From 17954f0c63499557eb93c162e300c3d86782a982 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?V=C3=ADctor?= Date: Sun, 10 Sep 2023 19:20:08 +0200 Subject: [PATCH] updated advanced probability and analysis --- .../Real_and_functional_analysis.tex | 8 +- .../Advanced_probability.tex | 164 ++++++++++++++++++ ...topics_in_functional_analysis_and_PDEs.tex | 55 ++++++ preamble_formulas.sty | 2 + 4 files changed, 225 insertions(+), 4 deletions(-) create mode 100644 Mathematics/5th/Advanced_probability/Advanced_probability.tex create mode 100644 Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex diff --git a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex index 3f39133..dc87479 100644 --- a/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex +++ b/Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex @@ -1087,7 +1087,7 @@ \begin{theorem}[Marcinkiewicz interpolation theorem] Let $T$ be a sublinear operator. Then: \begin{enumerate} - \item $\displaystyle\m{\{x\in\RR^n:\abs{Tf(x)}>t\}}\leq \frac{A}{t}\norm{f}_1$ + \item $\displaystyle\abs{\{x\in\RR^n:\abs{Tf(x)}>t\}}\leq \frac{A}{t}\norm{f}_1$ \item $\norm{Tf}_\infty\leq A_\infty\norm{f}_\infty$ \end{enumerate} \end{theorem} @@ -1188,13 +1188,13 @@ Let $X,Y\subseteq \RR^n$ be measurable spaces and $K\in L^2(X\times Y)$. We define the \emph{Hilbert-Schmidt operator with kernel $K$} as the operator $T:L^2(Y)\rightarrow L^2(X)$ defined by: $$Tf(x)\almoste{=}\int_YK(x,y)f(y)\dd{y}$$ \end{definition} \begin{proposition}\label{RFA:fredholm} - Let $X$, $Y$ be compact metric spaces and $K\in\mathcal{C}(X\times Y)$. The Fredholm operator $T$ with kernel $K$ is compact and satisfies $\norm{T}\leq\norm{K}_{X\times Y}\m{Y}$ + Let $X$, $Y$ be compact metric spaces and $K\in\mathcal{C}(X\times Y)$. The Fredholm operator $T$ with kernel $K$ is compact and satisfies $\norm{T}\leq\norm{K}_{X\times Y}\abs{Y}$ \end{proposition} \begin{sproof} It is a direct application of \mnameref{RFA:arzela}. The proof of the equicontinuity follows from the inequality - $$\abs{Tf(a)-Tf(b)}\leq \norm{f}\sup_{y\in Y}\{\abs{K(a,y)-K(b,y)}\}\m{Y}$$ + $$\abs{Tf(a)-Tf(b)}\leq \norm{f}\sup_{y\in Y}\{\abs{K(a,y)-K(b,y)}\}\abs{Y}$$ and the fact that $\norm{f}\leq 1$ and that $K$ is uniformly continuous. The pointwise boundedness follows from: - $$\abs{Tf(a)}\leq\sup_{(x,y)\in X\times Y}\{\abs{K(x,y)}\}\m{Y}$$ + $$\abs{Tf(a)}\leq\sup_{(x,y)\in X\times Y}\{\abs{K(x,y)}\}\abs{Y}$$ because $\norm{f}\leq 1$. And from here the inequality of the norm is clear. \end{sproof} diff --git a/Mathematics/5th/Advanced_probability/Advanced_probability.tex b/Mathematics/5th/Advanced_probability/Advanced_probability.tex new file mode 100644 index 0000000..dda4a32 --- /dev/null +++ b/Mathematics/5th/Advanced_probability/Advanced_probability.tex @@ -0,0 +1,164 @@ +\documentclass[../../../main_math.tex]{subfiles} + +\begin{document} +\changecolor{AP} +\begin{multicols}{2}[\section{Advanced probability}] + These summaries aims to review the basic notions of probability theory in a more abstract setting. We will not prove any result here as most of them are from previous courses. + \subsection{Basics of measure theory and integration} + \begin{definition}[$\sigma$-algebra] + Let $E$ be a set. A \emph{$\sigma$-algebra} $\mathcal{E}$ on $E$ is a collection of subsets of $E$ such that: + \begin{enumerate} + \item $\varnothing\in\mathcal{E}$. + \item $\forall A\in\mathcal{E}$, $A^c\in\mathcal{E}$. + \item $\forall (A_n)_{n\in\NN}\subseteq \mathcal{E}$, $\bigcup_{n\in\NN}{A_n}\in\mathcal{E}$. + \end{enumerate} + The pair $(E,\mathcal{E})$ is called a \emph{measurable space}. + \end{definition} + \begin{definition} + Let $E$ be a set and $\mathcal{F}$ be a collection of subsets of $E$. The \emph{$\sigma$-algebra generated by $\mathcal{F}$} is the smallest $\sigma$-algebra containing $\mathcal{F}$, i.e.: + $$ + \sigma(\mathcal{F}):=\bigcap_{\substack{\mathcal{E}\text{ is a }\sigma\text{-algebra}\\\mathcal{F}\subseteq \mathcal{E}}}{\mathcal{E}} + $$ + \end{definition} + \begin{definition} + Let $(E,\mathcal{E})$, $(F,\mathcal{F})$ be measurable spaces. A function $f:E\to F$ is said to be \emph{measurable} if $\forall A\in\mathcal{F}$, $f^{-1}(A)\in\mathcal{E}$. + \end{definition} + \begin{definition}[Measure] + Let $(E,\mathcal{E})$ be a measurable space. A function $\mu:\mathcal{E}\to [0,\infty]$ is said to be a \emph{measure} if: + \begin{enumerate} + \item $\mu(\varnothing)=0$. + \item $\mu$ is $\sigma$-additive, i.e. $\forall {(A_n)}_{n\in\NN}\subseteq \mathcal{E}$ pairwise disjoint, we have: + $$ + \mu\left(\bigcup_{n\in\NN}{A_n}\right)=\sum_{n\in\NN}{\mu(A_n)} + $$ + \end{enumerate} + The triple $(E,\mathcal{E},\mu)$ is called a \emph{measurable space}. + \end{definition} + \begin{definition} + Let $(E,\mathcal{E},\mu)$ be a measurable space and $f:E\to [0,\infty]$ be a measurable function. We define the \emph{integral of $f$ with respect to $\mu$} as: + $$ + \int_E{f\dd{\mu}}:=\sup\left\{\int_E{g\dd{\mu}}:g\leq f, g\text{ simple}\right\} + $$ + \end{definition} + \begin{definition} + Let $(E,\mathcal{E},\mu)$ be a measurable space and $f:E\to \RR$ be a measurable function. Suppose that $\int_E{\abs{f}\dd{\mu}}<\infty$. Then, we define the \emph{integral of $f$ with respect to $\mu$} as: + $$ + \int_E{f\dd{\mu}}:=\int_E{f^+\dd{\mu}}-\int_E{f^-\dd{\mu}} + $$ + \end{definition} + \begin{theorem}[Monotone convergence theorem] + Let $(E,\mathcal{E},\mu)$ be a measurable space and ${(f_n)}_{n\in\NN}$ be a sequence of measurable functions $f_n:E\to [0,\infty]$ such that $\forall n\in\NN$, $f_n\leq f_{n+1}$. Then: + $$ + \int_E{\lim_{n\to\infty}{f_n}\dd{\mu}}=\lim_{n\to\infty}{\int_E{f_n\dd{\mu}}} + $$ + \end{theorem} + \begin{theorem}[Fatou's lemma] + Let $(E,\mathcal{E},\mu)$ be a measurable space and ${(f_n)}_{n\in\NN}$ be a sequence of measurable functions $f_n:E\to [0,\infty]$. Then: + $$ + \int_E{\liminf_{n\to\infty}{f_n}\dd{\mu}}\leq \liminf_{n\to\infty}{\int_E{f_n\dd{\mu}}} + $$ + \end{theorem} + \begin{theorem}[Dominated convergence theorem] + Let $(E,\mathcal{E},\mu)$ be a measurable space and ${(f_n)}_{n\in\NN}$ be a sequence of measurable functions $f_n:E\to \RR$ such that $\forall n\in\NN$, $\abs{f_n}\leq g$ for some $g:E\to [0,\infty]$ integrable. Then: + $$ + \int_E{\lim_{n\to\infty}{f_n}\dd{\mu}}=\lim_{n\to\infty}{\int_E{f_n\dd{\mu}}} + $$ + \end{theorem} + \begin{proposition} + Let $(E,\mathcal{E},\mu)$ be a measurable space and $f:E\times I\rightarrow \RR$ be a measurable function, where $I\subseteq \RR$ is an interval. Assume that $\forall \lambda\in I$, $f(\cdot,\lambda)$ is integrable and that for some $k\in \NN\cup\{0\}$ and $\forall x\in E$ we have $f(x,\cdot)\in \mathcal{C}^k(I)$ and $\abs{\partial_\lambda^k f(x,\lambda)}\leq g(x)$ for some $g:E\to [0,\infty]$ integrable. Then, the function $F:I\to \RR$ defined by: + $$ + F(\lambda):=\int_E{f(x,\lambda)\dd{\mu(x)}} + $$ + is in $\mathcal{C}^k(I)$ and $\forall j\in\{0,\ldots,k\}$ we have: + $$ + \partial_\lambda^j F(\lambda)=\int_E{\partial_\lambda^j f(x,\lambda)\dd{\mu(x)}} + $$ + \end{proposition} + \begin{definition}[Product measure] + Let $(E,\mathcal{E},\mu)$ and $(F,\mathcal{F},\nu)$ be two measurable spaces. We define the \emph{product measure} $\mu\times\nu$ on $(E\times F,\mathcal{E}\otimes\mathcal{F})$ as: + $$ + \forall A\in\mathcal{E}, B\in\mathcal{F}, \quad \mu\times\nu(A\times B):=\mu(A)\nu(B) + $$ + \end{definition} + \begin{definition} + Let $(E,\mathcal{E},\mu)$ be a measurable space. We say that $\mu$ is \emph{$\sigma$-finite} if there exists a sequence ${(E_n)}_{n\in\NN}\subseteq \mathcal{E}$ such that $\forall n\in\NN$, $\mu(E_n)<\infty$ and $\bigcup_{n\in\NN}{E_n}=E$. + \end{definition} + \begin{theorem}[Fubini] + Let $(E,\mathcal{E},\mu)$ and $(F,\mathcal{F},\nu)$ be two $\sigma$-finite measurable spaces and $f:E\times F\to \RR$ be a measurable function. Then, the following are equivalent: + \begin{enumerate} + \item $f$ is integrable with respect to $\mu\times\nu$. + \item $\displaystyle\int_E{\left(\int_F{\abs{f(x,y)}\dd{\nu(y)}}\right)\dd{\mu(x)}}<\infty$. + \item $\displaystyle\int_F{\left(\int_E{\abs{f(x,y)}\dd{\mu(x)}}\right)\dd{\nu(y)}}<\infty$. + \end{enumerate} + And if any of the above holds, then: + \begin{align*} + \int_{E\times F}{f\dd{(\mu\times\nu)}} & =\int_E{\left(\int_F{f(x,y)\dd{\nu(y)}}\right)\dd{\mu(x)}} \\ + & =\int_F{\left(\int_E{f(x,y)\dd{\mu(x)}}\right)\dd{\nu(y)}} + \end{align*} + \end{theorem} + \begin{definition} + Let $(E,\mathcal{E},\mu)$ be a measurable space and $f:E\to [0,\infty]$ be a measurable function. We define the measure $\nu$ on $(E,\mathcal{E})$ as $\forall A\in\mathcal{E}$: + $$ + \nu(A):=\int_A{f\dd{\mu}} + $$ + In that case, we say that $f$ is the \emph{density} of $\nu$ with respect to $\mu$, also denoted by $\dv{\nu}{\mu}=f$. + \end{definition} + \begin{definition} + Let $(E,\mathcal{E},\mu)$ be a measurable space. A measure $\nu$ on $(E,\mathcal{E})$ is said to be \emph{absolutely continuous} with respect to $\mu$ if $\forall A\in\mathcal{E}$ such that $\mu(A)=0$, we have $\nu(A)=0$. + \end{definition} + \begin{theorem}[Radon-Nikodym] + Let $\mu$, $\nu$ be two $\sigma$-finite on a measurable space $(E,\mathcal{E})$ such that $\nu$ is absolutely continuous with respect to $\mu$. Then, $\nu$ admits a density $f$ with respect to $\mu$. + \end{theorem} + \subsection{Probability spaces and random variables} + \begin{definition} + A \emph{probability space} is a triple $(\Omega,\mathcal{F},\Prob)$ where $\Omega$ is a set, $\mathcal{F}$ is a $\sigma$-algebra on $\Omega$ and $\Prob$ is a measure on $(\Omega,\mathcal{F})$ such that $\Prob(\Omega)=1$. In this context, the elements if $\mathcal{F}$ are called \emph{events}. + \end{definition} + \begin{definition}[Random variable] + Let $(\Omega,\mathcal{F},\Prob)$ be a probability space and $(E,\mathcal{E})$ be a measurable space. An \emph{$E$-valued random variable} is a measurable function from $(\Omega,\mathcal{F})$ to $(E,\mathcal{E})$\footnote{When $E$ is not specified, we will assume that $E=\RR$.}. + \end{definition} + \begin{definition}[Expectation] + Let $(\Omega,\mathcal{F},\Prob)$ be a probability space and $X$ be a random variable. We define the \emph{expectation of $X$} as: + $$ + \Exp(X):=\int_\Omega{X\dd{\Prob}} + $$ + \end{definition} + \begin{definition} + Let $(\Omega,\mathcal{F},\Prob)$ be a probability space, $(E,\mathcal{E})$ be a measurable space and $X$ be a $E$-valued random variable. We define the \emph{law of $X$} as the measure image on $E$, defined for all $A\in\mathcal{E}$ as: + $$ + \mathcal{L}^X(A):=\Prob\circ X^{-1}(A)=\Prob(X\in A) + $$ + \end{definition} + \begin{proposition} + Let $(\Omega,\mathcal{F},\Prob)$ be a probability space, $X$ be a random variable and $h:\RR\to \RR$ be a measurable function such that $h(X)$ is integrable. Then: + $$ + \Exp(h(X))=\int_\RR{h(x)\dd{\mathcal{L}^X(x)}} + $$ + In particular, if the law of $X$ admits a density $f$ with respect to the Lebesgue measure, then: + $$ + \Exp(h(X))=\int_\RR{h(x)f(x)\dd{x}} + $$ + \end{proposition} + \begin{definition} + Let $(\Omega,\mathcal{F},\Prob)$ be a probability space and $X$ be a random variable. We define the \emph{$\sigma$-algebra generated by $X$} as the smallest $\sigma$-algebra containing $X$, i.e.: + $$ + \sigma(X):=\sigma(X^{-1}(A):A\in\mathcal{E}) + $$ + \end{definition} + \begin{proposition} + Let $X$ be a $(E,\mathcal{E})$-valued random variable and $Y$ be a $\sigma(X)$-measurable random variable. Then, there exists a measurable function $f:E\to \RR$ such that $Y=f(X)$. + \end{proposition} + \begin{proposition}[Jensen's inequality] + Let $(\Omega,\mathcal{F},\Prob)$ be a probability space, $X$ be a random variable and $h:\RR\to \RR$ be a convex function. Then: + $$ + h(\Exp(X))\leq \Exp(h(X)) + $$ + as long as the expectations are well-defined. + \end{proposition} + \begin{proposition} + Let $(\Omega,\mathcal{F},\Prob)$ be a probability space, $X$ be a random variable and $h:\RR\to \RR$ be a non-decreasing positive function. Then: + $$ + \Prob(X\geq t)\leq \frac{\Exp(h(X))}{h(t)} + $$ + \end{proposition} +\end{multicols} +\end{document} \ No newline at end of file diff --git a/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex b/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex new file mode 100644 index 0000000..4e87267 --- /dev/null +++ b/Mathematics/5th/Advanced_topics_in_functional_analysis_and_PDEs/Advanced_topics_in_functional_analysis_and_PDEs.tex @@ -0,0 +1,55 @@ +\documentclass[../../../main_math.tex]{subfiles} + +\begin{document} +\changecolor{ATFAPDE} +\begin{multicols}{2}[\section{Advanced topics in functional analysis and PDEs}] + \subsection{Sobolev spaces} + \begin{definition}[Sobolev spaces] + Let $\Omega\subseteq \RR^d$ be an open set, $m\in\NN$ and $1\leq p\leq \infty$. We define the \emph{Sobolev spaces} $W^{m,p}$ as: + $$ + W^{m,p}(\Omega)\!:=\!\{f\in L^p(\Omega): \forall\alpha\in\NN^d, \abs{\alpha}\leq m, \partial^\alpha f\in L^p(\Omega)\} + $$ + Moreover we define the associate norm $\norm{\cdot}_{W^{m,p}(\Omega)}$ as: + $$ + \norm{f}_{W^{m,p}(\Omega)}:={\left(\sum_{\abs{\alpha}\leq m}{\norm{\partial^\alpha f}_p}^p\right)}^{1/p} + $$ + If $p=2$, we denote $H^m(\Omega):=W^{m,2}(\Omega)$. + \end{definition} + \begin{theorem} + Let $\Omega\subseteq \RR^d$ be an open set. Then, for all $m\in\NN$ and all $1\leq p\leq \infty$, $(W^{m,p}(\Omega),\norm{\cdot}_{W^{m,p}(\Omega)}$ is Banach. Moreover, if $p<\infty$, it is separable and if $1