### Advanced Analysis, Notes 19: The holomorphic functional calculus II (definition and basic properties)

In this post we continue our discussion of the holomorphic functional calculus for elements of a Banach algebra (or operators). The beginning of this discussion can be found in Notes 18.

#### 1. Definition and main theorem

Recall that $\mathcal \mathcal{O}(a)$ denotes the algebra of functions defined and holomorphic in some neighborhood of $\sigma(a)$ (the spectrum of $a$). When working with $\mathcal{O}(a)$ some care is needed because it is not a usual algebra of functions, in that its elements are not functions with the same domain. Especially when $\sigma(a)$ is a point, the restriction $f \big|_{\sigma(a)}$ of a function $f \in \mathcal{O}(a)$ to $\sigma(a)$ loses almost all information on $f$. Let us show how multiplication is defined in this algebra. If $f,g \in \mathcal{O}(a)$, then there are neighborhoods $U,V$ of $\sigma(a)$ such that $f$ is analytic in $U$ and $g$ is analytic in $V$. Then $U \cap V$ is a neighborhood of $\sigma(a)$, and $f g$ is defined and holomorphic in the neighborhood of $U \cap V$ of $\sigma(a)$, so it is in $\mathcal{O}(a)$. All algebraic matters are handled in a similar matter. Analytic matters (such as defining a topology on $\mathcal{O}(a)$) are more involved and we will treat them only at the most basic level. We will require the following elementary result.

Exercise A: Let $K$ be a compact subset of the plane, and let $U$ be an open set containing $K$. Then there exists a bounded open set $V$ such that $K \subset V \subset \overline{V} \subset U$ and such that $\partial V$ is a finite union of $C^1$ curves. (In fact, one can also find such a $V$ such that its boundary is smooth, but we shall not require that). The following definition and theorem jointly comprise the holomorphic functional calculus. We shall spend the rest of this post proving it.

Definition (holomorphic functional calculus): Let $A$ be a unital Banach algebra and let $a \in A$. For every $f \in \mathcal{O}(a)$, we define an element $f(a) \in A$ in the following way: Let $U \supset \sigma(a)$ be an open set where $f$ is defined and holomorphic, and let $V$ be an open bounded set such that $\sigma(a) \subset V \subset \overline{V} \subset U$ and such that $\partial V$ is a finite union of $C^1$ curves; define

$f(a) = \int_{\partial V} f(z) (z-a)^{-1} dz .$

Theorem (holomorphic functional calculus): The mapping $f \mapsto f(a)$ has the following properties:

1. (Well definedness) $f(a)$ is well defined, and in particular does not depend on the choice of $U$ and $V$
2. (Homomorphism) The mapping $f \mapsto f(a)$ is a homomorphism.
3. (Naturality) If $f$ is a polynomial, a rational function with poles off $\sigma(a)$, or a function with a power series representation that converges in a neighborhood of $\sigma(a)$, then the definition of $f(a)$ coincides with the natural definitions given in the previous lecture (Sections 2,3 and 4 in the previous post). If $\zeta$ denotes the entire function $\zeta(z) = z$, one has $\zeta(a) = a$
4. (Spectral mapping theorem) For all $f \in \mathcal{O}(a)$, it holds that $\sigma(f(a)) = f(\sigma(a))$
5. (Conitnuity) If $f_n,f$ are holomorphic in $U \supset \sigma(a)$, and $f_n \rightarrow f$ uniformly on compact subsets of $U$, then $f_n(a) \rightarrow f(a)$

As a sanity check, note that if $A = \mathbb{C}$, then $\sigma(a) = \{a\}$, and by Cauchy’s formula we recover the usual definition of evaluating $f$ at the point $a$.

#### 2. Well definedness

First we fix $U$, and show that $f(a)$ is independent of $V$. Let $V,V'$ be as in the definition of the functional calculus. Let $W$ be bounded open set containing $\overline{V \cup V'}$ such that $\overline{W} \subset U$ and such that $\partial W$ consists of finitely many $C^1$ curves. $W \setminus \overline{V}$ is open and has piecewise $C^1$ boundary. Moreover, $z \mapsto f(z)(z-a)^{-1}$ is analytic in a neighborhood of the closure of this set. Therefore, from Theorem 8(2) in the previous post, we have that

$\int_{\partial W} f(z) (z-a)^{-1} dz = \int_{\partial V} f(z) (z-a)^{-1} dz .$

(Reader: make sure you understand this. In particular, why does the LHS equal the RHS, and not minus the RHS). Similarly,

$\int_{\partial W} f(z) (z-a)^{-1} dz = \int_{\partial V'} f(z) (z-a)^{-1} dz .$

Now let $U$ and $U'$ be two sets playing the role of $U$ in the theorem. Denote by $f_U(a)$ and $f_{U'}(a)$ the corresponding elements obtained. We know that $f_U(a)$ is independent of the subset $V$ on which boundary we integrate. Choose a bounded $V \supset \sigma(a)$ with piecewise $C^1$ boundary such that $\overline{V} \subset U \cap U'$. The boundary of this set $V$ can be used to compute both $f_U(a)$ and $f_{U'}(a)$. Thus $f_U(a)$ and $f_{U'}(a)$ are obtained as the same $A$ valued integral, hence $f_U(a) = f_{U'}(a)$.

#### 3. The holomorphic functional calculus is a homomorphism

It is clear that $f + g \mapsto f(a) + g(a)$ since the $A$-valued integral is linear. The $A$ valued integral is not multiplicative (even the scalar valued integral is not), so it is really not clear that $f g \mapsto f(a) g(a)$. However, it turns out that this is indeed the case. The proof is a little tricky. Let $f,g \in\mathcal{O}(a)$. Suppose that both functions are in holomorphic in an open set $U \subset \sigma(a)$. Let $V,W$ be bounded open sets such that

$\sigma(a) \subset V \subset \overline{V} \subset W \subset \overline{W} \subset U.$

Now,

(*) $f(a) g(a) = (2\pi i)^{-2} \int_{w\in \partial W} f(w)(w-a)^{-1} dw \int_{v \in \partial V}g(v) (v-a)^{-1} dv,$

while, on the other hand,

$fg(a) = \frac{1}{2\pi i} \int_{\partial V} f(v)g(v)(v-a)^{-1}dv .$

For every $v \in \partial V$ we write $f(v) = \frac{1}{2\pi i} \int_{\partial W} f(w)(w-v)^{-1} dw$. It is not hard to show that for continuous functions a Fubini type theorem holds, thus we get

(**)$fg(a) = (2 \pi i)^{-2} \int_{\partial V} \left(\int_{\partial W} f(w)(w-v)^{-1}dw \right) g(v) (v-a)^{-1} dv$

We want to show that the right hand sides of (*) and (**) are equal. We add and subtract a term in (*) and rewrite the left hand side of (*) as a sum $I_1 + I_2$, where

$I_1 = (2\pi i)^{-2} \int_{\partial V} \int_{\partial W} \frac{f(w)}{w-v}g(v) (v-a)^{-1} dw dv$,

and

$I_2= (2 \pi i)^{-2} \int_{\partial V} \int_{\partial W} f(w) [(w-a)^{-1} - (w-v)^{-1}]g(v)(v-a)^{-1} dw dv$.

The first integral $I_1$ is equal to (**), so to finish we must show that the second integral $I_2$ vanishes. Using the identity

$(w-a)^{-1} - (w-v)^{-1} = (v-a)(v-w)^{-1} (w-a)^{-1}$,

we rewrite the $2 \pi i I_2$ as

$\int_{\partial V} \int_{\partial W} f(w) (v-a) (v-w)^{-1}(w-a)^{-1} g(v) (v-a)^{-1} dw dv =$

$= \int_{\partial W} f(w) (w-a)^{-1} \left(\int_{\partial V}\frac{g(v)}{v-w} dv\right) dw .$

But the inner integral is, for every fixed $w \in \partial W$, the integral of a function analytic in a neighborhood of $\overline{V}$, hence it is zero by Cauchy’s formula. This establishes the equality $f(a)g(a) = fg(a)$, and completes the proof.

#### 4. Continuity

Crashing through with the triangle inequality for $A$-valued integrals, it is easy to see that if $\overline{V} \subset U$ and $f_n \rightarrow f$ uniformly on $\partial V$, then

$2 \pi i f_n(a) = \int_{\partial V} f_n(z) (z-a)^{-1} dz \rightarrow \int_{\partial V} f(z) (a-z)^{-1} dz = 2 \pi i f(a),$

as required.

#### 5. Naturality

Because the functional calculus is a continuous homomorphism, it suffices to prove that $\zeta(a) = a$ (explain why). Since $\zeta$ is entire, we may define

$\zeta(a) = \frac{1}{2\pi i} \int_C z(z-a)^{-1} dz$

where $C$ is a circle with radius bigger than the norm of $a$. But then we have

$\frac{1}{2\pi i} \int_C z(z-a)^{-1} dz = \frac{1}{2\pi i}\int_C (1-a/z)^{-1} dz =$

$= \frac{1}{2 \pi i} \sum_{k=0}^\infty \int_C a^k z^{-k} dz = a$,

because $\int_C a^k z^{-k} dz$  is equal $a^k \int_C z^{-k} dz$ is equal to $a 2 \pi i$ if $k=1$ and $0$ else.

#### 6. Spectral mapping theorem

Denote by $B$ a commutative unital Banach subalgebra of $A$ containing $a$ and $1$. Denote by $\Gamma : B \rightarrow C(sp(B))$ the Gelfand transform.

Lemma: For every $f \in \mathcal{O}(a)$, $f(a) \in B$ and

$\Gamma(f(a)) = f (\Gamma(a))$,

where $f(\Gamma(a))$ denote the application of the functional calculus to $\Gamma(a) \in C(sp(B))$

Remark:  Note that

Proof: It is clear that $f \in B$. If $V$ is an appropriately chosen open set,

$f(\Gamma(a)) = \frac{1}{2\pi i}\int_{\partial V} f(z)(z-\Gamma(a))^{-1}dz .$

Now if $\phi \in sp(B)$, then (denoting by $\rho_\phi$ the evaluation functional) we have

$f(\Gamma(a))(\phi) =\rho_\phi(f(\Gamma(a))) = \frac{1}{2 \pi i} \int_{\partial V} \rho_\phi(f(z) (z-a)^{-1}) dz =$

$= \frac{1}{2 \pi i}\int_{\partial V} f(z) (z-\rho_\phi(\Gamma(a)))^{-1} dz = \frac{1}{2\pi i} \int_{\partial V} f(z) (z - \phi(a))^{-1} dz.$

On the other hand,

$\Gamma(f(a))(\phi) = \phi(f(a)) = \frac{1}{2 \pi i}\int_{\partial V} \phi ( f(z) (z-a)^{-1}) dz =$

$\frac{1}{2 \pi i} \int_{\partial V} f(z) (z-\phi(a))^{-1} dz.$

This concludes the proof of the lemma.

The spectral mapping theorem in $B$ follows easily from the lemma, since by Gelfand’s theorem

$\sigma_B(f(a)) = \sigma(\Gamma(f(a))) = \sigma(f(\Gamma(a)))$

But the right hand side is the range of the function $f(\Gamma(a))$, which is $f$ applied to the range of $\Gamma(a)$, i.e., $f(\sigma(\Gamma(a)))$. But on the other hand

$f(\sigma_B(a)) = f(\sigma(\Gamma(a)))$,

thus the spectral mapping theorem holds in any commutative Banach subalgebra $B \subseteq A$ containing $a$ and the unit. We are not done yet since $A$ might be non-commutative.

Finally, $\sigma_A(f(a)) = f(\sigma_A(a))$ follows directly from the following exercise.

Exercise B: There exists a commutative Banach subalgebra $B \subseteq A$ containing $a$ and the unit such that

1. $B$ is inverse closed: if $x \in B$ and $x \in A^{-1}$ then $x^{-1} \in B$.
2. $B$ is the smallest commutative inverse closed subalgebra of $A$ containing $A$ and the unit.
3. $\sigma_B(a) = \sigma_A(a)$ (what about $f(a)$?).

(Hint: consider all elements of the form $f(a)$, where $f \in Rat(a)$. Alternatively, consider the double commutant $B''$ of $B$ in $A$).

#### 7. Application to invariant subspaces

There are many applications of the holomorphic functional calculus. We show just one application to the notorious invariant subspace problem.

Exercise C: Let $E$ be a Banach space, and let $T \in B(E)$. If $\sigma(T)$ is disconnected, then $T$ has a non-trivial invariant subspace.