Added the continuous functional calculus.

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Bokuan Li
2026-07-03 15:21:33 -04:00
parent 683b822e7e
commit 35e9550ff2
7 changed files with 144 additions and 16 deletions

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(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 $1 - yx \in G(A)$.
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\end{align*}
\end{proof}

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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}

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\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.
\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.
(Uniqueness): By the \hyperref[Stone-Weierstrass Theorem]{theorem:complex-stone-weierstrass}.
\end{proof}

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\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}

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(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}

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\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}

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$\Omega(A)$ & 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} \\