Proof of Proposition 3.1

We first prove that, for each $n\in {\Bbb N}$ ,

(3.2) $$ \begin{align} P(X, \sigma, \mathcal G)\leq \frac{1}{n}P(X^{(n)},\sigma^n, g_n). \end{align} $$

To see this, fix $n\in {\Bbb N}$ and let $\mu $ be an equilibrium measure for the potential ${\mathcal G}$ . Then

$$ \begin{align*} P(X, \sigma, \mathcal G)&=h_\mu(\sigma)+\mathcal G_*(\mu)\\ &\leq h_\mu(\sigma)+\frac{1}{n}\int g_n\, d\mu\quad\mbox{(by (2.5))} \nonumber \\&=\frac{1}{n}\bigg( h_\mu(\sigma^n)+\int g_n\, d\mu\bigg)\\ &\leq \frac{1}{n}P(X^{(n)},\sigma^n, g_n), \end{align*} $$

where in the last inequality, we use the fact that $\mu \in \mathcal M(X^{(n)}, \sigma ^n)$ and the classical variational principle for the topological pressure of additive potentials. This proves (3.2).

In what follows we prove that

(3.3) $$ \begin{align} P(X, \sigma, \mathcal G)\geq \limsup_{n\to \infty}\frac{1}{n}P(X^{(n)},\sigma^n, g_n). \end{align} $$

Clearly (3.2) and (3.3) imply (3.1). To prove (3.3), by the classical variational principle we can take a subsequence $\{n_i\}$ of natural numbers and $\nu _{n_i}\in \mathcal M(X^{(n_i)}, \sigma ^{n_i})$ such that

(3.4) $$ \begin{align} \limsup_{n\to \infty}\frac{1}{n}P(X^{(n)},\sigma^n, g_n)=\lim_{i\to \infty}\frac{1}{n_i} \bigg(h_{\nu_{n_i}}(\sigma^{n_i})+\int g_{n_i}\, d\nu_{n_i}\bigg). \end{align} $$

Set $\mu ^{(i)}=({1}/{n_i})\sum _{k=0}^{n_i-1}\nu _{n_i}\circ \sigma ^{-k}$ for each i. Taking a subsequence if necessary, we may assume that $\mu ^{(i)}$ converges to an element $\mu \in \mathcal M(\mathcal A^{\Bbb N})$ in the weak* topology. By Lemma 3.3, $\mu \in {\mathcal M}({\mathcal A}^{\Bbb N}, \sigma )$ and, moreover,

(3.5) $$ \begin{align} \limsup_{i\to\infty}\frac{1}{n_i}\int g_{n_i}(x)\,d\nu_{n_i}(x)\leq \mathcal G_*(\mu). \end{align} $$

Next, we show that $\mu $ is supported on X. For this we adopt some arguments from the proof of [Reference Kenyon and Peres30, Theorem 1.1]. Notice that, for each i, $\mu ^{(i)}$ is $\sigma $ -invariant supported on

$$ \begin{align*} \bigcup_{k=0}^{n_i-1}\sigma^k X^{(n_i)}=\bigcup_{k=1}^{n_i}\sigma^{n_i-k} X^{(n_i)}. \end{align*} $$

Hence, $\mu $ is supported on

$$ \begin{align*} \bigcap_{N=1}^\infty \overline{ \bigcup_{i=N}^\infty \bigcup_{k=1}^{n_i} \sigma^{n_i-k} X^{(n_i)}}. \end{align*} $$

If x is in this set, then, for each $N\geq 1$ , there exist integers $i(N)\geq N$ and $k(N)\in [1, n_{i(N)}]$ for which $d(x, \sigma ^{n_{i(N)}-k(N)}X^{(n_{i(N)})})<1/N$ (i.e., $d(x,z)<1/N$ for some $z\in \sigma ^{n_{i(N)}-k(N)}X^{(n_{i(N)})}$ ), hence

(3.6) $$ \begin{align} d(x, X)\leq d(x,z)+d(z, X)< \frac{1}{N}+ \sup_{y\in \sigma^{n_{i(N)}-k(N)}X^{(n_{i(N)})} } d(y, X) \leq \frac{1}{N}+2^{-k(N)} \end{align} $$

and

(3.7) $$ \begin{align} d(\sigma^{k(N)}x, X)&\leq d(\sigma^{k(N)}x,\sigma^{k(N)}z)+d(\sigma^{k(N)}z, X)\nonumber\\ &\leq 2^{k(N)}d(x,z)+d(\sigma^{k(N)}z, X) \nonumber \\&<\frac{2^{k(N)}}{N}+\sup_{y\in \sigma^{n_{i(N)}-k(N)}X^{(n_{i(N)})} } d(\sigma^{k(N)}y, X)\nonumber\\ &\leq \frac{2^{k(N)}}{N}+2^{-n_{i(N)}}\quad \mbox{(since}\ \sigma^{k(N)}y\in \sigma^{n_{i(N)}}X^{(n_{i(N)})}=X^{(n_{i(N)})}\mbox{)}. \end{align} $$

If the values $k(N)$ are unbounded as $N\to \infty $ , then (3.6) yields $x\in X$ , while if they are bounded then some value of k recurs infinitely often as $k(N)$ , which implies that $\sigma ^kx\in X$ by (3.7). Thus $\mu $ is supported on

$$ \begin{align*} \bigcup_{k=0}^\infty\sigma^{-k}X. \end{align*} $$

Since $\sigma X\subset X$ , the set $(\sigma ^{-1} X)\backslash X $ is wandering under $\sigma ^{-1}$ (i.e., its preimages under powers of $\sigma $ are disjoint), so it must have zero $\mu $ -measure. Consequently, $\mu \in \mathcal M(X,\sigma )$ .

Notice that $h_{\mu ^{(i)}}(\sigma )=({1}/{n_i})h_{\nu _{n_i}}(\sigma ^{n_i})$ (see Lemma 2.4). By the upper semi-continuity of the entropy map,

$$ \begin{align*} h_\mu(\sigma)\geq \limsup_{i\to \infty}h_{\mu^{(i)}}(\sigma)=\limsup_{i\to \infty}\frac{1}{n_i}h_{\nu_{n_i}}(\sigma^{n_i}), \end{align*} $$

which, together with (3.5), yields that

$$ \begin{align*} h_\mu(\sigma)+\mathcal G_*(\mu)&\geq \limsup_{i\to \infty}\frac{1}{n_i} \bigg(h_{\nu_{n_i}}(\sigma^{n_i})+\int g_{n_i} \, d\nu_{n_i}\bigg)\\ &=\limsup_{n\to \infty}\frac{1}{n}P(X^{(n)},\sigma^n, g_n). \end{align*} $$

Applying Theorem 2.2, we obtain (3.3). This completes the proof of the proposition.

Proof. The result is well known. For the reader’s convenience, we include a proof.

Define, for $n\in {\Bbb N}$ ,

$$ \begin{align*}\mbox{var}_nf=\sup\{|f(x)-f(y)|: x_i=y_i \mbox{ for all } 1\leq i\leq n\}.\end{align*} $$

Since f is uniformly continuous, $\mbox {var}_nf\to 0$ as $n\to \infty $ . It follows that

$$ \begin{align*}\lim_{n\to \infty} \frac{1}{n}\sum_{i=1}^n \mbox{var}_if=0.\end{align*} $$

This concludes the result of the lemma since

$$ \begin{align*}\sup\{|S_nf(x)-S_nf(y)|: x_i=y_i \mbox{ for all } 1\leq i\leq n\}\end{align*} $$

is bounded above by $\sum _{i=1}^n \mbox {var}_if$ .

Proof of Proposition 3.4

Set

$$ \begin{align*} \Theta_r=\sum_{I\in \mathcal A_r}\sup_{x\in [I]\cap X} \exp(S_{|I|}g(x)),\quad r\in (0, r_0). \end{align*} $$

Let $\epsilon>0$ . It is enough to show that

(3.9) $$ \begin{align} r^{-t+\epsilon}\leq \Theta_r\leq r^{-t-\epsilon} \end{align} $$

for sufficiently small r.

To this end, set, for $0<r<r_0$ ,

$$ \begin{align*} m(r)=\min\{|I|: I\in \mathcal A_r\}, \quad M(r)=\max\{|I|: I\in \mathcal A_r\}. \end{align*} $$

From the definition of $\mathcal A_r$ and the negativity of h, it follows that there exist two positive constants $a, b$ such that

(3.10) $$ \begin{align} a\log(1/r)\leq m(r) \leq M(r)\leq b \log (1/r)\quad \text{for all } r\in (0, r_0). \end{align} $$

Define

$$ \begin{align*} \Gamma_r=\sum_{I\in \mathcal A_r}\sup_{x\in [I]\cap X} \exp(S_{|I|}(g+th)(x)),\quad r\in (0, r_0). \end{align*} $$

By Lemma 3.5 and (3.10), it is readily checked that

(3.11) $$ \begin{align} r^{t+\epsilon/2} \Theta_r\leq \Gamma_r\leq r^{t-\epsilon/2} \Theta_r \quad \mbox{for sufficiently small }r, \end{align} $$

Hence, to prove (3.9), it suffices to prove that $r^{\epsilon /2}\leq \Gamma _r\leq r^{-\epsilon /2}$ for small r.

We first prove $\Gamma _r>r^{\epsilon /2}$ when r is small. Suppose to the contrary that this is not true. Then by (3.10) we can find some $r\in (0, r_0)$ and $\unicode{x3bb}>0$ such that $Z(r, \unicode{x3bb} )<1$ , where

(3.12) $$ \begin{align} Z(r, \unicode{x3bb}):=\sum_{I\in \mathcal A_r}\exp(\unicode{x3bb} |I|) \sup_{x\in [I]\cap X} \exp(S_{|I|}(g+th)(x)). \end{align} $$

Observe that $\{[I]: I\in \mathcal A_r\}$ is a cover of X. From [Reference Bowen7] it follows that $P(X, \sigma , g+th)\leq -\unicode{x3bb} $ , contradicting the fact that $P(X, \sigma , g+th)=0$ . Hence, we have $\Gamma _r>r^{\epsilon /2}$ when r is sufficiently small.

Next, we prove the inequality $\Gamma _r\leq r^{-\epsilon /2}$ for small r. To do this, fix $\unicode{x3bb} \in (0, \epsilon /(2b))$ , where b is the constant in (3.10). We claim that there exists $0<r_1<r_0$ such that

(3.13) $$ \begin{align} Z(r, -\unicode{x3bb})< 1\quad \mbox{for all } r\in (0, r_1), \end{align} $$

where Z is defined as in (3.12). Since $\unicode{x3bb} \in (0, \epsilon /(2b))$ , it follows from (3.10) that, for any $I\in \mathcal A_r$ ,

$$ \begin{align*}\exp(-\unicode{x3bb}|I|)\geq \exp(-\unicode{x3bb} b\log(1/r))=r^{b\unicode{x3bb}}\geq r^{\epsilon/2}.\end{align*} $$

Hence, (3.13) implies that $\Gamma _r\leq r^{-\epsilon /2}$ for $0<r<r_1$ .

Now it remains to prove (3.13). Since $P(X, \sigma , g+th)=0$ , by definition we have

$$ \begin{align*} \lim_{n\to \infty} \frac{1}{n}\log \bigg(\sum_{I\in X^*: \; |I|=n} \sup_{x\in [I]\cap X} \exp(S_{|I|}(g+th)(x))\bigg)=0. \end{align*} $$

Hence, there exists a large N such that $e^{-\unicode{x3bb} N/2}<1-e^{-\unicode{x3bb} /2}$ and, for any $n>N$ ,

$$ \begin{align*} \gamma_n:=\sum_{I\in X^*: \; |I|=n} \exp(-\unicode{x3bb} |I|) \sup_{x\in [I]\cap X} \exp(S_{|I|}(g+th)(x))\leq \exp(-\unicode{x3bb} n/2). \end{align*} $$

Take a small $r_1\in (0, r_0)$ so that $m(r)\geq N$ for any $0<r<r_1$ . By the definition of $m(r)$ , for any $0<r<r_1$ we have $\mathcal A_r\subset \{I\in X^*:|I|\geq N\}$ and so

$$ \begin{align*} Z(r, -\unicode{x3bb})\leq \sum_{n=N}^\infty \gamma_n\leq \sum_{n=N}^\infty \exp(-\unicode{x3bb} n/2)=\frac{e^{-\unicode{x3bb} N/2}}{1-e^{-\unicode{x3bb}/2}}<1. \end{align*} $$

This proves (3.13).

Proof. The result was implicitly proved in [Reference Zhang40, Lemma 3] by using an idea of [Reference Douady and Oesterlé12]. For the convenience of the reader, we provide a detailed proof.

Set

$$ \begin{align*} \gamma=\min_{y\in E}\alpha_d(D_yf). \end{align*} $$

Then

(4.1) $$ \begin{align} B(0, \gamma)\subset D_yf(B(0,1)) \quad\mbox{for all }y\in E. \end{align} $$

Since f is $C^1$ , non-degenerate on U and E is compact, it follows that $\gamma>0$ . Take $\epsilon =(2\sqrt {d}-1)/2$ . Then there exists a small $r_0>0$ such that, for $u, v, w\in V_{2r_0}(E):=\{x: d(x, E)<2r_0\}$ ,

(4.2) $$ \begin{align} |f(u)-f(v)-D_vf(u-v)|\leq \epsilon \gamma |u-v| \quad \mbox{ if }|u-v|\leq r_0, \end{align} $$

and

(4.3) $$ \begin{align} D_vf(B(0,1))\subset ((1+\epsilon)D_wf)(B(0,1))\quad \mbox{ if }|v-w|\leq r_0. \end{align} $$

Now let $y\in E$ and $z\in B(y, r_0)$ . For any $0<r<r_0$ and $x\in B(z, r)$ , taking $u=x$ and $v=z$ in (4.2) gives

$$ \begin{align*} f(x)-f(z)-D_zf(x-z)\in B(0, \epsilon \gamma r), \end{align*} $$

so by (4.3) and (4.1),

$$ \begin{align*} f(x)-f(z) &\in D_zf(B(0,r))+B(0, \epsilon\gamma r)\\ &\quad\subset ((1+\epsilon) D_yf)(B(0,r))+B(0, \epsilon \gamma r) \quad (\mbox{by (4.3)})\\ &\quad\subset D_yf(B(0,(1+\epsilon)r)+ B(0,\epsilon r)) \;\quad (\mbox{by (4.1)})\\ &\quad\subset D_yf(B(0,(1+2\epsilon)r))\\ &\quad = D_yf(B(0,2\sqrt{d}r)), \end{align*} $$

where $A+A':=\{u+v: u\in A, \; v\in A'\}$ . Therefore,

$$ \begin{align*} f(B(z, r))\subset f(z)+ D_yf(B(0,2\sqrt{d}r)). \end{align*} $$

That is, $f(B(z, r))$ is contained in an ellipsoid which has principle axes of lengths $4\sqrt {d}\alpha _i(D_yf)r$ , $i=1,\ldots , d$ . Hence, $f(B(z, r))$ is contained in a rectangular parallelepiped of side lengths $2\sqrt {d}\alpha _i(D_yf)r$ , $i=1,\ldots , d$ . Now we can divide such a parallelepiped into at most

$$ \begin{align*} \bigg(\prod_{i=1}^{k+1}\frac{4d\alpha_i(D_yf)}{\alpha_{k+1}(D_yf)}\bigg)\cdot (4d)^{d-k-1} \leq (4d)^d \cdot\frac{\phi^k(D_yf)}{(\alpha_{k+1}(D_yf))^k} \end{align*} $$

cubes of side $({2}/{\sqrt {d}})\cdot \alpha _{k+1}(D_yf)r$ . Therefore, this parallelepiped (and $f(B(z, r))$ as well) can be covered by

$$ \begin{align*}(4d)^d \cdot\frac{\phi^k(D_yf)}{(\alpha_{k+1}(D_yf))^k}\end{align*} $$

balls of radius $\alpha _{k+1}(D_yf)r$ .

Proof. Since $\{f_i\}_{i=1}^\ell $ is a $C^1$ IFS, there exists an open set $U\supset K$ such that each $f_i$ extends to a $C^1$ diffeomorphism $f_i: U\to f_i(U)$ . Applying Lemma 4.1 to the mappings $f_i$ , we see that there exists $r_0>0$ such that, for any $y\in K$ , $z\in B(y, r_0)$ , $0<r<r_0$ and $i\in \{1,\ldots , \ell \}$ , the set $f_i(B(z, r))$ can be covered by

$$ \begin{align*}\theta(y,i):=\frac{C\phi^k(D_{y}f_{i})}{\alpha_{k+1}(D_{y}f_{i})^k}\end{align*} $$

balls of radius $\alpha _{k+1}(D_{y}f_{i})r$ .

Since $f_1,\ldots , f_\ell $ are contracting on U, there exists $\gamma \in (0,1)$ such that

$$ \begin{align*} |f_i(x)-f_i(y)|\leq \gamma|x-y| \quad\mbox{for all }x, y\in U,\; i\in \{1,\ldots, \ell\}. \end{align*} $$

This implies that $\alpha _{1}(D_yf_i)\leq \gamma $ for any $y\in K$ and $i\in \{1,\ldots , \ell \}$ . Take a large integer $n_0$ such that

$$ \begin{align*}\gamma^{n_0}\max\{1,{\textrm{diam}}(K)\}<r_0/2.\end{align*}$$

Clearly there exists a large number $C_1$ so that the conclusion of the proposition holds for any positive integer $n\leq n_0$ and $\mathbf {i}\in \Sigma $ , that is, the set $f_{\mathbf {i}|n}(K)$ can be covered by $C_1\prod _{p=0}^{n-1}G(\sigma ^p \mathbf {i})$ balls of radius $\prod _{p=0}^{n-1}H(\sigma ^p \mathbf {i})$ . Below we show by induction that this holds for all $n\in {\Bbb N}$ and $\mathbf {i}\in \Sigma $ .

Suppose, for some $m\geq n_0$ , that the conclusion of the proposition holds for any positive integer $n\leq m$ and $\mathbf {i}\in \Sigma $ . Then, for ${\textbf {i}}\in \Sigma $ , $f|_{(\sigma {\textbf {i}})|m}(K)$ can be covered by $C_1\prod _{p=0}^{m-1}G(\sigma ^{p+1} \mathbf {i})$ balls of radius $\prod _{p=0}^{m-1}H(\sigma ^{p+1} \mathbf {i})$ . Let $B_1,\ldots B_N$ denote these balls. We may assume that $B_j\cap f|_{(\sigma {\textbf {i}})|m}(K)\neq \emptyset $ for each j. Since

$$ \begin{align*}\prod_{p=0}^{m-1}H(\sigma^{p+1} \mathbf{i})\leq \gamma^m\leq \gamma^{n_0}<r_0/2\end{align*} $$

and

$$ \begin{align*}d(\Pi\sigma\mathbf{i}, B_j\cap f_{(\sigma {\textbf{i}})|m}(K))\leq {\textrm{diam}}(f_{(\sigma {\textbf{i}})|m}(K))\leq \gamma^{n_0}{\textrm{diam}}(K)<r_0/2,\end{align*} $$

the center of $B_j$ is in $B(\Pi \sigma {\textbf {i}}, r_0)$ . Therefore, $f_{i_1}(B_j)$ can be covered by $\theta (\Pi \sigma \mathbf {i}, i_1)=G(\bf {i})$ balls of radius

$$ \begin{align*}H({\textbf{i}})\cdot \prod_{p=0}^{m-1}H(\sigma^{p+1} \mathbf{i})=\prod_{p=0}^{m}H(\sigma^{p} \mathbf{i}).\end{align*} $$

Since $f_{\mathbf {i}|{(m+1)}}(K)\subset \bigcup _{j=1}^N f_{i_1}(B_j)$ , it follows that $f_{\mathbf {i}|{(m+1)}}(K)$ can be covered by

$$ \begin{align*}G({\textbf{i}})N\leq G({\textbf{i}}) \cdot C_1\prod_{p=0}^{m-1}G(\sigma^{p+1} {\textbf{i}})=C_1\prod_{p=0}^{m}G(\sigma^{p} {\textbf{i}})\end{align*} $$

balls of radius $\prod _{p=0}^{m}H(\sigma ^{p} \mathbf {i})$ . Thus the proposition also holds for $n=m+1$ and all ${\textbf {i}}\in \Sigma $ , as desired.

Proof. Write $g=\log G$ and $h=\log H$ for short. Define

$$ \begin{align*} r_{\min}=\min_{x\in \Sigma}\alpha_{k+1}(D_{\Pi\sigma x}f_{x_1}),\quad r_{\max}=\max_{x\in \Sigma}\alpha_{k+1}(D_{\Pi\sigma x}f_{x_1}). \end{align*} $$

Then $0<r_{\min } \leq r_{\max }<1$ . For $0<r<r_{\min }$ , define

(4.5) $$ \begin{align} \mathcal A_r=\big\{i_1\ldots i_n\in \Sigma^*: \sup_{x\in [i_1\ldots i_n]}S_nh(x)<\log r\leq \sup_{y\in [i_1\ldots i_{n-1}]}S_{n-1}h(y)\big\}; \end{align} $$

clearly $\{[I]: I\in \mathcal A_r\}$ is a partition of $\Sigma $ . By Proposition 4.2, there exists a constant $C_1>0$ such that, for each $0<r<r_{\min }$ , every $I\in \mathcal A_r$ and $x\in [I]$ , $f_I(K)$ can be covered by

$$ \begin{align*} C_1\exp(S_{|I|}g(x)) \leq C_1\exp\big(\sup_{y\in [I]}S_{|I|}g(y)\big) \end{align*} $$

balls of radius

$$ \begin{align*}\exp(S_{|I|}h(x))\leq \exp\big(\sup_{y\in [I]}S_{|I|}h(y)\big)<r.\end{align*} $$

It follows that K can be covered by

$$ \begin{align*} C_1\sum_{I\in \mathcal A_r}\exp\big(\sup_{y\in [I]}S_{|I|}g(y)\big) \end{align*} $$

balls of radius r. Hence, by Proposition 3.4,

$$ \begin{align*} \overline{\dim}_BK\leq \limsup_{r\to 0}\frac{\log\big(\sum_{I\in \mathcal A_r}\exp\big(\sup_{y\in [I]}S_{|I|}g(y)\big)\big)}{\log (1/r)}= t. \end{align*} $$

This completes the proof of the proposition.

Proof. The result is obtained by applying Proposition 4.3 to the IFS $\{f_{I}: I\in X^*_n\}$ instead of $\{f_i\}_{i=1}^\ell $ , where $X_n^*$ stands for the collection of words of length n allowed in X.

Proof of Theorem 1.1

Write $s=\dim _SX$ . We may assume that $s<d$ ; otherwise we have nothing left to prove. Set $k=[s]$ , that is, k is the largest integer less than or equal to s. Let $\mathcal U=\{u_n\}_{n=1}^\infty $ be the subadditive potential on $\Sigma $ defined by

$$ \begin{align*}u_n(x)=\log \phi^s(D_{\Pi\sigma^n x} f_{x|n}).\end{align*} $$

Then $P(X, \sigma , {\mathcal U})=0$ by the definition of $\dim _SX$ . Hence, by Proposition 3.1,

$$ \begin{align*} \lim_{n\to \infty}\frac{1}{n}P(X^{(n)}, \sigma^n, u_n)=\inf_{n\geq 1}\frac{1}{n}P(X^{(n)}, \sigma^n, u_n)=0. \end{align*} $$

It follows that, for each $\epsilon>0$ , there exists $N_\epsilon>0$ such that

(4.7) $$ \begin{align} 0\leq P(X^{(n)}, \sigma^n, u_n)\leq n\epsilon \quad\mbox{for }\; n>N_\epsilon. \end{align} $$

For $n\in {\Bbb N}$ , let $s_n$ be the unique real number such that $P(X^{(n)}, \sigma ^n, v_n)=0$ , where $v_n$ is a continuous function on $\Sigma $ defined by

$$ \begin{align*} v_n(x)&= \log G_n(x)+ s_n \log H_n(x)\\ &=\log ( C\phi^k(D_{\Pi\sigma^nx}f_{x|n})\alpha_{k+1}(D_{\Pi\sigma^nx}f_{x|n})^{s_n-k}), \end{align*} $$

where $G_n, H_n$ are defined in (4.6) and $C=(4d)^d$ . By Proposition 4.4,

$$ \begin{align*}\overline{\dim}_B\Pi(X)\leq \overline{\dim}_B\Pi(X^{(n)})\leq s_n.\end{align*} $$

If $s\geq s_n$ for some n, then $\overline {\dim }_B\Pi (X)\leq s_n\leq s$ and we are done. In what follows, we assume that $s<s_n$ for each n. Then, for each $x\in \Sigma $ ,

$$ \begin{align*} u_n(x)-v_n(x)&= -\log C +(s-s_n)\log \alpha_{k+1}(D_{\Pi\sigma^nx}f_{x|n})\\ &\geq -\log C +(s-s_n)\log \alpha_1(D_{\Pi\sigma^nx}f_{x|n})\\ &\geq -\log C+ n(s-s_n) \log \theta, \end{align*} $$

where

$$ \begin{align*} \theta:=\max_{y\in \Sigma} \alpha_1(D_{\Pi\sigma y}f_{y_1})<1. \end{align*} $$

Hence,

$$ \begin{align*} P(X^{(n)}, \sigma^n, u_n)&=P(X^{(n)}, \sigma^n, u_n)-P(X^{(n)}, \sigma^n, v_n)\\ & \geq \inf_{x\in \Sigma} (u_n(x)-v_n(x))\\ &\geq -\log C+ n(s-s_n) \log \theta, \end{align*} $$

where in the second inequality, we used [Reference Walters39, Theorem 9.7(iv)]. Combining this with (4.7) yields that, for $n\geq N_\epsilon $ ,

$$ \begin{align*} n\epsilon\geq -\log C + n(s-s_n) \log \theta, \end{align*} $$

so

$$ \begin{align*} s\geq s_n + \frac{\epsilon+ n^{-1} \log C}{\log \theta}\geq \overline{\dim}_B\Pi(X)+\frac{\epsilon+ n^{-1} \log C}{\log \theta}. \end{align*} $$

Letting $n\to \infty $ and then $\epsilon \to 0$ , we obtain $s\geq \overline {\dim }_B\Pi (X)$ , as desired.

Proof. It is known (see, for example, [Reference Arnold1, Theorem 3.3.3]) that, for m-a.e. x,

$$ \begin{align*} \lim_{p\to \infty}\frac{1}{p} \log (\alpha_i (D_{\Pi \sigma^p x}f_{x|p}))=\unicode{x3bb}_i,\quad i=1,\ldots, d, \end{align*} $$

and

$$ \begin{align*} \lim_{p\to \infty}\frac{1}{p} \log (\phi^i (D_{\Pi \sigma^p x}f_{x|p}))=\unicode{x3bb}_1+\cdots+\unicode{x3bb}_i,\quad i=1,\ldots, d. \end{align*} $$

For $i\in \{1,\ldots , d\}$ , $x\in \Sigma $ and $p\in {\Bbb N}$ , set

$$ \begin{align*} \begin{split} w_p^{(i)}(x)&= \log \phi^i(D_{\Pi \sigma^p x}f_{x|p}), \\ v_p^{(i)}(x)&= \log \alpha_i(D_{\Pi \sigma^p x}f_{x|p}). \end{split} \end{align*} $$

Then, by the definition of $\phi ^i$ ,

(5.3) $$ \begin{align} w_p^{(i)}=v_p^{(1)}+\cdots+v_p^{(i)}. \end{align} $$

Since $\phi ^i$ is submultiplicative on ${\Bbb R}^{d\times d}$ (cf. [Reference Falconer13, Lemma 2.1]), $\{w^{(i)}_p\}_{p=1}^\infty $ is a subadditive potential satisfying

$$ \begin{align*} |w_p^{(i)}(x)|\leq pC,\quad p\in {\Bbb N}, x\in \Sigma, \end{align*} $$

for some constant $C>0$ . Set ${\mathcal C}_p:=\{B\in {\mathcal B}(\Sigma ): \sigma ^{-p}B=B\ \mathrm {almost\ everywhere}\}$ . Then, by Lemma 2.8, for m-a.e. x,

(5.4) $$ \begin{align} \lim_{p\to \infty}E\bigg(\frac{w^{(i)}_p}{p} \bigg| {\mathcal C}_p \bigg)(x) =\lim_{p\to \infty} \frac{1}{p}w^{(i)}_p(x)=\unicode{x3bb}_1+\cdots+\unicode{x3bb}_i, \quad i=1,\ldots, d, \end{align} $$

and so by (5.3),

(5.5) $$ \begin{align} \lim_{p\to \infty}E\bigg(\frac{v^{(i)}_p}{p} \bigg| {\mathcal C}_p \bigg)(x) =\unicode{x3bb}_i, \quad i=1,\ldots, d. \end{align} $$

Let $p\in {\Bbb N}$ . Applying Proposition 4.2 to the IFS $\{f_I: I\in \mathcal A^p\}$ , we see that there exists a positive number $C_1(p)$ such that, for any $x\in \Sigma $ and $n\in {\Bbb N}$ , the set $\Pi ([x|np])=f_{x|np}(K)$ can be covered by $C_1(p)\prod _{i=0}^{n-1}G_p(\sigma ^{pi} x)$ balls of radius $\prod _{i=0}^{n-1}H_p(\sigma ^{ip} x)$ , where

(5.6) $$ \begin{align} G_p(x):=\frac{C\phi^k(D_{\Pi \sigma^px }f_{x|p})}{\alpha_{k+1}(D_{\Pi \sigma^p x}f_{x|p})^k},\quad H_p(x):=\alpha_{k+1}(D_{\Pi \sigma^px}f_{x|p}), \end{align} $$

with $C=(4d)^d$ . By Birkhoff’s ergodic theorem, for m-a.e. $x\in \Sigma $ ,

(5.7) $$ \begin{align} \lim_{n\to \infty}\frac{1}{np}\log \bigg(\!C_1(p)\prod_{i=0}^{n-1}G_p(\sigma^{pi} x)\!\bigg)&=\frac{\log C}{p}+E\bigg(\frac{w^{(k)}_p}{p} \bigg| {\mathcal C}_p \!\bigg)(x) -kE\bigg(\frac{v^{(k+1)}_p}{p} \bigg| {\mathcal C}_p \!\bigg)(x) \end{align} $$

and

(5.8) $$ \begin{align} \lim_{n\to \infty}\frac{1}{np}\log \bigg( \prod_{i=0}^{n-1}H_p(\sigma^{ip} x)\bigg)=E\bigg(\frac{v^{(k+1)}_p}{p} \bigg| {\mathcal C}_p \bigg)(x). \end{align} $$

Let $\epsilon>0$ . By (5.4), (5.5), (5.7), (5.8), for m-a.e. x there exists a positive integer $p_0(x)$ such that, for any $p\geq p_0(x)$ ,

(5.9) $$ \begin{align} \prod_{i=0}^{n-1}H_p(\sigma^{ip} x)\leq (u+\epsilon)^{np}\quad \mbox{for large enough }n, \end{align} $$

and

(5.10) $$ \begin{align} \frac{1}{np}\log \bigg( C_1(p)\prod_{i=0}^{n-1}G_p(\sigma^{pi} x)\bigg) \leq \unicode{x3bb}_1+\cdots +\unicode{x3bb}_k-k\unicode{x3bb}_{k+1}+\epsilon \quad \mbox{for large enough }n. \end{align} $$

Fix such an x and let $p\geq p_0(x)$ . By (5.9),

$$ \begin{align*} N_{(u+\epsilon)^{np}}(\Pi([x|pn]))\leq C_1(p)\prod_{i=0}^{n-1}G_p(\sigma^{pi} x) \quad \mbox{for large enough }n. \end{align*} $$

Notice that there exists a constant $C_2=C_2(d)>0$ such that a ball of radius $(u+\epsilon )^{np}$ in ${\Bbb R}^d$ can be covered by $C_2 (1+\epsilon /u)^{dnp}$ balls of radius $u^{np}$ . It follows that, for large enough n,

$$ \begin{align*} N_{u^{np}}(\Pi([x|pn]))&\leq C_2 (1+\epsilon/u)^{dnp} N_{(u+\epsilon)^{np}}(\Pi([x|pn]))\\ &\leq C_1(p) C_2 (1+\epsilon/u)^{dnp} \prod_{i=0}^{n-1}G_p(\sigma^{pi} x). \end{align*} $$

Hence, by (5.10),

$$ \begin{align*} \limsup_{n\to \infty}\frac{1}{n}\log N_{u^{n}}(\Pi([x|n]))&=\limsup_{n\to \infty} \frac{1}{np}\log N_{u^{np}}(\Pi([x|pn]))\\ &\leq d\log(1+\epsilon/u)+ \limsup_{n\to \infty} \frac{1}{np}\log \bigg(\prod_{i=0}^{n-1}G_p(\sigma^{pi} x)\bigg)\\ &\leq d\log(1+\epsilon/u)+\epsilon+\unicode{x3bb}_1+\cdots+\unicode{x3bb}_k-k\unicode{x3bb}_{k+1}, \end{align*} $$

where the first equality follows from the fact that, for $pn\leq m<p(n+1)$ ,

$$ \begin{align*} N_{u^{m}}(\Pi([x|m]))\leq N_{u^{m}}(\Pi([x|pn]))&\leq 4^d (u^{pn-m})^d N_{u^{np}}(\Pi([x|pn]))\\ &\leq 4^d u^{-pd} N_{u^{np}}(\Pi([x|pn])), \end{align*} $$

using the fact that, for $R>r>0$ , a ball of radius R in ${\Bbb R}^d$ can be covered by $(4R/r)^d$ balls of radius r. Letting $\epsilon \to 0$ yields the desired inequality (5.2).

Proof. The formulation and the proof of the above lemma are adapted from an argument given by Jordan [Reference Jordan25]. A similar idea was also employed in the proof of [Reference Rossi36, Theorem 2.2].

For $n\in {\Bbb N}$ , let $\Lambda _n$ denote the set of the points $x=(x_n)_{n=1}^\infty \in \Sigma $ such that

$$ \begin{align*} m\circ\Pi^{-1} (B(\Pi x, 2\rho^n))< (1-\epsilon)^n\frac{m([x_1\ldots x_n])}{N_{\rho^n}(\Pi([x_1\ldots x_n]))}. \end{align*} $$

To prove that (5.11) holds almost everywhere, by the Borel–Cantelli lemma it suffices to show that

(5.12) $$ \begin{align} \sum_{n=1}^\infty m(\Lambda_n)<\infty. \end{align} $$

For this purpose, let us estimate $m(\Lambda _n)$ . Fix $n\in {\Bbb N}$ and $I\in \mathcal A^n$ . Notice that $\Pi ([I])$ can be covered by $N_{\rho ^n}(\Pi ([I]))$ balls of radius $\rho ^n$ . As a consequence, there exists $L\leq N_{\rho ^n}(\Pi ([I]))$ such that $\Pi (\Lambda _n\cap [I])$ can be covered by L balls of radius $\rho ^n$ , say, $B_1,\ldots , B_L$ . We may assume that $\Pi (\Lambda _n\cap [I])\cap B_i\neq \emptyset $ for each $1\leq i\leq L$ . Hence, for each i, we may pick $x^{(i)}\in \Lambda _n \cap [I]$ such that $\Pi x^{(i)} \in B_i$ . Clearly $B_i\subset B(\Pi x^{(i)}, 2 \rho ^n)$ . Since $x^{(i)}\in \Lambda _n \cap [I]$ , by the definition of $\Lambda _n$ we obtain

$$ \begin{align*}m\circ\Pi^{-1} (B(\Pi x^{(i)}, 2\rho^n))< (1-\epsilon)^n\frac{m([I])}{N_{\rho^n}(\Pi([I]))}.\end{align*} $$

It follows that

$$ \begin{align*} m(\Lambda_n\cap [I])&\leq m\circ \Pi^{-1}(\Pi(\Lambda_n\cap [I]))\\ &\leq m\circ \Pi^{-1}\bigg(\bigcup_{i=1}^LB_i\bigg)\\ &\leq m\circ \Pi^{-1}\bigg(\bigcup_{i=1}^LB(\Pi x^{(i)}, 2\rho^n)\bigg)\\ &\leq L (1-\epsilon)^n\frac{m([I])}{N_{\rho^n}(\Pi([I]))}\\ &\leq (1-\epsilon)^nm([I]). \end{align*} $$

Summing over $I\in \mathcal A^n$ yields that $m(\Lambda _n)\leq (1-\epsilon )^n$ , which implies (5.12).

Proof of Theorem 1.2

We may assume that $s:=\dim _{L}m<d$ ; otherwise there is nothing left to prove. Set $k=[s]$ . Let $\unicode{x3bb} _i$ , $i=1,\ldots , d$ , be defined as in (5.1). Then, by Definition 2.6,

$$ \begin{align*} h_m(\sigma)+\unicode{x3bb}_1+\cdots+\unicode{x3bb}_k+(s-k)\unicode{x3bb}_{k+1}=0. \end{align*} $$

Let $u=\exp (\unicode{x3bb} _{k+1})$ and $\epsilon \in (0,1)$ . Applying Lemma 5.2 yields that, for m-a.e. $x=(x_n)_{n=1}^\infty \in ~\Sigma $ ,

(5.13) $$ \begin{align} m\circ\Pi^{-1} (B(\Pi x, 2u^n))\geq (1-\epsilon)^n\frac{m([x_1\ldots x_n])}{N_{u^n}(\Pi([x_1\ldots x_n]))} \quad \mbox{for large enough }n. \end{align} $$

It follows that, for m-a.e. $x=(x_n)_{n=1}^\infty \in \Sigma $ ,

$$ \begin{align*} &\overline{d}(m\circ \Pi^{-1}, \Pi x)\\ &\quad=\limsup_{n\to \infty} \frac{\log(m\circ\Pi^{-1} (B(\Pi x, 2u^n)))}{n\log u} \\ &\quad\leq \frac{\log(1-\epsilon)}{\log u}+\limsup_{n\to \infty}\bigg(\frac{\log m([x_1\ldots x_n])}{n \log u} -\frac{\log N_{u^n}(\Pi([x_1\ldots x_n]))}{n\log u}\bigg)\\ &\quad\leq \frac{\log(1-\epsilon)-h_m(\sigma)-(\unicode{x3bb}_1+\cdots+\unicode{x3bb}_k)+k\unicode{x3bb}_{k+1}}{\log u}\\ &\quad= \frac{\log(1-\epsilon)+s\unicode{x3bb}_{k+1}}{\unicode{x3bb}_{k+1}}, \end{align*} $$

where in the third inequality, we used the Shannon–McMillan–Breiman theorem (cf. [Reference Walters39, p. 93]) and Lemma 5.1 (keeping in mind that $\log u=\unicode{x3bb} _{k+1}<0$ ). Letting $\epsilon \to 0$ yields the desired result.

Proof. This can be done by routinely modifying the proof of Lemma 4.1 and using similar arguments to the proof of [Reference Zhang40, Corollary 1].

Proof of Proposition 6.4

The proof is adapted from that of Proposition 4.2. For the reader’s convenience, we provide the full details.

First, we construct a local inverse $f_{i,j}$ of $\psi $ for each pair $(i,j)$ with $ij\in \Sigma _{A,2}$ . To do so, notice that $\psi (R_{ij})=R_j$ for each $ij\in \Sigma _{A,2}$ . Since $\psi $ is a diffeomorphism restricted on a small neighborhood of $R_i$ , we can find open sets $\widetilde {R}_{ij}$ and $\widetilde {R_j}$ such that $\widetilde {R}_{ij} \supset R_{ij}$ , $\widetilde {R_j}\supset R_j$ , $\psi (\widetilde {R}_{ij})=\widetilde {R_j}$ and $\psi : \widetilde {R}_{ij}\to \widetilde {R_j}$ is diffeomorphic. Then we take $f_{i,j}: \widetilde {R_j}\to \widetilde {R}_{ij}$ to be the inverse of $\psi : \widetilde {R}_{ij}\to \widetilde {R_j}$ , and the construction is done.

For any ${\textbf {i}}=(i_n)_{n=1}^\infty \in \Sigma _A$ , we see that $\Pi \sigma {\textbf {i}}\in R_{i_2}\subset \widetilde {R_{i_2}}$ and $(\psi \circ f_{i_1, i_2})|_{\widetilde {R_{i_2}}}$ is the identity restricted on $\widetilde {R_{i_2}}$ . Since $\psi (\Pi {\textbf {i}})=\Pi \sigma {\textbf {i}}$ , it follows that $f_{i_1, i_2}(\Pi \sigma {\textbf {i}})=\Pi {\textbf {i}}$ . Differentiating $\psi \circ f_{i_1, i_2}$ at $\Pi \sigma {\textbf {i}}$ and applying the chain rule, we get

$$ \begin{align*} (D_{\Pi {\textbf{i}}} \psi)(D_{\Pi\sigma {\textbf{i}}}f_{i_1, i_2})=\mbox{Identity}, \end{align*} $$

so

(6.5) $$ \begin{align} D_{\Pi\sigma {\textbf{i}}}f_{i_1, i_2}=(D_{\Pi {\textbf{i}}} \psi)^{-1}. \end{align} $$

According to Lemma 6.3, there exists $r_0>0$ such that, for each $ij\in \Sigma _{A,2}$ , $y\in R_j$ , $z\in B(y, r_0)$ and $0<r<r_0$ , the set $f_{i,j}(B(z, r))$ can be covered by

$$ \begin{align*} C_{\pmb M}\cdot\frac{\phi^k(D_yf_{i,j})}{(\alpha_{k+1}(D_yf_{i,j}))^k} \end{align*} $$

balls of radius $\alpha _{k+1}(D_yf_{i,j})r$ .

Since $\psi $ is expanding on $\Lambda $ , there exists $\gamma \in (0,1)$ such that $\sup _{{\textbf {i}}\in \Sigma _A}\alpha _1 ((D_{\Pi \mathbf {i}} \psi )^{-1})<\gamma $ . Then $\sup _{{\textbf {i}}\in \Sigma _A}H({\textbf {i}})<\gamma $ and

(6.6) $$ \begin{align} \mbox{diam}(R_{{\textbf{i}}|(n+1)})\leq \gamma \mbox{diam}(R_{(\sigma{\textbf{i}})|n}) \end{align} $$

for all ${\textbf {i}}\in \Sigma _A$ and $n\in {\Bbb N}$ . Take a large integer $n_0$ such that

(6.7) $$ \begin{align} \gamma^{n_0-1} \max\{1, \mbox{diam}(\Lambda)\}<r_0/2. \end{align} $$

By (6.6)–(6.7), $\mbox {diam}(R_{{\textbf {i}}|n})<r_0/2$ for all ${\textbf {i}}\in \Sigma _A$ and $n\geq n_0$ .

Clearly there exists a large number $C_1$ so that the conclusion of the proposition holds for any positive integer $n\leq n_0$ and $\mathbf {i}\in \Sigma _A$ , that is, the set $R_{\mathbf {i}|n}$ can be covered by $C_1\prod _{p=0}^{n-1}G(\sigma ^p \mathbf {i})$ balls of radius $\prod _{p=0}^{n-1}H(\sigma ^p \mathbf {i})$ . Below we show by induction that this holds for all $n\in {\Bbb N}$ and $\mathbf {i}\in \Sigma _A$ .

Suppose, for some $m\geq n_0$ , that the conclusion of the proposition holds for any positive integer $n\leq m$ and $\mathbf {i}\in \Sigma _A$ . Then, for given ${\textbf {i}}=(i_n)_{n=1}^\infty \in \Sigma _A$ , $R_{(\sigma {\textbf {i}})|m}$ can be covered by $C_1\prod _{p=0}^{m-1}G(\sigma ^{p+1} \mathbf {i})$ balls of radius $\prod _{p=0}^{m-1}H(\sigma ^{p+1} \mathbf {i})$ . Let $B_1,\ldots , B_N$ denote these balls. We may assume that $B_j\cap R_{(\sigma {\textbf {i}})|m}\neq \emptyset $ for each j. Since

$$ \begin{align*}\prod_{p=0}^{m-1}H(\sigma^{p+1} \mathbf{i})\leq \gamma^m\leq \gamma^{n_0}<r_0/2\end{align*} $$

and

$$ \begin{align*}d(\Pi\sigma\mathbf{i}, B_j\cap R_{(\sigma {\textbf{i}})|m})\leq {\textrm{diam}}(R_{(\sigma {\textbf{i}})|m})<r_0/2,\end{align*} $$

so the center of $B_j$ is in $B(\Pi \sigma {\textbf {i}}, r_0)$ . Therefore, by Lemma 6.3 and (6.5), $f_{i_1, i_2}(B_j)$ can be covered by

$$ \begin{align*} C_{\pmb M}\cdot\frac{\phi^k(D_{\Pi \sigma {\textbf{i}}}f_{i_1, i_2})}{(\alpha_{k+1}(D_{\Pi \sigma {\textbf{i}}}f_{i_1, i_2}))^k}=G({\textbf{i}}) \end{align*} $$

balls of radius

$$ \begin{align*}\alpha_{k+1}(D_{\Pi\sigma{\textbf{i}}}f_{i_1,i_2})\cdot\prod_{p=0}^{m-1}H(\sigma^{p+1} \mathbf{i})= H({\textbf{i}})\cdot \prod_{p=0}^{m-1}H(\sigma^{p+1} \mathbf{i})=\prod_{p=0}^{m}H(\sigma^{p} \mathbf{i}).\end{align*} $$

Since $\psi (R_{\mathbf {i}|{(m+1)}})\subset R_{(\sigma {\textbf {i}})|m}$ , it follows that

$$ \begin{align*}R_{\mathbf {i}|{(m+1)}}\subset f_{i_1, i_2} (R_{(\sigma {\textbf{i}})|m})\subset \bigcup_{j=1}^N f_{i_1, i_2}(B_j),\end{align*} $$

hence $R_{\mathbf {i}|{(m+1)}}$ can be covered by

$$ \begin{align*}G({\textbf{i}})N\leq G({\textbf{i}}) \cdot C_1\prod_{p=0}^{m-1}G(\sigma^{p+1} {\textbf{i}})=C_1\prod_{p=0}^{m}G(\sigma^{p} {\textbf{i}})\end{align*} $$

balls of radius $\prod _{p=0}^{m}H(\sigma ^{p} \mathbf {i})$ . Thus the proposition also holds for $n=m+1$ and all ${\textbf {i}}\in \Sigma _A$ , as desired.

Proof. Here we use similar arguments to that in the proof of Proposition 4.3. Write $g=\log G$ and $h=\log H$ . Define

$$ \begin{align*} r_{\min}=\min_{{\textbf{i}}\in \Sigma_A}h({\textbf{i}}),\quad r_{\max}=\max_{{\textbf{i}}\in \Sigma_A}h({\textbf{i}}). \end{align*} $$

Then $0<r_{\min }\leq r_{\max }<1$ . For $0<r<r_{\min }$ , define

(6.8) $$ \begin{align} \kern-8pt\mathcal A_r=\big\{i_1\ldots i_n\in \Sigma^*_A: \sup_{x\in [i_1\ldots i_n]\cap \Sigma_A}S_nh(x)<\log r\leq \sup_{y\in [i_1\ldots i_{n-1}]\cap \Sigma_A}S_{n-1}h(y)\big\}, \end{align} $$

where $\Sigma _A^*$ denotes the set of all finite words allowed in $\Sigma _A$ . Clearly $\{[I]: I\in \mathcal A_r\}$ is a partition of $\Sigma _A$ . By Proposition 6.4, there exists a constant $C_1>0$ such that, for each $0<r<r_{\min }$ , every $I\in \mathcal A_r$ and $x\in [I]$ , $R_I$ can be covered by

$$ \begin{align*} C_1\exp(S_{|I|}g(x)) \leq C_1\exp\big(\sup_{y\in [I]}S_{|I|}g(y)\big) \end{align*} $$

balls of radius

$$ \begin{align*}\exp(S_{|I|}h(x))\leq \exp\big(\sup_{y\in [I]}S_{|I|}h(y)\big)<r.\end{align*} $$

It follows that $\Lambda $ can be covered by

$$ \begin{align*} C_1\sum_{I\in \mathcal A_r}\exp\big(\sup_{y\in [I]}S_{|I|}g(y)\big) \end{align*} $$

balls of radius r. Hence, by Proposition 3.4,

$$ \begin{align*} \overline{\dim}_B\Lambda\leq \limsup_{r\to 0}\frac{\sum_{I\in \mathcal A_r}\exp\big(\sup_{y\in [I]}S_{|I|}g(y)\big)}{\log (1/r)}= t. \end{align*} $$

This completes the proof of the proposition.

Proof of Theorem 1.3

We follow the proof of Theorem 1.1 with slight modifications. Write $s=\dim _{S^*}(\Lambda )$ . We may assume that $s<d$ ; otherwise we have nothing left to prove. Set $k=[s]$ . Let $\mathcal G^s=\{g_n^s\}_{n=1}^\infty $ be the subadditive potential on $\Lambda $ defined by

$$ \begin{align*}g_n^s(z)=\log \phi^s((D_{z} \psi^n)^{-1}).\end{align*} $$

Then $P(\Lambda , \psi , {\mathcal G}^s)=0$ by the definition of $\dim _{S^*}(\Lambda )$ . Let $\widehat {\mathcal G}^s:=\{\widehat {g}_n^s\}_{n=1}^\infty $ , where $\widehat {g}_n^s\in C(\Sigma _A)$ is defined by

$$ \begin{align*} \widehat{g}^s_n({\textbf{i}})=g_n^s(\Pi{\textbf{i}})=\log \phi^s((D_{\Pi{\textbf{i}}} \psi^n)^{-1}),\quad {\textbf{i}}\in \Sigma_A. \end{align*} $$

Clearly, $\widehat {\mathcal G}^s$ is a subadditive potential on $\Sigma _A$ and

$$ \begin{align*} (\widehat{\mathcal G}^s)_*(m)=({\mathcal G}^s)_*(m\circ \Pi^{-1}), \quad m\in {\mathcal M}(\Sigma_A, \sigma). \end{align*} $$

Since the factor map $\Pi :\Sigma _A\to \Lambda $ is onto and finite-to-one, by Lemma 2.3, $m\to m\circ \Pi ^{-1}$ is a surjective map from ${\mathcal M}(\Sigma _A, \sigma )$ to $\mathcal M(\Lambda , \psi )$ and, moreover, $h_m(\sigma )=h_{m\circ \Pi ^{-1}}(\psi )$ for $m\in \mathcal M (\Sigma _A, \sigma )$ . By the variational principle for the subadditive pressure (see Theorem 2.2),

$$ \begin{align*} P(\Lambda, \psi, \mathcal{G}^{s}) &= \sup \{h_{\mu}(\psi)+(\mathcal{G}^{s})_{*}(\mu) : \mu \in \mathcal{M}(\Lambda, \psi)\}\\ &= \sup \{h_{m \circ \Pi^{-1}}(\psi)+(\mathcal{G}^{s})_{*}(m \circ \Pi^{-1}) : m \in \mathcal{M}(\Sigma_{A}, \sigma)\} \\ &= \sup \{h_{m}(\sigma)+(\widehat{\mathcal{G}}^{s})_{*}(m) : m \in \mathcal{M}(\Sigma_{A}, \sigma)\} \\ &= P(\Sigma_{A},\sigma,\widehat{\mathcal{G}}^{s}). \end{align*} $$

It follows that $P(\Sigma _{A},\sigma ,\widehat {\mathcal {G}}^{s})=0$ .

By [Reference Ban, Cao and Hu2],

$$ \begin{align*} \lim_{n\to \infty}\frac{1}{n}P(\Sigma_A, \sigma^n, \widehat{g}^s_n)=\inf_{n\geq 1}\frac{1}{n}P(\Sigma_A, \sigma^n, \widehat{g}^s_n)=P(\Sigma_A, \sigma, \widehat{\mathcal G}^s). \end{align*} $$

Since $P(\Sigma _{A},\sigma ,\widehat {\mathcal {G}}^{s})=0$ , the above equalities imply that, for each $\epsilon>0$ , there exists $N_\epsilon>0$ such that

(6.10) $$ \begin{align} 0\leq P(\Sigma_A, \sigma^n, \widehat{g}^s_n)\leq n\epsilon \quad\mbox{for }\; n>N_\epsilon. \end{align} $$

For $n\in {\Bbb N}$ , let $s_n$ be the unique real number such that $P(\Sigma _A, \sigma ^n, v_n^{s_n})=0$ , where $v_n^{s_n}$ is a continuous function on $\Sigma _A$ defined by

$$ \begin{align*} v_n^{s_n}(x)&= \log G_n(x)+ s_n \log H_n(x)\\ &=\log ( C\phi^k((D_{\Pi x}\psi^n)^{-1})\alpha_{k+1}((D_{\Pi x}\psi^n)^{-1})^{s_n-k}), \end{align*} $$

where $G_n, H_n$ are defined in (6.9) and $C=C_{\pmb M}$ . By Proposition 6.7,

$$ \begin{align*}\overline{\dim}_B\Lambda\leq s_n.\end{align*} $$

If $s\geq s_n$ for some n, then $\overline {\dim }_B\Lambda \leq s_n\leq s$ and we are done. In what follows, we assume that $s<s_n$ for each n. Then, for each $x\in \Sigma _A$ ,

$$ \begin{align*} \widehat{g}^s_n(x)-v_n^{s_n}(x)&= -\log C +(s-s_n)\log \alpha_{k+1}((D_{\Pi x}\psi^n)^{-1})\\ &\geq -\log C +(s-s_n)\log \alpha_1((D_{\Pi x}\psi^n)^{-1})\\ &\geq -\log C+ n(s-s_n) \log \theta, \end{align*} $$

where

$$ \begin{align*} \theta:=\max_{x\in \Sigma_A} \alpha_1((D_{\Pi x}\phi)^{-1})<1. \end{align*} $$

Hence,

$$ \begin{align*} P(\Sigma_A, \sigma^n,\widehat{g}^s_n)&=P(\Sigma_A, \sigma^n,\widehat{g}^s_n)-P(\Sigma_A, \sigma^n, v_n^{s_n})\\ &\geq \inf_{x\in \Sigma_A} ( \widehat{g}^s_n(x)-v_n^{s_n}(x))\\ &\geq -\log C+ n(s-s_n) \log \theta. \end{align*} $$

where in the second inequality, we used [Reference Walters39, Theorem 9.7(iv)]. Combining this with (6.10) yields that, for $n\geq N_\epsilon $ ,

$$ \begin{align*} n\epsilon\geq -\log C + n(s-s_n) \log \theta, \end{align*} $$

so

$$ \begin{align*} s\geq s_n + \frac{\epsilon+ n^{-1} \log C}{\log \theta}\geq \overline{\dim}_B\Lambda+\frac{\epsilon+ n^{-1} \log C}{\log \theta}. \end{align*} $$

Letting $n\to \infty $ and then $\epsilon \to 0$ , we obtain $s\geq \overline {\dim }_B\Lambda $ , as desired.

Proof of Theorem 1.4

Here we adapt the proof of Theorem 1.2. We may assume that $s:=\dim _{L^*}\mu <d$ ; otherwise there is nothing left to prove. Since $\Pi : \Sigma _A\to \Lambda $ is surjective and finite-to-one, by Lemma 2.3, there exists a $\sigma $ -invariant ergodic measure m on $\Sigma _A$ so that $m\circ \Pi ^{-1}=\mu $ and $h_m(\sigma )=h_\mu (\psi )$ .

Set $k=[s]$ . Let $\unicode{x3bb} _i$ , $i=1,\ldots , d$ , be defined as in (6.11). Then, by Definition 6.2,

$$ \begin{align*} h_m(\sigma)+\unicode{x3bb}_1+\cdots+\unicode{x3bb}_k+(s-k)\unicode{x3bb}_{k+1}=0. \end{align*} $$

Let $u=\exp (\unicode{x3bb} _{k+1})$ and $\epsilon \in (0,1)$ . Applying Lemma 6.9 (in which we take $\rho =u$ ) yields that, for m-a.e. $x=(x_n)_{n=1}^\infty \in \Sigma _A$ ,

(6.14) $$ \begin{align} \mu (B(\Pi x, 2u^n))\geq (1-\epsilon)^n\frac{m([x_1\ldots x_n])}{N_{u^n}(R_{x_1\ldots x_n})} \quad \mbox{for large enough }n. \end{align} $$

It follows that, for m-a.e. $x=(x_n)_{n=1}^\infty \in \Sigma _A$ ,

$$ \begin{align*} \overline{d}(\mu, \Pi x) &=\limsup_{n\to \infty} \frac{\log(\mu (B(\Pi x, 2u^n)))}{n\log u} \\[6pt] &\leq \frac{\log(1-\epsilon)}{\log u}+\limsup_{n\to \infty}\bigg(\frac{\log m([x_1\ldots x_n])}{n \log u} -\frac{\log N_{u^n}(\Pi([x_1\ldots x_n]))}{n\log u}\bigg)\\[6pt] &\leq \frac{\log(1-\epsilon)-h_m(\sigma)-(\unicode{x3bb}_1+\cdots+\unicode{x3bb}_k)+k\unicode{x3bb}_{k+1}}{\log u}\\[3pt] &= \frac{\log(1-\epsilon)+s\unicode{x3bb}_{k+1}}{\unicode{x3bb}_{k+1}}, \end{align*} $$

where in the third inequality we used the Shannon–McMillan–Breiman theorem (cf. [Reference Walters39, p. 93]) and Lemma 6.8 (keeping in mind that $\log u=\unicode{x3bb} _{k+1}<0$ ). Letting $\epsilon \to 0$ yields the desired result.