Let $X$ be a compact metric space and $f : X\rightarrow X$ such that $d (x,y)\le d (f(x),f(y))$ for all $x,y\in X$. Prove that $f$ is an isometry. I am getting stuck on this question. Can any one help me ?

  • $\begingroup$ I haven't solved it but a possible idea: The function $F: X \times X \rightarrow [0,\infty)$ given by $$F(x,y) = d(f(x),f(y)) - d(x,y)$$ is continuous, with $X \times X$ compact. If its image is not $\{0\}$, there is an absolute maximum $M > 0$. $\endgroup$ – D_S Jun 19 '16 at 13:39
  • $\begingroup$ I think that this idea does not work. $\endgroup$ – Omega Jun 19 '16 at 14:08
  • $\begingroup$ Is $f$ continuous? $\endgroup$ – Disintegrating By Parts Jun 20 '16 at 19:26
  • $\begingroup$ f is not continuous. $\endgroup$ – Omega Jun 21 '16 at 9:20

It is enough to prove that $f$ is bijective.

$f$ is injective for if not, there are $x\neq y$ s.t. $f(x)=f(y)\Rightarrow d(f(x),f(y))=0.$

If $f$ is continuous I think we can argue as follows: $f(X)$ is compact, and closed.

Now, assume $X\neq f(X)$. Then then is a point $x\in X\setminus f(X)$ s.t. $d(x,f(X))=a>0$. This follows because $f(X)$ is closed and $x\notin f(X)$.

Consider the sequence defined by $x_1=x$ and $x_n=f(x_{n-1})\in f(X)$ and extract a convergent subsequence $x_{n_k}\to y$. Of course, $y\in f(X)$ because $f(X)$ is closed.

But then, for all integers $k$, we have, for some integer $m$,

$d(f(x_{n_k}),f(x_{n_{k-1}}))\geq d(x_{n_k},x_{n_{k-1}})= d(f(x_{n_k-1}),f(x_{n_{k-1}-1}))\geq \cdots= \cdots \geq d(f(x_m),x)>a$, which implies that $f(x_{n_{k}})\not\to f(y)$, which contradicts the continuity of $f$.

  • $\begingroup$ I think "$=$" should be "$\ge$" in a few places. But more important, the fact that $d(x_n,x_{n+1})\ge1$ does not imply that $(x_n)$ has no convergent subsequence. Consider the sequence $0,1,0,1,\dots$ in $\Bbb R$. $\endgroup$ – David C. Ullrich Jun 19 '16 at 15:58
  • $\begingroup$ Yes of course thanks. I had another bash at it. I would appreciate your comments. $\endgroup$ – Matematleta Jun 19 '16 at 16:48
  • $\begingroup$ The $x_{n_{k-1}}$ is a typo for $x_{n_k-1}$. Hence I don't see why that chain of inequalities shows that $x_{n_k}\not\to y$. ??? $\endgroup$ – David C. Ullrich Jun 19 '16 at 17:34
  • 1
    $\begingroup$ Or rather why it shows that $f(x_{n_k})\not\to f(y)$. Typo. $\endgroup$ – David C. Ullrich Jun 19 '16 at 17:56
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    $\begingroup$ Btw, looking at the link provided in the other answer, you can make this work without continuity. Let $Y=f(X)$. The argument you give shows that $Y$ is dense. There is a function $g:Y\to X$ with $g(f(x))=x$. Since $g$ decreases distances it is uniformly continuous. Hence $g$ extends to a map from $X$ to $X$, and now $g$ is a surjective weak contraction, hence $g$ is an isometry. $\endgroup$ – David C. Ullrich Jun 19 '16 at 23:32

The answer can be found in Theorem 1.6.15 of "A Course in Metric Geometry", by Burago, Burago and Ivanov.


Here is another approach that uses only very basic ideas:

Pick $x,y\in X$ and take $x_0=x, y_0=y$ and then $x_n=f(x_{n-1}), y_n=f(y_{n-1})$.

Then, for all integers $n$, we have

$\tag1 d(x,y)\leq \cdots \leq d(x_n,y_n)\leq d(x_{n+1},y_{n+1})\leq \cdots $

Now, there are convergent subsequences $x_{n_{k}}, y_{n_{k}}$ and these are of course, Cauchy.

Let $\epsilon >0$. Then, if $k$ is large enough,

$\tag2 d(x_{n_{k+1}-n_{k}},x)\leq d(x_{n_{k+1}},x_{n_{k}})<\epsilon$

and $\tag3 d(y_{n_{k+1}-n_{k}},y)\leq d(y_{n_{k+1}},y_{n_{k}})<\epsilon$

where we have used $(1)$ successively.

But now, using $(1), (2), (3)$ and the triangle inequality, we have

$\tag4 d(f(x),f(y))=d(x_1,y_1)\leq \cdots \leq d(x_{n_{k+1}-n_{k}},y_{n_{k+1}-n_{k}})\leq d(x_{n_{k+1}-n_{k}},x)+d(x,y)+d(y_{n_{k+1}-n_{k}},y)<d(x,y)+2\epsilon$

so $d(f(x),f(y))\leq d(x,y)$ and we are done.


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