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 $(f^{\circ n})_{n=1}^{\infty}$ is $1$-Lipschitz, hence is uniformly equicontinuous. So by the Arzelà–Ascoli theorem, there is a uniformly convergent subsequence $(f^{\circ n_k})_{k=1}^{\infty}$. 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 $k\to\infty$ for $i = 1, 2$. Then 

\begin{align*} d(f(y_1), f(y_2)) &= \lim_{k\to\infty} d(f^{\circ(1+n_k)}(x_1), f^{\circ(1+n_k)}(x_2)) \\ &\geq \lim_{k\to\infty} d(f^{\circ n_{k+1}}(x_1), f^{\circ n_{k+1}}(x_2)) \\ &= d(y_1, y_2) \end{align*}

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

 

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