Isometry of a compact metric space

이번 포스팅에서는 뜬금없이 해석학 한 문제 다뤄보겠습니다.

 

Theorem. Let $(X,d)$ be a compact metric space, and let $f:X\to X$ be a surjective $1$-Lipschitz function. Then $f$ is an isometry.

Proof. Let $f^{\circ n}$ denote the $n$-fold composition of $f$, that is, $f^{\circ 0}={\mathrm{id}}_X$ and $f^{\circ (n+1)}=f\circ f^{\circ n}$. By the assumption, the family $(f^{\circ n}{)}_{n=1}^{\mathrm{\infty }}$ is $1$-Lipschitz, hence is uniformly equicontinuous. So by the Arzelà–Ascoli theorem, there is a uniformly convergent subsequence $(f^{\circ n_k}{)}_{k=1}^{\mathrm{\infty }}$. Let $g$ denote the corresponding subsequential limit, and let $Y=g(X)$. We claim that

 

1. $f{|}_Y$ is an isometry from $Y$ to $X$, and
2. $Y=X$.

Indeed, let $y_1,y_2\in Y$, and choose $x_1,x_2\in X$ so that $f^{\circ n_k}(x_i)\to y_i$ as $k\to \mathrm{\infty }$ for $i=1,2$. Then 

$$\begin{array}{rl}d(f(y_1),f(y_2))& =\underset{k\to \mathrm{\infty }}{lim}d(f^{\circ (1+n_k)}(x_1),f^{\circ (1+n_k)}(x_2))\\ & \ge \underset{k\to \mathrm{\infty }}{lim}d(f^{\circ n_{k+1}}(x_1),f^{\circ n_{k+1}}(x_2))\\ & =d(y_1,y_2)\end{array}$$

The opposite direction is obvious, and so, $f$ restricted to $Y$ is an isometry.

 

For the second claim, assume otherwise that $Y\ne X$. Then there exists $\epsilon >0$ such that the $\epsilon$-neighborhood $Y^{\epsilon }$ of $Y$ is not all of $X$. However, since $f^{\circ n_k}$ converges uniformly to $g$, there exists $k$ such that $\Vert f^{\circ n_k}-g{\Vert }_{sup}<\epsilon$. Then $f^{\circ n_k}(X)\subseteq Y^{\epsilon }$ and hence $f^{\circ n_k}$ is not surjective, contradicting the fact that $f$ is surjective. $◻$