updated improvements dyn syst

Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex

@@ -405,7 +405,7 @@
   \begin{remark}
     The converse is not true.
   \end{remark}
-  \begin{theorem}
+  \begin{theorem}\label{ADS:invariance_rho}
     Let $F,G\in\Homeoplus(\TT^1)$ be conjugate by $H\in \Homeoplus(\TT^1)$. Then, $\rho(F)=\rho(G)$.
   \end{theorem}
   \begin{proof}
@@ -441,7 +441,7 @@
     Let $F\in \Homeoplus(\TT^1)$ and $\mu\in \mathcal{M}(\TT^1)$. We define the \emph{push-forward measure} of $F$ as $F_*\mu(\varphi):=\mu(\varphi\circ F)$.
   \end{definition}
   \begin{definition}
-    We say that a measure $\mu\in\mathcal{M}(\TT^1)$ is \emph{invariant} by $F\in\Homeoplus(\TT^1)$ (or \emph{$F$-invariant}) if $F_*\mu=\mu$. We will denote by $\mathcal{M}_F(\TT^1)$ the set of $F$-invariant probability measures.
+    We say that a measure $\mu\in\mathcal{M}(\TT^1)$ is \emph{invariant} by $F\in\Homeo(\TT^1)$ (or \emph{$F$-invariant}) if $F_*\mu=\mu$. We will denote by $\mathcal{M}_F(\TT^1)$ the set of $F$-invariant probability measures.
   \end{definition}
   \begin{proposition}
     Let $F\in \Homeoplus(\TT^1)$, $x\in\TT^1$ and $n\in\NN$.
@@ -568,8 +568,8 @@
   \end{proposition}
   \begin{proof}
     \begin{itemizeiff}
-      For some $y\in Y\subseteq X$ closed, invariant and non-empty, we have:
-      $$Y\subseteq X = \overline{\mathcal{O}(y)}\subseteq \overline{\mathcal{O}(Y)}\subseteq \overline{Y}=Y$$
+      Let $Y\subseteq X$ be closed, invariant and non-empty and take $y\in Y$. We have:
+      $$Y\subseteq X = \overline{\mathcal{O}(y)}\subseteq \overline{Y}=Y$$
       \item Let $x\in X$. Since $\overline{\mathcal{O}(x)}\subseteq X$ is closed, invariant and non-empty, we have that $\overline{\mathcal{O}(x)}=X$.
     \end{itemizeiff}
   \end{proof}
@@ -581,7 +581,7 @@
     \end{enumerate}
   \end{theorem}
   \begin{proof}
-    First we assume $q=1$ and $p=0$. Let $f\in \mathcal{D}^0(\TT^1)$ be a lift of $F$. By \mcref{ADS:characterisation_rot_number}, we have that $\exists x\in \RR$ with $f(x)=x$. So $\Fix(f)\ne \varnothing$, and it is closed and invariant by translations. Now we write $\RR\setminus\Fix(f)$ as union of open intervals. Let $(a,b)$ be one of such connected components. Inside it, we must have either $f(x)>x$ or $f(x)<x$. In the first case we have that $(f^n(x))$ is strictly increasing $\forall x\in (a,b)$ and so $\omega(x)=\{b\}\in\Fix(f)$ and $\alpha(x)=\{a\}\in\Fix(f)$ $\forall x\in(a,b)$. The second case is exactly the opposite.
+    First we assume $q=1$ and $p=0$. Let $f\in \mathcal{D}^0(\TT^1)$ be a lift of $F$. By \mcref{ADS:characterisation_rot_number}, we have that $\exists x\in \RR$ with $f(x)=x$. So $\Fix(f)\ne \varnothing$, and it is closed and invariant by integer translations. Now we write $\RR\setminus\Fix(f)$ as union of open intervals. Let $(a,b)$ be one of such connected components. Inside it, we must have either $f(x)>x$ or $f(x)<x$. In the first case we have that $(f^n(x))$ is strictly increasing $\forall x\in (a,b)$ and so $\omega(x)=\{b\}\in\Fix(f)$ and $\alpha(x)=\{a\}\in\Fix(f)$ $\forall x\in(a,b)$. The second case is exactly the opposite.
 
     Now we do the general case. Assume $\rho(f)=\frac{p}{q}$. Then, again by \mcref{ADS:characterisation_rot_number}, we have that $\exists x\in \RR$ with $f^q(x)=x+p$. Assume we have $x'\in\RR$ and $p',q'\in\ZZ$ with $q'\geq 1$ such that $f^{q'}(x')=x'+p'$. By \mcref{ADS:characterisation_rot_number}, we have that $\frac{p}{q}=\frac{p'}{q'}$ and so $\exists k\in\NN$ such that $q'=kq$ and $p'=kp'$ because $\frac{p}{q}$ is irreducible. Now let $g=f^q-p$. Then, an easy calculation shows that $g^k(x')=x'$. But $\rho(g)=0$ and in the previous case we have seen that the periodic points are only fixed points, so $k=1$. For the second part, we proceed as in the previous case with the function $g=f^q-p$.
   \end{proof}
@@ -606,26 +606,60 @@
     $$
   \end{definition}
   \begin{remark}
-    Note that $\supp\mu $ is a closed set.
+    Note that $\supp\mu $ is a closed set and $\mu(\TT^1\setminus \supp\mu) = 0$.
+  \end{remark}
+  \begin{remark}
+    If $F\in \Homeoplus(\TT^1)$ and $\mu\in \mathcal{M}_F(\TT^1)$, then $\supp\mu$ is invariant.
   \end{remark}
   \begin{remark}
     $\supp\text{Leb}=\TT^1$ and $\supp\delta_{x}=\{x\}$.
   \end{remark}
   \begin{proposition}
-    Let $\mu\in \mathcal{M}(\TT^1)$ and $F\in\Homeoplus(\TT^1)$. $\mu$ is invariant by $F$ if and only if $\forall A\subseteq \TT^1$ Borel set, $\mu(A)=\mu(F^{-1}(A))$.
+    Let $\mu\in \mathcal{M}(\TT^1)$ and $F\in\Homeo(\TT^1)$. $\mu$ is invariant by $F$ if and only if $\forall A\subseteq \TT^1$ Borel set, $\mu(A)=\mu(F^{-1}(A))$.
   \end{proposition}
+  \begin{proof}
+    \begin{itemizeiff}
+      Let $A\subseteq \TT^1$ be Borel. We have:
+      \begin{multline*}
+        \mu(A) =\!\inf_{\substack{U\text{ open} \\A\subseteq U}}\sup_{\substack{\psi\text{ cont.}\\\varphi\leq \indi{U}}}\!\mu(\varphi)=\!\inf_{\substack{U\text{ open} \\A\subseteq U}}\sup_{\substack{\psi\text{ cont.}\\\varphi\leq \indi{U}}}\!\mu(\varphi\circ F)=\\
+        =\inf_{\substack{U\text{ open} \\A\subseteq U}}\sup_{\substack{\psi\text{ cont.}\\\psi\leq \indi{F^{-1}(U)}}}\!\mu(\psi)=\!\inf_{\substack{V\text{ open} \\F^{-1}(A)\subseteq V}}\sup_{\substack{\psi\text{ cont.}\\\psi\leq \indi{V}}}\mu(\psi)=\\=\mu(F^{-1}(A))
+      \end{multline*}
+      % \item Now, let $\varphi\in \mathcal{C}(\TT^1)$. Note that:
+    \end{itemizeiff}
+  \end{proof}
+  \begin{remark}
+    If $F\in \Homeoplus(\TT^1)$ and $\mu\in\mathcal{M}_F(\TT^1)$, then $\mu(F^n(A))=\mu(A)$ $\forall n\in\ZZ$ and $A\subseteq \TT^1$ Borel.
+  \end{remark}
   \begin{lemma}
     Let $\mu\in \mathcal{M}(\TT^1)$. We have a lift to a measure $\mu$ on $\RR$ invariant by integer translations: $\mu(A+k)=\mu(A)$ $\forall k\in\ZZ$ and $A\subseteq \mathcal{B}(\RR)$.
   \end{lemma}
   \begin{definition}
-    Let $\mu\in\mathcal{M}(\TT^1)$. We define $h_\mu:[0,1]\to [0,1]$ as the function with $h_\mu(0)=0$ and $h_\mu(x)=\mu([0,x))$ for $0<x\leq 1$. This definition extends to a non-decreasing function $h_\mu:\RR\to\RR$ such that $h_\mu(x+k)=h_\mu(x)+k$ $\forall k\in\ZZ$
+    Let $\mu\in\mathcal{M}(\TT^1)$. We define $h_\mu:[0,1]\to [0,1]$ as the function with $h_\mu(0)=0$ and $h_\mu(x)=\mu([0,x))$ for $0<x\leq 1$. This definition extends to a non-decreasing function $h_\mu:\RR\to\RR$ such that $h_\mu(x+k)=h_\mu(x)+k$ $\forall k\in\ZZ$.
+  \end{definition}
+  \begin{definition}
+    Let $\mu\in\mathcal{M}(\TT^1)$. We say that $\mu$ has atoms if $\exists x\in\TT^1$ such that $\mu(\{x\})>0$.
   \end{definition}
-  \begin{lemma}
-    Let $\mu\in\mathcal{M}(\TT^1)$. We say that $\mu$ has atoms if $\exists x\in\TT^1$ such that $\mu(\{x\})>0$
-  \end{lemma}
   \begin{lemma}\label{ADS:lemma_atom}
-    Let $\mu\in\mathcal{M}(\TT^1)$. $h_\mu$ is continuous if and only if $\forall x\in\RR$, $\mu(\{x\})=0$, that is $\mu$ has no atoms.
+    Let $\mu\in\mathcal{M}(\TT^1)$. $h_\mu$ is continuous if and only if $\forall x\in\RR$, $\mu(\{x\})=0$, that is if $\mu$ has no atoms.
   \end{lemma}
+  \begin{proof}
+    \begin{itemizeiff}
+      Let $x_n=x+\frac{1}{n}\to x$. Then, $[0,x_n)\supset [0,x_{n+1})$ and so by the continuity of $h_\mu$ and \mcref{RFA:decresingseq} we have:
+      \begin{multline*}
+        \mu([0,x))=\lim_{n\to\infty}\mu([0,x_n))=\mu\left(\bigcap_{n\in\NN}[0,x_n)\right)=\\=\mu([0,x])
+      \end{multline*}
+      which implies that $\mu(\{x\})=0$.
+      \item Let $x\in\RR$ and $x_n\to x$. Since $(x_n)$ is bounded, we can extract a monotone subsequence $(x_{n_k})$. Then:
+      \begin{multline*}
+        \lim_{n\to \infty}h(x_n)=\lim_{k\to \infty}\mu([0,x_{n_k}))=\\
+        =\begin{cases}
+          \mu\left(\bigcup_{k\in\NN}[0,x_{n_k})\right) = \mu([0,x)) & \text{if } x_{n_k}\nearrow x \\
+          \mu\left(\bigcap_{k\in\NN}[0,x_{n_k})\right) = \mu([0,x]) & \text{if } x_{n_k}\searrow x
+        \end{cases}
+      \end{multline*}
+      The first case is fine, and for the second one, since $\mu(\{x\})=0$, we have that $\mu([0,x])=\mu([0,x))$.
+    \end{itemizeiff}
+  \end{proof}
   \begin{definition}
     A subset $C\subseteq \RR$ is a \emph{Cantor set} if it is closed, it has no isolated points and it has empty interior.
   \end{definition}
@@ -637,26 +671,26 @@
     \end{enumerate}
   \end{theorem}
   \begin{proof}
-    Let $\mu\in\mathcal{M}_F(\TT^1)$ and consider $h_\mu:\RR\to\RR$ as defined above. Now assume $x\in \TT^1$ is such that $\mu(\{x\})=c>0$, then by invariance $\mu(A_n)=c>0$, where $A_n:=\{F^n(x)\}$. Note that since $\mu\leq 1$, $(A_n)$ cannot be disjoint. So $\exists n,m\in\NN$ with $n<m$ such that $F^n(x)=F^m(x)$. But then $F^{m-n}(x)=x$ and so $x$ is periodic, which is not possible since $\rho(F)\in\quot{\QQ}{\ZZ}$ by \mcref{ADS:characterisation_rot_number}. Thus, $\mu$ has no atoms and so $h_\mu$ is continuous by \mcref{ADS:lemma_atom}. Now, define $H:\TT^1\to \TT^1$ as the projection of $h_\mu$ to $\TT^1$, which is continuous and surjective. Let $f\in\mathcal{D}^0(\TT^1)$ be a lift of $F$. Then:
+    Let $\mu\in\mathcal{M}_F(\TT^1)$ and consider $h:=h_\mu:\RR\to\RR$ as defined above. Now assume $x\in \TT^1$ is such that $\mu(\{x\})=c>0$, then by invariance $\mu(A_n)=c>0$, where $A_n:=\{F^n(x)\}$. Note that since $\mu\leq 1$, $(A_n)$ cannot be disjoint. So $\exists n,m\in\NN$ with $n<m$ such that $F^n(x)=F^m(x)$. But then $F^{m-n}(x)=x$ and so $x$ is periodic, which is not possible since $\rho(F)\in\quot{\QQ}{\ZZ}$ by \mcref{ADS:characterisation_rot_number}. Thus, $\mu$ has no atoms and so $h$ is continuous by \mcref{ADS:lemma_atom}. Now, define $H:\TT^1\to \TT^1$ as the projection of $h$ to $\TT^1$, which is continuous and surjective. Let $f\in\mathcal{D}^0(\TT^1)$ be a lift of $F$. Then:
     $$
       h(f(x))-h(f(0))=\mu([f(0),f(x)))=\mu([0,x))=h(x)
     $$
-    where we have used the invariance of $\mu$. Thus, $h\circ f=R_{h(f(0))}\circ h$ and necessarily we need $h(f(0))=\rho(R_{h(f(0))})=\rho(f)$, by the invariance of the rotation number by semi-conjugacy. This gives $H\circ F=R_{\rho(F)}\circ H$. Now, we can express the dichotomy as follows: either $\supp\mu=\TT^1$ or $\supp\mu=:X\subsetneq \TT^1$. The first case is equivalent to $h$ being strictly increasing and so $h$ is a homeomorphism. Then, $H$ conjugates $F$ and $R_{\rho(F)}$ and so $F$ is minimal because $R_{\rho(F)}$ is minimal. In the second case, we have that $X$ is a nonempty closed invariant set that has no isolated points because $\mu$ has no atoms. To show that $X$ is minimal, let $\TT^1=X\sqcup U$ with $U$ open, and so it can be written as a countable union of open intervals. Let $D\subseteq X$ be the set containing the endpoints of those intervals and let
+    where we have used the invariance of $\mu$. Thus, $h\circ f=R_{h(f(0))}\circ h$ and necessarily we need $h(f(0))=\rho(R_{h(f(0))})=\rho(f)$. This gives $H\circ F=R_{\rho(F)}\circ H$. Now, we can express the dichotomy as follows: either $\supp\mu=\TT^1$ or $\supp\mu=:X\subsetneq \TT^1$. The first case is equivalent to $h$ being strictly increasing and so $h$ is a homeomorphism. Then, $H$ conjugates $F$ and $R_{\rho(F)}$ and so $F$ is minimal because $R_{\rho(F)}$ is minimal. In the second case, we have that $X$ is a nonempty closed invariant set that has no isolated points because $\mu$ has no atoms. To show that $X$ is minimal, let $\TT^1=X\sqcup U$ with $U$ open, and so it can be written as a countable union of open intervals. Let $D\subseteq X$ be the set containing the endpoints of those intervals and let
     \begin{equation}\label{ADS:eq1}
       Y:=\{y\in\TT^1:H^{-1}(\{y\})\text{ is a closed interval}\}
     \end{equation}
-    $Y$ is countable. Now take $M\subseteq X$ be nonempty, closed and invariant. We want to prove that $M=X$. Then, $H(M)\subseteq \TT^1$ is nonempty, closed and invariant by $R_{\rho(F)}$. So $H(M)=\TT^1$ because $R_{\rho(F)}$ is minimal. Now, since $H$ restricted to $X\setminus D$ is injective, then $M\supseteq X\setminus D$. Indeed, by contradiction let $x\in X\setminus D$ such that $x\notin M$. We have $H(X\setminus D)=\TT^1\setminus Y$. Thus, $H^{-1}(H(X\setminus D))=H^{-1}(\TT^1\setminus Y)=X\setminus D$. Then, $H(x)\in H(X\setminus D)\subseteq \TT^1=H(M)$. So $\exists y\in M$ such that $H(x)=H(y)$, and so$y\in H^{-1}(H(X\setminus X))=X\setminus D$. But $H|_{X\setminus D}$ is injective, so $x=y\in M$, which is a contradiction because $x\notin M$. Thus, $M\supseteq X\setminus D$, which implies:
+    $Y$ is countable ($H^{-1}(Y)=U\cup D$). Now take $M\subseteq X$ be nonempty, closed and invariant. We want to prove that $M=X$. We have that $H(M)\subseteq \TT^1$ is nonempty, closed (in fact it's compact because it is the image of a compact set) and invariant by $R_{\rho(F)}$. So $H(M)=\TT^1$ because $R_{\rho(F)}$ is minimal. Now, since $H$ restricted to $X\setminus D$ is injective ($h$ is strictly increasing in $X\setminus D$ thought inside $[0,1]$), then $M\supseteq X\setminus D$ because if not there would exist $x\in X\setminus D$ such that $x\notin M$. We have $H(X\setminus D)=\TT^1\setminus Y$. Thus, $H^{-1}(H(X\setminus D))=H^{-1}(\TT^1\setminus Y)=X\setminus D$. Then, $H(x)\in H(X\setminus D)\subseteq \TT^1=H(M)$. So $\exists y\in M$ such that $H(x)=H(y)$, and so $y\in H^{-1}(H(X\setminus X))=X\setminus D$. But $H|_{X\setminus D}$ is injective, so $x=y\in M$, which is a contradiction because $x\notin M$. Thus, $M\supseteq X\setminus D$, which implies:
     $$
       M=\overline{M}\supseteq \overline{X\setminus D}=\overline{X} = X
     $$
-    So $M=X$ and, thus, $X$ is minimal. Moreover, $X$ has empty interior. Indeed, if that wasn't the case, we would have $\Fr{X}=\overline{X}\setminus\Int(X)=X\setminus \Int(X)\subsetneq X$ and so $\Fr{X}$ would be a nonempty closed invariant set, which is not possible because $X$ is minimal. Finally, to prove $X=\Omega(F)$, by minimality it suffices to show that $\Omega(F)\subseteq X$. Let $x\in U$, where $\TT^1=X\sqcup U$, with $U=\bigsqcup_{i=1}^\infty I_i$ invariant and $I_i$ intervals. We need to see that $x$ is wandering. Let $I$ be one of such intervals. We may have either $F^n(I)=I$ for some $n\geq 1$ or $F^n(I)\cap I=\varnothing$ for all $n\geq 1$. But in the first case, we would have that the extremities of $I$ are periodic points, which is not possible because $\rho(F)\notin\quot{\QQ}{\ZZ}$. So we must have the second case, which implies that $I$ is a wandering domain, and thus so is $U$.
+    So $M=X$ and, thus, $X$ is minimal. Moreover, $X$ has empty interior. Indeed, if that wasn't the case, we would have $\Fr{X}=\overline{X}\setminus\Int(X)=X\setminus \Int(X)\subsetneq X$ and so $\Fr{X}$ would be a nonempty closed invariant set, which is not possible because $X$ is minimal. Finally, to prove $X=\Omega(F)$, by minimality it suffices to show that $\Omega(F)\subseteq X$. Let $x\in U$, where $\TT^1=X\sqcup U$, with $U=\bigsqcup_{i=1}^\infty I_i$ invariant and $I_i$ intervals. Note that $F$ maps connected components of $U$ to connected components of $U$. We need to see that $x$ is wandering. Let $I$ be one of such intervals. We may have either $F^n(I)=I$ for some $n\geq 1$ or $F^n(I)\cap I=\varnothing$ for all $n\geq 1$. But in the first case, we would have that the extremities of $I$ are periodic points, which is not possible because $\rho(F)\notin\quot{\QQ}{\ZZ}$. So we must have the second case, which implies that $I$ is a wandering domain, and thus so is $U$.
   \end{proof}
   \subsubsection{Unique ergodicity}
   \begin{definition}
     A homeomorphism $F:\TT^1\to\TT^1$ is \emph{uniquely ergodic} if it has a unique invariant probability measure.
   \end{definition}
   \begin{lemma}
-    If $\alpha\notin \QQ$, then $R_\alpha$ is uniquely ergodic and $\mathcal{M}_{R_\alpha}=\{\text{Leb}\}$.
+    If $\alpha\notin \QQ$, then $R_\alpha$ is uniquely ergodic and $\mathcal{M}_{R_\alpha}(\TT^1)=\{\text{Leb}\}$.
   \end{lemma}
   \begin{proof}
     Let $\mu\in\mathcal{M}_{R_\alpha}$. We want to see that $\forall \varphi\in \mathcal{C}(\TT^1)$:
@@ -673,27 +707,27 @@
     Let $F\in\Homeoplus(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, $F$ is uniquely ergodic.
   \end{proposition}
   \begin{proof}
-    Let $H$ be such that $H\circ F=R_\rho\circ H$ (by \mcref{ADS:theorem_irrational_rotation_number}) and so $F^{-1}(H^{-1}(A))=H^{-1}({R_\rho}^{-1}(A))$. Define:
+    Let $H$ be such that $H\circ F=R_{\rho(F)}\circ H$ (by \mcref{ADS:theorem_irrational_rotation_number}) and so $F^{-1}(H^{-1}(A))=H^{-1}({R_{\rho(F)}}^{-1}(A))$. Take $\mu\in\mathcal{M}_F(\TT^1)$ and define $H_*\mu$ as:
     $$
       H_*\mu(A):=\mu(H^{-1}(A))\qquad \forall A\subseteq \TT^1\text{ Borel}
     $$
     We have:
     \begin{multline*}
-      H_*\mu(A)=\mu(H^{-1}(A))=\mu(F^{-1}(H^{-1}(A)))=\\=\mu(H^{-1}({R_\rho}^{-1}(A)))=H_*\mu({R_\rho}^{-1}(A))
+      H_*\mu(A)=\mu(H^{-1}(A))=\mu(F^{-1}(H^{-1}(A)))=\\=\mu(H^{-1}({R_{\rho(F)}}^{-1}(A)))=H_*\mu({R_{\rho(F)}}^{-1}(A))
     \end{multline*}
-    where the second equality is due to the invariance of $\mu$. Hence, $H_*\mu$ is invariant by $R_\rho$, and so $H_*\mu=\text{Leb}$. That, is $\mu(H^{-1}(A))=\text{Leb}(A)$. Recall again the set $Y$ of \mcref{ADS:eq1} and $\TT^1=X\sqcup U$, with $H^{-1}(Y)=\overline{U}$. Since $Y$ is countable, $0=\text{Leb}(Y)=\mu(H^{-1}(Y))$. Now since $H|_X$ is a homeomorphism (??is it true??), we have that $\mu(B)=\text{Leb}(H(B))$, and so $\mu$ is uniquely determined.
+    where the second equality is due to the invariance of $\mu$. Hence, $H_*\mu$ is invariant by $R_{\rho(F)}$, and so $H_*\mu=\text{Leb}$. That, is $\mu(H^{-1}(A))=\text{Leb}(A)$. Recall again the set $Y$ of \mcref{ADS:eq1} and $\TT^1=X\sqcup U$, with $H^{-1}(Y)=\overline{U}$. Since $Y$ is countable, $0=\text{Leb}(Y)=\mu(H^{-1}(Y))$. Now since $H|_{X\setminus D}:X\setminus D\to \TT^1\setminus Y$ is a homeomorphism, we have that $\mu(B)=\text{Leb}(H(B))$, and so $\mu$ is uniquely determined.
   \end{proof}
   \begin{proposition}\label{ADS:birkov_sum_converge}
-    Let $F:\TT^1\to\TT^1$ be a homeomorphism. Then, $F$ is uniquely ergodic if and only if $\forall \varphi\in\mathcal{C}(\TT^1)$, $\exists c_\varphi\in\RR$ such that $\frac{1}{n}\sum_{i=0}^n\varphi\circ F^i$ converge uniformly to $c_\varphi$.
+    Let $F\in \Homeo(\TT^1)$. Then, $F$ is uniquely ergodic if and only if $\forall \varphi\in\mathcal{C}(\TT^1)$, $\exists c_\varphi\in\RR$ such that $\frac{1}{n}\sum_{i=0}^{n-1}\varphi\circ F^i$ converge uniformly to $c_\varphi$. In that case, $c_\varphi=\mu(\varphi)$, where $\mathcal{M}_F(\TT^1)=\{\mu\}$.
   \end{proposition}
   \begin{proof}
-    Assume first that $\mathcal{M}_F=\{\mu\}$ and argue by contradiction. That, is $\exists \varepsilon>0$, $(n_k)\in\NN$ with $n_k\nearrow +\infty$ and $(x_k)\in\TT^1$ such that $\forall k\geq 0$:
+    Assume first that $\mathcal{M}_F(\TT^1)=\{\mu\}$ and argue by contradiction. That, is $\exists \varepsilon>0$, $(n_k)\in\NN$ with $n_k\nearrow +\infty$ and $(x_k)\in\TT^1$ such that $\forall k\geq 0$:
     \begin{equation}\label{ADS:nuk_mu}
       \abs{\frac{1}{n_k}\sum_{i=0}^{n_k-1}\varphi\circ F^i(x_k)-\int_{\TT^1}\varphi\dd{\mu}}=\abs{\int_{\TT^1}\varphi\dd{\nu_k}-\int_{\TT^1}\varphi\dd{\mu}}>\varepsilon
     \end{equation}
     $$
     $$
-    where $\nu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}F_*^i\delta_{x_k}$. Note that $\nu_k\in\mathcal{\TT^1}$ and since $\mathcal{M}(\TT^1)$ is compact, after extracting a subsequence, $(\nu_k)$ converges to $\nu\in\mathcal{M}(\TT^1)$. Now, $\nu$ is invariant. Indeed:
+    where $\nu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}F_*^i\delta_{x_k}$. Note that $\nu_k\in\mathcal{M}(\TT^1)$ and since $\mathcal{M}(\TT^1)$ is compact, after extracting a subsequence, $(\nu_k)$ converges to $\nu\in\mathcal{M}(\TT^1)$. Now, $\nu$ is invariant. Indeed:
     $$
       F_*\nu_k-\nu_k=\frac{1}{n_k}(F_*^{n_k}\delta_{x_k}-\delta_{x_k})\overset{k\to\infty}{\longrightarrow} 0
     $$