MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 4 deleted 6 characters in body

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] "weak" contraction 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 $\|f(x)-f(y)\| < \|x-y\|$. )

show/hide this revision's text 3 m

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] "weak" contraction mapping $f: K\to K$ with no fixed point? ( i.e. i.e. a mapping $f$ such that [edit] for all $x,y\in x\neq y\in K$ one has $\|f(x)-f(y)\|\leq \lambda \|f(x)-f(y)\| < \|x-y\|$ for some $\lambda<1$)|x-y\|$. )

show/hide this revision's text 2 added 22 characters in body

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 question is: is it possible to construct a contraction mapping $f: K\to K$ with no fixed point? (i.e. a mapping $f$ such that for all $x,y\in K$ one has $\|f(x)-f(y)\|\leq \lambda \|x-y\|$ for some $\lambda<1$)

show/hide this revision's text 1