Added more facts about the positive square root.

document.tex

@@ -6,6 +6,8 @@
 
 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. 
+
 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. 
 
 \input{./src/cat/index}

src/op/banach/invertible.tex

@@ -52,6 +52,15 @@
     (3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}. 
 \end{proof}
 
+\begin{corollary}
+\label{corollary:invertible-boundary-explode}
+    Let $A$ be a unital Banach algebra, $x \in A \setminus G(A)$, and $r > 0$, then $\normn{y^{-1}}_A > 1/r$ for all $y \in B(x, r) \cap G(A)$. 
+\end{corollary}
+\begin{proof}
+    Let $y \in B(x, r)$, then $B(y, \normn{y^{-1}}_A^{-1}) \subset G(A)$ by \autoref{proposition:banach-algebra-inverse}. Since $x \not\in G(A)$, $r > \norm{x - y}_A \ge \normn{y^{-1}}_A^{-1}$, and $1/r < \normn{y^{-1}}_A$. 
+\end{proof}
+
+
 \begin{proposition}
 \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 $1 - yx \in G(A)$. 
@@ -74,5 +83,3 @@
     \end{align*}
     
 \end{proof}
-
-

src/op/banach/spectrum.tex

@@ -146,4 +146,29 @@
     The above holds for $x$ and $y$ with respect to $\sigma_A$. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:spectrum-subalgebra-gymnastics}
+    Let $A$ be a unital Banach algebra, $B \subset A$ be a closed subalgebra containing $1$, and $x \in B$, then:
+    \begin{enumerate}
+        \item $\sigma_A(x) \subset \sigma_B(x)$. 
+        \item $\partial \sigma_B(x) \subset \sigma_A(x)$. 
+        \item $\sigma_B(x)$ is the union of $\sigma_A(x)$ and some bounded components of $\complex \setminus \sigma_A(x)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): $G(B) \subset G(A)$. 
+
+    (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"]
+\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}
+\begin{proof}[Proof, {{\cite[Theorem 4.9]{MarcouxNotes}}}. ]
+    By construction, $P \subset \complex \setminus \sigma_B(x)$. In addition, for any polynomial $p \in \complex[z]$, $p(x) \in B$. Thus for every rational function $f \in \complex(z) \cap H(\complex_\infty \setminus (P \cup \bracs{\infty}); \complex)$, $f(x) \in B$. 
+
+    By \hyperref[Runge's Theorem]{theorem:runge}, $H(\complex_\infty \setminus (P \cup \bracs{\infty}); \complex)$ is dense in $H(\sigma_A(x); \complex)$. The continuity of the \hyperref[holomorphic functional calculus]{definition:holomorphic-functional-calculus} then implies that $f(x) \in B$ for all $f \in H(\sigma_A(x); \complex)$. In particular, $(\lambda - x)^{-1} \in B$ for all $\lambda \in \complex \setminus \sigma_A(x)$. Therefore $\sigma_B(x) \subset \sigma_A(x)$, and $\sigma_B(x) = \sigma_A(x)$ by \autoref{proposition:spectrum-subalgebra-gymnastics}.
+\end{proof}
+
 

src/op/c-star/cont.tex

@@ -0,0 +1,121 @@
+\section{The Continuous Functional Calculus}
+\label{section:continuous-functional-calculus}
+
+\begin{theorem}[Spectral Theorem for $C^*$-Algebras]
+\label{theorem:spectral-c-star}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then the mapping
+    \[
+    \Phi: \Omega(A[x]) \to \sigma_A(x) \quad \Phi(\psi) = \Gamma_{A[x]}(x)(\psi) = \psi(x)
+    \]
+
+    is a homeomorphism. 
+\end{theorem}
+\begin{proof}
+    Firstly, $A[x]$ is commutative by \autoref{proposition:generated-subalgebra-dense}. Thus \autoref{corollary:c-star-algebra-preserve-spectrum} and (3) of \autoref{proposition:gelfand-transform-gymnastics} imply that
+    \[
+    \Phi(\Omega(A[x])) = \Gamma_{A[x]}(\Omega(A[x])) = \sigma_{A[x]}(x) = \sigma_A(x)
+    \]
+
+    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. 
+
+    Finally, since $\Omega(A[x])$ is equipped with the weak* topology and $\Phi$ is the evaluation map at $x$, it is continuous. 
+
+    By \autoref{proposition:compact-hausdorff-homeomorphism}, $\Phi$ is a homeomorphism. 
+\end{proof}
+
+\begin{definition}[Continuous Functional Calculus]
+\label{definition:continuous-functional-calculus}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then there exists a unique continuous unital *-homomorphism
+    \[
+    C(\sigma_A(x); \complex) \to A[x] \quad f \mapsto f(x)
+    \]
+
+    such that:
+    \begin{enumerate}
+        \item $\one(x) = 1_A$. 
+        \item $\text{Id}(x) = x$.
+        \item $\overline{\text{Id}}(x) = x^*$.
+    \end{enumerate}
+
+    Moreover,
+    \begin{enumerate}[start=3]
+        \item Under the identification $\Omega(A[x]) = \sigma_A(x)$, for each $f \in C(\sigma_A(x); \complex)$, $f(x) = \Gamma^{-1}_{A[x]}(f)$.
+        \item The mapping $f \mapsto f(x)$ is a unital *-isomorphism.  
+        \item For each $U \subset \complex$ open and $f \in C(U; \complex)$, the mapping
+        \[
+        \bracs{x \in A|\sigma_A(x) \subset U} \to A \quad x \mapsto f(x)
+        \]
+
+        is continuous. 
+    \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}. 
+
+    (2): The identification 
+    \[
+    \Phi: \Omega(A[x]) \to \sigma_A(x) \quad \Phi(\psi) = \Gamma_{A[x]}(x)(\psi) = \psi(x)
+    \]
+
+    given by the Spectral Theorem implies that $\Gamma_{A[x]}(x) = \text{Id}$.
+
+    (3): The mapping $f \mapsto f(x)$ is a *-homomorphism. 
+
+    (6): Fix $x \in A$ and let $K \in \cn_U(\sigma_A(x); \complex)$ be compact. By \autoref{proposition:spectrum-continuous}, there exists $r > 0$ such that $\sigma_A(y) \subset K$ for all $y \in B_A(x, r)$. 
+
+    Let $\eps > 0$. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $p \in \complex[z, \ol z]$ such that $\sup_{\lambda \in K}|p(\lambda) - f(\lambda)| < \eps$. In which case, 
+    \begin{align*}
+    \norm{f(x) - f(y)}_A &\le \norm{f(x) - p(x)}_A + \norm{f(y) - p(y)}_A + \norm{p(x) - p(y)}_A \\
+    &\le \norm{p(x) - p(y)}_A + 2\eps
+    \end{align*}
+
+    for all $y \in B_A(x, r)$. Since $p$ is a continuous map on $A$, there exists $\delta \in (0, r)$ such that $\norm{p(x) - p(y)}_A < \eps$ for all $y \in B_A(x, \delta)$. Therefore $\norm{f(x) - f(y)}_A < 3\eps$ for all $y \in B_A(x, \delta)$. 
+
+    (Uniqueness): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
+\end{proof}
+
+\begin{theorem}[Spectral Mapping Theorem (Continuous)]
+\label{theorem:spectral-mapping-continuous}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item For every $f \in C(\sigma_A(x); \complex)$, $\sigma_A(f(x)) = f(\sigma_A(x))$. 
+        \item For every $f \in C(\sigma_A(x); \complex)$ and $g \in C(\sigma_A(f(x)); \complex)$, $(f \circ g)(x) = f(g(x))$.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Since $f(x) \in A[x]$, by \autoref{proposition:gelfand-transform-gymnastics} and definition of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    \sigma_A(f(x)) = \sigma_{A[x]}(f(x)) = \Gamma_{A[x]}(f(x))(\sigma_A(x)) = f(\sigma_A(x))
+    \]
+
+    (2): Firstly, the theorem holds directly if $f, g \in \complex[z, \ol z]$ are polynomials in $z$ and $\ol z$. 
+
+    Suppose that $g \in \complex[z, \ol z]$ but $f$ is arbitrary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to f$ uniformly on $\sigma_A(x)$. By property (6) of the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, 
+    \[
+    (g \circ f)(x) = \limv{n}(g \circ f_n)(x) = \limv{n}g(f_n(x)) = \limv{n}g(f(x))
+    \]
+
+    Finally, suppose that both $f$ and $g$ are arbitary. By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}, there exists $\seq{p_n} \subset \complex[z, \ol z]$ such that $p_n \to g$ uniformly on $\sigma_A(f(x))$. By continuity of the continuous functional calculus, 
+    \[
+    (g \circ f)(x) = \limv{n}(g_n \circ f)(x) = \limv{n}g_n(f(x)) = \limv{n}g(f(x))
+    \]
+\end{proof}
+
+
+
+\begin{corollary}
+\label{corollary:normal-spectrum-identity}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then:
+    \begin{enumerate}
+        \item $x$ is self-adjoint if and only if $\sigma(x) \subset \real$. 
+        \item $x$ is unitary if and only if $\sigma(x) \subset \partial B_\complex(0, 1)$. 
+        \item $x$ is a projection if and only if $\sigma(x) \subset \bracs{0, 1}$. 
+    \end{enumerate}
+\end{corollary}
+
+
+
+

src/op/c-star/index.tex

@@ -2,8 +2,10 @@
 \label{chap:c-star-algebras}
 
 \input{./involution.tex}
+\input{./sub.tex}
 \input{./unitary.tex}
 \input{./sa.tex}
-\input{./order.tex}
 \input{./homomorphism.tex}
-\input{./gelfand.tex}
+\input{./gelfand.tex}
+\input{./cont.tex}
+\input{./order.tex}

src/op/c-star/involution.tex

@@ -41,17 +41,4 @@
     (3): For any $x \in A$, $(x^{-1})^*x^* = (x^{-1}x)^* = 1 = (xx^{-1})^* = x^*(x^{-1})^*$.
 \end{proof}
 
-\begin{definition}[*-Homomorphism]
-\label{definition:star-homomorphism}
-    Let $A, B$ be $C^*$-algebras and $\phi: A \to B$, then $\phi$ is a \textbf{*-homomorphism} if:
-    \begin{enumerate}
-        \item $\phi$ is a homomorphism of Banach algebras.
-        \item For every $x \in A$, $\phi(x^*) = \phi(x)^*$. 
-    \end{enumerate}
-
-    If in addition, $\phi(1) = 1$, then $\phi$ is a \textbf{unital *-homomorphism}. 
-\end{definition}
-
-
-
 

src/op/c-star/order.tex

@@ -6,5 +6,54 @@
     Let $A$ be a $C^*$-algebra and $x \in A$, then $x$ is \textbf{positive} if there exists $y \in A$ such that $x = y^*y$. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:positive-spectrum}
+    Let $A$ be a unital $C^*$-algebra and $x \in A$ be normal, then $x$ is positive if and only if $\sigma_A(x) \subset [0, \infty)$. 
+\end{proposition}
+\begin{proof}
+    Using the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $x$ is positive if and only if $\Gamma_{A[x]}(x) = \text{Id}$ is positive in $C(\sigma_A(x); \complex)$, if and only if $\sigma_A(x) = \Gamma_{A[x]}(x)(\Omega(A[x])) \subset [0, \infty)$ by \autoref{proposition:gelfand-transform-gymnastics}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:positive-norm-inequality}
+    Let $A$ be a unital $C^*$-algebra and $x \in A_{sa}$, then the following are equivalent:
+    \begin{enumerate}
+        \item $x$ is positive.
+        \item $\sigma_A(x) \subset [0, \infty)$. 
+        \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}}}. ]
+    (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
+    \[
+    \norm{\lambda - x}_A = [\lambda - x]_{sp} = \sup\bracsn{\lambda - \mu|\mu\in \sigma_A(x)}
+    \]
+
+    which is bounded above by $\lambda$ if and only if $\sigma_A(x) \subset [0, \infty)$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:positive-ordering}
+    Let $A$ be a unital $C^*$-algebra. For each $x, y \in A$, denote $x \ge y$ if $x - y$ is positive, then $(A, \le)$ is an ordered vector space. 
+\end{corollary}
+\begin{proof}
+    By definition, the ordering is reflexive, antisymmetric, translation-invariant, and invariant under scaling by positive constants. It remains to show that $\le$ is transitive, or equivalently, the sum of two positive elements is positive. 
+
+    Let $x, y \in A$ be positive, then $x + y$ is self-adjoint. Thus there exists $\lambda \ge \norm{x}_A$ and $\mu \ge \norm{y}_A$ such that $\norm{\lambda - x}_A \le \lambda$ and $\norm{\mu - y}_A \le \mu$, so $\norm{(\lambda + \mu) - (x + y)}_A \le \lambda + \mu$, and $x + y$ is positive by \autoref{proposition:positive-norm-inequality}. 
+\end{proof}
+
+
+\begin{definition}[Positive Square Root]
+\label{definition:positive-square-root}
+    Let $A$ be a $C^*$-algebra and $x \in A$ be positive, then there exists a unique positive element $y \in A$ such that $y^2 = x$. The element $y$ is the \textbf{positive square root} of $x$, denoted $\sqrt{x}$. 
+\end{definition}
+\begin{proof}
+    Since $x$ is positive, $\sigma_A(x) \subset [0, \infty)$ by \autoref{proposition:positive-norm-inequality}. Therefore the square root function $f(t) = \sqrt{t}$ is defined and continuous on $\sigma_A(x)$. Using the \hyperref[continuous functional calculus]{definition:continuous-functional-calculus}, $f(x)$ is a positive element of $A$ such that $f(x)^2 = x$. 
+
+    Let $y \in A$ such that $y^2 = x$, then by the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-continuous}, $y = f(y^2) = f(x)$, so the square root is unique. 
+\end{proof}
+
 
 

src/op/c-star/sub.tex

@@ -0,0 +1,41 @@
+\section{Subalgebras}
+\label{section:sub-c-star-algebras}
+
+\begin{definition}[Generated Subalgebra]
+\label{definition:generated-subalgebra}
+    Let $A$ be a unital $C^*$-algebra and $S \subset A$, then $A[S]$ is the smallest $C^*$-subalgebra of $A$ containing $1$ and $S$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:generated-subalgebra-dense}
+    Let $A$ be a unital $C^*$-algebra, $S \subset A$, and $\mathcal{S} = S \cup \bracs{x^*|x \in S}$, then
+    \begin{enumerate}
+        \item The linear span
+        \[
+        \text{span}\bracs{\prod_{j = 1}^n x_j \bigg | \seqf{x_j} \subset \mathcal{S}}
+        \]
+
+        is dense in $A[S]$. 
+        \item If for any $x, y \in \mathcal{S}$, $xy = yx$, then $A[S]$ is commutative.
+        \item For any normal element $x \in A$, $A[x]$ is commutative. 
+    \end{enumerate}
+\end{proposition}
+% Obvious so omitted. 
+
+\begin{proposition}
+\label{proposition:c-star-algebra-preserve-gl}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra containing $1$, then $G(B) = G(A) \cap B$. 
+\end{proposition}
+\begin{proof}
+    Let $x \in G(A) \cap B$, then $x^*x \in G(A) \cap B$ as well. In particular, $0 \not\in \sigma_A(x^*x)$. Since $x^*x \in A_{sa}$, $\sigma_A(x^*x) \subset \real$ by \autoref{proposition:self-adjoint-spectrum}. By \autoref{proposition:spectrum-subalgebra-gymnastics}, $\partial \sigma_B(x^*x) \subset \sigma_A(x^*x) \subset \real$. Thus $\sigma_B(x^*x) \subset \real$ as well, which means that $\partial \sigma_B(x^*x) = \sigma_B(x^*x) = \sigma_A(x^*x)$. Therefore $0 \not\in \sigma_A(x^*x) = \sigma_B(x^*x)$, $x^*x \in G(B)$, and $x \in G(B)$. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:c-star-algebra-preserve-spectrum}
+    Let $A$ be a unital $C^*$-algebra, $B \subset A$ be a $C^*$-subalgebra containing $1$, and $x \in B$, then $\sigma_A(x) = \sigma_B(x)$. 
+\end{corollary}
+\begin{proof}
+    By \autoref{proposition:c-star-algebra-preserve-gl}. 
+\end{proof}
+
+

src/op/notation.tex

@@ -14,6 +14,7 @@
     $\Omega(A)$ & Space of multiplicative functionals on $A$. & \autoref{definition:multiplicative-functional} \\
     $\cm(A)$ & 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} \\ 
 
     $M_n(\complex)$ & 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} \\