Added the holomorphic functional calculus.

refs.bib

@@ -122,3 +122,32 @@
   year={1978},
   publisher={Springer}
 }
+@book{Zhu,
+  title={An Introduction to Operator Algebras},
+  author={Zhu, K.},
+  isbn={9780849378751},
+  lccn={93007172},
+  series={Studies in Advanced Mathematics},
+  url={https://books.google.ca/books?id=XHLj7bz8hOIC},
+  year={1993},
+  publisher={Taylor \& Francis}
+}
+@book{FollandHarmonic,
+  title={A Course in Abstract Harmonic Analysis},
+  author={Folland, G.B.},
+  isbn={9781498727150},
+  series={Textbooks in Mathematics},
+  url={https://books.google.ca/books?id=z-GYCgAAQBAJ},
+  year={2016},
+  publisher={CRC Press}
+}
+@book{Takesaki1,
+  title={Theory of Operator Algebras I},
+  author={Takesaki, M.},
+  isbn={9783540422488},
+  lccn={79013655},
+  series={Encyclopaedia of Mathematical Sciences},
+  url={https://books.google.ca/books?id=38QIwQEACAAJ},
+  year={2001},
+  publisher={Springer Berlin Heidelberg}
+}

src/dg/complex/runge.tex

@@ -3,10 +3,17 @@
 
 \begin{proposition}
 \label{proposition:existence-curves}
-    Let $K \subset \complex$ be compact and $U \in \cn_\complex(K)$, then there exists closed rectifiable curves $\seqf{\gamma_j}$ such that for any separated locally convex space $E$, $f \in H(U; E)$, and $z_0 \in K$, 
+    Let $K \subset \complex$ be compact and $U \in \cn_\complex(K)$, then there exists closed rectifiable curves $\seqf{\gamma_j}$ such that for any separated locally convex space $E$, 
-    \[
+    \begin{enumerate}
-    f(z) = \sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
+        \item For every $f \in H(U; E)$ and $z_0 \in K$, 
-    \]
+        \[
+        f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
+        \]
+        \item For every $V \in \cn_\complex(U)$, $f \in H(V; E)$, and $z_0 \in V \setminus U$, 
+        \[
+        \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = 0
+        \]
+    \end{enumerate}
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ]
     Via \hyperref[fattening]{proposition:distance-compact}, let $V \in \cn_\complex^o(K)$ precompact with $\ol V \subset U$. Identify $\complex = \real^2$, then since $V$ is precompact, there exists $\delta > 0$ and $\seqf{(x_j, y_j)} \subset U$ such that:
@@ -26,8 +33,8 @@
 
     and $\gamma_j = \gamma_{j, \downarrow} \cdot \gamma_{j, \leftarrow} \cdot \gamma_{j, \uparrow} \cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_j$, 
     \[
-    \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases}
+    \frac{1}{2\pi i}\int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases}
-        f(z) &z \in R_j^o \\
+        f(z_0) &z \in R_j^o \\
         0 &z \in U \setminus R_j
     \end{cases}
     \]
@@ -39,7 +46,7 @@
 
     then for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
     \[
-    f(z) = \sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz
     \]
 
     From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_j, \mu_k)$ is \textit{redundant} if for each $t \in [0, 1]$, $\mu_j(t) = \mu_k(1 - t)$. If $(\mu_j, \mu_k)$ are redundant, then
@@ -52,7 +59,7 @@
 
     By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_j, \mu_k)$ is redundant. By (5), for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
     \[
-    f(z) = \sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz
     \]
 
     Let $1 \le j \le N$ such that $\mu_j([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_\complex(K)$, there exists $1 \le k \le N$ such that $(\mu_j, \mu_k)$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^m \mu_j([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^n R_j^o$ is dense in $K$, the above also holds for all $z \in K$. 
@@ -60,6 +67,7 @@
     Finally, let $1 \le j \le m$. Since $\mu_j$ does not form a redundant pair, $\mu_j(1)$ intersects at most two distinct squares by (3). In which case, there must exist $1 \le k \le m$ such that $\mu_k(0) = \mu_j(1)$. Therefore $\seqf[m]{\mu_j}$ forms a collection of closed rectifiable curves. 
 \end{proof}
 
+
 \begin{lemma}
 \label{lemma:rational-curve-approximation}
     Let $\gamma \in C([a, b]; \complex)$ be a rectifiable curve, $K \subset \complex$ such that $K \cap \gamma([a, b]) = \emptyset$, $f \in C(\gamma([a, b]); \complex)$, and $\eps > 0$, then there exists $R \in \complex(z)$ such that:
@@ -145,20 +153,20 @@
 \begin{proof}
     Let $U \in \cn_\complex(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_0 \in K$, 
     \[
-    f(z_0) = \sum_{j = 1}^n \int_{\gamma_j}\frac{f(z)}{z - z_0}dz
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j}\frac{f(z)}{z - z_0}dz
     \]
 
     Let $T$ be the union of the images of $\seqf{\gamma_j}$, then by \autoref{lemma:rational-curve-approximation}, there exists $\seqf{R_j} \subset \complex(z) \cap H(\complex \setminus T; \complex)$ such that for each $z_0 \in K$ and $1 \le j \le n$, 
     \[
-    \abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
+    \abs{\frac{1}{2\pi i}\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
     \]
 
-    so
+    Thus
     \[
     \abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps
     \]
     
-    for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$. 
+    for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P; \complex)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$. Therefore $|S(z_0) - f(z_0)| < 2\eps$ for all $z_0 \in K$.
 \end{proof}
 
 \begin{corollary}

src/op/banach/definitions.tex

@@ -15,4 +15,14 @@
     Let $A$ 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, $1$ is the unique \textbf{multiplicative identity} of $A$. 
 \end{definition}
 
+\begin{definition}[Homomorphism]
+\label{definition:banach-algebra-homomorphism}
+    Let $A, B$ be Banach algebras and $\phi: A \to B$, then $\phi$ is a \textbf{homomorphism} if:
+    \begin{enumerate}
+        \item $\phi \in L(A; B)$. 
+        \item For each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
+    \end{enumerate}
+\end{definition}
+
+
 

src/op/banach/fc.tex

@@ -0,0 +1,96 @@
+\section{The Holomorphic Functional Calculus}
+\label{section:holomorphic-functional-calculus}
+
+
+\begin{definition}[Holomorphic Functional Calculus]
+\label{definition:holomorphic-functional-calculus}
+    Let $A$ be a unital Banach algebra and $x \in A$, then there exists a unique continuous homomorphism
+    \[
+    H(\sigma_A(x); \complex) \to A \quad f \mapsto f(x)
+    \]
+
+    such that:
+    \begin{enumerate}
+        \item $1(x) = 1$. 
+        \item $\text{Id}(x) = x$. 
+    \end{enumerate}
+
+    Moreover, 
+    \begin{enumerate}[start=2]
+        \item For each $U \in \cn_\complex(\sigma_A(x))$ and closed rectifiable curves $\seqf{\gamma_j}$ on $U \setminus \sigma_A(x)$ such that $f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} f(z)/(z - z_0)dz$ for all $f \in H(U; \complex)$ and $z_0 \in \sigma_A(x)$, 
+        \[
+        f(x) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(\lambda)}{\lambda - x}d\lambda
+        \]
+
+        for all $f \in H(U; \complex)$. 
+    \end{enumerate}
+
+    The mapping $f \mapsto f(x)$ is the \textbf{holomorphic functional calculus} of $x$. 
+\end{definition}
+\begin{proof}[Proof, {{\cite[Proposition I.2.7]{Takesaki1}}}. ]
+    (Definition): Let $U, V \in \cn_\complex(\sigma_A(x))$ such that $\ol V \subset U$, then by \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_A(x)$ such that 
+    \begin{enumerate}[label=(\alph*)]
+        \item For all $f \in H(V; \complex)$ and $z_0 \in \sigma_A(x)$, 
+        \[
+        f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
+        \]
+
+        \item For all $f \in H(U; \complex)$ and $z_0 \in U \setminus V$, 
+        \[
+        \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = 0
+        \]
+    \end{enumerate}
+    
+    For each $f \in H(U; \complex)$, define
+    \[
+    f(x) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(\lambda)}{\lambda - x}d\lambda
+    \]
+
+    then by \hyperref[Cauchy's Theorem]{theorem:cauchy-homotopy}, the above definition is independent of the choice of curves satisfying (a). 
+
+    (Linearity): By \autoref{proposition:rs-bound}, the mapping $f \mapsto f(z)$ is a continuous linear map from $H(\sigma_A(x); \complex)$ to $A$. 
+
+    (Homomorphism): Let $T$ be the union of the image of $\seq{\gamma_j}$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf[m]{\mu_j}$ on $U \setminus \ol V$ such that $g(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^m \int_{\mu_j} g(z)/(z - z_0)dz$ for all $f \in H(U; \complex)$ and $z_0 \in \ol V$. Now, 
+    \begin{align*}
+        f(x)g(x) &= \frac{1}{(2\pi i)^2}\sum_{j = 1}^n \sum_{k = 1}^m \braks{\int_{\gamma_j} \frac{f(z)}{z - x}dz} \cdot \braks{\int_{\mu_k}\frac{g(w)}{w - x}dw} \\
+        &= \frac{1}{(2\pi i)^2}\sum_{j = 1}^n \sum_{k = 1}^m \int_{\gamma_j}\int_{\mu_k} \frac{f(z)g(w)}{(z - x)(w - x)}dwdz
+    \end{align*}
+    
+    For each $z, w \in U \setminus \sigma_A(x)$ with $z \ne w$, by the \hyperref[resolvent equation]{lemma:resolvent-equation}, 
+    \[
+        \frac{f(z)g(w)}{(z - x)(w - x)} = \frac{f(z)g(w)}{w - z}\braks{\frac{1}{z - x} - \frac{1}{w - x}}
+    \]
+
+    By assumptions on $\seqf[m]{\mu_j}$, for each $1 \le j \le n$, 
+    \[
+        \frac{1}{2\pi i }\int_{\gamma_j} \sum_{k = 1}^m \int_{\mu_k} \frac{f(z)g(w)}{(w - z)(z - x)}dwdz = \int_{\gamma_j} \frac{f(z)g(z)}{z - x}dz
+    \]
+
+    By assumption (b) and \hyperref[Fubini's Theorem]{theorem:rs-fubini}, for each $1 \le k \le m$, 
+    \[
+        \sum_{j = 1}^n\int_{\gamma_j}\int_{\mu_k}  \frac{f(z)g(w)}{(w - z)(z - x)}dwdz = 0
+    \]
+
+    Therefore
+    \[
+        f(x)g(x) = \frac{1}{2\pi i }\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)g(z)}{z - x}dz = (fg)(x)
+    \]
+
+    and the mapping $f \mapsto f(x)$ is a homomorphism. 
+
+    (1): Since the constant $1$ function is the identity in $H(\sigma_A(x); \complex)$, $1(x) = 1$ by the homomorphism property. 
+
+    (2): Let $R > 0$ such that $\sigma_A(x) \subset B_\complex(0, R)$, then by \autoref{proposition:rs-complete} and \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}, 
+    \[
+        \text{Id}(x) = \frac{1}{2\pi i }\int_{\omega_{0, R}} \frac{1}{z - x}dz = \frac{1}{2\pi i }\sum_{n = 0}^\infty\int_{\omega_{0, R}}  z^{-n-1}x^n dz = 1
+    \]
+
+    and
+    \[
+        \text{Id}(x) = \frac{1}{2\pi i }\int_{\omega_{0, R}} \frac{z}{z - x}dz = \frac{1}{2\pi i }\sum_{n = 0}^\infty\int_{\omega_{0, R}}  z^{-n}x^n dz = x
+    \]
+
+    (Uniqueness): By (2), the homomorphism extends uniquely to $\complex(z) \cap H(\sigma_A(x); \complex)$. By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, it extends uniquely to $H(\sigma_A(x); \complex)$ by continuity. 
+
+\end{proof}
+

src/op/banach/igroup.tex

@@ -0,0 +1,34 @@
+\section{The Index Group}
+\label{section:index-group}
+
+\begin{definition}[Identity Component]
+\label{definition:identity-component}
+    Let $A$ be a unital Banach algebra, then the connected component $G_0(A)$ of $G(A)$ containing $1$ is the \textbf{identity component} of $G(A)$, and:
+    \begin{enumerate}
+        \item $G_0(A)$ is an open, closed, and normal subgroup of $G(A)$. 
+        \item The cosets of $G_0(A)$ are the connected components of $G(A)$. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}[Proof, {{\cite[Theorem 2.4]{Zhu}}}. ]
+    (1): Since $G_0(A)$ is connected, it is open and closed. By \autoref{proposition:locally-path-connected-properties}, $G_0(A)$ coincides with the path component of $G(A)$ containing $1$. 
+
+    Let $x, y \in G_0(A)$, then there exists paths $f, g \in C([0, 1]; G_0(A))$ such that $f(0) = g(0) = 1$, $f(1) = x$, and $g(1) = y$. The concatenation of $f$ and $xg$ then is a path from $1$ to $xy$, so $xy \in G_0(A)$. In addition, $t \mapsto f(t)^{-1}$ is a path from $1$ to $x^{-1}$, so $x^{-1} \in G_0(A)$ as well. Therefore $G_0(A)$ is a subgroup of $G(A)$. 
+
+    Finally, let $x \in G(A)$, then $x^{-1}G_0(A)x$ is a connected subset of $G(A)$ containing $1$, so $x^{-1}G_0(A)x \subset G_0(A)$ and $G_0(A) \subset xG_0(A)x^{-1}$. Since the above holds for all $x \in G(A)$, $x^{-1}G_0(A)x = G_0(A)$. 
+
+    (2): For each $x \in G(A)$, $xG_0(A)$ is connected, closed, and open, so it is a connected component by \autoref{lemma:union-connected-components}. 
+\end{proof}
+
+% Note: this setup appears to work in general topological groups. There does not seem to be any payoffs as of now.
+% However, should there be any, the proof should be improved to avoid the path argument. 
+
+
+\begin{definition}[Index Group]
+\label{definition:index-group}
+    Let $A$ be a unital Banach algebra, then the \textbf{index group} $I(A)$ is the quotient $G(A)/G_0(A)$, which is a discrete group. 
+\end{definition}
+\begin{proof}
+    By \autoref{definition:identity-component}, the components of $G(A)$ are the cosets of $G_0(A)$, so the quotient group is discrete. 
+\end{proof}
+
+

src/op/banach/index.tex

@@ -1,5 +1,8 @@
-\chapter{$C*$-Algebras}
+\chapter{Banach Algebras}
 \label{chap:banach-algebras}
 
 \input{./definitions.tex}
-\input{./invertible.tex}
+\input{./invertible.tex}
+\input{./igroup.tex}
+\input{./spectrum.tex}
+\input{./fc.tex}

src/op/banach/invertible.tex

@@ -4,15 +4,17 @@
 
 \begin{definition}[Invertible]
 \label{definition:banach-algebra-invertible}
-    Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is \textbf{invertible} if there exists $x^{-1} \in A$ such that $xx^{-1} = x^{-1}x = 1$. The set $G(A)$ denotes the collection of all invertible elements in $A$. 
+    Let $A$ be a unital Banach algebra and $x \in A$, then $x$ is \textbf{invertible} if there exists $x^{-1} \in A$ such that $xx^{-1} = x^{-1}x = 1$. The set $G(A)$ denotes the group of all invertible elements in $A$. 
 \end{definition}
 
-\begin{lemma}
+\begin{lemma}[Neumann Series]
 \label{lemma:neumann-series}
     Let $A$ be a unital banach algebra and $x \in B_A(1, 1)$, then $x \in G(A)$ with
     \[
     x^{-1} = \sum_{n = 0}^\infty (1 - x)^n
     \]
+
+    and $\normn{x^{-1}}_A \le (1 - \norm{x})_A^{-1}$. 
 \end{lemma}
 \begin{proof}
     Since $\norm{1 - x}_A < 1$, the series converges absolutely. Let $y = \sum_{n = 0}^\infty (1 - x)^n$, then
@@ -50,3 +52,27 @@
     (3): Since the inversion map is locally a power series, it is $C^\infty$ by \autoref{theorem:termwise-differentiation}. 
 \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 $1 - yx \in G(A)$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Proposition I.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 \\
+        &-yxy(1 - xy)^{-1}x - yx \\
+        &= y[(1 - xy)^{-1} - xy(1 - xy)^{-1}]x + 1 - yx \\
+        &= yx + 1 - yx = 1
+    \end{align*}
+    
+    Similarly, 
+    \begin{align*}
+        [y(1 - xy)^{-1}x + 1](1 - yx) &=  y(1 - xy)^{-1}x + 1 \\
+        &-y(1 - xy)^{-1}xyx - yx \\
+        &= y[(1 - xy)^{-1} - (1 - xy)^{-1}xy]y + 1 - yx \\
+        &= yx + 1 - yx = 1
+    \end{align*}
+    
+\end{proof}
+
+

src/op/banach/spectrum.tex

@@ -0,0 +1,120 @@
+\section{The Spectrum}
+\label{section:spectrum}
+
+\begin{definition}[Spectrum]
+\label{definition:spectrum}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    \sigma(x) = \sigma_A(x) = \bracs{\lambda \in \complex| \lambda - x \not\in G(A)}
+    \]
+
+    is the \textbf{spectrum} of $x$ in $A$, and its complement is the \textbf{resolvent set} of $x$. 
+\end{definition}
+
+\begin{definition}[Spectral Radius]
+\label{definition:spectral-radius}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    [x]_{sp} = \sup\bracs{|\lambda|: \lambda \in \sigma_A(x)}
+    \]
+
+    is the \textbf{spectral radius} of $x$. 
+\end{definition}
+
+\begin{definition}[Resolvent]
+\label{definition:resolvent}
+    Let $A$ be a unital Banach algebra and $x \in A$, then
+    \[
+    R_x: \complex \setminus \sigma_A(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}
+    \]
+
+    is the \textbf{resolvent} function of $x$, which is holomorphic on $\complex \setminus \sigma_A(x)$.
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:banach-algebra-inverse}, the mapping $y \mapsto y^{-1}$ is smooth on $G(A)$. Hence $R_x: \complex \setminus \sigma_A(x) \to A$ is holomorphic. 
+\end{proof}
+
+\begin{lemma}[Resolvent Equation]
+\label{lemma:resolvent-equation}
+    Let $A$ be a unital Banach algebra, $x \in A$, and $\lambda, \mu \in \complex \setminus \sigma_A(x)$, then
+    \[
+    R_x(\lambda) - R_x(\mu) = (\mu - \lambda)R_x(\lambda) R_x(\mu)
+    \]
+\end{lemma}
+\begin{proof}
+    \begin{align*}
+        [R_x(\lambda) - R_x(\mu)](\mu - x) &= (\lambda - x)^{-1}(\mu - x) - 1 \\
+        (\lambda - x)[R_x(\lambda) - R_x(\mu)](\mu - x)&= (\mu - x) - (\lambda - x) = \mu - \lambda \\
+        R_x(\lambda) - R_x(\mu) &= (\mu - \lambda)R_x(\lambda)R_x(\mu)
+    \end{align*}
+\end{proof}
+
+
+
+\begin{proposition}
+\label{proposition:spectrum-non-empty}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_A(x) \ne \emptyset$. 
+\end{proposition}
+\begin{proof}
+    Assume for contradiction that $\sigma_A(x) = \emptyset$, then by \autoref{definition:resolvent}, the resolvent
+    \[
+    R_x: \complex \setminus \sigma_A(x) \to A \quad \lambda \mapsto \frac{1}{\lambda - x}
+    \]
+
+    is an entire function. Since
+    \[
+    R_x(\lambda) = \frac{1}{\lambda - x} = \frac{\lambda^{-1}}{1 - \lambda^{-1}x}
+    \]
+
+    which tends to $0$ as $|\lambda| \to \infty$, $R_x \in H(\complex; A) \cap C_0(\complex; A)$. By \hyperref[Liouville's Theorem]{theorem:liouville}, $R_x = 0$, which is impossible. 
+\end{proof}
+
+\begin{theorem}[Gelfand-Naimark]
+\label{theorem:gelfand-naimark}
+    Let $A$ be a unital Banach algebra. If every non-zero element of $A$ is invertible, then $A$ is isometrically isomorphic to $\complex$. 
+\end{theorem}
+\begin{proof}
+    Let $x \in A$. By \autoref{proposition:spectrum-non-empty}, there exists $\lambda \in \sigma_A(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:spectral-radius-hadamard}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $[\cdot]_{sp} = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Theorem 1.8]{FollandHarmonic}}}. ]
+    Let $r = \limsup_{n \to \infty}\normn{x^n}_A^{1/n}$, then for any $\lambda \in \complex$ with $|\lambda| > r$, the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ converges absolutely, and to the inverse of $(x - \lambda)$. Therefore $r \ge [\cdot]_{sp}$. 
+
+    Let $D = \bracs{\lambda \in \complex|\ |\lambda| > [x]_{sp}}$, then since the series $\sum_{n = 0}^\infty \lambda^{-n-1}x^n$ is the expansion of $R_x$ at infinity, which is defined on $D$, it must converge on $D$. Let $\phi \in A^*$, then by the preceding discussion, the series $\sum_{n = 0}^\infty \lambda^{-n-1}\dpn{x^n, \phi}{A}$ converges on $D$. Thus for any $\lambda \in D$, 
+    \[
+    \sup_{n \in \natz}|\lambda^{-n-1} \dpn{x^n, \phi}{A}|^{1/n} < \infty
+    \]
+
+    By the \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness}, 
+    \[
+    \sup_{n \in \natz}|\lambda|^{-n-1} \cdot \normn{x^n}_A < \infty
+    \]
+
+    Therefore $\limsup_{n \to \infty}\norm{x^n}_A^{1/n} \le [x]_{sp}$. 
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:spectrum-compact}
+    Let $A$ be a unital Banach algebra and $x \in A$, then $\sigma_A(x)$ is a compact subset of $B_\complex(0, \norm{x}_A)$. 
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:banach-algebra-inverse}, $G(A)$ and hence the resolvent set of $x$ is open, so $\sigma_A(x)$ is closed. By \autoref{proposition:spectral-radius-hadamard}, $[x]_{sp} \le \norm{x}_A$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:spectrum-product-gymnastics}
+    Let $A$ be a unital Banach algebra and $x, y \in A$, then:
+    \begin{enumerate}
+        \item $\sigma(xy) \cup \bracs{0} = \sigma(yx) \cup \bracs{0}$. 
+        \item $[xy]_{sp} = [yx]_{sp}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $\lambda \in \complex \setminus \bracs{0}$, then by \autoref{proposition:swap-invertible}, $\lambda - xy \in G(A)$ if and only if $\lambda - yx \in G(A)$. 
+\end{proof}
+

src/op/notation.tex

@@ -5,4 +5,10 @@
     \textbf{Notation} & \textbf{Description} & \textbf{Source} \\
     \hline
     $1$ & Identity element of a unital algebra. & \autoref{definition:unital-banach-algebra} \\
+    $G(A)$ & Invertible group of a unital algebra. & \autoref{definition:banach-algebra-invertible} \\
+    $G_0(A)$ & The identity component of $G(A)$. & \autoref{definition:identity-component} \\
+    $I(A)$ & The index group of $A$. & \autoref{definition:index-group} \\
+    $\sigma_A(x) = \sigma(x)$ & The spectrum of $x$ in $A$. & \autoref{definition:spectrum} \\
+    $R_x(\lambda)$ & The resolvent of $x$. & \autoref{definition:resolvent} \\ 
+    $[x]_{sp}$ The spectral radius of $x$. & \autoref{definition:spectral-radius}
 \end{tabular}