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?



A variation of the Banach fixed-point theorem
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 $m=0$).


