The Concept of Functional Calculus

Consider the Banach space X :=C[0,1] of continuous functions on the unit in- terval with values in the complex numbersC Each functiona∈X determines a bounded linear operator

The multiplication operator \( M_a = (g \rightarrow ag) : C[0,1] \rightarrow C[0,1] \) is defined on \( X \) and has a spectrum \( \sigma(M_a) \) that corresponds to the range \( a[0,1] \) of the function \( a \) For any continuous function \( f : \sigma(M_a) \rightarrow \mathbb{C} \), one can define the multiplication operator \( M_{f \circ a} \) associated with \( f \circ a \), leading to an algebra homomorphism \( \Phi = (f \rightarrow M_{f \circ a}) : C(\sigma(M_a)) \rightarrow L(X) \), which maps into the algebra of bounded linear operators on \( X \) Notably, \( \Phi(z) = M_a^\dagger \) and \( \Phi((\lambda - z)^{-1}) = R(\lambda, M_a) \) for \( \lambda \in \sigma(M_a) \), indicating that the operator \( \Phi(f) \) represents multiplication by \( f(a(z)) \), effectively ‘inserting’ the operator.

M a into the function f and writes f(M a ) := Φ(f) Generalising this example to the Banach space X = C 0 (R) and continuous functions a ∈ C(R), one realises that boundedness of the operators is not an essential requirement.

† We simply write z := ( z −→ z ) for the coordinate function on C Hence the symbols f ( z ) and f are used interchangeably.

2 Chapter 1 Axiomatics for Functional Calculi

Functional calculus intuitively relates to the concept that each closed operator A on a Banach space X corresponds to an algebra of complex-valued functions defined on its spectrum, allowing for the operator A to be effectively represented within this framework.

In the context of functional calculus for a set A, it is essential that the mapping f → f(A) maintains expected meanings, such as ensuring that if λ belongs to A, then (λ−z) − 1 (A) equals R(λ, A), or if A generates a semigroup T, then tz(A) equals T(t) While these represent minimal requirements, additional reasonable criteria may also exist However, a comprehensive formalization of this concept has yet to be established, with current progress limited to case-by-case constructions.

In certain cases, if an operator A is fundamentally akin to a multiplication operator, it becomes easy to develop a functional calculus According to a specific interpretation of the spectral theorem, this relationship holds true under certain conditions.

A normal operator on a Hilbert space serves as a foundational concept in functional calculus, with the Fourier transform on L²(R) being a prime example, where the operator A is represented as id/dt Unlike some mathematical constructs, there is no standard method for diagonalizing a normal operator, which means that the functional calculus derived from it is influenced by the selected unitary equivalence Consequently, further justification is required to confirm the independence of this construction, as outlined in Theorem D.6.1.

To achieve a general understanding, it is essential to focus on intrinsic definitions Riesz and Dunford proposed constructing the function f(A) using a Cauchy-type integral, expressed as f(A) = 1.

The idea behind this is readily sketched Let us assume that we are given abounded operatorA and an open supersetU of the spectrum K := σ(A) The so-called

‘homology version’ of the Cauchy Integral Formula says the following One can choose a generalised contour (a ‘cycle’) Γ with the following properties:

2) each point ofC\Γ ∗ has index either 0 or 1 with respect to Γ;

3) each point ofK has index 1 with respect to Γ.

In the context of the contour Γ, the function f(a) can be expressed as f(a) = (1/(2πi)) ∮_Γ f(z) / (z - a) dz for each point a in K and for any holomorphic function defined on U, as demonstrated in [48, Proposition VIII.1.1] By substituting A for a in this formula and ensuring that 1/(z - A) is equivalent to R(z, A), we derive the equation (1.1).

The Dunford–Riesz calculus, a specific form of functional calculus, is a subset of a broader framework within Banach algebras This construction establishes an algebra homomorphism, denoted as Φ, which maps holomorphic functions from the algebra of germs on σ(A) to the bounded linear operators on a Banach space X.

Let us continue with a second example Consider the Banach space X : C[0,1] again, and thereon theVolterra operator V ∈ L(X) defined by

As is well known, V is a compact, positive contraction withσ(A) ={0} Given a holomorphic function f defined in a neighbourhood of 0, simply choose ε >0 small enough and define f(V) := 1

SinceV n = 1/n! for alln∈N, one can equally replacezbyV in the power-series expansion of f.

The operator V is injective but not invertible, and the function z - 1 is not holomorphic at 0 However, it is reasonable to express V - 1 as (z - 1)(V), indicating a potential extension of the functional calculus to a broader algebra that includes the function z - 1 In the specific case of the Volterra operator, this extension is straightforward, particularly when considering functions that are holomorphic in a neighborhood around 0 and lack essential singularities at that point, implying that the principal part of the Laurent series can be represented as a polynomial in z - 1.

V as in Appendix A.7 into the principal part and as above into the remainder. Finally add the results.)

In many cases, it is not possible to revert to power or Laurent series, necessitating the use of an abstract extension procedure This will be the focus of the upcoming section.

An Abstract Framework

The Extension Procedure

The extension procedure begins with a Banach space \( X \) and a commutative algebra \( M \) that includes a subalgebra \( E \subset M \) (with \( 1 \in E \) typically) This framework is defined by a homomorphism \( \Phi: E \rightarrow L(X) \) The combination of the elements \( (E, M, \Phi) \) is referred to as an abstract functional calculus (afc) over \( X \), though sometimes the focus is solely on the pair \( (E, M) \) without explicitly mentioning the homomorphism An afc is considered non-degenerate or proper if the associated set meets specific criteria.

The set Reg(E) is defined as the collection of elements e in E for which the function Φ(e) is injective, and it is not empty Each element within Reg(E) is referred to as a regulariser For a function f in M, if there exists an element e in Reg(E) such that the product ef is also in E, then f is considered regularisable by E, with e serving as a regulariser for f It is important to note that the element 1 is regularisable if and only if the associated affine combination (afc) is proper.

M r :={f ∈ M |f is regularisable} clearly is a subalgebra ofMwhich containsE.

In the context of a proper affine function (E, M, Φ), we define the mapping Φ(f) as Φ(e) − 1 Φ(ef), where e is a regularizer for f within the set of regularizers Reg(E) For convenience, we often denote this mapping as f • instead of Φ(f) This notation implies that the expression can be rewritten as f • := e − • 1 (ef) • The subsequent lemma demonstrates that this definition remains consistent regardless of the regularizer selected.

Lemma 1.2.1 Let (E,M,Φ)be a proper afc Then by (1.3)a closed operator on

X is well defined and the so-defined mapping Φ = (f −→Φ(f)) :M r −→ {closed operators on X} extends the original mapping Φ :E −→ L(X).

Proof Leth∈Reg(E) be a second regulariser forf, and define A:= (e • ) − 1 (ef) • and B := (h • ) − 1 (hf) • Because e • h • = (eh) • = (he) • = h • e • , inverting yields (e • ) − 1 (h • ) − 1 = (h • ) − 1 (e • ) − 1 Now it follows that

This shows that f • is the same whichever regulariser one chooses.

We need to demonstrate that the new mapping Φ serves as an extension of the previous one Given that the affine connection is proper, we have E as a subset of M r For elements e and f in E with an injective operation, it follows that (e • ) − 1 (ef) • equals (e • ) − 1 e • f • f • This confirms that the mapping on M r effectively extends the original.

Sometimes we call the original mapping Φ :E −→ L(X) theprimary(func- tional) calculus (in short: pfc) and the extension defined above the extended

(functional) calculus The algebraM r is called thedomainof the afc (E,M,Φ).

Properties of the Extended Calculus

We collect some basic properties.

Proposition 1.2.2 establishes key properties of a proper abstract functional calculus (E, M, Φ) over a Banach space X Firstly, if an operator T in L(X) commutes with every element e in E, it will also commute with all elements f in Mr Additionally, the identity element 1 is included in Mr, and the operation 1• corresponds to the identity operator I For functions f and g in Mr, their sum and product operations satisfy f• + g• being a subset of (f + g)• and f• g• being a subset of (fg)•, with the domains D(f• g•) equating to the intersection of D((fg)•) and D(g•) Furthermore, if f and g in Mr satisfy fg = 1, then f• is injective, and its inverse is given by g• Lastly, for a function f in Mr and a subspace F of D(f•), if there exists a sequence (e_n) in E such that e_n• converges strongly to I and R(e_n•) is contained in F for all n, then F serves as a core for f•.

6 Chapter 1 Axiomatics for Functional Calculi

In the proof, we consider a linear operator T that commutes with every regulariser e in a set E For a function f in M_r and a regulariser e, we establish that T(f •) equals f • T, indicating a relationship between T and the regularised function Given that the algebraic function class (afc) is proper, we can find an injective regulariser e that also regularises the constant function 1 Furthermore, for functions f and g in M_r with regularisers e_1 and e_2, the product e = e_1 e_2 serves as a regulariser for both f and g, and consequently for their sum f + g Additionally, the product ef g can be expressed as the product of the regularised functions, confirming that e remains a valid regulariser for the product f g, leading to the conclusion that f • + g • can be expressed in terms of the regularised functions.

To prove the assertions concerning the domains, let x∈D((f g) • )∩D(g • ) Since (e 1 f) • commutes with e 2 • it also commutes with e − 2 • 1 By assumption, y : (e 2 g) • x∈D(e − 2 • 1 ), whence also (e 1 f) • y∈D(e − 2 • 1 ) and

(e 1 f) • g • x= (e 1 f) • e − 2 • 1 y=e − 2 • 1 (e 1 f) • y=e − 2 • 1 (ef g) • x∈D(e − 1 • 1 ) by assumption and the identity e − • 1 =e − 1 • 1 e − 2 • 1 Consequently, g • x∈D(f • ), and hencex∈D(f • g • ). d) Supposef, g∈ M r withf g= 1 By b) and c) we haveg • f • ⊂(f g) • = 1 • =I andD(g • f • ) =D(I)∩D(f • ) =D(f • ) Interchangingfandgproves the statement. e) Letx∈D(f • ) and definey :=f • x Withx n :=e n • xandy n :=e n • y we have x n →x,x n ∈F and f • x n =f • e n • x=e n • f • x=y n →y

In general one cannot expect equality in c) of Proposition 1.2.2 However, if we define

Corollary 1.2.3 LetE,M,Φ, X be as above. a) Forf ∈ M r , g∈ M b one has f • +g • = (f+g) • andf • g • = (f g) • b) The setM b is a subalgebra with 1 ofM, and the map

(f −→f • ) :M b −→ L(X) is a homomorphism of algebras with 1 c) If f ∈ M b is such that f • is injective, then

Proof a) By c) of Proposition 1.2.2 we havef • +g • ⊂(f+g) • and (f+g) • −g • ⊂

The relationship between the functions f and g can be expressed as (f + g − g) • = f •, leading to the conclusion that D((f + g) •) = D(f •) since g • is bounded This indicates that f • + g • equals (f + g) • Furthermore, as established in Proposition 1.2.2, the assertion follows directly from the fact that D(g •) = X Additionally, we can identify an element e ∈ E that regularizes both f and g By performing calculations, we find that f • − 1 g • f • simplifies to g •, demonstrating the interrelationship between these functions.

Here we used b) and the identitye − • 1 f • − 1 =f • − 1 e − • 1 , which is true sincee • andf • are both bounded and injective

Using this new information we can improve d) of Proposition 1.2.2.

Corollary 1.2.4 Let E,M,Φ, X be as above Suppose that f ∈ M r , g∈ M such that f g= 1 Then g∈ M r ⇐⇒ f • is injective.

In this case, we have g • =f • − 1

Proof One direction of the equivalence is simply d) of Proposition 1.2.2 Suppose thatf • is injective and lete∈ Ebe a regulariser forf Thenf e∈ E,(f e)g=e∈ E and (f e) • =f • e • is injective This means thatf eis a regulariser forg

In this article, we explore the extension procedure through the example of the Volterra operator We define M as the algebra of germs of functions that are holomorphic in a pointed neighborhood of 0, while E represents the subalgebra of germs that are holomorphic in a whole neighborhood of 0 The foundational calculus is established using the Cauchy integral (1.1) or by inserting into the power series Notably, since (z)(V) = V, this algebraic function is proper, allowing for a natural extension to M r, which can be easily demonstrated.

M r is exactly the algebra of germs of meromorphic functions at 0 Iff is such a germ with z n f holomorphic at 0, we havef(V) =V − n (z n f)(V) In particular,(z − n )(V) =V − n for eachn∈N.

Generators and Morphisms

Let us return to the abstract treatment We start with a proper afc (E,M,Φ) over the Banach space X A subalgebraD ⊂ M b is called admissible if the set

{f ∈ D |f • is injective} is not empty In this case (D,M,Φ) is another proper afc overX Let us denote by

D:={f ∈ M |there is d∈ Dsuch that df∈ Dandd • is injective} the regularisable elements of this ‘sub-afc’.

8 Chapter 1 Axiomatics for Functional Calculi

Proposition 1.2.5 Let (E,M,Φ)be a proper afc over the Banach spaceX, and let

D be an admissible subalgebra ofM b If f ∈ M andg ∈ Dare such that g • is injective and gf∈ D, then alreadyf ∈ D.

Proof By assumption there are regularising elements d 1 , d 2 ∈ D for g, f g, re- spectively Letting d := d 1 d 2 we see that d regularises both g and f g Since

(d • ) − 1 (dg) • =g • is injective, the operator (dg) • also is Hence dg regularisesf, whence f ∈ D

The proposition indicates that Mr is equal to Eb, highlighting that a generator of the afc is an admissible subalgebra D, where D equals Mr The focus is on identifying small generators To verify if a specific admissible subalgebra acts as a generator, it is sufficient to encompass any existing generator, as demonstrated by the following corollary.

Corollary 1.2.6 Let (E,M,Φ) be a proper afc over the Banach space X Let

D,D be admissible subalgebras of M b such that D ⊂ D ThenD ⊂ D In particular, an admissible subalgebraDof M b is a generator of the afc if and only if E ⊂ D.

Proof This follows immediately from Proposition 1.2.5

Abstract functional calculi over the Banach space X form a category, where we detail the morphisms within this framework For two proper abstract functional calculi (E, M, Φ) and (E', M', Φ'), a morphism θ: (E, M, Φ) → (E', M', Φ') consists of a homomorphism of algebras θ: M → M' that satisfies θ(E) ⊆ E' and θ(e) • = e' • for all e ∈ E The extension procedure is inherently functorial, as demonstrated by the subsequent proposition.

Proposition 1.2.7 Let θ: (E,M,Φ)−→(E ,M ,Φ )be a morphism of proper afc on the Banach spaceX Thenθ(M r )⊂(M ) r withθ(f) • =f • for everyf ∈ M r

Proof Letf ∈ M r and lete∈Reg(E) be a regulariser forf Sinceθ(e) • =e • is injective and θ(e)θ(f) =θ(ef) ∈θ(E) ⊂ E , the elementθ(e) is a regulariser for θ(f) Moreover, θ(f) • = [θ(e) • ] − 1 [θ(e)θ(f)] • = (e • ) − 1 θ(ef) • = (e • ) − 1 (ef) • =f •

At first glance, the lemma may appear abstract, but it will be utilized multiple times later, with M representing a superalgebra of M and θ as the inclusion mapping According to Proposition 1.2.7, when extending an afc consistently, one only needs to focus on the primary calculus, ensuring that the consistency of the extended calculus follows automatically.

1 The correct plural of the word calculus is calculi, whence the plural of afc should be again afc

Remark 1.2.8 The extension procedure actually makes use only of the multi- plicative structure of M Except from those statements which explicitly involve addition — the first part of Proposition 1.2.2 c) and Corollary 1.2.3 a) — every- thing remains true when we merely assume that M is a multiplicative monoid extending the multiplicative structure of E ⊂ M In such a setting the above- mentioned statements involving addition can be appropriately modified In fact, call h∈ Ma sumof f, g∈ Mifef +eg=ehfor all e∈ E such that ef, eg∈ E, and if such elementseexist Then all statements remain true, replacing ‘f+g’ by

In the context of algebra, it is essential to adjust the concept of morphism to reflect an algebra homomorphism at the level of E, while recognizing it as a homomorphism of monoids at the level of M.

Meromorphic Functional Calculi

Rational Functions

Recall that forany operatorA with non-empty resolvent set there is a definition of ‘r(A)’ whereris a rational function onCwith all its poles outside ofσ(A) (see

In the context of injective operators, the polynomial p(A) is defined for any polynomial p in the variables z and z − 1 This concept is further explored in Appendix A.7 The subsequent results demonstrate that these general definitions align with any meromorphic functional calculus applied to the operator A.

Proposition 1.3.3 states that if A is a closed operator on a Banach space X with an empty spectrum, and (E(Ω), M(Ω), Φ) represents a meromorphic functional calculus for A, then any rational function r, whose poles are entirely within the spectrum of A, belongs to M(Ω) A Consequently, the expression 'r(A)' is understood in the standard context as outlined in Appendix A.6.

To prove the statement for polynomials \( p \in \mathbb{C}[z] \), we will use induction on the degree of \( p \) For polynomials of degree 0 or 1, the assertion is straightforward Therefore, we assume that the degree of \( p \) is \( n + 1 \), and we can express \( p \) as \( p = zq + a \), where \( a \) is a scalar and the degree of \( q \) is \( n \).

The induction hypothesis implies that q∈ M(Ω) A and D(q(A)) =D(A n ) Since

M(Ω) A is an algebra, p ∈ M(Ω) A and we have B := p(A) ⊃ q(A)A+à By

Theorem 1.3.2 c) we also have D(B)∩D(A) =D(q(A)A) =D(A n+1 ) Hence to prove the assertion we only have to show D(B)⊂ D(A) So let x ∈ D(B) and choose λ ∈ (A) Employing a) and f) of Theorem 1.3.2 we have R(λ, A)B ⊂ BR(λ, A), whenceR(λ, A)x∈D(A)∩D(B) This impliesR(λ, A)x∈D(A n+1 ), from which we readily infer that x∈D(A n ).

In the context of rational functions, Theorem 1.3.2 indicates that for any λ within the set A, the expression (λ−z) − 1 is included in H(A), and (λ−z) − 1 (A) equals R(λ, A) If q is a polynomial in C[z] with all its zeros contained in A, then q − 1 is also in H(A), leading to the conclusion that q − 1 (A) equals q(A) − 1 Consequently, we find that r(A) can be expressed as (p/q)(A) = p(A)q − 1 (A) = p(A)q(A) − 1, which aligns with the definition of r(A) provided in Appendix A.6 The discussion then shifts to the topic of injective operators.

Proposition 1.3.4 Let A be an injective, closed operator on the Banach space X such that(A) =∅, and let(E(Ω),M(Ω),Φ)be a meromorphic functional calculus forA Letp k ∈Z a k z k ∈C[z, z − 1 ]be a polynomial inz, z − 1 Thenp∈ M(Ω) A andp(A) k ∈Z a k A k

According to Theorem 1.3.2 and the assumptions, both z and z - 1 belong to M(Ω) A, which implies that C[z, z - 1] is a subset of M(Ω) A For a polynomial p in C[z, z - 1] that is not in C[z], we can express p(z) as a sum of terms m k= -n a k z k, where a -n is not zero and n is greater than or equal to 1 Theorem 1.3.2 indicates that p(A) can be represented as τ(A) - n p(A)τ(A) n = Λ n (τ n p)(A), with τ(z) defined as z/(λ - z)² and Λ as τ(A) - 1 Consequently, we derive that p(A) = Λ n (τ n p)(A) = Λ n.

12 Chapter 1 Axiomatics for Functional Calculi and this is contained in (λ−A) 2n m+n k=0 a k − n A k

A − n (λ−A) − 2n , by Lemma A.7.2 Applying Lemma A.7.1 and Lemma A.6.1 together with the previous Propo- sition 1.3.3, we see that this is equal to m+n k=0 a k − n A k

Taking into account also other pointsλ∈Csuch thatλ−Ais injective leads us to the following.

Theorem 1.3.5 LetAbe a closed operator on the Banach space X with(A) =∅. Let r be any rational function on C with no pole of r being an eigenvalue of A. Then r(A)is uniquely defined by any meromorphic functional calculus for A.

Proof Let (E(Ω),M(Ω),Φ) be a meromorphic functional calculus for A Then r ∈ M(Ω) A since M(Ω) A is an algebra and z, 1 ,(λ−z) − 1 ∈ M(A) A for all λ /∈ P σ(A) (cf Theorem 1.3.2 b) and f)) Letr =p/q, and fix λ∈(A) Take n >degp,degq Then r(A) =p q(A) (λ−z) n q p

(λ−z) n (A), where we have again used Theorem 1.3.2 Now Proposition 1.3.3 shows thatr(A) is independent of the chosen functional calculus

The previous result has to be complemented by the following remark The discussion of Appendix A.6 shows that any operator A with (A) = ∅ has a meromorphic functional calculus on the whole ofC, with

In the context of primary calculus, we consider rational functions p and q within the set C[z], where the degree of p is less than or equal to the degree of q, and the condition {q=0} is included in the domain A The extended calculus encompasses rational functions r on C that do not have poles, which are also eigenvalues of A Theorem 1.3.5 demonstrates that this calculus is minimal in a specific sense.

An Abstract Composition Rule

The purpose of functional calculus extends beyond merely creating new operators from existing ones; it also enables effective computations with these operators In addition to the fundamental rules outlined in Theorem 1.3.2, possessing the composition rule, expressed as f(g(A)) = (f◦g)(A), is crucial for successful application.

Examples are formulae like e log A =A, (A α ) β =A αβ , log(A α ) =αlog(A) etc.

To understand equation (1.4), it's essential to have a functional calculus for both A and g(A), ensuring that (f◦g) is meaningful Without more information about the functional calculi involved, we cannot fully prove (1.4) in a general context However, in specific cases, such as with sectorial and strip-type operators, we have a clearer framework, as outlined in Theorem 2.4.2 and Theorem 4.2.4 Generally, any issues arise at the level of the primary functional calculus.

Proposition 1.3.6 Let Ω,Ω ⊂Cbe two open subsets of the plane Suppose that

In a Banach space X, let A be a closed operator and (E(Ω), M(Ω), Φ) represent a meromorphic functional calculus for A If g: Ω → Ω is a meromorphic function that belongs to M(Ω) A, and (E(Ω), M(Ω), Φ) also serves as a meromorphic functional calculus for g(A), then it follows that f ◦ g is in M(Ω) A Moreover, the relationship (f ◦ g)(A) = f(g(A)) is valid for all functions f in M(Ω) g(A), provided it holds true for all f in E(Ω).

Note that sinceg(Ω)⊂Ω , the function gis actually holomorphic on Ω.

Proof Define B :=g(A) Letf ∈ M(Ω ) g(A) and e∈ E(Ω ) be a regulariser for f Thenef ∈ E(Ω ) ande(B) is injective By assumptione◦g,(ef)◦g∈ M(Ω) A and (e◦g)(A) =e(B) as well as

This shows that e◦g ∈ H(A) is a regulariser for f ◦g Employing Proposition

The significance of the previous proposition lies in its demonstration that examining primary functional calculi is sufficient for establishing effective composition rules However, this process can still be quite laborious, as highlighted in Theorem 2.4.2.

Multiplication Operators

In this section we make sure that the axiomatics developed in Section 1.2 meet our intuition in the case of multiplication operators.

Consider a σ-finite measure space (Ω,Σ, à) and thereon the scale of spaces

L p (Ω, à), where 1 ≤ p ≤ ∞ We usually omit reference to the underlying σ- algebra Σ Let M(Ω, à;C) be the space of (à-equivalence classes of) measurable

14 Chapter 1 Axiomatics for Functional Calculi complex-valued functions Eacha∈M(Ω, à;C) determines amultiplication oper- ator

M a := (f →af) on each spaceL p (Ω, à), with the maximal domain

It is easy to see thatM a is closed.

Lemma 1.4.1 The operator M a is bounded if and only if a ∈ L ∞ (Ω, à) The operator M a is injective if and only if à(a= 0) = 0 If1 ≤p 0 \) such that the set \( A_c := \{ |a| > c \} \) is not a null set Within \( A_c \), we can identify a subset \( A \) where \( 0 < \mu(A) < \infty \) We define the function \( f \) as \( f := \mu(A) - \frac{1}{p} \cdot 1_A \), where \( 1_A \) is the characteristic function of set \( A \) Consequently, it holds that \( f^p = 1 \).

Hence c≤ M a Taking the supremum overcyieldsa L ∞ ≤ M a

If \( a(a = 0) > 0 \), there exists a set \( A \) such that \( 0 < a(A) < \infty \) and \( a = 0 \) on \( A \), leading to \( M_a 1_A = 0 \) and \( 1_A = 0 \) in \( L^p \) Conversely, if \( M_a \) is not injective, there exists a function \( f \in L^p \) such that \( af = 0 \), indicating that \( \{f = 0\} \subset \{a = 0\} \) For \( 1 \leq p < \infty \) and \( f \in L^p(\Omega, a) \), the sequence \( f 1_{\{|a| \leq n\}} \) converges pointwise to \( f \) as \( n \to \infty \), and by Lebesgue's theorem, this convergence holds in \( L^p \) Furthermore, it is evident that \( f 1_{\{|a| \leq n\}} \in D(M_a) \) for all \( n \in \mathbb{N} \).

We set up a proper abstract functional calculus for M a on X :=L p (Ω, à). Let

The essential range of a function \( a \), denoted as \( K := \{ \lambda \in \mathbb{C} | a \in U > 0 \text{ for each neighbourhood } U \text{ of } \lambda \text{ in } \mathbb{C} \} \), is a closed subset of \( \mathbb{C} \) where \( a(\omega) \in K \) for almost all \( \omega \in \Omega \) We define the measure \( a[\mu] \) on \( K \) such that \( a[\mu](B) = \mu(a \in B) \) for all Borel subsets \( B \subset K \) For \( e \in L^\infty(K, a[\mu]) \), we establish \( \Phi(e) := M_e \circ a \), creating an algebra homomorphism \( \Phi: L^\infty(K, a[\mu]) \to L(X) \) with \( \Phi(1) = I \) The Riemann sphere is denoted as \( C^\infty \), and we embed \( L^\infty(K, a[\mu]) \) into the set \( M(K, a[\mu]; C^\infty) = \{ f: K \to C^\infty | f \text{ is } a[\mu]\text{-measurable} \} \), which forms a multiplicative monoid, not an algebra This leads to a proper abstract functional calculus as noted in Remark 1.2.8.

Theorem 1.4.2 Let (Ω, à),a∈ M(Ω, à;C), and K be as above Then the set K equals the spectrumσ(M a )of the operator M a (cf AppendixA.3) Forλ∈C\K one has

A function \( f \in M(K, a[\infty]; C^\infty) \) is regularizable within the abstract functional calculus if and only if the measure of the set where \( f = \infty \) is zero In this case, it holds that \( f(M_a) = M(f \circ a) \) The first assertion is derived from the equivalence \( \lambda \notin K \) if and only if \( (\lambda - a)^{-1} \in L^\infty \), as stated in Lemma 1.4.1 The second assertion follows from the identity \( f_{L^\infty} = \sup\{ |\lambda| \,|\, \lambda \in \text{essran} f \} \), which is straightforward to establish For a regularizable function \( f \), there exists a measurable and bounded function \( e: K \to \mathbb{C} \) such that \( ef \) is bounded and \( e(A) = M(e \circ a) \) is injective Consequently, the set where \( f = \infty \) is contained within the set where \( e = 0 \), leading to the conclusion that the measure of the set where \( e = 0 \) is zero Conversely, if this condition is satisfied, one can define \( e = \min(1, 1/|f|) \), which is clearly measurable and bounded, ensuring that \( |ef| \leq 1 \).

{e= 0}={f =∞}, whence from the hypothesis and Lemma 1.4.1 it follows that e(A) =M e ◦ a is injective Henceeis a regulariser forf

Note that if λ ∈ K\P σ(A) then, by Lemma 1.4.1, the operatorf(M a ) is independent of the value off at the pointλ.

Concluding Remarks

The findings from earlier sections reveal that once the primary calculus is established, extending it to a broader class of functions becomes relatively simple The key to developing a functional calculus for an operator A on a Banach space X is to first establish this primary calculus, which often necessitates a deep understanding of the operator itself and can be quite complex For those who prefer the Cauchy integral approach, this understanding primarily involves recognizing the growth behavior of the resolvent in relation to the spectrum.

The Volterra operator V, previously discussed, may initially seem to conclude the exploration of its functional calculus However, this perception is misleading, as the resolvent reveals further complexities and insights into the operator's behavior.

16 Chapter 1 Axiomatics for Functional Calculi

R(λ, V) behaves nicely forλ →0 when λis restricted to the left half-plane In fact, the inverse operatorV − 1 is justd/dtwith domain

D(V − 1 ) ={f ∈ C 1 [0,1]|f(0) = 0}, and one can show that both operatorsA=V, V − 1 satisfy a resolvent estimate of the form

In Section 8.5, it is established that both V − 1 and V are sectorial operators with an angle of π/2 This indicates that as one approaches the spectral point 0 along a straight line at an angle ϕ, where π ≥ |argϕ| > π/2, the resolvent of V exhibits optimal growth behavior, remaining bounded Specifically, the growth rate is approximately 1/|λ|² when approaching 0 along the imaginary axis, while it tends to be more exponential when approaching from the right.

Now suppose that one is given a function f, holomorphic on some open sector S ϕ :={z ∈C\ {0} |0 < |argz| < ϕ}, where ϕ∈ (π/2, π) and such that

|f(z)|=O(|z| α ) asz →0, for someα > 0 Then the Cauchy integral (1.1) still makes sense, Γ being a contour as in Figure 1 below.

In Chapter 2, we establish a foundational calculus that includes germs of functions holomorphic at zero, as well as those holomorphic within a specific sector S ϕ, exhibiting favorable behavior at zero Notably, we can define fractional powers V α (where Reα > 0) and the logarithm log V through this functional calculus, as discussed in Chapter 3.

The future of the Volterra operator remains uncertain, as a precise formulation of a 'maximal functional calculus' for individual operators has yet to be established Currently, progress has been made in developing primary calculi for specific classes of operators, including bounded operators, those with particular resolvent growth conditions, and semigroup generators.

Comments

This chapter is primarily based on the introduction of abstract functional calculus as presented in [107] This approach systematically formalizes the constructions commonly found in existing literature Consequently, Fundamental Theorem 1.3.2 encapsulates the core principles that an unbounded functional calculus must fulfill.

In this discussion, we focus on a 'purely commutative' approach, specifically considering algebras of scalar functions However, with some straightforward modifications, it is possible to establish an abstract framework for 'operator-valued' functional calculi, which accommodates a degree of non-commutativity To illustrate this, we consider the mapping Φ: E → L(X) under the condition that supλR(λ, V) < ∞.

Figure 1: The contour Γ surrounds the region of rapid growth of R ( , V ). a homomorphism of algebras and that E ⊂ M as before, but E (and hence M) need not be commutative Define thecentreofE to be

An element \( f \in M \) is defined as regularisable if there exists \( e \in Z(E) \) such that \( e \cdot f \) is injective and \( e f \in E \) This adjustment means that regularisers must now be selected from \( Z(E) \) Most results continue to hold true, except for those that explicitly require commutativity to be included in the assumptions, as noted in Corollary 1.2.4.

In later chapters, readers can explore additional insights on abstract functional calculi, specifically in Section 2.6.1, which covers dual functional calculi, and Section 2.6.3, focusing on approximations of functional calculi.

A comprehensive historical overview of the origins and evolution of functional calculus will be presented in Section 2.8 of the following chapter, following the development of the functional calculus specifically for sectorial operators.

The Functional Calculus for Sectorial Operators

Section 2.1 explores the fundamental theory of sectorial operators, highlighting key examples and the concept of sectorial approximation In Section 2.2, we establish notation for specific spaces of holomorphic functions defined on sectors A functional calculus for sectorial operators is developed in Section 2.3, following the abstract framework outlined in Chapter 1, where essential properties such as the composition rule are demonstrated Section 2.5 extends the functional calculus to broader function spaces, particularly when the operator is bounded and/or invertible, creating a comprehensive overview of functional calculi Mixed topics, including adjoints and restrictions of sectorial operators, along with significant boundedness results and initial approximation findings, are addressed in Section 2.6 Finally, Section 2.7 presents a spectral mapping theorem.

Sectorial Operators

Examples

Let us browse through a list of examples.

In a σ-finite measure space (Ω, Σ, μ), let a be a measurable function in Mes(Ω, μ; C), and define A as the multiplication operator on L^p(Ω, μ) According to Theorem 1.4.2, the essential range K is equal to the spectrum σ(A) We assert that the operator A is sectorial with an angle ω in the range [0, π) if and only if K is a subset of S_ω This condition is necessary, and we further explore its implications.

R(λ, A) = 1 dist(λ, K) ≤ 1 dist(λ, S ω ) for allλ∈C\S ω by Theorem 1.4.2 So the pure location of the spectrum already implies the resolvent growth condition (This is not true for general operators.)

Let −A generate a bounded semigroup (T(t)) t>0 in the sense of Appendix A.8. Then withM := sup t>0 T(t) one has

Reλ (Reλ >0) (2.2) by (the easy part of) the Hille–Yosida theorem (Theorem A.8.6) This shows that

An estimate of the form (2.2) is insufficient to guarantee that -A generates a semigroup, whether M = 1, as seen with the Volterra operator on C[0,1], or if A is densely defined However, there exists a strong correlation between the generators of bounded analytic semigroups and sectorial operators with angles strictly less than π/2 For further details, refer to Section 3.4.

Normal Operators and Numerical Range Conditions

Let X =H be a Hilbert space, and let A be a normal operator on H By the Spectral Theorem D.5.1 the operatorAis similar to a multiplication operator on someL 2 -space Hence by our first example,Ais sectorial of angleω if and only if σ(A)⊂S ω

If an operator A is not normal but its numerical range W(A) is contained within Sω for some ω in the range [0, π/2], then A is considered sectorial with angle ω if and only if there exists a resolvent point outside the sector This concept is elaborated in Appendix C.3, while a more comprehensive discussion on Hilbert space theory is presented in Chapter 7.

For more concrete examples see Chapter 8.

Sectorial Approximation

The present section formalises an important approximation tool in the theory of sectorial operators.

A uniformly sectorial sequence \((A_n)_{n \in \mathbb{N}}\) of angle \(\omega\) is defined as a sectorial approximation on \(S_\omega\) for the operator \(A\) if \(\lambda \in (A)\) and \(R(\lambda, A_n) \to R(\lambda, A)\) in \(L(X)\) for some \(\lambda \notin S_\omega\) According to Proposition A.5.3, this condition holds for all \(\lambda \notin S_\omega\), indicating that the operator \(A\) itself is sectorial with angle \(\omega\).

If (A n ) n is a sectorial approximation for A on S ω , we write A n → A(S ω ) and speak ofsectorial convergence.

Proposition 2.1.3 a) If A n → A(S ω ) and all A n as well as A are injective, thenA − 1 →A − 1 (S ω ). b) If A n → A(S ω ) and A ∈ L(X), then A n ∈ L(X) for large n ∈ N, and

If \( A_n \) converges to \( A \) in norm and \( 0 \in A \), then for sufficiently large \( n \), \( 0 \) will also belong to \( A_n \) Furthermore, if the sequence \( (A_n) \) is uniformly sectorial with angle \( \omega \) and converges to \( A \) in norm, it follows that \( A_n \) converges to \( A(S_\omega) \) Additionally, if \( A \) is in the sector \( Sect(S_\omega) \), then \( (A + \epsilon) \) for \( \epsilon > 0 \) serves as a sectorial approximation for \( A \) on \( S_\omega \) Lastly, if \( A \) is in \( Sect(S_\omega) \), then the sequence \( (A_\epsilon) \) for \( 0 < \epsilon \leq 1 \) is also relevant.

A ε := (A+ε) (1 +εA) − 1 , is a sectorial approximation for Aon S ω

1 Obviously this concept extends to nets However, in all relevant situations in this book everything can be reduced to sequences.

The proof begins with the assertion that if \( A_n \) converges to \( A(S \omega) \) and \( A \) belongs to \( L(X) \), then the sequence \( (1 + A_n)^{-1} \) converges to \( (1 + A)^{-1} \) in norm Since the set of bounded invertible operators on \( X \) is open and the inversion mapping is continuous, it follows that \( (1 + A_n)^{-1} \) is eventually invertible, confirming that \( (1 + A_n) \) converges to \( (1 + A) \) in norm Furthermore, by defining \( B_n := A^{-1} \) and \( B := A^{-1} \), we recognize that both \( B_n \) and \( B \) may be multi-valued operators However, applying the Spectral Mapping Theorem for the resolvent and relevant lemmas, we establish that \( (1 + B)^{-1} \) and \( (1 + B_n)^{-1} \) are contained in \( L(X) \).

(1 + B n ) − 1 → (1 +B) − 1 in norm With the same argument as in the proof of b) we conclude thatB n ∈ L(X) for largen∈N. d) Suppose that (A n ) n is uniformly sectorial with A n → A in norm Then

(1 +A n ) → (1 +A) in norm and sup n (1 +A n ) − 1 < ∞ This implies that

The expression (1 + A) − 1 is an element of L(X), and it has been established that (1 + A^n) − 1 converges to (1 + A) − 1 in norm Furthermore, if A is sectorial, the uniform sectoriality of (A + ε) for ε > 0 has been demonstrated in the proof of Proposition 2.1.1 f) It is evident that (1 + ε + A) − 1 also converges to (1 + A) − 1 in norm, which follows from Proposition 2.1.1 f) and the relevant identity.

Remark 2.1.4 Although we have assumed throughout the present section thatA is single-valued, thedefinitionof sectoriality makes perfect sense even ifAis multi- valued Admittedly, we deal mostly with single-valued operators, but sometimes it is quite illuminating to have the multi-valued case in mind Therefore, we speak of a ‘multi-valued, sectorial operator’ whenever it is convenient Note that the fundamental identity (2.1) still holds in the multi-valued case and readily implies that the inverse of a sectorial operator is sectorial, with the same angle One has x∈D(A) ⇔ lim t →∞ t(t+A) − 1 x=xin the multi-valued case as well (cf c) of Proposition 2.1.1) This showsA0∩D(A) = 0 Most statements of this section remain true in the multi-valued case, at least after adapting notation a little As a rule, one has to replace expressions of the formB(B+λ) − 1 byI−λ(B+λ) − 1 For example we obtain that A ε = 1/ε−

(1 +εA) − 1 is a sectorial approximation ofAby bounded and invertible operators (cf Proposition 2.1.3 f)). Note also that part k) of Proposition 2.1.1 holds even without the assumption that

Ais densely defined In this case the adjointA is a multi-valued sectorial operator.

Spaces of Holomorphic Functions

In this section, we develop a functional calculus specifically for sectorial operators, following the methodology outlined in Chapter 1 This primary calculus is established using a Cauchy integral approach, leveraging the unique properties of the spectrum associated with sectorial operators.

In the study of holomorphic functions, the operator is typically unbounded, necessitating integration along infinite lines, specifically the boundary of a sector This process is feasible only for a limited set of functions, prompting the introduction of specific notation to address these functions effectively.

In Chapter 1, we denote O(Ω) as the space of all holomorphic functions and M(Ω) as the space of all meromorphic functions within the open set Ω ⊂ C Let A represent a sectorial operator with angle ω on a Banach space X Our goal is to define operators of the form f(A) := 1.

The integral \( 2\pi i \int_{\Gamma} f(z) R(z, A) dz \) is defined for functions \( f \) belonging to the space \( O(S_{\phi}) \), where \( \phi \) is in the range \( (\omega, \pi] \) The contour \( \Gamma \) encircles the sector \( S_{\omega} \) in a positive orientation, indicating that it intersects the point at infinity on the Riemann sphere To ensure the integral is well-defined, the function \( f \) must exhibit rapid decay as it approaches infinity, leading to the concept of a polynomial limit.

Let ϕ be in the range (0, π], and consider a function f belonging to the class M(S ϕ ) We define f to have a polynomial limit c in the complex numbers at 0 if there exists a positive α such that the difference f(z) - c behaves like O(|z|^α) as z approaches 0 Additionally, we say that f has a polynomial limit of infinity at 0 if the reciprocal function 1/f has a polynomial limit of 0 at that point Furthermore, f is said to have a polynomial limit d at infinity if the transformed function f(z - 1) has a polynomial limit d at 0.

We say thatf has afinite polynomial limitat 0 (at∞) if there isc∈Csuch that f has polynomial limit c at 0 (at ∞) If f has polynomial limit 0 at 0 (at

∞), we call f regularly decayingat 0 (at∞).

Clearly, if f, g both have finite polynomial limits at 0 (at ∞) also f g and f +g do so.

For a function \( f \) to possess a polynomial limit at infinity, it must have a limit within the complex plane \( \mathbb{C} \) that is approached at least polynomially fast It is important to note that if this limit is infinity, it does not necessarily mean that \( f \) is polynomially bounded at infinity For example, when considered as a function on \( S_\phi \) with \( \phi \in [0, \pi/2) \), the function \( e^z \) has a polynomial limit of infinity at infinity, yet it is not polynomially bounded.

2) If f is meromorphic at 0, then f has a polynomial limit at 0 and this limit is finite if and only iff is holomorphic at 0 The same remark applies to the point∞.

The Cauchy integral highlights the significance of sectoriality in relation to the function f, which ensures integrability at infinity, particularly when the contour Γ becomes eventually straight This property is equally applicable when f exhibits regular decay at infinity Consequently, it is pertinent to examine the Dunford–Riesz class on S ϕ, which is defined under these conditions.

H 0 ∞ (S ϕ ) :={f ∈H ∞ (S ϕ )|f is regularly decaying at 0 and at∞}, where

H ∞ (S ϕ ) :={f∈ O(S ϕ )|f is bounded} is the Banach algebra of allbounded, holomorphic functionsonS ϕ , endowed with the norm f ∞ =f ∞ ,S ϕ = sup{|f(z)| |z∈S ϕ }.

ObviouslyH 0 ∞ (S ϕ ) is an algebra ideal in the algebra H ∞ (S ϕ ) With each f(z) also the function f(1/z) is contained in H 0 ∞ (S ϕ ) The following description is often helpful.

Lemma 2.2.2 Let ϕ∈(0, π], and let f :S ϕ −→Cbe holomorphic The following assertions are equivalent:

(ii) There isC≥0ands >0such that|f(z)| ≤Cmin

(iii) There isC≥0 ands >0 such that|f(z)| ≤C | z | s

(iv) There isC≥0 ands >0 such that|f(z)| ≤C

Proof The proof is easy and we omit it

It is obvious that neither the rational function (1 +z) − 1 nor the constant function1is contained inH 0 ∞ (S ϕ ) We therefore define

⊕ 1 called the extended Dunford–Riesz class This set is in fact a subalgebra of

Lemma 2.2.3 Let ϕ∈(0, π], and let f :S ϕ −→Cbe holomorphic The following assertions are equivalent:

(ii) The function f is bounded and has finite polynomial limits at0 and∞.

Proof The implication (i)⇒(ii) is obvious To prove the converse, suppose that f ∈H ∞ (S ϕ ) has finite polynomial limits at 0 and∞ Then the function g(z) : f(z)−f(∞)−[f(0)−f(∞)]/(1 +z) is contained inH 0 ∞ (S ϕ )

We remark that both algebrasH 0 ∞ (S ϕ ) andE(S ϕ ) are invariant under inver- sion, i.e., withf(z) alsof(z − 1 ) belongs to the set This is clear forH 0 ∞ (S ϕ ) For the larger algebra it follows from

If the precise sector S ϕ is understood from the context or unimportant, we simply writeH 0 ∞ instead ofH 0 ∞ (S ϕ ) in the sequel The same applies to the other function spaces.

In this example, we consider a function \( f \) that belongs to the space \( H^\infty(S_\phi) \), which is regularly decaying at infinity and has a holomorphic extension in the vicinity of zero This implies that there exists a constant \( C \) such that the difference \( |f(z) - f(0)| \) is bounded by \( C|z| \) for values of \( z \) close to zero Consequently, this behavior indicates that \( f \) has a finite polynomial limit at the point zero, establishing that \( f \) is indeed an element of \( E(S_\phi) \).

An important special case is the functione − z , provided one takesϕ∈(0, π/2).

Example 2.2.5 Let 0 0 We denote by Γ ϕ :=∂S ϕ the boundary of the sector

S ϕ , oriented in the positive sense, i.e., Γ ϕ :=−R+ e iϕ ⊕R+ e − iϕ

2πi Γ ω f(z)R(z, A)dz, (2.5) where ω ∈ (ω, ϕ) is arbitrary A standard argument using Cauchy’s integral theorem shows that this definition is actually independent ofω (Figure 3 below illustrates the definition of f(A).)

Lemma 2.3.1 Let A∈Sect(ω), and let ϕ∈(ω, π) Then the following assertions hold. a) The mapping (h→h(A)) :H 0 ∞ (S ϕ )−→ L(X)is a homomorphism of alge- bras. b) If B is a closed operator commuting with the resolvents ofA, thenB com- mutes withf(A) In particular,f(A)commutes withAand withR(λ, A)for all λ∈(A). c) We have R(λ, A)f(A) = ((λ−z) − 1 f)(A)for each λ /∈S ϕ

Proof a) is a simple application of Fubini’s theorem and the resolvent identity. b) is trivial. c) Defineg:= (λ−z) − 1 f and Γ := Γ ω Then

2πi Γ g(z)dz=f(A) since the latter summand equals 0 by Cauchy’s theorem Γ = Γ ω f ( A ) = 1

Figure 3: Integration into 0 is possible due to rapid decay of f

We expand the definition of \( f(A) \) from \( f \in H_0^\infty(S_\phi) \) to encompass all \( f \in E(S_\phi) \) by defining \( g(A) := f(A) + c(1 + A) - 1 + d \), where \( g = f + c(1 + z) - 1 + d \) with \( f \in H_0^\infty(S_\phi) \) and \( c, d \in \mathbb{C} \) This approach establishes an algebra homomorphism \( \Phi_A := (g \mapsto g(A)) : E(S_\phi) \to L(X) \), leading to a meromorphic functional calculus \( (E(S_\phi), M(S_\phi), \Phi) \) as outlined in Section 1.3, although further work is required to fully validate this result.

Let us (for the moment) introduce the notation

H (0) ∞ (S ϕ ) :={f ∈H ∞ (S ϕ )| f is regularly decaying at∞ and holomorphic at 0}.

We saw in Example 2.2.4 thatH (0) ∞ (S ϕ )⊂ E(S ϕ ) The next lemma shows that for f ∈H (0) ∞ (S ϕ ) the value of f(A) can also be computed by a Cauchy integral.

Lemma 2.3.2 Let f ∈ H (0) ∞ (S ϕ ), ω ∈ (ω, ϕ), and let δ >0 be small enough so that f is holomorphic in a neighbourhood of B δ (0) Then f(A) = 1

2πi Γ f(z)R(z, A)dz whereΓ = Γ ω ,δ is the positively oriented boundary of S ω ∪B δ (0).

Figure 4 below gives an impression of what is going on.

To prove the statement, we start with the assumption that if \( f \in H(0) \infty (S \phi) \cap H_0 \infty (S \phi) \), we can shrink the path around the point 0 to 0 without altering the integral's value A function \( f \in H(0) \infty \) can be expressed as \( f = g + c(1 + z)^{-1} \), where \( g \in H(0) \infty \cap H_0 \infty \) Our focus is on demonstrating the claim for \( f(z) = (1 + z)^{-1} \) We introduce the contour \( \Gamma := -\Gamma \omega, R \) with \( R > 1 \) According to Cauchy's theorem, the integral over \( \Gamma (1 + z)^{-1} R(z, A)dz \) equals 0 By shifting the path left without changing the integral's value, we find that it approaches 0 Adding the integrals over both \( \Gamma \) and the initial integral results in cancellations, leaving a simple closed curve around the singularity at -1 Thus, by Cauchy's theorem, we conclude that \( -R(-1, A) = (1 + A)^{-1} \) as the final outcome.

Theorem 2.3.3 LetA∈Sect(ω)onX, and let ϕ∈(ω, π) The mapping Φ A := (g−→g(A)) :E(S ϕ )−→ L(X) defined by (2.6) is a homomorphism of algebras Moreover, it has the following properties: a) z(1 +z) − 1

In the context of complex analysis, the contour of integration avoids the singularity at zero when the function \( f \) is analytic in that region Additionally, if \( B \) is a closed operator that commutes with the resolvents of \( A \), it follows that \( B \) also commutes with \( f(A) \) for any function \( f \) in the space \( E(S_\phi) \); notably, this means that \( f(A) \) commutes with \( A \) Furthermore, for any vector \( x \) in the null space \( N(A) \) and for any function \( f \) in \( E(S_\phi) \), the relationship \( f(A)x = f(0)x \) holds true Lastly, if \( B \) represents the injective part of \( A \), then the range \( Y = R(A) \) remains invariant under the action of each \( f(A) \), leading to the conclusion that \( f(B) = f(A)|_Y \).

In this article, we examine two functions \( g_i = f_i + c_i(1 + z)^{-1} + d_i \) for \( i = 1, 2 \) within the space \( E \), where \( f_i \in H_{0}^{\infty} \) and \( c_i, d_i \in \mathbb{C} \) Given that \( \Phi \) is linear, we must consider all mixed products The involvement of a constant \( d_i \ requires no additional proof, while the products \( f_1 f_2 \) and \( f_i (1 + z)^{-1} \) are addressed using Lemma 2.3.1 The assertion for the remaining product \( (1 + z)^{-1}(1 + z)^{-1} = (1 + z)^{-2} \) follows from Lemma 2.3.2, Fubini's theorem, and the resolvent identity Consequently, we establish that \( (z(1 + z)^{-1})(A) = 1(A) - (1 + z)^{-1}(A) I - (1 + A)^{-1} A(1 + A)^{-1} \) The proof for part b is straightforward, and for part c, for \( x \in N(A) \) and \( z \in (A) \), it holds that \( R(z, A)x = (1/z)x \), leading to the conclusion that \( f(A)x = 1 \).

2πi Γ f(z) z dz x= 0ãx= 0 forf ∈H 0 ∞ , by Cauchy’s theorem The rest follows. d) follows from the fact thatY isR(λ, A)-invariant, withR(λ, A)| Y =R(λ, B) for allλ∈(A)

We call the algebra homomorphism Φ A = (f −→f(A)) :E(S ϕ )−→ L(X) (2.7) theprimary functional calculus(in short: pfc) onS ϕ forAas a sectorial operator.

Remark 2.3.4 Note that the definition of this primary calculus is perfectly mean- ingful even if A is multi-valued Theorem 2.3.3 holds true (after reformulating part a) appropriately) even in this more general case We shall not use this fact except for Section 3.4 when we treat holomorphic semigroups.

We close this section with a result similar to Lemma 2.3.2.

Corollary 2.3.5 Let A∈Sect(ω),ϕ∈(ω, π), and let f ∈ O(S ϕ )be holomorphic at both points 0 and∞ Thenf ∈ E(S ϕ )andf(A)is given by f(A) =f(∞) + 1

2πi Γ f(z)R(z, A)dz where the contour Γ = Γ ω ,δ,R is as is shown in Figure5 (with ω ∈(ω, ϕ), δ >0 small and R >0 large enough).

To prove the result, we define \( g(z) = f(z) - f(\infty) \), which belongs to the space \( H(0)_{\infty} \) This allows us to express \( f(A) \) as \( f(\infty) + g(A) \) Utilizing Lemma 2.3.2, we compute \( g(A) \) as \( 2\pi i \int_{\Gamma_{\omega, \delta}} g(z) R(z, A) dz \) By applying Cauchy’s theorem, we can substitute the infinite integration path \( \Gamma_{\omega, \delta} \) with a sufficiently large finite path \( \Gamma_{\omega, \delta, R} \) A further application of Cauchy’s theorem leads to the cancellation of the constant \( f(\infty) \), completing the proof.

The Natural Functional Calculus

As before, letA∈Sect(ω) andϕ∈(ω, π) By Theorem 2.3.3 we have established an abstract functional calculus

(E(S ϕ ),M(S ϕ ),Φ A ), which is proper since (1 +z) − 1 (A) = (1 +A) − 1 is injective In fact, by Theorem 2.3.3 a) the function (1 +z) − 1 regulariseszand z(A) = (1 +A)(z(1 +z) − 1 )(A) = (1 +A)A(1 +A) − 1 =A.

Figure 5: The function f is holomorphic outside the dashed line, including the point ∞

Hence in the terminology of Section 1.3 our abstract functional calculus is even a meromorphicfunctional calculus forAonS ϕ We write

In the context of the abstract functional calculus, for a function \( f \) belonging to the class \( M(S ϕ) \), there exists an injective operator \( e \) from the set \( E(S ϕ) \) such that \( e(A) \) is also in \( E(S ϕ) \) The operator \( f(A) \) is defined as \( f(A) := e(A) - 1 (ef)(A) \), where \( e \) serves as an arbitrary regularizer for \( f \) This framework is essential for understanding the regularizability of functions within this calculus.

Lemma 2.3.6 Let A∈Sect(ω),ϕ∈(ω, π), andf ∈ M(S ϕ ) A Then one can find a regulariser e∈ E(S ϕ )for f with e(∞) = 0 If A is injective, one can even find a regulariser e∈H 0 ∞ (S ϕ ).

Proof If d is any regulariser for f then e := (1 +z) − 1 d is a regulariser as well which moreover satisfies e(∞) = 0 If A is injective, thene :=z(1 +z) − 2 dis a regulariser forf withe∈H 0 ∞

Since we have a meromorphic functional calculus forA, it follows from The- orem 1.3.2 that

(à−z)(A) =à−A, (λ−z) − 1 (A) =R(λ, A) (2.8) for all à ∈C andλ∈ (A) More generally,r(A) has its usual meaning (as ex- plained in Appendix A.6) for every rational functionrwith poles in (A) (Propo- sition 1.3.3).

We can select ϕ to be arbitrarily close to ω, allowing us to view M(S ϕ 1 ) as a subalgebra of M(S ϕ 2 ) when ϕ 1 > ϕ 2 > ω Consequently, we also have E(S ϕ 1 ) ⊂ E(S ϕ 2 ) These relationships indicate that the corresponding functional calculi are consistent, leading to a morphism as outlined in Proposition 1.2.7.

Writing E[S ω ] : ϕ ∈ (ω,π) E(S ϕ ) andM[S ω ] : ϕ ∈ (ω,π) M(S ϕ ) we thus (as an

‘inductive limit’) obtain a meromorphic functional calculus

(E[S ω ],M[S ω ],Φ) which we call the natural functional calculus for A as a sectorial operator on

S ω Note that again we may viewM[S ω ] as a subalgebra ofM(S ω ) so that this calculus is even a meromorphic functional calculus forAonS ω (in the terminology of Section 1.3) The set

M(S ϕ ) A is called thedomainof this calculus As in Chapter 1 we write

To enhance clarity, we often refer to the sector simply as ‘M A’ rather than ‘M[S ω] A’, as long as there are no ambiguities Additionally, we define the abbreviation O[S ω] A as O[S ω] ∩ M A, along with similar notations for other function classes.

Remarks 2.3.7 1) One should keep in mind that a functionfbelonging toE[S ω ] or M[S ω ] A , is actually defined on a larger sectorS ϕ for some ϕ∈(ω, π].

2) As in Section 1.3 we warn the reader not to forget that our notationsM(S ϕ ) A ,

The accuracy of M[S ω ] A and H(A) is limited, as these sets are influenced not only by A but also by the specific domain of the primary calculus It is important to keep this in mind as we explore additional primary calculi in subsequent sections.

3) To distinguish different functional calculi which may be defined for the same operatorA (even on the same set Ω) we may add the words ‘as a sectorial operator’ to the notion of ‘natural functional calculus’ If for example the operator A is also bounded, then one has in addition the usual Dunford calculus at hand (see the Preface) This calculus could be named the ‘natural functional calculus forAas a bounded operator’ If Ais not invertible then z 1/2 is in the domain of the natural funcional calculus forA as a sectorial, but not as a bounded operator Similarly, the functione 1/(r − z) forr > r(A) is in the domain of the natural functional calculus forA as a bounded, but not as a sectorial operator.

The general properties of the so-defined ‘natural functional calculus’ forAas a sectorial operator are summarised in Theorem 1.3.2 with Ω :=S ω (and, in fact,

M[S ω ] A instead ofM(S ω ) A ) We often shall use them without further mention.

Which functionsf are contained inM A ? The next lemma shows that a limit behaviour at 0 is necessary ifAis not injective.

Lemma 2.3.8 Let A∈ Sect(ω),f ∈ M A , and λ∈S ω Then at least one of the following assertions holds.

1) There isc∈Candα >0 such thatf(z)−c=O(|z| α )asz→λ.

If A is not injective, thenf(A)x=f(0)xfor allx∈N(A).

Proof Let e ∈ E be a regulariser for f Then also ef ∈ E Suppose that f fails 1), and that λ = 0 Then λ is a pole of f, whence e(λ) = 0 Hence also h(z) := e(z)/(z−λ) ∈ E and e(A) ⊃ h(A)(A−λ) Since e(A) is injective and h(A)∈ L(X),A−λmust be injective.

To complete the proof, suppose λ= 0 and Ais not injective Sincee(A)x e(0)xfor everyx∈N(A) (Theorem 2.3.3) ande(A) is injective,e(0) = 0 Hence f(z)−f(0) = [e(z)f(z)−e(0)f(0)]−f(0)[e(z)−e(0)] e(z) forz near 0, thereby showing the assertion

Corollary 2.3.9 Let X be reflexive, and let A∈Sect(ω)with N(A) = 0 Denote by B the injective part of A Iff ∈ M A , then also f ∈ M B with

The function f(0) is well-defined because the operator A is not injective A regularizer e for f in the functional calculus of A is introduced, ensuring that both e and ef belong to the set E, with e(0) equal to zero Notably, e(A) is injective According to Theorem 2.3.3, for any elements x and y from the direct sum of N(A) and R(A), the expression e(A)(x⊕y) equals e(0)x⊕e(B)y, which implies that e(B) is also injective Consequently, it is established that f is in M B, and the relationship f(A) = e(A) − 1 (ef)(A) = e(0) − 1 ⊕ e(B) − 1 [(ef)(0)⊕(ef)(B)] can be expressed in a suggestive notation as f(0)⊕f(B).

Functions of Polynomial Growth

To determine subclasses of M A , one looks for natural regularisers Taking ele- mentary rational functions leads to the definition

and A[S ω ] : ϕ ∈ (ω,π) A(S ϕ ) If ϕ is understood or not important, we write simplyAinstead ofA(S ϕ ).

Lemma 2.3.10 Forf ∈ O[S ω ]the following assertions are equivalent:

(ii) The function f has the following two properties:

2) f has a finite polynomial limit at0.

(iii) There isc∈C,n∈N, andF∈H 0 ∞ such that f(z) =c+ (1 +z) n F(z).

If f is bounded, one can taken= 1 in(iii).

In particular,A[S ω ] is an algebra of functions containing every rational func- tion with poles outside ofS ω

Proof (i)⇒(ii) Ifg(z) :=f(z)(1 +z) − n ∈ E, then clearly condition 1) is satisfied. Because g has a finite polynomial limit at 0, also f = (1 +z) n g has a finite polynomial limit at 0.

(ii)⇒(iii) Chooseαas in (ii), and letn > α Then (f(z)−f(0))(1 +z) − n ∈H 0 ∞

Proposition 2.3.11 Let A ∈ Sect(ω) Then A[S ω ] ⊂ M A For f ∈ A[S ω ] the following assertions hold. a) If A is bounded, so isf(A). b) One has [(z−à)f(z)](A) = (A−à)f(A)for allà∈C. c) If D(A) =X andf(z)(1 +z) − n ∈ E, thenD(A n )is a core for f(A).

Proof a) It is immediate thatA[S ω ]⊂ M A Let f ∈ A, and choosensuch that

F :=f(1 +z) − n ∈ E IfA is bounded, thenf(A) = (1 +A) n F(A)∈ L(X). b) By (2.8) and Lemma A.6.1 we have

= (A−à)(1 +A) n F(A) = (A−à)f(A). c) Note that the operator F(A) commutes with (1 +A) − n , whence D(A n ) is

F(A)-invariant This gives D(A n ) ⊂ D(f(A)) For arbitrary x ∈ D(f(A)) we haveT t (x) := (t(t+A) − 1 ) n x→xast→ ∞ As a matter of fact,T t (x)∈D(A n ) and — by Proposition 2.1.1 c) — alsof(A)T t x=T t f(A)x→f(A)x

Remark 2.3.12 Interestingly enough, polynomial growth at∞is not the best one can achieve for an arbitrary sectorial operator In fact, we shall see in Section 3.4 that ifω A < π/2 thene − A :=e − z (A) is injective (cf Proposition 3.4.4.) Hencee − z is a regulariser which compensates even exponential growth at ∞ Ifω A happens to be larger than π/2, one can use the functions e − z α with suitable α ∈ (0,1).Therefore we see that O[S ω ] A is much larger than A[S ω ] However, a property like b) of the last proposition cannot be expected for more general functions since its proof rests on the fact that the regulariser is the inverse of a polynomial.

Injective Operators

This section examines an injective, sectorial operator A within the Banach space X, where reflexivity ensures that D(A) = R(A) = X Given that A is injective, the function τ(z) := z(1 + z) − 2 belongs to H₀ ∞ and serves as a regulariser, as τ(A) = A(1 + A) − 2 is also injective We define its inverse as Λ_A := τ(A)⁻¹ = (1 + A)² A⁻¹ = (1 + A) A⁻¹ (1 + A) = (2 + A + A⁻¹) The domain of Λ_A is D(Λ_A) = D(A) ∩ R(A), and the class of functions regularised by powers of τ is established.

B(S ϕ ) :={f :S ϕ −→C|there isn∈N:τ(z) n f(z)∈H 0 ∞ (S ϕ )}, where ϕ ∈(0, π] As usual, we write B :=B[S ω ] : ϕ ∈ (ω,π) B(S ϕ ) Obviously,

B(S ϕ ) is an algebra of functions andA(S ϕ )⊂ B(S ϕ ) A holomorphic functionf onS ϕ belongs toB if and only iff has at most polynomial growth at 0 and at∞

(and is bounded in between) In particular,H ∞ (S ϕ ) is a subalgebra ofB(S ϕ ).

Proposition 2.3.13 Let A∈Sect(ω)be injective Then B[S ω ]⊂ M A Moreover, the following assertions hold. a) If X is reflexive, thenD(A) =R(A) =D(A n )∩R(A n )for every n∈N. b) If D(A) =X =R(A) andf(z)τ n ∈H 0 ∞ , then D(A n )∩R(A n ) is a core for f(A).

Proof a) Let x ∈ X Then A n (t+A) − n (1 +tA) − n x ∈ D(A n )∩R(A n ) By Proposition 2.1.1 we have lim t → 0 A n (t+A) − n (1 +tA) − n x=xifx∈D(A)∩R(A). b) is proved in the same way as c) of Proposition 2.3.11

Remark 2.3.14 A law of the form (zf(z))(A) = Af(A) cannot hold in general for all f ∈ B (cf Proposition 2.3.11 b)) Indeed, by Theorem 1.3.2 f) we have(z − 1 )(A) =A − 1 , but (zz − 1 )(A) = (1)(A) =I=AA − 1 =A[(z − 1 )(A)] in general.

We conclude this section with an important example.

Example 2.3.15 Let (Ω,Σ, à) be aσ-finite measure space, and leta∈M(Ω, à;C) be as in Sections 1.4 and 2.1.1 Let A :=M a be the multiplication operator on

L p (Ω, à) andK := essrana=σ(A) its spectrum We suppose that A∈Sect(ψ), i.e.,K⊂S ψ For simplicity we also suppose thatAis injective, i.e.,à(a= 0) = 0 (Lemma 1.4.1) Takeϕ ∈(ψ, π) Then we have defined two abstract functional calculi

(L ∞ (K, a[à]),M(K, a[à];C ∞ ),Φ1) and (E(S ϕ ),M(S ϕ ),Φ2) where Φ 1 (f) =M f ◦ a and Φ 2 (f) =f(A) Since the set{a= 0} is aà-null set by assumption,M(S ϕ ) ‘embeds’ naturally intoM(K, a[à];C ∞ ) More precisely, one considers the mapping θ: f −→[f

We claim that this mapping is a morphism of abstract functional calculi in the sense of Section 1.2.3.

Proof The only thing to show is that f(A) = M f ◦ a for f ∈ H 0 ∞ (S ϕ ) Take x∈ L p (Ω, à) andx ∈ L p (Ω, à), where p is the dual exponent Using Fubini’s and Cauchy’s theorem we compute x , f(A)x= 1

Sincex was arbitrary, we obtainf(A) =M f ◦ a as desired

The consequence of this fact is that also for the extended calculi one has compatibility, i.e., f(A) =M f ◦ a holds forallf ∈ M(S ϕ ) A

If A is not injective, all the above statements remain true, for the formal argument one however has to replaceM(S ϕ ) by its subalgebra of functions with limits at 0.

The Composition Rule

If \( A \in Sect(\omega) \) is injective, then \( A^{-1} \) is also sectorial and shares the same angle (Proposition 2.1.1) A natural conjecture arises regarding the identity \( f(z - 1)(A) = f(A - 1) \) for \( f \in M[S_\omega] A^{-1} \), provided that the left-hand side is defined In fact, a stronger statement holds true.

Proposition 2.4.1 Let A ∈ Sect(ω) be injective, and let f ∈ M(S ϕ ) for some ϕ∈(ω, π) Then f ∈ M(S ϕ ) A −1 ⇐⇒ f(z − 1 )∈ M(S ϕ ) A and in this casef(A − 1 ) =f(z − 1 )(A).

Proof We employ Proposition 1.3.6 Namely, we have meromorphic functional calculi (H 0 ∞ (S ϕ ),M(S ϕ ),Φ A ) and (H 0 ∞ (S ϕ ),M(S ϕ ),Φ A −1) andg(z) :=z − 1 maps

To prove the assertion for all functions \( f \in H_0^\infty(S_\phi) \), it suffices to apply Proposition 1.3.6, which can be demonstrated through a straightforward change of variables in the defining Cauchy integral This process utilizes the fundamental identity (2.1) and exemplifies the composition rule.

The current rule lacks clarity without additional hypotheses Specifically, it is essential that A is sectorial, g(A) is both defined and sectorial, and that g effectively maps one sector into another.

3) For every ϕ ∈(ω , π) there isϕ∈(ω, π) withg∈ M(S ϕ ) andg(S ϕ )⊂S ϕ Under these requirements obviouslyg(S ω )⊂S ω

Theorem 2.4.2 (Composition Rule) Let the operatorAand the function g satisfy the conditions 1), 2), and3) above Thenf ◦g∈ M[S ω ] A and

When considering the case where g = c is a constant, we find that g(A) = c If c equals 0, the situation simplifies due to Cauchy’s theorem In this scenario, where g(A) = 0, the function f ∈ M[S ω ] g(A) must exhibit 'nice' behavior at 0, as stated in Lemma 2.3.8 Consequently, the composition f◦g is well-defined, resulting in the constant f(0), and according to Lemma 2.3.8, the composition rule is upheld.

Hence in the following we may suppose without loss of generality that g is not a constant Given this, by the Open Mapping Theorem one hasg(S ω )⊂S ω and the stronger property

3) For everyϕ ∈(ω , π) there isϕ∈(ω, π) withg∈ M(S ϕ ) andg(S ϕ )⊂S ϕ

We now appeal to theabstractcomposition rule Proposition 1.3.6 from Chap- ter 1 with the data

Hence it suffices to prove the assertions of Theorem 2.4.2 only for f ∈ E[S ω ] A

Lemma 2.4.3 Let A and g be as in Theorem 2.4.2, and let f ∈ E[S ω ] Then f ◦g∈H ∞ [S ω ] A

In this proof, we assume that the function g is not constant, which leads us to conclude that the composition f◦g belongs to the space H ∞ [S ω ] If the function A is injective, there is nothing further to prove However, if A is not injective, g has a finite polynomial limit g(0) at 0, as stated in Lemma 2.3.8 Additionally, the function f has a finite polynomial limit at f(0) If g(0) equals 0, this conclusion follows from the holomorphy of f at c Consequently, the composition f◦g also has a finite polynomial limit at 0, ensuring that (f◦g)(A) is well-defined.

Lemma 2.4.4 LetA andg be as above, and letf ∈ E[S ω ] Then the composition rule f(g(A)) = (f ◦g)(A)holds.

In this proof, we assume that g is not a constant and consider a function f belonging to the space E[Sω] It can be expressed as a combination of constants c, d, and a function f1 from H0∞[Sω], specifically f = c + d/(1 + z) + f1 If f equals c, there is nothing further to prove; if f equals d/(1 + z), the result follows from Theorem 1.3.2 f) Thus, we can assume without loss of generality that f is in H0∞ For λ not in Sω, the expression (λ - g(z))^(-1) is both bounded and holomorphic on Sϕ Consequently, we conclude that f(g(A)) equals 1.

1 for suitableω 1 ∈(ω , π) We chooseϕ ∈(ω , ω 1 ) and according to 3) we findϕ∈(ω, π) such thatg(S ϕ )⊂S ϕ We consider two cases:

The first case is easier to handle If A is injective we can useτ(z) :=z(1 +z) − 2

2.4 The Composition Rule 43 as a regulariser, let Γ surroundS ω within S ϕ , and compute f(g(A)) = Λ A τ(A)f(g(A)) = Λ A 1

(Recall the definition Λ A :=τ(A) − 1 ) Equality (1) is an application of Cauchy’s integral theorem Before, one has to interchange the order of integration To justify this, note that the function f(λ) (λ−g(z))(1 +z) 2 = λ λ−g(z) f(λ) λ

(1 +z) 2 is product integrable on Γ×Γ since the first factor is uniformly bounded.

In the scenario where the function \( g \) is not injective, it possesses a finite polynomial limit \( c = g(0) \) at zero We define \( g_1(z) = g(z) - g(0) \) and select a regularizer \( e \in E \) such that \( e(\infty) = 0 \) Consequently, \( e g_1 \) belongs to \( H_{0}^{\infty} \) Additionally, for \( \lambda \notin S_{\phi} \), the expression \([(λ - g(z)) - 1 - (λ - c)^{-1}]\) is also regularized by \( e \) This leads us to compute \( f(g(A)) = e(A)^{-1} e(A) f(g(A)) = e(A)^{-1} 1 \).

The second summand equals f(c) by Cauchy’s theorem The first satisfies e(A) − 1 1

=e(A) − 1 ((f ◦g)(A)−f(c))e(A) = (f◦g)(A)−f(c) where we used Fubini’s theorem in (1) and Cauchy’s theorem in (2) To justify the application of Fubini’s theorem in (1) one has, after estimating the resolvent, to consider the function

(λ−c)(λ−g(z))z and prove its product integrability The representation

(g 1 e)(z) z shows thatc= 0 is harmless since λ/(λ−g(z)) is uniformly bounded because of the conditionsg(z)∈S ϕ and λ∈Γ (Recall that (eg 1)∈H 0 ∞ ) Ifc= 0 (hence g 1=g) we write

In the context where g has no poles within the region S ϕ, we select α from the interval (0,1) to ensure the first factor maintains integrability Consequently, the middle term remains uniformly bounded It is also evident that the expression eg α belongs to the space H 0 ∞, which confirms that F is integrable.

This concludes the proof of Theorem 2.4.2 We shall encounter several ap- plications of the composition rule throughout the remaining parts of this book, in particular in Chapter 3.

Extensions According to Spectral Conditions

Invertible Operators

Let A be an invertible element in Sect(ω) The ball of radius r centered at 0 is contained within the spectrum of A, where the inverse of r is equal to the spectral radius of A inverse When establishing a functional calculus for A on a sector defined by the angle ϕ, which lies between ω and π, the behavior of functions at the origin is irrelevant.

For A ∈ Sect(ω) satisfying 0∈ (A) andf ∈ E ∞ (S ϕ ) (ϕ∈ (ω, π]) we define as usual Φ(f) :=f(A) := 1

2πi Γ f(z)R(z, A)dz, where the path Γ bounds a sector S ω (with ω ∈ (ω, ϕ)), except for the region near 0 where it avoids 0 at small distance, cf Figure 6 below.

As a matter of fact, one can prove in a similar way a theorem analogous to Theorem 2.3.3 Hence a meromorphic functional calculus (E ∞ (S ϕ ),M(S ϕ ),Φ) for

A onS ϕ is defined As in Section 2.3 one can form the inductive limit asϕ→ω and obtains a meromorphic functional calculus

(E ∞ [S ω ],M[S ω ],Φ) on Ω :=S ω We call it thenatural functional calculusonS ω forAas an invertible, sectorial operator It clearly forms a consistent extension of the natural functional calculus for Aas a sectorial operator (defined in Section 2.3).

Remark 2.5.1 The construction of the calculus for invertible sectorial operators can of course be refined Indeed, one can pass to functionsf which are not even defined near the origin Since this would make necessary a lot more notation we omit it.

In order to prove a composition rule as in Theorem 2.4.2, we have to face several cases, namely the combinations

The proofs are similar to the proof of Theorem 2.4.2 However, in cases 1) and 3) one needs the additional assumption 0∈/g(S ϕ ) on the functiong. Γ f ( A ) = 1

Figure 6: Cauchy integral in the case of invertibility of A

Bounded Operators

Now we are concerned with the ‘dual’ situation, namely with a boundedoperator

A∈Sect(ω) It is then clear that for the expressionf(A) to make sense, the limit behaviour of f at∞is irrelevant Define

ForA∈Sect(ω)∩ L(X) andf ∈ E 0(S ϕ ) (ϕ∈(ω, π]) we define as usual Φ(f) :=f(A) := 1

The integral expression \(2\pi i \Gamma f(z)R(z, A)dz\) describes a contour \(\Gamma\) that bounds a sector \(S_\omega\) (where \(\omega \in (\omega, \phi)\), while avoiding the region near infinity and the spectrum \(\sigma(A)\) If the operator \(A\) is injective, we can derive a meromorphic functional calculus \((E_0(S_\phi), M(S_\phi), \Phi)\) for \(A\) on the sector \(S_\omega\) However, in cases where \(A\) is not injective, as seen with general sectorial operators, further considerations are necessary.

— we have to include also functions which are holomorphic at 0 The easiest way to do this is to pass to the algebraE 0(S ϕ ) :=E 0(S ϕ )⊕C(1 +z) − 1 and define the

2.5 Extensions According to Spectral Conditions 47 primary calculus by Φ f + c

1 +z :=f(A) +c(1 +A) − 1 , where f ∈ E 0(S ϕ ) In any case, one ends up with a meromorphic functional calculus forA onS ϕ and by forming the inductive limit asϕ→ω one obtains a meromorphic functional calculus

(E 0[S ω ],M[S ω ],Φ) forAonS ω ; this we call thenatural functional calculusonS ω forAas a bounded, sectorial operator It clearly forms a consistent extension of the natural functional calculus for Aas a sectorial operator (defined in Section 2.3).

Remark 2.5.2 Also in this case the so-obtained calculus is not the most general.

In fact, one can extend it to functions f which are not even defined near ∞.

As above, to prove acomposition ruleas in Theorem 2.4.2, we have to face the following cases:

2) Ainvertible and sectorial,g(A) bounded and sectorial.

5) Abounded and sectorial,g(A) invertible and sectorial.

The proofs resemble those of Theorem 2.4.2; however, additional assumptions about the mapping behavior of g may be necessary Specifically, for cases 1) to 3), it is required that ∞ does not belong to g(S ω), while in case 5), it is essential that 0 is not included in g(S ω).

Bounded and Invertible Operators

If \( A \in \text{Sect}(\omega) \) is both bounded and invertible, the primary calculus can be applied without assumptions regarding the behavior of \( f \) at 0 or \( \infty \) The standard Dunford calculus can be utilized for \( A \), leading to a consistent extension of the natural functional calculus for \( A \) as either an invertible or bounded sectorial operator Furthermore, under appropriate conditions on the function \( g \), composition rules can be established that connect all previously introduced calculi.

We have successfully completed our systematic introduction of functional calculi for sectorial operators The interpretation of the symbol ‘f(A)’—where ‘f’ is a meromorphic function on a sector—depends on the spectral properties of the operator A This mapping, denoted as the natural functional calculus (nfc), applies to bounded and invertible sectorial operators The definition of ‘f(A)’ relies significantly on the primary calculus employed, leading to distinct interpretations for the nfc in the context of general sectorial operators versus bounded sectorial operators In subsequent chapters, we will primarily focus on the nfc for sectorial operators.

Miscellanies

Adjoints

In the context of Banach spaces, let A represent a multi-valued, sectorial operator Its adjoint operator is also a multi-valued, sectorial operator, as demonstrated in Proposition 2.1.1 k) and Remark 2.1.4 This raises the question of how the functional calculi of A and its adjoint A are interconnected, prompting an abstract exploration of their relationship.

Let (E,M,Φ) be a proper abstract functional calculus over the Banach space

X Then bye ∗ := Φ(e) = (e • ) a homomorphism Φ ∗ : (e−→e ∗ ) :E −→ L(X ) is defined Hence (E,F,Φ ∗ ) is an abstract functional calculus overX , the so- called dual (functional) calculus This afc is proper if and only if there ise∈ E such that R(e • ) = X It is now natural to ask, for which f ∈ M the identity f ∗ = (f • ) holds We obtain the following result.

Proposition 2.6.1 Let (E,M,Φ)be an afc with dual calculus (E,M,Φ ∗ ), and let f ∈ M Suppose that there is a Φ-regulariser e ∈ E for f with the following property: There exists (f n ) n ⊂ M b such that

Then the space R(e • )is dense in X, it is a core for f • , andf ∗ = (f • )

Proof It is immediately clear from 1) and 2) that R(e • ) is dense in X Hencee is a regulariser for f in the dual calculus Let x ∈ D(f • ) and y :=f • x Then x n :=f n • x→xandx n ∈R(e • )⊂D(f • ) Moreoverf • x n =f • f n • x=f n • y→y.

Sincex∈D(f • ) was arbitrary, we see thatR(e • ) is a core forf • This means that (ef) • e − • 1 =f • Using this we compute f ∗ = (e ∗ ) − 1 (ef) ∗ = [(e • ) ] − 1 [(ef) • ] (1) = [e − • 1 ] [(ef) • ]

We used b), k) and a) of Proposition A.4.2 for (1), (2), and (3), respectively

Let us return to sectorial operators Suppose A∈Sect(ω) and D(A) =X. This extra condition ensures that A is again single-valued, whence also A ∈

Sect(ω) Note that ifA is in addition bounded or invertible, the same is true for

A Hence the primary functional calculi forAandA have the same domain.

Lemma 2.6.2 Let A∈Sect(ω) with D(A) =X Then the pfc for A is the dual of the pfc for A, i.e., f(A ) = f(A) , whenever f(A) is defined by the primary functional calculus for A.

Recall that the hypothesis onf means: f ∈ E[S ω ] ifAis neither bounded nor invertible, f ∈ E ∞ [S ω ] ifA∈Sect(ω) is invertible andf ∈E 0[S ω ] ifA∈Sect(ω) is bounded.

Proof Letf be as required Then we may writef(A) = 2πi 1 Γ f(z)R(z, A)dzfor some contour Γ Hence f(A) 1 2πi Γ f(z)R(z, A)dz

Combining this result with Proposition 2.6.1 we can prove the identity f(A) =f(A ) for certain classes of functionsf.

Proposition 2.6.3 LetA∈Sect(ω)withD(A) =X, and letf ∈ M[S ω ] Then the identity f(A) =f(A ) holds in the following cases: a) f ∈ A; b) R(A) =X andf ∈ B.

Proof In the case a) f is regularisable by e := (1 +z) − m for some m Since

D(A) is dense, the functionsf n := [n(n+z) − 1 ] m satisfy conditions 1) and 2) of Proposition 2.6.1, whence the statement follows The case b) is treated similarly, with e = z m /(1 +z) 2m for some m ∈ N Note that the additional condition

R(A) =X ensures that both AandA are injective

Remarks 2.6.4 1) As a matter of course similar results hold for bounded sec- torial and for invertible sectorial operatorsA.

2) Even if nothing more than the sectoriality ofAis required, the theorem holds for a much larger class than A Indeed, suppose ω A < π/2 for simplicity.

Then, since D(A) is dense, e − (1/n)A → I strongly (cf Proposition 3.4.1).Therefore even subexponential growth is admissible.

Restrictions

LetAbe a sectorial operator on the Banach spaceX, and letY be another Banach space, continuously embedded into X Let A Y :=A

(cf Proposition A.2.8) In general, A Y is not necessarily sectorial However, if it is, one may ask howf(A Y ) is computed fromf(A) The answer, given in the next proposition, is not surprising.

Proposition 2.6.5 Let A ∈ Sect(ω) on the Banach space X, and let Y ⊂ X be another Banach space, continuously included in X If A Y ∈ Sect(ω) then the following assertions hold: a) If e∈ E[S ω ], then Y is invariant undere(A) ande(A Y ) =e(A)

Proof a) Since A Y is sectorial of angle ω on Y, the space Y must be invariant under all resolvents R(λ, A),λ /∈S ω , with R(λ, A Y ) =R(λ, A)

Y for theseλ If e∈ E[S ω ], then by Proposition A.2.8 we havee(A Y ) =e(A)

The function \( Y \) can be expressed as \( f(z) = d + c(1 + z) - 1 + h(z) \), where \( h(A) \) is defined through a Cauchy integral involving resolvents To find a regularizer for a general function \( f \), ensure that \( e \) and \( ef \) are elementary and that \( e(A) \) is injective Consequently, it follows that \( e(A Y) = e(A) Y = e(A) \).

Y is injective Furthermore, (x, y)∈f(A)∩Y ⊕Y ⇔ x, y∈Y, (ef)(A)x=e(A)y ⇔ x, y ∈Y, (ef)(A Y )x e(A Y )y ⇔ (x, y)∈f(A Y )

Remark 2.6.6 Proposition 2.6.5 has an obvious analogue for the class of invertible(bounded, bounded and invertible) sectorial operators.

Sectorial Approximation

Suppose that A ∈Sect(ω) on X and that (A n ) n is a sectorial approximation of

A on S ω By Proposition 2.1.3, if A is bounded and/or invertible, the same is true for eventually allA n HenceAand theA n eventually have the same primary functional calculus.

Lemma 2.6.7 Let A∈Sect(ω), and let (A n ) n be a sectorial approximation of A on S ω Then f(A n ) → f(A) in norm whenever f(A) is defined by the primary functional calculus for A.

Proof Letf be as required Then we can writef(A) = 2πi 1 Γ f(z)R(z, A)dz for a certain contour Γ Hence lim n f(A n ) = lim n

2πi Γ f(z)R(z, A)dz=f(A) by an easy application of Lebesgue’s Dominated Convergence theorem

Exploring whether f(A n ) converges to f(A) for more general functions f raises intriguing questions, yet it is a complex issue The definition of f(A n ) for large n is not guaranteed even if f(A) is defined Furthermore, if we assume f(A n ) is defined, the regularizer for f may vary with n, complicating the analysis Therefore, we will transition to a more abstract framework.

Let (E,M,Φ),(E,M,Φ n ),n∈N, be a family of proper afc over the Banach space X We call the sequence [(E,M,Φ n )] n convergentto the afc (E,M,Φ) if Φ n (e)→Φ(e) in norm for everye∈ E.

Proposition 2.6.8 Let [(E,M,Φ n )] n be a sequence of proper afc converging to the proper afc (E,M,Φ) Letf ∈ M, and let e be a uniform regulariser for f, i.e., e, ef ∈ E and all e • , e • n are injective If x n ∈ D(f • n ) such that x n → x and f • n x n →y, then(x, y)∈f •

In particular, if all f • n ∈ L(X)andf • n →T ∈ L(X)strongly, then f • =T.

In the given proof, we consider elements \( e, f, x_n, x, y \) as stated in the hypothesis It is established that \( (ef) \cdot x = \lim_{n} (ef) \cdot n x_n = e \cdot y \), which implies that \( (x, y) \in f \cdot \) Additionally, Proposition 2.6.8 highlights the necessity of uniform regularizers for achieving general convergence results in the context of sectorial operators, with the subsequent proposition offering a relevant example.

Proposition 2.6.9 Let A∈Sect(ω), and let(A n ) n be a sectorial approximation of

A onS ω Take f ∈ M[S ω ]and suppose either

2) f ∈ B and all operatorsA, A n are injective.

If x n ∈ D(f(A n )) with x n → x and f(A n )x n → y, then also x∈ D(f(A)) and y = f(A)x In particular, if f(A n ) ∈ L(X) with f(A n ) → T ∈ L(X) strongly, thenf(A) =T.

Proof In case 1) a uniform regulariser is (1 +z) − m for somem, in case 2) a power ofz/(1 +z) 2 can be used The rest is only the application of Proposition 2.6.8.

Boundedness

Understanding which families of functions result in uniformly bounded families of operators is crucial in functional calculus Chapter 5 extensively explores this topic, but here we present a few straightforward yet valuable findings.

Lemma 2.6.10 Let A∈Sect(ω), ϕ∈(ω, π), and (f α ) α ⊂H 0 ∞ (S ϕ ) Let there be numbers C, s, c α >0 such that

(2.10) for allz∈S ϕ and all indices α Then sup α f α (A) ≤(2C/sπ)M(A, ϕ).

Proof Chooseω ∈(ω, ϕ) Then, by a simple change of variables in the integral, one obtains the estimate f α (A) ≤M(A, ω ) C

Now letω tend toϕ, cf Remark 2.1.2

Proposition 2.6.11 Let ϕ∈ (0, π], and let f ∈ E(S ϕ ) Then there is a constant

C f >0such that sup t>0 f(tA) ≤ C f M(A, ϕ) for each sectorial operator A∈Sect(ω),ω∈(0, ϕ), on a Banach spaceX. Moreover, givenθ∈[0, ϕ−ω)one has f(λA) ≤C f M(A, ϕ−θ) for allλ∈C,|argλ| ≤θ.

Proof Writef =ψ+a/(1 +z) +b with ψ∈ H 0 ∞ (S ϕ ), a, b∈C We can choose

By applying a change of variable in the defining Cauchy integral, we establish that ψ(tA) = ψ(tz)(A) for all t > 0 Consequently, according to Lemma 2.6.10, it follows that ψ(tA) is less than or equal to 2C sπM(A, ϕ).

The Spectral Mapping Theorem

The Spectral Inclusion Theorem

We begin with two auxiliary results.

Lemma 2.7.1 LetA∈Sect(ω), and let λ∈Cbe such that λ−Ais injective Let e∈H(A), and let0 =c∈Cbe such that f(z) := e(z)−c λ−z ∈H(A).

Proof By Theorem 1.3.2, the inclusione(A)(λ−A) − 1 ⊂(λ−A) − 1 e(A) is always true To prove the converse, take x∈X such thate(A)x∈D((λ−A) − 1 ) Then there is z ∈ D(A) with e(A)x = (λ−A)z Since e = (λ−z)f +c we have (λ−A)z=cx+ (λ−A)f(A)x, whencecx= (λ−A)(z−f(A)x) Now, c= 0 by assumption, hencex∈R(λ−A) =D((λ−A) − 1 )

Lemma 2.7.2 Let A∈Sect(ω), and let f ∈ M[S ω ] A Given 0 =λ∈S ω there is a regulariser efor f satisfying e(λ) = 0.

Let \( g \) be any regularizer for \( f \), with both \( f \) and \( g \) belonging to the space \( E \) and \( g(A) \) being injective We define \( e = \frac{g}{(\lambda - z)^n} \), where \( n \) is the order of the zero \( \lambda \) of \( g \), which may be zero It follows that \( e(\lambda) = 0 \) and \( e \) is also in \( E \) Consequently, we have \( g(A) = (\lambda - A)^n e(A) \), indicating that \( e(A) \) is injective Moreover, since \( ef = \frac{f g}{(\lambda - z)^n} \) is also in \( E \), \( e \) serves as a regularizer for \( f \) with \( e(\lambda) = 0 \).

We now come to the main part.

Proposition 2.7.3 Let A ∈Sect(ω),f ∈ M[S ω ] A , and 0 =λ∈ S ω If f(λ) = 0 andf(A)is invertible, then λ∈(A).

Proof Choose a regulariser e∈ E forf with c :=e(λ) = 0 Then (ef)∈ E and (ef)(λ) = 0 This implies that alsoh:=ef /(λ−z)∈ E, whence h(A)(λ−A)⊂(ef)(A) =f(A)e(A).

Becausee(A) andf(A) are both injective, this shows thatλ−Amust be injective. Also,eis a regulariser forf /(z−λ).

It is obvious thatg:= (e−c)/(λ−z)∈ E By Lemma 2.7.1,e(A)(λ−A) − 1 (λ−A) − 1 e(A) Inverting both sides of this equation yields (λ−A)e(A) − 1 e(A) − 1 (λ−A) Hence f(A)

Sincef(A) is surjective,λ−Amust also be surjective, henceλ∈(A)

Theorem 2.7.4 (Spectral Inclusion Theorem) Let A ∈ Sect(ω), and let f ∈

Proof Let 0 = λ ∈ σ(A), and define à := f(λ) If à = ∞, Proposition 2.7.3 applied to the function à−f shows that à−f(A) cannot be invertible Hence à∈σ(f(A)).

Suppose that à = ∞∈/ σ(f˜ (A)) Then f(A) ∈ L(X) and there is λ 0 ∈ C such that f(A)−λ 0 is invertible Hence g := 1/(f−λ 0 ) ∈ M[S ω ] A with g(A) being invertible and g(λ) = 0 Another application of Proposition 2.7.3 yields λ∈(A), contradicting the assumption made onλ

It is natural to ask what happens at the ‘critical’ points 0 and ∞ Since ˜ σ(f(A)) is a compact subset of the Riemann sphere, we immediately obtain f(σ(A)\ {0}) C ∞ ⊂σ(f˜ (A)) (2.11)

The Volterra operator V on C[0,1] has a spectrum σ(V) that is equal to {0}, making the inclusion trivial Additionally, a subsequent result provides a sufficient condition related to the concept of polynomial limit.

Theorem 2.7.5 Let A ∈ Sect(ω), and let λ 0 ∈ {0,∞} If f ∈ M[S ω ] A has polynomial limit àatλ 0 andλ 0 ∈˜σ(A), then à∈σ(f˜ (A)).

Proof Forà∈Cwe can considerf −àinstead of f and in this way reduce the problem to the casesà= 0 andà=∞.

We begin by considering the case where \( a = 0 \) and \( \lambda_0 = \infty \), assuming that \( f(A) \) is invertible and that \( f \) approaches a polynomial limit of 0 as \( \lambda_0 \) approaches infinity If infinity is included in the spectrum of \( A \) excluding zero, the conclusion follows from equation (2.11) Alternatively, if this is not the case, there exists a radius \( R > 0 \) such that the spectrum of \( A \) is contained within the ball of radius \( R \) centered at zero.

In the context of the bounded projection Q defined by R(z, A)dz, where Γ = ∂B δ(0) for any R < δ, we establish that the complementary projection P is given by P := I - Q, and its range space is denoted as Y := R(P) Notably, since P commutes with A, it follows that Ay ∈ Y for every y in the intersection of Y and the domain of A, denoted as D(A) Furthermore, B := A|Y is identified as a sectorial operator of angle ω on Y, with the relationship (A) ⊂ (B) and R(λ, B) = R(λ, A)|Y holding true for all λ in (A).

Moreover, ˜σ(B)⊂ {∞} Indeed, it is easily verified that for eachà∈σ(A) one has

The integral 1 à−zR(z, A)dz on Y indicates that e(B) equals e(A) restricted to Y for every e in E[S ω] This leads to the conclusion that D(f(B)) is equal to D(f(A)) intersected with Y, and f(B)x equals f(A)x for all x in D(f(A)) intersected with Y Given that f(A) is assumed to be invertible, it follows that f(B) is also invertible, with f(B) − 1 equal to f(A) − 1 restricted to Y Since the function f approaches a polynomial limit of 0 at infinity, f is included in the domain of the primary functional calculus for B Furthermore, for sufficiently large n in N, the function g defined as zf^n retains this property, leading to g(B) equating to Bf(B)^n in L(X) As f(B) is invertible, B must be bounded Consequently, since the spectrum ˜σ(B) is a subset of {∞}, we conclude that ˜σ(B) is empty, which implies that Y equals 0 This ultimately demonstrates that A is bounded, establishing that ∞ is part of the spectrum σ(A).

In the case where \( a = 0 \) and \( \lambda_0 = 0 \), we assume that \( f(A) \) is invertible and that \( f \) has a polynomial limit of 0 at 0 If \( e \in E \) serves as a regularizer for \( f \), then \( ef \in E \), making \( (ef)(A) \) injective with \( (ef)(0) = 0 \) According to Lemma 2.3.8, this implies that \( A \) must be injective Defining \( B := A - 1 \) and \( g(z) := f(z - 1) \), we find that \( g \in M[S_\omega]^B \), \( g(B) = f(A) \) is invertible, and \( g \) has a polynomial limit of 0 at infinity Since \( B \) is bounded, it follows that \( A \) is invertible.

In the case where ∞ = à and ∞ is not an element of the spectrum σ(f˜(A)), it follows that f(A) must be bounded Consequently, there exists a complex number λ such that λ - f(A) is invertible According to Theorem 1.3.2, the function g defined as g = 1/(λ - f) is part of M[S ω] A, and g(A) is invertible with a polynomial limit of 0 at λ Therefore, we can conclude that λ is not in the spectrum of A.

Cf [108] for a slightly stronger result.

The Spectral Mapping Theorem

We are heading towards the final goal As usual, we start with an auxiliary result.

Lemma 2.7.6 LetA∈Sect(ω)andf ∈ M[S ω ] Suppose that f has finite polyno- mial limits at σ(A)˜ ∩ {0,∞} and all poles of f are contained in (A) Then the following assertions are true. a) If {0,∞} ⊂σ(A),˜ f(A)is defined by the nfc for sectorial operators. b) IfAis invertible,f(A)is defined by the nfc for invertible sectorial operators. c) If A is bounded,f(A)is defined by the nfc for bounded sectorial operators.

Proof Let f be as required, and letλ 0 be a pole off Then, for suitably large n 0 ∈N, the function f 1 := (λ 0 −z) n 0

(1 +z) n 0 f(z) also has finite polynomial limits at ˜σ(A)∩{0,∞}, but one pole less thanf More- over, letting r 0 := (λ 0 −z) n 0 /(1 +z) n 0 , we see that r 0 (A) is bounded and invert- ible.

In scenario a), the function f is limited to having only a finite number of poles Through induction, we can identify a bounded rational function r, which is both bounded and invertible, ensuring that rf belongs to the space E Consequently, this regularizes f to E, allowing us to express f(A) as r(A) - 1 (rf)(A), which is an element of L(X).

In various scenarios, only a finite number of poles of a function are located within the relevant section of its domain For instance, when matrix A is invertible, the poles may cluster around zero; however, the behavior of the function near zero is not significant for the normal form of A.

Proposition 2.7.7 Let A ∈Sect(ω), and let f ∈ M[S ω ] A have polynomial limits at the points σ(A)˜ ∩ {0,∞} Thenσ(f˜ (A))⊂f(˜σ(A)).

Assuming that f(λ) is defined for each λ in the spectrum of the operator A, consider a complex number a that is not in the range of f(˜σ(A)) The function (a - f)^{-1} belongs to M[Sω] and exhibits finite polynomial limits at the points {0, ∞} intersected with σ(A) Furthermore, all poles, which are the points ˜λ in Sω \ {0} where f(λ) equals a, are located within the resolvent set of A Therefore, one can apply the relevant lemma.

2.7.6 to conclude that (à−f) − 1 (A) is defined and bounded But this implies that à−f(A) is invertible, whence à /∈σ(f˜ (A)).

Now suppose à =∞∈/ f(˜σ(A)) This implies that the poles of f are con- tained in the resolvent ofA An application of Lemma 2.7.6 yields thatf(A) is a bounded operator, whence∞∈/σ(f˜ (A))

Theorem 2.7.8 (Spectral Mapping Theorem) Let A ∈ Sect(ω), and let f ∈

M[S ω ] A have polynomial limits at {0,∞} ∩σ(A) Then˜ f(˜σ(A)) = ˜σ(f(A)).

Proof Combine Proposition 2.7.7 and Theorem 2.7.5.

Convergence Lemma

Equivalent Descriptions and Uniqueness

Perturbation Results

Characterisations

Extrapolation Spaces

Homogeneous Interpolation

More Characterisations and Dore’s Theorem

Fractional Powers as Intermediate Spaces

Numerical Range Conditions

Group Generators on Hilbert Spaces

Similarity Theorems for Sectorial Operators

Operators Without Bounded H ∞ -Calculus

Rational Approximation Schemes

Maximal Regularity