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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.

Aden Dong
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  • I think this(6c) is a fun example, if you are into algebraic topology... http://math.stackexchange.com/questions/481001/hatchers-algebraic-topology-problem-0-6bc-is-this-proof-legit?rq=1 – 1LiterTears Sep 01 '13 at 21:56

1 Answers1

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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 $x\ne y)$.

Ittay Weiss
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  • thank you for your answer. can you perhaps shed some light on whether a weak contraction is a contraction when M is compact? – Aden Dong Feb 22 '13 at 02:16
  • You have a typo: you meant the inequality to be strict where you claim the weak contraction. – Julien Feb 22 '13 at 02:20
  • @AdenDong I just improved the answer slightly to answer that as well. – Ittay Weiss Feb 22 '13 at 02:20
  • @julien thank you! I corrected the typo. – Ittay Weiss Feb 22 '13 at 02:21
  • @IttayWeiss thank you! great counterexample. i always have trouble doing those questions that ask you to find counterexamples... – Aden Dong Feb 22 '13 at 02:22
  • Excuse me, please. Can we say that since there is fixed point, then it`s a contraction? –  Apr 22 '13 at 07:07
  • Let $\alpha=7/12, x=1/3, y=1/4$ –  Apr 22 '13 at 11:54
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    @guest: no, existence of fixed point does not imply a map is a contraction. – Willie Wong Apr 22 '13 at 14:06
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    Let me just add a comment to Ittay's counterexample: the main driving force is that the definition of weak contraction as stated in the problem is given on the set $M\times M\setminus D$, where $D = {(x,x)}$ is the diagonal set. Now while $M$ being compact would imply $M\times M$ is compact, to draw the conclusion along the line the OP imagined requires knowing that $M\times M\setminus D$ is compact, which is not guaranteed by our assumptions thus far. – Willie Wong Apr 22 '13 at 14:10