# A variation of the Banach fixed-point theorem

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?

-
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. –  Alex Degtyarev Mar 6 at 0:02
@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. –  Mehmet Ozan Kabak Mar 6 at 0:12
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. –  Alex Degtyarev Mar 6 at 0:23
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. –  Mehmet Ozan Kabak Mar 6 at 0:27
If $m=n$, then it works. –  Suvrit Mar 6 at 2:15

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 $m=0$).