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Let $\Omega$ and $M$ be compact $C^\infty$ manifolds, let $\theta \colon \Omega \to \Omega$ be a $C^\infty$ diffeomorphism, and let $\Theta \colon \Omega \times M \to \Omega \times M$ be a $C^\infty$ diffeomorphism such that $\Theta(\{\omega\} \times M)=\{\theta\omega\} \times M$ for all $\omega \in \Omega$.

Let $A \colon \Omega \to M$ be a continuous function such that $\Theta(\mathrm{graph}\,A)=\mathrm{graph}\,A$, and suppose that

$$ \limsup_{n \to \infty} \max_{\omega \in \Omega} \tfrac{1}{n} \log \left\| \left. \frac{\mathrm{d}\Theta^n(\omega,\tilde{x})}{\mathrm{d}\tilde{x}} \right|_{\tilde{x}=A(\omega)} \right\| \ < \ 0. $$

Is it necessarily the case that for every neighbourhood $U$ of $\mathrm{graph}\,A$, there is a neighbourhood $V$ of $\mathrm{graph}\,A$ such that $\Theta^n(V) \subset U$ for all $n \geq 0$?

If the answer is no, does the answer become yes when we add the additional requirement that there exists at least one $\theta$-ergodic probability measure of full support in $\Omega$?


Remark. By Corollary 1.11 of https://iopscience.iop.org/article/10.1088/0951-7715/13/1/306/meta (Sturman & Stark, 2000), a sufficient condition for the above negative-limsup condition is that for every $\theta$-ergodic probability measure $\mathbb{P}$ the corresponding maximal Lyapunov exponent $\lambda_\mathbb{P}$ given by $$ \lambda_\mathbb{P} \overset{\mathbb{P}\textrm{-a.e.}\ \omega}{=} \lim_{n \to \infty} \tfrac{1}{n} \log \left\| \left. \frac{\mathrm{d}\Theta^n(\omega,\tilde{x})}{\mathrm{d}\tilde{x}} \right|_{\tilde{x}=A(\omega)} \right\| $$ is strictly negative.

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    $\begingroup$ I've realised that this is probably trivial. Cover graph$A$ by finitely many charts. Fix sufficiently large $n$ that max log norm $\partial_x(\Theta^n)(\omega,A(\omega))$ over all the charts is strictly negative; then take a nbhd $O$ of graph$A$ contained in the union of the charts such that log norm $\partial_x(\Theta^n)$ has strictly negative supremum over all the charts intersected with $O$. Then $\Theta^n$ is uniformly Lipschitz w.r.t. its second input across the whole of $O$, and hence $\Theta^{tn}(O)\to$ graph$A$ as $t\to\infty$, and therefore $\Theta^t(O)\to$ graph$A$ as $t\to\infty$. $\endgroup$ Commented Sep 1, 2021 at 4:47

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