RETRACTION: wrong word used.

document.tex

@@ -14,6 +14,7 @@ Occasionally, I make up some definitions to play with. These definition blocks w
 \input{./src/topology/index}
 \input{./src/fa/index}
 \input{./src/measure/index}
+\input{./src/harmonic/index}
 \input{./src/dg/index}
 \input{./src/op/index}
 \input{./src/dyn/index}

src/dyn/top/def.tex

@@ -66,6 +66,12 @@
     Let $G$ be an infinite group, $G \curvearrowright X$ be a $G$-flow, and $x \in X$, then $x$ is \textbf{recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is infinite.
 \end{definition}
 
+\begin{definition}[Uniformly Recurrent]
+\label{definition:tds-uniformly-recurrent}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{uniformly recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is syndetic. 
+\end{definition}
+
+
 \begin{definition}[Syndetic]
 \label{definition:group-syndetic}
     Let $G$ be a group and $S \subset G$, then $S$ is \textbf{(left) syndetic} if there exists $F \subset G$ finite $FS = G$. 
@@ -94,7 +100,7 @@
     \begin{enumerate}
         \item $G \curvearrowright X$ admits no proper subflows.
         \item Every point in $X$ is topologically transitive. 
-        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. 
+        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. In particular, $x$ is uniformly recurrent. 
     \end{enumerate}
     
 
@@ -119,10 +125,25 @@
 \end{proof}
 
 
+\begin{definition}[Minimal Point]
+\label{definition:tds-minimal-point}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $G \curvearrowright \ol{Gx}$ is a minimal $G$-subflow.
+        \item $x$ is uniformly recurrent. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): See \autoref{definition:tds-minimal}. 
+
+    (2) $\Rightarrow$ (1): Let $Y = \ol{Gx}$ and $y \in Y$. For any $U \in \cn_X(x)$, since $X$ is a compact Hausdorff space, there exists $V \in \cn_X(x)$ with $\ol{V} \subset U$. As $N(x, V)$ is syndetic in $G$, there exists $F \subset G$ finite such that $G = F^{-1}N(x, V)$. 
+
+    Since $\ol{Gx} = Y$, there exists an ultrafilter $\fF \subset 2^G$ such that $\fF x \to y$. As $G = \bigcup_{f \in F}N(x, V)$, there exists $f \in F$ such that $f^{-1}N(x, V) \in \fF$. Thus $y = \lim \fF x \subset \ol{f^{-1}N(x, V)} \subset \ol{f^{-1}V}$. Equivalently, $fy \in \ol{V}$, and $Gy \cap U \supset Gy \cao f^{-1}\ol{V} \ne \emptyset$. As this holds for all $U \in \cn_X(x)$, $x \in \ol{Gy}$, and $\ol{Gy} = \ol{Gx} = Y$. Therefore every point in $Y$ is transitive, and $G \curvearrowright Y$ is a minimal subflow. 
+\end{proof}
 
 \begin{theorem}[Existence of Minimal Subflows]
 \label{theorem:tds-minimal-existence}
-    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. 
+    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. In which case, every point in $Y$ is a minimal. 
 \end{theorem}
 % Proof omitted. 
 

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 $\bracs{\rho < 1} \subseteq A \subseteq \bracs{\rho \le 1} \subseteq \ol A$, with respect to any vector space topology on $E$. 
+        \item If $A$ is circled, then $\bracs{[\cdot]_A < 1} \subseteq A \subseteq \bracs{[\cdot]_A \le 1} \subseteq \ol A$, with respect to any vector space topology on $E$. 
     \end{enumerate}
 
     In particular,

src/harmonic/index.tex

@@ -0,0 +1,5 @@
+\part{Harmonic Analysis}
+\label{part:harmonic}
+
+\input{./lcg/index.tex}
+\input{./notation.tex}

src/harmonic/lcg/algebra.tex

@@ -306,3 +306,28 @@ Unfortunately, $L^1(G)$ is not a $C^*$-algebra, and its approximate identity req
     and $f * \phi_U \to f$ in $L^p(G; \complex)$ by \hyperref[continuity of translation]{proposition:haar-translation}.
 \end{proof}
 
+\begin{theorem}
+\label{theorem:lcg-group-algebra-ideal}
+    Let $G$ be a locally compact group and $J \subset L^1(G; \complex)$ be a closed subspace, then $J$ is a left ideal if and only if $L_xJ \subset J$ for all $x \in G$. 
+\end{theorem}
+\begin{proof}[Proof, {{\cite[Theorem 2.45]{FollandHarmonic}}}. ]
+    ($\Rightarrow$): Suppose that $J$ is a left ideal. By \autoref{proposition:group-algebra-approximate-identity}, $L^1(G; \complex)$ admits an approximate identity $\angles{\phi_\alpha}_{\alpha \in A}$. In which case, for any $f \in J$ and $x \in G$,
+    \[
+    L_xf = \lim_{\alpha \in A}L_x(\phi_\alpha * f) = \lim_{\alpha \in A}(L_x\phi_\alpha) * f \in J
+    \]
+
+    ($\Leftarrow$): Suppose that $L_xJ \subset J$ for all $x \in G$. Let $f \in J$, $A \subset G$ be compact, and $\eps > 0$. By \autoref{proposition:haar-translation}, for each $y \in A$, there exists $U_y \in \cn_G(y)$ such that $\norm{L_yf - L_zf}_{L^1(G; \complex)} \le \eps$ for all $z \in U_y$. By compactness of $A$, there exists $F \subset A$ finite such that $A \subset \bigcup_{y \in F}U_y$. In which case,
+    \begin{align*}
+        (\one_A * f)(x) &= \int_A L_yf(x)dy = \sum_{y \in F}\int_{A \cap U_y} L_zf(x)dz \\
+        (\one_A * f)(x) - \sum_{y \in F}|A \cap U_y|L_yf(x) &= \sum_{y \in F}\int_{A \cap U_y} (L_zf - L_yf)(x)dz \\
+        \abs{(\one_A * f) - \sum_{y \in F}|A \cap U_y|L_yf} &\le \eps |A|
+    \end{align*}
+
+    where for any $B \in \cb_G$, $|B|$ denotes its measure. As the above holds for all $\eps > 0$, $\one_A * f \in \ol J = J$. 
+
+    For any Borel set $B \in \cb_G$ with finite measure and $\eps > 0$, by inner regularity of the Haar measure, there exists $A \subset G$ compact such that $\norm{\one_B - \one_A}_{L^1(G; \complex)} < \eps$. By continuity of the convolution product, $\one_B * f \in J$ as well. 
+
+    By linearity, $\phi * f \in J$ for any $\phi \in \Sigma(G; \complex) \cap L^1(G; \complex)$. Finally, by \hyperref[density of simple functions]{proposition:lp-simple-dense}, $g * f \in J$ for all $g \in L^1(G; \complex)$. Therefore $J$ is a left ideal. 
+\end{proof}
+
+

src/harmonic/notation.tex

@@ -0,0 +1,3 @@
+\chapter{Notations}
+\label{chap:harmonic-notations}
+

src/measure/index.tex

@@ -9,5 +9,4 @@
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}
 \input{./weak-integral/index.tex}
-\input{./lcg/index.tex}
 \input{./notation.tex}

src/op/banach/definitions.tex

@@ -12,7 +12,7 @@
 
 \begin{definition}[Unital Banach Algebra]
 \label{definition:unital-banach-algebra}
-    Let $(A, \norm{\cdot}_A)$ be a Banach algebra, then $A$ is \textbf{unital} if there exists $1 \in A$ such that for any $x \in A$, $x1 = 1x = x$. In which case, there exists an equivalent norm $\norm{\cdot}_{1}: A \to [0, \inftu)$ such that $\norm{1}_1 = 1$, and $A$ is always assumed to be equipped with this norm.  
+    Let $(A, \norm{\cdot}_A)$ be a Banach algebra, then $A$ is \textbf{unital} if there exists $1 \in A$ such that for any $x \in A$, $x1 = 1x = x$. In which case, there exists an equivalent norm $\norm{\cdot}_{1}: A \to [0, \infty)$ such that $\norm{1}_1 = 1$, and $A$ is always assumed to be equipped with this norm.  
 \end{definition}
 \begin{proof}[Proof, {{\cite[Proposition I.1.3]{Takesaki1}}}. ]
     For each $x \in A$, let $L_x \in L(A; A)$ be defined by $y \mapsto xy$, and let $\norm{x}_1 = \norm{L_x}_{L(A; A)}$, then $\norm{x}_1 \le \norm{x}_A$ and $\norm{1}_1 = 1$. On the other hand, $\frac{\norm{x}_A}{\norm{1}_A} \le \norm{x}_1$, so $\norm{\cdot}_1$ is equivalent to $\norm{\cdot}_A$.
@@ -82,7 +82,9 @@
     The algebra $\tilde A$ is the \textbf{unitisation} of $A$. 
 \end{definition}
 \begin{proof}
-    (U): For each $\lambda + x \in \tilde A$, let $\tilde \phi(\lambda + x) = \lamdba + \phi(x)$. 
+    (U): For each $\lambda + x \in \tilde A$, let $\tilde \phi(\lambda + x) = \lambda + \phi(x)$. 
 \end{proof}
 
 
+
+

src/op/c-star/gelfand.tex

@@ -59,10 +59,10 @@
 
 \begin{corollary}
 \label{corollary:stonean-commutative-algebra}
-    Let $A$ be a unital $C^*$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected. 
+    Let $A$ be a unital $C^*$-algebra, then $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremally disconnected. 
 \end{corollary}
 \begin{proof}
-    By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremely-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremely disconnected.
+    By \autoref{theorem:gelfand-naimark}, $A$ and $C(\Omega(A); \complex)$ are isomorphic as $C^*$-algebras. In particular, $A_{sa}$ and $C(\Omega(A); \real)$ are isomorphic as ordered vector spaces, so $A_{sa}$ is order complete if and only if $C(\Omega(A); \real)$ is order complete. Thus the \hyperref[Stone-Nakano Theorem]{theorem:stone-nakano-extremally-disconnected} implies that $A_{sa}$ is order complete if and only if $\Omega(A)$ is extremally disconnected.
 \end{proof}
 
 

src/op/c-star/involution.tex

@@ -42,3 +42,21 @@
 \end{proof}
 
 
+\begin{definition}[$M_n(A)$]
+\label{definition:matrix-cstar}
+    Let $A$ be a $C^*$-algebra, $M_n(A) = M_n(\complex) \otimes A$ be the space of $n \times n$ matrices over $A$, equipped with the product
+    \[
+    M_n(A) \times M_n(A) \quad (M \otimes a)(N \otimes b) = MN \otimes ab
+    \]
+    and the involution
+    \[
+    *: M_n(A) \to M_n(A) \quad M \otimes a \mapsto M^* \otimes a^*
+    \]
+
+    Let $H$ be a complex Hilbert space and $\pi: A \to B(H)$ be a faithful representation of $A$, then $\pi$ induces a faithful representation $M_n(A) \otimes A \to B(H^n)$. Under the identification that $M_n(A) \subset B(H^n)$, the inherited norm\footnote{Which is equal to the projective norm because $M_n(\complex)$ is finite-dimensional and hence nuclear.} makes $M_n(A)$ a $C^*$-algebra. 
+\end{definition}
+
+\begin{definition}[Completely]
+\label{definition:cstar-completely}
+    Let $A, B$ be $C^*$-algebras and $\phi: A \to B$ be a function, then $\phi$ induces mappings $\phi_n = \text{Id}_n \otimes \phi: M_n(A) \to M_n(B)$. A property of $\phi$ holds \textbf{completely} if it holds for $\phi_n$ for all $n \in \natp$. 
+\end{definition}

src/op/example/bb.tex

@@ -88,11 +88,11 @@
 
 
 \begin{corollary}
-\label{corollary:linfinity-extremely-disconnected}
+\label{corollary:linfinity-extremally-disconnected}
-    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremely disconnected. 
+    Let $(X, \cm, \mu)$ be a localisable measure space, then $\Omega(L^\infty(X))$ is extremally disconnected. 
 \end{corollary}
 \begin{proof}
-    By \autoref{theorem:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremely disconnected. 
+    By \autoref{theorem:l-infty-dedekind-complete}, $L^\infty(X; \real)$ is order complete. By \autoref{corollary:stonean-commutative-algebra}, $\Omega(L^\infty(X))$ is extremally disconnected. 
 \end{proof}
 
 \begin{lemma}

src/op/example/bounded.tex

@@ -6,6 +6,22 @@
     Let $H$ be a complex Hilbert space, then $B(H) = L(H; H)$ is the algebra of all bounded linear operators on $H$. 
 \end{definition}
 
+\begin{lemma}
+\label{lemma:unitary-ots}
+    Let $H$ be a complex Hilbert space, then the strong and weak operator topologies on $U(B(H))$ coincide. 
+\end{lemma}
+\begin{proof}
+    Let $S, T \in U(B(H))$ and $\xi \in H$, then
+    \begin{align*}
+        \norm{(S - T)\xi}_H^2 &= \dpn{(S - T)\xi, (S - T)\xi}{H} \\
+        &= \norm{S\xi}_H^2 + \norm{T\xi}_H^2 - 2\text{Re}\angles{S\xi, T\xi}_H \\
+        &= 2\norm{\xi}_H^2  - 2\text{Re}\angles{S\xi, T\xi}_H
+    \end{align*}
+
+    which tends to $0$ as $S \to T$ in the weak operator topology. 
+    
+\end{proof}
+
 
 % 1. Every non-trivial ideal of B(H) contains the finite-rank operators.
 % 2. If H is separable, then the only non-trivial closed idela of B(H) are the compact operators. 
@@ -142,6 +158,18 @@
     so $QU^n|_{H} = V^n$.
 \end{proof}
 
+\begin{definition}[Compression]
+\label{definition:bh-compression}
+    Let $H$ be a complex Hilbert space, $K \subset H$ be a closed subspace, $P \in B(H)$ be the orthogonal projection onto $K$, $S \in B(K)$, and $T \in B(H)$, then $S$ is a \textbf{compression} of $T$ if $S = PT|_K$. 
+\end{definition}
+
+\begin{definition}[Dilation]
+\label{definition:bh-dilation}
+    Let $H$ be a complex Hilbert space and $T \in B(H)$, then a \textbf{dilation} of $T$ is a Hilbert space $K \supset H$ and an operator $S \in B(K)$ such that $T = PS|_H$, where $P \in B(K)$ is the orthogonal projection onto $H$. If $T^n = PS^n|_H$ for all $n \in \natz$, then $S$ is a \textbf{power dilation} of $T$. 
+\end{definition}
+
+
+
 \begin{definition}[Isometric Dilation]
 \label{definition:isometric-dilation}
     Let $H$ be a complex Hilbert space, $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, $D_T = (1 - T^*T)^{1/2}$, and
@@ -177,7 +205,7 @@
 
 \begin{theorem}[Sz.-Nagy's Dilation Theorem]
 \label{theorem:sznagy-dilation}
-    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a Hilbert space $K \supset H$ and a unitary operator $U \in B(K)$ such that $T^n = P_HU^n|_H$ for all $n \in \natp$, where $P_H$ is the orthogonal projection onto $H$. 
+    Let $H$ be a complex Hilbert space and $T \in B(H)$ with $\norm{T}_{B(H)} \le 1$, then there exists a unitary power dilation of $T$.
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Theorem 1.1]{Paulsen}}}. ]
     Let $K = l^2(\natp; H)^2$ and identify $H = (H \oplus \bracsn{0} \oplus \cdot) \times \bracsn{0}$. Let $V \in B(l^2(\natp; H))$ be the isometric dilation of $T$, and $U \in B(l^2(\natp; H)^2)$ be the unitary dilation of $V$. Identify $l^2(\natp; H)$ as a subspace of $K$ as $l^2(\natp; H) \times \bracsn{0}$, and let $P_{l^2(\natp; H)}$ be the orthogonal projection onto $l^2(\natp; H)$, then for each $n \ge 0$, 

src/op/system/definitions.tex

@@ -1,9 +1,134 @@
 \section{Operator Systems}
 \label{section:operator-systems}
 
-\begin{definition}[Operator System]
+\begin{definition}[Concrete Operator System]
-\label{definition:operator-system}
+\label{definition:concrete-operator-system}
-    Let $A$ be a $C^*$-algebra and $S \subset A$, then $S$ is an \textbf{operator system} if $S$ is a linear subspace of $A$ and $S = S^*$. 
+    Let $A$ be a $C^*$-algebra and $S \subset A$, then $S$ is a \textbf{concrete operator system} if:
+    \begin{enumerate}
+        \item $1_A \in S$.
+        \item $S = S^*$. 
+        \item $S$ is a linear subspace of $A$. 
+    \end{enumerate}
+\end{definition}
+
+\begin{definition}[Self-Adjoint Part]
+\label{definition:operator-system-sa}
+    Let $A$ be a $C^*$-algebra and $S \subset A$ be an operator system, then $S_{sa} = \bracsn{x \in S|x = x^*}$ is the \textbf{self-adjoint part} of $S$. 
+\end{definition}
+
+\begin{definition}[Positive Cone]
+\label{definition:operator-system-positive-cone}
+    Let $A$ be a $C^*$-algebra and $S \subset A$ be an operator system, then $S_{+} = \bracsn{x \in S_{sa}|x \ge 0}$ is the \textbf{positive cone} of $S$. 
 \end{definition}
 
 
+\begin{proposition}
+\label{proposition:operator-system-positive-span}
+    Let $A$ be a $C^*$-algebra, $S \subset A$ be an operator system, and $x \in S_{sa}$, then there exists positive elements $p, q \in S_{sa}$ such that $x = p - q$. 
+\end{proposition}
+\begin{proof}
+    Let $p = (\norm{x}_A + x)/2$ and $q = (\norm{x}_A - x)/2$, then $q$ is positive by \autoref{corollary:spectrum-characterisation-iff}. 
+\end{proof}
+
+\begin{definition}[Positive]
+\label{definition:operator-system-positive-map}
+    Let $A, B$ be $C^*$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a linear map, then $\phi$ is \textbf{positive} if $\phi(S_+) \subset \phi(T_+)$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:positive-automatic-continuity}
+    Let $A, B$ be $C^*$-algebra, $S \subset A$ and $T \subset B$ be operator systems, and $\phi: S \to T$ be a positive map, then
+    \begin{enumerate}
+        \item For any $x \in S_{sa}$, $\norm{\phi(x)}_{B} \le \norm{\phi(1_A)}_B \cdot \norom{x}_A$.
+        \item $\norm{\phi}_{L(S; T)} \le 2\norm{\phi(1_A)}_B$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $x \in T_+$, then $0 \le x \le \norm{x}_A$, and $0 \le \phi(x) \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$, so $\norm{\phi(x)}_B \le \norm{x}_A \cdot \norm{\phi(1_A)}_B$. 
+
+    For any $x, y \in T_+$, $x - y \le x$ and $y - x \le y$. As such, 
+    \[
+    \norm{x - y}_{B} \le \norm{x}_B \vee \norm{y}_B
+    \]
+
+    In particular, for any $s \in S_{sa}$, 
+    \begin{align*}
+        \phi(s) &= \frac{1}{2}\braks{\phi(\norm{s}_A + s) - \phi(\norm{s}_A - s)} \\
+        \norm{\phi(s)}_{B} &\le \frac{1}{2}\braks{\norm{\phi(\norm{s}_A + s)}_B \vee \norm{\phi(\norm{s}_A - s)}_B} \\
+        &\le \norm{s}_A \norm{\phi(1_A)}_B
+    \end{align*}
+    
+    (2): For any general $s \in S$, 
+    \begin{align*}
+        \phi(s) &= \phi(\text{Re}(s)) + i\phi(\text{Im}(s)) \\
+        \norm{\phi(s)}_B &\le \norm{\phi(\text{Re}(s))}_B + \norm{\phi(\text{Im}(s))}_{B} \\
+        &\le \norm{\phi(1_A)}_B \cdot (\norm{\text{Re}(s)}_A + \norm{\text{Im}(s)}_A) \le 2\norm{\phi(1_A)}_B\norm{s}_A
+    \end{align*}
+\end{proof}
+
+\begin{theorem}
+\label{theorem:positive-commutative-norm}
+    Let $X$ be a compact Hausdorff space, $A$ be a unital $C^*$-algebra and $\phi: C(X; \complex) \to A$ be a positive map, then $\norm{\phi}_{L(C(X; \complex); B)} = \norm{\phi(\one)}_{B}$. 
+\end{theorem}
+\begin{proof}
+    Assume without loss of generality that $\norm{\phi(\one)}_{B} \le 1$. 
+
+    Let $\seqf{h_j} \subset C(X; [0, 1])$ be a partition of unity on $X$, and $\seqf{\lambda_j} \subset \ol{B_\complex(0, 1)}$. 
+
+    Since $A$ is a unital $C^*$-algebra, there exists a complex Hilbert space $H$ such that $A$ may be identified with a $C^*$-subalgebra of $B(H)$, with $1_A = 1_{B(H)}$. In which case, for any $\xi, \eta \in H$, 
+    \begin{align*}
+        \abs{\sum_{j = 1}^n \dpn{\lambda_j \phi(h_j)\xi, \eta}{H}} &\le \sum_{j = 1}^n \abs{\dpn{\phi(h_j)\xi, \eta}{H}} \\
+        &= \sum_{j = 1}^n\abs{\dpn{\phi(h_j)^{1/2}\xi, \phi(h_j)^{1/2}\eta}{H}} \\
+        &\le \braks{\sum_{j = 1}^n \normn{\phi(h_j)^{1/2}\xi}_H^2}^{1/2} \cdot \braks{\sum_{j = 1}^n\normn{\phi(h_j)^{1/2}\eta}_H^2}^{1/2}\\
+        &\le \braks{\sum_{j = 1}^n \dpn{\phi(h_j)\xi, \xi}{H}}^{1/2} \cdot \braks{\sum_{j = 1}^n\dpn{\phi(h_j)\xi, \xi}{H}}^{1/2} \\
+        &\le \braks{\angles{\sum_{j = 1}^n \phi(h_j)\xi, \xi}{H}}^{1/2} \cdot \braks{\angles{\sum_{j = 1}^n \phi(h_j)\eta, \eta}{H}}^{1/2} \\
+        &\le (\dpn{\xi, \xi}{H})^{1/2}(\dpn{\eta, \eta}{H})^{1/2} =  \norm{\xi}_H \norm{\eta}_H
+    \end{align*}
+
+    so $\normn{\sum_{j = 1}^n \lambda_j\phi(h_j)}_A \le 1$.
+
+    Let $f \in C(X; \complex)$ with $\norm{f}_u \le 1$ and $\eps > 0$, then there exists a partition of unity $\seqf{h_j} \subset C(X; [0, 1])$ such that for each $1 \le j \le n$, $\sup_{x, y \in \supp{h_j}}|f(x) - f(y)| < \eps$. For each $1 \le j \le n$, choose $x_j \in \supp{h_j}$, then $\norm{f - \sum_{j = 1}^n f(x_j)h_j}_u \le \eps$.
+    
+    By the previous case and \autoref{proposition:positive-automatic-continuity}, 
+    \[
+    \norm{\phi(f)}_A \le \norm{\sum_{j = 1}^n f(x_j)\phi(h_j)}_A + 2\norm{f - \sum_{j = 1}^n f(x_j)h_j}_u \le 1 + 2\eps
+    \]
+\end{proof}
+
+
+\begin{example}[Arveson]
+    \label{example:positive-automatic-continuity}
+
+    Let $\mathbb{T} = \bracsn{z \in \complex|\ |z| = 1}$, and $S = \text{span}\bracsn{1, z, \ol z}  \subset C(\mathbb{T}; \complex)$, then $S$ is an operator system in $C(\mathbb{T} ; \complex)$. Let
+    \[
+    \phi: S \to M_2(\complex) \quad (a + bz + c\ol z) = \begin{bmatrix}
+        a & 2b \\
+        2c & a
+    \end{bmatrix}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item $\phi$ is a positive map.
+        \item $\norm{\phi}_{L(C(\mathbb{T} ; \complex); M_2(\complex))} = 2$. 
+        \item $\phi$ admits no extension to a positive map on $C(\mathbb{T}; \complex)$. 
+    \end{enumerate}
+
+    As such, $M_2(\complex)$ is not injective with respect to operator systems and positive maps. 
+\end{example}
+\begin{proof}[Proof, {{\cite[Example 2.2]{Paulsen}}}. ]
+    (1): Let $a + bz + c \ol z \in S$, then the following are equivalent:
+    \begin{enumerate}
+        \item $a + bz + c\ol z \ge 0$. 
+        \item $c = \ol b$ and $a \ge 2\text{Re}(b)$. 
+        \item $c = \ol b$ and $a \ge 2|b|$. 
+    \end{enumerate}
+
+    In addition, since a $2 \times 2$ matrix is positive if and only if its diagonal entries and its determinant are non-negative. Thus $\phi$ is positive. 
+
+    (2): $\norm{\phi(z)}_{M_2(\complex)} = 2$. 
+    
+    (3): By \autoref{theorem:positive-commutative-norm}, any such extension would have norm $1$. 
+\end{proof}
+
+

src/op/vn/spec.tex

@@ -322,7 +322,7 @@
     The complexity of $\Omega$, that is, the number of copies of $\Omega(A)$ that it contains, depends on two factors: 
     \begin{enumerate}
         \item The complexity of the von Neumann algebra $B$: If $B$ is sufficiently complex, then $\mathscr{E}$ cannot be expressed as the $L^1$ space of a single measure on $\Omega(A)$. Instead, multiple copies of $\Omega(A)$ are needed to handle mutually singular measures with overlapping supports. For more details on this phenomenon, see \autoref{theorem:hilbert-measures-dual}.
-        \item The size of the Hilbert space $H$ relative to $B$: If $H$ is extremely large, then a large number of vectors are required for $B$ to cover it. As such, many copies of $\Omega(A)$ are required to handle the complexity of $H$. 
+        \item The size of the Hilbert space $H$ relative to $B$: If $H$ is extremally large, then a large number of vectors are required for $B$ to cover it. As such, many copies of $\Omega(A)$ are required to handle the complexity of $H$. 
     \end{enumerate}
 
     More concretely, (1) manifests as the size of the space $\mathscr{E}$, and (2) manifests as the size of the kernel of the mapping $L^1(\Omega) \to \mathscr{E}$. 

src/topology/main/compactify.tex

@@ -79,7 +79,7 @@
     U_A = \bracsn{\fU \in \beta X| A \in \fU}
     \]
 
-    Equip $\beta X$ with the topology generated by $\fU := \bracsn{U_A|\emptyset \ne A \subset X}$. For each $x \in X$, let $e(x)$ be the principal ultrafilter at $x$, then
+    Equip $\beta X$ with the topology generated by $\fB := \bracsn{U_A|\emptyset \ne A \subset X}$. For each $x \in X$, let $e(x)$ be the principal ultrafilter at $x$, then
     \begin{enumerate}
         \item $(\beta X, e)$ is a compactification of $X$. 
         \item[(U1)] For any $f \in C(X; [0, 1])$, there exists a unique $\beta f \in C(\beta X; [0, 1])$ such that the following diagram commutes:
@@ -94,18 +94,18 @@
     In other words, $(\beta X, e)$ as constructed above is the Stone-Čech compactification of $X$. 
 \end{theorem}
 \begin{proof}
-    (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B} = U_A \cap U_B$. Thus $\fU$ is a base for the topology that it generates. 
+    (1): Firstly, $\beta X = U_{X}$. For any $A, B \subset X$, $U_{A \cap B} = U_A \cap U_B$. Thus $\fB$ is a base for the topology that it generates. 
 
     Let $\mathfrak{U}, \mathfrak{V} \in \beta X$ be distinct, then there exists $A \subset X$ with $A \in \mathfrak{U}$ and $A^c \in \mathfrak{V}$, so $\mathfrak{U} \in U_A$, $\mathfrak{V} \in U_{A^c}$, and $U_A \cap U_{A^c} = \emptyset$. Thus $\beta X$ is Hausdorff. 
 
-    Let $\mathcal{U} \subset 2^{\beta X}$ be an open cover of $X$. Since $\fU$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S} \subset 2^X \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,
+    Let $\mathcal{U} \subset 2^{\beta X}$ be an open cover of $X$. Since $\fB$ is a base for $\beta X$, assume without loss of generality that $\mathcal{U}$ is of the form $\bracsn{U_A|A \in \mathcal{S}}$, where $\mathcal{S} \subset 2^X \setminus \bracsn{\emptyset}$. Suppose for contradiction that $\mathcal{U}$ admits no finite subcover, then for any $F \subset \mathcal{S}$ finite,
     \[
     \bigcap_{A \in F}U_{A^c} = \bigcap_{A \in F}U_A^c = \braks{\bigcup_{A \in F}U_A}^c \ne \emptyset
     \]
     
     In particular, $\bigcap_{A \in F}A^c \ne \emptyset$. As such, $\bracsn{A^c|A \in \mathcal{S}}$ generates an ultrafilter $\fU \in \beta X$. Since $\fU \not\in U_A$ for all $A \in \mathcal{S}$, $\mathcal{U}$ does not cover $X$, which is a contradiction. Thus $\beta X$ is compact. 
 
-    Finally, since $X$ is discrete, $e$ is automatically continuous. For any $\emptyset \ne A \subset X$, $e(x) \in U_A$ for all $x \in A$, so $U_A \cap e(X) \ne \emptyset$, and $e(X)$ is dense in $\beta X$. Therefore $e(X)$ is embedded as a dense subspace of $\beta X$. 
+    Finally, since $X$ is discrete, $e$ is automatically continuous. On the other hand, for each $x \in X$, $U_{\bracsn{x}} = e(\bracsn{x})$ is open in $\beta X$. In addition, for any $\emptyset \ne A \subset X$, $e(x) \in U_A$ for all $x \in A$, so $U_A \cap e(X) \ne \emptyset$, and $e(X)$ is dense in $\beta X$. Therefore $e(X)$ is embedded as a dense subspace of $\beta X$. 
 
     (U1): Let $f \in C(X; [0, 1])$, and let $\beta f: \beta X \to [0, 1]$ be defined by $\beta f(\fU) = \lim_{\fU}f$. Fix $\fU \in \beta X$ and $\eps > 0$, then there exists $A \in \fU$ such that $f(A) \subset B_{[0, 1]}(\lim_{\fU}f, \eps)$. Thus for any $\fV \in U_A$, 
     \[
@@ -143,7 +143,7 @@
     The pair $(X^*, \iota)$ is the \textbf{one-point compactification} of $X$.
 \end{definition}
 \begin{proof}
-    Let $\infty$ be a point not in $X$, $X^* = X \sqcup \bracs{\infty}$, and $\mathcal{T}^* \subset 2^{X^*}$ such that for each $U \in \mathcal{T}^*$, one of the following holds:
+    (1): Let $\infty$ be a point not in $X$, $X^* = X \sqcup \bracs{\infty}$, and $\mathcal{T}^* \subset 2^{X^*}$ such that for each $U \in \mathcal{T}^*$, one of the following holds:
     \begin{enumerate}[label=(\alph*)]
         \item $U \in \mathcal{T}$.
         \item $\infty \in U$ and $U^c \subset X$ is compact. 
@@ -153,7 +153,7 @@
     
     Let $\iota: X \to X^*$ be the inclusion map. For each $U \in \mathcal{T}^*$ satisfying (b), $\iota^{-1}(U) = U \cap X$. Since $U^c \subset X$ is compact, $U \cap X$ is open by \autoref{proposition:compact-closed}, so $\iota \in C(X; X^*)$. By (a), $\iota$ is an embedding. 
 
-    Finally, let
+    (U): Let
     \[
     \varphi^*: Y \to X^* \quad x \mapsto \begin{cases}
         \varphi^{-1}(x) &x \in \varphi(X) \\

src/topology/main/stonean.tex

@@ -1,18 +1,18 @@
 \section{Extremely Disconnected Spaces}
-\label{section:extremely-disconnected-space}
+\label{section:extremally-disconnected-space}
 
 \begin{definition}[Extremely Disconnected]
-\label{definition:extremely-disconnected}
+\label{definition:extremally-disconnected}
-    Let $X$ be a topological space, then $X$ is \textbf{extremely disconnected} if for every $U \subset X$ open, $\ol U$ is also open. 
+    Let $X$ be a topological space, then $X$ is \textbf{extremally disconnected} if for every $U \subset X$ open, $\ol U$ is also open. 
 \end{definition}
 
 
 \begin{theorem}[Stone-Nakano]
-\label{theorem:stone-nakano-extremely-disconnected}
+\label{theorem:stone-nakano-extremally-disconnected}
-    Let $X$ be a topological space. If $X$ is extremely disconnected, then $C(X; \real)$ is order complete.  Conversely, if $X$ is completely regular and $C(X; \real)$ is order complete, then $X$ is extremely disconnected. 
+    Let $X$ be a topological space. If $X$ is extremally disconnected, then $C(X; \real)$ is order complete.  Conversely, if $X$ is completely regular and $C(X; \real)$ is order complete, then $X$ is extremally disconnected. 
 \end{theorem}
 \begin{proof}
-    ($\Rightarrow$): Suppose that $X$ is extremely disconnected. Let $\cf \subset C(X; \real)$ and $F \in C(X; \real)$ be an upper bound of $\cf$. For each $q \in \real$, let
+    ($\Rightarrow$): Suppose that $X$ is extremally disconnected. Let $\cf \subset C(X; \real)$ and $F \in C(X; \real)$ be an upper bound of $\cf$. For each $q \in \real$, let
     \[
     U(q) = \overline{\bigcup_{f \in \cf}\bracs{f > q}} \subset \overline{\bracs{F > q}}
     \]