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Let $(X, d)$ be a compact metric space, and suppose $f : X → X$ satisfies $$d(f(x), f(y)) < d(x, y)$$ for all $x \neq y \in X$. Show that f has a unique fixed point.

All I've gotten it so far is that we need to somehow use another function $g(x)=(x,f(x))$.

Thanks

Ben Grossmann
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Gauss
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  • Intuitively you could seach for the fixed point by reapplying $f$ over and over. if you do this you get a sequence of some sort. Can you say anything about convergence of this sequence? – An.Ditlev Nov 28 '16 at 14:58
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    Those downvoting should provide some kind of feedback. The way I see it, this is a reasonable question with a reasonable amount of context given. – Ben Grossmann Nov 28 '16 at 15:00
  • @Omnomnomnom I agree. Moreover, it allows interesting answers. – Hello Nov 28 '16 at 15:07
  • Please also have a look at the following post: https://math.stackexchange.com/questions/3139704/prob-7-b-sec-28-in-munkres-topology-2nd-ed-a-shrinking-self-map-of-a-co – Saaqib Mahmood Apr 01 '20 at 12:32

2 Answers2

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Hint: Consider the function $$ g(x) = d(x,f(x)) $$ Note that this is a continuous function (why?) over a compact space, so it attains its minimum.

Suppose for contradiction that the minimimum of $g$ over $X$ is not $0$. That is, suppose that for every $x \in X$, $g(x) \geq \epsilon > 0$. By compactness, there is an $x^* \in X$ is such that $g(x^*) = \epsilon$. Reach a contradiction (how?) to conclude that $g$ must have a minimum of $0$.

Uniqueness is easy using the inequality.

Saaqib Mahmood
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Ben Grossmann
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  1. Show that $f$ is continuous.

  2. Take $x_0 \in X$ and define $x_{n+1}:=f(x_n)$ for $n \ge 1$

  3. By compactness of $X$ ,$(x_n)$ contains a convergent subsequence.

  4. Show that the limit of this convergent subsequence is a fixed point of $f$

  5. Show that this fixed point is unique.

Ben Grossmann
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Fred
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    Can it be shown that the whole sequence converges? If not, do you know a counter example? – Hello Nov 28 '16 at 15:09
  • @Hello it must, based on what's proven here. Each subsequence converges to a fixed point, but each fixed point is unique, which means that each subsequence converges to the same element. – Ben Grossmann Nov 28 '16 at 15:17
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    @Fred how would you show that the limit of the subsequence should be a fixed point? I can't see a quick way to do it. – Ben Grossmann Nov 28 '16 at 15:18
  • @Omnomnomnom well isn't it that each "convergent" subsequence converges to a fixed point? For your question to Fred, I think this follows from continuity. – Hello Nov 28 '16 at 15:19
  • @Omnomnomnom $x=\lim x_{n+1} = \lim f(x_{n}) = f(\lim x_n)=f(x)$ – Hello Nov 28 '16 at 15:20
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    @Hello you can use that to see that if the full sequence convereges, its limit is a fixed point. However, that line of reasoning can't be directly applied to the subsequence. – Ben Grossmann Nov 28 '16 at 15:21
  • @Omnomnomnom Good point! and actually, if the map is only non-expansive, the set of accumulation points might not contain any fixed points. However, I think it can be shown that $f$ acts transitively on this set. (such examples are known in nonlinear perron-frobenius theory) – Hello Nov 28 '16 at 15:23