updated advanced probability and analysis

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}

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}

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 $1<p<\infty$, it is reflexive. Finally, $H^m(\Omega)$ is a separable Hilbert space.
+  \end{theorem}
+  \begin{definition}
+    Let $\Omega\subseteq \RR^d$ be an open set, $m\in\NN$ and $1\leq p\leq \infty$. We define the space $W_0^{m,p}(\Omega):=\overline{\mathcal{C}_0^\infty(\Omega)}$, where the closure is taken with the norm of $W^{m,p}(\Omega)$. Similarly, we set $H_0^m(\Omega):=W_0^{m,2}(\Omega)$.
+  \end{definition}
+  \begin{remark}
+    Note that $W_0^{m,p}(\Omega)$ is also Banach (with the same norm as $W^{m,p}(\Omega)$) because it is a closed subspace of a Banach space.
+  \end{remark}
+  \begin{theorem}
+    For $1\leq p < \infty$ we have that $W_0^{m,p}(\RR^d)=W^{m,p}(\RR^d)$. In particular, $\mathcal{C}^\infty(\RR^d)\cap W^{m,p}(\RR^d)$ and $\mathcal{C}_0^{\infty}(\RR^d)$ are dense in $W^{m,p}(\RR^d)$ for $1\leq p<\infty$.
+  \end{theorem}
+  \begin{theorem}[Poincaré's inequality]\label{ATFAPDE:poincare_ineq}
+    Let $\Omega$ be a bounded open set in $\RR^d$ and let $1\leq p<\infty$. Then, there is a constant $C=C(\Omega,p)$ so that $\forall u\in W_0^{m,p}(\Omega)$ we have:
+    $$
+      \norm{u}_p\leq C\norm{\grad u}_p
+    $$
+  \end{theorem}
+  \begin{remark}
+    \mnameref{ATFAPDE:poincare_ineq} is also valid when $\Omega$ is unbounded in one direction.
+  \end{remark}
+  \begin{corollary}
+    If $\Omega$ is bounded, then the constant function $f(x)=C$ with $C\ne 0$ is not in $W_0^{1,p}$. Thus, we cannot approximate constant functions by $\mathcal{C}_0^\infty(\Omega)$ functions, with the $W^{1,p}(\Omega)$ norm.
+  \end{corollary}
+  \begin{definition}
+    Let $\Omega$ be a bounded set. We define the average of $u$ in $\Omega$ as:
+    $$
+      \fint_\Omega u:=\frac{1}{\abs{\Omega}}\int_\Omega u
+    $$
+  \end{definition}
+  \begin{theorem}[Poincaré-Wirtinger's inequality]
+    Let $\Omega\subseteq \RR^d$ be a bounded connected open set with $\mathcal{C}^1$ boundary, and let $1\leq p<\infty$. Then, there is a constant $C=C(\Omega,p)$ so that $\forall u\in W^{1,p}(\Omega)$ we have:
+    $$
+      \norm{u-\fint_\Omega u}_p\leq C\norm{\grad u}_p
+    $$
+  \end{theorem}
+\end{multicols}
+\end{document}

preamble_formulas.sty

@@ -89,6 +89,8 @@
       {PDE}{\apl}   % partial differential equations
       {RFA}{\ana}   % real and functional analysis
       {SP}{\sta}    % stochastic processes
+      {ATFAPDE}{\ana} % advanced topics in functional analysis and pdes
+      {AP}{\sta}    % advanced probability
   }{\col}%
 }
 \ExplSyntaxOff