Advanced Analysis, Notes 19: The holomorphic functional calculus II (definition and basic properties)
by Orr Shalit
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 denotes the algebra of functions defined and holomorphic in some neighborhood of (the spectrum of ). When working with 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 is a point, the restriction of a function to loses almost all information on . Let us show how multiplication is defined in this algebra. If , then there are neighborhoods of such that is analytic in and is analytic in . Then is a neighborhood of , and is defined and holomorphic in the neighborhood of of , so it is in . All algebraic matters are handled in a similar matter. Analytic matters (such as defining a topology on ) are more involved and we will treat them only at the most basic level. We will require the following elementary result.
Exercise A: Let be a compact subset of the plane, and let be an open set containing . Then there exists a bounded open set such that and such that is a finite union of curves. (In fact, one can also find such a 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 be a unital Banach algebra and let . For every , we define an element in the following way: Let be an open set where is defined and holomorphic, and let be an open bounded set such that and such that is a finite union of curves; define
Theorem (holomorphic functional calculus): The mapping has the following properties:
- (Well definedness) is well defined, and in particular does not depend on the choice of and .
- (Homomorphism) The mapping is a homomorphism.
- (Naturality) If is a polynomial, a rational function with poles off , or a function with a power series representation that converges in a neighborhood of , then the definition of coincides with the natural definitions given in the previous lecture (Sections 2,3 and 4 in the previous post). If denotes the entire function , one has .
- (Spectral mapping theorem) For all , it holds that .
- (Conitnuity) If are holomorphic in , and uniformly on compact subsets of , then .
As a sanity check, note that if , then , and by Cauchy’s formula we recover the usual definition of evaluating at the point .
2. Well definedness
First we fix , and show that is independent of . Let be as in the definition of the functional calculus. Let be bounded open set containing such that and such that consists of finitely many curves. is open and has piecewise boundary. Moreover, is analytic in a neighborhood of the closure of this set. Therefore, from Theorem 8(2) in the previous post, we have that
(Reader: make sure you understand this. In particular, why does the LHS equal the RHS, and not minus the RHS). Similarly,
Now let and be two sets playing the role of in the theorem. Denote by and the corresponding elements obtained. We know that is independent of the subset on which boundary we integrate. Choose a bounded with piecewise boundary such that . The boundary of this set can be used to compute both and . Thus and are obtained as the same valued integral, hence .
3. The holomorphic functional calculus is a homomorphism
It is clear that since the -valued integral is linear. The valued integral is not multiplicative (even the scalar valued integral is not), so it is really not clear that . However, it turns out that this is indeed the case. The proof is a little tricky. Let . Suppose that both functions are in holomorphic in an open set . Let be bounded open sets such that
while, on the other hand,
For every we write . It is not hard to show that for continuous functions a Fubini type theorem holds, thus we get
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 , where
The first integral is equal to (**), so to finish we must show that the second integral vanishes. Using the identity
we rewrite the as
But the inner integral is, for every fixed , the integral of a function analytic in a neighborhood of , hence it is zero by Cauchy’s formula. This establishes the equality , and completes the proof.
Crashing through with the triangle inequality for -valued integrals, it is easy to see that if and uniformly on , then
Because the functional calculus is a continuous homomorphism, it suffices to prove that (explain why). Since is entire, we may define
where is a circle with radius bigger than the norm of . But then we have
because is equal is equal to if and else.
6. Spectral mapping theorem
Denote by a commutative unital Banach subalgebra of containing and . Denote by the Gelfand transform.
Lemma: For every , and
where denote the application of the functional calculus to .
Remark: Note that
Proof: It is clear that . If is an appropriately chosen open set,
Now if , then (denoting by the evaluation functional) we have
On the other hand,
This concludes the proof of the lemma.
The spectral mapping theorem in follows easily from the lemma, since by Gelfand’s theorem
But the right hand side is the range of the function , which is applied to the range of , i.e., . But on the other hand
thus the spectral mapping theorem holds in any commutative Banach subalgebra containing and the unit. We are not done yet since might be non-commutative.
Finally, follows directly from the following exercise.
Exercise B: There exists a commutative Banach subalgebra containing and the unit such that
- is inverse closed: if and then .
- is the smallest commutative inverse closed subalgebra of containing and the unit.
- (what about ?).
(Hint: consider all elements of the form , where . Alternatively, consider the double commutant of in ).
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 be a Banach space, and let . If is disconnected, then has a non-trivial invariant subspace.