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The Banach fixed point theorem has the following statement

THEOREM ( Banach contraction principle). Let $(Y,d)$ be a complete metric space and $F:Y\to Y$ be contractive . Then $F$ has a uniqe fixed point $u$, and $F^n(y)\to u$ for each $y\in Y$.

This statement is from the excellent book Fixed Point Theory written by Andrzej Granas and James Dugundji. In this book several generalizations of this theorem are given. Several of them are to relax the hypothesis of classic contraction to another contraction hypothesis that is weaker. For example, $d(F (x) ,F (y)) \le \phi(d (x, y))$ for all $x,y\in Y$ with $ \lim_{n \to \infty}\phi^n (d (x, y )) = 0$ ( see p. 15).

QUESTION Let $ (M, d) $ be a complete metric space and $ (M\times M, \mathcal{M}\otimes\mathcal{M}, \mu) $ be a probability space. Suppose that $F:M\to M$ is continuous and a.s. contractive i.e. there is a $0<\lambda<1$ such that $$ \mu\left\{ (x,y) :\quad d(F (x) ,F (y)) \le \lambda \cdot d(x, y)\quad \right\}=1. $$ So is there a counterexample to the claim that $ F $ has a unique fixed point? If not, a reference in the form of a book that treats the subject would be well accepted answer.

My attempt. Let $\delta_{(x, y)} $ the measure of Dirac concentrated in $ (x, y) $. The above question has a positive answer to the dirac measure (equivalent to the deterministic case). I have tried to prove my assertion probability measures of the form $$ \mu=\dfrac{1}{1+\epsilon}\left(\delta_{(x,y)}+\epsilon\cdot \nu \right) $$ for a $ \epsilon>0 $ small enough and any probability measure $\nu$ on $(M\times M, \mathcal{M}\otimes\mathcal{M})$.

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    $\begingroup$ What about $\mu$ being Dirac measure? I'd at least ask $\mu$ to be positive on open sets. Still, for the Lebesgue measure on $[0,1]$ consider a contraction mapping with a fixpoint $x^*$ and change $F(x^*)$, then there is no fixpoint and $\lambda(\{x\}) = 0$. If you ask $F$ to be continuous a priori, and $\mu$ to be positive on open sets then I think your claim shall be correct. $\endgroup$ – Ilya Jul 24 '14 at 16:47
  • $\begingroup$ Ok, the Lebesgue measure example does not work, other things I think are true. $\endgroup$ – Ilya Jul 24 '14 at 17:10
  • $\begingroup$ @Ilya My mistake, I forgot to put the hypothesis of continuity. With hiótese the existence of continuity follows easily from other fixed point theorems. I'll add the hypothesis of the continuity issue. My interest is in the uniqueness of the fixed point. $\endgroup$ – MathOverview Jul 24 '14 at 17:43
  • $\begingroup$ What about avoiding Dirac measure $\mu$? $\endgroup$ – Ilya Jul 24 '14 at 18:29
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    $\begingroup$ @PhoemueX: if $L$ can depend upon $x,y$ and is not bounded away from $1$ then even if the inequality holds everywhere, $F$ fails to have a fixpoint in general. $\endgroup$ – Ilya Jul 24 '14 at 21:31

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