From dd7981f367831788f69fdbf0322149d75969cbc2 Mon Sep 17 00:00:00 2001 From: George Stamatiou Date: Sun, 17 Dec 2023 17:17:00 +0100 Subject: [PATCH 1/2] Filled proof of unique ergodicity for irrational rotations --- .../Advanced_dynamical_systems.tex | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex index 62554c5..c61f884 100644 --- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex +++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex @@ -663,11 +663,15 @@ $$ \int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x} $$ - An easy check shows that if $P_n=\sum_{k=-n}^na_k\exp{2\pi\ii k x}$ is a trigonometric polynomial, then $\int_{\TT^1}P_n(x)\dd{x}=a_0$. Moreover, if $k\ne 0$: + First we consider the case of trigonometric polynomials, i.e. $P_n(x) =\sum_{k=-n}^na_k\exp{2\pi\ii k x}$, where $a_k = \bar a_{-k}$, then $\int_{\TT^1}P_n(x)\dd{x}=a_0$. + Moreover, if $k\ne 0$: $$ \int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=\exp{2\pi\ii k\alpha}\int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}\implies \int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=0 $$ - where the equality is due to the invariance of $\mu$. So, we also have $\int_{\TT^1}P_n(x)\dd{\mu}=a_0$. Now consider the Féjer trigonometric polynomial that converge uniformly to $\varphi$ and use the \mnameref{RFA:domianted}. + where the equality is due to the invariance of $\mu$. + Thus $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$, and the result holds for trigonometric polynomials. + In the general case of a continuous function $\phi \in \mathcal{C}(\TT^1)$, we recall that the set of trigonometric polynomials is dense in $\mathcal{C}(\TT^1)$. + Thus the result for the case of trigonometric polynomials passes to the limit and we have that $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$. \end{proof} \begin{proposition}\label{ADS:uniquely_ergodic} Let $F\in\Homeoplus(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, $F$ is uniquely ergodic. From bad52d18a0782590e632bc8b54d022132eabb2c1 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Sun, 17 Dec 2023 20:10:42 +0100 Subject: [PATCH 2/2] modified thm --- .../2nd/Mathematical_analysis/Mathematical_analysis.tex | 2 +- .../Advanced_dynamical_systems.tex | 8 ++------ 2 files changed, 3 insertions(+), 7 deletions(-) diff --git a/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex b/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex index 67351e3..9657d3f 100644 --- a/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex +++ b/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex @@ -175,7 +175,7 @@ \begin{theorem} Let $(f_n)$ be a sequence of continuous functions defined on $D\subseteq\RR $. If $(f_n)$ converges uniformly to $f$ on $D$, then $f$ is continuous on $D$, that is, for any $x_0\in D$, it satisfies: $$\lim_{n\to\infty}\left(\lim_{x\to x_0} f_n(x)\right)=\lim_{x\to x_0} f(x)$$ \end{theorem} - \begin{theorem} + \begin{theorem}\label{MA:uniformintegral} Let $(f_n)$ be a sequence of functions defined on $I=[a,b]\subseteq\RR $. If $(f_n)$ are Riemann-integrable on $I$ and $(f_n)$ converges uniformly to $f$ on $I$, then $f$ is Riemann-integrable on $I$ and $$\int_a^b\lim_{n\to\infty} f_n(x) \dd{x}=\lim_{n\to\infty} \int_a^bf_n(x) \dd{x}$$ \end{theorem} \begin{theorem} diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex index c61f884..18a055d 100644 --- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex +++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex @@ -663,15 +663,11 @@ $$ \int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x} $$ - First we consider the case of trigonometric polynomials, i.e. $P_n(x) =\sum_{k=-n}^na_k\exp{2\pi\ii k x}$, where $a_k = \bar a_{-k}$, then $\int_{\TT^1}P_n(x)\dd{x}=a_0$. - Moreover, if $k\ne 0$: + An easy check shows that if $P_n=\sum_{k=-n}^na_k\exp{2\pi\ii k x}$ is a trigonometric polynomial, then $\int_{\TT^1}P_n(x)\dd{x}=a_0$. Moreover, if $k\ne 0$: $$ \int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=\exp{2\pi\ii k\alpha}\int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}\implies \int_{\TT^1}\exp{2\pi\ii kx}\dd{\mu}=0 $$ - where the equality is due to the invariance of $\mu$. - Thus $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$, and the result holds for trigonometric polynomials. - In the general case of a continuous function $\phi \in \mathcal{C}(\TT^1)$, we recall that the set of trigonometric polynomials is dense in $\mathcal{C}(\TT^1)$. - Thus the result for the case of trigonometric polynomials passes to the limit and we have that $\int_{\TT^1}\varphi(x)\dd{\mu}=\int_{\TT^1}\varphi(x)\dd{x}$. + where the equality is due to the invariance of $\mu$. So, we also have $\int_{\TT^1}P_n(x)\dd{\mu}=a_0$. Now consider the Féjer means, which converge uniformly to $\varphi$ (recall \mnameref{MA:fejerthm}) and use the \mcref{MA:uniformintegral}. \end{proof} \begin{proposition}\label{ADS:uniquely_ergodic} Let $F\in\Homeoplus(\TT^1)$ with $\rho(F)\notin\quot{\QQ}{\ZZ}$. Then, $F$ is uniquely ergodic.