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Let $K$ be a compact normed space and $f:K\rightarrow K$ such that $$\|f(x)-f(y)\|<\|x-y\|\quad\quad\forall\,\, x, y\in K, x\neq y.$$ Prove that $f$ has a fixed point.

Peter_Pan
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Roiner Segura Cubero
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4 Answers4

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The sequence $K \supset f(K) \supset f(f(K)) \supset \cdots$ is a nested sequence of compact sets. Then their intersection is non-empty.

Denote the intersection by $A$. Then since

$$A =\cap f^{(n)}(K)$$

it is easy to show that $f(A)=A$.

Note there is a typo in the problem, the inequality can only hold for $x \neq y$.

Since $A$ is non-empty and compact, by This Question you can find $a_1,a_2 \in A$ so that

$$\|a_1-a_2\| = \operatorname{diam}(A) \,.$$

I claim that $a_1=a_2$ which shows that $A$ consists of a single point.

Indeed, as $f(A)=A$, pick $x_1,x_2 \in A$ so that $f(x_i)=a_i$. If we assume by contradiction that $a_1 \neq a_2$ then

$$\operatorname{diam}(A)=\|a_1-a_2\| = \| f(x_1)-f(x_2) \| < \|x_1-x_2 \|$$

but this is a contradiction, as $x_1,x_2 \in A$.

As $A$ is a single point set and $f(A)=A$ we are done.

Community
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N. S.
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  • I like this proof, but the containment $A \subseteq f(A)$ is unclear to me when I work though it. How might I show that? – Matt R. May 26 '16 at 14:48
  • @MattR. If $x \in A$ then $x \in \cap_n f^n(A)$. Therefore, for all $n \geq 0$ there exists some $y_n \in A$ such that $f^n(y_n)=x$....Now pick a cluster point $z$ of $f^{n-1}(y_n)$ and show that $z \in A$ and $f(z)=x$. – N. S. Jun 05 '16 at 13:21
  • Very good, thank you. – Matt R. Jun 16 '16 at 01:25
  • @N.S. Can you explain why K⊃f(K)⊃f(f(K))⊃⋯ holds? – user1559897 Sep 30 '16 at 01:18
  • @user1559897 $f(K) \subset K$ from the definition. Now prove by induction that $f^{n}(K) \subset f^{n+1}(K)$. – N. S. Sep 30 '16 at 03:20
  • @N.S. We have f(K)⊂K, then why do we have f(f(K)⊂f(K)? We do not have f(A)⊂A for arbitrary A. K is the metric space. – user1559897 Sep 30 '16 at 22:49
  • @user1559897 But we do have in general that $f(A) \subset f(B)$ whenever $A \subset B$.... Set $A=f(K)$ and $B=K$ ;) – N. S. Sep 30 '16 at 23:03
  • @N.S. Maybe I am missing something. Where is f(A)⊂f(B) whenever A⊂B from? – user1559897 Sep 30 '16 at 23:04
  • @user1559897 For any sets $A,B$ if $y \in f(A)$ then there exist some $x \in A$ such that $f(x)=y$. Since $x \in A \subset B$ it follows that $x \in B$ and hence $y=f(x) \in f(B)$..... Now set $A=f(K) \subset K=B$ and see what you get. – N. S. Sep 30 '16 at 23:09
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    @user1559897 OR, if you don't want to worry about general set theory results, here is the proof that $f(f(K)) \subset f(K)$. ..... Let $y \in f(f(K))$. Then there exists $x \in f(K)$ such that $f(x)=y$. Now, since $f(K)\subset K$ you get that $x \in K$ and $y=f(x)$ therefore $y =f(x) \in f(K)$................................... To put it very simple any element $f(f(k))$ has the form $f(y)$ for $y=f(k) \in K$. – N. S. Sep 30 '16 at 23:11
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Edit: Note that $f$ is continuous. (Why?) Define $$g(x)=\lVert f(x)-x\rVert$$ for all $x\in K$. Use the compactness of $K$ to show that $g$ obtains a non-negative minimum. If $f$ has no fixed point, we can then derive a contradiction. (Apologies for my earlier, bogus approach.) You can further show that $f$ has a unique fixed point, if you like.

Cameron Buie
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  • Nice solution :) – N. S. Jun 03 '13 at 03:31
  • @N.S.: Yours, too (+1). – Cameron Buie Jun 03 '13 at 03:32
  • Cameron how to get the contradiction? – Roiner Segura Cubero Jun 03 '13 at 03:48
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    @Roiner: If $f$ has no fixed point, then the minimum of $g$ is positive. Suppose $0<g(x)\le g(y)$ for all $y,$ so that $f(x)\ne x,$ but then, considering $y=f(x),$ we can use the hypothesis to see that $$\begin{align}g(y) &= \lVert f(y)-y\rVert\ &= \lVert f(f(x))-f(x)\rVert\ &< \lVert f(x)-x\rVert\ &= g(x),\end{align}$$ contradicting our choice of $x.$ – Cameron Buie Jun 03 '13 at 10:22
  • This same proof works for compact metric spaces as well. I guess that's the fixed point theorem which applies here in the question you linked – leo Jun 06 '13 at 04:18
  • @leo: Yes, that's true, it does work there; and since this question is about a compact normed (so metric) space, it works here. However, I linked that one to show that Mhenni's answer incorrectly assumed a contraction map where there need be none (as you, yourself observed in your own comment on that answer). What exactly are you hinting at/asking? – Cameron Buie Jun 06 '13 at 04:26
  • Just noting that we earn a bit more of generality with your approach :-) – leo Jun 06 '13 at 04:34
  • @leo: Ah. For whatever reason, your comment sounded doubtful to me. Perhaps I should get some sleep! ^_^ – Cameron Buie Jun 06 '13 at 04:37
  • Hi, I'm wondering whether this fixed point is unique. Could you please show me how to prove the uniqueness? – lalala8797 Jun 14 '20 at 14:57
  • @JiuLian It is indeed unique. To prove it, suppose $x$ is a fixed point of $f,$ and take any $y\in K$ such that $y\ne x.$ Then we know that $$\lVert f(x)-f(y)\rVert<\lVert x-y\rVert.$$ Since $f(x)=x,$ what can we conclude? – Cameron Buie Jun 14 '20 at 16:28
  • Thanks so much! It is really helpful! – lalala8797 Jun 15 '20 at 01:03
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Here is a proof by contradiction:

First we note that if $f(x)=x$ and $f(x')=x'$, we must have $x=x'$ (otherwise $\|f(x')-f(x)\| =\|x'-x\| < \|x'-x\|$, which is a contradiction). Hence any fixed point is unique.

Choose $x_1 \in K$ and let $x_{n+1} = f(x_n)$. If $x_{n+1} = x_n$ for some $n$, we have a fixed point, so assume that $x_{n+1} \neq x_n$ for all $n$. Note that $\|x_{n+2}-x_{n+1}\| < \|x_{n+1} - x_{n}\|$ for all $n$.

Since $K$ is compact, we can find a subsequence such that $x_{n_k} \to x$ and $x_{n_k+1} \to x'$. If $x' = x$ we have found a fixed point, so suppose $x \neq x'$. Then we have $\|f(x')-f(x) \| < \|x'-x\|$. Let $\lambda = \frac{\|f(x')-f(x) \|}{\|x'-x\|}$, and note that $\lambda <1$.

By continuity, for some $\beta \in (\lambda,1)$ we have $\|f(y')-f(y) \| \le \beta \|y'-y\|$ for $y'$ close to $x'$ and $y$ close to $x$.

Let $d_n = x_{n+1}-x_n$. Then we have $\|d_{n+1} \| < \|d_n\|$ for all $n$, and $\|d_{n_{k+1}}\| \le \beta \|d_{n_k} \|$ infinitely often. Hence $d_n \to 0$, which is a contradiction.

copper.hat
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  • Why is $d_n \to 0$ a contradiction? Note that it is possible to have $|x_{n+1}-x_n| \to 0$ and $x_n$ to have two cluster points.... – N. S. Jun 03 '13 at 12:33
  • Because at that stage of the proof I have assumed $x' \neq x$. If $d_n \to 0$, then $x'=x$. – copper.hat Jun 03 '13 at 14:33
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You can use Banach fixed point theorem by noting that, Your space $K$ is complete, since compactness implies completeness, and your map is a contraction.

Mhenni Benghorbal
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