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Let $K\subset \mathbb{R}^{n}$ be a compact non-empty set and $f: K \to K$ such that $|f(x) - f(y)| \geq |x-y|$, for all $x,y \in K$. Consider the sequnce $\{f_{n}\}$ given by $f_{n} = f^{n} = f\circ \cdots \circ f$. For $a,b \in K$, let $a_{n} = f_{n}(a)$ and $b_{n} = f_{n}(b)$.

(a) Show that for any $\epsilon > 0$, there is $k \in \mathbb{N}$ such that $$|a-a_{k}| < \epsilon, |b-b_{k}|<\epsilon.$$

(b) Show that $f(K)$ is dense in $K$.

(c) Show that $f$ is an isometry

My attempt

(a) I dont have any idea

(b) Given $x \in K$, take $x_{n} = f_{n}(x)$. By (a), for any $\epsilon > 0$, there is $k \in \mathbb{N}$ such that $|x - x_{k}| < \epsilon$. The sequence $\{x_{k}\}$ satisfying (a) has a convergent subsequence, since $K$ is compact. But this sequence converges to $x$.

(c) My idea was suppose that $|a - b| > |f(a) - f(b)|$ for some $a,b$ using (b), but I cannot conclude anything.


Can someone help me? Also, I'm not sure about (b). This not seems a hard problem, but I'm stuck on it.

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$(a) \quad$ Let $\varepsilon > 0$. For all $x \in K$, denote by $B_x$ the open ball centered in $x$, of radius $\varepsilon$. The family of open sets $(B_x \times B_y)_{x,y \in K}$ covers $K \times K$. Because $K \times K$ is compact, there exists a finite family $x_1, ..., x_n$ and $y_1, ..., y_n$ of points of $K$ such that $$(K \times K) \subset \bigcup_{1 \leq m \leq n} (B_{x_m} \times B_{y_m}) $$

Now consider the $n+1$ points $(a,b)$, $(a_1, b_1)$, ..., $(a_n, b_n)$ of $K \times K$. You have $n+1$ points, and $K\times K$ is covered by $n$ open sets, so two of these points are in the same open. So there exists $1 \leq m \leq n$, and $0 \leq i < j \leq n$ such that $a_i$ and $a_j$ are in $B_{x_m}$, and $b_i$ and $b_j$ are in $B_{y_m}$. In particular, $|a_i - a_j| < \varepsilon$, and $|b_i - b_j| < \varepsilon$. Let $k=j-i$. It is easy to see that, because $f$ is expansive, $|f^i(a)-f^j(a)| \geq |a-f^{j-i}(a)|$, i.e $|a_i-a_j| \geq |a-a_k|$. Similarly, $|b_i - b_j| \geq |b-b_k|$. So you deduce that $$|a-a_k| < \varepsilon \quad \text{and} \quad|b-b_k| < \varepsilon$$

$(b) \quad$ The first question shows that for every $\varepsilon > 0$ and $a \in K$, there exists $k \neq 0$ such that $|a-f^k(a)| < \varepsilon$. But $f^k(a)=f(f^{k-1}(a)) \in f(K)$, so you get a point of $f(K)$ at distance less than $\varepsilon$ of $a$. This shows that $f(K)$ is dense in $K$.

$(c) \quad$ Suppose that there exists $a \neq b$ such that $|f(a) - f(b)| > |a-b|$. Let $$\varepsilon = \frac{|f(a) - f(b)| - |a,b|}{2}$$

By the first question, there exists $k$ such that $|a-a_k| <\varepsilon$ and $|b-b_k| < \varepsilon$. So you get $$|a_k-b_k| \leq |a-a_k| + |a-b| + |b-b_k| < 2\varepsilon + |a-b| = |f(a)-f(b)|$$

This is impossible, because $f$ is expansive so you must have $|f^k(a)-f^k(b)| \geq |f(a)-f(b)|$.

This shows that $f$ is an isometry.

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Let $\epsilon>0$. Cover $K$ with finitely many open balls of radius $\epsilon/2$, as we may, since $K$ is compact. The sequence $(a_n)$ must visit at least one such ball, say $B$, infinitely many times. Therefore there is a subsequence $(a_{n_k})$ of $(a_n)$ that lives wholly in $B$. In particular, $|a_{n_k} - a_{n_{k+1}}| < \epsilon$ for all $k$, where we also used the triangle inequality. By the expansive property of $f$, and by induction, it follows that $|a - a_{n_k - n_{k+1}}| < \epsilon$ for all $k$, that is, $(\alpha_k) = (a_{n_k - n_{k+1}})$ is a subsequence of $(a_n)$ that lives wholly in the $\epsilon$-ball centred on $a$. Taking the corresponding subsequence of $(b_n)$, and applying the same argument using compactness of $K$ and expansiveness of $f$, we may pass to further corresponding subsequences of $(a_n)$ and $(b_n)$ that live wholly in the $\epsilon$-balls centred on $a$ and on $b$, respectively. In particular, there is an index $k$ such that $a_k$ and $b_k$ have the property required in part a) of the question.

For part b), let $c \in K$. We already showed in part a) that points in the sequence $f^n(c)$ may be found that are arbitrarily close to $c$. It follows that the range of $f$ is dense in $K$, as required.

For part c), let $a, b \in K$ and suppose, for a contradiction, that $|f(a) - f(b)| = |a - b| + \eta$, where $\eta > 0$. Then for all $k$, $|a_k - b_k| \geq |a - b| + \eta$. By a suitable choice of $\epsilon$, and by applying the triangle inequality, we can contradict the assertion in part a). Conclude that $|f(a) - f(b)| = |a - b|$.

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  • $\begingroup$ I didn't see TheSilverDoe's nice answer, until after I'd posted this. $\endgroup$ – Simon Mar 20 at 13:02

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