# Diffeomorphisms with Shadowable Measures
## 1 Introduction
: \(f\in\text{int}\mathcal{PS}\) _if and only if \(f\) satisfies both Axiom A and the strong transversality condition._  
The proof is based on the main result of [7], and the result is obtained by proving that \(\text{int}\mathcal{PS}\subset\text{int}\mathcal{S}\) (see Section 3). Denote by \(\mathcal{F}^{1}(M)\) the set of \(f\in\text{Diff}(M)\) such that there is a \(C^{1}\)-neighborhood \(\mathcal{U}(f)\) of \(f\) with property that every \(p\in P(g)\)\((g\in\mathcal{U}(f))\) is hyperbolic. It is proved by [PERSON] [10] that \(f\in\mathcal{F}^{1}(M)\) if and only if \(f\) satisfies both Axiom A and no-cycle condition. The proof of the second result is based on [PERSON], and the result is obtained by proving that \(\text{int}\mathcal{PIS}\subset\mathcal{F}^{1}(M)\) (see Section 4). **Theorem 2**. : \(f\in\text{int}\mathcal{PIS}\) _if and only if \(f\) satisfies both Axiom A and the no-cycle condition._  
Owing to the above two theorems, the relationship between uniform hyperbolicity and \(C^{1}\)-robustness of the existence of shadowable measures turns out to be clear. **2. Examples**  
Recall that if a dynamical system has the shadowing property, then any probability measure is shadowable. The dynamical systems considered in this section do not have the shadowing property. First two examples have a measure that is not shadowable, but the last example has a shadowable measure. **Example 1**. : _(The identity map of the unit interval)._  
Let \(I\) be the unit interval \([0,1]\), and let \(id_{I}:I\to I\) be the identity map, that is, \(id_{I}(x)=x\) for \(x\in I\). It is well-known that the identity map of \(I\) does not have the shadowing property. Let \(\mu\) be the Lebesgue probability measure on \(I\). Then we can see that there exists \(\epsilon_{0}>0\) such that for any \(n\in\mathbb{N}\) and \(\delta>0\), there is a subset  
\[\Gamma_{\delta}^{n}\subset\mathcal{PO}(\delta,n)\]  
with the property that \(\mu_{n}(\Gamma_{\delta}^{n})>0\) and non of any element in \(\Gamma_{\delta}^{n}\) can be \(\epsilon_{0}\)-shadowed. For any integer \(m>0\) and \(0\leq i\leq 3m-1\), denote by \(I_{i}^{m}=[\frac{i}{3m},\frac{i+1}{3m}]\). First we put \(J_{i}^{m}=I_{i}^{m}\) for \(0\leq i\leq 3m-1\), and then set \(J_{i}^{m}=I_{3m-1}^{m}\) for \(i\geq 3m\). It is easy to see that for any \(x_{i}\in J_{i}^{m}\)\((i\geq 0)\),  
\[d(id_{I}(x_{i}),x_{i+1})=d(x_{i},x_{i+1})\leq\frac{2}{3m}<\frac{1}{m}\]  
so that \(\{x_{i}\}_{i=0}^{\infty}\) is a \(\frac{1}{m}\)-pseudo-orbit of \(id_{I}\). Set  
\[\Gamma_{\delta}^{n}=\prod_{i=0}^{n-1}J_{i}^{m}\subset X^{n}.\]  
Then  
\[\mu_{n}\left(\mathcal{PO}\left(\frac{1}{m},n\right)\right)\geq\mu_{n}\left( \Gamma_{\delta}^{n}\right)=\mu_{n}\left(\prod_{i=0}^{n-1}|J_{i}^{m}|\right)= \left(\frac{1}{3m}\right)^{n}>0\]  
for \(n>0\). Since \(id_{I}\) is the identity map, it is not hard to show that there exists \(\epsilon_{0}>0\) (say, \(\epsilon_{0}=1/3\)) such that for any \(\frac{1}{m}\)-pseudo-orbit \(\{x_{i}\}_{i=0}^{\infty}\)\((m\gg 0)\) of \(id_{I}\), there is \(n\) such that \(\{x_{i}\}_{i=0}^{n-1}\) cannot be \(\epsilon_{0}\)-shadowed by any point \(y\in I\). **Example 2**. : _(Rotation maps of the unit circle)._Let \(S^{1}=\{e^{2\pi i\theta}:\theta\in\mathbb{R}\}\subset\mathbb{C}\), and define a metric on \(S^{1}\) by  
\[\|Df^{n}\|_{E^{s}_{x}}\|\leq C\lambda^{n}\ \text{ and }\ \|Df^{-n}\|_{E^{u}_{x}}\|\leq C\lambda^{n}\]0  
For \(\alpha\in\mathbb{R}\), let \(\rho_{\alpha}:S^{1}\to S^{1}\) be a rotation map defined by \(\rho_{\alpha}(e^{2\pi i\theta})=e^{2\pi i(\alpha+\theta)}\). Then the map \(\rho_{\alpha}\) does not have the shadowing property. In fact, the identity map \(id_{S^{1}}\) does not have the shadowing property. For any (small) \(\delta>0\), put \(x_{j}=e^{2\pi i\beta}\ (j\in\mathbb{N})\). Then since  
\[\|Df^{n}\|_{E^{s}_{x}}\|\leq C\lambda^{n}\ \text{ and }\ \|Df^{-n}\|_{E^{u}_{x}}\|\leq C\lambda^{n}\]1  
for any \(j\in\mathbb{N}\), the sequence \(\{x_{j}\}_{j=0}^{\infty}\) is a \(\delta\)-pseudo-orbit of \(id_{S^{1}}\).