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Let $X$ be a compact metric space with metric $d$ and let $f \in C (X, X)$ and such that $d (f(a ),f(b))\ge d (a, b)$ for all $a$ and $b$ in $X.$ Show that $d(f(a), f(b)) = d(a, b)$ for all $a$ and $b$ in $X.$

My attempt:

$X$ is compact implies that $f(X)$ is also compact, since $f$ is continuous. Also, $f$ is uniformly continuous.

On the contrary, let us assume that there exists a pair of points $(x,y) \in X^2$ such that $d (f(x ),f(y)) > d (x, y).$ Since, $X$ is compact, diameter of $X$ is finite, say $M.$ And there exits points $x_0,y_0 \in X$ such that $d(x_0,y_0)=M.$

Suppose that $(x_0,y_0)=(x,y),$ we are done with the contradiction that $d (f(x_0 ),f(y_0)) > d (x_0, y_0)>M,$ since diameter of $f(X) \subseteq X$ is atmost M.

Now we are left with the case where $d(x,y) < M.$

I am stuck here. Please help. Thanks in advance.

MathRookie2204
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2 Answers2

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Verify that $G(x)=d(f(x),x)$ is a continuous real valued function on the compact set X. Hence, G attains a maximum and a minimum. Let $x^*$ be the point of maximum of $G$. Assume that $G(x^*)>0$. By the property of the metric given in the question and the fact that $x^*$ is the point where $G$ attains the maximum, we must have $G(f(x^*))=G(x^*)$ and $G(x^*)>0$ also implies that the sequence $<f^t(x^*)>$ has distinct points in X, that is, $f^t(x^*)\neq f^{t+1}(x^*)$ . Since X is a compact metric space, it is sequentially compact and hence $<f^t(x^*)>$ has a convergent subsequence. In fact given that the distance between points of the sequence $t,t+1$ is at least as large as that of $t-1, t$ the sequence itself must converge. But the distance between any two consecutive points of this sequence is bounded below by $G(x^*)>0$, which leads to the desired contradiction. Hence, $G(x)=d(f(x),x)\leq G(x^*)=0$ implies $f(x)=x$ for all $x\in X$.

Rex Nolan
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  • Just had a confusion. It is very clear that $f^t(x^) \ne f^{t+1}(x^).$ Is it possible that $f^i(x^) \ne f^{j}(x^)$ for some $i,j \in \mathbb{N}$ ?? – MathRookie2204 Feb 29 '24 at 08:41
  • You are correct. The approach seemed intuitive to me, but it does not work the way it is written. In particular, the point about possible cycles cannot be ruled out and hence makes the proof incorrect. I have provided the another proof. Sorry about that. – Rex Nolan Mar 01 '24 at 05:08
  • Thank you very much for your interest in my question. – MathRookie2204 Mar 01 '24 at 07:27
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    Just want to add an explicit example. Let $D={z \in \mathbb{C}: |z| \le 1. }$ Induce the standard metric from $\mathbb{C}$ on $D.$ See that $D$ is a compact metric space. Consider the function $f:D \to D$ defined by $$f(z) = e^{i \frac{\pi}{4}}z.$$ Observe that $f(z)$ is essentially rotating $z$ by an angle $\frac{\pi}{4}$ radians about the origin. $f$ satisfies the given condition. But, that does not imply that $f(z)=z.$ – MathRookie2204 Mar 01 '24 at 07:34
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I am skipping some of the details that are easy to fill. Also, I use sup instead of max but it does not matter for this proof. The set $f(X)$ is a compact subset of X. Hence, $\sup_{(x,y)\in X^2} d(f(x),f(y))\leq \sup_{(x,y)\in X^2} d(x,y)$. Observe that $X^2$ is also compact and the left hand side is supremum taken over a subset of $X^2$, so the sup cannot dominate.

However, from the property of the metric in the question it also follows that $\sup_{(x,y)\in X^2} d(f(x),f(y))\geq \sup_{(x,y)\in X^2} d(x,y)$.

So, $\sup_{(x,y)\in X^2} d(f(x),f(y))= \sup_{(x,y)\in X^2} d(x,y)$ must hold.

In fact, for any closed subset $A$ of $X^2$ we must have, $\sup_{(x,y)\in A} d(f(x),f(y))= \sup_{(x,y)\in A} d(x,y)$ must hold.

If there is some $a,b \in X$ such that $d(f(a),f(b))> d(a,b)$ we simply define the closed set $A=\{(x,y):d(x,y)\leq d(a,b)\}$ and note that $\sup_{(x,y)\in A} d(x,y)=d(a,b)<d(f(a),f(b))\leq \sup_{(x,y)\in A} d(f(x),f(y))$. This contradiction completes the argument.

MathRookie2204
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Rex Nolan
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