# Density of periodic points and density of periodic measures

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 $\mathcal{M}_i(M)$
• The north-south map solves 2. – Anthony Quas Mar 11 '15 at 23:24
• @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. – Pengfei Mar 12 '15 at 2:58
• 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 $\mathcal{M}_p$? – Ian Morris Mar 12 '15 at 12:31
• @IanMorris this was a part of my question in a sense. But this simple equality you wrote answers it. – Dmitry Todorov Mar 12 '15 at 13:00

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 $1\in S^1$.
• 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 $\mathcal{M}_p$. – Dmitry Todorov Mar 12 '15 at 8:15
• 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 $1$. – Ian Morris Mar 12 '15 at 12:33
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.
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).