diff --git a/src/cat/tricks/product.tex b/src/cat/tricks/product.tex index 134ab99..1dd7a94 100644 --- a/src/cat/tricks/product.tex +++ b/src/cat/tricks/product.tex @@ -26,3 +26,34 @@ \] \end{proof} + +\begin{lemma} +\label{lemma:power-difference} + Let $p \in [1, \infty)$, then for each $a, b \ge 0$, + \[ + |a^p - b^p| \le p|a - b|(a \vee b)^{p - 1} + \] +\end{lemma} +\begin{proof} + Assume without loss of generality that $0 < a < b$, then + \[ + b^p - a^p = p\int_{a}^b t^{p - 1}dt \le p|a - b|(a \vee b)^{p - 1} + \] +\end{proof} + +\begin{lemma} +\label{lemma:power-sum} + Let $p \in [1, \infty)$, then for each $a, b \ge 0$, + \[ + (a + b)^p \le 2^{p-1}(a^p + b^p) + \] +\end{lemma} +\begin{proof} + Since $t \mapsto t^p$ is convex, + \begin{align*} + \braks{\frac{a + b}{2}}^p &\le \frac{a^p}{2} + \frac{b^p}{2} \\ + \frac{(a+b)^p}{2^p} &\le \frac{a^p}{2} + \frac{b^p}{2} \\ + (a + b)^p &\le 2^{-1}(a^p + b^p) + \end{align*} + +\end{proof} diff --git a/src/fa/lp/definition.tex b/src/fa/lp/definition.tex index 6b82ebc..c0e5922 100644 --- a/src/fa/lp/definition.tex +++ b/src/fa/lp/definition.tex @@ -169,36 +169,6 @@ \end{proof} - -\begin{theorem}[Markov's Inequality] -\label{theorem:markov-inequality} - Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then - \begin{enumerate} - \item For any $\alpha > 0$, - \[ - \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)} - \] - \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$, - \[ - \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)} - \] - \item For any $\alpha > 0$, - \[ - \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p - \] - \end{enumerate} -\end{theorem} -\begin{proof} - (1): For any $\alpha > 0$, - \begin{align*} - \mu\bracs{\norm{f}_E \ge \alpha} &= \int_{\bracs{\norm{f}_E \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |\norm{f}_Ed\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha} - \end{align*} - - (2): By \autoref{lemma:monotone-measurable}, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{\norm{f}_E \ge \alpha} = \bracs{\phi \circ \norm{f}_E \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ \norm{f}_E$. - - (3): Since $x \mapsto x^p$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^p$ yields the desired result. -\end{proof} - \begin{proposition} \label{proposition:lp-in-measure} Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty]$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$ such that $f_n \to f$ in $L^p$, then $f_n \to f$ in measure. @@ -244,3 +214,5 @@ \end{proof} + + diff --git a/src/fa/lp/index.tex b/src/fa/lp/index.tex index f1937f6..a13e9d9 100644 --- a/src/fa/lp/index.tex +++ b/src/fa/lp/index.tex @@ -2,5 +2,7 @@ \label{chap:lp} \input{./definition.tex} +\input{./ineq.tex} \input{./duality.tex} +\input{./ui.tex} \input{./seq.tex} diff --git a/src/fa/lp/ineq.tex b/src/fa/lp/ineq.tex new file mode 100644 index 0000000..3f9ab6f --- /dev/null +++ b/src/fa/lp/ineq.tex @@ -0,0 +1,50 @@ +\section{$L^p$ Inequalities} +\label{section:lp-inequalities} + + +\begin{theorem}[Markov's Inequality] +\label{theorem:markov-inequality} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then + \begin{enumerate} + \item For any $\alpha > 0$, + \[ + \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha}\norm{f}_{L^1(X; E)} + \] + \item For $\alpha > 0$ and strictly increasing function $\phi: [0, \infty) \to [0, \infty)$, + \[ + \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\phi(\alpha)}\norm{\phi \circ |f|}_{L^1(X; E)} + \] + \item For any $\alpha > 0$, + \[ + \mu\bracs{\norm{f}_E \ge \alpha} \le \frac{1}{\alpha^p}\norm{f}_{L^p(X; E)}^p + \] + \end{enumerate} +\end{theorem} +\begin{proof} + (1): For any $\alpha > 0$, + \begin{align*} + \mu\bracs{\norm{f}_E \ge \alpha} &= \int_{\bracs{\norm{f}_E \ge \alpha}} d\mu \le \frac{1}{\alpha}\int |\norm{f}_Ed\mu = \frac{\norm{f}_{L^1(X; E)}}{\alpha} + \end{align*} + + (2): By \autoref{lemma:monotone-measurable}, $\phi: [0, \infty) \to [0, \infty)$ is Borel measurable. Since $\phi$ is strictly increasing, $\bracs{\norm{f}_E \ge \alpha} = \bracs{\phi \circ \norm{f}_E \ge \phi(\alpha)}$, and the result holds by applying (1) to $\phi \circ \norm{f}_E$. + + (3): Since $x \mapsto x^p$ is strictly increasing on $[0, \infty)$, applying (2) to $\phi(x) = x^p$ yields the desired result. +\end{proof} + + +\begin{lemma} +\label{lemma:lp-p-diff} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty)$, and $f, g \in L^p(X; E)$, then + \[ + |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| \le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1} + \] +\end{lemma} +\begin{proof} + By \autoref{lemma:power-difference}, + \begin{align*} + |\norm{f}_{L^p(X; E)}^p - \norm{g}_{L^p(X; E)}^p| &\le p|\norm{f}_{L^p(X; E)} - \norm{g}_{L^p(X; E)}|\\ + &\times (\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1} \\ + &\le p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1} + \end{align*} + +\end{proof} diff --git a/src/fa/lp/ui.tex b/src/fa/lp/ui.tex new file mode 100644 index 0000000..bff6d89 --- /dev/null +++ b/src/fa/lp/ui.tex @@ -0,0 +1,140 @@ +\chapter{Uniform Integrability} +\label{chap:uniform-integrable} + +\begin{definition}[Uniform Integrability] +\label{definition:uniform-integrable} + Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\cf \subset L^p(X; E)$, then $\cf$ is \textbf{uniformly $p$-integrable} if + \[ + \lim_{M \to \infty}\sup_{f \in \cf} \int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu = 0 + \] +\end{definition} + +\begin{proposition} +\label{proposition:ui-gymnastics} + Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, $\cf \subset L^p(X; K)$, and let $|\cf|^p = \bracsn{\norm{f}^p| f \in \cf}$, then: + \begin{enumerate} + \item If $\cf$ is uniformly $p$-integrable, then $|\cf|^p$ is uniformly absolutely continuous with respect to $\mu$. + \item If $\sup_{f \in \cf}\norm{f}_{L^p(X; E)} < \infty$ and $|\cf|^p$ is uniformly absolutely continuous with respect to $\mu$, then $\cf$ is uniformly $p$-integrable. + \end{enumerate} +\end{proposition} +\begin{proof} + (1): Let $\eps > 0$, then there exists $M \ge 0$ such that + \[ + \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu < \eps/2 + \] + + Thus for any $A \in \cm$, + \begin{align*} + \sup_{f \in \cf}\int_{A}\norm{f}^p d\mu &\le \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \le M} \cap A} \norm{f}^p d\mu+ \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M} \cap A} \norm{f}^p d\mu \\ + &\le M \mu(A) + \eps/2 + \end{align*} + + so if $\mu(A) \le M\eps/2$, then $\sup_{f \in \cf}\int_{A}\norm{f}^p d\mu \le \eps$. + + (2): Let $\eps > 0$, then there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\sup_{f \in \cf}\int_A \norm{f}^p d\mu < \eps$. By \hyperref[Markov's inequality]{theorem:markov-inequality}, + \[ + \sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) \le \frac{\sup_{f \in \cf}\norm{f}_{L^p(X; E)}^p}{M^p} + \] + + Thus for sufficiently large $M$, $\sup_{f \in \cf}\mu(\bracs{\norm{f}_E \ge M}) < \delta$, and + \[ + \sup_{f \in \cf}\int_{\bracs{\norm{f}_E \ge M}} \norm{f}^p d\mu < \eps + \] +\end{proof} + + + +\begin{theorem}[Vitali Convergence Theorem] +\label{theorem:vitali-convergence} + Let $(X, \cm, \mu)$ be a measure space, $p \in [1, \infty)$, $E$ be a normed vector space over $K \in \RC$, and $\fF \subset 2^{L^p(X; E)}$ be a filter, then $\fF$ is Cauchy in $L^p(X; E)$ if and only if: + \begin{enumerate} + \item[(M)] $\fF$ is Cauchy in measure. + \item[(UI)] For each $\eps > 0$, there exists $M \ge 0$ and $F \in \fF$ such that + \[ + \sup_{f \in F}\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu < \eps + \] + \item[(T)] For each $\eps > 0$, there exists $A \in \cm$ and $F \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F}\int_{A^c}\norm{f}_E^p < \eps$. + \end{enumerate} + + % M stands for measure, UI stands for Uniform Integrability, T stands for tightness. +\end{theorem} +\begin{proof} + ($L^p$) $\Rightarrow$ (M): By \hyperref[Markov's inequality]{theorem:markov-inequality}. + + ($L^p$) $\Rightarrow$ (UI): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)} < \eps$. Fix $f \in F$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $M > 0$ such that $\int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^p d\mu < \eps$. For any $g \in F$, + \[ + \bracs{\norm{g}_E \ge 2M} \subset \bracs{\norm{f}_E \ge M} \cup \bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M} + \] + + so by \autoref{lemma:lp-p-diff}, + \begin{align*} + \int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_E^pd\mu &\le \int_{\bracs{\norm{f}_E \ge M}}\norm{g}_E^pd\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{g}_E^pd\mu \\ + &\le \int_{\bracs{\norm{f}_E \ge M}}\norm{f}_E^pd\mu + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_E^pd\mu \\ + &+ 2p\norm{f - g}_{L^p(X; E)}(\norm{f}_{L^p(X; E)} \vee \norm{g}_{L^p(X; E)})^{p-1} + \end{align*} + + Since $\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M} \subset \bracs{\norm{f - g}_E \ge M}$, by \hyperref[Markov's inequality]{theorem:markov-inequality}, + \[ + \int_{\bracs{\norm{g}_E \ge 2M, \norm{f}_E \le M}}\norm{f}_E^pd\mu \le M^p\mu\bracs{\norm{f - g}_E \ge M} \le \norm{f - g}_{L^p(X; E)}^p + \] + + Therefore + \[ + \sup_{g \in F}\int_{\bracs{\norm{g}_E \ge 2M}}\norm{g}_E^p d\mu \le 2p \eps (\norm{f}_{L^p(X; E)} + \eps)^{p - 1} + \eps + \eps^p + \] + + ($L^p$) $\Rightarrow$ (T): Let $F \in \fF$ such that for each $f, g \in F$, $\norm{f - g}_{L^p(X; E)} < \eps$. Fix $f \in F$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, there exists $A \in \cm$ such that $\mu(A) < \infty$ and $\norm{\one_{A^c}f}_{L^p(X; E)} < \eps$. In which case, for any $g \in F$, + \[ + \norm{\one_{A^c}g}_{L^p(X; E)} \le \norm{\one_{A^c}f}_{L^p(X; E)} + \norm{f - g}_{L^p(X; E)} + \le \norm{\one_{A^c}f}_{L^p(X; E)} + \eps + \] + + + + (M) + (UI) + (T) $\Rightarrow$ ($L^p$): Let $\eps > 0$. By (T), there exists $A \in \cm$ and $F_1 \in \fF$ with $\mu(A) < \infty$ and $\sup_{f \in F_1}\int_{A^c}\norm{f}_E^p < \eps^p$. Thus for every $f, g \in F_1$, + \begin{align*} + \norm{f - g}_{L^p(X; E)} &\le \norm{\one_A(f - g)}_{L^p(X; E)} + \norm{\one_{A^c}f}_{L^p(X; E)} + \norm{\one_{A^c}g}_{L^p(X; E)} \\ + &\le \norm{\one_A(f - g)}_{L^p(X; E)} + 2\eps + \end{align*} + + By (UI), there exists $M > 0$ and $F_2 \in \fF$ with $F_2 \subset F_1$ such that + \[ + \sup_{h \in F_2} \int_{\bracs{\norm{h}_E \ge M}}\norm{h}_E^p < \eps^p + \] + + Assume without loss of generality that $\mu(A) > 0$ and let $\delta = \eps\mu(A)^{-1/p}$. By (M), there exists $F_3 \in \fF$ with $F_3 \subset F_2$, such that for any $f, g \in F_3$, + \[ + \mu\bracsn{\norm{f - g}_E \ge \delta} \le \paren{\frac{\eps}{2M}}^p + \] + + In which case, + \begin{align*} + \norm{\one_{A}(f - g)}_{L^p(X; E)} &\le \normn{\one_{A \cap \bracs{\norm{f - g}_E \le \delta}}(f - g)}_{L^p(X; E)} \\ + &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} \\ + &\le \delta\mu(A)^{1/p} + \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} \\ + &\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} + \eps + \end{align*} + + then for any $f, g \in F_3$, + \begin{align*} + \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} &\le \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)} \\ + &+ \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)} + \end{align*} + + Now, + \begin{align*} + \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}f}_{L^p(X; E)} &\le \normn{\one_{\bracsn{\norm{f}_E \ge M}}f}_{L^p(X; E)} \\ + &+ \normn{\one_{\bracsn{\norm{f - g}_E \ge \delta, \norm{f}_E \le M}}f}_{L^p(X; E)} \\ + &\le \eps + \eps/2 = 3\eps/2 + \end{align*} + + Similarly, $\normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}g}_{L^p(X; E)} \le 3\eps/2$. Thus + \[ + \normn{\one_{A \cap \bracsn{\norm{f - g}_E \ge \delta}}(f - g)}_{L^p(X; E)} \le 3\eps + \] + + + Therefore for any $f, g \in F_3$, $\norm{f - g}_{L^p(X; E)} \le 6 \eps$, so $\fF$ is Cauchy in $L^p(X; E)$. + +\end{proof} + diff --git a/src/measure/measurable-maps/in-measure.tex b/src/measure/measurable-maps/in-measure.tex index df9f96a..5cc8d1b 100644 --- a/src/measure/measurable-maps/in-measure.tex +++ b/src/measure/measurable-maps/in-measure.tex @@ -3,18 +3,15 @@ \begin{definition}[Convergence in Measure] \label{definition:convergence-in-measure} - Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{in measure} if for every $\eps > 0$, + Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, and $f$ be a $(\cm, \cb_Y)$-measurable function, then $\fF \to f$ \textbf{in measure} if for each $\eps > 0$, \[ - \lim_{n \to \infty}\mu(\bracs{d(f_n, f) > \eps}) = 0 + \lim_{g, \fF}\mu(\bracs{d(f, g) > \eps}) = 0 \] \end{definition} \begin{definition}[Cauchy in Measure] \label{definition:cauchy-in-measure} - Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is \textbf{Cauchy in measure} if for every $\eps > 0$, - \[ - \mu(\bracs{d(f_m, f_n) > \eps}) \to 0 \quad \text{as}\ m, n \to \infty - \] + Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $A \in \fF$ such that $\mu(\bracs{d(f, g) > \delta}) < \eps$ for all $f, g \in A$. \end{definition} \begin{lemma} diff --git a/src/measure/vector/ac.tex b/src/measure/vector/ac.tex index 53cc99c..f487b6e 100644 --- a/src/measure/vector/ac.tex +++ b/src/measure/vector/ac.tex @@ -38,6 +38,8 @@ \begin{definition}[Uniformly Absolutely Continuous] \label{definition:u-ac} Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\mathcal{U} \subset M(X, \cm; E)$ be a family of $E$-valued vector measures on $(X, \cm)$, then $\mathcal{U}$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if for every $\eps > 0$, there exists $\delta > 0$ such that for all $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_E < \eps$ for all $\nu \in \mathcal{U}$. + + For any family $\cf$ of measurable functions, $\cf$ is \textbf{uniformly absolutely continuous} with respect to $\mu$ if the measures $\mathcal{U} = \bracs{fd\mu|f \in \cf}$ are uniformly absolutely continuous with respect to $\mu$. \end{definition} \begin{theorem}[Vitali-Hahn-Saks]