TITLE: Is every weak contraction a contraction?
QUESTION [8 upvotes]: A weak contraction is a function $f:M \to M$ such that for all $x \neq y$, $d(f(x), f(y)) < d(x, y)$. I don't think every weak contraction is a contraction, but I'm having a hard time finding a counterexample. Also, is it true that if $M$ is compact, then a weak contraction is a contraction? I think it might be true, but I'm not sure how to prove it.

REPLY [13 votes]: No, a weak contraction need not be a contraction, not even when the domain is compact. For instance, let $M=[0,1/2]$ with the usual metric, and let $f:M\to M$ be $f(x)=x^2$. Then $f$ is a weak contraction since $|x^2-y^2|=|x-y||x+y| < |x-y|$ when $x\ne y\in [0,1/2]$. However, it is not a contraction since for any $0 < \alpha<1$ the inequality $|f(x)-f(y)|\le \alpha |x-y|$ doesn't hold for all $x,y\in [0,1/2)$. This is so since for $x,y\in (1/2-\epsilon /2, 1/2)$, for $1>\epsilon >0$, holds that $|x^2-y^2|=|x-y|\cdot |x+y|>|x-y|\cdot (1-\epsilon)$ (assuming $x\ne y)$.