The Closing Lemma for Expanding Maps

In this post I discuss a variation on the closing lemma.

Set Up

The goal of this article is to prove a version of the closing lemma for expanding maps on the circle. We’ll first recall some key definitions:

(1) A \(C^1\) map \(f : S^1 \rightarrow S^1\) is said to be expanding if \(\mid f'(x) \mid > 1\) for every \(x \in S^1\).

(2) Let \((x_i)_{i=1}^\infty\) be a sequence of points in \(S^1\). We say that this is an \(\epsilon\)-orbit if for each \(k\) we have

\[d(f(x_k), x_{k+1}) < \epsilon.\]

Now recall (or see Theorem 6.4.15 in Katok and Hasselblatt) that the closing lemma states that if we have an Anosov diffeomorphism \(f : M \rightarrow M\), then there exists constants \(C, \epsilon_0 > 0\) so that for any \(\epsilon < \epsilon_0\) and any periodic \(\epsilon\)-orbit \((x_0, \ldots, x_m)\) there is an honest periodic point \(y \in M\) with \(f^m(y) = y\) so that for \(0 \leq k \leq m-1\) we have

\[d(f^k(y), x_k) < C \epsilon.\]

The proof can be found in either Katok and Hasselblatt or my notes. The gist is to use a contracting mapping principle argument to find the periodic point.

Result and Proof

Theorem: Let \(f : S^1 \rightarrow S^1\) be a \(C^1\) expanding map. There exists a \(C, \epsilon_0 > 0\) so that if \(\epsilon < \epsilon_0\) and \((x_0, \ldots, x_m)\) is a periodic \(\epsilon\)-orbit, then there exists a \(y \in S^1\) so that \(d(f^k(y), x_k) < \epsilon\). Furthermore this \(y\) is unique and periodic.

Proof: This basically will just be the solution to Problem 24 in this. As seen in Proposition 3.2.3 in Katok and Hasselblatt, there is a constant \(\delta\) sufficiently small so that

\[d(x,y) < \delta \implies d(f(x), f(y)) < D \delta,\]

where \(D\) is some constant depending on the expansion. Thus for sufficiently small \(\delta > 0\), we have

\[f(\overline{B_\delta(x)}) = \overline{B_{D\delta}(f(x))}.\]

Hence we have the property that if \(d(f(x), y) \leq D \delta\) then there is a \(z \in \overline{B_\delta(x)}\) with \(f(z) = y\). We’re going to first ignore the fact that \((x_0, \ldots, x_m)\) is periodic and just extend it to an infinite \(\epsilon\)-orbit. Fix very small \(\epsilon > 0\) – it will be clear in our proof how small we’ll need as we go. Let

\[A_n := \{z \in S^1 : f^i(z) \in \overline{B_\epsilon(x_i)} \text{ for } 0 \leq i \leq n\},\] \[A := \bigcap_{n \geq 0} A_n.\]

We check that the \(A_n\) are non-empty and closed. We can then deduce by the finite intersection property that \(A \neq \varnothing\). We will then show that \(A\) is just one point, and we’ll be done.

Let’s show that the \(A_n\) are non-empty. Let \(z_n = x_n\). By making \(2\epsilon\) small enough, we can use our observation above to deduce that there is some \(z_{n-1} \in \overline{B_{\epsilon/2}(x_{n-1})}\) with \(f(z_{n-1}) = x_n\). Now

\[d(f(x_{n-2}), z_{n-1}) \leq d(f(x_{n-2}), x_{n-1}) + d(x_{n-1}, z_{n-1}) < \epsilon (1 + 1/2) < 2 \epsilon.\]

Let \(\lambda_j = 2 - 2^{1-j}\). We can inductively construct \(z_{n-j}\) so that \(z_{n-j} \in \overline{B_{\lambda_j\epsilon/2}(x_{n-j})}\) and \(f(z_{n-j}) = z_{n-j+1}\). Notice this gives us a point \(z_0\) with

\[d(f^{n-j}(z_0), x_{n-j}) \leq \epsilon \text{ for } 0 \leq j \leq n.\]

So each \(A_n\) is non-empty. It’s easy to see that the continuity of \(f\) will give us that \(A_n\) is closed – just take sequences. Finally if \(x,y \in A\), then \(d(f^i(x), f^i(y)) \leq 2\epsilon\) for all \(i\), which if we take \(\epsilon\) small enough forces \(x = y\) since it’s expanding.

Suppose \(z\) is the point in \(A\). We have that \(f^m(z)\) and \(z\) will both lie in \(A\), so this uniqueness forces \(f^m(z) = z\), which means we have a periodic point. This finishes the proof.

The other thing to note here is that we can actually get a variation of Proposition 6.4.16 from Katok and Hasselblatt. Namely if \(x,y \in S^1\) satisfy the property that \(d(f^k(x), f^k(y)) < \delta\) for \(\delta\) sufficiently small and \(0 \leq k \leq n\) then we have that

\[d(f^k(x), f^k(y)) \leq C \alpha^{k, n-k} (d(x,y) + d(f^n(x), f^n(y))),\]

where \(\alpha\) comes from the Anosov exponents. To deal with this in just the expanding case, we have that as long as \(\delta\) is small we can get

\[d(f^k(x), f^k(y)) \leq \alpha^k d(x,y),\]

where here we have \(\alpha = \max_{x \in S^1} \mid f'(x) \mid.\) The same result then follows easily. We can get the so-called “improved closing lemma” as well. Let’s state this one, since it’s useful.

Theorem: If \(f : S^1 \rightarrow S^1\) is an expanding map with \(\alpha = \max_{x \in S^1} \mid f'(x) \mid\), then there exists a \(\delta, C > 0\) so that if \(d(f^k(x), x) < \delta\) for \(k = 0, \ldots, n\) then there is a periodic point \(y\) so that \(f^n(y) = y\) and

\[d(f^k(y), f^k(x)) < C \alpha^k d(f^n(x), x).\]

Proof: Notice that we can apply the closing lemma as long as \(\delta\) is small since

\[d(f^k(x), f^{k+1}(x)) \leq d(f^k(x), x) + d(x, f^{k+1}(x)) < 2 \delta.\]

This gives us a periodic point \(y\) with

\[d(f^k(x), f^k(y)) < \delta \text{ for } 0 \leq k \leq n.\]

Now as long as we make sure \(\delta\) is small here, we can apply the above result to get

\[d(f^k(x), f^k(y)) < \alpha^k (d(x,y) + d(f^n(x), y)).\]

At the cost of some constant \(C\) we get

\[d(f^k(x), f^k(y)) < C \alpha^k d(f^n(x),x),\]

as desired.