# Counterexample to a “modified” Banach Fixed Point Theorem?

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?

## 2 Answers

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

• 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?) – Marcus Aurelius Jan 4 '18 at 19:43
• Yes, you can say it fixes only the point at infinity. – Kavi Rama Murthy Jan 5 '18 at 9:33

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