# Anosov closing lemma: how to control the exact period

Question: I would like to state and prove Anosov's closing lemma in such a way that we know the exact period of the orbit.

Framework: $$\varphi : \mathbb R^n \longrightarrow \mathbb R^n$$ is a smooth diffeomorphism and $$\Lambda \subset \mathbb R^n$$ is a hyperbolic set. Here is how I was presented Anosov's closing lemma :

Take $$\epsilon, \delta > 0$$ as in the shadowing lemma. Suppose that for some $$x \in \Lambda$$ there exists an $$N \in \mathbb N^*$$ such that $$|| \varphi^N(x) - x|| < \delta$$ Then there exists a periodic orbit of $$\varphi$$ that is in $$\Lambda + B(0,\epsilon)$$.

Here is how I state Anosov's shadowing lemma:

There exists $$\epsilon_0 > 0$$ such that forall $$\epsilon \in (0, \epsilon_0)$$, there exists $$\delta > 0$$ such that forall $$(q_j)_{j \in \mathbb Z}$$ $$\delta$$-pesudo orbit of $$\varphi$$ that is in $$\Lambda + B(0,\delta)$$, there exists a unique $$\epsilon$$-shadowing orbit of $$(q_j)$$.

The idea of the proof of the closing lemma is to consider the $$\delta$$-pseudo orbit $$q_j = \varphi^{j ~\mathrm{mod} ~N}(x)$$ and apply the shadowing lemma. By uniqueness one can prove the resulting shadowing orbit is $$N$$-periodic.

Discussion: In the last idea of proof one don't prove that $$N$$ is the exact period. Considering $$x$$ fixed point of $$\varphi$$ and $$N > 1$$ it is clear we cannot ask for $$N$$ to be the period. I think I can achieve my goal asking $$N$$ to be minimal in $$\mathbb N^*$$. I don't see how to complete the proof of this last claim, in my opinion I don't have enough information about how the periodic point behaves as a function of the pseudo-orbit. In my course we have not covered the proof of the shadowing lemma. Any hint or reference to a proof including ideas on how to control the exact period are welcome.

Let $$A$$ be a hyperbolic set for $$f:U \rightarrow M$$. Then there exists an open neighborhood $$V$$ containing $$A$$ and $$C$$, $$\varepsilon > 0$$ such that for $$\varepsilon < \varepsilon_0$$ and any periodic $$\varepsilon$$-orbit $$(x_0,..., x_m) \in V$$, there is a point $$y\in U$$ such that $$f^m(y) = y$$ and $$dist(f^k(y), x_k) < C\varepsilon$$ for $$k = 0,..., m-1$$.
If $$\epsilon$$ is small with respect to the mutual distances $$d(x_i, x_j)$$, the $$x_j$$ being assumed to be mutually distinct, then since the point $$f^i(y)$$ is very close to $$x_i$$, it cannot be close to $$x_j$$ and so neither to $$f^j(y)$$. This guarantees that the smallest period of $$y$$ is $$m$$. This is better seen on a figure, with small disjoint balls around the $$x_i's$$ and the $$f^i(y)$$ belonging to these balls.