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The Banach theorem states that if a (self) map on a complete metric space is Lipschitz with ratio $< 1$, it has a unique fixed point. What about modifying the hypotheses to say that the only condition the map satisfies is that $d(x, y) > d(f(x), f(y))$ where $x, y$ are in the domain? (I think this condition is known as "weakly contracting" I think that this type of map would fix a point if the metric space were compact, but what about in the general case where the metric space is not compact? Is there a counterexample?

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$$ \frac{3x + \sqrt {x^2 + 1}}{4} $$

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$X=[1,\infty)$ with the usual metrix, $f(x)=x+e^{-x}$. Use Mean Value Theorem to verify that the hypothesis is satisfied.

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  • $\begingroup$ I like this example. It "almost" has a fixed point because it asymptotically approaches y=x, but the weakness of the contraction condition makes it not approach the fixed point anywhere on the standard real line. (But if the extended real numbers make a good metric space which still makes this map contracting, is it true that this map uniquely fixes a point at infinity?) $\endgroup$ Jan 4, 2018 at 19:43
  • $\begingroup$ Yes, you can say it fixes only the point at infinity. $\endgroup$ Jan 5, 2018 at 9:33

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