Theorem 29.6.4 (Riesz Decomposition).label 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:
- (1)
For each $K \in \text{Clop}(\sigma_{A}(x))$, $P(K)$ is idempotent and commutes with $x$.
- (2)
$P(\emptyset) = 0$ and $P(\sigma_{A}(x)) = 1$.
- (3)
For each $B, C \in \text{Clop}(\sigma_{A}(x))$, $P(B \cap C) = P(B)P(C)$.
- (4)
For pairwise disjoint sequence $\seqf{B_j}\in \text{Clop}(\sigma_{A}(x))$, $P(\bigsqcup_{j = 1}^{n} B_{j}) = \sum_{n = 1}^{n} P(B_{j})$.
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
- (5)
For any $B, C \in \text{Clop}(\sigma_{A}(x))$, $E_{B \cap C}= E_{B} \cap E_{C}$.
- (6)
For any $B, C \in \text{Clop}(\sigma_{A}(x))$, $E_{B \cup C}= E_{B} + E_{C}$.
- (7)
$E_{K}$ is a closed subspace of $E$.
- (8)
$E = E_{K} \oplus E_{K^c}$ as a sum of complementary closed subspaces.
- (9)
$E_{K}$ is $x$-invariant.
Finally, let $x_{k} = x|_{E_K}$, then
- (10)
$\sigma_{B(E_K)}(x_{K}) = K$.
- (11)
For any $f \in H(\sigma_{A}(x); \complex)$, $f(a_{K}) = f(a)|_{E_K}$.
Proof, [Corollary 3.15, Mar21]. (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
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),
and by (3),
so $E_{B \cup C}\supset E_{B} + E_{C}$. On the other hand,
(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 Spectral Mapping Theorem,
Since $E_{K}$ is $x$-invariant, $\sigma_{A}(\text{Id}\cdot f_{K}(x)) = \sigma_{B(E_K)}(x_{K})$.$\square$
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