Proof. Since $\phi \in C_{uc}(\Sigma )$ , we have that $\text {var}_m(\phi )\to 0$ as $m\to \infty $ . Let $m\in {\mathbb {N}}$ be large enough such that $\text {var}_m(\phi )<\infty $ . Define $\phi _m(x)=\sup \{\phi (y):y\in [x_1,\ldots , x_{m}]\}$ , where $x=(x_1,x_2,\ldots )$ . Note that $\phi _m$ is non-negative, $\text {var}_m(\phi _m)=0$ , and $\|\phi -\phi _m\|_0\le \text {var}_m(\phi )$ . In particular, $\|\phi -\phi _m\|_0\to 0$ as $m\to \infty $ . Let $(C_i^m)_i$ be an enumeration of the set of cylinders of length m. Define $A^m_k=\bigcup _{i=1}^k C_i^m$ . It follows that

$$ \begin{align*}\liminf_{n\to\infty}\int \phi_m\,d\mu_n\ge \lim_{n\to\infty}\int \phi_m 1_{A^m_k}\,d\mu_n=\int \phi_m 1_{A_k^m}\,d\mu.\end{align*} $$

Sending $k\to \infty $ , we get that $\liminf _{n\to \infty }\int \phi _m\,d\mu _n\ge \int \phi _m\,d\mu .$ Finally, send $m\to \infty $ and use that $|\!\int \phi \,d\nu -\int \phi _m\,d\nu |\le \|\phi -\phi _m\|_0$ for every $\nu \in {\mathcal {M}}_{\le 1}(\sigma )$ .

Proof. We will first prove that $P^{\mathrm {top}}_\infty (\phi ,a)=P^{\mathrm {top}}_\infty (\phi -tV,a)$ for every $t\ge 0$ . Observe that if $x\in \text {Per}_a(q,M,n)$ , then $S_nV(x)\le ({n}/{M})+({n}/{q})$ . It follows that

$$ \begin{align*}Z_n(\phi-tV,a;q,M)\ge Z_n(\phi,a;q,M)\exp\bigg(\!{-}\,nt\bigg(\frac{1}{M}+\frac{1}{q}\bigg)\bigg),\end{align*} $$

and therefore,

$$ \begin{align*}P^{\mathrm{top}}_\infty(\phi-tV,a;q,M)\ge P^{\mathrm{top}}_\infty(\phi,a;q,M)-t\bigg(\frac{1}{M}+\frac{1}{q}\bigg).\end{align*} $$

Sending q and M to infinity, we obtain $P^{\mathrm {top}}_\infty (\phi -tV,a)\ge P^{\mathrm {top}}_\infty (\phi ,a)$ . The other inequality follows from $\phi \ge \phi -tV$ . We conclude that $P_\infty ^{\mathrm {top}}(\phi ,a)=P_\infty ^{\mathrm {top}}(\phi -tV,a)$ .

Since $P^{\mathrm {top}}_\infty (\phi ,a)=P^{\mathrm {top}}_\infty (\phi -tV,a)\le P(\phi -tV)$ , we have $P^{\mathrm {top}}_\infty (\phi ,a)\le \lim _{t\to \infty } P(\phi -tV)$ . To complete the proof, we need to show that $\lim _{t\to \infty }P(\phi -tV)\le P^{\mathrm {top}}_\infty (\phi ,a)$ .

Fix $q,M\in {\mathbb {N}}$ . If $x\in \text {Per}_a^c(q,M,n)$ , then $\#\{1\le k\le n:x_k\le q\}> ({n}/{M})$ . In particular, $S_nV(x)> ({n}/{Mq}).$ Then,

$$ \begin{align*}\sum_{x\in \text{Per}_a^c(q,M,n)}\exp(S_n(\phi-tV)(x))\le \exp\bigg(\!{-}\,\frac{tn}{Mq}\bigg)\sum_{x\in \text{Per}_a^c(q,M,n)}\exp(S_n\phi(x)).\end{align*} $$

By definition of the pressure, there exists $n_0\in {\mathbb {N}}$ such that if $n\ge n_0,$ then $\sum _{x\in \text {Per}_a^c(q,M,n)} \exp (S_n\phi (x))\le \exp (n(P(\phi )+1)).$ Therefore,

$$ \begin{align*} \sum_{x\in \text{Per}_a^c(q,M,n)}\exp(S_n(\phi-tV)(x))\le \exp\bigg(n\bigg(P(\phi)+1-\frac{t}{Mq}\bigg)\bigg) \end{align*} $$

for every $n\ge n_0$ . Finally, for $n\ge n_0$ and $t>qM(P(\phi )+1+\kappa )$ , we have

$$ \begin{align*} Z_n(\phi-tV,a)&=\sum_{x\in \text{Per}_a(q,M,n)}\exp(S_n(\phi-tV)(x))+\sum_{x\in \text{Per}_a^c(q,M,n)}\exp(S_n(\phi-tV)(x))\\ &\le \sum_{x\in \text{Per}_a(q,M,n)}\exp(S_n\phi(x))+e^{-\kappa n} \end{align*} $$

for every $\kappa>0$ . We conclude that $P(\phi -tV)\le \max \{P^{\mathrm {top}}_\infty (\phi ,a;q,M),-\kappa \}$ and, therefore, $\lim _{t\to \infty }P(\phi -tV)\le P^{\mathrm {top}}_\infty (\phi ,a;q,M)$ .

Proof. Parts (1)–(5) follow easily from the definition. For part (6), note that $\lim _{n\to \infty } \int g\,d\mu _n=0$ for every sequence $(\mu _n)_n$ that converges on cylinders to the zero measure.

Proof. Since periodic measures are dense in ${\mathcal {M}}(\sigma )$ , there exists a sequence $(\nu _n)_n$ in ${\mathcal {M}}_\phi (\sigma )$ that converges on cylinders to the zero measure. By the variational principle, there exists a sequence $(\mu _n)_n$ in ${\mathcal {M}}_\phi (\sigma )$ such that $h_{\mu _n}(\sigma )+\int \phi \,d\mu _n\ge n^2$ . Define ${\eta _n=(1-({1}/{n}))\nu _n+({1}/{n})\mu _n}$ . The sequence $(\eta _n)_n$ converges to the zero measure and satisfies that $h_{\eta _n}(\sigma )+\int \phi \,d\eta _n\ge n$ .

Proof. Let $V=\sum _{k=1}^\infty ({1}/{k})1_{[k]}$ . Note that $\lim _{n\to \infty }\int Vd\mu _n=0$ . By the standard variational principle, we have that

$$ \begin{align*} \limsup_{n\to\infty}\bigg(h_{\mu_n}(\sigma)+\int\phi \,d\mu_n\bigg)=\limsup_{n\to\infty}\bigg(h_{\mu_n}(\sigma)+\int (\phi -tV)\,d\mu_n\bigg)\le P(\phi -tV). \end{align*} $$

In particular, the left-hand side is bounded from above by $\lim _{t\to \infty }P(\phi -tV)$ . Inequality (3) follows from Proposition 3.2.

Let us suppose now that $M:=P^{\mathrm {top}}_\infty (\phi )>-\infty $ . We will prove that inequality (3) is sharp. It follows from Proposition 3.2 that $P(\phi -tV)\ge M$ holds for all $t\in {\mathbb {R}}$ . The variational principle guarantees the existence of $\sigma $ -invariant probability measures $(\nu _n)_n$ such that

(4) $$ \begin{align} h_{\nu_n}(\sigma)+\int (\phi-nV) \,d\nu_n\ge P(\phi -nV)-\frac{1}{n},\end{align} $$

where $\int (\phi -nV)\,d\nu _n>-\infty $ . In particular, $\int \phi \,d\nu _n>-\infty $ . We claim that $(\nu _n)_n$ converges on cylinders to the zero measure. We argue by contradiction and suppose that  $\limsup _{n\to \infty }\nu _n(C)>0$ for some cylinder $C\subseteq \Sigma $ . In this case, we would have that $\limsup _{n\to \infty }n\int V\,d\nu _n=\infty $ , which is not possible because inequality (4) implies that $P(\phi )-n\int V\,d\nu _ n\ge M-1$ . However,

$$ \begin{align*} P^{\mathrm{top}}_\infty(\phi )=\lim_{n\to\infty}P(\phi -nV) &\le \liminf_{n\to\infty}\bigg(h_{\nu_n}(\sigma)+\int \phi\,d\nu_n-n\int V\,d\nu_n\bigg)\\ &\le \liminf_{n\to\infty} \bigg(h_{\nu_n}(\sigma)+\int \phi\,d\nu_n\bigg). \end{align*} $$

In conclusion, $(\nu _n)_n$ is a sequence in ${\mathcal {M}}_\phi (\sigma )$ that converges on cylinders to the zero measure and $P^{\mathrm {top}}_\infty (\phi )=\lim _{n\to \infty } h_{\nu _n}(\sigma )+\int \phi \,d\nu _n.$

Proof of Theorem 1.2

In Proposition 3.6, we proved that $P_\infty (\phi )\le P^{\mathrm {top}}_\infty (\phi )$ and that equality holds if $P^{\mathrm {top}}_\infty (\phi )>-\infty $ . Observe that inequality $P_\infty (\phi )\le P^{\mathrm {top}}_\infty (\phi )$ also implies that if $P^{\mathrm {top}}_\infty (\phi )=-\infty $ , then $P_\infty (\phi )=-\infty $ .

Proof. Since the pressure is convex whenever finite, it follows that $P(t\phi )<\infty $ for every $t\in (s_\infty (\phi ),1]$ . Consider $\epsilon>0$ small such that $P((1-\epsilon )\phi )<\infty $ , and note that

$$ \begin{align*}h_{\mu_n}+\int \phi\,d\mu_n\le P((1-\epsilon)\phi)+\epsilon\int\phi\,d\mu_n\end{align*} $$

for every $n\in {\mathbb {N}}$ .

Proof. We argue by contradiction and assume that $\liminf _{n\to \infty }\int \phi \,d\mu _n=-\infty $ . In particular, there exists a subsequence whose limit is $-\infty $ . The result now follows from Lemma 3.9.

Proof. Since $\nu $ has positive mass, there exists a cylinder $[c_i^{a,b}]$ such that $\nu ([c_i^{a,b}])$ is positive. If $i=1$ , there is nothing to prove. Assume that $i>1$ , in particular, $\tau (a,b)\ge 2$ . In this case, $\sigma ^{-(i-1)}[c_i^{a,b}]=[c_1^{a,b}, c_2^{a,b}]$ and, therefore, $\nu ([a,c_2^{a,b}])=\nu ([c_i^{a,b}])>0.$

Proof. Let $C\subseteq \widehat {\Sigma }$ be a cylinder. By Remark 5.4, there exists a cylinder $D\subseteq \Sigma $ such that $\widehat {\mu }(C)={1}/({\int {\tau }\,d\mu })\mu (D)$ for every $\mu \in {\mathcal {M}}(\sigma ;\tau )$ . Then,

We conclude that $(\widehat {\mu }_n)_n$ converges on cylinders to $({\unicode{x3bb} _1}/{\unicode{x3bb} _2})\widehat {\mu }$ .

Proof. Case 1. There exists a sequence $(\mu _n)_n$ in ${\mathcal {M}}(\sigma ;\tau )$ that converges to the zero measure and such that $\lim _{n\to \infty }(h_{\mu _n}(\sigma )- s_\tau \int \tau d\mu _n)=0$ . Then, by Abramov’s formula, $\lim _{n\to \infty }h_{\widehat {\mu }_n}=s_\tau $ and $(\widehat {\mu }_n)_n$ converges to the zero measure (see Lemma 5.5). It follows that $h_\infty (\widehat {\sigma })\ge {s}_{\tau }$ .

Let $(\nu _n)_n$ be a sequence of measures in ${\mathcal {M}}(\widehat {\sigma })$ that converges to the zero measure and such that $\lim _{n\to \infty }h_{\nu _n}(\widehat {\sigma })=h_\infty (\widehat {\sigma })$ (the existence of such a sequence is given by Theorem 2.3). Let $\eta _n$ be the measure in ${\mathcal {M}}(\sigma ;\tau )$ such that $\widehat {\eta }_n =\nu _n$ . Then,

$$ \begin{align*} \lim_{n\to\infty}\bigg(\frac{h_{\eta_n}(\sigma)-h_\infty(\widehat{\sigma})\int \tau\,d\eta_n}{\int \tau\,d\eta_n}\bigg)=0. \end{align*} $$

If $\limsup _{n\to \infty }\int \tau \,d\eta _n<\infty $ , then $(\eta _n)_n$ converges on cylinders to the zero measure (see Lemma 5.5) and $\lim _{n\to \infty }(h_{\eta _n}(\sigma )-h_\infty (\widehat {\sigma })\!\int \! \tau \,d\eta _n)=0.$ In particular, ${{P}_\infty (-h_\infty (\widehat {\sigma })\tau )\ge 0}$ and, therefore, ${s}_{\tau }\ge h_\infty (\widehat {\sigma })$ . If $\limsup _{n\to \infty }\int \tau \,d\eta _n=\infty $ , then there exists a subsequence for which the integral diverges. Maybe after passing to a subsequence can we assume that $\lim _{n\to \infty }\int \tau \,d\eta _n=\infty $ . Fix $\epsilon>0$ and note that

(11) $$ \begin{align} \limsup_{n\to\infty} \frac{h_{\eta_n}(\sigma)}{\int \tau\,d\eta_n}\le (s_\infty(-\tau)+\epsilon)+\limsup_{n\to\infty}\frac{P(-(s_\infty(-\tau)+\epsilon)\tau)}{\int \tau\,d\eta_n}=s_\infty(-\tau)+\epsilon.\end{align} $$

Therefore, $h_\infty (\widehat {\sigma })\le s_\infty (-\tau )$ . This is not possible since we already proved that $h_\infty (\widehat {\sigma })\ge {s}_{\tau }$ .

Case 2. Set $s=s_\infty (-\tau )$ . We claim that $h_\infty (\widehat {\sigma })\ge s$ . We assume that $s>0$ , otherwise there is nothing to prove. Fix $\epsilon>0$ small. Since $P(-(s-\epsilon )\tau )=\infty $ , there exists a sequence $(\mu _n)_n$ in ${\mathcal {M}}(\sigma ;\tau )$ such that

(12) $$ \begin{align} \lim_{n\to\infty}\bigg(h_{\mu_n}(\sigma)-(s-\epsilon)\int \tau\,d\mu_n\bigg)=\infty. \end{align} $$

In particular, the limit is greater than zero and, therefore, $\liminf _{n\to \infty }h_{\widehat {\mu }_n}(\widehat {\sigma })\ge s-\epsilon $ . It is enough to prove that $\limsup _{n\to \infty }\int \tau \,d\mu _n=\infty $ , in which case, $(\widehat {\mu }_n)_n$ has a subsequence that converges to the zero measure (see Lemma 5.5) and $\epsilon>0$ was arbitrary. Suppose otherwise that $\limsup _{n\to \infty }\int \tau \,d\mu _n<\infty $ and, therefore, there exists M such that $\int \tau \,d\mu _n\le M$ for all $n\in {\mathbb {N}}$ . It follows that $h_{\mu _n}(\sigma )\le M+P(-\tau )$ , which contradicts (12).

We now claim that $h_\infty (\widehat {\sigma })\le s$ . Let $(\nu _n)_n$ be a sequence in ${\mathcal {M}}(\widehat {\sigma })$ that converges to the zero measure and such that $\lim _{n\to \infty }h_{\nu _n}(\widehat {\sigma })=h_\infty (\widehat {\sigma })$ . Let $\eta _n$ be the measure in ${\mathcal {M}}(\sigma ;\tau )$ such that $\widehat {\eta }_n =\nu _n$ . Then,

$$ \begin{align*}\lim_{n\to\infty}\bigg(\frac{h_{\eta_n}(\sigma)-h_\infty(\widehat{\sigma})\int \tau \,d\eta_n}{\int \tau \,d\eta_n}\bigg)=0.\end{align*} $$

If $\limsup _{n\to \infty }\int \tau \,d\eta _n<\infty $ , then $\lim _{n\to \infty }(h_{\eta _n}(\sigma )-h_\infty (\widehat {\sigma })\int \tau \,d\eta _n)=0$ and $(\eta _n)_n$ converges to the zero measure (see Lemma 5.5). In particular, $P_\infty (-h_\infty (\widehat {\sigma })\tau )\ge 0$ . If $h_\infty (\widehat {\sigma })> s$ , then $\tau \in \Psi _1$ , which contradicts our assumption. We therefore have that $h_\infty (\widehat {\sigma })\le s$ . Now, suppose that $\limsup _{n\to \infty }\int \tau \,d\nu _n=\infty .$ Maybe after passing to a subsequence can we assume that $\lim _{n\to \infty }\int \tau \,d\nu _n=\infty $ . Inequality (11) implies that $h_\infty (\widehat {\sigma })\le s$ .

Proof of Theorem 1.1

Consider $M \in \mathbb {R}$ such that $\sup \phi < M$ . Define ${\psi = -(\phi - M)}$ . Note that $\psi \in \Psi $ . Define $\tau :\Sigma \to {\mathbb {R}}$ by $\tau (x) = \sup _{y \in [x_1, x_2]} \lceil \psi (y) \rceil $ , where $x = (x_1, x_2, \ldots )\in \Sigma $ . Since $\text {var}_2(\phi )=\text {var}_2(\psi )$ is finite, it follows that $\|\tau - \psi \|_0 < \infty $ . In particular, $P(-\tau )<\infty $ . Moreover, $\tau $ takes values in $\mathbb {N}$ and $\text {var}_2(\tau ) = 0$ .

Let $(\widehat {\Sigma }, \widehat {\sigma })$ be the CMS constructed at the beginning of this section for the pair $(\Sigma , \sigma )$ and $\tau $ . Note that $(\widehat {\Sigma },\widehat {\sigma })$ is transitive, because $(\Sigma ,\sigma )$ is transitive. Additionally, $h_{\textrm {top}}(\widehat {\sigma })$ is finite (see Remark 5.7). By Theorem 2.3, ${\mathcal {M}}_{\le 1}(\widehat {\sigma })$ is compact with respect to the cylinder topology. As before, we denote by $\widehat {\nu }$ the measure in ${\mathcal {M}}(\widehat {\sigma })$ associated to $\nu \in {\mathcal {M}}(\sigma ;\tau )$ . Maybe after passing to a subsequence can we assume that $(\widehat {\mu }_n)_n$ converges on cylinders to $\unicode{x3bb} \widehat {\mu }$ , where $\unicode{x3bb} \in [0,1]$ and $\mu \in {\mathcal {M}}(\sigma ;\tau )$ . Maybe after passing to a further subsequence can we assume that $A:=\lim _{n\to \infty }\int \tau d\mu _n$ is well defined and finite (the limsup is finite by assumption). Let C be a cylinder in $\Sigma $ . By Remark 5.4, there exists a cylinder $D\subseteq \widehat {\Sigma }$ such that $\widehat {\nu }(D)={1}/({\int \tau \,d\nu })\nu (C)$ for every $\nu \in {\mathcal {M}}(\sigma ;\tau )$ . Then, $\lim _{n\to \infty }\mu _n(C)= {\unicode{x3bb} A}/({\int \tau \,d\mu })\mu (C).$ We conclude that $(\mu _n)_n$ converges on cylinders to $\mu _0={\unicode{x3bb} A}/({\int \tau \,d\mu })\mu $ . Note that $\int \tau \,d{\mu _0}<\infty $ and, therefore, $\int \phi \,d{\mu }_0>-\infty $ (see Lemma 2.5).

Proof. Suppose that $\lim _{n\to \infty }\mu _n(C)=\mu (C)$ for every cylinder $C\subseteq \Sigma $ of length m. It follows by Theorem 1.1 (applied to the potential $-\tau $ ) that every subsequence of $(\mu _n)_n$ has a subsubsequence that converges on cylinders to a countably additive measure $\nu \in {\mathcal {M}}_{\le 1}(\sigma )$ . Since $\nu (C)=\mu (C)$ , for every cylinder of length m, it follows that $\nu =\mu $ . We conclude that $(\mu _n)_n$ converges on cylinders to $\mu $ . The forward implication follows by definition.

Proof. It is sufficient to prove the proposition in the case where $(\int \tau \,d\mu _n)_n$ is convergent (otherwise, we can consider a subsequence for which the limsup in (13) is a limit and then pass to a further subsequence for which the integral converges). By Lemma 2.5, we have that $\int \tau \,d\mu <\infty $ . Set $\unicode{x3bb} _0$ such that $\lim _{n\to \infty }\int \tau \,d\mu _n=\unicode{x3bb} _0\int \tau \,d(\unicode{x3bb} \mu )$ , which, by Lemma 2.5, also satisfies that $\unicode{x3bb} _0\ge 1$ .

Let $(\widehat {\Sigma },\widehat {\sigma })$ be the CMS associated with $(\Sigma ,\sigma )$ and $\tau $ constructed in §5. By Remark 5.7, $h_{\textrm {top}}(\widehat {\sigma })$ is finite. It follows by Lemma 5.6 that $(\hat {\mu }_n)_n$ converges on cylinders to ${1}/({\unicode{x3bb} _0})\hat {\mu }$ . We use Theorem 2.7 and Proposition 5.8 to conclude that

Then,

$$ \begin{align*}\limsup_{n\to\infty} h_{\mu_n}(\sigma) \le \unicode{x3bb} h_\mu(\sigma) +\bigg(\lim_{n\to\infty}\int \tau\,d\mu_n-\unicode{x3bb} \int \tau \,d\mu\bigg)h(\tau),\end{align*} $$

which is equivalent to inequality (13).

Proof. If $\|\phi -\phi _k\|<\infty $ , then $P(-t\phi )=\infty $ if and only if $P(-t\phi _k)=\infty $ . It follows that $s_\infty (-\phi )=s_\infty (-\phi _k)$ . By Lemma 3.4, we have also that $\lim _{k\to \infty } P_\infty (-t\phi _k)=P_\infty (-t\phi )$ .

Let us first consider the case where $\phi \in \Psi _1$ . Choose $\epsilon>0$ small such that $P_\infty (-(s_\phi -\epsilon )\phi )>0>P_\infty (-(s_\phi +\epsilon )\phi )$ . Then, $P_\infty (-(s_\phi -\epsilon )\phi _k)>0>P_\infty (-(s_\phi +\epsilon )\phi _k)$ for large enough k and, therefore, $|s_\phi -s_{\phi _k}|<\epsilon $ . It follows that for large enough k, we have that $\phi _k\in \Psi _1$ and that $\lim _{k\to \infty }s_{\phi _k}=s_\phi $ , which proves the claim in this case.

Suppose that $\phi \in \Psi _2$ . For every $\epsilon>0$ , we have that $P_\infty (-(s_\infty (-\phi )+\epsilon )\phi )<0$ . Therefore, for large enough k, we have that $P_\infty (-(s_\infty (-\phi )+\epsilon )\phi _k)<0$ and, thus, $s_{\phi _k}\le s_\infty (-\phi )+\epsilon $ . This implies that $\limsup _{n\to \infty } s_{\phi _k}\le s_\infty (-\phi )$ . Since $s_\infty (-\phi )=s_\infty (-\phi _k)$ and $h(\phi )=s_\infty (-\phi )$ , the claim follows.

Proof. Define $\tau _k(x)={1}/{k}\lceil k\tau (x)\rceil $ and observe that $\lim _{k\to 0}\|\tau -\tau _k\|_0=0$ . Set $f_k=k\tau _k$ and note that $h(f_k)={1}/{k}h(\tau _k)$ . The function $f_k$ takes values in ${\mathbb {N}}$ and ${\text {var}_2(f_k)=0}$ . Proposition 6.1 holds for the functions $f_k$ , which is equivalent to say that the proposition holds for $\tau _k$ . By Lemma 6.2, it follows that $\lim _{k\to \infty }h(\tau _k)=h(\tau )$ . Additionally, $|\!\int \tau _k\,d\nu -\int \tau \,d\nu |\le \|\tau -\tau _k\|_0$ for every $\nu \in {\mathcal {M}}(\sigma )$ . All of this together implies that Proposition 6.1 holds for $\tau \in \Psi _2$ .

Proof. Observe that $h_\mu (\sigma )=h_{\tilde {\mu }}(\sigma _m)$ and $\int \tau \,d\mu =\int \tau _m\,d\tilde {\mu }$ for every $\mu \in {\mathcal {M}}(\sigma )$ . It follows that $P^{\Sigma _m}(-t\tau _m)=P^{\Sigma }(-t\tau )$ and, therefore, $s_\infty (-\tau _m)=s_\infty (-\tau )$ .

We will prove that if $t>s_\infty (-\tau )$ , then $P_\infty ^{\Sigma _m}(-t\tau _m)=P^{\Sigma }_\infty (-t\tau )$ . It follows from this fact that $\tau \in \Psi _1$ if and only if $\tau _m\in \Psi _1$ , and that ${s}_\tau ={s}_{\tau _m}$ , which finishes the proof of the lemma, as we can conclude that $h(\tau )=h(\tau _m)$ .

If there exists a sequence $(\mu _n)_n$ in ${\mathcal {M}}(\sigma )$ that converges on cylinders to the zero measure and such that $\lim _{n\to \infty }(h_{\mu _n}(\sigma )-t\int \tau \,d\mu _n)=P_\infty (-t\tau )$ , then $(\tilde {\mu }_n)_n$ converges to the zero measure (see Remark 2.12) and $\lim _{n\to \infty }(h_{\tilde {\mu }_n}(\sigma _m)-t\int \tau _m\,d\tilde {\mu }_n)=P_\infty (-t\tau )$ . Therefore, we have $P_\infty ^{\Sigma _m}(-t\tau _m)\ge P^{\Sigma }_\infty (-t\tau )$ . If no such sequence exists, then we still have the inequality $P_\infty ^{\Sigma _m}(-t\tau _m)\ge P^{\Sigma }_\infty (-t\tau )$ . Note that if $P_\infty ^{\Sigma _m}(-t\tau _m)=-\infty $ , then $P_\infty ^{\Sigma }(-t\tau )=-\infty $ . We will now suppose that $P_\infty ^{\Sigma _m}(-t\tau _m)>-\infty $ .

Let $(\tilde {\mu }_n)_n$ be a sequence in ${\mathcal {M}}_{-\tau _m}(\sigma _n)$ that converges on cylinders to the zero measure and such that $P_\infty ^{\Sigma _m}(-t\tau _m)=\lim _{n\to \infty }(h_{\tilde {\mu }_n}(\sigma _m)-t\int \tau _m\,d\tilde {\mu }_n)$ . Since ${t>s_\infty (-\tau _m),}$ we have that $\limsup _{n\to \infty }\int \tau _m \,d\tilde {\mu }_n<\infty $ (see Lemma 3.10); equivalently, $\limsup _{n\to \infty }\int \tau \,d\mu _n<\infty $ . By Remark 6.4, $(\mu _n)_n$ converges to the zero measure and, therefore, $P_\infty ^{\Sigma _m}(-t\tau _m)\le P^{\Sigma }_\infty (-t\tau )$ . We conclude that $P_\infty ^{\Sigma _m}(-t\tau _m)=P^{\Sigma }_\infty (-t\tau )$ for every $t>s_\infty (-\tau )$ .

Proof. Let $(\mu _n)_n$ be a sequence in ${\mathcal {M}}_{-\tau }(\sigma )$ that converges on cylinders to $\unicode{x3bb} \mu $ , where ${\mu \in {\mathcal {M}}_{-\tau }(\sigma )}$ and $\unicode{x3bb} \in [0,1]$ . By assumption, $\limsup _{n\to \infty }\int \tau \,d\mu _n<\infty $ . Consider $(\Sigma _m,\sigma _m)$ and note that $\tau _m\in \Psi _1$ , as it only depends on the first coordinate of $\Sigma _m$ . We can apply Proposition 6.3 to the pair $(\Sigma _m,\sigma _m)$ and $\tau _m$ . Since $(\tilde {\mu }_n)_n$ converges on cylinders to $\unicode{x3bb} \tilde {\mu }$ (see Remark 2.12), we have that

$$ \begin{align*} \limsup_{n\to\infty}\bigg(h_{\tilde{\mu}_n}(\sigma_m)-h(\tau_m)\int \tau_m \,d{\tilde{\mu}_n}\bigg)\le \unicode{x3bb}\bigg(h_{\tilde{\mu}}(\sigma_m)-h(\tau_m)\int \tau_m \,d{\tilde{\mu}}\bigg). \end{align*} $$

Finally, recall that $h_\nu (\sigma )=h_{\tilde {\nu }}(\sigma _m)$ and $\int \tau \,d\nu =\int \tau _m\,d\tilde {\nu }$ for every $\nu \in {\mathcal {M}}(\sigma )$ , and that $h(\tau )=h(\tau _m)$ (see Lemma 6.5). We conclude that inequality (13) holds for $\tau $ .

Proof. Since $\tau \in \Psi $ , we have that $\text {var}_2(\tau )<\infty $ and that $\lim _{n\to \infty }\text {var}_n(\tau )=0$ . Fix ${m\ge 2}$ and define $\tau _m(x)=\sup \{\tau (y):y\in [x_1,\ldots , x_{m}]\}$ , where $x=(x_1,x_2,\ldots )$ . Note that $\|\tau -\tau _m\|_0\le \text {var}_m(\tau )$ and that $\text {var}_2(\tau _m)<\infty $ . Moreover, $\tau _m\in \Psi _m$ and by Proposition 6.6, we have that

$$ \begin{align*} \limsup_{n\to\infty}\bigg(h_{\mu_n}(\sigma)-h(\tau_m)\int \tau_m \,d\mu_n\bigg)\le \unicode{x3bb}\bigg(h_{\mu}(\sigma)-h(\tau_m)\int \tau_m\,d\mu\bigg). \end{align*} $$

Since $|\!\int \tau _m\,d\nu -\int \tau \,d\nu |\le \|\tau _m-\tau \|_0$ for every $\nu \in {\mathcal {M}}(\sigma )$ , and $\|\tau _m-\tau \|_0\to 0$ , we obtain the desired inequality (see Lemma 6.2).

Proof. Let $a:{\mathbb {N}}\to {\mathbb {R}}$ be given by $a(n)=\inf _{x\in [n]}\phi (x)$ . Define $\phi _1:{\mathbb {N}}^{\mathbb {N}}\to {\mathbb {R}}$ , where $\phi _1(x)=a(k)$ if $x\in [k]$ . Note that $\|\phi -\phi _1\|_0\le \text {var}_1(\phi )$ . In particular, $P(-\phi _1)=\log (\sum _{n=1}^\infty e^{-a(n)})$ is finite (see Example 2.9). We conclude that $\lim _{n\to \infty } a(n)=\infty $ .

Let us now suppose that $(\nu _n)_n$ converges on cylinders to the zero measure. We argue by contradiction and suppose that $\liminf _{n\to \infty }\int \phi\, d\nu _n<\infty $ . In this case, the inequality

$$ \begin{align*} 1-\nu_n\bigg(\bigcup_{k< n}[k]\bigg)=\nu_n\bigg(\bigcup_{k\ge n}[k]\bigg)\le \frac{\int \phi\,d\nu_n}{\inf \{\phi(x):x\in \bigcup_{k\ge n}[k]\}}=\frac{\int \phi\,d\nu_n}{\inf_{k\ge n} a(k)} \end{align*} $$

implies that $(\nu _n)_n$ does not converge on cylinders to the zero measure.

Proof. By Remark 6.9, the potential $\phi $ is bounded below. After adding a positive constant, we can assume that $\phi $ is bounded away from zero and, therefore, $\phi \in \Psi $ . By Remark 6.9, we also have that $\phi \in \Psi _2$ and, therefore, $h(\phi )=s_\infty (-\phi )$ . It follows by Theorem 6.7 that

$$ \begin{align*}\limsup_{n\to\infty}\bigg( h_{\mu_n}(\sigma)-s_\infty(-\phi)\int \phi\,d\mu_n\bigg)\le h_\mu(\sigma)-s_\infty(-\phi)\int \phi\,d\mu.\end{align*} $$

Lemma 2.5 implies that

$$ \begin{align*}\limsup_{n\to\infty}-(1-s_\infty(-\phi))\int \phi\,d\mu_n\le -(1-s_\infty(-\phi))\int \phi\,d\mu.\end{align*} $$

The result follows from adding these inequalities together.

Proof. The case $\unicode{x3bb} =0$ follows from the variational principle. Suppose that $\unicode{x3bb}>0$ . We will further suppose that $P(-\phi )=0$ (otherwise, we can subtract $P(-\phi )$ from both sides and consider the potential $\phi '=\phi +P(-\phi )$ ).