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Recently under the tutelage of a Mentor, we were told to complete this example where contractiveness plays a pivotal role in the Contraction mapping principle. I already have an example by taking $E=c_0$ and $B=\{x=\{x_n\}\in c_0:\,0\leq x_n\leq 1 \}$ and defining $T:B\to B$ by $$T(x_1,x_2,\cdots)=(1,x_1,x_2,\cdots).$$ Then, $B$ is non-empty, closed, bounded and convex. Also, $T$ maps $B$ into $B,\;$ and by what I showed here, $T$ is non-expansive. Yet, it fails to have a fixed point in $B.$

Question

The kind of example that is expected from me, is that from Contemporary Mathematics, Volume 18. Here is the hint that was given to me. Take $E=l_1$, $B=\{\bar{x}\in l_1 ;\,\|x\|\leq 1\}$ and define $T:B\to B$ by $\_\_\_\_\_\_\_\_\_\_$. Then, $T$ does not necessarily have a fixed point in $B.$

I have tried to get access to the book but couldn't. Is there anyone who knows this example? Kindly share with me. The right link to the book might also be of help. Thanks.

Omojola Micheal
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1 Answers1

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I don't know if this is the example that you are looking for. Consider the closed bounded convex set $B=\{x\in l_1 ;\,x_i\geq 0\;\; \forall i, \|x\|= 1\}$ then the map $T:B\to B$ defined by $$T(x_1,x_2,\cdots)=(0,x_1,x_2,\cdots)$$ is an isometry which fails to have a fixed point in $B$.

Robert Z
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  • Thanks, Robert Z! I also have this example but I would be glad if you can maintain the condition on $B$. Anyway, plus one! – Omojola Micheal Apr 22 '19 at 13:10
  • @OmojolaMicheal I revised my answer but $B$ is not the full unit ball yet. BTW I found a copy of Contemporary Mathematics 18 - Fixed Points and Nonexpansive Mappings-AMS (1983) but I did not find any counterexample related to your question. – Robert Z Apr 22 '19 at 14:53
  • You are indeed, a Mentor! Thank you for taking such great steps because of me. It makes me remember Jesus. Happy Easter. – Omojola Micheal Apr 22 '19 at 15:25
  • It could be that the person missed his reference. Maybe he missed the volume number. – Omojola Micheal Apr 22 '19 at 15:27
  • \begin{align}\Vert T(x)-T(y) \Vert &=\Vert (1-\sum_{i=1}^{\infty}x_i,x_1,x_2,\cdots)-(1-\sum_{i=1}^{\infty}y_i,y_1,y_2,\cdots) \Vert\ &=\Vert (\sum_{i=1}^{\infty}y_i-x_i,x_1-y_1,x_2-x_2,\cdots) \Vert\ &\leq \sum_{i=1}^{\infty}\Vert (y_i-x_i,x_1-y_1,x_2-x_2,\cdots)\Vert\ &= \sum_{i=1}^{\infty}\Vert y_i-x_i\Vert.\end{align} – Omojola Micheal Apr 22 '19 at 15:53
  • The above is what I'm getting. Can you please, throw more light as to how it is non-expansive? – Omojola Micheal Apr 22 '19 at 15:54
  • I am also trying to see that it has no fixed point. Suppose it does, then $x_i=1-\sum^{\infty}{i=1}x_i.$ So that $|1|\leq |x_i|+\sum^{\infty}{i=1}|x_i |=1+k$ where $k>0.$ Contradiction. I guess, I'm mixing up some things. Can you guide my reasoning? – Omojola Micheal Apr 22 '19 at 16:02
  • @OmojolaMicheal Sorry, my fault. I have to go back to my first version. – Robert Z Apr 22 '19 at 16:40
  • No problems at all. You have really tried for me. – Omojola Micheal Apr 22 '19 at 17:06