Added the GKZ Theorem.

src/dg/complex/derivative.tex

@@ -280,4 +280,10 @@
     (5) $\Rightarrow$ (3): By the equivalence of the prior points, for any $\phi \in E^*$, $\phi \circ f$ satisfies (3). By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $f$ also satisfies (3). 
 \end{proof}
 
+\begin{remark}
+\label{remark:weak-holomorphy}
+    Since weak holomorphy is equivalent to strong holomorphy, most common results of scalar-valued holomorphic functions may be transferred to the vector valued case with little effort. Since this chapter represents my first attempt at learning complex analysis, and I happen to require vector-valued results, most results here are stated and proven in the vector-valued case. However, this is strictly unnecessary if one already knows the scalar-valued results. 
+\end{remark}
+
+
 

src/dg/complex/entire.tex

@@ -40,5 +40,53 @@
     Let $p \in \complex[z]$ such that $p(z) \ne 0$ for all $z \in \complex$. Let $f = 1/p$, then $f \in H(\complex; \complex) \cap C_0(\complex; \complex)$, so $f$ is bounded. By \hyperref[Liouville's Theorem]{theorem:liouville}, $f$ and thus $p$ is constant. 
 \end{proof}
 
+\begin{proposition}
+\label{proposition:entire-logarithm}
+    Let $f \in H(\complex; \complex)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f(z) \ne 0$ for all $z \in \complex$. 
+        \item There exists $\ell \in H(\complex; \complex)$ such that $f = e^\ell$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): For each $z \in \complex$, let
+    \[
+    g(z) = \int_0^w \frac{f'(w)}{f(w)}dw
+    \]
+
+    then $g \in H(\complex; \complex)$ with $g' = f/f'$, and
+    \begin{align*}
+        \frac{d}{dz}\braks{f(z)e^{-g(z)}} &= f'(z)e^{-g(z)} - f(z)g'(z)e^{-g(z)} \\
+        &= f'(z)e^{-g(z)} - f'(z)e^{-g(z)} = 0
+    \end{align*}
+
+    Therefore $fe^{-g}$ is a non-zero constant. For any $\lambda \in \complex$ with $e^\lambda = fe^{-g}$, let $h = g + \lambda$, then $f = e^h$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:entire-constant}
+    Let $f \in H(\complex; \complex)$ such that:
+    \begin{enumerate}[label=(\alph*)]
+        \item $f(0) = 1$. 
+        \item $f'(0) = 0$. 
+        \item $0 < |f| < |\exp|$. 
+    \end{enumerate}
+
+    then $f = 1$. 
+\end{proposition}
+\begin{proof}[Proof, {{\cite[Lemma I.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
+    \[
+    g_r: \ol{B_\complex(0, r)} \setminus \bracs{0} \to \complex \quad z \mapsto \frac{r^2}{z^2} \frac{g(z)}{2r - g(z)}
+    \]
+
+    extends to a holomorphic function on $B_\complex(0, r)$ by \autoref{proposition:zero-finite-multiplicity}. Since $|g_r(z)| \le 1$ for all $z \in \partial B_\complex(0, r)$, $g_r(z) \le 1$ for all $z \in B_\complex(0, r)$ by the \hyperref[Maximum Modulus Theorem]{theorem:maximum-modulus-theorem}. 
+
+    Finally, since $|g_r(z)| \le 1$ for all $z \in \complex$ and $r > |z|$, $g = 0$ and $f = 1$. 
+\end{proof}
+
+
 
 

src/dg/complex/index.tex

@@ -3,6 +3,7 @@
 
 
 \input{./derivative.tex}
+\input{./zero.tex}
 \input{./sphere.tex}
 \input{./space.tex}
 \input{./log.tex}

src/dg/complex/zero.tex

@@ -0,0 +1,96 @@
+\section{Zeroes of Holomorphic Functions}
+\label{section:complex-zeroes}
+
+\begin{definition}[Zero]
+\label{definition:complex-zero}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $f \in H(U; E)$, and $z_0 \in U$, then $z_0$ is a \textbf{zero of $f$ of multiplicity $n \in \natp$} if there exists $g \in H(U; E)$ such that $f(z) = (z - a)^n g(z)$ for all $z \in \bracs{g \ne 0}$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:complex-zero-gymnastics}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open and connected, and $f \in H(U; E)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $f = 0$. 
+        \item There exists $z_0 \in U$ such that $D^nf(z_0) = 0$ for all $n \in \natp$. 
+        \item There exists $z_0 \in U$ such that $z_0 \in \ol{\bracs{f = 0} \setminus \bracs{z_0}}$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (3) $\Rightarrow$ (2): Let $r > 0$ and $\bracs{a_n}_0^\infty \subset E$ such that $f(z) = \sum_{n = 0}^\infty a_n (z - z_0)^n$ for all $z \in B(z_0, r)$, where the series has a radius of convergence of at least $r$. By continuity of $f$, $a_0 = f(0) = 0$. 
+    
+    Suppose inductively that $a_k = 0$ for each $0 \le k \le n$. Let $[\cdot]_E: E \to [0, \infty)$ be a continuous seminorm, then for each $z \in B(z_0, r) \cap \bracs{f = 0} \setminus \bracs{z_0}$, 
+    \begin{align*}
+        f(z) &= \sum_{k = n + 1}^\infty a_k(z - z_0)^k \\
+        [a_{n+1}(z - z_0)^{n+1}]_E &\le \sum_{k = n+2}^\infty [a_k(z - z_0)^k]_E = \sum_{k = n+2}^\infty |z - z_0|^k[a_k]_E  \\
+        [a_{n+1}]_E &\le |z - z_0| \sum_{k = 0}^\infty |z - z_0|^{k}[a_{k+n+2}]_E
+    \end{align*}
+
+    Since the series $\sum_{n = 0}^\infty a_n (z - z_0)^n$ has a radius of convergence of at least $r$, $\limsup_{n \to \infty}[a_n]_E^{1/n} \le 1/r$, so there exists $N \ge n+2$ such that $[a_k]_E^{1/k} \le 2/r$ for all $k \ge N$. For each $s \in (0, r/4)$, $B(z_0, s) \cap \bracs{f = 0} \setminus \bracs{z_0} \ne \emptyset$, thus
+    \[
+        [a_{n+1}]_E \le s \braks{\sum_{k = 0}^N \paren{\frac{r}{4}}^k[a_{k+n+2}]_E + \sum_{k > N}\frac{1}{2^k}}
+    \]
+
+    As the above holds for all $s \in (0, r/4)$, $[a_{n+1}]_E = 0$. Therefore $a_n = 0$ for all $n \in \natp$. 
+
+    (2) $\Rightarrow$ (1): By continuity, $\bracs{f = 0}$ is closed. By local power series expansion and (3), $\bracs{f = 0}$ is open. As $U$ is connected, $U = \bracs{f = 0}$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:zero-finite-multiplicity}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be connected, and $f \in H(U; E) \setminus \bracs{0}$, then for each $z_0 \in \bracs{f = 0}$, there exists $n \ge 1$ and $g \in H(U; \complex)$ such that:
+    \begin{enumerate}
+        \item $g(a) \ne 0$.
+        \item For each $z \in U$, $f(z) = (z - z_0)^ng(z)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    Since $f \ne 0$, $n = \min\bracs{n \in \natp| D^nf(z_0) \ne 0} < \infty$ by \autoref{proposition:complex-zero-gymnastics}. Let
+    \[
+    g: U \to E \quad z \mapsto \begin{cases}
+        \frac{f(z)}{(z - z_0)^n} &z \ne z_0 \\
+        \frac{1}{n!}D^nf(z_0) &z = z_0
+    \end{cases}
+    \]
+
+    then $g$ is continuous on $U$ and holomorphic on $U \setminus \bracs{z_0}$. As $f \in H(U; E)$, there exists $r > 0$ and $\bracs{a_k}_0^\infty \subset E$ such that $f(z) = \sum_{k = 0}^\infty a_k (z - z_0)^k$ for all $z \in B(z_0, r)$, where the series has a radius of convergence of at least $r$. By assumption on $n$, for each $z \in B(z_0, r)$, 
+    \begin{align*}
+        f(z) &= \sum_{k = 0}^\infty a_k (z - z_0)^k = \sum_{k = n}^\infty a_k(z - z_0)^k \\
+        g(z) &= \sum_{k = 0}^\infty a_{k+n}(z - z_0)^{k}
+    \end{align*}
+
+    where the series $\sum_{k = 0}^\infty a_{k+n}(z - z_0)^{k}$ has the same radius of convergence. Therefore $g$ is holomorphic at $z_0$. 
+\end{proof}
+
+\begin{theorem}[Maximum Modulus Theorem]
+\label{theorem:maximum-modulus-theorem}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open and connected, then
+    \begin{enumerate}
+        \item For each $f \in H(U; E)$ and continuous seminorm $[\cdot]_E: E \to [0, \infty)$, if there exists $z_0 \in U$ with
+        \[
+        [f(z_0)]_E = \sup_{z \in U}[f(z)]_E
+        \]
+
+        then $[f]_E$ is constant. 
+        \item For any $f \in H(U; \complex)$, if there exists $z_0 \in U$ with
+        \[
+        |f(z_0)| = \sup_{z \in U}|f(z)|
+        \]
+
+        then $f$ is constant. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    (1): Let $M = \sup_{z \in U}[f(z)]_E$. Since $[f]_E$ is continuous, $\bracsn{[f]_E = M}$ is closed. By \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, $\bracsn{[f]_E = M}$ is open. As $U$ is connected, $\bracsn{[f]_E = M} = U$. 
+
+    (2): By (1), $|f|$ is constant. After rescaling, assume without loss of generality that $\ol f = 1/f$. In this case, for each $z \in U$, 
+    \[
+    D\ol f(z) = \lim_{\substack{h \to 0 \\ h \in \real}}\frac{\ol{f(x + h) - f(x)}}{h} = \ol{Df}(z)
+    \]
+
+    and
+    \[
+    D\ol f(z) = \lim_{\substack{h \to 0 \\ h \in \real}}\frac{\ol{f(x + ih) - f(x)}}{ih} = -\ol{Df}(z)
+    \]
+
+    so $Df = 0$ and $f$ is constant. 
+\end{proof}

src/op/banach/definitions.tex

@@ -12,8 +12,11 @@
 
 \begin{definition}[Unital Banach Algebra]
 \label{definition:unital-banach-algebra}
-    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 $A$. 
+    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.  
 \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 $\norm{\cdot}_A$.
+\end{proof}
 
 \begin{definition}[Homomorphism]
 \label{definition:banach-algebra-homomorphism}

src/op/banach/fc.tex

@@ -103,7 +103,7 @@
 \begin{proof}
     (1): Let $\lambda \in \complex \setminus f(\sigma_A(x))$, then $1/(\lambda - f) \in H(\sigma_A(x); \complex)$. Since the holomorphic functional calculus is a homomorphism, $[1/(\lambda - f)](x) = (\lambda - f(x))^{-1}$. Thus $\sigma_A(f(x)) \subset f(\sigma_A(x))$. 
 
-    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$. Therefore by the homomorphism property, 
+    On the other hand, let $\lambda \in \sigma_A(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_A(x); \complex)$ by \autoref{proposition:zero-finite-multiplicity}. Therefore by the homomorphism property, 
     \[
     f(x) - f(\lambda) = (x - \lambda)h(x) = h(x)(x - \lambda)
     \]
@@ -120,7 +120,7 @@
     
     By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation}, there exists $h \in U \cap \complex(z)$. Thus $\norm{g(f(x)) - (g \circ f)(x)}_A < \eps$ as well. As this holds for all $\eps > 0$, $g(f(x)) = (g \circ f)(x)$ for any $f \in H(\sigma_A(x); \complex)$ and $g \in \complex(x) \cap H(f(\sigma_A(x)); \complex)$. 
     
-    By \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} again and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
+    Through a second application of \hyperref[Runge's Theorem]{corollary:runge-rational-approximation} again and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_A(x); \complex)$ and $g \in H(f(\sigma_A(x)); \complex)$. 
 \end{proof}
 
 \begin{proposition}

src/op/banach/ideal.tex

@@ -0,0 +1,59 @@
+\section{Ideals}
+\label{section:banach-algebra-ideals}
+
+\begin{definition}[Ideal]
+\label{definition:banach-algebra-ideal}
+    Let $A$ be a Banach algebra and $I \subset A$, then $I$ is a \textbf{left ideal} if:
+    \begin{enumerate}
+        \item $I$ is a subspace of $A$. 
+        \item For each $x \in A$, $aI \subset I$. 
+    \end{enumerate}
+
+    and $I$ is a \textbf{two-sided ideal} if in addition to the above, 
+    \begin{enumerate}[start=2]
+        \item For each $x \in A$, $Ia \subset I$. 
+    \end{enumerate}
+
+    \textit{In this document, the "sidedness" of ideals are omitted. Within each block, the statement should hold as long as the interpretation is consistent. }
+\end{definition}
+
+\begin{definition}[Proper Ideal]
+\label{definition:banach-algebra-proper-ideal}
+    Let $A$ be a Banach algebra and $I \subset A$ be an ideal, then $I$ is \textbf{proper} if $I \subsetneq A$.  
+\end{definition}
+
+\begin{definition}[Maximal Ideal]
+\label{definition:maximal-ideal}
+    Let $A$ be a Banach algebra and $I \subset A$ be an ideal, then $I$ is \textbf{maximal} if:
+    \begin{enumerate}
+        \item $I$ is proper.
+        \item For any proper ideal $J \subsetneq A$ with $J \supset I$, $I = J$. 
+    \end{enumerate}
+
+    The set $\cm(A)$ of maximal two-sided ideals of $A$ is the \textbf{maximal ideal space} of $A$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:ideal-gymnastics}
+    Let $A$ be a unital Banach algebra and $I \subset A$ be a proper ideal, then:
+    \begin{enumerate}
+        \item $I$ contains no invertible elements. 
+        \item $\ol I$ is also a proper ideal.
+        \item $I$ is contained in a maximal ideal.
+        \item If $I$ is maximal, then $I$ is closed. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}[Proof, {{\cite{FollandHarmonic}}}. ]
+    (1): If $I$ contains an invertible element, then $1 \in I$ and thus $A = I$. 
+
+    (2): By \autoref{proposition:banach-algebra-inverse}, $G(A)$ is open. Since $G(A) \cap I = \emptyset$, $G(A) \cap \ol I = \emptyset$ as well. 
+
+    (3): By Zorn's lemma. 
+
+    (4): By (2). 
+\end{proof}
+
+
+
+
+

src/op/banach/index.tex

@@ -4,5 +4,7 @@
 \input{./definitions.tex}
 \input{./invertible.tex}
 \input{./igroup.tex}
+\input{./ideal.tex}
 \input{./spectrum.tex}
 \input{./fc.tex}
+\input{./multiplicative.tex}

src/op/banach/multiplicative.tex

@@ -0,0 +1,108 @@
+\section{Multiplicative Functionals}
+\label{section:multiplicative-functional}
+
+\begin{definition}[Multiplicative Functional]
+\label{definition:multiplicative-functional}
+    Let $A$ be a unital Banach algebra and $\phi \in A^*$, then $\phi$ is \textbf{multiplicative} if $\phi \ne 0$ and for each $x, y \in A$, $\phi(xy) = \phi(x)\phi(y)$. 
+\end{definition}
+
+\begin{proposition}
+\label{proposition:multiplicative-unit}
+    Let $A$ be a unital Banach algebra and $\phi \in A^*$ be a multiplicative functional, then
+    \begin{enumerate}
+        \item For each $x \in A$, $|\phi(x)| \le [x]_{sp} \le \norm{x}_A$. 
+        \item $\norm{\phi}_{A^*} = 1$. 
+        \item $\phi(G(A)) \subset \complex \setminus \bracs{0}$. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (3): For each $x \in G(A)$, $1 = \phi(xx^{-1}) = \phi(x)\phi(x^{-1}) \ne 0$. 
+    
+    (1): By (3), for every $\lambda \in \complex$ with $|\lambda| > [x]_{sp}$, $\lambda x \in G(A)$ and $\lambda - \phi(x) \ne 0$. Therefore $\phi(x) \in \ol{B(0, [x]_{sp})}$. 
+
+    (2): For each $\lambda \in \complex$, $\phi(\lambda 1) = \lambda$, so $\norm{\phi}_{A^*} \le 1$. 
+\end{proof}
+
+\begin{definition}[Space of Multiplicative Linear Functionals]
+\label{definition:multiplicative-linear-functional-space}
+    Let $A$ be a unital Banach algebra, then $\Omega(A)$ is the \textbf{space of multiplicative linear functionals}, which is a compact Hausdorff space under the weak-* topology. 
+\end{definition}
+\begin{proof}
+    By the \hyperref[Banach-Alaoglu Theorem]{theorem:alaoglu}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:commutative-maximal-ideal-space}
+    Let $A$ be a commutative unital Banach algebra, then the mapping
+    \[
+    \Omega(A) \to \cm(A) \quad \phi \mapsto \ker(\phi)
+    \]
+
+    is a bijection. 
+\end{proposition}
+\begin{proof}
+    For each $\phi \in \cm(A)$, $\ker(\phi)$ is an ideal of codimension $1$, and must be maximal. 
+
+    On the other hand, for each $I \in \Omega(A)$, the quotient $A/I$ is a field, so by the \hyperref[Gelfand-Mazur Theorem]{theorem:gelfand-mazur}, it is isomorphic to $\complex$. Thus the canonical projection $A \to A/I \iso \complex$ induces a multiplicative functional on $A$. 
+\end{proof}
+
+\begin{theorem}[Gleason-Kahane-Żelazko]
+\label{theorem:gkz}
+    Let $A$ be a unital Banach algebra and $\phi \in A^*$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\phi$ is a multiplicative linear functional.
+        \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}}}. ]
+    (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 $\norm{\phi}_{A^*} = 1$. 
+
+    Let $x \in \ker \phi$ with $\norm{x}_A \le 1$, and let
+    \[
+    f: \complex \to \complex \quad \lambda \mapsto \phi[\exp(\lambda x)] = \sum_{n = 0}^\infty \frac{\phi[(\lambda x)^n]}{n!}
+    \]
+
+    then
+    \begin{enumerate}[label=(\alph*)]
+        \item $f(0) = \phi(1) = 1$. 
+        \item $Df(0) = \phi(x) = 0$. 
+        \item Since $\norm{x}_A \le 1$, $|f| \le |\exp|$. As $\exp(\lambda x) \in G(A)$ for all $\lambda \in \complex$, $f(\lambda) \ne 0$ for all $\lambda \in \complex$. 
+    \end{enumerate}
+
+    By \autoref{proposition:entire-constant}, $f = 1$, and $\phi(x^2) = 0$, so for any $x \in \ker\phi$, $x^2 \in \ker\phi$ as well. 
+
+    Now, for each $x, y \in A$, there exists $x_0, y_0 \in \ker \phi$ such that $x = x_0 + \phi(x)$ and $y = y_0 + \phi(x)$. In which case, 
+    \begin{align*}
+        \phi(xy) &= \phi(x_0y_0 + \phi(x)y_0 + \phi(y)x_0 + \phi(x)\phi(y)) \\
+        &= \phi(x_0y_0) + \phi(x)\phi(y)
+    \end{align*}
+
+    In particular, 
+    \[
+    \phi(x^2) = \phi(x_0^2) + \phi(x)^2 = \phi(x)^2
+    \]
+
+    To conclude, let $x \in \ker\phi$, $y \in A$, then
+    \begin{align*}
+        \phi((x + y)^2) &= (\phi(x + y))^2 = \phi(x)^2 + 2\phi(x)(y) + \phi(y)^2 \\
+        \phi(xy + yx) &= 2\phi(x)\phi(y)
+    \end{align*}
+
+    so $xy + yx \in \ker\phi$ as well. Finally, 
+    \begin{align*}
+        (xy + yx)^2 + (xy - yx)^2 &= 2(xyxy + yxyx) \\
+        &= 2[x(yxy) + y(xyx)] \in \ker \phi
+    \end{align*}
+
+    and $(xy - yx)^2 \in \ker\phi$ as well, and
+    \[
+    (\phi(xy - yx))^2 = \phi((xy - yx)^2) = 0
+    \]
+
+    Therefore $2xy = (xy + yx) - (xy - yx) \in \ker \phi$, $\ker \phi$ is an ideal, and $\phi$ is a homomorphism. 
+\end{proof}
+
+

src/op/notation.tex

@@ -10,5 +10,7 @@
     $I(A)$ & 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}
+    $[x]_{sp}$ & The spectral radius of $x$. & \autoref{definition:spectral-radius} \\
+    $\Omega(A)$ & Space of multiplicative functionals on $A$. & \autoref{definition:multiplicative-functional} \\
+    $\cm(A)$ & Maximal ideal space of $A$. & \autoref{definition:maximal-ideal}
 \end{tabular}