Let $(X, d)$ be complete metric space, $q \in [0, 1)$ be a real number, and $f$ be a map that satisfies $$d(f(x), f(y)) \leq q \cdot d(x, y)$$ for all $x, y \in X$. Then, Banach fixed-point theorem says that there is a unique fixed-point of $f$ in $X$.

I am interested in figuring out what happens if we relax the condition on $f$ to require, for all $x, y \in X$:

$$d(f^n(x), f^m(y)) \leq q \cdot d(x, y)\tag{1}$$ for some $n, m \geq 0.$

In general, there need not be a fixed-point when the constraint on $f$ is weakened to the above form. For example, points on a periodic orbit satisfy this weakened constraint, yet they never converge to a fixed-point. This means that, even though we conserve the notion of pairwise convergence, it is not enough to guarantee the existence of a fixed-point.

What can we say about maps that exhibit a fixed point under $(1)$? In other words, what properties do such maps have?

  • $\begingroup$ What exactly is the question? The way it is stated, the answer seems to be "maps satisfying (1) and having a fixed point". Just (1) guarantees, for example, that $f^{m+n}$ has a fixed point. $\endgroup$ Mar 6, 2014 at 0:02
  • $\begingroup$ @AlexDegtyarev: First of all, numbers $m$ and $n$ are not fixed. Each $x,y \in X$ might have a different $m$ and $n$. Also, I don't understand what you mean by "maps satisfying (1) and having a fixed point". It seems like you are answering the question circularly. $\endgroup$ Mar 6, 2014 at 0:12
  • $\begingroup$ That's exactly what I mean when I ask "what's the question". I answered the question as it was stated. (You can take it as a joke as well.) Speaking about $m$, $n$, then you should change the order of the quantifiers in (1), as otherwise it's misleading. $\endgroup$ Mar 6, 2014 at 0:23
  • $\begingroup$ I will rephrase (1) to make it more clear. Anyway, I am interested in finding out the properties of maps that exhibit a fixed point whenever (1) holds. $\endgroup$ Mar 6, 2014 at 0:27
  • $\begingroup$ If $m=n$, then it works. $\endgroup$
    – Suvrit
    Mar 6, 2014 at 2:15

1 Answer 1


If you really mean $m=m(x,y)$ and $n=n(x,y)$ (as per your comments), your assumption is equivalent to say that every pair of orbits have minimum distance $0$: for any $x\in X$ and $y\in X$ $$\inf_{ n\in\mathbb{N}\atop m\in\mathbb{N}}d\big((f^n(x),f^m(y)\big)=0\, .$$

A map $f$ satisfying this condition may be discontinuous. Even assuming continuity, it may have no fixed point: for instance, any irrational rotation on the unit circle (one can then fix $m=0$).

  • $\begingroup$ Yes. The condition requires all pairs of orbits to come arbitrarily close infinitely many times. I was curious to see how that restricted the possible long term behavior of the map, especially in relation to fixed points. It seems like this restriction is too weak to deduce any interesting results from. $\endgroup$ Mar 6, 2014 at 8:20

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