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Preliminars and notation:

Let $M$ be an $n$-dimensional compact manifold, $T\colon M\rightarrow M$ a diffeomorphism and $( x_n)_{n\in\mathbb{Z}}$ a dense orbit under $T$, ($x_n = T^n(x_0)$). Let $p\in M$ be another point and define, for $\delta>0$, $B_n(\delta) = B(p, e^{-n\delta})$.

Question: Is it true that for every $\delta>0$ the set

$A = $( $n\in\mathbb{N} | x_n \in B_n(\delta)$ )

has finite cardinality? How can it be proven?

Thank you for the answers!

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Does the exact choice of metric on the manifold affect this question? If not, do you know a purely topological statement (without invoking an arbitrary metric on the manifold)? – macbeth Apr 8 '10 at 15:08
You've used the symbol $n$ twice, once for the dimension and once to index a sequence. – Ian Morris Apr 16 '10 at 16:46
up vote 6 down vote accepted

This is called the shrinking target problem, and there is a reasonably large literature on it. For hyperbolic dynamical systems we can usually find quite a few pairs $x$, $p$ such that $A$ is infinite for all $\delta$. Indeed, I believe that there are results showing that in certain cases, for any point $z$ and positive real number $\delta>0$, the set of all $x$ such that $d(T^nx,z)<\exp(-n\delta)$ for infinitely many $n \geq 1$ has positive Hausdorff dimension. A good place to start would be the articles "Ergodic theory of shrinking targets" and "The shrinking target problem for matrix transformations of tori", both by Hill and Velani, but there are many results beyond this.

For illustration, here is a nice example in the case where $T$ is a smooth map of the circle which is not a diffeomorphism. I realise that this falls slightly outside the purview of your question, but it is possible to extend this argument to the case of toral diffemorphisms using the technical device of a Markov partition. (I will not attempt this here because it is very fiddly.) Let $X=\mathbb{R}/\mathbb{Z}$ be the circle, let $T \colon X \to X$ be given by $Tx = 2x \mod 1$, and let $d$ be a metric on $X$ which locally agrees with the standard metric on $\mathbb{R}$. Take $p=0 \in X$ and fix any $\delta>0$. Now, the orbit of $x$ is dense if and only if it enters every interval of the form $(k/2^n,(k+1)/2^n)$, if and only if every possible finite string of 0's and 1's occurs somewhere in the tail of its binary expansion. On the other hand, we have $d(T^nx,0)<2^{-\delta n}$ as long as the binary expansion of $x$ contains a string of zeroes starting at position $n$ and having length $\lceil \delta n \rceil$. I think that it is not difficult to see that we can construct an infinite binary expansion, and hence a point $x$, such that this condition is met for infinitely many $n$, whilst simultaneously meeting the condition that the orbit of $x$ is dense. In particular we can construct a point $x$ for which $A$ is infinite, even for all $\delta$ simultaneously if you like.

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Extremely good reference. Thanks for the article!!! I will enjoy a lot it! – Kaminoite Apr 16 '10 at 17:46

$x_n \in B(p,e^{-n\delta})$ iff $p \in B(x_n,e^{-n\delta})$. Thus we ask whether $$ \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty B(x_n,e^{-n\delta}) = \varnothing $$ But that is a countable intersection of dense open sets, so (by Baire category) is NOT empty.

(I hope my quantifiers are right...)

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This does not mean that $p$ is in the intersection but in its closure. – Kaminoite Apr 8 '10 at 19:55
Nice! We could even take a further intersection over a sequence of $\delta$'s tending to zero, proving the existence of points which work simultaneously for all $\delta$. – Ian Morris Apr 16 '10 at 16:52
Ian, I think that Gerald's proof is not correct? Am I wrong? – Kaminoite Apr 16 '10 at 17:25
Oops, I should have said for $\delta$ tending to infinity, not zero. I don't think that there's a problem with Gerald's proof: given any dense orbit $(x_n)$ and real number $\delta>0$, Baire's theorem shows that there exists $p \in \bigcap_{k=1}^\infty \bigcup_{n=k}^\infty B(x_n,e^{-n\delta})$, and by unravelling the meaning of this statement we find that this $p$ satisfies $d(x_n,p)<e^{-n\delta})$ for infinitely many $n$. – Ian Morris Apr 16 '10 at 20:19
I see.. The problem that I found is that it not assures that $p$ is in the intersection. Nevertheless, as you said, the Baire's theorem asserts that this intersection is dense but, by Borel-Cantelli lemma, that its measure is $0$. – Kaminoite Apr 17 '10 at 9:07

Take $M=\mathbb{R} / \mathbb{Z}$, $T(x)=x+\alpha$ for some $\alpha$. If $\alpha$ is irrational, all orbits will be dense. Set $p=0$, then the set $A$ can be made to be infinite by choosing an $\alpha$ that can be approximated well: choose $\alpha$ from $B_1(\delta-2\cdot 10^{-k_1})$ for some $k_1$, then modify it at most by $10^{-k_1}$to make $10^{k_1} \alpha -[10^{k_1} \alpha] \in B_{10^{k_1}}(\delta-2 \cdot 10^{-k_2})$ and repeat ad infinitum.

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The thing is that you can not modify the value $\delta$. The question states that it must be fixed. – Kaminoite Apr 8 '10 at 16:00
You mean $\alpha$, if I understand you correctly. I am giving you a construction of an $\alpha$ for which the dynamical system exhibits an infinite $A$, so there is no modification. I assume there could be an explicit $\alpha$ given, something along the lines of $\sum 10^{-10^{k^2}}$ - as long as it can be approximated superexponentially, the set $A$ will be infinite for all $\delta>0$. – Thorny Apr 9 '10 at 7:03

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