From 15ba221caa1f7be3a27c8325af89a6e4853eba19 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Sun, 17 Dec 2023 20:05:53 +0100 Subject: [PATCH] more improvs --- .../Advanced_dynamical_systems.tex | 56 ++++++++++--------- ...ntroduction_to_nonlinear_elliptic_PDEs.tex | 2 +- 2 files changed, 32 insertions(+), 26 deletions(-) diff --git a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex index 9eb6e18..20b0d76 100644 --- a/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex +++ b/Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex @@ -727,17 +727,21 @@ \end{equation} $$ $$ - where $\nu_k=\frac{1}{n_k}\sum_{i=0}^{n_k-1}F_*^i\delta_{x_k}$. Note that $\nu_k\in\mathcal{M}(\TT^1)$ 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: + 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 with the weak$^*$-topology\footnote{$\mathcal{M}(\TT^1)$ is closed in $E=(\mathcal{C}(\TT^1))^*$ and it's contained in the unit ball of $E$, which is compact for the weak$^*$-topology. Thus, $\mathcal{M}(\TT^1)$ is compact.}, after extracting a subsequence, $(\nu_k)$ converges weakly to $\nu\in\mathcal{M}(\TT^1)$. Now, $\nu$ is invariant. Indeed: $$ - F_*\nu_k-\nu_k=\frac{1}{n_k}(F_*^{n_k}\delta_{x_k}-\delta_{x_k})\overset{k\to\infty}{\longrightarrow} 0 + \norm{F_*\nu_k-\nu_k}=\norm{\frac{1}{n_k}({(F^{n_k})}_*\delta_{x_k}-\delta_{x_k})}\leq \frac{ + 2\norm{\varphi}}{n_k}\overset{k\to\infty}{\longrightarrow} 0 $$ - because $\forall \varphi\in\mathcal{C}(\TT^1)$, $(F_*^{n_k}\delta_{x_k}-\delta_{x_k})(\varphi)$ is bounded by $2\norm{\varphi}$. So $\nu =\mu$, but this is a contradiction with \mcref{ADS:nuk_mu}. Now we prove the converse. Let $\mu\in\mathcal{M}_F(\TT^1)$. Then: + So $\nu =\mu$, but this is a contradiction with \mcref{ADS:nuk_mu}. Now we prove the converse. Let $\mu\in\mathcal{M}_F(\TT^1)$. Then: \begin{multline*} c_\varphi =\int_{\TT^1}c_\varphi\dd{\mu} =\int_{\TT^1}\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\varphi\circ F^i\dd{\mu} =\\ =\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\int_{\TT^1}\varphi\circ F^i\dd{\mu}=\lim_{n\to\infty}\frac{1}{n}\sum_{i=0}^{n-1}\int_{\TT^1}\varphi\dd{\mu} =\int_{\TT^1}\varphi\dd{\mu} \end{multline*} where the third equality is due to the uniform convergence and the penultimate equality is due to the invariance of $\mu$. This implies that $\mu$ is uniquely determined. \end{proof} + \begin{remark} + The unique ergodicity property is preserved under conjugation. + \end{remark} \begin{definition} For $k\in\NN\cup\{0\}$ we define the set $\mathcal{D}^k(\TT^1)$ as: \begin{multline*} @@ -761,8 +765,7 @@ $$ \frac{1}{n}\log Df^n=\frac{1}{n}\sum_{i=0}^{n-1}\log(Df\circ f^i)=\frac{1}{n}\sum_{i=0}^{n-1}\log(Df\circ F^i) $$ - where in the last equality we have used the fact that $Df=1+D\varphi\in \mathcal{C}(\TT^1)$. By \mcref{ADS:uniquely_ergodic - ,ADS:birkov_sum_converge}, we have that $\frac{1}{n}\sum_{i=0}^{n-1}\log(Df\circ F^i)$ converges uniformly to $c:=\int_{\TT^1}\log(Df)\dd{\mu}$. Moreover, since $Df^n=1+D\varphi_n\in \mathcal{C}(\TT^1)$, we have that $\int_{\TT^1}Df^n\dd{x} =1$. Now assume without loss of generality that $c>0$. Then, for $n$ large enough we must have $Df^n(x)\sim e^{nc}$ and so: + where in the last equality we have used the fact that $Df=1+D\varphi\in \mathcal{C}(\TT^1)$. By \mcref{ADS:uniquely_ergodic,ADS:birkov_sum_converge}, we have that $\frac{1}{n}\sum_{i=0}^{n-1}\log(Df\circ F^i)$ converges uniformly to $c:=\int_{\TT^1}\log(Df)\dd{\mu}$. Moreover, since $Df^n=1+D\varphi_n\in \mathcal{C}(\TT^1)$, we have that $\int_{\TT^1}Df^n\dd{x} =1+\int_{\TT^1}D\varphi_n\dd{x}=1$. Now assume without loss of generality that $c>0$. Then, for $n$ large enough we must have $Df^n(x)\sim e^{nc}$ and so: $$ 1=\int_{\TT^1}Df^n\dd{x}\sim \int_{\TT^1}e^{nc}\dd{x}\overset{n\to\infty}{\longrightarrow} +\infty $$ @@ -775,6 +778,27 @@ $$ The constant $C$ is usually denoted as $\Var(\varphi)$. \end{definition} + \begin{remark} + If $\varphi\in\mathcal{C}^1(\TT^1)$, then $\Var(\varphi)=\norm{D\varphi}_{\mathcal{C}^0}$. + \end{remark} + \begin{lemma}\label{ADS:lema_pnqn} + Let $\alpha\notin\QQ$. Then, $\forall n\in\NN$, $\exists \frac{p_n}{q_n}\in\QQ$ such that: + \begin{enumerate} + \item $\displaystyle\abs{\alpha-\frac{p_n}{q_n}}<\frac{1}{{q_n}^2}$ + \item $q_n\overset{n\to\infty}{\longrightarrow}+\infty$ + \end{enumerate} + \end{lemma} + \begin{proof} + Let $Q\in\NN$. By the Pigeon-hole principle, there exist two elements among $0,\{\alpha\},\ldots,\{Q \alpha\}$ (here $\{\cdot\}$ denotes the fractional part) such that they are in one of the intervals among $\left[0,\frac{1}{Q}\right], \left[\frac{1}{Q},\frac{2}{Q}\right],\ldots,\left[\frac{Q-1}{Q},1\right]$. That is, $\exists q_1,q_2\in\QQ_{\geq 0}$ and $p\in\ZZ$ such that $\abs{q\alpha -p}\leq \frac{1}{Q}$ with $q:=q_2-q_1\leq Q$. Now apply this to $Q_n=n\geq 1$: $\exists \frac{p_n}{q_n}\in \QQ$ with $1\leq q_n\leq n$ such that: + $$ + \abs{q_n\alpha-p_n}<\frac{1}{n}\leq \frac{1}{q_n} + $$ + To prove that $q_n\overset{n\to\infty}{\longrightarrow}+\infty$, we argue by contradiction. Assume that $\exists M\in\NN$ such that $q_n\leq M$ for all $n\in\NN$. Then, $\exists n_0\in\NN$ such that $\forall n\geq n_0$ we have $q_n=q_{n_0}$. But then by the irrationality of $\alpha$ we have that $\exists c>0$ such that: + $$ + c\leq \abs{q_n\alpha-p_n}\leq \frac{1}{n} \overset{n\to\infty}{\longrightarrow} 0 + $$ + which is a contradiction. + \end{proof} \begin{lemma}\label{ADS:lema_alpha_i} Let $\alpha\notin\QQ$ and $\frac{p}{q}\in \QQ$ with $\abs{\alpha-\frac{p}{q}}<\frac{1}{q^2}$. Set $\alpha_i:=\{i\alpha\}$ for $1\leq i\leq q$, where $\{x\}$ denotes the fractional part of $x$. Then, each $\alpha_i$ belongs to a different interval of the form $\left(\frac{k_i}{q},\frac{k_i+1}{q}\right)$ with $k_i\in\{0,\ldots,q-1\}$. \end{lemma} @@ -786,9 +810,9 @@ We claim that the numbers $\{i \frac{p}{q}\}$ are all distinct for $1\leq i\leq q$. Indeed, if $\exists i,j$ with $i\frac{p}{q} -j\frac{p}{q}=k\in\ZZ^*$, then $\frac{p}{q}=\frac{k}{i-j}$, which is not possible because $\frac{p}{q}$ is irreducible and $i-j \leq q-1$. So we can write $\{i \frac{p}{q}\}=k_i\frac{p}{q}$ for some $k_i\in\{0,\ldots,q-1\}$. Finally, \mcref{ADS:ineq_alpha_i} implies that $\alpha_i\in \left(\frac{k_i}{q},\frac{k_i+1}{q}\right)$. \end{proof} \begin{proposition}[Denjoy-Koksma inequality]\label{ADS:denjoy_koksma} - Let $F\in\Homeoplus(\TT^1)$ with $\rho(F)=\alpha\notin \quot{\QQ}{\ZZ}$, $\mu\in\mathcal{M}_F(\TT^1)$ and $\frac{p}{q}\in \QQ$ with $\abs{\alpha-\frac{p}{q}}<\frac{1}{q^2}$. For any $\psi\in \mathcal{C}(\TT^1)$ with $\Var(\psi)<\infty$ we have $\forall x \in \TT^1$: + Let $F\in\Homeoplus(\TT^1)$ with $\alpha:=\rho(F)\notin \quot{\QQ}{\ZZ}$, $\mu\in\mathcal{M}_F(\TT^1)$ and $\frac{p}{q}\in \QQ$ with $\abs{\alpha-\frac{p}{q}}<\frac{1}{q^2}$. Then, $\forall\psi\in \mathcal{C}(\TT^1)$ with $\Var(\psi)<\infty$ we have: $$ - \abs{\sum_{i=0}^{q-1}\psi(F^i(x))-q\int_{\TT^1}\psi\dd{\mu}}\leq\Var(\psi) + \abs{\sum_{i=0}^{q-1}\psi(F^i(x))-q\int_{\TT^1}\psi\dd{\mu}}\leq\Var(\psi) \qquad \forall x\in \TT^1 $$ \end{proposition} \begin{proof} @@ -806,24 +830,6 @@ $$ because $H_*\mu = \text{Leb}$ and the invariance of $\mu$ (?), and at the end the supremum is reached at some point $t_i\in I_i$ because the intervals are closed. \end{proof} - \begin{lemma}\label{ADS:lema_pnqn} - Let $\alpha\notin\QQ$. Then, $\forall n\in\NN$, $\exists \frac{p_n}{q_n}\in\QQ$ such that: - \begin{enumerate} - \item $\displaystyle\abs{\alpha-\frac{p_n}{q_n}}<\frac{1}{{q_n}^2}$ - \item $q_n\overset{n\to\infty}{\longrightarrow}+\infty$ - \end{enumerate} - \end{lemma} - \begin{proof} - For $Q\geq 1$, by the Pigeon-hole principle, there exist two elements among $0,\{\alpha\},\ldots,\{Q \alpha\}$ such that they are in one of the intervals among $[0,\frac{1}{Q}], [\frac{1}{Q},\frac{2}{Q}],\ldots,[\frac{Q-1}{Q},1]$. That is, $\exists q_1,q_2\in\QQ_{\geq 0}$ and $p\in\ZZ$ such that $\abs{q\alpha -p}\leq \frac{1}{Q}$ with $q:=q_2-q_1$. Now apply this to $Q_n=n\geq 1$: $\exists \frac{p_n}{q_n}\in \QQ$ with $1\leq q_n\leq n$ such that: - $$ - \abs{q_n\alpha-p_n}<\frac{1}{n}\leq \frac{1}{q_n} - $$ - To prove that $q_n\overset{n\to\infty}{\longrightarrow}+\infty$, we argue by contradiction. Assume that $\exists M\in\NN$ such that $q_n\leq M$ for all $n\in\NN$. Then, $\exists n_0\in\NN$ such that $\forall n\geq n_0$ we have $q_n=q_{n_0}$. But then by the irrationality of $\alpha$ we have that $\exists c>0$ such that: - $$ - c\leq \abs{q_n\alpha-p_n}\leq \frac{1}{n} \overset{n\to\infty}{\longrightarrow} 0 - $$ - which is a contradiction. - \end{proof} \begin{lemma}\label{ADS:lema_var_log} Let $f\in \mathcal{D}^1(\TT^1)$. $Df$ has bounded variation if and only if $\log Df$ has bounded variation. \end{lemma} diff --git a/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex b/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex index e5e949a..5b32484 100644 --- a/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex +++ b/Mathematics/5th/Introduction_to_nonlinear_elliptic_PDEs/Introduction_to_nonlinear_elliptic_PDEs.tex @@ -536,7 +536,7 @@ \end{proposition} \begin{lemma}\label{INEPDE:optimization} - Let $X$ be a Banach space and $\Phi:X\to \RR$ be continuous and convex, then it is weakly sequentially lower semicontinuous, that is, if $u_n\rightharpoonup u$ in $X$, then $\Phi(u)\leq \liminf_{n\to\infty}\Phi(u_n)$. + Let $X$ be a Banach space and $\Phi:X\to \RR$ be continuous and convex, then it is weakly sequentially lower semicontinuous, that is, if $u_n\rightharpoonup u$ in $X$, then $\displaystyle\Phi(u)\leq \liminf_{n\to\infty}\Phi(u_n)$. \end{lemma} \begin{theorem} Let $(X,\norm{\cdot})$ be a reflexive Banach space and $\Phi:X\to \RR$ be continuous, convex and such that $\displaystyle \lim_{\norm{u}\to\infty}\Phi(u)=+\infty$. Then, $\Phi$ has a minimizer. This minimizer is unique if $\Phi$ is strictly convex.