33.6 The Holomorphic Functional Calculus

Proof, [Proposition I.2.7, Tak01]. (Definition), (3): Let $U, V \in \cn_{\complex}(\sigma_{A}(x))$ such that $\ol V$ is a compact subset of $U$, then by Proposition 31.7.1, there exists closed rectifiable curves $\seqf{\gamma_j}$ on $V \setminus \sigma_{A}(x)$ such that

1. (a)

   For all $f \in H(V; \complex)$ and $z_{0} \in \sigma_{A}(x)$,

   \[f(z_{0}) = \frac{1}{2\pi i}\sum_{j = 1}^{n} \int_{\gamma_j}\frac{f(z)}{z - z_{0}}dz\]

   
   
2. (b)

   For all $f \in H(U; \complex)$ and $z_{0} \in U \setminus V$,

   \[\frac{1}{2\pi i}\sum_{j = 1}^{n} \int_{\gamma_j}\frac{f(z)}{z - z_{0}}dz = 0\]

   
   

For each $f \in H(U; \complex)$, define

then by Cauchy’s Theorem, the above definition is independent of the choice of curves satisfying (a).

(Linearity): By Proposition 14.4.1, 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 Proposition 31.7.1, 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,

For each $z, w \in U \setminus \sigma_{A}(x)$ with $z \ne w$, by the resolvent equation,

By assumptions on $\seqf[m]{\mu_j}$, for each $1 \le j \le n$,

By assumption (b) and Fubini’s Theorem, for each $1 \le k \le m$,

Therefore

and the mapping $f \mapsto f(x)$ is a homomorphism.

(1), (2): Let $R > 0$ such that $\sigma_{A}(x) \subset B_{\complex}(0, R)$, then by Proposition 14.4.2 and Cauchy’s Integral Formula,

and

(Uniqueness): By (2), the homomorphism extends uniquely to $\complex(z) \cap H(\sigma_{A}(x); \complex)$. By Runge’s Theorem, it extends uniquely to $H(\sigma_{A}(x); \complex)$ by continuity.$\square$

Proof. (1): Let $\lambda \in \complex \setminus f(\sigma_{A}(x))$, then $1/(\lambda - f) \in H(\sigma_{A}(x); \complex)$. Since the holomorphic functional calculus is a homomorphism, $[1/(\lambda - f)](x) = (\lambda - f(x))^{-1}$. Thus $\sigma_{A}(f(x)) \subset f(\sigma_{A}(x))$.

On the other hand, let $\lambda \in \sigma_{A}(x)$, then the function $h(\mu) = [f(\mu) - f(\lambda)]/(\mu - \lambda)$ extends to an element of $H(\sigma_{A}(x); \complex)$ by Proposition 31.2.3. By the homomorphism property,

Since $(x - \lambda) \not\in G(A)$, $f(x) - f(\lambda) \not\in G(A)$ as well. Therefore $\sigma_{A}(f(x)) \supset f(\sigma_{A}(x))$.

(2): If additionally $f, g \in \complex(z)$, then the theorem holds by (1) and (2) of Definition 33.6.1.

Now suppose that $f \in H(\sigma_{A}(x); \complex)$ is arbitrary and $g \in \complex(z)$. For each $\eps > 0$, by Proposition 33.5.10 and continuity of the holomorphic functional calculus, there exists $U \in \cn_{H(\sigma_A(x); \complex)}(f)$ such that for each $h \in U$,

1. (a)

   $g \in H(\sigma_{A}(h(x)); \complex)$.

2. (b)

   $\norm{g(f(x)) - g(h(x))}_{A} < \eps$.

By Runge’s Theorem, there exists $h \in U \cap \complex(z)$. Thus $\norm{g(f(x)) - (g \circ f)(x)}_{A} < \eps$ as well. As this holds for all $\eps > 0$, $g(f(x)) = (g \circ f)(x)$ for any $f \in H(\sigma_{A}(x); \complex)$ and $g \in \complex(x) \cap H(f(\sigma_{A}(x)); \complex)$.

Through a second application of Runge’s Theorem and continuity of the holomorphic functional calculus, the above also holds for all $f \in H(\sigma_{A}(x); \complex)$ and $g \in H(f(\sigma_{A}(x)); \complex)$.$\square$

Proof. (1): By the power series representation of $\exp(x)$ and continuity of the holomorphic functional calculus.

(2): By the homomorphism property of the functional calculus, $\exp(x)^{-1}= \exp(-x)$. Thus $\exp(x)$ is connected to $1$ by the path $t \mapsto \exp(tx)$.

(3): Since the exponential is an entire function, the following series converges absolutely, and is eligible for arbitrary manipulations:

(4): By the Spectral Mapping Theorem.$\square$

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, $\sigma_{A}(\text{Id}\cdot f_{K}(x)) = \sigma_{B(E_K)}(x_{K})$.$\square$