Added nuclear operators.

document.tex

@@ -6,7 +6,7 @@
 
 Hi, welcome to my digital garden, where I collect math results that I learn. 
 
-Despite being presented in a linear order, I will frequently reference things between chapters and sections. 
+Despite the contents being presented in a linear order by the table of contents, I will frequently reference things between chapters and sections. 
 
 Occasionally, I make up some definitions to play with. These definition blocks will always have a * at the end of its title to indicate that it lives mostly in my head. These terms will always be referenced with a link to their definition block. 
 

src/dg/complex/entire.tex

@@ -74,7 +74,7 @@
 
     then $f = 1$. 
 \end{proposition}
-\begin{proof}[Proof, {{\cite[Lemma I.4.4]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Lemma 4.4]{Zhu}}}. ]
     By (c) and \autoref{proposition:entire-logarithm}, there exists $g \in H(\complex; \complex)$ such that $f = e^g$. Since $f(0) = 1$, $g(0) = 0$. As $f'(0) = g'(0)f(0) = 0$, $g'(0) = 0$ as well. 
 
     From (c), $|g(z)| \le e^{|z|}$ for each $z \in \complex$. Thus for every $r > 0$ and $z \in B_\complex(0, r)$, $|g(z)| \le |g(z) - 2r|$, and

src/fa/lc/convex.tex

@@ -159,7 +159,7 @@
         \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
         \item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
         \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
-        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$.
+        \item If $A$ is circled, then $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$, with respect to any vector space topology on $E$. 
     \end{enumerate}
 
     In particular,

src/fa/lc/hahn-banach.tex

@@ -26,10 +26,10 @@
 
     then for any $x \in F$ and $t > 0$,
     \begin{align*}
-        \phi(x + tx_0) &= \phi(x) + t\lambda = t\braks{\phi(t^{-1}x) + \lambda} \\
+        \phi_{x_0, \lambda}(x + tx_0) &= \phi(x) + t\lambda = t\braks{\phi(t^{-1}x) + \lambda} \\
         &\le t\braks{\rho(t^{-1}x + x_0) - \phi(t^{-1}x) + \phi(t^{-1}x)} \\
         &= t\rho(t^{-1}x + x_0) = \rho(x + tx_0) \\
-        \phi(x - tx_0) &= \phi(x) - t\lambda = t\braks{\phi(t^{-1}x) - \lambda} \\
+        \phi_{x_0, \lambda}(x - tx_0) &= \phi(x) - t\lambda = t\braks{\phi(t^{-1}x) - \lambda} \\
         &\ge t\braks{\rho(t^{-1}x - x_0) + \phi(t^{-1}x) - \phi(t^{-1}x)} \\
         &= t\rho(t^{-1}x - x_0) = \rho(x - tx_0)
     \end{align*}
@@ -40,7 +40,7 @@
     Let $E$ be a vector space over $K \in \RC$, $\rho: E \to \real$ be a sublinear functional, and $F \subsetneq E$ be a subspace, then
     \begin{enumerate}
         \item For any $\phi \in \hom(F; \real)$ with $\phi \le \rho|_F$, there exists $\Phi \in \hom(E; \real)$ such that $\Phi \le \rho$ and $\Phi|_F = \phi$.
-        \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; \complex)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; \complex)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
+        \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; K)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; K)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
     \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 5.6, 5.7]{Folland}}}. ]
@@ -60,7 +60,7 @@
 
     By Zorn's lemma, $\mathbf{F}$ admits a maximal element $\Phi$. If $\cd(\Phi) \subsetneq E$, then $\Phi$ is not maximal by the preceding discussion. Therefore $\cd(\Phi) = E$ and $\Phi$ is a desired extension.
 
-    (2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$.
+    (2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$. Assume without loss of generality that $K = \complex$. 
 
     Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \autoref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \autoref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
     \[

src/fa/lc/index.tex

@@ -13,3 +13,4 @@
 \input{./hahn-banach.tex}
 \input{./spaces-of-linear.tex}
 \input{./tensor.tex}
+\input{./nuclear.tex}

src/fa/lc/nuclear.tex

@@ -0,0 +1,158 @@
+\section{Nuclear Operators}
+\label{section:nuclear-operator}
+
+\begin{definition}[Nuclear Operator Between Banach Spaces]
+\label{definition:nuclear-operator-normed}
+    Let $E, F$ be Banach spaces, $E^*$ be the dual of $E$, equipped with the uniform topology, and $T \in L(E; F)$, then $T$ is \textbf{nuclear} if there exists $\seq{\phi_n} \subset E^*$ and $\seq{y_n} \subset F$ such that:
+    \begin{enumerate}
+        \item For each $x \in E$, $Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E}$. 
+        \item $\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} < \infty$. 
+    \end{enumerate}
+
+    The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$. For each $T \in N(E; F)$, let
+    \[
+    \norm{T}_{N(E; F)} = \inf\bracs{\sum_{n \in \natp}\norm{y_n}_F\norm{\phi_n}_{E^*} \bigg | Tx = \sum_{n = 1}^\infty y_n \dpn{x, \phi_n}{E} \forall x \in E}
+    \]
+
+    then $\norm{\cdot}_{N(E; F)}$ is a norm on $N(E; F)$, and $N(E; F)$ is a Banach space. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:nuclear-operator-normed-tensor}
+    Let $E, F$ be Banach spaces, $E^*$ be the dual of $E$, equipped with the uniform topology, then the mapping
+    \[
+    E^* \otimes F \to N(E; F) \quad \sum_{j = 1}^n \phi_j \otimes y_j \mapsto \sum_{j = 1}^n y_j\dpn{\cdot, \phi_j}{E}
+    \]
+
+    extends continuously into a surjective linear map $E^* \tilde \otimes_\pi F \to N(E; F)$. 
+\end{lemma}
+
+
+
+
+\begin{definition}[Nuclear Operator]
+\label{definition:nuclear-operator}
+    Let $E, F$ be separated locally convex spaces over $K \in \RC$ and $T \in L(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists convex and circled sets $U \in \cn_E(0)$ and $B \in B(F)$ such that:
+        \begin{enumerate}[label=(\alph*)]
+            \item The auxiliary space $F_B$ is a Banach space.
+            \item $T(U) \subset B$. 
+            \item The induced map $\wh E_U \to F_B$ is nuclear. 
+        \end{enumerate}
+        \item There exists an equicontinuous sequence $\seq{\phi_n} \subset E^*$, a convex, circled, and bounded subset $B \subset F$, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ such that
+        \begin{enumerate}[label=(\alph*)]
+            \item The auxiliary space $F_B$ is a Banach space. 
+            \item For each $x \in E$, $Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}$. 
+            \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+        \end{enumerate}
+    \end{enumerate}
+    
+    If the above holds, then $T$ is \textbf{nuclear}. The set $N(E; F)$ is the \textbf{space of nuclear operators} from $E$ to $F$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem III.7.1]{SchaeferWolff}}}. ]
+    (1) $\Rightarrow$ (2): Let $\pi: E \to E_U$ be the canonical projection map associated with $E_U$ and $\iota: F_B \to F$ be the canonical inclusion map associated with $F_B$. By assumption (1b), there exists an induced map $\hat T: E_U \to F_B$ such that the following diagram commutes:
+    \[
+    \xymatrix{
+    E \ar@{->}[r]^{T} \ar@{->}[d]_{\pi} & F \\
+    E_U \ar@{->}[r]_{\hat T} & F_B \ar@{->}[u]_{\iota}
+    }
+    \]
+
+    By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}, $E_U^* = (\wh E_U)^*$. Assume without loss of generality that $E_U$ is a Banach space, then (1c) implies that $\hat T \in L(E_U; F_B)$ is a nuclear operator. By \autoref{lemma:nuclear-operator-normed-tensor} and \autoref{theorem:metrisable-tensor-product}, there exists $\seq{\phi_n} \subset E_U^*$, $\seq{y_n} \subset F_B$, and $\seq{\lambda_n} \subset K$ such that:
+    \begin{enumerate}[label=(\roman*)]
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+        \item $\limv{n}\phi_n = 0$ and $\limv{n}y_n = 0$. 
+        \item For each $x \in E_U$, $\hat Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E_U}$. 
+    \end{enumerate}
+
+    By (ii), $\sup_{n \in \natp}\norm{\phi_n}_{E_U^*} < \infty$ and $\sup_{n \in \natp}\norm{y_n}_{F_B} < \infty$, so $\seq{\phi_n}$ is equicontinuous, and there exists $R > 0$ such that $\seq{y_n} \subset RB$. After rescaling, assume without loss of generality that $\seq{y_n} \subset B$. By unraveling the factorisation, (iii) shows that for each $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ \pi}{E}
+    \]
+
+    Therefore the decomposition using $\seq{\phi_n \circ \pi} \subset E^*$, $B \subset F$, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ given above satisfies (2).
+
+    (2) $\Rightarrow$ (1): Since $\seq{\phi_n}$ is equicontinuous, $U = \bigcap_{n \in \natp}\phi_n^{-1}(B_K(0, 1))$ is a convex and circled neighbourhood of $0$ in $E$. 
+    
+    (1b): Using assumption (2c) and rescaling, assume without loss of generality that $\sum_{n = 1}^\infty |\lambda_n| < 1$. Let $\rho: F_B \to [0, \infty)$ be the gauge of $B$, then for any $x \in U$, 
+    \begin{align*}
+        \rho(Tx) &= \rho\braks{\sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E}} \le \sum_{n \in \natp} |\lambda_n| \cdot \underbrace{|\dpn{x, \phi_n}{E}|}_{\le 1} \cdot \underbrace{\rho(y_n)}_{\le 1} \\
+        &\le \sum_{n \in \natp}|\lambda_n| < 1
+    \end{align*}
+
+    so $\rho(Tx) < 1$ and $Tx \in B$. Therefore $T(U) \subset B$. 
+
+    (1c): Let $\pi: E \to E_U$ be the canonical projection map associated with $E_U$ and $n \in \natp$. By construction, $U \subset \phi_n^{-1}(B_K(0, 1))$, so there exists $\hat \phi_n \in E_U^*$ such that the following diagram commutes: 
+    \[
+    \xymatrix{
+    E \ar@{->}[d]_{\pi} \ar@{->}[rd]^{\phi_n} &  \\
+    E_U \ar@{->}[r]_{\hat \phi_n} & K
+    }
+    \]
+
+    Thus for each $x \in E$, 
+    \[
+    Tx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{E} = \sum_{n = 1}^\infty y_n \dpn{\pi(x), \lambda_n \hat \phi_n}{E_U}
+    \]
+    
+    and the induced map $\hat T: \wh E_U \to F_B$ takes the form $\hat Tx = \sum_{n = 1}^\infty y_n \dpn{x, \lambda_n \hat \phi_n}{\wh E_U}$. Finally, for each $n \in \natp$, $U \subset \phi_n^{-1}(B_K(0, 1))$, and $\normn{\hat \phi_n}_{E_U^*} \le 1$. Similarly, since $y_n \in B$, $\norm{y_n}_{F_B} \le 1$ as well. Therefore
+    \[
+        \normn{\hat T}_{N(\wh E_U; F_B)} \le \sum_{n \in \natp}|\lambda_n| \cdot \norm{y_n}_{F_B} \cdot \normn{\hat \phi_n}_{E_U^*} \le \sum_{n \in \natp}|\lambda_n| < \infty
+    \]
+
+    and $\hat T: \wh E_U \to F_B$ is nuclear. 
+    
+    
+\end{proof}
+
+\begin{proposition}
+\label{proposition:nuclear-gymnastics}
+    Let $E, F, G, H$ be separated locally convex spaces and $S \in N(F; G)$, then:
+    \begin{enumerate}
+        \item $S$ is compact.
+        \item For any $T \in L(E; F)$, $S \circ T \in N(E; G)$. 
+        \item For any $R \in L(G; H)$, $R \circ S \in N(F; H)$. 
+        \item There exists a unique $\wh S \in L(\wh F; G)$ such that $\wh S|_{F} = S$. Moreover, $\wh S \in N(\wh F; G)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Corollary III.7.1.1-III.7.1.3]{SchaeferWolff}}}. ]
+    Let $\seq{\phi_n} \subset F^*$ be an equicontinuous sequence, $B \in B(G)$ be convex and circled, $\seq{y_n} \subset B$, and $\seq{\lambda_n} \subset K$ such that
+    \begin{enumerate}[label=(\alph*)]
+        \item The auxiliary space $G_B$ is a Banach space. 
+        \item For each $x \in F$, $Sx = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n}{F}$. 
+        \item $\sum_{n \in \natp}|\lambda_n| < \infty$. 
+    \end{enumerate}
+
+    (1): Let $U = \bigcap_{n \in \natp}\phi_n^{-1}(B_K(0, 1))$, then since $\seq{\phi_n}$ is equicontinuous, $U$ is a convex and circled neighbourhood of $0$ in $F$. Given that $G_B$ is complete, $S$ is the following composition of continuous maps:
+    \[
+    \begin{CD}
+    U @>{\prod_{n \in \natp} \phi_n}>> \overline{B_K(0,1)}^{\natp} @>{x \mapsto \sum_{n=1}^\infty \lambda_n x_n y_n}>> G_B @>>> G
+    \end{CD}
+    \]
+
+    By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $\overline{B_K(0,1)}^{\natp}$ is compact. Since $S(U)$ is contained in its image in the above diagram, $S(U)$ is relatively compact. 
+
+    (2): Since $T \in L(E; F)$, $\seq{\phi_n \circ T} \subset E^*$ is equicontinuous. Thus for any $x \in E$, 
+    \[
+    (S \circ T)x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \phi_n \circ T}{E}
+    \]
+
+    and $S \circ T \in N(E; G)$. 
+
+    (3): Using (1), assume without loss of generality that $B$ is also compact. In which case, $R(B)$ is a convex, circled, and compact set in $H$ containing $0$. Thus $H_{R(B)}$ is a Banach space. For each $x \in F$, 
+    \[
+    (R \circ S)x = \sum_{n = 1}^\infty \lambda_n R(y_n) \dpn{x, \phi_n}{F}
+    \]
+
+    and $R \circ S \in N(F; H)$. 
+
+    (4): By the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-tvs}, such an extension exists and is unique. Moreover, $\seq{\phi_n} \subset F^*$ extend into an equicontinuous family $\bracsn{\wh \phi_n}_1^\infty \subset \wh F^*$. Since $G_B$ is complete, the extension $\wh S \in L(\wh F; G)$ takes the form
+    \[
+    \wh S x = \sum_{n = 1}^\infty \lambda_n y_n \dpn{x, \wh \phi_n}{\wh F}
+    \]
+
+    Therefore $\wh S \in N(\wh F; G)$.
+\end{proof}
+
+

src/fa/lc/tensor.tex

@@ -135,7 +135,7 @@
     (5): By (6) of \autoref{definition:projective-tensor-product}. 
 \end{proof}
 
-\begin{theorem}[{{\cite[III.6.4]{SchaeferWolff}}}]
+\begin{theorem}
 \label{theorem:metrisable-tensor-product}
     Let $E, F$ be metrisable locally convex spaces over $K \in \RC$, then for any $z \in E \td{\otimes}_\pi F$, there exists $\seq{\lambda_n} \subset K$ and $\seq{(x_j, y_j)} \subset E \times F$ such that:
     \begin{enumerate}
@@ -146,7 +146,7 @@
 
     
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[III.6.4]{SchaeferWolff}}}.]
     Let $\seq{p_n}$ and $\seq{q_n}$ be increasing sequences of continuous seminorms that induce the topology on $E$ and $F$, respectively. For each $n \in \natp$, let $r_n = p_n \otimes q_n$, and $\td r_n$ be the continuous extension of $r_n$ to $E \td{\otimes}_\pi F$. 
 
     Let $u \in E \td{\otimes}_\pi F$, then there exists $\seq{u_n} \subset E \otimes_\pi F$ such that $\td r_n(u - u_n) < 2^{-n}/n^2$ for all $n \in \natp$. For each $N \in \natp$, let $v_N = u_{N+1} - u_N$, then

src/fa/norm/hilbert.tex

@@ -83,20 +83,23 @@
 
 \begin{definition}[Inner Product]
 \label{definition:inner-product}
-    Let $E$ be a vector space over $K$ and $\inp_E: E \times E \to K$, then $\inp_E$ is an \textbf{inner product} if:
+    Let $E$ be a vector space over $K$ and $\inp_E: E \times E \to K$, then $\inp_E$ is a \textbf{pseudo inner product} if:
     \begin{enumerate}[label=(H\arabic*)]
         \item For each $x, y, z \in E$, $\angles{x + y, z}_E = \dpn{x, z}{E} + \dpn{y, z}{E}$. 
         \item For any $x, y \in E$ and $\mu \in K$, $\dpn{\mu x, y}{E} = \mu \dpn{x, y}{E}$. 
         \item For every $x, y \in E$, $\dpn{x, y}{E} = \ol{\dpn{y, x}{E}}$. 
-        \item[(I)] For each $x \in E$, $\dpn{x, x}{E} \ge 0$, with equality if and only if $x = 0$. 
+        \item[(I)] For each $x \in E$, $\dpn{x, x}{E} \ge 0$. 
     \end{enumerate}
 
+    and an \textbf{inner product} if for each $x \in E$, $\dpn{x, x}{E} = 0$ if and only if $x = 0$. 
+    
+
 \end{definition}
 
 
 \begin{proposition}[Cauchy-Schwarz Inequality]
 \label{proposition:cauchy-schwarz}
-    Let $H$ be a vector space over $K \in \RC$ and $\inp_H: E \times E \to K$ be an inner product, then for any $x, y \in H$, $\dpn{x, y}{H} \le \norm{x}_H\norm{y}_H$. 
+    Let $H$ be a vector space over $K \in \RC$ and $\inp_H: E \times E \to K$ be a pseudo inner product, then for any $x, y \in H$, $\dpn{x, y}{H} \le \norm{x}_H\norm{y}_H$. 
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Theorem 5.19]{Folland}}}. ]
     Assume without loss of generality that $\dpn{x, y}{H} > 0$, then for each $t \in \real$, 

src/fa/norm/normed.tex

@@ -68,7 +68,7 @@
 \end{proof}
 
 
-\begin{theorem}[Successive Approximation]
+\begin{theorem}[Successive Approximations]
 \label{theorem:successive-approximation}
     Let $E, F$ be normed vector spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:
     \begin{enumerate}
@@ -84,7 +84,7 @@
 
     In particular, if $E$ is a Banach space, then for every $y \in F$, there exists $x \in E$ such that $\norm{x}_E \le C\norm{y}_F/(1 - \gamma)$ and $Tx = y$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof, learned from Anson Li (https://ansonli0.com/). ]
     Let $y_1 = y \in F$. Let $n \in \natp$ and suppose inductively that $\bracs{x_k| 0 \le k < n}$ and $\bracs{y_k| 0 \le k \le n}$ has been constructed. By (a) and (b), there exists $x_n \in E$ such that $\norm{x_k}_E \le C\norm{y_k}_F$ and $\norm{y_n - Tx_n}_F \le \gamma \norm{y_{n}}_F$. Let $y_{n+1} = y_n - Tx_n$, then $\norm{y_{n+1}}_F \le \gamma \norm{y_n}_F$.
 
     For each $n \in \nat$,

src/fa/notation.tex

@@ -26,6 +26,7 @@
     $E \otimes_\pi F$ & Projective tensor product of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $E \,\widetilde{\otimes}_\pi F$ & Projective completion of $E$ and $F$. & \autoref{definition:projective-tensor-product} \\
     $p \otimes q$ & Cross seminorm of $p$ and $q$. & \autoref{definition:cross-seminorm} \\
+    $N(E; F)$ & Nuclear mappings from $E$ to $F$. & \autoref{definition:nuclear-operator-normed} \\ 
     % ---- Order Structures ----
     $x \vee y$, $x \wedge y$ & $\sup$ and $\inf$ in vector lattice. & \autoref{definition:vector-lattice} \\
     $|x|$ & Absolute value $x \vee (-x)$ in a vector lattice. & \autoref{definition:order-absolute-value} \\

src/fa/order/index.tex

@@ -1,5 +1,7 @@
 \chapter{Order Structures}
 \label{chap:order-structure}
 
+\input{./order.tex}
+\input{./positive.tex}
 \input{./lattice.tex}
 \input{./norm.tex}

src/fa/order/lattice.tex

@@ -1,63 +1,6 @@
 \section{Vector Lattices}
 \label{section:vector-lattice}
 
-\begin{definition}[Ordered Vector Space]
-\label{definition:ordered-vector-space}
-    Let $E$ be a vector space over $\real$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
-    \begin{enumerate}
-        \item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
-        \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
-    \end{enumerate}
-
-    
-\end{definition}
-
-\begin{proposition}
-\label{proposition:ordered-vector-space-properties}
-    Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then 
-    \begin{enumerate}
-        \item $\sup(A + B) = \sup(A) + \sup(B)$.
-        \item $\sup(A) = -\inf (-A)$
-    \end{enumerate}
-
-    
-\end{proposition}
-\begin{proof}
-    (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
-
-    (2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$. 
-\end{proof}
-
-
-\begin{definition}[Interval]
-\label{definition:ordered-vector-space-interval}
-    Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
-    \[
-    [x, y] = \bracs{z \in E| x \le z \le y}
-    \]
-
-    is the \textbf{order interval} with endpoints $x$ and $y$.
-\end{definition}
-
-\begin{definition}[Order Bounded]
-\label{definition:ordered-vector-space-bounded}
-    Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
-\end{definition}
-
-\begin{definition}[Order Complete]
-\label{definition:order-vector-complete}
-    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
-\end{definition}
-
-\begin{definition}[Order Bounded Dual]
-\label{definition:order-bounded-dual}
-    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
-\end{definition}
-
-\begin{definition}[Order Dual]
-\label{definition:order-dual}
-    Let $(E, \le)$ be an ordered vector space and $\Phi^+ \in \hom(E; \real)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\Phi^+(x) \ge 0$. The subspace $E^+ \subset \hom(E; \real)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
-\end{definition}
 
 
 \begin{definition}[Vector Lattice]
@@ -249,7 +192,7 @@
 \end{proof}
 
 
-\begin{proposition}[{{\cite[V.1.4]{SchaeferWolff}}}]
+\begin{proposition}
 \label{proposition:order-vector-dual}
     Let $(E, \le)$ be an ordered vector space. If for any $x, y \in E$ with $x, y \ge 0$, $[0, x] + [0, y] = [0, x + y]$, then:
     \begin{enumerate}
@@ -264,7 +207,7 @@
 
     
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[V.1.4]{SchaeferWolff}}}. ]
     (1): Let $C = \bracs{x \in E|x \ge 0}$, $\Phi^+ \in E^b$, and
     \[
     \Phi^+: C \to [0, \infty) \quad x \mapsto \sup(f([0, x]))

src/fa/order/order.tex

@@ -0,0 +1,56 @@
+\section{Ordered Vector Spaces}
+\label{section:ovs}
+
+\begin{definition}[Ordered Vector Space]
+\label{definition:ordered-vector-space}
+    Let $E$ be a vector space over $K \in \RC$ and $\le$ be a partial order on $E$, then $(E, \le)$ is a \textbf{ordered vector space} if
+    \begin{enumerate}
+        \item[(LO1)] For any $x, y, z \in E$ with $x \le y$, $x + z \le y + z$.
+        \item[(LO2)] For any $x, y \in E$ and $\lambda > 0$, $x \le y$ implies that $\lambda x \le \lambda y$.
+    \end{enumerate}
+
+    The set $C = \bracs{x \in E|x \ge 0}$ is the \textbf{positive cone} of $E$. 
+\end{definition}
+
+\begin{definition}[Ordered Topological Vector Space]
+\label{definition:ordered-tvs}
+    Let $(E, \le)$ be an ordered vector space over $K \in \RC$, and $\topo$ be a vector space topology on $E$, then the triple $(E, \topo, \le)$ is an \textbf{ordered topological vector space} if the positive cone $C = \bracs{x \in E|x \ge 0}$ is closed. 
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:ordered-vector-space-properties}
+    Let $(E, \le)$ be an ordered vector space and $A, B \subset E$ such that $\sup(A)$ and $\sup (B)$ exist, then 
+    \begin{enumerate}
+        \item $\sup(A + B) = \sup(A) + \sup(B)$.
+        \item $\sup(A) = -\inf (-A)$
+    \end{enumerate}
+
+    
+\end{proposition}
+\begin{proof}
+    (1): For any $a \in A$ and $b \in B$, $a + b \le \sup(A) + \sup(B)$ by (LO1). Let $c \in E$ such that $c \ge a + b$ for all $a \in A$ and $b \in B$, then $c \ge a + \sup(B)$ for all $a \in A$, so $c \ge \sup(A) + \sup(B)$. Therefore $\sup(A) + \sup(B) = \sup(A + B)$.
+
+    (2): For any $a \in A$, $\sup(A) \ge a = -(-a)$, so $-\sup(A) \le \inf(-A)$, and $\sup(A) \ge -\inf(-A)$. On the other hand, for any $a \in A$, $-\inf(-A) \ge a$. Therefore $\sup(A) \le -\inf(-A)$. 
+\end{proof}
+
+
+\begin{definition}[Interval]
+\label{definition:ordered-vector-space-interval}
+    Let $(E, \le)$ be an ordered vector space and $x, y \in E$, then
+    \[
+    [x, y] = \bracs{z \in E| x \le z \le y}
+    \]
+
+    is the \textbf{order interval} with endpoints $x$ and $y$.
+\end{definition}
+
+\begin{definition}[Order Bounded]
+\label{definition:ordered-vector-space-bounded}
+    Let $(E, \le)$ be an ordered vector space and $A \subset E$, then $A$ is \textbf{order bounded} if there exists $x, y \in E$ such that $A \subset [x, y]$.
+\end{definition}
+
+\begin{definition}[Order Complete]
+\label{definition:order-vector-complete}
+    Let $(E, \le)$ be an ordered vector space, then $E$ is \textbf{order complete} if for any order bounded set $A \subset E$, $\sup (A)$ and $\inf (A)$ exist.
+\end{definition}

src/fa/order/positive.tex

@@ -0,0 +1,34 @@
+\section{The Order Dual}
+\label{section:order-dual}
+
+
+\begin{definition}[Order Bounded Dual]
+\label{definition:order-bounded-dual}
+    Let $(E, \le)$ be an ordered vector space, then the space $E^b$ consisting of all linear functionals on $E$ that are bounded on order bounded sets is the \textbf{order bounded dual} of $E$.
+\end{definition}
+
+
+\begin{definition}[Order Dual]
+\label{definition:order-dual}
+    Let $(E, \le)$ be an ordered vector space over $K \in \RC$ and $\Phi^+ \in \hom(E; K)$, then $\Phi^+$ is \textbf{positive} if for any $x \in E$ with $x \ge 0$, $\text{Re}\dpn{x, \Phi^+}{E} \ge 0$. The subspace $E^+ \subset \hom(E; K)$ generated by the positive linear functionals on $E$ is the \textbf{order dual} of $E$. 
+\end{definition}
+
+
+\begin{theorem}[Bauer-Namioka]
+\label{theorem:bauer-namioka}
+    Let $E$ be an ordered vector space over $K \in \RC$ with positive cone $C$, $\topo$ be a vector space topology on $E$\footnote{The order and the topology need not to be compatible. }, $F \subset E$ be a subspace, and $\phi \in (F, \topo)^*$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a continuous positive linear functional $\Phi \in (E, \topo)^*$ such that $\Phi|_F = \phi$. 
+        \item There exists $U \in \cn_\topo(0)$ convex such that
+        \[
+        \sup\bracs{\text{Re}\dpn{x, \phi}{F}|x \in F \cap (U - C)} < \infty
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[V.5.4]{SchaeferWolff}}}. ]
+    (1) $\Rightarrow$ (2): Let $U = \bracs{\text{Re}(\Phi) < 1}$, then $U \in \cn_\topo(0)$ is convex. Since $\Phi$ is positive, for any $x \in U$ and $y \in C$, $\text{Re}\dpn{x - y, \Phi}{E} \le \text{Re}\dpn{x, \Phi}{E} < 1$. 
+
+    (2) $\Rightarrow$ (1): Assume without loss of generality that $K = \real$. Let $\alpha > 0$ such that $F \cap (U - C) \subset \bracs{\phi < \alpha}$. Since $U$ is convex and open, and $C$ is convex, $U - C \subset E$ is an open convex set, disjoint from the convex set $\bracs{\phi = \alpha}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach-geometric-1}, there exists $\Phi \in E^*$ such that $U - C \subset \bracs{\Phi < \alpha}$, and $\bracs{\phi = \alpha} \subset \bracs{\Phi \ge \alpha}$. 
+
+    After rescaling, assume without loss of generality that $\bracs{\phi = \alpha} \subset \bracs{\Phi = \alpha}$, then $\Phi \in E^*$ is an extension of $\phi$. For each $x \in C$ and $\lambda > 0$, $-\lambda x \in U - C$, and $-\lambda\dpn{x, \Phi}{E} < \alpha$. As this holds for all $\lambda > 0$, $\dpn{x, \Phi}{E} \ge 0$, so $\Phi$ is the desired extension.
+\end{proof}

src/fa/tvs/complexify.tex

@@ -84,6 +84,18 @@
     
 \end{proof}
 
+\begin{definition}[Hermitian]
+\label{definition:hermitian-functional}
+    Let $E$ be a vector space over $\real$, $*: \complex(E) \to \complex(E)$ be the canonical complex conjugation map, and $\phi \in \hom(\complex(E); \complex)$, then the following are equivalent: 
+    \begin{enumerate}
+        \item $\phi|_E \in \hom(E; \real)$.
+        \item For each $x \in E$, $\dpn{x, \phi}{\complex(E)} = \ol{\dpn{x^*, \phi}{\complex(E)}}$. 
+    \end{enumerate}
+
+    If the above holds, then $\phi$ is \textbf{Hermitian}.     
+\end{definition}
+
+
 
 
 
@@ -107,8 +119,7 @@
     The pair $(\complex(E), \iota)$ is the \textbf{complexification} of $E$ as a topological vector space, and
     \begin{enumerate}[start=4]
         \item If $E$ is locally convex, then so is $\complex(E)$. 
-        \item If $E$ is normed, then $\complex(E)$ is normable, and there exists a norm $\norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty)$ such that $\iota: E \to \complex(E)$ is isometric. 
-        \item[(F)] For any vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
+        \item[(F)] For any topological vector space $F$ over $\real$ and continuous $\real$-linear map $T: E \to F$, there exists a unique continuous $\complex$-linear map $\complex(T): \complex(E) \to \complex(F)$ such that the following diagram commutes:
         \[
         \xymatrix{
         \mathbb{C}(E) \ar@{->}[r]^{\mathbb{C}(T)} & \mathbb{C}(F) \\
@@ -120,6 +131,8 @@
         \[
         \complex(T)(x + iy) = Tx + iTy
         \]
+
+        Moreover, if $E$ and $F$ are normed, then $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. 
     \end{enumerate}
     
 \end{definition}
@@ -129,25 +142,81 @@
     (U): By (U) of the \hyperref[complexification]{definition:complexification}, there exists a $\complex$-linear map $\complex(T): \complex(E) \to F$ such that the given diagram commutes. Since $T \circ \iota$ and $iT \circ \iota$ are continuous, $T$ is continuous by (U) of the \hyperref[direct sum]{definition:tvs-direct-sum}.
 
     (4): By \autoref{proposition:finite-lc-product}, the direct sum and product of finitely many locally convex spaces coincide. By \autoref{proposition:lc-projective-topology}, this topology is locally convex. 
+    
 
-    (5): Let $\norm{\cdot}_E: E \to [0, \infty)$ be the norm of $E$, and define
+    (F): Existence of $\complex(T)$ is given by (U) applied to $\iota \circ T$. 
+
+    For the isometry, 
+    \begin{align*}
+    \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \norm{\complex(T)}
+    \end{align*}
+    
+\end{proof}
+
+\begin{definition}[Complexification of Normed Spaces]
+\label{definition:complexification-of-normed-spaces}
+    Let $E$ be a normed vector space over $\real$, then
     \[
     \norm{\cdot}_{\complex(E)}: \complex(E) \to [0, \infty) \quad (x, y) \mapsto \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x + \sin(\theta)y}_E
     \]
 
-    then for any $\phi \in [0, 2\pi]$ and $x, y \in E$, 
+    is a norm on $\complex(E)$ such that the inclusion map $\iota: E \to \complex(E)$ is isometric. 
+    
+    Moreover, for any normed vector space $F$ over $\real$ and $T \in L(E; F)$, $\norm{\complex(T)}_{L(\complex(E); \complex(F))} = \norm{T}_{L(E; F)}$. 
+\end{definition}
+\begin{proof}
+    For any $\phi \in [0, 2\pi]$ and $x, y \in E$, 
     \begin{align*}
         \normn{e^{i \phi}(x, y)}_{\complex(E)} &= \normn{(\cos(\phi)x - \sin(\phi)y, \sin(\phi)x + \cos(\phi)y)}_{\complex(E)} \\
         &= \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta - \phi)x + \sin(\theta - \phi)y}_E \\
         &= \norm{(x, y)}_{\complex(E)}
     \end{align*}
 
-    so $\norm{(x, y)}_{\complex(E)}$ is a norm. For any $x \in E$, 
+    so $\norm{\cdot}_{\complex(E)}$ is a norm on $\complex(E)$. For any $x \in E$, 
     \[
     \norm{\iota x}_{\complex(E)} = \sup_{\theta \in [0, 2\pi]}\norm{\cos(\theta)x}_E = \norm{x}_E \\
     \]
 
     Therefore $\iota: E \to \complex(E)$ is isometric. 
 
-    (F): By (U) applied to $\iota \circ T$.
+    Now, let $F$ be a normed vector space over $\real$ and $T \in L(E; F)$, then
+    \begin{align*}
+        \norm{\complex(T)}_{L(\complex(E); \complex(F))} &= \sup\bracsn{\norm{\complex(T)(x,y)}_{\complex(F)}|(x, y) \in \complex(E), \norm{(x, y)}_{\complex(E)} = 1} \\
+        &\ge \sup\bracsn{\norm{Tx}|x \in E, \norm{x}_E = 1} = \norm{T}_{L(E; F)}
+    \end{align*}
+
+    On the other hand, let $(x, y) \in \complex(E)$, then there exists $\theta \in [0, 2\pi]$ such that
+    \begin{align*}
+    \normn{\complex(T)(x, y)}_{\complex(F)} &= \normn{(Tx, Ty)}_{\complex(F)} = \norm{\cos(\theta)Tx + \sin(\theta)Ty}_F \\
+    &\le \norm{T}_{L(E; F)} \cdot \norm{\cos(\theta)x + \sin(\theta)y}_E \\
+    &\le \norm{T}_{L(E; F)} \cdot \sup_{\phi \in [0, 2\pi]}\norm{\cos(\phi)x + \sin(\phi)y}_E \\
+    &= \norm{T}_{L(E; F)} \cdot \norm{(x, y)}_E
+    \end{align*}
+    
+    
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:hermitian-functional-norm}
+    Let $E$ be a normed vector space over $\complex$, $*: E \to E$ be a complex conjugation map such that $\norm{x}_E = \normn{x^*}_E$ for all $x \in E$, and $\phi \in E^*$ be a Hermitian functional, then
+    \[
+    \norm{\phi}_{E^*} = \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
+    \]
+\end{proposition}
+\begin{proof}
+    Since $\bracsn{x \in E|x = x^*} \subset E$, $\norm{\phi}_{E^*} \ge \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}$. 
+
+    On the other hand, let $x \in E$ with $\norm{x}_E = 1$. Assume without loss of generality that $\dpn{x, \phi}{E} \in \real$, then
+    \begin{align*}
+        \dpn{x, \phi}{E} &= \dpn{\text{Re}(x), \phi}{E} + \underbrace{i\dpn{\text{Im}(x), \phi}{E}}_{\in \real} =
+\dpn{\text{Re}(x), \phi}{E} \\
+&\le  \norm{\text{Re}(x)}_E \cdot \sup\bracsn{\dpn{y, \phi}{E}|y \in E, y = y^*, \norm{y}_E = 1}
+    \end{align*}
+
+    where $\norm{\text{Re}(x)}_E = \norm{{(x + x^*)}/{2}}_E \le \norm{x}_E$. As the above holds for all $x \in E$, 
+    \[
+    \norm{\phi}_{E^*} \le \sup\bracsn{\dpn{x, \phi}{E}|x \in E, x = x^*, \norm{x}_E = 1}
+    \]
+\end{proof}
+

src/measure/lcg/lcg.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Locally Compact Group]
 \label{definition:lcg}
-    Let $G$ be a topological group, then $G$ is \textbf{locally compact} if $G$ is a LCH space. 
+    Let $G$ be a topological group, then $G$ is \textbf{locally compact} if $G$ is an LCH space. 
 \end{definition}
 
 \begin{proposition}

src/measure/measure/regular.tex

@@ -28,7 +28,7 @@
 \label{theorem:sigma-compact-regular-measure}
     Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure. If:
     \begin{enumerate}
-        \item[(a)] $X$ is a LCH space.
+        \item[(a)] $X$ is an LCH space.
         \item[(b)] Every open set of $X$ is $\sigma$-compact.
         \item[(c)] For any $K \subset X$ compact, $\mu(K) < \infty$.
     \end{enumerate}

src/measure/measure/scaffold.tex

@@ -16,7 +16,7 @@
 \end{definition}
 
 \begin{example}
-    Let $X$ be a LCH space, $\mu$ be a Radon measure, and $\mathcal{K}$ be the collection of compact subsets of $X$, then $\mathcal{K}$ is a scaffold for $\mu$. 
+    Let $X$ be an LCH space, $\mu$ be a semifinite Radon measure, and $\mathcal{K}$ be the collection of compact subsets of $X$, then $\mathcal{K}$ is a scaffold for $\mu$. 
 \end{example}
 % Omitted
 

src/measure/radon/c0.tex

@@ -20,14 +20,14 @@
 
 \begin{definition}[Radon Measure]
 \label{definition:radon-measure-extended}
-    Let $X$ be a LCH space and $\mu$ be a Borel signed/vector measure on $X$, then $\mu$ is \textbf{Radon} if $|\mu|$ is Radon. 
+    Let $X$ be an LCH space and $\mu$ be a Borel signed/vector measure on $X$, then $\mu$ is \textbf{Radon} if $|\mu|$ is Radon. 
 \end{definition}
 
 
 
 \begin{definition}[Space of Finite Radon Measures]
 \label{definition:space-radon-measures}
-    Let $X$ be a LCH space and $E$ be a normed vector space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
+    Let $X$ be an LCH space and $E$ be a normed vector space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
 \end{definition}
 \begin{proof}
     Let $\mu, \nu \in M_R(X; E)$, then for any $A \in \cb_X$, $|\mu + \nu|(A) \le |\mu|(A) + |\nu|(A)$. Let $\eps > 0$, then by outer regularity and \autoref{proposition:radon-measurable-description}, there exists $K \subset A$ compact and $U \in \cn^o(A)$ such that $(|\mu| + |\nu|)(A \setminus K), (|\mu| + |\nu|)(U \setminus A) < \eps$. Therefore $|\mu + \nu|$ is regular on all Borel sets, and hence Radon. 
@@ -132,7 +132,7 @@
 
     then $T$ is maps $C_0(X; E)$ continuously into a subspace of $C_0(X \times B; K)$. 
 
-    Let $I \in C_0(X; E)^*$, then by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\ol{I} \in C_0(X \times B; K)^*$ such that $\ol I \circ T = I$. By \hyperref[Alaoglu's Theorem]{theorem:alaoglu}, $B$ is a compact Hausdorff space. Therefore $X \times B$ is a LCH space by \autoref{proposition:lch-product}. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(X \times B; K)$ such that for any $f \in C_0(X \times B; K)$, 
+    Let $I \in C_0(X; E)^*$, then by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\ol{I} \in C_0(X \times B; K)^*$ such that $\ol I \circ T = I$. By \hyperref[Alaoglu's Theorem]{theorem:alaoglu}, $B$ is a compact Hausdorff space. Therefore $X \times B$ is an LCH space by \autoref{proposition:lch-product}. By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, there exists $\mu \in M_R(X \times B; K)$ such that for any $f \in C_0(X \times B; K)$, 
     \[
     \dpn{f, \ol I}{C_0(X \times B; K)} = \int_{X \times B} f d\mu
     \]

src/measure/radon/radon.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Radon Measure]
 \label{definition:radon-measure}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure, then $\mu$ is a \textbf{Radon measure} if:
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure, then $\mu$ is a \textbf{Radon measure} if:
     \begin{enumerate}
         \item[(R1)] For any $K \subset X$ compact, $\mu(K) < \infty$.
         \item[(R2)] $\mu$ is outer regular on all Borel sets.
@@ -13,7 +13,7 @@
 
 \begin{proposition}
 \label{proposition:radon-measure-cc}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
     \begin{enumerate}
         \item For any $U \subset X$ open,
         \[
@@ -53,7 +53,7 @@
 
 \begin{proposition}
 \label{proposition:radon-regular-sigma-finite}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
     \begin{enumerate}
         \item $\mu$ is inner regular on all its $\sigma$-finite sets.
         \item If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
@@ -87,7 +87,7 @@
 
 \begin{proposition}
 \label{proposition:radon-measurable-description}
-    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
+    Let $X$ be an LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
     \begin{enumerate}
         \item For every $\eps > 0$, there exists $U \in \cn^o(E)$ and $F \subset E$ closed such that $\mu(U \setminus F) < \eps$.
         \item There exists a $F_\sigma$ set $A$ and a $G_\delta$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
@@ -111,7 +111,7 @@
 
 \begin{proposition}
 \label{proposition:finite-compact-regular}
-    Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure such that:
+    Let $X$ be an LCH space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure such that:
     \begin{enumerate}
         \item[(a)] For any $U \subset X$ open, $U$ is $\sigma$-compact.
         \item[(b)] For any $K \subset X$ compact, $\mu(K) < \infty$.
@@ -151,7 +151,7 @@
 
 \begin{lemma}
 \label{lemma:radon-compact-project}
-    Let $X$ be a LCH space, $Y$ be a compact Hausdorff space, $\mu$ be a finite Radon measure on $X \times Y$, and $\nu$ be a measure on $X$. If for each $A \in \cb_X$, $\nu(A) \le \mu(A \times Y)$, then $\nu$ is also a Radon measure. 
+    Let $X$ be an LCH space, $Y$ be a compact Hausdorff space, $\mu$ be a finite Radon measure on $X \times Y$, and $\nu$ be a measure on $X$. If for each $A \in \cb_X$, $\nu(A) \le \mu(A \times Y)$, then $\nu$ is also a Radon measure. 
 \end{lemma}
 \begin{proof}
     Let $A \in \cb_X$ and $\eps > 0$. By outer regularity of $\mu$, there exists $U \in \cn_{X \times Y}(A \times Y)$ such that $\mu(U \setminus (A \times Y)) < \eps$. By the \hyperref[Tube Lemma]{lemma:tube-lemma}, there exists $V \in \cn_X(A)$ such that $V \times Y \subset U$. In which case, 
@@ -172,7 +172,7 @@
 
 \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 vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
+    Let $X$ be an LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
     By \autoref{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.
@@ -187,7 +187,7 @@
 
 \begin{theorem}[Lusin]
 \label{theorem:lusin}
-    Let $X$ be a LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
+    Let $X$ be an LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
     \begin{enumerate}
         \item There exists $A \subset \bracs{f \ne 0}$ such that $f|_A$ is continuous and $\mu(\bracs{f \ne 0} \setminus A) < \eps$
         \item If $E = \complex$, then there exists $\phi \in C_c(X; E)$ such that $\mu\bracs{f \ne \phi} < \eps$.
@@ -228,7 +228,7 @@
 
 \begin{proposition}[Monotone Convergence Theorem (LSC)]
 \label{proposition:mct-radon}
-    Let $X$ be a LCH space, $\net{f}$ and $f: X \to [0, \infty]$ be non-negative lower semicontinuous functions such that $f_\alpha \upto f$, then for any Radon measure $\mu$ on $X$, 
+    Let $X$ be an LCH space, $\net{f}$ and $f: X \to [0, \infty]$ be non-negative lower semicontinuous functions such that $f_\alpha \upto f$, then for any Radon measure $\mu$ on $X$, 
     \[
     \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu
     \]

src/measure/radon/riesz.tex

@@ -8,7 +8,7 @@
 
 \begin{proposition}
 \label{proposition:positive-linear-functional-cc-property}
-    Let $X$ be a LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then
+    Let $X$ be an LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then
     \begin{enumerate}
         \item For any $f, g \in C_c(X; \real)$ with $f \le g$, $\dpb{f, I}{C_c(X; \real)} \le \dpb{g, I}{C_c(X; \real)}$.
         \item For any $K \subset X$ compact, there exists $C_K \ge 0$ such that for all $f \in C_c(X; \real)$ with $\supp{f} \subset K$, $|{\dpb{f, I}{C_c(X; \real)}}| \le \norm{f}_u$.
@@ -28,7 +28,7 @@
 
 \begin{theorem}[Riesz Representation Theorem]
 \label{theorem:riesz-radon}
-    Let $(X, \topo)$ be a LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that:
+    Let $(X, \topo)$ be an LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that:
     \begin{enumerate}
         \item For any $U \subset X$ open, $\mu(U) = \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}$.
         \item For any $K \subset X$ compact, $\mu(K) = \inf_{f \in C_c(X; [0, 1]), f \ge \one_K}\dpb{f, I}{C_c(X; \real)}$.

src/measure/vector/fin.tex

@@ -117,3 +117,46 @@ Despite not covering the full dual space, the bounded Borel functions still form
     (2) $\Rightarrow$ (1): By the \hyperref[Dominated Convergence Theorem]{theorem:dct-bochner-vector}. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:space-of-measures-extreme-points}
+    Let $X$ be an LCH space and $\cm \subset \overline{B_{M_R(X; \complex)}(0, 1)}$ be a compact convex set such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $\mu \in \cm$ and $A, B \in \cb_X$, let $\mu_A(B) = \mu(A \cap B)$, then $\mu_A \in \cm$.
+        \item For each $\mu \in \cm \setminus \bracs{0}$ and $t \in [0, 1/\norm{\mu}_{\text{var}}]$, $t\mu \in \cm$. 
+    \end{enumerate}
+
+    then for any $\mu \in \cm \setminus \bracs{0}$, the following are equivalent: 
+    \begin{enumerate}
+        \item $\norm{\mu}_{\text{var}} = 1$ and $\mu$ takes on exactly two distinct values. 
+        \item There exists $x \in X$ and $\lambda \in \partial B_\complex(0, 1)$ such that $\mu = \lambda \delta_x$. 
+        \item $\mu$ is an extreme point of $\cm$. 
+    \end{enumerate}
+
+    Moreover, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)} \cup\bracs{0}) \cap \cm$. 
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Assume without loss of generality that $\mu$ is positive and $\mu(\cb_X) = \bracs{0, 1}$. By inner regularity on open sets, there exists at least one compact set $K \subset X$ such that $\mu(K) = 1$. 
+
+    Let $\mathcal{F} = \bracs{K \subset X|K \text{ compact}, \mu(K) = 1}$, then $\mathcal{F}$ is a $\pi$-system that does not contain $\emptyset$, and as such satisfies the finite intersection property. Thus $A = \bigcap_{K \in \mathcal{F}}K \ne \emptyset$. 
+
+    Let $U \in \cn_X(A)$ and $K \in \cf$, then $K \setminus U$ is compact. Since $K \setminus U \cap A = \emptyset$, $K \setminus U \not\in \cf$, and $\mu(K \setminus U) = 0$. Thus $\mu(U) = \mu(K \cap U) = 1$. As this holds for all $U \in \cn_X(A)$, $\mu(A) = 1$ by outer regularity. 
+
+    Finally, let $x \in A$ and $U \in \cn_X(x)$, then $A \setminus U \subsetneq A$, so $A \setminus U \not\in \cf$. As such, $A \subset A \cap \ol U$ for all $U \in \cn_X(x)$. Since $\bigcap_{U \in \cn_X(x)}\ol{U} = \bracs{x}$, $A = \bracs{x}$, and $\mu = \delta_x$. 
+    
+    (2) $\Rightarrow$ (3): Assume without loss of generality that $\mu = \delta_x$. 
+    
+    Let $\nu, \rho \in \cm$ and $t \in (0, 1)$ such that $\mu = (1 - t)\nu + t\rho$, then $1 = \mu(\bracs{x}) = (1 - t)\nu(\bracs{x}) + t\rho(\bracs{x})$. Since $\mu(\bracs{x}) = 1$ and $|\nu(\bracs{x})|, |\rho(\bracs{x})| \le 1$, $\nu(\bracs{x}) = \rho(\bracs{x}) = 1$. As $\norm{\nu}_{\text{var}}, \norm{\rho}_{\text{var}} \le 1$, $\nu = \rho = \delta_x = \mu$. Therefore $\mu$ is an extreme point of $\cm$. 
+
+    (3) $\Rightarrow$ (1): If $\norm{\mu}_{\text{var}} \in (0, 1)$, then $\mu$ is a convex combination of $0$ and $\mu/\norm{\mu}_{\text{var}}$, so $\norm{\mu}_{\text{var}}$ must be $1$. 
+    
+    Suppose that $\mu$ takes on at least three distinct values, then there exists $A \in \cb_X$ such that $|\mu|(A), |\mu|(X \setminus A) > 0$. For each $B \in \cb_X$, let $\nu(B) = \mu(B \cap A)$ and $\rho(B) = \mu(B \setminus A)$, then $\mu = \nu + \rho$, $\nu, \rho \ne 0$, $\nu \perp \rho$, and $\norm{\nu}_{\text{var}} + \norm{\rho}_{\text{var}} = \norm{\mu}_{\text{var}}$. In which case, 
+    \[
+    \mu = \frac{\norm{\nu}_{\text{var}}}{\norm{\mu}_{\text{var}}} \cdot \underbrace{\frac{\norm{\mu}_{\text{var}} \cdot \nu}{\norm{\nu}_{\text{var}}}}_{\in \cm} + \frac{\norm{\rho}_{\text{var}}}{\norm{\mu}_{\text{var}}} \cdot \underbrace{\frac{\norm{\mu}_{\text{var}} \cdot \rho}{\norm{\rho}_{\text{var}}}}_{ \in \cm}
+    \]
+
+    is a convex combination of $\mu$ in terms of two other elements of $\cm$. 
+
+    Finally, by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $\cm$ is the closed convex hull of $(\bracsn{\lambda \delta_x|x \in X, \lambda \in \partial B_\complex(0, 1)} \cup\bracs{0}) \cap \cm$. 
+\end{proof}
+
+

src/op/banach/definitions.tex

@@ -29,9 +29,16 @@
 
 \begin{definition}[Unital Homomorphism]
 \label{definition:banach-algebra-unital-homomorphism}
-    Let $A, B$ be unital Banach algebras and $\phi: A \to B$ be a homomorphism, then $\phi$ is a \textbf{unital homomorphism} if $\phi(1) = 1$. 
+    Let $A, B$ be unital Banach algebras and $\phi: A \to B$ be a homomorphism, then $\phi$ is a \textbf{unital homomorphism} if $\phi(1_A) = 1_B$. 
 \end{definition}
 
+\begin{definition}[Representation]
+\label{definition:banach-algebra-representation}
+    Let $A$ be a Banach algebra, then a \textbf{representation} of $A$ is a pair $(E, \pi)$ where $E$ is a Banach space, and $\pi: A \to L(E; E)$ is a continuous homomorphism. 
+\end{definition}
+
+
+
 \begin{definition}[Unitisation]
 \label{definition:unitisation}
     Let $A$ be a Banach algebra over $\complex$, and $\tilde A = \complex \oplus A$ with

src/op/banach/invertible.tex

@@ -65,7 +65,7 @@
 \label{proposition:swap-invertible}
     Let $A$ be a unital Banach algebra and $x, y \in A$, then $1 - xy \in G(A)$ if and only if $1 - yx \in G(A)$. 
 \end{proposition}
-\begin{proof}[Proof, {{\cite[Proposition I.3.4]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Proposition 3.4]{Zhu}}}. ]
     If $1 - xy \in G(A)$, then
     \begin{align*}
         (1 - yx)[y(1 - xy)^{-1}x + 1] &= y(1 - xy)^{-1}x + 1 \\

src/op/banach/multiplicative.tex

@@ -37,7 +37,7 @@
     Let $A$ be a Banach algebra, then $\Omega(A)$ is the \textbf{space of multiplicative linear functionals}, and with respect to the weak-* topology, 
     \begin{enumerate}
         \item If $A$ is unital, then $\Omega(A)$ is a compact Hausdorff space. 
-        \item $\Omega(A) \cup \bracs{0}$ is a compact Hausdorff space, and $\Omega(A)$ is a LCH space.
+        \item $\Omega(A) \cup \bracs{0}$ is a compact Hausdorff space, and $\Omega(A)$ is an LCH space.
     \end{enumerate}
 \end{definition}
 \begin{proof}
@@ -69,7 +69,7 @@
         \item $\phi(1) = 1$ and $\phi(G(A)) \subset \complex \setminus \bracs{0}$. 
     \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem I.4.5]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 4.5]{Zhu}}}. ]
     (1) $\Rightarrow$ (2): \autoref{proposition:multiplicative-unit}. 
 
     (2) $\Rightarrow$ (1): Let $x \in A$ and $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, then $\lambda - x \in G(A)$ and $\phi(\lambda - x) \ne 0$. Therefore $\phi(x) \subset \ol{B(0, [x]_{sp})}$, and $\norm{\phi}_{A^*} = 1$. 

src/op/banach/spectrum.tex

@@ -128,7 +128,7 @@
 
 \begin{proposition}
 \label{proposition:commutative-spectrum-gymnastics}
-    Let $A$ be a commutative 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 $xy = yx$, then
     \begin{enumerate}
         \item $\sigma_A(x + y) \subset \sigma_A(x) + \sigma_A(y)$.
         \item $\sigma_A(xy) \subset \sigma_A(x)\sigma_A(y)$.
@@ -161,7 +161,7 @@
     (2): Let $\lambda \in \partial \sigma_B(x)$, then there exists $\seq{\lambda_n} \subset \complex \setminus \sigma_B(x)$ such that $\lambda_n - x \in G(B)$ for all $n \in \natp$, and $\lambda_n \to \lambda$ as $n \to \infty$. By \autoref{corollary:invertible-boundary-explode}, $\norm{(\lambda_n - x)^{-1}}_A \to \infty$ as $n \to \infty$. If $\lambda - x \in G(A)$, then $(\lambda_n - x)^{-1} \to (\lambda - x)^{-1}$ as $n \to \infty$. Thus $\norm{(\lambda - x)^{-1}}_A = \infty$, which is impossible. Therefore $\lambda - x \not\in G(A)$, and $\lambda \in \sigma_A(x)$. 
 \end{proof}
 
-\begin{theorem}["Runge's Theorem"]
+\begin{theorem}[Runge]
 \label{theorem:spectrum-subalgebra-sufficiency}
     Let $A$ be a unital Banach algebra, $x \in A$, $P \subset \complex \setminus \sigma_A(x)$ such that $P$ intersects every bounded component of $\complex \setminus \sigma_A(x)$, and $B \subset A$ be a closed algebra containing $1$, $x$, and $\bracsn{(\lambda - x)^{-1}|\lambda \in P}$, then $\sigma_A(x) = \sigma_B(x)$. 
 \end{theorem}

src/op/c-star/cont.tex

@@ -18,7 +18,7 @@
 
     and $\Phi$ is a surjection onto $\sigma_A(x)$. 
 
-    On the other hand, the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark} implies that $\Phi(x^*) = \ol{\Phi(x)}$, so since $A[x]$ is the smallest $C^*$-algebra containing $x$, any element $\psi \in \Omega(A[x])$ is uniquely determined by $\psi(x)$. Therefore $\Phi$ is injective. 
+    On the other hand, the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark} implies that $\psi(x^*) = \ol{\psi(x)}$ for any $\psi \in \Omega(A[x])$. Since $A[x]$ is the smallest $C^*$-algebra containing $x$, any element $\psi \in \Omega(A[x])$ is uniquely determined by $\psi(x)$. Therefore $\Phi$ is injective. 
 
     Finally, since $\Omega(A[x])$ is equipped with the weak* topology and $\Phi$ is the evaluation map at $x$, it is continuous. 
 
@@ -49,8 +49,8 @@
         \]
 
         is continuous. 
+        \item Let $B$ be a unital $C^*$-algebra and $\Phi: A \to B$ be a unital *-homomorphism, then for any $x \in A$ and $f \in C(\sigma_A(x); \complex)$, $\Phi(f(x)) = f(\Phi(x))$.
     \end{enumerate}
-    
 \end{definition}
 \begin{proof}
     (1), (4), (5): By the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, $\Omega(A[x])$ and $\sigma_A(x)$ may be identified. For each $f \in C(\sigma_A(x); \complex)$, define $f(x) = \Gamma^{-1}_{A[x]}(f)$, then the mapping $f \mapsto f(x)$ is a unital *-isomorphism by the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}. 
@@ -116,6 +116,30 @@
     \end{enumerate}
 \end{corollary}
 
-
+\begin{corollary}
+\label{corollary:sum-of-unitaries}
+    Let $A$ be a unital $C^*$-algebra, then
+    \begin{enumerate}
+        \item For any $x \in A_{sa}$, there exists a unitary element $u \in A$ such that $x = (u + u^*)/(2\norm{x}_A)$. 
+        \item For any $x \in A$, there exists unitary elements $u, v \in A$ such that $x = (u + u^*)/(\norm{x}_A) + i(v + v^*)/(2\norm{x}_A)$. 
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Assume without loss of generality that $\norm{x}_A \le 1$. In which case, $\sigma(x) \subset [-1, 1]$, and $f(\lambda) = \lambda + i\sqrt{1 - \lambda^2}$ is defined and continuous on $[-1, 1]$. Furthermore, $|f(\lambda)| = 1$ for all $\lambda \in [-1, 1]$. Thus \autoref{corollary:normal-spectrum-identity} implies that $f(x)$ is unitary. Finally, since $f + \ol f = 2\text{Id}$, $x = (f(x) + \ol{f(x)})/(2\norm{x}_A)$.
+\end{proof}
+
+\begin{corollary}
+\label{corollary:star-homomorphism-continuous}
+    Let $A, B$ be unital $C^*$-algebras and $\Phi: A \to B$ be an injective unital *-homomorphism, then for each $x \in A$,
+    \begin{enumerate}
+        \item $\sigma_B(\Phi(x)) = \sigma_A(x)$. 
+        \item $\norm{\Phi(x)}_B = \norm{x}_A$.
+    \end{enumerate}
+\end{corollary}
+\begin{proof}[Proof, {{\cite[10.7]{Zhu}}}. ]
+    (1): Since $\Phi(G(A)) \subset G(B)$, $\sigma_B(\Phi(x)) \subset \sigma_A(x)$. If $\sigma_B(\Phi(x)) \subsetneq \sigma_A(x)$, then \hyperref[Urysohn's Lemma]{lemma:urysohn} implies that there exists $C(\sigma_A(x); \complex)$ such that $f|_{\sigma_B(\Phi(x))} = 0$ but $f \ne 0$. In which case, by (7) of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(f(x)) = f(\Phi(x)) = 0$, which contradicts the fact that $\Phi$ is injective. 
+
+    (2): By \autoref{corollary:c-star-unique-norm}, $\Phi$ is isometric. 
+\end{proof}
 
 

src/op/c-star/gelfand.tex

@@ -10,7 +10,7 @@
 
     is a unital $C^*$-isomorphism. 
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem II.9.4]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 9.4]{Zhu}}}. ]
     By construction $\Gamma_A$ is a unital algebra homomorphism. 
 
     To see that $\Gamma_A$ preserves involutions, let $y \in A$ be self-adjoint. By \autoref{proposition:gelfand-transform-gymnastics} and \autoref{proposition:self-adjoint-spectrum}, $\Gamma_A(y)(\Omega(A)) = \sigma_A(y) \subset \real$, so $\Gamma_A(y) \in C(\Omega(A); \real)$. For any $x \in A$, write $x = \text{Re}(x) + i\text{Im}(x)$, where $\text{Re}(x)$ and $\text{Im}(x)$ are both self-adjoint, then

src/op/c-star/gns.tex

@@ -0,0 +1,176 @@
+\section{The GNS Construction}
+\label{section:gns}
+
+\begin{definition}[Cyclic Representation]
+\label{definition:cyclic-representation}
+    Let $A$ be a $C^*$-algebra, $(H, \pi)$ be a representation of $A$, and $\xi \in H$, then $\xi$ is a \textbf{cyclic vector} for $(H, \pi)$ if $\bracsn{\pi(x)(\xi)|x \in A}$ is dense in $H$. The representation $(H, \pi)$ is \textbf{cyclic} if it admits a cyclic vector. 
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:cstar-state-kernel}
+    Let $A$ be a $C^*$-algebra, $\phi \in S(A)$, and
+    \[
+    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For any $x, y \in A$ with $x \in N_\phi$ or $y \in N_\phi$, $\dpn{x, y}{A} = 0$. 
+        \item $N_\phi$ is a closed left ideal of $A$. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for any $x, y \in A$, 
+    \[
+    |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
+    \]
+
+    If $x \in N_\phi$ or $y \in N_\phi$, then the above inequality shows that $\dpn{x, y}{\phi} = 0$. 
+
+    (2): As the zero set of a continuous function on $A$, $N_\phi$ is closed.
+
+    For any $x, y \in N_\phi$, 
+    \begin{align*}
+        \dpn{x + y, x + y}{\phi} &= \dpn{x, x}{\phi} + \dpn{x, y}{\phi} + \dpn{y, x}{\phi} + \dpn{y, y}{\phi} \\
+        &= \dpn{x, y}{\phi} + \dpn{y, x}{\phi} 
+    \end{align*}
+
+    By (1), $\dpn{x, y}{\phi} = \dpn{y, x}{\phi} = 0$. Therefore $x + y \in N_\phi$. 
+
+    Finally, for each $x \in N_\phi$ and $y \in A$, 
+    \[
+        \dpn{yx, yx}{\phi} = \dpn{x^*y^*yx, \phi}{A} = \dpn{x^*(y^*yx), \phi}{A} = \dpn{y^*yx, x}{\phi} =0 
+    \]
+
+    by (1). 
+\end{proof}
+
+
+\begin{definition}[GNS Triple]
+\label{definition:gns-triple}
+    Let $A$ be a unital $C^*$-algebra, $\phi \in S(A)$, and
+    \[
+    N_\phi = \bracsn{x \in A| \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = 0}
+    \]
+
+    Let $H_\phi^0 = A/N_\phi$, $H_\phi$ be its completion with respect to $\dpn{\cdot, \cdot}{\phi}$, and
+    \[
+    \pi_\phi^0: A \to B(H_\phi^0) \quad \pi_\phi^0(x)(y + N_\phi) = xy + N_\phi
+    \]
+
+    For each $x \in A$, let $\pi_\phi(x)$ be the continuous extension of $\pi_\phi^0(x)$ to an element of $B(H_\phi)$, then:
+    \begin{enumerate}
+        \item $(H_\phi, \dpn{\cdot, \cdot}{\phi})$ is a Hibert space.
+        \item $(H_\phi, \pi_\phi)$ is a well-defined representation of $A$. 
+        \item $\xi_\phi = 1_A + N_\phi$ is a unit vector in $H_\phi$, and $\bracsn{\pi_\phi(x)\xi_\phi| x \in A}$ is dense in $H_\phi$. Moreover, for each $x, y \in A$,
+        \[
+        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
+        \]
+    \end{enumerate}
+    
+    The representation $(H_\phi, \pi_\phi)$ is the \textbf{cyclic representation of $A$ induced by $\phi$}, and the triple $(H_\phi, \pi_\phi, \xi_\phi)$ is the \textbf{Gelfand-Naimark-Segal (GNS) triple associated with $\phi$}. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition 14.2]{Zhu}}}. ]
+    (2): Fix $x \in A$, then for each $y_1, y_2 \in A$ with $y_1 - y_2 \in N_\phi$, $x(y_1 - y_2) \in N_\phi$ by \autoref{lemma:cstar-state-kernel}, so $\pi_\phi^0(x)$ is well-defined on $A/N_\phi$. 
+
+    By rescaling, assume without loss of generality that $\norm{x}_A \le 1$. In which case, for each $y \in A$, 
+    \[
+        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*y, \phi}{A} - \dpn{y^*x^*xy, \phi}{A} = \dpn{y^*(1 - x^*x)y, \phi}{A}
+    \]
+
+    Since $\sigma_A(x^*x) \subset [0, 1]$, $\sigma_A(1 - x^*x) \subset [0, 1]$ and is positive by \autoref{corollary:spectrum-characterisation-iff}. Thus there exists $z \in A$ positive such that $(1 - x^*x) = z^*z$, so
+    \[
+        \dpn{y, y}{\phi} - \dpn{xy, xy}{\phi} = \dpn{y^*z^*zy, \phi}{A} = \dpn{zy, zy}{\phi} \ge 0
+    \]
+
+    and $\dpn{y, y}{\phi} \ge \dpn{xy, xy}{\phi}$. Therefore $\pi_\phi^0(x)$ extends continuously into an element of $B(H_\phi)$ by the \hyperref[linear extension theorem]{theorem:linear-extension-theorem-normed}. 
+    
+
+    Now, let $x, y, z \in A$, then
+    \[
+    \pi_\phi^0(x)[\pi_\phi^0(y)(z + N_\phi)] = \pi_\phi^0(x)(yz + N_\phi) = xyz + N_\phi = \pi_\phi^0(xy)(z + N_\phi)
+    \]
+
+    and by uniqueness of continuous extensions, $\pi_\phi(x)\pi_\phi(y) = \pi_\phi(xy)$, so $\pi_\phi$ is a homomorphism. 
+
+    Finally, 
+    \[
+        \dpn{\pi_\phi^0(x^*)y, z}{\phi} = \dpn{z^*x^*y, \phi}{A} = \dpn{y, xz}{\phi} = \dpn{y, \pi_\phi^0(x)z}{\phi}
+    \]
+
+    By uniqueness of continuous extensions, $\pi_\phi(x^*) = \pi_\phi(x)^*$. Therefore $\pi_\phi$ is a *-homomorphism, and $(H_\phi, \pi_\phi)$ is a representation of $A$. 
+
+    (3): Since $\phi$ is a state, $\dpn{1_A, 1_A}{\phi} = 1$, and $1_A$ is a unit vector. As $H_\phi$ is the completion of $A/N_\phi$ and $A/N_\phi = \bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$, $\bracsn{\pi_\phi(x)(1_A + N_\phi)| x \in A}$ is dense in $H_\phi$. 
+
+    For each $x, y \in A$, $\dpn{x, y}{\phi} = \dpn{\pi_\phi(x)(1_A + N_\phi), \pi_\phi(y)(1_A + N_\phi)}{H_\phi}$ by well-definedness of the inner product on $H_\phi$.
+\end{proof}
+
+\begin{theorem}[Gelfand-Naimark-Segal]
+\label{theorem:gns}
+    Let $A$ be a unital $C^*$-algebra, then:
+    \begin{enumerate}
+        \item For each $\phi \in S(A)$, there exists a triple $(H_\phi, \pi_\phi, \xi_\phi)$ where $(H_\phi, \pi_\phi)$ is a representation of $A$, $\xi_\phi$ is a cyclic unit vector of $(H_\phi, \pi_\phi)$, and
+        \[
+        \dpn{x, y}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(y)\xi_\phi}{H_\phi}
+        \]
+        \item For each representation $(H, \pi)$ of $A$ with cyclic unit vector $\xi$, the mapping
+        \[
+        \phi: A \to \complex \quad x \mapsto \dpn{\pi(x)\xi, \xi}{H}
+        \]
+
+        is a state on $A$. Moreover, if $(H_\phi, \pi_\phi, \xi_\phi)$ is the GNS triple associated with $\phi$, then there exists a unitary equivalence $U: H \to H_\phi$ such that $U\xi = \xi_\phi$. 
+        \item For each $\mathcal{S} \subset S(A)$, the mapping
+        \[
+        \pi_{\mathcal{S}}: A \to B([l^2(\mathcal{S}); H_\phi]) \quad \pi_{\mathcal{S}}(x)(\eta)_\phi = \pi_{\phi}(x)(\eta_\phi)
+        \]
+
+        is a representation of $A$, which is injective if for every $x \in A$, there exists $\phi \in \mathcal{S}$ with $\dpn{x^*x, \phi}{A} \ne 0$. 
+
+        In particular, $A$ is isomorphic to a closed subalgebra of $B([l^2(P(A)); H_\phi])$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): By the \hyperref[GNS construction]{definition:gns-triple}. 
+
+    (2): For each $x \in A$, if $x$ is positive, then so is $\pi(x)$, so $\dpn{\pi(x)\xi, \xi}{H} \ge 0$. Since $\xi$ is a unit vector, $\dpn{\pi(1_A)\xi, \xi}{H} = \dpn{\xi, \xi}{H} = 1$, and $\phi$ is a state. 
+
+    Let $H^0 = \bracsn{\pi(x)\xi|x \in A}$ and $H_\phi^0 = \bracsn{\pi_\phi(x)\xi_\phi|x \in A}$. Define
+    \[
+    U: H^0 \to H_\phi^0 \quad \pi(x)\xi \mapsto \pi_\phi(x)\xi_\phi
+    \]
+
+    then for each $x, y \in A$ with $\pi(x - y)\xi = 0$,
+    \begin{align*}
+    0 &= \dpn{\pi(x - y)\xi, \pi(x - y)\xi}{H} = \dpn{(x - y)^*(x - y), \phi}{A} \\
+    &= \dpn{x - y, x- y}{\phi} = \dpn{\pi_\phi(x - y)\xi_\phi, \pi_\phi(x - y)\xi_\phi}{H_\phi}
+    \end{align*}
+    
+
+    and $\pi_\phi(x - y)\xi_\phi = 0$ as well. Thus $U$ is well-defined. Moreover, for each $x \in A$,
+    \[
+        \dpn{\pi(x)\xi, \pi(x)\xi}{H} = \dpn{x^*x, \phi}{A} = \dpn{x^*x, 1_A}{\phi} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x) \xi_\phi}{H_\phi}
+    \]
+
+    so $U$ is an isometry. For each $x, y \in A$,
+    \begin{align*}
+    U(\pi(x)[\pi(y)\xi]) &= U(\pi(xy)\xi) = \pi_\phi(xy)\xi_\phi \\
+    &= \pi_\phi(x)[\pi_\phi(y)\xi_\phi] = \pi_\phi(x)[U(\pi(y)\xi)]
+    \end{align*}
+    
+
+    so $U$ \hyperref[extends continuously]{theorem:linear-extension-theorem-normed} to a unitary equivalence between $(H, \pi)$ and $(H_\phi, \pi_\phi)$, with $U(\xi) = \xi_\phi$. 
+
+    (3): Suppose that for each $x \in A$, there exists $\phi \in \mathcal{S}$ such that $\dpn{x^*x, \phi}{A} \ne 0$. In which case,
+    \[
+    0 \ne \dpn{x, x}{\phi} = \dpn{x^*x, \phi}{A} = \dpn{\pi_\phi(x)\xi_\phi, \pi_\phi(x)\xi_\phi}{H_\phi}
+    \]
+
+    so $\pi_\phi(x) \ne 0$, and $\pi_{\mathcal{S}}(x) \ne 0$ as well. 
+
+    By \autoref{corollary:cstar-positive-weakstar-dense}, for each $x \in A$, there exists $\phi \in P(A)$ with $\dpn{x^*x, \phi}{A} \ne 0$, so $\pi_{P(A)}$ is injective. By \autoref{theorem:continuity-of-homomorphism-c-star}, $\pi_{P(A)}(A)$ is closed in $B([l^2(P(A)); H_\phi])$. 
+    
+\end{proof}
+
+
+

src/op/c-star/homomorphism.tex

@@ -12,7 +12,7 @@
 
     If $A$ and $B$ are unital, then $\phi$ is \textbf{unital} if:
     \begin{enumerate}
-        \item[(U)] $\phi(1_A) = \one_B$. 
+        \item[(U)] $\phi(1_A) = 1_B$. 
     \end{enumerate}
 \end{definition}
 
@@ -36,4 +36,42 @@
     
 \end{proof}
 
+\begin{theorem}
+\label{theorem:continuity-of-homomorphism-c-star}
+    Let $A, B$ be unital $C^*$-algebras and $\Phi: A \to B$ be a unital *-homomorphism, then $\Phi(A)$ is closed. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 11.1]{Zhu}}}. ]
+    Let $y \in \ol{\Phi(A)} \cap B_{sa}$, then there exists $x \in A_{sa}$ such that $\norm{y - \Phi(x)}_B \le \norm{y}_B/2$. Let
+    \[
+    f: \complex \to \complex \quad z \mapsto \begin{cases}
+        z & |z| \le 2\norm{y}_F \\
+        2\norm{y}_F \cdot \sgn z = 2\norm{y}_F \cdot \frac{z}{|z|} & |z| \ge 2\norm{y}_F
+    \end{cases}
+    \]
+
+    then $f \in C(\complex; \complex)$. Since $\norm{\Phi(x)}_B \le \norm{y}_B + \norm{y - \Phi(x)}_B \le 2\norm{y}_B$, $\sigma_B(\Phi(x)) \subset \ol{B_\complex(0, 2\norm{y}_B)}$, and $f|_{\sigma_B(\Phi(x))}$ is the identity. Thus by the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $\Phi(x) = f(\Phi(x)) = \Phi(f(x))$. By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $\sigma_A(f(x)) = f(\sigma_A(x))$. By \autoref{theorem:c-star-normal-spectral-radius}, $\norm{f(x)}_A = [f(x)]_{sp} \le \norm{f}_u = 2\norm{y}_F$. 
+
+    The above setup implies that for every $y \in \ol{\Phi(A)} \cap B_{sa}$, there exists $z \in A_{sa}$ such that $\norm{y - \Phi(z)}_{B} \le \norm{y}_B/2$, and $\norm{z}_A \le 2\norm{y}_B$. By the \hyperref[method of successive approximations]{theorem:successive-approximation}, $\phi(A_{sa}) = \ol{\Phi(A)} \cap B_{sa}$. Therefore $\Phi(A) = \ol{\Phi(A)}$. 
+\end{proof}
+
+\begin{definition}[Representation of $C^*$-Algebra]
+\label{definition:representation-cstar-algebra}
+    Let $A$ be a $C^*$-algebra, then a \textbf{representation} of $A$ is a pair $(H, \pi)$, where $H$ is a Hilbert space, and $\pi: A \to B(H)$ is a *-homomorphism. 
+\end{definition}
+
+\begin{definition}[Unitary Equivalence]
+\label{definition:representation-unitary-equivalent}
+    Let $A$ be a $C^*$-algebra and $(H_1, \pi_1), (H_2, \pi_2)$ be representations of $A$, then $(H_1, \pi_1)$ and $(H_2, \pi_2)$ are \textbf{unitarily equivalent} if there exists an isometry $U \in L(H_1; H_2)$ such that the following diagram commutes
+    \[
+    \xymatrix{
+    H_1 \ar@{->}[r]^{U} \ar@{->}[d]_{\pi_1(x)} & H_2 \ar@{->}[d]^{\pi_2(x)} \\
+    H_1 & H_2 \ar@{->}[l]^{U^*}
+    }
+    \]
+
+    for all $x \in A$. 
+\end{definition}
+
+
+
 

src/op/c-star/index.tex

@@ -8,4 +8,7 @@
 \input{./homomorphism.tex}
 \input{./gelfand.tex}
 \input{./cont.tex}
-\input{./order.tex}
+\input{./order.tex}
+\input{./positive.tex}
+\input{./state.tex}
+\input{./gns.tex}

src/op/c-star/order.tex

@@ -23,7 +23,7 @@
         \item There exists $\lambda \ge \norm{x}_A$ such that $\norm{\lambda - x}_A \le \lambda$. 
     \end{enumerate}
 \end{proposition}
-\begin{proof}[Proof, {{\cite[Lemma II.11.3]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Lemma 11.3]{Zhu}}}. ]
     (1) $\Leftrightarrow$ (2): \autoref{proposition:positive-spectrum}. 
 
     (2) $\Leftrightarrow$ (3): By assumption, $\sigma_A(x) \subset \real$, so \autoref{theorem:c-star-normal-spectral-radius} implies that
@@ -56,4 +56,32 @@
 \end{proof}
 
 
+\begin{definition}[Absolute Value]
+\label{definition:absolute-value-c-star}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$, then $|x| = \sqrt{x^*x}$ is the \textbf{absolute value} of $x$. 
+\end{definition}
+
+\begin{definition}[Positive and Negative Parts]
+\label{definition:positive-negative-cstar-algebra}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be self-adjoint, then there exists unique positive elements $x^+, x^- \in A$ such that
+    \begin{enumerate}
+        \item $x = x^+ - x^-$.
+        \item $x^+x^- = x^-x^+ = 0$. 
+    \end{enumerate}
+    
+    The pair $(x^+, x^-)$ are the \textbf{positive and negative parts} of $x$.
+\end{definition}
+\begin{proof}
+    Since $x$ is self-adjoint, $\sigma_A(x) \subset \real$ by \autoref{proposition:self-adjoint-spectrum}. Using the continuous functional calculus, existence is given by the functions $f^+(\lambda) = \lambda \vee 0$ and $f^-(\lambda) = \lambda \wedge 0$ and \autoref{proposition:positive-norm-inequality}. 
+
+    On the other hand, for each $p \in \real[z]$ with $p(0) = 0$, (2) implies that $p(x) = p(x^+) + p(-x^-)$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:stone-weierstrass}, $f(x) = f(x^+) + f(-x^-)$ for all $f \in C(\real; \real)$ with $f(0) = 0$. In particular, (1) then implies that $f^+(x) = f+(x^+) + f^+(-x^-) = f^+(x^+) = x^+$, and likewise $f^-(x) = x^-$. Therefore the decomposition is given uniquely by the continuous functional calculus. 
+\end{proof}
+
+\begin{remark}
+\label{remark:positive-negative-cstar-algebra}
+    The condition in the sign decomposition that $x^+x^- = x^-x^+ = 0$ is essential. Otherwise I may use silly decompositions like $0 = 1 - 1$. 
+\end{remark}
+
+
+
 

src/op/c-star/positive.tex

@@ -0,0 +1,41 @@
+\section{Positive Linear Functionals}
+\label{section:cstar-positive}
+
+\begin{definition}[Positive Linear Functional]
+\label{definition:cstar-positive-functional}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$, then $\phi$ is \textbf{positive} if $\dpn{x, \phi}{A} \ge 0$ for all positive elements $x \in A$. 
+\end{definition}
+
+\begin{theorem}
+\label{theorem:cstar-positive-algebraic}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in \hom(A; \complex)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is a positive linear functional.
+        \item $\phi \in A^*$ with $\normn{\phi}_{A^*} = \dpn{1, \phi}{A}$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For any $x \in A_{sa}$ with $\norm{x}_A \le 1$, $\sigma_A(1 - x) \subset 1 - [-1, 1] = [0, 2]$ by \autoref{proposition:commutative-spectrum-gymnastics}. Thus $1 - x \ge 0$ by \autoref{proposition:positive-spectrum}, and $\dpn{x, \phi}{A} \le \dpn{1, \phi}{A}$. By \autoref{proposition:hermitian-functional-norm}, $\norm{\phi}_{A^*} \le \dpn{1, \phi}{A}$, so $\norm{\phi}_{A^*} = \dpn{1, \phi}{A}$. 
+
+    (2) $\Rightarrow$ (1): For each $x \in A_{sa}$, $\sigma_A(x) = \sigma_{A[x]}(x)$ by \autoref{corollary:c-star-algebra-preserve-spectrum}, so $x \ge 0$ in $A$ if and only if $x \ge 0$ in $A[x]$ by \autoref{proposition:positive-spectrum}. In addition, the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark} implies that $x \ge 0$ if and only if $\Gamma_{A[x]}(x) \ge 0$. Moreover, since $1 \in A[x]$, the norm of $\phi$ alongside (2) is preserved by restricting to $A[x]$. By considering each commutative subalgebra, assume without loss of generality that $A$ is commutative. 
+    
+    By the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon-c0}, $\phi$ takes the form of a complex Radon measure $\mu$ on $\Omega(A)$, and $\norm{\phi}_{A^*} = \norm{\mu}_{\text{var}}$. For each Borel set $E \in \cb_{\Omega(A)}$, 
+    \begin{align*}
+        \norm{\mu}_{\text{var}} &= \int_{\Omega(A)} 1 d\mu = \mu(E) + \mu(\Omega(A) \setminus E) \\
+        &\le |\mu(E)| + |\mu(\Omega(A) \setminus E)| \le \norm{\mu}_{\text{var}}
+    \end{align*}
+
+    which is only possible if $\mu(E), \mu(\Omega(A) \setminus E) \ge 0$. As this holds for all $E \in \cb_{\Omega(A)}$, $\mu$ is positive, and $\phi$ then is a positive linear functional. 
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:positive-linear-functional-extension}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a closed subspace with $1_A \in B$, and $\phi \in B^*$ with $\norm{\phi}_{B^*} = \dpn{1_A, \phi}{B}$, then there exists a positive linear functional $\Phi \in A^*$ such that $\Phi|_B = A$. 
+\end{corollary}
+\begin{proof}
+    By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Phi \in A^*$ such that $\Phi|_B = A$ and $\norm{\Phi}_{A^*} = \norm{\phi}_{B^*}$. In which case, $\norm{\Phi}_{A^*} = \dpn{1_A, \Phi}{A}$, and $\Phi$ is also positive by \autoref{theorem:cstar-positive-algebraic}. 
+\end{proof}
+
+
+

src/op/c-star/sa.tex

@@ -6,7 +6,7 @@
     Let $A$ be an involutive algebra over $\complex$ and $x \in A$, then $x$ is \textbf{self-adjoint} if $x = x^*$. The space $A_{sa} = \bracs{x \in A| x = x^*}$ is the \textbf{self-adjoint part} of $A$, and:
     \begin{enumerate}
         \item $A_{sa}$ is a $\real$ subspace of $A$. 
-        \item $A = \complex(A_{sa})$ as a vector space.
+        \item $A = \complex(A_{sa})$, with equivalent norms. 
         \item For each $x \in A$, let
         \[
         \text{Re}(x) = \frac{x + x^*}{2} \quad \text{Im}(x) = \frac{x - x^*}{2i}
@@ -35,7 +35,7 @@
 \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}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 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 \\
@@ -64,7 +64,7 @@
     \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}, $|\lambda| = [x]_{sp} = \norm{x}_A$. 
+    (1): Since $\sigma_A(x)$ is compact, there exists $\lambda \in \sigma_A(x)$ 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}, 
     \[

src/op/c-star/state.tex

@@ -0,0 +1,173 @@
+\section{States}
+\label{section:cstar-states}
+
+
+\begin{definition}[State]
+\label{definition:cstar-state}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$, then $\phi$ is a \textbf{state} if $\phi$ is positive and $\dpn{1, \phi}{A} = 1$. 
+
+    The set of states $S(A) \subset A^*$ of $A$ equipped with the weak* topology is the \textbf{state space} of $A$.
+\end{definition}
+
+\begin{definition}
+\label{definition:cstar-state-pseudo-inner-product}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in A^*$ be a positive linear functional, then the mapping
+    \[
+    A \times A \to \complex \quad (x, y) \mapsto \dpn{x, y}{\phi} := \dpn{y^*x, \phi}{A}
+    \] 
+
+    is a pseudo inner product. In particular, for any $x, y \in A$, 
+    \[
+    |\dpn{y^*x, \phi}{A}|^2 = |\dpn{x, y}{\phi}|^2 \le \dpn{x, x}{\phi} \cdot \dpn{y, y}{\phi}
+    \]
+
+    The pairing $\dpn{\cdot, \cdot}{\phi}$ is the \textbf{pseudo inner product associated with $\phi$}. 
+\end{definition}
+\begin{proof}
+    By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}. 
+\end{proof}
+
+
+
+\begin{definition}[Pure State]
+\label{definition:pure-state}
+    Let $A$ be a unital $C^*$-algebra and $\phi \in S(A)$, then $\phi$ is a \textbf{pure state} if $\phi$ is an extreme point of $S(A)$. The set $P(A)$ is the collection of all pure states of $A$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:state-space-compact-convex}
+    Let $A$ be a unital $C^*$-algebra, then $S(A)$ is a compact convex set, and $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+\end{proposition}
+\begin{proof}
+    Since the evaluation map is weak* continuous and
+    \[
+    S(A) = \bracs{\phi \in A^*|\dpn{1, \phi}{A} = 1} \cap \bigcap_{\substack{x \in A \\ x \ge 0}}\bracs{\phi \in A^*|\dpn{x, \phi}{A} \ge 0}
+    \]
+
+    the state space is an intersection of convex and weak*-closed sets, so it is closed and convex. 
+
+    By \autoref{theorem:cstar-positive-algebraic}, $S(A) \subset \ol{B_{A^*}(0, 1)}$, which is weak* compact by the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}. Therefore $S(A)$ is compact by \autoref{proposition:compact-extensions}. 
+
+    By the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, $S(A)$ is the weak*-closed convex hull of $P(A)$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:multiplicative-pure-state}
+    Let $A$ be a unital $C^*$-algebra, then:
+    \begin{enumerate}
+        \item $\Omega(A) \subset P(A)$. 
+        \item If $A$ is commutative, then $\Omega(A) = P(A)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\phi \in \Omega(A)$. By \autoref{proposition:multiplicative-unit}, $\norm{\phi}_{A^*} = \dpn{1, \phi}{A} = 1$. Thus $\phi$ is a state by \autoref{theorem:cstar-positive-algebraic}, and $\Omega(A) \subset S(A)$. 
+
+    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\phi = (1 - t)\psi + t\rho$, then for each $x \in \ker(\phi)$, $x^*x \in \ker(\phi)$ as well. As $t \ne 0$, $x^*x \in \ker(\psi)$ and $x^*x \in \ker(\rho)$. By the \hyperref[Cauchy-Schwarz inequality]{definition:cstar-state-pseudo-inner-product}, 
+    \[
+    |\dpn{x, \psi}{A}|^2 = |\dpn{1^*x, \psi}{A}|^2 \le \dpn{1, \psi}{A} \cdot \dpn{x^*x, \psi}{A} = 0
+    \]
+
+    Likewise, $\dpn{x, \rho}{A} = 0$ as well. Hence $\ker(\psi), \ker(\rho) \supset \ker(\phi)$. Thus there exist scalars $\alpha, \beta \in \complex$ such that $\phi = \alpha \psi = \beta \rho$. However, since $\phi, \psi, \rho \in S(A)$, $\alpha = \beta = 1$, and $\phi = \psi = \rho$. Therefore $\phi$ is a pure state. 
+
+    (2): Using the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, identify $A$ with $C(\Omega(A); \complex)$ and $S(A)$ as Radon probability measures on $\Omega(A)$. 
+    
+    Let $\cm = \bracs{t\mu|\mu \in S(A), t \in [0, 1]}$. By \autoref{proposition:space-of-measures-extreme-points}, the extreme points of $\cm$ are the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, and possibly $0$. For any $\mu \in S(A)$, $\nu, \rho \in \cm$, and $t \in (0, 1)$, $\mu = (1 - t)\nu + t\rho$ implies that $\nu(\Omega(A)) = \rho(\Omega(A)) = 1$, and $\nu, \rho \in S(A)$ as well. Thus the extreme points of $S(A)$ are exactly the delta masses $\bracs{\delta_x|x \in \Omega(A)}$, which correspond to $\Omega(A)$ itself. 
+\end{proof}
+
+
+\begin{theorem}[Extension of States]
+\label{theorem:cstar-pure-state-extension}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra with $1_A \in B$, and $\phi \in S(B)$, then
+    \begin{enumerate}
+        \item There exists $\Phi \in S(A)$ such that $\Phi|_B = \phi$.
+        \item If $\phi \in P(B)$, then there exists $\Phi \in P(A)$ such that $\Phi|_B = \phi$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): By \autoref{theorem:cstar-positive-algebraic}, $\norm{\phi}_{B^*} = \dpn{1_A, \phi}{B}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\Phi \in A^*$ such that $\Phi|_B = \phi$ and $\norm{\Phi}_{A^*} = \norm{\phi}_{B^*} = \dpn{1_A, \Phi}{A}$. Thus \autoref{theorem:cstar-positive-algebraic} implies that $\Phi \in S(A)$.
+
+    (2): Let $E(\phi) = \bracs{\Phi \in S(A)|\Phi|_B = \phi}$ be the collection of all extensions of $\phi$, then $E(\phi)$ is a weak*-closed convex subset of $S(A)$. By (1), $E(\phi)$ is non-empty, and as such admits an extreme point $\Phi$ by the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}. 
+
+    Let $\psi, \rho \in S(A)$ and $t \in (0, 1)$ such that $\Phi = (1 - t)\psi + t\rho$. In which case, $\phi = (1 - t)\psi|_B + t\rho|_B$. Since $\phi \in P(B)$, $\phi = \psi|_B = \rho|_B$, so $\psi, \rho \in E(\phi)$. As $\Phi$ is an extreme point of $E(\phi)$, $\Phi = \psi = \rho$. Therefore $\Phi \in P(A)$. 
+\end{proof}
+
+
+\begin{corollary}
+\label{corollary:cstar-positive-property-probe}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then\footnote{The crude bound seems kind of tragic, but it wouldn't be true otherwise. }
+    \begin{align*}
+    \sigma_A(x) &\subset \bracs{\dpn{x, \phi}{A}|\phi \in P(A)} \\
+    &\subset \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} = \ol{\text{Conv}}(\sigma_A(x))
+    \end{align*}
+    
+
+    In particular, there exists $\phi \in P(A)$ such that $\norm{x}_A = |\dpn{x, \phi}{A}|$. 
+\end{corollary}
+\begin{proof}
+    Let $\lambda \in \sigma_A(x)$. By \autoref{proposition:gelfand-transform-gymnastics}, there exists $\phi \in \Omega(A[x])$ such that $\dpn{x, \phi}{A[x]} = \lambda$. By \autoref{proposition:multiplicative-pure-state}, $\phi \in P(A[x])$. The \hyperref[pure state extension theorem]{theorem:cstar-pure-state-extension} implies that there exists $\Phi \in P(A)$ such that $\Phi|_{A[x]} = \phi$. Thus $\Phi$ is a pure state with $\dpn{x, \Phi}{A} = \lambda$, and $ \sigma_A(x) \subset \bracs{\dpn{x, \Phi}{A}|\Phi \in P(A)}$. 
+
+    Let $\Phi \in S(A)$ and $\phi = \Phi|_{A[x]}$, then $\phi \in S(A[x])$ as well. By the \hyperref[Gelfand-Naimark Theorem]{theorem:gelfand-naimark}, the \hyperref[Spectral Theorem]{theorem:spectral-c-star}, and the \hyperref[Riesz Representation Theorem]{theorem:riesz-radon}, $\phi$ takes the form of a Radon probability measure $\mu$ on $\sigma_A(x)$. In which case,
+    \[
+    \dpn{x, \Phi}{A} = \dpn{x, \phi}{A[x]} = \int_{\sigma_A(x)}\lambda \mu(d\lambda) \in \ol{\text{Conv}}(\sigma_A(x))
+    \]
+
+    Finally, since $S(A)$ is compact and convex by \autoref{proposition:state-space-compact-convex}, 
+    \begin{align*}
+        \bracs{\dpn{x, \phi}{A}|\phi \in S(A)} &= \ol{\text{Conv}}(\bracs{\dpn{x, \phi}{A}|\phi \in P(A)}) \\
+        &\subset \ol{\text{Conv}}(\sigma_A(x))
+    \end{align*}
+
+    by \autoref{proposition:compact-extensions} and \autoref{proposition:closure-of-image}. 
+\end{proof}
+
+\begin{theorem}
+\label{theorem:cstar-state-existence}
+    Let $A$ be a unital $C^*$-algebra, $x \in A$, and $\lambda \in \sigma_A(x)$, then there exists $\phi \in S(A)$ such that $\dpn{x, \phi}{A} = \lambda$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 13.7]{Zhu}}}. ]
+    Let $B = \text{span}\bracs{x, 1}$. For each $\alpha x + \beta \in B$, let $\dpn{\alpha x + \beta, \phi_0}{B} = \alpha \lambda + \beta$. Since $\sigma_A(1) = \bracs{1}$, $\phi_0 \in B^*$ is a well-defined linear functional with $\dpn{x, \phi_0}{B} = \lambda$ and $\dpn{1, \phi_0}{B} = 1$. 
+
+    In addition, for each $\alpha x + \beta \in B$, $\alpha \lambda + \beta \in \sigma_A(\alpha x + \beta)$ by \autoref{proposition:commutative-spectrum-gymnastics}, and
+    \[
+    |\alpha \lambda + \beta| \le [\alpha x + \beta]_{sp} \le \norm{\alpha x + \beta}_A
+    \]
+
+    Thus $\norm{\phi_0}_{B^*} = \dpn{1, \phi_0}{B} = 1$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\phi \in A^*$ such that $\phi|_B = \phi_0$ and $\norm{\phi}_{A^*} = \norm{\phi_0}_{B^*}$. In which case, $\dpn{x, \phi}{A} = \lambda$ and $\dpn{1, \phi}{A} = \norm{\phi}_{A^*} = 1$. By \autoref{theorem:cstar-positive-algebraic}, $\phi$ is positive and hence a state. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:cstar-positive-weakstar-dense}
+    Let $A$ be a unital $C^*$-algebra, then:
+    \begin{enumerate}
+        \item For each $x \in A$, $x = 0$ if and only if $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. 
+        \item The linear span of $P(A)$ is weak*-dense in $A^*$. 
+    \end{enumerate}
+
+    Moreover, for any $x \in A$, 
+    \begin{enumerate}[start=2]
+        \item $x$ is self-adjoint if and only if $\dpn{x, \phi}{A} \in \real$ for all $\phi \in P(A)$. 
+        \item $x$ is positive if and only if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$. 
+    \end{enumerate}
+    
+\end{corollary}
+\begin{proof}[Proof, {{\cite[Theorem 13.9]{Zhu}}}. ]
+    (1): Let $x \in A$ such that $\dpn{x, \phi}{A} = 0$ for all $\phi \in P(A)$. First suppose that $x$ is self-adjoint. By \autoref{theorem:cstar-state-existence}, $\sigma_A(x) = \bracs{0}$, and $\norm{x}_A = [x]_{sp} = 0$ by \autoref{theorem:c-star-normal-spectral-radius}. 
+
+    Now suppose that $x$ is arbitrary. In this case, for each $\phi \in P(A)$, 
+    \[
+    0 = \text{Re}(\dpn{x, \phi}{A}) = \dpn{\text{Re}(x), \phi}{A}
+    \]
+
+    because $\phi$ is Hermitian. Similarly, $\dpn{\text{Im}(x), \phi}{A} = 0$ as well. Thus $\text{Re}(x) = \text{Im}(x) = 0$, and $x = 0$ as well. 
+
+    (2): Since the linear span of $P(A)$ separates points in $A$, it is weak*-dense in $A^*$ by \autoref{lemma:duality-dense}. 
+
+    (3): Let $\phi \in P(A)$, then $\phi$ is Hermitian. If $x$ is self-adjoint, then $\dpn{x, \phi}{A} \in \real$. 
+
+    On the other hand, if $\dpn{x, \phi}{A} \in \real$, then $\dpn{x, \phi}{A} = \dpn{x^*, \phi}{A}$, and $\dpn{x - x^*, \phi}{A} =0 $. If this holds for all $\phi \in P(A)$, then $x - x^* = 0$ by (1), and $x$ is self-adjoint. 
+
+    (4): Let $\phi \in P(A)$, then $\phi$ is positive. Thus if $x$ is positive, $\dpn{x, \phi}{A} \ge 0$. 
+
+    On the other hand, if $\dpn{x, \phi}{A} \ge 0$ for all $\phi \in P(A)$, then $x$ is self-adjoint by (3). By \autoref{corollary:cstar-positive-property-probe}, $\sigma_A(x) \subset [0, \infty)$. As such, $x$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
+\end{proof}
+

src/op/c-star/unitary.tex

@@ -42,7 +42,7 @@
 \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}}}. ]
+\begin{proof}[Proof, {{\cite[Proposition 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)}

src/op/example/bc.tex

@@ -17,7 +17,7 @@
 
     is a homeomorphism. Under the identification $\beta X = \Omega(BC(X; \complex))$, $\Gamma_{BC(X; \complex)} = \beta$. 
 \end{theorem}
-\begin{proof}[Proof, {{\cite[Theorem I.6.4]{Zhu}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 6.4]{Zhu}}}. ]
     Let $\phi \in BC(X; \complex)^* \setminus \ol{E(X)}$, then there exists $\seqf{f_k} \subset BC(X; \complex)$ and $\eps > 0$ such that for every $x \in X$, 
     \[
     f(x) = \sum_{k = 1}^n |f_k(x) - \dpn{f_k, \phi}{BC(X; \complex)}|^2 \ge \eps^2

src/op/example/c0.tex

@@ -3,12 +3,12 @@
 
 \begin{definition}[$C_0(X)$]
 \label{definition:vanishing-infinity-algebra}
-    Let $X$ be a LCH space, then $C_0(X; \complex)$ equipped with pointwise operations and the uniform norm is a $C^*$-algebra. 
+    Let $X$ be an LCH space, then $C_0(X; \complex)$ equipped with pointwise operations and the uniform norm is a $C^*$-algebra. 
 \end{definition}
 
 \begin{theorem}
 \label{theorem:vanishing-infinity-multiplicative-functional}
-    Let $X$ be a LCH space, then the mapping
+    Let $X$ be an LCH space, then the mapping
     \[
     E: X \to \Omega(C_0(X)) \quad E(x)(f) = f(x)
     \]

src/op/example/matrix.tex

@@ -26,3 +26,61 @@
 \begin{proof}
     (1), (2): By \autoref{proposition:matrix-algebra-spectrum}, $\sigma(x)$ is finite. By \autoref{proposition:log-identity-component}, there exists $y \in M_n(\complex)$ such that $x = \exp(y)$. In which case, $x \in G_0(M_n(\complex))$. 
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:matrix-algebra-state-space}
+    Let $n \in \natp$. For each $y \in M_n(\complex)$, let
+    \[
+    \phi_y: M_n(\complex) \to \complex \quad x \mapsto \dpn{x, y}{F} = \text{tr}(y^*x)
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi_y \in S(M_n(\complex))$. 
+        \item $y \ge 0$ and $\text{tr}(y) = 1$.
+    \end{enumerate}
+\end{proposition}
+% "Obvious" so won't prove. 
+
+\begin{proposition}
+\label{proposition:matrix-algebra-pure-state}
+    Let $n \in \natp$. For each $y \in M_n(\complex)$, let
+    \[
+    \phi_y: M_n(\complex) \to \complex \quad x \mapsto \dpn{x, y}{F} = \text{tr}(y^*x)
+    \]
+
+    then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi_y$ is a pure state of $M_n(\complex)$. 
+        \item $y$ is a projection operator with rank $1$. 
+        \item There exists $v \in \complex^n$ with $\norm{v}_{\complex^n} = 1$ such that $\dpn{x, \phi_y}{M_n(\complex)} = \dpn{xv, v}{\complex^n}$ for all $x \in M_n(\complex)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2): By \autoref{proposition:matrix-algebra-state-space}, $y \ge 0$ with $\text{tr}(y) = 1$. Via an orthogonal change of coordinates, assume without loss of generality that $y$ is diagonal. In which case, $y$ corresponds to an extreme point of $S(M_n(\complex))$ if and only if it is of rank $1$. As $\text{tr}(y) = 1$, $y$ is a projection.
+
+    (2) $\Rightarrow$ (3): Let $v \in \complex^n$ be a unit eigenvector of $y$, then for each $x \in M_n(\complex)$,
+    \[
+        \dpn{x, \phi_y}{M_n(\complex)} = \text{tr}(y^*x) = \text{tr}(yx) = \dpn{xv, v}{\complex^n}
+    \]
+\end{proof}
+
+\begin{example}
+\label{proposition:spectrum-pure-state-counterexample}
+    Let
+    \[
+    A = \begin{bmatrix}
+        1 & 0 \\
+        0 & -1
+    \end{bmatrix}
+    \]
+
+    then $A$ is a self-adjoint element of $M_2(\complex)$. By \autoref{proposition:matrix-algebra-pure-state}, the mapping $T \mapsto \dpn{Tv, v}{\complex^2}$ is a pure state on $M_2(\complex)$ for every unit vector $v \in \complex^2$. In particular, if $v = (\sqrt{2}, \sqrt{2})/2$, then $\dpn{Tv, v}{\complex^2} = 0 \not\in \sigma_{M_2(\complex)}(A)$. Therefore
+    \[
+    \sigma_{M_2(\complex)}(A) \subsetneq \bracsn{\dpn{T, \phi}{M_2(\complex)}|\phi \in P(M_2(\complex))}
+    \]
+\end{example}
+
+
+

src/op/notation.tex

@@ -15,6 +15,10 @@
     $\cm(A)$ & Maximal ideal space of $A$. & \autoref{definition:maximal-ideal} \\
     $\Gamma = \Gamma_A$ & The Gelfand transform on $A$. & \autoref{definition:gelfand-transform} \\ 
     $A[S]$ & $C^*$-subalgebra of $A$ generated by $S \subset A$. & \autoref{definition:generated-subalgebra} \\ 
+    $S(A)$ & State space of a $C^*$-algebra $A$. & \autoref{definition:cstar-state} \\ 
+    $P(A)$ & Pure state space of a $C^*$-algebra $A$. & \autoref{definition:pure-state} \\ 
+    $\dpn{x, y}{\phi}$ & Defined as $\dpn{y^*x, \phi}{A}$, the pseudo inner product associated to a positive linear functional. & \autoref{definition:cstar-state-pseudo-inner-product} \\ 
+    $(H_\phi, \pi_\phi, \xi_\phi)$ & GNS triple associated with $\phi \in S(A)$. & \autoref{definition:gns-triple} \\ 
 
     $M_n(\complex)$ & Algebra of $n \times n$ matrices over $\complex$.  & \autoref{definition:matrix-algebra} \\
     $B(H)$ & Algebra of bounded operators on a Hilbert space. & \autoref{definition:hilbert-endomorphism} \\

src/topology/main/c0.tex

@@ -17,7 +17,7 @@ For details regarding the complex-valued cased, in particular its properties as
     \begin{enumerate}
         \item $C_0(X; E) \subset BC(X; E)$.
         \item $C_0(X; E)$ is a closed subspace of $BC(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $C_0(X; E)$.
-        \item If $X$ is a LCH space, then $C_c(X; E)$ is a dense subspace of $C_0(X; E)$ with respect to the uniform topology.
+        \item If $X$ is an LCH space, then $C_c(X; E)$ is a dense subspace of $C_0(X; E)$ with respect to the uniform topology.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
@@ -44,7 +44,7 @@ For details regarding the complex-valued cased, in particular its properties as
 
 \begin{proposition}
 \label{proposition:c0-tensor}
-    Let $X$ be a LCH space and $E$ be a locally convex space over $K \in \RC$. Identify $C_0(X; K) \otimes E$ as a subspace of $C_0(X; E)$ under the natural map
+    Let $X$ be an LCH space and $E$ be a locally convex space over $K \in \RC$. Identify $C_0(X; K) \otimes E$ as a subspace of $C_0(X; E)$ under the natural map
     \[
     C_0(X; K) \otimes E \to C_0(X; E) \quad \sum_{j = 1}^n \phi_j \otimes x_j \mapsto \sum_{j = 1}^n x_j \cdot \phi_j
     \]

src/topology/main/compact.tex

@@ -19,9 +19,7 @@
     U_J = \bigcup_{j \in J}E_j^c
     \]
 
-    then $U_J \subset X$ is open. For any $J, J' \subset I$, $U_J \cup U_{J'} = U_{J \cup J'}$.
-
-    Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
+    then $U_J \subset X$ is open. Suppose for contradiction that $\bigcap_{i \in I}E_i = \emptyset$, then
     \[
     \mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
     \]

src/topology/main/compactify.tex

@@ -73,7 +73,7 @@
 
 \begin{lemma}
 \label{lemma:lch-compactification-open}
-    Let $X$ be a LCH space and $(Y, \varphi)$ be a compactification of $X$, then $\varphi(X) \subset Y$ is open. 
+    Let $X$ be an LCH space and $(Y, \varphi)$ be a compactification of $X$, then $\varphi(X) \subset Y$ is open. 
 \end{lemma}
 \begin{proof}
     For each $x \in X$, let $U \in \cn_X(x)$ be a compact neighbourhood. Since $Y$ is a compact Hausdorff space, $\varphi(U)$ is closed by \autoref{proposition:compact-closed}. As $\varphi \in C(X; Y)$ is an embedding, there exists $V \in \cn_Y(\varphi(x))$ such that $\varphi(U) = \varphi(X) \cap V$. Given that $\varphi(X)$ is dense in $Y$, $\varphi(U) = \ol{\varphi(X) \cap V} \supset V$. Therefore $\varphi(U) \in \cn_{Y}(\varphi(x))$, and $\varphi(X)$ is open in $Y$. 
@@ -83,7 +83,7 @@
 
 \begin{definition}[One-Point Compactification]
 \label{definition:alexandroff-compactification}
-    Let $(X, \mathcal{T})$ be a LCH space, then there exists a pair $(X^*, \iota)$ such that:
+    Let $(X, \mathcal{T})$ be an LCH space, then there exists a pair $(X^*, \iota)$ such that:
     \begin{enumerate}
         \item $(X^*, \iota)$ is a compactification of $X$. 
         \item[(U)] For any pair $(Y, \varphi)$ satisfying (1), there exists a unique $\varphi^* \in C(Y; X^*)$ such that the following diagram commutes:

src/topology/main/lch.tex

@@ -20,7 +20,7 @@
 
 \begin{lemma}
 \label{lemma:lch-compact-neighbour}
-    Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exits $V \in \cn^o(K)$ relatively compact such that $K \subset V \subset \ol{V} \subset U$.
+    Let $X$ be an LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exits $V \in \cn^o(K)$ relatively compact such that $K \subset V \subset \ol{V} \subset U$.
 \end{lemma}
 \begin{proof}
     For each $x \in K$, there exists $V_x \in \cn^o(x)$ be relatively compact such that $x \in V_x \subset \overline{V_x} \subset U$ by (3) of \autoref{definition:lch}. Since $K$ is compact, there exists $\seqf{x_j} \subset K$ such that
@@ -38,7 +38,7 @@
 
 \begin{lemma}[Urysohn's Lemma (LCH)]
 \label{lemma:lch-urysohn}
-    Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $F \in C_c(X; [0, 1])$ such that $\supp{F} \subset U$.
+    Let $X$ be an LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $F \in C_c(X; [0, 1])$ such that $\supp{F} \subset U$.
 \end{lemma}
 \begin{proof}[Proof, {{\cite[Lemma 4.32]{Folland}}}. ]
     By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ relatively compact such that
@@ -61,7 +61,7 @@
 
 \begin{theorem}[Tietze Extension Theorem (LCH)]
 \label{theorem:lch-tietze}
-    Let $X$ be a LCH space, $K \subset X$ be compact, $U \in \cn^o(K)$, and $f \in C(K; \real)$, then there exists $F \in C_c(U; \real)$ such that $F|_K = f$.
+    Let $X$ be an LCH space, $K \subset X$ be compact, $U \in \cn^o(K)$, and $f \in C(K; \real)$, then there exists $F \in C_c(U; \real)$ such that $F|_K = f$.
 \end{theorem}
 \begin{proof}
     By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ relatively compact such that $K \subset V \subset \ol{V} \subset U$. As $\ol{W}$ is compact, it is normal by \autoref{proposition:compact-hausdorff-normal}. Since $X$ is Hausdorff, $K \subset \ol{W}$ is closed by \autoref{proposition:compact-closed}.
@@ -79,7 +79,7 @@
 
 \begin{proposition}
 \label{proposition:lch-compactly-generated}
-    Let $X$ be a LCH space, then:
+    Let $X$ be an LCH space, then:
     \begin{enumerate}
         \item $X$ is compactly generated. 
         \item For any uniform space $Y$, $C(X; Y) \subset Y^X$ is closed with respect to the compact-open topology. 
@@ -94,7 +94,7 @@
 
 \begin{proposition}
 \label{proposition:lch-product}
-    Let $\seqi{X}$ be a family of LCH spaces. If $X_i$ is compact for all but finitely many $i \in I$, then $X = \prod_{i \in I}X_i$ is a LCH space. 
+    Let $\seqi{X}$ be a family of LCH spaces. If $X_i$ is compact for all but finitely many $i \in I$, then $X = \prod_{i \in I}X_i$ is an LCH space. 
 \end{proposition}
 \begin{proof}
     By \autoref{proposition:product-hausdorff}, $\prod_{i \in I}X_i$ is Hausdorff. Let $x \in \prod_{i \in I}X_i$ and $i \in I$. If $X_i$ is not compact, let $U_i \in \cn_{X_i}(\pi_i(x))$ be compact. Otherwise, let $U_i = X_i$. Let $U = \prod_{i \in I}U_i$, then since $U_i \ne X_i$ for only finitely many $i \in I$, $U \in \cn_X(x)$. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff}, $U$ is compact. Therefore $X$ is locally compact. 
@@ -108,7 +108,7 @@
 
 \begin{proposition}
 \label{proposition:lch-sigma-compact}
-    Let $X$ be a LCH space, then the following are equivalent:
+    Let $X$ be an LCH space, then the following are equivalent:
     \begin{enumerate}
         \item $X$ is $\sigma$-compact.
         \item There exists an exhaustion of $X$ by compact sets.
@@ -134,7 +134,7 @@
 
 \begin{proposition}
 \label{proposition:lch-partition-of-unity}
-    Let $X$ be a LCH space, $K \subset X$ be compact, and $\seqi{U}$ be an open cover of $K$, then there exists a $C_c$ partition of unity on $K$ subordinate to $\seqi{U}$.
+    Let $X$ be an LCH space, $K \subset X$ be compact, and $\seqi{U}$ be an open cover of $K$, then there exists a $C_c$ partition of unity on $K$ subordinate to $\seqi{U}$.
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Proposition 4.41]{Folland}}}. ]
     Since $K$ is compact, assume without loss of generality that $\seqi{U} = \seqf{U_j}$.
@@ -160,7 +160,7 @@
 
 \begin{lemma}
 \label{lemma:lch-locally-finite-relatively-compact-refine}
-    Let $X$ be a LCH space and $\ce \subset 2^X$ be a locally finite relatively compact open cover of $X$, then there exists locally finite relatively compact open covers $\bracs{F_E}_{E \in \ce}, \bracs{G_E}_{E \in \ce} \subset 2^X$ such that for each $E \in \ce$, $G_E \subset \ol{G_E} \subset E \subset \ol{E} \subset F_E$.
+    Let $X$ be an LCH space and $\ce \subset 2^X$ be a locally finite relatively compact open cover of $X$, then there exists locally finite relatively compact open covers $\bracs{F_E}_{E \in \ce}, \bracs{G_E}_{E \in \ce} \subset 2^X$ such that for each $E \in \ce$, $G_E \subset \ol{G_E} \subset E \subset \ol{E} \subset F_E$.
 \end{lemma}
 \begin{proof}
     $(\bracs{F_E}_{E \in \ce})$: For each $E \in \ce$, $\bracsn{F \in \ce|F \cap \ol E \ne \emptyset}$ is finite by \autoref{lemma:locally-finite-compact}. Let
@@ -210,7 +210,7 @@
 
 \begin{theorem}
 \label{theorem:lch-paracompact}
-    Let $X$ be a LCH space, then the following are equivalent:
+    Let $X$ be an LCH space, then the following are equivalent:
     \begin{enumerate}
         \item $X$ is paracompact.
         \item There exists a locally finite relatively compact open cover $\cf$ of $X$.

src/topology/main/semicontinuity.tex

@@ -64,7 +64,7 @@
 
 \begin{proposition}
 \label{proposition:semicontinuous-lch}
-    Let $X$ be a LCH space and $f: X \to [0, \infty]$ be lower semicontinuous, then
+    Let $X$ be an LCH space and $f: X \to [0, \infty]$ be lower semicontinuous, then
     \[
     f = \sup_{\substack{\phi \in C_c(X) \\ 0 \le \phi \le f}}\phi
     \]

src/topology/main/stonean.tex

@@ -40,7 +40,7 @@
     so $g \le F$. As this holds for all upper bounds of $\cf$, $g$ is the supremum of $\cf$ in $C(X; \real)$. 
     
 
-    ($\Leftarrow$, \cite[Theorem II.9.6]{Zhu}): Suppose that $X$ is completely regular and $C(X; \real)$ is order complete. Let $U \subset X$ be open and
+    ($\Leftarrow$, \cite[Theorem 9.6]{Zhu}): Suppose that $X$ is completely regular and $C(X; \real)$ is order complete. Let $U \subset X$ be open and
     \[
     \cf = \bracs{f \in C(X; [0, 1])| 0 \le f \le \one_U}
     \]

src/topology/uniform/equicontinuous.tex

@@ -57,7 +57,7 @@
         \item $\cf$ is a relatively compact subset of $C(X; Y)$ with respect to the uniform structure of compact convergence. 
     \end{enumerate}
 
-    Conversely, if $X$ is a LCH space, then (C3) implies (E1) + (E2).
+    Conversely, if $X$ is an LCH space, then (C3) implies (E1) + (E2).
 \end{theorem}
 \begin{proof}
     (E1) $\Rightarrow$ (C1): It is sufficient to show that (ii) is finer than (iii).
@@ -80,7 +80,7 @@
 
     (E1) + (E2) $\Rightarrow$ (C3): Using (C2), assume without loss of generality that $\cf$ is closed in $Y^X$ with respect to the product topology. In which case, $\cf$ is a closed subset of $\prod_{x \in X}\ol{\cf(x)}$ with respect to the product topology. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff} and \autoref{proposition:compact-extensions}, $\cf$ is compact in the product topology. By (C1), $\cf$ is also compact in the compact uniform topology.
 
-    (C3) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
+    (C3) $\Rightarrow$ (E1): Assume that $X$ is an LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
     \[
     (g(x), f_j(x)), (f_j(x), f_j(y)), (f_j(y), g(y)) \in U \circ U \circ U
     \]