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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  "weak" contraction nonexpansive mapping $f: K\to K$ with no fixed point? ( i.e. a mapping $f$ such that  for all $x\neq y\in K$ one has $\|f(x)-f(y)\| < \|x-y\|$. )

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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  "weak" contraction mapping $f: K\to K$ with no fixed point? ( i.e. i.e. a mapping $f$ such that  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\|$. ) 2 added 22 characters in body I am interested in constructing the following "counter-example" to the Banach's fixed point theorem. Let$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\$)

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