# Isometry composition, (f^n (x)) - increasing sequence

• May 4th 2013, 05:19 AM
wilhelm
Isometry composition, (f^n (x)) - increasing sequence
Let $\displaystyle f: [0,1] \rightarrow [0,1]$ be an isometry in a metric space $\displaystyle ([0,1], d_{Eucl})$. Suppose $\displaystyle f(0)=0, \ \ \ x\in [0,1]$

Could you explain to me why the sequence $\displaystyle (f^n(x))_{n\geq1}$ is monotone and why does it converge with respect to the usual topology of $\displaystyle [0,1]$ and also with respect to $\displaystyle d$ and, in particular, it satisfies the condition of Cauchy?

I know that since isometry is a continuous and bijective function, it must be monotone. We put $\displaystyle f(0)=0$, so it must be increasing. But I can't figure out why $\displaystyle f^n(x) \le f^{n+1}(x)$

Could you help me with that? I would really appreciate a thorough explanation, because I can't find anything about isometry composition anywhere.

Thank you.
• May 4th 2013, 05:40 AM
emakarov
Re: Isometry composition, (f^n (x)) - increasing sequence
Quote:

Originally Posted by wilhelm
I know that since isometry is a continuous and bijective function, it must be monotone. We put $\displaystyle f(0)=0$, so it must be increasing. But I can't figure out why $\displaystyle f^n(x) \le f^{n+1}(x)$

If you know that f is monotone, then it's trivial: apply n times the function f to both sides of $\displaystyle x\le f(x)$. And $\displaystyle f^n(x)$ converges by the monotone convergence theorem.
• May 4th 2013, 06:20 AM
wilhelm
Re: Isometry composition, (f^n (x)) - increasing sequence
Thank you. Could you tell me why $\displaystyle x \le f(x)$?
• May 4th 2013, 07:46 AM
emakarov
Re: Isometry composition, (f^n (x)) - increasing sequence
Quote:

Originally Posted by wilhelm
Could you tell me why $\displaystyle x \le f(x)$?

Good question. I was a little hasty. But does not it follow that f(x) = x? Indeed, since f is an isometry and f(0) = 0, for every 0 <= x <= 1 we have |f(x)| = |f(x) - f(0)| = |x - 0| = |x| = x. But since f(x) ∈ [0, 1], f(x) >= 0, so |f(x)| = f(x) = x.
• May 4th 2013, 06:16 PM
wilhelm
Re: Isometry composition, (f^n (x)) - increasing sequence
Am I missing something or does it follow straight from $\displaystyle d(x-0) = d(f(x)-f(0))$ that this isometry is identity?
• May 5th 2013, 05:34 AM
wilhelm
Re: Isometry composition, (f^n (x)) - increasing sequence
I'm sorry there should be commas instead of "-" signs.
• May 5th 2013, 10:29 AM
emakarov
Re: Isometry composition, (f^n (x)) - increasing sequence
Quote:

Originally Posted by wilhelm
Am I missing something or does it follow straight from $\displaystyle d(x-0) = d(f(x)-f(0))$ that this isometry is identity?

In addition to this fact, one needs f(0) = 0 and f(x) ∈ [0, 1]. Otherwise, $\displaystyle f_1(x) = 1 - x$ and $\displaystyle f_2(x)= \begin{cases}x&x\in\mathbb{Q}\\ -x&x\notin\mathbb{Q} \end{cases}$ satisfy the quoted property but are not identities. With these additional properties, a proof that f is the identity is in post #4.
• May 9th 2013, 09:19 AM
wilhelm
Re: Isometry composition, (f^n (x)) - increasing sequence
I hope I'm not bothering you. I have one more question. Do you know how to prove that the sequence is increasing for an arbitrary metric consistent with natural topology of $\displaystyle [0,1]$?
• May 9th 2013, 11:35 AM
emakarov
Re: Isometry composition, (f^n (x)) - increasing sequence
Do I understand right that now we have some metric on [0, 1] that generates the natural topology and that f is still an isometry w.r.t. this metric?
• May 9th 2013, 10:52 PM
wilhelm
Re: Isometry composition, (f^n (x)) - increasing sequence
Yes, that's right.