Cleanup

src/fa/lp/definition.tex

@@ -7,6 +7,7 @@
     \[
     \norm{f}_{L^p(X; E)} = \norm{f}_{L^p(\mu; E)} = \norm{f}_{L^p(X, \cm, \mu; E)} = \braks{\int \norm{f}_E^p d\mu}^{1/p} < \infty
     \]
+
     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}
 
@@ -16,6 +17,7 @@
     \[
     \norm{f}_{L^\infty(X; E)} = \norm{f}_{L^\infty(\mu; E)} = \norm{f}_{L^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
     \]
+
     In which case, $\norm{f}_{L^\infty(X; E)}$ is the \textbf{essential supremum} of $f$.
 \end{definition}
 
@@ -25,6 +27,7 @@
     \[
     \frac{1}{p} + \frac{1}{q} = 1
     \]
+
 \end{definition}
 
 \begin{lemma}
@@ -43,16 +46,19 @@
     \[
     \int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
     \]
+
 \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}),
+
+    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 \hyperref[Young's inequality]{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}}}]
@@ -61,6 +67,7 @@
     \[
     \norm{f + g}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)}
     \]
+
 \end{theorem}
 \begin{proof}
     If $p = 1$, then the theorem holds directly.
@@ -70,6 +77,7 @@
     \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} \\
@@ -85,10 +93,11 @@
     \[
     L^p(X, \cm, \mu; E) = \mathcal{L}^p(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
     \]
+
     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}).
+    By \hyperref[Minkowski's Inequality]{theorem:minkowski}.
 \end{proof}
 
 \begin{proposition}
@@ -96,11 +105,12 @@
     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 $n \to \infty$.
+    Let $f \in L^p(X; E)$. By \autoref{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}),
+    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 \hyperref[Dominated Convergence Theorem]{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}