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.
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 $B$.