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Let $\varphi:\mathbb{R}^d\to\mathbb{R}^d$ be an injective continuous function. Denote by $\varphi_n$ the $n$-th iterate of $\varphi$, i.e. $\varphi_n(x)=\varphi_{n-1}(\varphi(x))$ for all $x\in\mathbb{R}^d$. Consider two properties of $\varphi$:

1) For every compact set $K\subset\mathbb{R^d}$ there exists $n\in\mathbb{N}$ such that $\varphi_n(K)\cap K=\emptyset$.

2) For every compact set $K\subset\mathbb{R^d}$ there exists $n_0\in\mathbb{N}$ such that for every $n\geq n_0$ we have $\varphi_n(K)\cap K=\emptyset$.

The question is the following: is 1 equivalent to 2?

Remark 1. It is obvious that 2 implies 1.

Remark 2. If $d=1$, then the answer is YES and it is a simple consequence of the fact that an injective function must be monotonic.

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  • $\begingroup$ Your question seems intersting. I suggest you add more tags to the question, for instance, "ca.analysis-and-odes" and "gn.general-topology" $\endgroup$ – Bin Yu Oct 24 '13 at 3:47
  • $\begingroup$ I have added some tags. The question is related to functional analysis because the conditions can be translated into properties (like transitivity) of the composition operator $f\mapsto f\circ \varphi$ on the Frechet space of continuous functions on $\mathbb R^d$. $\endgroup$ – Jochen Wengenroth Oct 24 '13 at 7:10
  • $\begingroup$ I wonder what injective continuous self maps on $\mathbb R^n$ with no fixed points look like. Is there a characterization? $\endgroup$ – Jason Rute Oct 24 '13 at 15:34
  • $\begingroup$ Actually, the answer is "No" (at least in $\mathbb R^3$, dimension $2$ may be special). I'll try to post my example later, unless somebody beats me to it. $\endgroup$ – fedja Oct 28 '13 at 0:54
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This is an observation, along with a sketch of a potential construction where (1) does not imply (2) for $\mathbb{R}^3$.

Observation. I want to point out an example of a closed set $X \subset \mathbb{R}^2$ and a continuous injective function $f: X \rightarrow X$ such that $f$ has property (1) but not (2).

To construct this example, start with the sequence of points $a_n = (n - 10^{\lfloor log_{10}{n} \rfloor}, 10^{- \lfloor log_{10}{n} \rfloor})$. Informally: "rows" of points approaching the $x$-axis, where the size of the rows increases exponentially (9, 90, 900, etc.), and their distance to the $x$-axis decreases exponentially (1, 1/10, 1/100, etc.).

Let $X = \{ a_1, a_2, a_3, ... \} \cup \{ (0,0), (1,0), (2,0), ... \}$

Define $f$ by $f(a_n) = a_{n+1}$, and $f(x, 0) = (x + 1, 0)$.

This dynamical system breaks into two pieces: translation on non-negative lattice points on the $x$-axis, and an orbit, $\{a_n\}$, that accumulates on the origin (and subsequent lattice points) at exponentially increasing intervals. On the rows of $\{a_n\}$, points shift rightward until they reach the end of the row, in which case they "wrap around" to the first position in the next row down.

The function $f$ is continuous and injective. It does not have property (2), since if $K$ is a closed ball around the origin, it contains some of the $\{a_n\}$, whose orbit will accumulate on the origin and intersect $K$ infinitely often.

The function $f$ does have property (1). Let $K$ be a compact set, and let $M$ be the maximum $x$-coordinate of any point in $K$. Let $10^m$ be the smallest power of ten greater than $M$. Then $f^{5 \cdot 10^m} (K) \cap K = \emptyset$. To see this, note that all points in the initial rows of size less than or equal to $10^m$ will be shifted to the row of size $9 \cdot 10^{m}$ and on this row will be shifted more than $10^m$ to the right of the $x$-axis, so won't lie in $K$. Points in other rows will also shift far enough to the right so they don't hit $K$, but their rows are long enough that they won't "wrap" to come back to the $x$-axis.

Notes and solution sketch

Going back to the original problem, this example shows that a proof that (1) implies (2) will have to use some fact about $\mathbb{R}^d$, rather than just pure dynamical considerations.

It also suggests a path to construct an example where (1) holds but not (2) for a dynamical system on $\mathbb{R}^d$. A sketch of a potential construction might go something like this:

First, we can certainly extend $f$ to a homeomorphism $g$ on all of $\mathbb{R}^2$. (For the "wrapping" from one row to the next, imagine making a loop around the top; there could be a repelling fixed point somewhere on the positive $y$-axis.) Below the $x$-axis, we can just set $g$ to be a translation one unit to the right.

All the extensions I can think of for $g$ will lose property (1). To save the situation, we move to $\mathbb{R}^3$.

Let $d(x, X)$ be the distance from the point $x$ to the set $X$. For $x \in \mathbb{R}^2$ and $z \in \mathbb{R}$, define $$\varphi(x, z) = (g(x), z - d(x, X))$$

In other words, under the action of $\varphi$ points move steadily downwards, with the exception of points lying directly above or below $X \times \{0\}$. Edit: fedja points out that a stronger downward component might be useful. (For points near $X$ one could shift by $d(x, X)^{1/2}$ for example.)

The map $\varphi$ doesn't have property (2) since it is the same as $f$ restricted to $X \times \{0\}$. On the other hand, with careful attention to the steady downward movement away from $X$, one might be able to show that for a compact $K$, for high enough $n$, iterates $\varphi_n(K)$ could only intersect $K$ very close to $X \times \{0\}$, and property (1) would hold for the same reasons as in the construction of $f$.

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    $\begingroup$ That's pretty much what I did except I used a flow instead of a single homeomorphism and made the downward part a bit stronger than just the distance to $X$. I guess anyone can fill the details in now, so there is no point in posting a second version of the same argument :-). $\endgroup$ – fedja Oct 28 '13 at 11:45
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    $\begingroup$ Thanks! In light of your comment, I added your good point about a stronger downward component and made the wording overall a bit less tentative. $\endgroup$ – Martin M. W. Oct 28 '13 at 14:11
  • $\begingroup$ Thanks a lot, Martin. You will certainly get the 100 reps, but I wait till Adam accepts your answer. $\endgroup$ – Jochen Wengenroth Oct 29 '13 at 16:05
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    $\begingroup$ Thanks a lot Martin for this answer. The idea is really nice and I have a feeling that it works at least for the case d>2. Unfortunately, I am not able to produce a proof that the constructed map satisfies (1), so if it is possible please post a more detailed proof of this fact $\endgroup$ – adamp Oct 29 '13 at 18:46
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Martin's negative answer leaves open the case of $n=2$, and it seems this is for good reason.

Indeed, I think that, for a homeomorphism $\phi$ of $\mathbb{R}^2$, the two properties are indeed equivalent.

Firstly, note that $\phi$ can be assumed to be orientation-preserving (otherwise, replace $\phi$ by its second iterate).

Let $\phi$ be an orientation-preserving homeomorphism of $\mathbb{R}^2$ without fixed points. The Brouwer plane translation theorem states that every point is contained in a domain of translation (which is disjoint from both its image and preimage). See

http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=2143652

So you can cover your compact set $K$ with finitely many such translation domains. It should follow (though I have not checked the details) that $K$ is disjoint from all of its sufficiently large iterates.

(I noticed that you did not say that $\phi$ should be a homeomorphism, but only injective and continuous. It appears plausible that any counterexample in this setting would also give rise to a homeomorphic one, but one should always be careful with such topological questions - as indeed Brouwer himself discovered in the course of the proof of his theorem, I believe.)

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