A Note on the Relation Between the Metric Entropy and the Generalized Fractal Dimensions of Invariant Measures

Abstract

We investigate in this work some situations where it is possible to estimate or determine the upper and the lower q-generalized fractal dimensions $D^{\pm }_{\mu }(q)$, $q\in {\mathbb {R}}$, of invariant measures associated with continuous transformations over compact metric spaces. In particular, we present an alternative proof of Young’s Theorem by Young (Ergod. Theory Dyn. Syst. 2(1):109–124, 1982) for the generalized fractal dimensions of the Bowen-Margulis measure associated with a $C^{1+\alpha }$-Axiom A system over a two-dimensional compact Riemannian manifold M. We also present estimates for the generalized fractal dimensions of an ergodic measure for which Brin-Katok’s Theorem is pointwise satisfied, in terms of its metric entropy. Furthermore, for expansive homeomorphisms (like $C^1$-Axiom A systems), we show that the set of invariant measures such that $D_\mu ^+(q)=0$ ($q\ge 1$), under a hyperbolic metric, is generic (taking into account the weak topology). We also show that for each $s\in [0,1)$, $D^{+}_{\mu }(s)$ is bounded above, up to a constant, by the topological entropy, also under a hyperbolic metric. Finally, we show that, for some dynamical systems, the metric entropy of an invariant measure is typically zero, settling a conjecture posed by Sigmund (Trans. Am. Math. Soc. 190:285–299, 1974) for Lipschitz transformations which satisfy the specification property.

6 Appendix

Proof (Proposition 1.1)

The second inequalities in (2) and (3) are obvious. The equalities in (2) and (3) are just Proposition 1.1 in Carvalho and Condori (2021b). A proof of the first inequality in (2) and the last inequality in (3) is presented in Rudnicki (2002). We provide the proof of the last inequality in (2) and of the first inequality in (3).

We begin proving that, for each $q>1$, $D_{\mu }^+(q)\le \mu -{\text {ess inf}}\overline{d}_\mu (x)$. For each $\nu \in (0,1)$ and each $\varepsilon >0$, set

$$\begin{aligned}A_\nu (\varepsilon )=\{x\in X\mid \mu (B(x,\varepsilon ))\ge \varepsilon ^{a+\nu }\}.\end{aligned}$$

It is easy to check that

$$\begin{aligned}A_\nu :=\liminf _{\varepsilon \downarrow 0}A_\nu (\varepsilon )\supset B_\nu :=\{x\in X\mid \overline{d}_\mu (x)\le a+\nu /2\}\,. \end{aligned}$$

It follows from the definition of $\mu -{\text {ess inf}}$ that, for each $\nu >0$, there exists $\varepsilon _0>0$ such that, for each $\varepsilon \in (0,\min \{\varepsilon _0,1\})$, $2\mu (A_\nu (\varepsilon ))\ge \mu (A_\nu )\ge \mu (B_\nu )>0$. Now, for each $\varepsilon \in (0,\min \{\varepsilon _0,1\})$, one has

$$\begin{aligned}I_\mu (q,\varepsilon )= & {} \int \mu (B(x,\varepsilon ))^{q-1}\, d\mu (x)\ge \int _{A_\nu (\varepsilon )}\mu (B(x,\varepsilon ))^{q-1}\,d\mu (x)\\\ge & {} \varepsilon ^{(q-1)(a+\nu )}\frac{\mu (A_\nu )}{2}\,,\end{aligned}$$

which yields $D^+_\mu (q)\le a+\nu$. Since $\nu >0$ is arbitrary, one gets $D^+_\mu (q)\le \mu -{\text {ess inf}}\overline{d}_\mu (x)$.

The proof that, for each $s\in (0,1)$, $D_{\mu }^-(s)\ge \mu -{\text {ess sup}}\underline{d}_\mu (x)$ is completely analogous; namely, let $\mathfrak {h}=\mu -{\text {ess sup}}\underline{d}_\mu (x)$, assume that $\mathfrak {h}>0$ (otherwise, there is nothing to prove) and for each $\nu \in (0,\mathfrak {h})$ and each $\varepsilon >0$, set

$$\begin{aligned} \tilde{A}_\nu (\varepsilon ):=\{x\in X\mid \mu (B(x,\varepsilon ))\le \varepsilon ^{\mathfrak {h}-\nu }\}\,. \end{aligned}$$

Then,

$$\begin{aligned} \tilde{A}_\nu :=\liminf _{\varepsilon \downarrow 0}\tilde{A}_\nu (\varepsilon )\supset \tilde{B}_\nu :=\{x\in X\mid \underline{d}_\mu (x)\ge \mathfrak {h}-\nu /2\}\,, \end{aligned}$$

so for each $\nu >0$, there exists $\varepsilon _0>0$ such that, for each $\varepsilon \in (0,\min \{\varepsilon _0,1\})$, $2\mu (\tilde{A}_\nu (\varepsilon ))\ge \mu (\tilde{A}_\nu )\ge \mu (\tilde{B}_\nu )>0$. Now, for each $\varepsilon \in (0,\min \{\varepsilon _0,1\})$, one has

$$\begin{aligned}I_\mu (s,\varepsilon )\ge \int _{\tilde{A}_\nu (\varepsilon )}\mu (B(x,\varepsilon ))^{s-1}\,d\mu (x)\ge \varepsilon ^{(s-1)(\mathfrak {h}-\nu )}\frac{\mu (\tilde{A}_\nu )}{2}\,,\end{aligned}$$

which yields $D^-_\mu (s)\ge \mathfrak {h}-\nu$. Since $\nu >0$ is arbitrary, one gets $D^-_\mu (s)\ge \mu -{\text {ess sup}}\underline{d}_\mu (x)$. $\square$