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/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/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/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/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

@@ -54,4 +54,24 @@
     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

@@ -10,3 +10,5 @@
 \input{./cont.tex}
 \input{./order.tex}
 \input{./positive.tex}
+\input{./state.tex}
+\input{./gns.tex}

src/op/c-star/positive.tex

@@ -29,4 +29,13 @@
 \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/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/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/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
     \[
     \mathbf{U} = \bracs{U_J|J \subset I \text{ finite}}
     \]