# Counterexamples to Banach Fixed Point (Banach's Contraction) Theorem with relaxed inequalities?

Banach Fixed Point theorem states: Let $(X,d)$ be a complete metric space. Suppose that $f:X→X$ is a strong contraction, i.e. there exists $q ∈ [0, 1)$ such that

$d(f(x),f(y))$ $\le$ $q$ $d(x,y)$, then there is a unique point $x_0∈X$ s.t. $f(x_0)=x_0$

My questions are:

1- If we allow $q$ to be equal to $1$, does the theorem fail? Could someone provide an example?

2- If we substitute the strong contraction condition with the following condition: $d(f(x),f(y))$ $<$ $d(x,y)$, does the theorem fail? example?

• For the first question, consider the map $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x + 1$. This map has no fixed points, but clearly $d(f(x),f(y)) = d(x,y)$. for any $x$ and $y$. Aug 14, 2015 at 20:13
• @Bryan Brown, great! Thank you.
– MATH
Aug 14, 2015 at 20:18
• In order for the result to hold under your second condition you need to assume $X$ is compact and not merely complete. Aug 14, 2015 at 20:24
• @Santiago Canez, That's helpful, thank you.
– MATH
Aug 14, 2015 at 20:57

1. Yes, it fails. Consider $f : [0, \infty) \to [0, \infty)$ given by

$$f(x) = \sqrt{1+x^2}.$$

Since

$$f'(\xi) = \frac{\xi}{\sqrt{1+\xi^2}} < 1 \text{ for each } \xi \in [0, \infty),$$

by the mean value theorem

$$|f(x) - f(y)| = f'(\xi) |x-y| < |x-y| \text{ for every } x, y \in [0, \infty).$$

But $f$ clearly has no fixed points.

– MATH
Aug 14, 2015 at 20:19
• If you want an example where $f$ is a surjective map onto $X$, I believe the function $f(x) = x + 1 - x/\sqrt{1 + x^2}$ (with $X = \mathbb{R}$) works by the same logic. Aug 14, 2015 at 20:27
• @ Michael Seifert Thank you Michael.
– MATH
Aug 14, 2015 at 21:02
• But how do we know that there does not exist a number $c$ between $f'(\xi)$ and $1$, such that $|f(x)-f(y)|<c|x-y|$? May 31, 2017 at 13:32
• If it existed, the Banach fixed point theorem would imply that $f$ had a fixed point, which it doesn't. Alternatively, one can prove that $$\lim_{\xi \to \infty} \frac{\xi}{\sqrt{1+\xi^2}} = 1,$$ so no such $c$ can exist. Jun 1, 2017 at 21:39