how the latter implies that $C$ is a diameter set". I think that It means we need $K$ has normal structure?? – kimtahe6 Feb 13 '14 at 4:44
• I'm trying to show that "If $C \ne \left \{ \text{one point} \right \}$ implies that $C$ is a diameter set." But I still have no solution!!! :( – kimtahe6 Feb 13 '14 at 7:48
• @ Daniel Fischer: I think that we need to show that: $\text{diam}C=r(C)$. It means $$Z(C)=C$$ We have $C=\overline{\text{conv}}T(C)$. And $r(C,Tx) \le r(C) \implies Tx \in Z(C), \forall x \in Z(C)$??? – kimtahe6 Feb 13 '14 at 8:34
• I suppose you have a general theorem that $Z(C)$ is convex and closed? (If not, it's not difficult to prove.) Then all you need to show is what you did show, $T(Z(C)) \subset Z(C)$. Yes, correct, very good. – Daniel Fischer Feb 13 '14 at 14:48