## On C^1 closing lemma

May 25, 2010

Let $f: M \rightarrow M$ be a diffeomorphism. A point $p$ is non-wandering if for all neighborhood $U$ of $p$, there is increasing sequence $(n_k) \subseteq \mathbb{N}$ where $U \cap f^{n_k}(U) \neq \phi$. We write $p \in \mathcal{NW}(f)$.

Closing lemma: For any diffeomorphism $f: M \rightarrow M$, for any $p \in \mathcal{NW}(f)$. For all $\varepsilon>0$ there exists diffeomorphism $g$ s.t. $||f-g||_{C^1} < \varepsilon$ and $g^N(p) = p$ for some $N \in \mathbb{N}$.

Suppose $p \in \mathcal{NW}(f)$, $\overline{\mathcal{O}(p)}$ is compact, then for any $\varepsilon>0$, there exists $x_0 \in B(p, \varepsilon)$, $k \in \mathbb{N}$ s.t. $f^k(x) \in B(p, \varepsilon)$.

First we apply a selection process to pick an appropriate almost-orbit for the closing. Set $x_i = f^i(x_0), \ 0 \leq i \leq k$.

If there exists $0 < j < k$ where

$\min \{ d(x_0, x_j), d(x_j, x_k) \} < \sqrt{\frac{2}{3}}d(x_0, x_k)$

then we replace the origional finite sequence by $(x_0, x_1, \cdots, x_j)$ or $(x_j, \cdots, x_k)$. Iterate the above process. since the sequence is at least one term shorter after each shortening, the process stops in finite time. We obtain final sequence $(p_0, \cdots, p_n)$ s.t. for all $0 < i < n$,

$\min \{ d(p_0, p_i), d(p_i, p_n) \} \geq \sqrt{\frac{2}{3}}d(p_0, p_n)$.

Since the process is applied at most $k$ times, $x_0, x_k \in B(p, \varepsilon)$, after the first shortening, $d(p, x_{i_1}) \leq \max \{d(p, x_0), d(p, x_k) \} + \sqrt{\frac{2}{3}}d(x_0, x_k)$ $\leq \varepsilon + 2 \sqrt{\frac{2}{3}} \varepsilon$.

i.e. both initial and final term of the sequence is at most $(\frac{1}{2}+ \sqrt{\frac{2}{3}}) 2 \varepsilon$. Along the same line, we have, at the $i$-th shortening, the distance between the initial and final sequence and $p$ is at most $(\frac{1}{2} + \sqrt{\frac{2}{3}} + (\sqrt{\frac{2}{3}})^2 + \cdots (\sqrt{\frac{2}{3}})^i) 2 \varepsilon$. Hence for the final sequence $p_0, p_n \in B(p, 1+2 \sqrt{\frac{2}{3}}/(1-\sqrt{\frac{2}{3}}) \varepsilon) \subseteq B(p, 10 \varepsilon)$.

There is a rectangle $R \subseteq M$ where $p_0, p_n \in \sqrt{\frac{3}{4}}R$
(i.e. shrunk $R$ by a factor of $\sqrt{\frac{3}{4}}$ w.r.t. the center) and for all $0 < i < n, \ p_i \notin R$.

Next, we perturb $f$ in $R$ i.e. find $h: M \rightarrow M$ with $||h||_{C^1} < \delta$ and $h|_{M \backslash R} =$ id. Hence $||h \circ f - f ||_{C^1} < \delta$.

Suppose $R = I_1 \times I_2; L_1, L_2$ are the lengths of $I_1, I_2$, $L_1 < L_2$.
By main value theorem, for all $x \in M, \ d(x, h(x)) < \delta L_1$.
On the other hand, since $p_0 \in \sqrt{\frac{3}{4}}R$, it's at least $\frac{1}{2}(1-\sqrt{\frac{3}{4}})L_1$ away from the boundary of $R$. i.e. there exists bump function $h$ satisfying the above condition and $d(p_0, h(p_0)) > \frac{\delta}{8}(1-\sqrt{\frac{3}{4}})L_1$.

Hence in order to move a point by a distance $L_1$, we need about $1/ \delta$ such bump functions, to move a distance $L_2$, we need about $\frac{L_2}{\delta L_1}$ bumps.

For simplicity, we now suppose $M$ is a surface. By starting with an $\varepsilon$ (and hence $R$) very small, we have for all $0 \leq i \leq N+M, \ f^i(R)$ is contained in a small neighbourhood of $p_i$. Hence on $f^i(B), f^i$ is $C^1$ close to the linear map $p_i + Df^i(p_0)(x-p_0)$. Hence mod some details we may reduce to the case where $f$ is linear in a neighborhood of $\mathcal{O}(p_0)$.

By choosing appropiate coordinate system in $R$, we can have $f$ preserving the horizontal and vertical foliations and the horizontal vectors eventually grow more rapidly than the vertical vectors.

It turns out to be possible to choose $R$ to be long and thin such that for all $i \leq 40 / \delta$, $f^i(R)$ has height greater than width. (note that $M = \lfloor 40/ \delta \rfloor$ bumps will be able to move the point by a distance equal to the width of the original rectangle $R$. Since horizontal vectors eventually grow more rapidly than the vertical vectors, there exists $N$ s.t. for all $N \leq i \leq N+M$, $f^i(R)$ has width greater than its height.
For small enough $\epsilon$, the boxes $f^i(R)$ are disjoint for $0 \leq i \leq N+40/ \delta$. Construct $h$ to be identity outside of

$\displaystyle \bigsqcup_{i=0}^M f^i(R) \sqcup \bigsqcup_{i=N}^{N + M} f^i(R)$

For the first $M$ boxes, we let $h$ preserve the horizontal foliation and move along the width so that $g = h \circ f$ has the property that $g^M(p_n)$ lies on the same vertical fiber as $f^M(p_0)$.

On the boxes $f^{N+i}(R), \ 0 \leq i \leq M$, we let $h$ pushes along the vertical direction so that

$g^{N+M}(p_n) = f^{N+M}(p_0)$

Since iterates of the rectangle are disjoint, for $N+M \leq i \leq n, \ h(p_i) = p_i$, $g(p_i) = f(p_i)$.

Hence $g^n(p_n) = g^{n-(N+M)} \circ g^{N+M}(p_n)$ $= g^{n-(N+M)} f^{N+M}(p_0) = g^{n-(N+M)} (p_{N+M}) = p_n$.

Therefore we have obtained a periodic point $p_n$ of $g$.

Since $p_n \in B(p, 10 \varepsilon)$, we may further perturb $g$ to move $p_n$ to $p$. This takes care of the linear case on surfaces.

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