TITLE: Contraction mapping with no fixed point
QUESTION [5 upvotes]: 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 $\|f(x)-f(y)\| <  \|x-y\|$. )

REPLY [4 votes]: 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 $x_j>y_j$.