I am interested in constructing the following "counter-example" to the Banach's fixed point theorem.


Let $K=$ {$ g\in L\_1: \|g\|=1, g(\cdot)\ge0 $}.
 Clearly, $K$ is not a compact and $K$ is  ~~not~~  closed.


~~My~~ A question is: is it possible to construct a [*edit*] *nonexpansive* mapping $f: K\to K$ with no fixed point? ( i.e. a mapping $f$ such that [*edit*] for all $x\neq y\in K$ one has $\|f(x)-f(y)\| < \|x-y\|$. )



Contraction mapping with no fixed point
There need not be a fixed point. First note that by composing with a conditional expectation onto the closed span of indicator functions of disjoint sets it is sufficient to build an example on $W:=\{x\in \ell\_1 : x\_i \ge 0, \sum x\_i =1\}$. Given $x\in W$, define $y=Tx \in W$ by $y\_1=0$, $y\_2 = x\_1/2$, and, for $k\ge 2$, $y\_{k+1} = y\_k/2 + x\_{k+1}/2$. It is obvious that $T$ is nonexpansive. The inequality $\|Tx-Ty\|<\|x-y\|$ when $x\not= y$ follows from the fact that if $x\not= y$ there are coordinates $i$ and $j$ s.t. $x\_i<y\_i$ and $x\_j>y\_j$.


