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 $latex mathcal mathcal{O}(a)$ denotes the algebra of functions defined and holomorphic in some neighborhood of $latex sigma(a)$ (the spectrum of $latex a$). When working with $latex mathcal{O}(a)$ 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 $latex sigma(a)$ is a point, the restriction $latex f big|_{sigma(a)}$ of a function $latex f in mathcal{O}(a)$ to $latex sigma(a)$ loses almost all information on $latex f$. Let us show how multiplication is defined in this algebra. If $latex f,g in mathcal{O}(a)$, then there are neighborhoods $latex U,V$ of $latex sigma(a)$ such that $latex f$ is analytic in $latex U$ and $latex g$ is analytic in $latex V$. Then $latex U cap V$ is a neighborhood of $latex sigma(a)$, and $latex f g$ is defined and holomorphic in the neighborhood of $latex U cap V$ of $latex sigma(a)$, so it is in $latex mathcal{O}(a)$. All algebraic matters are handled in a similar matter. Analytic matters (such as defining a topology on $latex mathcal{O}(a)$) are more involved and we will treat them only at the most basic level. We will require the following elementary result.
Exercise A: Let $latex K$ be a compact subset of the plane, and let $latex U$ be an open set containing $latex K$. Then there exists a bounded open set $latex V$ such that $latex K subset V subset overline{V} subset U$ and such that $latex partial V$ is a finite union of $latex C^1$ curves. (In fact, one can also find such a $latex V$ 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 $latex A$ be a unital Banach algebra and let $latex a in A$. For every $latex f in mathcal{O}(a)$, we define an element $latex f(a) in A$ in the following way: Let $latex U supset sigma(a)$ be an open set where $latex f$ is defined and holomorphic, and let $latex V$ be an open bounded set such that $latex sigma(a) subset V subset overline{V} subset U$ and such that $latex partial V$ is a finite union of $latex C^1$ curves; define
$latex f(a) = int_{partial V} f(z) (z-a)^{-1} dz .$
Theorem (holomorphic functional calculus): The mapping $latex f mapsto f(a)$ has the following properties:
- (Well definedness) $latex f(a)$ is well defined, and in particular does not depend on the choice of $latex U$ and $latex V$.
- (Homomorphism) The mapping $latex f mapsto f(a)$ is a homomorphism.
- (Naturality) If $latex f$ is a polynomial, a rational function with poles off $latex sigma(a)$, or a function with a power series representation that converges in a neighborhood of $latex sigma(a)$, then the definition of $latex f(a)$ coincides with the natural definitions given in the previous lecture (Sections 2,3 and 4 in the previous post). If $latex zeta$ denotes the entire function $latex zeta(z) = z$, one has $latex zeta(a) = a$.
- (Spectral mapping theorem) For all $latex f in mathcal{O}(a)$, it holds that $latex sigma(f(a)) = f(sigma(a))$.
- (Conitnuity) If $latex f_n,f$ are holomorphic in $latex U supset sigma(a)$, and $latex f_n rightarrow f$ uniformly on compact subsets of $latex U$, then $latex f_n(a) rightarrow f(a)$.
As a sanity check, note that if $latex A = mathbb{C}$, then $latex sigma(a) = {a}$, and by Cauchy’s formula we recover the usual definition of evaluating $latex f$ at the point $latex a$.
2. Well definedness
First we fix $latex U$, and show that $latex f(a)$ is independent of $latex V$. Let $latex V,V’$ be as in the definition of the functional calculus. Let $latex W$ be bounded open set containing $latex overline{V cup V’}$ such that $latex overline{W} subset U$ and such that $latex partial W$ consists of finitely many $latex C^1$ curves. $latex W setminus overline{V}$ is open and has piecewise $latex C^1$ boundary. Moreover, $latex z mapsto f(z)(z-a)^{-1}$ is analytic in a neighborhood of the closure of this set. Therefore, from Theorem 8(2) in the previous post, we have that
$latex int_{partial W} f(z) (z-a)^{-1} dz = int_{partial V} f(z) (z-a)^{-1} dz .$
(Reader: make sure you understand this. In particular, why does the LHS equal the RHS, and not minus the RHS). Similarly,
$latex int_{partial W} f(z) (z-a)^{-1} dz = int_{partial V’} f(z) (z-a)^{-1} dz .$
Now let $latex U$ and $latex U’$ be two sets playing the role of $latex U$ in the theorem. Denote by $latex f_U(a)$ and $latex f_{U’}(a)$ the corresponding elements obtained. We know that $latex f_U(a)$ is independent of the subset $latex V$ on which boundary we integrate. Choose a bounded $latex V supset sigma(a)$ with piecewise $latex C^1$ boundary such that $latex overline{V} subset U cap U’$. The boundary of this set $latex V$ can be used to compute both $latex f_U(a)$ and $latex f_{U’}(a)$. Thus $latex f_U(a)$ and $latex f_{U’}(a)$ are obtained as the same $latex A$ valued integral, hence $latex f_U(a) = f_{U’}(a)$.
3. The holomorphic functional calculus is a homomorphism
It is clear that $latex f + g mapsto f(a) + g(a)$ since the $latex A$-valued integral is linear. The $latex A$ valued integral is not multiplicative (even the scalar valued integral is not), so it is really not clear that $latex f g mapsto f(a) g(a)$. However, it turns out that this is indeed the case. The proof is a little tricky. Let $latex f,g inmathcal{O}(a)$. Suppose that both functions are in holomorphic in an open set $latex U subset sigma(a)$. Let $latex V,W$ be bounded open sets such that
$latex sigma(a) subset V subset overline{V} subset W subset overline{W} subset U.$
Now,
(*) $latex f(a) g(a) = (2pi i)^{-2} int_{win partial W} f(w)(w-a)^{-1} dw int_{v in partial V}g(v) (v-a)^{-1} dv,$
while, on the other hand,
$latex fg(a) = frac{1}{2pi i} int_{partial V} f(v)g(v)(v-a)^{-1}dv .$
For every $latex v in partial V$ we write $latex f(v) = frac{1}{2pi i} int_{partial W} f(w)(w-v)^{-1} dw$. It is not hard to show that for continuous functions a Fubini type theorem holds, thus we get
(**)$latex fg(a) = (2 pi i)^{-2} int_{partial V} left(int_{partial W} f(w)(w-v)^{-1}dw right) g(v) (v-a)^{-1} dv$
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 $latex I_1 + I_2$, where
$latex I_1 = (2pi i)^{-2} int_{partial V} int_{partial W} frac{f(w)}{w-v}g(v) (v-a)^{-1} dw dv$,
and
$latex 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 $latex I_1$ is equal to (**), so to finish we must show that the second integral $latex I_2$ vanishes. Using the identity
$latex (w-a)^{-1} – (w-v)^{-1} = (v-a)(v-w)^{-1} (w-a)^{-1}$,
we rewrite the $latex 2 pi i I_2$ as
$latex int_{partial V} int_{partial W} f(w) (v-a) (v-w)^{-1}(w-a)^{-1} g(v) (v-a)^{-1} dw dv = $
$latex = int_{partial W} f(w) (w-a)^{-1} left(int_{partial V}frac{g(v)}{v-w} dvright) dw .$
But the inner integral is, for every fixed $latex w in partial W$, the integral of a function analytic in a neighborhood of $latex overline{V}$, hence it is zero by Cauchy’s formula. This establishes the equality $latex f(a)g(a) = fg(a)$, and completes the proof.
4. Continuity
Crashing through with the triangle inequality for $latex A$-valued integrals, it is easy to see that if $latex overline{V} subset U$ and $latex f_n rightarrow f$ uniformly on $latex partial V$, then
$latex 2 pi i f_n(a) = int_{partial V} f_n(z) (z-a)^{-1} dz rightarrow int_{partial V} f(z) (a-z)^{-1} dz = 2 pi i f(a),$
as required.
5. Naturality
Because the functional calculus is a continuous homomorphism, it suffices to prove that $latex zeta(a) = a$ (explain why). Since $latex zeta$ is entire, we may define
$latex zeta(a) = frac{1}{2pi i} int_C z(z-a)^{-1} dz $
where $latex C$ is a circle with radius bigger than the norm of $latex a$. But then we have
$latex frac{1}{2pi i} int_C z(z-a)^{-1} dz = frac{1}{2pi i}int_C (1-a/z)^{-1} dz = $
$latex = frac{1}{2 pi i} sum_{k=0}^infty int_C a^k z^{-k} dz = a$,
because $latex int_C a^k z^{-k} dz$ is equal $latex a^k int_C z^{-k} dz$ is equal to $latex a 2 pi i$ if $latex k=1$ and $latex 0$ else.
6. Spectral mapping theorem
Denote by $latex B$ a commutative unital Banach subalgebra of $latex A$ containing $latex a$ and $latex 1$. Denote by $latex Gamma : B rightarrow C(sp(B))$ the Gelfand transform.
Lemma: For every $latex f in mathcal{O}(a)$, $latex f(a) in B$ and
$latex Gamma(f(a)) = f (Gamma(a))$,
where $latex f(Gamma(a))$ denote the application of the functional calculus to $latex Gamma(a) in C(sp(B))$.
Remark: Note that
Proof: It is clear that $latex f in B$. If $latex V$ is an appropriately chosen open set,
$latex f(Gamma(a)) = frac{1}{2pi i}int_{partial V} f(z)(z-Gamma(a))^{-1}dz .$
Now if $latex phi in sp(B)$, then (denoting by $latex rho_phi$ the evaluation functional) we have
$latex 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 = $
$latex = frac{1}{2 pi i}int_{partial V} f(z) (z-rho_phi(Gamma(a)))^{-1} dz = frac{1}{2pi i} int_{partial V} f(z) (z – phi(a))^{-1} dz.$
On the other hand,
$latex Gamma(f(a))(phi) = phi(f(a)) = frac{1}{2 pi i}int_{partial V} phi ( f(z) (z-a)^{-1}) dz = $
$latex frac{1}{2 pi i} int_{partial V} f(z) (z-phi(a))^{-1} dz.$
This concludes the proof of the lemma.
The spectral mapping theorem in $latex B$ follows easily from the lemma, since by Gelfand’s theorem
$latex sigma_B(f(a)) = sigma(Gamma(f(a))) = sigma(f(Gamma(a)))$
But the right hand side is the range of the function $latex f(Gamma(a))$, which is $latex f$ applied to the range of $latex Gamma(a)$, i.e., $latex f(sigma(Gamma(a)))$. But on the other hand
$latex f(sigma_B(a)) = f(sigma(Gamma(a)))$,
thus the spectral mapping theorem holds in any commutative Banach subalgebra $latex B subseteq A$ containing $latex a$ and the unit. We are not done yet since $latex A$ might be non-commutative.
Finally, $latex sigma_A(f(a)) = f(sigma_A(a))$ follows directly from the following exercise.
Exercise B: There exists a commutative Banach subalgebra $latex B subseteq A$ containing $latex a$ and the unit such that
- $latex B$ is inverse closed: if $latex x in B$ and $latex x in A^{-1}$ then $latex x^{-1} in B$.
- $latex B$ is the smallest commutative inverse closed subalgebra of $latex A$ containing $latex A$ and the unit.
- $latex sigma_B(a) = sigma_A(a)$ (what about $latex f(a)$?).
(Hint: consider all elements of the form $latex f(a)$, where $latex f in Rat(a)$. Alternatively, consider the double commutant $latex B”$ of $latex B$ in $latex A$).
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 $latex E$ be a Banach space, and let $latex T in B(E)$. If $latex sigma(T)$ is disconnected, then $latex T$ has a non-trivial invariant subspace.
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