Contraction mapping with no fixed point

2012-02-27T14:35:22Z

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 $\|f(x)-f(y)\| < \|x-y\|$. )

Answer by Bill Johnson for Contraction mapping with no fixed point

2012-02-28T01:22:06Z

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 $x_j>y_j$.

Contraction mapping with no fixed point - MathOverflow

Contraction mapping with no fixed point

Answer by Bill Johnson for Contraction mapping with no fixed point