Added localisable order completeness.

src/measure/measure/localisable.tex

@@ -199,4 +199,42 @@
     \end{enumerate}
 \end{proof}
 
+\begin{corollary}
+\label{corollary:l-infty-dedekind-complete}
+    Let $(X, \cm, \mu)$ be a localisable measure space, then $L^\infty(X; \real)$ is order complete. 
+\end{corollary}
+\begin{proof}
+    Let $\seqi{f} \subset L^\infty(X; \real)$ and $M \in \real$ such that $f_i \le M$ almost everywhere for all $i \in I$. 
 
+    Fix $A \in \cm$ with $\mu(A) < \infty$, and let
+    \[
+    \mathcal{S}_A = \bracs{g \in L^\infty(A; \real)| f_i|_A \le g \text{ almost everywhere }\forall i \in I}
+    \]
+
+    then since $f_i \le M$ almost everywhere for all $i \in I$, $\mathcal{S}_A \ne \emptyset$, and $m_A = \inf_{g \in \mathcal{S}_A}\int g d\mu \in \real$.
+    
+    Let $\seq{g_{A, n}} \subset \mathcal{S}_A$ such that $\seq{g_{A, n}}$ is decreasing pointwise and $\limv{n}\int_A g_{A, n} d\mu \downto m_A$. Take $g_A = \limv{n}g_{A, n}$, then by the \hyperref[Dominated Convergence Theorem]{theorem:dct}, $\int g_A d\mu = m_A$. 
+
+    For each $i \in I$, since $g_{A, n} \ge f_i|_A$ almost everywhere for all $n \in \natp$, $g_A \ge f_i|_A$ almost everywhere as well. Thus $g_A \in \mathcal{S}_A$. For any $h \in \mathcal{S}_A$, $g_A \wedge h \in \mathcal{S}_A$ with
+    \[
+    m_A \le \int_A g_A \wedge h d\mu \le \int_A g_A d\mu = m_A 
+    \]
+
+    Thus $g_A \wedge h = g_A$ almost everywhere, so $g_A \le h$ almost everywhere, and $g_A$ is an essential supremum of $\bracsn{f_i|_A}_{i \in I}$. 
+
+    Now, let $A, B \in \cm$ with $\mu(A), \mu(B) < \infty$, then $\one_{A \cap B} g_B + \one_{A \setminus B}M \in \mathcal{S}_A$, and
+    \[
+    m_A \le \int_A g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M)d\mu \le \int_A g_A d\mu = m_A
+    \]
+
+    Thus $g_A \wedge (\one_{A \cap B} g_B + \one_{A \setminus B}M) = g_A$ almost everywhere, so $g_A|_{A \cap B} \le g_B|_{A \cap B}$ almost everywhere. As the argument is symmetric, $g_A|_{A \cap B} = g_B|_{A \cap B}$ almost everywhere. 
+
+    By the \hyperref[gluing lemma for measurable functions]{lemma:gluing-measurable}, there exists a measurable function $g:X \to \real$ such that $g|_A = g_A$ for all $A \in \cm$ with $\mu(A) < \infty$. 
+
+    Let $h \in L^\infty(X; \real)$ with $h \ge f_i$ almost everywhere for all $i \in I$, then for any $A \in \cm$ with $\mu(A) < \infty$, 
+    \[
+    \mu(\bracs{h < g} \cap A) \le \mu(\bracs{h|_A < g_A} \cup \bracs{g|_A \ne g_A}) = 0
+    \]
+    
+    As $\mu$ is semifinite, $\mu(\bracs{h < g}) = 0$. Finally, since $g_A \le M$ almost everywhere for all $A \in \cm$ with $\mu(A) < \infty$, $g \le M$ almost everywhere. Therefore $g \in L^\infty(X; \real)$ is indeed the essential supremum of $\seqi{f}$. 
+\end{proof}

src/measure/measure/scaffold.tex

@@ -3,10 +3,12 @@
 
 \begin{definition}[Scaffold*]
 \label{definition:measure-scaffold}
-    Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \bracs{A \in \cm|\mu(A) < \infty}$ be an ideal, then $\cf$ is a \textbf{scaffold} for $\mu$ if for all $E \in \cm$, 
+    Let $(X, \cm, \mu)$ be a measure space and $\cf \subset \cm$, then $\cf$ is a \textbf{scaffold} for $\mu$ if:
-    \[
+    \begin{enumerate}[label=(S\arabic*)]
-    \mu(E) = \sup\bracs{\mu(E \cap A)|A \in \cf}
+        \item For each $A \in \cf$, $\mu(A) < \infty$. 
-    \]
+        \item For all $E \in \cm$, $\mu(E) = \sup\bracs{\mu(E \cap A)|A \in \cf}$. 
+        \item For any $A, B \in \cf$, $A \cup B \in \cf$. 
+    \end{enumerate}
 
     and the quadruple $(X, \cm, \cf, \mu)$ is a \textbf{scaffolded measure space}. 
 
@@ -21,10 +23,11 @@
 
 \begin{lemma}[Gluing Lemma for Measures]
 \label{lemma:gluing-measure}
-    Let $(X, \cm)$ be a measurable space, $\cf \subset \cm$ be an ideal, and $\bracsn{\mu_A}_{A \in \cf}$ such that:
+    Let $(X, \cm)$ be a measurable space, $\cf \subset \cm$, and $\bracsn{\mu_A}_{A \in \cf}$ such that:
     \begin{enumerate}[label=(\alph*)]
         \item For each $A \in \cf$, $\mu_A$ is a finite measure on $A$.
         \item For each $A, B \in \cf$ and $E \in \cm$, $\mu_A(E \cap A \cap B) = \mu_B(E \cap A \cap B)$. 
+        \item For each $A, B \in \cf$, $A \cup B \in \cf$. 
     \end{enumerate}
 
     Let

src/measure/vector/rn-draft.tex

@@ -50,7 +50,7 @@
     &= \sup_{A \in \cf}\braks{\int_{A \cap E}f_A d\mu + \nu_s^A(A \cap E)}
     \end{align*}
 
-    As $\cf$ is an ideal, for any $A, B \in \cf$, $A \cup B \in \cf$, and
+    By (S3), for any $A, B \in \cf$, $A \cup B \in \cf$, and
     \begin{align*}
         \int_{A \cap E}f_A d\mu \vee \int_{B \cap E}f_B d\mu &\le \int_{(A \cup B) \cap E}f_{A \cup B} d\mu \\
         \nu_s^A(A \cap E) \vee \nu_s^B(B \cap E) &\le \nu_s^{A \cup B}((A \cup B) \cap E)

src/op/banach/spectrum.tex

@@ -77,7 +77,7 @@
     Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. 
 \end{proof}
 
-\begin{proposition}[Spectral Radius Formula]
+\begin{proposition}[Beurling's Spectral Radius Formula]
 \label{proposition:spectral-radius-hadamard}
     Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. 
 \end{proposition}
@@ -128,7 +128,7 @@
 
 \begin{proposition}
 \label{proposition:commutative-spectrum-gymnastics}
-    Let $A$ be a unital Banach algebra and $x, y \in A$ with $x = y$, then
+    Let $A$ be a commutative unital Banach algebra and $x, y \in A$ with $x = y$, then
     \begin{enumerate}
         \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
         \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.

src/op/c-star/index.tex

@@ -2,5 +2,6 @@
 \label{chap:c-star-algebras}
 
 \input{./involution.tex}
+\input{./unitary.tex}
 \input{./sa.tex}
 \input{./order.tex}

src/op/c-star/sa.tex

@@ -31,6 +31,91 @@
     If the above holds, then $x$ is \textbf{normal}. 
 \end{definition}
 
+\begin{theorem}
+\label{theorem:c-star-normal-spectral-radius}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $\norm{x}_A = [x]_{sp}$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem II.8.1]{Zhu}}}. ]
+    First suppose that $x$ is self-adjoint. In this case, 
+    \begin{align*}
+        \normn{x^2}_A &= \normn{xx^*}_A = \norm{x}_A^2 \\
+        \normn{x^{2^n}}_A &= \norm{x}_A^{2^n}
+    \end{align*}
+
+    for all $n \in \natp$. Thus by the \hyperref[spectral radius formula]{proposition:spectral-radius-hadamard},
+    \[
+    [x]_{sp} = \limsup_{n \to \infty}\norm{x^{n}}_A^{1/n} \ge \limsup_{n \to \infty}\normn{x^{2^n}}_A^{1/2^n} = \norm{x}_A
+    \]
+
+    Now suppose that $x$ is only normal. Since $x$ and $x^*$ commute, $[xx^*]_{sp} \le [x]_{sp}[x^*]_{sp}$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus
+    \[
+    \norm{x}^2_A = \normn{xx^*}_A = [xx^*]_{sp} \le [x]_{sp}[x^*]_{sp} = [x]_{sp}^2
+    \]
+
+    by (5) of \autoref{proposition:c-star-algebra-gymnastics}. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:c-star-normal-spectral-radius-corollary}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item There exists $\lambda \in \sigma_A(x)$ such that $|\lambda| = \norm{x}_A$. 
+        \item If there exists $n \in \natp$ such that $x^n = 0$, then $x = 0$ as well. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Since $\sigma_A(x)$ is compact, there exisst $\lambda \in \sigma_A(x)$ such that $|\lambda| = [x]_{sp}$. By \autoref{theorem:c-star-normal-spectral-radius}, $|\lambda| = [x]_{sp} = \norm{x}_A$. 
+
+    (2): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, 
+    \[
+    \bracs{\lambda^n| \lambda \in \sigma_A(x)} = \bracs{0}
+    \]
+
+    Thus $\sigma_A(x) = \bracs{0}$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{x}_A = [x]_{sp} = 0$. 
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:c-star-unique-norm}
+    Let $A$ be a unital $C^*$-algebra over $\complex$, then for each $x \in A$, 
+    \[
+    \norm{x}_A^2 = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
+    \]
+
+    In particular, there exists at most one norm on $A$ making it a $C^*$-algebra. 
+\end{corollary}
+\begin{proof}
+    Since $x^*x$ is self-adjoint, \autoref{theorem:c-star-normal-spectral-radius} implies that
+    \[
+    \norm{x}_A^2 = \norm{x^*x}_A = \sup\bracs{|\lambda|\ | \lambda \in \sigma_A(x^*x)}
+    \]
+
+    which depends only on the algebraic structure of $A$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:self-adjoint-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then $\sigma_A(x) \subset \real$. 
+\end{proposition}
+\begin{proof}
+    Let
+    \[
+    y = \exp(ix) = \sum_{n = 0}^\infty \frac{i^nx^n}{n!}  
+    \]
+
+    then
+    \[
+    y^*= \sum_{n = 0}^\infty \frac{(-i)^n (x^*)^n}{n!} = \exp(-ix^*) 
+    \]
+
+    Since $x$ is normal, $y$ is also normal. By \autoref{proposition:functional-calculus-exp},,
+    \[
+    y^*y = \exp(-ix^* + ix) = \exp(-ix + ix) = 1
+    \]
+    
+    so $y$ is unitary. By \autoref{proposition:unitary-spectrum} and the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, $\exp(i\sigma_A(x)) = \sigma_A(y) \subset \partial B_\complex(0, 1)$. Thus $i\sigma_A(x) \subset \bracs{\text{Re} = 0}$, and $\sigma_A(x) \subset \real$. 
+\end{proof}
+
 
 
 

src/op/c-star/unitary.tex

@@ -0,0 +1,55 @@
+\section{Unitary Elements}
+\label{section:unitary-c-star}
+
+\begin{definition}[Unitary]
+\label{definition:unitary-element}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $x$ is \textbf{unitary} if $x \in G(A)$ and $x^* = x^{-1}$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:unitary-unit}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\norm{x}_A = 1$. 
+\end{lemma}
+\begin{proof}
+    $\normn{x^2}_A = \norm{xx^*}_A = \norm{1}_A = 1$. 
+\end{proof}
+
+\begin{definition}[Unitarily Equivalent]
+\label{definition:unitary-equivalent}
+    Let $A$ be a unital $C^*$-algebra and $x, y \in A$, then $x$ and $y$ are \textbf{unitarily equivalent} if there exists a unitary element $u \in A$ such that $x = uyu^*$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:unitary-equivalent-same-stuff}
+    Let $A$ be a unital $C^*$-algebra and $x, y \in A$ be unitarily equivalent, then:
+    \begin{enumerate}
+        \item $\norm{x}_A = \norm{y}_A$. 
+        \item $\sigma_A(x) = \sigma_A(y)$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): Let $u \in A$ be unitary such that $x = uyu^*$, then $\norm{x}_A \le \norm{u}_A\norm{x}_A\normn{u^*}_A$. By \autoref{lemma:unitary-unit} and (1) of \autoref{proposition:c-star-algebra-gymnastics}, $\norm{u}_A = \normn{u^*}_A = 1$, so $\norm{x}_A \le \norm{y}_A$. As the argument is symmetric, $\norm{x}_A = \norm{y}_A$. 
+
+    (2): Let $\lambda \in \complex$, then
+    \[
+    u(y - \lambda)u^* = uyu^* - \lambda uu^* = uyu^* - \lambda = x - \lambda
+    \]
+
+    Since $u, u^* \in G(A)$, $x - \lambda \in G(A)$ if and only if $y - \lambda \in G(A)$. Therefore $\sigma_A(x) = \sigma_A(y)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:unitary-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be unitary, then $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition II.8.2]{Zhu}}}. ]
+    By \autoref{lemma:unitary-unit}, $\norm{x}_A = 1$, so $\sigma_A(x) \subset \ol{B_\complex(0, 1)}$. Thus
+    \[
+    \bracsn{\ol{\lambda}|\lambda \in \sigma_A(x)} = \sigma_A(x^*) = \sigma_A(x^{-1}) \subset \ol{\complex \setminus B_\complex(0, 1)}
+    \] 
+
+    by (4) of \autoref{proposition:c-star-algebra-gymnastics}. For any $\lambda \in \complex$, $\lambda \in \ol{B_\complex(0, 1)}$ and $\ol \lambda \in \ol{\complex \setminus B_\complex(0, 1)}$ if and only if $|\lambda| = 1$. Therefore $\sigma_A(x) \subset \partial B_\complex(0, 1)$. 
+\end{proof}
+
+
+