Proof. The case $d=1$ is immediate. Assume $d>1$ , fix $K\in \mathcal K_{\mathcal F}$ , and set

$$ \begin{align*} K^{*}:=\bigcup_{K'\in\mathcal K_{\mathcal F}\setminus\{K\}}K'. \end{align*} $$

Since the minimal invariant sets are pairwise disjoint and compact,

$$ \begin{align*} \delta:=\operatorname{dist}(K,K^{*})>0. \end{align*} $$

By definition, $\mathbb {S}^1\setminus D_K$ is the union of the components $I$ of $\mathbb {S}^1\setminus K$ that meet $K^{*}$ . Each such component has length at least $\delta $ . Hence, since these components are pairwise disjoint and $\mathbb {S}^1$ has finite length, there are only finitely many of them. Therefore, $D_K$ is a finite union of closed intervals.

Proof. Consider $f\in \mathcal {F}$ and $K\in \mathcal {K}_{\mathcal {F}}$ . Since $K\subset D_K$ , by the $\mathcal {F}$ -invariance of K,

$$ \begin{align*} f(x_{K,j}),f(y_{K,j})\in f(K)\subset K\subset D_K \end{align*} $$

for all $j=1,\ldots ,\ell _K$ . If $\ell _K=1$ , the conclusion is immediate with $k_1=r_1=1$ .

Assume $\ell _K>1$ . For $j=1,\ldots ,\ell _K$ , there exist $k_j^-,k_j^+\in \{1,\ldots ,\ell _K\}$ such that $f(x_{K,j})\in I_{K,k_j^-}$ and $f(y_{K,j})\in I_{K,k_j^+}$ . By definition, for all

$$ \begin{align*} a,b\in\{x_{K,1},y_{K,1}, x_{K,2},y_{K,2},\ldots, x_{K,\ell_K},y_{K,\ell_K}\}, \end{align*} $$

with $b\neq y_{K,j}$ , $a\neq x_{K,j}$ , we have that both intervals $[a,x_{K,j}]$ and $[y_{K,j},b]$ intersect $\bigcup _{K'\in \mathcal {K}_{\mathcal {F}}\backslash \{K\}}K$ . Indeed, for some $i=1,\ldots ,\ell _K$ , we have $[y_{K,i},x_{K,j}]\subset [a,x_{K,j}]$ and $[y_{K,j},x_{K,i}]\subset [y_{K,j},b]$ . By $\mathcal {F}$ -invariance, the intervals $f([a,x_{K,j}])$ and $f([y_{K,j},b])$ also intersect $\bigcup _{K'\in \mathcal {K}_{\mathcal {F}}\setminus \{K\}}K'$ . In particular, for all $I\in \{I_{K,j}\}_{j=1}^{\ell _K}$ , $f([a,x_{K,j}])\not \subset I$ and $f([y_{K,j},b])\not \subset I$ . Therefore, for all $i,j\in \{1,\ldots ,\ell _K\}$ , $i\neq j$ ,

$$ \begin{align*} k_i^+,k_i^-\notin\{k_j^+,k_j^-\}, \end{align*} $$

and so,

$$ \begin{align*} \{k_j^-\colon j=1,\ldots,\ell_K\}=\{k_j^+\colon j=1,\ldots,\ell_K\}=\{1,\ldots,\ell_K\}. \end{align*} $$

So, we must necessarily have equality $k_j^-=k_j^+$ for all $j=1,\ldots ,\ell _K$ . Thus, for each j, the image $f(I_{K,j})$ is contained in some component $I_{K,k_j}$ . Since the map $j\mapsto k_j$ is a permutation of $\{1,\ldots ,\ell _K\}$ , for every j, there exists $r_j$ such that $k_{r_j}=j$ . Hence, $f(I_{K,r_j})\subset I_{K,j}$ .

Proof. Consider $K_1,K_2\in \mathcal {K}_{\mathcal {F}}$ , $y_1\in K_1$ , and $x_2\in K_2$ be such that $(y_1,x_2)\cap K=\emptyset $ for all $K\in \mathcal {K}_{\mathcal {F}}$ . Let us proceed by contradiction. Given $f\in \mathcal {F}$ , if $f([y_1,x_2])\cap K\neq \emptyset $ for some $K\in \mathcal {K}_{\mathcal {F}}\backslash \{K_1,K_2\}$ , then there exists a connected component I of $D_K$ (that is, ${I\in \{I_{K,j}\}_{j=1}^{\ell _K}}$ ) such that $I\subset f([y_1,x_2])$ . By Lemma 2, there exists $J\in \{I_{K,j}\}_{j=1}^{\ell _K}$ such that $f(J)\subset I\subset f([y_1,x_2])$ . Since f is a homeomorphism, $J\subset [y_1,x_2]$ . Hence, $(y_1,x_2)\cap K\neq \emptyset $ , which contradicts the hypotheses. The lemma is proven.

Proof of Proposition 1.

Let $x\in \mathbb {S}^1$ . If $x\in K$ , for some $K\in \mathcal {K}_{\mathcal {F}}$ , it is clear that for all $\omega \in \mathcal {F}^{\mathbb {N}}$ , $f_{\omega }^n(x)\in K$ for all $n\in \mathbb {N}$ . Hence, the proposition follows by taking ${K_1=K_2=K}$ . If $x\notin \bigcup _{K\in \mathcal {K}_{\mathcal {F}}} K$ , there exist $K_1$ , $K_2\in \mathcal {K}_{\mathcal {F}}$ and $y_1\in K_1$ , $x_2\in K_2$ , such that ${(y_1,x)\cap K=(x,x_2)\cap K=\emptyset }$ for all $K\in \mathcal {K}_{\mathcal {F}}$ . By Lemma 3, we get that for all ${f\in \mathcal {F}}$ , $f(y_1,x_2)\cap K=\emptyset $ for all $K\in \mathcal {K}_{\mathcal {F}}\backslash \{K_{1},K_{2}\}$ . Therefore, the cluster set of the sequence $(f_\omega ^n(x))_{n\in \mathbb {N}}$ does not intersect any $K\in \mathcal {K}_{\mathcal {F}}\backslash \{K_{1},K_{2}\}$ . The proposition is proven.

Proof. Let us show the $\mathcal {F}$ -invariance of $\{u_i=1\}$ . For $\{u_i=0\}$ , the proof is analogous. We need to prove that for all $f\in \mathcal {F}$ ,

(5) $$ \begin{align} f(\{x\in\mathbb{S}^1\colon u_i(x)=1\})\subset \{x\in\mathbb{S}^1\colon u_i(x)=1\}. \end{align} $$

In fact, using (4), for x such that $u_i(x)=1$ , we get

$$ \begin{align*} 1=u_i(x)&=\int_{\mathcal{F}} u_i(f(x))\,d\nu(f). \end{align*} $$

Since $0\leq u_i\leq 1$ , we get $u_i(f(x))=1$ for $\nu $ -almost every $f\in \mathcal {F}$ . Using the continuity of $u_i$ and the fact that the support of $\nu $ is exactly $\mathcal {F}$ , we get $u_i(f(x))=1$ for all $f\in \mathcal {F}$ and so (5) holds.

Proof. When $d=1$ , the result is evident by taking $i_1=i_2=1$ for all $x\in \mathbb {S}^1$ because $u_1\equiv 1$ . Now, consider $d\geq 2$ . Since the collection $\mathcal {K}_{\mathcal {F}}=\{K_1,\ldots , K_{ d }\}$ of the closed $\mathcal {F}$ -minimal invariant sets is a finite set, we can apply Proposition 1 to get that for every $x\in \mathbb {S}^1$ , there exist $i_1,i_2\in \{1,\ldots ,d\}$ such that for all $\omega \in \mathcal {F}^{\mathbb {N}}$ , the cluster set of the sequence $(f_\omega ^n(x))_{n\in \mathbb {N}}$ does not intersect any element of $\mathcal {K}_{\mathcal {F}}\backslash \{K_{i_1},K_{i_2}\}$ . In particular, we can find $x_1\in K_{i_1}$ , $x_2\in K_{i_2}$ , such that $x\in [x_1,x_2]$ and $[x_1,x_2]\cap K_j=\emptyset $ for all $j\in \{1,\ldots ,d\}\backslash \{i_1,i_2\}$ . By (3), using the definition of the maps $u_i$ , we get that $u_{i_1}(x)+u_{i_2}(x)=1$ (case $i_1\neq i_2$ ) or $u_{i_1}(x)=1$ (case $i_1= i_2$ ). Hence, $u_j(x)=0$ for all ${j\in \{1,\ldots ,d\}\backslash \{i_1,i_2\}}$ .

Proof of Theorem 4.

When $d=1$ , the result is immediate, since $ \{u_1=1\}=\mathbb {S}^1. $ Thus, assume that $d\geq 2$ .

Fix $i\in \{1,\ldots ,d\}$ . Recall that $u_i:\mathbb {S}^1\to [0,1]$ is continuous. Hence, the proof that $ \{u_i=1\} $ is a finite union of closed intervals is analogous to the proof of Lemma 1. This proves item (a).

To prove item (b), we use the $\mathcal {F}$ -invariance of $\{u_i=1\}$ established in Lemma 4. The conclusion then follows by arguing analogously to the proof of Lemma 2.

Finally, item (c) follows from Lemma 3 and the fact that every connected component of $\{u_i=1\}$ intersects $K_i$ .

To prove the second part of Theorem 4, consider $i,j\in \{1,\ldots ,d\}$ , $i\neq j$ and ${I=[x_i,y_i]\subset \{u_i=1\}}$ , $J=[x_j,y_j]\subset \{u_j=1\}$ such that

$$ \begin{align*} (y_i,x_j)\cap \bigcup_{r=1}^d K_r=\emptyset. \end{align*} $$

By Proposition 2, for $x\in [y_i,x_j]$ , we have $u_i(x)+u_j(x)=1$ . Let us show that $u_i$ is a monotone decreasing function on $[y_i,x_j]$ . Note that for $f\in \mathcal {F}$ , there exist ${I(f)=[x_i(f),y_i(f)]\subset \{u_i=1\}}$ , $J(f)=[x_j(f),y_j(f)]\subset \{u_j=1\}$ such that ${f(I)\subset I(f)}$ , $f(J)\subset J(f)$ , and

$$ \begin{align*} (y_i(f),x_j(f))\cap \bigcup_{r=1}^d K_r=\emptyset\quad\mathrm{or}\quad (y_j(f),x_i(f))\cap \bigcup_{r=1}^d K_r=\emptyset. \end{align*} $$

The last fact will depend on whether f preserves orientation or not. Therefore, for ${x\in [y_i,x_j]}$ and $\omega \in \mathcal {F}^{\mathbb {N}} $ such that $\mathrm {cl}((f_\omega ^n(x))_{n \geq 0})=K_i$ , we get that for all $y\in [y_i,x]$ , we also have $\mathrm {cl}((f_\omega ^n(y))_{n \geq 0})=K_i$ . Hence,

(6) $$ \begin{align} \{\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(x))_{n \geq 0})=K_i\}\subset \{\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(y))_{n \geq 0})=K_i\} \end{align} $$

for every $x\in [y_i,x_j]$ and $y\in [y_i,x]$ . Consequently, for $x,y\in [y_i,x_j]$ , $( y_i,x,y,x_j)$ is circularly ordered, we have that

$$ \begin{align*} u_i(x)\leq u_i(y). \end{align*} $$

This implies the desired conclusion. The monotonicity of $u_j$ is an immediate consequence of $u_i+u_j=1$ on $[y_i,x_j]$ .

Proof. Consider $i\in \{1,\ldots ,d\}$ and $I=[a,b]$ such that $u_i|_I\equiv C$ for some $C\in [0,1]$ . By Lemma 4, when $C\in \{0,1\}$ , it is enough to consider $c_f=C$ for all $f\in \mathcal {F}$ .

Assume $C\in (0,1)$ . Since $C<1$ , $I\cap K_j=\emptyset $ for all $j\in \{1,\ldots ,d\}$ . Take $i_1,i_2\in \{1,\ldots ,d\}$ , $i_1\neq i_2$ , $x_1\in K_{i_1}$ , $x_2\in K_{i_2}$ such that $I\subset [x_1,x_2]$ , and for all $j\in \{1,\ldots ,d\}$ ,

$$ \begin{align*} (x_1,x_2)\cap K_j=\emptyset. \end{align*} $$

Since $C>0$ , either $i_1=i$ or $i_2=i$ . Consider $c_1,c_2>0$ such that $c_1+c_2=1$ , ${u_{i_1}|_I=c_1}$ , and $u_{i_2}|_I=c_2$ . By the invariance of the weight maps in (4), we get for $x,y\in I$ , $x\leq y$ ,

(7) $$ \begin{align} &\int_{\mathcal{F}} \nu^{\mathbb{N}}(\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(f(x)))_{n \geq 0})=K_{i_1})\,d\nu(f)\nonumber\\ & \quad =u_{i_1}(x)=c_{1}=u_{i_1}(y) \nonumber \\ & \quad =\int_{\mathcal{F}} \nu^{\mathbb{N}}(\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(f(y)))_{n \geq 0})=K_{i_1})\,d\nu(f). \end{align} $$

By the continuity of the maps in $\mathcal {F}$ and the argument leading to (6), for all $f\in \mathcal {F}$ , we have

(8) $$ \begin{align} \{\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(f(y)))_{n \geq 0})=K_{i_1}\}\subset\{\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(f(x)))_{n \geq 0})=K_{i_1}\}. \end{align} $$

By (7) and (8), we get that for $\nu $ -almost every $f\in \mathcal {F}$ ,

$$ \begin{align*} \nu^{\mathbb{N}}(\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(f(x)))_{n \geq 0})=K_{i_1})= \nu^{\mathbb{N}}(\omega\in\mathcal{F}^{\mathbb{N}}\colon \mathrm{cl}((f_\omega^n(f(y)))_{n \geq 0})=K_{i_1}), \end{align*} $$

that is, $u_{i_1}(f(x))=u_{i_1}(f(y))$ . By the continuity of $u_{i_1}$ and the fact that $\nu $ is supported on $\mathcal {F}$ , we conclude $u_{i_1}(f(x))=u_{i_1}(f(y))$ for all $f\in \mathcal {F}$ . Since x and y are arbitrary in I, we get $u_{i_1}|_{f(I)}=c_{1,f}$ for some $c_{1,f}\in [0,1]$ . Analogously, we can show that for all $f\in \mathcal {F}$ , we have $u_{i_2}|_{f(I)}=c_{2,f}=1-c_{1,f}$ . To conclude this demonstration, it only remains to consider $c_f=c_{k,f}$ for all $f\in \mathcal {F}$ and $k\in \{1,2\}$ such that $i=i_k$ .

Proof. If $d=1$ , the corollary follows easily because $u_1$ is constant, implying that ${V(u_1)=0}$ . Assume $d\geq 2$ and take $i\in \{1,\ldots ,d\}$ . By Theorem 4, we can find a circularly ordered tuple $( a_0,a_1,\ldots ,a_m)$ , with $m=4k-1$ for some $k\in \mathbb {N}$ such that for ${j=1,\ldots ,k}$ :

• • $u_i=1$ on $[a_{4j-4},a_{4j-3}]$ ;

• • $u_i$ is monotone decreasing on $[a_{4j-3},a_{4j-2}]$ ;

• • $u_i=0$ on $[a_{4j-2},a_{4j-1}]$ ; and

• • $u_i$ is monotone increasing on $[a_{4j-1},a_{4j}]$ , where $a_{4k}=a_0$ .

Therefore, for any $P\in \mathcal {P}$ , the sum

$$ \begin{align*} \sum_{j=0}^{n_P-1}\vert u_i(x_j)-u_i(x_{j+1})\vert \end{align*} $$

can be separated into $m+1=4k$ sums with half of them being 0 and the other half (corresponding to the monotonicity intervals of $u_i$ ) being telescopic sums. Thus, using $0\leq u_i\leq 1$ ,

$$ \begin{align*} \sum_{j=0}^{n_P-1}\vert u_i(x_j)-u_i(x_{j+1})\vert\leq 2k. \end{align*} $$

Since k does not depend on the partition P, we can conclude the bounded variation of  $u_i$ .

Proof. Let $K_i=\operatorname {supp}(\mu _i)$ , $i=1,\ldots ,d$ . We first prove the inclusion

(10) $$ \begin{align} \mathcal{S}_{\nu}\subset \{t\mu_i+(1-t)\mu_j\colon t\in[0,1],\,(i,j)\in\mathcal{E}\}. \end{align} $$

Take $x\in \mathbb {S}^1$ . By [Reference MalicetMal17, Proposition 4.11], the sequence in (1) converges in the weak- $*$ topology to $ m_x=\sum _{i=1}^d u_i(x)\mu _i. $ Since for every $n\geq 0$ , the support of the n-step transition measure starting from x is contained in $\mathcal O_{\mathcal {F}}(x)$ , we must have

$$ \begin{align*} \operatorname{supp}(m_x)\subset \overline{\mathcal O_{\mathcal{F}}(x)}. \end{align*} $$

If $u_i(x)>0$ , then $K_i\subset \operatorname {supp}(m_x)$ . Consequently, $ K_i\subset \overline {\mathcal O_{\mathcal {F}}(x)}. $ Thus, every minimal set $K_i$ with positive weight $u_i(x)$ intersects $\overline {\mathcal O_{\mathcal {F}}(x)}$ . By Proposition 1, the set $\overline {\mathcal O_{\mathcal {F}}(x)}$ can intersect at most two minimal sets. Therefore, at most two coefficients in the convex decomposition of $m_x$ are non-zero. It follows that, for each $x\in \mathbb {S}^1$ ,

$$ \begin{align*} m_x=t\mu_i+(1-t)\mu_j \end{align*} $$

for some $t\in [0,1]$ and some pair $(i,j)\in \mathcal {E}$ . Hence,

$$ \begin{align*} m_x\in \{t\mu_i+(1-t)\mu_j\colon t\in[0,1],\,(i,j)\in\mathcal{E}\}. \end{align*} $$

Since, by definition, every element of $\mathcal {S}_{\nu }$ is of the form $m_x$ for some $x\in \mathbb {S}^1$ , we obtain the inclusion in (10).

We now prove the reverse inclusion. Let $(i,j)\in \mathcal {E}$ . If $i=j$ , by the Birkhoff ergodic theorem, we have $\mu _i\in \mathcal {S}_{\nu }$ . Let us assume $i\neq j$ .

By the definition of $\mathcal {E}$ , there exists a point $x\in \mathbb {S}^1$ such that $ \overline {\mathcal O_{\mathcal {F}}(x)} $ intersects both $K_i$ and $K_j$ . Moreover, by Proposition 1, there is no $l\in \{1,\ldots ,d\}\backslash \{i,j\}$ such that $ \overline {\mathcal O_{\mathcal {F}}(x)} $ intersects $K_l$ . Take $x_i\in K_i$ and $x_j\in K_j$ such that $x\in (x_i,x_j)$ and $(x_i,x_j)\cap K=\emptyset $ for $K\in \{K_1,\ldots ,K_d\}\backslash \{K_i,K_j\}$ . By the continuity of the weight maps, we must have for all $t\in [0,1]$ ,

$$ \begin{align*} t\mu_i+(1-t)\mu_j\in \mathcal{S}_{\nu}. \end{align*} $$

Combining both inclusions, we obtain the desired equality.

Proof of (12).

Assume $d\geq 2$ . For $i=1,\ldots ,d$ , let $\mathcal A_i$ be the collection of connected components of $D_{K_i}$ . Fix $i=1,\ldots ,d$ . Note that there exists $j\neq i$ such that

$$ \begin{align*} \mathrm{dist}\bigg(\bigcup_{I\in\mathcal{A}_i}I,\bigcup_{J\in\mathcal{A}_j}J\bigg)=\min_{k\neq i}\mathrm{dist}\bigg(\bigcup_{I\in\mathcal{A}_i}I,\bigcup_{J\in\mathcal{A}_k}J\bigg). \end{align*} $$

From the continuity of the weight maps, we can conclude that $j\in V_i$ . Hence, $\# V_i\geq 1$ .

Now, let us show that $\# V_i\leq 2$ . Take $j\in V_i$ and consider $I'=[a_i',b_i']\in \mathcal {A}_i$ and ${J'=[a_j',b_j']\in \mathcal {A}_j}$ such that:

• • $(b_i',a_j')\cap A=\emptyset $ for all $A\in \bigcup _{k}\mathcal {A}_k$ ; or

• • $(b_j',a_i')\cap A=\emptyset $ for all $A\in \bigcup _{k}\mathcal {A}_k$ .

In any case, by item (c) of Theorem 4, we have that for every $I=[a_i,b_i]\in \mathcal {A}_i$ , there exists $f\in \mathcal {F}$ and $J=[a_j,b_j]\in \mathcal {A}_j$ such that $f(I')\subset I$ and $f(J')\subset J$ . From Lemma 3, we get:

• • $(b_i,a_j)\cap A=\emptyset $ for all $A\in \bigcup _{k}\mathcal {A}_k$ ; or

• • $(b_j,a_i)\cap A=\emptyset $ for all $A\in \bigcup _{k}\mathcal {A}_k$ .

We have just shown that for all $j\in V_i$ and $I=[a_i,b_i]\in \mathcal {A}_i$ , there exists ${J=[a_j,b_j]\in \mathcal {A}_j}$ such that the two items above are satisfied. Using the topological structure of $\mathbb {S}^1$ , we conclude that $\# V_i\leq 2$ since each interval $I\in \mathcal {A}_i$ has only two intervals in $\bigcup _{k\neq i}\mathcal {A}_k$ as neighbors.

Proof. Assume $d\geq 3$ . For $d<3$ , the conclusion is clear. We argue by contradiction. Take $i,j,k,a,b,c\in \{1,\ldots ,d\}$ , with $i\neq j$ , $j\neq k$ , and $k\neq i$ , such that $[a,i,a]$ , $[b,j,b]$ , and $[c,k,c]$ are $\mathcal {F}$ -paths, that is, $\#V_i=\#V_j=\#V_k=1$ . Using (12), we can conclude that ${\#V_a=2}$ . In fact, if $\#V_a=1$ , then $[i,a,i,a,i]$ is an $\mathcal {F}$ -path and, in particular, for any $n\in \mathbb {N}$ , any n-tuple of the form $[i,a,i,a,\ldots ,i]$ is an $\mathcal {F}$ -path, which is not possible by Remark 3 and the fact $d\geq 3$ . Analogously, we have $\#V_a=\#V_b=\#V_c=2$ . Using the circle structure, we get that for $n\in \mathbb {N}$ and any two $\mathcal {F}$ -paths $[i_1,\ldots ,i_n]$ and $[j_1,\ldots ,j_n]$ with $i_1=j_n=a$ and $i_n=j_1=b$ , we have $i_m=j_{n-m+1}$ for $m=1,\ldots ,n$ . Further, we have $i_m=k$ for some $m=1,\ldots ,n$ . Since we must conclude the same for $\mathcal {F}$ -paths starting at a (or b) and ending at c, we can conclude that $a=b$ . However, this contradicts that $i\neq j$ , since $\#V_a=2$ . Therefore, at least one of the triples $[a,i,a]$ , $[b,j,b]$ , or $[c,k,c]$ is not an $\mathcal {F}$ -path.

Proof of (13) in Lemma 6.

Since $\#V_i=1$ implies that $[j,i,j]$ is an $\mathcal {F}$ -path for some $j\neq i$ , we can apply Lemma 7 to get (13).

Proof of Theorem 2.

Proposition 2 and Lemma 6 imply the first part of the theorem. Item (i) follows from Lemma 4 and Theorem 4. The second part of Theorem 4 implies item (ii). Item (iii) was already established in Corollary 2.

Proof. Using that $S_I^+$ is a semigroup, we get that for $g\in S_I^+$ ,

$$ \begin{align*} g(\hat a)&= g(\sup\{f(a)\colon f\in S_I^+\})\\ &=\sup\{g(f(a))\colon f\in S_I^+\}\\ &\leq \sup\{f(a)\colon f\in S_I^+\}=\hat a, \end{align*} $$

that is, $g(\hat a)\leq \hat a$ . Analogously, for $g\in S_I^+$ , we get $g(\hat b)\geq \hat b$ .

Case $\hat a>\hat b$ . In this case, it is easy to show the proximal action of $G_{\mathcal {F}}$ on I. In fact, take $g\in S_I^+$ such that $\hat b<g(a)\leq g(b)\leq b$ . Then, consider a sequence $(g_n)_{n\in \mathbb {N}}$ in $ S_I^+$ such that $g_n(b) $ converges to $\hat b$ as $n\to \infty $ . Hence, for every $x\in I$ , $\hat b\leq g_n(\hat b)\leq g_n(g(a))\leq g_n(g(x))\leq g_n(g(b))\leq g_n(b)$ and so the sequence $(g_n(g(x)))_{n\in \mathbb {N}}$ converges to $\hat b$ so that the sequence $(g_n\circ g)_{n\in \mathbb {N}}$ in $S_I^+\subset G_{\mathcal {F}}$ degenerates the interval I to $\{\hat b\}$ , which implies what is desired.

Case $\hat a=\hat b$ . For all $g\in S_I^+$ , we get

$$ \begin{align*} g(\hat b)\geq \hat b=\hat a\geq g(\hat a), \end{align*} $$

that is, $c=\hat a=\hat b\in [a,b]$ is a fixed point of every map in $S_I^+$ . Now, let us show that for any $\varepsilon>0$ , there exists $g\in S_I^+$ such that $g([a,b])\subset (c-\varepsilon ,c+\varepsilon )$ . By definition of supremum and infimum, given $\varepsilon>0$ , there exist $f\in S_I^+$ and a sequence $(f_n)_{n\in \mathbb {N}}$ such that

$$ \begin{align*} c\leq f(b)<c+\varepsilon \end{align*} $$

and

$$ \begin{align*} \lim_{n\to\infty}f_n(a)=c. \end{align*} $$

By continuity,

$$ \begin{align*} \lim_{n\to\infty}f\circ f_n(a)=f(c)=c. \end{align*} $$

By definition of c, it is clear that $c\leq f_n(b)\leq b$ for all $n\in \mathbb {N}$ so that for N large enough, we have

$$ \begin{align*} f\circ f_N(a)\in (c-\varepsilon,c]\quad\mathrm{and}\quad f\circ f_N(b)\in [c,f(b)]\subset [c,c+\varepsilon). \end{align*} $$

Consequently, $g([a,b])\subset (c-\varepsilon ,c+\varepsilon )$ for $g=f\circ f_N$ . Using the arbitrariness of $\varepsilon $ , we can conclude that the semigroup $S_I^+$ (and so $G_{\mathcal {F}}$ ) acts proximally on $I=[a,b]$ .

Case $\hat a<\hat b$ . By definition of $\hat a$ and $\hat b$ , it is clear that $S_I^+$ acts proximally on $[a,\hat a]$ and $[\hat b,b]$ . Since $S_I^+\subset S_I\subset G_{\mathcal {F}}$ , we have that $G_{\mathcal {F}}$ acts proximally on $[a,\hat a]$ and $[\hat b,b]$ . Now, let us show that $f( a),f( b)\in [a,\hat a]\cup [\hat b,b]$ for all $f\in S_I$ . Take $f\in S_I$ and note ${f^2\in S_I^+}$ . If f is not orientation-preserving, then $f(a)>\hat b$ , because

$$ \begin{align*} a\leq f(b) <f(a)<\hat b \end{align*} $$

implies

$$ \begin{align*} f(\hat b)<f^2(a) <f^2(b)\leq f(a)<\hat b, \end{align*} $$

which is a contradiction to the definition of $\hat b$ . Analogously, $f(b)<\hat a$ . Therefore, for ${f\in S_I}$ ,

$$ \begin{align*} f(a)\in [a,\hat a],\quad f(b)\in[\hat b,b] \end{align*} $$

whenever f preserves orientation, or

$$ \begin{align*} f(a)\in[\hat b,b],\quad f(b)\in [a,\hat a] \end{align*} $$

whenever f does not preserve orientation. Using Theorem 4, for $f\in G_{\mathcal {F}}$ such that ${f(a)\in [a,b]}$ , we conclude that $f\in S_I$ . Since $u_i(a)=1$ , for any $x\in K_i$ , we can find $f\in G_{\mathcal {F}}$ such that $f(a)$ is arbitrarily close to x. Hence, we must have $(\hat a,\hat b)\cap K_i=\emptyset $ .

Moreover, $\hat a,\hat b\in K_i$ . Let us show that $\hat a\in K_i$ ; the proof of $\hat b\in K_i$ is analogous. Use the finiteness of the collection $\mathcal {A}_i$ , the definition of the weight map $u_i$ , and that ${u_i(a)=u_i(b)=1}$ to conclude $K_i\cap [a,\hat a]\neq \emptyset $ . Hence, if $\hat a\notin K_i$ , then take ${x\in [a,\hat a)\cap K_i}$ such that $(x,\hat a)\cap K_i=\emptyset $ . Now, by $\mathcal {F}$ -invariance of $K_i$ , for all $f\in S_I^{+}$ , $f(x)\in K_i$ and $f(a)\leq f(x)\leq \hat a$ . However, by definition of x, $f(x)\leq x$ so that $f(a)\leq x<\hat a$ for all $f\in S_I^+$ , which contradicts the definition of $\hat a.$

The lemma is proven.

Proof of Theorem 5.

Consider $i\in \{1,\ldots ,d\}$ and $\mathcal {A}_i=\{I_1,\ldots ,I_m\}$ as in Theorem 4. Given $I_j=[a_j,b_j]\in \mathcal {A}_i$ , consider $A_j=[a_j,\hat a_j]$ and $B_j=[\hat b_j,b_j]$ , where $\hat a_j$ and $\hat b_j$ are defined analogously to $\hat a$ and $\hat b$ in Lemma 8 taking $a=a_j$ and $b=b_j$ . In the case $\hat a_j\geq \hat b_j$ , set $\mathcal {B}_{i,j}=\{I_j\}$ . In the case $\hat a_j<\hat b_j$ , set $\mathcal {B}_{i,j}=\{A_j,B_j\}$ . Consider

$$ \begin{align*} \mathcal{B}_i=\bigcup_{j=1}^m\mathcal{B}_{i,j}. \end{align*} $$

By definition of $\mathcal {B}_i$ , item (c) holds. Items (b) and (e) are consequences of items (i) and (ii) of Lemma 8. Item (d) is a consequence from item (b) of Theorem 4 and items (iii), (iv) of Lemma 8.

To see item (a), it is enough to show that $\# \mathcal {B}_{i,j}=\# \mathcal {B}_{i,k}$ for all $j,k\in \{1,\ldots ,m\}$ . Assume that there exist $j,k$ such that $\# \mathcal {B}_{i,j}=2$ and $\# \mathcal {B}_{i,k}=1$ . Since $\# \mathcal {B}_{i,j}=2$ , we have $\hat a_j<\hat b_j$ . Set $\delta =\mathrm {dist}(\hat a_j, \hat b_j)>0$ . Fix $x\in [a_j,\hat a_j]\subset [a_j,b_j]$ and $y\in [\hat b_j,b_j]\subset [a_j,b_j]$ . By item (iv) of Lemma 8, we get $\mathrm {dist}(g(x),g(y))\geq \delta>0$ for all g preserving $I_j$ . By item (b) of Theorem 4, there exists $f\in G_{\mathcal {F}}$ such that $f(I_j)\subset I_k$ . From the proximal action of $G_{\mathcal {F}}$ on $I_k$ , we can find a sequence $(g_n)_{n\in \mathbb {N}}$ in $G_{\mathcal {F}}$ such that

$$ \begin{align*} \lim_{n\to\infty}\mathrm{dist}(g_n(f(x)),g_n(f(y)))=0. \end{align*} $$

Without loss of generality, let us assume that the sequences $(g_n(f(x)))_{n\in \mathbb {N}}$ and $(g_n(f(y)))_{n\in \mathbb {N}}$ converge to a point $z\in I_r$ for some $r\in \{1,\ldots ,m\}$ . Take $h\in \mathcal {F}$ such that $h(I_r)\subset I_j$ . Consider $V_z=h^{-1}(h(z)-\varepsilon ,h(z)+\varepsilon )$ for some $\varepsilon \in (0,\delta /4)$ . Take N such that $g_N(f(x)),g_N(f(y))\in V_z$ . Hence, $g=h\circ g_N\circ f$ preserves $I_j$ . Further, $g(x),g(y)\in (h(z)-\varepsilon ,h(z)+\varepsilon )$ , which is a contradiction because $\mathrm {dist}(g(x), g(y))>\delta >2\varepsilon $ . Therefore, there is no $i,j$ such that $\# \mathcal {B}_{i,j}\neq \# \mathcal {B}_{i,k}$ , which implies the desired.

Proof of Corollary 4.

Following the notation in the proof of Theorem 5, by Remark 5, for all i and j, we get that $\mathcal {B}_{i,j}=\{I_j\}$ . Therefore, for all $i=1,\ldots ,d$ , we have $\mathcal {B}_i=\mathcal {A}_i$ . Applying Theorem 5, we conclude the desired.

Proof of Theorem 1.

For each i, write

$$ \begin{align*} D_i := D_{K_i}. \end{align*} $$

We repeat the proof of Theorem 5, replacing the set $\{u_i=1\}$ by the domain $D_i$ . This gives a finite collection $\widehat {\mathcal B}_i$ of pairwise disjoint closed intervals with the same properties as the collection $\mathcal B_i$ in Theorem 5. In particular,

$$ \begin{align*} K_i \subset \bigcup_{I\in\widehat{\mathcal B}_i} I \subset D_i. \end{align*} $$

Moreover, the intervals in $\widehat {\mathcal B}_i$ are mapped into each other by elements of $\mathcal F$ , and the semigroup $G_{\mathcal F}$ acts proximally on each of them.

We now define

$$ \begin{align*} L_i := \bigcup_{I\in\widehat{\mathcal B}_i} I. \end{align*} $$

Then, $L_i$ is a finite union of pairwise disjoint closed intervals, contains $K_i$ , is contained in $D_i$ , and satisfies all the required properties. This proves the theorem.

Proof of Lemma 9.

Assume $d_+=n+2$ . By Lemma 6, then

$$ \begin{align*} \#\{i\in \{1,\ldots,d_+\}\colon \#V_i=2\}\geq n. \end{align*} $$

Hence, there are n distinct points $i_1,\ldots ,i_n,$ in $\{1,\ldots ,d_+\}$ such that $V_{i_k}=2$ for ${k=1,\ldots ,n}$ . Since $d_+=n+2>n$ , we have that any n-cycle is not an $\mathcal {F}$ -path. (An n-cycle is a path of the form $[j_0,j_1,\ldots ,j_n]$ with $j_0=j_n$ and $j_k\neq j_{\ell }$ for ${k,\ell =1,\ldots ,n}$ and $\ell \neq k$ .) Therefore, we can find $i_{n+1},j_1,\ldots ,j_n,j_{n+1}\in \{1,\ldots ,d_+\}$ with $(i_1,j_1),\ldots ,(i_{n+1},j_{n+1})\in \mathcal {E}$ such that for $k\neq \ell $ ,

$$ \begin{align*} i_k\neq j_k\quad \mathrm{and}\quad\{i_k,j_k\}\neq \{i_\ell,j_\ell\}. \end{align*} $$

For $k=1,\ldots ,n+1$ , consider the set $C_{i_k,j_k}$ as defined above Remark 6. Let $C_k=C_{i_k,j_k}$ . Note that for $k\neq \ell $ , $C_k\neq C_\ell $ and so

$$ \begin{align*} \bigg(\bigcup_{I\in C_k}I\bigg)\cap\bigg(\bigcup_{J\in C_\ell}J\bigg)=\emptyset. \end{align*} $$

By Remark 6, for $k=1,\ldots ,n+1$ , there exists a closed $\mathcal {F}^-$ -minimal set $F_k\subset C_k$ . Therefore, there exist at least $n+1$ closed $\mathcal {F}^-$ -minimal sets and so $d_-\geq n+1$ .

Proof of Lemma 10.

If $d_+=1$ , it is clear that $d_-\geq d_+$ . Assume $d_+\geq 2$ . For all ${i=1,\ldots ,d_+}$ , take $j_i\in \{1,\ldots ,d_+\}\backslash \{i\} $ such that $[i,j_i]$ is an $\mathcal {F}$ -path and so $(i,j_i)\in \mathcal {E}$ . As we discussed above, we have for each $\mathcal {F}$ -path $[i,j_i]$ (that is, $(i,j_i)\in \mathcal {E}$ ), there exists a closed $\mathcal {F}^-$ -minimal set $F_i\subset C_{[i,j_i]}$ . Hence, there exist at least $d_+$ closed $\mathcal {F}^-$ -minimal sets and so $d_-\geq d_+$ .

Proof of Theorem 3.

By Lemma 9, $d_-\geq d_+-1$ . Using that $(\mathcal {F}^-)^-=\mathcal {F}$ , we also get ${d_+\geq d_--1}$ . Consequently, $d_++1\geq d_-\geq d_+-1$ and, hence, $|d_+-d_-|\leq 1$ .

Now, assume $\mathcal {F}\subset \mathrm {Homeo}^+(\mathbb {S}^1 )$ . Since $(\mathcal {F}^-)^-=\mathcal {F}$ , by Lemma 10, $d_-\geq d_+$ and ${d_+\geq d_-}$ . Therefore, $d_-=d_+$ .