Proof. (i). Suppose \((f_n,T_n,g_n)\to (f,T,g)\) in \(\mathcal{G}\) and let \(x_n\in \Lambda_{f_n,T_n,g_n}\) be such that \(x_n\to x\). Let us show \(x\in \Lambda_{f,T,g}\). Indeed, for each integer \(l\ge 0\) and each \(n\ge 1\), \(T_n^l(x_{n})\in E_{f_{n}, T_n, g_{n}}\), so \[g_{n}(T_n^l(x_{n}))+f_{n}(T_n^l(x_{n}))= g_{n}(T_n^{l+1}(x_{n}))+\beta(f_{n},T_{n}),\] and hence \[g(T^l(x))+f(T^l(x))=g(T^{l+1}(x))+\beta(f,T).\] This implies that \(x\in \Lambda_{f,T,g}\).

(ii). Since \(T(\Lambda_{f,T,g})=\Lambda_{f,T,g}\) for all \(g\in G_{f,T}\), we have \(T(K(f,T))= K(f,T)\). To show that \(K(f,T)\) is compact, let \(\{x_n\}\subset K(f,T)\), so for each \(n\), \(x_n\in \Lambda_{f,T,g_n}\) for some \(g_n\in G_{f,T}\). Fixing an arbitrary \(y\in \mathbb{R}/\mathbb{Z}\), let \(\tilde{g}_n=g_n-g_n(y)\). Then \(\tilde{g}_n\in G_{f,T}\) and \(\Lambda_{f,T, g_n}=\Lambda_{f,T,\tilde{g}_n}\). Since \(\mathrm{Lip}(\tilde{g}_n)\) is bounded in \(n\), passing to a subsequence, we may assume that \(\tilde{g}_n\) converges to some \(g\) in the \(C^0\) topology. Then \(g\in G_{f,T}\). By (i), any accumulation point of \(\{x_n\}\) belongs to \(\Lambda_{f,T,g}\subset K(f,T)\). Thus \(K(f,T)\) is compact.

(iii). Suppose \((f_n,T_n)\to (f,T)\) in \(\mathcal{C}^1\times\mathcal{E}^1\) and \(x_n\in K(f_n,T_n)\), \(x_n\to x\). Then for each \(n\) there exists \(g_n\in G_{f_n,T_n}\) such that \(x_n\in \Lambda_{f_n,T_n, g_n}\). As in the proof of (ii), \(x_n\in \Lambda_{f_n,T_n,\tilde{g}_n}\) for \(\tilde{g}_n=g_n-g_n(y)\), and passing to a subsequence, we may assume that \(\tilde{g}_n\) converges to some \(g\) in the \(C^0\) topology. Again \(g\in G_{f,T}\). By (i), we obtain \(x\in \Lambda_{f,T,g} \subset K(f,T)\).

(iv). Any \(\mu\in \mathcal{M}_{\max}(f,T)\) is supported on \(\Lambda_{f,T,g}\) for any continuous sub-action \(g\) of \(f\). So \(\mu\) is supported on \(K(f,T)\), which implies that \(\mu\) is an invariant Borel probability measure of \(T|K(f,T)\).

For the other direction, it suffices to show each ergodic Borel probability measure \(\mu\) of \(T|K(f,T)\) satisfies \(\int f \,\mathrm{d}\mu=\beta(f,T)\). By the ergodicity of \(\mu\), there exists \(x\in K(f,T)\) such that \[\int f\,\mathrm{d}\mu=\lim_{n\to\infty} \frac{1}{n}\sum_{i=0}^{n-1} f(T^i(x)).\] Since \(x\in K(f,T)\), there exists \(g\in G_{f,T}\) with \(x\in \Lambda_{f,T,g}\). Thus for each \(n\ge 1\) \[\sum_{i=0}^{n-1} f(T^i(x))= n\beta(f,T)+ g(T^n(x))-g(x),\] which implies that \(\int f\,\mathrm{d}\mu=\beta(f,T)\). ◻

Proof. Let \(S_j\), \(j=1,2,\dots\) be a sequence of finite-dimensionally shy subsets of \(\mathcal{V}\). So for each \(j\), there exists \(\{\varphi_n^j\}_{n=1}^{N_j}\subset\mathcal{V}\setminus \left\lbrace 0\right\rbrace\), such that for each \(f\in \mathcal{V}\), \[m_{N_j}\left(\left\{(t_n)_{n=1}^{N_j}\in [0,1]^{N_j}: f+\sum_{n=1}^{N_j} t_n \varphi_n^j\in S_j\right\}\right)=0.\] Put \(\psi_n^j=\frac{1}{2^jN_j}\varphi_n^j/\|\varphi_n^j\|\). Let \(\mathcal{N}=\{(n,j): j\in \mathbb{N}_1, n\in \{1,2,\dots, N_j\}\}\). Then \(\sum_{(n,j)\in \mathcal{N}} \|\psi_n^j\|=1\). For each \(k=1,2,\dots\), \[m_\infty\left(\left\{(t_n^j)_{n,j}\in [0,1]^\mathcal{N}: f + \sum_{n,j} t_n^j \psi_n^j\in S_k\right\}\right)=0\] by Fubini’s theorem. Thus \[m_\infty\left(\left\{(t_{n}^j)_{n,j}\in [0,1]^\mathcal{N}: f + \sum_{n,j} t_n^j \psi_n^j \in \bigcup_k S_k \right\}\right)=0.\] ◻

Proof. Let \((\mathcal{W}_n)_{n=1}^\infty\) be a countable topological basis for \(\mathcal{V}\). Let \(E\) be the collection of \(n\ge 1\) such that \(\mathcal{W}_n\cap S\) is a countable union of finite-dimensionally shy subsets of \(\mathcal{H}\). Then \[S=\bigcup_{n\in E} (\mathcal{W}_n\cap S),\] and the conclusion follows. ◻

Proof. For each \(T\)-periodic orbit \(\mathcal{O}\), let \(\mu_{\mathcal{O}}\) denote the averaged Dirac measure on \(\mathcal{O}\), and let \[\mathcal{X}_{\mathcal{O}}(T)=\{f\in \mathcal{X}: \mu_{\mathcal{O}}\text{ is }(f,T)\text{-maximizing}\}.\] To prove (i), since \(T\) has only countably many periodic orbits, it suffices to show that for each periodic orbit \(\mathcal{O}\) of \(T\), \(\mathcal{X}_{\mathcal{O}}(T)\setminus \mathcal{X}_{\#}(T)\) is a \(1\)-dimensionally shy subset of \(\mathcal{X}\). To prove (ii), it suffices to show that \(\mathcal{X}_{\#}(T)\cap \mathcal{X}_{\mathcal{O}}(T)\) is dense in \(\mathcal{X}_{\mathcal{O}}(T)\) for each \(\mathcal{O}\).

To this end, let us follow an argument in [15]. Fix an arbitrary \(\mathcal{O}\), and let \(\varphi: \mathbb{R}/\mathbb{Z}\to \mathbb{R}\) be a trigonometric polynomial such that \(\varphi|\mathcal{O}=0\) and \(\varphi(x)<0\) for each \(x\in (\mathbb{R}/\mathbb{Z})\setminus \mathcal{O}\).

Now let \(f_0\in\mathcal{X}\) be arbitrary and define \(f_t=f_0+t \varphi\) for \(t\in\mathbb{R}\). Then \(f_t\in \mathcal{X}\), and let us show that \[\{t\in \mathbb{R}:f_t\in \mathcal{X}_{\mathcal{O}}(T)\setminus \mathcal{X}_{\#}(T)\}\] contains at most one point. Arguing by contradiction, assume that there exist \(t_1<t_2\) such that \(f_{t_1}, f_{t_2}\in \mathcal{X}_{\mathcal{O}}(T)\setminus \mathcal{X}_{\#}(T).\) Then \[\beta(f_{t_1}, T)=\beta(f_{t_2}, T)=\frac{1}{\#\mathcal{O}}\sum_{p\in \mathcal{O}} f_0(p),\] where \(\#\mathcal{O}\) denotes the cardinality of \(\mathcal{O}\). Let \(g\) be an arbitrary continuous sub-action of \((f_{t_1},T)\). Then for each \(x\in \mathbb{R}/\mathbb{Z}\), \[g(T(x))+\beta(f_{t_1}, T)\ge g(x)+f_{t_1}(x).\] Since \(\beta(f_{t_2}, T)=\beta(f_{t_1}, T)\) and \(f_{t_2}\le f_{t_1}\), \(g\) is also a sub-action of \((f_{t_2},T)\). Also note that \(f_{t_2}(x)<f_{t_1}(x)\) if \(x\notin \mathcal{O}\). It follows that \[\Omega(f_{t_2},T)\subset E(f_{t_2}, T, g)\subset\mathcal{O}.\] As \(\Omega(f_{t_2},T)\) is \(T\)-invariant and non-empty, this implies that \(\Omega(f_{t_2},T)=\mathcal{O}\) is a periodic orbit of \(T\), so \(f_{t_2}\in \mathcal{X}_{\#}(T)\), a contradiction. This shows that \(\mathcal{X}_{\mathcal{O}}(T)\setminus \mathcal{X}_{\#}(T)\) is \(1\)-dimensionally shy.

To show that \(\mathcal{X}_{\#}(T)\cap \mathcal{X}_{\mathcal{O}}(T)\) is dense in \(\mathcal{X}_{\mathcal{O}}(T)\), we observe that if \(f_0\in \mathcal{X}_{\mathcal{O}}(T)\), then \(f_t\in \mathcal{X}_{\#}(T)\cap \mathcal{X}_{\mathcal{O}}(T)\) for each \(t>0\). ◻

Proof. It suffices to show that for any (Lipschitz) calibrated sub-action \(g\) of \((f,T)\), we have \(\Omega(f,T)\supset \Lambda_{f,T,g}\). For case (i), since \(\Omega(f,T)\) is \(T\)-minimal, it is \(g\)-irreducible (see [15]), so by [15], the inclusion follows. For case (ii), it follows from [15]. ◻

Proof. Let \((f_0,T_0)\in \mathcal{C}^1\times\mathcal{E}^1\) be such that \(K_0=\Omega(f_0,T_0)\) is a \(T_0\)-periodic orbit. By Lemma 5, \(K_0=K(f_0, T_0)\). Fix a small neighborhood \(V\) of \(K_0\) in \(\mathbb{R}/\mathbb{Z}\). By the upper semi-continuity of \((f,T)\mapsto K(f,T)\), there is a neighborhood \(\mathcal{W}\) of \((f_0,T_0)\) in \(\mathcal{C}^1\times \mathcal{E}^1\) such that \(K(f,T)\subset V\) for \((f,T)\in \mathcal{W}\). Shrinking \(V\) and \(\mathcal{W}\) if necessary, we have that \(\bigcap_{n=0}^\infty T^{-n}(V)\) is a periodic orbit of \(T\). Since \(K(f,T)\) is \(T\)-invariant, \(K(f,T)\subset \bigcap_{n=0}^\infty T^{-n}(V)\). Thus \(K(f,T)\) is a \(T\)-periodic orbit. Since \(\Omega(f,T)\subset K(f,T)\), \(\Omega(f,T)\) is a periodic orbit of \(T\). ◻

Proof of Theorem 2 assuming Theorem 3. Let \(\mathcal{X}=\mathcal{C}^r\), \(r=1,2,\dots,\infty\) and let \[\mathcal{E}_*^r=\{T\in\mathcal{E}^r: \mathcal{X}_\#(T) \text{ is open and dense in }\mathcal{X}\}.\] It suffices to show that \(\mathcal{E}_*^r\supset \mathcal{E}^\omega\) and that \(\mathcal{E}_*^r\) is a residual subset of \(\mathcal{E}^r\). Note that by Lemma 6, \(\mathcal{X}_\#(T)\) is open in \(\mathcal{X}\) for each \(T\in \mathcal{X}\). Therefore, \[\mathcal{E}_*^r=\{T\in\mathcal{E}^r: \mathcal{X}_\#(T) \text{ is dense in }\mathcal{X}\}.\]

Let us first show that \(\mathcal{E}^\omega\subset \mathcal{E}_*^r\). Let \(T\in \mathcal{E}^\omega\). Let \(\mathcal{H}=\mathcal{C}^\omega_{1,1}\) be as defined in 4 , which is dense in \(\mathcal{X}\). By Theorem 3, \(\mathcal{H}_{\#}(T)\) is a prevalent subset of \(\mathcal{H}\), and hence it is dense in \(\mathcal{H}\). Thus \(\mathcal{X}_\#(T)\supset \mathcal{H}_\#(T)\) is dense in \(\mathcal{X}\). Therefore \(T\in \mathcal{E}_*^r\).

To prove that \(\mathcal{E}_*^r\) is a residual subset of \(\mathcal{E}^r\), choose \(T_n\in \mathcal{E}^\omega\), \(n=1,2,\dots\), such that \(\{T_n\}_{n=1}^\infty\) is dense in \(\mathcal{E}^r\). Then for each \(n=1,2,\dots\), \(\mathcal{X}_\#(T_n)\) is an open and dense subset of \(\mathcal{X}\). Thus \(\mathcal{D}_\#=\bigcap_{n=1}^\infty\mathcal{X}_\#(T_n)\) is a residual subset of \(\mathcal{X}\). Let \(\{f_m\}_{m=1}^\infty\) be a dense subset of \(\mathcal{D}_\#\). By Lemma 6, for each \(m\), there is an open subset \(\mathcal{U}_m\) of \(\mathcal{E}^r\) containing \(\{T_n\}_{n=1}^\infty\) such that \(f_m\in\mathcal{X}_\#(T)\) for each \(T\in\mathcal{U}_m\). Let \(\mathcal{U}=\bigcap_{m=1}^\infty \mathcal{U}_m\). Then \(\mathcal{U}\) is a residual subset of \(\mathcal{E}^r\). Now let us show that \(\mathcal{U}\subset \mathcal{E}_*^r\). Indeed, for each \(T\in \mathcal{U}\), \(\{f_m\}_{m=1}^\infty\subset\mathcal{X}_\#(T)\), and hence \(\mathcal{X}_\#(T)\) is dense in \(\mathcal{X}\). Thus \(T\in \mathcal{E}_*^r\), which completes the proof. ◻

Proof. For each \(((f_{\mathbf{t}}), T)\in \mathcal{F}\times \mathcal{E}^r\), let \[\mathbf{L}((f_{\mathbf{t}}), T)=\{\mathbf{t}\in [0,1]^q: f_{\mathbf{t}} \in \mathcal{X}_{\#}(T)\}.\] By Lemma 6, for any compact subset \(K\) of \([0,1]^q\), \[\{(\mathbf{f},T)\in \mathcal{F}\times \mathcal{E}^r: \mathbf{L}(\mathbf{f}, T)\supset K\}\] is an open subset of \(\mathcal{F}\times \mathcal{E}^r\).

Now suppose that \(m_q\big(\mathbf{L}(\mathbf{f}, T))>\varepsilon\) for some \((\mathbf{f}, T)\in \mathcal{F}\times \mathcal{E}^r\). Then there exists a compact set \(K\subset \mathbf{L}(\mathbf{f}, T)\) such that \(m_q(K)>\varepsilon\). Thus if \((\tilde{\mathbf{f}}, \tilde{T})\) is sufficiently close to \((\mathbf{f}, T)\) in \(\mathcal{F}\times \mathcal{E}^r\), then \(\mathbf{L}(\tilde{\mathbf{f}}, \tilde{T}) \supset K\) and hence \(m_q(\mathbf{L}(\tilde{\mathbf{f}}, \tilde{T}))>\varepsilon\). The conclusion follows. ◻

Proof of Theorem 1 assuming Theorem 3. For each integer \(k\ge 1\) and each \(T\in \mathcal{E}^r\), let \[\mathcal{F}_k(T)=\left\{(f_{\mathbf{t}})\in \mathcal{F}: m_q\big(\{\mathbf{t}\in [0,1]^q: f_{\mathbf{t}} \in \mathcal{X}_{\#}(T) \}\big)>1-\tfrac{1}{k}\right\},\] and let \[\mathcal{F}_*(T)=\left\{(f_{\mathbf{t}})\in \mathcal{F}: m_q \big(\{\mathbf{t}\in [0,1]^q: f_{\mathbf{t}} \in \mathcal{X}_{\#}(T) \} \big) = 1 \right\}.\] Let \[\mathcal{E}_k^r=\{T\in\mathcal{E}^r: \mathcal{F}_k(T) \text{ is a residual subset of } \mathcal{F}\}\] for each \(k\), and let \[\mathcal{E}_*^r=\{T\in\mathcal{E}^r: \mathcal{F}_*(T) \text{ is a residual subset of } \mathcal{F}\}.\] Then for each \(T\in\mathcal{E}^r\), \[\mathcal{F}_*(T)=\bigcap_{k=1}^\infty \mathcal{F}_k(T).\] As a result, \[\mathcal{E}_*^r= \bigcap_{k=1}^\infty \mathcal{E}_k^r.\] So it suffices to show that for each \(k\), \(\mathcal{E}_k^r\supset \mathcal{E}^\omega\) and \(\mathcal{E}_k^r\) is a residual subset of \(\mathcal{E}^r\).

Let us first show that \(\mathcal{E}_k^r \supset \mathcal{E}^\omega\). Given \(T\in \mathcal{E}^\omega\), it suffices to show that \(\mathcal{F}_k(T)\) is open and \(\mathcal{F}_*(T)\) is dense in \(\mathcal{F}\). By Lemma 8, \(\mathcal{F}_k(T)\) is open. It remains to show that \(\mathcal{F}_*(T)\) is dense in \(\mathcal{F}\). To this end, let \(\mathcal{H}=\mathcal{C}^\omega_{1,1}\) be as defined in 4 . By Theorem 3, there exists \(\{\varphi_n\}_{n=1}^\infty\subset \mathcal{H}\), such that \(\sum_{n=1}^\infty \|\varphi_n\|_{\mathcal{H}}<\infty\) and such that for each \(f\in \mathcal{H}\), \[m_\infty \left(\left\{(t_n)_{n=1}^\infty \in [0,1]^{\mathbb{N}_1}: f+\sum_{n=1}^\infty t_n\varphi_n\not\in \mathcal{H}_\#(T)\right\}\right)=0.\] Given any \(C^r\) family \((f_{\mathbf{s}})\in \mathcal{F}\), we may approximate it in the \(C^r\) topology by a family \((\tilde{f}_{\mathbf{s}})\in \mathcal{F}\) with \(\tilde{f}_{\mathbf{s}}\in \mathcal{H}\). Then for each \(\mathbf{s}\in [0,1]^q\), \[m_\infty\left(\left\{(t_n)_{n=1}^\infty \in [0,1]^{\mathbb{N}_1}: \tilde{f}_{\mathbf{s}} + \sum_{n=1}^\infty t_n\varphi_n \not\in \mathcal{H}_{\#}(T)\right\}\right)=0.\] By Fubini’s theorem, for \(m_\infty\)-a.e. \((t_n)_{n=1}^\infty\), \[m_q\left(\left\{\mathbf{s}\in [0,1]^q: \tilde{f}_{\mathbf{s}} +\sum_{n=1}^\infty t_n\varphi_n\not\in \mathcal{H}_{\#}(T)\right\}\right)=0.\] For any such \((t_n)_{n=1}^\infty\), \((\tilde{f}_{\mathbf{s}}+ \sum_{n=1}^\infty t_n \varphi_n)\) is a family in \(\mathcal{F}_*(T)\). Since the topology on \(\mathcal{H}\) is stronger than the \(C^r\) topology, this implies that \(\mathcal{F}_*(T)\) is dense in \(\mathcal{F}\).

Now let us turn to the proof that \(\mathcal{E}_k^r\) is a residual subset of \(\mathcal{E}^r\). The argument is similar to the corresponding part in the proof of Theorem 2. To begin with, let \(\{T_n\}_{n=1}^\infty\subset \mathcal{E}^\omega\) be dense in \(\mathcal{E}^r\). For each \(n\), since \(T_n\in \mathcal{E}^\omega\), \(\mathcal{F}_k(T_n)\) is a residual subset of \(\mathcal{F}\). Thus \[\mathcal{F}_k:=\bigcap_{n=1}^\infty \mathcal{F}_k(T_n)\] is a residual subset of \(\mathcal{F}\). Let \(\{\mathbf{f}_m\}_{m=1}^\infty\) be a dense subset of \(\mathcal{F}_k\). For each \(m\), consider \[\mathcal{E}^r_{k,m}:=\{T\in \mathcal{E}^r: \mathbf{f}_m \in \mathcal{F}_k(T)\}.\] By Lemma 8, \(\mathcal{E}^r_{k,m}\) is open in \(\mathcal{E}^r\). On the other hand, since \(T_n\in \mathcal{E}^r_{k,m}\) for each \(n\), and since \(\{T_n\}_{n=1}^\infty\) is dense in \(\mathcal{E}^r\), \(\mathcal{E}^r_{k,m}\) is dense in \(\mathcal{E}^r\). Therefore, \(\bigcap_{m=1}^\infty \mathcal{E}^r_{k,m}\) is a residual subset of \(\mathcal{E}^r\). For each \(T\in \bigcap_{m=1}^\infty \mathcal{E}^r_{k,m}\), \(\mathcal{F}_k(T)\) is open in \(\mathcal{F}\) and contains \(\{\mathbf{f}_m\}_{m=1}^\infty\), so \(\mathcal{F}_k(T)\) is a residual subset of \(\mathcal{F}\). Thus \[\mathcal{E}^r_k\supset \bigcap_{m=1}^\infty \mathcal{E}^r_{k,m}\] is a residual subset of \(\mathcal{E}^r\). ◻

Proof. Since \(g\) is a sub-action of \(f\), for each \(N\ge 1\) we have: \[g(\pi(y))\ge \sum_{n=1}^N \widehat{f}(\tau_{\mathbf{i},n}(y)) + g(\pi(\tau_{\mathbf{i},N}(y))).\] Moreover, since \(\mathbf{i}\in \mathcal{CL}(x,K)\), we have: \[g(\pi(x)) = \sum_{n=1}^N \widehat{f}(\tau_{\mathbf{i},n}(x)) + g(\pi(\tau_{\mathbf{i},N}(x))).\] Subtracting the two displayed expressions above first and then letting \(N\to \infty\), the conclusion follows. ◻

Proof. Arguing by contradiction, assume that \(K\) is not Sturmian-like. Then there exists \(x_0\in \pi^{-1}(K)\) such that it is a critical value of \(K\) but not regular. As a result, there exist two sequences \(x_n, y_n\in \pi^{-1}(K)\), both converging to \(x_0\) from the same side (say from the right hand side), and distinct \(i_1, j_1\in \mathbb{A}\), such that for each \(n\), \[\tau_{i_1}(x_n), \tau_{j_1}(y_n)\in \pi^{-1}(K).\] As \(T(K)=K\), there exist \[\mathbf{i}^n=(i^n_k)_{k\ge 1}\in \mathcal{CL}(x_n,K) \quad\text{and}\quad \mathbf{j}^n=(j^n_k)_{k\ge 1}\in \mathcal{CL}(y_n, K)\] such that \(i^n_1=i_1\ne j_1=j^n_1\). Passing to a subsequence, we may assume \(\mathbf{i}^n\to\mathbf{i}\) and \(\mathbf{j}^n\to\mathbf{j}\). By continuity, \(\mathbf{i},\mathbf{j}\in \mathcal{CL}(x_0,K)\). By Lemma 12, \[\int_{x_0}^{x_n} \mathrm{D}h^f_{\mathbf{i}^n}(t)\,\mathrm{d}t\ge g(\pi(x_n))-g(\pi(x_0))\ge \int_{x_0}^{x_n} \mathrm{D}h^f_{\mathbf{j}}(t) \,\mathrm{d}t,\] \[\int_{x_0}^{y_n} \mathrm{D}h^f_{\mathbf{j}^n}(t) \,\mathrm{d}t\ge g(\pi(y_n))-g(\pi(x_0))\ge \int_{x_0}^{y_n} \mathrm{D}h^f_{\mathbf{i}}(t)\,\mathrm{d}t.\] Dividing the first inequality by \(x_n-x_0\) and then letting \(n\to\infty\) gives \[\mathrm{D}h^f_{\mathbf{i}}(x_0)\ge \mathrm{D}h^f_{\mathbf{j}}(x_0).\] Here we used the fact that as \(n\to\infty\), \(\mathrm{D}h^f_{\mathbf{i}^n}(t)\) converges to \(\mathrm{D}h^f_{\mathbf{i}}(t)\) locally uniformly in \(t\). Similarly, the second gives \[\mathrm{D}h^f_{\mathbf{j}}(x_0)\ge \mathrm{D}h^f_{\mathbf{i}}(x_0).\] Thus \[\mathrm{D}h^f_{\mathbf{j}}(x_0)= \mathrm{D}h^f_{\mathbf{i}}(x_0).\] Since \(i_1\ne j_1\), this implies that \(x_0\) is an \(f\)-supercritical value, a contradiction. ◻

Proof. By Lemma 5, \(K(f)=\Omega(f,T)\), so for each continuous sub-action \(g\) of \((f,T)\), \(K(f)\subset E_{f,T,g}\). Alternatively, since \((f,T)\) has a unique maximizing measure, it is well-known that there exists a unique calibrated sub-action \(g\) of \((f,T)\) up to an additive constant (see [6] or [33]), so \(K(f)=\Lambda_{f,T,g}\subset E_{f,T,g}\) for this \(g\). Thus the conclusion follows from Proposition 13 applied to \(K=K(f)\). ◻

Proof. Suppose that \(f_n\to f_0\) in \(\mathcal{C}^1\) and for each \(n\), \(K(f_n)\) has an \(f_n\)-supercritical value \(x_n\in \pi^{-1}(K(f_n))\). We need to show that \(f_0\) is supercritical. Without loss of generality, we may assume \(x_n\in [0,1)\). After passing to a subsequence, we may assume \(x_n\to x_0\). By the upper semi-continuity of \(K(\cdot)\), we have \(\pi(x_0)\in K(f_0)\). Let \(\mathbf{i}^n=(i^n_k)_{k=1}^\infty, \mathbf{j}^n =(j_k^n)_{k=1}^\infty\) be elements in \(\mathcal{CL}(x_n, K(f_n))\) such that \(i^n_1\not=j^n_1\) and \[\label{eqn:superfn} \mathrm{D}h^{f_n}_{\mathbf{i}^n}(x_n)=\mathrm{D}h^{f_n}_{\mathbf{j}^n}(x_n).\tag{13}\] Passing to a further subsequence, we may assume that \(\mathbf{i}^n\to \mathbf{i}=(i_k)_{k=1}^\infty\) and \(\mathbf{j}^n\to (j_k)_{k=1}^\infty\) in \(\mathbb{A}^{\mathbb{N}_1}\). Then \(i_1\not=j_1\). Again by the upper semi-continuity of \(K(\cdot)\), we have \(\mathbf{i},\mathbf{j}\in \mathcal{CL}(x_0, K(f_0))\). Letting \(n\to\infty\) in (13 ), we obtain \[\mathrm{D}h^f_\mathbf{i}(x_0)=\mathrm{D}h^f_{\mathbf{j}}(x_0),\] i.e., \(x_0\) is an \(f_0\)-supercritical value of \(K(f_0)\). ◻

Proof. By [23], for each \(f\in \mathcal{H}\), either each \(\mu\in \mathcal{M}(T)\) is a maximizing measure of \(f\), or each maximizing measure has entropy zero. In the former case, \(f\in \mathcal{H}_{\mathrm{per}}\); in the latter case, by Proposition 1 and the variational principle, \(h_{\mathrm{top}} (T|K(f))=0\). So for each \(f_0\in \mathcal{H}_{\mathrm{sc}}\setminus \mathcal{H}_{\mathrm{per}}\), \(h_{\mathrm{top}} (T|K(f_0))=0\). Applying Theorem 16 to \(\mathcal{D}=\mathcal{H}\), for \(f_0\) as above we can find an open neighborhood \(\mathcal{U}\) of \(f_0\) in \(\mathcal{C}^2\) such that \(\mathcal{U}\cap (\mathcal{H}_{\mathrm{sc}}\setminus \mathcal{H}_{\mathrm{per}})\) is finite-dimensionally shy. Then by Proposition 3 (taking \(\mathcal{V}=\mathcal{C}^2\)), the corollary follows. ◻

Proof. Direct calculation yields that \[|\mathrm{D}h^f_{\mathbf{i}}(x) - \mathrm{D}h^f_{\mathbf{i}}(y)| \le \sum_{n=1}^{|\mathbf{i}|} \sup_{z\in \mathbb{R}} |(\widehat{f}\circ\tau_{\mathbf{i},n})''(z)| \cdot |x-y|,\] and \[|(\widehat{f}\circ\tau_{\mathbf{i},k})''(z)| \le \|\mathrm{D}f\|_{C^0}\cdot|\tau_{\mathbf{i},n}''(z)| + \|\mathrm{D}^2 f\|_{C^0}\cdot|\tau_{\mathbf{i},n}'(z)|^2.\] On the other hand, by 2 , \(|\tau_{\mathbf{i},n}'|\le (C_*\lambda_*^n)^{-1}\), and the conclusion follows once we show that \(|\tau_{\mathbf{i},n}''|\le C_0 \lambda_*^{-n}\) for some \(C_0 >1\) dependent only on \(T\). To estimate \(|\tau_{\mathbf{i},n}''|\), denote \(\mathbf{i}=i_1i_2\dots\) and \(\tau_{\mathbf{i},0}=\mathrm{id}\). Then by 11 and the chain rule, \[\tau_{\mathbf{i},n}' = \prod_{m=1}^n \tau_{i_m}'\circ \tau_{\mathbf{i},m-1},\] and hence \[\tau_{\mathbf{i},n}'' = \sum_{m=1}^n \frac{\tau_{\mathbf{i},n}'}{\tau_{i_m}'\circ \tau_{\mathbf{i},m-1}}\cdot \left(\tau_{i_m}''\circ \tau_{\mathbf{i},m-1}\right) \cdot \tau_{\mathbf{i},m-1}'.\] Since \(|\tau_{i_m}'|\) is bounded from below, \(|\tau_{i_m}''|\) is bounded from above and \(|\tau_{\mathbf{i},k}'|\le (C_*\lambda_*^k)^{-1}\), the desired upper bound of \(|\tau_{\mathbf{i},n}''|\) follows. ◻

Proof. Fix \(\theta>h_{\mathrm{top}} (T|K(f_0))\) and \(\delta>0\). For each \(n\), let \(\mathcal{A}_n\) denote the subset of \(\mathbb{A}^n\) consisting of \(i_1i_2\dots i_n\) such that \[\tau_{i_n}\circ \dots \circ\tau_{i_2}\circ\tau_{i_1} ((0,1))\cap \pi^{-1}(K(f_0))\not=\emptyset.\] Then as \(n\to\infty\), \[\frac{1}{n}\log \#\mathcal{A}_n\to h_{\mathrm{top}} (T|K(f_0))<\theta,\] so there exists a positive integer \(N\) such that ?? holds, and such that for \[V:=\bigcup_{i_1i_2\dots i_N\in \mathcal{A}_N} \tau_{i_N}\circ\dots\circ \tau_{i_2}\circ\tau_{i_1}((0,1)),\] each connected component of \(V\) has length less than \(\delta\). Then \(\pi(V)\) is an open neighborhood of \(K(f_0)\), and by the upper semi-continuity of \(f\mapsto K(f)\) (Proposition 1), there exists \(\varepsilon_0\in (0,1)\) such that if \(\|f-f_0\|_{C^1}<\varepsilon_0\), then \(K(f)\subset \pi(V)\).

Now suppose \(f\) is as in the lemma and \(x_0\in (0,1)\cap \pi^{-1}(K(f))\) is an \(f\)-supercritical value of \(K(f)\). Let \(\mathbf{i}=(i_n)_{n\ge 1}, \mathbf{j}=(j_n)_{n\ge 1}\in \mathbb{A}^{\mathbb{N}_1}\) be such that \(i_1\not=j_1\), \[\tau_{\mathbf{i},n}(x_0), \tau_{\mathbf{j},n}(x_0) \in \pi^{-1}(K(f)) for alln\ge 1,\] and such that \[\mathrm{D}h^f_{\mathbf{i}}(x_0)=\mathrm{D}h^f_{\mathbf{j}}(x_0).\] It follows that \(i_{k+1}i_{k+2}\dots i_{k+N}, j_{k+1}j_{k+2}\dots j_{k+N}\in \mathcal{A}_N\) for all \(k\ge 0\).

Let \(m\ge 1\) be given. Take \(\mathbf{u}=i_1i_2\dots i_{mN}\) and \(\mathbf{v}=j_1j_2\dots j_{mN}\). Then \(u_1\ne v_1\), ?? holds for \(\mathbf{u},\mathbf{v}\) and \[\left|\mathrm{D}h^f_{\mathbf{u}}(x_0)-\mathrm{D}h^f_{\mathbf{v}}(x_0)\right| \le \sum_{k=mN+1}^\infty \left|(\widehat{f}\circ\tau_{\mathbf{i},k})'(x_0) - (\widehat{f}\circ\tau_{\mathbf{j},k})'(x_0)\right|\le C\lambda_*^{-mN}\] for some constant \(C=C(f_0)\ge 1\). On the other hand, since \(\pi(x_0)\in K(f)=T(K(f))\subset \pi(V)\), there exists \(\mathbf{w}\in \mathbb{A}^{mN}\) such that ?? holds, ?? holds for \(\mathbf{w}\) and \[|\tau_{\mathbf{w},mN}(0)-x_0| \le (C_*\lambda_*^{mN})^{-1}.\] Combining the last inequality with \(\|\mathrm{D}f\|_{C^1}<\|\mathrm{D}f_0\|_{C^1}+1\) and Lemma 18 yields that \[|\mathrm{D}h^f_{\mathbf{u}}(\tau_{\mathbf{w},mN}(0))-\mathrm{D}h^f_{\mathbf{v}}(\tau_{\mathbf{w},mN}(0))|\le |\mathrm{D}h^f_{\mathbf{u}}(x_0)-\mathrm{D}h^f_{\mathbf{v}}(x_0)|+\tilde{C}\lambda_*^{-mN}\] for some \(\tilde{C}=\tilde{C}(f_0)\), which proves ?? and completes the proof of the lemma. ◻

Proof. Fix an integer \(M\ge 2\) with \[\lambda_*^M \ge \frac{8C_T}{C_*(\lambda_*-1)} \quad\text{for}\quad C_T:=\sup_{x\in \mathbb{R}/\mathbb{Z}}|T'(x)|>1.\] Then there exists \(\delta>0\) such that \(\bigcup_{z\in \widehat{K}}[z-4\delta,z+4\delta]\subset[0,1]\) and such that the following hold for each \(i\in \mathbb{A}\) and each \(y\in\widehat{K}\):

• \(\tau_i([y-4\delta,y+4\delta])\cap \tau_j([y-4\delta,y+4\delta])=\emptyset\) for any \(j\in\mathbb{A}\setminus\{i\}\);

• \(\tau_i([y-4\delta,y+4\delta])\cap\tau_{\mathbf{j},m}([y-4\delta,y+4\delta])=\emptyset\) for any word \(\mathbf{j}\) and any \(2\le m\le \min\{M,|\mathbf{j}|\}\).

Note that in the last item we used the assumption that \(K(f_0)\) contains no periodic point. By the choice of \(\delta\), for each \(i\in\mathbb{A}\) and each \(y\in \widehat{K}\), we can easily find \(\psi_{i,y}\in \mathcal{C}^1\) such that for \(\widehat{\psi}_{i,y}=\psi_{i,y}\circ\pi\) we have:

• \(|(\widehat{\psi}_{i,y})'|\le 2C_T\) on \(\mathbb{R}\);

• \((\widehat{\psi}_{i,y}\circ\tau_i)'=1\) on \([y-2\delta,y+2\delta]\);

• \((\widehat{\psi}_{i,y})'=0\) on \([0,1]\setminus \big(\tau_i([y-4\delta,y+4\delta])\big)\).

For each \(i\in\mathbb{A}\) and each \(y\in \widehat{K}\), by our choices of \(\delta\) and \(\psi_{i,y}\), when \(|x-y|\le 2\delta\), for any word \(\mathbf{j}\) we have \[(\widehat{\psi}_{i,y}\circ\tau_{\mathbf{j},m})'(x)=0, \quad 2\le m\le \min\{M,|\mathbf{j}|\},\] and therefore \[\big|\mathrm{D}h^{\psi_{i,y}}_{\mathbf{j}}(x)-(\widehat{\psi}_{i,y}\circ\tau_{j_1})'(x)\big| =\Big|\sum_{M<m\le|\mathbf{j}|} (\widehat{\psi}_{i,y}\circ\tau_{\mathbf{j},m})'(x)\Big|\le 2 C_T\sum_{m>M} (C_*\lambda_*^m)^{-1} \le \frac{1}{4}.\] That is to say,

\[\big|\mathrm{D}h^{\psi_{i,y}}_{\mathbf{j}}(x)-1\big| \le \frac{1}{4} \;\;if\;\;j_1=i\,;\quad \big|\mathrm{D}h^{\psi_{i,y}}_{\mathbf{j}}(x)\big| \le \frac{1}{4} \;\;if\;\;j_1\ne i.\] By compactness, we can choose \(x_q\in \widehat{K},1\le q\le Q\) such that assertion (i) holds. Since \(\mathcal{D}\) is dense in \(\mathcal{C}^1\), for each \(i\in \mathbb{A}\) and each \(q=1,\dots,Q\), we may choose \(\varphi_{i,q}\in \mathcal{D}\) sufficiently close to \(\psi_{i,x_q}\) in \(C^1\) topology such that both ?? and ?? hold. ◻

Proof of Theorem 16. Let \(f_0\) be as in the statement. Let \(\delta\) and \(\varphi_{i,q}\) be given by Lemma 20. Consider the following family of functions: \[\varphi_{\mathbf{t}}=\sum_{i,q} t_{i,q} \varphi_{i,q},\quad \mathbf{t}=(t_{i,q})_{i\in\mathbb{A}, 1\le q\le Q}\in\mathbb{R}^{d_TQ}.\] Fix \(f_0\) and \(\delta\) as above and fix \(\theta>0\) such that \[h_{\mathrm{top}} (T|K(f_0))<\theta<\tfrac{1}{3}\log \lambda_*.\] Let \(C_0,\varepsilon_0,N,\mathcal{A}_N\) be given by Lemma 19 accordingly. Let \(\mathcal{U}\) be an open neighborhood of \(f_0\) in \(\mathcal{C}^2\) such that for each \(f\in \mathcal{U}\), \[\|f-f_0\|_{C^1}< \varepsilon_0 \quad\text{and}\quad \|f-f_0\|_{C^2}< 1.\]

Our goal is to show that the set \[\mathbf{S}(F):=\{\mathbf{t}\in [0,1]^{d_TQ}: F+ \varphi_{\mathbf{t}} \in \mathcal{U} \text{ is supercritical}\}\] has Lebesgue measure zero for any \(F\in \mathcal{C}^2\). To this end, fix \(F\in \mathcal{C}^2\) and denote \(F_{\mathbf{t}}=F+ \varphi_{\mathbf{t}}\) below. We make the following three observations.

Firstly, if \(\mathbf{t}\in \mathbf{S}(F)\), then \[\|F_{\mathbf{t}}-f_0\|_{C^1}<\varepsilon_0 \quad\text{and}\quad \|F_{\mathbf{t}}-f_0\|_{C^2}<1,\] and hence all the statements in Lemma 19 hold for \(F_{\mathbf{t}}\) instead of \(f\). In particular, for any \(m\ge 1\), there exist \(\mathbf{u}, \mathbf{v},\mathbf{w}\) (depending on \(\mathbf{t}\)) as in Lemma 19 such that \[\label{eqn:bc1} \left| \mathrm{D}h^{F_{\mathbf{t}}}_{\mathbf{u}}(\tau_{\mathbf{w},mN}(0))-\mathrm{D}h^{F_{\mathbf{t}}}_{\mathbf{v}}(\tau_{\mathbf{w},mN}(0))\right|\le C_0 \lambda_* ^{-mN}.\tag{14}\] Moreover, by ?? and Lemma 20 (i), \(|\tau_{\mathbf{w},mN}(0)-x_q|<2\delta\) holds for some \(q\).

Secondly, if \(1\le q\le Q\) and \(\mathbf{u},\mathbf{v},\mathbf{w}\in \mathbb{A}^{mN}\) satisfy \(u_1 \ne v_1\) and \(|\tau_{\mathbf{w},mN}(0)-x_q|\le 2\delta\), then by Lemma 20 (ii), \[\left| \mathrm{D}h^{\varphi_{u_1,q}}_{\mathbf{u}}(\tau_{\mathbf{w},mN}(0))-\mathrm{D}h^{\varphi_{u_1,q}}_{\mathbf{v}}(\tau_{\mathbf{w},mN}(0))\right|\ge \frac{1}{3}.\] Also note that \(\mathrm{D}h^{f}_{\bullet}(\cdot)\) is a linear functional of \(f\) and \(F_{\mathbf{t}}\) is affine in \(\mathbf{t}\). Then by Fubini’s theorem, it follows that \[\{\mathbf{t}\in [0,1]^{d_TQ}: \text{\eqref{eqn:bc1} holds for \mathbf{t}}\}\] has Lebesgue measure bounded from above by \(6C_0\lambda_* ^{-mN}\).

Thirdly, by ?? , the number of possible choices of the triple \((\mathbf{u},\mathbf{v},\mathbf{w})\) satisfying ?? is bounded from above by \[(\#\mathcal{A}_N)^{3m} < e^{3\theta mN}.\]

From the three observations above we conclude that \(\mathbf{S}(F)\) has Lebesgue measure bounded by \(6C_0(e^{3\theta}\lambda_*^{-1})^{mN}\). Combining this with \(\theta<\frac{\log\lambda_*}{3}\) and letting \(m\to\infty\), it follows that \(\mathbf{S}(F)\) is of Lebesgue measure zero. ◻

Proof of Theorem 3. By Proposition 2, it suffices to show that \(\mathcal{H}\setminus \mathcal{H}_{\#}\) is a countable union of finite-dimensionally shy subsets of \(\mathcal{H}\). To this end, let us write \[\mathcal{X}_1=\mathcal{H}\setminus \mathcal{H}_{\mathrm{u}}, \quad \mathcal{X}_2=(\mathcal{H}_{\mathrm{u}}\cap \mathcal{H}_{\mathrm{st}})\setminus \mathcal{H}_{\mathrm{per}}, \quad \mathcal{X}_3=\mathcal{H}_{\mathrm{u}}\setminus(\mathcal{H}_{\mathrm{st}}\cup\mathcal{H}_{\mathrm{sc}}),\] \[\mathcal{X}_4=\mathcal{H}_{\mathrm{sc}}\setminus \mathcal{H}_{\mathrm{per}},\quad \mathcal{X}_5=\mathcal{H}_{\mathrm{per}}\setminus \mathcal{H}_\#.\] Then \[\mathcal{H}\setminus \mathcal{H}_{\#} \subset \bigcup_{i=1}^5\mathcal{X}_i,\] and it remains to prove that for \(1\le i\le 5\), \(\mathcal{X}_i\) is a countable union of finite-dimensionally shy subsets of \(\mathcal{H}\). This is verified as follows.

• For \(i=1\), in [18] it was implicitly proved that \(\mathcal{H}\setminus \mathcal{H}_{\mathrm{u}}\) is a countable union of \(1\)-dimensionally shy subsets of \(\mathcal{H}\). More specifically, given \(g\in\mathcal{H}\), let \[\mathcal{P}_g=\{f\in\mathcal{H}: s\mapsto \beta(f+sg,T) \text{ is differentiable at 0} \}.\] By the convexity of \(\beta(\cdot,T)\), the set \(\{t\in[0,1]: f+tg \notin \mathcal{P}_g\}\) is countable for any \(f\in\mathcal{H}\), so \(\mathcal{H}\setminus \mathcal{P}_g\) is a \(1\)-dimensionally shy subset of \(\mathcal{H}\). On the other hand, Morris [18] showed that if \(\{g_n\}_{n=1}^\infty\subset \mathcal{H}\) and it is dense in \(\mathcal{C}^0\), then \(\mathcal{H}_u=\cap_{n=1}^\infty \mathcal{P}_{g_n}\).

• For \(i=2\), this follows from Theorem 21.

• For \(i=3\), in fact \(\mathcal{X}_3=\emptyset\). This follows from Corollary 14.

• For \(i=4\), this follows from Corollary 17.

• For \(i=5\), this follows from Theorem 4.

The proof is done. ◻

Proof. For each \(x\in \mathbb{R}/\mathbb{Z}\), let \(\widehat{x}\) denote the unique point in \(\pi^{-1}(x)\cap [0,1)\). For each \(x\in K(f)\setminus \mathrm{Crit}(f)\), let \(i(x)\) denote the unique element in \(\mathbb{A}\) such that \[\tau_{i(x)}(\widehat{x})\in \pi^{-1}(K(f)).\] Then \(x\mapsto i(x)\) is a continuous map from \(K(f)\setminus \mathrm{Crit}(f)\) to \(\mathbb{A}\), and hence locally constant. Moreover, since \(K(f)\) is Sturmian-like, there exists \(N\) such that for each \(U\in \mathcal{P}_N\), \(\#(U\cap \mathrm{Crit}(f))\le 1\), and for each component \(B\) of \(U\setminus \mathrm{Crit}(f)\), either \(B\cap K(f)=\emptyset\) or \(x\mapsto i(x)\) is constant on \(B\cap K(f)\).

For each \(n\ge N\), let \(V(n)\) denote the union of elements of \(\mathcal{P}_n\) which intersect \(K(f)\). For each component \(U\) of \(V(n)\), define \(U'\subset T^{-1}(U)\) as follows.

• If \(U\cap \mathrm{Crit}(f)=\emptyset\), then take \(U'\) as the unique component of \(T^{-1}(U)\) that intersects \(K(f)\).

• If \(U\cap \mathrm{Crit}(f)=\{c\}\) for some \(c\in \mathrm{Crit}(f)\), let \(U_+\) (resp. \(U_-\)) be the right (resp. left) component of \(U\setminus\{c\}\). For \(\alpha\in \{+,-\}\), let \(U_\alpha'\) be a component of \(T^{-1}(U)\) such that \(T^{-1}(U_\alpha\cap K(f))\subset U_\alpha'\) and such that \(U_\alpha'\cap K(f)\cap T^{-1}(c)\not=\emptyset\). Note that if \(U_\alpha\cap K(f)\not=\emptyset\), then \(U_\alpha'\) is uniquely determined, and otherwise there are multiple choices of \(U_\alpha'\). Finally, we take \(U'=U_+'\cup U_-'\) (\(U_+'\) and \(U_-'\) may coincide).

Let \(V'(n)\) be the union of all \(U'\) defined as above. The frame \((V(n), V'(n))\) is obviously marked so that the resulting marked frame is \(f\)-compatible.

Let \(s(n)\) denote the complexity of the frame \((V(n), V'(n))\). Fix \(\varepsilon>0\). It remains to show that \(s(n)<\varepsilon\) when \(n\) is large enough. Let \(q(n)\) denote the number of components of \(V(n)\). Since \(T|K(f)\) has topological entropy zero by [23], \(q(n)< e^{n\varepsilon/2}\) provided that \(n\) is large enough. For each \(k\ge n\), let \[V^k(n)=\{x\in V'(n): T(x), T^2(x), \dots, T^{k}(x)\in V(n)\}.\] Write \(k=\ell n+r\) with \(\ell\ge 1\) and \(0\le r<n\). Then \(V^k(n)\) is contained in \[W:=\{x\in V(n): T^{jn}(x)\in V(n) \text{ for } 1\le j\le \ell\},\] and each component of \(W\) contains at most \(d_T^r\) components of \(V^{k}(n)\). Thus \[\#\{\text{components of }V^k(n)\} \le d_T^r\cdot \#\{\text{components of } W\}\le d_T^n\cdot q(n)^{\ell+1}.\] Combining this with \(q(n)< e^{n\varepsilon/2}\) and letting \(k\to\infty\), it follows that \(s(n)<\varepsilon\) provided that \(n\) is large enough. ◻

Proof. Suppose \(f_n,f\in \mathcal{H}^{\mathcal{M}}\) with \(f_n\to f\) in \(C^1\) topology. For each \(1\le \nu\le p\), it suffices to show that \(c_\nu(f)\) is the unique limit point of \(\{c_\nu(f_n)\}\). For simplicity of notation, we may further assume that \(\{c_\nu(f_n)\}\) itself converges. Let \(x_{n,+}\) and \(x_{n,-}\) be two distinct points in \(T^{-1}(c_\nu(f_n))\cap K(f_n)\) (these points may lie outside \(V'\)). Passing to a subsequence, we may assume that \(x_{n,+}\to x_+\) and \(x_{n,-}\to x_-\) simultaneously for some \(x_+,x_-\). Since \(T(x_{n,+})=T(x_{n,-})\in V_\nu\), \(T(x_+)=T(x_-)\) is contained in the closure of \(V_\nu\). By the upper semi-continuity of \(K(\cdot)\), we have \(x_+, x_-\in K(f)\). Since \(T\) is locally injective, we have \(x_+\ne x_-\). Note also that \(T(K(f))=K(f)\) does not meet the boundary of \(V\). It follows that \(T(x_+)=T(x_-)\in \mathrm{Crit}(f)\cap V_\nu\) and hence \(T(x_+)=T(x_-)=c_\nu(f)\). Consequently, \(c_\nu(f_n)\to c_\nu(f)\). ◻

Proof. Take \(\nu_0\in \{1,2,\dots, p\}\) such that \(c=c_{\nu_0}(f_0)\in \mathrm{Crit}(f_0)\) is not in the forward orbit of any \(c'\in \mathrm{Crit}(f_0)\) under iteration of \(T\). Such a \(\nu_0\) exists for otherwise, \(K(f_0)\) would contain a periodic orbit. Take a backward orbit \((z_{-n})_{n=0}^\infty\) in \(K(f_0)\) with \(z_0=c\) and \(z_{-1}=\lim_{x\to c^+}\tau^{f_0}(x)\). Then \(z_{-n}\not\in \mathrm{Crit}(f_0)\) for each \(n\ge 1\). We claim that the closure of \(\{z_{-n}\}_{n=0}^\infty\) contains \(\mathrm{supp}(\mu_{f_0})\). Indeed, any accumulation point of \(\frac{1}{n}\sum_{k=0}^{n-1} \delta_{z_{-k}}\) in the weak-* topology is a maximizing measure of \(f_0\). Since \(\mu_{f_0}\) is the only maximizing measure of \(f_0\), the statement follows. We may then choose \(n_1, n_2\ge 1\) and \(a_i(f_0)=z_{-n_i}\), \(i=1,2\), such that the desired properties hold for \(f=f_0\) (since \(\mu_{f_0}\) has no atom, we can guarantee that \((a_1({f_0}),a_2({f_0}))\cap \mathrm{supp}(\mu_{{f_0}})\ne\emptyset\)). By continuity and Lemma 24, the construction can be extended to a \(C^1\) neighborhood of \(f_0\) in \(\mathcal{H}^{\mathcal{M}}\). More precisely, we can define a sequence of \(C^1\) neighborhoods \(\mathcal{U}_1\supset \mathcal{U}_2\supset\cdots\) of \(f_0\) in \(\mathcal{H}^{\mathcal{M}}\) and continuous maps \(b_n:\mathcal{U}_n \to V\) with \(b_n(f_0)=z_{-n}\) for every \(n\ge 1\) as follows: first let \(b_1(f)=\lim_{x\to c_{\nu_0}(f)^+}\tau^{f}(x)\), which is well-defined and continuous on some \(\mathcal{U}_1\) by Lemma 24 and the construction of \(\tau^f\); then define \(b_{n+1}(f)=\tau^f(b_n(f))\) inductively on \(n\) (shrinking the domain if necessary in this process). As a consequence, \(a_i(f):=b_{n_i}(f)\) is continuous on \(\mathcal{U}_{n_i}\) for \(i=1,2\), and all the desired properties hold by choosing \(\mathcal{U}\) small enough. ◻

Proof. Since \(T\) is expanding and \(\tau^f\) is injective, \(|W_f^n|\) is exponentially small in \(n\), which implies that \(\psi_f\in L^q(\mathbb{R}/\mathbb{Z})\) for each \(1\le q<\infty\).

Now let us show \(\|\psi_f\|_{\infty}=\infty\). Since \(\mu_f((a_1(f),a_2(f)))>0\) and \(\mu_{f}\) has no atom, by the Poincaré recurrence theorem, there exists \(x_0\in (a_1(f),a_2(f))\cap K(f)\) such that \(x_0\) returns to \((a_1(f),a_2(f))\) infinitely often under iteration of \(T\), and \(T^n(x_0)\not\in \mathrm{Crit}(f)\) for each \(n\). To complete the proof, it suffices to show that for each positive integer \(N\), there exists a neighborhood \(U_N\) of \(x_0\) such that \(\psi_f> N\) on \(U_N\). To this end, let \(0=n_0<n_1<n_2<\cdots\) be nonnegative integers such that \(T^{n_j}(x_0)\in (a_1(f), a_2(f))\), \(\forall j\ge 0\). By continuity, there exists an open arc \(U_N\) containing \(x_0\) such that \(T^k(U_N)\subset V\setminus \mathrm{Crit}(f)\) for each \(0\le k\le n_N\), and \(T^{n_i}(U_N)\subset (a_1(f), a_2(f))\) for each \(0\le i\le N\). Consequently, \(U_N\subset W_f^{n_i}\) for each \(0\le i\le N\). Thus \(\psi_f>N\) on \(U_N\). ◻

Proof. Let \(g\) be a Lipschitz sub-action of \(f\). Note that by Lemma 5 (ii), \(K(f)\subset E_{f,T,g}\). We first prove \[\label{eqn:gableft} g(a_2(f))-g(a_1(f))=\int_{\mathbb{R}/\mathbb{Z}} f'(x)\psi_f(x) \,\mathrm{d}x.\tag{17}\] To this end, for each open set \(U\) of \(\mathbb{R}/\mathbb{Z}\) with \(U\not=\mathbb{R}/\mathbb{Z}\), let us write it as a disjoint union of open arcs \(J_j=(u_j,v_j)\), and define \[\mathbf{O}_g(U)=\sum_j (g(v_j)-g(u_j)).\] Provided \(\partial U\subset K(f)\) (\(\partial U\) stands for the boundary of \(U\) in \(\mathbb{R}/\mathbb{Z}\)) and \(U\subset V\), \[\mathbf{O}_g(U)=\mathbf{O}_g(\tau^f(U\setminus \mathrm{Crit}(f)))+ \int_{\tau^f(U\setminus \mathrm{Crit}(f))} f'(x) \,\mathrm{d}x.\] To check this, it suffices to consider the case \(U\) is connected and it is a routine task.

Therefore, for each \(n=1,2,\dots\), \[g(a_2(f))-g(a_1(f))=\mathbf{O}_g(W_f^n)+\sum_{i=1}^n \int_{W_f^i} f'(x) \,\mathrm{d}x.\] As \(n\to\infty\), since \(g\) is Lipschitz and \(T\) is expanding, \[\mathbf{O}_g(W_f^n)=O(|W_f^n|)\to 0;\] at the same time, \[\sum_{i=1}^n 1_{W_f^i} \to \psi_f ~~\text{in}~~ L^1(\mathbb{R}/\mathbb{Z}).\] Thus 17 follows.

On the other hand, since the orbits of \(a_1(f)\) and \(a_2(f)\) are contained in the \(g\)-action set \(E_{f,T,g}\) and since \(T^{n_1}(a_1(f))=T^{n_2}(a_2(f))\), we have \[g(a_2(f))-g(a_1(f))=\sum_{k=0}^{n_1-1} f(T^k(a_1(f)))-\sum_{k=0}^{n_2-1} f(T^k(a_2(f)))+(n_2-n_1)\int f \,\mathrm{d}\mu_f.\] Together with 17 , this implies the identity ?? . ◻

Proof. For \(j=1,2\), let \(W_j^0=(a_1(f_j), a_2(f_j))\) and for each \(k\ge 0\), let \[W_{j}^{k+1}=\tau^{f_j}(W_j^k\setminus \mathrm{Crit}(f_j)).\] Then \[\label{eqn:varphif} \|\psi_{f_1}-\psi_{f_2}\|_1\le \sum_{k=0}^\infty |W_1^k\triangle W_2^k|.\tag{18}\] Let \(A\) denote the union of grand orbits of points in \(\mathrm{Crit}(f_1)\cup \mathrm{Crit}(f_2)\) under the iteration of \(T\). That is to say, \(x\in A\) if and only if there exists \(y\in \mathrm{Crit}(f_1)\cup \mathrm{Crit}(f_2)\) and integers \(m,n\ge 0\) such that \(T^m(x)=T^n(y)\). By definition, \(A\) is a countable set. For \(j=1,2\) and \(k\ge 0\), let \(F_j^k=W_j^k\setminus A\), so that \(\tau^{f_1}\) and \(\tau^{f_2}\) are both well-defined on \(F_j^k\), and \(|F_j^k|=|W_j^k|\). For \(1\le \nu\le p\), let \(I_\nu\subset V_\nu\) be the (possibly empty) open arc with endpoints \(c_\nu(f_1),c_\nu(f_2)\). Since \(\tau^{f_1}=\tau^{f_2}\) on \(F_2^k\setminus(\cup_{\nu=1}^p I_\nu)\), \[|\tau^{f_1}(F_2^k)\triangle \tau^{f_2}(F_2^k)|\le \sum_{\nu=1}^p |\tau^{f_1}(I_\nu)\cup \tau^{f_2}(I_\nu)|\le pC\delta,\] where \(C=2/\min_{x\in \mathbb{R}/\mathbb{Z}}|T'(x)|\). It follows that \[|W_1^{k+1}\triangle W_2^{k+1}|\le |\tau^{f_1}(F_1^k\triangle F_2^k)|+|\tau^{f_1}(F_2^k)\triangle \tau^{f_2}(F_2^k)|\le |\tau^{f_1}(F_1^k\triangle F_2^k)|+ pC\delta.\] Consequently, for each \(k\ge 0\) we have: \[|W_1^k\triangle W_2^k|\le k p C\delta + |(\tau^{f_1})^k (F_1^0\triangle F_2^0)|.\] Also note that by 16 , \[|F_1^0\triangle F_2^0|=|W_1^0\triangle W_2^0|\le 2C_0\delta.\] Combining the two displayed lines above with 2 we have: \[|W_1^k\triangle W_2^k|\le (kpC+2C_0(C_*\lambda_*^k)^{-1})\delta,\] and hence there exists \(C'>0\) independent of \(\delta\) such that \[\sum_{\lambda_*^{-k} \ge \delta} |W_1^k\triangle W_2^k| \le C' \delta (\log \delta)^2.\] On the other hand, \[|W_1^k\triangle W_2^k|\le |W_1^k|+|W_2^k|=|(\tau^{f_1})^k(F_1^0)|+|(\tau^{f_2})^k(F_2^0)|\le 2(C_*\lambda_*^k)^{-1},\] and hence there exists \(C''>0\) independent of \(\delta\) such that \[\sum_{\lambda_*^{-k} < \delta} |W_1^k\triangle W_2^k| \le C'' \delta .\] Combining the two estimates above with 18 , the lemma follows. ◻

Proof. Since \(\mathcal{H}\) is dense in \(\mathcal{C}^1\), it suffices to find \(\varphi\in \mathcal{C}^1\) satisfying ?? . To prove this, we first pick a Borel set \(J\subset \mathbb{R}/\mathbb{Z}\) with positive Lebesgue measure such that \(\psi_{f_0}|J\) is bounded from above by a constant \(C>0\). Such a set exists because \(\|\psi_{f_0}\|_1<\infty\). Since \(\|\psi_{f_0}\|_\infty=\infty\) and \(\psi_{f_0}\ge 0\), the set \[E=\{x\in \mathbb{R}/\mathbb{Z}: \psi_{f_0}(x) \ge 4L+C+8\}\] has positive Lebesgue measure. Let \[\rho=\tfrac{1}{4|E|}\cdot 1_E- \tfrac{1}{4|J|} \cdot 1_J \in L^\infty(\mathbb{R}/\mathbb{Z}).\] Then \[\int_{\mathbb{R}/\mathbb{Z}}\rho \,\mathrm{d}x =0\,, \quad \|\rho\|_1=\frac{1}{2}\,, \quad \text{and} \quad \int_{\mathbb{R}/\mathbb{Z}} \rho \psi_{f_0} \,\mathrm{d}x \ge L+2\,.\] Now let \(\{\rho_n\}_{n=1}^\infty\) to be a sequence in \(\mathcal{C}^0\) converging to \(\rho\) in \(L^2(\mathbb{R}/\mathbb{Z})\). Let \(\varphi_n\) be the function in \(\mathcal{C}^1\) such that \(\varphi_n'=\rho_n-\int_{\mathbb{R}/\mathbb{Z}}\rho_n \,\mathrm{d}x\) and \(\varphi_n(\pi(0))=0\). Then \[\|\varphi_n'-\rho\|_1 \le \|\varphi_n'-\rho\|_2 \le \|\rho_n-\rho\|_2 + \left|\int_{\mathbb{R}/\mathbb{Z}}\rho_n \,\mathrm{d}x\right| \to \left|\int_{\mathbb{R}/\mathbb{Z}}\rho \,\mathrm{d}x\right|= 0.\] It follows that \[\|\varphi_n\|_\infty \le \|\varphi_n'\|_1 \le \|\varphi_n'-\rho\|_1 + \|\rho\|_1 \to \frac{1}{2},\] and since \(\|\psi_{f_0}\|_2<\infty\), \[\int_{\mathbb{R}/\mathbb{Z}} \varphi_n' \psi_{f_0} \,\mathrm{d}x \ge \int_{\mathbb{R}/\mathbb{Z}} \rho \psi_{f_0} \,\mathrm{d}x - \|(\varphi_n'-\rho)\psi_{f_0}\|_1 \ge L+2 - \|\varphi_n'-\rho\|_2\cdot \|\psi_{f_0}\|_2 \to L+2.\] Therefore, ?? holds by taking \(\varphi=\varphi_n\) for a sufficiently large \(n\). ◻

Proof of Theorem 29. Fix \(L=2(n_1+n_2+1)\) and let \(\varphi\in \mathcal{H}\) be as in ?? . By Lemma 24, Lemma 28 and the upper semi-continuity of \(K(\cdot)\), there exists an open neighborhood \(\mathcal{U}_{f_0}\subset \mathcal{U}\) of \(f_0\) in \(\mathcal{C}^1\) and \(C_2\ge 1\) such that for any \(f\in \mathcal{U}_{f_0}\cap \mathcal{H}^{\mathcal{M}}\), we have: \[\label{eqn:integralphipsi} \int_{\mathbb{R}/\mathbb{Z}} \varphi'(x)\psi_f(x) \,\mathrm{d}x \ge L,\tag{19}\] and \[\label{eqn:f39bounded} \|f'\|_\infty\le C_2.\tag{20}\]

To proceed, take an arbitrary \(F\in \mathcal{H}\), and let \(F_t=F+t\varphi\in \mathcal{H},\,t\in\mathbb{R}\).

Finally, let us show that the set \[S:=\{t\in [0,1] : F_t\in \mathcal{U}_{f_0}\cap \mathcal{H}^{\mathcal{M}}\}\] has Lebesgue measure zero.

Let \((V, V')\) be the underlying frame of \(\mathcal{M}\). Let \[V^n=\{x\in V': T(x), T^2(x), \dots, T^n(x)\in V\}\] for each \(n\). Let \(\mathcal{R}^n\) denote the collection of rectangles \(U_1\times U_2\times \dots\times U_p\), where \(U_1, U_2, \dots, U_p\) are connected components of \(V^n\). As the complexity of \((V,V')\) is less than \(\frac{1}{p}\log \lambda_*\), there exist \(1<\lambda_2<\lambda_1<\lambda_*\) such that the number of elements in \(\mathcal{R}^n\) is at most \(\lambda_2^n\), and each of them has diameter at most \(\lambda_1^{-n}\), provided that \(n\) is large enough. Thus \[\label{eqn:Rn} \sum_{R\in \mathcal{R}^n} \mathop{\mathrm{diam}}(R)(\log \mathop{\mathrm{diam}}(R))^2\to 0\tag{26}\] as \(n\to\infty\).

Let \(\mathbf{c}(F_t)=(c_\nu(F_t))_{\nu=1}^p\) for \(t\in S\). Note that \(\mathbf{c}(F_t)\) is contained in a rectangle in \(\mathcal{R}^n\). On the other hand, by ?? , if \(\mathbf{c}(F_{t_1}), \mathbf{c}(F_{t_2})\) are contained in the same element \(R\) of \(\mathcal{R}^n\), then \[|t_1-t_2|\le C_3 \cdot \mathop{\mathrm{diam}}(R)(\log \mathop{\mathrm{diam}}(R))^2.\] Combining this with 26 , the statement follows. ◻