Contraction mapping with no fixed point - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T08:38:39Z http://mathoverflow.net/feeds/question/89670 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89670/contraction-mapping-with-no-fixed-point Contraction mapping with no fixed point Oleg 2012-02-27T14:35:22Z 2012-02-28T08:32:51Z <p>I am interested in constructing the following "counter-example" to the Banach's fixed point theorem.</p> <p>Let $K=$ {$g\in L_1: \|g\|=1, g(\cdot)\ge0$}. Clearly, $K$ is not a compact and $K$ is <s> not </s> closed.</p> <p><s>My</s> A question is: is it possible to construct a [<em>edit</em>] <em>nonexpansive</em> mapping $f: K\to K$ with no fixed point? ( i.e. a mapping $f$ such that [<em>edit</em>] for all $x\neq y\in K$ one has $\|f(x)-f(y)\| &lt; \|x-y\|$. )</p> http://mathoverflow.net/questions/89670/contraction-mapping-with-no-fixed-point/89716#89716 Answer by Bill Johnson for Contraction mapping with no fixed point Bill Johnson 2012-02-28T01:22:06Z 2012-02-28T01:22:06Z <p>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 <code>$W:=\{x\in \ell_1 : x_i \ge 0, \sum x_i =1\}$</code>. 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 $\|Tx-Ty\|&lt;\|x-y\|$ when $x\not= y$ follows from the fact that if $x\not= y$ there are coordinates $i$ and $j$ s.t. <code>$x_i&lt;y_i$</code> and $x_j>y_j$.</p>