From b16666e74eaf4fd544bff68efde7f202a148fac3 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 26 Jan 2026 17:47:10 -0500 Subject: [PATCH] Added basic theory of Lp spaces, alongside some integral stuff. --- src/cat/tricks/index.tex | 1 + src/cat/tricks/product.tex | 25 +++++ src/fa/index.tex | 1 + src/fa/lp/definition.tex | 106 ++++++++++++++++++++++ src/fa/lp/index.tex | 4 + src/measure/bochner-integral/index.tex | 4 + src/measure/bochner-integral/strongly.tex | 27 ++++++ src/measure/index.tex | 1 + src/measure/measure/radon.tex | 12 ++- 9 files changed, 180 insertions(+), 1 deletion(-) create mode 100644 src/cat/tricks/product.tex create mode 100644 src/fa/lp/definition.tex create mode 100644 src/fa/lp/index.tex create mode 100644 src/measure/bochner-integral/index.tex create mode 100644 src/measure/bochner-integral/strongly.tex diff --git a/src/cat/tricks/index.tex b/src/cat/tricks/index.tex index 93248ad..e4a04fd 100644 --- a/src/cat/tricks/index.tex +++ b/src/cat/tricks/index.tex @@ -2,3 +2,4 @@ \label{chap:tricks} \input{./src/cat/tricks/dyadic.tex} +\input{./src/cat/tricks/product.tex} diff --git a/src/cat/tricks/product.tex b/src/cat/tricks/product.tex new file mode 100644 index 0000000..0615593 --- /dev/null +++ b/src/cat/tricks/product.tex @@ -0,0 +1,25 @@ +\section{Product Inequalities} +\label{section:product-inequalities} + +\begin{lemma}[Young's Inequality] +\label{lemma:young-inequality} + Let $p, q \in (1, \infty)$ such that $1/p + 1/q = 1$, then for any $a, b > 0$, + \[ + ab \le \frac{a^p}{p} + \frac{b^q}{q} + \] + and for any $\eps > 0$, + \[ + ab \le \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q + \] +\end{lemma} +\begin{proof} + Since $x \mapsto \exp(x)$ is convex, + \begin{align*} + ab &= \exp(\ln(a) + \ln(b)) = \exp\paren{\frac{1}{p}\ln(a^p) + \frac{1}{q}\ln(b^q)} \\ + &\le \frac{1}{p}\exp(\ln(a^p)) + \frac{1}{q}\exp(\ln(a^q)) = \frac{a^p}{p} + \frac{b^q}{q} + \end{align*} + For any $\eps > 0$, applying the above inequality to $(\eps p)^{1/p}a$ and $b/(\eps p)^{1/p}$ yields + \[ + ab = (\eps p)^{1/p}a \cdot \frac{b}{(\eps p)^{1/p}} \le \eps a^p + \frac{b^q}{q}(\eps p)^{-(1/p)q} = \eps a^p + \frac{1}{q}(\eps q)^{-q/p}b^q + \] +\end{proof} diff --git a/src/fa/index.tex b/src/fa/index.tex index 8bd6220..d81fc4d 100644 --- a/src/fa/index.tex +++ b/src/fa/index.tex @@ -6,3 +6,4 @@ \input{./src/fa/lc/index.tex} \input{./src/fa/norm/index.tex} \input{./src/fa/rs/index.tex} +\input{./src/fa/lp/index.tex} diff --git a/src/fa/lp/definition.tex b/src/fa/lp/definition.tex new file mode 100644 index 0000000..100f79d --- /dev/null +++ b/src/fa/lp/definition.tex @@ -0,0 +1,106 @@ +\section{Basic Properties} +\label{section:lp-basic} + +\begin{definition}[$\mathcal{L}^p$ Spaces] +\label{definition:lp-unequivalence} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $f: X \to E$ be strongly measurable, and $p \in [1, \infty)$, then $f$ is \textbf{$p$-integrable} if + \[ + \norm{f}_{L^p(X; E)} = \norm{f}_{L^p(\mu; E)} = \norm{f}_{L^p(X, \cm, \mu; E)} = \braks{\int \norm{f}_E^p d\mu}^{1/p} < \infty + \] + The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $p$-integrable functions on $X$. +\end{definition} + +\begin{definition}[Essential Supremum] +\label{definition:esssup} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be strongly measurable, then $f$ is \textbf{essentially bounded} if + \[ + \norm{f}_{L^\infty(X; E)} = \norm{f}_{L^\infty(\mu; E)} = \norm{f}_{L^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty + \] + In which case, $\norm{f}_{L^\infty(X; E)}$ is the \textbf{essential supremum} of $f$. +\end{definition} + +\begin{definition}[Hölder conjugates] +\label{definition:holder-conjugates} + Let $p, q \in (1, \infty)$, then $p$ and $q$ are \textbf{Hölder conjugates} if + \[ + \frac{1}{p} + \frac{1}{q} = 1 + \] +\end{definition} + +\begin{lemma} +\label{lemma:holder-conjugate-gymnastics} + Let $p, q \in (1, \infty)$ be Hölder conjugates, then: + \begin{enumerate} + \item $q = p/(p - 1)$. + \item $p = q(p - 1)$. + \end{enumerate} +\end{lemma} + + +\begin{theorem}[Hölder's Inequality, {{\cite[6.2]{Folland}}}] +\label{theorem:holder} + Let $(X, \cm, \mu)$ be a measure space, $E, F$ be a normed vector spaces, $p, q \in [1, \infty]$. If $p, q$ are Hölder conjugates or if $p = 1$ and $q = \infty$, then for any $f \in \mathcal{L}^p(X; E)$ and $g \in \mathcal{L}^q(X; F)$, + \[ + \int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)} + \] +\end{theorem} +\begin{proof} + First suppose that $p = 1$ and $q = \infty$. In this case, + \[ + \int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)} + \] + Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By Young's inequality (\ref{lemma:young-inequality}), + \[ + \int \norm{f}_E \norm{g}_F d\mu \le \int \frac{\norm{f}_E^p}{p} + \frac{\norm{g}_F^q}{q} d\mu = \frac{1}{p}\int \norm{f}_E d\mu + \frac{1}{q}\int \norm{g}_F^q d\mu = 1 + \] +\end{proof} + +\begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}] +\label{theorem:minkowski} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^p(X; E)$, then + \[ + \norm{f + g}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)} + \] +\end{theorem} +\begin{proof} + If $p = 1$, then the theorem holds directly. + + If $p = \infty$, then for any $\alpha > \norm{f}_{L^\infty(X; E)}$ and $\beta > \norm{g}_{L^\infty(X; E)}$, $\alpha + \beta \ge f + g$ $\mu$-almost everywhere, so + \[ + \norm{f + g}_{L^\infty(X; E)} \le \norm{f}_{L^\infty(X; E)} + \norm{g}_{L^\infty(X; E)} + \] + + Now suppose that $p \in (1, \infty)$, then $p = q(p - 1)$, and + \begin{align*} + \norm{f + g}_E^p &\le (\norm{f}_E + \norm{g}_E)\norm{f + g}_E^{p - 1} \\ + \int\norm{f + g}_E^pd\mu &\le (\norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{q(p - 1)}d\mu}^{1/q} \\ + \int\norm{f + g}_E^pd\mu &\le (\norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)})\braks{\int \norm{f + g}_E^{p}d\mu}^{1/q} \\ + \braks{\int\norm{f + g}_E^pd\mu}^{1/p} &\le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)} + \end{align*} +\end{proof} + +\begin{definition}[$L^p$ Space] +\label{definition:lp} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty]$, then $\norm{\cdot}_{L^p(X; E)}$ is a seminorm on $\mathcal{L}^p(X; E)$. The quotient + \[ + L^p(X, \cm, \mu; E) = \mathcal{L}^p(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}} + \] + is a normed vector space, known as the $E$-valued \textbf{$L^p$ space} on $(X, \cm, \mu)$. +\end{definition} +\begin{proof} + By Minkowski's Inequality (\ref{theorem:minkowski}). +\end{proof} + +\begin{proposition} +\label{proposition:lp-simple-dense} + Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. +\end{proposition} +\begin{proof} + Let $f \in L^p(X; E)$. By \ref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$. + + For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the Dominated Convergence Theorem (\ref{theorem:dct}), + \[ + \limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0 + \] + Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$. +\end{proof} diff --git a/src/fa/lp/index.tex b/src/fa/lp/index.tex new file mode 100644 index 0000000..af714ae --- /dev/null +++ b/src/fa/lp/index.tex @@ -0,0 +1,4 @@ +\chapter{$L^p$ Spaces} +\label{chap:lp} + +\input{./src/fa/lp/definition.tex} diff --git a/src/measure/bochner-integral/index.tex b/src/measure/bochner-integral/index.tex new file mode 100644 index 0000000..adc96a8 --- /dev/null +++ b/src/measure/bochner-integral/index.tex @@ -0,0 +1,4 @@ +\chapter{Bochner Integral} +\label{chap:bochner-integral} + +\input{./src/measure/bochner-integral/strongly.tex} diff --git a/src/measure/bochner-integral/strongly.tex b/src/measure/bochner-integral/strongly.tex new file mode 100644 index 0000000..75c8080 --- /dev/null +++ b/src/measure/bochner-integral/strongly.tex @@ -0,0 +1,27 @@ +\section{Strongly Measurable Functions} +\label{section:strongly-measurable} + +\begin{definition}[Strongly Measurable Function] +\label{definition:strongly-measurable} + Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent: + \begin{enumerate} + \item For each $\phi \in E^*$, $\phi \circ f$ is $(\cm, \cb_K)$-measurable and $f(X) \subset E$ is separable. + \item $f$ is $(\cm, \cb_E)$ measurable and $f(X) \subset E$ is separable. + \item There exists a sequence $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that + \begin{enumerate} + \item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$. + \item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$. + \end{enumerate} + \end{enumerate} +\end{definition} +\begin{proof} + (1) $\Rightarrow$ (2): TODO + + (2) $\Rightarrow$ (3): By \ref{proposition:measurable-simple-separable-norm}. + + (3) $\Rightarrow$ (1): For each $\phi \in E^*$, $\phi \circ f = \limv{n}\phi \circ f_n$ is measurable by \ref{proposition:limit-measurable}. Since + \[ + f(X) \subset \ol{\bigcup_{n \in \natp}f_n(X)} + \] + and each $f_n$ is finitely-valued, $f(X)$ is separable. +\end{proof} diff --git a/src/measure/index.tex b/src/measure/index.tex index 8161296..f9de750 100644 --- a/src/measure/index.tex +++ b/src/measure/index.tex @@ -5,3 +5,4 @@ \input{./src/measure/measure/index.tex} \input{./src/measure/measurable-maps/index.tex} \input{./src/measure/lebesgue-integral/index.tex} +\input{./src/measure/bochner-integral/index.tex} diff --git a/src/measure/measure/radon.tex b/src/measure/measure/radon.tex index 8527744..14dd44a 100644 --- a/src/measure/measure/radon.tex +++ b/src/measure/measure/radon.tex @@ -131,7 +131,17 @@ and $\mu = \nu$. Since $X$ is $\sigma$-compact, $\mu$ is $\sigma$-finite, so $\mu$ is regular by \ref{proposition:radon-regular-sigma-finite}. +\end{proof} + +\begin{proposition} +\label{proposition:radon-cc-dense} + Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$. +\end{proposition} +\begin{proof} + By \ref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure. + + Let $A \in \cb_X$ and $\eps > 0$. By \ref{proposition:radon-regular-sigma-finite}, there exists $U \in \cn^o(A)$ and $K \subset A$ compact such that $\mu(U \setminus A), \mu(A \setminus K) < \eps/2$. By Urysohn's lemma (\ref{lemma:lch-urysohn}), there exists $f \in C_c(X; [0, 1])$ such that $f|_K = 1$ and $\supp{f} \subset U$. In which case, for any $y \in E$, \[ - \color{blue}{\int}\llap{\color{green}{\int}} + \norm{x \cdot \one_A - x \cdot f}_{L^p(X; E)} \le \norm{x}_E \mu(\bracs{f \ne \one_A})^{1/p} \le \norm{x}_E\mu(U \setminus K)^{1/p} < \eps^{1/p}\norm{x}_E \] \end{proof}