Added continuity of homomorphisms.

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/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/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}
+
+
 
 
 

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 a 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/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

@@ -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/c-star/cont.tex

@@ -136,7 +136,7 @@
         \item $\norm{\Phi(x)}_B = \norm{x}_A$.
     \end{enumerate}
 \end{corollary}
-\begin{proof}[Proof, {{\cite[II.10.7]{Zhu}}}. ]
+\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. 

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/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 +36,22 @@
     
 \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}
+
 

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

src/op/c-star/sa.tex

@@ -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 \\

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/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
     \[
     \cf = \bracs{f \in C(X; [0, 1])| 0 \le f \le \one_U}
     \]