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

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The Banach contraction mapping theorem only requires the space to be complete, not compact. en.wikipedia.org/wiki/Banach_fixed-point_theorem. –  Pablo Shmerkin Feb 27 '12 at 14:41
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@Oleg: that $g_n$ is not Cauchy, so it does not disprove $K$ closed. In fact, $K$ is closed (in the $L_1$ metric). –  Gerald Edgar Feb 27 '12 at 14:51
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To save the question, the OP could ask for a "weak contraction" mapping with no fixed points $f:K\to K$ on the same set $K$ defined above, but now only satisfying $\|f(x)-f(y)\| < \|x -y\|$ for all $x\neq y$ in $K$. (On a compact metric space $K$ such maps do have fixed points). –  Pietro Majer Feb 27 '12 at 16:26
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I took the liberty to edit the question and modify it, in order to keep it alive. –  Pietro Majer Feb 27 '12 at 16:38
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@Yemon: yes, for $K:=\mathbb{R}$ it's OK but I meant the $K$ defined in the OP. –  Pietro Majer Feb 27 '12 at 20:50
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1 Answer 1

up vote 4 down vote accepted

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 $x_j>y_j$.

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I meant nonexpansive in its usual sense ($\|Tx - Ty \| \le \|x-y\|$) rather than the way Pietro used it. The main problem was to make the inequality strict when $x\not= y$. –  Bill Johnson Feb 28 '12 at 14:32
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