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Why is the following true?

If $(X,d)$ is a compact metric space and $f: X \rightarrow X$ is non-expansive (i.e $d(f(x),f(y)) \leq d(x,y)$) and surjective then $f$ is an isometry.

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4 Answers 4

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Consider the map $f\times f: X\times X \mapsto X\times X$. It is continuous (because f is) and surjective (because f is). Consider the orbit of $(x,y)$ under $f$, i.e. the sequence $p_n=(f^n(x), f^n(y))$. Because $X\times X$ is compact, it has a convergent subsequence, hence for any $\epsilon$ (sorry, I hoped to do no epsilons, but can't), there is an $N_0, k>0$, such that $f^{N_0}(x)$ is $\epsilon$ close to $f^{k+N_0}(x)$ and $f^{N_0}(y)$ is epsilon close to $f^{k+N}(y)$, hence $d(f^{k+N_0}(x), f^{k+N_0}(y))$ is $2\epsilon$ close to $d(f^{N_0}(x), f^{N_0}(y))$. Then because $f$ does not expand distances, induction shows the same is true for all $N \geq N_0$. But we want more. We want this to be true for some fixed $N$ (and varying $k$) for all $(x,y)$. That is achievable by compactness as follows:

As we showed (replacing $\epsilon$ by $\epsilon/2$) we have $d(f^{k+N_0}(x), f^{k+N_0}(y))$ is $\epsilon$ close to $d(f^{N_0}(x), f^{N_0}(y))$. Because $f$ does not expand distances (and triangle inequality), this means that for any $(v,w)$ with $d(v,x)<\epsilon/4$ and $d(w,y)<\epsilon/4$ the we have $d(f^{k+N_0}(v), f^{k+N_0}(w))$ is $2\epsilon$ close to $d(f^{N_0}(v), f^{N_0}(w))$. Now cover $X$ with $\epsilon/4$ balls and pick a finite subcover. For any pair of centers $(x_i,y_i)$ of the balls we have some $N_0$ which gives $\epsilon$-closeness, and hence gives $2\epsilon$ closeness on the ball, hence the maximum $M$ of these $N_0$'s gives $2\epsilon$ closeness everywhere.

Now we can prove $f$ is an isometry. To get contradiction, assume you have $(x,y)$ with $d(f(x), f(y))< d(x,y)$, hence for some $\epsilon$ also $d(f(x), f(y))< d(x,y)-2\epsilon$. Then for any $k>0$, we have $d(f^k(x), f^k(y))< d(x,y)-2\epsilon$. Finally we use surjectivity. Take $(a,b)$ with $f^M(a)=x, f^M(b)=y$. Then $d(f^{M+k}(a), f^{M+k} (b))= d(f^k(x), f^k(y))$ is $2\epsilon$ close to $d(f^M (a), f^M(b))=d(x,y)$ for some $k$, contradiction.

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  • $\begingroup$ I don't know if you are interested, but I have found a non-epsilon approach here. $\endgroup$
    – Akira
    Commented Feb 28, 2023 at 0:52
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In fact more is true. For a compact, metric space, $f$ locally non-expansive $\implies$ $f$ local isometry.

This is theorem 4.2 in the paper linked below.

Link

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    $\begingroup$ Thanks for the reference to the stronger result. Still, I wonder if one can extract an easier proof for the OP's precise question? $\endgroup$ Commented Nov 29, 2010 at 6:01
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Here's a different argument. Recall that an $\epsilon$-covering $S$ of a metric space is a set such that for every point $x$ there exists $s\in S$ with $d(x,s)\leq \epsilon.$

Lemma. Let $X$ be a metric space with a finite $\epsilon/4$-cover. If $f:X\to X$ is a non-expansive surjection then $d(f(x),f(y))\geq d(x,y)-\epsilon$ for all $x,y\in X.$

Proof: Set $D=d(x,y)-\epsilon/2.$ Let $S$ be an $\epsilon/4$-cover minimizing the quantity $N(S)=|\{(s_1,s_2)\in S\mid d(s_1,s_2)\geq D\}|.$ Note that $f(S)$ is also an $\epsilon/4$-cover, and whenever $d(s_1,s_2)< D$ we have $d(f(s_1),f(s_2))< D.$ But $N(f(S))\geq N(S),$ so no pair can go from distance $\geq D$ to $<D:$ whenever $d(s_1,s_2)\geq D$ with $s_1,s_2\in S,$ we must have $d(f(s_1),f(s_2))\geq D.$

Pick $s_1,s_2\in S$ with $d(s_1,x),d(s_2,y)\leq \epsilon/4$; this gives $d(s_1,s_2)\geq d(x,y)-\epsilon/2,$ which we argued implies $d(f(s_1),f(s_2))\geq d(x,y)-\epsilon/2.$ So $d(f(x),f(y))\geq d(x,y)-\epsilon.$ $\Box$

To get the result in the question, note that a compact space is totally bounded i.e. has a finite $\epsilon$-covering for each $\epsilon.$

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For any $x,y\in X$, since $f$ is surjective we can find a sequence $x_{-n}$ such that $f(x_{-n})=x_{-n+1}$, $x_0=x$. As $X$ is compact one can find a convergent subsequence $x_{-n_k}$ (with similar notation one can assume that $y_{-n_k}$ converges as well). By assumption we have \begin{equation*} d(x_{-n_k},x_{-n_{k+1}})\geq d(x,x_{-n_{k+1}+n_k})\to0 \end{equation*} whence $x_{-n_{k+1}+n_k}\to x$. If the set $V:=\{n_{n+1}-n_n\}$ is finite, we have $v\in V$ that is attained infinitely often whence $d(x,x_{-v}))=d(y,y_{-v}))=0$, i.e. $x_{-v}=x$ and $y_{-v}=y$. Then \begin{equation*} d(x,y)=d(f^v(x_{-v}),f^v(y_{-v}))\leq d(x_{-v},y_{-v})=d(x,y). \end{equation*} If $V$ is infinite, let $V\ni v_n\nearrow\infty$, then $u_n:=d(x_{-v_n},y_{-v_n})\nearrow d(x,y)$ by hypothesis of $f$, then \begin{equation*} d(x,y)\leq u_n\nearrow d(x,y). \end{equation*} Say one always have $d(x,y)=d(x_{-1},y_{-1})$ and that is the same as $d(x,y)=d(f(x),f(y))$.

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  • $\begingroup$ That's very nice. $\endgroup$ Commented Oct 19, 2023 at 4:36

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