From 99d772d1c8422f6fcbed255fcdbccca8638aa7a5 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Fri, 5 Jun 2026 21:57:02 -0400 Subject: [PATCH] Added the Riesz decomposition. --- refs.bib | 12 +++++++ src/op/banach/fc.tex | 67 +++++++++++++++++++++++++++++++++++++++ src/op/example/matrix.tex | 16 ---------- 3 files changed, 79 insertions(+), 16 deletions(-) diff --git a/refs.bib b/refs.bib index 1aea105..7a5e46a 100644 --- a/refs.bib +++ b/refs.bib @@ -152,6 +152,18 @@ publisher={Springer Berlin Heidelberg} } +@unpublished{MarcouxNotes, + author = {Laurent W. Marcoux}, + title = {An Introduction to Banach Algebras and Operator Algebras}, + year = {2021}, + month = apr, + url = {https://www.math.uwaterloo.ca/~lwmarcou/notes/pmath810.pdf}, + note = {Unpublished lecture notes, Department of Pure Mathematics, + University of Waterloo. Draft dated April 30, 2021. + Potential victim of Google search scraping. }, + institution = {University of Waterloo}, +} + @techreport{aronszajn1964interpolation, diff --git a/src/op/banach/fc.tex b/src/op/banach/fc.tex index 88eb925..a141452 100644 --- a/src/op/banach/fc.tex +++ b/src/op/banach/fc.tex @@ -154,4 +154,71 @@ (4): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}. \end{proof} +\begin{theorem}[Riesz Decomposition] +\label{theorem:riesz-decomposition} + Let $A$ be a unital Banach algebra, $x \in A$, and $\text{Clop}(\sigma_A(x))$ be the family of all relatively open and closed subsets of $\sigma_A(x)$, then there exists a mapping $P: \text{Clop}(\sigma_A(x)) \to A$ such that: + \begin{enumerate} + \item For each $K \in \text{Clop}(\sigma_A(x))$, $P(K)$ is idempotent and commutes with $x$. + \item $P(\emptyset) = 0$ and $P(\sigma_A(x)) = 1$. + \item For each $B, C \in \text{Clop}(\sigma_A(x))$, $P(B \cap C) = P(B)P(C)$. + \item For pairwise disjoint sequence $\seqf{B_j} \in \text{Clop}(\sigma_A(x))$, $P(\bigsqcup_{j = 1}^n B_j) = \sum_{n = 1}^n P(B_j)$. + \end{enumerate} + + Let $E$ be a Banach space such that $A \subset B(E)$. For each $K \in \text{Clop}(\sigma_A(x))$, let $E_K = P(E)(K)$, then + \begin{enumerate}[start=4] + \item For any $B, C \in \text{Clop}(\sigma_A(x))$, $E_{B \cap C} = E_B \cap E_C$. + \item For any $B, C \in \text{Clop}(\sigma_A(x))$, $E_{B \cup C} = E_B + E_C$. + \item $E_K$ is a closed subspace of $E$. + \item $E = E_K \oplus E_{K^c}$ as a sum of complementary closed subspaces. + \item $E_K$ is $x$-invariant. + \end{enumerate} + + Finally, let $x_k = x|_{E_K}$, then + \begin{enumerate}[start=9] + \item $\sigma_{B(E_K)}(x_K) = K$. + \item For any $f \in H(\sigma_A(x); \complex)$, $f(a_K) = f(a)|_{E_K}$. + \end{enumerate} +\end{theorem} +\begin{proof}[Proof, {{\cite[Corollary 3.15]{MarcouxNotes}}}. ] + (1)-(4): Since $K$ is relatively open and closed, $\sigma_A(x) \setminus K$ is also closed. Therefore there exists $U \in \cn_{\complex}(K)$ and $V \in \cn_{\complex}(\sigma_A(x) \setminus K)$ with $U \cap V = \emptyset$. Let + \[ + f_K: U \sqcup V \to \complex \quad \lambda \mapsto \begin{cases} + 1 &\lambda \in U \\ + 0 &\lambda \in V + \end{cases} + \] + + then $f_K \in H(U \sqcup V; \complex)$. Let $P(K) = f_K(x)$, then by properties of the holomorphic functional calculus, $P$ satisfies (1)-(4). + + (6): By (4), + \[ + P(B \cup C) = P(B \setminus C) + P(B \cap C) + P(C \setminus B) + \] + + and by (3), + \begin{align*} + E_{B \cup C} &\supset [P(B \setminus C) + P(B \cap C) + P(C \setminus B)](E_{B}) \\ + &= [P(B \setminus C) + P(B \cap C) + P(C \setminus B)]P(B)(E_{B}) \\ + &= [P(B \setminus C) + P(B \cap C)]P(B)(E_{B}) = E_B + \end{align*} + + so $E_{B \cup C} \supset E_B + E_C$. On the other hand, + \begin{align*} + E_{B \cup C} &= [P(B \setminus C) + P(B \cap C) + P(C \setminus B)](E_{B \cup C}) \\ + &\subset P(B \setminus C)(E_{B \cup C}) + P(B \cap C)(E_{B \cup C}) + P(C \setminus B)(E_{B \cup C}) \\ + &\subset E_B + E_C - E_{B \cap C} \subset E_B + E_C + \end{align*} + + (7), (8): By (2) and (3), $P(K^c)(E_K) = P(K)P(K^c)(E_K) = \bracs{0}$, so $E_K \subset \ker P(K^c)$. By (5) and (6), $E = E_K \oplus E_{K^c}$ as an algebraic direct sum, so $E_K = \ker P(K^c)$, and is closed. + + (9): By (1), $x(E_K) = xP(K)(E_K) = P(K)(xE_K) \subset E_K$. + + (10): By the \hyperref[Spectral Mapping Theorem]{theorem:spectral-mapping-holomorphic}, + \[ + \sigma_A(\text{Id} \cdot f_K(x)) = \text{Id} \cdot f_K(x)(\sigma_A(x)) = K + \] + + Since $E_K$ is $x$-invariant, $\sigma_A(\text{Id} \cdot f_K(x)) = \sigma_{B(E_K)}(x_K)$. +\end{proof} + diff --git a/src/op/example/matrix.tex b/src/op/example/matrix.tex index a485969..ead52ea 100644 --- a/src/op/example/matrix.tex +++ b/src/op/example/matrix.tex @@ -26,19 +26,3 @@ \begin{proof} (1), (2): By \autoref{proposition:matrix-algebra-spectrum}, $\sigma(x)$ is finite. By \autoref{proposition:log-identity-component}, there exists $y \in M_n(\complex)$ such that $x = \exp(y)$. In which case, $x \in G_0(M_n(\complex))$. \end{proof} - -\begin{proposition} -\label{proposition:matrix-algebra-ideals} - Let $n \in \natp$, then - \begin{enumerate} - \item $M_n(\complex)$ admits no nontrivial two-sided ideals. - \item $M_n(\complex)$ admits no multiplicative functionals. - \end{enumerate} - -\end{proposition} -\begin{proof} - Let $x = (x_{ij}) \in M_n(\complex) \setminus \bracs{0}$, then there exists $1 \le i, j \le n$ such that $x_{ij} \ne 0$. In which case, for any $1 \le k, l \le n$ and $\lambda \in \complex$, there exists $y_k, z_l \in M_n(\complex)$ such that $y_kxz_l$ is the matrix with $\lambda$ on its $(k, l)$ entry and $0$ everywhere else. Therefore every non-trivial two-sided ideal of $M_n(\complex)$ is $M_n(\complex)$. -\end{proof} - - -