updated probability and functional analysis

Mathematics/3rd/Probability/Probability.tex

@@ -125,10 +125,10 @@
   \begin{proposition}
     Let $(\Omega,\mathcal{A},\Prob)$ be a probability space such that $\Omega$ is finite and all its elements are equiprobable. Let $A\in\mathcal{A}$ be an event. Then: $$\Prob(A)=\frac{|A|}{|\Omega|}$$
   \end{proposition}
-  \begin{theorem}[Continuity from below]
+  \begin{theorem}[Continuity from below]\label{P:continuity_below}
     Let\\ $(\Omega,\mathcal{A},\Prob)$ be a probability space and $(A_n)\subset\mathcal{A}$ be an increasing sequence of events, that is: $$A_1\subseteq A_2\subseteq\cdots\subseteq A_n\subseteq\cdots$$ Let $A:=\bigcup_{n=1}^\infty A_n$. Then: $$\Prob(A)=\lim_{n\to\infty}\Prob(A_n)$$
   \end{theorem}
-  \begin{corollary}[Continuity from above]
+  \begin{corollary}[Continuity from above]\label{P:continuity_above}
     Let $(\Omega,\mathcal{A},\Prob)$ be a probability space and $(A_n)\subset\mathcal{A}$ be a decreasing sequence of events, that is: $$A_1\supseteq A_2\supseteq\cdots\supseteq A_n\supseteq\cdots$$ Let $A:=\bigcap_{n=1}^\infty A_n$. Then: $$\Prob(A)=\lim_{n\to\infty}\Prob(A_n)$$
   \end{corollary}
   \begin{proposition}[Countable subadditivity]
@@ -1035,10 +1035,27 @@
     Let $(\Omega,\mathcal{A},\Prob)$ be a probability space and $(A_n)\subset\mathcal{A}$ be a sequence of events such that: $$\sum_{n=1}^\infty\Prob(A_n)<\infty$$
     Then, $\displaystyle\Prob\left(\limsup_{n\to\infty} A_n\right)=0$.
   \end{lemma}
+  \begin{proof}
+    Let $B_n:=\bigcup_{k\geq n}A_k$ and note that $B_{n+1}\subseteq B_n$. Thus, using the definition of $\limsup$ and \mnameref{P:continuity_above} we have that
+    $$
+      \Prob\left(\limsup_{n\to\infty} A_n\right)=\lim_{n\to\infty}\Prob(B_n)\leq \lim_{n\to\infty}\sum_{k\geq n}\Prob(A_n)=0
+    $$
+    because it is the tail of a convergent sequence.
+  \end{proof}
   \begin{lemma}[Second Borel-Cantelli lemma]
     Let $(\Omega,\mathcal{A},\Prob)$ be a probability space and $(A_n)\subset\mathcal{A}$ be a sequence of independent events such that: $$\sum_{n=1}^\infty\Prob(A_n)=\infty$$
     Then, $\displaystyle\Prob\left(\limsup_{n\to\infty} A_n\right)=1$.
   \end{lemma}
+  \begin{proof}
+    We will prove that $\displaystyle\Prob\left({\left[\limsup_{n\to\infty} A_n\right]}^c\right)=0$. From \mnameref{P:continuity_below}, if $B_n:= \bigcap_{k\geq n}{A_k}^c$ we have:
+    $$
+      \Prob\left({\left[\limsup_{n\to\infty} A_n\right]}^c\right)=\lim_{n\to\infty}\Prob(B_n)
+    $$
+    Now, $\forall N\geq n$ we have $\Prob(B_n)\leq \Prob\left(\bigcap_{k=n}^N{A_n}^c\right)$. Using the independence and the inequality $1+x\leq \exp{x}$, we get:
+    $$
+      \Prob(B_n)\leq \prod_{k=n}^N(1-\Prob(A_n))\leq \exp{-\sum_{k=n}^{N}}\overset{N\to\infty}{\longrightarrow}0
+    $$
+  \end{proof}
   \subsubsection{Convergence in mean}
   \begin{definition}
     Let $(\Omega,\mathcal{A},\Prob)$ be a probability space, $p\geq 1$, $(X_n)$ be a sequence of random variables such that $\Exp({|X_n|}^p)<\infty$ and $X$ be a random variable such that $\Exp({|X|}^p)<\infty$. We say that $(X_n)$ \emph{converges in the $p$-th mean} to $X$, and we denote it by $X_n\overset{L^p}{\longrightarrow} X$, if  $$\lim_{n\to\infty}\Exp({|X_n-X|}^p)=0$$

Mathematics/4th/Harmonic_analysis/Harmonic_analysis.tex

@@ -449,6 +449,19 @@
   \begin{proof}
     The first property follows from its definition. For the second one, if $\displaystyle f(x)\overset{L^2}{=}\lim_{n\to\infty}f_n(x)$, by \mnameref{HA:plancherel} we have $\norm{f_n}_2=\norm{\widehat{f}}_2$. Now use the continuity of the norm. For the other properties, take the function given in the proof of \mcref{HA:preDefFTinL2} and use the \mnameref{RFA:monotone}.
   \end{proof}
+  \begin{proposition}[Jensen's inequality]
+    Let $J:\RR\to\RR$ be a convex function, $f$ be a measurable function, and $\mu:\Omega\to\RR$ be measurable with $\int_\Omega\dd{\mu}=1$. Then:
+    $$
+      \int_\Omega J(f)\dd{\mu}\geq J\left(\int_\Omega f\dd{\mu}\right)
+    $$
+  \end{proposition}
+  \begin{sproof}
+    We assume differentiability on $J$ for simplicity. Since $J$ is convex we have that $\forall a,b\in\RR$:
+    $$J(b)\geq J(a)+J'(a)(b-a)$$
+    Taking $a=\int_\Omega f\dd{\mu}$ and $b=f(x)$, we have:
+    $$J(f(x))\geq J\left(\int_\Omega f\dd{\mu}\right)+J'\left(\int_\Omega f\dd{\mu}\right)\!\!\left(f(x)-\int_\Omega f\dd{\mu}\right)$$
+    Multiplying by $\dd{\mu}$ and integrating, yields the result.
+  \end{sproof}
   \begin{lemma}[Generalized Hölder's inequality]\label{HA:holderGeneralized}
     Let $E\subseteq\RR^n$ be a measurable set, $1\leq p_1,\ldots,p_n\leq \infty$ be such that $\sum_{i=1}^n\frac{1}{p_i}=1$ and ${f_i}\in L^{p_i}(E)$. Then:
     $$\norm{\prod_{i=1}^{n}f_i}_1\leq\prod_{i=1}^{n}\norm{{f_i}}_{p_i}$$

Mathematics/4th/Real_and_functional_analysis/Real_and_functional_analysis.tex

@@ -481,7 +481,7 @@
   \end{proposition}
   \begin{important}
     \begin{theorem}[Dominated convergence theorem]\label{RFA:dominated}
-      Let $E\subseteq\RR^n$ be a measurable set, $f$ be a measurable function over $E$ such that $\exists (f_m)\geq 0$ with $f_m\almoste{\rightarrow} f$ and $\abs{f_m(x)}\almoste\leq g(x)$ on $E$ with $g\in \mathcal{L}^1(E)$ $\forall m\in\NN$. Then, $f, f_m\in \mathcal{L}^1(E)$ $\forall m\in\NN$ and: $$\int_Ef(x)\dd{x}=\lim_{m\to\infty}\int_Ef_m(x)\dd{x}$$
+      Let $E\subseteq\RR^n$ be a measurable set, $f$ be a measurable function over $E$ such that $\exists (f_m)$ measurable with $f_m\almoste{\rightarrow} f$ and $\abs{f_m(x)}\almoste\leq g(x)$ on $E$ with $g\in \mathcal{L}^1(E)$ $\forall m\in\NN$. Then, $f, f_m\in \mathcal{L}^1(E)$ $\forall m\in\NN$ and: $$\int_Ef(x)\dd{x}=\lim_{m\to\infty}\int_Ef_m(x)\dd{x}$$
     \end{theorem}
   \end{important}
   \begin{proposition}
@@ -843,6 +843,20 @@
     $$\int_E\abs{f(x)g(x)}\dd{x}\leq\int_E\left(\frac{\abs{f(x)}^p}{p}+\frac{\abs{g(x)}^q}{q}\right)\dd{x}=\frac{1}{p}+\frac{1}{q}=1$$
     The equality follows from the equality in \mnameref{RFA:young}.
   \end{proof}
+  \begin{corollary}[Hölder's inequality]
+    Let $E\subseteq\RR^n$ be a measurable set, $1\leq p,q, r\leq \infty$ be such that $\frac{1}{p}+\frac{1}{q}=\frac{1}{r}$ and ${f}\in L^p(E)$, ${g}\in L^q(E)$. Then, $fg\in L^r(E)$ and:
+    $$\norm{{fg}}_r\leq\norm{{f}}_p\norm{{g}}_q$$
+  \end{corollary}
+  \begin{sproof}
+    Use \mnameref{RFA:holder} with $F:=\abs{f}^r\in L^{\frac{p}{r}}(E)$ and $G:=\abs{g}^r\in L^{\frac{q}{r}}(E)$, noting that $p/r$ and $q/r$ are Hölder conjugates.
+  \end{sproof}
+  \begin{corollary}[Interpolation inequality]
+    Let $E\subseteq\RR^n$ be a measurable set, $1\leq p_1\leq p_2\leq \infty$ and $f\in L^{p_1}(E)\cap L^{p_2}(E)$. Then, $\forall p\in[p_1,p_2]$ we have $f\in L^p(E)$ and:
+    $$
+      \norm{f}_p\leq {\norm{f}_{p_1}}^\alpha{\norm{f}_{p_2}}^{1-\alpha}
+    $$
+    with $\alpha\in[0,1]$ such that $\frac{1}{p}=\frac{\alpha}{p_1}+\frac{1-\alpha}{p_2}$.
+  \end{corollary}
   \begin{proposition}\label{RFA:lpnorm}
     Let $E\subseteq\RR^n$ be a measurable set and $1\leq p<\infty$. The set $L^p(E)$ is a normed vector space with the norm: $$\norm{{f}}_p:={\left(\int_E\abs{{f}}^p\right)}^{1/p}\qquad\forall{f}\in L^p(E)$$
     And the set $L^\infty(E)$ is also a normed vector space with the norm: $$\norm{{f}}_\infty=\inf \{M:\abs{{f}(x)}\almoste{\leq} M, x\in E\}\qquad\forall{f}\in L^\infty(E)$$