2
$\begingroup$

There are many results (usually connected to specification-like properties) about density of periodic measures in the space of all invariant ones. However some questions that seem to be easy (at first glance at least) always puzzle me.

Consider dynamical system generated by a continuous self-map $f$ of a connected compact topological manifold $M$.

Denote by $\mathcal{M}_i(M)$ the set of $f$-invariant probability measures.

Denote by $\mathcal{M}_p(M)$ the set of measures of the form $(1/k)\sum_{i=0}^{k-1} \delta_{f^i (x)}$ where $x$ is a $k$-periodic orbit of $f$.

Denote by $\mathcal{M}'_p(M)$ the set of measures of the form $\sum_{i=0}^{k-1} \alpha_i \delta_{f^i (x)}$ where $x$ is a $k$-periodic orbit of $f$ and $\alpha_i > 0$ and $\sum_{i=0}^{k-1} \alpha_i = 1$.

My question: is there an example of $f$ such that one of the following holds?

  1. the set of periodic points is dense in $M$ but $\mathcal{M}_p(M)$ is not dense in $\mathcal{M}_i(M)$
  2. $\mathcal{M}_p(M)$ is dense in $\mathcal{M}_i(M)$ but the set of periodic points is not dense in $M$
  3. $\mathcal{M}'_p(M)$ is dense in $\mathcal{M}_i(M)$ but $\mathcal{M}_p(M)$ is not dense in $\mathcal{M}_i(M)$
Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ The north-south map solves 2. $\endgroup$
    – Anthony Quas
    Commented Mar 11, 2015 at 23:24
  • $\begingroup$ @AnthonyQuas (A completely trivial comment)Depends on what $\mathcal{M}_p$ is, just on a single orbit, or convex combinations. Or equivalently, use the set of ergodic measures instead of invariant measures. $\endgroup$
    – Pengfei
    Commented Mar 12, 2015 at 2:58
  • 1
    $\begingroup$ Do you really mean to define $\mathcal{M}_p'$ as above? As formulated above we have $\mathcal{M}_p' \cap \mathcal{M}_i=\mathcal{M}_p$. Did you actually mean $\mathcal{M}_p'$ to be the set of finite linear combinations of elements of $\mathcal{M}_p$? $\endgroup$
    – Ian Morris
    Commented Mar 12, 2015 at 12:31
  • $\begingroup$ @IanMorris this was a part of my question in a sense. But this simple equality you wrote answers it. $\endgroup$
    – Dmitry Todorov
    Commented Mar 12, 2015 at 13:00

2 Answers 2

Reset to default
2
$\begingroup$

For (1) and (3), as mentioned by Christian, consider the identity map $I:S^1\to S^1$.

For (2), consider the map $f:S^1\to S^1$ fixing $1\in S^1$, and $f^n x\to 1$ (as $n\to\pm\infty$) for all other points. Then $\mathcal{M}_p=\mathcal{M}_i=\{\delta_1\}$, but there is no other periodic point beside $1\in S^1$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Concerning (2) -- I think I miss something. As I see you can not have only one fixed point like this on a circle by topological reasons. So you will have a repeller as well and finally the map will be north pole -- south pole as was suggested by Anthony Quas. And due to your comment to the Anthony's comment, this concerns the set $\mathcal{M}'_p$ whereas I was asking about $\mathcal{M}_p$. $\endgroup$
    – Dmitry Todorov
    Commented Mar 12, 2015 at 8:15
  • 1
    $\begingroup$ You can certainly have only one fixed point on the circle: consider $f \colon [0,1] \to [0,1]$ defined by $f(x):=x^2$ and pass to the circle by identifying $0$ with $1$. $\endgroup$
    – Ian Morris
    Commented Mar 12, 2015 at 12:33
0
$\begingroup$

Situations (1) and (3) are not possible (I assume you are using the weak $*$ topology on the measures). If the periodic points are dense and $\mu$ is invariant, then we can approximate $\mu$ by a finite sum $\nu=\sum g_j \delta_{p_j}$ for certain periodic points $p_j$. Then, since $f\mu=\mu$ by assumption (where I write $f\mu$ for the image measure), another approximation of the same quality is given by $f^N\nu$, and if $N$ is a multiple of all periods that are involved, then this is of the desired form.

(3) is ruled out in the same way.

Edit: On rereading this, I now discover that this answers a slightly different question, namely, are the convex combinations of periodic measures dense. Obviously, the periodic measures themselves are not dense (consider $1/2(\mu_1+\mu_2)$ for two different periodic measures for, say, a rational rotation on the circle).

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ I'm trying to understand your solution for (1), namely that measures which are finite convex combinations of periodic measures are weak-* dense in the space of all invariant measures (with the assumption that periodic points are dense) You start by approximating $\mu$ with $\nu$ which is supported on periodic points. Then you consider the measure $f^N\nu$ which is indeed of the desired form. However, $N$ may be very large, so what guarantees that $f^N\nu$ is still a good approximation for $\mu$? In addition, isn't it true that $f^N \nu=\nu$? Did you mean to take an average instead? $\endgroup$
    – Itamar Vigdorovich
    Commented Jan 26, 2022 at 10:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .