Added the Riesz decomposition.

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+@unpublished{MarcouxNotes,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+  note         = {Unpublished lecture notes, Department of Pure Mathematics,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,

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}
+
 

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}
-
-
-