Added the holomorphic functional calculus.

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/index.tex

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

src/op/banach/spectrum.tex

@@ -34,6 +34,22 @@
     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}