Prove that $\exists k\in \mathbb{N}^*$ such that $\|a-a_k\|<\varepsilon$

Question

Let $E$ be a normed linear space, $C$ compact and $f:C\to C$ a function such that $\|f(x)-f(y)\|\geq \|x-y\|$ for all $x,y\in C$. Then $f$ is an isometry.

Note: I'm having trouble trying to prove it! I feel stuck. My approach: Some hint says define: $a_0=a$, $a_n=f\circ ...\circ f(a)=f^{n}(a)$ and similar $b_n=f^{n}(b)$. Added: prove that for all $\varepsilon>0$ there is $k\in \mathbb{N}^*$ such that $\|a-a_k\|<\varepsilon$ and $\|b-b_k\|<\varepsilon$

How can I deduce from this that $\|a_1-b_1\|\leq \|a-b\|+2\varepsilon$? Because in that case I know that $\|f(a)-f(b)\|\leq \|a-b\|$ and since $a,b$ were arbitrary we are done!

Please any help in proving this two things!!

My attempt: $a_n$ have (by compacity) a subsequence which converges, say $a_{n_k}\to z\in C$. Can I say that $z=a$?? What I know is that for some $x_1,...x_p\in C$ we have that $C\subseteq \bigcup_{i=1}^p B(x_i,\varepsilon)$.

Thanks I really feel stuck!

@TheOscillator you can't, because it doesn't necessarily have one — Ben Grossmann, Jun 23 '15 at 01:03
@TheOscillator even with continuity we might not have a fixed point. Take $x \mapsto -x$ on ${-1,1}$, which is continuous. — Ben Grossmann, Jun 23 '15 at 01:08
No the continuity of $f$ is not assumed! Just the inequality above and the Compacity of $f$. — Valent, Jun 23 '15 at 01:14
Is that your hint, or part of the question as it was presented? — Ben Grossmann, Jun 23 '15 at 01:19
See my example; we can't necessarily prove the result in your hint — Ben Grossmann, Jun 23 '15 at 01:21
Is a hint in the problem! But I cannot prove it, and even don't know why I can deduce from this hint that $|a_1-b_1|\leq |a-b|+2\varepsilon$. — Valent, Jun 23 '15 at 01:24
Your example works, to see that there is not a fix point, but your $f$ is an isometry, and then I cannot say the problem is wrong :( — Valent, Jun 23 '15 at 01:35

score 3 · Accepted Answer · edited Jun 24 '15 at 15:06

Let $a_n=f^n(a)$ then we must prove that for arbitrary $\epsilon$ there exist natural K such that |$a-a_K$|<$\epsilon$

So assume that it's not true i.e. there exist $\epsilon_0$ s.t. |$a-a_k$|$\ge\epsilon_0$ for arbitrary natural $k$

Since C is compact , there are a finite number of points $x_1 , ... , x_n$ s.t. C $\subset$ $\cup D(x_i,\frac{\epsilon_0}2)$ where $D$ means disk

Then the above statement tells us that $a$ $\in$ $D(x_l,\frac{\epsilon_0}2)$ for some $l$ and $a_k$ $\in$ $D(a_i,\frac{\epsilon_0}2)$ where $i\ne l$

So there exists some K s.t. $a_1 , a_K$ are in same disk (without loss of generality) but then it is a contradiction to

$\epsilon_0$ $\le$ |$a-a_{K-1}$| $\le$ |$a_1-a_K$| $\le$ |$a_1-x_{i_0}$|+|$x_{i_0}-a_K$| $<$ $\frac{\epsilon_0}2+\frac{\epsilon_0}2$ = $\epsilon_0$ where $i_0$ is same disk's center

Is there any problem? Please check this.

To conclude I just note $|f(a)-f(b)|\leq |a_k-b_k|\leq |a-a_k|+|a-b|+|b-b_k|\leq |a-b|+2\varepsilon$. — Valent, Jun 23 '15 at 07:35

Prove that $\exists k\in \mathbb{N}^*$ such that $\|a-a_k\|<\varepsilon$

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