The Functional Calculus for Sectorial Operators

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The Functional Calculus for Sectorial Operators Markus Haase

Preliminary Version 19th February 2005

Author’s Address:

Abt. Angewandte Analysis Helmholtzstraße 18 89069 Ulm

e-mail: [email protected]

c by Markus Haase

Contents

Preface 1 Axiomatics for Functional Calculi 1.1 The Concept of Functional Calculus . . . . 1.2 An Abstract Framework . . . . . . . . . . . 1.2.1 The Extension Procedure . . . . . . 1.2.2 Properties of the Extended Calculus 1.2.3 Generators and Morphisms . . . . . 1.3 Meromorphic Functional Calculi . . . . . . 1.3.1 Rational Functions . . . . . . . . . . 1.3.2 An Abstract Composition Rule . . . 1.4 Multiplication Operators . . . . . . . . . . . 1.5 Concluding Remarks . . . . . . . . . . . . . 1.6 Comments . . . . . . . . . . . . . . . . . . .

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1 1 3 4 5 7 8 10 12 12 14 16

2 The Functional Calculus for Sectorial Operators 2.1 Sectorial Operators . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Sectorial Approximation . . . . . . . . . . . . . . . 2.2 Spaces of Holomorphic Functions . . . . . . . . . . . . . . 2.3 The Natural Functional Calculus . . . . . . . . . . . . . . 2.3.1 Primary Functional Calculus via Cauchy Integrals 2.3.2 The Natural Functional Calculus . . . . . . . . . . 2.3.3 Functions of Polynomial Growth . . . . . . . . . . 2.3.4 The Case when A is Injective . . . . . . . . . . . . 2.4 The Composition Rule . . . . . . . . . . . . . . . . . . . . 2.5 Extensions According to Spectral Conditions . . . . . . . 2.5.1 The Case when 0 ∈ ̺(A) . . . . . . . . . . . . . . . 2.5.2 The Case when A ∈ L(X) . . . . . . . . . . . . . . 2.5.3 The Case when A Is Bounded and Invertible . . . 2.6 Miscellanies . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Adjoints . . . . . . . . . . . . . . . . . . . . . . . . 2.6.2 Restrictions . . . . . . . . . . . . . . . . . . . . . .

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vi

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3 Fractional Powers and Related Topics 3.1 Fractional Powers with Positive Real Part . 3.2 Fractional Powers with Arbitrary Real Part 3.3 Holomorphic Semigroups . . . . . . . . . . . 3.4 The Logarithm and the Imaginary Powers . 3.5 Comments . . . . . . . . . . . . . . . . . . .

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5 The Boundedness of the H ∞ -calculus 5.1 Convergence Lemma . . . . . . . . . . . . . . . . . . . . . . . . . 5.1.1 Convergence Lemma for Sectorial Operators. . . . . . . . 5.1.2 Convergence Lemma for Strip Type Operators. . . . . . . 5.2 A Fundamental Approximation Technique . . . . . . . . . . . . . 5.3 Bounded H ∞ -Calculus: Equivalent Descriptions and Uniqueness 5.3.1 Adjoints . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.2 Logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.3 Boundedness on Subalgebras of H ∞ . . . . . . . . . . . . 5.3.4 Uniqueness . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 The Minimal Angle . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 An Operator without a Bounded H ∞ -Calculus . . . . . . . . . . 5.6 Perturbation Results . . . . . . . . . . . . . . . . . . . . . . . . . 5.6.1 Resolvent Growth Conditions . . . . . . . . . . . . . . . . 5.6.2 A Theorem of Pr¨ uss and Sohr . . . . . . . . . . . . . . . . 5.7 Connections to Harmonic Analysis . . . . . . . . . . . . . . . . . 5.8 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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91 91 91 93 93 97 97 97 98 100 101 103 104 105 109 111 112

2.7

2.8

2.6.3 Sectorial Approximation . . . . . 2.6.4 Boundedness . . . . . . . . . . . The Spectral Mapping Theorem . . . . . 2.7.1 The Spectral Inclusion Theorem 2.7.2 The Spectral Mapping Theorem Comments . . . . . . . . . . . . . . . . .

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4 Strip Type Operators and the Logarithm 4.1 Strip Type Operators . . . . . . . . . . . . . . . 4.2 The Natural Functional Calculus . . . . . . . . . 4.3 The Spectral Height of the Logarithm . . . . . . 4.4 Monniaux’s Theorem and the Inversion Problem 4.5 A Counterexample . . . . . . . . . . . . . . . . . 4.6 Comments . . . . . . . . . . . . . . . . . . . . . .

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6 Interpolation Spaces 113 6.1 Real Interpolation Spaces . . . . . . . . . . . . . . . . . . . . . . . 113 6.2 Characterizations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 6.2.1 A First Characterization . . . . . . . . . . . . . . . . . . . . 116

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120 122 123 124 125 128 130 130 132 134 134 135 136 137 137 138 140 141 143 146

7 The Functional Calculus on Hilbert Spaces 7.1 Numerical Range Conditions . . . . . . . . . . . . . . . . . 7.1.1 Accretive and ω-accretive Operators . . . . . . . . . 7.1.2 Normal Operators . . . . . . . . . . . . . . . . . . . 7.1.3 Functional Calculus for m-accretive Operators . . . 7.1.4 Mapping Theorems for the Numerical Range . . . . 7.1.5 The Crouzeix-Delyon Theorem . . . . . . . . . . . . 7.2 Group Generators on Hilbert Spaces . . . . . . . . . . . . . 7.2.1 Liapunov’s Direct Method for Groups . . . . . . . . 7.2.2 A Decomposition Theorem for Group Generators . . 7.2.3 A Characterization of Group Generators . . . . . . . 7.3 Similarity Theorems for Sectorial Operators . . . . . . . . . 7.3.1 The Theorem of McIntosh . . . . . . . . . . . . . . . 7.3.2 Interlude: Operators Defined by Sesquilinear Forms 7.3.3 Similarity Theorems . . . . . . . . . . . . . . . . . . 7.3.4 A Counterexample . . . . . . . . . . . . . . . . . . . 7.4 Cosine Function Generators . . . . . . . . . . . . . . . . . . 7.5 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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147 149 149 151 152 154 155 158 159 161 163 167 167 170 174 177 179 183

6.3

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6.7 6.8

6.2.2 A Second Characterization . . . . . . . . . . . . 6.2.3 Examples . . . . . . . . . . . . . . . . . . . . . . Extrapolation Spaces . . . . . . . . . . . . . . . . . . . . 6.3.1 An Abstract Method . . . . . . . . . . . . . . . . 6.3.2 Extrapolation for Injective Sectorial Operators . 6.3.3 The Homogeneous Fractional Domain Spaces . . Homogeneous Interpolation . . . . . . . . . . . . . . . . 6.4.1 Some Intermediate Spaces . . . . . . . . . . . . . 6.4.2 . . . are actually Real Interpolation Spaces . . . . More Characterizations and Dore’s Theorem . . . . . . 6.5.1 A Third Characterization (Injective Operators) . 6.5.2 A Fourth Characterization (Invertible Operators) 6.5.3 Dore’s Theorem Revisited . . . . . . . . . . . . . Fractional Powers as Intermediate Spaces . . . . . . . . 6.6.1 Density of Fractional Domain Spaces . . . . . . . 6.6.2 The Moment Inequality . . . . . . . . . . . . . . 6.6.3 Reiteration and Komatsu’s Theorem . . . . . . . 6.6.4 The Complex Interpolation Spaces and BIP . . . Characterizing Growth Conditions . . . . . . . . . . . . Comments . . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents

A Linear Operators A.1 The Algebra of Multivalued Operators . . . . . . . . A.2 Resolvents . . . . . . . . . . . . . . . . . . . . . . . . A.3 The Spectral Mapping Theorem for the Resolvent . . A.4 Adjoints . . . . . . . . . . . . . . . . . . . . . . . . . A.5 Convergence of Operators . . . . . . . . . . . . . . . A.6 Polynomials and Rational Functions of an Operator A.7 Injective Operators . . . . . . . . . . . . . . . . . . . A.8 Semigroups and Generators . . . . . . . . . . . . . .

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189 189 192 195 197 199 201 203 205

B Interpolation Spaces B.1 Interpolation Couples . . . . . . . . B.2 Real Interpolation by the K-Method B.3 The Trace Method . . . . . . . . . . B.4 The Reiteration Theorem . . . . . . B.5 Complex Interpolation . . . . . . . .

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C Operator Theory on Hilbert Spaces C.1 Sesquilinear Forms . . . . . . . . . . . . . . . . . C.2 Adjoint Operators . . . . . . . . . . . . . . . . . C.3 The Numerical Range . . . . . . . . . . . . . . . C.4 Symmetric Operators . . . . . . . . . . . . . . . . C.5 Equivalent Scalar Products and the Lax-Milgram C.6 Weak Integration . . . . . . . . . . . . . . . . . . C.7 Accretive Operators . . . . . . . . . . . . . . . . C.8 The Theorems of Plancherel and Gearhart . . . . D The D.1 D.2 D.3 D.4 D.5 D.6

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Spectral Theorem Multiplication Operators . . . . . . . . . . . . . . . . . . . Commutative C ∗ -Algebras. The Cyclic Case . . . . . . . . Commutative C ∗ -Algebras. The General Case . . . . . . . The Spectral Theorem: Bounded Normal Operators . . . The Spectral Theorem: Unbounded Selfadjoint Operators The Functional Calculus . . . . . . . . . . . . . . . . . . .

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E Approximation by Rational Functions

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Preface Here will be the text of the Preface.

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Preface

Chapter 1

Axiomatics for Functional Calculi During the course of the book the reader will encounter several examples of functional calculi. This chapter on the one hand is intended to convey the fundamental intuition behind the concept “functional calculus” (in Section 1.1). On the other hand, we present a formalization of certain ideas common to many functional calculus constructions (Section 1.2). In particular we introduce a method of extending an elementary functional calculus to a larger algebra. In Section 1.3 we introduce the notion of a meromophic functional calculus for a closed operator A on a Banach space and specialize the abstract results from Section 1.2. As an important example we deal with multiplication operators in Section 1.4. Finally, in Section 1.5 we motivate by example the construction of the functional calculus for sectorial operators which will be the topic of Chapter 2.

1.1 The Concept of Functional Calculus Consider the Banach space X := C[0, 1] of complex valued continuous functions on the unit interval. Each function a ∈ X determines a bounded linear operator Ma = (g 7−→ ag) : C[0, 1] −→ C[0, 1] on X called the multiplication operator associated with a. Its spectrum is simply the range a[0, 1] of a. Given any other continuous function f : σ(Ma ) −→ C one can consider the multiplication operator Mf ◦a associated with f ◦ a. This yields an algebra homomorphism Φ = (f 7−→ Mf ◦a ) : C(σ(Ma )) −→ L(X). Since one has Φ(z) = Ma and Φ((λ − z)−1 ) = R(λ, Ma ) for λ ∈ ̺(Ma ), and since the operator Φ(f ) is simply multiplication by f (a(z)), one says that Φ(f ) is obtained by “inserting” the operator Ma into the function f and writes f (Ma ) := Φ(f ). Generalizing this example to the Banach space X = C0 (R) one realizes that boundedness of the operators is not an essential requirement. The intuition of functional calculus now consists, roughly, in the idea that to every closed operator A on a Banach space X there corresponds an algebra of

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Chapter 1. Axiomatics for Functional Calculi

complex-valued functions on its spectrum in which the operator A can somehow be “inserted” in a reasonable way. Here “reasonable” means at least that f (A) should have the expected meaning if one expects something, e.g., if λ ∈ ̺(A) then one expects (λ−z)−1 (A) = R(λ, A) or if A generates a semigroup T then etz (A) = T (t). (This is just a minimal requirement. There may be other reasonable criteria.) In summary we think of a mapping f 7−→ f (A) which we (informally) call a functional calculus for A. Unfortunately, up to now there is no overall formalization of this idea. The best thing achieved so far is a case by case construction. We turn back to examples. If one knows that the operator A is “essentially” (i.e., is similar to) a multiplication operator, then it is straightforward to construct a functional calculus. By one version of the spectral theorem, this is the case if A is a normal operator on a Hilbert space. (For the convenience of the reader, we have put an account of the spectral theorem into Appendix D.) As is well-known, the Fourier transform on L2 (R) is an instance of this, hence provides one of the earliest examples of a non-trivial functional calculus. (The operator A in that case is just id/dt.) In general, there is no canonical “diagonalization” of the normal operator A, and so the resulting functional calculus depends — at least a priori — on the chosen unitary equivalence. Hence an additional argument is needed to ensure independence of the construction (compare Theorem D.6.1). Having this in mind as well as for the sake of generality, one looks for intrinsic definitions. It was the idea of Riesz and Dunford to base the construction of f (A) on a Cauchy-type integral Z 1 f (A) = f (z)R(z, A) dz. (1.1) 2πi Γ The idea behind this is readily sketched. Let us suppose we are given a bounded operator A and an open superset U of the spectrum K := σ(A). The so-called “homology version” of the Cauchy Integral Formula says the following. One can choose a generalized contour (a “cycle”) Γ with the following properties: 1) Sp Γ ⊂ U \ K. 2) Every point of C \ Sp Γ has index either 0 or 1 with respect to Γ. 3) Each point of K has index 1 with respect to Γ. With respect to such a contour Γ, one has Z 1 1 f (a) = f (z) dz 2πi Γ z−a

(1.2)

for each a ∈ K and for every holomorphic function defined on U . (See [19, Ch. VIII, Prop.1.1] for a proof.) Replacing a by A on both sides in the formula (1.2) 1 and insisting that “ z−A ” should be the same as R(z, A), we arrive at (1.1). The so obtained functional calculus is called the Dunford-Riesz calculus, see [39, Section VII.3] for the mathematics and [39, Section VII.11] for some historical remarks. (It turns out to be only a special case of a general construction in Banach

1.2. An Abstract Framework

3

algebras an account of which can be found in [20, Chapter VII, §4].) Actually, it yields an algebra homomorphism Φ := (f 7−→ f (A)) : A −→ L(X), where A is the algebra of germs of holomorphic functions on σ(A). Let us continue with a second example. Consider the Banach space X := C[0, 1] and the Volterra operator V ∈ L(X) defined by Z t (V f )(t) := f (s) ds (t ∈ [0, 1]). 0

As is well-known, V is a compact, positive contraction with σ(A) := {0}. Given a holomorphic function f defined in a neighborhood of 0, simply choose ε > 0 small enough and define Z 1 f (V ) = f (z)R(z, V ) dz. 2πi ∂Bε (0) 1 for all n, one can equally replace z by V in the power-series Since kV n k = n! expansion of f . Now note that V is injective but not invertible, and z −1 is not holomorphic in 0. Nevertheless it seems reasonable to write V −1 = (z −1 )(V ). This suggests an extension of the functional calculus to a larger algebra of germs of holomorphic functions which contains the function z −1 . In the concrete example of the Volterra operator this extension is easy. (Consider functions holomorphic in a pointed neighborhood of 0 which have no essential singularity in 0. This means that the principal part of the Laurent series is a polynomial in z −1 . Now insert V as in Appendix A.7 into the principal part and as above into the remainder. Finally add the results.)

In more general situations one cannot go back to power or Laurent series. Luckily, there is a general extension procedure which will be described in the next section.

1.2 An Abstract Framework In this section we will describe abstractly how to extend a certain basic functional calculus to a wider class of functions. It is a little “Bourbakistic” in spirit and the reader who is not so fond of axiomatic treatment may skip it on first reading and come back to it when needed. To have a model in mind, assume we are given an operator A on a Banach space X and a basic class of functions E on the spectrum of A. Assume further we have a “method” Φ = (f 7−→ f (A)) : E −→ L(X)

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Chapter 1. Axiomatics for Functional Calculi

of “inserting” the operator into functions from E, i.e., E is an algebra and Φ is a homomorphism of algebras. Moreover the mapping Φ should have something to do with the operator A and so we assume for the moment that for some λ ∈ ̺(A) the function (λ − z)−1 is contained in E and Φ((λ − z)−1 ) = R(λ, A). Then one can think of defining f (A) := (λ − A)n Φ[f (z)(λ − z)−n ] = Φ[(λ − z)−n ]−1 Φ[f (z)(λ − z)−n ] for each function f such that f (z)(λ − z)−n is contained in E for some n ∈ N. This would clearly yield a closed operator and one only has to make sure that the definition is independent of the chosen n. Hence one has defined “f (A)” for f taken from a larger algebra. To proceed even further assume now that e is any function from E such that e(A) is injective and f is a function with ef ∈ E. Then, as above where we had e = (λ − z)−n , one can define f (A) = e(A)−1 (ef )(A) = Φ(e)−1 Φ(ef ).

Of course one has to make sure that the use of different e’s does not lead to different f (A)’s. This will shortly be proved, after we will have cast all this motivation into an abstract axiomatic framework.

1.2.1 The Extension Procedure The starting point of what we will call the extension procedure is a Banach space X, a commutative algebra M with 1 together with a subalgebra E ⊂ M (with 1∈ / E in general) and a homomorphism Φ : (e −→ Φ(e)) : E −→ L(X). Let us call the triple (E, M, Φ) an abstract functional calculus (in short: AFC) over X. Sometimes we will denote this object by just the pair (E, M) and suppress the explicit reference to the homomorphism Φ : E −→ L(X). We say that the AFC is non-degenerate or proper, if the set Reg(E) := {e ∈ E | Φ(e) is injective} is not empty. The members of Reg(E) are called regularizers. Take f ∈ M. If there is e ∈ Reg(E) such that also ef ∈ E, we call f regularizable by E and e a regularizer for f . Note that 1 is regularizable if and only if the AFC is proper. In this case Mr := {f ∈ M | f is regularizable}

clearly is a subalgebra of M which contains E. Let (E, M, Φ) be a proper AFC. We define

Φ(f ) := Φ(e)−1 Φ(ef )

(1.3)

for f ∈ Mr and e ∈ Reg(E) being a regularizer for f . However, we will often write f• instead of Φ(f ) to shorten the computations. (Hence, (1.3) reads f• := e−1 • (ef )• with this convention.)

1.2. An Abstract Framework

5

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 7−→ Φ(f )) : Mr −→ {closed operators on X} extends the original mapping Φ : E −→ L(X).

Proof. Let h ∈ Reg(E ) be a second regularizer for f and define A := (e• )−1 (ef )• and B := (h• )−1 (hf )• . Because e• h• = (eh)• = (he)• = h• e• , we have (e• )−1 (h• )−1 = (h• )−1 (e• )−1 . Now it follows that A = (e• )−1 (ef )• = (e• )−1 (h• )−1 h• (ef )• = (h• )−1 (e• )−1 (hef )• = (h• )−1 (e• )−1 e• (hf )• = (h• )−1 (hf )• = B.

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

We are left to show that the new Φ extends the old one. Now, since the AFC is proper, E ⊂ Mr . If e, f ∈ E with e• injective, one has (e• )−1 (ef )• = (e• )−1 e• f• = f• , whence it is shown that the map on Mr is indeed an extension of the original.

Sometimes we will call the original mapping Φ : E −→ L(X) the primary (functional) calculus (in short: PFC) and the extension defined above the extended (functional) calculus. The algebra Mr is called the domain of the AFC (E, M, Φ).

1.2.2 Properties of the Extended Calculus We collect some basic properties. Proposition 1.2.2. Let (E, M, Φ) be a proper abstract functional calculus over the Banach space X. Then the following assertions hold. a) If T ∈ L(X) commutes with each e• , e ∈ E, then it commutes with each f• , f ∈ Mr . b) One has 1 ∈ Mr and 1• = I. c) Given f, g ∈ Mr , one has

f• + g• ⊂ (f + g)• f• g• ⊂ (f g)•

with D(f• g• ) = D((f g)• ) ∩ D(g• ).

d) If f, g ∈ Mr such that f g = 1, then f• is injective with (f• )−1 = g• .

e) Let f ∈ Mr and F be a subspace of D(f• ). Assume that there is a sequence (en )n ⊂ E such that en• → I strongly as n → ∞ and R(en• ) ⊂ F for all n ∈ N. Then F is a core for f• .

Proof. Ad a). Let T ∈ L(X) commute with every e• , e ∈ E. Take f ∈ Mr and a regularizer e ∈ E for f . Then we have T f• = T (e• )−1 (ef )• ⊂ (e• )−1 T (ef )• = (e• )−1 (ef )• T = f• T . Ad b). Since the AFC is proper, there exists e ∈ E such that e• is injective. Clearly, e −1 regularizes 1. Moreover, we have 1• = e−1 • (e1)• = e• e• = I. Ad c). Take f, g ∈ Mr and let e1 , e2 ∈ E be regularizers for f, g, respectively. Then e := e1 e2 is a regularizer for both f and g, hence for f + g. Also ef g = (e1 f )(e2 g) ∈ E,

6

Chapter 1. Axiomatics for Functional Calculi

−1 −1 whence e is also a regularizer for f g. We have f• + g• = e−1 • (ef )• + e• (eg)• ⊂ e• [(ef )• + −1 −1 −1 −1 (eg)• ] = e−1 (e(f + g)) = (f + g) and f g = e (e f ) e (e g) ⊂ e e (e f • • • • 1 • 2• 2 • 1 )• (e2 g)• = • 1• 1• 2• (ef g) = (f g) . [e2• e1• ]−1 (ef g)• = e−1 • • • To prove the assertions concerning the domains, let x ∈ D((f g)• ) ∩ D(g• ). Note that since −1 (e1 f )• commutes with e2• it also commutes with e−1 2• . By assumption, y := (e2 g)• x ∈ D(e2• ), −1 whence also (e1 f )• y ∈ D(e2• ) and −1 −1 −1 (e1 f )• g• x = (e1 f )• e−1 2• y = e2• (e1 f )• y = e2• (ef g)• x ∈ D(e1• ) −1 −1 by assumption and the identity e−1 • = e1• e2• . Consequently, g• x ∈ D(f• ), whence x ∈ D(f• g• ). Ad d). Assume f, g ∈ Mr with f g = 1. By b) and c) we have g• f• ⊂ (f g)• = 1• = I and D(g• f• ) = D(I) ∩ D(f• ) = D(f• ). Interchanging f and g proves the statement. Ad e). Let x ∈ D(f• ) and define y := f• x. With xn := en• x and yn := en• y we have xn → x, xn ∈ F and f• xn = f• en• x = en• f• x = yn → y.

In general one cannot expect equality in c) of Proposition 1.2.2. However, if we define Mb := {f ∈ Mr | f• ∈ L(X)}, we obtain the following. Corollary 1.2.3. Let E, M, Φ, X be as above. a) For f ∈ Mr , g ∈ Mb the identities f• + g• = (f + g)• ,

f• g• = (f g)•

hold. b) The set Mb is a subalgebra with 1 of M and the map (f 7−→ f• ) : Mb −→ L(X) is a homomorphism of algebras with 1. c) If f ∈ Mb is such that f• is injective, then (f• )−1 g• f• = g• holds for all g ∈ Mr . Proof. Ad a). By c) of Proposition 1.2.2 we have f• + g• ⊂ (f + g)• and (f + g)• − g• ⊂ (f + g − g)• = f• . Since g• is assumed to be bounded, D((f + g)• ) = D(f• ). This readily implies f• + g• = (f + g)• . The second assertion is immediate from c) of Proposition 1.2.2 since we have D(g• ) = X. The statement in b) follows from a). Ad c). We can find e ∈ E which regularizes both f and g. Now we compute −1 −1 −1 −1 −1 f•−1 g• f• = f•−1 e−1 • (eg)• f• = f• e• f• (eg)• = e• f• f• (eg)• = e• (eg)• = g• . −1 Here we used b) and the identity e−1 = f•−1 e−1 which is true since e• and f• are both • f• • bounded and injective.

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

1.2. An Abstract Framework

7

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

In this case, we have g• = f•−1 .

Proof. One direction of the biimplication is simply d) of Proposition 1.2.2. Assume that f• is injective and let e ∈ E be a regularizer for f . Then f e ∈ E, (f e)g = e ∈ E and (f e)• = f• e• is injective. This means that f e is a regularizer for g!

Although the reader will encounter several examples of the extension procedure during the following chapters, we will illustrate it here in the case of the Volterra operator. We let M be the algebra of germs of functions holomorphic in a pointed neighborhood of 0 and E be the subalgebra of germs of functions holomorphic in a whole neighborhood of 0. The primary calculus is given by the Cauchy integral (1.1) (or, alternatively, by insertion into the power series). Since (z)(V ) = V , this AFC is proper and there is a natural extension to Mr . It is easy to see that Mr is exactly the algebra of germs meromorphic in 0. If f is such a germ with z n f holomorphic in 0, we have f (V ) = V −n (z n f )(V ). In particular, (z −n )(V ) = V −n for each n ∈ N.

1.2.3 Generators and Morphisms Let us return to the abstract treatment. We start with a proper AFC (E, M, Φ) on the Banach space X. Call a subalgebra D ⊂ Mb admissible if the set {f ∈ D | f• is injective} is not empty. In this case (D, M, Φ) is another proper AFC. Let us denote hDi := {f ∈ M | there is d ∈ D s.t. df ∈ D and d• is injective} the regularizable elements of this AFC. We call hDi the domain generated by D and D its generator. In this terminology, Mr = hEi. The next lemma shows in particular that hDi is always contained in Mr whence nothing is gained by passing from E to Mb . Proposition 1.2.5. Let (E, M, Φ) be a proper AFC over the Banach space X and let D be an admissible subalgebra of Mb . If f ∈ M and g ∈ hDi such that g• is injective and gf ∈ hDi, then already f ∈ hDi.

Proof. By assumption there are regularizing elements d1 , d2 ∈ D for g, f g, respectively. Letting d := d1 d2 we see that d regularizes both g and f g. Since (d• )−1 (dg)• = g• is injective, the operator (dg)• also is. Hence, dg regularizes f , whence f ∈ hDi.

A generator of the AFC is an admissible subalgebra D such that hDi = Mr . Corollary 1.2.6. Let (E, M, Φ) be a proper AFC over the Banach space X. Let D, D′ be admissible subalgebras of Mb such that D ′ ⊂ hDi. Then hD ′ i ⊂ hDi. In particular, an admissible subalgebra D of Mb is a generator of the AFC if and only if E ⊂ hDi.

8

Chapter 1. Axiomatics for Functional Calculi

Proof. This follows immediately from Proposition 1.2.5.

It should be clear that abstract functional calculi over the Banach space X are the objects of a category. We describe the morphisms of this category. Let (E, M, Φ) and (E ′ , M′ , Φ′ ) be proper AFCs over the Banach space X. A morphism θ : (E, M, Φ) −→ (E ′ , M′ , Φ′ ) between these AFCs consists of a homomorphism of algebras θ : M −→ M′ with θ(E) ⊂ E ′ and θ(e)• = e• for all e ∈ E. Not surprisingly, the extension procedure is functorial, as the next proposition shows. Proposition 1.2.7. Let θ : (E, M, Φ) −→ (E ′ , M′ , Φ′ ) be a morphism of proper AFCs on the Banach space X. Then θ(Mr ) ⊂ (M′ )r with θ(f )• = f• for every f ∈ Mr .

Proof. Let f ∈ Mr and let e ∈ Reg(E) be a regularizer for f . Since θ(e)• = e• is injective and θ(e)θ(f ) = θ(ef ) ∈ θ(E) ⊂ E ′ , the element θ(e) is a regularizer for θ(f ). Moreover, θ(f )• = [θ(e)• ]−1 [θ(e)θ(f )]• = (e• )−1 θ(ef )• = (e• )−1 (ef )• = f• .

Remark 1.2.8. The above lemma may seem to be another bit of abstract nonsense on the first glance. However, it will be applied several times in the sequel, where M′ is in fact a superalgebra of M and θ is just the inclusion mapping. In such a situation Proposition 1.2.7 implies that for a consistent extension of an AFC one has to take care only of the primary calculus, whereas consistency of the extended calculus is then automatic. Remark 1.2.9. The attentive reader might have noticed that the extension procedure actually only makes use of the multiplicative 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)), everything remains true when we merely assume M to be a multiplicative monoid extending the multiplicative structure of E ⊂ M. In such a setting, the abovementioned statements involving addition can be appropriately modified. In fact, call h ∈ M a of f, g ∈ M if ef + eg = eh for all e ∈ E such that ef, eg ∈ E, and if such elements e exist. Then all statements remain true, replacing “f + g” by “h”. As a matter of fact, one has to adapt the notion of morphism which should be an algebra homomorphism on the level of E, but only a homomorphims of monoids on the level of M.

1.3 Meromorphic Functional Calculi We want to apply the results of the previous section to functional calculi for operators. Let Ω ⊂ C be an open subset of the complex plane. We denote by O(Ω)

and

M(Ω)

the algebras of holomorphic and meromorphic functions on the set Ω. (Note that M(Ω) is in fact a field.)

1.3. Meromorphic Functional Calculi

9

Let E(Ω) be a subalgebra of M(Ω) and Φ : E(Ω) −→ L(X) be an algebra homomorphism where X is a Banach space. Hence we are given an abstract functional calculus (E(Ω), M(Ω), Φ) in the terminology of the previous section. Assume that the following holds: 1. The function z := (w 7−→ w) ∈ M(Ω) is regularizable by E(Ω) whence the operator A := Φ(z) is well-defined by the extension procedure. 2. An operator T ∈ L(X) which commutes with A also commutes with each Φ(e), e ∈ E(Ω). Then we call the AFC (E(Ω), M(Ω), Φ) a (meromorphic) functional calculus for A. We write M(Ω)A := M(Ω)r and f (A) := Φ(f )

(f ∈ M(Ω)A )

in order to use a more common notation. Also we introduce the alternative notation H(A) := {f ∈ M(Ω)A | f (A) ∈ L(X)} for the set which was called M(Ω)b in our abstract setting of Section 1.2. Remark 1.3.1. Note that the terminology M(Ω)A is inaccurate in that this set heavily depends on the primary functional calculus (E(Ω), Φ). This is even more true for the notation H(A) where also the reference to the set Ω has to be understood from the context. We can now rephrase the properties of AFC’s in terms of meromorphic functional calculi. Theorem 1.3.2. (Fundamental Theorem of Functional Calculus) Let A be a closed operator on the Banach space X and (E(Ω), M(Ω), Φ) be a meromorphic functional calculus for A. Let f ∈ M(Ω)A . Then the following assertions hold. a) If T ∈ L(X) commutes with A, it also commutes with f (A). If f (A) ∈ L(X), i.e. f ∈ H(A), then f (A) commutes with A. b) 1(A) = I and (z)(A) = A.

c) Let also g ∈ M(Ω)A . Then f (A) + g(A) ⊂ (f + g)(A)

and

f (A) g(A) ⊂ (f g)(A).

Furthermore, D((f g)(A)) ∩ D(g(A)) = D(f (A)g(A)) and one has equality in the above relations if g ∈ H(A).

d) The mapping (f 7→ f (A)) : H(A) −→ L(X) is a homomorphism of algebras. e) One has f (A) = g(A)−1 f (A)g(A) if g ∈ H(A) and g(A) is injective.

10

Chapter 1. Axiomatics for Functional Calculi f ) Let λ ∈ C. Then 1 ∈ M(Ω)A λ − f (z)

⇐⇒

λ − f (A) is injective.

In this case (λ − f (z))−1 (A) = (λ − f (A))−1 . In particular, λ ∈ ̺(f (A)) if and only if (λ − f (z))−1 ∈ H(A). Proof. This is more or less a restatement of Proposition 1.2.2 and Corollaries 1.2.3 and 1.2.4. Note that in b) the statement (z)(A) = A is just a reformulation of the hypothesis.

Our definition of meromorphic functional calculus for an operator A immediately raises the question of uniqueness. Unfortunately we are unable to prove a positive result at this level of abstraction, knowing too little e.g. about the space E(Ω). We will present a uniqueness result for the functional calculus for sectorial operators in Section 5.3.4, where additional topological assumptions are needed. The next results show that for rational functions there is no problem.

1.3.1 Rational Functions Recall that for any operator A with nonempty resolvent set there is a definition of r(A) where r is a rational function on C with all its poles outside of σ(A) (see Appendix A.6). Also, for any injective operator A there is a definition of p(A) where p ∈ C[z, z −1] is a polynomial in the variables z and z −1 (see Appendix A.7). The following results shows that we meet these general definitions with any meromorphic functional calculus for A. Proposition 1.3.3. Let A be a closed operator on the Banach space X such that ̺(A) 6= ∅ and (E(Ω), M(Ω), Φ) be a meromorphic functional calculus for A. Let r be a rational function with all its poles being contained in ̺(A). Then r ∈ M(Ω)A and r(A) has its standard meaning as defined in Appendix A.6. Proof. Let us first deal with polynomials p ∈ C[z]. We prove the statement by induction on deg p. If deg p = 0 or deg p = 1, the assertion is trivial. So assume that deg p = n+1 and write p = zq+µ where µ is a scalar and deg q = n. The induction hypothesis implies that q ∈ M(Ω)A and D(q(A)) = D(An ). Since M(Ω)A is an algebra, p ∈ M(Ω)A and we have B := p(A) ⊃ q(A)A+µ. By Theorem 1.3.2, c) we also have D(B) ∩ D(A) = D(q(A)A) = D(An+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), whence R(λ, A)x ∈ D(A) ∩ D(B). This implies R(λ, A)x ∈ D(An+1 ) from which we readily infer x ∈ D(An ).

Turning to rational functions we observe that whenever λ ∈ ̺(A), one has (λ−z)−1 ∈ H(A) and (λ − z)−1 (A) = R(λ, A). This follows from Theorem 1.3.2. Hence if q ∈ C[z] is such that all zeros are contained in ̺(A) we obtain q −1 ∈ H(A) and q −1 (A) = q(A)−1 . Then r(A) = pq (A) = p(A)q −1 (A) = p(A)q(A)−1 , again by Theorem 1.3.2. The last expression is exactly how r(A) is defined in Appendix A.6.

Next we turn to injective operators. Proposition 1.3.4. Let A be an injective closed operator on the Banach space X such that ̺(A) 6= ∅ and (E(Ω), M(Ω), Φ) be a meromorphic functional calculus for

1.3. Meromorphic Functional Calculi

11

P A. Let p = k∈Z ak z k ∈ C[z, z −1 ] be a polynomial in z, z −1. Then p ∈ M(Ω)A P and p(A) = k∈Z ak Ak .

Proof. We note first that it follows from Theorem 1.3.2 and the assumptions that z, z −1 ∈ M(Ω)A . Hence C[z, z −1 ] ⊂ M(Ω)A . Now take p ∈ C[z, z −1 ] asPin the hypothesis of the k statement. We can assume that p ∈ / C[z]. Hence we can write p(z) = m k=−n ak z where a−n 6= −n n 0 and n ≥ 1. It follows from Theorem 1.3.2 that p(A) = τ (A) p(A)τ (A) = Λn (τ n p)(A), z where τ (z) := (λ−z) 2 . Writing Λ := τ (A) we obtain ! ! m+n m+n X X n n n k −2n n k Λ (τ p)(A) = Λ ( ak−n z )(λ − z) (A) = Λ ak−n z (A)(λ − A)−2n k=0

k=0

!

m+n X

= (λ − A)2n A−n ( m+n X

⊂ (λ − A)2n ( m+n X

=( !

=

k=0 m X

k=−n

k=0

k=0

ak−n Ak )(λ − A)−2n

ak−n Ak )A−n (λ − A)−2n m+n X

ak−n Ak )(λ − A)2n (λ − A)−2n A−n = ( ak Ak =

m X

(ak z k )(A) ⊂ (

k=−n

m X

ak−n Ak )A−n

k=0

ak z k )(A) = p(A)

k=−n

Here we have used Lemma A.7.2, Lemma A.7.1, Lemma A.6.1, and c) of Theorem 1.3.2. Of course we also have used implicitly the previous Proposition 1.3.4.

Taking into account also other points λ ∈ C such that λ − A is injective, leads us to the following. Theorem 1.3.5. Let A be a closed operator on the Banach space X with ̺(A) 6= ∅. 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 )). Let r = p/q and fix λ ∈ ̺(A). Take n > deg p, deg q. Then „ « „ « „ « p (λ − z)n p (λ − z)n p r(A) = (A) = (A) = (A) (A) n n q q (λ − z) q (λ − z) „ « „ « q p = (A)−1 (A) (λ − z)n (λ − z)n Where we have again used Theorem 1.3.2. Now Proposition 1.3.3 shows that r(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) 6= ∅ has a meromorphic functional calculus on the whole of C, with p E(C) = { | p, q ∈ C[z], deg(p) ≤ deg(q), {q = 0} ⊂ ̺(A)} q as the domain of the primary calculus. The domain of the extended calculus consists exactly of the rational functions r on C with no pole of r being an eigenvalue of A. Theorem 1.3.5 shows that this calculus is minimal in a sense.

12

Chapter 1. Axiomatics for Functional Calculi

1.3.2 An Abstract Composition Rule The intention behind using a functional calculus consists not only in generating new operators from old ones but also to be able to perform computations with them. Beside the basic rules given in Proposition 1.3.2 it is also very important to have at ones disposal the so-called composition rule by which we mean an identity of the form f (g(A)) = (f ◦ g)(A). (1.4) Examples are formulae like elog A = A,

(Aα )β = Aαβ ,

log(Aα ) = α log(A)

etc.

To make sense of (1.4) one needs not only a functional calculus for A but also one for g(A), and (f ◦ g) should be meaningful. As with the uniqueness question above, we are unable to prove (1.4) in full generality without knowing more about the involved functional calculi. In concrete cases, as for sectorial and strip type operators, we are in a much better situation (see Theorem 2.4.2 and Theorem 4.2.4). All we can say in general is that the problem — if there is one — is already in the primary functional calculus. Proposition 1.3.6. Let Ω, Ω′ ⊂ C be two open subsets of the plane. Assume A is a closed operator on the Banach space X and (E(Ω), M(Ω), Φ) is a meromorphic functional calculus for A. Suppose further g ∈ M(Ω)A , g : Ω −→ Ω′ and (E(Ω′ ), M(Ω′ ), Φ′ ) is a meromorphic functional calculus for g(A). Then the statement f ◦ g ∈ M(Ω)A , and (f ◦ g)(A) = f (g(A)) holds for all f ∈ M(Ω′ )g(A) provided it holds for all f ∈ E(Ω′ ). Note that by assumption f has to be holomorphic on Ω. Proof. Define B := g(A). Let f ∈ M(Ω′ )g(A) and e ∈ E(Ω′ ) be a regularizer for f . Then ef ∈ E(Ω′ ) and e(B) is injective. By assumption e ◦ g, (ef ) ◦ g ∈ M(Ω)A and (e ◦ g)(A) = e(B) as well as [(e ◦ g)(f ◦ g)](A) = [(ef ) ◦ g](A) = (ef )(B).

This shows that e ◦ g ∈ H(A) is a regularizer for f ◦ g. Employing Proposition 1.2.5 we conclude that f ◦ g ∈ M(Ω)A and (f ◦ g)(A) = f (B).

The previous proposition is important in that it shows that for nice composition rules one only has to look at the primary functional calculi. Nevertheless, this still can be a tedious task, cp. Theorem 2.4.2.

1.4 Multiplication Operators In this section we make sure that the axiomatics developed in Section 1.2 meets the intuition in the case of multiplication operators.

1.4. Multiplication Operators

13

Consider a σ-finite measure space (Ω, Σ, µ) and thereon the scale of spaces Lp (Ω, µ), where 1 ≤ p ≤ ∞. (We will usually omit reference to the underlying σ-algebra Σ.) Let M(Ω, µ; C) the space of (µ-equivalence classes of) measurable C-valued functions. Each a ∈ M(Ω, µ; C) determines a multiplication operator Ma := (f 7→ af ) on each space Lp (Ω, µ) with the maximal domain D(Ma ) = {f ∈ Lp | af ∈ Lp (Ω, µ)}. It is easy to see that Ma is closed. Moreover, the equivalence Ma ∈ L(X)

⇐⇒

a ∈ L∞ (Ω, µ)

holds. (One implication is trivial, a proof of the other can be found in [89, Theorem 1.3.4].) In this case, one has kMa k = kakL∞ . Lemma 1.4.1. The operator Ma is injective if, and only if µ(a = 0) = 0. If 1 ≤ p < ∞, the operator Ma is densely defined.

Proof. If µ(a = 0) > 0 one can find a set A such that 0 < µ(A) < ∞ and a = 0 on A. Clearly Ma 1A = 0 and 1A 6= 0 in Lp . On the other hand, if Ma is not injective, one finds 0 6= f ∈ Lp such that af = 0. Then {f 6= 0} ⊂ {a = 0}. Now let 1 ≤ p < ∞ and f ∈ Lp (Ω, µ). Then f 1{|a|≤n} → f pointwise as n → ∞, whence in Lp by Lebesgue’s theorem. But it is clear that f 1{|a|≤n} ∈ D(Ma ) for all n ∈ N.

We will set up a proper abstract functional calculus for Ma on X := Lp (Ω, µ). Let K := essran a := {λ ∈ C | µ(a ∈ U ) > 0, for each

neighborhood U of λ in C}

be the essential range of the function a. It is a closed subset of C with a(ω) ∈ K for µ-almost all ω ∈ Ω. On K we consider the measure a[µ] defined by a[µ](B) = µ(a ∈ B) for all B ⊂ K Borel. For e ∈ L∞ (K, a[µ]) we define Φ(e) := Me◦a . This determines an algebra homomorphism Φ : L∞ (K, a[µ]) −→ L(X) with Φ(1) = I. Let C∞ denote the Riemann sphere. We embed L∞ (K, a[µ]) into the set M(K, a[µ]; C∞ ) := {f : K −→ C∞ | f is a[µ]-measurable}. This set is not an algebra, but certainly a multiplicative monoid. Taking into account Remark 1.2.9 therefore yields a proper abstract functional calculus (L∞ (K, a[µ]), M(K, a[µ]; C∞ ), Φ) .

(1.5)

14

Chapter 1. Axiomatics for Functional Calculi

Theorem 1.4.2. Let (Ω, µ), a ∈ M(Ω, µ; C) and K as above. Then the set K equals the spectrum σ(Ma ) of the operator Ma (cp. Appendix A.3). For λ ∈ C \ K one has 1 . kR(λ, Ma )k = dist(λ, K) A function f ∈ M(K, a[µ]; C∞ ) is regularizable within the abstract functional calculus (1.5) if, and only if µ(a ∈ {f = ∞}) = 0. In this case we have f (Ma ) = Mf ◦a Proof. The first assertion follows from the equivalence λ∈ /K

⇐⇒

(λ − a)−1 ∈ L∞

and the fact that for each function b one has Mb ∈ L(X) if and only if b ∈ L∞ (see above). The second assertion follows from the identity kf kL∞ = sup{|λ| | λ ∈ essran f } which is straightforward to prove. Now take f ∈ M(K, a[µ]; C∞ ) which is regularizable. By definition, there is e : K −→ C measurable and bounded such that also ef is bounded and e(A) = Me◦a is injective. Hence {f = ∞} ⊂ {e = 0} and µ(a ∈ {e = 0}) = µ{e ◦ a = 0} = 0 by Lemma 1.4.1. This implies µ(a ∈ {f = ∞}) = 0. If conversely this condition holds, define e := min(1, |f1 | ). Clearly, e is measurable and bounded (by 1) and one has |ef | ≤ 1. Moreover, {e = 0} = {f = ∞}, whence from the hypothesis and Lemma 1.4.1 it follows that e(A) = Me◦a is injective. Hence, e is a regularizer for f .

Note that if λ ∈ K \ P σ(A) then, by Lemma 1.4.1, the operator f (Ma ) is independent of the value f (λ).

1.5 Concluding Remarks The considerations of the last sections lead to a somewhat surprising result: Once you have the primary calculus, the extension to a larger class of functions is easy or at least straightforward. The crucial point in constructing a functional calculus for an operator A on a Banach space X lies in setting up the primary calculus. This often requires a good knowledge of the operator in question (and even some fancy sometimes). If one favours the approach via Cauchy-integrals (as we do in this treatment), this knowledge consists mostly of insight into the growth behaviour of the resolvent near the spectrum. To illustrate this, let us turn back to the Volterra operator V introduced above. On first sight one can think that the functional calculus for V obtained so far is the end of the story. However, this is false. The reason is that the resolvent R(λ, V ) behaves nicely for λ → 0 when λ is restricted to the left halfplane. In fact, the inverse operator V −1 is just d/dt with domain D(V −1 ) = {f ∈ C1 [0, 1] | f (0) = 0}

1.5. Concluding Remarks

15

and so the operator −V −1 generates the bounded (nilpotent) right translation semigroup on X = C[0, 1]. Consequently, V −1 and V are both sectorial operators of angle π/2 (see Section 2.1 below). This means that in approaching the spectral point 0 (for V ) on a straight line with an angle ϕ such that π ≥ |arg ϕ| > π/2 the resolvent shows optimal growth behaviour, i.e. kλR(λ, V )k stays bounded. (One can show that the growth is like 1/ |λ|2 when appoaching 0 on the imaginary axis, and is more or less exponential when coming from the right.)

sup kλR(λ, V )k < ∞

ϕ

Γ

Figure 1: The contour Γ surrounds the region of rapid growth of R(., V ).

Now suppose one is given a function f , holomorphic on some open sector Sϕ := {z ∈ C \ {0} | 0 < |arg z| < ϕ}, where π/2 < ϕ < π and such that α |f (z)| = O(|z| ) for z → 0 and some α > 0. Then the Cauchy-integral (1.1) still makes sense when Γ is a contour as in Figure 1. As we will see in Chapter 2, in this way a primary calculus is set up which contains (germs of) functions holomorphic in 0 as well as (germs of) functions holomorphic on some sector Sϕ and with a good decay in 0. In particular, the fractional powers V α (Re α > 0) and the logarithm log V can be defined (see Chapter 3) via this functional calculus. After these remarks it seems hard to judge if this is now the end of the story for the Volterra operator. This situation is somehow characteristic for the state of the art in the general theory of functional calculus. Although there is a basic intuition of a kind of a “maximal functional calculus” for each single operator, this has not been made precise yet. The best achieved so far is a bundle of construc-

16

Chapter 1. Axiomatics for Functional Calculi

tions of primary calculi for certain classes of operators, like bounded operators or operators described by certain resolvent growth conditions (like sectorial operators). In the course of this text the reader will encounter more examples of such operator classes (strip type, parabola type) each coming along with a somehow natural functional calculus.

1.6 Comments The contents of this Chapter mainly rests on [54], where the notion of abstract functional calculus was introduced. . . .

Chapter 2

The Functional Calculus for Sectorial Operators In Section 2.1 the basic theory of sectorial operators is developed including examples and the concept of sectorial approximation. In Section 2.2 we introduce some notation for certain spaces of holomorphic functions on sectors. A functional calculus for sectorial operators is constructed in Section 2.3 along the lines of the abstract framework of Chapter 1. Fundamental properties including the composition rule are proved. In Section 2.5 we give natural extensions of the functional calculus to larger function spaces in case the operator is bounded and/or invertible. In this way a panorama of functional calculi is developed. In Section 2.6 some mixed topics are discussed, e.g., adjoints and restrictions of sectorial operators and some fundamental boundedness and approximation results. (This matter is continued in Chapter 5.) Finally, a (fairly general) spectral mapping theorem is proved in Section 2.7.

2.1 Sectorial Operators In the following, X always denotes a (nontrivial) Banach space and A a (singlevalued linear) operator on X. (Note the “Agreement” on page 189 in Appendix A.) For 0 ≤ ω ≤ π let ( {z ∈ C | z 6= 0 and |arg z| < ω} 0 < ω ≤ π Sω := (0, ∞) ω = 0. Hence if ω > 0, Sω denotes the open sector symmetric about the positive real axis with opening angle ω. An operator A is called sectorial of angle ω < π (in short: A ∈ Sect(ω)) if 1) σ(A) ⊂ Sω and 2) M (A, ω ′ ) := sup{kλR(λ, A)k | λ ∈ / Sω′ } < ∞ for all ω ′ ∈ (ω, π).

Figure 2 on page 18 illustrates this notion. A family of operators (Aι )ι is called uniformly sectorial of angle ω, if Aι ∈ Sect(ω) for each ι, and supι M (Aι , ω ′ ) < ∞ for all ω ′ ∈ (ω, π). Finally, we call ωA := min{0 ≤ ω < π | A ∈ Sect(ω)}

18

Chapter 2. The Functional Calculus for Sectorial Operators

sup kλR(λ, A)k < ∞ π − ω′

σ(A)

∂Sω

Figure 2: Spectrum of a sectorial operator.

the spectral angle (or sectoriality angle ) of A. Proposition 2.1.1. Let A be a closed operator on a Banach space

X.

a) If (−∞, 0) ⊂ ̺(A) and M (A) := M (A, π) := supt>0 t(t + A)−1 < ∞, 1 then M (A) ≥ 1 and A ∈ Sect(π − arcsin( M(A) )). Let A ∈ Sect(ω) for some ω ∈ [0, π). b) If A is injective, then A−1 ∈ Sect(ω), and the fundamental identity −1 −1 1 1 λ λ + A−1 =I− +A (2.1) λ λ holds for all 0 6= λ ∈ C. In particular, M (A−1 , ω ′ ) ≤ 1 + M (A, ω ′ ) for all ω ′ ∈ (ω, π).

c) Let n ∈ N and x ∈ X. Then one has x ∈ D(A) x ∈ R(A)

⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒

limt→∞ tn (t + A)−n x = x limt→∞ An (t + A)−n x = 0, and limt→0 An (t + A)−n x = x limt→0 tn (t + A)−n x = 0.

2.1. Sectorial Operators

19

d) We have N(A) ∩ R(A) = 0. If R(A) = X, then A is injective. e) The identity N(An ) = N(A) holds for all n ∈ N.

f ) The family of operators (A + δ)δ≥0 is uniformly sectorial of angle ω. In fact, M (A + δ, ω ′ ) ≤ c(ω ′ )M (A, ω ′ ) (δ > 0, ω ′ ∈ (ω, π)) where c(ω ′ ) = sin1ω′ if ω ′ ∈ (0, π2 ] and c(ω ′ ) = 1 if ω ′ ∈ [ π2 , π). The family of operators (εA)ε≥0 is uniformly sectorial of angle ω. In fact M (εA, ω ′ ) = M (A, ω ′ ) for all ω ′ ∈ (ω, π) and all ε > 0. The family of operators {(A + δ)(A + ε + δ)−1 | ε > 0, δ ≥ 0} is uniformly sectorial of angle ω.

g) Let ε > 0, n, m ∈ N and x ∈ X. Then we have n A(A + ε)−1 x ∈ D(Am ) ⇐⇒

x ∈ D(Am ).

h) If the Banach space X is reflexive, one has D(A) = X and X = N(A) ⊕ R(A). i) If 0 6= λ ∈ C with |arg λ| + ω < π, then λA ∈ Sect(ω + |arg λ|) and M (λA, ω ′ ) ≤ M (A, ω ′ − |arg λ|) for each ω ′ ∈ (|arg λ| + ω, π]. j) If D(A) = X then A′ ∈ Sect(ω, X ′ ) with M (ω ′ , A′ ) = M (ω, A) for all ω ′ ∈ (ω, π].

Proof. Ad a). Set M := M (A). For each x ∈ D(A) and λ > 0 we have

1 M kAxk + M kxk . λ This implies kxk ≤ M kxk for all x ∈ D(A). If M < 1, then D(A) = 0, but this is impossible, since X 6= 0 and ̺(A) 6= ∅. |λ | Let λ0 < 0 and M ′ > M . For each µ ∈ C with |µ − λ0 | ≤ M0′ we have µ ∈ ̺(A) and P n n+1 R(µ, A) = n (µ − λ0 ) R(λ0 , A) (Neumann series, see Proposition A.2.3). Hence it follows that „ « |µ| X |µ − λ0 |n n+1 |µ| X M n |µ| kR(µ, A)k ≤ M ≤ M |λ0 | n |λ0 |n |λ0 | n M′ „ « „ « |µ − λ0 | M 1 MM′ (M ′ + 1)M ≤ 1+ ≤ 1+ = . M ′ ′ |λ0 | M M −M M′ − M 1 − M′ kxk ≤

Now, if we choose ω ′ > π − arcsin

1 M

and define M ′ :=

1 , sin(π−ω ′ )

then it is clear that M ′ > M . |λ |

Choose µ ∈ C such that π ≥ |arg µ| ≥ ω ′ and define λ0 := Re µ. Then λ0 < 0 and |µ − λ0 | ≤ M0′ , whence kµR(µ, A)k ≤ (M ′ + 1)M/(M ′ − M ). Ad b). The identity (2.1) is true for all operators and all λ 6= 0 (cf. Lemma A.2.1 in the Appendix). The statement follows. Ad c). Consider the stated equivalence x ∈ D(A) ⇐⇒ limt→∞ tn (t + A)−n x = x. One implication is clear. For the other implication, let first x ∈ D(A) and consider the identity x = t(t + A)−1 x + 1t [t(t + A)−1 ]Ax. After repeatedly inserting this identity into itself one arrives at n 1 X x = [t(t + A)−1 ]n x + [t(t + A)−1 ]k Ax. t k=1

20

Chapter 2. The Functional Calculus for Sectorial Operators

This shows limt→∞ [t(t + A)−1 ]n x = x for x ∈ D(A). By the uniform boundedness of the operators ([t(t + A)−1 ]n )t>0 this is eventually true for all x ∈ D(A). The other equivalences are treated similarly. Ad d). This is an immediate consequence of c). Ad e). Evidently, N(A) ⊂ N(An ). But if x ∈ N(An ) and n ≥ 2, then in particular x ∈ D(An−1 ) and one has 0 = (t + A)−1 An x = A(t + A)−1 An−1 x for all t > 0. Because of An−1 x ∈ R(A) one can apply c) to obtain An−1 x = 0. By repeating this argument one finally arrives at Ax = 0. Ad f ). Define (for the moment) Aδ := A + δ, and let ω ′ ∈ (ω, π). From λ ∈ Sπ−ω ′ it follows that λ + δ ∈ Sπ−ω ′ , and because of λ (δ + λ)(A + δ + λ)−1 δ+λ ˛ ˛ ˛ λ ˛ we obtain M (Aδ , ω ′ ) ≤ M (A, ω ′ ) supλ∈S Hence, the family (Aδ )δ≥0 is uniformly ′ ˛ λ+δ ˛. λ(λ + Aδ )−1 =

π−ω

sectorial of angle ω and M (Aδ , ω ′ ) ≤ c(ω ′ )M (A, ω ′ ) with c(ω ′ ) is as described. The uniform sectoriality of the family (εA)ε≥0 is clear from the fact that sectors are invariant under dilations with positive factors. λ To prove the third statement, choose ε > 0 . Because λ ∈ Sπ−ω ′ we have 1+λ ε + δ ∈ Sπ−ω ′ . For these λ one obtains „ « λ λ + Aδ (Aδ + ε)−1 = (λ(Aδ + ε) + Aδ )(Aδ + ε)−1 = (λ + 1) Aδ + ε (Aδ + ε)−1 . 1+λ Hence the operator λ + Aδ (Aδ + ε)−1 is invertible and λ[λ + Aδ (Aδ + ε)−1 ]−1

= = =

λ λ (Aδ + ε)[Aδ + ε]−1 1+λ 1+λ λ λ ε λ (Aδ + ε+ )[Aδ + ε]−1 1+λ 1+λ 1+λ 1+λ " „ «−1 # λ 1 λε λε I+ Aδ + . 1+λ 1+λ 1+λ 1+λ

The uniform sectoriality of the family (Aδ )δ now gives the first statement of f ). Ad g). We consider first the reverse implication. To prove it we select x ∈ D(Am ). Then A(A + ε)−1 x = x − ε(A + ε)−1 x ∈ D(Am ), and an iteration of this argument yields (A(A + ε)−1 )n x ∈ D(Am ). The proof of the direction “⇒” can also be reduced to the case n = 1 which is proved by induction on m. Ad h). Let x ∈ X and X be reflexive. The sequence (n(n + A)−1 x)n∈N is bounded, hence it has a weakly convergent subsequence nk (nk + A)−1 x ⇀ y. This means that A(nk + A)−1 x ⇀ x − y. Now, (nk + A)−1 x → 0 even strongly. Because the graph of A is closed and a linear subspace of X ⊕ X, it is weakly closed, whence x − y = 0. But this means that x lies in the weak closure of D(A) which is the same as the strong closure, for D(A) is a subspace of X. Altogether it follows that x ∈ D(A). Let x ∈ N(A) ∩ R(A). Then we have 0 = Ax = limt→0 (t + A)−1 Ax = limt→0 A(t + A)−1 x = x by c). Therefore, the sum is direct. For arbitrary x ∈ X, by the reflexivity of X on can find a sequence tn ց 0 and a y such that tn (tn + A)−1 x ⇀ y. But we have tn A(tn + A)−1 x → 0. The graph of A is weakly closed, hence y ∈ N(A). This implies that A(tn + A)−1 x ⇀ x − y. Therefore, x − y is in the weak closure of R(A) which is identical to the strong closure R(A), for R(A) is a subspace of X. It follows that x ∈ N(A) + R(A). Ad i). If µ ∈ / Sω ′ , then clearly µλ−1 ∈ / Sω ′ −|arg λ| . Hence µR(µ, λA) = (µλ−1 )R(µλ−1 , A) is uniformly bounded by M (A, ω ′ − |arg λ|) for such µ. Ad k). This follows since ̺(A) = ̺(A′ ) and kR(λ, A)k = kR(λ, A)′ k = kR(λ, A′ )k for all λ ∈ ̺(A), see Proposition A.4.2 and Corollary A.4.3.

2.1. Sectorial Operators

21

Remark 2.1.2. Let A ∈ Sect(ω) be a sectorial operator. Then M (A, ϕ) =

inf

ω ′ ∈(ω,ϕ)

M (A, ω ′ )

for each ϕ ∈ (ω, π]. This follows from the proof of Proposition 2.1.1, a). Actually, the function (ϕ 7−→ M (A, ϕ)) : (ω, π] −→ [1, ∞) is continuous, see [89, Prop. 1.2.2]. Let A ∈ Sect(ω) on the Banach space X. We define Y := R(A) and denote by B the part of A in Y , i.e., D(B) := D(A) ∩ R(A)

with By := Ay

(y ∈ D(B)).

It is easy to see that B ∈ Sect(ω) on Y with M (ω ′ , B) ≤ M (ω ′ , A) for all ω < ω ′ ≤ π. Moreover, B is injective. Therefore, we call B the injective part of A. The identity D(B n ) = D(An ) ∩ Y (n ∈ N) is easily proved by induction. Proposition 2.1.1 yields N(A)∩R(A) = N(A)∩Y = 0, and by a short argument one obtains R(λ, A)(x ⊕ y) =

1 x ⊕ R(λ, B)y λ

for all x ∈ N(A), y ∈ Y .

2.1.1 Examples We now want to browse shortly through a list of examples. Multiplication Operators Let (Ω, Σ, µ) be a σ-finite measure space, and a ∈ Mes(Ω, µ; C) as in Section 1.4. Let A := Ma be the multiplication operator on Lp (Ω, µ). Theorem 1.4.2 shows that K = essran a equals the spectrum σ(A). We claim that A is sectorial of angle ω ∈ [0, π) if and only if K ⊂ Sω . This clearly is a necessary condition, but we also have 1 1 kR(λ, A)k = ≤ dist(λ, K) dist(λ, Sω ) for all λ ∈ C \ Sω 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).

22

Chapter 2. The Functional Calculus for Sectorial Operators

Generators of Semigroups Let −A generate a bounded semigroup (T (t))t>0 in the sense of Appendix A.8. Then with M := supt>0 kT (t)k one has

(λ + A)−1 = kR(λ, −A)k ≤ M Re λ

(Re λ > 0)

(2.2)

by the (easy part of the) Hille-Yosida Theorem A.8.6. This shows that A is sectorial of angle π2 . It is well-known that an estimate of the form (2.2) does not suffice to ensure that −A generates a semigroup, not even if A is densely defined (see [?]). However, there is a perfect correspondence between generators of bounded analytic semigroups and sectorial operators of angle strictly less than π2 . See Section 3.3 for more information. Normal Operators and Numerical Range Conditions Let X = H be a Hilbert space and A be a normal operator on H. By the Spectral Theorem D.5.1, the operator A is similar to a multiplication operator on some L2 -space. Hence by our first example, A is sectorial of angle ω if and only if σ(A) ⊂ Sω . If A is not normal but has its numerical range W (A) contained in Sω for some ω ∈ [0, π2 ] then A is sectorial of angle ω if and only if A has some resolvent point outside the sector. This follows from Appendix C.3. The Hilbert space theory is dealt with more deeply in Chapter 7. Other, more concrete examples will appear in Section ??. One can also look in [89, Chapter 1 and 2].

2.1.2 Sectorial Approximation The present section formalizes an important approximation tool in the theory of sectorial operators. A uniformly sectorial sequence (An )n∈N of angle ω is called a sectorial approximation on Sω for the operator A, if λ ∈ ̺(A)

and R(λ, An ) → R(λ, A) in L(X).

(2.3)

for some λ ∈ / Sω . From Proposition A.5.3 it follows that in this case (2.3) is true for all λ ∈ / Sω . Moreover, A itself is sectorial of angle ω. If (An )n is a sectorial approximation for A on Sω , we write An → A (Sω ) and speak of sectorial convergence. Proposition 2.1.3. a) If An → A (Sω ) and all An as well as A are injective, then A−1 → A−1 (Sω ).

2.2. Spaces of Holomorphic Functions

23

b) If An → A (Sω ) and A ∈ L(X), then An ∈ L(X) for large n and An → A in norm. c) If An → A (Sω ) and 0 ∈ ̺(A), then 0 ∈ ̺(An ) for large n.

d) If (An )n ⊂ L(X) is uniformly sectorial of angle ω and if An → A in norm, then An → A (Sω ). e) If A ∈ Sect(Sω ), then (A + ε)ε>0 is a sectorial approximation for A on Sω . f ) If A ∈ Sect(Sω ), then (Aε )0<ε≤1 , where

Aε := (A + ε) (1 + εA)−1 ,

is a sectorial approximation for A on Sω . Proof. Assertion a) follows from b) in Proposition 2.1.1. If An → A (Sω ) and A ∈ L(X), then (1 + An )−1 → (1 + A)−1 in norm. But the set of bounded invertible operators on X is open with the inversion mapping being continuous, hence eventually (1 + An )−1 is invertible, and −1 . Then (1 + An ) → (1 + A) in norm. This gives b). To prove c) define Bn := A−1 n and B := A Bn and B are possibly multivalued operators. However, by the Spectral Mapping Theorem for the resolvent (Proposition A.3.1) and Lemma A.2.1 we have (1 + B)−1 , (1 + Bn )−1 ∈ L(X) and (1+Bn )−1 → (1+B)−1 in norm. With the same argument as in the proof of b) we conclude that Bn ∈ L(X) for large n. For d) suppose that ‚(An )n is uniformly sectorial with An → A in norm. ‚ Then (1 + An ) → (1 + A) in norm and supn ‚(1 + An )−1 ‚ < ∞. This implies (1 + A)−1 ∈ L(X) and (1 + An )−1 → (1 + A)−1 in norm. Finally, we prove e) and f ). Let A be sectorial. The uniform sectoriality of (A + ε)ε>0 has been shown in the proof of part f ) of Prop. 2.1.1. But it is clear that (1 + ε + A)−1 → (1 + A)−1 in norm, whence e) follows. Assertion f ) results from Prop. 2.1.1, part f ) and the identity (A + ε)(1 + εA)−1 = ε−1 (A + ε)(A + ε +

1−ε2 −1 ) . ε

Remark 2.1.4. Although we have assumed throughout the section that A is singlevalued, the definition of sectoriality makes perfect sense even if A is multivalued. Although we deal mostly with single-valued operators, sometimes it is quite illuminating to have the multivalued case in mind. Therefore, we will speak of a “multivalued sectorial operator” whenever it is convenient. Note that the fundamental identity (2.1) still holds in the multivalued case and implies readily that the inverse of a sectorial operator is sectorial with the same angle. One has x ∈ D(A) ⇔ limt→∞ t(t + A)−1 x = x in the multivalued case as well (cf. c) of Prop. 2.1.1). This shows A0 ∩ D(A) = 0. Most statements of this section remain true in the multivalued case, at least after adapting notation a bit. As a rule, one −1 −1 has to replace expressions ofthe form B(B + λ) by I − λ(B + λ) . For example 1 1−ε2 −1 we obtain that Aε = ε − (1 + εA) is a sectorial approximation of A ε by bounded and invertible operators (cf. f ) of Proposition 2.1.3). Note also that part k) of Proposition 2.1.1 holds even without the assumption that A is densely defined. In this case the adjoint A′ will be a multivalued sectorial operator.

2.2 Spaces of Holomorphic Functions In the next section we will construct a functional calculus for sectorial operators. This will be done by proceeding along the lines of Chapter 1 with the primary

24

Chapter 2. The Functional Calculus for Sectorial Operators

calculus constructed by means of a Cauchy integral. Since the spectrum of a sectorial operator in general is unbounded, one has to integrate along infinite lines (here: the boundary of a sector). As a matter of fact, this is only possible for a restricted collection of functions. Dealing with these functions requires some notation which we now introduce. As already stated in Chapter 1 we write O(Ω) to denote the space of all holomorphic functions and M(Ω) to denote the space of all meromorphic functions on the open set Ω ⊂ C. Suppose A denotes a sectorial operator of angle ω on a Banach space X. We want to define operators of the form Z 1 f (z)R(z, A) dz (2.4) f (A) := 2πi Γ where f ∈ O(Sϕ ) (ϕ > ω), and the path Γ “surrounds” the sector Sω in the positive sense. This means in particular that — considered as a curve on the Riemann sphere — Γ passes through the point ∞. To give meaning to the above integral, the function f should have a quick decay at ∞. We therefore introduce the notion of polynomial limit. Polynomial Limits Let ϕ ∈ (0, π], f ∈ M(Sϕ ) and c ∈ C. We say that f has polynomial limit c at α 0 if there is α > 0 such that f (z) − c = O(|z| ) as z → 0. We say that f has polynomial limit ∞ at 0 if 1/f has polynomial limit 0 at 0. Similarly, we say that f has polynomial limit d ∈ C∞ at ∞ if f (z −1 ) has polynomial limit d at 0. We say that f has a finite polynomial limit at 0 (at ∞) if there is c ∈ C such that f has polynomial limit d at 0 (at ∞). If f has polynomial limit 0 at 0 (at ∞), we call f regularly decaying at 0 (at ∞). Clearly, if f, g both have finite polynomial limits at 0 (at ∞) also f g and f + g do so. Remarks 2.2.1. 1) For a function f to have a polynomial limit at ∞ means that f has a limit (within C∞ ) and this limit is approached at least polynomially fast. Note that if this limit is ∞ this does not imply that f is polynomially bounded at ∞. In fact, the function ez has polynomial limit ∞ on Sϕ for every 0 < ϕ < π2 but is far from being polynomially bounded. 2) If f is meromorphic in 0, then f has a polynomial limit at 0 and this limit is finite if and only if f is holomorphic in 0. The same remark applies to the point ∞.

Now look again at the Cauchy integral (2.4). By the sectoriality of A, the function f being regularly decaying at ∞ guarantees integrability at infinity, at least if Γ is eventually straight. The same holds if f is regularly decaying at ∞. It is therefore natural to consider the so-called Dunford-Riesz class on Sϕ , defined by H0∞ (Sϕ ) := {f ∈ H ∞ (Sϕ ) | f is regularly decaying at 0 and at ∞},

2.2. Spaces of Holomorphic Functions

25

where H ∞ (Sϕ ) := {f ∈ O(Sϕ ) | f is bounded} is the Banach algebra of all bounded, holomorphic functions on Sϕ . Obviously, H0∞ (Sϕ ) is an algebra ideal in the algebra H ∞ (Sϕ ). With each f (z) also the function f (1/z) is a member of H0∞ (Sϕ ).The following description is often helpful. Lemma 2.2.2. Let ϕ ∈ (0, π] and let f : Sϕ −→ C be holomorphic. The following assertions are equivalent: (i) The function f belongs to H0∞ (Sϕ ).

n o s −s (ii) There is C ≥ 0 and s > 0 such that |f (z)| ≤ C min |z| , |z| for all z ∈ Sϕ . (iii) There is C ≥ 0 and s > 0 such that |f (z)| ≤ C (iv) There is C ≥ 0 and s > 0 such that |f (z)| ≤ C Proof. The proof is easy and we omit it.

|z|s 1+|z|2s

for all z ∈ Sϕ . s |z| for all z ∈ Sϕ . 1+|z|2

−1

It is obvious that neither the rational function (1 + z) function 1 is contained in H0∞ (Sϕ ). We therefore define

E(Sϕ ) := H0∞ (Sϕ ) ⊕ (1 + z)−1 ⊕ h1i

nor the constant

called the extended Dunford-Riesz class. This set is in fact a subalgebra of H ∞ (Sϕ ), due to the identity 1 z 1 = − . 2 (1 + z) 1 + z (1 + z)2

Lemma 2.2.3. Let ϕ ∈ (0, π] and f : Sϕ −→ C holomorphic. The following assertions are equivalent: (i) The function f belongs to E(Sϕ ).

(ii) The function f is bounded and has finite polynomial limits at 0 and ∞.

Proof. The implication (i) ⇒ (ii) is obvious. To prove the converse assume f ∈ H ∞ (Sϕ ) has f (0)−f (∞) finite polynomial limits at 0 and ∞. Then f (z)−f (∞)− is contained in H0∞ (Sϕ ). 1+z

We remark that both algebras H0∞ (Sϕ ) and E(Sϕ ) are invariant under inversion, i.e., with f (z) also f (z −1 ) belongs to the set. This is clear for H0∞ (Sϕ ). For the larger algebra it follows from 1 z 1 = =1− . −1 1+z 1+z 1+z If the precise sector Sϕ is understood from the context, we will simply write H0∞ instead of H0∞ (Sϕ ) in the sequel. The same applies to the other function

26

Chapter 2. The Functional Calculus for Sectorial Operators

spaces. Given ω ∈ [0, π) we let H0∞ [Sω ] :=

[

H0∞ (Sϕ )

ϕ∈(ω,π)

and use a similar notation for the other function spaces, e.g., E[Sω ], O[Sω ], M[Sω ] . . . . Note that all these spaces can be viewed as algebras in a natural way. In the following we will present some instances of functions from the classes we have introduced. Example 2.2.4. Let f ∈ H ∞ (Sϕ ) be regularly decaying at ∞ and have a holomorphic extension to a neighborhood of 0. Then f ∈ E(Sϕ ). In fact, by holomorphy, there is a constant C such that |f (z) − f (0)| ≤ C |z| for z near 0. Hence f has a finite polynomial limit at 0. An important special case is the function e−z , provided one takes ϕ ∈ (0, π2 ). Example 2.2.5. Let 0 < Re α < Re β. Then zα 1 ∈ H0∞ (Sϕ ), ∈ E(Sϕ ), β (1 + z) (1 + z)α

and

zα ∈ E(Sϕ ) (1 + z)α

for all ϕ ∈ (0, π).

Proof. The function (1+z)−α clearly is regularly decaying at ∞ and has a holomorphic extension round 0. Hence by Example 2.2.4 it lies in E. This implies readily that z β (1 + z)−α ∈ H0∞ . Since (zw)α = z α w α whenever z, w, zw ∈ Sπ , we have „ «α „ «α 1 1 z zα = = = , −1 α −1 (1 + z ) 1+z 1+z (1 + z)α also the last assertion is clear.

Next, we consider an important construction. Example 2.2.6. Let ψ ∈ H0∞ (Sϕ ) and define Z 1 Z ∞ ds ds h(z) := ψ(sz) and g(z) := ψ(sz) s s 0 1 for z ∈ Sϕ . Then h, g ∈ E(Sϕ ), h(0) = g(∞) = 0, and Z ∞ ds h(z) + g(z) = ψ(s) s 0

is constant.

n o Proof. Choose C, α > 0 such that |ψ(z)| ≤ C min |z|α , |z|−α . Then Z 1 ds C |h(z)| ≤ C |sz|α = |z|α . s α 0

Analogously, |g(z)| ≤ C |z|−α . This shows not only that g, h are well defined, but also that h α behaves well at 0 and g behaves well at ∞. By Morera’s theorem, g and h are also holomorphic. The function Z ∞ ds c(z) := ψ(sz) s 0 is constant on (0, ∞) (by change of variables) hence on the whole sector Sϕ (by holomorphy). Thus we can write h − c = g which shows g, h ∈ E(Sϕ ) and g(0) = h(∞) = c.

2.3. The Natural Functional Calculus

27

2.3 The Natural Functional Calculus In this section A always denotes a sectorial operator of angle ω on a Banach space X. We pursue our idea of defining a functional calculus by means of a Cauchy integral. Following the abstract setting of Section 1.2 and Section 1.3, we start with the primary calculus.

2.3.1 Primary Functional Calculus via Cauchy Integrals Let ϕ ∈ (0, π) und δ > 0. We call Γϕ = ∂Sϕ the boundary of the sector Sϕ , oriented in the positive sense, i.e., Γϕ := −R+ eiϕ ⊕ R+ e−iϕ . For ϕ ∈ (ω, π) and f ∈ H0∞ (Sϕ ) we define Z 1 f (A) := f (z) R(z, A) dz 2πi Γω′

(2.5)

where ω ′ ∈ (ω, ϕ) is arbitrary. A standard argument using Cauchy’s integral theorem shows that this definition is actually independent of ω ′ . (Figure 3 on page 28 illustrates the definition of f (A).) Lemma 2.3.1. . Let A ∈ Sect(ω) and ϕ ∈ (ω, π). Then the following assertions hold. a) The mapping (h → 7 h(A)) : H0∞ (Sϕ ) −→ L(X) is a homomorphism of algebras. b) If B is a closed operator commuting with the resolvents of A, then B also commutes with f (A). In particular, f (A) commutes with A and R(λ, A) for all λ ∈ ̺(A). c) We have R(λ, A)f (A) = ((λ − z)−1 f )(A) for each λ ∈ / Sϕ .

Proof. Ad a). This is a simple application of Fubini’s theorem and the resolvent identity. Ad b). This is trivial. Ad c). Let g := (λ − z)−1 f and Γ = Γω ′ . We compute Z Z 1 1 g(z)(λ − A)R(z, A) dz = g(z)(λ − z)R(z, A) + I dz (λ − A)−1 g(A) = 2πi Γ 2πi Γ Z Z 1 1 = f (z)R(z, A) dz + g(z) dz = f (A) 2πi Γ 2πi Γ since the latter summand equals 0 by Cauchy’s theorem.

We extend the definition of f (A) from f ∈ H0∞ (Sϕ ) to all f ∈ E(Sϕ ). This can be done by defining g(A) := f (A) + c(1 + A)−1 + d

(2.6)

28

Chapter 2. The Functional Calculus for Sectorial Operators

Γω ′

f (A) =

1 2πi

R

Γ

...

σ(A)

∂Sϕ ∂Sω Figure 3: Integration into 0 is possible due to rapid decay of f .

2.3. The Natural Functional Calculus

29

whenever g = f + c(1 + z)−1 + d with f ∈ H0∞ (Sϕ ) and c, d ∈ C. We will show that this constitutes an algebra homomorphism Φ := (g 7−→ g(A)) : E(Sϕ ) −→ L(X) which gives rise to a meromorphic functional calculus (E(Sϕ ), M(Sϕ ), Φ) in the sense of Section 1.3. To establish this fact, there is still some work to do. Let us introduce the notation ∞ H(0) (Sϕ ) := {f ∈ H ∞ (Sϕ ) | f is regularly decaying at ∞

and holomorphic around 0}

∞ We saw in Example 2.2.4 that H(0) (Sϕ ) ⊂ E(Sϕ ). The next lemma shows that for ∞ f ∈ H(0) (Sϕ ) the value of f (A) can also be computed by a Cauchy integral. ∞ Lemma 2.3.2. Let f ∈ H(0) (Sϕ ), ω ′ ∈ (ω, ϕ), and let δ > 0 be small enough so

that f is holomorphic in a neighborhood of Bδ (0). Then Z 1 f (z)R(z, A) dz f (A) = 2πi Γ

where Γ = Γω′ ,δ is the positively oriented boundary of Sω′ ∪ Bδ(0) . Figure 4 on page 30 gives an impression what is going on. ∞ (S ) ∩ H ∞ (S ) then the statement is true, since one can shrink the path Proof. If f ∈ H(0) ϕ ϕ 0 ∞ can be written around 0 to 0 without changing the value of the integral. As a general f ∈ H(0)

∞ ∩ H ∞ , we are left to show the claim for f (z) = (1 + z)−1 . as f = g + c(1 + z)−1 with g ∈ H(0) 0 The idea is to introduce the contour Γ′ := −Γω ′ ,R with R > 1. Cauchy’s theorem shows that R −1 R(z, A) dz = 0. In fact, without changing the value of the integral, we can shift the Γ′ (1 + z) path to the left which makes Rthe value of the integral tend to 0. However, if we add the integral R . . . to the initial integral Γ ′ . . . , some parts cancel and only a simple curve surrounding Γ′ ω ,δ

the singularity −1 remains. Applying Cauchy’s theorem again yields −R(−1, A) = (1+A)−1 .

Now we reach our goal. Theorem 2.3.3. Let A ∈ Sect(ω) on X and ϕ ∈ (ω, π). The mapping ΦA := (g 7−→ g(A)) : E(Sϕ ) −→ L(X) defined by (2.6) is a homomorphism of algebras. Morover, it has the following properties: a) z(1 + z)−1 (A) = A(1 + A)−1 .

b) If B is a closed operator commuting with the resolvents of A, then B also commutes with f (A). In particular, f (A) commutes with A. c) If x ∈ N(A) and f ∈ E(Sϕ ), then f (A)x = f (0)x.

30

Chapter 2. The Functional Calculus for Sectorial Operators Γω′ ,δ

f (A) =

1 2πi

R

Γ

...

σ(A)

∂Sω Figure 4: The contour of integration avoids 0 with f being analytic there.

d) Let B denote the injective part of A. Then Y := R(A) is invariant under the action of f (A), and one has f (B) = f (A)|Y . Proof. Take two functions gi = fi + ci (1 + z)−1 + di (i = 1, 2) in E with fi ∈ H0∞ and ci , di ∈ C. Since Φ is linear one has to care for all mixed products. When a constant di is involved there is nothing to show, the products f1 f2 and fi (1 + z)−1 are dealt with using Lemma 2.3.1. The assertion for the remaining product (1 + z)−1 (1 + z)−1 = (1 + z)−2 follow from Lemma 2.3.2, Fubini’s theorem and the resolvent identity. Ad a) We use what we have already proven and compute (z(1 + z)−1 )(A) = 1(A) − (1 + z)−1 (A) = I − (1 + A)−1 A(1 + A)−1 . Ad b). This is trivial. Ad c). For x ∈ N(A) and z ∈ ̺(A) one has R(z, A)x = z1 x. Hence one has Z 1 f (z) f (A)x = dzx = 0 2πi Γ z

for f ∈ H0∞ , by Cauchy’s theorem. The rest follows. Ad d). The statement follows from the fact that Y is R(λ, A)-invariant, with R(λ, A)|Y = R(λ, B) for all λ ∈ ̺(A).

We call the algebra homomorphism ΦA = (f 7−→ f (A)) : E(Sϕ ) −→ L(X)

(2.7)

2.3. The Natural Functional Calculus

31

the primary functional calculus (in short: PFC) on Sϕ for A as a sectorial operator. Remark 2.3.4. Note that the definition of this primary calculus is perfectly meaningful even when A is multivalued. Theorem 2.3.3 holds true (after reformulating part d) appropriately) even in this more general case. We will not use this fact except for Section 3.3 when dealing with holomorphic semigroups.

2.3.2 The Natural Functional Calculus We continue to have A ∈ Sect(ω) and ϕ ∈ (ω, π). By Theorem 2.3.3 we have established an abstract functional calculus (E(Sϕ ), M(Sϕ ), ΦA ) which is proper, since (1 + z)−1 (A) = (1 + A)−1 is injective. In fact, by a) of Theorem 2.3.3 the function (1 + z)−1 regularizes z and z(A) = (1 + A)(z(1 + z)−1 )(A) = (1 + A)A(1 + A)−1 = A Hence, in the terminology of Section 1.3 our abstract functional calculus is even a meromorphic functional calculus for A on Sϕ . We write M(Sϕ )A := {f ∈ M(Sϕ ) | ∃ e ∈ E(Sϕ ) : e(A) is injective and ef ∈ E(Sϕ )} to denote the class of functions which are regularizable within this abstract functional calculus. For f ∈ M(Sϕ ), the operator f (A) is defined by f (A) := e(A)−1 (ef )(A) with e ∈ E(Sϕ ) being an arbitrary regularizer of f . The next lemma is quite useful. Lemma 2.3.5. Let A ∈ Sect(ω), ϕ ∈ (ω, π) and f ∈ M(Sϕ )A . Then one can find a regularizer e ∈ E(Sϕ ) for f with e(∞) = 0. If A is injective, one can even find a regularizer e ∈ H0∞ (Sϕ ).

Proof. If d is any regularizer for f then also e := (1 + z)−1 d is one which satisfies e(∞) = 0. If A is injective, then e := z(1 + z)−2 d is a regularizer for f with e ∈ H0∞ .

Since we have a meromorphic functional calculus for Am it follows from Theorem 1.3.2 that (µ − z)(A) = µ − A,

(λ − z)−1 (A) = R(λ, A)

(2.8)

for all µ ∈ C and λ ∈ ̺(A). More general, r(A) has its usual meaning (as explained in Appendix A.6) for every rational function r with poles in ̺(A) (Proposition 1.3.3). Now note that we can choose ϕ arbitrarily close to ω. For ϕ1 > ϕ2 > ω we can view M(Sϕ1 ) in a natural way as a subalgebra of M(Sϕ2 ) and we have

32

Chapter 2. The Functional Calculus for Sectorial Operators

likewise E(Sϕ1 ) ⊂ E(Sϕ2 ). The corresponding functional calculi are consistent, which mean that these inclusions give rise to a morphism (see Proposition 1.2.7) (E(Sϕ1 ), M(Sϕ1 ), ΦA ) ֒→ (E(Sϕ2 ), M(Sϕ2 ), ΦA ). S S Writing E[Sω ] := ω<ϕ<π E(Sϕ ) and M[Sω ] := ω<ϕ<π M(Sϕ ) we thus (as an “inductive limit”) obtain a meromorphic functional calculus (E[Sω ], M[Sω ], Φ) which we call the natural functional calculus for A as a sectorial operator on Sω . Note that again we can view M[Sω ] as a subalgebra of M(Sω ) so that this calculus is in fact a meromorphic functional calculus for A on Sω (in the terminology of Section 1.3). The set [ M[Sω ]A = M(Sϕ )A ϕ∈(ω,π)

is called the domain of this calculus. Finally, we write H(A) := {f ∈ M[Sω ]A | f (A) ∈ L(X)} as in Chapter 1. For simplicity, we will frequently omit explicit reference to the sector and simply write “MA ” instead of “M[Sω ]A ”, provided ambiguities are excluded. Also we will use the abbreviation O[Sω ]A := O[Sω ] ∩ MA and similar ones for other function classes. Remarks 2.3.6. 1) One should keep in mind that a function f belonging to E[Sω ] or M[Sω ]A , is in fact defined on a larger sector Sϕ for some ϕ > ω. 2) As in Section 1.3 we warn the reader not to forget that our notations M(Sϕ )A , M[Sω ]A and H(A) are somewhat inaccurate. In fact, these sets do not only depend on A but also on the domain of the primary calculus. Since we are going to encounter other primary calculi in later sections, this remark should always be recalled. 3) To distinguish different functional calculi which may be defined for the same operator A (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 for A as a bounded operator”. If A is not invertible then 1 z 2 is in the domain of the n.f.c for A as a sectorial, but not as a bounded 1 operator. Similarly, the function e r−z for r > r(A) is in the domain of the n.f.c for A as a bounded, but not as a sectorial operator. The general properties of the so defined “natural functional calculus” for A ∈ Sect(ω) as a sectorial operator are summarized in Theorem 1.3.2 with Ω := Sω

2.3. The Natural Functional Calculus

33

(and, in fact, M[Sω ]A instead of M(Sω )A ). We will often use them without further mention. Which functions f are contained in MA ? The next lemma shows that a limit behaviour in 0 is necessary if A is not injective. Lemma 2.3.7. Assume A ∈ Sect(ω), f ∈ MA and λ ∈ Sω . Then (at least) one of the following assertions holds. α

1) There is c ∈ C and α > 0 such that f (z) − c = O(|z| ) for z → λ. 2) The operator λ − A is injective.

If A is not injective, then f (A)x = f (0)x for all x ∈ N(A).

Proof. Let e ∈ E be a regularizer for f . Then also ef ∈ E. Suppose that f fails 1) and λ 6= 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 − λ must be injective. To complete the proof, assume λ = 0 and A is not injective. Since e(A)x = e(0)x for every x ∈ N(A) (Theorem 2.3.3) and e(A) is injective, e(0) 6= 0. Hence f (z) − f (0) =

[e(z)f (z) − e(0)f (0)] − f (0)[e(z) − e(0)] e(z)

for z near 0, which shows the assertion.

Corollary 2.3.8. Let X be reflexive and A ∈ Sect(ω) with N(A) 6= 0, and denote by B the injective part of A. If f ∈ MA , then also f ∈ MB with D(f (A)) = N(A)⊕D(f (B)) and f (A)(x⊕y) = f (0)x⊕f (B)y for all x⊕y ∈ N(A)⊕D(f (B)).

Proof. Note first that f (0) makes sense since A is not injective. Let e be a regularizer for f in the functional calculus for A. Then e, ef ∈ E with e(0) 6= 0 (since A is not injective) and e(A) is injective. By Theorem 2.3.3, e(A)(x ⊕ y) = e(0)x ⊕ e(B)y for all x ⊕ y ∈ N(A) ⊕ R(A), whence e(B) is also injective. This shows that f ∈ MB and f (A) = e(A)−1 (ef )(A) = e(0)−1 ⊕ e(B)−1 [(ef )(0) ⊕ (ef )(B)] = f (0) ⊕ f (B)

in suggestive notation.

2.3.3 Functions of Polynomial Growth The search for regularizers immediately leads one to think of resolvents, like (1 + z)−1 , (1 + z)−2 , . . . . Therefore we define A(Sϕ ) := {f : Sϕ −→ C | ∃ n ∈ N : f (z)(1 + z)−n ∈ E(Sϕ )}, S and A[Sω ] := ϕ∈(ω,π) A(Sϕ ). If ω is understood, we write simply A instead of A[Sω ]. Lemma 2.3.9. For f ∈ O[Sω ] the following assertions are equivalent: (i) The function f belongs to A.

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

1) f (z) = O(|z| ) (z → ∞) for some α ∈ R, and 2) f has a finite polynomial limit at 0.

34

Chapter 2. The Functional Calculus for Sectorial Operators (iii) There is c ∈ C, n ∈ N, and F ∈ H0∞ such that f (z) = c + (1 + z)n F (z).

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

In particular, A[Sω ] is an algebra of functions containing every rational function with poles outside of Sω . Proof. (i)⇒(ii): If g(z) := f (z)(1 + z)−n ∈ E, then clearly condition 1) is satisfied. Because g has finite polynomial limit at 0, also f = (1 + z)n g must have one as well. (ii)⇒(iii): Let c := f (0), choose α as in (ii), and let n > α. Then clearly (f (z) − c)(1 + z)−n ∈ H0∞ . The implication (iii)⇒(i) is trivial.

Proposition 2.3.10. Let A ∈ Sect(ω). Then A[Sω ] ⊂ MA . If f ∈ A[Sω ], the following assertions hold. a) If A is bounded, so is f (A). b) One has [(z − µ)f (z)](A) = (A − µ)f (A) for all µ ∈ C.

c) If D(A) = X and f (z)(1 + z)−n ∈ E, then D(An ) is a core for f (A). Proof. It is immediate that A[Sω ] ⊂ MA . Let f ∈ A and choose n such that F := f (1 + z)−n ∈ E. If A is bounded, then f (A) = (1 + A)n F (A) ∈ L(X). This proves a). To prove b), note that « „ f (z)(z − µ) ((z − µ)f (z))(A) = (1 + A)n+1 (A) = (1 + A)n+1 [(A − µ)(1 + A)−1 ]F (A) (1 + z)n+1 = (A − µ)(1 + A)n+1 (1 + A)−1 F (A) = (A − µ)(1 + A)n F (A) = (A − µ)f (A).

Here we have used (2.8) and Lemma A.6.1. Finally, we prove c). Note that the operator F (A) commutes with (1 + A)−n , whence D(An ) is F (A)-invariant. This gives D(An ) ⊂ D(f (A)). For arbitrary x ∈ D(f (A)) we have Tt (x) := (t(t + A)−1 )n x → x with t → ∞. As a matter of fact, Tt (x) ∈ D(An ) and, by c) of Proposition 2.1.1. also f (A)Tt x = Tt f (A)x → f (A)x.

Remark 2.3.11. Interesting enough, polynomial growth in ∞ is not the best one can achieve for an arbitrary sectorial operator. In fact, we will see in Section 3.3 that if ωA < π2 then e−A := e−z (A) is injective. (See Proposition 3.3.3.) Hence e−z is a regularizer which compensates even exponential growth in ∞. If α ωA happens to be larger than π2 , one can use the functions e−z with suitable 0 < α < 1. Therefore we see that O[Sω ]A is much larger than only A[Sω ]. However, a property like b) of the last proposition can not be expected for more general functions, for its proof rests on the fact that the regularizer is the inverse of a polynomial.

2.3.4 The Case when A is Injective We now consider an injective sectorial operator A ∈ Sect(ω) on the Banach space X. (If X is reflexive, then automatically D(A) = R(A) = X.) With A being injective, also the function τ (z) := z(1 + z)−2 ∈ H0∞ (and each of its powers) is a

2.3. The Natural Functional Calculus

35

good regularizer, since τ (A) = A(1 + A)−2 is injective. We will denote its inverse by ΛA := τ (A)−1 = (1 + A)2 A−1 = (1 + A)A−1 (1 + A) = (2 + A + A−1 ) with D(ΛA ) = D(A) ∩ R(A). The class of functions regularized by powers of τ is B(Sϕ ) := {f : Sϕ −→ C | there is n ∈ N : τ (z)n f (z) ∈ H0∞ (Sϕ )},

S where ϕ ∈ (0, π]. As usual, we write B := B[Sω ] := ϕ∈(ω,π) B(Sϕ ). Obviously, B(Sϕ ) is an algebra of functions which contains A(Sϕ ). A holomorphic function f on Sϕ belongs to B if and only if f has at most polynomial growth at 0 and at ∞ (and is bounded in between). In particular, H ∞ (Sϕ ) is a subalgebra of B(Sϕ ). Proposition 2.3.12. Let A ∈ Sect(ω) be injective. Then B[Sω ] ⊂ MA . Moreover, the following assertions hold. a) If X is reflexive, D(A) = R(A) = D(An ) ∩ R(An ) for every n ∈ N.

b) If D(A) = X = R(A) and f (z)τ n ∈ H0∞ , then D(An ) ∩ R(An ) is a core for f (A).

Proof. Ad a). Let x ∈ X. Then An (t+A)−n (1+tA)−n x ∈ D(An )∩R(An ). Now, by Proposition 2.1.1, we have limt→0 An (t + A)−n (1 + tA)−n x = x, if x ∈ D(A) ∩ R(A). Assertion b) is proved in the same way as c) of Proposition 2.3.10.

Remark 2.3.13. A law of the form (zf (z))(A) = Af (A) can not hold for all f ∈ B (cf. part b) of Proposition 2.3.10). For, by f ) of Theorem 1.3.2, we certainly have (z −1 )(A) = A−1 . Therefore, (zz −1 )(A) = (1)(A) = I 6= AA−1 = A(z −1 )(A) in general. We end this section with an important example. Example 2.3.14. Let (Ω, Σ, µ) a σ-finite measure space and a ∈ M(Ω, µ; C) as in Sections 1.4 and 2.1.1. Let A := Ma be the multiplication operator on Lp (Ω, µ) and K := essran a = σ(A) its spectrum. We assume A ∈ Sect(ψ), i.e., K ⊂ Sψ . For simplicity we assume A to be injective, i.e., µ(a = 0) = 0 (Lemma 1.4.1). Take ϕ ∈ (ψ, π). Then we have defined two abstract functional calculi (L∞ (K, a[µ]), M(K, a[µ]; C∞ ), Φ1 ) and (E(Sϕ ), M(Sϕ ), Φ2 ) where Φ( f ) = Mf ◦a and Φ2 (f ) = f (A). Since the set {a = 0} is a µ-nullset by assumption, M(Sϕ ) “embeds” naturally into M(K, a[µ]; C∞ ). More precisly, one considers the mapping θ := f 7−→ [f K ] : M(Sϕ ) −→ M(K, a[µ]; C∞ ). We claim that this mapping is a morphism of abstract functional calculi in the sense of Section 1.2.3.

36

Chapter 2. The Functional Calculus for Sectorial Operators

Proof. The only thing to show is that f (A) = Mf ◦a for f ∈ H0∞ (Sϕ ). Take x ∈ Lp (Ω, µ) and ′ x′ ∈ Lp (Ω, µ) where p′ is the dual exponent. Then we compute using Fubini’s and Cauchy’s theorem Z Z Z ˙ ¸ ˙ ′ ¸ 1 1 x(ω) x , f (A)x = f (z) x′ , R(z, A)x dz = f (z) x′ (ω) µ(dω) dz 2πi Γ 2πi Γ z − a(ω) Ω Z Z 1 f (z) = x′ (ω)x(ω) dz µ(dω) 2πi z − a(ω) Ω Γ Z ˙ ¸ = x′ (ω)x(ω)f (a(ω)) µ(dω) = x′ , Mf ◦a x . Ω

Since

x′

was arbitrary, we obtain f (A) = Mf ◦a as desired

The consequence of this fact is, that also for the extended calculi one has compatibility, i.e. f (A) = Mf ◦a holds for all f ∈ M(Sϕ )A . If A is not injective, all the above statements remain true. However, for the formal arguing one has to replace M(Sϕ ) by its subalgebra of functions with limits at 0.

2.4 The Composition Rule Let A ∈ Sect(ω) be injective. Then we know from Proposition 2.1.1 that also A−1 is sectorial, of same angle. A straightforward conjecture now would be the validity of an identity f (z −1 )(A) = f (A−1 ) for f ∈ M[Sω ]A−1 , at least if also the left side is defined. In fact, even more is true. Proposition 2.4.1. Let A ∈ Sect(ω) be injective and f ∈ M[Sω ]. Then f ∈ M[Sω ]A−1

⇐⇒

f (z −1 ) ∈ M[Sω ]A

and in this case f (A−1 ) = f (z −1 )(A). Proof. Let ϕ ∈ (ω, π). We employ the abstract composition rule Proposition 1.3.6. Namely, we have meromorphic functional calculi (H0∞ (Sϕ ), M(Sϕ ), ΦA ) and (H0∞ (Sϕ ), M(Sϕ ), ΦA−1 ) and g(z) := z −1 sends Sϕ to itself. Proposition 1.3.6 shows that one only has to show the assertion for f ∈ H0∞ (Sϕ ). But this folows from an easy change of variables in the defining Cauchy integral, using the fundamental identity (2.1).

The foregoing was an instance of the so-called composition rule (f ◦ g)(A) = f (g(A)). (see Section 1.3). As a matter of fact, the rule as it stands does not make sense unless we require some additional hypotheses. Basically, we make the following requirements: 1) A is sectorial.

2.4. The Composition Rule

37

2) g(A) is defined and also sectorial. 3) g maps a sector into another sector. More precisely, we obtain the following result. Theorem 2.4.2. Let A ∈ Sect(ω) with ω ∈ [0, π), and let ϕ ∈ (ω, π), g ∈ M(Sϕ )A . Assume that g(A) ∈ Sect(ω ′ ) with g(Sϕ ) ⊂ Sω′ ∪ {∞}. If f ∈ M[Sω′ ]g(A) then f ◦ g ∈ M(Sϕ )A and the composition rule (f ◦ g)(A) = f (g(A))

(2.9)

holds true. Note that, by the open mapping theorem, if g is not constant and has the required mapping property (namely, g(Sϕ ) ⊂ Sω′ ∪ {∞}), then g(Sϕ ) ⊂ Sω′ . Hence g is always holomorphic. Moreover, unless g = 0, g cannot have zeroes or poles within Sϕ . Hence if g 6= 0 then g and g −1 are both holomorphic on Sϕ . and g(Sϕ ) ⊂ Sϕ′ for every ω ′ < ϕ′ < π. We are going to prove Theorem 2.4.2 in several steps. At first, let us deal with a trivial case. Lemma 2.4.3. Theorem 2.4.2 holds true if g = 0. Proof. If g = 0 then g(A) = 0. Hence if f ∈ M(Sω ′ )g(A) then by Lemma 2.3.7 f must have a well-defined finite limit at 0 which is approached polynomially fast. Therefore, f ◦ g = f (0) is well defined. It is easy to see that f (g(A)) = f (0) whence the lemma is proved.

From now on we will assume that g 6= 0. As remarked above, this implies g(Sϕ ) ⊂ Sϕ′ for each ω ′ < ϕ′ < π. Lemma 2.4.4. Let ω, ϕ, A, and g 6= 0 be as above. If f ∈ M[Sω′ ] then f ◦ g ∈ M(Sϕ ). If f ∈ E[Sω′ ], then f ◦ g ∈ H ∞ (Sϕ )A .

Proof. The first statement is clear by the remarks preceeding this lemma. Obviously also f ◦ g ∈ H ∞ (Sϕ ) holds for every f ∈ E[Sω ′ ]. If A is injective, there is nothing more to show. Suppose A is not injective. Then by Lemma 2.3.7 the function g must have a limit at 0 with g(z) − g(0) = O(|z|α ) for some α > 0 as z → 0. Since f ∈ E, the function f ◦ g will have a similar limit behaviour in 0. Hence we have f ◦ g ∈ A, whence the lemma is proved.

Lemma 2.4.5. Let ω, ϕ, A, g be as above. Then the composition rule f (g(A)) = (f ◦ g)(A) is true for every f ∈ E[Sω′ ].

Proof. Let f ∈ E[Sω ′ ]. Then there are constants c, d ∈ C and f1 ∈ H0∞ [Sω ′ ] such that f = d d c1 + 1+z + f1 . Now for f = c1 there is nothing to prove. If f = 1+z then the assertion is contained in f ) of Theorem 1.3.2. Hence we can assume without restriction that f ∈ H0∞ . For λ∈ / Sω ′ the function (λ − g(z))−1 certainly is bounded and holomorphic on Sϕ . Now „ « Z Z 1 1 1 f (g(A)) = f (λ)R(λ, g(A)) dλ = f (λ) (A) dλ 2πi Γ′ 2πi Γ′ λ − g(z) where Γ′ surrounds the sector Sω ′ . We consider two cases: 1)

A is injective.

38 2)

Chapter 2. The Functional Calculus for Sectorial Operators A is not injective.

The first case is easier to handle. If A is injective we can use z(1 + z)−2 as a regularizer, let Γ surround Sω within Sϕ and compute Z 1 f (g(A)) = ΛA τ (A)f (g(A)) = ΛA f (λ)τ (A)R(λ, g(A)) dλ 2πi Γ′ „ « Z 1 z = ΛA f (λ) (A) dλ 2πi Γ′ (1 + z)2 (λ − g(z)) Z Z 1 z = ΛA f (λ) R(z, A) dz dλ (2πi)2 Γ′ Γ (1 + z)2 (λ − g(z)) Z 1 f (g(z)) z (1) = ΛA R(z, A) dz 2πi Γ (1 + z)2 = ΛA (f (g(z)) τ (z)) (A) = (f ◦ g)(A).

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 λ f (λ) 1 f (λ) = (λ − g(z))(1 + z)2 λ − g(z) λ (1 + z)2

is product integrable on Γ × Γ′ since the first factor is uniformly bounded. To cover the second case, assume that A is not injective. Then by Lemma 2.3.7, g must have a finite limit c := g(0) in 0 which is approached at least polynomially. Define g1 (z) := g(z)− g(0) and choose a regularizer e ∈ E for g1 with e(∞) = 0. Then eg1 ∈ H0∞ . Furthermore, for λ ∈ / Sω ′ the function [(λ − g(z))−1 − (λ − c)−1 ] is also regularized by e. Thus we compute Z 1 f (g(A)) = e(A)−1 e(A)f (g(A)) = e(A)−1 f (λ)e(A)R(λ, g(A)) dλ 2πi Γ′ „ « Z 1 e(z) f (λ) (A) dλ = e(A)−1 2πi Γ′ λ − g(z) „ « Z 1 g1 (z)e(z) e(z) = e(A)−1 f (λ) + (A) dλ 2πi Γ′ (λ − g(z))(λ − g(0)) λ−c „ « Z 1 (g1 e)(z) f (λ) = e(A)−1 f (λ) (A) + e(A) dλ 2πi Γ′ (λ − g(z))(λ − c) λ−c „ « Z 1 (g1 e)(z) = e(A)−1 f (λ) (A) dλ 2πi Γ′ (λ − g(z))(λ − c) Z 1 f (λ) + e(A)−1 dλ e(A) 2πi Γ′ λ − c The second summand equals f (c) by Cauchy’s theorem. The first satisfies „ « Z 1 (g1 e)(z) e(A)−1 f (λ) (A) dλ 2πi Γ′ (λ − c)(λ − g(z)) Z Z 1 1 (g1 e)(z) = e(A)−1 f (λ) R(z, A) dz dλ 2πi Γ′ 2πi Γ (λ − c)(λ − g(z)) » – « Z „ Z 1 1 1 1 (1) = e(A)−1 f (λ) − dλ e(z)R(z, A) dz 2πi Γ 2πi Γ′ λ − g(z) λ−c Z (2) −1 1 = e(A) (f (g(z)) − f (c)) e(z)R(z, A) dz 2πi Γ (3)

= e(A)−1 [(f (g(z)) − f (c))e(z)] (A)

= e(A)−1 ((f ◦ g)(A) − f (c)) e(A) = (f ◦ g)(A) − f (c)

2.5. Extensions According to Spectral Conditions

39

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 F (λ, z) :=

f (λ)(g1 e)(z) (λ − c)(λ − g(z))z

and prove its product integrability. The representation „ «„ «„ « f (λ) λ (g1 e)(z) F (λ, z) = λ (λ − c)(λ − g(z)) z

shows that c 6= 0 is harmless, since λ/(λ − g(z)) is uniformly bounded because of the conditions g(z) ∈ Sω ′ and λ ∈ Γ′ . (Recall that (eg1 ) ∈ H0∞ .) If c = 0 (hence g1 = g) we write „ «„ «„ « f (λ) 1 (ge)(z) F (λ, z) = λ λ − g(z) z „ «„ α «„ « f (λ) λ g(z)1−α g(z)α e(z) = . 1+α λ λ − g(z) z

(Recall that g has no poles in Sϕ .) Here, 0 < α < 1 is chosen in such a way that the first factor remains integrable. Then, the middle term is still uniformly bounded. It is easily seen that also eg α ∈ H0∞ , whence F is integrable.

To complete the proof of Theorem 2.4.2, we invoke Proposition 1.3.6 from Chapter 1 with the data (E(Sϕ ), M(Sϕ ), ΦA )

(meromorphic f.c. for A on Sϕ ),

(E(Sϕ′ ), M(Sϕ′ ), Φg(A) ) (meromorphic f.c. for g(A) on Sϕ′ ), g : Sϕ −→ Sϕ′ , where ϕ′ ∈ (ω ′ , π) is fixed. Lemma 2.4.5 shows that the hypothesis of Proposition 1.3.6 holds, namely that the Composition Rule is true for functions f in the domain of the primary calculus, i.e. for f ∈ E(Sϕ′ ). Hence it is true for every f ∈ M(Sϕ′ )g(A) . Since ϕ′ ∈ (ω ′ , π) was arbitrary, Theorem 2.4.2 is completely proved.

2.5 Extensions According to Spectral Conditions In this section we treat the special cases of sectorial operators which are invertible and/or bounded. The additional hypothesis makes it possible to set up a new primary functional calculus extending the E-calculus. Strictly speaking this leads to completely new abstract functional calculi extending the one from last section.

2.5.1 The Case when 0 ∈ ρ(A) Let A ∈ Sect(ω) and assume that A is invertible. Then Br (0) ⊂ ̺(A) where 1 −1 −1 is the spectral radius of A . It should be clear that in setting up a r = rA functional calculus for A on a sector Sϕ with ϕ > ω, the behaviour of the functions in 0 does not matter. Define −α

E ∞ (Sϕ ) := {f ∈ O(Sϕ ) | f = O(|z|

) (z → ∞) for some α > 0}

40

Chapter 2. The Functional Calculus for Sectorial Operators

Γ

f (A) =

1 2πi

R

Γ

f (z) R(z, A) dz

σ(A)

∂Sϕ ∂Sω Figure 5: Cauchy integral in the case of invertibility.

2.5. Extensions According to Spectral Conditions

41

For A ∈ Sect(ω) satisfying 0 ∈ ̺(A) and f ∈ E ∞ (Sϕ ) (ϕ > ω) we define as usual Z 1 f (z)R(z, A) dz Φ(f ) := f (A) := 2πi Γ where the path Γ bounds a sector Sω′ (with ω ′ ∈ (ω, ϕ)), except for the region near 0 where it avoids 0 at small distance, cp. Figure 5 on page 40. As a matter of fact, one can prove in a similar way a theorem which is analogous to Theorem 2.3.3. Hence, a meromorphic functional calculus (E ∞ (Sϕ ), M(Sϕ ), Φ) for A on Sϕ is defined. As in Chapter 2.3 one can form the inductive limit as ϕ → ω and obtains a meromorphic functional calculus (E ∞ [Sω ], M[Sω ], Φ) on Ω := Sω . This we call the natural functional calculus on Sω for A as an invertible, sectorial operator. It clearly forms a consistent extension of the natural functional calculus for A as a sectorial operator (defined in Section 2.3). Remark 2.5.1. The construction of the calculus for invertible sectorial operators could of course be refined. Namely on could even go over to functions f which are not even defined near the origin. Since this would make necessary a lot more notation we omit it. We now realize that in order to prove a composition rule as in Theorem 2.4.2, we have to face several cases, namely the combinations 1) A sectorial, g(A) invertible and sectorial. 2) A invertible and sectorial, g(A) sectorial. 3) Both A, g(A) invertible and sectorial. 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(Sϕ ) on the function g.

2.5.2 The Case when A ∈ L(X) Now we are concerned with the “dual” situation, namely with a bounded operator A ∈ Sect(ω). It is then clear that for the expression f (A) to make sense, the limit behaviour of f in ∞ will be irrelevant. Define E 0 (Sϕ ) := {f ∈ O(Sϕ ) | f = O(|z|α ) (z → 0) for some α > 0} For A ∈ Sect(ω) ∩ L(X) and f ∈ E 0 (Sϕ ) (ϕ > ω) we define as usual Z 1 Φ(f ) := f (A) := f (z)R(z, A) dz 2πi Γ where the path Γ bounds a sector Sω′ (with ω ′ ∈ (ω, ϕ)), except for the region near ∞ where it avoids ∞ and stays away from σ(A). If A is injective we can

42

Chapter 2. The Functional Calculus for Sectorial Operators

leave it like this, obtaining a meromorphic functional calculus (E 0 (Sϕ ), M(Sϕ ), Φ) for A on Sω . If A is not injective — like in the case of a general sectorial operator — we have to include also functions which are holomorphic in 0. The easiest way 1 and define the to do this is to go over to the algebra Ee 0 (Sϕ ) := E 0 (Sϕ ) ⊕ C 1+z primary calculus by c Φ(f + ) := f (A) + c(1 + A)−1 1+z where f ∈ E 0 (Sϕ ). In any case, one ends up with a meromorphic functional calculus for A on Sϕ and by forming the inductive limit as ϕ → ω obtains a meromorphic functional calculus (Ee 0 [Sω ], M[Sω ], Φ)]

for A on Sω which we call the natural functional calculus on Sω for A as a bounded, sectorial operator. It clearly forms a consistent extension of the natural functional calculus for A as a sectorial operator (defined in Chapter 2.3). Remark 2.5.2. Also in this case the so obtained calculus is not the most general. In fact, one could go over to functions f which are not even defined near ∞. As above, to prove a composition rule as in Theorem 2.4.2, we have to face the cases 1) 2) 3) 4) 5)

A sectorial, g(A) bounded and sectorial. A invertible and sectorial, g(A) bounded and sectorial. Both A, g(A) bounded and sectorial. A bounded and sectorial, g(A) sectorial. A bounded and sectorial, g(A) invertible and sectorial.

Again, the proofs are similar to the proof of Theorem 2.4.2, but sometimes one needs an additional assumption on the mapping behaviour of g: In cases 1) - 3) one needs ∞ ∈ / g(Sϕ ) and in case 5) one needs 0 ∈ / g(Sϕ ).

2.5.3 The Case when A is bounded and invertible If A ∈ Sect(ω) is bounded and invertible then for a primary calculus no assumptions on the behaviour of f in 0 or ∞ are needed. In fact, one can use the ususal Dunford calculus for A. Easy arguments show that this yields a consistent extension of either the natural functional calculus for A as a invertible sectorial operator and the natural functional calculus for A as a bounded sectorial operator. Again, composition rules (under suitable assumptions on the function g) could be proved linking all calculi introduced up to now.

Final Remarks Our systematic introduction of functional calculi for sectorial operators has now come to an end. Depending on how “good” the operator A is in terms of cer-

2.6. Miscellanies

43

tain spectral conditions we have given meaning to the symbol “f (A)” where “f ” denotes a meromorphic function on a sector. As introduced above, the resulting mapping (f 7−→ f (A)) is called the natural functional calculus (in short: NFC) for A as a (bounded, invertible, bounded and invertible) sectorial operator A. Recall that the definition of f (A) depends heavily on the used primary calculus. Hence, the assertions “f (A) is defined by the NFC for sectorial operators” and “f (A) is defined by the NFC for bounded sectorial operators” have different meanings. However, in the following chapters we use mostly the NFC for sectorial operators.

2.6 Miscellanies In this section we examine the relation between the functional calculi for A and its adjoint A′ . Then we look at the part AY of A in a second Banach space Y ⊂ X. After this we examine the beehaviour of the functional calculus with respect to sectorial approximation and prove a fundamental boundedness result.

2.6.1 Adjoints Let A be a m.v. sectorial operator on the Banach space X. Then its adjoint is again a m.v. sectorial operator, see k) of Proposition 2.1.1 and Remark 2.1.4). How do the functional calculi of A and A′ relate? We begin with an abstract consideration. Let (E, M, Φ) be a proper abstract functional calculus over the Banach space X. Then by e∗ := Φ(e)′ = (e• )′ a homomorphism Φ∗ : (e 7−→ e∗ ) : E −→ L(X ′ ) is defined. Hence, (E, F, Φ∗ ) is an abstract functional calculus over X ′ , the socalled dual (functional) calculus. This AFC is proper if and only if there is e ∈ 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 there is a Φ-regularizer e ∈ E for f with the following property: There exists (fn )n ⊂ Mb such that 1) fn• → I strongly on X. 2) R(fn• ) ⊂ R(e• ).

Then R(e• ) is dense in X, is a core for f• , and f∗ = (f• )′ . Proof. It is immediately clear from 1) and 2) that R(e• ) is dense in X. Hence, e is a regularizer for f in the dual calculus. Let x ∈ D(f• ) and y := f• x. Then xn := fn• x → x and xn ∈ R(e• ) ⊂ D(f• ). Moreover f• xn = f• fn• x = fn• y → y. Since x ∈ Df• was arbitrary, we see that R(e• ) is a core for f• . This means that (ef )• e−1 • = f• . Using this we compute

h i′ ˆ ˜ (1) ′ ′ (2) −1 ′ (3) f∗ = (e∗ )−1 (ef )∗ = [(e• )′ ]−1 [(ef )• ]′ = [e−1 = (ef )• e−1 = (f• )′ . • • ] [(ef )• ] = (ef )• e•

44

Chapter 2. The Functional Calculus for Sectorial Operators

Here 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 if A is in addition bounded or invertible, the same is true for A′ . Hence, the primary functional calculi for A and A′ 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 on f means: f ∈ E[Sω ] if A is neither bounded nor invertible, f ∈ E ∞ [Sω ] if A ∈ Sect(ω) is invertible and f ∈ Ee 0 [Sω ] if A ∈ Sect(ω) is bounded. R 1 Proof. Let f be as required. Without restriction we can write f (A) = 2πi f (z)R(z, A) dz for Γ a certain Γ. Then «′ „ Z Z Z 1 1 1 f (A)′ = f (z)R(z, A) dz = f (z)R(z, A)′ dz = f (z)R(z, A′ ) dz = f (A′ ). 2πi Γ 2πi Γ 2πi Γ

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

Proposition 2.6.3. Let A ∈ Sect(ω) with D(A) = X. Then the identity f (A)′ = f (A′ ) holds in the following cases: a) f ∈ A[Sω ];

b) R(A) = X and f ∈ B[Sω ].

Proof. In the case a), f is regularizable by e := (1 + z)−m for some m. Since D(A) is dense, the functions fn := [n(n + z)−1 ]m satisfy conditions 1) and 2) of Proposition 2.6.1, whence the statement follows. Part b) is treated similarly with e = τ m for some m. Note that the additional condition R(A) = X ensures that both A and A′ are injective.

Remarks 2.6.4. 1) As a matter of course, one could prove a similar result for bounded sectorial and invertible sectorial operators A. 2) Even if nothing more than sectoriality for A is required, the theorem holds for a much larger class than A. Assume ωA < π2 for simplicity. Then, since 1 D(A) is dense, e− n A → I strongly (see Proposition 3.3.1). Therefore, even subexponential growth is admissible.

2.6. Miscellanies

45

2.6.2 Restrictions Let A be a sectorial operator on the Banach space X, and let Y be another Banach space, continuously embedded into X. Let AY := A Y be the part of A in Y , i.e., D(AY ) := {x ∈ Y ∩ D(A) | Ax ∈ Y },

AY y = Ay

(y ∈ D(AY )).

(cf. Proposition A.2.8). In general AY will not be sectorial any more. However, if it is, one can ask how f (AY ) is computed from f (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 AY ∈ Sect(ω) then the implication f ∈ M[Sω ]A

=⇒

f ∈ M[Sω ]AY

and

f (AY ) = f (A)Y

holds. Proof. If f is a function of the primary functional calculus for A, then by Proposition A.2.8 we have f (AY ) = f (A)Y . In fact, f can be decomposed as f (z) = d + c(1 + z)−1 + e(z) where e(A) is defined by the Cauchy integral. If F is more general, on finds a regularizer e, i.e., e, (ef ) are elementary and e(A) is injective. But then also e(AY ) = 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 )(AY )x = e(AY )y ⇔ (x, y) ∈ f (AY ).

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

2.6.3 Sectorial Approximation Suppose A ∈ Sect(ω) on X and (An )n is a sectorial approximation of A on Sω . By Proposition 2.1.3, if A is bounded and/or invertible, the same is true for almost all An . Hence A and the An eventually have the same primary functional calculus. Lemma 2.6.7. Let A ∈ Sect(ω) and (An )n a sectorial approximation of A on Sω . Then f (An ) → f (A) in norm whenever f (A) is defined by the primary functional calculus for A. R 1 Proof. Let f be as required. Without restriction we can write f (A) = 2πi Γ f (z)R(z, A) dz for a certain Γ. Then „ « Z Z 1 1 lim f (An ) = lim f (z)R(z, An ) dz = f (z) lim R(z, An ) dz n n n 2πi Γ 2πi Γ Z 1 = f (z)R(z, A) dz = f (A) 2πi Γ by an easy application of Lebesgue’s Dominated Convergence theorem.

It is certainly interesting whether f (An ) → f (A) (in a sense) for more general functions f . However, this is a delicate matter since it is not even clear that f (An ) is defined for large n whenever f (A) is. And even if one assumes this, the

46

Chapter 2. The Functional Calculus for Sectorial Operators

regularizer for f will in general depend on n. Let us quickly skip to the abstract setting. Assume there is a family (E, M, Φ), (E, M, Φn ) of proper AFC over the Banach space X. We call the sequence [(E, M, Φn )]n convergent to the AFC (E, M, Φ) if Φn (e) → Φ(e) in norm for every e ∈ E. Proposition 2.6.8. Let [(E, F, Φn )]n be a sequence of proper AFCs converging to the proper AFC (E, M, Φ). Let f ∈ M and let e be a uniform regularizer for f , i.e., e, ef ∈ E and all e• , e•n are injective. If xn ∈ D(f•n ) such that xn → x and f•n xn → y, then (x, y) ∈ f• . In particular, if all f•n ∈ L(X) and f•n → T ∈ L(X) strongly, then f• = T . Proof. Assume that f, e, xn , x, y are as in the hypothesis. We have (ef )• x = limn (ef )•n xn = limn e•n f•n xn = e• y. This yields (x, y) ∈ f• .

Coming back to sectorial operators, Proposition 2.6.8 shows that for a general convergence result one needs uniform regularizers. The next proposition provides an example. Proposition 2.6.9. Let A ∈ Sect(ω) and (An )n a sectorial approximation of A on Sω . Suppose either 1) f ∈ A[Sω ], or 2) f ∈ B[Sω ] and all A, An are injective.

If xn ∈ D(f (An )) with xn → x and f (An )xn → y, then also x ∈ D(f (A)) and y = f (A)x. In particular, if f (An ) ∈ L(X) with f (An ) → T ∈ L(X) strongly, then f (A) = T . Proof. In case 1) a uniform regularizer is (1 + z)−m for some m, in case 2) a power of τ (z) can be used. The rest is only the application of Proposition 2.6.8

2.6.4 Boundedness In the theory of functional calculus it is of ultimate importance to know, which families of functions lead to uniformly bounded families of operators. The whole Chapter 5 is dedicated to that question. At this point we provide only some simple although useful results. Lemma 2.6.10. Let A ∈ Sect(ω), ϕ ∈ (ω, π), and (fα )α ⊂ H0∞ (Sϕ ). Assume that there are numbers C, s, cα > 0 such that n o s −s |fα (z)| ≤ C min |cα z| , |cα z| (z ∈ Sϕ ) (2.10)

for all α. Then

sup kfα (A)k ≤ α

2C M (A, ϕ). sπ

2.7. The Spectral Mapping Theorem

47

Proof. Choose ω ′ ∈ (ω, ϕ). Then, by a simple n change ofovariables in the integral, one obtains the R |dz| C min |z|s , |z|−s = 2C M (A, ω ′ ). Now let ω ′ tend estimate kfα (A)k ≤ M (A, ω ′ ) 2π Γω ′ |z| sπ to ϕ, cf. Remark 2.1.2.

Proposition 2.6.11. Let ϕ ∈ (0, π] and f ∈ E(Sϕ ). Then there is a constant Cf > 0 such that sup kf (tA)k ≤ Cf M (A, ϕ) t>0

holds for each sectorial operator A ∈ Sect(ω), ω ∈ (0, ϕ) on a Banach space X. Moreover, given θ ∈ [0, ϕ − ω) one has kf (λA)k ≤ Cf M (A, ϕ − θ) for all λ ∈ C, |arg λ| ≤ θ. Proof. Let f = ψ +

a 1+z

+ b with ψ ∈ H0∞ (Sϕ ), a, b ∈ C. Choose C, s > 0 such that n o |ψ(z)| ≤ C min |z|s , |z|−s

(z ∈ Sϕ ).

By a change of variable in the defining Cauchy integral (or, if you wish, an application of the composition rule), we obtain ψ(tA) = ψ(tz)(A) for all t > 0. Hence by Lemma 2.6.10 we have kψ(tA)k ≤

2C M (A, ϕ). sπ

This implies kf (tA)k ≤

» – 2C 2C M (A, ϕ) + |a| M (A) + |b| ≤ + |a| + |b| M (A, ϕ), sπ sπ

since 1 ≤ M (A) ≤ M (A, ϕ) (see Proposition 2.1.1, a)). To prove the second assertion, write λ = te−iµ with |µ| ≤ θ. Note that eiµ A ∈ Sect(ω + θ). Since ω + θ < ϕ, we can apply the (already proved) first part of the Proposition and obtain ‚ ‚ kf (λA)k = ‚f (teiµ A)‚ ≤ Cf M (eiµ A, ϕ) ≤ Cf M (A, ϕ − θ).

Remark 2.6.12. Let us note that Proposition 2.6.11 also remains meaningful and true when A is a multivalued operator (cp. Remark 2.3.4). This will play a role in Section 3.3. For more results on boundedness and on approximation we refer to Section 5.2.

2.7 The Spectral Mapping Theorem In this section we will prove a spectral mapping theorem for the NFC for sectorial operators. The first step is a spectral inclusion theorem.

48

Chapter 2. The Functional Calculus for Sectorial Operators

2.7.1 The spectral inclusion f (σ(A) \ {0}) ⊂ σ ˜ (f (A)) We begin with two auxiliary results. Lemma 2.7.1. Let A ∈ Sect(ω), e ∈ H(A), λ ∈ C with λ − A being injective. Assume there is 0 6= c ∈ C such that f (z) :=

e(z) − c ∈ H(A). λ−z

Then e(A)(λ − A)−1 = (λ − A)−1 e(A). Note that by Theorem 1.3.2 the inclusion e(A)(λ − A)−1 ⊂ (λ − A)−1 e(A) is always true. Proof. We only have to prove the inclusion e(A)(λ − A)−1 ⊃ (λ − A)−1 e(A). Take x ∈ X such that e(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, whence cx = (λ − A)(z − f (A)x). Now, c 6= 0 by assumption, hence x ∈ R(λ − A) = D((λ − A)−1 .

Lemma 2.7.2. Let A ∈ Sect(ω) and f ∈ M[Sω ]A . Given 0 6= λ ∈ Sω there is a regularizer e for f satisfying e(λ) 6= 0. Proof. Let g be any regularizer for f , i.e., g, f g ∈ E, g(A) is injective. Define e := g/(λ − z)n where n ∈ N is the order of the zero λ of g. (This order n may be zero.) Then e(λ) 6= 0. Furthermore, e ∈ E and g(A) = (λ − A)n e(A), whence e(A) is injective. Clearly, also ef = f g/(λ − z)n ∈ E. Hence e is a regularizer for f with e(λ) 6= 0.

We now come to main part. Proposition 2.7.3. Let A ∈ Sect(ω), f ∈ M[Sω ]A and 0 6= λ ∈ Sω . If f (λ) = 0 and f (A) is invertible, then λ ∈ ̺(A). Proof. Choose a regularizer e ∈ E for f with c := e(λ) 6= 0. Then (ef ) ∈ E and (ef )(λ) = 0. This implies that also h := ef /(λ − z) ∈ E, whence h(A)(λ − A) ⊂ (ef )(A) = f (A)e(A). Because e(A) and f (A) are both injective, this shows that λ − A must be injective. Also, e is a regularizer for f /(z − λ). e−c It is obvious that g := λ−z ∈ E. Lemma 2.7.1 implies that e(A)(λ−A)−1 = (λ−A)−1 e(A). Inverting both sides of this equation gives (λ − A)e(A)−1 = e(A)−1 (λ − A). Now, we compute „ « „ « „ « f ef ef f (A) = (λ − z) (A) = e(A)−1 (λ − z) (A) = e(A)−1 (λ − A) (A) λ−z λ−z λ−z „ « „ « f ef = (λ − A)e(A)−1 (A) = (λ − A) (A). λ−z λ−z Since f (A) is surjective, λ − A must also be surjective, whence we arrive at λ ∈ ̺(A).

Theorem 2.7.4. (Spectral Inclusion Theorem) Let A ∈ Sect(ω) and f ∈ M[Sω ]A . Then f (σ(A) \ {0}) ⊂ σ ˜ (f (A)).

2.7. The Spectral Mapping Theorem

49

Proof. Let 0 6= λ ∈ σ(A) and define µ := f (λ). If µ 6= ∞, Proposition 2.7.3 applied to the function µ − f shows that µ − f (A) cannot be invertible. Hence µ ∈ σ(f (A)). Assume now that µ = ∞ ∈ / σ ˜ (f (A)). Then f (A) ∈ L(X) and there is λ0 ∈ C such that 1 f (A) − λ0 is invertible. Hence, g := f −λ ∈ M[Sω ]A with g(A) being invertible and g(λ) = 0. 0 Another application of Proposition 2.7.3 yields λ ∈ ̺(A) which contradicts the assumption made on λ.

It is natural to ask what happens in the “critical” points 0 and ∞. Since σ ˜ (f (A)) is a compact subset of the Riemann sphere, we immediately obtain C∞

f (σ(A) \ {0})

⊂σ ˜ (f (A)).

(2.11)

However, consider the Volterra operator V on C[0, 1] (cf. Chapter 1). We know that σ(V ) = {0}, whence the above inclusion is trivial. The next result gives a sufficient condition, involving the notion of polynomial limit (see page 24). Theorem 2.7.5. Let A ∈ Sect(ω) and λ0 ∈ {0, ∞}. If f ∈ M[Sω ]A has polynomial limit µ at λ0 and λ0 ∈ σ ˜ (A) then µ ∈ σ ˜ (f (A)). Proof. For µ ∈ C we can consider f − µ instead of f and in this way reduce the problem to the cases µ = 0, ∞. Let us start with the case µ = 0, λ0 = ∞. Therefore, we assume that f (A) is invertible C∞

and f has polynomial limit 0 at λ0 = ∞. If ∞ ∈ σ(A) \ {0} then the assertion follows from R 1 (2.11). Otherwise there is R > 0 such that σ(A) ⊂ BR (0). Define Q := 2πi Γ R(z, A) dz where Γ = ∂Bδ (0) for (any) R < δ. It is easy to see that Q is a bounded projection. Let P := I − Q the complementary projection and Y := R(P ) be its range space. Since P commutes with A, Ay ∈ Y for every y ∈ Y ∩ D(A). It is also easy to see that B := A|Y is a sectorial operator of angle ω on Y with ̺(A) ⊂ ̺(B) and R(λ, B) = R(λ, A)|Y for all λ ∈ ̺(A). R 1 1 R(z, A) dz on Moreover, σ ˜ (B) ⊂ {∞}. In fact, for each µ ∈ σ(A), R(µ, B) = 2πi Γ µ−z Y as is easily verified. Clearly we have e(B) = e(A)|Y for each e ∈ E[Sω ]. From this follows that D(f (B)) = D(f (A)) ∩ Y and f (B)x = f (A)x ∈ Y for all x ∈ D(f (A)) ∩ Y . Since f (A) is invertible by assumption, also f (B) is with f (B)−1 = f (A)−1 |Y . Since f has polynomial limit 0 at ∞, clearly f is in the domain of the primary functional calculus for B! Moreover, for large n ∈ N the function g := zf n still has this property. This gives g(B) = Bf (B)n ∈ L(X). But f (B) is invertible, whence B must be bounded. Since σ ˜ (B) ⊂ {∞}, we conclude that σ ˜ (B) = ∅, whence Y = 0. This shows that A is bounded, i.e., ∞ 6∈ σ ˜ (A). We now consider the case µ = 0, λ0 = 0. Assume again that f (A) is invertible and f has polynomial limit 0 at 0. Let e ∈ E be a regularizer for f . Then ef ∈ E, (ef )(A) is injective and (ef )(0) = 0. Lemma 2.3.7 now shows that A must be injective. Let B := A−1 , g(z) := f (z −1 ). By Proposition 2.4.1, g ∈ M[Sω ]B , g(B) = f (A) is invertible and g has polynomial limit 0 at ∞. By what we have shown above, B is bounded, whence A is invertible. Finally, we deal with the case µ = ∞. Assume that ∞ = µ ∈ / σ ˜ (f (A)). By definition, f (A) must be bounded. Hence we can find λ ∈ C such that λ − f (A) is invertible. By Theorem 1 1.3.2, g := λ−f ∈ M[Sω ]A , g(A) is invertible and g has polynomial limit 0 at λ0 . By the results achieved so far, we can conclude that λ0 ∈ /σ ˜ (A).

2.7.2 The Spectral Mapping Theorem We are heading towards the final goal. As usual, we start with an auxiliary result.

50

Chapter 2. The Functional Calculus for Sectorial Operators

Lemma 2.7.6. Let A ∈ Sect(ω) and f ∈ M[Sω ]. Assume f has finite polynomial limits at σ ˜ (A)∩{0, ∞} and all poles of f are contained in ̺(A). Then the following assertions are true. 1) In case {0, ∞} ⊂ σ ˜ (A), f (A) is defined by the NFC for sectorial operators. 2) In case A is invertible, f (A) is defined by the NFC for invertible sectorial operators. 3) In case A is bounded, f (A) is defined by the NFC for bounded sectorial operators. In either case, f (A) ∈ L(X).

Proof. Assume f is as required and λ0 is a pole of f . Then, for suitably large n0 ∈ N, the function (λ0 − z)n0 f1 := f (z) (1 + z)n0 also has finite polynomial limits in σ ˜ (A) ∩ {0, ∞}, but one pole less than f . Moreover, letting (λ0 −z)n0 r0 := (1+z) n0 , we see that r0 (A) is bounded and invertible. Assume that we are in the situation of a). Then f can have only finitely many poles. By induction we find a bounded rational function r such that r(A) is bounded and invertible and rf ∈ E. Hence r regularizes f to E and f (A) = r(A)−1 (rf )(A) ∈ L(X). (Confer also Proposition 1.2.5.) A similar reasoning applies in the other cases. Namely, only finitely many poles of f lie in the “relevant” part of the domain of f . For example, if A is invertible, the poles of f may accumulate in 0, but for the NFC for A, behaviour of f near 0 is irrelevant.

Proposition 2.7.7. Let A ∈ Sect(ω) and f ∈ M[Sω ]A having polynomial limits in σ ˜ (A) ∩ {0, ∞}. Then σ ˜ (f (A)) ⊂ f (˜ σ (A)).

Proof. Note that by assumption f (λ) is defined for each λ ∈ σ ˜ (A). Take µ ∈ C such that µ∈ / f (˜ σ (A)). Then the function (µ− f )−1 ∈ M[Sω ] has finite polynomial limits in {0, ∞} ∩ σ ˜ (A) and all of its poles (namely, the points λ ∈ Sω \{0} where f (λ) = µ) are contained in the resolvent of A. Hence one can apply Lemma 2.7.6 to conclude that (µ − f )−1 (A) is defined and bounded. But this implies that µ − f (A) is invertible, whence µ ∈ /σ ˜ (f (A)). Now assume µ = ∞ ∈ / f (˜ σ (A)). This implies that the poles of f are contained in the resolvent of A. An application of Lemma 2.7.6 yields that f (A) is a bounded operator, whence ∞∈ /σ ˜ (f (A)).

In summary, we have the following theorem. Theorem 2.7.8. (Spectral Mapping Theorem) Let A ∈ Sect(ω) and f ∈ M[Sω ]A having polynomial limits in {0, ∞} ∩ σ ˜ (A). Then f (˜ σ (A)) = σ ˜ (f (A)). Proof. Put together Proposition 2.7.7 and Theorem 2.7.5.

2.8 Comments 2.1 Sectorial Operators. Most of the material is adapted from the monograph [89, Section 1.2] where also the standard examples are presented. The notions

2.8. Comments “uniform sectoriality” and “sectorial in the literature. E.g., the family of in [109, Section 8.1] and [78, Section [101]. Basics on multivalued sectorial

51

approximation” unify methods well-known operators Aε = (A + ε)(1 + εA)−1 is used 2] and is called “Nollau approximation” in operators can be found in [88, Section 2].

2.2 Spaces of Holomorphic Functions. The name “Dunford-Riesz class” together with the notation H0∞ is taken from [123, Section 1.3.3.1 ] and [124, Section 2]. Other notations for the same class are used in the literature, e.g., Ψ(Sϕ ) in [21] and H0∞ (Sϕ ) in [78]. 2.3 The Natural Functional Calculus. As for the historical roots of the functional calculus developed here, we have already given a short account in Chapter 1. The “mother of all functional calculi” so to speak, is provided by the Fourier transform or, more generally, by the spectral theorem for bounded normal operators on a Hilbert space (see Appendix D). The aim to obtain a similar tool for general bounded operators on a Banach space lead to the so called Dunford-Riesz calculus, see [39, Section VII.3] for the mathematics and [39, Section VII.11] for some historical remarks. This turned out to be only a special case of a general construction in Banach algebras an account of which can be found in [20, Chapter VII, §4], cf. also Chapter 1. However, in the Banach space situation, so far all appearing operators had to be bounded. The first approach to a (Banach space) calculus for unbounded operators was to reduce it to the bounded case by an application of a resolvent/elementary rational function, cf. [39, Section VII.9] or [3, Lecture 2]. This functional calculus is sometimes called Taylor calculus. Here as in the bounded case, only functions are used which are holomorphic in a neighborhood of the spectrum, where ∞ has to be considered a member of the spectrum if A is unbounded (see Section A.2). With the Hilbert space (i.e. multiplication operator) examples in mind it became clear that boundedness of operators should not play an essential role in a general framework, neither as a requirement for the original operator A nor as a restriction for the result f (A). That f (A) is always bounded in the Taylor setting can be regarded as a mere coincidence due to the particular functions used. The decisive step to abandon boundedness was to realize how to involve functions which are singular in certain boundary points of the spectrum. Despite later development this was first done for strip type operators (see Section 4.1) by Bade in [12]. He introduces the “regularization trick” which we have formalized in the extension procedure of Section 1.2. However, since ∞ is kind of a “hidden” spectral point, it was not obvious what was going on and in fact became fully clear only after McIntosh in [95] constructed the setup for sectorial operators. More or less systematic accounts of the natural functional calculus for sectorial operators can be found, e.g., in [3], [123] or [126]. Our presentation differs from these approaches in two decisive points. First, they mostly consider only injective operators, with the obvious disadvantage that the usual fractional powers

52

Chapter 2. The Functional Calculus for Sectorial Operators

of an arbitrary sectorial operator (see Chapter 3) including the bounded ones can not be treated by their methods. The second main difference between our and other treatments lies in that we do not make any density assumptions on either the domain or the range of the operator. This is motivated by the fact that there are important examples of operators which are not densely defined, like the Dirichlet Laplacian on C(Ω) where Ω is some open and bounded subset of Rn , compare [6, Section 6.1]. Moreover, it seems to be advantageous to discard density assumptions in a systematic treatment because then one is not tempted to prove things by approximation and closure arguments (which usually are a tedious matter). The focus on injective operators (with dense range and dense domain) in the literature is of course not due to unawareness but a matter of convenience. As one can see, e.g., in the proof of Lemma 2.4.5, working only with injective operators often makes life easier. And McIntosh’s seminal paper [95] outlines how to treat the matter without the injectivity assumption. 2.4 The Composition Rule. Most of the main properties of the natural functional calculus as enumerated, e.g., in Theorem 1.3.2 are folklore (even if e) and f ) have not yet appeared in the literature in this generality). The composition rule (Theorem 2.4.2) however has been underestimated for a long time. Our account is based on [54]. The only previous account I know is in [126] but there only a very special case is treated. We believe that only the composition rule opens the door for a fruitful use of the natural functional calculus beyond the mere definition. 2.5 Extensions According to Spectral Conditions. This section supplements the two previous ones. To sum up, we have defined four compatible “natural” functional calculi for certain classes of (sectorial) operators. This is by far not the end of the story. For example, suppose that A is sectorial but there is ε > 0 such that the vertical line {Re z = ε} separates the spectrum of A. Then by using appropriate contours, a new primary calculus can be defined, leading to a natural NFC for such operators. Obviously one can think of infinitely many modifications which all show the same pattern: The shape of the spectrum together with resolvent growth conditions enables one to set up an AFC via Cauchy-type integrals. The properties of the resulting calculi are more or less the same as in Theorem 1.3.2. Moreover, as long as only such calculi are involved, one can expect composition rules to hold. These however have to be proved case by case, which certainly is a tedious task. Even worse, lacking a unifying terminology makes it necessary to create an absurd amount of notation (as this section convincingly illustrates). Both are strong reasons to search for a “good” definition of the primary functional calculus for an operator A. 2.6 Miscellanies The results on adjoints can also be found in [54]. The definition of the numbers C(f, ω ′ ) as well as Proposition 2.6.11 were created in order to unify

2.8. Comments

53

certain arguments in the literature. The payoff will appear in Section 3.3. 2.7 The Spectral Mapping Theorem. Spectral mapping theorems (SMTs) have a long tradition and are ultimately important. In semigroup theory for example a spectral mapping theorem σ(etA ) = etσ(A) has a wealth of consequences regarding the asymptotics of the semigroup. (One can find a thorough discussion in [40, Chapter IV, Section 3].) It is well known that SMT holds for the Dunford-Riesz calculus, see [39, Section VII.3] or [20, Chapter VII, §4]. Also, there is a SMT for the Hirsch functional calculus, see [89, Chapter 4]. Moreover, the unpublished notes [37] of Dore contain basic ingredients of Theorem 2.7.8, with the key notion of polynomial limits, although implicit. However, our presentation is more streamlined, making use of the elegance of our abstract framework. One can find even more general results in [55]. Other Functional Calculi. Although not particularly important for our purposes, we mention that there are (a lot of) other functional calculi around in the literature. The common pattern is this: Suppose you are given an operator A and a function f for which you would like to define f (A). Take some (usually: integral) representation of f in terms of other functions g for which you already “know” g(A). Then plug in A into the known parts and hope that the formulas still make sense. Obviously, our Cauchy integral-based natural calculus is of this type (the known parts are resolvents). Other examples are: 1) The Hirsch functional calculus, based on a representation Z z f (z) = a + µ(dt) R+ 1 + tz where µ is a suitable complex measure on R+ . Details can be found in [89, Chapter 4]. 2) The Phillips calculus, based on the Laplace transform Z f (z) = e−zt µ(dt) R+

where µ is a finite complex measure on R+ . (Here, the “known” part is e−zt (A) = e−tA , i.e., −A is assumed to generate a bounded C0 -semigroup.) The main reference for this is [59, Chapter XV]. 3) The Mellin transform calculus as developed in [110] and [124]. 4) A functional calculus based on the Poisson integral formula, see [16] and [32]. (Clearly this list of functional calculi is far from being complete.) Of course, each of these calculi can be extended by the “regularization trick”. For the Phillips calculus, this is done in [123, Section 1.3.4]. Let us mention that there are first attempts to define a functional calculus for multivalued sectorial operators, see [1] and [88].

54

Chapter 2. The Functional Calculus for Sectorial Operators

Concluding Remarks. The reader might have noticed that that we have never defined formally what a functional calculus is(In fact we have only defined the terms “abstract functional calculus”, “primary functional calculus”, “extended f.c.” and “natural f.c.”.) The reason is that we feel that the notion of functional calculus is basically intuitive and should remain so — at least for the moment — in order to keep its full power. In fact there are formal definitions in the literature, as, e.g., in [15], but these attempts are ad hoc and one can usually find a functional calculus (in the intuitive sense) which does not meet the given formal definition. This indicates that the time for a “good” definition has not yet come and the theory developed here may be seen as a contribution to the strive for finding one some time in the future. Refusing to give a formal definition of a functional calculus has an important consequence: the uniqueness question does not appear, it can not even be formulated. It is our point of view that this is an advantage or at least no drawback. In Section 5.3 the uniqueness question will be treated in a quite special setting, and it will turn out that uniqueness is delicate (even if a formal definition is at hand) and seems not to be handable if situations become more general (missing density, continuity . . . ). All in all, we do not consider the uniqueness problem very important. (We are in good company, with almost no paper elaborating on the matter.) This is because — philosophically speaking — there is only one functional calculus. Though plausible, this remains intuitive reasoning until the “Great Unifying Theory” will be found. In the meantime it is therefore recommendable to give a name to each newly constructed functional calculus, at least if this calculus is widely used. Since the “ansatz” via Cauchy integrals seems natural, we advocate the term “natural functional calculus” for so constructed calculi.

Chapter 3

Fractional Powers and Related Topics In this chapter, the basic theory of fractional powers Aα of a sectorial operator A is presented. This is done (quite elegantly) by making use of the functional calculus developed in Chapter 2. In Section 3.1 we introduce fractional powers with positive real part. The section comprises proofs for the scaling property, the power laws, the spectral mapping theorem, and the Balakrishnan representation. Furthermore, we examine the behaviour of (A + ε)α for variable ε and of Aα x with variable α. In Section 3.2 we generalize the results from Section 3.1 to fractional powers with arbitrary real part. (Here, the operator A has to be injective.) A bit of a detour, the definition and the fundamental properties of holomorphic semigroups are presented in Section 3.3. The usual generator/semigroup correspondence is extended to the multivalued case. Once again, the account is elegant by using the theory of functional calculus. In Section 3.4 the logarithm of an injective sectorial operator A is defined. We prove an important spectral result originally due tue Nollau and consider the connection of log A to the family of imaginary powers (Ais )s∈R of A.

3.1 Fractional Powers with Positive Real Part In this section X always denotes a Banach space and A a sectorial operator of angle ω on X. (Recall the general agreement on terminology, see page 189.) Observe that for each α ∈ C with Re α > 0 the function f (z) = z α is contained in the class A(Sϕ ) for any ϕ ∈ (0, π]. In fact, one has z α (1 + z)−n ∈ H0∞ if n > Re α. Hence the operator zα α α n A := (z )(A) = (1 + A) (A) (0 < Re α < n) (1 + z)n is defined by the natural functional calculus for sectorial operators. We call Aα the fractional power with exponent α of A. Proposition 3.1.1. Let A be a sectorial operator on the Banach space X. Then the following assertions hold. a) If A is bounded, then also Aα is, and the mapping (α 7−→ Aα ) : {Re α > 0} −→ L(X)

56

Chapter 3. Fractional Powers and Related Topics is holomorphic. b) If n > Re α > 0, then D(An ) ⊂ D(Aα ), and the mapping (α 7−→ Aα x) : {0 < Re α < n} −→ X is holomorphic for each x ∈ D(An ).

c) (First Power Law) The identity

Aα+β = Aα Aβ holds for all Re α, Re β > 0. In particular, D(Aγ ) ⊂ D(Aα ) for 0 < Re α < Re γ. d) One has N(Aα ) = N(A) for all Re α > 0. e) If is A injective, then (A−1 )α = (Aα )−1 . If 0 ∈ ̺(A), then also 0 ∈ ̺(Aα ). f ) If T ∈ L(A) commutes with A, then it also commutes with Aα .

g) Let Re α, Re β > 0, x ∈ X, and ε > 0. Then α A(A + ε)−1 x ∈ D(Aβ ) ⇐⇒

x ∈ D(Aβ ).

h) If A is densely defined, then D(An ) is a core for Aα , where 0 < Re α < n. i) If the Banach space X is reflexive, then Aα (x ⊕ y) = 0 ⊕ B α y for all x ∈ N(A), y ∈ R(A), where B := A|R(A) is the injective part of A. j) (Spectral Mapping Theorem) σ(Aα ) = {µα | µ ∈ σ(A)} for all Re α > 0. Proof. Part a) easily follows from the definition and the Dominated Convergence Theorem. The same applies to b), if one notes that Aα x = (z α /(1 + z)n )(A)(1 + A)n x for x ∈ D(An ) and 0 < Re α < n. The power law is a little more involved. From the general statements on the functional calculus (Theorem 1.3.2) we know that Aα Aβ ⊂ Aα+β , with D(Aα Aβ ) = D(Aα+β ) ∩ D(Aβ ). Let n > Re α, Re β be fixed. We define Φα := (z α /(1 + z)n )(A) ∈ L(X) and Φβ analogously. Let x ∈ D(Aα+β ). Then Φα Φβ x =

z α+β (A)x (1+z)2n

An (1 + A)−2n Φβ x =

∈ D(A2n ). From this it follows that

z n+β z n−α (A)x = (A)Φα Φβ x ∈ D(A2n ). 3n (1 + z) (1 + z)n

Applying Proposition 2.1.1, part g), we obtain (1 + A)−n Φβ x ∈ D(A2n ). But this gives Φβ x ∈ D(An ), hence x ∈ D(Aβ ). Assertion d) follows from An x = An−α Aα x and part e) of Proposition 2.1.1. We prove e). Note that f (z) := z −α ∈ B. Hence by Proposition 2.4.1 we have (A−1 )α = f (z −1 )(A−1 ) = f (A) = (f −1 (A))−1 = (Aα )−1 . The second assertion follows from the first

3.1. Fractional Powers with Positive Real Part

57

because of a). Assertion f ) is a special case of part a) of Theorem 1.3.2. We prove g). Recall that A(A + ε)−1 is bounded and sectorial and commutes with A. Hence, by f ), (A(A + ε)−1 )α commutes with A, whence it commutes with Aβ by another application of f ). This gives one implication. Let (A(A + ε)−1 )α x ∈ D(Aβ ) and choose n > Re α, Re β. Applying the implication just proved with α replaced by n − α, we obtain (A(A + ε)−1 )n x ∈ D(Aβ ). From this we conclude that (A(A + ε)−1 )n Φβ x ∈ D(An ), where Φβ := (z β /(1 + z)n )(A). By Proposition 2.1.1, part g) this implies Φβ x ∈ D(An ), whence x ∈ D(Aβ ). Part h) follows from Proposition 2.3.10. The statement in i) is obvious. Consider the function f (z) = z α . Clearly, f has polynomial limits both at 0 and at ∞ (cf. Section 2.7). By Theorem 2.7.8 we therefore have f (˜ σ (A)) = σ ˜ (f (A)) = σ ˜ (Aα ). But µα = ∞ if and only if µ = ∞, whence j) is proved.

Sαω

Consider an operator A ∈ Sect(ω) and 0 < α < ωπ . Then σ(Aα ) = σ(A)α ⊂ by part j) of the previous proposition. But there is more to say in this case.

Proposition 3.1.2. (Scaling Property) Let A ∈ Sect(ω) for some 0 < ω < π on the Banach space X. Let 0 < α < Then Aα ∈ Sect(αω). In particular, we have ωAα = αωA .

π ω.

Proof. Let λ ∈ / Sαω and define fλ (z) := (λ − z α )−1 . By f ) of Theorem 1.3.2, λ ∈ ̺(Aα ) if and only if fλ ∈ H(A). Define 1

ψλ (z) :=

1

λ |λ| α λz + |λ| α z α − = . 1 1 λ − zα z + |λ| α (λ − z α )(z + |λ| α )

Obviously, ψλ ∈ H0∞ (Sθ ) for each 0 < θ < π. Then 1 1 fλ (z) = = λ − zα λ

1

|λ| α

1

z + |λ| α

!

+ ψλ (z)

∈ H(A).

So we have shown that λ ∈ ̺(Aα ) and R(λ, Aα ) = fλ (A). To prove sectoriality it suffices to show supλ∈S / αϕ kψλ (A)k < ∞ for ω < ϕ < π/α. By the sectoriality of A there is a constant c such that Z |dz| kψλ (A)k ≤ c |ψλ (z)| . |z| Γ Employing the scaling property ψtα λ (tz) = ψλ (z) for t > 0 we see that it is sufficient to prove the uniform boundedness of ψλ (A) f¨ ur λ ∈ / Sαϕ , |λ| = 1. This is easily done.

Corollary 3.1.3. Let (An )n ⊂ Sect(ω) be uniformly sectorial for some ω ∈ (0, π). Let 0 < α < ωπ . Then (Aα n )n ⊂ Sect(αω) is also uniformly sectorial. Furthermore, α from An → A (Sω ) it follows that Aα n → A (Sω ). The next result is an immediate consequence of Proposition 3.1.2 and the general composition rule Theorem 2.4.2. Proposition 3.1.4. Let A ∈ Sect(ω) for some ω. Assume 0 < α < ωπ and ω < π ϕ< α . If f ∈ H0∞ (Sαϕ ) (f ∈ A(Sαϕ )/ f ∈ M[Sαω ]Aα ), the function f (z α ) is in ∞ H0 (Sϕ ) (in A(Sϕ )/ M[Sω ]A ), and the identity f (Aα ) = (f (z α ))(A). holds.

58

Chapter 3. Fractional Powers and Related Topics

Corollary 3.1.5. (Second Power Law) Let A ∈ Sect(ω) with 0 < ω < π, and let 0 < α < ωπ . Then (Aα )β = Aαβ for all Re β > 0. Corollary 3.1.6. Assume A ∈ Sect(ω), Re γ > 0, and x ∈ D(Aγ ). Then the mapping (α 7−→ Aα x) : {0 < Re α < Re γ} −→ X is holomorphic. Proof. We can assume γ > 0 without restriction. Choose n > γ. Then Aα x = B αn/γ x and x ∈ D(B n ) where B := Aγ/n . Now the claim follows from b) of Proposition 3.1.1.

We now compare the operators Aα and (A + ε)α for ε > 0. Proposition 3.1.7. Let 0 < Re α < 1 and ε > 0. Then Tε := ((z + ε)α − z α )(A) ∈ L(X). Moreover, we have Aα + Tε = (A + ε)α and kTε k ≤ CεRe α , where C = C(α, ϕ, M (A, ϕ)) for ω < ϕ < π. Proof. Let ψ1 (z) := (z + 1)α − z α − (z + 1)−1 . A short computation shows that ψ1 ∈ H0∞ . Hence ψε (z) := εα ψ1 (z/ε) = (z + ε)α − z α − εα (ε−1 z + 1)−1 is also contained in H0∞ . This shows that (z + ε)α − z α ∈ H(A). By part c) of Theorem 1.3.2 and the composition rule we obtain Aα + Tε = (A + ε)α . Furthermore, we have ε−α Tε = ε−α ψε (A) − (1 + ε−1 A)−1 = ψ1 (ε−1 A) − (1 + ε−1 A)−1 . ‚ ‚ ‚ ‚ The claim now follows from supε>0 ‚(1 + ε−1 A)−1 ‚ = M (A) and ‚ψ1 (ε−1 A)‚ ≤ M (A, ϕ)C(ψ1 , ϕ) for ω < ϕ < π (apply Proposition 2.6.11 with ψ = 0).

Remark 3.1.8. With the help of the Balakrishnan representation (see below) the last proposition can be improved with respect to the constant C. Namely, one can explicitly determine a constant which only depends on M (A), Re α, and |sin απ|, see [89, Proposition 5.1.14]. The last proposition implies in particular that D((A + ε)α ) = D(Aα ) for ε > 0 and 0 < Re α < 1. However, this is true for all Re α > 0 as it is shown by the next result. Proposition 3.1.9. Let Re α > 0 and ε > 0. Then the following assertions hold. a) D(Aα ) = D((A + ε)α ). α b) Aα ((A + ε)−1 )α = A(A + ε)−1 . c) limε→0 (A + ε)α x = Aα x for each x ∈ D(Aα ).

Proof. Apply the (z) := (z + ε)−1 and g(z) := z α to obtain ` composition ´α rule to the functions f−α (z + ε)−α (A) = (A + ε)−1 ∈ L(X). Hence (z + ε) ∈ H(A) and „ « α z (A(A + ε)−1 )α = (A) = z α (A)(z + ε)−α (A) = Aα ((A + ε)−1 )α (z + ε)α by the composition rule again, whence b) is proved. Furthermore, we have ` ´α (A + ε)−1 = (z + ε)−α (A) = ((z + ε)α (A))−1 = ((A + ε)α )−1

3.1. Fractional Powers with Positive Real Part

59

by the composition rule and f ) of Theorem 1.3.2. Together with b) this shows that D((A+ε)α ) = R(((A + ε)−1 )α ) ⊂ D(Aα ). Let us prove the other inclusion of a). Choose x ∈ D(Aα ) and n > Re α. Applying the first power law and b) we obtain D(An ) ∋ (A + ε)−n Aα x = Aα (A + ε)−n x = Aα ((A + ε)−1 )α ((A + ε)−1 )n−α x = (A(A + ε)−1 )α ((A + ε)−1 )n−α x.

This gives ((A + ε)−1 )n−α x ∈ D(An ) by g) of Proposition 3.1.1. From this it follows that x = (A + ε)n−α ((A + ε)−1 )n−α x ∈ D((A + ε)α ). We are left to show c). The family of operators (A + ε)(A + 1)−1 is a sectorial approximation of A(A + 1)−1 (apply c) of Proposition 2.1.3 together with f ) of Proposition 2.1.1). Lemma 2.6.7 together with b) implies that (A + ε)α ((A + 1)−1 )α = ((A + ε)(A + 1)−1 )α → (A(A + 1)−1 )α = Aα ((A + 1)α )−1 in norm. In particular we have limεց0 (A + ε)α x = Aα x for all x ∈ D((A + 1)α ) = D(Aα ).

Remark 3.1.10. The last result together with the rule (Aα )−1 = (A−1 )α (if A is injective) reduces the definition of Aα for (general) sectorial operators A to the one for operators A ∈ L(X) with 0 ∈ ̺(A) where the usual Dunford calculus is at hand. Therefore, the last result can be viewed as an “interface” to the literature where often the fractional powers are defined in a different way. Instead of proving a statement for fractional powers directly with recourse to the definition one can proceed in three steps: 1) The validity of the statement is proved for A ∈ L(X) with 0 ∈ ̺(A). 2) One shows that in case A is injective the statement for A follows from the statement for A−1 . 3) One shows that the statement is true for A if it is true for all A + ε (0 < ε small ). In fact, many proofs follow this scheme. Corollary 3.1.11. One has D(Aα ) ⊂ D(A) and R(Aα ) ⊂ R(A) for each Re α > 0 and each sectorial operator A. Proof. Assume first that A is bounded, i.e., A ∈ L(X). Then with Γ being an appropriate finite path, we have Z Z 1 1 Aα = z α R(z, A) dz = z α−1 AR(z, A) dz 2πi Γ 2πi Γ

by Cauchy’s theorem. This gives R(Aα ) ⊂ R(A) if A ∈ L(X). For arbitrary A we apply this to the operator A(A + 1)−1 and obtain R(Aα ) = R[Aα ((A + 1)α )−1 ] = R[(A(A + 1)−1 )α ] ⊂ R(A(A + 1)−1 ) = R(A).

(Here we used b) of the last proposition.) Finally we conclude from this and a) of Proposition 3.1.9 that D(Aα ) = D[(A + 1)α ] = R[((A + 1)−1 )α ] ⊂ R((A + 1)−1 ) = D(A). This proves the statement.

We now turn to an integral representation which historically was one of the first approaches to fractional powers.

60

Chapter 3. Fractional Powers and Related Topics

Proposition 3.1.12. (Balakrishnan-Representation) Let A ∈ Sect(ω) on the Banach space X and let 0 < Re α < 1. Then sin απ A x= π α

Z

∞

tα−1 (t + A)−1 Ax dt

(3.1)

0

for all x ∈ D(A). More general, we have Aα x =

Γ(m) Γ(α)Γ(m − α)

Z

∞

tα−1 [A(t + A)−1 ]m x dt

(3.2)

0

where 0 < Re α < n ≤ m and x ∈ D(An ). Proof. Assume first that 0 < Re α < 1. For x ∈ D(A) we have Aα x =

1 2πi

Z

z α−1 R(z, A)Ax dz,

(3.3)

Γϕ

where ω < ϕ < π. In fact, we can compute „ α « „ α « „ « z z zα Aα x = (A)(ε + A)x = (A)Ax + ε (A)(1 + A)x z+ε z+ε (z + ε)(z + 1) „ « « Z Z „ 1 z ε zα = z α−1 R(z, A)Ax dz + R(z, A)(1 + A)x dz, 2πi Γϕ z+ε 2πi Γϕ (z + ε)(z + 1) where ε > 0. Letting ε ց 0, the second summand vanishes and we obtain (3.3). Note that the function z 7−→ z α−1 R(z, A)Az is integrable on the boundary Γ of the sector: R(z, A)Az is bounded at 0 and O(|z|−1 ) for z → ∞. The functions z 7−→ (z/z + ε) are bounded on Γϕ uniformly in ε, hence the Dominated Convergence Theorem is applicable. Starting from (3.3) we “deform” the path Γϕ onto the negative real axis. This means that the opening angle ϕ of Γϕ is enlarged until the angle π is reached. Cauchy’s theorem ensures that the integral does not change its value during this deforming procedure. Lebesgue’s theorem shows that the limit is exactly (3.1). So we have proved the first part of the Proposition. For the second we assume m = n and n − 1 < Re α < n. Then we write Z sin(α − n + 1)π ∞ α−n t (t + A)−1 An x dt Aα x = Aα−(n−1) An−1 x = π 0 Z ∞ sin((α − n + 1)π)(n − 1)! Int.b.p. = tα−1 (t + A)−n An x dt π(α − n + 1)(α − n + 2) . . . (α − 1) 0 Z ∞ Γ(n) = tα−1 (t + A)−n An x dt, Γ(α)Γ(n − α) 0 where x ∈ D(An ). Here we used the standard formulae for the Gamma function zΓ(z) = Γ(z +1) and sin(πz)/π = 1/(Γ(z)Γ(1−z)). An holomorphy argument allows us to replace the assumption n − 1 < Re α by 0 < Re α < n = m. Thus we have proved that (3.2) holds for n = m. To prove the general statement we use induction on m. The assertion is already known for n = m. Define Z ∞ Γ(m) cm := and Im := tα−1 [A(t + A)−1 ]m x dt. Γ(m − α)Γ(α) 0

3.1. Fractional Powers with Positive Real Part

61

Then we have Im

Z tα m ∞ α [A(t + A)−1 ]m x|∞ t [A(t + A)−1 ]m (t + A)−1 x dt 0 + α α 0 Z m ∞ α = t [A(t + A)−1 ]m (t + A)−1 x dt α 0 Z ∞ m tα−1 ([A(t + A)−1 ]m x − [A(t + A)−1 ]m+1 )x dt = α 0 m = (Im − Im+1 ). α I.b.p.

This means that

=

m I m−α m+1

m = Im . Since cm m−α = cm+1 , the induction is complete.

Corollary 0 < α < n. Then supt>0 (t(t + A)−1 )α ≤ M (A)n and

3.1.13. Let supt>0 (A(t + A)−1 )α ≤ (M (A) + 1)n . Proof. Let 0 < α < 1. Then we have

Z ` ´α sin πα ∞ α−1 (t + A)−1 = s (t + A)−1 [s + (t + A)−1 ]−1 ds π 0 Z sin πα ∞ α−1 = s (st + 1 + A)−1 ds, whence π 0 Z ∞ ‚` ´ ‚ 1 ‚ (t + A)−1 α ‚ ≤ sin πα sα−1 (st + 1)−1 dtM (A) = α M (A). π t 0

The general statement follows with the help of an easy induction argument. The proof of the second assertion is similar, see [89, Remark 5.1.2].

Corollary 3.1.14. Let A ∈ Sect(ω) and 0 < Re α < 1. Then Aα x =

sin πα π

Z

∞

t−α (t + A−1 )−1 x dt

0

for x ∈ D(A). (The operator A−1 may be multivalued, see Remark 2.1.4.) Proof. Starting from the Balakrishnan representation (3.1) we obtain Z Z sin πα ∞ α ds t=1/s sin πα ∞ −α 1 dt Aα x = s (s + A)−1 Ax = t ( + A)−1 Ax π s π t t 0 0 Z Z sin πα ∞ −α 1 1 dt sin πα ∞ −α −1 −1 −1 dt = t (I − ( + A) )x = t t(t + A ) x π t t t π t 0 0 Z ∞ sin πα −α −1 −1 = t (t + A ) x dt π 0

by the fundamental identity (2.1).

Proposition 3.1.15. Assume A ∈ Sect(ω), x ∈ D(A), and 0 ≤ ϕ < following assertions hold. a) x ∈ R(A) ⇐⇒ limα→0, α∈Sϕ Aα x = x. b) Ax ∈ D(A)

⇐⇒

limα→1, α∈1−Sϕ Aα x = Ax.

π 2.

Then the

62

Chapter 3. Fractional Powers and Related Topics

Proof. We prove a). The implication “⇐” is immediate from Corollary 3.1.11. To prove the reverse direction we use the Balakrishnan representation 3.1 and write ‚ ‚ Z ‚ ‚ sin απ ∞ α−1 1 t ((t + A)−1 Ax − x)dt‚ kAα x − Axk = ‚ ‚ ‚ π t+1 0 ˛ ˛Z L ˛ sin απ ˛ ‚ ‚ ˛ ≤ ˛˛ tRe α−1 ‚(t + A)−1 Ax − x‚ + tRe α kxk dt π ˛ 0 ˛ ˛ ‚ ‚ ˛ sin απ ˛ Z ∞ Re α−2 ‚ ‚ ˛ ‚t(t + A)−1 Ax − t x‚ dt + ˛˛ t ‚ π ˛ L 1+t ‚ ˛ ˛ ‚ ‚ ˛ sin απ ˛ |α| Re α ˛ LRe α sup ‚(t + A)−1 Ax − x‚ + LRe α+1 kxk ≤ ˛˛ απ ˛ Re α Re α + 1 t≤L « Re α + LRe α−1 (M kAxk + kxk) . 1 − Re α Note that sin απ/απ is continuous at 0 and that |α| / Re α is bounded by (cos ϕ)−1‚. Due to ‚ x ∈ R(A) we can choose the number L in such a way that supt≤L ‚(t + A)−1 Ax − x‚ is small. For a fixed L the other summands tend to zero as Re α → 0. The proof of b) requires similar arguments, see [89, p.62].

Remark 3.1.16. Let A ∈ Sect(ω) and 0 < α < 1. Then Aα ∈ Sect(αω) as we know from Proposition 3.1.2. By applying the same technique as in the proof of Proposition 3.1.12 one can obtain a Balakrishnan-type representation for the resolvent of Aα , i.e., Z − sin απ ∞ tα R(λ, Aα ) = (t + A)−1 dt α iπα α e−iπα ) π (λ − t e )(λ − t 0 for |arg λ| > απ. One can deduce M (Aα ) ≤ M (A) from this, see [89, (5.24) and (5.25)] and [120, (2.23)]. We will discuss the so-called moment inequality in connection with interpolation spaces in Section 6.6.2.

3.2 Fractional Powers with Arbitrary Real Part To introduce fractional powers with arbitrary real part, i.e., in order to define Aα := (z α )(A)

(α ∈ C).

we have to assume that the sectorial operator A is injective, because z α ∈ B(Sϕ ) for all α ∈ C and each 0 < ϕ < π. Proposition 3.2.1. Let A ∈ Sect(ω) be injective. Then the following assertions hold. a) The operator Aα is injective with (Aα )−1 = A−α = (A−1 )α for each α ∈ C. b) We have Aα Aβ ⊂ Aα+β with D(Aβ ) ∩ D(Aα+β ) = D(Aα Aβ ) for all α, β ∈ C.

3.2. Fractional Powers with Arbitrary Real Part

63

c) If D(A) = X = R(A), then Aα+β = Aα Aβ for all α, β ∈ C.

d) If 0 < Re α < 1 then

−α

A for all x ∈ R(A).

sin πα x= π

e) If α ∈ R satisfying |α| <

π ω,

Z

∞

t−α (t + A)−1 x dt

0

then Aα ∈ Sect(|α| ω) and one has (Aα )β = Aαβ

for all β ∈ C.

f ) Let Re α0 , Re α1 > 0. Then D(Aα1 ) ∩ R(Aα0 ) ⊂ D(Aα ) for each α with − Re α0 < Re α < Re α1 . The mapping (α 7−→ Aα x) : {α | − Re α0 < Re α < Re α1 } −→ X is holomorphic for each x ∈ D(Aα1 ) ∩ R(Aα0 ). Proof. Part a) follows from f ) of Theorem 1.3.2 in the same way as e) of Proposition 3.1.1. Part b) is immediate from c) of Theorem 1.3.2. We prove c). For 1 < n ∈ N we define τn (A) := n(n + A)−1 −

1 1 ( + A)−1 = (n + A)−1 n n

„

n−1 n

«

A(

1 + A)−1 . n

Then it is easy to see that for k ∈ N the following assertions hold. ‚ ‚ a) supn ‚τn (A)k ‚ < ∞. b)

limn→∞ τn (A)k x = x for x ∈ D(A) ∩ R(A).

c) R(τn (A)k ) ⊂ D(Ak ) ∩ R(Ak ), and D(Ak ) ∩ R(Ak ) = D(A) ∩ R(A). We now choose k ∈ N such that D(Ak ) ∩ R(Ak ) ⊂ D(Aβ ). Let x ∈ D(Aα+β ) and define xn := τn (A)k x. Then xn ∈ D(Aα+β ) ∩ D(Aβ ) ⊂ D(Aα Aβ ) and xn → x. Furthermore, Aα Aβ xn = τn (A)k Aα+β x → Aα+β x. This shows the claim. Part d) follows from Corollary 3.1.14 by replacing A by A−1 ; part e) follows from a) and Proposition 3.1.2 together with the composition rule (Theorem 2.4.2). To prove f ), let Re α0 , Re α1 > 0, x ∈ D(Aα1 ) ∩ R(Aα0 ), and − Re α0 < α < Re α1 . If α∈ / iR, then it is clear from Proposition 3.1.1 that x ∈ D(Aα ). Assume α = is ∈ iR. Choose n ∈ N such that Re α0 , Re α1 < n and define B := A1/n . Then x ∈ D(B) ∩ R(B), hence x ∈ D(B ins ) = D(Ais ) = D(Aα ). Since Aα x = Aα+α0 A−α0 x the second statement follows from Corollary 3.1.6.

For actual computations the following integral representations are of great importance. Proposition 3.2.2. (Komatsu-Representation)

64

Chapter 3. Fractional Powers and Related Topics

Let A ∈ Sect(ω) be injective. The identities Z 1 sin πα 1 1 α x− A−1 x + tα+1 (t + A)−1 A−1 x dt A x = π α 1+α 0 Z ∞ α−1 −1 + t (t + A) Ax dt

(3.4)

1

Z 1 sin πα 1 x+ t−α (1 + tA)−1 Ax dt π α 0 Z 1 − tα (1 + tA−1 )−1 A−1 x dt

=

(3.5)

0

hold for |Re α| < 1 and x ∈ D(A) ∩ R(A).

Note that the second formula is symmetric in A and A−1 whence it can again be seen that A−α x = (A−1 )α x for x ∈ D(A) ∩ R(A).

Proof. We assume first that 0 < Re α < 1. Starting from the Balakrishnan-representation (3.1) we obtain Z ∞ Z 1 Z ∞ π Aα x = tα−1 (t + A)−1 Ax dt = tα−1 (t + A)−1 Ax dt + tα−1 (t + A)−1 Ax dt sin απ 0 0 1 Z 1 Z ∞ = tα−1 (1 + tA−1 )−1 x dt + tα−1 (t + A)−1 Ax dt 0

= = =

Z

1

1

α−1

t 0

1 x− α 1 x− α

Z

(1 − (1 + tA 1

−1 −1

)

tA

−1

)x dt +

tα (1 + tA−1 )−1 A−1 x dt +

0

Z

Z

Z

∞

tα−1 (t + A)−1 Ax dt

1 ∞

tα−1 (t + A)−1 Ax dt

(∗)

1

1

tα (1 + tA−1 )−1 A−1 x dt +

0

Z

1

t−α (1 + tA)−1 Ax dt

0

for x ∈ D(A) ∩ R(A), where in the last step we have replaced t by t−1 in the second integral. This last formula makes sense even for −1 < Re α < 1. Hence, by holomorphy, we obtain (3.5) for |Re α| < 1. The representation (3.4) now follows from (∗) with the help of the identity t−1 (t + A)−1 = t−1 A−1 − (t + A)−1 A−1 .

A particularly important subclass of the injective sectorial operators are the invertible ones. Proposition 3.2.3. Let A ∈ Sect(ω) and assume 0 ∈ ̺(A). Then the mapping (α 7−→ A−α ) : {Re α > 0} −→ L(X) is holomorphic. For every 0 < ϕ < π2 ,

sup{ A−α | 0 < Re α < 1, |arg α| ≤ ϕ} < ∞.

Moreover, we have

x ∈ D(A)

⇐⇒

lim

α→0,|arg α|≤ϕ

A−α x = x.

(3.6)

3.3. Holomorphic Semigroups

65

Proof. The first assertion follows from A−α = (A−1 )α and a) of Proposition 3.1.1. To prove (3.6) we employ d) of Proposition 3.2.1 and obtain Z ∞ ‚ −α ‚ |sin πα| ‚A x‚ ≤ K |sin πα| t− Re α (t + 1)−1 dt kxk = K kxk , π sin(π Re α) 0 ‚ ‚ ‚ ‚ ˛ πα ˛ π Re α |α| ˛ where K := supt>0 ‚(t + 1)(t + A)−1 ‚ < ∞. Hence ‚A−α ‚ ≤ K ˛ sinπα . If sin(π Re α) Re α

A−α x → x then x ∈ D(A) by Corollary 3.1.11. The second implication follows from a) of Proposition 3.1.15.

Proposition 3.2.4. Let −A be the generator of an exponentially stable semigroup T = (T (t))t≥0 . Then A ∈ Sect( π2 ) and 0 ∈ ̺(A). Furthermore, −α

A

1 = Γ(α)

Z

∞

tα−1 T (t) dt

(3.7)

0

for all Re α > 0. Proof. If −A generates a bounded semigroup, it is sectorial of angle π/2. This follows from Proposition A.8.1. By exponential stability there are M ≥ 1, ε > 0 such that kT (t)k ≤ M e−εt for all t ≥ 0. The resolvent of −A can be computed via the Laplace transform of the semigroup, i.e., Z ∞

(u + A)−1 x =

e−ut T (s)x dt

0

for each x ∈ X and each u > −ε. In particular, 0 ∈ ̺(A). To establish the identity (3.7) for all Re α > 0, it suffices to prove it for 0 < α < 1 (by holomorphy). Using d) of Proposition 3.2.1 we obtain for x ∈ X Z Z Z sin πα ∞ −α sin πα ∞ −α ∞ −ut A−α x = u (u + A)−1 x du = u e T (t)x dt du π π 0 0 0 « Z ∞ „Z ∞ Fub. sin πα u−α e−ut du T (t)x dt = π 0 0 Z ∞Z ∞“ ” Z ∞ s −α −s sin πα s=u/t sin πα = e ds t−1 T (t)x dt = Γ(1 − α) tα−1 T (t)x dt. π t π 0 0 0 Now, the well known formula Γ(α)Γ(1 − α) =

π sin απ

(valid for all α ∈ C \ Z) yields (3.7).

Note that in the last proposition we did not assume the semigroup to be strongly continuous at 0.

3.3 Holomorphic Semigroups In this section we survey the basic properties of (bounded) holomorphic semigroups. The material is (almost) standard and is only included because it fits nicely into our functional calculus framework. Let A be a sectorial operator of angle 0 ≤ ω < π2 . In contrast to other sections we allow the operator A to be multivalued. For 0 6= λ ∈ C with |arg λ| < π2 − ω the ∞ function e−λz is in H(0) (Sϕ ) for each ϕ ∈ (ω, π2 − |arg λ|), i.e., is in E(Sϕ ), decays regularly at ∞ and is holomorphic round 0 (see Example 2.2.4 and Lemma 2.3.2).

66

Chapter 3. Fractional Powers and Related Topics By Remark 2.3.4, this allows us to define e−λA := (e−λz )(A) ∈ L(X).

(3.8)

The family (e−λA )|arg λ|< π2 −ω is called the holomorphic semigroup generated by −A (see below for a general introduction of this terminology). We summarize some basic properties. Proposition 3.3.1. Let A be a multivalued sectorial operator of angle ω with ω < Then the following assertions hold. a) e−λA e−µA = e−(λ+µ)A for all λ, µ ∈ S π2 −ω .

π 2.

b) The mapping

(λ 7−→ e−λA ) : S π2 −ω −→ L(X) is holomorphic. c) Let 0 < θ <

π 2

− ω. Then sup{ke−λA k | |arg λ| ≤ θ} < ∞.

More precisely, for each choice of ω ′ ∈ (ω, π2 −θ) there is C = C(e−z , ω ′ +θ) with ke−λA k ≤ CM (A, ω ′ ) for all |arg λ| ≤ θ.

d) The identity

(µ + A)−1 =

Z

∞

e−µt e−tA dt

0

holds true for all Re µ > 0.

e) For all λ ∈ S π2 −ω we have R(e−λA ) ⊂ f ) If x ∈ D(A) then

lim

λ→0,|arg λ|≤ϕ

for each 0 < ϕ <

π 2

T

n∈N

D(An ).

e−λA x = x

− ω.

Proof. Assertion a) follows from the multiplicativity of the E-functional calculus (Theorem 2.3.3). The statement in b) is easily proved iun writeing e−λA as a Cauchy integral (Lemma 2.3.2) and applying the Dominated Convergence Theorem. We obtain dn −λA (e ) = ((−z)n e−λz )(A). dλn Assertion c) is a special case of Proposition 2.6.11 (writing ϕ := ω ′ + θ) in noting that e−λA = (e−z )(λA) (which is an instance of the composition rule). To prove d), given Re µ > 0 we compute Z ∞ Z ∞ Z Z Z ∞ 1 1 e−µt e−tA dt = e−µt e−tz R(z, A) dz = e−µt e−tz dt R(z, A) dz 2πi Γ 2πi Γ 0 0 0 Z 1 1 = R(z, A) dz = (µ + A)−1 2πi Γ µ + z by Lemma 2.3.2. (Here, Γ is an appropriate path avoiding 0, see Section 2.3.) To prove e), take n ∈ N and λ ∈ S π −ω . Since both functions (1 + z)−n and (1 + z)n e−λz are 2

3.3. Holomorphic Semigroups

67

contained in E we obtain e−λA = (1 + A)−n ((1 + z)n e−λz )(A) by multiplicativity. We prove f ). Choose ϕ ∈ (0, π/2 − ω). Then (e−λz (1 + z)−1 )(A) → (1 + A)−1 in norm for Sϕ ∋ λ → 0 by the Dominated Convergence Theorem. Hence e−λA x = (e−λz (1+z)−1 )(A)z → x for x ∈ D(A) and z ∈ (1 + A)x. By the uniform boundedness proved in c) we obtain e−λA x → x even for all x ∈ D(A). This completes the proof.

Proposition 3.3.2. Let A ∈ Sect(ω) for ω < π2 and assume that A is single valued. Then we have f (A)e−λA ∈ L(X) for all λ ∈ S π2 −ω and f ∈ A[Sω ]. Moreover, let 0 < ϕ < π2 − ω and Re α > 0. Then n o Re α sup |λ| kAα e−λA k | arg λ ≤ ϕ < ∞.

More precisely, for each choice of ω ′ ∈ (ω, π2 − θ) there is C = C(z α e−z , ω ′ + θ) − Re α such that kAα e−λA k ≤ C M (A, ω ′ )e(Im α arg λ) |λ| for all |arg λ| ≤ θ.

Proof. Take f ∈ A[Sω ]. Then f (z)e−λz is regularly decaying at ∞ and f (z)e−λz − f (0) is regularly decaying at 0, hence f (z)e−λz ∈ E. By a standard functional calculus argument we obtain f (A)e−λA = (f (z)e−λz )(A) ∈ L(X). The last statement follows from Proposition 2.6.11 in noting that λα Aα e−λA = (z α e−z )(λA) (composition rule).

Let θ ∈ (0, π2 ]. A mapping T : Sθ −→ L(X) is called a bounded holomorphic (degenerate) semigroup (of angle θ) if it has the following properties: 1) The semigroup law T (λ)T (µ) = T (λ + µ) holds for all λ, µ ∈ Sθ . 2) The mapping T : Sθ −→ L(X) is holomorphic. 3) The mapping T satisfies supλ∈Sϕ kT (λ)k < ∞ for each 0 < ϕ < θ.

(Note that, by a), b) and c) of Proposition 3.3.1, (e−λA )λ∈Sθ is a bounded holomorphic semigroup if A ∈ Sect(ω) and θ := π2 − ω > 0.) By holomorphy, T is uniquely determined by its values on (0, ∞). Moreover, if the semigroup law holds for real values, then it holds for all λ (see the proof of Proposition 3.7.2 in [6]). Also by holomorphy and the semigroup law, the space NT := N(T (λ)) is independent of λ ∈ Sθ . Hence either each or none of the operators T (λ) is injective. If we restrict a bounded holomorphic semigroup T to the positive real axis (0, ∞), we obtain a bounded semigroup as defined in Section A.8. This semigroup has a generator B which is defined via its resolvent by Z ∞ R(λ, B) := e−λt T (t) dt 0

for Re λ > 0. Let A := −B. Then we have A0 = NT by (A.1) on page 206. Hence, A is single-valued if and only if T (w) is injective for one/all w ∈ Sθ . The m.v. operator A is sectorial (at least of angle π2 ), since

λ(λ + A)−1 = kλR(λ, B)k ≤ (sup kT (t)k) |λ| Re λ t>0

for all Re λ > 0. But even more is true: The m.v. operator A is sectorial of angle π −λA for all λ ∈ Sθ . 2 − θ, and we have T (λ) = e

68

Chapter 3. Fractional Powers and Related Topics

Proof. Let ϕ ∈ (−θ, θ) and consider the bounded holomorphic semigroup T (eiϕ ·) on Sθ−|ϕ| . We know already that there is a m.v. operator Aϕ ∈ Sect(π/2) such that (λ + Aϕ )−1 = R ∞ −λt e T (teiϕ ) dt for all Re λ > 0. Let Γ = (0, ∞)eiϕ . By Cauchy’s theorem, 0 Z ∞ Z Z ∞ iϕ (s+A)−1 = e−st T (t) dt = e−sz T (z) dz = eiϕ e−se t T (teiϕ ) dt = eiϕ (seiϕ +Aϕ )−1 0

Γ

0

for all s > 0. Hence (s + A)−1 = (s + e−iϕ Aϕ )−1 for all s > 0. This gives eiϕ A = Aϕ . Since ϕ ranges between −θ and θ and each Aϕ is sectorial of angle π/2 we obtain that A is sectorial of angle π/2 − θ. Employing d) of Proposition 3.3.1 and the injectivity of the Laplace transform we conclude T (t) = e−tA for all t > 0. By holomorphy, this implies T (λ) = e−λA for all λ ∈ Sθ .

Combining the above considerations with Proposition 3.3.1 we can summarize: Proposition 3.3.3. There is a one-one correspondence between m.v. sectorial operators A of angle ω < π2 and bounded holomorphic semigroups T on S π2 −ω given by the relations Z ∞ T (z) = e−zA (z ∈ S π2 −ω ), (λ + A)−1 = e−λt T (t) dt (Re λ > 0). 0

The operator A is single-valued if and only if T (z) is injective for one/all z ∈ S π2 −ω . The last result establishes a fact pointed at in Remark 2.3.11. Namely, if A is sectorial and single-valued and ωA < π2 then e−tz for t > 0 is a regularizer which compensates exponential growth at ∞. Remark 3.3.4. An exponentially bounded holomorphic semigroup of angle θ ∈ (0, π2 ] is a holomorphic mapping T : Sθ −→ L(X) with the semigroup property and such that {T (λ) | λ ∈ Sϕ , |λ| ≤ 1} is bounded for each ϕ ∈ (0, θ). Given such a semigroup for each 0 < ϕ < θ one can find Mϕ ≥ 1, wϕ ≥ 0 such that kT (λ)k ≤ Mϕ ewϕ Re λ for λ ∈ Sϕ . In particular, T |(0,∞) is an exponentially bounded semigroup in the sense of Section A.8. Thus it has a generator −A, which is characterized by Z ∞ (λ + A)−1 = e−λt T (t) dt 0

which holds for sufficiently large Re λ. Given ϕ ∈ (0, θ) and wϕ as above, we have e−wϕ λ T (λ) = e−λ(A+wϕ ) for λ ∈ Sϕ . Thus, all statements on exponentially bounded holomorphic semigroups can be reduced to statements on bounded holomorphic semigroups. For example, given ϕ ∈ (−θ, θ), the mapping (t 7−→ T (eiϕ t)) is an exponentially bounded semigroup with generator −eiϕ A. (See the proof before Proposition 3.3.3.)

3.4. The Logarithm and the Imaginary Powers

69

From Corollary A.8.3 we know that the space of strong continuity of T |(0,∞) is exactly D(A). Employing Proposition 3.3.3 and part f ) of Proposition 3.3.1 we see that even lim

Sϕ ∋λ→0

T (λ)x = x

for x ∈ D(A) and each 0 ≤ ϕ < θ. Let A ∈ Sect(ω) on the Banach space X and assume that 0 ∈ ̺(A). We know from Proposition 3.2.3 that (A−z )Re z>0 is an exponentially bounded holomorphic semigroup of angle π2 , with D(A) as its space of strong continuity. In the next section we will identify its generator.

3.4 The Logarithm and the Imaginary Powers We return to our terminological agreement that “operator” always is to be read as “single-valued operator” (see the agreement on p. 189). In fact, we will work with an injective, single-valued operator A ∈ Sect(ω), where ω < π. The function log z has subpolynomial growth at 0 and at ∞, whence it belongs to the class B(Sϕ ) for each 0 < ϕ ≤ π. Since A is assumed to be injective, the operator log A := (log z)(A) is defined and is called the logarithm of the operator A. Because of log(z −1 ) = − log z we have log(A−1 ) = − log A. The next result is quite fundamental. Lemma 3.4.1. (Nollau) Let A ∈ Sect(ω) be injective. If |Im λ| > π then λ ∈ ̺(log A) and R(λ, log A) =

Z

0

Hence kR(λ, log A)k ≤

∞

−1 (t + A)−1 dt. (λ − log t)2 + π 2

(3.9)

M(A)π |Im λ|−π .

We call the formula (3.9) the Nollau representation of the resolvent of log A.

Proof. Assume first that A, A−1 ∈ L(X). We choose ω < ϕ < π, a > 0 small and b > 0 large

70

Chapter 3. Fractional Powers and Related Topics

enough and denote by Γ the positively oriented boundary of the bounded sector Sϕ (a, b). Then Z 1 1 (λ − log z)−1 (A) = R(z, A) dz 2πi Γ λ − log z Z b Z ϕ 1 e−iϕ 1 beis = R(te−iϕ , A) dt + R(beis , A) ds 2πi a λ − log t + iϕ 2π −ϕ λ + log b − is Z b Z ϕ 1 eiϕ 1 aeis − R(teiϕ , A) dt − R(aeis , A) ds 2πi a λ − log t − iϕ 2π −ϕ λ + log a − is Z b Z π 1 beis −1 (1) −1 (t + A) dt + R(beis , A) ds = 2 + π2 (λ − log t) 2π λ + log b − is −π a Z π aeis 1 R(aeis , A) ds − 2π −π λ + log a − is Z ∞ −1 (2) = (t + A)−1 dt =: J(A), (λ − log t)2 + π 2 0

where we have let ϕ → π in (1) and a → 0, b → ∞ in (2). Hence we have proved the claim in the special case of A and A−1 being bounded. In the general case, define Aε := (A + ε)(1 + εA)−1 . Then (Aε )ε>0 is a sectorial approximation of A (see Proposition 2.1.3). Let f (z) := (λ−log z)−1 . Obviously f ∈ H ∞ (Sϕ ) for ϕ > ω. We have already shown that f (Aε ) = J(Aε ) ∈ L(X). It follows from the Dominated Convergence Theorem that Z ∞ −1 f (Aε ) → J(A) = (t + A)−1 dt. (λ − log t)2 + π 2 0

in norm. Applying Proposition 2.6.9 we obtain f (A) = J(A). From f ) of Theorem 1.3.2 we conclude that in fact f (A) = (λ − log A)−1 . Having shown this we compute Z ∞ Z M (A) dt M (A) kR(λ, log A)k ≤ = ds 2 2 |(λ − log t)2 + π 2 | t 0 R |(i Im λ − s) + π | Z M (A) = p ds (s2 − ((Im λ)2 − π 2 )2 + 4s2 (Im λ)2 R Z M (A)π M (A) M (A)π ≤ ds = p ≤ 2 2 2 |Im λ| − π (Im λ)2 − π 2 R (s + ((Im λ) − π ) where we have used the inequalities

(s2 − ((Im λ)2 − π 2 ))2 + 4s2 (Im λ)2 ≥ (s2 − ((Im λ)2 − π 2 ))2 + 4s2 ((Im λ)2 − π 2 )

which are easy to verify.

= (s2 + ((Im λ)2 − π 2 ))2 , q (Im λ)2 − π 2 ≥ |Im λ| − π

and

Proposition 3.4.2. Let A ∈ Sect(ω) be injective. If |Im λ| > ω then λ ∈ ̺(log A) and for each ϕ > ω there is a constant Mϕ such that kR(λ, log A)k ≤ π

Mϕ |Im λ| − ϕ

for |Im λ| > ϕ. In fact Mϕ = πM (A ϕ ). Furthermore, D(log A) ⊂ D(A) ∩ R(A).

Proof. Let α := π/ϕ. Then Aα ∈ Sect(αω) is injective. The composition rule yields log(Aα ) = α log(A). Applying Nollaus’s Lemma 3.4.1 we obtain kR(µ, α log A)k ≤ M (Aα )π(|Im µ| − π)−1 for |Im µ| > π. Letting λ = µ/α we arrive at kR(λ, log A)k ≤ M (Aα )π(|Im λ| − ϕ)−1 for |Im λ| > π/α = ϕ. From the Nollau-representation (3.9) for R(·, log A) it is immediate that D(log A) ⊂ D(A). But − log A = log A−1 , hence also D(log A) ⊂ D(A−1 ) = R(A).

3.4. The Logarithm and the Imaginary Powers

71

As promised, we can now identify the generator of the holomorphic semigroup (A−z )Re z>0 , where A is a sectorial and invertible operator. Proposition 3.4.3. Let A ∈ Sect(ω) with 0 ∈ ̺(A). Then − log A is the generator of the holomorphic semigroup (A−z )Re z>0 . In particular, D(log A) = D(A). Proof. By semigroup theory there is c > 0 suchR that (e−ct A−t )t>0 is a bounded semigroup. Clearly it suffices to show that (λ + log A)−1 = 0∞ e−λt A−t dt for some λ with Re λ > c and |Im λ > π|. So choose such a λ and let 0 < a < b < ∞. Then Z b Z b Z Z Z b 1 1 e−λt A−t dt = e−λt z −t R(z, A) dz dt = e−λt z −t dtR(z, A) dz 2πi Γ 2πi Γ a a a Z −λa −a 1 e z − e−λb z −b = R(z, A) dz 2πi Γ λ + log z = e−λa A−a (λ + log A)−1 − e−λb (λ + log A)−1 ,

where Γ is an appropriate path avoiding 0 (see Section 2.5). The last equality is due to the fact −1 −1 ∈ L(X) by Nollau’s Lemma 3.4.1. Since Re λ > c we that (λ ‚ + log z)‚ (A) = (λ + log A) have ‚e−λb A−b ‚ → 0 for b → ∞. From D(log A) ⊂ D(A) and Proposition 3.2.3 we see that e−λa A−a (λ + log A)−1 → (λ + log A)−1 strongly. This finishes the proof.

Remark 3.4.4. Suppose that A is a bounded sectorial operator. As a matter of fact, the fractional powers (Az )Re z>0 of A form a quasi-bounded holomorphic semigroup as well. What is its generator? If A is injective, the answer is of course log A. But if A is not injective, the generator is not single-valued and can not be obtained by our functional calculus. Since also in this case the generator should be log A in a sense, we feel the necessity of a sophisticated functional calculus for multivalued operators. However, we do not further pursue this matter here. We end our investigation of the logarithm with the following nice observation. Consider the function f (z) := log z − log(z + ε) = log(z(z + ε)−1 ). A short computation reveals that f is holomorphic at ∞ with f (∞) = 0. Hence, if A ∈ Sect(ω) and 0 ∈ ̺(A), then f (A) ∈ L(X), i.e., f ∈ H(A). From the usual rules of functional calculus we see that log(A + ε) is a bounded perturbation of log A. In particular, D(log A) = D(log(A + ε)). We now turn to the imaginary powers Ais of an injective operator A ∈ Sect(ω). Proposition 3.4.5. Let A ∈ Sect(ω) be injective and let 0 6= s ∈ R. Then the following assertions hold. a) If Ais ∈ L(X) then D(Aα ) ⊂ D(Aα+is ) for all α ∈ C. Conversely, if D(Aα ) ⊂ D(Aα+is ) for α ∈ {−1, 1}, then Ais ∈ L(X). b) If 0 ∈ ̺(A) and ε > 0, then

(A + ε)is = (1 + εA−1 )is Ais . In particular, D((A + ε)is ) = D(Ais ). Moreover, (A + ε)is x → Ais x for ε → 0 and x ∈ D(Ais ).

72

Chapter 3. Fractional Powers and Related Topics c) The space D := D((A + ε)is ) is independent of ε > 0. If x ∈ D and limε→0 (A + ε)is x =: y exists, then x ∈ D(Ais ) with y = Ais x. is d) If Ais ∈ L(X), then

also (A + ε) ∈ L(X)is for allis ε > 0. Moreover, is

sup0≤ε≤1 (A + ε) < ∞ and limε→0 (A + ε) x = A x for all x ∈ R(A).

is e) Let 0 ∈ ̺(A). If A ∈ L(X), then sup0<α<1 A−α+is < ∞ and

lim A−α+is x = Ais x

αց0

for all x ∈ D(A). Conversely, if sup0<α<1 A−α+is < ∞ and D(A) = X, then Ais ∈ L(X).

Proof. Ad a). Assume Ais ∈ L(X) and let α ∈ C. Then Ais Aα ⊂ Aα+is , hence D(Aα ) = D(Ais Aα ) ⊂ D(Aα+is ). Conversely, assume D(A) ⊂ D(A1+is ) and R(A) ⊂ D(A−1+is ). Since always D(A1+is ) ∩ D(A) = D(Ais A), we have D(A) = D(Ais A). Hence R(A) ⊂ D(Ais ). Similarly we obtain D(A) = R(A−1 ) ⊂ D(Ais ). But D(A) + R(A) = X, whence Ais ∈ L(X). Ad b). Since A(A + ε)−1 is bounded and invertible, we have (A(A + ε)−1 )is ∈ L(X). But „ «is z (1 + εA−1 )−is = (A(A + ε)−1 )is = (A) z+ε by the composition rule. Hence (z(z + ε)−1 )is ∈ H(A). Therefore, " «is # „ z (A + ε)is (1 + εA−1 )−is = (z + ε)is (A) = (z is )(A) = Ais z+ε (see Theorem 1.3.2). From this we conclude (A + ε)is = (1 + εA−1 )is (A + ε)is (1 + εA−1 )−is = (1 + εA−1 )is Ais by e) of Theorem 1.3.2. The last statement follows from the fact that (1 + εA−1 )is → I in norm for ε → 0 (apply Lemma 2.6.7). Ad c). The first statement is immediate from b). Now assume x ∈ D and limεց0 (A + ε)is x =: y exists. Now ! „ «is z (τ (z)z is )(A) = τ (z) (A + ε)is x → τ (A)y (ε ց 0) z+ε by Lemma 5.1.2. But this is just to say that Ais x = y. Ad d). Assume Ais ∈ L(X). Then

D(A + ε) = D(A) ⊂ D(A1+is ) = D((A + ε)1+is )

by a) and Proposition 3.1.9. Since (A + ε)−1+is is bounded, we also have D((A + ε)−1 ) = X ⊂ D((A + ε)−1+is ). Applying a) again we obtain (A + ε)is ∈ L(X). From i h ˆ ˜1 1 1 1 (A + ε)is x = (A + ε) 2 +is − A 2 +is (A + ε)− 2 x + Ais A(A + ε)−1 2 x ‚ ‚ we conclude sup0≤ε≤1 ‚(A + ε)is ‚ < ∞ with the help of Proposition 3.1.7 and Corollary 3.1.13. 1

1

1

Now, let x = A 2 y ∈ R(A 2 ). Then the first summand is O(ε 2 ). The second summand can

be written as (A(A + ε)−1 )Ais (A + 1 ε) 2

1 A2

1 ε) 2

1

y and this tends to Ais A 2 y = Ais x for ε ց 0, since

Ais ∈ L(X), (A + y→ y (by Proposition 3.1.9), and A(A + ε)−1 → I strongly on R(A) (by Proposition 2.1.1). Ad e). The statements follow almost immediately from Proposition 3.2.3.

3.4. The Logarithm and the Imaginary Powers

73

Example 7.3.3 in [89] which goes back to Komatsu yields a bounded sectorial operator A on a Banach space X such that Ais ∈ / L(X) for all 0 6= s ∈ R. This shows in particular that in general D(Ais ) 6= D((A + ε)is ). The next result shows that — in a way — the operator i log A can be considered the “generator” of the operator family (Ais )s∈R . Recall the notation −2 Λ−1 for an injective sectorial operator A and note that (Ais Λ−1 A = A(1 + A) A )s∈R is a strongly continuous family of bounded operators on X. Proposition 3.4.6. Let A ∈ Sect(ω) be injective. For each θ > ω there is a constant Cθ ≥ 0 such that

is −1

A Λ ≤ M (A, θ)Cθ e|s|θ A for all s ∈ R. Moreover, one has

(λ − i log A)−1 = ΛA

Z

∞

0

e−λs Ais Λ−1 A ds

(3.10)

for all Re λ > ω. Proof. Choose θ > ω. Then Ais Λ−1 A Z ‚ ‚ M (A, θ) ‚ is −1 ‚ ‚A ΛA (A)‚ ≤ 2π Γθ For 0 < a < b we compute Z b Z e−λs Ais Λ−1 ds = A a

b

= (z −is τ )(A), whence ˛ is ˛ ˛z ˛

e−λs

a

1 = 2πi

Z

1 = 2πi

Z

|1 + z|2

1 2πi

Z

Γ

z (1 + z)2

Γ

e−λa = 2πi

1

Γ

Z

|dz| ≤ e|s|θ

M (A, θ) 2π

Z

1 Γθ

|1 + z|2

|dz| .

z is z R(z, A) dz (1 + z)2 „Z b « z is e−λs ds R(z, A) dz a

z e−(λ−i log z)a − e−(λ−i log z)b R(z, A) dz 2 (1 + z) λ − i log z

Γ

z ia z R(z, A) dz− (1 + z)2 (λ − i log z) Z e−λb z ib z R(z, A) dz 2πi Γ (1 + z)2 (λ − i log z)

−λb ib −1 = e−λa Aia Λ−1 A ΛA R(λ, i log A) A R(λ, i log A) − e

The second summand tends to 0 in norm for b → ∞, since Re λ > ω. The first summand tends to Λ−1 A R(λ, i log A) strongly for a → 0 (by f ) of Proposition 3.2.1). Hence we obtain Z ∞ −1 e−λs Ais Λ−1 A ds = ΛA R(λ, i log A) 0

which immediately implies (3.10).

Corollary 3.4.7. Let A ∈ Sect(ω) be injective. The following assertions are equivalent. (i) D(A) ∩ R(A) = X and Ais ∈ L(X) for all s ∈ R.

(ii) The operators (Ais )s∈R form a C0 -group of bounded operators on X.

74

Chapter 3. Fractional Powers and Related Topics (iii) The operator i log A generates a C0 -group (T (s))s∈R of bounded operators on X.

In this case we have T (s) = Ais for all s ∈ R.

Proof. (i) ⇒ (ii). Define T (s) := Ais for s ∈ R. Obviously, (T (s))s∈R is a group. For every x ∈ D(A) ∩ R(A), T (·)x is continuous. This follows from Proposition 3.2.1. Since D(A) ∩ R(A) = X, we have that for each x ∈ X the trajectory T (·)x is at least strongly measurable. From [59, Theorem 10.2.3] we infer that T is strongly continuous on (0, ∞), but this implies readily that T is strongly continuous on the whole real line. (ii) ⇒ (iii). This follows from Proposition 3.4.6, since the hypothesis allows us to put the operator Λ−1 A in front of the integral. (iii) ⇒ (i). Assume that i log A generates the C0 -group T . In particular, D(log A) must be dense in X. Now, D(log A) ⊂ D(A) ∩ R(A) = D(A) ∩ R(A) by Proposition 3.4.2. Thus, we are left to show T (s) = Ais for all s ∈ R. Employing Proposition 3.4.6 we obtain Z ∞ Z ∞ −1 e−λs Ais Λ−1 e−λs Λ−1 A ds = ΛA R(λ, i log A) = A T (s) ds 0

0

−1 for Re λ large. The Uniqueness Theorem for the Laplace transform implies Ais Λ−1 A = ΛA T (s) for all s > 0. Multipying both sides of this equation with ΛA yields Ais = T (s) for s > 0. From this we infer A−is = (Ais )−1 = T (s)−1 = T (−s) for all s > 0, and the proof is complete.

If A satisfies the equivalent conditions in Corollary 3.4.7 we say that A has bounded imaginary powers and write A ∈ BIP(X). Note that D(A) ∩ R(A) = X is included in this definition; there exist operators A such that Ais ∈ L(X) for all

is

s and even sups∈R A < ∞, but D(A) ∩ R(A) 6= X (see Example 7.3.1 in [89]). If A ∈ BIP(X), by semigroup theory, we know that

θA := inf{θ ≥ 0 | ∃ C : Ais ≤ C e|s|θ (s ∈ R)} < ∞. (3.11)

We will write A ∈ BIP(X, θ) if A ∈ BIP(X) and θ ≥ θA . Note that this notation slightly differs from the terminology used in [89, Definition 8.1.1]. The first part of the celebrated Pr¨ uss-Sohr theorem states that always θA ≥ ωA if A ∈ BIP(X). We will reprove (and in fact generalize) this result in Section 4.3, see Corollary 4.3.4. Remark 3.4.8. Suppose that (Ais )s>0 ⊂ L(X) is strongly continuous and norm bounded at 0. This means that T (s) := Ais is an exponentially bounded semigroup in the terminology of Section A.8. It follows from Proposition 3.4.6 that the operator i log A is the generator of T even in this situation. The same conclusion holds (with an appropriate notion of “generator”) in case that (s 7−→ Ais ) is continuous only with respect to a “very weak” topology, like those considered in [76] and [75]. In any case, Proposition 3.4.6 implies that (Ais Λ−1 A )s∈R is an exponentially bounded Λ−1 -regularized group with generator i log A, see [33] and A compare also [101, Addendum] and [100].

3.5 Comments Literature on Fractional Powers, past and present. Beginning with the seminal papers by Krasnoselskii and Sobolevski [72], Balakrishnan [13], Yosida

3.5. Comments

75

[129], Kato [61], around 1960, fractional powers have been the subject of extensive research. A first attempt to exhaustively present the theory was undertaken by Komatsu in a series of papers [66], [67], [69], [68], [70] since around 1965. Although the fractional powers were widely used in the theory of evolution equations (an early example is Krein’s book [74], see also [105]), the first monograph to appear on the theory of fractional powers was [89] in 2001. Without doubt, this book is a milestone. However, it has some considerable drawbacks. The authors are quite fond of proving results in such a way that as much as possible remains valid on general sequentially complete locally convex spaces, and on these spaces “non-negativity” of an operator is not the same as “sectoriality” (as it is the case on Banach spaces, see Proposition 2.1.1, a)). To this aim they base their construction on the Balakrishnan-representation, which is a real integral. As a consequence, they are forced to give a definition of fractional powers “by cases” as is sketched in Remark 3.1.10. This leads to fuzzy proofs, compare their proof of the scaling property. Apart from elegance, another drawback is that their construction is not embedded in a large functional calculus (although they make constant use of an extension of the Hirsch functional calculus). The powerful composition rule therefore is not at hand. 3.1 Fractional Powers with Positive Real Part. The results presented in this section are standard and in fact all included in [89, Chapter V]. The proof of Proposition 3.1.2 is from [10, Proposition 3.5]. After proving Proposition 3.1.7 we learned that the same already had been done in [81, Lemma 3.3] The proof of the Balakrishnan-representation and its corollaries is from [89]. The Spectral Mapping Theorem for fractional powers (our part j of Proposition 3.1.1) was originally proved by Balakrishnan in [13] (there is a “reprint” in [89, p. 121]). In [89, Thm. 5.2.1] Martinez and Sanz give their own proof based on the Spectral Mapping Theorem for the Hisch functional calculus. 3.2 Fractional Powers with Arbitrary Real Part. Like in Section 3.1, the results are well-known. We took Proposition 3.2.3 and Proposition 3.2.4 from [4, Section III.4.6]. 3.3 Holomorphic Semigroups. This section is inspired by [86, ]. The generalization to the multivalued case seems to be new as it stands. However, there are similar results in [42, Chapter III]. 3.4 The Logarithm and the Imaginary Powers. The first reference for Lemma 3.4.1 is Nollau’s paper [99]. Our proof is an adaptation of [89, Lemma 10.1.5] and [101, Lemma 5.1] (for the norm estimate). Proposition 3.4.2 is stated and proved in the Hilbert space case in [89, Theorem 10.1.6] (without using any Hilbert space properties). Our proof is slightly different. The basic facts on the imaginary powers collected in Proposition 3.4.5 are extracted from [89, Chapter 7]. Part d)

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Chapter 3. Fractional Powers and Related Topics

is a perturbation result, see also Section 5.6.2. Proposition 3.4.6 is new, as far as we know. Corollary 3.4.7 is essentially in [89, Theorem 10.1.3 and Theorem 10.1.4]. An earlier account can be found in [101, Addendum] and [100]. From [123, Proposition 2.2.31] have we learned the argument in the proof of the implication (i) ⇒ (ii).

Chapter 4

Strip Type Operators and the Logarithm In Section 4.1 strip type operators are introduced. They are a natural abstraction of logarithms of injective sectorial operators (by Nollau’s theorem). Since the resolvent of a strip type operator by definition is bounded outside a horizontal strip, a “natural” functional calculus based on Cauchy integrals can be set up. This is done in Section 4.2 by means of the abstract extension procedure introduced in Chapter 1. Moreover, a composition rule is proved. Section 4.3 is devoted to prove Theorem 4.3.1 which states equality between the spectral angle of an injective sectorial opertor A and the spectral height of the strip type operator log A. As a corollary one obtains a theorem ¨ ss and Sohr that in case A ∈ BIP, the group type of (Ais )s∈R is always larger of Pru than the spectral angle of A. In Section 4.4 the problem of “inversion” is discussed, namely the question, which strip type operators are actually logarithms of sectorial operators. Here a theorem of Monniaux is generalized. In Section 4.5 we construct the example of an injective sectorial operator A ∈ BIP on a UMD space which is such that the group type of (Ais )s∈R is larger than π.

4.1 Strip Type Operators We are going to introduce a new class of operators, in order to have an abstract concept at hand when dealing with logarithms of sectorial operators. For ω > 0 we denote by Hω := {z ∈ C | |Im z| < ω} the horizontal strip of height ω which is symmetric with respect to the real axis. In the case ω = 0 we define H0 := R. An operator B on a Banach space X is said to be a strip type operator of height ω (in short: B ∈ Strip(ω)), if 1) σ(B) ⊂ Hω and 2) L(B, ω ′ ) := sup{kR(λ, B)k | |Im λ| ≥ ω ′ } < ∞ for all ω < ω ′ .

It is clear that B ∈ Strip(ω) if and only if −B ∈ Strip(ω) with L(B, ω ′ ) = L(−B, ω ′ ) for each ω ′ > ω. We call ωst (B) := min{ω ≥ 0 | B ∈ Strip(ω)}

78

Chapter 4. Strip Type Operators and the Logarithm

sup kR(λ, B)k < ∞ iω ′

σ(B)

−iω

sup kR(λ, B)k < ∞

Figure 6: Spectrum of a strip type operator.

the spectral height of B. Figure 6 on page 78 illustrates the notion of a strip type operator. Let us say that an operator B is a strong strip type operator of height ω (in short: B ∈ SST(ω)) if for each ϕ > ω there is Lϕ such that kR(λ, B)k ≤

Lϕ |Im λ| − ϕ

for all |Im λ| > ϕ. Such operators obviously have the property that for each ϕ > ω both operators ϕ − iB and ϕ + iB are sectorial of angle π2 . Examples 4.1.1. We describe three classes of strip type operators which arise in a natural manner. 1) Let 0 ≤ ω < π and A ∈ Sect(ω) be injective. Then B := log A is a strong strip type operator of height ω, as we learn from Nollau’s theorem (Proposition 3.4.2). We will prove below (see §4.3) that in fact ωA = ωst (log A). 2) Let iB generates a C0 -group T . Then B is a strong strip type operator of height θ(T ), where θ(T ) is the group type of T . In general, it can occur that ωst (B) < θ(T ). However, Gearhart’s Theorem C.8.2 implies that ωst (B) = θ(T ) if X is a Hilbert space. 3) Let H be a Hilbert space and B a selfadjoint operator on H. Then σ(B) ⊂ R and 1 kR(λ, B)k ≤ (λ ∈ / R) |Im λ|

4.2. The Natural Functional Calculus

79

(see Proposition C.4.2). In particular, B is a strong strip type operator of height 0. Of course, this is a special case of 2), since iB generates a unitary group on H, by Stone’s Theorem C.7.4. Remark 4.1.2. As a matter of fact, instead of dealing with horizontal strips we could have defined all notions for vertical strips. (The stripe type operators in the horizontal and vertical case correspond to each other via the mapping (B 7→ iB).) This in fact seems more natural for generators of groups and therefore was done in [49] and [52]. Since the logarithm of a sectorial operator is our guiding example, we will stick to horizontal strips in this exposition.

4.2 The Natural Functional Calculus We want to define a functional calculus for a strip type operator B ∈ Strip(ω) analogously to the sectorial case. Given ϕ > 0 we define −α

F (Hϕ ) := {f ∈ O(Hϕ ) | f (z) = O(|Re z|

) (|z| → ∞) for some α > 1}

and define γϕ := ∂Hϕ (oriented in the positive sense). Then for B ∈ Strip(ω), ω < ω ′ < ϕ, and f ∈ F(Hϕ ), the Cauchy integral Z 1 f (B) := f (z)R(z, B) dz ∈ L(X) 2πi γω′ exists, since R(·, B) is bounded on the path γω′ , see Figure 7 on page 80. By Cauchy’s theorem, the definition of f (B) is independent of the actual choice of ω ′ . The following result is not surprising. Proposition 4.2.1. a) The mapping (f 7−→ f (B)) : F (Hϕ ) −→ L(X) is a homomorphism of algebras. 1 b) For λ, µ ∈ / Hϕ the identity (λ−z)(µ−z) (B) = R(λ, B)R(µ, B) holds.

c) If f ∈ F(Hϕ ) and λ ∈ / Hϕ , then (f (z)(λ − z)−1 )(B) = R(λ, B)f (B) = f (B)R(λ, B).

d) If C is a closed operator commuting with the resolvents of B, then C also commutes with f (B), where f ∈ F(Hϕ ). In particular, f (B) commutes with B. Proof. a), c) and d) are proved analogously to the corresponding statements in Lemma 2.3.1. The only difficulty is in b). We give a sketch of the proof, leaving details to the reader. Let f (z) := (λ − z)−1 (µ − z)−1 with λ, µ ∈ / Hϕ . Let us assume that λ, µ lie on the same side of Hϕ , e.g., above. f (B) is given by a Cauchy integral on the contour (R − iδ1 ) ⊖ (R + iδ2 ) where δ1 , δ2 ∈ (ϕ, ∞) are close to ϕ. Since the integrand has no singularity within the lower halfplane, we shift the lower path (as δ1 → ∞) towards Im z = −∞ without changing its value. This value must therefore be equal to 0. When increasing δ2 , nothing happens in the beginning. When crossing λ we obtain a residue, which is (µ − λ)−1 R(λ, B), and when crossing µ we obtain in

80

Chapter 4. Strip Type Operators and the Logarithm

iϕ γω ′

σ(B)

−iω

γω ′

Figure 7: Cauchy integral for strip type operators. addition (λ − µ)−1 R(µ, B). Then the path wanders off to Im z = +∞ without changing the value any more. To sum up, we obtain f (B) =

R(λ, B) R(µ, B) + = R(λ, B)R(µ, B) µ−λ λ−µ

by the resolvent identity. (When λ = µ a similar reasoning applies, also in the case that λ, µ are on different sides of the strip.)

S S We define F [Hω ] := ϕ>ω F (Hϕ ) and M[Hω ] := ϕ>ω M(Hϕ ), and call the mapping ΦB := (f 7→ f (B)) : F [Hω ] −→ L(X)

the primary functional calculus (in short: PFC) for the strip type operator B. Obviously, F [Hω ] is a subalgebra of M[Hω ] ⊂ M(Hω ). Hence, an abstract functional calculus (F [Hω ], M(Hω ), ΦB )

in the sense of Chapter 1 is established which is proper, by part b) of Proposition 4.2.1. By the abstract results of Section 1.2, the primary functional calculus has a natural extension to a map M[Hω ]B −→ {closed operators}, where M[Hω ]B := {f ∈ M[Hω ] | ∃e ∈ F(Hω ) : e(B) is injective and ef ∈ F[Hω ]}

is the set of regularizable elements of M[Hω ]. The operator f (B) is then defined as f (B) = e(B)−1 (ef )(B)

4.2. The Natural Functional Calculus

81

whenenver f ∈ M(Hω )B , and e ∈ F[Hω ] is such that e(B) is injective and ef ∈ F[Hω ]. The so obtained abtract functional calculus (F [Hω ], M(Hω ), ΦB ) is even a meromorphic functional calculus for B (see Section 1.3), as the following Lemma shows. Lemma 4.2.2. Let B ∈ Strip(ω). Then (z)(B) = B.

Proof. Let λ ∈ / Hω . Then (λ − z)−3 (B) = R(λ, B)3 , by Proposition 4.2.1, b) and c). Hence (λ − z)−3 regularizes z. Now, z(λ − z)−3 = λ(λ − z)−3 − (λ − z)−2 , whence z(λ − z)−3 (B) = λR(λ, B)3 − R(λ, B)2 = BR(λ, B)3 . This yields finally z(B) = (λ − B)3 (z(λ − z)−3 )(B) = (λ − B)3 BR(λ, B)3 = B(λ − B)3 R(λ, B)3 = B.

As a consequence, we have at hand all the results on meromorphic functional calculi from Section 1.3, in particular Theorem 1.3.2. It is now natural to ask which meromorphic functions can be guaranteed to be members of M[Hω ]B .

Lemma 4.2.3. Let B ∈ Strip(ω). a) If f ∈ O[Hω ] is polynomially bounded at ∞, then f ∈ M[Hω ]B . In particular, f (B) is defined for every f ∈ H ∞ [Sω ].

b) Assume ω < π and let λ ∈ C \ {0} with ω < |arg λ| ≤ π. Then (λ − ez )−1 ∈ H ∞ [Hω ] and the operator 1 (B) λ − ez is injective.

c) Assume ω < π and let λ, µ ∈ C \ {0} with ω < |arg λ| , |arg µ| ≤ π. Then the operator ez T (λ, µ) := (B) z (λ − e )(µ − ez ) is injective. Note that the function f (z) =

ez (λ−ez )(µ−ez )

is indeed a member of F [Hω ].

Proof. Since ϕ(z) := (µ2 + z 2 )−1 ∈ F [Hω ] and (µ2 + z 2 )−1 (B) = −R(iµ, B)R(−iµ, B) for each µ > ω, on can use powers of ϕ as regularizers for polynomially bounded holomorphic functions. Hence the assertion of a) is clear. To prove b), let f ∈ F [Hω ] be such that f (B) is injective. Then, for every H ∞ -function g, f is a regularizer for g and ker g(B) = ker(f g)(B). Let ω < |arg λ| , |arg µ| ≤ π. Since λf (z) µf (z) (µ − λ) µf (z) = + λ − ez µ − ez µ(λ − ez ) µ − ez it follows that ker(µ − ez )−1 (B) ⊂ ker(λ − ez )−1 (B), hence equality by symmetry. Now, let λf (z) λ → −∞ to obtain λ−ez (B) → f (B) in norm (use the definition by Cauchy integrals). Hence ker(λ − ez )−1 (B) = ker f (B) = 0. 1 In a totally analogous way one can prove that the operator ( µ − e−z )−1 (B) is injective. The assertion in c) now follows from −f (B)2 T (λ, µ) =

λf (z) µez f (z) λf (z) (B) (B) = (B) z λ−e µ − ez λ − ez

1 f (z) µ 1 − e−z µ

(B).

82

Chapter 4. Strip Type Operators and the Logarithm

The last result shows that for an arbitrary operator B ∈ Strip(ω) and f ∈ O[Hω ] such that f = O(eα|Re z| ) as |Re z| → ∞ for some α < π/ω, the operator f (B) is defined. In particular, if ω < π then eB := (ez )(B) is defined. This operator is injective and satisfies C \ Sω ∩ P σ(eB ) = ∅. Proof. Let λ ∈ C \ Sω . Then Theorem 1.3.2. Since the operator

eB

is injective.

1 λ−ez

∈ H ∞ [Hω ] ⊂ M[Hω ]B whence λ − eB is injective by f ) of

(λ − eB )−1 (µ − eB )−1 eB ⊂ T (λ, µ)

By Proposition 3.4.2 we know that log A ∈ Strip(ω) if A ∈ Sect(ω) is injective. So we expect elog A = A in this case. In fact, this is an instance of the following composition rule. Theorem 4.2.4. (Composition Rule) Let ϕ ∈ (ω, π) and let A ∈ Sect(ω), g ∈ M(Sϕ )A . Assume that g(A) ∈ Strip(ω ′ ) and g(Sϕ ) ⊂ Hω′ ∪ {∞}. If f ∈ M[Sω′ ]B then f ◦ g ∈ M[Sω ]A and (f ◦ g)(A) = f (g(A)). Analogous statements hold in case A ∈ Strip(ω), g ∈ M(Hϕ )A , g(A) ∈ Sect(ω ′ ) or g(A) ∈ Strip(ω ′ ) (with f in the appropriate function class and g satisfying the appropriate mapping condition). As in the case of Theorem 2.4.2 the requirement on the range of g ensures that g cannot have poles in Sϕ , whence actually g is holomorphic and g(Sϕ ) ⊂ Hψ for every ω ′ < ψ. Proof. We prove only the first statement of the theorem. This will be done in several steps which are analogous to the proof of Theorem 2.4.2. First, if f ∈ M[Hω ′ ] then clearly f ◦g ∈ M(Sϕ ). Moreover, if f ∈ F [Hω ′ ] then f ◦g ∈ H ∞ (Sϕ )A . (Compare Lemma 2.4.4.) The next step is to reduce the assertion to functions f ∈ F [Hω ′ ]. This is done by an application of Proposition 1.3.6. Finally, one has to prove the assertion for f ∈ F [Hω ′ ]. Since we are mainly interested in the case g(z) = log(z) we assume for simplicity that A is injective. We choose a strip H between Hω ′ and the domain of f , as well as a sector S between Sω and Sϕ . Then we can compute (quite canonically) Z 1 f (g(A)) = f (λ)R(λ, g(A)) dλ 2πi ∂H Z Z 1 1 τ (z) = ΛA f (λ) R(z, A) dz dλ 2πi ∂H 2πi ∂S λ − g(z) – Z » Z 1 1 f (λ) = ΛA dλ τ (z)R(z, A) dz 2πi ∂S 2πi ∂H λ − g(z) Z 1 = ΛA τ (z)(f ◦ g)(z) R(z, A) dz = ΛA (τ (f ◦ g)) (A) = (f ◦ g)(A), 2πi ∂S where we have used the abbreviation τ (z) := z(1 + z)−2 and ΛA := τ (A)−1 . The application of Fubini’s Theorem is justified since the function f (λ) is product integrable on ∂H × ∂S.

1 τ (z) λ − g(z) z

4.3. The Spectral Height of the Logarithm

83

Here is an immediate application. Corollary 4.2.5. Let ω ∈ [0, π) and A ∈ Sect(ω) be injective. Then elog A = A

and

f (log A) = (f (log z))(A)

for all f ∈ H ∞ (Sϕ ) and all ϕ ∈ (ω, π). In particular we have

1 λ − ez

ez (λ − ez )(µ − ez )

(log A) = R(λ, A)

and

(log A) = AR(λ, A)R(µ, A)

for all λ, µ ∈ / Sω . Corollary 4.2.6. An injective sectorial operator on a Banach space X is uniquely determined by its logarithm, i.e., if A and B are injective sectorial operators on X with log A = log B, then A = B. Remark 4.2.7. 1) Suppose B ∈ Strip(ω) is bounded. Then one can set up the ususal Dunford calculus for B. It is easy to see, that if f ∈ O[Hω ]B then f (B) (defined by the NFC for strip type operators) is bounded and coincides with f (B) defined by the Dunford calculus. Hence the Dunford calculus is an extension of the NFC for strip type operators in this case. 2) By methods analogous to those developed in Section 2.7 one can prove a spectral inclusion theorem and a spectral mapping theorem for the NFC for strip type operators. Theorem 4.2.8. (Spectral Mapping Theorem) Let A ∈ Sect(ω) be injective. Then the identity σ ˜ (log A) = log (˜ σ (A))

(4.1)

holds. Proof. The Spectral Inclusion Theorem 2.7.4 yields the inclusion log (σ(A) \ {0}) ⊂ σ(log A). Now, by Remark 4.2.7 one has also a spectral inclusion theorem for strip type operators. Hence we can infer eσ(log A) ⊂ σ(elog A ) \ {0} = σ(A) \ {0}, whence σ(log A) ⊂ log(σ(A)) \ {0}. On the other hand it is obvious that log(A) is bounded if and only if A is bounded and invertible. This concludes the proof.

4.3 The Spectral Height of the Logarithm Let A be an injective sectorial operator on the Banach space X. Recall that Nollau’s theorem (Proposition 3.4.2) says that log A is a strip type operator with ωst (log A) ≤ ωA . The purpose of this section is to prove the following.

84

Chapter 4. Strip Type Operators and the Logarithm

Theorem 4.3.1. Let X be a Banach space and A ∈ Sect(ω) be injective. If log A ∈ Strip(ω ′ ) for some ω ′ ≥ 0, then A ∈ Sect(ω ′ ). In particular, we have ωst (log A) = ωA . We let B := log A. Without restriction we can assume ω ′ < ω. Let us consider the operator family ez (B) ∈ L(X) T (λ, µ) := (λ − ez )(µ − ez ) defined for λ, µ ∈ / Sω′ . (Note that (ez (λ − ez )−1 (µ − ez )−1 ) ∈ F(Hϕ ) for ω ′ < ϕ < |arg λ| , |arg µ|.) From Corollary 4.2.5 we learn that T (λ, µ) = AR(λ, A)R(µ, A)

(4.2)

for ω < |arg λ| , |arg µ|. Lemma 4.3.2. Fix µ with |arg µ| > ω ′ . Then the mapping (λ 7−→ T (λ, µ)) : C \ Sω′ −→ L(X) is holomorphic. Proof. It suffices to show that the function ez (λ −

ez )(µ

− ez )

R(z, B)

is integrable on horizontal lines uniformly in λ ∈ K, for each compact K ⊂ C \ Sω ′ . Choose ω ′ < ϕ < |arg µ|. Define Cν := dist(ν, Sϕ ) for |arg ν| > ϕ. Let |arg λ| > ϕ and define fλ (z) := ez (λ − ez )−1 (µ − ez )−1 . Then |fλ (z)| ≤

eRe z Cµ Cλ

and

|fλ (z)| ≤

e− Re z |λ| |µ| C 1 C 1 λ

µ

for z ∈ Hϕ . Hence there is C(λ), locally bounded in λ, such that |fλ (z)| ≤ C(λ)e−|Re z| . This proves the claim.

We need another Lemma. Lemma 4.3.3. Assume ϕ ∈ (ω ′ , π). Then the set {k(µ − λ)T (λ, µ)k | |arg λ| ≥ ϕ, µ = − |λ|} is bounded. Proof. Choose ω1 ∈ (ω ′ , ϕ). We write (µ − λ)T (λ, µ) as a Cauchy integral along the path γ := ∂Hω1 . We obtain ‚ ‚ Z Z ‚ 1 ‚ (µ − λ)ez |(µ − λ)ez | |dz| ‚ ≤ L(B, ω1 ) k(µ − λ)T (λ, µ)k = ‚ R(z, B) dz ‚ ‚ 2πi z z z z 2π γ (λ − e )(µ − e ) γ |λ − e )| |(µ − e )|

4.3. The Spectral Height of the Logarithm

85

We estimate the integral over the path R + iω1 by ˛ ˛ Z ∞ Z ∞ |µ − λ| ˛er+iω1 ˛ |µ − λ| t=er dr = dt r+iω r+iω 1 | |µ − e 1| |λ − teiω1 | |µ − teiω1 | −∞ |λ − e 0 ˛ ˛ ˛ λ ˛ Z ∞ Z ∞ ˛ ˛1 + |λ| |µ − λ| |λ| µ=−|λ| ˛ ˛ = dt dt = iω1 | |µ − |λ| teiω1 | ˛ ˛ λ |λ − |λ| te iω 0 0 ˛ |λ| − te 1 ˛ |1 + teiω1 | Z ∞ 2 ˛ ˛ ≤ dt. ˛ λ ˛ 0 ˛ |λ| − teiω1 ˛ |1 + teiω1 |

The last term is uniformly bounded, since ϕ ≤ |arg λ| ≤ π. Needless to say that the second integral can be treated analogously.

We are now able to complete the proof of Theorem 4.3.1. By elementary calculations we obtain from (4.2) the identity λR(λ, A) = (µ − λ)T (λ, µ) + µR(µ, A)

(4.3)

which holds for all λ, µ with |arg λ| , |arg µ| > ω. Keeping µ fixed we see that the right hand side of this equation is defined even for |arg λ| > ω ′ , and is in fact holomorphic as a function of λ (Lemma 4.3.2). Since the norm of the resolvent blows up if one approaches a spectral value, we see that no λ with |arg λ| > ω ′ can belong to σ(A). Furthermore, if we choose ω ′ < ϕ < π and let µ = µλ := − |λ| in (4.3), we arrive at

kλR(λ, A)k ≤ k(µλ − λ)T (λ, µλ )k + |λ| (|λ| + A)−1 .

This is bounded uniformly for |arg λ| ≥ ϕ, by Lemma 4.3.3 and the sectoriality of A.

We state two important corollaries. Recall that, if iB generates a group T on a Banach space X, one always has θ(T ) ≥ ωst (B) by the Hille -Yosida Theorem A.8.6. If X = H is a Hilbert space, even θ(T ) = ωst (B) holds, by Gearhart’s Theorem C.8.2. Now, if A is sectorial and A ∈ BIP, we know that i log A generates the group (Ais )s∈R . Employing the identity ωA = ωst (log A) we obtain the following two results. Corollary 4.3.4. (Pr¨ uss-Sohr I) Let A be an injective and sectorial operator on a Banach space X, having dense domain and dense range. Assume that A ∈ BIP(X). Then ωA ≤ θA , i.e., the group type of the group (Ais )s∈R is always larger than the spectral angle of A. Corollary 4.3.5. (McIntosh) Let H be a Hilbert space and A an injective sectorial operator on H which has bounded imaginary powers. Then θA = ωA , i.e., the group type of (Ais )s∈R equals the spectral angle of A. We will see in Section 7.3.3 that even more is true in the Hilbert space situation.

86

Chapter 4. Strip Type Operators and the Logarithm

4.4 Monniaux’s Theorem and the Inversion Problem The logarithm of an injective sectorial operator A is a strip type operator, and in fact elog A = A holds, by the composition rule (Corollary 4.2.5). It is therefore natural to ask, whether eB is sectorial whenever B ∈ Strip(ω) with ω < π. Let us give name to that question and call it the inversion problem for strip type operators. Unfortunately, the answer is “not always”. d d Example 4.4.1. Let X := L1 (R) and let B := i dt , where dt denotes the usual d 1,1 derivative with domain D( dt ) = W (R). Then iB generates a strongly continuous isometric group. In particular, B ∈ SST(0). Consider the function f (z) = 1 1+ez . Since f is bounded and holomorphic on each Hϕ (0 < ϕ < π), one can compute the operator f (B). If A := eB is sectorial then clearly f (B) = (1 + A)−1 is a bounded operator. It is easily seen that this amounts to the fact, that the function f is a Fourier multiplier on L1 (R). But this is false.

Proof. A bounded Fourier multiplier on L1 is the Fourier tranform µ ˆ of a bounded measure µ ∈ M(R). With the help of Fubinis theorem and Lebesgue’s Dominated Convergence Theorem R one easily proves the fact that the Cesaro limits limT →±∞ T1 0T µ ˆ(x) dx both exist and are equal to µ{0}. In particular, if µ ˆ has limits in −∞ and ∞ these limits must be equal. This is not the case for f (x) = (1 + ex )−1 .

Similarly or with the help of the next Lemma 4.4.2 it follows that C \ [0, ∞) ⊂ σ(eB ). Hence, σ(eB ) = C. Therefore, one can ask for additional sufficient conditions to ensure that A := eB is sectorial. A very small step is in the following lemma. Lemma 4.4.2. Let B ∈ Strip(ω) with ω < π and A := eB . If λ ∈ ̺(A) for some λ with |arg λ| > ω then this is true for all such λ. Proof. This is due to the formula

λ λ−ez

z

= (µ − λ) (λ−eze)(µ−ez ) +

µ µ−ez

.

Unfortunately this lemma does not give estimates for t(t + A)−1 as λ → 0 and λ → ∞. In fact, it is an open problem if existence of the resolvent already implies sectoriality of eB . However, there is a remarkable result by Monniaux from [98]. Theorem 4.4.3. (Monniaux) Let X be a Banach space with the UMD property and let iB generate a strongly continuous group on X. If ωst (B) < π, then eB is sectorial. (See [6, Section 3.12] for the definition of UMD spaces.) Proof. In [98, Theorem 4.3] the theorem is proved under the additional assumption that θ(U ) < π, where U denotes the group generated by iB and θ(U ) is its group type, cf. page 210. Using this result, we deal with the general case. If θ := θ(U ) ≥ π then one can choose 0 < α < 1 such that θα < π. Then iαB generates the group (U (αs))s∈R which has group type αθ. Furthermore, αB ∈ Strip(αω). Applying [98, Theorem 4.3] we can find an injective sectorial operator C on X such that C is = U (αs) for all s ∈ R. Theorem 4.3.1 yields that ωC ≤ αω. If we define A := C 1/α we know from Proposition

4.5. A Counterexample

87

3.1.2 and Proposition 3.2.1 that A is also an injective sectorial operator and Ais = U (s) for all s ∈ R. Hence i log A generates U , whence log A = B.

It would lead us too far astray to give a complete proof of Monniaux’s original result from [98, Theorem 4.3]. However, in the next section we will apply Theorem 4.4.3 to obtain an interesting counterexample.

4.5 A Counterexample In this section we want to point out that Theorem 4.3.1 sheds light on a longstanding problem in the field. Question: Is there a Banach space X and a sectorial operator A ∈ BIP(X) such that θA ≥ π? On UMD spaces, the question stated above is intimately connected with the failing of Gearhart’s theorem (Theorem C.8.2) for C0 -groups. We state this as a proposition. Proposition 4.5.1. Let X be a Banach space with the UMD property. Then the following assertions are equivalent. (i) There is a sectorial operator A ∈ BIP(X) such that θA ≥ π.

(ii) There is a sectorial operator A ∈ BIP(X) such that ωA < θA .

(iii) There is an operator B on X which generates a C0 -semigroup T such that s0 (B) < ω0 (T ) and T is a group. Recall the definition (C.7) of s0 (B) on page 239. Proof. The implication (i) ⇒ (ii) is trivial. To prove (ii) ⇒ (i) simply consider a scaled operator Aα for som α > 1 in case that θA < π. We prove (ii) ⇒ (iii). Assume that A ∈ BIP(X) with ωA < θA . Then i log A generates T1 := (Ais )s≥0 and −i log A generates T2 := (A−is )s≥0 . But max{s0 (i log A), s0 (−i log A)} = ωst (log A) = ωA . < θA = max{ω0 (T1 ), ω0 (T2 )} Hence either B := i log A or B := −i log A will do the job. Finally we prove the implication (iii) ⇒ (ii). Assume that (U (s))s∈R is a C0 -group on X, the operator B generates the semigroup T := (U (s))s≥0 , and s0 (B) < ω0 (T ). Changing B to α(B + λ) with suitable λ, α > 0 we can assume that sups≤0 kU (s)k < ∞ and 0 ≤ s0 (B) < ω0 (T ) < π. By Monniaux’s theorem 4.4.3 there is a sectorial operator A ∈ BIP(X) such that Ais = U (s) for all s ∈ R. Hence i log A = B by the uniqueness of generators. Obviously, ωst (log A) = s0 (B) < ω0 (T ) = θ(U ) = θA . Since ωA = ωst (log A) by Theorem 4.3.1, assertion (ii) readily follows.

We are now going to give an example of a Banach space X with the UMD property such that (iii) of Proposition 4.5.1 holds. Let 1 < p < 2 < q < ∞ and a > q such that p2 < aq . We define the weight w : R → [0, ∞) by ax e x≤0 w(x) := 1 x≥0

88

Chapter 4. Strip Type Operators and the Logarithm

Now we let X := Lp (R, e2x dx) ∩ Lq (R, w(x)dx) with the norm kf kX := kf kp + kf kq , where kf kp =

Z

∞ −∞

p 2x

|f (x)| e

p1 dx

kf kq =

and

Z

∞

−∞

q1 |f (x)| w(x) dx . q

It can be shown that X has in fact the UMD property. The space Cc (R) of compactly supported continuous functions is dense in X. Proof. Let f ∈ X. Then 1[−n,n] f → f in X for n → ∞. Hence we can assume that f is compactly supported. Now we approximate f in Lq (R) by functions fn ∈ Cc (R) such that supp(fn ) ⊂ [a, b] where a < b do not depend on n. Since Lq ((a, b), w(x)dx) ∼ = Lq (a, b) ֒→ Lp (a, b) ∼ = Lp ((a, b), e2x dx)

the sequence fn tends to f even in the norm of X.

On X we consider the left shift group (T (t))t∈R defined by (x ∈ R, t ∈ R)

[T (t)f ](x) := f (x + t) We have the norm inequalities 2

kT (t)f kX ≤ e− p t kf kp + kf kq ≤ kf kX , 2

a

kT (−t)f kX ≤ e p t kf kp + e q t kf kq for f ∈ X and t ≥ 0.

−2t

Proof. Obviously, kT (t)f kp = e p kf kp for t ∈ R. If t ≥ 0, we have Z ∞ kT (t)f kqq = |f (x)|q w(x − t) dt −∞

=e

−at

Z

−∞

≤ e−at

Z

−∞

|f (x)|q eax dx + e−at |f (x)|q eax dx +

The computation for kT (−t)f kqq is similar.

Z

∞

0

Z

t

0

|f (x)|q eax dx +

Z

∞ t

|f (x)|q dx

|f (x)|q dx ≤ kf kqq .

In particular, it follows that kT (t)k ≤ 1 for all t ≥ 0 and that T is a group. Since Cc (R) is dense in X and (t 7−→ T (t)f ) : R −→ X is continuous for each f ∈ Cc (R), we conclude that T is in fact a C0 -group. Claim. We have kT (t)k = 1 for all t ≥ 1. Proof. Let t0 ≥ 0. Choose t1 > t0 arbitrary. Since Lp ((t0 , t1 ), e2x dx) ∼ = Lp (t0 , t1 ) 6֒→ Lq (t0 , t1 ), there is no inequality of the form kf kq ≤ C kf kp with f ∈ Cc (t0 , t1 ). Hence there is a sequence gn ∈ Cc (t0 , t1 ) such that kgn kq ≥ n kgn kp for all n. Letting fn := gn / kgn kq we have kfn kq = 1, Thus, kT (t0 )fn kX =

−2t e p 0

kfn k ≤

1 , n

and

fn ≡ 0 on (−∞, t0 ]

kfn kp + kfn kq ≥ 1 and kfn kX ≤

all n ∈ N. This proves the claim.

1 n

+ 1. Hence, kT (t0 )k ≥

1 1 1+ n

for

4.6. Comments

89

We now determine the generator A of T . We claim that D(A) = {f ∈ X | f ′ ∈ X} and Af = f ′ , where f ′ denotes the (distributional) derivative of f . Proof. Let us denote the derivative operator on distributions by ∂. If f ∈ D(A) there is a g ∈ X such that 1t (T (t)f − f ) → g in X for t ց 0. Since X ֒→ D(R)′ we see that g = f ′ . Hence, D(A) ⊂ D := {f ∈ X | f ′ ∈ X} and Af = f ′ for f ∈ D(A). Now, I − A is bijective (since T is a contraction semigroup) and I − ∂ : D −→ X is injective (since f ′ = f ∈ X implies that f is more or less the exponential function). This implies the claim.

Claim. We have s(A) := sup{Re λ | λ ∈ σ(A)} ≤ − p2 . Proof. Let Xp := Lp (R, e2x dx) with the norm kf kp defined above, and denote by Ap the genera−2t

tor of the left shift group on Xp . Since kT (t)f kp = e p kf kp for all t ∈ R and all f ∈ Xp , we have s0 (Ap ) ≤ − p2 . By the same reasoning as above one can show that D(Ap ) = {f ∈ Xp | f ′ ∈ Xp } with Ap f = f ′ for f ∈ Ap . Next, we claim that D(Ap ) ֒→ X. Actually, by the Closed Graph The2·

orem, only inclusion has to be shown. If f ∈ D(Ap ) we let g := e p f and note that g ∈ Lp (R, dx) and g ′ =

2 g p

2·

+ e p f ′ ∈ Lp (R, dx). Hence g ∈ W1,p (R) and since W1,p (R) ֒→ C0 (R) there

is a constant C > 0 such that |f (x)| ≤ Ce

2x −p

ax eq

for x“ ∈ R.” This immediately implies that a−2

x

f ∈ Moreover, we have |f (x)| ≤ Ce q p for all x ∈ R, and since aq > p2 q ax we conclude that f ∈ L ((−∞, 0), e dx). Altogether we obtain f ∈ X. Now take λ ∈ C with − p2 < Re λ. Since (λ − A)f = (λ − Ap )f for f ∈ D(A) ⊂ D(Ap ) we see that λ − A is injective. If f ∈ X then f ∈ Xp and g := R(λ, Ap )f ∈ D(Ap ) ⊂ X. But this implies that g ∈ D(A), since g ′ = λg − f ∈ X. Hence λ − A is also surjective. This proves that λ ∈ ̺(A). Lq ((0, ∞), dx).

We take the last step. Obviously, the space X is not only a Banach space but even a Banach lattice and the semigroup T is positive, i.e., T (t)f ≥ 0 for all 0 ≤ f ∈ X and all t ≥ 0. By [6, Theorem 5.3.1] we conclude that s0 (A) = s(A) ≤ − p2 < 0 = ω0 (T ), whence we are done. Corollary 4.5.2. There is a Banach space X with the UMD property and an injective sectorial operator A on X such that A ∈ BIP(X) and θA > π.

4.6 Comments 4.1/4.2 Strip Type Operators and their Natural Functional Calculus. The natural functional calculus for strip type operators appears first in [12]. It is used in [15] and companion papers. However, with all the work done in Chapters 1 and 2 we could give a streamlined account fitting into a general pattern. In particular, the composition rule is at hand. Suppose one is given an operator B with spectrum in a horizontal strip Hω and such that the resolvent is bounded on some horizontal lines R±iϕ with ϕ > ω. Then one can construct the “functional calculus” like we did in Section 4.2, since the basic Cauchy integrals converge. However, it can happen that this “functional

90

Chapter 4. Strip Type Operators and the Logarithm

calculus” is the zero mapping. If one requires that the calculus behaves well for rational functions, i.e., is really a functional calculus for B, then B has to be strip type. This was shown in [49]. 4.3 The Spectral Height of the Logarithm. Theorem 4.3.1 is contained in [49]. ¨ss The Pr¨ uss-Sohr I result (Corollary 4.3.4) is part of a celebrated theorem of Pru and Sohr given in [110, Theorem 3.3]. (See Corollary 5.6.12 for the second part.) Its original proof rests on the Mellin transform calculus for C0 -groups, cf. also [123, Proposition 3.19], [124], and [89, Chapter 9]. The content of Corollary 4.3.5 is originally due to McIntosh (in [95]). 4.4 Monniaux’s Theorem and the Inversion Problem. For the proof of Theorem 4.4.3 in [98], Monniaux utilizes the theory of analytic generators of C0 -groups. In the same paper, she gives our Example 4.4.1 to show that the conclusion fails when one discards the UMD assumption. The term “UMD” is an abbreviation for “Unconditional Martingale Differences” and was introduced by Burkholder. It was shown by Burkholder, McConnel and Bourgain that the UMD property is equivalent to the boundedness of the Hilbert transform, see [89, Section 8.4] and the references therein. Up to now, Monniaux’s theorem seems to be the only positive result on the inversion problem. 4.5 A Counterexample. The example of a group with differing growth bound and abszissa of uniform boundedness of the resolvent is an adaptation of an example given by Wolff in [127]. We are indebted to Charles Batty for showing this result to us. Kalton recently has given an example of a sectorial operator with bounded H ∞ -calculus whose sectoriality angle differs from the H ∞ -angle, see [60].

Chapter 5

The Boundedness of the H ∞-Calculus In this chapter we deal with a circle of questions described in short by “boundedness of the H ∞ -calculus”. This notion was originally developed by McIntosh and his coworkers and has a strong relation to interpolation theory and harmonic analysis. We start with two technical tools, the so-called Convergence Lemma (Section 5.1) and some fundamental boundedness and approximation results (Section 5.2). Afterwards, in Section 5.3, we prove equivalence of boundedness of F -functional calculi for different subalgebras F of H ∞ . Then we introduce and study the H ∞ -angle (Section 5.4). In Section 5.5, we describe the construction of an operator on a Hilbert space with no bounded H ∞ -calculus. Permanence results with respect to additive perturbations are presented in Section 5.6. Last, we provide a technical Lemma which indicates the connections to Harmonic Analysis (Section 5.7).

5.1 Convergence Lemma In this section we will prove a basic approximation result, called the Convergence Lemma. Whereas in Section 2.6.3 we only dealt with sectorial approximation, we now consider the approximation of functions. More precisely, our hypothesis will be pointwise convergence of holomorphic functions, thereby bringing us in need of the following fact. Proposition 5.1.1. (Vitali) Assume Ω ⊂ C is open and connected, and (fα )α is a locally bounded net of holomorphic functions on Ω. If the set {z ∈ Ω | fα (z) converges} has a limit point in Ω, then fα converges to a holomorphic function uniformly on compact subsets of Ω. Proof. For sequences, an elegant proof can be found in [6, Theorem A.5]. However, one can easily modify the proof given there to obtain the result for nets.

5.1.1 Convergence Lemma for Sectorial Operators. The next lemma, although very elementary, is most fundamental. Lemma 5.1.2. Let A ∈ Sect(ω) and ϕ ∈ (ω, π). Let (fα )α ⊂ H0∞ (Sϕ ) be a net of functions converging pointwise to a function f . Assume that there are C, s > 0

Chapter 5. The Boundedness of the H ∞ -calculus

92 such that

n o s −s |fα (z)| ≤ C min |z| , |z|

(z ∈ Sϕ )

(5.1)

independent of α. Then f ∈ H0∞ (Sϕ ) and kfα (A) − f (A)k → 0. Proof. Because of (5.1) the family (fα )α is locally bounded. Vitali’s theorem implies that f is holomorphic, hence f ∈ H0∞ (Sϕ ). Moreover, fα → f uniformly on compact sets. Now, the claim follows from a version of the Dominated Convergence Theorem.

Now we are heading for the main result. Note that one has a) D(A) ⊂ D(f (A)) for all f ∈ H ∞ (Sϕ ) ∩ A, and

b) D(A) ∩ R(A) ⊂ D(f (A)) for all f ∈ H ∞ (Sϕ ), if A is injective.

Proof. If f ∈ H ∞ ∩ A we have f (z)(1 + z)−1 ∈ f (0)(1 + z)−1 + H0∞ . This gives f (A)(1 + A)−1 ∈ L(X), whence D(A) ⊂ D(f (A)) follows. In the second case consider f (z)z(1 + z)−2 .

Proposition 5.1.3. (Convergence Lemma) Let A ∈ Sect(ω), ϕ ∈ (ω, π), and (fα )α ⊂ H ∞ (Sϕ ). Assume supα kfα k∞ < ∞ and that the limit f (z) := limα fα (z) exists pointwise on Sϕ . Suppose that fα (A) and f (A) are defined by the natural functional calculus for sectorial operators. Then fα (A)x → f (A)x for all x ∈ D(A) ∩ R(A). Moreover the following assertions hold: a) If A is injective, fα (A) ∈ L(X), and fα (A) → T ∈ L(X) strongly, then f (A) = T . b) If A is densely defined with dense range and supn kfα (A)k < ∞, then f (A) ∈ L(X) and fα (A) → f (A) strongly.

Proof. Vitali’s theorem implies that f ∈ H ∞ (Sϕ ). Define g(z) := f (z)z/(1 + z)2 and gα (z) := fα (z)z/(1 + z)2 . The net (gα )α ⊂ H0∞ (Sϕ ) obviously satisfies the hypotheses of Lemma 5.1.2. Hence fα (A)A(1 + A)−2 = gα (A) → g(A) = f (A)A(1 + A)−2 in norm. So fα (A)x → f (A)x for all x ∈ D(A) ∩ R(A). Assume that A is injective, fα (A) ∈ L(X), and fα (A) → T ∈ L(X) strongly. Let x ∈ X. Then gα (A)x =: yα ∈ D(ΛA ) for each α and yα → y := g(A)x. Now, ΛA yα = fα (A)x → T x, by assumption. Since the operator ΛA is closed, we have y ∈ D(ΛA ) and ΛA y = T x. But this means x ∈ D(f (A)) with f (A)x = T x. The proof of assertion b) is now trivial.

Remark 5.1.4. One can wonder whether a) remains true if we drop the hypothesis that A is injective. (In this case we must of course assume fα , f ∈ A.) Then we can still conclude that f (A) ∈ L(X). However, T = f (A) is not true in general, but only T = f (A) on R(A) and R(T − f (A)) ⊂ N(A). Proof. Take x ∈ R(A). Then (1 + A)−1 x ∈ D(A) ∩ R(A). Hence, yα := (fα (z)(1 + z)−1 )(A)x = fα (A)(1 + A)−1 x → f (A)(1 + A)−1 x = (f (z)(1 + z)−1 )(A)x =: y, by Proposition 5.1.3. Since fα (A) ∈ L(X), yα ∈ D(A) and (1 + A)yα = fα (A)x → T x. Because (1 + A) is closed, we have y ∈ D(A) and (1 + A)y = T x. This means that x ∈ D(f (A)) and f (A)x = T x. Since always D(A) ⊂ D(f (A)) for f ∈ A ∩H ∞ and X = D(A) + R(A) we obtain X ⊂ D(f (A)), whence f (A) ∈ L(X). From above we see that T = f (A) on R(A). Let x ∈ D(A). Then f (A)x, T x ∈ D(A) as well and A(f (A)x − T x) = f (A)Ax − T Ax = 0.

5.2. A Fundamental Approximation Technique

93

Example 5.1.5. The above conclusion is in fact the best one can expect. Let, e.g., A be a sectorial operator on a reflexive space X such that N(A) 6= 0 and let P : X −→ R(A) the projection along N(A) (see h) of Proposition 2.1.1). With f := 1 and fn (z) := z( n1 + z)−1 we have fn (A) → P strongly, but f (A) = I 6= P .

5.1.2 Convergence Lemma for Strip Type Operators. It should be clear that in case of a strip type operator B ∈ Strip(ω) a result similar to Proposition 5.1.3 holds. Proposition 5.1.6. (Convergence Lemma on the Strip) Let X be a Banach space and B ∈ Strip(ω) on X. Let ϕ > ω and (fα )α a net of holomorphic functions on Hϕ converging pointwise to a function f on Hϕ . β a) If β > 1 and supα supz∈Hϕ |fα (z)| (1 + |Re z| ) < ∞, then f ∈ F(Hϕ ) and fα (B) → f (B) in norm. b) If fα ∈ H ∞ (Hϕ ), fα (B) ∈ L(X) for all α, and sup sup |f (z)| < ∞

sup kfα (B)k < ∞,

as well as

α z∈Hϕ

α

then at least fα (B)x → f (B)x for all x ∈ D(B 2 ). If in addition fα (B) → T ∈ L(X) strongly, then T = f (B).

Proof. The proof is anlogous to the proof of Proposition 5.1.3.

Note that if B ∈ Strip(ω) is densely defined, then D(B 2 ) = D(B) = X.

5.2 A Fundamental Approximation Technique In this section we provide auxiliary results which will not be used until the next chapter. (So the reader might skip this section o first reading.) The main part is an approximation method which has been used extensively by McIntosh and his co-workers. We start with an auxiliary result. Lemma 5.2.1. Let [a, b] ⊂ R, F : [a, b] × Sϕ −→ C continuous such that F (t, .) : Sϕ −→ C is holomorphic for each t ∈ [a, b]. Assume there are C, s > 0 such that n o s −s |F (t, z)| ≤ C min |z| , |z| for all t ∈ [a, b], z ∈ Sϕ . We define f (z) := assertions hold. a) f ∈ H0∞ (Sϕ );

Z

b

F (t, z) dt. Then the following

a

b) (t 7−→ F (t, A)) : [a, b] −→ L(X) is continuous.

Chapter 5. The Boundedness of the H ∞ -calculus

94 c) f (A) =

Z

b

F (t, A) dt.

a

Proof. Assertion a) is clear and b) is an immediate consequence of Lemma 5.1.2. Furthermore we have Z b Z bZ (2πi) F (t, A) = F (t, z) R(z, A) dz dt a

a

Fub.

=

Γ

Z „Z Γ

b

a

« Z F (t, z) dt R(z, A) dz = f (z)R(z, A) dz Γ

= (2πi)f (A).

This proves c).

Let us introduce the following notation. For ψ ∈ E(Sϕ ) and t > 0 we write ψt := (z 7−→ ψ(tz))

(z ∈ Sϕ )

(5.2)

It is then clear that again ψt ∈ E(Sϕ ) and ψ(tA) = ψt (A) for all t > 0. Theorem 5.2.2. Let A ∈ Sect(ω), ϕ ∈ (ω, π), and ψ, θ ∈ H0∞ (Sϕ ). Then the following statements hold. a) For each f ∈ H ∞ (Sϕ ) the mapping (t 7−→ (f ψt )(A)) : (0, ∞) −→ L(X) is continuous. Moreover, there is a constant C = C(ψ) with sup k(f ψt )(A)k ≤ C M (A, ϕ) kf k∞,ϕ t>0

b) One has sup t>0

Z

0

∞

kθ(tA)ψ(rA)k

(f ∈ H ∞ (Sϕ )).

dr < ∞. r

Proof. Choose C, s > 0 with |ψ(z)| ≤ C min{|z|s , |z|−s }. Let [a, b] ⊂ (0, ∞) and define F (t, z) := f (z)ψ(tz) on [a, b] × Sϕ . Then we have n o ˘ ¯ |F (t, z)| ≤ kf k∞ C min |z|s , |z|−s max ts , t−s . which shows that F satisfies the hypothesis of Lemma 5.2.1. This implies continuity. To prove the boundedness, simply apply Lemma 2.6.10. This is possible since we have |(f ψt )(z)| ≤ C kf k∞,ϕ min{|tz|s , |tz|−s } for all t > 0, z ∈ Sϕ . Assertion b) is a little more involved. Choose ω ′ ∈ (ω, ϕ) and define Γ := Γω ′ . Z ∞Z Z ∞Z Z ∞ ˛ ˛ |dz| dr dr ˛θ(z)ψ(rt−1 z)˛ |dz| dr kθ(tA)ψ(rA)k . |θ(tz)ψ(rz)| = r |z| r |z| r 0 0 Γ 0 Γ „Z ∞ ˛ « Z Z ∞ Z ˛ ˛ ˛ |dz| |θ(z)| ˛ i arg z ˛ dr εiω ′ ˛ dr ˛ ≤ ψ(re ) |θ(z)| ≤ |dz| max )˛ . ˛ψ(re ε=±1 r |z| r Γ 0 Γ |z| 0 The constant hidden in the symbol “.” is of course M (A, ω ′ )/2π.

5.2. A Fundamental Approximation Technique

95

Let A ∈ Sect(ω), ϕ ∈ (ω, π], ψ ∈ H0∞ (Sϕ ) and define ψa,b (z) :=

Z

b

a

ψ(tz)

dt t

(0 < a < b < ∞, z ∈ Sϕ ).

(5.3)

One knows from Example 2.2.6 that ψa,b (z) → c pointwise as (a, b) → (0, ∞), where c is a constant. The interesting question now is, for which x ∈ X one has ψa,b (A)x → cx as

(a, b) → (0, ∞).

The first step towards a general result is the following. Lemma 5.2.3. Let A ∈ Sect(ω), ϕ ∈ (ω, π], and ψ ∈ H0∞ (Sϕ ). Define ψa,b as in (5.3). Then the following assertions hold. a) ψa,b ∈ H0∞ (Sϕ ). b) supa,b kψa,b kϕ < ∞.

c) ψa,b (z) Z ∞→ c as (a, b) → (0, ∞) uniformly on compact sets of Sϕ , where dt c := ψ(t) . t 0 Z b dt d) ψa,b (A) = ψ(tA) . t a e) supa,b kψa,b (A)k < ∞.

Proof. We write Z b Z a Z 1 Z 1 dt dt dt dt ψa,b (z) = ψ(tz) − ψ(tz) = ψ(tbz) − ψ(taz) = h(bz) − h(az), t t t t 0 0 0 0 R where h(z) = 01 ψ(tz) dt as in Example 2.2.6. Using the results shown there, a)-c) follow t immediately. To prove assertion d) we apply Lemma 5.2.1 to the function F (t, z) := ψ(tz). Assertion e) is immediate from ψa,b (A) = h(bA) − h(aA) and Proposition 2.6.11.

Proposition 5.2.4. Let A ∈ Sect(ω), ϕ ∈ (ω, π], ψ ∈ H0∞ (Sϕ ). Define h(z) :=

Z

0

1

ψ(tz)

dt , t

g(z) :=

Z

∞

ψ(tz)

1

as in Example 2.2.6. Let also ψa,b (z) := the following assertions hold. Z 1 dt a) If kψ(tA)xk < ∞, then t 0

Z

b

ψ(tz)

a

h(A)x = lim ψa,1 (A)x = aց0

dt , t

Z

c :=

Z

∞

0

ψ(t)

dt t

dt as above. Then for x ∈ X t

0

1

ψ(tA)x

dt . t

Chapter 5. The Boundedness of the H ∞ -calculus

96 Z

∞

dt < ∞ and A is injective, then t 1 Z ∞ dt g(A)x = lim ψ1,b (A)x = ψ(tA)x . bր∞ t 1 Z ∞ dt ∞ < ∞ and c) If x ∈ R(θ(A)) for some θ ∈ H0 (Sϕ ), then kψ(tA)xk t 0 b) If

kψ(tA)xk

lim

a→0,b→∞

ψa,b (A)x =

lim

a→0,b→∞

Z

a

b

dt ψ(tA)x = t

Z

∞

ψ(tA)x

0

dt = c x. t

Proof. Find C, s > 0 such that |ψ(z)| ≤ C min{|z|s , |z|−s }. Ad a). It is clear that ψa,1 (z) → h(z) pointwise, with Z ∞ Z ∞ dt 2C dt ≤C min{ts , t−s } = . |ψa,1 (z)| ≤ |ψ(tz)| t t s 0 0 Also, |ψa,1 (z)| ≤

Z

1 0

|ψ(tz)|

dt C ≤ |z|s t s ψ

(z)

and obtain (1 + A)−1 ψa,1 (A) → Hence, one can apply Lemma 5.1.2 to the functions a,1 1+z (1 + A)−1 h(A) in norm as a → 0. On the other hand, by assumption and d) of Lemma 5.2.3 one has Z 1 Z 1 dt dt (1 + A)−1 ψa,1 (A)x = (1 + A)−1 ψ(tA)x → (1 + A)−1 ψ(tA) t t a 0 R as a ց 0. By injectivity of (1 + A)−1 , it follows that h(A)x = 01 ψ(tA) dt . t Ad b). This is proved in the same way as a), where one has to regularize at 0 with a function −1 −1 z(1 + z) . Note that one needs A(1 + A) to be injective in the end of the argument. Ad c). The first assertion is straightforward from Theorem 5.2.2, part b). For the second, let x = θ(A)y and ‚ ‚ consider the functions ψa,b (z)θ(z) which converge to the function cθ(z). Since supa,b ‚ψa,b ‚∞ < ∞ by Lemma 5.2.3, one can apply Lemma 5.1.2 and obtains ψa,b (A)x = ψa,b (A)θ(A)y → cθ(A)y = cx.

Remark 5.2.5. Part b)R of Proposition 5.2.4 is false in general if A is not injective. ∞ More precisely, if c = 0 ψ(t) dt/t 6= 0 and 0 = 6 x ∈ N(A) then ψ(tA)x = 0 for all t > 0 but g(A)x = cx 6= 0. The final result of this section is just a corollary of the foregoing considerations. Theorem 5.2.6. (McIntosh Approximation) R ∞ Let A ∈ Sect(ω), and ψ ∈ H0∞ [Sω ] such that 0 ψ(t) dt/t = 1. Then Z

b

a

dt ψ(tA)x = t

Z

for all x ∈ D(A) ∩ R(A).

a

b

dt ψ(tz) t

!

(A)x

a→0,b→∞

−→

Z

0

∞

ψ(tA)x

dt = x t

5.3. Bounded H ∞ -Calculus: Equivalent Descriptions and Uniqueness

97

5.3 Equivalent Descriptions and Uniqueness Let A ∈ Sect(ω) be a sectorial operator on the Banach space X and let ϕ ∈ (ω, π). Suppose we are given a subalgebra F ⊂ H ∞ (Sϕ ) such that f (A) is defined by the natural functional calculus for each f ∈ F. (This is a restriction only if A is not injective.) We say that the natural F -calculus for A is bounded if f (A) ∈ L(X) for all f ∈ F and kf (A)k ≤ C kf kϕ (f ∈ F) (5.4) for some constant C ≥ 0. Here, kf kϕ is shorthand for kf kϕ := kf k∞,Sϕ = sup{|f (z)| | z ∈ Sϕ }

(5.5)

We call inf{C ≥ 0 | (5.4) holds} the bound of the natural F -calculus.

If F is a closed subalgebra of H ∞ (Sϕ ) and A is injective, the Closed Graph Theorem together with part a) of the Convergence Lemma (Proposition 5.1.3) yields the existence of a constant C satisfying (5.4) if only f (A) ∈ L(X) for all f ∈ F.

5.3.1 Adjoints The next, in no way surprising result shows that boundedness is preserved when taking adjoints. Proposition 5.3.1. Let A ∈ Sect(ω) with dense domain and dense range and ϕ ∈ (ω, π). Then f (A) ∈ L(X) ⇐⇒ f (A′ ) ∈ L(X) for every f ∈ H ∞ (Sϕ ). Moreover, the natural H ∞ (Sϕ )-calculus for A is bounded with bound C if and only if the natural H ∞ (Sϕ )-calculus for A′ is bounded with bound C. Proof. By Proposition 2.6.3 we have f (A)′ = f (A′ ). If f (A) is bounded, then also f (A)′ is, by Proposition A.4.2. Moreover, the norms are the same. If f (A′ ) = f (A)′ is bounded, then the same reasoning shows that f (A)′′ ∈ L(X ′′ ). However, the hypotheses imply that f (A) is densely defined in X. Because f (A) = f (A)′′ ∩ (X ⊕ X), it follows from Lemma A.1.2 that f (A) ∈ L(X). The rest is clear.

5.3.2 Logarithms It is clear that the definition of boundedness of the natural H ∞ -calculus applies as well in the case of strip type operators. The next result shows in a way that via the logarithm one can switch back and forth from the sectorial to the strip type case.

Chapter 5. The Boundedness of the H ∞ -calculus

98

Proposition 5.3.2. Let A ∈ Sect(ω) be injective, let B := log A and ϕ ∈ (ω, π]. If the natural H ∞ (Sϕ )-calculus for A is bounded with bound C, i.e., one has f (A) ∈ L(X)

and

kf (A)k ≤ C kf kSϕ

for all f ∈ H ∞ (Sϕ ).

(5.6)

and

kg(B)k ≤ C kgkHϕ

for all g ∈ H ∞ (Hϕ ).

(5.7)

then also g(B) ∈ L(X)

Conversely, if B ∈ Strip(ω) such that (5.7) holds for some ϕ ∈ (ω, π], then A := eB is sectorial and (5.6) holds. Proof. The first statement is clear from the composition rule Theorem 4.2.4, since z 7−→ log z : Sϕ −→ Hϕ is biholomorphic with inverse z 7−→ ez . Assume that B ∈ Strip(ω) and (5.7) holds with ω < ϕ ≤ π. Then (t + ez )−1 (B) is bounded by M/t for some constant M and all t > 0. Hence A := eB is sectorial. By Theorem 4.3.1, ωA = ωst (B) ≤ ω. The rest follows again from the composition rule.

5.3.3 Boundedness on Subalgebras of H ∞ In the remaining part of this section we will examine how — for different subalgebras F ⊂ H ∞ — the boundedness of the natural F -calculus and the corresponding bounds interrelate. Here is the “Omnibus Theorem” in case that the operator has dense domain and range. Proposition 5.3.3. Let A ∈ Sect(ω) with dense domain and dense range. Let ϕ ∈ (ω, π) and C ≥ 0. The following assertions are equivalent. (i) The natural H0∞ (Sϕ )-calculus for A is bounded with bound C.

(ii) The natural H ∞ (Sϕ )-calculus for A is bounded with bound C. (iii) The natural H ∞ (Sϕ ) ∩ C0 (Sϕ )-calculus for A is bounded with bound C. (iv) The natural R∞ 0 (Sϕ )-calculus for A is bounded with bound C. (v) The natural R∞ (Sϕ )-calculus for A is bounded with bound C.

Proof. Let us first note that the implications (ii) ⇒ (i) and (ii) ⇒ (v) ⇒ (iv) are trivial. By ∞ Proposition E.0.5 the space R∞ 0 (Sϕ ) is uniformly dense in H (Sϕ ) ∩ C0 (Sϕ ). Applying part a) of the Convergence Lemma (Proposition 5.1.3) yields the implication (iv) ⇒ (iii). (Note that here only injectivity of A is needed.) To show the implication (iii) ⇒ (i) let f ∈ H0∞ (Sϕ ). Define 1 fn (z) := f (z + n ). Obviously, fn ∈ H ∞ (Sϕ ) ∩ C0 (Sϕ ), kfn kϕ ≤ kf kϕ and fn → f pointwise on Sϕ . Now (i) follows from another application of the Convergence Lemma. Finally, we show (i) ⇒ (ii). Let f ∈ H ∞ (Sϕ ). Define ψs (z) := z s /(1 + z)2s for s > 0. Then ψs , f ψs ∈ H0∞ (Sϕ ). Moreover, kf ψs kϕ ≤ kf kϕ K s where K := kψ1 kϕ . Using (i) gives k(f ψs )(A)k ≤ C kf k∞ K s . Now we apply the Convergence Lemma (Prop. 5.1.3) and infer that f (A) ∈ L(X) with (f ψs )(A) → f (A) strongly if s ց 0. Since lims→0 K s = 1 we obtain kf (A)k ≤ C kf k∞ .

5.3. Bounded H ∞ -Calculus: Equivalent Descriptions and Uniqueness

99

Note that in reflexive spaces, injective sectorial operators automatically have dense domain and dense range. The questions become more difficult if we drop the density assumptions. Recall that for a sectorial operator A we can always form the uniformly sectorial family Aε := (ε + A)(1 + εA)−1 which is a sectorial approximation of A consisting of bounded and invertible operators (see Proposition 2.1.3). Lemma 5.3.4. Let A ∈ Sect(ω), ϕ ∈ [ω, π), and C ≥ 0. The following assertions are equivalent. (i) The natural R∞ (Sϕ )-calculus for A is bounded with bound C.

(ii) The natural R∞ (Sϕ )-calculus for Aε is bounded with bound C, for each ε > 0. Proof. Let rε (z) = (ε + z)(1 + εz)−1 ∈ R∞ (Sϕ ). Then Aε = rε (A). If r ∈ R∞ (Sϕ ) then r ◦ rε ∈ R∞ (Sϕ ) with kr ◦ rε kϕ ≤ krkϕ , since rε maps Sϕ into Sϕ . The composition rule yields r(Aε ) = (r ◦ rε )(A), whence we have proved the implication (i) ⇒ (ii). The reverse implication follows from the fact that r(Aε ) → r(A) in norm, by Lemma 2.6.7. (Note that R∞ (Sϕ ) ⊂ E(Sϕ ).)

One should note that ϕ = ω is allowed in Lemma 5.3.4. Proposition 5.3.5. Let A ∈ Sect(ω), ϕ ∈ (ω, π) and C ≥ 0. We define K to be the closure of Sϕ in C∞ . The following assertions are equivalent. (i) The natural A(Sϕ ) ∩ C(K)-calculus for A is bounded with bound C.

(ii) The natural R∞ (Sϕ )-calculus for A is bounded with bound C.

If A is injective, one can replace A(Sϕ ) by H ∞ (Sϕ ) in a).

Proof. The implication (i) ⇒ (ii) is obviously true. To prove the reverse implication, take f ∈ A(Sϕ ) ∩ C(K). By Proposition E.0.5 there is a sequence (rn )n ⊂ R∞ (Sϕ ) such that krn − f kϕ → 0. By Lemma 5.3.4, the hypothesis (ii) implies that the natural R∞ (Sϕ )-calculus for Aε is bounded with bound C, for each ε > 0. Since each Aε is bounded and invertible, we can apply Proposition 5.3.3 to conclude that the natural H ∞ (Sϕ )-calculus for Aε is bounded with bound C for each ε > 0. This implies that rn (Aε ) → f (Aε ) uniformly in ε > 0. From Lemma 2.6.7 we see that rn (Aε ) → rn (A) as ε ց 0 for each n. A standard argment from functional analysis now implies that there is T ∈ L(X) with f (Aε ) → T . An application of Proposition 2.6.9 yields T = f (A). The same proof works in the case where A is injective and f ∈ H ∞ (Sϕ ) ∩ C(K).

Proposition 5.3.6. Let A ∈ Sect(ω), ϕ ∈ [ω, π) and C ≥ 0. Assume that the natural R∞ 0 (Sϕ )-calculus for A is bounded with bound C. If A is densely defined and ϕ ≤ π2 then the natural R∞ (Sϕ )-calculus for A is bounded with bound C.

n Proof. Let rn := n+z . Then krn kϕ = 1 since ϕ ≤ π2 . Moreover, rn (A)x → x for n → ∞ (by Proposition 2.1.1), since D(A) is dense in X. Thus for a function r ∈ R∞ (Sϕ ) we have

kr(A)xk = lim kr(A)rn (A)xk = lim k(rrn )(A)xk ≤ C krrn kϕ kxk ≤ C krkϕ kxk n

for all x ∈ X.

n

100

Chapter 5. The Boundedness of the H ∞ -calculus

We do not know if in the second part of Proposition 5.3.6 one can omit the assumption D(A) = X or ϕ ≤ π2 .

5.3.4 Uniqueness Let us turn (for the moment) to a more general situation. Assume that A is a closed operator on a Banach space X, Ω ⊂ C is open and F ⊂ H ∞ (Ω) is a subalgebra containing the rationals rλ (z) = (λ − z)−1 for λ ∈ / Ω. We say that a mapping Φ : F → L(X) is a bounded F -calculus for A if the following conditions are satisfied: 1) 2) 3) 4)

The mapping Φ is a homomorphism of algebras. We have Φ(rλ ) = R(λ, A) for all λ ∈ / Ω. If 1 ∈ F then Φ(1) = I. There is C ≥ 0 such that kΦ(f )k ≤ C kf kΩ for all f ∈ F.

Here, kf kΩ denotes the supremum norm of f on Ω. We say that a sequence (fn )n ⊂ F converges boundedly and pointwise on Ω to a function f if fn → f pointwise on Ω and supn kfn kΩ < ∞. We say that Φ is continuous with respect to bounded and pointwise convergence (in short: b.p.-continuous), if it has the following property. 5) If fn , f ∈ F such that fn → f boundedly and pointwise on Ω, then Φ(fn ) → Φ(f ) strongly on X. Note that if 1 ∈ / F we can always extend Φ to F ′ = F ⊕ C1 satisfying 3) and keeping all other properties. Lemma 5.3.7. Let A be a closed operator on the Banach space X. Let ω ∈ (0, π) and assume that Φ is a bounded H ∞ (Sω )-calculus for A with bound C ≥ 0. Then A is sectorial with ωA ≤ ω. Take a sector Sϕ with ωA < ϕ and ω ≤ ϕ and denote by K the closure of Sϕ in C∞ . Then the following assertions hold. a) Φ(f ) = f (A) for all f ∈ A(Sϕ ) ∩ C(K). b) If A is injective, then Φ(f ) = f (A) for f ∈ H ∞ (Sϕ ) ∩ C(K).

c) If A has dense domain and dense range then the natural H ∞ (Sϕ )-calculus for A is bounded with bound C. λ . By Proposition E.0.5, Proof. The sectoriality of A follows by applying Φ to the rationals λ−z we have Φ(f ) = f (A) for f ∈ R∞ (Sω ). The assertions a) and b) now follow from Proposition 5.3.5. Part c) is immediate from Proposition 5.3.3.

Proposition 5.3.8. Let A be a closed operator on a Banach space X and let ω ∈ (0, π). Then there is at most one bounded and b.p.-continuous H ∞ (Sω )-calculus Φ for A. If such a Φ exists, the operator A is sectorial with ωA ≤ ω and has dense domain and dense range. Moreover, Φ coincides with the natural functional calculus on H ∞ (Sϕ ), where ϕ > ωA with ϕ ≥ ω.

5.4. The Minimal Angle

101

Proof. Uniqueness follows from Proposition E.0.6. Define fn (z) := n(n + z)−1 . Then fn → 1 pointwise on Sω and supn kfn kω < ∞. Since Φ is b.p.-continuous, n(n + A)−1 = Φ(fn ) → 1 Φ(1) = I strongly whence A is densely defined. The same argument with fn (z) = z( n + z)−1 yields that A has dense range. Take ϕ > ωA with ϕ ≥ ω. From Lemma 5.3.7 it follows that the natural H ∞ (Sϕ )-calculus for A is bounded. By uniqueness, the natural calculus must coincide with Φ.

Remark 5.3.9. Let A ∈ Sect(ω) with dense domain and range and let ϕ > ω. Proposition 5.3.3 implies that if A has some bounded H ∞ (Sϕ )-calculus with bound C, then the natural H ∞ (Sϕ )-calculus for A is bounded with bound C. However, the question remains if there really can be two different bounded H ∞ -calculi for the same operator.

5.4 The Minimal Angle Suppose we are given an injective sectorial operator A such that the natural H ∞ (Sϕ )-calculus for A is bounded for some ϕ > ωA . Hence we can ask for the minimal angle ϕ such that this is true. In fact, in general such a minimal angle does not exist but we can define the H ∞ -angle ωH ∞ (A) := inf{ϕ > ωA | the nat. H ∞ (Sϕ )-calc. for A is bdd.} which is understood to have the value ∞ if the set on the right hand side is empty. It is clear that if ωH ∞ (A) < ∞, then the imaginary powers Ais of A are bounded operators and for every ωH ∞ (A) < µ < π there is a constant Cµ such that

is

A ≤ Cµ eµ|s| (s ∈ R). Although the imaginary powers do not need to form a C0 -group (since D(A)∩R(A) need not be dense in X) the definition of the type θA from (3.11) is still meaningful: n o

θA = inf θ > 0 | ∃ M ≥ 1 : Ais ≤ M eθ|s| (s ∈ R) .

So with this notation we obtain θA ≤ ωH ∞ (A). We will see below (Corollary 5.4.3) that one actually has θA = ωH ∞ (A) if ωH ∞ (A) < ∞. Since the proof anyway works with strips instead of sectors, we first formulate the result in the strip version. Theorem 5.4.1. Let B ∈ Strip(ω), ω < α and assume that the natural H ∞ (Hα )calculus

is bounded. Let 0 < µ < θ < α and assume that there is a constant C with eisB ≤ Ceµ|s| for all s ∈ R. Then the natural H ∞ (Hθ )-calculus for B is bounded. For the proof we utilize the following auxiliary result.

Lemma 5.4.2. Let 0 < µ < θ < α. Then there is a constant cµ,θ,α such that for every f ∈ H ∞ (Hθ ) there exists (fk )k∈Z ⊂ H ∞ (Hα ) with

Chapter 5. The Boundedness of the H ∞ -calculus

102 P

eµ|k| kfk (z)kH ∞ (Hα ) ≤ cµ,θ,α kf kHθ P 2. f (z) = k∈Z eikz fk (z) uniformly for z ∈ Hµ . 1.

k∈Z

Proof. Choose ϕ ∈ D(R) with the following properties: supp ϕ ⊂ [−1, 1], 0 ≤ ϕ ≤ 1, and P n∈Z ϕ(x − n) = 1 for all x ∈ R. By the Paley-Wiener Theorem [114, Thm. 7.22] one knows that the inverse Fourier transform ϕ ˇ of ϕ is an entire function which decreases rapidly along horizontal lines, uniformly within bounded horizontal strips. Hence for n ∈ Z the function Z fn (z) := ϕ(z ˇ − t)f (t)e−int dt R

is entire. Choose ε ∈ (−1, 1) such that εn ≤ 0. We can shift the contour of integration to obtain Z fn (z) = ϕ(z ˇ − t − iεθ)f (t + iεθ)e−int eεnθ dt. R

This (with |ε| → 1 ) yields the estimate |ϕ ˇn (z)| ≤ Cθ,α e−|n|θ kf kHθ for z ∈ Hα , where Cθ,α = R sup|y|≤α+θ R |ϕ(t ˇ + iy)| dt. Hence we obtain X eµ|n| kfn kHα ≤ cµ,θ,α kf kHθ . n∈Z

So we know that the sum n∈Z einz fn (z) converges uniformly within the strip Hµ to a bounded function g. However, in the distributional sense, we have X X` X ´ f = δ0 ∗ f = (ϕ(· − n))∨ ∗ f = ein· ϕ ˇ ∗f = ein· fn (·) P

n∈Z

n∈Z

n∈Z

on R. Hence we have g = f on R and therefore on the whole of Hµ .

Let us now turn to the proof of Theorem 5.4.1. Proof of Theorem 5.4.1. Choose f ∈ H ∞ (Hθ ) and employ Lemma 5.4.2 to find (fn )n∈Z ⊂ H ∞ (Hα ) with properties 1) and 2). Denote by C ′ the bound of the H ∞ (Hα )-calculus for B. Then we have ‚ X ‚` X‚ X ‚ ´ ‚ inB ‚ ‚ einz fn (z) (B)‚ = fn (B)‚ ≤ C C ′ eµ|n| kfn kHα ≤ C C ′ cµ,θ,α kf kHθ . ‚e n∈Z

n∈Z

n∈Z

` ´ Hence the sum n∈Z einz fn (z) (B) converges uniformly in L(X) towards an operator, say, T . By the Convergence Lemma for strip type operators (Proposition 5.1.6) we see that T = f (B) and, in fact, kf (B)k ≤ CC ′ cµ,θ,α kf kHθ . P

Corollary 5.4.3. (Cowling, Doust, McIntosh, Yagi) Let A ∈ Sect(ω) be injective with ωH ∞ (A) < ∞. Then ωH ∞ (A) = θA . Note that we do not suppose A to be densely defined or to have dense range. Proof. We already know that the inequality θA ≤ ωH ∞ (A) holds. Choose θA < µ < θ < α ≤ π such that the natural H ∞ (Sα )-calculus for A is bounded. Let B := log(A). By Proposition 5.3.2, the operator B satisfies the hypotheses of Theorem 5.4.1. Applying again Proposition 5.3.2 we obtain that the natural H ∞ (Sθ )-calculus for A is bounded. Since θ can be arbitrarily close to θA , we are done.

Remark 5.4.4. In Section ?? we will present an example of an injective sectorial operator A on X := Lp (R) (1 < p < ∞, p 6= 2) such that A ∈ BIP(X, 0) but ωH ∞ (A) = ∞.

5.5. An Operator without a Bounded H ∞ -Calculus

103

5.5 An Operator without a Bounded H ∞ -Calculus In the present section we will present the “construction” of an operator A on a separable Hilbert space H such that the following conditions are satisfied: 1) The operator A is invertible and sectorial of angle 0. 2) The semigroup generated by −A is immediately compact (see [40, II.4.23] for the definition). 3) The natural H ∞ (S π2 )-calculus of A is not bounded. By the McIntosh-Yagi Theorem 7.3.1 below, this implies that A ∈ / BIP(H) (see Section 3.4 for the definition) and the natural H ∞ (Sϕ )-calculus for A is not bounded for any 0 < ϕ ≤ π. The construction uses the notion of a conditional basis. This is a sequence (en )n≥1 in H with the following properties: 1) P For every x ∈ H there is a unique sequence (xn )n ⊂ C such that x = ∞ n=1 xn en in H. 2) P There is a sequence (θn )nP⊂ {−1, 1} and a vector x ∈ H such that x = ∞ n=1 xn en but the series n θn xn en does not converge.

This means that (en )n is a Schauder basis of H which is not unconditional, see [82, Part I, Ch. I] for definitions. By [82, Part I, Prop. 2.b.11] each separable Hilbert space has a conditional basis. If (en )n is a Schauder basis of H, one can define the projections Pn :

X∞

k=1

xk ek 7−→

Xn

k=1

xk ek

and obtains Pn ∈ L(H) for each n with M0 := supn kPn k being finite. The number M0 is called the basis constant of the basis (en )n . Given a scalar sequence a = (an )n ⊂ C we define X k| ak| := lim sup |an | + |an+1 − an | n

n≥1

which may be infinite. With a we associate an operator A on H by X X D(A) := {x = xk ek ∈ H | ak xk ek converges} k

k

X

and Ax = ak xk ek for x ∈ D(A). An easy partial summation argument — k see [125, Lemma 2.2] — yields that A ∈ L(H) if k| ak| < ∞, and in this case we have kAk ≤ M0 k| ak| . Furthermore, if in addition we have limn an = 0, then kPn A − Ak → 0 whence A is compact. Assume now that (an )n is a strictly increasing sequence of positive real numbers with a1 > 0 and limn an = ∞. Given λ ∈ / [a1 , ∞) we form the sequence

Chapter 5. The Boundedness of the H ∞ -calculus

104

1 )n and compute rλ := ( λ−a n

k| rλ k| =

Z ak+1 Z ∞ X 1 1 dt dt − = ≤ 2. 2 k≥1 λ − ak+1 k≥1 a λ − ak (λ − t) a1 |λ − t| k

X

Hence, the operator associated to rλ is bounded (even compact) and it is easily seen that this operator equals R(λ, A). In particular, σ(A) ⊂ [a1 , ∞). For λ = |λ| eiϕ , 0 < |ϕ| ≤ π, we obtain kλR(λ, A)k ≤ M0 |λ| k| rλ k| ≤ M0

Z

0

∞

|λ| dt

2

|λ − t|

= M0

Z

0

∞

dt

2.

|eiϕ − t|

This shows that A ∈ Sect(0) and A has compact resolvent. Since the semigroup generated by −A is holomorphic (see Section 3.3), it is immediately compact, by [40, Theorem II.4.29]. Let D := span{en | n ∈ N} ⊂ D(A). Since the projections Pn commute with A, we easily see that Xn Xn f (A) xn en = f (an )xn en k=1

k=1

for each n ∈ N, (xk )nk=1 ∈ Cn and each f ∈ H ∞ [S0 ]. Given ϕ > 0 we conclude that X∞ the natural H ∞ (Sϕ )-calculus for A is bounded if and only if f (an )xn en k=1 X∞ converges whenever xn en does. k=1

We now specialize the sequence a to an := 2n . Since the basis (en )n is assumed to be conditional, we pick a sequence (θn )n ⊂ {−1, 1} and a vector X∞ X∞ x = xn en ∈ H such that θn xn en does not converge. Employing k=1 k=1 a result of Carleson (from [46, Section VII.1]) one can find f ∈ H ∞ (S π2 ) such that f (2n ) = θn for all n. By the remarks above, we obtain that f (A) ∈ / L(H).

5.6 Perturbation Results In this section we examine permanence properties of the functional calculus for sectorial operators under additive perturbations, i.e., we start with a sectorial operator A and a linear operator B : D(A) −→ X and consider the operator A + B with domain D(A + B) = D(A). We first remark that in order to have λ0 + A + B being sectorial for some λ0 > 0, it is neccessary that B ∈ L(D(A), X). Proof. If λ0 + A + B is sectorial, then A + B is closed. Hence also A(1 + A)−1 + B(1 + A)−1 = (A + B)(1 + A)−1 is closed. From this it follows that B(1 + A)−1 is closed and fully defined, hence bounded.

5.6. Perturbation Results

105

5.6.1 Resolvent Growth Conditions In this section we examine some conditions under which the perturbed operator A + B is (quasi-)sectorial and preserves boundedness of the H ∞ -calculus. A “Domain Condition” The next lemma describes a class of well-behaving perturbation operators. Lemma 5.6.1. Let A ∈ Sect(ω), B ∈ L(D(A), X) and 0 < θ. Then the following assertions are equivalent.

(i) supt≥1 tθ B(t + A)−1 < ∞. (ii)

θ

sup{|λ| kBR(λ, A)k | |λ| ≥ 1, ω ′ ≤ |arg λ| ≤ π} < ∞

for each ω ′ ∈ (ω, π).

In this case there is λ0 ∈ R such that λ0 + A + B ∈ Sect(ω).

‚ ‚ Proof. The implication (ii) ⇒ (i) is trivial. Suppose (i) holds and set d := supt≥1 ‚tθ B(t + A)−1 ‚. Let ω < ω ′ < π and ω ′ ≤ arg λ ≤ π, |λ| ≥ 1. Then ‚ ‚ ‚ ‚ |λ|θ kBR(λ, A)k ≤ |λ|θ ‚BR(λ, A) + B(|λ| + A)−1 ‚ + |λ|θ ‚B(|λ| + A)−1 ‚ ≤ |λ|θ kB (R(λ, A) − R(− |λ| , A))k + d

= |λ|θ k−B(|λ| + λ)R(− |λ| , A)R(λ, A)k + d ≤ 2dM (A, ω ′ ) + d.

To prove the additional assertion, we write (λ − (A + B)) = (I − BR(λ, A))(λ − A) ‚ ‚ for λ ∈ / Sω ′ . Since lim|λ|→∞ ‚B(λ + A)−1 ‚ = 0 by (ii) we see that λ − (A + B) is invertible for large |λ| and R(λ, A + B) = R(λ, A)(I + BR(λ, A))−1 . The rest follows.

Remark 5.6.2. We will prove in Section 6.7 that B : D(A) −→ X satisfies the equivalent conditions (i), (ii) from Lemma 5.6.1 with θ ∈ (0, 1] if and only if B extends to a bounded operator B : (X, D(A))1−θ,1 −→ X (see Proposition 6.7.3). So the condition (i) is more or less a condition on the domain of the perturbing operator B. This motivates the headline of this section. Let us formulate the main result. Proposition 5.6.3. Let A ∈ Sect(ω) be injective and ϕ ∈ (ω, π). Assume that B has the equivalent properties (i), (ii) from Lemma 5.6.1 for some θ > 0. Assume furthermore that A+B ∈ Sect(ω) and A+B is invertible. Then there are constants K1 , K2 such that kf (A + B)k ≤ K1 kf (A)k + K2 kf kϕ for all f ∈ H ∞ (Sϕ ) such that f (A) ∈ L(X).

106

Chapter 5. The Boundedness of the H ∞ -calculus

Proof. Take f ∈ H ∞ (Sϕ ) such that f (A) ∈ L(X) and a contour Γ = Γω ′ with ω < ω ′ < ϕ. By definition, Z f (z)z 1 R(z, A + B) dz. f (A + B) = (2 + (A + B) + (A + B)−1 ) 2πi Γ (1 + z)2 Since 2 + (A + B)−1 is a bounded operator we can estimate ‚ ‚ Z ‚ ‚ f (z)z ‚(2 + (A + B)−1 ) 1 ‚ R(z, A + B) dz ‚ ‚ 2πi Γ (1 + z)2 Z ‚ ‚ 1 |dz| ≤ kf kϕ M (A + B, ω ′ ) ‚2 + (A + B)−1 ‚ . 2π Γ |1 + z|2

As to the remaining part, writing R(z, A + B) = R(z, A + B)BR(z, A) + R(z, A) within the integral, we have to estimate the two summands Z 1 f (z)z (1) (A + B) R(z, A) dz and 2πi Γ (1 + z)2 Z 1 f (z)z (2) (A + B) R(z, A + B)BR(z, A) dz. 2πi Γ (1 + z)2 Since f (A) is known to be a bounded operator we can write the first term as (A + B)A(1 + A)−2 f (A) and estimate ‚ ‚ ‚ ‚ ‚(A + B)A(1 + A)−2 ‚ ≤ (M + 1)2 + (M + 1) ‚B(1 + A)−1 ‚

where M = M (A). We split the second summand into two parts by decomposing the contour Γ into Γ ∩ {|z| ≤ 1} ⊕ Γ ∩ {|z| ≥ 1} and obtain ‚ ‚ Z ‚ ‚ f (z)z ‚(A + B) 1 ‚ R(z, A + B)BR(z, A) dz ‚ ‚ 2πi Γ (1 + z)2 Z ‚ ‚ 1 |f (z)| ≤ k(A + B)R(z, A + B)k ‚B(1 + A)−1 ‚ kz(A + 1)R(z, A)k |dz| 2π Γ,|z|≤1 |1 + z|2 Z |f (z)| |z|1−θ 1 k(A + B)R(z, A + B)k |z|θ kBR(z, A)k |dz| + 2π Γ,|z|≥1 |1 + z|2 Z ‚ ‚ 1 |dz| ≤ kf kϕ (M (A + B, ω ′ ) + 1)(2M (A, ω ′ ) + 1) ‚B(1 + A)−1 ‚ 2π Γ,|z|≤1 |1 + z|2 ! Z 1 |z|1−θ |dz| . + kf kϕ (M (A + B, ω ′ ) + 1) sup |z|θ kBR(z, Ak 2π Γ,|z|≥1 |1 + z|2 |arg z|=ω ′ ,|z|≥1

Remark 5.6.4. The statement is false without the assumption that A + B is invertible. This follows from the fact that there is a bounded, injective operator A (on a Hilbert space) with unbounded natural H ∞ -calculus (see Section 5.3). However, A can be written as A = (A + 1) + (−1), and A + 1 does have a bounded H ∞ -calculus, since it is bounded and invertible. In passing by, we note the following corollary. Corollary 5.6.5. Let A be an injective operator on a Banach space X and B ∈ L(X). Assume that A, A+ B ∈ Sect(ω) and 0 ∈ ̺(A+ B). If, for some ϕ ∈ (ω, π), the natural H ∞ (Sϕ )-calculus for A is bounded, then this also holds for A + B.

5.6. Perturbation Results

107

A “Range Condition” The next lemma introduces another class of perturbations. Essentially, the required property is a condition on the range of the perturbing operator (see Remark 5.6.7 below). Lemma 5.6.6. Let A ∈ Sect(ω), B ∈ L(D(A), X) and 0 < θ. Then the following assertions are equivalent.

(i) supt≥1 tθ A(t + A)−1 B L(D(A),X) < ∞. (ii)

θ

sup{|λ| kAR(λ, A)BkL(D(A),X) | |λ| ≥ 1, ω ′ ≤ |arg λ| ≤ π} < ∞ for each ω ′ ∈ (ω, π).

In this case there is λ0 ∈ R such that λ0 + A + B ∈ Sect(ω).

‚ ‚ Proof. The implication (ii) ⇒ (i) is trivial. Suppose (i) holds and set d := supt≥1 ‚tθ A(t + A)−1 B ‚. Let ω ′ ∈ (ω, π) and ω ′ ≤ arg λ ≤ π, |λ| ≥ 1. Then ‚ ‚ ‚ ‚ |λ|θ kAR(λ, A)Bk ≤ |λ|θ ‚AR(λ, A)B + A(|λ| + A)−1 B ‚ + |λ|θ ‚A(|λ| + A)−1 B ‚ ≤ |λ|θ kA (R(λ, A) − R(− |λ| , A)) Bk + d

= |λ|θ k−(|λ| + λ)R(λ, A)AR(− |λ| , A)Bk + d ≤ 2dM (A, ω ′ ) + d

(all the norms in L(D(A), X)). To prove the additional assertion, we write (λ − (A + B)) = (λ − A)(I − R(λ, A)B) for λ ∈ / Sω ′ . Choose σ ∈ (0,‚θ] with σ < 1. Then our assumptions and (ii) imply ac‚ tually sup|λ|≥1 λσ ‚(λ + A)−1 B ‚D(A)−→D(A) < ∞. (Note that B ∈ L(D(A), X)). Hence ‚ ‚ lim|λ|→∞ ‚(λ + A)−1 B ‚L(D(A)) = 0, and so (I + (λ + A)−1 B) is an isomorphism on D(A) for large |λ|. Therefore, for |λ| large enough, we see that λ − (A + B) is invertible and R(λ, A + B) = (I + BR(λ, A))−1 R(λ, A). The rest follows.

Remark 5.6.7. We will prove in Section 6.7 that B : D(A) −→ X satisfies the equivalent conditions (i), (ii) from Lemma 5.6.1 with θ ∈ (0, 1] if and only if B : D(A) −→ X is bounded and R(B) ⊂ (X, D(A))θ,∞ (see Proposition 6.7.1). Proposition 5.6.8. Let A ∈ Sect(ω) be injective and ϕ ∈ (ω, π). Assume that B has the equivalent properties (i), (ii) from Lemma 5.6.6 for some θ > 0. Assume furthermore that A+B ∈ Sect(ω) and A+B is invertible. Then there are constants K1 , K2 such that kf (A + B)k ≤ K1 kf (A)k + K2 kf kϕ for all f ∈ H ∞ (Sϕ ) such that f (A) ∈ L(X).

Proof. The proof follows the same lines as the proof of Proposition 5.6.3. The main difference is in writing R(z, A + B) = R(z, A) − R(z, A)BR(z, A + B) instead of R(z, A + B) = R(z, A) + R(z, A + B)BR(z, A).

108

Chapter 5. The Boundedness of the H ∞ -calculus

A Degenerate Case We turn to a different class of operators, somehow “dual” to the first one. Assume that A ∈ Sect(ω) has dense domain. Let B : D(A) −→ X be an operator such that

sup tθ (t + A)−1 B < ∞ (5.8) t≥1

for some 0 < θ < 1. If we replace A by 1 + A for ε > 0, condition (5.8) still holds. So in the following we assume A to be invertible. The problem now is that in general the operator A+B with D(A+B) := D(A) will not be quasi-sectorial any more, and not even a closed operator. However, it always has a quasi-sectorial extension. To prove this we introduce a suitable extrapolation space of X. Namely, let

Y := (X, A−1 · )∼ ,

i.e., Y is the completion of X with respect to the norm kxkY := A−1 x X . Then, by density of D(A) in X, the operator A extends in a unique way to an isometric isomorphism AY : X −→ Y . Moreover, by (5.8), one has

kBxkY = A−1 Bx X ≤ (A + 1)A−1 L(X) (1 + A)−1 Bx X

≤ c (A + 1)A−1 L(X) kxkX (x ∈ D(A)) for some c > 0. This implies that B extends uniquely to a bounded operator BY : X −→ Y . Using these extensions, condition (5.8) becomes

sup tθ AY (t + AY )−1 BY L(X,Y ) < ∞. (5.9) t≥1

Considering the operator AY as an operator on Y with domain D(AY ) = X, we realize that AY is sectorial of angle ω, isometrically similar to the operator A on X. Moreover, BY : D(AY ) −→ Y is a perturbation of AY of the type which we considered in Lemma 5.6.6. Hence we conclude that for some λ0 > 0 the operator λ0 + AY + BY is also sectorial (of angle ω) and invertible (on the Banach space Y ). If we take the part in X of this operator we have our sectorial extension, as announced above. Proposition 5.6.9. Let A, B, λ0 as above. Let C := (AY + BY ) X be the part of AY + BY in X. Then λ0 + C is an invertible sectorial operator of angle ω in X. The operator C extends A+ B whence this operator is closable. Also, the following two assertions are equivalent: (i) C = A + B. (ii) λ0 + A + B has dense range. Moreover, for every ϕ ∈ (ω, π), f ∈ H ∞ (Sϕ ) such that f (A) ∈ L(X), also f (λ0 + C) ∈ L(X) with an estimate of the form kf (λ0 + C)k ≤ K1 kf (A)k + K2 kf kϕ .

5.6. Perturbation Results

109

Proof. Since D(AY + BY ) = X, the space X ˛is invariant under the resolvent of AY + BY . Also it is clear that R(λ, C) = R(λ, AY + BY )˛X for all λ ∈ ̺(AY + BY ). This proves that σ(λ0 + C) ⊂ Sω . To prove the sectoriality estimate, we simply write (λ + AY + BY )−1 = (I + (λ + AY )−1 BY )−1 (λ + AY )−1 .

By restriction this gives (λ + C)−1 = (I + (λ + AY )−1 BY )−1 (λ + A)−1 . Now one combines the sectoriality of A with condition (5.9) (and the resolvent identity) to obtain the sectoriality of λ0 + C. Let us prove the second assertion. If C = A + B we have X = R(λ0 + C) = R(λ0 + A + B) ⊂ R(λ0 + A + B) whence λ0 + A + B has dense range. For the converse, assume this. From A + B ⊂ C ist follows (λ0 + C)′ ⊂ (λ0 + A + B)′ = (λ0 + A + B)′ . The latter is injective, as an immediate cconsequence of the assumption. On the other hand, (λ0 + C)−1 is even invertible, and this is only possible if one has equality (λ0 + C)′ = (λ0 + A + B)′ . Taking adjoints and intersecting with X ∩ X yields λ0 + C = λ0 + A + B. To prove the third assertion, take ϕ ∈ (ω, π) and f ∈ H ∞ (Sϕ ) such that f (A) ∈ L(X). Then clearly f (AY ) ∈ L(Y ) with kf (AY )kL(Y ) = kf (A)kL(X) . By Proposition 5.6.8 we therefore have f (λ0 + AY + BY ) ∈ L(Y ) with kf (λ0 + AY + BY )kL(Y ) ≤ K1 kf (AY )kL(Y ) + K2 kf kϕ for some constants K1 , K2 . Now note that, by Proposition 2.6.5, we have f (λ0 + C) = f ((λ0 + AY + BY )X ) = f (λ0 + AY + BY )X . Moreover, X = D(λ0 + AY + BY ) which means on the one hand that k(λ0 + AY + BY )xkY provides an equivalent norm on X. On the other hand, the space X is invariant under f (λ0 + AY + BY ) with kf (λ0 + C)kL(X) = kf (λ0 + AY + BY )X kL(X) . kf (λ0 + AY + BY )kL(Y )

This concludes the proof.

One can show that each of the two equivalent conditions appearing in Proposition 5.6.9 is also equivalent to the identity (A + B)′ = A′ + B ′ .

5.6.2 A Theorem of Pr¨ uss and Sohr Recall that, by Proposition 3.4.5, one has Ais ∈ L(X)

⇒

(A + ε)is ∈ L(X)

for each injective sectorial operator A and all s ∈ R, ε > 0. The reverse implication is valid if A is invertible. Unfortunately, this perturbation result does not say anything about the group types θA and θA+ε (in case D(A) and R(A) are dense). In this section we want to prove the following theorem. Theorem 5.6.10. Let A be an injective sectorial operator on the Banach

space

X (no density assumptions!). Assume there is a constant K such that Ais ≤ Keθ|s| (s ∈ R) for some θ > ωA . Then there is K ′ such that

(A + ε)is ≤ K ′ eθ|s| (s ∈ R) uniformly in ε > 0.

Chapter 5. The Boundedness of the H ∞ -calculus

110

To reach this goal, we have to investigate in more detail a special case of Corollary 5.6.5. Proposition 5.6.11. Let A ∈ Sect(ω) be injective. Let ϕ ∈ (ω, π) and f ∈ H ∞ (Sϕ ). If f (A) ∈ L(X) then f (A + 1) ∈ L(X) and kf (A + 1)k ≤ K1 kf (A)k + K2 kf kϕ where K1 = 1 + M (A) and K2 = (2 + M (A) + M (A, ϕ))M (A, ϕ)C(ϕ) where C(ϕ) only depends on ϕ. Proof. Take ω ′ ∈ (ω, ϕ) and let Γ = ∂Sω ′ . One has to estimate Z f (z)z 1 R(z, A + 1) dz. f (A + 1) = (2 + (A + 1) + (A + 1)−1 ) 2π Γ (1 + z)2 As in the proof of Proposition 5.6.3 we split this expression into two summands, namely Z 1 f (z)z f (A + 1) = (2 + (A + 1)−1 ) R(z, A + 1) dz 2πi Γ (1 + z)2 Z 1 f (z)z + (A + 1) R(z, A + 1) dz 2πi Γ (1 + z)2 The first summand is estimated by Z 1 |dz| (2 + M (A)) kf kϕ M (A + 1, ω ′ ) = kf kϕ (2 + M (A))M (A + 1, ω ′ )T (ω ′ ) 2π Γ |1 + z|2 R |dz| 1 where T (ω ′ ) = 2π Γ |1+z|2 . The second is written as the difference f (z)z R(z, A + 1) dz = (1 + z)2 Z Z 1 f (z)z 1 f (z)z (A + 1) R(z, A) dz − (A + 1) R(z, A)R(z, A + 1) dz 2 2πi Γ (1 + z) 2πi Γ (1 + z)2 ‚ ‚ Here we obtain the bounds ‚A(A + 1)−1 f (A)‚ ≤ (1 + M (A)) kf (A)k and Z |dz| 1 = kf kϕ M (A, ω ′ )M (A + 1, ω ′ )T (ω ′ ). kf kϕ M (A, ω ′ )M (A + 1, ω ′ ) 2π Γ |1 + z|2 (A + 1)

1 2πi

Z

Γ

Hence we obtain ˆ ˜ kf (A + 1)k ≤ (1 + M (A)) kf (A)k + (2 + M (A) + M (A, ω ′ ))M (A + 1, ω ′ )T (ω ′ ) kf kϕ .

We now let ω ′ → ϕ and apply f ) of Proposition 2.1.1) to obtain a constant C(ϕ) depending only on ϕ such that kf (A + 1)k ≤ (1 + M (A)) kf (A)k + [(2 + M (A) + M (A, ϕ))M (A, ϕ)C(ϕ)] kf kϕ . This proves the statement.

We will now prove Theorem 5.6.10. Assume A is sectorial and injective and θ > ωA . Choose ω < ϕ < min(θ, π) and suppose Ais ≤ Keθ|s| (s ∈ R). Given

5.7. Connections to Harmonic Analysis

111

ε > 0 and s ∈ R we apply Proposition 5.6.11 to the operator ε−1 A and the function f (z) = z is and obtain

−1

(ε A + 1)is ≤ K1 (ε−1 A)is + K2 z is ϕ

for all s ∈ R. A closer look on the shape of the constants K1 , K2 in Proposition 5.6.11 −is reveals

that

they do not depend on ε (due to i) of Proposition 2.1.1). Now ε = 1, z is = eϕ|s| , and (ε−1 A + 1)is = (ε−1 (A + ε))is = ε−is (A + ε)is ϕ

by a (trivial) application of the composition rule. Similarly, (ε−1 A)is = ε−is Ais . Altogether this yields

(A + ε)is ≤ K1 Ais + K2 eϕ|s| ≤ (K1 K + K2 )eθ|s| (s ∈ R) uniformly in ε ≥ 0, whence Theorem 5.6.10 is proved.

We state a corollary which summarizes our considerations. Corollary 5.6.12. (Pr¨ uss-Sohr II) Let A be sectorial with dense domain and dense range and let θ ≥ 0. If A ∈ BIP(X, θ) then A + ε ∈ BIP(X, θ) for all ε > 0. In fact, if θ > ωA , there is K ′ such that (A + ε)is ≤ K ′ eθ|s| (s ∈ R) uniformly in ε ≥ 0.

Remark 5.6.13. We have labelled the last corollary “Pr¨ uss-Sohr II”, since the state¨ss and Sohr from [110]. However, ment very much resembles a theorem of Pru this is only resemblance and not complete conformity. First, we do not have to assume neither θ > 0 nor θ < π, and in so far Corollary 5.6.12 is a generalization of the original result. On the other hand, “A ∈ BIP(X, θ)” in [110] means that there is K such that Ais ≤ Keθ|s| (s ∈ R).

(Compare our definition on page 74.) Therefore, in case one knows that Ais ≤ Keθ|s| with θ = ωA , our result Theorem 5.6.10 is weaker than the original Pr¨ uss-Sohr theorem.

5.7 Connections to Harmonic Analysis The next result, well known in Harmonic Analysis, is of great importance and shows that a bounded H0∞ -calculus implies a large amount of “unconditionality”. Lemma 5.7.1. Let ω < ϕ < π and f ∈ H0∞ (Sϕ ). Then there is a constant D > 0 such that the following holds. If A ∈ Sect(ω) such that the natural H0∞ (Sϕ )calculus for A is bounded with bound C, then

X

k ak f (t2 A) ≤ CD kak∞ (5.10)

k∈Z

for all t > 0 and all finite sequences a = (ak )k∈Z ⊂ C.

This result will be called Unconditionality Lemma in the sequel.

Chapter 5. The Boundedness of the H ∞ -calculus

112

|z|s

Proof. Since f ∈ H0∞ (Sϕ ) there is C ′ > 0 and s > 0 such that |f (z)| ≤ C ′ 1+|z|2s for all

z ∈ Sϕ . (See Lemma 2.2.2.) Let a = (ak )k∈Z be a finite sequence of complex numbers such that kak∞ ≤ 1 and let t > 0. We estimate ‚ ‚ ‚ ‚ ˛ ˛ ! ‚X ‚ ‚ X ‚ ˛X ˛ ‚ ˛ ˛ ‚ ‚ ‚ ak f (t2k A)‚ = ‚ ak f (t2k z) (A)‚ ≤ C sup ˛ ak f (t2k z)˛ ‚ ‚ ‚ ‚ ‚ ˛ ˛ z∈S ϕ k k k ˛ k ˛s ˛ ˛ ˛ X ˛˛ X 2 z ˛ ≤ C sup ˛f (2k z)˛ ≤ CC ′ sup ˛ ˛2s z∈Sϕ z∈Sϕ 1 + ˛2k z ˛ k

′

= CC sup

Thus with D :=

C′

sup1≤t≤2

P

t>0

X k

(2k t)s k 1+(2k t)2s

k

X (2k t)s (2k t)s = CC ′ sup k 2s k 2s 1 + (2 t) 1≤t≤2 k 1 + (2 t) the lemma is proved.

Remark 5.7.2. One can in fact show that an inequality (5.10) for certain functions f already implies the boundedness of the natural H0∞ -calculus. See [?] for a proof.

5.8 Comments

Chapter 6

Interpolation Spaces The present chapter is devoted to the connections between the theories of functional calculus and interpolation spaces. Section 6.1, kind of a “starter”, presents two central ideas. On the one hand we obtain a model how to describe the real interpolation spaces (X, D(A))θ,p using the functional calculus and on the other we prove a theorem of Dore which roughly says that an invertible sectorial operator A has bounded H ∞ calculus when it is restricted to its interpolation spaces. In Section 6.2 we examine the first of these ideas, proving several representation results for the spaces (X, D(Aα ))θ,p . Afterwards, in Section 6.3 we introduce extrapolation spaces for injective operators. With the help of these spaces we derive in Section 6.4 two important results (Theorem 6.4.3 and Theorem 6.4.6). These lead to some more characterizations of interpolation spaces by functional calculus (Section 6.5.1) and a generalization of Dore’s theorem (Section 6.5.3). Afterwards, in Section 6.6, we establish all the common properties of fractional domain spaces as intermediate spaces: density, the moment inequality, reiteration and a result by Komatsu (Theorem 6.6.8). Moreover, we prove the intrigueing fact that for an operator A ∈ BIP(X) the fractional domain spaces equal the complex interpolation spaces (Theorem 6.6.9).‚ In the end of the chapter we chacterize growth ‚ conditions like supt≥1 ‚tθ B(t + A)−1 ‚ < ∞ (which appeared in our considerations on perturbation) in terms of interpolation spaces (Section 6.7).

6.1 Real Interpolation Spaces In the following sections we assume the reader is familiar with the basic theory of real interpolation spaces. A short account and references can be found in Appendix B. For convenience we recall our notation Lp∗ ((a, b); X) to denote the Bochner space of (equivalence classes of) measurable functions f : (a, b) −→ X which are p-integrable with respect to the measure dt/t. Also, we define Γ ⊂ [0, 1] × [1, ∞] by ( θ ∈ (0, 1) ∧ 1 ≤ p < ∞ or (θ, p) ∈ Γ :⇐⇒ θ ∈ [0, 1] ∧ p = ∞. The present section should be viewed as an “appetizer” for a more deeper study of the connections of functional calculus and real interpolation spaces. Let A ∈ Sect(ω) on the Banach space X. In the following, we will use the

114

Chapter 6. Interpolation Spaces

common notation DA (θ, p) := (X, D(A))θ,p for (θ, p) ∈ Γ, with norm denoted by k·kθ,p . Using only elementary arguments (see [87, Prop.3.1.1]) one can derive the following description. Lemma 6.1.1. Let A ∈ Sect(ω) on the Banach space X. Then DA (θ, p) := {x ∈ X | tθ A(t + A)−1 x ∈ Lp∗ ((0, ∞); X)}

(6.1)

with equivalence of norms

kxkθ,p ∼ kxkX + tθ A(t + A)−1 x Lp ((0,∞);X) . ∗

If A is invertible, one can even drop the k·kX -part and has the norm equivalence

kxkθ,p ∼ tθ A(t + A)−1 x Lp ((0,∞);X) . ∗

We do not prove the lemma here, since it will follow from more general descriptions we give in the next section. However, we point out that the result is in fact a characterization of DA (θ, p) in terms of functional calculus. Namely, letting ψ(z) := z(1 + z)−1 after a change of parameter t 7→ t−1 , the description (6.1) becomes (X, D(A))θ,p = {x ∈ X | t−θ ψ(tA)x ∈ Lp∗ ((0, ∞); X)}. This gives the role model of our characterizations Theorems 6.2.1, 6.2.9 and 6.5.3 below. One of the central features of real interpolation spaces is that the operator A when restricted to DA (θ, p) improves some of its functional calculus properties. This heuristic statement is made precise by a result of Dore from [36] which we now want to prove. Let again A ∈ Sect(ω) on X. Clearly, the spaces DA (θ, p) are invariant under application of the resolvent of A, i.e., R(λ, A)DA (θ, p) ⊂ DA (θ, p) with kR(λ, A)xkθ,p ≤ kR(λ, A)kL(X) kxkθ,p

(6.2)

for all λ ∈ ̺(A), (θ, p) ∈ Γ, and x ∈ DA (θ, p). Restricting the operator A to the space DA (θ, p) therefore yields again a sectorial operator. The following proposition is stated for convenience. Its proof follows from Proposition 2.6.5. Proposition 6.1.2. Let A ∈ Sect(ω), 0 < θ < 1 and 1 ≤ p ≤ ∞. Denote by Aθ,p the part of A in DA (θ, p), i.e., (x, y) ∈ Aθ,p

⇐⇒

x ∈ D(A) and y = Ax ∈ DA (θ, p)

for all x, y ∈ X. Then the following assertions hold.

6.1. Real Interpolation Spaces a) ̺(A) ⊂ ̺(Aθ,p ) with

115

R(λ, Aθ,p ) = R(λ, A) D

A (θ,p)

for all λ ∈ ̺(A).

b) Aθ,p ∈ Sect(ω) in the Banach space DA (θ, p). c) If A is injective or invertible, so is Aθ,p .

d) If f ∈ M[Sω ]A , then f ∈ M[Sω ]Aθ,p with f (Aθ,p ) being the part of f (A) in DA (θ, p), i.e., ( x ∈ D(f (A)) ∩ DA (θ, p), (x, y) ∈ f (Aθ,p ) ⇐⇒ y = f (A)x ∈ DA (θ, p) for all x, y ∈ X. Now we can state Dore’s Theorem. Note that we know already from Section 5.5 that there are sectorial operators without a bounded H ∞ -calculus. Theorem 6.1.3. (Dore) Let A be an invertible sectorial operator on the Banach space X, and let θ ∈ (0, 1) and p ∈ [1, ∞. Then for each ϕ ∈ (ωA , π), the natural H ∞ (Sϕ )-calculus for Aθ,p is bounded. In particular, DA (θ, p) ⊂ D(f (A)) for all f ∈ H ∞ [Sω ]. Proof. We first treat the case p = ∞. Employing the resolvent identity and the second part of Lemma 6.1.1 one sees that for each ω ′ ∈ (ωA , π) there is a constant c(ω ′ ) such that ‚ ‚ ‚ ‚ sup ‚z θ AR(z, A)x‚ ≤ c(ω ′ ) kxkθ,∞ z∈∂Sω′

for all x ∈ DA (θ, ∞). Take ϕ ∈ (ωA , π), f ∈ H ∞ (Sϕ ) and x ∈ DA (θ, p). Then « „ Z Z f (z) 1 f (z) 1 f (z) (A)x = R(z, A)x dz = A−1 [z θ AR(z, A)x] dz 1+z 2πi Γ 1 + z 2πi Γ (1 + z)z θ

where Γ = ∂Sω ′ for some ω ′ ∈ (ω, ϕ). This shows that (f /(1 + z))(A)x ∈ D(A) whence x ∈ D(f (A)). Moreover, ‚Z ‚ ‚ ‚ ‚ 1 ‚ f (z)tθ θ ‚θ ‚ ‚ ‚ z AR(z, A)x dz ‚t A(t + A)−1 f (A)x‚ = ‚ X 2π ‚ Γ (t + z)z θ X Z 1 |f (z)| tθ |dz| ≤ c(ω ′ ) kxkθ,∞ 2π Γ |t + z| |z|θ Z 1 |f (tz)| |dz| = c(ω ′ ) kxkθ,∞ 2π Γ |1 + z| |z|θ Z 1 |dz| c(ω ′ ) kxkθ,∞ ≤ kf kϕ 2π Γ |1 + z| |z|θ

for each t > 0. Hence f (A)x ∈ DA (θ, ∞) with kf (A)xkθ,∞ ≤ C kf kϕ kxkθ,∞ for some constant C independent of x. Let us turn to the case 1 ≤ p < ∞. By the Reiteration Theorem B.4.2 one can view DA (θ, p) as a real interpolation space between DA (θ, ∞) and DA (β, ∞), where θ < β < 1. Combining this with Proposition 6.1.2 concludes the proof.

116

Chapter 6. Interpolation Spaces

Example 6.1.4. The hypothesis of invertibility can not be dropped from Theorem 6.1.3. Let A be an unbounded, invertible sectorial operator on a Hilbert space H such that the natural H ∞ -calculus for A is not bounded. Such an operator was constructed in Section 5.5. From the composition rule it is clear that A−1 cannot have a bounded H ∞ -calculus. Form the diagonal operator A 0 −1 A := diag(A, A ) := 0 A−1 on the space H ⊕ H. Then it is clear that DA (θ, p) = DA (θ, A) ⊕ H with the induced operator Aθ,p being the diagonal operator diag(Aθ,p , A−1 ). Since obviously f (A) = diag(f (A), f (A−1 )), A cannot have a bounded H ∞ -calculus on DA (θ, p) for any pair (θ, p). Corollary 6.1.5. Let A ∈ Sect(ω) such that D(Aα ) = DA (Re α, p) (with equivalent norms) for some 0 < Re α < 1 and some 1 ≤ p ≤ ∞. Then for each ε > 0 and each ω < ϕ ≤ π the natural H ∞ (Sϕ )-calculus for A + ε is bounded.

Proof. Since (A + ε)−α : X −→ D(Aα ) is an isomorphism, the assertion follows from Theorem 6.1.3.

We will prove below that the identity D(Aα ) = DA (α, p) for one α implies already that it holds for all α. Also we will give a much more general form of Dore’s Theorem, and a proof of it which does not rely on the Reiteration Theorem.

6.2 Characterizations In the present section we will use the functional calculus to give several descriptions of the real interpolation spaces associated with a sectorial operator A on a Banach space X. Given Re α > 0 the space D(Aα ) will be considered endowed with the norm kxkX + kAα xkX which turns it into a Banach space continuously embedded in X. Hence, the pair (D(Aα ), D(Aβ )) is an interpolation couple whenever α, β ∈ {z | Re z > 0} ∪ {0}. Note that, since Aα (A + 1)−α = (A(A + 1)−1 )α is bounded, one has D(Aα ) = D((A + 1)α ) not only as sets but also with equivalent norms. (Apply the closed graph theorem.) Therefore, in considerations involving only the Banach space D(Aα ) one can always assume that A is invertible. Given this, the homogeneous norm kxkAα := kAα kX is an equivalent norm on D(Aα ).

6.2.1 A First Characterization In this section we consider a description of the real interpolation spaces as follows. Theorem 6.2.1. Let A ∈ Sect(ω), ϕ ∈ (ω, π), Re α > 0. Assume one is given a function ψ ∈ O(Sϕ ) with the following properties: a) ψ, ψz −α ∈ E(Sϕ ). b) limz→0 ψ(z)z −α 6= 0.

6.2. Characterizations c) ψ(z) 6= 0 for all z ∈ Sϕ . Then if sup z∈Sϕ ,s≥1

one has the identity

117

ψ(sz) sα ψ(z) < ∞,

(6.3)

(X, D(Aα ))θ,p = {x ∈ X | t−θ Re α ψ(tA)x ∈ Lp∗ ((0, ∞); X)

(6.4)

with the equivalence of norms kxk(X,D(Aα ))θ,p

∼

for all (θ, p) ∈ Γ.

kxkX + t−θ Re α ψ(tA)x Lp ((0,∞);X) ∗

The assumptions on ψ may seem rather strong. This is due to the fact that the theorem also holds in the extremal case θ = 1, p = ∞. In Section 6.2.2 we weaken the assumtions on ψ while loosing this etremal case (cf. Theorem 6.2.9). Remarks 6.2.2. 1) By Proposition 2.6.11, the function t 7−→ ψ(tA) is bounded, whence t−θ Re α ψ(tA) ∈ Lp∗ ((1, ∞); L(X)) for all (θ, p) ∈ Γ. Hence one can replace (6.4) by (X, D(Aα ))θ,p = x ∈ X | t−θ Re α ψ(tA)x ∈ Lp∗ ((0, 1); X) (6.5) with the equivalence of norms kxk(X,D(Aα ))θ,p

∼

kxkX + t−θ Re α ψ(tA)x Lp ((0,1);X) ∗

for all (θ, p) ∈ Γ. 2) If A is invertible, one can discard the k·kX -part and has the simpler norm equivalence

kxk(X,D(Aα ))θ,p ∼ t−θ Re α ψ(tA)x Lp ((0,∞);X) . ∗

This is proved in Section 6.5.2, see Corollary 6.5.5.

Before we give the proof of Theorem 6.2.1, we formulate an important special case. Corollary 6.2.3. Let A ∈ Sect(ω), ϕ ∈ (ω, π), Re α > 0. Assume one is given a function ψ ∈ O(Sϕ ) which satisfies a) ψ, ψz −α ∈ E(Sϕ ). b) limz→0 ψ(z)z −α 6= 0.

c) ψ(z) 6= 0 for all z ∈ Sϕ .

d) ψ(∞) 6= 0. Then the conclusion of Theorem 6.2.1 holds.

118

Chapter 6. Interpolation Spaces

Proof. One only has to show that also (6.3) is satisfied, at least on a slightly smaller sector. Choose ϕ′ ∈ (ω, ϕ). Then ˛ ˛ ˛ ˛ ˛ ˛ ˛ ˛ ψ(sz) ˛ ˛ ψ(sz)(sz)−α ˛ ‚ −α ‚ ˛ 1 ˛ ˛ ˛ ˛ ‚ ‚ ˛ ˛ ˛ sα ψ(z) ˛ = ˛ ψ(z)z −α ˛ ≤ ψz ˛ ϕ ψ(z)z −α ˛ and

˛ ˛ ˛ ˛ ˛ ψ(sz) ˛ ˛ ˛ ˛ ˛ ≤ kψk ˛ 1 ˛ ϕ˛ ˛ sα ψ(z) ˛ ψ(z) ˛

independent of s ≥ 1. The first is bounded for z ∈ Sϕ′ , |z| ≤ 1, the second for z ∈ Sϕ′ , |z| ≥ 1.

Let us now turn to the proof of Theorem 6.2.1. We begin with noticing that one of the desired inclusions is easy, and in fact requires only condition a). Lemma 6.2.4. Let A ∈ Sect(ω), ϕ ∈ (ω, π), Re α > 0. Let ψ, ψ˜ := ψz −α ∈ E(Sϕ ). Then kψ(tA)xk ≤ C K(tRe α , x, X, D(Aα )) (x ∈ X, t > 0), where C = max(Cψ , Cψ˜ ), cf. Proposition 2.6.11. Consequently, there is a continuous inclusion (X, D(Aα ))θ,p ⊂ {x ∈ X | t−θ Re α ψ(tA)x ∈ Lp∗ ((0, ∞); X) for all (θ, p) ∈ Γ.

α b. This ˜ Proof. Let x = a + b where a ∈ X, b ∈ D(Aα ). Then ψ(tA)x = ψ(tA)a + tα ψ(tA)A yields kψ(tA)xk ≤ Cψ kakX + tRe α Cψ˜ kAα bkX ≤ C(kakX + tRe α kbkD(Aα ) ).

Taking the infimum gives the first statement. The second follows readily.

The other inclusion is not so easy to establish. In fact, we need several auxiliary results. Lemma 6.2.5. Let 0 6= ψ ∈ E(Sϕ ) and Re α > 0. Then there is f ∈ H0∞ (Sϕ ) such that Z ∞ ds (f ψ)(sz) = 1 (z ∈ Sϕ ) s 0

and z α f ∈ H0∞ (Sϕ ).

2 z Proof. Let ψ(z) := ψ(z) and τ (z) := (1+z) for all t > 0. Choose 2 . Then ψ(t)ψ(t) = |ψ(t)| m > Re α. Since 0 6= ψ, Z ∞ ds c := τ (s)m ψ(s)ψ(s) > 0. s 0

Then f := c−1 τ m ψ is a valid choice.

The next lemma states the classical Hardy-Young inequality (which is ubiquitous in interpolation theory). Lemma 6.2.6. (Hardy-Young Inequality) Let σ > 0, p ∈ [1, ∞] and f : (0, ∞) −→ [0, ∞).

6.2. Characterizations

119

a) If t−σ f ∈ Lp∗ (0, ∞) then f ∈ L1∗ (0, T ) for every T ∈ (0, ∞) and with Z t ds g(t) := f (s) s 0 one has t−σ g ∈ Lp∗ (0, ∞) and kt−σ gkLp∗ (0,∞) ≤

1 σ

kt−σ f kLp∗ (0,∞) .

b) If tσ f ∈ Lp∗ (0, ∞), then f ∈ L1∗ (T, ∞) for every T ∈ (0, ∞) and with Z ∞ ds g(t) := f (s) s t one has tσ g ∈ Lp∗ (0, ∞) and ktσ gkLp∗ (0,∞) ≤

1 σ

ktσ f kLp∗ (0,∞) .

Note that the constant appearing in the norm inequality is independent of the length of the interval (0, T ) and of the parameter p. Proof. The second assertion follows from the first by a change of parameter t 7→ t−1 . The first assertion can be proved easily by Riesz-Thorin interpolation since the case p = ∞ is trivial and the case p = 1 is just the Fubini theorem. (Only positive operators are involved here, so interpolation is elementary, cp. [51].) Of course one can also look into a book, e.g., in [58, p.245-246].

Returning to the proof of Theorem 6.2.1, we look for a possibility to write the constant 1 as a sum of appropriate functions in E. This will be furnished by the next lemma. Lemma 6.2.7. Let A ∈ Sect(ω), ϕ ∈ (ω, π), Re α > 0, and assume that ψ ∈ E(Sϕ ) satisfies all the hypotheses of Theorem 6.2.1. Then there are functions f ∈ H0∞ (Sϕ ), g ∈ E(Sϕ ) such that Z 1 ds (f ψ)(sz) + g(z)ψ(z)z −α = 1 (z ∈ Sϕ ). (6.6) s 0 Proof. We apply Lemma 6.2.5 to find a function f ∈ H0∞ such that f˜ := z α f ∈ H0∞ and R∞ R∞ ds ds 0 (f ψ)(sz) s = 1 for all z ∈ Sϕ . Let h(z) := 1 (f ψ)(sz) s . From Example 2.2.6 it follows ˛ h(z) that h ∈ E and h(0) = 1. Define g(z) := ψ(z)z −α . Since by assumption ψ(z)z −α ˛z=0 6= 0, g is “good” at 0. To see that g is also “good” at ∞, we use the crucial condition (6.3) and write ˛ ˛ ˛Z ˛ Z ∞˛ ˛ ds ˛ h(z) ˛ ˛ ∞ ψ(sz) ds ˛˛ ˛˜ ˛ α ˛=˛ |g(z)| = ˛˛ (sz) f (sz) ≤ C ˛f (sz)˛ ψ(z)z −α ˛ ˛ sα ψ(z) s ˛ s 1

the latter being good at ∞, cf. Example 2.2.6.

1

Now we take the last step. Proposition 6.2.8. Let A ∈ Sect(ω), ϕ ∈ (ω, π), Re α > 0, and assume that ψ, ψz −α ∈ E(Sϕ ). Assume furthermore that there are functions f, g ∈ E(Sϕ ) with (f ψ) ∈ H0∞ (Sϕ ) and Z 1 ds (f ψ)(sz) + g(z)ψ(z)z −α = 1 (z ∈ Sϕ ). s 0 Then the conclusion of Theorem 6.2.1 holds.

120

Chapter 6. Interpolation Spaces

Proof. As already noted, one inclusion is clear from Lemma 6.2.4. Let Z 1 ds h1 (z) := (f ψ)(sz) , and h2 (z) := g(z)ψ(z)z −α . s 0 Clearly, h1 , h2 ∈ E, cf. Example 2.2.6. Moreover, for every x ∈ X, t > 0 we have h2 (tA)x ∈ D(Aα ) with Aα h2 (tA)x = t−α g(tA)ψ(tA)x, hence kh2 (tA)xkD(Aα ) ≤ t− Re α Cg kψ(tA)xk+Ch2 kxk. (See Proposition 2.6.11 for the meaning of Cg . . . .) This yields K(tRe α , x, X, D(Aα )) ≤ kh1 (tA)xk + Cg kψ(tA)xk + tRe α Ch2 kxk .

Since also K(tRe α , x, X, D(Aα )) ≤ kxk, we can enlarge the constants a bit to obtain “ ” K(tRe α , x, X, D(Aα )) ≤ C kh1 (tA)xk + kψ(tA)xk + min(tRe α , 1) kxk .

(6.7)

Lp ∗ (0, ∞).

t−θ Re α

Now, take (θ, p) ∈ Γ, θ 6= 0 and x ∈ X such that kψ(tA)xk ∈ Then also t−θ Re α k(f ψ)(tA)xk ∈ Lp older’s inequality ∗ (0, ∞), since k(f ψ)(tA)xk ≤ Cf kψ(tA)xk. By H¨ and θ Re α > 0, one has [s 7→ (f ψ)(stA)x] ∈ L1 ∗ ((0, 1); X) for each t > 0. An application of Proposition 5.2.4, a) yields ‚Z 1 ‚ Z t ‚ ds ds ‚ ‚≤ k(f ψ)(sA)xk =: σ(t) kh1 (tA)xk = ‚ (f ψ)(stA)x ‚ s ‚ s 0 0

Here we employ the Hardy-Young inequality (Lemma 6.2.6) to obtain t−θ Re α σ(t) ∈ Lp ∗ (0, ∞) with ‚ ‚ ‚ ‚ 1 ‚ −θ Re α ‚ ‚ ‚ σ(t)‚ p ≤ Cf ‚t−θ Re α ψ(tA)x‚ p . ‚t L∗ L∗ (0,∞) θ Re α From (6.7) we can now infer that

t−θ Re α K(tRe α , x, X, D(Aα )) ∈ Lp ∗ (0, ∞) ‚ −θ Re α ‚ ‚t and estimate the Lp -norm of this function in terms of kxk and ψ(tA)x‚Lp . ∗ ∗

Combining Lemma 6.2.7 and Proposition 6.2.8 completes the proof of Theorem 6.2.1.

6.2.2 A Second Characterization In this section we formulate and prove a result which is more general than Theorem 6.2.1, at least when the interpolation parameter θ is restricted to the open interval (0, 1). Theorem 6.2.9. Let A ∈ Sect(ω), ϕ ∈ (ω, π) and Re α > 0. Take any function ψ ∈ O(Sϕ ) such that 0 6= ψ, ψz −α ∈ E(Sϕ ). Then one has (X, D(Aα ))θ,p = x ∈ X | t−θ Re α ψ(tA)x ∈ Lp∗ ((0, ∞); X) with the equivalence of norms kxk(X,D(Aα ))θ,p for all θ ∈ (0, 1), p ∈ [1, ∞].

∼

kxkX + t−θ Re α ψ(tA)x Lp ((0,∞);X) ∗

6.2. Characterizations

121

We will give an even more general result for the case of an injective operator A in Section 6.5.1 below. As before, it follows from Lemma 6.2.4 that we only have to care for the inclusion {x | t−θ Re α kψ(tA)xk ∈ Lp∗ (0, ∞)} ⊂ (X, D(Aα ))θ,p . The main ingredients of the proof are already known. There are just some further technicalities. Lemma 6.2.10. Let A ∈ Sect(ω), ϕ ∈ (ω, π), f ∈ H0∞ (Sϕ ) and Re α > 0. Define h(z) :=

Z

∞

s−α f (sz)

1

Then h ∈ H0∞ (Sϕ ) and h(A) =

Z

∞

s−α f (sA)

1

ds . s

ds . s

Proof. If one chooses 0 < ε < Re α small enough such that z ±ε f ∈ H ∞ , one also has z ±ε h ∈ H ∞ . This shows h ∈ H0∞ . The rest is simply Fubini’s theorem and the definition the H0∞ -functional calculus by the Cauchy integral. Note that the function s 7−→ s−α f (sA) is in L1∗ ((1, ∞); L(X)). We return to our main task. Proof of Theorem 6.2.9. As in the proof of Theorem 6.2.1 we try to write the constant function 1 as a sum of appropriate functions. Hence we start as before and choose a function f ∈ H0∞ (Sϕ ) such that f˜ := z α f ∈ H0∞ (Sϕ ) and Z ∞ ds (f ψ)(s) =1 s 0 (Lemma 6.2.5). Then we define Z 1 ds h1 (z) := (f ψ)(sz) s 0

and

h2 (z) :=

Z

1

∞

(f ψ)(sz)

ds . s

By Example 2.2.6, h1 , h2 ∈ E(Sϕ ) and h1 + h2 = 1. Moreover, Z ∞ Z ∞ ds ds z α h2 (z) = z α f (sz)ψ(sz) = s−α (f˜ψ)(sz) s s 1 1 is — by Lemma 6.2.10 — a function in H ∞ (Sϕ ). Moreover, Lemma 6.2.10 yields Z ∞ ds Aα h2 (tA) = t−α (z α h2 )(tA) = t−α s−α (f˜ψ)(sz) (tA) s 1 Z ∞ Z ∞ ds ds = (st)−α (f˜ψ)(stA) = s−α (f˜ψ)(sA) s s 1 t for all t > 0. In particular, h2 (tA) maps X into D(Aα ). Now, let θ ∈ (0, 1), p ∈ [1, ∞] and take x ∈ X such that t−θ Re α kψ(tA)xk ∈ Lp ∗ (0, ∞). By the above, x = h1 (tA)x + h2 (tA)x and h2 (tA)x ∈ D(Aα ), hence K(tRe α , x, X, D(Aα )) ≤ kh1 (tA)xk + tRe α kh2 (tA)xk + tRe α kAα h2 (tA)xk

122

Chapter 6. Interpolation Spaces

for t > 0. In the middle term we estimate kh2 (tA)xk ≤ Ch2 kxk and — enlarging the constant — arrive at the inequality K(tRe α , x, X, D(Aα )) ≤ » Z “ ” C kh1 (tA)xk + min tRe α , 1 kxk + tRe α

∞

t

s− Re α kψ(sA)xk

– ds . s

Since we want to have t−θ Re α K(tRe α , x, X, D(Aα )) ∈ Lp ∗ (0, ∞), the middle term is obviously good. The first term is dealt with exactly as in the proof of Proposition 6.2.8. For the third we define g(s) := s− Re α kψ(sA)xk and observe that s(1−θ) Re α g(s) = sθ Re α kψ(sA)xk ∈ Lp ∗ (0, ∞)

by assumption. Hence we can apply part b) of the Hardy-Young inequality (Lemma 6.2.6), and we are done.

Let us note the following corollary to the proof. Corollary 6.2.11. Let A ∈ Sect(ω), ϕ ∈ (ω, π), ψ ∈ E(Sϕ ) and Re α > 0. Then the implication Z 1 ds s− Re α kψ(sA)xk < ∞ =⇒ x ∈ D(Aα ) s 0 holds. Proof. Let f, f˜, h1 , h2 be as before. Then x = h1 (A)x + h2 (A)x and h2 (A)x ∈ D(Aα ). By R assumption and Proposition 5.2.4, h1 (A)x = 01 f (sA)ψ(sA)x ds . Since the operator D(Aα ) is s closed and Aα f (sA)ψ(sA)x = s−α f˜(sA)ψ(sA)x is in L1 ((0, 1); X), one has also h1 (A)x ∈ D(Aα ) ∗ and Z ∞ ds Aα x = f˜(sA)[s−α ψ(sA)x] . s 0

6.2.3 Examples In this section we apply Theorem 6.2.9 to obtain various descriptions of the real interpolation spaces. The Komatsu Spaces Let A be a sectorial operator on the Banach space X and let Re α > 0. We consider the function zα ψ(z) := (1 + z)α which, by Example 2.2.5, satisfies the hypotheses of Corollary 6.2.3. Applying that result yields, after a change of variable t 7→ t−1 , the description (X, D(Aα ))θ,p = x ∈ X tθ Re α [A(t + A)−1 ]α x ∈ Lp∗ (0, ∞)

for all (θ, p) ∈ Γ. If 0 < θ < 1, this description is connected with the name of Komatsu, see [67], [71] and compare also [89]. The case α = 1 is the description (X, D(A))θ,p = x ∈ X tθ [A(t + A)−1 ]x ∈ Lp∗ (0, ∞) which we started with in Lemma 6.1.1.

6.3. Extrapolation Spaces

123

Description by Holomorphic Semigroups Let A be a sectorial operator on the Banach space X with ωA < π2 . Choose ωA < ϕ < π2 . The function f (z) := e−z belongs to E(Sϕ ) and f (tA) = e−tA is the bounded holomorphic semigroup generated by A, see Section 3.3. Let Re α > 0. The function ψ(z) := z α e−z does not satisfy the conditions of Corollary 6.2.3. However, it satisfies the more general condition (6.3), since ψ(sz) e−sz = = e−(s−1)z , sα ψ(z) e−z and this is uniformly bounded for z ∈ Sϕ , s ≥ 1. Hence Theorem 6.2.1 applies and yields the description n o (X, D(Aα ))θ,p = x ∈ X t(1−θ) Re α Aα e−tA x ∈ Lp∗ (0, ∞) for all (θ, p) ∈ Γ.

Another Characterization Let again A be a sectorial operator on the Banach space X and consider the function zα ψ(z) := (1 + z)β where 0 < Re α < Re β. Then ψ satisfies the conditions of Theorem 6.2.9 (but not those of Theorem 6.2.1). Hence, after changing variables, we obtain n o (X, D(Aα ))θ,p = x ∈ X tRe β−(1−θ) Re α [Aα (t + A)−β ]x ∈ Lp∗ (0, ∞)

for all (θ, p) ∈ Γ. Specializing, e.g., α = 1, β = 2 yields

(X, D(A))θ,p = x ∈ X t1+θ A(t + A)−2 x ∈ Lp∗ (0, ∞) .

6.3 Extrapolation Spaces

From now on, we only consider injective sectorial operators. We will see in Theorem 6.5.3 that they allow a much more flexible description of interpolation spaces and we will shed new light on Dore’s theorem (Section 6.5.3). In order to do this we need what is called extrapolation spaces.

124

Chapter 6. Interpolation Spaces

6.3.1 An Abstract Method Suppose one is given a Banach space X and a bounded linear operator T ∈ L(X) which is

injective. The space R(T ) is then a Banach space endowed with the norm

T −1 x , and T : X −→ R(T ) is an isometric isomorphism. Now one considers X the commuting diagram id /X XO T

X

T

/ R(T )

and renames some of its components: X−1 O

T−1

/X

ι

X

T

/ R(T )

The map ι is bounded and injective, i.e., it am is embedding and after some settheoretical work, we can view X−1 as a proper superspace of X. We arrive at the commuting diagram X−1

X

T−1

T

/X

/ R(T )

T

/ R(T 2 ) which shows that T−1 X = T . Hence no confusion can arise in renaming T−1 by T . When we define X1 := R(T ), X2 := R(T 2 ) . . . we obtain R(T )

X−1

X

X1

T

/X

T

/ X1

T

/ X2

with T being an isometric isomorphism on every stage. Iterating this procedure yields an upwards directed series of new spaces (X−n )n∈N , i.e., / X−1 / X−n / X−2 / ... / ... X = X0

6.3. Extrapolation Spaces

125

where as before we can always view the embeddings as proper inclusions. Moreover, we have (compatible) isometric isomorphisms T : X−n −→ X−n−1 . As a last extension we finally construct the algebraic inductive limit of the directed family (X−n )n∈N [ U := X−∞ := limn∈N X−n = X−n . −→ n∈N

The space U may be called the universal extrapolation space corresponding to T . On U we define the following notion of convergence. Let (xα )α∈Λ ⊂ U be a net and x ∈ U . Then xα → x in U :⇐⇒ ∃n ∈ N, α0 ∈ Λ :

x, xα ∈ X−n (α ≥ α0 ) and

kxα − xkX−n → 0.

(This will not give a proper topology on U but is well adapted to our purposes, cf. Remark 6.3.1 below.) One easily sees that the limit of a net in U is unique, and that sum and scalar multiplication are “continuous” with respect to the so-defined notion of convergence. Finally, since the operator T is defined on each space X−n , n ∈ N, it is a forteriori defined on the whole of U . The so-defined mapping T : U −→ U is obviously surjective, whence it is an algebraic isomorphism, continuous with respect to the notion of convergence defined above. Remarks 6.3.1. 1) Our construction of extrapolation spaces can be generalized. Let X be a Banach space and T ⊂ L(X) a set of pairwise commuting, injective bounded linear operators on X, closed under multiplication. To each T ∈ T one can trivially construct a 1-step extrapolation space XT ⊃ X together with an extension of T to an isometric isomorphism T : XT −→ T . For two elements S, T ∈ T , XST can be regarded a common superspace of XT , XS , since ST = T S. Hence the set of spaces (XT )T ∈T is upward directed, and the inductive limit U := − lim →T XT is an extrapolation space, on which all operators T ∈ T become isomorphisms. 2) Our notion of convergence on the universal space U is adapted to our purposes but is very unlikely to be induced by a proper vector space topology on U . Of course one thinks of the inductive limit topology as an alternative, but apart from very special cases (e.g., where one has weakly compact embeddings X−n ֒→ X−(n+1) ) one can not guarantee that this topology is Hausdorff. This is so unpleasant a feature that we wanted to avoid it at any cost.

6.3.2 Extrapolation for Injective Sectorial Operators We now apply this abstract procedure to the case of an injective sectorial operator A on X. The operator T which is constitutive for the extrapolation method is defined by T := A(1 + A)−2 = (1 + A)−1 (1 + A−1 )−1 .

126

Chapter 6. Interpolation Spaces

Then T is injecitve, since A is. Moreover, R(T ) = D(A) ∩ R(A). By the abstract method described above, we obtain a sequence of nested spaces X = X0 ⊂ X−1 ⊂ X−2 ⊂ · · · ⊂ X−n ⊂ . . . U with U being the “universal” space. Recall that T extends to a “topological” isomorphism on U . Let us now extend the operator A to the whole of U . Using the isometric isomorphism T : X−1 −→ X, one can just transfer the operator A to X−1 by defining A−1 := T −1 AT with D(A−1 ) := T −1 D(A). By construction, A−1 is an injective sectorial operator on X−1 , isometrically similar to A. Clearly, X ⊂ D(A−1 ). Moreover, A is the part of A−1 in X, i.e., D(A) = {x ∈ X | A−1 x ∈ X} and A−1 x = Ax

(x ∈ D(A)).

(This is due to the fact that the operator T (considered as an operator on X) commutes with A in the sense that T A ⊂ AT .) The inverse of A−1 is given by [A−1 ]−1 = T −1 A−1 T with the appropriate domain D(A−1 −1 ) = R(A−1 ) = T −1 R(A). Iterating this procedure we obtain a sequence of isometrically similar sectorial operators A−n on X−n , with each A−n being the part of A−(n+1) in X−n . Since X−n ⊂ D(A−(n+1) ), by abuse of notation we obtain an extension of A to the whole of U which we again denote by A. This extension has the pleasant feature that it is invertible, i.e., A : U −→ U is an isomorphism, continuous with respect to the notion of convergence we introduced above. We now can define within U an array of spaces as follows. Already within X we have the following natural spaces: • D := D1 := D(A) with norm kxkD := k(1 + A)xk.

• R := R1 := R(A) with norm kxk := (1 + A−1 )x . R

• D ∩ R := D(A) ∩ R(A) with the norm

kxkD∩R = (2 + A + A−1 )x = T −1 x .

• D2 := D(A2 ) with the norm kxkD2 = (1 + A)2 x .

• R2 := R(A2 ) = D(A−2 ) with the norm kxkR2 = (1 + A−1 )2 x .

These can be arranged into a diagram.

XF FF xx FF x xx FF x x FF xx R DF FF x AAA } x } F x AA FF }} xx FF AA }} xx F }} xx D2 R2 D∩R

?

1+A

A

/

6.3. Extrapolation Spaces

D−1

127

X−1 GG GG ww w GG w GG ww w G ww

R−1 vv v T−1 vv vv vv u X III u II u uu II uu II u u I u D HH v R DDD y HH vv DD yy HH v v y HH v DD y v y H v D H yy vv D2 R2 D∩R HH HH HH HH H

Figure 8: The first extrapolation space.

Here a downward meeting of two lines means intersection and an upward meeting of two lines means sum of the spaces. E.g., X = D + R and D = D2 + (D ∩ R). The operator A + 1 acts as an isometric isomorphism in the ր-direction,

A+1 / D A+1 / X or D ∩ R A+1 / R . (One can replace A + 1 by A + λ for i.e. D2 each λ > 0, but then the isomorphisms seize to be isometric.) Furthermore, the A / R operator A acts as an isometric isomorphism in the →-direction, e.g., D

A / or D2 D ∩ R. Via the isometric ismorphism T : X−1 −→ X this diagram can be transported to X−1 . That is, we form the spaces

• D−1 := T −1 (D) with the norm kukD−1 = kT ukD and • R−1 := T −1 (R) with the norm kukR−1 = kT ukR and obtain a situation as shown in Figure 8. Note that we already know D−1 = D(A−1 ), R−1 = R(A−1 ). By applying T −1 again and again, we generate the spaces D−2 , D−3 . . . and R−2 , R−3 , . . . . Now, since A is an isomorphism on U , we define

X (n) := A−n (X) with norm kxkX (n) := A−n x X . We will frequently write

D(n) := X (n)

and R(n) := X (−n)

for n ∈ N, and D˙ := D(1) and R˙ := R(1) in the case n = 1. These spaces are called the homogeneous spaces associated with the injective sectorial operator A. Figure 9 on page 128 illustrates the situation. As before, T , A, (A + 1) and (A−1 + 1)

128

Chapter 6. Interpolation Spaces U =

X−2

D−2

R−2

...

/

...

D (2)

A

D−1

/

˙ D

...

X−1

...

R−1

A

/

X

D

A

/

R˙

A

/

R(2)

R

D∩R Figure 9: An array of extrapolation spaces.

act as isometric isomorphisms in the directions ↓, →, ր, տ, respectively. If D(A) is dense in X, then all inclusions in the ր-direction are dense. If R(A) is dense then all inclusions in the տ-direction are dense.

6.3.3 The Homogeneous Fractional Domain Spaces Having constructed the universal space U we look at the behaviour of the functional calculus. Assume A ∈ Sect(ω) and ϕ ∈ (ω, π]. Recall the definition of the function algebra n o s −s B(Sϕ ) := f ∈ O(Sϕ ) | ∃C, s > 0 : |f (z)| ≤ C max(|z| , |z| ) . Since A is injective, B(Sϕ ) ⊂ M(Sϕ )A in the notation of Chapter 2, and in fact f (A) is defined for f ∈ B(Sϕ ) in each of the spaces X−n . These definitions are consistent, i.e., f (A−(n+1) ) = f (A−n ) (n ∈ N). X−n

/

6.3. Extrapolation Spaces

129

By definition of the class B, for each f ∈ B(Sϕ ) there is m ∈ N such that τ m f ∈ z E(Sϕ )), where τ (z) = (1+z) 2 . This implies, that actually f (A) : X−n → X−(n+m)

is bounded for each n ∈ N. To sum up, one can say that f (A) is actually a continuous fully-defined operator on the whole of U . Thus, with a slight abuse of notation, we have turned our unbounded functional calculus into a proper algebra homomorphism (f 7−→ f (A)) : B(Sϕ ) −→ L(U ). The following lemma is important and straightforward to prove. Lemma 6.3.2. Let f ∈ B(Sϕ ). Then

D(f (A)) = {x ∈ X | f (A)x ∈ X},

i.e., the operator f (A) considered as an operator in X is the part in X of f (A) considered as an operator on U . Now note that the function z α is contained in B for each α ∈ C. The resulting operators Aα are isomorphisms on U , and in fact (α 7−→ Aα ) : C −→ L(U ) is a representation of the abelian group (C, +) on U . Consequently, we can generalize the definition of the homogeneous spaces: X (α) := A−α (X),

with norm

kxkX (α) = kAα xkX

for all α ∈ C. These are the homogeneous fractional domain spaces. Similar to the last section we write D(α) := X (α) and R(α) := X (−α) for Re α > 0. Obviously we have X (α) ⊂ X−1 whenever |Re α| < 1. Lemma 6.3.3. Let A be an injective sectorial operator on the Banach space X. Then for Re α > 0 one has a) D(α) ∩ X = D(Aα ) and kAα xk + kxk ∼ k(1 + A)α xk. b) R(α) ∩ X = R(Aα ) and kA−α xk + kxk ∼ k(1 + A)α A−α xk.

c) D(α) ∩ R(α) = D(Aα ) ∩ R(Aα ) with kAα xk + kA−α xk ∼ (1 + A)2α A−α x .

d) D(Aα ) + R(Aα ) = X.

Proof. a) The identity D (α) ∩ X = D(Aα ) follows from Lemma 6.3.2 and the equivalence of norms is just the well known fact D(Aα ) = D((A + 1)α ) (Proposition 3.1.9). b) is the same as a), with A replaced by A−1 . One has to use the identity (A + 1)α A−α = (1 + A−1 )α which follows from the composition rule. c) Let x ∈ D (α) ∩ R(α) . Then y := A−α x, Aα x = A2α x ∈ X, whence y ∈ D(A2α ). But as is well-known, A2α = Aα Aα even as operators in X, and so x = Aα y ∈ X. This shows D (α) ∩ R(α) ⊂ X and together with a) and b) this implies the stated set equality. From a) we have the norm equivalence ‚ 2α ‚ ‚ ‚ ‚A x‚ + kxk ∼ ‚(1 + A)2α x‚ (which holds for x ∈ D(A2α )). Now we replace x by A−α x, x ∈ D(Aα ) ∩ R(Aα ) and we are done. d) Take n > Re α and expand I = [A(1 + A)−1 + (1 + A)−1 ]2n .

130

Chapter 6. Interpolation Spaces

6.4 Homogeneous Interpolation We make the overall assumption that A ∈ Sect(ω) is an injective sectorial operator on the Banach space X. In this section we will characterize the real interpolation spaces (D(α) , R(β) )θ,p between the homogeneous fractional domain spaces by means of functional calculus. To achieve this we freely use the extrapolation space U constructed in Section 6.3.

6.4.1 Some Intermediate Spaces Fix 1 ≤ p ≤ ∞, θ ∈ R, and 0 6= ψ ∈ O[Sω ] with z −θ ψ ∈ H0∞ [Sω ]. Then we define Xθ,ψ,p := {x ∈ U | t−θ ψ(tA)x ∈ Lp∗ ((0, ∞); X)}. For x ∈ Xθ,ψ,p we let

kxkθ,ψ,p := t−θ ψ(tA)x Lp ∗

(Note that it can well be that x ∈ Xθ,ψ,p but x ∈ / X.)

Remark 6.4.1. This construction goes back to Auscher, McIntosh and Nahmod in the Hilbert space setting ([10, Section 4]). There, no extrapolation space was made explicit, and one can in fact work with completions since injective sectorial operators on Hilbert spaces are densely defined and have dense range. It was the absence of such density assumptions, which motivated our construction of the universal extrapolation space U . Proposition 6.4.2. Let 1 ≤ p ≤ ∞, θ ∈ R, and ψ as above. Then the following statements hold. a) The operator Aθ (defined on U ) induces an isometric isomorphism Aθ : Xθ,ψ,p −→ X0,z−θ ψ,p b) The space Xθ,ψ,p is continuously included in A−θ X−1 . c) The space Xθ,ψ,p is a Banach space. Proof. Assertion a) is clear from t−θ ψ(tA)x = (tA)−θ ψ(tA)Aθ x = (z −θ ψ)(tA)Aθ x for each x ∈ U . To prove b) it suffices to consider the case θ = 0 (by a)). Therefore ψ ∈ H0∞ , by assumption. Take x ∈ X0,ψ,p , i.e., x ∈ U such that ψ(tA)x ∈ Lp ∗ (X). Apply Lemma R 6.2.5 with α = 0 to find a function f ∈ H0∞ with 0∞ (f ψ)(t) dt = 1. By Theorem 5.2.2, the t function t 7→ τ (A)f (tA) is bounded and absolutely integrable in L(X). (We use the abbreviation ′

τ (z) := z(1 + z)−2 as usual.) In particular, by H¨ older’s inequality, it is contained in Lp ∗ (L(X)), ˜ where p is the conjugated exponent to p. Hence we obtain τ (A)(ψψ)(tA)x ∈ L1 ∗ (X) with Z ∞‚ ‚ dt ‚ ‚ ‚ ‚ ‚ ‚ ˜ ˜ ≤ kxkX0,ψ,p · ‚τ (A)ψ(tA) ‚τ (A)(ψψ)(tA)x ‚ ‚ p′ X t L∗ (L(X)) 0 R∞ dt Applying Proposition 5.2.4 we obtain 0 (f ψ)(tA)x t = x in U (or, if you wish, in some space X−m ). Consequently, τ (A)x ∈ X, i.e., x ∈ X−1 , with ‚ ‚ ‚ ‚ ˜ kxkX−1 = kτ (A)xkX ≤ kxkX0,ψ,p · ‚τ (A)ψ(tA) . ‚ p′ L∗ (L(X))

6.4. Homogeneous Interpolation

131

The proof of c) is now easy. Again, it suffices to consider the case θ = 0. Let (xn )n ⊂ X0,ψ,p be a Cauchy sequence. By b) there is an x ∈ X−1 with xn → x in X−1 . Then ψ(tA)xn → ψ(tA)x in X−1 uniformly in t. On the other hand there is f ∈ Lp ∗ (X) with ψ(tA)xn → f (t) in the p Lp ∗ (X)-norm, hence a forteriori in the L∗ (X−1 )-norm. This shows that f (t) = ψ(tA)x almost everywhere, whence we are done.

We can now prove the main result. Theorem 6.4.3. The spaces (Xθ,ψ,p , k.kθ,ψ,p ) are independent of the chosen ψ.

Theorem 6.4.3 allows us to discard ψ from the definition and to write simply Xθ,p instead of Xθ,ψ,p .

Proof. By a) of Proposition 6.4.2 it suffices to prove the theorem in the case θ = 0. We choose ϕ ∈ (ω, π), 0 6= ψ, γ ∈ H0∞ (Sϕ ), and x ∈ X0,γ,p . Then we apply Lemma 6.2.5 (with α = 0) to R find f ∈ H0∞ (Sϕ ) such that 0∞ (γf )(s) ds = 1. By Theorem 5.2.2, the numbers s Z ∞ dt E := sup kψ(sA)f (tA)k L(X) and t s>0 0 Z ∞ ds F := sup kψ(sA)f (tA)k L(X) s t>0 0 are both finite. Also, Proposition 5.2.4 shows that for each s > 0 Z ∞ dt ψ(sA)x = (γf )(tA)ψ(sA)x (6.8) t 0 as a convergent integral in some space X−m . Now, for s, t > 0 we have (γf )(tA)ψ(sA)x = [ψ(sA)f (tA)] [γ(tA)x],

and considered as a product of functions in t this is integrable within X, since ′

p 1 ψ(sA)f (tA) ∈ L∞ ∗ (L(X)) ∩ L∗ (L(X)) ⊂ L∗ (L(X))

γ(tA)x ∈

Lp ∗ (X)

by Proposition 5.2.4 and the choice of x. Hence we actually have Z ∞ dt ∈ X with ψ(sA)x = (γf )(tA)ψ(sA)x t 0 Z ∞ dt kψ(sA)xkX ≤ kψ(sA)f (tA) γ(tA)xkX ≤ kψ(sA)f (·A)k p′ kxk0,γ,p . L∗ (L(X)) t 0 Using (6.8) together with H¨ older’s inequality we see that the function h(s) := (s 7−→ ψ(sA)x) : (0, ∞) −→ X is bounded. Moreover, similar estimates show that the continuous(!) functions »Z b – Z b dt dt ha,b (s) := (f γ)(tA)ψ(sA)x = ψ(sA) f (tA)γ(tA)x t t a a converge uniformly to h as a ց 0, b ր ∞. Hence the function h is in fact continuous. In particular, it is measurable. Now, in case p = ∞ we are already done, obtaining x ∈ X0,ψ,∞ with In case p < ∞ we compute Z kψ(sA)xkX ≤

kxk0,ψ,∞ ≤ E kxk0,γ,∞ .

∞

0

≤

„Z

≤E

∞ 0

1 p′

kψ(sA)f (tA)k

„Z

0

∞

dt t

kψ(sA)f (tA)γ(tA)xk dt t

«

1 p′

„Z

0

∞

kψ(sA)f (tA)k kγ(tA)xkp

kψ(sA)f (tA)k kγ(tA)xkp

dt t

«1

p

.

dt t

«1

p

132

Chapter 6. Interpolation Spaces

The last factor considered as a function in s lies in Lp ∗ (0, ∞), since Z ∞Z ∞ dt ds kψ(sA)f (tA)k kγ(tA)xkp t s 0 0 Z ∞Z ∞ ds dt kψ(sA)f (tA)k = kγ(tA)xkp s t 0 0 „ «Z ∞ Z ∞ ds dt ≤ sup kψ(sA)f (tA)k kγ(tA)xkp = F kxkp0,γ,p . s t t>0 0 0 Therefore we end up with 1

kxk0,ψ,p ≤ E p′ F

and the theorem is completely proved.

1 p

kxk0,γ,p

(6.9)

In the following we will simply write Xθ,p instead of Xθ,ψ,p . Note that one always has the (continuous) inclusions Xθ,1 ⊂ X (θ) ⊂ Xθ,∞ . Namely, from Proposition 5.2.4 the inclusion Xθ,1 ⊂ Xθ,p is immediate. The proof of Theorem 6.4.3 also shows that Xθ,p ⊂ Xθ,∞ . In fact it was proved that if x ∈ Xθ,ψ,p , the function (t 7→ t−θ ψ(tA)x) : (0, ∞) −→ X is actually continuous and uniformly bounded. Corollary 6.4.4. The operators Ais , s ∈ R act as topological isomorphisms on each of the spaces Xθ,p , θ ∈ R, p ∈ [1, ∞].

Proof. It suffices to prove the statement for θ = 0. But if 0 6= ψ ∈ H0∞ (Sϕ ) also 0 6= z is ψ ∈ H0∞ (Sϕ ) and since ψ(tA)Ais x = t−is (z is ψ)(tA)x ˛ −is ˛ ˛ ˛ and t ≤ 1 we have x ∈ X0,(z is ψ),p if and only if Ais x ∈ X0,ψ,p .

6.4.2 . . . are actually Real Interpolation Spaces

We are now going to show that the spaces Xθ,p are nothing than real interpolation spaces between homogeneous fractional domain spaces. Proposition 6.4.5. Let Re α, Re β > 0 and define θ :=

Re α Re α+Re β

∈ (0, 1). Then

(D(α) , R(β) )θ,p = X0,p with equivalent norms, for each p ∈ [1, ∞].

Proof. Choose ϕ ∈ (ωA , π) and ψ ∈ H0∞ (Sϕ ) with z −α ψ, z β ψ ∈ E(Sϕ ). For x ∈ U , x = a + b with a ∈ D (α) , b ∈ R(β) we have ψ(tA)x = tα (tA)−α ψ(tA)Aα a + tβ (tA)−β ψ(tA)A−β b

which yields

kψ(tA)xkX ≤ tRe α M1 kakD (α) + t− Re β M2 kbkR(β) for all t > 0 and some constants M1 , M2 . Taking the infimum with respect to the representations x = a + b yields kψ(tA)xk ≤ M tRe α K(t−(Re α+Re β) , x, D (α) , R(β) )

(t > 0)

6.4. Homogeneous Interpolation

133

for some constant M . This proves the continuous inclusion (D (α) , R(β) )θ,p ⊂ X0,p . R For the reverse inclusion choose a function ψ ∈ H0∞ (Sϕ ) with 0∞ ψ(t) dt = 1 and define t h(z) :=

Z

1

ψ(tz)

0

dt t

g(z) :=

Z

∞

1

ψ(tz)

dt t

as in Example 2.2.6 and Proposition 5.2.4. One can assume without restriction that the decay of ψ at 0 and at ∞ is fast enough to ensure ψ1 := z α g, ψ2 := z −β h ∈ H0∞ (Sϕ ). Take x ∈ Xp and write x = g(tA)x + h(tA)x (t > 0) Now observe that g(tA)x ∈ D (α) , since Aα g(tA)x = t−α (tA)α g(tA)x = t−α ψ1 (tA)x ∈ X Analogously, h(tA)x ∈ R(β) with A−β h(tA)x = tβ (tA)−β h(tA)x = tβ ψ2 (tA)x. Therefore, for each s > 0 one obtains K(s, x, D (α) , R(β) ) ≤ t− Re α kψ1 (tA)xkX + stRe β kψ2 (tA)xkX . Letting s := t−(Re α+Re β) yields tRe α K(t−(Re α+Re β) , x, D (α) , R(β) ) ≤ kψ1 (tA)xk + kψ2 (tA)xk The right hand side (as a function of t) is in

Lp ∗.

(t > 0).

This finishes the proof.

We remark that in the proof we made essential use of Theorem 6.4.3. Namely, it was important that one can choose every H0∞ -function to describe the space X0,p . With the help of Proposition 6.4.5 we are now able to show that each real interpolation space between homogeneous spaces is in fact a space Xθ,p . Theorem 6.4.6. Let A be an injective, sectorial operator on the Banach space X and let α, β ∈ C with Re α 6= Re β. Then, for all θ ∈ (0, 1), p ∈ [1, ∞], X (α) , X (β)

θ,p

= X(1−θ) Re α+θ Re β,p

with equivalent norms. Proof. Without restriction we can assume Re α > Re β (replace θ by 1 − θ if necessary). Define δ := (1 − θ)α + θβ. Then α′ := α − δ = θ(α − β) and β ′ := δ − β = (1 − θ)(α − β). Hence α′ Re α′ , Re β ′ > 0 and θ = Re αRe ′ +Re β ′ . By Proposition 6.4.5 we have “

X (α−δ) , X (β−δ)

”

θ,p

“ ” ′ ′ = X (α ) , X (−β )

θ,p

“ ” ′ ′ = D (α ) , R(β )

θ,p

= X0,p .

Applying A−δ = A− Re δ A−i Im δ to this identity finishes the proof. (Note that by Corollary 6.4.4 the operator A−i Im δ is an isomorphism on X0,p .)

Corollary 6.4.7. Let θ ∈ (0, 1), p ∈ [1, ∞]. Then ˙ R) ˙ θ,p = X1−2θ,p . (D,

134

Chapter 6. Interpolation Spaces

6.5 More Characterizations and Dore’s Theorem In this section we revisit Dore’s theorem and give evidence to the heuristic statement that the functional calculus properties of an injective sectorial operator A improve in its interpolation spaces. First however, we complement the characterizations of the interpolation spaces obtained in Section 6.2.

6.5.1 A Third Characterization (Injective Operators) With Theorem 6.4.6 we have established a powerful description of real interpolation spaces in terms of functional calculus. However, the space (X (α) , X (β) )θ,p are quite uncommon and we look for spaces which are naturally included in the original space X. Lemma 6.5.1. Let A be an injective sectorial operator on the Banach space X. Then (X, D(Aα ) ∩ R(Aα ))θ,p = (X, D(α) )θ,p ∩ (X, R(α) )θ,p , (X, D(Aα ))θ,p = (X, D(α) )θ,p ∩ X,

(D(Aα ), R(Aα ))θ,p = Aα (1 + A)−α (D(α) , X)θ,p for all Re α > 0, θ ∈ [0, 1], p ∈ [1∞].

Proof. The last assertion follows from the fact that isomorphism A−α (1 + A)α sends D(Aα ) to D (α) and R(Aα ) to X, see Lemma 6.3.3. For the first two assertions we use Proposition B.2.7 from the Appendix. Namely, the second assertion follows directly from (B.4) and D (α) ∩ X = D(Aα ) (see Lemma 6.3.3). To prove the first, let us denote for the moment Y := D (α) , Z := R(α) . Then we have Y ∩ Z = D(α) ∩ R(α) = D(Aα ) ∩ R(Aα ) ⊂ X and (X ∩ Y ) + (X ∩ Z) = (D(α) ∩ X) + (R(α) ∩ X) = D(Aα ) + R(Aα ) = X by Lemma 6.3.3. Hence we obtain (X, Y )θ,p ∩ (X, Z)θ,p ⊂ (X + Y ) ∩ (X + Z) = X. This yields ˆ ˜ ˆ ˜ (X, Y )θ,p ∩ (X, Z)θ,p = (X, Y )θ,p ∩ X ∩ (X, Z)θ,p ∩ X = (X, X ∩ Y )θ,p ∩ (X, X ∩ Z)θ,p = (X, X ∩ Y ∩ X ∩ Z)θ,p = (X, Y ∩ Z)θ,p

which proves the result.

Remark 6.5.2. In the first statement of the last lemma one writes “vertical” interpolation spaces as intersections of “horizontal” ones. As a matter of fact, one can also go the other way round. E.g., one has the equality (D, R)θ,p = (D−1 , D)1−θ,p ∩ (R−1 , R)θ,p which can be deduced from an interpolation identity which is in the spirit of Proposition B.2.7 (see [53, Theorem 1]). Combining Lemma 6.5.1 with Theorem 6.4.6 we arrive at a third representation theorem.

6.5. More Characterizations and Dore’s Theorem

135

Theorem 6.5.3. Let A be an injective sectorial operator on a Banach space X, Let ϕ ∈ (ωA , π), Re α > 0, θ ∈ (0, 1), and p ∈ [1, ∞]. a) One has (X, D(Aα ))θ,p = x ∈ X | t−θ Re α ψ(tA)x ∈ Lp∗ ((0, ∞); X) with equivalence of norms

kxk(X,D(Aα ))θ,p ∼ kxkX + t−θ Re α ψ(tA)x Lp ((0,∞);X) ∗

whenever 0 6= ψ ∈ O(Sϕ ) such that z

b) One has

−θα

ψ∈

H0∞ (Sϕ ).

(X, D(Aα )∩R(Aα ))θ,p = x ∈ X | t−θ Re α ψ1 (tA)x, tθ Re α ψ2 (tA)x ∈ Lp∗ ((0, ∞); X)

with equivalence of norms

kxk(X,D(Aα )∩R(Aα ))θ,p ∼ t−θ Re α ψ1 (tA)x Lp + tθ Re α ψ2 (tA)x Lp ∗

∗

whenever 0 6= ψ1 , ψ2 ∈ O(Sϕ ) such that z

−θα

ψ1 , z

θ Re α

ψ2 ∈

H0∞ (Sϕ ).

` ´ Proof. a). Let us specialize β = 0 and Re α > 0 in Theorem 6.4.6. Thus we obtain X, X (α) θ,p = Xθ Re α,p . Now we intersect both sides with X and employing Lemma 6.5.1 we arrive at (X, D(Aα ))θ,p = Xθ Re α,p ∩ X. The rest follows from Theorem 6.4.3. The proof of b) is similar.

When A is even invertible, some simplifactions apply. This is the content of the next section.

6.5.2 A Fourth Characterization (Invertible Operators) Let us specialize the given characterizations to the case of an invertible sectorial operator A. If A is invertible, the array of extrapolation spaces becomes “halftrivial”, since we obtain D(α) = D(Aα ),

R(Aα ) = X

for all Re α > 0. Moreover, D−1 = X and R−1 = X−1 . Consequently, one has (X, D(Aα ))θ,p = Xθ Re α,p by Theorem 6.4.6. Comparing this with the characterization Theorem 6.5.3, part a), this means that in case A is invertible, one can leave out the k·kX -part and simply has a norm equivalence

kxk(X,D(Aα ))θ,p ∼ t−θ Re α ψ(tA)x Lp ((0,∞);X) . ∗

The next proposition reproves this fact and moreover shows that the statement holds also in the extremal case θ = 1, p = ∞ when ψ is chosen as in Theorem 6.2.9.

136

Chapter 6. Interpolation Spaces

Proposition 6.5.4. Let A be an invertible sectorial operator on X, ϕ ∈ (ωA , π) and 0 6= ψ ∈ B(Sϕ ). Given Re α > 0, θ ∈ (0, 1], p ∈ [1, ∞] there is a constant c such that

kxkX ≤ c t−θ Re α ψ(tA)x Lp ((0,∞);X) ∗

for all x ∈ X.

Note that ψ(tA)x ∈ U is always defined but the right-hand side maybe infinite, in which case the inequality holds trivially. Proof. Without harm we can replace α by θα and assume θ = 1. Choose f ∈ H0∞ (Sϕ ) such R that f1 := z α+1 f ∈ H0∞ , ψf ∈ H0∞ and 0∞ (f ψ)(t) dt = 1 (Lemma 6.2.5). Define as usual t h(z) :=

Z

1

(f ψ)(tz)

0

dt , t

and

g(z) :=

Z

∞

(f ψ)(tz)

1

dt . t

Then h + g = 1 (Example 2.2.6). Take x ∈ X with t−θ Re α ψ(tA)x ∈ Lp ∗ ((0, ∞); X). Then (f ψ)(tA)x = A−(α+1) [t−1 f1 (tA)][t−α ψ(tA)x] ∈ L1 ∗ ((1, ∞); X) since by assumption the last factor is in Lp ∗ , the first factor is bounded and the second factor in ′ Lp ∗ . Hence by Proposition 5.2.4 we obtain ‚ ‚Z ∞ ‚ ‚ ‚ dt ‚ ‚ ‚ ≤ c1 ‚ kg(A)xk = ‚ (f ψ)(tA)x ‚t− Re α ψ(tA)x‚ p ‚ ‚ L∗ ((1,∞);X) t 1 ‚ −1 ‚ ‚ −(α+1) ‚ ‚ ‚ ‚ ‚ with c1 = t A supt>1 kf1 (tA)k. On the other hand we have p′ L∗ (1,∞)

(f ψ)(tA)x = [tα f (tA)][t−α ψ(tA)x] ∈ L1 ∗ ((0, 1); X) ′

since the first factor is in Lp ∗ ((0, 1), X). Again by Proposition 5.2.4 we obtain ‚Z 1 ‚ ‚ ‚ ‚ dt ‚ ‚ ‚ ≤ c2 ‚ kh(A)xk = ‚ (f ψ)(tA) ‚t− Re α ψ(tA)x‚ p ‚ ‚ L∗ ((0,1);X) t 0 ‚ Re α ‚ ‚ ‚ with c2 = t supt<1 kf (tA)k. Thus, the proof is complete. p′ L∗ (0,1)

Corollary 6.5.5. Let A be an invertible sectorial operator on a Banach space X. Then one has equivalence of norms

kxk ∼ t−θ Re α ψ(tA)x p α (X,D(A ))θ,p

L∗ ((0,∞);X)

in Theorem 6.2.1 and Theorem 6.5.3.

6.5.3 Dore’s Theorem Revisited Dore’s theorem (Theorem 6.1.3) says that an invertible operator has a bounded H ∞ -calculus in each of its real interpolation spaces (X, D(A))θ,p , whenever θ ∈ (0, 1). We have seen in Example 6.1.4 that the assumption of invertibility cannot be dropped. This phenomenon can now be fully understood, since by the results of the last section we have the following theorem.

6.6. Fractional Powers as Intermediate Spaces

137

Theorem 6.5.6. Let A be an injective sectorial operator on the Banach space X, and let α, β ∈ C with Re α 6= Re β. Then for each ϕ ∈ (ωA , π), the operator A has a bounded H ∞ (Sϕ )-calculus in each of the spaces X (α) , X (β) θ,p

(θ ∈ (0, 1), p ∈ [1, ∞]).

Proof. Employing Theorem 6.4.6 and Proposition 6.4.2 we are reduced to the spaces X0,p . Let x ∈ X0,p and choose two functions 0 6= ψ, γ ∈ H0∞ (Sϕ ). By Theorem 6.4.3 we have X0,p = X0,γ,p = X0,(ψγ),p . Now (ψγ)(tA)f (A)x = (ψt f )(A)γ(tA)x for all t > 0. But supt>0 k(f ψt )(A)kL(X) ≤ C(ψ)M (A, ϕ) kf kSϕ by Theorem 5.2.2.

We obtain an immediate corollary. Corollary 6.5.7. Let A be an injective sectorial operator on the Banach space X, and let ϕ ∈ (ωA , π). Then A has a bounded H ∞ (Sϕ )-calculus in each of the spaces (D(Aα ), R(Aα ))θ,p , (Re α > 0, θ ∈ (0, 1), p ∈ [1, ∞]). Proof. By Lemma 6.5.1, A−α (1 + A)α : (D(Aα ), R(Aα ))θ,p −→ (X (α) , X (0) )θ,p is an isomorphism which commutes with A. Now the assertion follows from Theorem 6.5.6.

We now apply Lemma 6.5.1 again to pass from the “horizontal” interpolation spaces to the “vertical ones”. Corollary 6.5.8. Let A be an injective sectorial operator on the Banach space X, and let ϕ ∈ (ωA , π). Then A has a bounded H ∞ (Sϕ )-calculus in each of the spaces (X, D(Aα ) ∩ R(Aα ))θ,p , (Re α > 0, θ ∈ (0, 1), p ∈ [1, ∞]). Assuming A to be invertible and taking α = 1 gives back Dore’s Theorem 6.1.3. Note that this new proof does not use the Reiteration Theorem.

6.6 Fractional Powers as Intermediate Spaces In this section we examine the domains D(Aα ) as intermediate spaces. Here A is any sectorial operator on a Banach space X, not necessarily injective.

6.6.1 Density of Fractional Domain Spaces If A is a sectorial operators A with non-dense domain, then for no pair α, β with 0 < Re α < Re β the space D(Aβ ) will be dense in D(Aα ). However, within the real interpolation spaces we have a quite different behaviour. Theorem 6.6.1. Let A be a sectorial operator on the Banach space X and let 0 < Re α < Re β, θ ∈ (0, 1), p ∈ [1, ∞). Then the space D(Aβ ) is dense in (X, D(Aα ))θ,p .

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Chapter 6. Interpolation Spaces

Proof. One can assume without restriction that A is invertible. Choose ϕ ∈ (ωA , π) and 0 6= ψ ∈ O(Sϕ ) such that z −θα ψ ∈ H0∞ (Sϕ ). Now, choose 0 6= τ ∈ H0∞ (Sϕ ) such that (τ ψ) ∈ H0∞ (Sϕ ). Then also z −θα (τ ψ) ∈ H0∞ (Sϕ ) and both ψ, τ ψ can be used to describe (X, D(Aα ))θ,p (see Corollary 6.5.5). We choose f ∈ H0∞ (Sϕ ) in such a way that f ψ, z β f ψ ∈ H0∞ (Sϕ ) and R∞ ds 0 (f ψ)(s) s = 1 (Lemma 6.2.5). Define Z ∞ ds gn (z) := (f ψ)(sz) . 1 s n Then gn ∈ E(Sϕ ) as was shown in Example 2.2.6. We will show (i) gn (A)(X) ⊂ D(Aβ ) and (ii) gn (A)x → x within (X, D(Aα ))θ,p for x ∈ (X, D(Aα ))θ,p . Claim (i) follows from Z ∞ ds z β gn (z) = (z β f ψ)(sz)s−β 1 s n which is a function in H0∞ (Sϕ ). To prove Claim (ii) we take x ∈ (X, D(Aα ))θ,p . We have to show t−θ Re α (τ ψ)(tA)gn (A)x → t−θ Re α (τ ψ)(tA)x

in Lp ∗ ((0, ∞); X). The Convergence Lemma (Prop. 5.1.3, see also c) of Proposition 5.2.4) shows that the convergence is at least pointwise. Employing the Dominated Convergence Theorem we have to prove domination. Now Z ∞ ds t−θ Re α (τ ψ)(tA)gn (A)x = (f ψ)(sz)τ (tz) (A) t−θ Re α ψ(tA)x 1 s n

and, by Proposition 5.2.4 and Theorem 5.2.2, ‚Z ‚ ‚Z ‚ ‚ ∞ ‚ ‚ ∞ ds ds ‚ ‚ ‚ ‚ ‚ sup ‚ (f ψ)(sz)τ (tz) (A)‚ = sup ‚ (f ψ)(sA)τ (tA) ‚ ‚ t>0 ‚ 1 s s ‚ t>0 ‚ 1 n n Z ∞ ds ≤ sup k(f ψ)(sA)τ (tA)k =: c < ∞. s t>0 0 ‚ ‚ ‚ ‚ This yields ‚t−θ Re α (τ ψ)(tA)gn (A)x‚X ≤ c ‚t−θ Re α ψ(tA)x‚X which is in Lp ∗ (0, ∞) by assumption.

6.6.2 The Moment Inequality We first establish, that — in the language of interpolation theory — the space D(Aβ ) is always of class Jθ ∩Kθ between X and D(Aα ), provided 0 < Re β < Re α and θ is defined as θ := Re β/ Re α. The following result is of auxiliary nature. Proposition 6.6.2. Let A ∈ Sect(ω), ϕ ∈ (ω, π) and ψ ∈ E(Sϕ ). Then the following assertions hold. a) Let Re α > 0 and assume that also z −α ψ ∈ E(Sϕ ). Then

x ∈ D(Aβ ) =⇒ sup t− Re β ψ(tA)x < ∞. t>0

for every β with Re β ∈ (0, Re α) or β ∈ {0, α}. Z 1 dt b) Let Re α > 0. Then t− Re α kψ(tA)xkX <∞ t 0

X

=⇒

x ∈ D(Aα ).

6.6. Fractional Powers as Intermediate Spaces

139

Proof. Assume as in a) that x ∈ D(Aβ ) where β is in the given range. From the hypotheses it β x, whence the assertion follows from ˜ is clear that ψ˜ := ψz −β ∈ E. Now, ψ(tA)x = tβ ψ(tA)A Proposition 2.6.11. R To prove b), choose δ > Re α and f ∈ H0∞ such that z δ f ∈ H0∞ and 0∞ (f ψ)(t) dt = 1, t cf. Lemma 6.2.5. Define Z 1 Z ∞ dt dt h(z) := (f ψ)(tz) and g(z) := (f ψ)(tz) t t 0 1 as in Example 2.2.6. Then 1 = h + g. Moreover, z α g ∈ H0∞ , since gz δ is still a bounded R function. Therefore g(A)X ⊂ D(Aα ). Now take x ∈ X such that 01 t− Re α kψ(tA)xk dt < ∞. t α ∞ ˜ Let f := z f ∈ H0 . Then clearly also Z 1 Z 1‚ ‚ dt dt ‚˜ ‚ kf (tA)ψ(tA)xk < ∞ and < ∞. ‚f (tA)t−α ψ(tA)x‚ t t 0 0 Proposition 5.2.4 implies that in fact

h(A)x =

Z

1

f (tA)ψ(tA)x 0

dt , t

and since f (tA)ψ(tA)x ∈ D(Aα ) for all t ∈ (0, 1) and the operator Aα is closed, we have h(A)x ∈ D(Aα ) with Z 1 dt Aα h(A)x = t−α f˜(tA)ψ(tA)x . t 0 Thus we arrive at x = h(A)x + g(A)x ∈ D(Aα ).

Specializing ψ to a function which describes the real interpolation spaces as in Theorem 6.2.9, yields the following. Corollary 6.6.3. Let Re γ < Re β < Re α with Re γ > 0 or γ = 0. Then (D(Aγ ), D(Aα ))θ,1 ⊂ D(Aβ ) ⊂ (D(Aγ , D(Aα ))θ,∞

(6.10)

where θ ∈ (0, 1) is such that Re β = (1 − θ) Re γ + θ Re α.

Proof. Since (1+A)−γ : X −→ D(Aγ ) which carries D(Aβ−γ ) to D(Aβ ) and D(Aα−γ ) to D(Aα ) we can assume γ = 0 without restriction. Then the statement follows from the characterization Theorem 6.2.9 by taking an appropriate function ψ.

It is well known that the left hand inclusion in (6.10) implies (and is in fact

1−θ θ equivalent to) an inequality of the form Aβ x ≤ C kAγ xk kAα xk . The usual interpolation-theoretic proof requires a description of the interpolation spaces different from the K-method, cp. [14, Thm.3.5.2] or [86, Thm 1.2.15]. However, our methods allow also a direct proof. Proposition 6.6.4. (Moment Inequality) Let A be a sectorial operator on the Banach space X. Let α, β, γ ∈ C such that Re γ < Re β < Re α and Re γ > 0 or γ = 0. Then there is a constant C such that

β

A x ≤

C kAγ xk1−θ kAα xkθ θ(1 − θ)

where θ is defined as θ :=

Re β−Re γ Re α−Re γ .

(x ∈ D(Aα )),

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Chapter 6. Interpolation Spaces

Proof. Choose any ψ ∈ H0∞ such that ψz α , ψz −α are still bounded functions. Define h(z) := R1 R∞ ds ds −α h and z α g are bounded func0 ψ(sz) s , g(z) := 1 ψ(sz) s as in Example 2.2.6. Then z −(α−β) β−γ ∞ ˆ tions, whence h := z h and gˆ := z g are both in H0 . For x ∈ D(Aα ) we have α ˆ Aβ x = h(tA)Aβ x + g(tA)Aβ x = tα−β h(tA)A x + t−(β−γ) gˆ(tA)x.

This yields ‚ ‚ ‚ β ‚ α − Re αθ Cgˆ kAγ xk ‚A x‚ ≤ tRe α(1−θ Ch ˆ kA xk + t

(see Proposition 2.6.11). Taking the infimum with respect to t > 0 we arrive at "„ « „ «1−θ # ‚ ‚ 1−θ θ θ ‚ β ‚ + kAγ xk1−θ kAα xkθ , ‚A x‚ ≤ C θ 1−θ where C := max{Cgˆ , Ch ˆ }. The term in brackets can be estimated by

1 . θ(1−θ)

Remark 6.6.5. A more detailed analysis would reveal the kind of dependence of the constant C on A and on α, β, γ. The classical estimate is [89, Lemma 3.1.7]. Corollary 6.6.6. Let A ∈ Sect(ω) and 0 ≤ Re γ < Re β < Re α, where either γ = 0 or Re γ > 0. Then D(Aβ ) ∈ Jθ (D(Aγ ), D(Aα ))

and

D(Aβ ) ∈ Kθ (D(Aγ ), D(Aα )),

where θ ∈ (0, 1) is such that Re β = (1 − θ) Re γ + θ Re α.

6.6.3 Reiteration and Komatsu’s Theorem The next result could be obtained from Corollary 6.6.6 by the Reiteration Theorem B.4.2 from Interpolation Theory. However we can give a direct proof. Proposition 6.6.7. (Reiteration) Let A be a sectorial operator on a Banach space X. Then the following assertions hold. a) If 0 < Re β ≤ Re α, then (X, D(Aα ))θ Re β ,p = (X, D(Aβ ))θ,p Re α

for all θ ∈ (0, 1), p ∈ [1, ∞].

b) If 0 < Re γ < Re β ≤ Re α, σ ∈ (0, 1), p ∈ [1, ∞], and x ∈ X, then x ∈ (X, D(Aα ))θ,p

⇐⇒

x ∈ D(Aγ )

and

Aγ x ∈ (X, D(Aβ−γ ))σ,p ,

Re γ Re β where θ := (1 − σ) Re α + σ Re α .

c) If α, β, γ, p, σ, θ are as in b), then (X, D(Aα ))θ,p = (D(Aγ ), D(Aβ ))σ,p .

6.6. Fractional Powers as Intermediate Spaces

141

Proof. To prove a), simply choose a function ψ ∈ E such that z −α ψ ∈ H0∞ . Then also ψz −β ∈ H0∞ and one can apply Theorem 6.2.9 twice. For the proof of b), choose ψ˜ ∈ E such that z −α ψ˜ ∈ E and use it to describe (X, D(Aα ))θ,p . Observe that also ψ := z −γ ψ˜ ∈ E and z −(β−γ) ψ = z −β ψ˜ ∈ E. Hence we can use ψ to describe (X, D(Aβ−γ ))σ,p . (Note that θ, σ ∈ (0, 1), whence Theorem 6.2.9 is applicable.) Now, since θ Re α − Re γ > 0, we have Z 1 Z 1 ‚ ‚ ds ‚ ‚ ds ‚˜ ‚ ‚ ‚ ˜ s− Re γ ‚ψ(sA)x = sθ Re α−Re γ ‚s−θ Re α ψ(sA)x <∞ ‚ ‚ s s 0 0 if x ∈ (X, D(Aα ))θ,p . Applying b) of Proposition 6.6.2 we see that (X, D(Aα ))θ,p ⊂ D(Aγ ). Moreover, ˜ t−σ Re(β−γ) ψ(tA)Aγ x = t−θ Re α ψ(tA)x

for all t > 0. This finishes the proof of the stated equivalence. The statement c) follows immediately from b) since here one can assume without restriction that A is invertible.

Part of the next result is due to Komatsu. Theorem 6.6.8. (Komatsu) Let A be a sectorial operator on a Banach space X, and let θ ∈ (0, ), p ∈ [1, ∞]. Suppose the identity (X, D(Aα ))θ,p = D(Aθα )

(6.11)

holds for some Re α > 0. Then it holds for all such α. Moreover, if A is invertible, its natural H ∞ (Sϕ )-calculus is bounded on X for each ϕ ∈ (ωA , π). Proof. We can without restriction assume that A is invertible (replace A by A + 1). Then, by Dore’s Theorem, A has bounded H ∞ -calculus on D(Aθα ), whence on X, since A−θα maps X isomorphically onto D(Aθα ). In particular, A ∈ BIP and the spaces D(Aβ ) do only depend on Re β (see Proposition 3.4.5). We show first that for fixed θ we have D(Aβθ ) = (X, D(Aβ ))θ,p for arbitrary Re β > 0. To obtain this we can assume Re α 6= Re β without restriction. Assume that Re β > Re α. Then we let γ := β − α and find by Proposition 6.6.7: (X, D(Aβ ))θ,p = (D(Aθγ ), D(Aβ ))θ1 ,p

and

(X, D(Aβ−γ ))θ,p = (X, D(Aβ−θγ ))θ1 ,p ,

Re β−Re γ where θ1 := θ Re . Now A−θγ : (X, D(Aβ−θγ ))θ1 ,p −→ (D(Aθγ ), D(Aβ ))θ1 ,p is an β−θ Re γ isomorphism. Employing the assumption we obtain

(X, D(Aβ ))θ,p = A−θγ (X, D(Aα ))θ,p = A−θγ D(Aθα ) = D(Aθβ ) The case Re β < Re α is proved analogously. θ θ1

=

To complete the proof, let Re γ > 0 and 0 < θ1 < 1 with θ 6= θ1 . Find Re β > 0 such that Re γ . Then by Proposition 6.6.7 we have (X, D(Aγ ))θ1 ,p = (X, D(Aβ ))θ,p = D(Aθβ ) = Re β

D(Aθ1 γ ). The theorem is completely proved.

6.6.4 The Complex Interpolation Spaces and BIP So far we have only encountered real interpolation spaces. We have seen, that these spaces allow characterizations in terms of the functional calculus and that the operator improves when it is restricted to these spaces. However, Komatsu’s Theorem 6.6.8 shows that an identity of the form (X, D(Aα ))θ,p = D(Aθα ) can

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Chapter 6. Interpolation Spaces

only happen when the operator A (or a translate of it) has bounded H ∞ -calculus in the orginal space X. On the other hand, using complex interpolation spaces we have the following intrigueing result. Theorem 6.6.9. Let A be sectorial and densely defined. Assume that A + 1 ∈ BIP(X). Then, for 0 < θ < 1, we have [D(Aα ), D(Aβ )]θ = D(A(1−θ)α+θβ ) for all 0 ≤ Re α < Re β with either Re α > 0 or α = 0. Proof. Replacing A by A + 1 we can assume without restriction that A is invertible. Moreover, applying the isomorphism Aα on both sides of the identity, reduces the whole statement to the identity [X, D(Aα )]θ = D(Aθα (Re α > 0, θ ∈ (0, 1)).

From A ∈ BIP also follows that D(Aα ) does only depend on the real part Re α, whence we can assume α > 0. By assumption, D(A) is dense in X and (A−is )s∈R is a C0 -group on X, whence one can find M ≥ 1, η ≥ 0 such that ˛ is ˛ ˛A ˛ ≤ M eη|s| (s ∈ R).

Now, two inclusions have to be shown. To prove the first, we take x ∈ D(Aαθ ) and consider the function 2 f (z) := e(z−θ) A−αz Aθα x (z ∈ S. ‚ ‚ Clearly f (θ) = x and f : S −→ X is holomorphic. Since the norms ‚Ais ‚ of the imaginary powers grow as s → ∞ at most exponentially fast. Hence supz∈S kf (z)kX < ∞. From Proposition 3.4.5 it follows that f : S −→ X is continuous, and ‚ ‚ 2 2 ‚ ‚ kf (1 + is)kD(Aα ) = kAα f (1 + is)kX = eδ(1−θ) e−δs M eη|s| ‚Aθα x‚ X

So we see that in fact f ∈ F (X, D(Aα )) and this implies x ∈ [X, D(Aα )]θ . The reverse inclusion [X, D(Aα )]θ ⊂ D(Aα ) is shown by using the fact that D(Aα ) is dense in [X, D(Aα )]θ (see Theorem B.5.3). Take x ∈ D(Aα ), and let fP∈ V(X, D(Aα )) be such that f (θ) = x (cf. Proposition B.5.4). Then f is of the form f (z) = n j=1 ϕj (z)aj for certain ϕj ∈ F 0 (C, C) and aj ∈ D(Aα ). Define 2

g(z) := eδ(z−θ) Aαz f (z) = eδ(z−θ)

2

n X

j=1

ϕj (z)Aαz aj

(z ∈ S).

Then g ∈ F 0 (X, X) with g(θ) = Aθα x. Now we apply the Three Line Theorem: ‚ ‚ ˘ ¯ ‚ ‚ kxkD(Aθα ) = ‚Aθα x‚ = kf (θ)kX ≤ sup kg(is)kX , kg(1 + is)kX , X

s∈R

but for each s ∈ R we have

‚ 2 2 ‚ 2 2 kg(is)kX ≤ eδθ e−δs ‚Aiαs f (is)‚X ≤ M eδθ e−δs eη|s| kf (is)kX ‚ 2 2 ‚ 2 2 kg(1 + is)kX ≤ eδθ e−δs ‚Aα Aiαs f (1 + is)‚X ≤ M eδθ e−δs eη|s| kf (1 + is)kD(Aα ) .

Hence we can find a constant c > 0 (not depending on f and x of course) such that kxkD(Aθα ) ≤ c kf kF . Taking the infimum with respect to f (and applying Proposition B.5.4) we find kxkD(Aθα ) ≤ c kxk[X,D(Aα )]θ . Now density cares for the rest.

6.7. Characterizing Growth Conditions

143

Remark 6.6.10. It is unknown whether the equality [X, D(A)]θ = D(Aθ ) for some 0 < θ < 1 in general is already sufficient to have A + 1 ∈ BIP. It is true in Hilbert space due to the fact that (X, D(A))θ,2 = [X, D(A)]θ for all 0 < θ < 1 (see [87, Cor. 4.3.12]). However, this is a very special case. In the following we give an example of an injective, densely defined sectorial operator A on a Banach space such that (X, D(A))θ,p 6== [X, D(A)]θ for all θ ∈ (0, 1) and all p ∈ [1, ∞]. However, the operator will have a bounded H ∞ -calculus. In particular, we have (X, D(A))θ,p 6= D(Aθ ) for all θ ∈ (0, 1), p ∈ [1, ∞]. Example 6.6.11. Consider the negative Laplacian −∆p on X := Lp (Rn ) where 1 < p < ∞ and p 6= 2. It has domain D(−∆p ) = W2,p (Rn ) and is an injective sectorial operator with spectral angle 0. Since Lp is reflexive, −∆p has dense domain and dense range. Moreover, by applying Mikhlin’s multiplier theorem (see [6, Appendix E]) one can easily prove that the natural H ∞ -calculus for A is bounded. In particular, −∆p ∈ BIP(X), whence H 2θ,p (Rn ) = [Lp (Rn ), W2,p (Rn )]θ = D((−∆p )θ ) for all 0 < θ < 1, see [87, Ex. 2.1.12] However, the real interpolation spaces 2θ (Lp (Rn ), W2,p (Rn ))θ,q are the so-called Besov spaces Bp,q (see [87, Ex. 1.3.10]) s s,p n and it is well known that Bp,q 6= H (R ) for no s, p, q (see [121]). It should be clear from these considerations that an identity of the form (X, D(A))θ,p = D(Aθ ) is — apart from the Hilbert space case — extremely seldom.

6.7 Characterizing Growth Conditions When we examined perturbations B of a sectorial operator A in Section 5.6, an important condition proved to be an estimate of the form

sup tθ B(t + A)−1 < ∞ t>0

(see Lemma 5.6.1 and Proposition 5.6.3). We remarked (in Remark 5.6.2) that one can describe such a condition in terms of interpolation spaces. The present section is to give evidence to this assertion. Proposition 6.7.1. Let A be a sectorial operator on the Banach space X, let B : Y −→ X be a bounded operator, where Y is another Banach space, and let θ ∈

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Chapter 6. Interpolation Spaces

[0, 1]. Then the equivalences

R(B) ⊂ (X, D(A))θ,∞ ⇐⇒ sup λθ (λ + A)−1 B Y →D(A) < ∞ λ>1

⇐⇒ sup λθ (λ + A)−1 B Y →D˙ < ∞ λ>1

hold, with the latter applying only when A is injective. Furthermore, if A is injective, then also the equivalences

R(B) ⊂ (X, R(A))1−θ,∞ ⇐⇒ sup λθ (λ + A)−1 B Y →X < ∞ 0<λ<1

⇐⇒ sup λθ (λ + A)−1 B Y →D(A) < ∞ 0<λ<1

hold true.

Proof. Let us consider the first assertion with its two biimplications. The case θ = 0 is trivial, so let θ ∈ (0, 1]. We observe that, by the sectoriality of A, the second biimplication holds trivially. The first is derived immediately from Lemma 6.1.1 (see also the Komatsu representation in Section 6.2.3) and the Closed Graph theorem. 1 : The second statement is obtained from the first by substituting µ := λ „ «−1 1 1 λθ (λ + A)−1 = µ1−θ +A = µ1−θ A−1 (µ + A−1 )−1 µ µ

and replacing A by A−1 .

Combining both statements yields the following. Corollary 6.7.2. Let A be an injective sectorial operator on the Banach space X, let B : Y −→ X be a bounded operator, where Y is another Banach space, and let θ ∈ [0, 1]. Then the conditon

sup λθ (λ + A)−1 B Y →D(A) < ∞ 0<λ<∞

is equivalent to R(B) ⊂ (D(A), R(A))1−θ,∞ .

Proof. Combining both conditions from Proposition 6.7.1 yields R(B) ⊂ (X, D(A))θ,∞ ∩ (X, R(A))1−θ,∞ . Now observe that X = D(A) + R(A) and apply (B.7) from Proposition B.2.7.

Let us turn to the “dual” conditions. Proposition 6.7.3. Let A be a sectorial operator on the Banach space X, let C : D(A) −→ Y a bounded operator, where Y is a second Banach space, and let θ ∈ [0, 1]. Then the condition

sup λ1−θ C(λ + A)−1 X→Y < ∞ (6.12) λ>1

is trivially satisfied if θ = 1. For 0 < θ < 1 it is equivalent to the boundedness of C : (X, D(A))θ,1 −→ Y

on the space D(A). In case θ = 0 it is equivalent to the fact that C (defined on D(A)) is bounded for the norm k·kX .

6.7. Characterizing Growth Conditions

145

Proof. Let us start with the case θ ∈ (0, 1). We use again the representation of (X, D(A))θ,1 from Lemma 6.1.1. If C : (X, D(A))θ,1 −→ X is bounded, we can estimate ‚ ‚ ‚ ‚ ‚ 1−θ ‚ C(t + A)−1 x‚ ≤ ct1−θ ‚(t + A)−1 x‚(X,D(A)) ‚t θ,1 Z ∞‚ ‚ ds ‚ ‚ ‚ θ ‚ 1−θ ‚ −1 ‚ 1−θ ≤ ct (t + A) x + ct ‚s A(s + A)−1 (t + A)−1 x‚ s 0 ‚ Z ∞‚ ‚ ‚ 1 1 ‚sθ−1 A1−θ (1 + A)−1 t−θ Aθ (1 + A)−1 x‚ ds ≤ ct−θ M (A) + c ‚ ‚ s s t 0 Z ∞ ‚ ‚ ‚ϕ1/s (A)ψ1/t (A)x‚ ds = ct−θ M (A) + c s 0

for x ∈ X, t > 0 and some constant c, where we have used the notation ϕ(z) := z 1−θ /1 + z and ψ(z) := z θ /1 + z. By Theorem 5.2.2, part b) one sees that the second summand is uniformly bounded in t > 0. Assume now that (6.12) holds, again under the assumption θ ∈ (0, 1). We want to show the boundedness of C : (X, D(A))θ,1 −→ Y , and since D(A) = D(A + 1), we can assume without restriction that A is invertible and ‚ ‚ ‚ ‚ c′ := sup ‚t1−θ C(t + A)−1 ‚ < ∞. t>0

z From Theorem 6.6.1 we know that D(A2 ) is dense in (X, D(A))θ,1 . Let τ (z) := (1+z) 2 and R R∞ ∞ −1 −1 A)x ds for all x ∈ D(A) (Proposition c := 0 τ (s−1 ) ds > 0. Then we have x = c τ (s 0 s s 5.2.4). So, for x ∈ D(A2 ) this integral converges in D(A), whence Z ∞ ‚ ‚ ‚Cτ (s−1 A)x‚ ds kCxk ≤ c−1 s 0 Z ∞ ‚ ‚ −1 −1 −1 ‚C(1 + s A) s−1 A(1 + s−1 A)−1 x‚ ds =c s 0 Z ∞‚ ‚ ‚ 1−θ −1 −1 θ −1 ‚ ds =c C(s + A) s A(s + A) x‚ ‚s s 0 Z ∞‚ ‚ ds ‚ ‚ ≤ c′ c−1 ≤ c′ c−1 kxk(X,D(A))θ,1 ‚sθ A(s + A)−1 x‚ s 0

for x ∈ D(A2 ). Since D(A2 ) is dense in (X, D(A))θ,1 , we obtain the desired result.

We are left to deal with the cases θ = 0, 1. It is easily seen that condition (6.12) with θ = 1 is equivalent to kCkD(A)→Y < ∞. So let θ = 0. If C is bounded for the norm k·kX then clearly ‚ ‚ (6.12) holds, by sectoriality of A. For the converse assume ‚λC(λ + A)−1 x‚Y ≤ γ kxk for each x ∈ X and λ > 1. Let x ∈ D(A). Then ‚ ‚ ‚ ‚ 1 kCxk = ‚λC(λ + A)−1 x‚ + ‚C(λ + A)−1 Ax‚ ≤ γ kxk + kAxk λ for all λ > 1. As λ → ∞ we obtain kCxk ≤ γ kxk for all x ∈ D(A).

Proposition 6.7.4. Let A be an injective sectorial operator on the Banach space X, let C : D(A) −→ Y a bounded operator, where Y is a second Banach space, and let θ ∈ [0, 1]. Then the condition

sup λ1−θ C(λ + A)−1 X→Y (6.13) 0<λ<1

is trivially satisfied for θ = 0. In case 0 < θ < 1 it is equivalent to the boundedness of CA−1 : (X, R(A))1−θ,1 −→ Y

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Chapter 6. Interpolation Spaces

on R(A). In case θ = 1, it is equivalent to the boundedness of C : D˙ −→ Y on D(A). 1 Proof. Assume 0 < θ < 1. Writing µ = λ yields λ1−θ C(λ + A)−1 = λ−θ Cλ(λ + A)−1 = θ −1 −1 θ −1 −1 −1 µ C[I − µ(µ + A ) = µ [CA ](µ + A ) . Now one can apply Proposition 6.7.3 with A replaced by A−1 and θ replaced by 1 − θ. Let us look at the remaining choices for θ.‚ The case θ‚ = 0 being easy, θ = ‚ we assume ‚ ˙ −→ Y is bounded, one has ‚C(λ + A)−1 x‚ ≤ kCk ˙ ‚A(λ + A)−1 x‚ ≤ 1. If C : D D→Y ‚ Y ‚ kCkD→Y (M (A) + 1), hence (6.13). Conversely, suppose ‚C(λ + A)−1 x‚ ≤ γ kxk for all x ∈ X ˙ and all 0 < λ < 1. Then ‚ ‚ ‚ ‚ kCxk ≤ λ ‚C(λ + A)−1 x‚ + ‚C(λ + A)−1 Ax‚ ≤ λγ kxk + γ kAxk

for x ∈ D(A). Hence, if we let λ → 0 we obtain kCxk ≤ γ kxkD˙ .

Again, we combine the last two propositions. Corollary 6.7.5. Let A be an injective sectorial operator on the Banach space X, let C : D(A) −→ Y be a bounded operator, where Y is another Banach space, and let θ ∈ [0, 1]. Then the condition

sup λ1−θ C(λ + A)−1 X→Y < ∞ 0<λ<∞

is equivalent to the boundedness of   C : X −→ Y, ˙ θ,1 −→ Y, C : (X, D)   C : D˙ −→ Y,

if if if

θ=0 0<θ<1 θ=1

on D(A).

Proof. In the cases θ = 0, 1 this is just a combination of Propositions 6.7.3 and 6.7.4. Assume 0 < θ < 1. Since A is injective, it induces a topological isomorphism ˙ D(A)) −→ (X, R(A)) A : (D, of Banach couples. Hence the boundedness of CA−1 : (X, R(A))1−θ,1 −→ Y is then equiv˙ D(A))1−θ,1 −→ Y . This yields the boundedness of alent to the boundedness of C : (D, ˙ D(A))1−θ,1 −→ Y as a characterization. However, (B.6) of ProposiC : (X, D(A))θ,1 + (D, ˙ D(A))1−θ,1 = (X, D) ˙ θ,1 which tion B.2.7 applies and we obtain the identity (X, D(A))θ,1 + (D, finishes the proof.

6.8 Comments

Chapter 7

The Functional Calculus on Hilbert Spaces We start in providing necessary background information on the functional calculus on Hilbert spaces. In Section 7.1 we prove several results which infer a bounded H ∞ calculus (on sectors or strips) from a corresponding numerical range condition. In the sector case this is essentially von Neumann’ inequality (Section 7.1.3). In the strip case it is a result by Crouzeix and Delyon which we prove in detail (Section 7.1.5)). From von Neumann’s inequality we obtain certain “mapping theorems for the numerical range” (Section 7.1.4). Section 7.2 is devoted to C0 -groups on Hilbert spaces. We discuss Liapunov’s direct method (Section 7.2.1) from Linear Systems Theory and apply it to obtain a remarkable decomposition and similarity result for group generators (Section 7.2.2). This allows to prove the deLaubenfels-Boyadzhiev Theorem 7.2.1 on the boundedness of the H ∞ -calculus on strips for such operators. A different approach to this result yielding also a characterization of group generators is presented in Section 7.2.3. Section 7.3 is devoted mainly to the connection of similarity and functional calculus on Hilbert spaces. The main tool is McIntosh’s fundamental result on the boundedness of the H ∞ -calculus (Section 7.3.1). The similarity questions we are interested in deal with operators defined by sesquilinear forms. We introduce these operators in Section 7.3.2 and obtain in Section 7.3.3 a characterization of such operators up to similarity. Afterwards several theorems on similarity are proved, touching also the famous square root problem. Then (Section 7.3.4) we give an example of a C0 -semigroup which is not similar even to a quasi-contractive semigroup. Finally, in Section 7.4 the foregoing results are applied to generators of cosine functions. We show that after a similarity those operators always have numerical range in a horizontal parabola.

Preliminaries on Functional Calculus Before we can come to more substantial things we have to do some preliminary work. Let U ⊂ C be open and invariant under complex conjugation, e.g., U = Sω or U = Hω for some ω. Let f : U −→ C∞ be meromorphic. Then the function f ∗ , defined by f ∗ := z 7−→ f (z) : U −→ C∞

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is called the conjugate of the function f . Obviously, f ∗ is meromorphic again. Moreover, the spaces H0∞ (Sω ), H ∞ (Sω ), . . . and F (Hω ), H ∞ (Hω ), . . . are invariant under conjugation. The next proposition gives some background information on the functional calculus for sectorial operators on Hilbert spaces. Proposition 7.0.1. Let A ∈ Sect(ω) be a sectorial operator on a Hilbert space H. Then the following assertions hold. a) The operator A is densely defined and H = N(A) ⊕ R(A). In particular, A is injective if and only if R(A) is dense in H. b) The operator A∗ is also sectorial of angle ω with M (A, ω ′ ) = M (A∗ , ω ′ ) for all ω ′ ∈ (ω, π]. In particular, we have ωA = ωA∗ .

c) The operator A is injective (invertible, bounded) if and only if A∗ is injective (invertible, bounded). d) The identity f (A∗ ) = [f ∗ (A)]∗ holds whenever f ∈ A[Sω ] or, if A is injective, f ∈ B[Sω ]. e) The identity

(Aα )∗ = (A∗ )α

(7.1)

is true for all Re α > 0, and even for all α ∈ C in case that A is injective.

f ) If A ∈ BIP(H) then also A∗ ∈ BIP(H) with θA = θA∗ .

g) Let ϕ > ω. If the natural H0∞ (Sϕ )-calculus for A is bounded, the same is true for A∗ and the bounds are the same. Proof. Ad a). This follows from Proposition 2.1.1, since a Hilbert space is reflexive. Ad b). This follows from Corollary C.2.2. Ad c) If A is injective, then R(A) is dense. By part e) of Proposition C.2.1, this implies N(A∗ ) = 0. The reverse implication follows from A∗∗ = A. Ad d). Let f ∈ H0∞ (Sϕ ). Then „ «∗ Z Z 1 −1 [f ∗ (A)]∗ = f ∗ (z)R(z, A) dz = f (z)R(z, A)∗ dz 2πi Γ 2πi Γ Z Z −1 1 ! = f (z)R(z, A∗ ) dz = f (z)R(z, A∗ ) dz = f (A∗ ), 2πi Γ 2πi Γ since Γ is such that (z 7−→ z) just reverses the orientation of Γ. Since g ∗ = g if g(z) = (1+z)−1 we thus obtain f ∗ (A)∗ = f (A∗ ) for all f ∈ E(Sϕ ). Now, if f ∈ A such that F (z) := (1+z)−n f (z) ∈ E we have (1)

(2)

(3)

f (A∗ ) = (1 + A∗ )n F (A∗ ) = [(1 + A)n ]∗ [F ∗ (A)]∗ = [F ∗ (A)(1 + A)n ]∗ = [f ∗ (A)]∗ Here we have used Proposition C.2.3 in (1) and Proposition C.2.1 in (2). Equation (3) is justified by Proposition C.2.1 and the fact that D(An ) is a core for f ∗ (A) by c) of Proposition 2.3.10. The proof of the statement in case that A is injective and f ∈ B is similar. One has to use the ∗ n identity [Λn A ] = ΛA∗ which holds for each n. The statements e), f ) and g) are consequences of d).

Clearly one could prove a similar result for strip type operators. We do not go into the details.

7.1. Numerical Range Conditions

149

7.1 Numerical Range Conditions In this section we will examine sectorial and strip type operators A on a Hilbert space H with the additional property, that the numerical range W (A) is contained in the relevant set (sector or strip). The ultimate goal is to obtain boundedness of the natural H ∞ -calculus for such operators.

7.1.1 Accretive and ω-accretive Operators Let us recall that an operator A on a Hilbert space is called accretive if its numerical range W (A) := {( Ax | x) | x ∈ D(A), kxk = 1} (see Appendix C.3) is contained in the closed right halfplane S π2 . Since any restriction of an accretive operator is likewise accretive, an accretive operator need not be closed. However, if assume in addition that −1 ∈ ̺(A) we obtain a so-called m-accretive operator. We have given an account of basic facts on accretive operators in Appendix C.7. By Proposition C.7.2 and Theorem C.7.3 the following assertions are equivalent for a closed operator A on H. (i) A is m-accretive, i.e. W (A) ⊂ S π2 and R(A + 1) is dense in H.

(ii) −A generates a C0 -semigroup (T (t))t≥0 with kT (t)k ≤ 1 for all t ≥ 0.

(iii) {Re z < 0} ⊂ ̺(A) and (λ + A)−1 ≤ (Re λ)−1 for all Re λ > 0.

In particular, A ∈ Sect( π2 ).

As a matter of fact, one can consider also operators whose numerical range is contained in a smaller sector. Let 0 ≤ ω ≤ π2 . An operator A on a Hilbert space H is called ω-accretive if W (A) ⊂ Sω . Note that this means |Im ( Au | u )| ≤ (tan ω) Re ( Au | u )

(u ∈ D(A)).

The operator A is called m-ω-accretive if it is ω-accretive and R(A + 1) is dense in H. Hence, an operator is (m-)accretive if and only if it is (m-) π2 -accretive. Furthermore, each m-ω-accretive operator is m-accretive. A 0-accretive operator is symmetric (since our Hilbert spaces are complex). An operator is positive if and only if it is m-0-accretive. The following proposition gives a useful characterization. Proposition 7.1.1. Let A be an operator on the Hilbert space H and let 0 < ω < and θ := π2 − ω. The following assertions are equivalent.

π 2

(i) The operator A is m-ω-accretive.

(ii) The operators e±iθ A are m-accretive. (iii) The operator −A generates a holomorphic C0 -semigroup (T (z))z∈Sθ on Sθ such that kT (z)k ≤ 1 for all z ∈ Sθ .

If one of these equivalent conditions is satisfied then A ∈ Sect(ω).

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Note that (ii) does not imply (i) if ω = 0. For the precise meaning of (iii) see § 3.3 of Chapter 3. Proof. The equivalence of (i) and (ii) is clear from Corollary C.3.2. Assume (ii). From Proposition C.3.1 we infer that A ∈ Sect(ω). Since ω < π2 we conclude that −A generates a bounded holomorphic semigroup T : Sθ −→ L(H). For each ϕ ∈ (−θ, θ) the operator −eiϕ A generates the ‚semigroup (T (eiϕ t))t>0 (see Remark 3.3.4). But clearly eiϕ A is m-accretive, whence ‚ ‚T (eiϕ t)‚ ≤ 1 for all t > 0. This proves (iii). Assume (iii). As above, the operator −eiϕ A generates the semigroup (T (eiϕ t))t>0 , for each ϕ ∈ (−θ, θ). Since this is a contraction semigroup we conclude that eiϕ A is m-accretive. Letting ϕ tend to ±θ yields that eϕθ A is m-accretive. The final statement follows from Proposition C.3.1.

Parallel to sectorial operators we want to study strip type operators. Here we can say the following. Proposition 7.1.2. Let A be an operator on a Hilbert space H and let ω > 0. Then the following assertions are equivalent. (i) W (A) ⊂ Hω and R(A ± i(ω + 1)) are dense in H.

(ii) The operators ω ± iA are both m-accretive.

(iii) iA generates a C0 -group (T (s))s∈R such that kT (s)k ≤ eω|s| for all s ∈ R.

(iv) σ(A) ⊂ Hω and kR(λ, A)k ≤

1 |Im λ|−ω

for all λ ∈ / Hω .

(v) There are selfadjoint operators B and C with −ω ≤ C ≤ ω and A = B + iC. (Note that the operator C in (iv) is necessarily bounded.) If these equivalent conditions are satisfied, A ∈ SST(ω), D(A) = D(A∗ ) and A∗ = B − iC. An operator A on H is called m-Hω -accretive if it satisfies the equivalent conditions of Proposition 7.1.2. Note that ω = 0 is allowed in Proposition 7.1.2 in which case the operator A is selfadjoint. Proof. The equivalence of (i) and (ii) readily follows from the definition of “m-accretive”, while (ii) ⇔ (iii) is due to the Lumer-Philips Theorem C.7.3 and Proposition A.8.9. Assertions (iv) and (ii) are equivalent by Proposition C.7.2. Assume (v). Then clearly W (A) ⊂ Hω . Moreover, iA is a bounded perturbation of the generator of a C0 -group, whence by itself generates a C0 group, by Proposition A.8.8. In particular, A has enough resolvent to conclude (ii). Finally, assume that (i) − (iv) are satisfied. Applying the generalized Cauchy-Schwarz 1 ( Au | v ) − ( u | Av )) Inequality (Proposition C.1.2) to the form c(u, v) := Im[( A · | · )](u, v) = 2i on V := D(A) we obtain |( Au | v ) − ( u | Av )| ≤ 2ω kuk kvk

for u, v ∈ D(A). Since D(A) is dense in H, the form c extends to a continuous symmetric form on H. Hence we can find a bounded symmetric operator C ∈ L(H) such that ( Cx | y ) = 1 (( Ax | y ) − ( x | Ay )). Obviously, −ω ≤ C ≤ ω. Now we define B := A − iC. Then ( Bu | u ) = 2i ( Au | u ) − i Im ( Au | u ) = Re ( Au | u ) for all u ∈ D(A) = D(B), whence W (B) ⊂ R. Since

7.1. Numerical Range Conditions

151

A generates a C0 -group by (iii), also B does (Proposition A.8.8) and this implies that B is skewadjoint, by Stone’s Theorem C.7.4. The remaining assertions are clear, since A∗ = (B + iC)∗ = B ∗ − iC ∗ = B − iC , by j of Proposition C.2.1.

7.1.2 Normal Operators Let us start with a discussion of normal operators. By the Spectral Theorem (see Appendix D), normal operators are unitarily equivalent to multiplication operators on L2 spaces. Proposition 7.1.3. Let (Ω, µ) be a standard measure space and let a : Ω → C be a continuous function with a(Ω) ⊂ Sω , where ω ∈ [0, π). Denote by A := Ma the multiplication operator on H := L2 (Ω, µ). Then A is m-ω-accretive and f (A) = Mf ◦a whenever f ∈ O[Sω ]A . Proof. Let ψ ∈ D(A). Then

( Aψ | ψ )L2 =

Z

Ω

aψψ dµ =

Z

Ω

a |ψ|2 dµ.

Now a(s) |ψ(s)|2 ∈ Sω for every s ∈ Ω by hypothesis and Sω is closed and convex. Hence also R 2 Ω a |ψ| ∈ Sω . This shows that W (A) ⊂ Sω . To prove the second assertion, asssume first that f ∈ H0∞ [Sω ]. Then f (A) is given by the integral R 1 f (A) = 2πi Γ f (z)R(z, A) dz which converges in the operator norm topology. By Proposition D.1.1, R(z, A) = M(z−a)−1 for each z ∈ / Sω . Moreover, the mapping (g 7−→ Mg ) : Cb (Ω) −→ L(L2 (Ω, µ)) is an isometric embedding. Therefore, f (A) = Mg for some bounded and continuous function on Ω. Since evaluation at a point s ∈ Ω is a continuous functional on Cb (Ω), we obtain » – Z Z 1 1 g(s) = f (z)(z − a(·))−1 dz (s) = f (z)(z − a(s))−1 dz = f (a(s)) 2πi Γ 2πi Γ by Cauchy’s theorem. This shows f (A) = Mf ◦a . The same proof (with a different Γ) works in the case f ∈ H0∞ 0 [Sω ]. So we know the statement for f ∈ E[Sω ]. Let f ∈ O[Sω ]A and let e ∈ E[Sω ] be a regularizer for f , i.e., ef ∈ E and e(A) is injective. Since e(A) = Me◦a , the set {e ◦ a = 0} is locally µ-null. Therefore, (x, y) ∈ f (A) ⇐⇒ (ef )(A)x = e(A)y ⇐⇒ (e ◦ a)(f ◦ a)x = (e ◦ a)y µ − a.e. ⇐⇒ (f ◦ a)x = y µ − a.e.. This yields f (A) = Mf ◦a .

Remark 7.1.4. Proposition 7.1.3 might look as a repetition of Example 2.3.14. But note that there we assumed the measure space to be σ-finite and applied Fubini’s theorem. One could in fact justifty this reasoning also in the case of a standard measure space (which is in general not σ-finite) but it is also instructive to see how continuity helps to avoid Fubini’s theorem. In the strip case, we obtain the analogous result (with almost the same proof). Proposition 7.1.5. Let (Ω, µ) be a standard measure space and let a : Ω → C be a continuous function with a(Ω) ⊂ Hω , where 0 ≤ ω. Denote by A := Ma the multiplication operator on H := L2 (Ω, µ). Then W (A) ⊂ Hω , A ∈ Strip(ω) and f (A) = Mf ◦a whenever f ∈ O[Hω ]A . Corollary 7.1.6. Let A be a selfadjoint operator on a Hilbert space H.

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a) If A ≥ 0, then A ∈ Sect(0) and for each ϕ > 0 the natural H0∞ (Sϕ )-calculus is bounded. In fact kf (A)k ≤ kf k(0,∞) for all f ∈ E[S0 ]. If in addition A is injective, then for each ϕ ∈ (0, π] the natural H ∞ (Sϕ )calculus for A is bounded (with bound 1). b) We have iA ∈ Sect( π2 ) and kf (iA)k ≤ kf kiR for all f ∈ E[S π2 ].

c) A ∈ Strip(0) and for each ϕ > 0 the natural H ∞ (Hϕ )-calculus for A is bounded (with bound 1).

Proof. By the spectral theorem D.5.1 there is a standard measure space (Ω, µ) and a continuous, real-valued function a ∈ C(Ω) such that (H, A) is unitarily equivalent to (L2 (Ω, µ), Ma ). Now the assertions follow from Proposition 7.1.3, Proposition 7.1.5 and Proposition D.1.1.

Corollary 7.1.6 shows that selfadjoint operators behave well with respect to functional calculus. Moreover, our restriction to selfadjoint operators is only for the sake of simplicity. Clearly the same proof works for normal operators. However, these results were obtained using the Spectral Theorem, which actually gives a much stronger result, namely a bounded Borel functional calculus (see Theorem D.6.1). It is interesting, that in the context of holomorphic functional calculi, sole conditions on the numerical range of A suffice. This will be shown in the next subsection.

7.1.3 Functional Calculus for m-accretive Operators We immediately start with the main result Theorem 7.1.7. Let A ∈ Sect( π2 ) on the Hilbert space H. Then A is m-accretive if and only if kf (A)k ≤ kf k π (7.2) 2

for all f ∈ E[S π2 ]. If A is injective, then this implies that (7.2) even holds for all f ∈ H ∞ [S π2 ].

We will give two proofs of this fact. The first relies on the following theorem, which we state without proof. Recall that A is m-accretive if and only if −A generates a contraction semigroup on H (Theorem C.7.3). Theorem 7.1.8. (Sz¨okefalvi-Nagy) Let (T (t))t≥0 be a contraction semigroup on a Hilbert space H. Then there exists a Hilbert space K, an isometric embedding ι : H −→ K and a unitary C0 -group (U (t))t∈R on K such that P ◦ U (t) ◦ ι = ι ◦ T (t) for all t ≥ 0. Here, P : K −→ ι(H) denotes the orthogonal projection on the closed subspace ι(H) of K. The triple (K, U, ι) is called a dilation of the contraction semigroup T . For a proof see, e.g., [29, Chapter 6, Section 3]. We now give our first proof of Theorem 7.1.7.

7.1. Numerical Range Conditions

153

‚ ‚ First Proof of Theorem 7.1.7. If we take f (z) = z−1 in (7.2) we obtain ‚(A − 1)(A + 1)−1 ‚ ≤ z+1 1. Then we apply (iv) of Proposition C.7.2 to conclude that A is m-accretive. The converse is proved by means of the Sz.-Nagy Theorem. Assume that A is m-accretive. Then −A generates a contraction semigroup T . By Theorem 7.1.8 there is a dilation (K, U, ι) of T . By Stone’s theorem C.7.4 the generator of U is of the form −iB where B is a selfadjoint operator on K. We claim that P R(λ, iB)ι = ιR(λ, A) for each Re λ < 0. In fact, Z ∞ Z ∞ P R(λ, iB)ι = −P (−λ, −iB)ι = − eλs P U (s)ι ds = − eλs ιT (s) ds 0

0

= −ιR(−λ, −A) = ιR(λ, A)

Now choose

π 2

< ϕ ≤ π and f ∈ H0∞ (Sϕ ). Then Z Z 1 1 f (z)P R(z, iB)ι dz = f (z)ι R(z, A) dz P f (iB)ι = 2πi Γ 2πi Γ Z 1 =ι f (z)R(z, A) dz = ιf (A) 2πi Γ

The same arguments apply if f ∈ H0∞ 0 . Hence we obtain P f (iB)ι = ιf (A) for all f ∈ E. Since ι is isometric, by b) of Corollary 7.1.6 we have kf (A)xkH = kιf (A)xkK = kP f (iB)ιxkK ≤ kf (iB)ιxkK ≤ kf k π kιxkK = kf k π kxkH 2

for every x ∈ H. This finishes the proof.

2

This proof loks nice but does not really illuminate how the numerical range condition comes into play. We therefore give a second proof. Second Proof of Theorem 7.1.7. By Propositions 5.3.5 and 5.3.6 combined with Lemma 5.3.4 it suffices to show (7.2) for f ∈ R0 (Sϕ ) for ϕ ∈ ( π2 , π) and A replaced by its sectorial approximation Aε := (ε + A)(1 + εA)−1 . Note that if A is m-accretive, Aε is bounded and strictly accretive, i.e., Re ( Aε x | x ) ≥ ε kxk2 for all x ∈ H. Hence in the following we can assume A to be bounded and strictly accretive. Given f ∈ R0 (Sϕ ) one can compute f (A) by the usual Cauchy integral, but since A is bounded and invertible, one can shift the path of integration onto the imaginary axis (from +i∞ to −i∞). Now, on iR we have z = −z, whence R(z, A)∗ = R(−z, A∗ ). We therefore can compute Z 1 f (A) = f (z)R(z, A) dz 2πi −iR Z Z 1 1 = f (z) (R(z, A) + R(z, A)∗ ) dz + f (z)R(z, A)∗ dz 2πi −iR 2πi −iR Z Z −1 1 f (z)R(z, A)∗ (A + A∗ )R(z, A) dz + f (z)R(−z, A∗ ) dz. = 2πi −iR 2πi −iR The latter integral equals zero, since f (z)R(−z, A∗ ) is holomorphic in the right halfplane and we can shift the the path of integration to Re z = +∞. Hence we obtain Z 1 f (A) = f (z)S(z) dz 2πi iR with S(z) := R(z, A)∗ (A + A∗ )R(z, A). Note that each S(z) is a strictly positive selfadjoint n and letting n tend to ∞, we obtain operator. By inserting f (z) := n+z Z 1 I= S(z) dz. 2πi iR

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Chapter 7. The Functional Calculus on Hilbert Spaces

Thus we have ‚ ‚ ‚ ‚ Z Z ‚ ‚ ‚ ‚ 1 ‚ ≤ kf k ‚ 1 ‚ = kf k , kf (A)k = ‚ f (it)S(it) dt S(it) dt ∞‚ ∞ ‚ 2π ‚ ‚ 2π R R where we have used Corollary C.6.4.

This second proof reveals a basic scheme: The numerical range condition implies positivity of certain selfadjoint operators. The integral over this family has a norm independent of the individual operator A. Then, Proposition C.6.3 furnishes the final estimate. We will encounter the same scheme below in the proof of Theorem 7.1.14. Remark 7.1.9. Let us sketch another proof of Theorem 7.1.7. Since A is maccretive, the Cayley transform T := (A − 1)(A + 1)−1 satisfies kT k ≤ 1. Now the von Neumann inequality says that kp(T )k ≤ kpk∞ for every polynomial p ∈ C[z], where kpk∞ is the uniform norm of p on the unit disc. This readily yields kr(T )k ≤ krk∞ for every rational function r with poles outside of {z | |z| ≤ 1}. Hence we can conclude that the natural R∞ (S π2 )-calculus for A is bounded with bound 1. Now apply Proposition 5.3.5. (In turn, one can derive von Neumann’s inequality from Theorem 7.1.7.)

7.1.4 Mapping Theorems for the Numerical Range As an application of Theorem 7.1.7 we prove what is called “mapping theorems for the numerical range”, a name somewhat inaccurate but not totally inappropriate. Here is the main result. Theorem 7.1.10. Let A be an m-accretive operator on the Hilbert space H and f ∈ O[S π2 ]A . Then W (f (A)) ⊂ convf (S π2 ).

Proof. If the right-hand side is the whole plane, then there is nothing to prove. If not, then there must be proper closed halfplane containing it, and it is in fact the intersection of all such halfplanes. Let P be any closed halfplane which contains f (S π ). Then there are numbers a, b ∈ C 2 such that a + bP equals the right halfplane {Re z ≥ 0}. Therefore, ˛ ˛ ˛ a + bf − 1 ˛ ˛ ˛ ˛ a + bf + 1 ˛ ≤ 1.

By Theorem 7.1.7 and some functional calculus arguments the operator a+bf (A)+1 is invertible and ‚ ‚ ‚(a + bf (A) + 1)(a + f (A) + 1)−1 ‚ ≤ 1 whence a + bf (A) is m-accretive. This means that a + bW (f (A)) = W (a + bf (A)) ⊂ {Re z ≥ 0} which implies W (f (A)) ⊂ P .

Corollary 7.1.11. Let A be m-accretive and let α ∈ (0, 1). Then the operator Aα is m-α π2 -accretive, i.e., |Im ( Aα x | x)| ≤ tan for all x ∈ D(Aα ).

απ Re ( Aα x | x) 2

7.1. Numerical Range Conditions

155

Proof. Let f (z) := z α . Then f (S π ) = Sα π and the assertion follows from Theorem 7.1.10. 2

2

We remark that the converse of Corollary 7.1.11 is not true, as Simard has shown. In fact there is α ∈ (0, 1) and an operator A on the two-dimensional Hilbert space which is not accretive, but Aα is α π2 -accretive (see [?]). (Surprisingly, the converse does hold modulo similarity which was proved by LeMerdy [?].) This example shows that Theorem 7.1.10 is of restricted value when the original operator is not m-accretive bu satisfies other numerical range conditions. Sometimes hoowever one is lucky, like in the following example. Corollary 7.1.12. Let A be an injective m-ω-accretive operator for some 0 ≤ ω ≤ Then log A is m-Hω -accretive.

π 2.

Proof. Theorem 7.1.10 shows that log A is m-H π -accretive whenever A is m-accretive. Define θ :=

π 2

2

− ω and apply the foregoing to e±iθ . Since log(e±iθ A) = ±iθ + log(A) we obtain ±iθ + W (log(A)) = W (log(e±iθ A)) ⊂ H π . 2

This proves te claim.

For completeness, we state another mapping theorem for the numerical range. Corollary 7.1.13. Let δ > 0 and A − δ be m-accretive. Then Aα − δ α is m-accretive for each 0 < α < 1. Proof. Without restriction we can assume δ = 1. Define B := A − 1. Then B is m-accretive and the claim is that also (B + 1)α − 1 is m-accretive. Theorem 7.1.10 it suffices to show that f (z) := (z+1)α −1 maps the right halfplane to itself. This can be done by using the Balakrishnan representation (3.1) on page 60 with A replaced by z + 1. However, one can use this formula also directly with the operator A as in [120, Lemma 2.3.6].

7.1.5 The Crouzeix-Delyon Theorem We now turn to the strip case, i.e., to the case of an m-Hω -accretive operator for ω > 0. (The case of ω = 0 has been settled in c) of Corollary 7.1.6.) The next result was originally published in [25]. Theorem 7.1.14. (Crouzeix-Delyon) Let ω > 0 and let A be a m-Hω -accretive operator on the Hilbert space H. Then 2 kf (A)k ≤ (2 + √ ) kf kHω 3 for all f ∈ H ∞ [Hω ].

Proof. We begin with some reducing steps. First, we can assume that f ∈ F [Hω ]. (Use Proposi“ ”2 n tion 5.1.6 together with the approximation fn (z) := f (z) n+ω+iz .) Next we want to reduce the theorem to bounded operators being Hεω -accretive for some 0 < ε < 1. To this aim, consider 1 the operator An := (1 − n )[n(n + ω + iA)−1 ]∗ A[n(n + ω + iA)−1 ]. Clearly, An is bounded and H(1−1/n)ω -accretive and An x → Ax for all x ∈ D(A). Using Proposition 7.1.2 we see that kR(λ, An )k is bounded uniformly in n ∈ N and λ on horizontal lines. This implies R(λ, An )x − R(λ, A)x = R(λ, An )[An − A]R(λ, A)x → 0

156

Chapter 7. The Functional Calculus on Hilbert Spaces

for each x ∈ H. Hence f (An ) → f (A) strongly for every f ∈ F [Hω ].

We thus can assume A ∈ L(X) and Hω ′ -accretive for some 0 < ω ′ < ω. Write A = B + iC with B, C bounded and selfadjoint and −ω ′ ≤ C ≤ ω ′ . For f ∈ F [Hω ] we can shift the path of integration to ∂Hω to obtain Z 1 f (A) = f (z)R(z, A) dz 2πi ∂Hω Z Z 1 1 = f (z) (R(z, A) − R(z, A)∗ ) dz + f (z)R(z, A∗ ) dz 2πi ∂Hω 2πi ∂Hω Z Z 1 = f (z)T (z) dz + f (z)R(z, A∗ ) dz, (7.3) 2πi ∂Hω ∂Hω with  1 1 ≤ 0 for Im z ≥ ω (R(z, A) − R(z, A)∗ ) = R(z, A)∗ (C − Im z)R(z, A) ≥ 0 for Im z ≤ −ω 2πi π  −T (z) Im z ≥ ω ] We now deal with the first summand. Let us define T (z) := . Then T (z) Im z ≤ −ω ] T (z) dz = T (z) |dz| on ∂Hω . Hence, Z Z ] S := T (z) dz = T (z) |dz| T (z) :=

∂Hω

∂Hω

is a positive operator. Moreover, by Corollary C.6.4 we can estimate ‚Z ‚ ‚ ‚ ‚ ≤ kf k ‚ f (z)T (z) dz Hω kSk ‚ ‚

(7.4)

∂Hω

so we have to specify kSk. To this end we write „ « „ « „ « 1 1 1 1 1 1 1 T (z) = R(z, A) − + − + − R(z, A∗ ) 2πi z 2πi z z 2πi z 1 R(z, A)A 1 R(z, A∗ )A∗ − Im z − = + 2πi z 2πi z π |z|2 R When we form S = ∂Hω T (z) dz the first and the last summand vanish, by Cauchy’s theorem (deform the path according to ω → ∞). Hence we arrive at Z Z Z ω −ω − Im z S= dz = dr − dr = 2 (7.5) 2 2 2 R π |r − iω| R π |r + iω| ∂Hω π |z| We now treat the second summand, namely Z 1 f (z) R(z, A∗ ) dz 2πi ∂Hω Z Z 1 1 = f (r − iω) R(r + iω, A∗ ) dz + f (r + iω) R(r − iω, A∗ ) dz 2πi R 2πi R Z Z 1 1 = f (z)R(z + 2iω, A) dz − f (z)R(z + 2iω, A) dz 2πi Im z=−ω 2πi Im z=ω Z Z 1 (∗) 1 = f (z)R(z + 2iω, A) dz − f (z)R(z + 2iω, A) dz 2πi R 2πi R Z 1 = f (z) [R(z + 2iω, A∗ ) − R(z − 2iω, A∗ )] dz 2πi R Z −2ω = f (t)R(t + 2iω, A∗ ) R(t − 2iω, A∗ ) dt, π R

7.1. Numerical Range Conditions

157

where in (∗) we have used Cauchy’s theorem. Now write ` ´−1 R(t+2iω, A∗ )R(t − 2iω, A∗ ) = (t − A∗ )2 + 4ω 2 ` ´−1 ` ´−1 = (t − B + iC)2 + 4ω 2 = (t − B)2 + 4ω 2 − C 2 + i[2tC − CB − BC] −1 2

= (Mt + iNt )−1 = Mt

1 −2

(I + iRt )−1 Mt

−1

−1

2 2 with Mt := (t − B)2 + 4ω 2 − C 2 , Nt := 2tC − BC − CB and ‚ Rt := Mt ‚ Nt Mt . Note that all operators Mt , Nt , Rt are selfadjoint and Mt ≥ 3ω 2 . Now ‚(I + iRt )−1 ‚ ≤ 1 for all t since Tt is selfadjoint. Employing Proposition C.6.3 we obtain ‚ ‚ ‚ ‚Z Z 1 ‚ 1 ‚ ‚ 2ω ‚ −2 −1 −1 2 ‚ ‚ ‚ f (z) R(z, A∗ ) dz ‚ ‚ 2πi ‚ = π ‚ f (t)Mt (I + iRt ) Mt dt‚ ∂Hω R ‚Z ‚ ‚ 2ω ‚ ‚ ≤ kf kHω Mt−1 dt‚ (7.6) ‚ ‚. π R ` ´−1 Now, Mt ≥ (t − B)2 + 3ω 2 whence Mt−1 ≤ (t − B)2 + 3ω 2 by Proposition C.4.7. This gives Z Z ` ´−1 Mt−1 dt ≤ (t − B)2 + 3ω 2 dt. (7.7)

R

R ` ´−1 Claim: R (t − B)2 + 3ω 2 dt =

R

√π . 3ω

Proof of Claim. The assertion is readily seen to be true √ by using the Spectral Theorem. Without this powerful tool one can argue as follows (with η := 3ω): Z Z ` ´−1 1 (t − B)2 + η2 dt = (R(t − iη, B) − R(t + iη, B)) dt 2iη R R « „ « „ « Z „ 1 1 1 1 1 R(t − iη, B) − + − − R(t + iη, B) − dt = 2iη R t − iη t − iη t + iη t + iη Z Z 1 R(t − iη, B)B R(t + iη, B)B dt = − dt + 2 + η2 2iη R t − iη t + iη t R Z 1 R(z, B)B π π = dz + = , 2iη ∂Hη z η η by Cauchy’s theorem (make, e.g. a change of variable w := 1z in the integral and note that 1 1 R( w , B) is holomorphic in 0). w ‚R ‚ ‚ ‚ The Claim together with (7.7) shows ‚ R Mt−1 dt‚ ≤ √π . Combining this with (7.3), 3ω (7.4), (7.5), and (7.6) gives „ « 2 2ω π kf (A)k ≤ kf kHω 2 + kf kHω √ = 2+ √ kf kHω π 3ω 3

which concludes the proof.

As a corollary we obtain a statement for ω-accretive operators. Corollary 7.1.15. Let A be m-ω-accretive for some 0 ≤ ω ≤ 2 kf (A)k ≤ 2 + √ kf kSω 3

π 2.

Then (7.8)

for all f ∈ R(Sω ). If A is injective, then (7.8) even holds for all f ∈ H ∞ [Sω ].

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Chapter 7. The Functional Calculus on Hilbert Spaces

√ Proof. Let B := log A and c := 2 + 2/ 3. Then B is Hω -accretive by Corollary 7.1.12. Hence kf (A)k = kf (ez )(B)k ≤ c kf (ez )kHω = c kf kSω by the composition rule.

Remark 7.1.16. Let us consider the following statement, which is considerably weaker than the assertion of Theorem 7.1.14: If A is Hω -accretive and ϕ > ω, then the natural H ∞ (Hϕ )-calculus for A is bounded. This proposition can be proved much easier than Theorem 7.1.14. Namely, one can view A = B + iC as a bounded perturbation of a selfadjoint operator and use an argument similar to the proof of Proposition 5.6.3. This has been done in [52].

7.2 Group Generators on Hilbert Spaces We have seen in the last section, that an m-Hω -accretive operator A on a Hilbert space H has bounded natural H ∞ (Hϕ )-calculi for all ϕ > ω. Due to the numerical range condition, the bounds are even independent of ϕ. In this section we want to weaken the assumptions made on A and nevertheless retain the conclusion (with bounds now being dependent on ϕ.) More precisely, we want to prove the following theorem result from [15]. Theorem 7.2.1. (Boyadzhiev-deLaubenfels) Let iA generate a C0 -group T on the Hilbert space H. Then, for each α > θ(T ), the natural H ∞ (Hα )-calculus for A is bounded. Remark 7.2.2. Let us remark, that Monniaux’s Theorem 4.4.3 in the Hilbert space case is an immediate corollary of Theorem 7.2.1. Indeed, if iB generates the group T then ωst(B) = θ(T ) by Gearharts theorem. If this is smaller than π, eB must be sectorial by Proposition 5.3.2 and Theorem 7.2.1. We will prove Theorem 7.2.1 in two different ways. The first employs Theorem 7.1.14 (or Remark 7.1.16) by an elegant similarity result. The second approach yields even a characterization of group generators on Hilbert spaces. The key idea for our first proof of Theorem 7.2.1 is that the conclusion is not dependent on any particular scalar product on H. So if we can find, for fixed α > θ(T ), an equivalent scalar product ( . | .)◦ on H with respect to which the operator A is Hα -accretive, we are done by the Crouzeix-Delyon Theorem 7.1.14. This method indeed works, see Theorem 7.2.8 below. However, we have to make a little detour before.

7.2. Group Generators on Hilbert Spaces

159

7.2.1 Liapunov’s Direct Method for Groups Recall the classical Liapunov theorem for linear dynamical systems in Cn . Theorem 7.2.3. (Liapunov) Let A ∈ Mat(n, C) with σ(A) ⊂ {z | Re z > 0}. Then there is a Hilbert norm k·k◦ on Cn and ε > 0 such that

−tA

e

≤ e−εt ◦ for all t ≥ 0.

The theorem has two components. First it states that the spectral condition s(−A) := sup{Re λ | λ ∈ σ(−A)} < 0

for the generator −A of a semigroup T (t) = e−tA on the Hilbert space Cn implies exponential stability of the semigroup. Second it states that an exponentially stable semigroup on (Cn , k·k2 ) is similar to a contraction semigroup. One can wonder about infinite-dimensional analogues. Concerning the first part, it is well-known that a sole condition on the position of the spectrum does not imply exponential stability of the semigroup on a general Hilbert space. (See [6, Example 5.3.2] or [26, Example 5.1.4].) Concerning the second part, we will see in Section 7.3.4 below that there is a Hilbert space H and a C0 -semigroup T on H that is not similar to a quasi-contractive semigroup, i.e., for no ω ∈ R one can find an equivalent Hilbert norm k·k◦ on H such that kT (t)k◦ ≤ eωt for all t ≥ 0. So if one looks for infinite dimensional versions of Liapunov’s theorem, one has to impose further conditions. One guiding principle is to look at the proof in finite dimensions. The central point is to find a Liapunov function for the dynamical system given by u˙ + (A − ε)u = 0, i.e., a function which decreases along the orbits (t 7−→ eεt T (t)x). In the finite dimensional situation such a Liapunov function is given by Z ∞

εt

2

e T (t)x 2 dt kxk◦ := (7.9) 0

εt

for x ∈ H. (Note that (e T (t))t≥0 is still exponentially stable.) This is sometimes called Liapunov’s direct method. The problem is that in general (7.9) does not define an equivalent Hilbert norm (see the remark after Proposition 7.2.7 below). (It always defines a continuous norm, though, as it is easily seen.) However, this norm is equivalent if T is a group. Proposition 7.2.4. Let A be the generator of an exponentially stable C0 -semigroup T on H. Then the operator Q defined by Z ∞ Q := T (t)∗ T (t) dt 0

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Chapter 7. The Functional Calculus on Hilbert Spaces

is a bounded, positive, and injective operator on H. By Z ∞ ( x | y )◦ := ( Qx | y ) = ( T (t)x | T (t)y ) dt 0

a continuous scalar product is defined on H. The semigroup T is contractive with respect to k·k◦ and Q satisfies the Liapunov inclusion QA ⊂ −I − A∗ Q.

(7.10)

Equivalently, QD(A) ⊂ D(A∗ ) and QAx + A∗ Qx = −x for all x ∈ D(A). If T is a group then Q is invertible and ( · | ·)◦ is an equivalent scalar product on H. Moreover, one has QA = −I − A∗ Q

and

A = −Q−1 − A◦ ,

(7.11)

where A◦ denotes the adjoint of A with respect to the scalar product ( · | ·)◦ .

Proof. Boundedness and positivity of Q are clear from the definition. Since we have ( Qx | x ) = R∞ kT (t)xk2 dt, the operator Q is injective and ( · | · )◦ is in fact a scalar product (not only 0 a semi-scalar product, see Proposition C.5.1). A simple change of variable yields the formula kT (t0 )xk◦ ≤ kxk◦ for all x ∈ H and t0 ≥ 0. If T is a group, we find M ≥ 1 and ω > 0 such that kT (t)k ≤ M eω|t| for all t ∈ R. Thus, Z ∞ Z ∞ 1 kxk2 kxk2◦ = ( Qx | x ) = kT (t)xk2 dt ≥ M −2 e−2ωt dt kxk2 = 2 2ωM 0 0 for all x ∈ H. Hence in this case the new scalar product is equivalent and Q is invertible (Proposition C.5.1). For the proof of the Liapunov equation note that since T is assumed to be exponentially stable, the resolvent of its generator A is given by Z ∞ −A−1 = R(0, A) = T (s) ds 0

(see Proposition A.8.1). Therefore, Z ∞ Z ∞ Z ∞ Z ∞ Q(−A−1 ) = T (t)∗ T (t) T (s) ds dt = T ∗ (t) T (s) ds dt 0 0 0 t « Z ∞ „Z s = T (t)∗ dt T (s) ds. 0

0

Similarly, −(A∗ )−1 Q =

Z

0

∞

T (t)∗

Z

∞

T (s)∗ T (s) ds dt =

0

Z

0

∞

„Z

∞ s

« T (t)∗ dt T (s) ds.

Adding the two identities we obtain −(QA−1 + (A∗ )−1 Q) =

Z

0

∞

Z

∞

T (t)∗ dt T (s) ds = (A∗ )−1 A−1 .

0

This yields QA−1 = −(A∗ )−1 (A−1 + Q), whence 7.10 holds. Suppose T is a group. Then Q is invertible and A◦ = Q−1 A∗ Q is the adjoint with respect to the new scalar product (see Lemma C.5.2). Multiplying the Liapunov inclusion from the left by Q−1 yields A ⊂ −A◦ − Q−1 . But −A◦ − Q−1 is a bounded perturbation of a C0 -semigroup generator, whence it is also a generator of a C0 -semigroup. This implies readily A = −A◦ − Q−1 . Multiplying by Q from the left yields (7.11).

7.2. Group Generators on Hilbert Spaces

161

Corollary 7.2.5. (Liapunov’s theorem for groups) Let T be an exponentially stable C0 -semigroup on the Hilbert space H. Assume that T is a group. For each 0 < ε < −ω0 (T ) there is an equivalent Hilbert norm k·k◦ on H such that kT (t)k◦ ≤ e−εt

for all t ≥ 0.

Remarks 7.2.6. 1) Since the “Liapunov inclusion” QA ⊂ −I − A∗ Q is not an equation in general, we do not want to call it “Liapunov equation” (which is the name in the finite dimensional setting). However, one can reformulate the inclusion as a system of equations ( Ax | Qy ) + ( Qx | Ay ) = − ( x | y ) with x, y ∈ D(A). In [26, p.160 and p.217] this system is called “Liapunov equation”. 2) We will prove an infinite-dimensional Liapunov theorem for holomorphic semigroups in Corollary 7.3.8.

The next proposition shows in which cases the “direct method” works. Consult [52] for a proof. Proposition 7.2.7. Let A be the R ∞ generator of an exponentially stable C0 -semigroup T on H and define Q := 0 T (t)∗ T (t) dt. Then the following assertions are equivalent. (i) The semigroup T is a group.

(ii) The operator Q is invertible and T (t) has dense range for some t > 0. (iii) The operator Q is invertible and T ∗ (t) is injective for some t > 0. R ˜ := ∞ T (t)T ∗(t) dt are invertible. (iv) Both operators Q and Q 0

(v) The operator Q is invertible and no left halfplane is contained in the residual spectrum of A.

As a consequence we obtain that the direct method works in the case of holomorphic semigroups if and only if the generator is bounded.

7.2.2 A Decomposition Theorem for Group Generators We come back to our original goal. Let A be the generator of a C0 -group T on the Hilbert space H. Recall the definition of the group type θ(T ) in Appendix A on page 210. We fix ω > θ(T ) and define Z ( x | y )◦ := ( T (t)x | T (t)y ) e−2ω|t| dt Z ∞R Z ∞ = ( T (t)x | T (t)y ) e−2ωt dt + ( T (−t)x | T (−t)y ) e−2ωt dt (7.12) 0

0

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Chapter 7. The Functional Calculus on Hilbert Spaces

for x, y ∈ H, i.e., we apply the Liapunov method simultaneously to the rescaled ’forward’ and ’backward’ semigroups obtained from the group T . From Proposition 7.2.4 it is immediate that ( · | ·)◦ is an equivalent scalar product on H. The following theorem summarizes its properties. Theorem 7.2.8. Let iA be the generator of a C0 -group T on a Hilbert space H and let ω > θ(T ). With respect to the (equivalent) scalar product ( · | ·)◦ defined by (7.12) the operator A is m-Hω -accretive. If one writes A = B + iC with B, C selfadjoint and −ω ≤ C ≤ ω with respect to ( . | .)◦ , then D(A) = D(A◦ ) is C-invariant and [B, C] = BC − CB has an extension to a bounded operator skewadjoint with respect to ( . | .)◦ . Note that the decomposition A = B + iC exists by Proposition 7.1.2.

Proof. One has Z kT (t)T (s)xk2 e−2ω|t| dt = kT (t + s)xk2 e−2ω|t| dt R R Z Z = kT (t)xk2 e−2ω|t−s| dt = kT (t)xk2 e−2ω|t| e2ω(|t|−|t−s|) dt

kT (s)xk2◦ =

Z

R

R

≤ e2ω|s| kxk20

(s ∈ R, x ∈ H)

since |t| − |t − s| ≤ |s| for all s, t ∈ R by the triangle inequality. This gives kT (t)k ◦ ≤ eω|t| for all t ∈ R, whence iA is m-Hω -accretive by Proposition 7.1.2. Now, let Z ∞ Z ∞ Q⊕ := T (t)∗ T (t)e−2ωt dt, Q⊖ := T (−t)∗ T (−t)e−2ωt dt, 0

0

and

Q := Q⊕ + Q⊖ =

Z

T (t)∗ T (t)e−2ω|t| dt. R

Then ( x | y )◦ = ( Qx | y ) for all x, y ∈ H. The Liapunov equations for Q⊕ and Q⊖ read Q⊕ (iA − ω) = −I − (−iA∗ − ω)Q⊕

Q⊖ (−iA − ω) = −I − (iA∗ − ω)Q⊖

(7.13) (7.14)

Subtracting the second from the first yields Q−1

QA ⊂ A∗ Q + 2iω(Q⊕ − Q⊖ ).

If we multiply by we arrive at A ⊂ A◦ + 2iωQ−1 (Q⊕ − Q⊖ ). Since D(A) = D(A◦ ) by Proposition 7.1.2, we actually have equality and C = ωQ−1 (Q⊕ − Q⊖ ).

The C-invariance of D(A) is clear from the formula C = ωQ−1 (Q⊕ − Q⊖ ) and D(A) = D(A◦ ) = Q−1 D(A∗ ). Furthermore, employing (7.13), (7.14) and the fact that A◦ = A − 2iC, we compute iCA = ωQ−1 (Q⊕ iA − Q⊖ iA)

= ωQ−1 (−1 + 2ωQ⊕ + iA∗ Q⊕ − 1 + 2ωQ⊖ − iA∗ Q⊖ ) = ωQ−1 (−2 + 2ωQ + iA∗ (Q⊕ − Q⊖ ))

= −2ωQ−1 + 2ω 2 + iωA◦ Q−1 (Q⊕ − Q⊖ ) = −2ωQ−1 + 2ω 2 + iA◦ C

= −2ωQ−1 + 2ω 2 I + i(A − 2iC)C = −2ωQ−1 + 2ω 2 + 2C 2 + iAC.

This shows, that [B, C] = [A, C] has an extension to a bounded operator which is skewadjoint with respect to ( · | · )◦ .

7.2. Group Generators on Hilbert Spaces

163

A simple renaming of terms and forgetting the new scalar product yields the following stunning corollary. Corollary 7.2.9. Let A generate a C0 -group T on a Hilbert space H. Then there exists a bounded operator C such that B := A − C generates a bounded C0 -group. Moreover, the operator C can be chosen in such a way that D(A) is C-invariant and the commutator [A, C] = AC − CA has an extension to a bounded operator on H. Remark 7.2.10. Let A be an Hω -accretive operator on H. By Proposition 7.1.2, A = B + iC with B, C selfadjoint and −ω ≤ C ≤ ω. In general however, D(A) is not C-invariant. In fact, let H := L2 (R) and B = d/dt the generator of the shift group. Furthermore, let C := (f 7→ ωmf ) where m(x) = sgn x is the sign function. Then C is bounded and selfadjoint and A := B +C generates a C0 -group T with kT (t)k ≤ eω|t| . Obviously, D(A) = D(B) = W1,2 (R) is not invariant with respect to multiplication by m. This shows that the additional statement in Theorem 7.2.8 is not a matter of course and is due to the particular way of renorming. Theorem 7.2.8 together with the Crouzeix-Delyon Theorem 7.1.14 immediately imply Theorem 7.2.1.

7.2.3 A Characterization of Group Generators We now turn to a second proof of Theorem 7.2.1. In fact, we will prove a more general result characterizing generators of C0 -groups on Hilbert spaces. In this section we consider a strip type operator A ∈ Strip(ω) on the Hilbert space H. Let ϕ1 , ϕ2 ∈ R \ [−ω, ω]. Then we have (t 7→ R(t + iϕ1 , A)x) ∈ L2 (R, H)

⇐⇒

(t 7→ R(t + iϕ2 , A)x) ∈ L2 (R, H).

for each x ∈ H. (Use the resolvent identity and the fact, that the resolvent is uniformly bounded on the horizontal lines R + iϕ1 and R + iϕ2 ). We say that A allows quadratic estimates, if for every ϕ ∈ R \ [−ω, ω] there is c = c(A, ϕ) such that Z

R

2

kR(t + iϕ, A)xk dt ≤ c(A, ϕ) kxk

2

(x ∈ H).

From the remarks above and the Closed Graph Theorem it follows that A allows quadratic estimates if and only if there exists ϕ ∈ R \ [−ω, ω] such that R(· + iϕ, A)x ∈ L2 (R, H) for every x ∈ H. If A allows quadratic estimates then also −A and A + λ do, for each λ ∈ C.

164

Chapter 7. The Functional Calculus on Hilbert Spaces

Example 7.2.11. Let A ∈ Strip(ω) and assume that iA is the generator of a C0 semigroup T on H. We claim that A allows quadratic estimates. In fact, we can find constants M, ω0 such that kT (t)k ≤ M eω0 t for all t ≥ 0. Then Z ∞ R(λ, A)x = e−λt T (t)x dt 0

for all Re λ > ω0 . An application of the Plancherel Theorem C.8.1 now yields Z Z Z ∞

−ϕs

2 2 2

e kR(t − iϕ, A)xk dt = kR(it, A − ϕ)xk dt = 2π T (s)x ds R

R

0

πM 2 2 ≤ kxk ϕ − ω0

for all ϕ > ω, ω0 and all x ∈ H. Since also iA∗ = −(iA)∗ generates a C0 -semigroup, the operator A∗ also allows quadratic estimates. We can now state the main result. Theorem 7.2.12. Let H be a Hilbert space and A ∈ Strip(ω) a strip type operator on A, where ω ≥ 0. Assume that A is densely defined. Then the following assertions are equivalent. (i) The natural H ∞ (Hα )-calculus for A is bounded, for one / all α > ω. (ii) The operator iA generates a C0 -group. (iii) The operator iA generates a C0 -semigroup. (iv) The operators A and A∗ both allow quadratic estimates. If iA generates the C0 -group T , then ω(T ) ≤ ω. Note that “Gearhart’s theorem for groups” is implicit in Theorem 7.2.12. Let us turn to the Proof of Theorem 7.2.12. The implication (ii) ⇒ (iii) is obvious, and (iii) ⇒ (iv) is in Example 7.2.11. To prove (i) ⇒ (ii) one only has to note that the boundedness of the H ∞ -functional calculus for A on some horizontal strip Hα implies the Hille-Yosida conditions for iA and −iA (see Theorem A.8.6). Hence iA generates a C0 -group T . Futhermore, we see that in this case the group type of T is as most as large as α. To establish the implication (iv) ⇒ (i) needs a little more effort. We fix ω < α < µ. We will need the auxiliary function ψ defined by ψ(z) :=

(µ2

c + z 2 )2

where c is chosen such that Z Z ψ(t) dt = R

R

(z ∈ Hα )

c dt = 1. (µ2 + t2 )2

7.2. Group Generators on Hilbert Spaces

165

(One can easily compute c = (4µ3 )/π). For a given f ∈ H ∞ (Hα ) we now define the approximants fn by Z n Z n (f ψt )(z) dt = f (z) ψ(z + t) dt (z ∈ Hα ), (7.15) fn (z) := −n

−n

where here and in the following for a function g on the strip Hα we will denote by gt the function gt := (z 7−→ g(t + z)) : Hα −→ C. The next lemma collects the properties of these approximants. Lemma 7.2.13. Let f ∈ H ∞ (Sα ) and let the sequence (fn )n be defined by (7.15). Then the following holds. a) fn ∈ F(Hα ) for all n. b) supn kfn k∞ < ∞.

c) fn → f pointwise on Sα .

d) The function (t 7→ (f ψt )(A)) : R −→ L(H) is continuous and Z n fn (A) = (f ψt )(A) dt ∈ L(H). −n

e) supn kfn (A)k < ∞. Proof. Ad a). By elementary Complex Analysis it is clear that fn is holomorphic on Hα for each n. We can choose d > 0 such that |ψ(z)| =

c |µ2 + z 2 |2

d

≤

(1 + |Re z|2 )2

(z ∈ Hα ).

For fixed n ∈ N one can find dn > 0 such that 1 1 + |Re z + t|2

≤

dn 1 + |Re z|2

(z ∈ Hα , |t| ≤ n).

With the help of this we can compute |fn (z)| ≤ kf k∞

Z

n

−n

|ψ(z + it)| dt ≤ kf k∞ ≤ kf k∞

n

d dt (1 + |Re z + t|2 )2 „Z « d dn ddn π dt = kf k∞ 2 2 1 + |Re z| 1 + |Re z|2 R 1+t

Z

−n

for z ∈ Hα . This proves a). Ad b). Let n ∈ N and z ∈ Hα . Then Z |fn (z)| ≤ kf k∞ |ψ(z + t)| dt R Z Z d d ≤ kf k∞ dt = kf k dt. ∞ 2 2 2 2 R (1 + |Re z + t| ) R (1 + t ) Ad c). By b) and Vitali’s Theorem is suffices to show that fn (z) → f (z) for all z ∈ R. But this is obvious from (7.15). Ad d). This is immediate from the Convergence Lemma (Proposition 5.1.6).

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Chapter 7. The Functional Calculus on Hilbert Spaces

Ad e). We let η(z) := 1/(µ2 + z 2 ). Hence, we have ψ = cη2 . Choose ω < ω1 < α and let γ := γω1 = ∂Hω1 . Now we fix t ∈ R and compute Z Z 1 ds d |z| = 2 2 2 2 2 γ |µ + (z + t) | R |µ + (s + iω1 ) | Z ds = 2 R |(s + iω1 + iµ)(s + iω1 − iµ)| „Z « 1 „Z «1 2 2 ds ds ≤ 2 2 2 |s + iω + iµ| |s + iω − iµ| R R 1 1 „Z « 1 „Z «1 2 2 ds ds = 2 2 2 2 2 R (µ − ω1 ) + s R (µ + ω1 ) + s 2π = q . µ2 − ω12 Using this we can estimate k(f ηt )(A)k for each t ∈ R by ‚ ‚ Z ‚ 1 ‚ f (z) k(f ηt )(A)k = ‚ R(z, A) dz ‚ ‚ 2πi ‚ 2 2 γ µ + (z + t) Z L(A, ω1 ) 1 ≤ kf k∞ d |z| 2 2 2π γ |µ + (z + t) | L(A, ω1 ) 2π kf k∞ q 2π 2 µ − ω12

≤

L(A, ω1 ) q kf k∞ . µ2 − ω12

=

Thus, for arbitrary x, y ∈ H one has ˛Z n ˛ ˛ ˛ ( (f ψt )(A)x | y ) dt˛˛ |( fn (A)x | y )| = ˛˛ −n ˛Z n ˛ ˛ ˛ ˛ = c˛ ( ηt (A)(f ηt )(A)x | y ) dt˛˛ −n ˛Z n ˛ ˛ ˛ ˛ = c˛ ( (f ηt )(A)R(iµ − t, A)x | R(iµ − t, A∗ )y ) dt˛˛ −n Z c L(A, ω1 ) ≤ q kf k∞ kR(t + iµ, A)xk kR(t + iµ, A∗ )yk dt R µ2 − ω12 ≤

c

q

µ2

− ω12

„Z

R

≤

c

q

L(A, ω1 ) kf k∞

« 1 „Z «1 2 2 kR(t + iµ, A)xk2 dt kR(t + iµ, A∗ )yk2 dt

µ2 − ω12

R

∗

L(A, ω1 ) c(A, µ) c(A , µ) kf k∞ kxk kyk .

In particular, this shows that

c kfn (A)k ≤ kf k∞ q L(A, ω1 ) c(A, µ) c(A∗ , µ) µ2 − ω12

for each n. Thus e) is completely proved.

7.3. Similarity Theorems for Sectorial Operators

167

It is now easy to complete the proof of Theorem 7.2.12. We simply apply the Convergence Lemma (Proposition 5.1.6) to the sequence (fn (A))n∈N . Hence, we obtain the boundedness of the functional calculus and, more explicitly, 4µ3 kf (A)k ≤ kf k∞ p L(A, ω1 ) c(A, µ) c(A∗ , µ) 2 2 π µ − ω1

(7.16)

for all f ∈ H ∞ (Hα ).

Corollary 7.2.14. (Liu) Let A be the generator of a C0 -semigroup T on the Hilbert space H. If the resolvent of A exists and is uniformly bounded on a left halfplane, then T is a group.

7.3 Similarity Theorems for Sectorial Operators In the last section we have seen how the boundedness of an H ∞ -calculus could be proved by first changing the scalar product appropriately. In the current section we will do it the other way round. We will see, how the boundedness of H ∞ -calculus for a sectorial operator leads to an interesting similarity theorem.

7.3.1 The Theorem of McIntosh The result which is the objective of the present section is among the most important results in the theory of functional calculus. Theorem 7.3.1. (McIntosh) Let A be an injective sectorial operator on the Hilbert space H. Then the following assertions are equivalent: (i) For some/each 0 6= f ∈ H0∞ [SωA ] there are constants C1 (f ), C2 (f ) > 0 such that Z ∞ 2 2 dt 2 C1 (f ) kxk ≤ kf (tA)xk ≤ C2 (f ) kxk (7.17) t 0 for all x ∈ H.

(ii) For one/all ϕ ∈ (ωA , π), the natural H ∞ (Sϕ )-calculus for A is bounded.

(iii) A ∈ BIP(H).

The implication (ii) ⇒ (iii) is obvious. To prove the converse, we go over to B := log A. If A ∈ BIP(H), iB generates the C0 -group T := (Ais )s∈R . By the Boyadzhiev-deLaubenfels Theorem 7.2.1, B has bounded H ∞ -calculus on every strip Hα , with α > θ(T ) = θA . But θA = ωA (Corollary 4.3.5), hence (ii) follows from Proposition 5.3.2. Reformulating statement (i) yields H = {x ∈ H | t−1/2 f (tA)x ∈ L2∗ ((0, ∞); H)}

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Chapter 7. The Functional Calculus on Hilbert Spaces

with equivalent norms kxkH ∼ t−1/2 f (tA)x L2 ((0,∞);H) (we use the terminology ∗ of Chapter 6). Employing Theorem 6.4.6 and some density arguments, this is equivalent to the statement ˙ R) ˙ 1 H = (D, (7.18) 2 ,2 where D˙ is the homogeneous domain and R˙ is the homogeneous range space (see Section 6.3.3). Hence the implication (i) ⇒ (ii) follows from Theorem 6.5.6. So only the implication (ii) ⇒ (i) remains to be proved. (Alternatively, we could prove the implication (iii) ⇒ (i). A sketch of this will be given in Remark 7.3.2.) We have to introduce a new concept. The Cantor group is the set G := {−1, 1}Z, i.e., the Z-fold direct product of the multiplicative discrete group Z2 ∼ = {−1, 1}. By Tychonoff’s theorem, G is a compact topological group. We denote by µ the normalized Haar measure on G. The projections εk := ((gn )n 7−→ gk ) : G −→ {−1, 1} (k ∈ Z) are called Rademacher functions. As the Rademachers obviously are continuous characters of the compact group G, they form an orthonormal set in L2 (G, µ), i.e., Z εn εm dµ = δnm G

for all n, m ∈ Z. (One can show that the set of Rademachers actually generates the character group of G.) Given any Banach space X, the space Rad(X) is defined by Rad(X) := span{εn ⊗ x | n ∈ Z, x ∈ X} ⊂ L2 (G, X)

with the norm induced by the norm of L2 (G, X)1 . If X = H is a Hilbert space, we have the important identity

X

2 X

kxn k2H = εn ⊗ xn (7.19) n

n

Rad(H)

for every finite two-sided sequence (xn )n∈Z ⊂ H. In fact, Z X

X

2

2

εn ⊗ xn = εn (g)xn µ(dg)

n n Rad(H) H G Z X = εn (g)εm (g) ( xn | xm ) µ(dg) n,m G X X 2 = δnm ( xn | xm ) = kxn kH . n,m

n

Let us return to the main theme.

1 Actually,

the Khintchine-Kahane inequality asserts, that the norms on Rad(X) induced by the different embeddings Rad(X) ⊂ Lp (G, X) for 1 ≤ p < ∞ are all equivalent, see [82, Part I, Theorem 1.e.13].

7.3. Similarity Theorems for Sectorial Operators

169

Proof of the implication (ii) ⇒ (i). Let C be such that kg(A)k ≤ C kgkSϕ for all g ∈ H0∞ (Sϕ ) and let f ∈ H0∞ (Sϕ ). We have Z

2N 2−N

kf (tA)xk2

Z 2k+1 Z 2‚ N N ‚2 dt X X dt dt ‚ ‚ = kf (tA)xk2 = ‚f (t2k A)x‚ k t t t 2 1 k=−N k=−N

‚2 Z 2 ‚X N ‚ ‚2 dt X ‚ ‚ N dt ‚ ‚ ‚ = εk ⊗ f (t2k A)x‚ ‚f (t2k A)x‚ ‚ ‚ −N t 1 1 k=−N Rad(H) t ‚2 Z 2 ‚X ‚ ‚ N dt ‚ ≤ εk ⊗ f (t2k A)‚ kxk2 . ‚ ‚ −N 1 Rad(L(H)) t

=

Z

2

At this point we use the Unconditionality Lemma 5.7.1 to conclude that ‚ ‚ ‚ XN ‚ ‚ εk (g)f (t2k A)‚ ≤ DC ‚ ‚ −N

L(H)

for all g ∈ G, t > 0, and N ∈ N, where D is a constant which does only depend on f and ϕ. This implies Z ∞ dt ≤ C 2 D 2 (log 2) kxk2 kf (tA)xk2 t 0 for all x ∈ H. So we are left to show the second inequality. For this we observe that by Proposition 7.0.1 the operator A∗ also satisfies the hypothesis of the proposition, even with the same constant C. By what we have shown, Z ∞ dt kf (tA∗ )xk2 ≤ (log 2)C 2 D 2 kxk2 t 0 for all x ∈ H. Consider the function g := f ∗ f . Obviously, g ∈ H0∞ (Sϕ ) and g(t) := |f (t)|2 for t > 0. Since f 6= 0 and f is holomorphic, we have Z ∞ dt α := |f (t)|2 > 0. t 0 By Theorem 5.2.6 we can compute ˛ « Z „Z ∞ dt ˛˛ x = α kxk2 = ( αx | x ) = g(tA)x t ˛ 0

=

Z

∞

0

( f (tA)x | f (tA∗ )x ) 1

≤ CD(log 2) 2 kxk

„Z

0

∞

dt ≤ t

„Z

∞

0

kf (tA)xk2

∞ 0

( f ∗ (tA)f (tA)x | x )

kf (tA)xk2

dt t

«1

dt t

« 1 „Z 2

0

dt t

∞

kf (tA∗ )xk2

dt t

«1 2

2

1

for all x ∈ D(A)∩R(A). We can divide by (log 2) 2 CD kxk and approximate an arbitrary element of H by elements from D(A) ∩ R(A) to finish the proof.

Remark 7.3.2. We sketch a second proof of the implication (iii) ⇒ (i) using 1 interpolation spaces. Namely, since A, A−1 ∈ BIP, one has D(A 2 ) = [H, D(A)] 12 1

and R(A 2 ) = [H, R(A)] 12 by Theorem 6.6.9. On the other hand, it is known that on Hilbert spaces real interpolation spaces (with p = 2) and complex interpolation spaces are equal (see [87, Cor. 4.3.12]). Hence, we have 1

(H, D(A)) 12 ,2 = D(A 2 )

1

and (H, D(A−1 )) 12 ,2 = R(A 2 ).

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Chapter 7. The Functional Calculus on Hilbert Spaces

Now we can compute 1

1

1

A 2 (1 + A)−1 (H) = D(A 2 ) ∩ R(A 2 ) = (H, D) 12 ,2 ∩ (H, R) 12 ,2 ˙ 1 ,2 = A 12 (1 + A)−1 X ( 12 ) , X (− 12 ) = (D, R) 12 ,2 = (1 + A)−1 (X, R) 1 2

and this gives H = D(1/2) , R we wanted to prove.

(1/2)

2 ,2

1 2 ,2

and this by Theorem 6.4.6 is exactly what

7.3.2 Interlude: Operators Defined by Sesquilinear Forms This section has a preparatory character. We briefly review the construction of operators by means of elliptic forms. This construction provides us with two interesting questions concerning similarity, which will be answered in the subsequent section with the help of the McIntosh’s theorem. The following scheme is an abstraction of the standard way for obtaining L2 realizations of elliptic differential operators in divergence form, see [41, Chapter 6]. Let H be a Hilbert space, V ⊂ H a dense subspace and a ∈ Ses(V ) a sesquilinear form on V . For λ ∈ C we define the form aλ ∈ Ses(V ) by aλ (u, v) := a(u, v) + λ ( u | v )H for u, v ∈ V . The form a is called elliptic if there is λ0 ≥ 0 such that the following two conditions hold. 1) The form Re aλ0 = (Re a) + λ0 ( . | .)H is a scalar product on V which turns V into a Hilbert space such that the inclusion mapping V ⊂ H is continuous. 2) The form a is continuous with respect to this scalar product on V . With an elliptic form a on V ⊂ H, we associate an operator A in the following way. We define a norm on V by 2

2

kukV := Re a(u) + λ0 kukH for u ∈ V . Condition 1 then implies that V is a Hilbert space. Since V is continuously embedded in H (also by Condition 1) there is a continuous embedding of H ∗ into V ∗ (injectivity follows since V is dense in H). If we identify H with its antidual by means of the scalar product ( · | ·)H of H (Riesz-Fr´echet Theorem), we obtain a sequence of continuous embeddings V ⊂ H(∼ = H ∗) ⊂ V ∗.

In doing this, H becomes a dense subspace of V ∗ . Now we define the mapping A : V −→ V ∗ by A := (u 7−→ a(u, .)) : V −→ V ∗

By means of the identifications above we obtain (A +λ0 )(u) = a(u, .) + λ0 ( u | .)H . The operator B := A +λ0 : V −→ V ∗ is an isomorphism.

7.3. Similarity Theorems for Sectorial Operators

171

Proof. Consider the form aλ0 . Then Re aλ0 is the scalar product of V , hence a is coercive (see (C.5) on page 235). By Condition 1, a is also continuous, hence satisfies the hypotheses of the Lax-Milgram Theorem C.5.3.

Employing the embedding H ⊂ V ∗ we define D(A)

:= =

{u ∈ V | there is y ∈ H s.t. a(u, ·) = ( y | ·)H } {u ∈ V | A u ∈ H}

and Au := A u regarded as an element of H. Then the two fundamental identities a(u, v) = hA u, vi (u ∈ V ) a(u, v) = ( Au | v )H (u ∈ D(A))

(7.20) (7.21)

hold for v ∈ V . The operator A is called associated with the form a (notation: a ∼ A). (Note that the definition of A is actually independent of the chosen λ0 .) An operator A on H is called Kato-sectorial if there is a dense subspace V ⊂ H and an elliptic form a ∈ Ses(V ) on V such that a ∼ A. Clearly, if A is Katosectorial, then A + λ is Kato-sectorial for each λ ∈ C. Furthermore, also A∗ is Kato-sectorial (if A ∼ a then A∗ ∼ a). Remarks 7.3.3. 1) We use the name “Kato-sectorial” for the following reason. Originally, these operators were treated extensively by Kato. In [65] he calls these operators just “sectorial” but meanwhile this name is mainly used as we defined it in Chapter 1. 2) Very often the starting point is a bit different in that V already carries a (natural) Hilbert space structure such that V is densely and continuously embedded in H. We denote this given scalar product on V by ( · | ·)V . One can then replace the Conditions 1) and 2) by 2

3) Re aλ (u) ≥ δ kukV for some δ > 0 and all u ∈ V . 4) |a(u, v)| ≤ M kukV kvkV for some M ≥ 0 and all u, v ∈ V .

Here, Condition 4) says that a is continuous on V and this together with Condition 3) implies that Re aλ is an equivalent scalar product on V , hence conditions 1) and 2). The following characterization of Kato-sectorial operators is well-known. We include the proof for convenience of the reader. Proposition 7.3.4. Let A ∼ a where a : V × V −→ H is an elliptic form. Then there is 0 ≤ ω < π2 such that A + λ0 is an invertible m-ω-accretive operator on H. (Here, λ0 is such that the conditions 1) and 2) above hold.) Conversely, let A be an m-ω-accretive operator on H for some Let 0 ≤ ω < π2 . Then A is Kato-sectorial; more precisely, there is a dense subspace V ⊂ H and an elliptic form a ∈ Ses(V ) such that A ∼ a and Re a is positive.

172

Chapter 7. The Functional Calculus on Hilbert Spaces

Proof. Assume that A ∼ a and a : V × V −→ H is elliptic, and λ0 is chosen as above. Then the operator B := A + λ0 : D(A) −→ H is bijective and B −1 is a bounded operator on H, since B is the restriction of B : V −→ V ∗ to the range H. Because H is dense in V ∗ , D(A) = D(B) is dense in V , hence a forteriori in H. This means that B is a densely defined closed operator in H with 0 ∈ ̺(B). Claim: The form aλ0 is continuous with respect to the scalar product Re aλ0 .

Proof of Claim: Since the embedding V ⊂ H is continuous, there is a constant M1 such that kuk2H ≤ M1 kuk2V for all u ∈ V . The continuity of the form a (Condition 2) yields the existence of a constant M2 such that |a(u, v)| ≤ M2 kuk2V . Putting the two inequalities together we obtain ˛ ˛ ˛aλ (u, v)˛ ≤ (M2 + λ0 M1 ) kuk2 .] V 0

Now a short glance on Proposition C.1.3 shows that the form aλ0 is sectorial of some angle ω which depends on the continuity constants of aλ0 with respect to Re aλ0 . Since ( Bu | u )H = aλ0 (u, u) for u ∈ D(B) the operator B is m-ω-accretive. Conversely, let A be m-ω-accretive. On D(A) we define the sesquilinear form a by a(u, v) := ( Au | v )H

and a scalar product (!) by ( u | v )V := (Re a)(u, v) + ( u | v )H . The form a is continuous with respect to this scalar product. In fact, since a is sectorial of angle ω by Proposition C.1.3 we have p p |a(u, v)| ≤ (1 + tan ω) Re a(u) Re a(v) ≤ (1 + tan ω) kukV kvkV

for all u, v ∈ D(A). Obviously, the embedding (D(A), k.kV ) ⊂ H is continuous. Hence it has a continuous extension ι : V −→ H, where V is the (abstract) completion of D(A) with respect to k.kV Claim: The mapping ι is injective.

Proof of Claim. Let x ∈ V and ιx = 0. This means that there is (un )n ⊂ D(A) such that kun − um kV → 0 and kun kH → 0. Hence for all n, m ∈ N we have Re a(un )

= = ≤ ≤

Re ( Aun | un )H = Re ( Aun | un − um )H + Re ( Aun | um )H Re[a(un , un − um )] + Re ( Aun | um )H ˛ ˛ |a(un , un − um )| + ˛( Aun | um ) ˛ H

M kun kV kun − um kV + kAun kH kum kH .

Since (un ) is V -Cauchy, C := supn kun kV < ∞. Hence Re a(un ) ≤ M C lim sup kun − um kV → 0 m

as n → ∞.

This shows that kun kV → 0, whence x = 0.

We therefore can regard V as being continuously and densely embedded into H. The form a has a unique extension to a continuous sesquilinear form on V (again denoted by a). Clearly, the form a is elliptic (with λ0 = 1). Hence it remains to show that A is associated with a. We denote by B the operator which is associated with a and choose u ∈ D(A). The equation a(u, v) = ( Au | v )H holds for all v ∈ V since it holds for all v ∈ D(A), D(A) is dense in V , a is continuous on V , and V is continuously embedded in H. This shows u ∈ D(B) and Bu = Au, whence A ⊂ B. On the other hand, we have −1 ∈ ̺(B) by construction and −1 ∈ ̺(A) since A is m-ω-accretive. This implies that A = B.

Combining Proposition 7.3.4 with Proposition 7.1.1 yields the following corollary.

7.3. Similarity Theorems for Sectorial Operators

173

Corollary 7.3.5. An operator A on H is Kato-sectorial if and only if there is λ ∈ R such that A + λ is m-ω-accretive for some ω < π2 if and only if −A generates a holomorphic semigroup T on some sector Sθ such that kT (z)k ≤ eλ Re z

(z ∈ Sθ )

for some λ ∈ R. Note that the corollary says that A is Kato-sectorial if and only if A is “quasi-m-ω-accretive” for some 0 ≤ ω < π2 . The construction of operators by means of sesquilinear forms, though quite powerful, raises two highly nontrivial questions. The first is due to the fact, that contrary to many other concepts, the notion of “Kato-sectoriality” relies heavily on the particular chosen scalar product. That is, an operator may be Katosectorial with respect to the original but not Kato-sectorial with respect to some new (although equivalent) scalar product on H. (A concrete example is given in [7].) In fact, a result of Matolcsi says that if A is not a bounded operator, one always can find a scalar product on H such that A is not Kato-sectorial (not even quasi-m-accretive, see [90]). It is therefore reasonable to ask for a characterization of Kato-sectorial operators modulo similarity. Such a characterization can indeed be given with the help of the functional calculus (see Corollary 7.3.10 below.) To understand the second problem, we have to cite a stunning theorem from [62]. Theorem 7.3.6. (Kato) Let A be m-accretive and 0 ≤ α < 21 . Then the following assertions hold: 1) D(Aα ) = D(A∗α ) =: Dα . 2) kA∗α uk ≤ tan π(1+2α) kAα uk for all u ∈ Dα . 4 α ∗α 3) Re ( A u | A u ) ≥ (cos πα) kAα uk kA∗α uk for all u ∈ Dα .

Proof. See [62, Theorem 1.1]. The proof proceeds in two steps. First the statements are proved under the additional hypothesis of A being bounded with Re ( Au | u ) ≥ δ kuk2 for all u ∈ H and some δ > 0. (This is the difficult part of the proof.). The second (easy) step uses “sectorial approximation” (in our language) and Proposition 3.1.9.

Kato’s theorem is remarkable, also in that it fails in the case α = 21 . Kato was not aware of this when he wrote the article [62] but only shortly afterwards, Lions in [83] produced a counterexample. For some time it had been an open question if at least for m-ω-accretive operators A it is true that 1

1

D(A 2 ) = D(A 2 ∗ ).

(7.22)

If A is associated with the form a ∈ Ses(V ) and, say, we have λ0 = 0, then (7.22) is 1 equivalent to V = D(A 2 ), as Lions in [83] and Kato in [63] have shown. Finally, McIntosh gave a counterexample in [91]. However, this was not the end of the story. In fact, even if the statement is false in general it might be true for particular

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Chapter 7. The Functional Calculus on Hilbert Spaces

operators such as second order elliptic operators on L2 (Rn ) in divergence form. For these operators, the problem became famous under the name Kato’s Square Root Problem and has stimulated a considerable amount of research which lead to the discovery of deep results connecting Operator Theory and Harmonic Analysis (see also the comments in Section 7.5). Let us call an operator A on the Hilbert space H square root regular if A + λ 1 1 is sectorial and D((A + λ) 2 ) = D((A + λ) 2 ∗ ) for large λ ∈ R.

Note that whether an operator is square root regular does also depend on the particular chosen scalar product. (If the scalar product changes, A stays the same while the adjoint A∗ changes.) Hence it is reasonable to ask the following: Assume that A is Kato-sectorial with respect to some scalar product. Is there an equivalent scalar product on H such that A is Kato-sectorial and square root regular with respect to the new scalar product? This is indeed the case (see Corollary 7.3.10 below).

7.3.3 Similarity Theorems After so much propedeutical work we can now bring in the harvest. The first similarity result actually has a very interesting historical background (see the comments in Section 7.5 of this chapter). Theorem 7.3.7. (Callier-Grabowski-LeMerdy) Let −A be the generator of a bounded holomorphic C0 -semigroup T on the Hilbert space H. Then T is similar to a contraction semigroup if and only if B ∈ BIP(R(A)), where B is the injective part of A. Recall that the injective part of A is the part of A in K := R(A). Clearly, A is m-ω-accretive if and only if B is. Proof. For abbreviation we define K := R(A). Assume that T is similar to a contraction semigroup. Changing the scalar product we can assume that A is m-accretive. Hence B ∈ BIP(K) by Theorem 7.1.7. To prove the converse, assume that B ∈ BIP(K). Since −A generates a bounded holomorphic semigroup, A is sectorial with ωA < π2 . By Theorem 7.3.1 we can change the norm on R 1 1 K to ( 0∞ kf (tA)xk2 dt/t) 2 where f ∈ H0∞ [SωA ] is arbitrary. If we choose f (z) := z 2 e−z we obtain the new norm Z ∞‚ Z ∞‚ ‚2 dt ‚2 1 ‚ ‚ ‚ 21 ‚ kxk2new = = ‚(tA) 2 e−tA x‚ ‚A T (t)x‚ dt t 0 0 on K. Employing the semigroup property it is easy to see that each T (t) is contractive on K with respect to this new norm. However, on N(A) each operator T (t) acts like the identity. Since H = N(A) ⊕ K we can choose the new scalar product on H in such a way that it is the old one on N(A), the one just constructed on K and N(A) ⊥ K. With respect to this new scalar product, the semigroup T is contractive.

For a different proof of Theorem 7.3.7 using interpolation theory, see the Comments in Section 7.5 of this Chapter. As an immediate corollary we obtain the announced Liapunov theorem for holomorphic semigroups.

7.3. Similarity Theorems for Sectorial Operators

175

Corollary 7.3.8. (Liapunov’s theorem for holomorphic semigroups) Let A be an operator on a Hilbert space H. Assume that −A generates a holomorphic C0 -semigroup T and s(−A) < 0. If A + λ ∈ BIP(H) for some λ > 0 then, for every 0 < ε < −s(−A) there is an equivalent Hilbert norm k·k◦ on H such that kT (t)k◦ ≤ e−εt for all t ≥ 0. Proof. Choose 0 < ε < δ := −s(−A) and consider the operator B := A − ε. We claim that B is still sectorial with ωB < π2 . For this, it suffices to show that {λ | Re λ < 0} ⊂ ̺(B) with kλR(λ, B)k being uniformly bounded for Re λ < 0. Since 0 ∈ ̺(B) weponly have to consider |λ| ≥ R for some radius R > 0. We choose ω < ω ′ < π2 and let R := ε2 (1 + tan2 ω ′ ). Now λ λR(λ, B) = λ+ε [(λ + ε)R(λ + ε, A)] and the factor λ/(λ + ε) is uniformly bounded for |λ| ≥ R. The second factor is uniformly bounded for Re λ ≤ −ε hence we have to check that sup{k(λ + ε)R(λ + ε, A)k | − ε ≤ Re λ ≤ 0, |λ| ≥ R} < ∞. √ But |λ| ≥ R together with −ε ≤ Re λ ≤ 0 implies |Im(λ + ε)| = |Im λ| ≥ R2 − ε2 = ε tan ω ′ , whence |arg(λ + ε)| ≥ ω ′ . Since ω ′ > ωA , the claim is proved. So −B generates a bounded holomorphic C0 -semigroup. Moreover, B + λ ∈ BIP(H) for some λ > 0. By b) of Proposition 3.4.5, B is ∈ L(H) for all s ∈ R. Applying Corollary 3.4.7 we obtain B ∈ BIP(H). We can now apply the Callier-Grabowski-LeMerdy Theorem 7.3.7. This yields an equivalent scalar product such that (eεt T (t))t≥0 becomes contractive.

Recall that one can write down the new scalar product in Proposition 7.3.8. In fact, it follows from the argument in the proof of Theorem 7.3.7 that Z ∞

2 1

2 kxk◦ =

(A − ε) 2 eεt T (t)x dt 0

2

for x ∈ H. The function k·k◦ is a Liapunov function for the dynamical system given by u˙ + (A − ε)u = 0. Now we can state our main result.

Theorem 7.3.9. (Similarity Theorem) Let A be a sectorial operator on the Hilbert space H satisfying ωA < π2 . Assume that A satisfies the equivalent properties (i) to (iv) of Theorem 7.3.1. Then, for each ωA < ϕ < π2 there is an equivalent scalar product ( · | ·)◦ on H with the following properties. 1) N(A) ⊥ R(A) with respect to ( · | ·)◦ . 2) The operator A is m-ϕ-accretive with respect to ( · | ·)◦ . π 3) One has D(Aα ) = D(A◦α ) for 0 ≤ α ≤ 4ϕ . Here, A◦ denotes the adjoint of A with respect to ( · | ·)◦ . 4) One has kf (A)k◦ ≤ kf kϕ for all f ∈ E(Sϕ ) ∩ H ∞ (Sϕ ). Note that

π 4ϕ

1

1

> 21 . Hence in particular D(A 2 ) = D(A◦ 2 ).

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π β ′ Proof. We choose ωA =: ω < ϕ′ < ϕ and define β := 2ϕ ′ and B := A . Then B ∈ Sect(ω ), ω π π ′ where ω = βω = ϕ′ 2 < 2 . Hence −B generates a bounded holomorphic C0 -semigroup. By hypothesis, A satisfies the condition (i) of Theorem 7.3.1. Applying Proposition 3.1.4 we see that B has the same property. By Theorem 7.3.7 there is an equivalent scalar product ( · | · )◦ 1

which makes B m-accretive and such that N(A) ⊥ R(A). Now, A = B β , whence by Corollary π 7.1.11, A is m- 2β -accretive with respect to the new scalar product. Moreover, Kato’s theorem 7.3.6 says that D(Aαβ ) = D(B α ) = D(B ◦α ) = D(A◦αβ ) for all 0 ≤ α < 21 . Thus, Assertion 3) π π π follows since 0 < αβ < β2 = 4ϕ ′ and 4ϕ′ > 4ϕ . If we take a bounded f ∈ E(Sϕ ) then ‚ ‚ ‚ 1 1 ‚ ‚ ‚ ‚ ‚ kf (A)k◦ = ‚f (z β )(B)‚ ≤ ‚f (z β )‚ ◦

Sπ 2

= kf kS

π 2β

= kf kS

ϕ′

≤ kf kSϕ

by Proposition 3.1.4 and Theorem 7.1.7.

As a corollary, we obtain a simultaneous solution to the similarity problems posed in Section 7.3.2 (see page 173). Corollary 7.3.10. Let A be a closed operator on a Hilbert space. Then A is Katosectorial with respect to some equivalent scalar product if and only if −A generates a holomorphic C0 -semigroup and A + λ ∈ BIP(H) for large λ ∈ R. In this case, the scalar product can be chosen such that A is Kato-sectorial and square root regular. We state as another corollary a theorem which (historically) was the starting point for our investigations. Corollary 7.3.11. (Franks-LeMerdy) Let A be an injective sectorial operator on the Hilbert space H. Assume that A ∈ BIP(H). Then for each ϕ ∈ (ωA , π) there is an equivalent scalar product on H with respect to which the natural H ∞ (Sϕ )-calculus for A is contractive. Proof. If ωA <

π 2

one can just apply the Similarity Theorem 7.3.9 in combination with Propo-

sition 5.3.3. In case ωA ≥

π 2

1

we apply this to the operator A 2 and use Proposition 3.1.4.

Finally we give an application of the Similarity Theorem to derive a dilation theorem for groups on Hilbert spaces. Corollary 7.3.12. (Dilation theorem for groups) Let T be a C0 -group on the Hilbert space H. For each ω > θ(T ) there is a Hilbert space K, a (not necessarily isometric) embedding ι : H −→ K onto a closed subspace of K, and a normal C0 -group U on K with kU (s)k ≤ eω|s| for all s ∈ R such that P ◦ U (s) ◦ ι = ι ◦ T (s) for all s ∈ R. Here, P : K −→ ι(H) denotes the orthogonal projection of K onto ι(H). Proof. Choose α > 0 such that αω = π/2. We consider the group T (α·) which has group type αθ(T ). By Monniaux’s Theorem 4.4.3, cf. also Remark 7.2.2) we find an injective sectorial operator A on H such that Ais = T (αs) for all s ∈ R. Then ωA = αθ(T ) < π/2 by Gearhart’s Theorem C.8.2 and Theorem 4.3.1. By the Callier-Grabowski-LeMerdy Theorem

7.3. Similarity Theorems for Sectorial Operators

177

7.3.7 we can change the scalar product on H in order to have −A generating a contraction semigroup (e−tA )t≥0 . The Sz.-Nagy Theorem 7.1.8 yields a new Hilbert space K and an isomorphic embedding ι : H −→ K (which is isometric with respect to the new scalar product) and a unitary group (W (t))t∈R on K such that P W (t)ι = ιe−tA for all t ≥ 0. Let −B be the generator of W . In the proof of Proposition 7.1.7 we showed that P f (B)ι = ιf (A) for all f ∈ E[S π ]. By 2

applying the Convergence Lemma (Proposition 5.1.3) to the sequence fn (z) := P B is ι

ιAis

z is zn 1 )(n+z) (z+ n

we

obtain = = ιT (αs) for s ∈ R. Thus we have arrived at the desired dilation when we ‚ ‚ π π define U (s) := B is/α . Since ‚B is ‚ ≤ e|s| 2 and ω = 2α we clearly have kU (s)k ≤ eω|s| for all s ∈ R.

7.3.4 A Counterexample Let if −A generate a bounded C0 -semigroup T and assume that A is injective. The Callier-Grabowski-LeMerdy Theorem 7.3.7 states that if A ∈ BIP(H) and T is even bounded holomorphic, T is similar to a contraction semigroup. What if we drop the assumption on holomorphy? This question is answered by the following theorem. Theorem 7.3.13. There exists a Hilbert space H and an operator A on H such that the following conditions hold. 1) The operator −A generates a bounded C0 -semigroup T on H. 2) The operator A is invertible and A ∈ BIP(H). 3) The semigroup T is not similar to a quasi-contractive semigroup. Here, a C0 -semigroup T on H is called quasi-contractive, if there is ω ∈ R such that kT (t)k ≤ eωt for all t ≥ 0. We will combine a counterexample given by LeMerdy with a technique due to Chernoff. Based on an example given by Pisier in [106], LeMerdy showed in [79, Proposition 4.8] that there exists a Hilbert space H, a bounded C0 -semigroup T on H with injective generator −A ∈ BIP(H) such that T is not similar to a contraction semigroup. Let α = (αn )n be a scalar sequence with αn > 0 for all n. We consider the space H := ℓ2 (H) and the operator A defined on H by D(A) := {x = (xn )n ∈ H | xn ∈ D(A) ∀n ∈ N}

Ax := (αn Axn )n .

Denote by k| ·k| the (Hilbert)-norm on H. Lemma 7.3.14. The following assertions hold. a) The operator −A generates the bounded C0 -semigroup T defined by T(t)x := (T (αn t))n

(x = (xn )n ∈ H, t ≥ 0).

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b) We have

−1

̺(A) = {λ | λα−1 < ∞}, n ∈ ̺(A) and sup (λ − αn A) n

with R(λ, A)x = ((λ − αn A)−1 xn )n for x = (xn )n ∈ H.

c) The operator A is injective.

d) Let ϕ > π2 and f ∈ H0∞ (Sϕ ). Then f (A)x = (f (αn A)xn )n for each x = (xn ) ∈ H. Moreover k| f (A)k| ≤ supn kf (αn A)k

Proof. Since T is a bounded semigroup, all operators T(t) are well defined bounded operators on H, and even uniformly bounded in t. Obviously, the semigroup law holds. Since the space of finite H-sequences is dense in H and each T (αn ·) is strongly continuous on H, T is a C0 -semigroup on H. Denote its generator by B and let x = (xn )n ∈ D(B). Then limtց0 1t (T(t)x−x) → Bx. This implies limtց0 1t (T (αn t)x − x) → [Bx]n for each n, whence xn ∈ D(A) and [Bx]n = −αn Axn for each n. Hence we have B ⊂ −A. Therefore, by a resolvent argument, a) will be proved as soon as we will have established b). Obviously, the inclusion “⊃” holds in b). Assume λ ∈ ̺(A). Then ((xn ) 7−→ ((λ − αn A)xn )) : D(A) −→ H

is bijective. By composing this mapping with suitable injections and projections we see that ‚ ‚ (λ − αn A) : D(A) −→ H is bijective for each n and that supn ‚(λ − αn A)−1 ‚ < ∞. The assertion c) follows from the fact that A is injective and each αn > 0. Part d) is immediate from b) and the representation of f (A) as a Cauchy integral.

Since A ∈ BIP(H) we know from the McIntosh-Yagi Theorem 7.3.1 that the natural H ∞ (Sϕ )-calculus is bounded for each ϕ > π2 . Fix ϕ > π2 and C ≥ 0 such that kf (A)k ≤ C kf kϕ for all f ∈ H ∞ (Sϕ ). If f ∈ H0∞ (Sϕ ), part d) of Lemma 7.3.14 yields kf (A)kL(H) ≤ C supn kf (αn ·)kϕ = C kf kϕ . Hence, also A has a bounded H ∞ (Sϕ )-calculus (see Proposition 5.3.3), whence A ∈ BIP(H). We turn to Chernoff’s observation from [18]. We now specialize αn := n1 in the above construction. Lemma 7.3.15. (Chernoff ) If T is similar to a quasi-contractive semigroup, then T is similar to a contractive semigroup. Proof. We denote by ( · | · ) and h·, ·i the scalar products on H and H, respectively. Assume that there is an equivalent scalar product h·, ·inew on H and a scalar ω ≥ 0 such that k| T(t)xk| new ≤ eωt k| xk| new for all x ∈ H and all t ≥ 0. Define ( x | y )n := hιn x, ιn yinew for x, y ∈ H and n ∈ N, where ιn : H −→ H is the natural inclusion mapping onto the n-th coordinate. Obviously, each ( · | · )n is an equivalent scalar product on H. More precisely, if c > 0 such that c−1 k| xk| 2 ≤ k| xk| new ≤ c k| xk| for all x ∈ H then c−1 kxk2 ≤ ( x | x )n ≤ c kxk2 for all n ∈ N and all x ∈ H. In particular, for each pair of vectors x, y ∈ H, the sequence (( x | y )n )n is bounded by c kxk kyk. Now, choose some free ultrafilter U on N and define ( x | y )new := U − lim ( x | y )n

for x, y ∈ H. Obviously, this yields a positive sesquilinear form on H. Moreover, we have c−1 kxk2 ≤ ( x | x )new ≤ c kxk2

for all x ∈ H, whence ( · | · )new is an equivalent scalar product on H. Since t

t

kT (t)xk2n = k| ιn T (t)xk| 2new = k| T(t/n)ιn xk| 2new ≤ e2ω n k| ιn k| 2new = e2ω n kxk2n ,

7.4. Cosine Function Generators

179

we obtain kT (t)xk2new ≤ kxk2new for all x ∈ H, t ≥ 0. Hence T is contractive with respect to ( · | · )new .

Since we started with a bounded semigroup which is not similar to a contraction semigroup, T is not similar to a quasi-contraction semigroup. Obviously this remains true also for each rescaled semigroup (e−εt T(t))t≥0 . Hence for each ε > 0 the operator A + ε on the Hilbert space H has the properties required in Theorem 7.3.13. Remark 7.3.16. Actually a statement slightly stronger than Theorem 7.3.13 is valid. In LeMerdy’s example the operator A even has a bounded R∞ (S π2 )calculus. By construction, this immediately carries over to the operator A, and this is strictly stronger than to say that A ∈ BIP(H).

7.4 Cosine Function Generators A cosine function on a Banach space X is a strongly continuous mapping Cos : R+ −→ L(X) such that Cos(0) = I, 2 Cos(t) Cos(s) = Cos(t + s) + Cos(t − s) (t, s ∈ R). In the following we cite some basic results of the theory of cosine functions from [6, Sections 3.14-3.16]. Given a cosine function, one can take its Laplace transform and define its generator B by Z ∞

λ R(λ2 , B)x =

e−λt Cos(t)x dt

0

for x ∈ X and Re λ sufficiently large. Then, for each pair (x, y) ∈ X 2 the function Z t u(t) := Cos(t)x + Cos(s)y ds 0

is the unique mild solution of the second order abstract Cauchy problem  ′′   u (t) = B u(t) (t ≥ 0), u(0) = x,   ′ u (0) = y

(cf. [6, Corollary 3.14.8]). If B generates a cosine function, then it also generates an exponentially bounded holomorphic semigroup of angle π/2 (cf. [6, Theorem 3.14.17]). Proposition 7.4.1. [6, Theorem 3.14.11] Let A generate a cosine function on the Banach space X. Let the operator A on X × X be defined by x 0 I x y D(A) := D(A) × X, A = = . y A 0 y Ax

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Chapter 7. The Functional Calculus on Hilbert Spaces

Then there exists a unique Banach space V such that D(A) ֒→ V ֒→ X and the part B of A in V × X generates a C0 -semigroup. The space V × H is called the phase space associated with A. If A generates a cosine function and λ ∈ C, then A + λ generates a cosine function with the same phase space (cf. [6, Corollary 3.14.13]). The connection to the theory of C0 -groups is given by the following: If an operator A generates a C0 -group (U (t))t∈R on the Banach space X, then A2 generates a cosine function Cos with phase space D(A) × X, where Cos(t) = (U (t) + U (−t))/2 (t ≥ 0) (cf. [6], Example 3.14.15). Moreover, a remarkable theorem of Fattorini states the (partial) converse. Theorem 7.4.2. (Fattorini) Let A be the generator of a cosine function on an UMD-space X. If −A is sectorial, 1 then C := i(−A) 2 generates a strongly continuous group and C 2 = A. A proof can be found in [6, Theorem 3.16.7]. Altogether this suggests to consider squares of group generators. Theorem 7.4.3. Assume that A is m-Hω -accretive for some ω ≥ 0. Then, for every 0 ≤ |ϕ| < π/2 the operator ! 2 ω iϕ 2 e (7.23) +A cos ϕ is m-accretive. The operator −A2 generates a holomorphic semigroup (S(z))Re z>0 of angle π/2 such that 2

kS(z)k ≤ e( cos ϕ ) ω

Moreover, one has

Re z

(|arg z| ≤ ϕ <

π ). 2

1 D (ω 2 + A2 ) 2 = D(A).

Proof. The case ω = 0 is trivial since then the group is unitary and iA is skewadjoint. This implies that A2 is selfadjoint with −A2 ≤ 0, and the assertions of the theorem are immediate. Assume ω > 0, let 0 ≤ |ϕ| < π/2 and fix ε > 0. Define α = ω tan ϕ, i.e., z := ω − iα = (ω/ cos ϕ)e−iϕ . By assumption and (ii) of Proposition C.7.2, the operators z − iA and z + iA are m-accretive. Applying (iv) of Proposition C.7.2) we obtain ‚ ‚ ‚ ‚ ‚ iA − (z − ε) ‚ ‚ iA + (z − ε) ‚ ‚ ‚,‚ ‚ ‚ iA − (z + ε) ‚ ‚ iA + (z + ε) ‚ ≤ 1. iA+λ (Here and in the following we write iA+µ instead of (iA+λ)(iA+µ)−1 to make the computations more perspicious.) Hence, ‚ 2 ‚ ‚„ «„ «‚ ‚ A + (z − ε)2 ‚ ‚ iA − (z − ε) iA + (z − ε) ‚ ‚ ‚=‚ ‚ ≤ 1. ‚ A2 + (z + ε)2 ‚ ‚ iA − (z + ε) iA + (z + ε) ‚

Now,

A2 + (z − ε)2 A2 + (z 2 + ε2 ) − 2zε eiϕ [A2 + (z 2 + ε2 )] − 2ε |z| = 2 = iϕ 2 . A2 + (z + ε)2 A + (z 2 + ε2 ) + 2zε e [A + (z 2 + ε2 )] + 2ε |z|

7.4. Cosine Function Generators

181

We can apply Proposition C.7.2 again (note that 2ε |z| > 0) to conclude that ω 2 −iϕ eiϕ [(z 2 + ε2 ) + A2 ] = ( ) e + ε2 eiϕ + eiϕ A2 cos ϕ ω is m-accretive. Letting ε ց 0 we see that ( cos )2 e−iϕ + eiϕ A2 and finally that ϕ

eiϕ ((

ω 2 ω 2 iϕ ) + A2 ) = ( ) e + eiϕ A2 cos ϕ cos ϕ

is m-accretive. (Note that eiϕ differs from e−iϕ only by a purely imaginary number and this does not affect m-accretivity by (ii) of Proposition C.7.2.) This finishes the proof of the first part of the theorem. The second part follows from Proposition 7.1.1, cf. also [6, Chapter 3.4 and Chapter 3.9]. To prove the last assertion, note first that the operator ω 2 + A2 is m-accretive, hence sectorial. Thus the square root is well defined. Since iA generates a group, −A2 generates a cosine function with phase space D(A) × H. By general cosine function theory (see the remarks at the beginning of this section), −(A2 + ω 2 ) also generates a cosine function with the same 1 phase space. Fattorini’s Theorem 7.4.2 implies that B := i(ω 2 + A2 ) 2 generates a group and 2 2 2 B = −(A + ω ). Then D(B) = D(A) follows from the uniqueness of the phase space (see Proposition 7.4.1).

Combining this result with the Similarity Theorem 7.2.8 for groups yields the following corollary: Corollary 7.4.4. Let −B be the generator of a cosine function on a Hilbert space H. Then B is Kato-sectorial and square root regular with respect to some equivalent scalar product. In particular, λ + B ∈ BIP(H) for large λ ∈ R.

Proof. First, one can find β such that B + β is sectorial. Since −(B + β) generates a cosine function as well, we can apply Fattorini’s Theorem 7.4.2. Thus, the operator iA := i(β + B)1/2 generates a strongly continuous group T on H. Choose ω > θ(T ). By Theorem 7.2.8 we can find a new scalar product ( · | · )◦ making A m-Hω -accretive and such that D(A) = D(A◦ ) holds. Apply now Theorem 7.4.3 together with Corollary 7.3.5 to conclude that A2 = B + β is Katosectorial. This implies that B is Kato-sectorial. Finally, we apply Theorem 7.4.3 to the operators A and A◦ and obtain 1

1

D((ω 2 + β + B) 2 ) = D((ω 2 + A2 ) 2 ) = D(A) = D(A◦ ) 1

1

= D((ω 2 + A◦2 ) 2 ) = D((ω 2 + β + B)◦ 2 ).

This completes the proof.

Let us turn to a second corollary. Namely, Theorem 7.4.3 can be seen as a mapping theorem for the numerical range (cf. also Section 7.1.4). Corollary 7.4.5. Let A be an Hω -accretive operator on the Hilbert space H for some ω ≥ 0. Then W (A2 ) ⊂ {z 2 | z ∈ Hω }, i.e., the numerical range of B := ω 2 + A2 is contained in the vertical parabola √ Πω = {z ∈ C | Re z ≥ 0 and |Im z| ≤ 2ω Re z}. This is equivalent to say that B is m-accretive and 1

|Im ( Bu | u )| ≤ 2ω Re ( Bu | u ) 2 kuk for all u ∈ D(B) = D(A2 ).

(7.24)

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Chapter 7. The Functional Calculus on Hilbert Spaces

Proof. By Theorem 7.4.3 we obtain that eiϕ ((ω/ cos ϕ)2 + A2 ) is m-acccretive for every 0 ≤ |ϕ| < π/2. Specializing ϕ = 0 yields that B = ω 2 + A2 is m-accretive. Now we write eiϕ (

ω2 ω2 + A2 ) = eiϕ (B + ( − ω 2 )) = eiϕ (B + ω 2 tan2 ϕ). cos2 ϕ cos2 ϕ

Since this operator is m-accretive and tan2 ϕ = tan2 (−ϕ) we obtain W (B + ω 2 tan2 ϕ) ⊂ S π −ϕ 2 by Proposition 7.1.1, i.e., |Im ( Bu | u )| ≤ tan(

π − ϕ)(Re ( Bu | u ) − ω 2 tan2 ϕ kuk2 ) 2

(u ∈ D(B))

for 0 < ϕ < π2 . Now tan( π2 − ϕ) = (tan ϕ)−1 and if we parametrize τ := tan ϕ with 0 < τ < ∞ we obtain |Im ( Bu | u )| ≤ τ1 Re ( Bu | u ) − ω 2 τ kuk2 for every u ∈ D(B). The right hand side q has 2 Re ( Bu | u ) ω 2 kuk2 as its minimum value, whence we arrive at (7.24).

Corollary 7.4.6. Let −B be the generator of a cosine function on the Hilbert space H. Then, with respect to an equivalent scalar product, B has numerical range in a horizontal parabola λ + Πω for some λ ∈ R, ω > 0. Proof. Apply Fattorini’s theorem 7.4.2, change the scalar product according to 7.2.8, then apply Corollary 7.4.5.

Let us call an operator m-Πω -accretive if its numerical range is contained in Πω and C \ Πω is contained in its resolvent set. The last corollary immediately raises the following question. Suppose −B is m-Πω -accretive for some ω > 0. Does it follow that B already generates a cosine function? This is in fact true, since Crouzeix has proved in [24] the following result which we quote without proof. Theorem 7.4.7. (Crouzeix) Let A be an m-Πω -accretive operator on a Hilbert space H. Then there is a constant c = c(ω) such that kf (A)k ≤ c kf kΠω for all f ∈ R∞ (Πω ). The proof follows in principle the lines of the proof of Theorem 7.1.14. However, it is much harder in detail. Corollary 7.4.8. Let B be an operator on a Hilbert space H. Then −B generates a cosine function if and only if there is λ ∈ R and equivalent scalar product on H with respect to which λ + B is m-Πω -accretive. Proof. One direction is Corollary 7.4.6. Assume that λ + B is m-Πω -accretive. Without restriction we can assume that B is m-Πω -accretive and invertible. Then we can set up a holomorphic functional calculus for B on horizontal parabolas Πα for α > ω. Using Crouzeix’ Theorem 7.4.7 together with similar arguments as in Chapter 6, §5.3 we conclude that the natural H∞ (Πα )1

calculus is bounded for every α > ω. Plugging in this calculus the functions etiz 2 for t ∈ R we 1 obtain the C0 -group generated by −B 2 . (The density of D(B) in H follows from sectoriality.) 1

Hence, −B = (iB 2 )2 generates a cosine function.

7.5. Comments

183

7.5 Comments 7.1 Numerical Range Conditions. The material on normal or selfadjoint operators is standard. The proof is usually given via a spectral measure respresentation, cp. [87, Section 4.3.1]. 7.1.3 Functional Calculus for m-accretive Operators. The fundamental Theorem 7.1.7 has been folklore for a long time, due to its connection to von Neumann’s inequality (cf. Remark 7.1.9 and below). The classical reference for the dilation theorem is the book [118] by Sz.-Nagy and Foias. Therein one can find also the construction of a functional calculus for so-called completely non-unitary contractions. The proof of Theorem 7.1.7 via the dilation theorem can also be found in [78, Theorem 4.5]. Our second proof is taken from [3], where it is attributed to Franks. This proof is in the spirit of Bernard and Fran¸cois Delyon’s proof of the von Neumann inequality in [35]. Other proofs of von Neumann’s inequality can be found in [104, Corollary 2.7] and [108, Chapter 1]. 7.1.4 Numerical Range Mapping Theorems. To establish “mapping theorems for the numerical range” Theorem 7.1.7 seems to be at the core of the theory, since it ties closely contractivity and m-accretivity to the functional calculus. Kato in [64] uses this connection to prove more general results. Based on that, Crouzeix and Delyon in [25] essentially prove Corollary 7.1.12. In [62] Kato gives a proof of Corollary 7.1.11 different from ours see also [120, Lemma 2.3.6]. Corollary 7.1.13 is proved in [120] by directly applying the Balakrishnan representation for Aα . We just wanted to show that in principle one can reduce the question to complex numbers by means of Theorem 7.1.7. In general, “mapping theorems for the numerical range” are a delicate matter. For example, Simard in [117] constructed an example of an m-H π2 -accretive operator on H = C2 such that eA is not m-accretive. 7.1.5 The Crouzeix-Delyon Theorem. The Crouzeix-Delyon Theorem 7.1.14 from [25] can be considered a major breakthrough in deriving boundedness of functional calculi from numerical range conditions. For a long time, the case of m-accretive operators was more or less the only example of this connection. The new insights came when a new proof of von Neumann’s inequality was found by directly manipulating the corresponding Cauchy integrals (see above). Although this has been developed further by Crouzeix to obtain similar results for operators with numerical range in a parabola (see Theorem 7.4.7 and [24]) and more general convex domains. 7.2 Group Generators on Hilbert Spaces. The Boyadzhiev-deLaubenfels Theorem 7.2.1 appears first in [15, Theorem 3.2], although in a little different form. The original proof proceeds in two steps. First, assuming ω < π/2 without loss of

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generality, the authors construct the operator eB (see Chapter 4; their construction however relies on the theory of regularized semigroups). Then they show that this operator is sectorial and has a bounded H ∞ -functional calculus on a sector. The statement then follows by means of some special case of the composition rule. Our first proof for Theorem 7.2.1 is just a combination of the Decomposition Theorem 7.2.8 and the Crouzeix-Delyon Theorem 7.1.14. Since the proof of the latter is quite involved, this proof cannot be regarded as simple. However, there is no need of the full power of the Crouzeix-Delyon theorem. In fact, the weaker version mentioned in Remark 7.1.16 suffices. Regarding the spectral theorem as known, this can be proved by a simple perturbation argument (see [52]). So we arrive at a proof considerably shorter and more perspicious than the proof in [15] and even more elegant than our second proof in Section 7.2.3 7.2.1 Liapunov’s Direct Method. In the case when A is bounded, the Liapunov method is used in Chapter I of the book [27]. There the operator equation QA + A∗ Q = −I is directly linked to the problem of finding a Liapunov function for the semigroup (which is sometimes called ‘Liapunov’s direct method’). For the unbounded case, the relevant facts are included in [26, Theorem 5.1.3], where a characterization of exponential stability of the semigroup is given in terms of the existence of an operator Q satisfying the Liapunov equation(s). (Extensions of this result can be found in [48], [8] and [2].) It is shown in [130] that this method in fact gives an equivalent scalar product if the semigroup is a group. In the paper [85] on exact controllability Liu and Russell establish almost the same results (and others). The Liapunov theorem for holomorphic semigroups (Proposition 7.3.8) is [7, Theorem 4.1]. In that paper, Liapunov type theorems are also established for hyperbolic holomorphic semigroups and quasi-compact holomorphic semigroups. Moreover, Proposition 7.3.8 is applied to semilinear equations. 7.2.2 The Decomposition Theorem. The following well-known theorem by Sz.Nagy from [31] can be regarded as the “limit case” in Theorem 7.2.8: Every generator of a bounded group is similar to a skew-adjoint operator. This result cannot be deduced directly from Theorem 7.2.8. However, Zwart in [130] gives a proof using the Liapunov renorming. In [34, Theorem 2.4] it is proved that, given a C0 -group T on a Hilbert space, one has kT (t)k◦ ≤ eω|t| for some equivalent scalar product ( · | ·)◦ and some ω strictly larger than the group type θ(T ). (This is covered by part a) of our Theorem 7.2.8.) While the proof in [34] is based on the boundedness of the H ∞ calculus and on Paulsen’s theorem (see below) our approach is more direct and considerably shorter. One can wonder, if one starts with a group T such that kT (s)k ≤ Keθ|s| (s ∈ R), whether one can always take ω = θ in Theorem 7.2.8. However, in [117] it is shown that this is not possible in general.

7.5. Comments

185

7.2.3 A Characterization of Group Generators. The idea to the second proof for Theorem 7.2.1 (from [49]) is to somehow adapt McIntosh’s methods from [95] for sectorial operators to strip type operators. In the sectorial case, the multiplicative group of positive real numbers acts on the sector by dilation, in the strip case the additive group R acts on the strip by translation. Combining this structural property with the right notion of quadratic estimates yields a proof which is quite analogous to McIntosh’s method in the sectorial case. As a byproduct, this second proof of the Boyadzhiev-deLaubenfels theorem yields a characterization of group generators and as Corollary 7.2.14 the result [84, Theorem 1] by Liu. Actually, Liu has an additional assumption which however is easily seen to be redundant. Another characterization of C0 -group generators appearing as Theorem 2 in [84] can also be reproved by functional calculus methods, see [49, Corollary 5.4]. In [49, Corollary 5.3] we gave a short argument proving Zwart’s generalization [130, Theorem 2.2] of Liu’s first result. 7.3.1 McIntosh”s Theorem. We do not know which people exactly contributed to the results stated in McIntosh’s Theorem 7.3.1. Our choice of name reflects the history of the ideas, originating in the 1984 paper [128] by Yagi and subsequently developed by McIntosh in [95] and both of them in [97]. Note that norm inequalities like kxkf . kxk or kxk . kxkf for x ∈ D(A) ∩ R(A) are called quadratic estimates or square function estimates in the terminology of those authors. The importance of quadratic estimates pertains when studying boundedness of H ∞ functional calculus on Lp -spaces, see [21] and [80]. The connection to interpolation theory is made explicit in [10] (see also Chapter 6 and the Comments there). The proof of the implication (ii) ⇒ (i) in Theorem 7.3.1 via the Unconditionality Lemma 5.7.1 is taken from [81, Theorem 4.2], although it seems probable that it has been known before. Let us point out that in Theorem 7.3.1 the equivalence of (iii) and the other statements is proved without making use of interpolation theory (cf. Remark 7.3.2. 7.3.2 Sesquilinear Forms. The classical reference for operators constructed via sesquilinear forms is Kato’s early paper [62] as well as his book [65]. This is the reason why we called this operators “Kato-sectorial”. Proposition 7.1.1 essentially is Kato’s “First Representation Theorem” [65, Chapter VI, Theorem 2.1]. It is also included as Theorem 1.2 in [7]. The applications to PDE in the literature are numerous and we omit further references. In [7, Example 3.2] an example of two similar operators on a Hilbert space is given, one variational but not the other. As mentioned already, by Matolcsi’s result in [90], this is not a pathological case. It would be interesting if there is a proof for Kato’s Theorem 7.3.6 which differs essentially from Kato’s original proof. Apart from [120, Lemma 2.3.8] and [87, Theorem 4.3.4] which more or less copy Kato’s arguments, we do not know of any other account. The Square Root Problem (cf. Remark 7.3.2) has been solved recently by Auscher, Hoffman, Lacey, Lewis, McIntosh, and

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Tchamitchian in [9]. Surveys and a deeper introduction to these matters as well as the connection to Calderon’s conjecture for Cauchy integrals on Lipschitz curves are in [96] [93], [94], and [11]. A counterexample to Kato’s original question is presented in [11, Section 0, Theorem 6]. 7.3.3 Similarity Theorems. Similarity problems have a long tradition in operator theory. In 1947 Sz.-Nagy had observed in [31] that a bounded and invertible operator T on a Hilbert space H is similar to a unitary operator if and only if the discrete group (T n )n∈Z is uniformly bounded. His question was whether the same is true if one discards the invertibility of T . In [119] he showed that the answer is yes if T is compact, but Foguel disproved the general conjecture by giving a counterexample in [43]. Von Neumann’s inequality shows that if T is a contraction, then T is not only power-bounded, but polynomially bounded, i.e., sup{kp(T )k | p ∈ C[z], kpk∞ ≤ 1} < ∞ where kpk∞ denotes the uniform norm on the unit disc. So Halmos asked in [57] if polynomial boundedness in fact characterizes the bounded operators on H which are similar to a contraction. This question remained open only until recently when Pisier found a counterexample, see [106] and [28]. Meanwhile, Paulsen had shown in [103] that T is similar to a contraction if and only if T is completely polynomially bounded. His characterization is a special case of a general similarity result [104, Theorem 8.1] for completely bounded homomorphisms of operator algebras (see [104] for definitions and further results). We will address this result as “Paulsen’s theorem” in the following. One can set up a semigroup analogue of Sz.-Nagy’s question, namely if every bounded C0 -semigroup on a Hilbert space is similar to a contraction semigroup. The corresponding result for groups is true as was also proved by Sz.-Nagy in the very same article [31]. The question is a bit more special than the original one, since not every power-bounded operator is the Cayley transform of a C0 semigroup generator. It was answered in the negative by Packel in [102]. By using this result, Chernoff provided an example of a bounded operator with the generated C0 -semigroup not being similar to a contraction one. Furthermore, he constructed a C0 -semigroup which is not even similar to a quasi-contractive semigroup. It is in this construction where Lemma 7.3.15 appears.2 Being probably unaware of the history of this problem, Callier and Grabowski in an unpublished research report [47] proved Theorem 7.3.7 in the case where the semigroup is exponentially stable. They did this using two facts from interpolation theory: first, both the (complex) interpolation space [H, D(A)] 12 and the 1

(real) interpolation space (H, D(A)) 12 ,2 are equal to D(A 2 ) if A ∈ BIP(H) (see 2 Our setting up this historical panorama is mainly based on [79, Introduction]. More on similarity problems and their connection to the theory of operator algebras can be found in [107].

7.5. Comments

187

Theorem 6.6.9) and second, this real interpolation space is given by Z ∞

12 −tA 2 dt (H, D(A)) 12 ,2 = {x ∈ H | x < ∞}

t Ae t 0

(see [86, Proposition 2.2.2]). Our proof of Theorem 7.3.7 uses an observation of LeMerdy in [81, Theorem 4.2]. Theorem 7.3.7 as it stands is also a consequence of Corollary 7.3.11 and is stated as such in [79, Theorem 4.3]. LeMerdy proves Corollary 7.3.11 with the help of Paulsen’s theorem (see above). The same is done by Franks in [44, Section 4]. Note that — by using the scaling technique and the results of Kato on accretive operators — the Franks-LeMerdy and the Callier-Grabowski-LeMerdy theorems are actually equivalent, and that Paulsen’s theorem is not necessary to prove the Similarity Theorem 7.3.9. It has been noted in [7, Theorem 3.3] that the Franks-LeMerdy result solves the first similarity problem posed on page 173. However, from their proof it is not clear if also the second problem concerning the square roots can be solved. That it actually can sheds an interesting light on the orginal square root problem. It also complements a result of Yagi in [128, Theorem B] which says that a sectorial invertible operator A on a Hilbert space has bounded imaginary powers if D(Aα ) = D(A∗α ) for all α contained in a small interval [0, ε). It is consequence of Theorem 7.3.9 combined with the scaling technique that the converse holds modulo similarity. The following problem seems to be open until now. Problem: Is there an m-accretive operator A on a Hilbert space H such that 1

1

D(A 2 ) 6= D(A◦ 2 ) for every equivalent scalar product ( · | ·)◦ ? (Note that, by Theorem 7.3.9, such an operator necessarily has to satisfy the relation ωA = π2 .) 7.4 Cosine Function Generators. The mapping theorem for the numerical range established in Theorem 7.4.3 and Corollary 7.4.5 is from [52] and [50]. McIntosh has shown in [92] that an operator B satisying the conclusion of Corollary 7.4.5 also 1 1 has the square root property D(B 2 ) = D(B ∗ 2 ). Employing this one can eliminate the cosine function theory from the proof of Corollary 7.4.4. Crouzeix’s result Theorem 7.4.7 from [24] is a milestone. This author is up to prove a general theorem on bounded H ∞ -calculus for operators with numerical range in a convex subset of the plane, cf. [23]. For more information on cosine functions consult [6] and the references therein.

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Appendix A

Linear Operators This chapter is supposed to be a “reminder” of some operator theory, including elementary spectral theory and approximation results, rational functional calculus and semigroup theory. There is a slight deviation from the standard literature on operator theory in that we deal with multivalued operators right from the start.

A.1 The Algebra of Multivalued Operators Let X, Y, Z be Banach spaces. A linear operator from X to Y is a linear subspace of the direct sum space X ⊕ Y . A linear operator may fail to be the graph of a mapping. The subspace A0 := {x ∈ X | (0, x) ∈ A} ⊂ X is a measure for this failure. In case A0 = 0, the relation A ⊂ X ⊕ Y is functional, i.e., the operator A is the graph of a mapping, and it is called single-valued. Since in the main text we deal with single valued linear operators almost exclusively (not without significant exceptions, of course), we make the following Agreement: Unless otherwise stated, the term “operator” always is to be understood as “single-valued linear operator”. We call an operator multivalued (in short “m.v.”) if we want to stress that it is not necessarily single-valued (but it may be). The image of a point x under the m.v. operator A is the set Ax := {y ∈ Y | (x, y) ∈ A}. This set is either empty (this means, the m.v. operator A is “undefined” at x) or it is an affine subspace of Y in the “direction” of the space A0.

190

Appendix A. Linear Operators With a m.v. operator A ⊂ X ⊕ Y we associate the spaces kernel N(A) domain D(A) range R(A)

:= := :=

{x ∈ X | (x, 0) ∈ A}, {x ∈ X | there is y ∈ Y such that (x, y) ∈ A}, {y ∈ Y | there is x ∈ X such that (x, y) ∈ A}.

The m.v. operator A is called injective if N(A) = 0 and surjective if R(A) = Y . If D(A) = X, then A is called fully defined. Let A, B ⊂ X ⊕ Y and C ⊂ Y ⊕ Z be m.v. operators and λ ∈ C a scalar. We define the sum A + B, the scalar multiple λA, the inverse A−1 and the composite CA by A + B := {(x, y + z) ∈ X ⊕ Y | (x, y) ∈ A, (x, z) ∈ B}, λA := {(x, λy) ∈ X ⊕ Y | (x, y) ∈ A},

A−1 := {(y, x) ∈ Y ⊕ X | (x, y) ∈ A}, CA := {(x, z) ∈ X ⊕ Z | ∃ y ∈ Y : (x, y) ∈ A ∧ (y, z) ∈ C}.

Then we have the following identities: D(A + B) = D(A) ∩ D(B), D(A−1 ) = R(A), D(λA) = D(A),

D(CA) = {x ∈ D(A) | ∃ y ∈ D(C) : (x, y) ∈ A}. The zero-operator is 0 := {(x, 0) | x ∈ X} and I := {(x, x) | x ∈ X} is the identity operator. (This means that in general we have only 0A ⊂ 0 but not 0A = 0.) Proposition A.1.1. Let X be a Banach space and A, B, C ⊂ X ⊕ X m.v. linear operators on X. a) The set of m.v. operators on X is a semigroup with respect to composition, i.e., the Law of Associativity A(BC) = (AB)C holds. The identity operator I is a (the) neutral element in this semigroup. Moreover, the Law of Inversion (AB)−1 = B −1 A−1 holds. b) The set of m.v. operators on X is an abelian semigroup with respect to sum, with the zero operator 0 as its neutral element. c) For λ 6= 0 one has

λA = (λI)A = A(λI).

d) The m.v. operators A, B, C satisfy the following Laws of Monotonicity: A ⊂ B =⇒ AC ⊂ BC, CA ⊂ CB

A ⊂ B =⇒ A + C ⊂ B + C, λA ⊂ λB.

A.1. The Algebra of Multivalued Operators

191

e) The following Distributivity Inclusions hold: (A + B)C ⊂ AC + BC, CA + CB ⊂ C(A + B),

with equality if C is single-valued; and with equality if R(A) ⊂ D(C).

In particular, there is equality in both cases if C ∈ L(X) (see below). A m.v. operator A ⊂ X ⊕ Y is called closed if it is closed in the natural topology on X ⊕ Y . If A is closed, then the m.v. operators λA (for λ 6= 0) and A−1 are closed as well. Furthermore, the spaces N(A) and A0 are closed. Sum and composition of closed m.v. operators are not necessarily closed. For each m.v. operator A one can consider its closure A in X ⊕ Y . A single-valued operator is called closable if A is again single-valued. If A is closed, then every subspace V ⊂ D(A) such that {(x, y) ∈ A | x ∈ V } = A is called a core for A. A single-valued operator A is called continuous if there is c ≥ 0 such that kAxk ≤ c kxk for all x ∈ D(A). Every continuous operator is closable. An operator is called bounded if it is continuous and fully defined. We let L(X, Y ) := {A ⊂ X ⊕ Y | A is a bounded operator} be the set of all bounded operators from X to Y . If X = Y we write just L(X) in place of L(X, X). For a single-valued operator A there is a natural norm on D(A), namely the graph norm kxkA := kxk + kAxk (x ∈ D(A)). Note that A is closed if and only if (D(A), k.kA ) is complete.

Lemma A.1.2. Let X and Y be Banach spaces. A single-valued operator A ⊂ X ⊕Y is continuous if and only if D(A) is a closed subspace of X. Proof. The continuity of the operator is equivalent to the fact that the graph norm is equivalent to the original norm. Closedness of the operator is equivalent to the fact that D(A) is complete with respect to the graph norm. Therefore, the assertion follows from the open mapping theorem.

Lemma A.1.3. Let A be a closed m.v. operator on the Banach space X, and let T ∈ L(X). Then AT is closed.

Proof. Let (xk , yk ) ∈ AT with xk → x and yk → y. Then there is zk such that (xk , zk ) ∈ T and (zk , yk ) ∈ A. Since T ∈ L(X), one has zk = T xk → T x. The closedness of A implies that z ∈ D(A) and (z, y) ∈ A. But this means exactly that (x, z) ∈ AT .

A m.v. operator A ⊂ X ⊕ Y is called invertible, if A−1 ∈ L(Y, X). We denote by L(X)× := {T | T, T −1 ∈ L(X)} the set of bounded invertible operators on X.

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Appendix A. Linear Operators

Lemma A.1.4. Let A be a closed m.v. operator on the Banach space X, and let T be an invertible m.v. operator. Then T A is closed. Proof. Suppose (xn , yn ) ∈ T A with xn → x and yn → y. Then there are zn such that (xn , zn ) ∈ A and (zn , yn ) ∈ T . Since T −1 ∈ L(X), zn = T −1 yn → T −1 y. The closedness of A implies (x, T −1 y) ∈ A, whence (x, y) ∈ T A.

Note that the last result in general is false without the assumption that T is invertible.

A.2 Resolvents In this section A denotes a m.v. linear operator on the Banach space X. The starting point for the spectral theory is the following lemma. Lemma A.2.1. The identity I − [I + λA−1 ]−1 = λ(λ + A)−1 holds for all λ ∈ C. Proof. For λ = 0 the assertion is (almost) trivial. Therefore, let λ 6= 0. If x, y ∈ X and z := (1/λ)y, one obtains (x, y) ∈ λ(λ + A)−1 ⇔ (x, z) ∈ (λ + A)−1 ⇔ (z, x) ∈ (λ + A) ⇔ (z, x − λz) ∈ A ⇔ (x − λz, z) ∈ A−1 ⇔ (x − λz, λz) ∈ λA−1

⇔ (x − λz, x) ∈ I + λA−1 ⇔ (x, x − λz) ∈ [I + λA−1 ]−1

⇔ (x, λz) ∈ I − [I + λA−1 ]−1 ⇔ (x, y) ∈ I − [I + λA−1 ]−1 . This shows the claim to be true.

We call the mapping (λ 7−→ R(λ, A) := (λ − A)−1 ) : C −→ {m.v. operators on X} the resolvent of A. The set ̺(A) := {λ ∈ C | R(λ, A) ∈ L(X)} is called the resolvent set and σ(A) := C \ ̺(A) the spectrum of A. Corollary A.2.2. For all λ, µ ∈ C the identity I − [I + (λ − µ)R(µ, A)]−1 = (λ − µ)R(λ, A). holds true. Proof. Just replace λ by (λ − µ) and A by µ − A in Lemma A.2.1.

A.2. Resolvents

193

Proposition A.2.3. Let A be a closed m.v. linear operator on the Banach space X. The resolvent set ̺(A) is an open subset of C. More precisely, for µ ∈ ̺(A) one −1 has dist(µ, σ(A)) ≥ kR(µ, A)k and R(λ, A) =

∞ X

k=0

(µ − λ)k R(µ, A)k+1

(|λ − µ| < kR(µ, A)k−1 ).

The resolvent mapping R(., A) : ̺(A) −→ L(X) is holomorphic and the resolvent identity R(λ, A) − R(µ, A) = (µ − λ)R(µ, A)R(λ, A) holds for all λ, µ ∈ ̺(A). case A ∈ L(X) one has ∅ 6= σ(A) ⊂ {z ∈ C | |z| ≤ kAk} where R(λ, A) = P∞ In −(k+1) Ak for all |λ| > kAk. k=o λ If A ∈ L(X) we call

rA := r(A) := inf{r > 0 | σ(A) ⊂ Br (0)} the spectral radius of A. Proof. As an abbreviation we write R(λ) instead of R(λ, A). Take µ ∈ ̺(A) and λ ∈ C such that |λ − µ| < kR(µ)k−1 . Then, a well known result from the theory of bounded operators states that I + (λ − µ)R(µ) is invertible with X (I + (λ − µ)R(µ))−1 = (µ − λ)k R(µ)k ∈ L(X) k≥0

being its inverse. Combined with Corollary A.2.2 this gives R(λ) ∈ L(X) and X 1 1 X R(λ) = (I − (µ − λ)k R(µ)k ) = (µ − λ)k R(µ)k λ−µ µ − λ k≥1 k≥0 X = (µ − λ)k R(µ)k+1 . k≥0

Let λ, µ ∈ ̺(A), and a ∈ X. Set x := R(λ)a. Then, (x, a) ∈ (λ − A), hence (x, a + (µ − λ)x) ∈ (µ − A). This implies R(λ)a = x = R(µ)(a + (µ − λ)x) = (R(µ) + (µ − λ)R(µ)R(λ))a. The proofs of the remaining statements are well known.

Each mapping R : Ω −→ L(X) with ∅ 6= Ω ⊂ C such that the resolvent identity R(λ) − R(µ) = (µ − λ)R(λ)R(µ) (λ, µ ∈ Ω) holds, is called a pseudo resolvent. Since we allow m.v. operators, we obtain the following proposition which fails to be true for single-valued operators. Proposition A.2.4. Let R : Ω −→ L(X) be a pseudo resolvent. Then there is one and only one m.v. operator A on X such that Ω ⊂ ̺(A) and R(λ) = R(λ, A) for all λ ∈ Ω.

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Appendix A. Linear Operators

Proof. If R(λ) = R(λ, A), then A = λ − R(λ)−1 . This shows that the operator A is uniquely determined by each single R(λ). Thus we define Aλ := λ − R(λ)−1 . What we have to show is that all the Aλ are equal, i.e., that λ + R(µ)−1 = µ + R(λ)−1 for all λ, µ ∈ Ω. By interchanging the roles of µ and λ it is clear that we are done as soon as we know one inclusion. Let (x, y) ∈ µ + R(λ)−1 . This means that x = R(λ)a where a := y − µx. Then R(µ)(y − λx) = R(µ)(a + (µ − λ)x)

= R(µ)(I + (µ − λ)R(λ))a = R(λ)a = x.

But this gives (x, y − λx) ∈

R(µ)−1 ,

whence (x, y) ∈ λ + R(µ)−1 .

Corollary A.2.5. Let R1 : Ω1 −→ L(X) and R2 : Ω2 −→ L(X) be pseudo resolvents. If R1 (z) = R2 (z) for some z ∈ Ω1 ∩ Ω2 , then R1 (z) = R2 (z) for all such z.

A, if

Let T ∈ L(X). We say that the operator T commutes with the m.v. operator (x, y) ∈ A

=⇒

(T x, T y) ∈ A

for all x, y ∈ X. This is equivalent to the condition T A ⊂ AT . Obviously, T commutes with A if and only if T commutes with A−1 . An immediate consequence of this fact is the following proposition. Proposition A.2.6. Let T ∈ L(X) and ̺(A) 6= ∅. The following assertions are equivalent. (i) [T, R(λ, A)] := T R(λ, A) − R(λ, A)T = 0 for some λ ∈ ̺(A).

(ii) [T, R(λ, A)] = 0 for all λ ∈ ̺(A).

(iii) T A ⊂ AT .

Let A, B closed m.v. operators, and suppose ̺(A) 6= ∅. We say that B commutes with the resolvents of A if B commutes with R(λ, A) for each λ ∈ ̺(A). If ̺(B) 6= ∅, it is sufficient that this is the case for a single λ. Moreover, it follows that A commutes with the resolvents of B. One should note that each single-valued operator A with ̺(A) 6= ∅ commutes with its own resolvents. (This is due to the identity B −1 Bx = x + N(B) for x ∈ D(B) which is true for all operators B.) Proposition A.2.7. Let A be a m.v. operator such that ̺(A) 6= ∅, and let T ∈ L(X) be injective. If T commutes with A, then T −1 AT = A. Proof. Because of AT ⊃ T A we have T −1 AT ⊃ T −1 T A = A. The reverse inclusion T −1 AT ⊂ A is equivalent to the following statement: If (T x, T y) ∈ A, then (x, y) ∈ A, for all x, y ∈ X. Let λ ∈ ̺(A). From (T x, T y) ∈ A it follows that (T x, T (y − λx)) = (T x, T y − λT x) ∈ (A−λ). This gives R(λ, A)T (λx−y) = T x. Now, T commutes with R(λ, A), hence T R(λ, A)(λx− y) = T x. By injectivity of T we obtain R(λ, A)(λx − y) = x, whence (x, y) ∈ A. This completes the proof.

A.3. The Spectral Mapping Theorem for the Resolvent

195

Restrictions Let A be a m.v. operator on the Banach space X, and let Y be another Banach space, continuously embedded into X. We denote by AY := A Y the part of A in Y , i.e., AY := A ∩ (Y ⊕ Y ). In case A is single-valued this means

D(AY ) = {x ∈ Y ∩ D(A) | Ax ∈ Y },

and AY y = Ay

(y ∈ D(AY )).

Here are the important facts. Proposition A.2.8. Let A be any operator on X and Y ⊂ X be another Banach space with continuous inclusion. Then the following assertions hold. a) If A is closed, then also AY is closed. b) [A−1 ]Y = [AY ]−1 . c) λ − AY = (λ − A)Y holds for all λ ∈ C.

d) If λ ∈ ̺(A) then the two assertions (i) λ ∈ ̺(AY )

(ii) Y is R(λ, A)-invariant. are equivalent. In this case, R(λ, AY ) = R(λ, A)Y . Proof. a) Assume that A is closed. Let (yn , zn ) ∈ AY , (yn , zn ) → (y, z) within Y ⊕ Y . Since the inclusion Y ⊂ X is continuous, (yn , zn ) → (y, z) within X ⊕ X, and since A is closed, it follows that (y, z) ∈ A ∩ Y ⊕ Y = AY . b) and c) are trivial. d) follows from a), b) and c).

Corollary A.2.9. Let X, Y, A as above. If D(A) ⊂ Y , then ̺(A) ⊂ ̺(AY ).

A.3 The Spectral Mapping Theorem for the Resolvent We define C∞ := C ∪ {∞} to be the one-point compactification of C (Riemann sphere) with the usual conventions for computation 1 . For a m..v. operator A we call the set σ(A) if A ∈ L(X) σ ˜ (A) := σ(A) ∪ {∞} if A ∈ / L(X) the extended spectrum of A.

Proposition A.3.1. (Spectral Mapping Theorem) Let A be a closed m.v. operator on the Banach space X. We have σ ˜ (λA) = λ˜ σ (A),

σ ˜ (A + λ) = σ ˜ (A) + λ,

for all λ ∈ C. In particular, σ ˜ (R(λ, A)) = 1 like,

1 λ−˜ σ(A)

and

σ ˜ (A−1 ) =

for all λ ∈ C.

e.g., ∞ + λ = ∞, λ · ∞ = ∞ for 0 6= λ ∈ C, and 0 = 0 · ∞ = 1/∞.

1 σ ˜ (A)

196

Appendix A. Linear Operators

This will follow from Corollary A.3.3 below. Let λ ∈ C∞ . The eigenspace of A at λ is defined by N(A − λ) for λ ∈ C N(λ, A) := A0 for λ = ∞. Similarly, we define the range space R(A − λ) R(λ, A) := D(A)

λ∈C λ=∞

f¨ ur f¨ ur

of A at λ. Using this notation we define the extended point spectrum, approximate point spectrum, residual spectrum, and surjectivity spectrum by Pσ ˜ (A) := {λ ∈ C∞ | N(λ, A) 6= 0},

A˜ σ (A) := {λ ∈ C∞ | N(λ, A) 6= 0 or R(λ, A) is not closed},

R˜ σ (A) := {λ ∈ C∞ | R(λ, A) 6= X}, Sσ ˜ (A) := {λ ∈ C∞ | R(λ, A) 6= X}.

Clearly we have σ ˜ (A) = P σ ˜ (A) ∪ S σ ˜ (A) = A˜ σ (A) ∪ R˜ σ (A). The classical spectra are obtained by intersecting the extended spectra with the complex plane, i.e., P σ(A) = C ∩ P σ ˜ (A),

Aσ(A) = C ∩ A˜ σ (A),

....

By introducing the extended spectra we have obtained, for example, that singlevaluedness of an operator is a spectral condition. Lemma A.3.2. Let µ ∈ C and λ ∈ C∞ . Then N(λ, µ − A) = N(µ − λ, A)

and

R(λ, µ − A) = R(µ − λ, A)

and

1 N(λ, A−1 ) = N( , A), λ 1 R(λ, A−1 ) = N( , A). λ

as well as

Proof. We only show the first and the last equality. The other two are proved similarly. Let y ∈ X. We have y ∈ N(λ, µ − A)

⇔

⇔ and

∨

⇔

y ∈ R(λ, A−1 )

∨ ∨

⇔ ∨ This shows

R(λ, A−1 )

⊂

⇒

λ = ∞, (0, y) ∈ A

(µ − λ) ∈ C, (y, 0) ∈ A − (µ − λ)

∨

(µ − λ) = ∞, (0, y) ∈ A

0 6= λ ∈ C, ∃x : (x, y) ∈ A−1 − λ

∨

λ = 0, ∃x : (x, y) ∈ A−1

y ∈ N(µ − λ, A),

⇔ ⇔

λ ∈ C, (y, 0) ∈ (µ − A) − λ = (µ − λ) − A

λ = ∞, ∃x : (y, x) ∈ A−1

0 6= λ ∈ C, ∃x : (y + λx, x) ∈ A λ = ∞, ∃x : (x, y) ∈ A

∨

λ = 0, ∃x : (y, x) ∈ A

0 6= λ−1 ∈ C, ∃x : (y + λx, −λ−1 y) ∈ A − λ−1 λ−1 = ∞, ∃x : (y, x) ∈ A −1

y ∈ R(λ

R(λ−1 , A).

∨

λ−1 = 0, ∃x : (x, y) ∈ A

, A).

The reverse inclusion follows by symmetry.

A.4. Adjoints

197

Corollary A.3.3. Let µ ∈ C and λ ∈ C∞ . Then X˜ σ (µ − A) = µ − X˜ σ (A)

und

X˜ σ (A−1 ) =

1 X˜ σ (A)

for X ∈ {P, A, R, S}. We give in addtion a characterization of the approximate point spectrum in terms of approximate eigenvectors. Proposition A.3.4. Let A be a closed m.v. operator on X and let λ ∈ C. The follwing assertions are equivalent. (i) λ ∈ Aσ(A).

(ii) There is (xn , yn ) ∈ λ − A such that kxn k = 1 for all n and yn → 0 as n → ∞.

Proof. On can assume without restriction that B := λ − A is injective. Since B is closed one ˜ : D(B) −→ X/B0. Clearly B has a closed can consider the induced (single-valued) operator B ˜ has. Now, 0 ⊕ B0 ⊂ B, ˜ whence B ˜ is a closed operator B ˜ ⊂ X ⊕ (X/B0). range if and only if B ˜ −1 : X/B0 −→ X is therefore a single-valued closed operator. By Lemma A.1.2 The inverse B ˜ — is closed. This means this operator is continuous if and only if its domain — which is R(B) ˜ X/B0 ≥ c kxk for all that B has a closed range if and only if there is a constant c such that kBxk x ∈ D(B). From this readily follows the stated equivalence.

Corollary A.3.5. Let A be a closed m.v. operator on X then ∂ σ ˜ (A) ⊂ A˜ σ (A). The boundary ∂ σ ˜ (A) has to be regarded as a boundary in C∞ . Proof. Let λ ∈ ∂ σ ˜ (A). Assume first that λ ∈ C. Then there λn → λ with λn ∈ ̺(A). Since σ ˜ (A) is closed, λ ∈ σ(A). By Proposition A.2.3 one has kR(λn , A)k → ∞. From this follows that there is a sequence (zn )n such that kzn k = 1 and kR(λn , A)zn k → ∞. Let xn := kR(λn , A)zn k−1 R(λn , A)zn and wn := kR(λn , A)zn k−1 zn . Then kxn k = 1, wn → 0 and (xn , wn ) ∈ λn − A. Defining yn := wn + (λ − λn )xn we obtain (xn , yn ) ∈ λ − A and yn → 0. Hence λ ∈ Aσ(A), by Proposition A.3.4. If λ = ∞ we use the spectral mapping theorem to apply the above to 0 ∈ ∂ σ ˜ (A−1 ). Then we use Corollary A.3.3.

A.4 Adjoints We denote the dual space of a Banach space X by X ′ . Note that there is a canonical identification of (X ⊕ X)′ and X ′ ⊕ X ′ which we will often employ without explicit mention. The canonical duality between X and X ′ is denoted by h·, ·i : X × X ′ −→ C. Given A ⊂ X and B ⊂ X ′ we define M ⊥ := {x′ ∈ X ′ | hx, x′ i = 0 ∀ x ∈ M }

N ⊤ := { x ∈ X | hx, x′ i = 0 ∀ x′ ∈ N }

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Appendix A. Linear Operators

Identifying canonically X with a closed subspace of X ′′ we therefore have N ⊥ ∩X = N ⊤. Let A ⊂ X ⊕ X a m.v. operator. The adjoint of A, usually denoted by A′ is defined by (x′ , y ′ ) ∈ A′

:⇔

hv, x′ i = hu, y ′ i for all (u, v) ∈ A.

If we define J := ((u, v) 7−→ (−v, u)) : X ⊕ X −→ X ⊕ X we can write A′ = [JA]⊥ . It is clear by definition that A′ is always a closed operator. The following lemma is an easy consequence of the Hahn-Banach Theorem. Lemma A.4.1. Let A be a m.v. operator on the Banach space X. Then D(A′ ) = {x′ ∈ X ′ | x′ A is s.v. and continuous} A′ x = {y ′ ∈ X ′ | y ′ ⊃ x′ A} (x ∈ D(A′ ))

and

We enumerate the basic properties of adjoints. Proposition A.4.2. Let A, B be m.v. linear operators on X. Then the following statements hold. a) A′ = (A)′ is w∗ -closed. b) (A−1 )′ = (A′ )−1 . c) (λA)′ = λA′ , for 0 6= λ ∈ C.

d) A′′ ∩ (X ⊕ X) = A, where we identify X ⊕ X canonically with a closed subspace of X ′′ ⊕ X ′′ . e) N(A′ ) = R(A)⊥ and N(A) = R(A′ )⊤ . f ) A′ 0 = D(A)⊥ and A0 = D(A′ )⊤ . g) D(A′ ) ⊂ (A0)⊥ and R(A′ ) ⊂ N(A)⊥ .

h) If A ∈ L(X), then A′ ∈ L(X ′ ) and kA′ k = kAk. i) A ⊂ B

⇒

B ′ ⊂ A′ .

j) A′ + B ′ ⊂ (A + B)′ with equality if A ∈ L(X).

k) A′ B ′ ⊂ (BA)′ with equality if B ∈ L(X). If A ∈ L(X) and B is closed, one has A′ B ′

w∗

= (BA)′ .

′

⊥

Proof. Ad a). Since J is a topological isomorphism, A = (JA)⊥ = JA = (JA)⊥ . Ad b). This follows from (JA)−1 = J(A−1 ) and (A−1 )⊥ = (A⊥ )−1 . Ad c). We have (x′ , y ′ ) ∈ (λA)′ ⇔ h−λv, xi + ( u | y ′ ) = 0 ∀(u, v) ∈ A ⇔ h−v, λx′ i + hu, y ′ i = 0 ∀(u, v) ∈ A ⇔ (λx′ , y ′ ) ∈ A′ ⇔ (λx′ , λy ′ ) ∈ λA′ ⇔ (x′ , y ′ ) ∈ A′ . Ad d). A′′ ∩ (X ⊕ X) = (J(JA)⊥ )⊤ = (JJA)⊥⊤ = A⊥⊤ = A. Ad e). We have x′ ∈ N(A′ ) ⇔ (x′ , 0) ∈ A′ ⇔ h−v, x′ i = 0 ∀v ∈ R(A) ⇔ x′ ∈ R(A)⊥ . Using this and d) , we obtain N(A) = N(A′′ ∩ (X ⊕ X)) = N(A′′ ) ∩ X = R(A′ )⊥ ∩ X = R(A′ )⊤ .

A.5. Convergence of Operators

199

Ad f ). This follows from e) and b). Ad g). If (x′ , y ′ ) ∈ A′ and v ∈ A0 then (0, v) ∈ A and hv, x′ i = h0, y ′ i = 0, whence x ∈ (A0)⊥ . The second statement follows from this and b). Ad h). Let A ∈ L(H). Then A∗ is closed and single-valued by f ). If x′ ∈ X ′ onee easily sees that (x′ , x′ ◦A) ∈ A′ , whence A′ x = x′ ◦A. Thus, D(A′ ) = X ′ and A′ ∈ L(X ′ ). By the HahnBanach Theorem, kA′ k = supkx′ k≤1 kA′ x′ k = supkx′ k≤1, kxk≤1 |hx, A′ x′ i| = supkx′ k≤1, kxk≤1 |hAx, x′ i| = supkxk≤1 kAxk = kAk. Ad i). We have A ⊂ B ⇒ JA ⊂ JB ⇒ (JB)⊥ ⊂ (JA)⊥ . Ad j). Let (x′ , y ′ ) ∈ A′ , (x′ , z ′ ) ∈ B ∗ . The generic element of J(A + B) is (−v − w, u), where (u, v) ∈ A and (u, w) ∈ B. So (x′ , y ′ ) ⊥ (−v, u) and (x′ , z ′ ) ⊥ (−w, u), hence (x′ , y ′ +z ′ ) ⊥ (−v − w, u). If A ∈ L(X), we write B = (A + B) − A and note that A′ ∈ L(X ′ ) by h). Ad k). Let (x′ , y ′ ) ∈ A′ B ′ . Then there is z ′ such that (x′ , z ′ ) ∈ B ′ and (z ′ , y ′ ) ∈ A′ . If (u, v) ∈ BA, one has (u, w) ∈ A and (w, v) ∈ B for some w. Hence hv, x′ i = hw, z ′ i = hu, y ′ i. Since (u, v) ∈ BA was arbitrary, we conclude that (x′ , y ′ ) ∈ (BA)∗ . Assume now B ∈ L(X) and (x′ , y ′ ) ∈ (BA)′ . Define z ′ := B ′ x′ . It suffices to show that (z ′ , y ′ ) ∈ A′ . Take (u, w) ∈ A and define v := Bw. Hence, (u, v) ∈ BA. Therefore ¸ ˙ ¸ ˙ ¸ ˙ ¸ ˙ ¸ ˙ u, y ′ = v, x′ = Bw, x′ = w, B ′ x′ = w, z ′ ,

whence (z ′ , y ′ ) ∈ A′ . Finally, assume that B is closed and A ∈ L(X). Since (BA)′ is w ∗ -closed and A′ B ′ ⊂ w∗

(BA)′ we have A′ B ′ ⊂ (BA)′ . To prove the reverse inclusion, by the Hahn-Banach Theorem ′ ′ it suffices to show (A B )⊤ ⊂ (BA)′⊤ . Now, (BA)′⊤ = J(BA)⊥⊤ = J(BA) = J(BA) since BA is closed. Let (u, v) ∈ (A′ B ′ )⊤ , i.e., hu, x′ i + hv, y ′ i = 0 for all (x′ , y ′ ) ∈ A′ B ′ . This is equivalent to hu, x′ i + hAv, z ′ i = 0 ∀(x′ , z ′ ) ∈ B ′ . This yields (−Av, u) ∈ B ′′ ∩ (X ⊕ X) = B = B since B is closed. Hence, J(u, v) ∈ BA and the statement is proved.

Corollary A.4.3. Let A be a m.v. linear operator on X. Then (λ − A)′ = (λ − A)′

and

R(λ, A)′ = R(λ, A′ )

for every λ ∈ C. For each λ ∈ C∞ we have N(λ, A′ ) = R(λ, A)⊥ . In particular, Pσ ˜ (A′ ) = R˜ σ (A). If A is closed, one has ̺(A) = ̺(A′ ). Let A be a single-valued operator on X. From part f ) of Proposition A.4.2 we see that A′ is densely defined if and only if A is closable, i.e., A is still single-valued; and A′ is single-valued if and only if A is densely defined. Remark A.4.4. Using [17, Thm. II.15] one can prove the Closed Range Theorem for m.v. operators in the same way as for single-valued operators, compare also [22, Thm. III.4.4].

A.5 Convergence of Operators In this section we consider the following situation: TLet (An )n be a sequence of m.v. operators on X and λ0 ∈ C such that λ0 ∈ n ̺(An ) for all n. Suppose further that the sequence (R(λ0 , An ))n converges to an operator Rλ0 ∈ L(X) (in norm or in the strong sense). We know from Proposition A.2.4 that there is a unique m.v. operator A with Rλ0 = (λ0 − A)−1 . Since Rλ0 ∈ L(X), we have

200

Appendix A. Linear Operators

λ0 ∈ ̺(A). We are interested in the question, whether there is convergence of the resolvents at other common resolvent points of the operators An . To answer this question, we let Ω := {λ ∈ C | ∃N ∈ N : λ ∈ ̺(An ) ∀n ≥ N and sup kR(λ, An )k < ∞} n≥N

It is immediate from the proof of Proposition A.2.3 that Ω is an open subset of C and that the mapping n(λ) := min{n ∈ N | λ ∈ ̺(Ak ) ∀ k ≥ n} locally can only decrease on Ω. Lemma A.5.1. Let λ ∈ C. a) If λ ∈ ̺(An ) for almost all n and if R(λ, An )x → y, then (x, y) ∈ R(λ, A). b) If λ ∈ Ω and (x, y) ∈ R(λ, A), then R(λ, An )x → y.

Proof. Define Qn := (λ − λ0 )R(λ0 , An ) and Q := (λ − λ0 )R(λ0 , A). The resolvent identity implies (I + Qn )R(λ, An ) = R(λ0 , An ). But R(λ0 , An ) → R(λ0 , A) strongly, hence Qn → Q strongly as well. Thus, we have (I + Q)y = R(λ0 , A)x. This implies readily that (x, y) ∈ R(λ, A). Let λ ∈ Ω and (x, y) ∈ R(λ, A). By Corollary A.2.2 one has I − (I + Q‚n )−1 = (λ −‚ λ0 )R(λ, An ). From λ ∈ Ω it follows that (I+Qn )−1 ∈ L(X) (n ≥ n0 ) and supn≥n0 ‚(I + Qn )−1 ‚ < ∞ for some n0 ∈ N. From (x, y) ∈ R(λ, A) it follows that (I + Q)y = R(λ0 , A)x. Therefore, (I + Qn )(R(λ, An )x − y) = R(λ0 , An )x − (I + Qn )y → R(λ0 , A)x − (I + Q)y = 0.

Applying (I + Qn )−1 yields R(λ, An )x − y → 0.

Corollary A.5.2. For λ ∈ Ω the m.v. operator (λ − A) is injective and has a closed range, i.e., λ ∈ / Aσ(A). In particular, σ(A) ∩ Ω is an open subset of C. Furthermore, we have Ω ∩ ̺(A) = {λ | λ ∈ ̺(An ) almost all n, (R(λ, An ))n strongly convergent}. Proof. Let λ ∈ Ω and K ≥ 0 such that kR(λ, An )k ≤ K for almost all n. By the last lemma, if (x, y) ∈ R(λ, A), then R(λ, An )x → y. Hence kyk ≤ K kxk. This shows that R(λ, A) is singlevalued and has a closed domain. From Corollary A.3.5 we know that ∂σ(A) ⊂ Aσ(A). Hence ∂σ(A) ∩ Ω = ∅. For λ ∈ Ω we have λ ∈ ̺(A) if and only if R(λ, A) is fully defined and this is the case if and only if (R(λ, An ))n is strongly convergent by Lemma A.5.1

The interesting question now is whether Ω = ̺(A) holds. It is true for the norm topology as the following proposition shows. Proposition A.5.3. Let An , A and Ω as above. Suppose R(λ0 , An ) → R(λ0 , A) in norm. Then Ω = ̺(A), and R(λ, A) → R(λ, A) in norm for all λ ∈ ̺(A).

Proof. Let λ ∈ ̺(A), and let Q, Qn be defined as in the proof of Lemma A.5.1. Then (I +Q)−1 ∈ L(X). Because of (I + Qn ) → (I + Q) in norm, we have (I + Qn )−1 ∈ L(X) for almost all n and (I + Qn )−1 → (I + Q)−1 in norm. This yields R(λ, An ) → R(λ, A) in norm. In particular, λ ∈ Ω. Conversely, ‚ ‚ let λ ∈ Ω. Then, Tn := I + Qn is invertible for all large n with K := ‚ ‚ supn≥n0 ‚Tn−1 ‚ < ∞. But Tn → T := I + Q in norm. Thus ‚ −1 ‚ ‚ ‚ −1 ‚ ‚ −1 −1 ‚ ‚ ≤ K 2 kTm − Tn k , ‚Tn − Tm ≤ ‚Tn ‚ kTm − Tn k ‚Tm

hence (Tn−1 )n is a norm-Cauchy sequence. Obviously, this implies that T is invertible, whence λ ∈ ̺(A).

A.6. Polynomials and Rational Functions of an Operator

201

A.6 Polynomials and Rational Functions of an Operator In this section A always denotes a single-valued operator on X. The sequence of natural powers (An )n∈N is defined recursively by A0 := I,

An+1 := An A (n ≥ 0).

A simple induction argument shows the validity of the power law An+m = An Am for all n, m ∈ N.2 In particular, An+1 = A An , and this shows that the sequence of domains D(An ) is decreasing, i.e., D(An+1 ) ⊂ D(An ). Another consequence is the inclusion An (D(Am )) ⊂ D(Am−n ) for m ≥ n. P Let p(z) = k≥0 ak z k ∈ C[z] be a polynomial. The operator p(A) :=

X

ak Ak

k≥0

is well-defined (by associativity of operator sums) with domain D(p(A)) = D(An ), where n = deg(p) in case that p 6= 0 and n = 0 in case that p = 0. (Note that this is not a definition but a conclusion.) Lemma A.6.1. Let p, q ∈ C[z]. The following statements hold. a) If p 6= 0, then p(A)q(A) = (pq)(A). In particular, if p 6= 0 and x ∈ X: x ∈ D(Adeg(p) ) and p(A)x ∈ D(An )

⇐⇒

x ∈ D(An+deg(p) ).

b) If p(A) is injective and q 6= 0, then D(p(A)−1 )∩D(q(A)) ⊂ D(p(A)−1 q(A))∩ D(q(A)p(A)−1 ) and p(A)−1 q(A)x = q(A)p(A)−1 x for each x ∈ D(p(A)−1 ) ∩ D(q(A)).

c) One has p(A) + q(A) ⊂ (p + q)(A), and equality holds in case deg(p + q) = max(deg(p), deg(q)). d) If T ∈ L(X) commutes with A, then it also commutes with p(A). In particular we have p(A)q(A) = q(A)p(A) for p, q 6= 0, and p(A) commutes with the resolvents of A. Proof. We prove a). The assertion is obviously true, if p or q are just scalars. Hence, by the Fundamental Theorem of Algebra and the associativity of operator multiplication, we can reduce the problem to the case deg(p), deg(q) = 1. This means that we have to establish the identity A2 − (µ + λ)A + µλ = (A − λ)(A − µ) 2 This can also be seen as follows: The set of all linear operators is a subsemigroup of the semigroup of all binary relations on X, with respect to the usual composition of relations. The power law now is actually true in every semigroup.

202

Appendix A. Linear Operators

for µ, λ ∈ C. Let y ∈ X be arbitrary. Because of D(A2 ) ⊂ D(A) we have y ∈ D(A) und (A − µ)y ∈ D(A) if and only if y ∈ D(A2 ). Thus, a) is proved. From a) it follows that p(A)q(A) = q(A)p(A) whenever p, q 6= 0. Now, a short argument gives b). The statement in c) is trivial. To prove d), one first shows T An ⊂ An T for n ∈ N by induction (Step: Because of T A ⊂ AT one has T An+1 ⊂ AT An ⊂ AAn T = An+1 T ). This yields T p(A) ⊂ p(A)T almost immediately.

Proposition A.6.2. Let A be an operator on a Banach spaces X such that ̺(A) 6= ∅. Then p(A) is a closed operator for each polynomial p ∈ C[z]. Furthermore, the spectral mapping theorem σ(p(A)) = p(σ(A)) holds. Proof. We prove the first statement by induction on n := deg p. The case n = 1 follows from the fact that the norms kxk + kAxk and kxk + k(A − µ)xk are equivalent norms on D(A). Now let n ≥ 1, deg p = n + 1, and λ ∈ ̺(A). If one defines µ := p(λ), then there is r ∈ C[z] such that deg r = n and p = (x − λ)r + µ. Let (xk )k be a sequence in D(An+1 ) converging to x (in X) such that p(A)xk → y (in X as well). Then we have r(A)xk → (A − λ)−1 (y − µx). By the induction hypothesis, r(A) is closed, hence x ∈ D(An ) with r(A) = (A − λ)−1 (y − µx) ∈ D(A). From a) and c) in Lemma A.6.1 it follows that x ∈ D(An+1 ) and y − µx = (A − λ)r(A)x = (p − µ)(A)x = p(A)x − µx. To prove the spectral mapping theorem it suffices to show that p(A) is invertible if and only if the roots of p are contained in the resolvent set of A. Therefore, let λ1 , . . . , λn ∈ C, and let T := (A − λ1 )(A − λ2 ) . . . (A − λn ) : D(An ) −→ X

be invertible. Then clearly A − λ1 is surjective and A − λn is injective. But all the A − λj commute with each other, by part a) of Lemma A.6.1. Hence all A − λj are bijective. On the other hand, if we assume λj ∈ ̺(A) for all j, then obviously T is invertible.

We denote by RA := {

p | p, q ∈ C[z] and {λ : q(λ) = 0} ⊂ ̺(A)} q

the set of all rational functions having their poles contained in ̺(A). For r = p/q ∈ RA we define r(A) := p(A)q(A)−1 . (This is independent of the special choice of p and q by Lemma A.6.1.) Note that there is some arbitrariness in this definition. For example, one could have equally defined r(A) := q(A)−1 p(A) (this is essentially the same operator but with a smaller domain). With our definition, the domain of r(A) is deg(p) − deg(q) if deg(p) ≥ deg(q), D(r(A)) = D(Am ) where m = 0 else. Proposition A.6.3. Let A be an operator on a Banach space X such that ̺(A) 6= ∅. For 0 6= r = p/q, r˜ = p˜/˜ q ∈ RA the following assertions hold. a) r(A) is a closed operator. b) r(˜ σ (A)) ⊂ σ ˜ (r(A)).

c) r(A)˜ r (A) ⊂ (r˜ r )(A), and equality holds, e.g., if (deg(p) − deg(q))(deg(˜ p) − deg(˜ q ) ≥ 0.

A.7. Injective Operators

203

d) r(A) + r˜(A) ⊂ (r + r˜)(A), and equality holds, e.g., if deg(p˜ q + p˜q) = max{deg(p˜ q ), deg(˜ pq)}. e) If T ∈ L(X) commutes with A, then it commutes also with r(A). Proof. Assertion a) is trivial, and assertions c) and d) follow from Lemma A.6.1. To prove b) we note first that r(A) − λI = [(p − λq)/q](A) for λ ∈ C. Hence we are left to show that r(A) is invertible, if {z : p(z) = 0} ⊂ ̺(A). Because of q(A)−1 p(A) ⊂ r(A) = p(A)q(A)−1 , the operator r(A) is invertible if and only if p(A) is. Assertion e) is a consequence of Lemma A.6.1 and of T r(A) = T p(A)q(A)−1 ⊂ p(A)T q(A)−1 = p(A)q(A)−1 T = r(A)T .

Here is an interesting corollary. Corollary A.6.4. Let A be a closed operator on a Banach space X. Let r˜ = p˜/˜ q∈ RA such that deg(˜ p) = deg(˜ q ). Then r(A)˜ r (A) = r˜(A)r(A) for every r ∈ RA . Proof. Just apply c) of Proposition A.6.3 twice. Another proof rests on the fact that r˜ can be written as a product of operators of the form α − βR(λ, A) for some numbers λ ∈ ̺(A), α, β ∈ C such that α 6= 0. For such operators we have x ∈ D(An )

⇐⇒

(α − βR(λ, A))x ∈ D(An )

for all n and all x ∈ X.

We end the section with a result for adjoints. Proposition A.6.5. Let A be a densely defined (single-valued) operator on X with ̺(A) 6= ∅. For each polynomial p ∈ C[z] we have p(A)′ = p(A′ ). The same statement holds if p is a rational function with poles inside ̺(A) = ̺(A′ ). Proof. Let r = p/q be a rational function with poles inside the set ̺(A′ ). Assume first that deg p ≤ deg q.QHence, r(A′ ) and r(A) are bounded operators. The function r can be written as a product r = j rj where each rj is either of the form α(λ − z)−1 or of the form α(λ − z)−1 + β. Now the claimed formula [r(A)]′ = r(A′ ) follows from part k) of Proposition A.4.2 and Corollary A.4.3. From this and part b) of Proposition A.4.2 we can infer that p(A)′ = p(A′ ) holds for all polynomials p having their roots inside ̺(A′ ). Now suppose deg p > deg q. Since ̺(A) 6= ∅ we can find a polynomial q1 having its roots inside ̺(A′ ) with deg q + deg q1 = deg p. Define r˜ := p/(q1 q). Then (1)

(2)

r(A′ ) = q1 (A′ )˜ r (A′ ) = q1 (A)′ r˜(A)′ = [˜ r(A)q1 (A)]′ = [q1 (A)˜ r (A)]′ = [r(A)]′ , where we have used k) from Proposition A.4.2 in (1) and Corollary A.6.3 in (2).

A.7 Injective Operators In this section we consider an injective single-valued operator A on a Banach space X. This enables us to extend the insertion mapping p 7−→ p(A) to the set of polynomials in z and z −1 . We begin with a surprising fact.

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Appendix A. Linear Operators

Lemma A.7.1. Let A be injective with ̺(A) 6= 0. Then p(A−1 )R(λ, A) = R(λ, A)p(A−1 ) for all λ ∈ ̺(A) and all polynomials p ∈ C[z].

Proof. Because R(λ, A) ∈ L(X) both distributivity inclusions (see Proposition A.1.1) are actually equalities, hence we can reduce the proof to the case p(z) = z. We have R(λ, A)A−1 ⊂ A−1 R(λ, A), since R(λ, A) commutes with A, hence with A−1 . Let x ∈ D(A−1 R(λ, A)). Then R(λ, A)x = Ay for some y ∈ D(A). But R(λ, A)x ∈ D(A), whence y ∈ D(A2 ), and we can apply λ − A on both sides to obtain x = (λ − A)Ay = A(λ − A)y ∈ R(A). This implies x ∈ D(A−1 ) = D(R(λ, A)A−1 ).

Let p(z) =

P

k∈Z

ak z k ∈ C[z, z −1] a polynomial. The operator p(A) :=

X

ak Ak

k∈Z

is well-defined (associativity for operator sums and injectivity of A). Its domain is D(p(A)) = D(Am ) ∩ D(An ) where n = min{k | (z k p(z))|z=0 6= ∞} and m = min{k | (z −k p(z))|z=∞ 6= ∞}. One can write p ∈ C[z, z −1] in a unique way as p(z) = q(z)z −n

n ∈ Z, q ∈ C[z], q(0) 6= 0.

Then we have p(A) = q(A)A−n . (This is clear for n ≤ 0. In the other case only equality of domains is to be shown, which is easy.) Note that a nice product law like in part a) of Lemma A.6.1 for polynomials can not hold in this situation. Simply look at A−1 A ⊂ I ⊃ AA−1 where the inclusions are strict in general. This example also shows that a general law of commutativity can not be expected. However, this is not the end of the story. Lemma A.7.2. Let p, q ∈ C[z] with q(0) 6= 0. Then p(A−1 )q(A) ⊂ q(A)p(A−1 ). Proof. We can assume p(A−1 ) = A−n without restriction. If x ∈ D(q(A)) with q(A)x ∈ D(A−n ), then there is y ∈ D(An ) such that q(A)x = An y. Because q(0) 6= 0 this implies x ∈ R(A), say, x = Ax1 . But then Aq(A)x1 = An y, and this yields q(A)x1 = An−1 y by inectivity of A. Inductively, it follows that x ∈ R(An ), say, x = An x0 . Hence we finally arrive at q(A)A−n x = q(A)x0 = y = A−n q(A)x. This proves the statement.

A.8. Semigroups and Generators

205

The simple example A−1 (1 + A) 6= (1 + A)A−1 shows that the inclusion in the last lemma is strict in general. Corollary A.7.3. Let p, q ∈ C[z] with p(0)q(0) 6= 0. Then p(A−1 )q(A) = (p(z −1 )q(z))(A) = q(A)p(A−1 ). Proof. The first equality is immediate from Lemma A.7.2. To prove the second, we can assume deg(p) = deg(q) = 1, employing the Fundamental Theorem of Algebra and Proposition A.6.3. The operator identity (A−1 + µ)(A + λ) = (1 + µλ) + µA + λA−1 then reduces to an almost trivial comparison of domains.

Proposition A.7.4. Let A be an injective operator and p, q ∈ C[z, z −1 ]. Then the following inclusions hold. p(A)q(A) ⊂ (pq)(A) and p(A) + q(A) ⊂ (p + q)(A). If T ∈ L(X) commutes with A, then it also commutes with p(A). If ̺(A) 6= ∅, p(A) is a closed operator.

Proof. We leave the proof to the reader.

A.8 Semigroups and Generators In this section we review the basic facts on semigroup theory. Our exposition is different to others in that we do not restrict the approach to strongly continuous semigroups. Let X be a Banach space. A (degenerate) semigroup on X is a strongly continuous mapping T : (0, ∞) −→ L(X) that possesses the semigroup property T (t)T (s) = T (t + s)

(t, s > 0).

The semigroup T is called bounded if sup kT (t)k < ∞.

0
If the semigroup is just bounded in 0, i.e., if supt≤1 kT (t)k < ∞, then there are constants M ≥ 1, ω ∈ R such that kT (t)k ≤ M eωt for all t > 0, compare [40, Chapter I, Proposition 5.5]. Hence such a semigroup is called exponentially bounded. Given an exponentially bounded semigroup T , the number ω0 (T ) := inf{ω ∈ R | ∃ M : kT (t))k ≤ M eωt (t > 0} is called the growth bound of T . The semigroup T is said to be exponentially stable if ω0 (T ) < 0. A semigroup T satisfying kT (t)k ≤ 1 for each t > 0 is called

206

Appendix A. Linear Operators

contractive or a contraction semigroup. It is called quasi-contractive if there is ω ≥ 0 such that the semigroup eω· T (·) is contractive. The space {x ∈ X | lim T (t)x = x} tց0

is called the space of strong continuity of the semigroup T . If it is the whole space X, the semigroup is called strongly continuous or a C0 -semigroup. Let T be an exponentially bounded semigroup and choose constants ω, M such that kT (t)k ≤ M eωt for t > 0. Then the Laplace transform of T exists at least in the halfplane {Re λ > ω}, i.e., Z ∞ Tˆ(λ)x := e−λt T (t)x dt (x ∈ X) 0

defines a bounded operator on X for every λ with Re λ > ω. One can show that in fact Tˆ(·) is a pseudo-resolvent, compare [6, p.114]. Hence there is a unique m.v. operator A such that Z (λ − A)−1 =

∞

e−λt T (t) dt.

0

We call A the generator of the semigroup T . By the injectivity of the Laplace transform the semigroup is uniquely determined by its generator. The semigroup T is said to be non-degenerate if A is single-valued. We obtain \ A0 = N(T (t)) = N(R(λ, A)) (λ ∈ ̺(A)) (A.1) t>0

again by the injectivity of the Laplace transform. Obviously, a C0 -semigroup is non-degenerate. If µ ∈ C, then A + µ generates the semigroup t 7−→ eµt T (t). Hence an operator generates a quasi-bounded semigroup if and only if there is ω ∈ R such that A − ω generates a bounded semigroup. Proposition A.8.1. Let T be a semigroup on the Banach space X satisfying kT (t)k ≤ M eωt for all t > 0. Then Z ∞ 1 R(λ, A)n = tn−1 e−λt T (t) dt and (A.2) (n − 1)! 0 M kR(λ, A)n k ≤ (A.3) (Re λ − ω)n for all n ∈ N and all Re λ > ω.

Proof. The proof is the same as in the strongly continuous case, compare [40, Chapter I, Corollary 1.11].

Note that each operator T (t) commutes with the resolvent of A, whence D(A) is left invariant by the semigroup T .

A.8. Semigroups and Generators

207

Proposition A.8.2. (Fundamental Identity for Semigroups) Let T be a quasi-bounded semigroup with generator A on the Banach space X. Define Z 1 1 t Dt := (T (t) − I) and Vt := T (s) ds t t 0 for t > 0. Then

(λR(λ, A) − I)Vε = Dε R(λ, A)

(Vε x, Dε x) ∈ A

and

for all ε > 0, Re λ > ω0 (T ), and all x ∈ X.

Proof. We compute

(λR(λ) − I)

Z

ε

0

Z ε Z ε λe−λt T (t) T (s) ds dt − T (s) ds 0 0 0 « Z ∞„ Z ε Z ε = λe−λt T (t + s) ds − λe−λt T (s) ds dt

T (s) ds =

Z

=

Z

∞

0

0

∞

λe

−λt

0

=

Z

=

0

t+ε

··· −

t

∞

λe−λt

0

I.p.

„Z „Z

ε

Z

0

∞

ε+t

··· −

Z

ε

...

0

Z

0

t

...

«

«

dt dt

e−λt (T (t + ε) − T (t)) dt = (T (ε) − I)R(λ).

Dividing by ε completes the proof of the first statement. Using this we obtain Vε = R(λ)(λVε − Dε ). This shows that (Vε x, λVε x − Dε x) ∈ (λ − A) for every x ∈ X.

Corollary A.8.3. Let T be a quasi-bounded semigroup on the Banach space X and let x ∈ X. The following assertions are equivalent. (i) x ∈ D(A).

(ii) T (t)x → x as t ց 0. Rε (iii) 1ε 0 T (s)x ds → x as ε ց 0. (iv) λR(λ, A)x → x as λ → ∞.

In particular, D(A) is the space of strong continuity of T . One has the inclusion T (t)X ⊂ D(A) for each t > 0.

Proof. Obviously we have (iv) ⇒ (i) and (ii) ⇒ (iii). Since R(Vε ) ⊂ D(A) by Proposition A.8.2, we also obtain (iii) ⇒ (i). Let x ∈ D(A) and pick y ∈ (λ − A)x where λ > ω0 (T ). Then T (ε)x − x = (T (ε) − I)R(λ)y = εVε (λR(λ) − I)y → 0 as ε → 0. Since supt≤1 kT (t)k < ∞, we obtain (i) ⇒ (ii). To prove the remaining implication (i) ⇒ (iv) we note first that kλR(λ)k is uniformly bounded for λ > ω where ω > ω0 (T ). This follows from Proposition A.8.1. Given x ∈ D(A) we choose µ ∈ ̺(A) and y ∈ (µ − A)x to obtain λR(λ)x = λR(λ)R(µ)y =

as λ → ∞.

λ (R(µ)y − R(λ)y) → R(µ)y = x λ−µ

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Appendix A. Linear Operators

Corollary A.8.4. Let x ∈ X. Then we have x ∈ D(A), Ax ∩ D(A) 6= ∅

⇔

1 lim (T (t)x − x) =: y t

tց0

exists

and in this case {y} = Ax ∩ D(A). In particular, A0 ∩ D(A) = 0, whence XT := A0 ⊕ D(A) is a closed subspace of X. It follows from Corollary A.8.4 that the part B := A ∩ (D(A) ⊕ D(A)) of A in D(A) is single-valued. Proposition A.8.5. Let T be a quasi-bounded semigroup with generator A on the Banach space X. Let Y := D(A). The space Y is left invariant by the semigroup T . The semigroup T restricts to a strongly continuous semigroup on Y which has B = A ∩ (Y ⊕ Y ) as its generator. The next result is one of the cornerstones of the theory of C0 -semigroups. (See [40, Chapter II, Section 3] or [6, Theorem 3.3.4] for proofs.) Theorem A.8.6. (Hille-Yosida) Let A be a (single-valued) linear operator on the Banach space X. Assume that A has dense domain and there is M ≥ 1 and ω ∈ R such that (ω, A) ⊂ ̺(A) and kR(λ, A)n k ≤

M (λ − ω)n

(A.4)

for all n ∈ N and all λ > ω. Then A generates a C0 -semigroup satisfying kT (t)k ≤ M eωt for t ≥ 0. Remark A.8.7. Unfortunately there is no similar characterization for generators of general exponentially bounded semigroups. The resolvent condition (A.3) guarantees that A generates a so-called integrated semigroup S(·), see [6, Theorem Rt 3.3.1]. For x ∈ D(A) one has S(t)x = 0 T (s)x ds, where T is the semigroup generated by the part B of A in D(A). Then A generates an exponentially bounded semigroup if and only if S(·)x ∈ C 1 ((0, ∞), X) for each x ∈ X. Employing [5, Theorem 6.2] one can show that this is always true if the Banach space X has the Radon-Nikodym property. Let us note an important result on bounded perturbations. For a proof see [40, Theorem III.1.3]. Proposition A.8.8. Let A be the generator of a C0 -semigroup on the Banach space X and let B ∈ L(X) be a bounded operator. Then A+B generates a C0 -semigroup on X. Finally we deal with the important case of groups. Proposition A.8.9. Let T be an exponentially bounded semigroup with generator A. The following assertions are equivalent:

A.8. Semigroups and Generators

209

(i) There exists t0 > 0 such that T (t0 ) is invertible. (ii) Each T (t) is invertible and the mapping T˜ : R → L(X)× , defined by  (t > 0)  T (t) I (t = 0) T˜(t) :=  T (−t)−1 (t < 0) is a strongly continuous group homomorphism.

(iii) The operator −A generates an exponentially bounded semigroup and A is single-valued. Proof. (i) ⇒ (ii) ⇒ (iii). Assume that T (t0 ) is invertible. Then the semigroup property readily implies that each T (t) is invertible. Moreover, since R(T (t)) ⊂ D(A) for each t > 0, we have that A is densely defined. Hence A is single valued and we are in the standard (C0 -) case. So we can refer to [40, subsection 3.11], or [105, Section 1.6] for the remaining arguments. (iii) ⇒ (i). We denote by S the semigroup generated by −A. Choose ω > ω0 (T ), ω0 (S). Let Y := D(A) and let B := A ∩ (Y ⊕ Y ) the part of A in Y . From Proposition A.8.5 we know that B generates the C0 -semigroup obtained by restricting T to Y . Analogously, −B generates the C0 -semigroup obtained by restricting S to Y . (Note that D(−A) = D(A).) The Theorem from [40, Section 3.11] now yields that T (t)S(t)y = S(t)T (t)y = y for all y ∈ Y . Let t0 > 0 and suppose that T (t0 )x = 0 for some x ∈ X. Then Z t0 R(λ, A)x = e−λt T (t)x dt =: f (λ) 0

for Re λ > ω. Obviously, f has a holomorphic continuation to an entire function which is bounded on every right halfplane. Claim: f (λ) = R(λ, A) for all Re λ < −ω.

Proof of Claim. Consider the function F : C → X ⊕X/A defined by F (λ) := (f (λ), λf (λ)− x) + A. Then F is entire and F (λ) = 0 for Re λ > ω, since (f (λ), x) ∈ λ − A for these λ. By the uniqueness theorem for holomorphic functions, F (λ) = 0 for all λ ∈ C. Hence (f (λ), x) ∈ λ − A even for Re λ < −ω. However, these λ are contained in the resolvent set of A, whence the claim is proved. From the claim we follow that f is in fact bounded also on some left halfplane. Hence it is constant. However, f (λ) → 0 if Re λ → ∞. Thus, f (λ) = 0 for all λ ∈ C. This implies R(λ, A)x = 0 for a lot of λ, whence x = 0, since A is assumed to be single-valued. So we have shown that T (t0 ) is injective. We obtain that T (t0 ) : X −→ Y is an isomorphism. But since T (t0 ) : Y −→ Y is also an isomorphism, we must have X = Y .

One usually writes T again instead of T˜ and calls it a C0 -group. In general one cannot omit the assumption “A is single-valued” from (iii). Indeed, let S be a C0 -group on a Banach space Y and let X := Y ⊕ C with T (t)(y, λ) := (S(t), 0) for all t ∈ R. If B generates S, then A = {((y, 0), (By, λ)) | y ∈ D(B), λ ∈ C} generates (T (t))t≥0 and −A generates (T (−t))t≥0 . Given a C0 -group (T (t))t∈R we call θ(T ) := inf{θ > 0 | ∃ M ≥ 1, kT (t)k ≤ M eθ|t| (t ∈ R)}

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Appendix A. Linear Operators

the group type of T . Let us call the semigroups T⊕ and T⊖ , defined by T⊕ (t) := T (t) and T⊖ (t) := T (−t) (t ≥ 0), the forward semigroup and the backward semigroup corresponding to the group T . Then we obviously have θ(T ) = max{ω0 (T⊕ ), ω0 (T⊖ )}.

References The basic results on m.v. operators, covered by Section A.1 and Section A.2, are contained, e.g., in [42, Chapter I] or the monograph [22]. Proposition A.2.4 and its corollary on pseudo-resolvents is not included in these books and may be new. The same is true for Propositions A.2.7 and parts of Section A.3. An exhaustive treatment of adjoints of multivalued operators can be found in [22, Chapter III]. The results on convergence in Section A.5 are generalizations of well-known facts for single-valued operators which can be found, e.g., in [65, Theorem IV.2.23 and Chapter VIII,§1]. Polynomials of operators (Section A.6) are studied in [39, Chapter VII.9] including the spectral mapping theorem (Proposition A.6.2). By [22, Theorem VI.5.4], the spectral mapping theorem for polynomials remains valid for multivalued operators. We provided the results for rational functions and on injective operators (Section A.7) without direct reference, but it is quite likely that these facts have been published somewhere. The results of [22, Chapter 6] let expect that there is a generalization of Sections A.6 and A.7 to multivalued operators. Section A.8 is an adaptation from the standard monographs in semigroup theory, like [105], [40], and [6]. However, if multivalued operators are involved (like in Proposition A.8.2 and Proposition A.8.9), we do not know of a direct reference.

Appendix B

Interpolation Spaces In this appendix we will shortly survey basic definitions and results of interpolation theory of Banach spaces.

B.1 Interpolation Couples Let X, Y be two normed spaces. A closed linear (single-valued) and injective operator ι ⊂ X ⊕ Y is called an (interpolation) coupling between X and Y . A pair of normed spaces (X, Y ) together with an interpolation coupling ι is called an interpolation couple. If (X, Y ) is an interpolation couple and both X, Y are Banach spaces then (X, Y ) is called a Banach couple Lemma B.1.1. Let X, Y be normed spaces. If there is a Hausdorff topological vector space Z and continuous linear injections ιX : X −→ Z and ιY : Y −→ Z, then (X, Y ) is an interpolation couple with respect to the coupling ι defined by (x, y) ∈ ι

⇐⇒

ιX (x) = ιY (y)

(B.1)

for x ∈ X, y ∈ Y . Conversely, if ι is some coupling between X and Y then there is a normed space Z and continuous injections ιX : X −→ Z and ιY : Y −→ Z such that (B.1.1) holds. Proof. The first assertion is straightforward to prove. Assume that ι is some coupling between X and Y . Then the operator −ι is a closed subspace of the normed space X ⊕ Y . Define the normed space Z as the quotient space Z := X ⊕ Y / − ι and denote by [x, y] the image of (x, y) under the canonical projection of X ⊕ Y onto Z. Then let ιX := (x 7−→ [x, 0]) : X −→ Z and ιY := (y 7−→ [0, y]) : Y −→ Z. Clearly, ιX and ιY are linear and continuous. Moreover, they are injective. To see this, consider x ∈ X such that ιX (x) = 0. This means that (x, 0) ∈ (−ι). Since ι is injective, x = 0. The injectivity of ιY follows similarly from the single-valuedness of ι. Now, for x ∈ X, y ∈ Y , we have ιX (x) = ιY (y) ⇐⇒ [x, 0] = [0, y] ⇐⇒ (x, −y) ∈ (−ι) ⇐⇒ (x, y) ∈ ι. This proves (B.1.1).

Given an interpolation couple (X, Y ) (with coupling ι) the normed space X + Y := X ⊕ Y /(−ι)

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Appendix B. Interpolation Spaces

defined in the proof of Lemma B.1.1 is called the sum of the couple (X, Y ). Furthermore, the space X ∩ Y := D(ι)

with the norm kxkX +kι(x)kY is called the intersection of the couple (X, Y ). If the coupling ι is obtained by (B.1.1) based on imbeddings ιX , ιY into some Hausdorff topological vector space Z, then X ∩ Y and X + Y embed canonically in Z. In fact, the most common situation is when X and Y are already subspaces of some space Z with continuous inclusion mappings. Then the coupling is natural and reference to it is often omitted. Also, the spaces X ∩ Y and X + Y appear naturally as subspaces of Z. Viewing X und Y as subspaces of X + Y , one has kxkX+Y = inf{kakX + kbkY | a ∈ X, b ∈ Y, a + b = x} kzkX∩Y = kzkX + kzkY

(B.2) (B.3)

for the norms on X + Y and X ∩ Y , respectively.

If both X and Y are complete, i.e., if (X, Y ) is a Banach couple, then X + Y and X ∩ Y are also complete.

The interpolation couples form the objects of a category whose morphisms are the pairs (T1 , T2 ) of continuous linear mappings ˜ and T2 : Y −→ Y˜ T1 : X −→ X such that ˜ι ◦ T1 = T2 ◦ ι. This condition ensures that one can unambiguously define an operator ˜ + Y˜ T : X + Y −→ X ˜ + Y˜ ) and T |X∩Y ∈ with T |X = T1 and T |Y = T2 . One then has T ∈ L(X + Y, X ˜ ˜ L(X ∩ Y, X ∩ Y ). Hence sum and intersection are (covariant) functors from the category of interpolation couples into the category of normed spaces. One usually abuses notation a bit and simply writes T instead of T1 , T2 . ˜ and T ∈ L(X, X). ˜ Lemma B.1.2. Let Y ⊂ X, Y˜ ⊂ X If T (Y ) ⊂ Y˜ , then ˜ T ∈ L(Y, Y ). Proof. This is an easy application of the Closed Graph Theorem.

˜ Y˜ ) is already characterized By Lemma B.1.2, a morphism T : (X, Y ) −→ (X, by the following properties. a) T ∈ L(X + Y, X ′ + Y ′ ) und b) T (X) ⊂ X ′ , T (Y ) ⊂ Y ′

A functor F from the category of interpolation couples into the category of normed spaces is called an interpolation functor if the following assertions hold. 1) X ∩ Y ⊂ F(X, Y ) ⊂ X + Y for all interpolation couples (X, Y ). ˜ Y˜ ) of interpolation 2) F (T ) = T |F(X,Y ) for every morphism T : (X, Y ) −→ (X, couples. 3) If (X, Y ) is a Banach couple then also F (X, Y ) is a Banach space.

B.2. Real Interpolation by the K-Method

213

B.2 Real Interpolation by the K-Method Let (X, Y ) be an interpolation couple. We define K(t, x) := K(t, x, X, Y ) := inf{kakX + t kbkY | x = a + b, a ∈ X, b ∈ Y } for x ∈ X + Y and t > 0. Lemma B.2.1. The following assertions hold true. a) For every t > 0 the mapping x 7−→ K(t, x) is an equivalent norm on X + Y . More precisely, min(t, 1) kxkX+Y ≤ K(t, x) ≤ max(t, 1) kxkX+Y for all x ∈ X + Y .

b) For fixed x ∈ X + Y the mapping t 7−→ K(t, x) is monotone increasing and concave, whence continuous. c) K(t, x, X, Y ) = tK( 1t , x, Y, X) for all x ∈ X + Y, t > 0.

d) K(t, x) ≤ st K(s, x) for 0 < s < t, i.e., for fixed x the mapping t 7−→ 1 t K(t, x) is monotone decreasing. e) For x ∈ X ∩ Y one has K(t, x) ≤ min(1, t) kxkX∩Y .

f ) If Y ⊂ X, then K(t, x) ≤ kxkX for all t > 0, x ∈ X.

Proof. The proofs are elementary.

Let us denote by Lp∗ (a, b) the space of p-integrable functions on the interval (a, b) ⊂ (0, ∞) with respect to the measure dt/t. We abbreviate the positive real coordinate (t 7→ t) simply by t. Now we define (X, Y )θ,p := {x ∈ X + Y | (t 7−→ t−θ K(t, x)) ∈ Lp∗ (0, ∞)} for 0 ≤ θ ≤ 1 and 1 ≤ p ≤ ∞, and endow this space with the norm

kxkθ,p = t−θ K(t, x) Lp (0,∞) . ∗

We call the space (X, Y )θ,p the real interpolation space with parameters θ, p. Here are basic properties. Proposition B.2.2. Let (X, Y ) be an interpolation couple, 0 ≤ θ ≤ 1 and 1 ≤ p ≤ ∞. Then the following statements hold. a) (X, Y )θ,p = (Y, X)1−θ,p with equality of norms. b) (X, Y )θ,p = 0 in case p < ∞ and θ ∈ {0, 1}.

c) There is c = c(θ, p) such that K(t, x) ≤ c tθ kxkθ,p fotr all t > 0 and x ∈ (X, Y )θ,p . In particular, (X, Y )θ,p ⊂ (X, Y )θ,∞ .

d) One has X ∩ Y ⊂ (X, Y )θ,p ⊂ X + Y , except for p < ∞ and θ ∈ {0, 1}.

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Appendix B. Interpolation Spaces

e) For x ∈ X + Y the following assertions are equivalent: (i) x ∈ (X, Y )0,∞

(ii) there is (xn )n ⊂ X with supn kxn kX < ∞ and kxn − xkX+Y → 0. In particular, (X, Y )0,∞ ⊂ X.

f ) Let 0 < θ < 1 and x ∈ (X, Y )θ,∞ . Then there are sequences (xn )n ⊂ X and (yn )n ⊂ Y with the following properties: a) kx − xn kX+Y , kx − yn kX+Y → 0 and b) kxn kX = O(nθ ), kyn kY = O(n1−θ ). In particular, (X, Y )θ,∞ ⊂ X ∩ Y .

Proof. Ad a). This follows from the equality K(t, x, X, Y ) = tK(1/t, x, Y, X) which holds for all x ∈ X + Y and t > 0. Ad b). By a) it suffices to consider the case θ = 0. So assume K(., x) ∈ Lp ∗ . Because of R kxkX+Y ≤ K(t, x) for t ≥ 1, we have 1∞ kxkpX+Y t−1 dt < ∞. This implies kxkX+Y = 0. Ad c). In case p = ∞ one can choose c = 1 by definition. Now assume 1 ≤ p < ∞. If θ ∈ {0, 1}, then by b) every c does the job. Hence we can suppose θ ∈ (0, 1) for the following. Then we write „Z ∞ «1 „Z ∞ «1 1 1 1 ds p ds p s−θp K(t, x) ≤ (θp) p s−θp K(s, x)p ≤ (θp) p kxkθ,p t−θ K(t, x) = (θp) p s s t t for x ∈ (X, Y )θ,p . Ad d). We let t = 1 in c) and obtain kxkX+Y ≤ c(θ, p) kxkθ,p . For x ∈ X ∩ Y we have K(t, x) ≤ min(1, t) kxkX∩Y . Now the assertion follows from the implication p = ∞ ∨ θ ∈ (0, 1)

=⇒

t−θ min(1, t) ∈ Lp ∗ (0, ∞).

Ad e). Assume first x ∈ (X, Y )0,∞ , i.e. there is c > 0 such that K(t, x) < c for all t > 0. For each n ∈ N choose x = xn + yn ∈ X + Y with kxn kX + n kyn kY < c. Then supn kxn kX < c and kx − xn kX+Y ≤ kykY ≤ cn → 0. On the other hand, if (ii) is assumed, then K(t, x)n ≤ kxn kX and K(t, xn ) → K(t, x), since each K(t, ·) is a norm on X + Y . Hence (i) follows. Ad f ). This is similar to the proof of part e).

The next result shows in particular that the mapping (X, Y ) 7−→ (X, Y )θ,p is an interpolation functor whenever 1 ≤ p ≤ ∞, θ ∈ [0, 1], except for the case p < ∞ and θ ∈ {0, 1}. ˜ Y˜ ) be interpolation couples. The following Theorem B.2.3. Let (X, Y ) and (X, assertions hold. a) If (X, Y ) is a Banach couple, then (X, Y )θ,p is a Banach space for all 1 ≤ p ≤ ∞, θ ∈ [0, 1]. b) One has always

(X, Y )θ,p ⊂ (X, Y )θ,q for 1 ≤ p ≤ q ≤ ∞ und θ ∈ [0, 1].

B.2. Real Interpolation by the K-Method

215

˜ Y˜) be a morphism of interpolation couples. Then it c) Let T : (X, Y ) −→ (X, ˜ Y˜ )θ,p such that restricts to a bounded linear mapping T : (X, Y )θ,p −→ (X, 1−θ θ kT k(X,Y )θ,p →(X, ˜ Y˜ )θ,p ≤ kT kX→X ˜ kT kY →Y˜

for all 1 ≤ p ≤ ∞ und θ ∈ [0, 1]. Proof. Ad a). This is [86, p.9/10]. Ad b). By c) of Proposition B.2.2 we have (X, Y )θ,p ⊂ (X, Y )θ,∞ , whence kxkθ,∞ ≤ c kxkθ,p for all x ∈ (X, Y )θ,p . If p < q < ∞, then kxkθ,q =

„Z

∞

≤

„Z

∞

t−θq K(t, x)q

0

0

dt t

dt t−θp K(t, x)p t

«1 q

«1 q

=

„Z

∞

t−θp K(t, x)p (t−θ K(t, x))(q−p)

0

sup(t−θ K(t, x))

q−p q

t>0

p

dt t

«1 q

q−p

q q = kxkθ,p kxkθ,∞ ≤ c kxkθ,p .

Ad c). This is [86, p.10].

Corollary B.2.4. Let (X, Y ) be an interpolation couple, 1 ≤ p ≤ ∞ and θ ∈ (0, 1). Then there is c = c(p, θ) such that 1−θ

θ

kxkθ,p ≤ c kxkX kxkY f¨ ur alle x ∈ X ∩ Y .

˜ Y = Y˜ and T = I in c) of Theorem B.2.3. Proof. Let X = X,

Remark B.2.5. Let us briefly consider the special case Y ⊂ X. Then clearly (X, Y ) is an interpolation couple. Since in this case K(t, x) is bounded, we certainly have t−θ K(t, x) ∈ Lp∗ (1, ∞) for all θ ∈ (0, 1] and all p ∈ [1, ∞] (and also for θ = 0 and p = ∞). Hence one only has to check the condition at 0, and

−θ

t K(t, x) p L (0,1) ∗

is an equivalent norm on (X, Y )θ,p (in the relevant cases). A similar remark holds in the case X ⊂ Y , where only the behaviour of K(t, x) at ∞ is relevant, and so kt−θ K(t, x)kLp∗ (1,∞) is an equivalent norm on (X, Y )θ,p (in the relevant cases). Proposition B.2.6. Let Y ⊂ X and 0 < θ < σ < 1. Then Y = (X, Y )1,∞ ⊂ (X, Y )σ,∞ ⊂ (X, Y )θ,1 ⊂ (X, Y )0,∞ = X. Proof. This is [86, p.9].

The following result is often useful.

216

Appendix B. Interpolation Spaces

Proposition B.2.7. Let (X, Y ) be a Banach couple, θ ∈ [0, 1], p ∈ [1, ∞]. Then the following identities hold. (X + Y, Y )θ,p ∩ X = (X, Y )θ,p ∩ X = (X, X ∩ Y )θ,p

(X + Y, X)θ,p ∩ (X + Y, Y )θ,p = (X + Y, X ∩ Y )θ,p (X, X ∩ Y )θ,p + (Y, X ∩ Y )θ,p = (X + Y, X ∩ Y )θ,p

(X + Y, X)1−θ,p ∩ (X + Y, Y )θ,p = (X, Y )θ,p (X, X ∩ Y )θ,p + (Y, X ∩ Y )1−θ,p = (X, Y )θ,p

(B.4) (B.5) (B.6) (B.7) (B.8)

Proof. In the first assertion the chain of inclusions “⊃” holds trivially. Now let x ∈ (X+Y, Y )θ,p ∩ X. Then K(t, x, X, X ∩ Y ) ≤ kxkX and we only have to account for the case 0 < t ≤ 1. Let x = a + b + c, where a ∈ X and b, c ∈ Y . Then b + c = x − a ∈ X ∩ Y , whence K(t, x, X, X ∩ Y ) ≤ kakX + t kb + ckX∩Y = kakX + t kb + ckY + t kx − akX ≤ kakX + kbkY + t kxkY + t kxkX + kakX ˆ ˜ ≤ 2 kakX + kbkY + t kckY + t kxkX

Taking the infimum first over a, b (with c fixed) and afterwards over c yields K(t, x, X, X ∩ Y ) ≤ 2K(t, x, X + Y, Y ) + t kxkX . This finishes the proof of the first statement. For the remaining arguments we are in need of the so-called “modular law”: B⊂C

=⇒ (A + B) ∩ C = (A ∩ C) + B

which holds for all subspaces A, B, C of a common superspace. The two identities (B.5) and (B.6) are proved by using (B.4) and the modular law (several times): (X + Y,X ∩ Y )θ,p ⊂ (X + Y, X)θ,p ∩ (X + Y, Y )θ,p

= [(X + Y, X)θ,p ∩ (X + Y )] ∩ [(X + Y, Y )θ,p ∩ (X + Y )] ˘ ¯ ˘ ¯ = X + [(X + Y, X)θ,p ∩ Y ] ∩ Y + [(X + Y, Y )θ,p ∩ X] ˘ ¯ ˘ ¯ = X + (Y, X ∩ Y )θ,p ∩ Y + (X, X ∩ Y )θ,p ˘ ¯ = X ∩ [Y + (X, X ∩ Y )θ,p ] + (Y, X ∩ Y )θ,p = (X ∩ Y ) + (X, X ∩ Y )θ,p + (Y, X ∩ Y )θ,p

= (X, X ∩ Y )θ,p + (Y, X ∩ Y )θ,p ⊂ (X + Y, X ∩ Y )θ,p .

This kind of proof also works for the last two identities:

(X,Y )θ,p ⊂ (X, X + Y )θ,p ∩ (X + Y, Y )θ,p ˆ ˜ ˆ ˜ = (X, X + Y )θ,p ∩ (X + Y ) ∩ (X + Y, Y )θ,p ∩ (X + Y ) ˆ ` ´˜ ˆ ` ´˜ = X + (X, X + Y )θ,p ∩ Y ∩ Y + (X + Y, Y )θ,p ∩ X ˆ ˜ ˆ ˜ = X + (X ∩ Y, Y )θ,p ∩ Y + (X, X ∩ Y )θ,p ˆ ` ´˜ = (X ∩ Y, Y )θ,p + X ∩ Y + (X, X ∩ Y )θ,p ˆ ˜ = (X ∩ Y, Y )θ,p + (X ∩ Y )θ,p + (X, X ∩ Y )θ,p ⊂ (X ∩ Y, Y )θ,p + (X, X ∩ Y )θ,p ⊂ (X, Y )θ,p .

B.3 The Trace Method The trace method yields an equivalent description of the real interpolation spaces. As a tool we need the Hardy-Young inequalities.

B.3. The Trace Method

217

Proposition B.3.1. (Hardy-Young Inequality) Let 0 < a ≤ ∞ and ϕ : (0, a) −→ [0, ∞] be a positive measurable function. Let also α > 0 and 1 ≤ p < ∞. Then the following inequalities hold. p Z t ds t−α ϕ(s) s 0 0 Z a Z a p ds α t ϕ(s) s 0 t Z

a

! p1

1 ds p s 0 p1 Z a p1 1 dt α p ds ≤ [s ϕ(s)] t α s 0

dt t

≤

1 α

Z

a

[s−α ϕ(s)]p

(B.9) (B.10)

Proof. Using Jensen’s inequality yields „ «p „ «p Z t Z t Z t α ϕ(s) α ϕ(s) α−1 α ϕ(s)p α−1 ds = s ds ≤ α s ds. α α α t s t s t sαp 0 0 0 Hence αp

«p Z a Z t ϕ(s) dt α ϕ(s)p α−1 dt ds ≤ s ds α s t t sαp t 0 0 0 « Z p a α dt a` ´ ϕ(s)p α−1 ϕ(s) s ds = s−α − a−α sα−1 ds α+1 αp αp t s s 0 0 s Z a Z a ϕ(s)p ϕ(s)p ds ≤ s−α αp sα−1 ds = . s sαp s 0 0

Z

a

t−αp q 0 Z α „Z =

„Z

t

The second inequality is proved similarly.

Let 1 ≤ p ≤ ∞, 0 ≤ a < ∞, 0 < α < 1 and f such that tα f (t) ∈ Lp∗ (0, a). Then immediately f ∈ L1 (0, a′ ) for all a′ ≤ a. This follows from Z

0

a′

|f (t)| dt =

Z

0

a′

α

t |f (t)| t

1−α

dt ≤ ktα f kLp∗ t

Z

a′

t 0

(1−α)p′

dt t

! 1′ p

with an according modification in the case p = 1. However, the Hardy-Young inequality yields more. Corollary B.3.2. Let 1 ≤ p ≤ ∞, 0 < α < 1 and 0 < a ≤ ∞. If tα f (t) ∈ Lp∗ (0, a), then if Z 1 t g(t) := f (s) ds t 0

also tα g ∈ Lp∗ (0, a) and ktα gkLp∗ (0,a) ≤

1 1−α

ktα f kLp∗

Proof. Replace ϕ(s) in (B.9) by sϕ(s) and α by 1 − α.

We can now turn to the trace method of interpolation. Let again (X, Y ) be an interpolation couple, 0 < θ < 1 and 1 ≤ p ≤ ∞. For simplicity, in the rest of this section we assume the spaces X, Y to be Banach spaces. We consider functions 1,p u ∈ Wloc ((0, ∞); X + Y )

218

Appendix B. Interpolation Spaces

such that uθ vθ

:= :=

t1−θ u(t) t1−θ u′ (t)

∈ Lp∗ ((0, ∞); Y ) ∈ Lp∗ ((0, ∞); X)

The space of all such functions is denoted by V (p, θ, X, Y ) and it is naturally endowed with the norm kukV (p,θ) := kuθ kLp∗ (0,∞) + kvθ kLp∗ (0,∞) Obviously, u ∈ C((0, ∞); X + Y ). Moreover, since Z

1

Z

1

1

1

t(1−θ)− p ku′ (r)kX tθ−1+ p dt 0

1

θ−1+ p1 ≤ t1−θ− p u′ p

t

p′ L (0,1;X) L (0,1)

1 −1 ′ − p′ 1−θ ′ ≤ (p θ) t u Lp (0,∞;X) = (p′ θ) p′ kvθ kLp∗ (0,∞;X) ,

ku′ (r)kX dr =

0

∗

we have also u′ ∈ L1 ((0, 1); X). Hence u(t) − u(s) =

Z

s

t

u′ (r) dr ∈ X

f¨ ur

0 < s < t.

Note that in general one does not have u(s) ∈ X. However, the limit u(0) := lim u(s) = u(t) − sց0

Z

t

0

u′ (r) dr ∈ X + Y

exists. This yields a continuous linear mapping (“trace”) (u 7−→ u(0)) : V (p, θ, X, Y ) −→ X + Y 1

− p1′

with ku(0)kX+Y ≤ (p(1 − θ))− p kuθ kLp∗ ((0,∞);Y ) + (p′ θ)

kvθ kLp∗ ((0,∞);X) .

R [By u(0) = u(t) − 0 u′ (r) dr one has ku(0)kX+Y ≤ ku(t)kY + 01 ku′ (r)kX dr. Multiply ‚ ‚ p both sides by t1−θ and take the L∗ (0, 1)-norm. The assertion follows from ‚t1−θ ‚Lp (0,1) = 1 −p

((1 − θ)p)

Rt

∗

.]

Theorem B.3.3. Let (X, Y ) be a Banach couple, 0 < θ < 1 and 1 ≤ p ≤ ∞. Then (X, Y )θ,p = {x | ∃ u ∈ V (θ, p, X, Y ) : x = u(0)} if on the right-hand space we choose the norm tr

kxkθ,p := inf{kukV (p,θ,X,Y ) | u ∈ V (p, θ, X, Y ), u(0) = x}.

B.3. The Trace Method

219

R Proof. We prove “⊃”. Let x = u(0) for u ∈ V (p, θ, X, Y ). Then x = − 0t u′ (r) dr + u(t), hence „ Z t « ‚ ′ ‚ 1 ‚u (r)‚ dr + t1−θ ku(t)k t−θ K(t, x) ≤ t1−θ for t > 0. Y X t 0 “ R ” By Corollary B.3.2 we have t1−θ 1t 0t ku′ (r)kX dr ∈ Lp ∗ (0, ∞) and

Since

1 θ

‚ „ Z ‚ 1−θ 1 ‚t ‚ t

t

0

«‚ ‚ ′ ‚ ‚ ‚u (r)‚ dr ‚ ‚ X

> 1, it follows x ∈ (X, Y )θ,p with kxkθ,p ≤

1 θ

‚ 1‚ ‚ 1−θ ′ ‚ u‚ p . ‚t L∗ (0,∞) θ

≤

p L∗ (0,∞)

kxktr θ,p .

To prove the reverse inclusion “⊂” we have to identify x ∈ (X, Y )θ,p as a trace u(0). Without restriction one can assume x 6= 0. For each n ∈ N there exists an , a ˜n ∈ X, bn , ˜bn ∈ 1 1 Y such that x = an + bn = a ˜n + ˜ bn and kan kX + n kbn kY ≤ 2K( n , x), and additionally k˜ an kX + nk˜bn kY ≤ 2K(n, x). Now we define b(t) :=

a(t) :=

∞ X

n=1 ∞ X

bn+1 1( an+1 1(

n=1

1 , 1 ] (t) n+1 n

+

1 , 1 ] (t) n+1 n

+

∞ X

n=1 ∞ X

˜ bn 1(n,n+1] (t) a ˜n 1(n,n+1] (t)

n=1

Then x = a(t) + b(t) for all t > 0. Furthermore, ∞ X

t1−θ kb(t)kY = ≤2

≤2

∞ X

n=1

t1−θ kbn+1 kY 1(

t1−θ (n + 1)K(

n=1

∞ X

t1−θ

n=1

1 , 1 ] (t) n+1 n

+

∞ X

n=1 ∞ X

t1−θ k˜bn kY 1(n,n+1] (t)

1 1 , x)1( 1 , 1 ] (t) + 2 t1−θ K(n, x)1(n,n+1] (t) n+1 n n+1 n n=1

∞ X n+1 n+1 1 nK( , x)1( 1 , 1 ] (t) + 2 t−θ K(n, x)1(n,n+1] (t) n+1 n n n n n=1

≤ 4t−θ K(t, x).

Y ). By Corollary B.3.2 also t1−θ u(t) ∈ Lp Hence we have t1−θ b(t) ∈ Lp ∗ (0, ∞; ∗ (0, ∞; Y ) holds, R where u(t) is defined by u(t) := 1t 0t b(s) ds. Clearly we have ka(t)k X = ≤2

∞ X

n=1

∞ X

n=1

K(

kan+1 kX 1(

1 , 1 ] (t) n+1 n

+

∞ X

n=1 ∞ X

k˜ an kX 1(n,n+1] (t)

1 , x)1( 1 , 1 ] (t) + 2 K(n, x)1(n,n+1] (t) ≤ 2K(t, x). n+1 n n+1 n=1

Now u(t) + v(t) = x and u′ (t) = −v′ (t) = − 1t v(t) + 1t a(t) on (0, ∞). For the norms we obtain ‚ ‚ ‚ 1−θ ′ ‚ u (t)‚ ≤ t−θ sup ka(s)kX + kt−θ a(t)kX ≤ 4t−θ K(t, x). ‚t X

t1−θ u′ (t)

Lp ∗ (0, ∞; X).

0<s
∈ Because of (X, Y )θ,p ⊂ (X, Y )θ,∞ obviously t−θ K(t, x) is bounded. This implies kan kX = O(n−θ ), whence in particular kan kX → 0. It follows that limtց0 kv(t)k X = 0. This shows limt→0 u(t) = x in X + Y and the proof is complete. Hence

220

Appendix B. Interpolation Spaces

Remark B.3.4. If one has found some function u ∈ V (θ, p, X, Y ) with u(0) = x ∈ (X, Y )θ,p , then one can even find a much Rmore regular one. Namely, choose a test ∞ function ϕ ∈ D(0, ∞) with the property 0 ϕ(s)/s ds = 1 and define Z ∞ Z ∞ ds ds v(t) := ϕ(t/s)u(s) = ϕ(s)u(t/s) . s s 0 0 (The function v is the multiplicative convolution v = ϕ ⋆ u.) This function enjoys the following properties. 1) v ∈ C∞ ((0, ∞); Y ), v ′ ∈ C∞ ((0, ∞);

X ∩Y)

2) t1−θ v ∈ Lp∗ ((0, ∞); Y ) with t1−θ v Lp ((0,∞);Y ) ≤ c1 t1−θ u Lp ((0,∞);Y )

∗

∗ 3) t2−θ v ′ ∈ Lp∗ ((0, ∞); Y ) with t2−θ v Lp ((0,∞);Y ) ≤ c2 t1−θ u Lp ((0,∞);Y )

∗

∗ 4) t1−θ v ′ ∈ Lp∗ ((0, ∞); X) with t1−θ v ′ Lp ((0,∞);X) ≤ c3 t1−θ u′ Lp ((0,∞);X) ∗ ∗ 5) v(0) = u(0) und limt→∞ kv(t)kY = 0.

The constants ci depend only on ϕ, θ, p, but not on u.

[The assertions follow from the inequalities “ ” “ ” t1−θ v(t) = s1−θ ϕ(s) ⋆ s1−θ u(s) Z ∞ ds = [s1−θ u′ (s)] ⋆ [ϕ(s)s−θ ] (t) t1−θ v′ (t) = [(t/s)1−θ u′ (t/s)][s−θ ϕ(s)] s 0 Z ∞ ds = [s1−θ u(s)] ⋆ [ϕ′ (s)s2−θ ] (t) t2−θ v′ (t) = [s1−θ u(s)][ϕ′ (t/s)(t/s)2−θ ] s 0 and some general facts on convolutions. ]

As a matter of fact, similar statements hold for the higher derivatives of v. Corollary B.3.5. Let (X, Y ) be a Banach couple, 1 ≤ p < ∞ and 0 < θ < 1. Then X ∩ Y is dense in (X, Y )θ,p .

Proof. Let x ∈ (X, Y )θ,p such that x = v(0) and v ∈ V (θ, p, X, Y ) as in the previous remark. For ε > 0 we let gε := v(ε)1[0,ε) +v1[ε,∞) . As iss easily seen, gε ∈ V (θ, p, X, Y ) and gε (0) = v(ε) ∈ Y . Hence one has zε := gε − v = (v(ε) − v)1(0,ε) ∈ V (θ, p, X, Y ). Since zε′ = −v′ 1(0,ε) , one has ‚ ‚ ‚ ‚ limεց0 ‚t1−θ zε′ (t)‚Lp ((0,∞);X) = 0. To estimate ‚t1−θ zε (t)‚Lp ((0,∞);Y ) we employ the Hardy∗ ∗ Young inequality (Proposition B.3.1) with α = 1 − θ, a = ∞, ϕ(s) = s1(0,ε) (s) kv′ (s)kY and obtain „Z ∞ ‚ ‚ ‚ ‚p ds « p1 ‚ 1−θ ‚ ‚ 1−θ ‚ zε (t)‚ p = (z(ε) − z(t))1(0,ε) (s)‚ ‚t ‚t L∗ ((0,∞);Y ) Y s 0 „Z ∞ ‚ ‚p ds « p1 ‚ 1−θ ‚ ≤ (z(ε) − z(t))1(0,ε) (s)‚ ‚t Y s 0 „Z ∞ „ « «1 Z ∞ ‚ ‚ ds p ds p ≤ t1−θ s1(0,ε) (s) ‚v′ (s)‚Y s s 0 t „Z ∞ «1 ‚ ‚ ‚ p 1 ds 1 ‚ ‚ 2−θ ′ ‚ ≤ [s2−θ 1(0,ε) (s) ‚v′ (s)‚Y = v (s)‚ p . ‚s L∗ ((0,ε):Y ) 1−θ s 1−θ 0 This vanishes with ε. Hence we have shown that Y ∩ (X, Y )θ,p is dense in (X, Y )θ,p . By Proposition B.2.7 and part a) of Proposition B.2.2 one has Y ∩ (X, Y )θ,p = (Y, X ∩ Y )1−θ,p . By what we have shown, X ∩ Y is dense in that space.

B.4. The Reiteration Theorem

221

Let us point out that the statement of the corollary even holds in the case if the spaces x, Y are not complete, cf. [14, Thm. 3.4.2].

B.4 The Reiteration Theorem Reiteration is one of the most important techniques, making interpolation theory the powerful tool that it is. Let (X, Y ) be a Banach couple and let E be an intermediate space, i.e., a space for which the continuous embeddings X ∩ Y ⊂ E ⊂ X + Y hold. We say, E is of class Jθ in (X, Y ), if there is a c ≥ 0 such that 1−θ

θ

kxkE ≤ c kxkX kxkY

(x ∈ X ∩ Y )

(0 ≤ θ ≤ 1). We write E ∈ Jθ (X, Y ) if this holds. Lemma B.4.1. Let X ∩ Y ⊂ E ⊂ X + Y and 0 < θ < 1. Then the following assertions are equivalent. (i) E ∈ Jθ (X, Y ).

(ii) (X, Y )θ,1 ⊂ E. Note that for θ ∈ {0, 1} this characterization must fail, since in this case (X, Y )θ,1 = 0. Proof. The implication (ii) → (i) follows directly from Corollary B.2.4. We show (i) → (ii). Let x ∈ (X, Y )θ,1 . Applying the trace method one finds u ∈ V (θ, 1, X, Y ) with x = u(0). Now we pass over R to a more regular function v (as in Remark B.3.4). Then, since v(∞) = 0, we have x = − 0∞ v′ (s) ds in X + Y . Because of v′ (t) ∈ X ∩ Y for t > 0 one can write using (i): ‚ ‚1−θ ‚ ‚ ‚ ‚ ‚ ′ ‚ ‚ ‚ ‚ ‚ 2−θ ′ ‚θ ‚v (t)‚ ≤ c ‚v′ (t)‚1−θ ‚v′ (t)‚θ = ct−1 ‚ v (t)‚ . ‚t1−θ v′ (t)‚ ‚t E X Y X

Y

‚ ‚1−θ ‚ ‚θ 1/(1−θ) 1/θ Now ‚t1−θ v′ (t)‚X ∈ L∗ and ‚t2−θ v′ (t)‚Y ∈ L∗ . By H¨ older we thus obtain t kv′ (t)kE ∈ 1 ′ 1 L∗ , i.e., v ∈ L ((0, ∞); E). It follows x ∈ E with ‚ ‚1−θ ‚ ‚ ‚ ‚ ‚ 2−θ ′ ‚θ kxkE ≤ c ‚t1−θ v′ ‚ 1 v‚ 1 ≤ c˜ kukV (θ,1,X,Y ) . ‚t L∗ ((0,∞);X)

Passing to the infimum we obtain kxkE ≤

L∗ ((0,∞);Y )

c˜ kxktr θ,1 .

We say, E is of class Kθ in (X, Y ), if there is a k ≥ 0 such that K(t, x) ≤ ktθ kxkE

(x ∈ E, t > 0)

(0 ≤ θ ≤ 1). One writes E ∈ Kθ (X, Y ) if this holds. In the case when 0 < θ < 1, then E ∈ Kθ (X, Y ) if, and only if E ⊂ (X, Y )θ,∞ . We now start with an Banach couple (X, Y ) and two intermediate Banach spaces X ∩ Y ⊂ E1 , E2 ⊂ X + Y . Then (E0 , E1 ) is another interpolation couple. The reiteration theorem deals with this situation. Theorem B.4.2. (Reiteration Theorem) Let (X, Y ) be a Banach couple, 0 ≤ θ0 < θ1 ≤ 1 and 0 < θ < 1.

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Appendix B. Interpolation Spaces

a) If Ei ∈ Kθi (X, Y ) for i = 0, 1, then (E0 , E1 )θ,p ⊂ (X, Y )ω,p

(1 ≤ p ≤ ∞).

b) If Ei ∈ Jθi (X, Y ) and Ei is complete for i = 0, 1, then (X, Y )ω,p ⊂ (E0 , E1 )θ,p

(1 ≤ p ≤ ∞).

Here, ω := (1 − θ)θ0 + θθ1 ∈ (0, 1).

Proof. We prove a). Choose constants ki in such a way that K(t, x) ≤ ki tθi kxkEi holds for x ∈ Ei , t > 0, i = 0, 1. Then for x = a + b ∈ E0 + E1 and t > 0 the inequality K(t, x, X, Y ) ≤ K(t, a, X, Y )+K(t, b, X, Y ) ≤ k0 tθ0 kakE0 +k1 tθ1 kbkE1 ≤ max(k0 , k1 )tθ0 (kakE0 +tθ1 −θ0 kbkE1 holds. Passing to the infimum yields K(t, x, X, Y ) ≤ ctθ0 K(tθ1 −θ0 , x, E0 , E1 ), where c := max(k0 , k1 ). This implies t−ω K(t, x, X, Y ) ≤ c[tθ1 −θ0 ]θ K(tθ1 −θ0 , x, E0 , E1 ), and hence the assertion. 1−θ θ We show b). Choose constants ci in such a way that kxkEi ≤ ci kxkX i kxkYi for all x ∈ X ∩ Y . Let x ∈ X ∩ Y and choose (by Theorem B.3.3) some u ∈ V (ω, p, X, Y ) with u(0) = x. Applying again Remark B.3.4 yields a function v = ϕ ⋆ u with v(0) = x and R the other properties listed in this remark. In particular v′ (t) ∈ X ∩Y for all t > 0 and v(t) = − t∞ v′ (s) ds. One can show that the function w = (t 7−→ v(t1/θ1 −θ0 )) is contained in V (θ, p, E0 , E1 ), and in fact prove a norm estimate kwkV (θ,p,E0 ,E1 ) ≤ K kukV (p,ω,X,Y ) (cf. [87, p.32/33]). After this is done, a density argument finishes the proof.

B.5 Complex Interpolation The complex method of interpolation is an abstraction of Thorin’s proof of the classical Riesz Interpolation theorem. Let us denote (for the moment) by S the vertical strip S := {z ∈ C | 0 < Re z < 1}. We consider vector-valued functions f : S −→ X, where X is some Banach space. For 0 ≤ θ ≤ 1 we denote by fθ the function fθ : R −→ X,

fθ (t) := f (θ + it) (t ∈ R),

i.e., the restriction of f to the vertical line {Re z = θ}. The following lemma is at the heart of the whole construction. Lemma B.5.1. (Three Line Lemma) Let X be a Banach space, f : S −→ X continuous, uniformly bounded and holomorphic on S. Then 1−θ

kf (θ)kX ≤ kf0 k∞

θ

kf1 k∞

(0 ≤ θ ≤).

In particular, kf k∞,S ≤ max(kf0 k∞ , kf1 k∞ ) (Maximum Modulus Principle). Proof. See [113, Theorem 12.8].

B.5. Complex Interpolation

223

Let now (X, Y ) be an interpolation couple. We define F (X, Y ) := {f ∈ Cb (S; X + Y ) ∩ O(S; X + Y ) | f0 ∈ Cb (R; X), f1 ∈ Cb (R; Y )} and a subspace F 0 (X, Y ) := {f ∈ F(X, Y ) | f0 ∈ C0 (R; X), f1 ∈ C0 (R; Y )}. On F (X, Y ) a norm(!) is defined by kf kF := max(kf0 kX,∞ kf1 kY,∞ ) (f ∈ F(X, Y )). It is easy to see that F (X, Y ) is a Banach space if (X, Y ) is a Banach couple. Furthermore, F 0 (X, Y ) is a closed subspace of F (X, Y ). The following technical lemma is quite important. Lemma B.5.2. The set M := {eδz

2

+λz

⊗ a | δ > 0, λ ∈ R, a ∈ X ∩ Y }

is total in F 0 (X, Y ).

Proof. This in fact needs an intricate argument a more detailed account of which can be found in 2 [73, Ch. IV, Thm. 1.1] or [14, Lemma 4.2.3].) We only sketch the proof. Note that |eδz +λz | ≤ 2 −δ|Im z| δ+|λ| ce for c = e . Hence the considered set is in fact contained in F 0 (X, Y ) and consists of very rapidly decreasing functions. Let f ∈ F 0 be arbitrary. One has to approximate f by finite sums of elements of M . The proof of this fact consists of several steps. 2 1) One has eεz f → f in F 0 as ε ց 0 for each f ∈ F 0 . 2) By P Step 1) one can assume without restriction that f decays rapidly. For such f consider fn (z) := k∈Z f (z + i(2πkn)). (The sum converges absolutely, uniformly for z in compacts of S.) This yields a 2πi-periodic function fn ∈ F . 3) Let bnk (s) be the k-th Fourier coefficient of the 2πn-periodic function fn (s + i·). An k

intricate argument using Cauchy’s theorem shows that ank (s) := e− n s bnk (s) is independent of s ∈ (0, 1). By continuity, ank (0) = ank (1) ∈ X ∩ Y . 4) Fejer’s theorem yields that fn (z) can be (uniformly in z) approximated by sums of k

2

2

terms of the form ank e n z . Multiplying by eδz shows that each function eδz fn is in the closed span of M . 5) Finally, fn → f uniformly on compacts, hence with δ > 0, n ∈ N to be chosen appro2 priately, keδz fn − f kF can be made arbitrarly small.

We are now in the position to define the complex interpolation spaces. Let (X, Y ) be an interpolation couple and let 0 < θ < 1. We define [X, Y ]θ := {f (θ) | f ∈ F(X, Y )} with the norm kxkθ := inf{kf kF | f ∈ F(X, Y ), f (θ) = x}. 2

In passing to functions of the form eδ(z−θ) f one can see that in the definition of [X, Y ]θ one can replace the space F by the space F 0 . We obtain the following theorem.

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Appendix B. Interpolation Spaces

˜ Y˜ ) be interpolation couples. The following Theorem B.5.3. Let (X, Y ) and (X, assertions hold. a) The space [X, Y ]θ is an intermediate space, i.e., X ∩ Y ⊂ [X, Y ]θ ⊂ X + Y. b) The space X ∩ Y is dense in [X, Y ]θ for each θ ∈ (0, 1).

c) If (X, Y ) is a Banach couple, then [X, Y ]θ is a Banach space for all θ ∈ [0, 1]. ˜ Y˜) be a morphism of interpolation couples. Then it d) Let T : (X, Y ) −→ (X, ˜ Y˜ ]θ such that restricts to a bounded linear mapping T : [X, Y ]θ −→ [X, 1−θ

θ

kT k[X,Y ]θ →[X, ˜ Y˜ ]θ ≤ kT kX→X ˜ kT kY →Y˜ for all θ ∈ (0, 1). Proof. Ad a). The first inclusion is clear by considering constant functions. The second follows from the Maximum Modulus Principle. Ad b). This is immediate from Lemma B.5.2. Ad c). This follows from abstract nonsense, since [X, Y ]θ can be written as the factor space F (X, Y )/ N , where N is the kernel of the evaluation mapping f 7−→ f (θ). Ad d). This is straightforward.

Proposition B.5.4. Let (X, Y ) be a Banach couple. Define V(X, Y ) := span{ϕ ⊗ x | ϕ ∈ F 0 (C, C), x ∈ X ∩ Y } Then kak[X,Y ]θ = inf{kf kF(X,Y ) | f ∈ V(X, Y )} for any a ∈ X ∩ Y, θ ∈ (0, 1).

Proof. We reproduce the proof from [87, Remark 2.1.5]. Take any g ∈ F 0 (X, Y ) such that f (θ) = a. Then define r(z) :=

z−θ , z+θ

2

and

h(z) :=

g(z) − e(z−θ) a . r(z)

Then |r(z)| ≤ 1 for z ∈ S with |r(z)| → 1 as |z| → ∞. Hence h ∈ F 0 (X, Y ). By Lemma B.5.2 one can find a function k ∈ M such that kh − f kF(X,Y ) is arbitrarily small. Now define 2

f (z) := e(z−θ) a + r(z)k(z) which is obviously contained in V(X, Y ) and satisfies f (θ) = a. Moreover, ‚ ‚ 2 ‚ ‚ kg − f kF (X,Y ) = ‚rh + e(z−θ) a − f ‚ = kr(h − k)kF(X,Y ) ≤ kh − kkF (X,Y ) F (X,Y )

Now the statement follows.

An important feature of the complex interpolation spaces is their connection to the real interpolation spaces.

B.5. Complex Interpolation

225

Proposition B.5.5. Let (X, Y ) be an interpolation couple and 0 < θ < 1. Then [X, Y ]θ is of the class K(θ) and of the class J(θ), i.e., the inclusion (X, Y )θ,1 ⊂ [X, Y ]θ ⊂ (X, Y )θ,∞ holds true. ˜ Y = Y˜ and T = I in c) of Theorem B.5.3. This gives [X, Y ]θ ∈ Jθ (X, Y ), Proof. Choose X = X, by definition. For the proof of the second inclusion we refer to [87, Prop. 2.1.10]. The key step is, by applying the conformal mapping eπiz and the Poisson kernel for the half-plane, to write each f ∈ F (X, Y ) can be written on the open strip S as a sum f = f0 + f1 , where f0 , f1 are given by integrals of f on the axis Re = 0 and Re = 1, respectively. (see [87, Lemma 2.1.9] or [73, Ch. IV.1.2]).

Let us remark that there is also a reiteration theorem for the complex method, see [14] or [73].

References Our main references are [86] and [87]. Other expositions are [14], [73] and [122]. However, the latter books contain much more material than is needed for our purposes. On the other hand, although [87] fits best to our needs, it is just a collection of Lecture Notes and seems not to be overall available. So we decided to give at least parts of the proofs in detail, whenever it seemed necessary or helpful.

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Appendix B. Interpolation Spaces

Appendix C

Operator Theory on Hilbert Spaces This chapter provides some facts on linear operators on Hilbert spaces, including adjoints (of multivalued operators), numerical range, symmetric and accretive operators, and the Lax-Milgram Theorem. The main difference to standard texts lies in the fact that we have avoided to employ the spectral theorem while dealing with spectral theory of selfadjoint operators (Proposition C.4.3 – Corollary C.4.6). We take for granted the basic Hilbert space theory as can be found in [20, Chapter I], [111, Chapter II], or [113, Chapter 4]. During the whole chapter the letter H denotes some complex Hilbert space. The scalar product on H is denoted by ( · | ·).

C.1 Sesquilinear Forms Let V be a vector space over the field of complex numbers. We denote by Ses(V ) := {a | a : V × V −→ C sesquilinear} the space of sesquilinear forms on V . Given a ∈ Ses(V ), the adjoint form a is defined by a(u, v) := a(v, u) (u, v ∈ V ). The real part and the imaginary part of the form a ∈ Ses(V ) are defined by Re a :=

1 (a + a) and 2

Im a :=

1 (a − a), 2i

respectively. Hence a = (Re a)+i(Im a) for every a ∈ Ses(V ). Using the shorthand notation a(u) := a(u, u) for a ∈ Ses(V ) and u ∈ V , we obtain (Re a)(u) = Re(a(u))

(u ∈ V ).

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Appendix C. Operator Theory on Hilbert Spaces

Given a ∈ Ses(V ) we have

1 (a(u + v) + a(u − v)) = a(u) + a(v) 2

(C.1)

1 (a(u + v) − a(u − v) + i(a(u + iv) − a(u − iv))) (C.2) 4 which equations are called Parallelogram Law and Polarization Identity, respectively. (The importance of the Polarization Identity lies in the consequence that each sesquilinear form a is already determined by the associated quadratic form a(·).) A form a is called real if a(u) ∈ R for all u ∈ V . It is called symmetric if a = a. Note that Re a and Im a are always symmetric forms. A sesquilinear form a ∈ Ses(V ) is called positive (or monotone) if Re a(u) ≥ 0 for all u ∈ V . a(u, v) =

Lemma C.1.1. A form a ∈ Ses(V ) is real if and only if it is symmetric. Proof. Obviously, a symmetric form is real. Let a be real. We have 1 a(u, v) + a(v, u) = (a(u + v) − a(u − v)) ∈ R 2 for all u, v ∈ V . Replacing u and v by iu and iv, respectively, we obtain also ia(u, v) − ia(v, u) ∈ R

for all u, v ∈ V . Combining these informations we arrive at Im(a(u, v)) = − Im(a(v, u)) and Re(a(u, v)) = Re(a(v, u)). But this is nothing else than a(u, v) = a(v, u).

Proposition C.1.2. (Generalized Cauchy-Schwarz Inequality) Let a, b ∈ Ses(V ) be symmetric, and assume there is c ≥ 0 such that |a(u)| ≤ c b(u)

(u ∈ V ).

(Note that this implies b to be positive.) Then p p |a(u, v)| ≤ c b(u) b(v)

for all u, v ∈ V .

Proof. The simple proof can be found in [115, Chapter XII, Lemma 3.1].

A positive form a is sometimes called a semi-scalar product. It is called a scalar product if it is even positive definite, i.e., if it is positive and if a(u) = 0 implies u = 0 for each u ∈ V . If a ∈ Ses(V ) is a semi-scalar product on V , then by p kxka := a(u)

a seminorm on V is defined. The form a is continuous with respect to this seminorm. (This follows from the Cauchy-Schwarz Inequality.) Let 0 ≤ ω <

π 2.

A form a ∈ Ses(V ) is called sectorial of angle ω if

a(u) 6= 0 ⇒ |arg a(u)| ≤ ω

(u ∈ V ).

The form a is called sectorial if it is sectorial of some angle ω < a is sectorial, then Re a is positive.

π 2.

Obviously, if

C.2. Adjoint Operators

229

Proposition C.1.3. Let a ∈ Ses(V ) such that Re a ≥ 0. The following assertions are equivalent. (i) The form a is sectorial. (ii) The form a is continuous with respect to the seminorm induced by the semi-scalar product Re a. p p More precisely: If |a(u, v)| ≤ M Re a(u) Re a(v) for all u, v ∈ V , then a is sectorial of angle arccos M −1 . Conversely, if a is sectorial of angle ω, then p p |a(u, v)| ≤ (1 + tan ω) Re a(u) Re a(v)

for all u, v ∈ V .

p p Proof. Suppose |a(u, v)| ≤ M Re a(u) Re a(v) for all u, v ∈ V . Then, letting u = v, we have |a(u)| ≤ M Re a(u), whence M ≥ 1 and a(u) 6= 0 ⇒ |arg a(u)| ≤ arccos M −1 . for all u ∈ V . Conversely, if a is sectorial of angle ω < π2 , then |Im a(u)| ≤ (tan ω) Re a(u) for all u ∈ V . The generalized Cauchy-Schwarz Inequality, applied to Re a and Im a, yields p p |a(u, v)| ≤ |(Re a)(u, v)| + |(Im a)(u, v)| ≤ (1 + tan ω) Re a(u) Re a(v) for all u, v ∈ V .

C.2 Adjoint Operators In this section we provide the theory of Hilbert space adjoints of m.v. linear operators. The results are more or less the same as for general adjoints, compare Section A.4. Let A ⊂ H ⊕ H be a multivalued linear operator on H. The Hilbert space adjoint of A, usually denoted by A∗ , is defined by (x, y) ∈ A∗

:⇐⇒

( v | x) = ( u | y ) for all (u, v) ∈ A.

(C.3)

If for the moment we define J := ((u, v) 7−→ (−v, u)) : H ⊕ H −→ H ⊕ H, then we can write A∗ = [JA]⊥ where the orthogonal complement is taken in the Hilbert space H ⊕ H. Hence A∗ is always a closed operator. We enumerate the basic properties. Proposition C.2.1. Let A, B be m.v. linear operators on H. Then the following statements hold. a) A∗ = (A)∗ b) (A−1 )∗ = (A∗ )−1 . c) (λA)∗ = λA∗ , for 0 6= λ ∈ C.

d) A∗∗ = A.

e) N(A∗ ) = R(A)⊥ and N(A) = R(A∗ )⊥ .

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Appendix C. Operator Theory on Hilbert Spaces

f ) A∗ 0 = D(A)⊥ and A0 = D(A∗ )⊥ . g) D(A∗ ) ⊂ (A0)⊥ and R(A∗ ) ⊂ N(A)⊥ .

h) If A ∈ L(H), then A∗ ∈ L(H) and kA∗ k = kAk = i) A ⊂ B

⇒

B ∗ ⊂ A∗ .

p kA∗ Ak.

j) A∗ + B ∗ ⊂ (A + B)∗ with equality if A ∈ L(H).

k) A∗ B ∗ ⊂ (BA)∗ with equality if B ∈ L(H). If A ∈ L(H) and B is closed, one has A∗ B ∗ = (BA)∗ . ∗

⊥

Proof. Ad a). Since J is a topological isomorphism, A = (JA)⊥ = JA Ad b). This follows from (JA)−1 = J(A−1 ) and (A−1 )⊥ = (A⊥ )−1 .

Ad c). We have (x, y) ∈ (λA)∗ ⇔ ( −λv | x ) + ( u | y ) = 0 ∀(u, v) ∈ A ⇔

= (JA)⊥ . “

˛ ” ˛ −v ˛ λx + ( u | y ) =

0 ∀(u, v) ∈ A ⇔ (λx, y) ∈ A∗ ⇔ (λx, λy) ∈ λA∗ ⇔ (x, y) ∈ A∗ . Ad d). A∗∗ = (J(JA)⊥ )⊥ = (JJA)⊥⊥ = A⊥⊥ = A. Ad e). We have x ∈ N(A∗ ) ⇔ (x, 0) ∈ A∗ ⇔ ( −v | x ) = 0 ∀v ∈ R(A) ⇔ x ∈ R(A)⊥ . The second satement follows from the first together with d). Ad f ). We have y ∈ A∗ 0 ⇔ (0, y) ∈ A∗ ⇔ ( u | y ) = 0 ∀u ∈ D(A) ⇔ y ∈ D(A)⊥ . The second statement follows from the first together with d). Ad g). If (x, y) ∈ A∗ , v ∈ A0, and u ∈ N(A), then ( −v | x ) = 0 and ( u | y ) = 0 by (C.3). Ad h). Let A ∈ L(H). Then A∗ is closed and single-valued by f ). We show that D(A∗ ) = H. In fact, let x ∈ H. Then (u 7−→ ( Au | x )) is a continuous linear functional on H. By the RieszFr´ echet Theorem [113, Theorem 4.12] there is y ∈ H such that ( Au | x ) = ( u | y ) for all u ∈ H. But this means exactly that (x, y) ∈ A∗ . The equation kAk = kA∗ k is easily proved by using the identity kT k = sup{|( T u | v )| | kuk = kvk = 1}, which holds for every bounded operator on H. This implies kA∗ Ak ≤ kA∗ k kAk = kAk2 . But if x ∈ H is arbitrary, we have kAxk2 = ( Ax | Ax ) = ( A∗ Ax | x ) ≤ kA∗ Ak kxk2 by the Cauchy-Schwarz inequality. Hence kAk2 ≤ kA∗ Ak. Ad i). We have A ⊂ B ⇒ JA ⊂ JB ⇒ (JB)⊥ ⊂ (JA)⊥ . Ad j). Let (x, y) ∈ A∗ , (x, z) ∈ B ∗ . The generic element of J(A + B) is (−v − w, u), where (u, v) ∈ A and (u, w) ∈ B. So (x, y) ⊥ (−v, u) and (x, z) ⊥ (−w, u), hence (x, y+z) ⊥ (−v−w, u). If A ∈ L(H), we write B = (A + B) − A and note that A∗ ∈ L(H) by h). Ad k). Let (x, y) ∈ A∗ B ∗ . Then there is z such that (x, z) ∈ B ∗ and (z, y) ∈ A∗ . If (u, v) ∈ BA, one has (u, w) ∈ A and (w, v) ∈ B for some w. Hence ( v | x ) = ( w | z ) = ( u | y ). Since (u, v) ∈ BA was arbitrary, we conclude that (x, y) ∈ (BA)∗ . Assume now B ∈ L(H) and (x, y) ∈ (BA)∗ . Define z := B ∗ x. It suffices to show that (z, y) ∈ A∗ . Take (u, w) ∈ A and define v := Bw. Hence, (u, v) ∈ BA. Therefore ( u | y ) = ( v | x ) = ( Bw | x ) = ( w | B ∗ x ) = ( w | z ) , whence (z, y) ∈ A∗ by (C.3). Finally, assume that B is closed and A ∈ L(H). The assertions already proved yield (A∗ B ∗ )∗ = B ∗∗ A∗∗ = (B)(A) = BA. Hence A∗ B ∗ = (BA)∗ .

Corollary C.2.2. Let A be a m.v. linear operator on H. Then (λ − A)∗ = (λ − A∗ )

and

R(λ, A)∗ = R(λ, A∗ )

for every λ ∈ C. In particular, we have ̺(A∗ ) = {λ | λ ∈ ̺(A)}. Let A be a single-valued operator on H. From part f ) of Proposition C.2.1 we see that A∗ is densely defined if and only if A is closable, i.e., A is still single-valued; and A∗ is single-valued if and only if A is densely defined.

C.3. The Numerical Range

231

Proposition C.2.3. Let A be a densely defined (single-valued) operator on H with ̺(A) 6= ∅. For a polynomial p ∈ C[z] define p∗ (z) := p(z). (Hence, p∗ is obtained from p by conjugating all coefficients.) Then we have [p∗ (A)]∗ = p(A∗ ). The same statement holds if p is a rational function with poles inside ̺(A) = ̺(A∗ ). Proof. Let r = p/q be a rational function with poles inside the set ̺(A∗ ). Assume first that deg p ≤ deg q. Q Hence, r(A∗ ) and r ∗ (A) are bounded operators. The function r can be written as a product r = j rj where each rj is either of the form α(λ − z)−1 or of the form α(λ − z)−1 + β. Now the claimed formula [r ∗ (A)]∗ = r(A∗ ) follows from part k) of Proposition C.2.1 and Corollary C.2.2. From this and part b) of Proposition C.2.1 we can infer that p∗ (A)∗ = p(A∗ ) holds for all polynomials p having their roots inside ̺(A∗ ). Now suppose deg p > deg q. Since ̺(A) 6= ∅ we can find a polynomial q1 having its roots inside ̺(A∗ ) with deg q + deg q1 = deg p. Define r˜ := p/(q1 q). Then (1)

(2)

r(A∗ ) = q1 (A∗ )˜ r (A∗ ) = [q1∗ (A)]∗ [˜ r ∗ (A)]∗ = [˜ r ∗ (A)q1∗ (A)]∗ = [q1∗ (A)˜ r ∗ (A)]∗ = [r ∗ (A)]∗ , where we have used k) from Proposition C.2.1 in (1) and Corollary A.6.3 in (2).

C.3 The Numerical Range From now on, all considered operators are single-valued. We therefore will follow the general terminological agreement made on page 189. Given an operator A on H we call W (A) := {( Au | u ) | u ∈ D(A), kuk = 1} ⊂ C the numerical range of A. By [115, Chapter XII, Theorem 5.2] the numerical range W (A) is always a convex subset of the plane. Proposition C.3.1. Let A be a closed operator. Then P σ(A) ⊂ W (A) and Aσ(A) ⊂ W (A). Furthermore, one has kR(λ, A)k ≤

1 dist(λ, W (A))

for λ ∈ ̺(A) \ W (A). If A ∈ L(H), we have σ(A) ⊂ W (A).

Proof. If λ ∈ P σ(A), there is u ∈ D(A), kuk = 1 such that Au = λu. This gives ( Au | u ) = ( λu | u ) = λ kuk2 = λ. Hence λ ∈ W (A). Assume λ ∈ / W (A) and define δ := dist(λ, W (A)). By definition, |( Ax | x ) − λ| ≥ δ for all x ∈ D(A) with kxk = 1. Hence, ˛ ˛ ˛ ˛ |( (A − λ) x | x )| = ˛( Ax | x ) − λ kxk2 ˛ ≥ δ kxk2 for all x ∈ D(A). But |( (A − λ)x | x )| ≤ k(A − λ)xk kxk, whence k(λ − A) xk ≥ δ kxk

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Appendix C. Operator Theory on Hilbert Spaces

for all x ∈ D(A). Since A is closed, this implies that (λ − A) is injective and has closed range, i.e., λ ∈ / Aσ(A). Moreover, it shows that kR(λ, A)k ≤ δ−1 if λ ∈ ̺(A). If A ∈ L(X), then W (A∗ ) = {λ | λ ∈ W (A)}. Now, if λ ∈ Rσ(A), then clearly λ ∈ P σ(A∗ ), whence λ ∈ W (A).

Corollary C.3.2. Let A be a single-valued closed operator on H. Let U ⊂ C\ W (A) be open and connected. If U ∩ ̺(A) 6= 0, then U ⊂ ̺(A). Proof. The statement follows from Proposition C.3.1 and the fact that kR(λ, A)k blows up if λ approaches a spectral value (cf. Proposition A.2.3).

C.4 Symmetric Operators We begin with a lemma. Lemma C.4.1. For an operator A on H the following assertions are equivalent: (i) W (A) ⊂ R.

(ii) ( Au | v ) = ( u | Av ) for all u, v ∈ D(A).

(iii) A ⊂ A∗ .

If this is the case, the operator A is called symmetric. Proof. Define the form a on V := D(A) by a(u, v) := ( Au | v ). The proof is now an easy consequence of Lemma C.1.1 and the definition of the adjoint (see (C.3)).

An operator A on H is called selfadjoint if A∗ = A. If A is symmetric/selfadjoint and injective, then A−1 is symmetric/selfadjoint, by part b) of Proposition C.2.1. Proposition C.4.2. Let A be an operator on H. Then A is self-adjoint if and only if A is symmetric, closed, and densely defined, and R(A ± i) is dense. In this case we have σ(A) ⊂ R and 1 kR(λ, A)k ≤ |Im λ| for all λ ∈ C \ R.

Proof. Assume that A = A∗ . Then A is closed since A∗ is. Morover, A∗ is single-valued, since A is, and this implies that A is densely defined by part f ) of Proposition C.2.1. Since A = A∗ we have W (A∗ ) = W (A) ⊂ R. Hence N(A∗ ± i) = 0 by Proposition C.3.1. This gives R(A ± i) = H by part e) of Proposition C.2.1. Now suppose that A is symmetric, closed, and densely defined with R(A ± i) being dense. By Proposition C.3.1 and its corollary we conclude that σ(A) ⊂ R and N(A∗ ± i) = 0. But A − i : D(A) −→ H is bijective and A − i ⊂ A∗ − i, whence A − i = A∗ − i. This proves A = A∗ . The norm inequality for the resolvent follows from Proposition C.3.1.

Let A be an operator on H and α ∈ R. We write α ≤ A if A is selfadjoint and W (A) ⊂ [α, ∞), and we write A ≤ α if −α ≤ −A. The operator A is called positive if 0 ≤ A. We obtain the following characterization.

C.4. Symmetric Operators

233

Proposition C.4.3. A closed and densely defined operator A on H is positive if and only if W (A) ⊂ [0, ∞) and A + λ is surjective for some/each λ > 0. In this case, {Re λ < 0} ⊂ ̺(A) and

(λ + A)−1 ≤ 1 Re λ

for all Re λ > 0. Moreover, 0 ≤ t(t + A)−1 ≤ 1 and 0 ≤ A(t + A)−1 ≤ 1 for all t > 0.

Proof. Assume that A is single-valued, closed and densely defined with W (A) ⊂ [0, ∞). Since C \ [0, ∞) is open and connected, the stated equivalence is a consequence of Proposition C.3.1 and its corollary. Moreover it follows that in this case {Re λ < 0} ⊂ ̺(A) and the stated norm inequality holds. Since t+A is selfadjoint, we know that (t+A)−1 also is. Furthermore 0 ≤ A(t+ A)−1 ≤ 1 ⇔ 0 ≤ t(t + A)−1 ≤ 1 ⇔ ( tu | (t + A)u ) ≤ ( (t + A)u | (t + A)u ) ∀u ∈ D(A). However, this is true if and only if kAuk2 + t ( Au | u ) ≥ 0 which is always the case.

Lemma C.4.4. Let A be a closed and densely defined operator on H. a) If −α ≤ A ≤ α for some α ≥ 0, then A ∈ L(H) and kAk ≤ α.

b) If A ∈ L(H) is selfadjoint, then kAk = sup{|λ| | λ ∈ W (A)}. In particular, W (A) is a bounded subset of R. c) If 0 ≤ A ≤ 1 then 0 ≤ A2 ≤ A.

Proof. Ad a). Define a(u, v) := ( Au | v ) and b(u, v) := ( u | v ) on V := D(A). The hypothesis implies that |a(u)| ≤ αb(u) for all u ∈ H. Moreover, a is symmetric. An application of the generalized Cauchy-Schwarz inequality (Proposition C.1.2) yields |( Au | v )| ≤ α kuk kvk for all u, v ∈ D(A). Since A is densely defined, this inequality holds for all v ∈ H, whence we have kAuk ≤ α kuk for all u ∈ D(A). Since A is closed, this implies that D(A) is closed. Hence A ∈ L(H) and kAk ≤ α. Ad b). Let A ∈ L(H) be selfadjoint. Then |( Au | u )| ≤ kAk kuk2 for all u ∈ H. Hence sup |W (A)| ≤ kAk. If sup |W (A)| ≤ α, then |( Au | u )| ≤ α kuk2 for all u. Since W (A) ⊂ R, this is equivalent to −α ≤ A ≤ α. Now a) implies that kAk ≤ α. Ad c). We have A − A2 = A2 (1 − A) + A(1 − A)2 , and both summands are positive since A and 1 − A are.

The following result is usually proved with the help of the spectral theorem. Proposition C.4.5. Assume A ≥ α for some α ∈ R and define α0 := inf W (A). Then α0 ∈ σ(A). If α0 ∈ W (A), then even α0 ∈ P σ(A). Proof. Without restriction we can assume α0 = 0. Hence A ≥ 0. Let Q := A(1 + A)−1 . Then 0 ≤ Q ≤ 1. Moreover, Q ≤ A in the obvious sense, since for x ∈ D(A) we have ˛ ´ ˛ ´ ` ` A(A + 1)−1 x ˛ x − ( Ax | x ) = A((A + 1)−1 − I)x ˛ x ˛ ˛ ` ´ ` ´ = − A(A + 1)−1 Ax ˛ x = − (A + 1)−1 Ax ˛ Ax ≤ 0

Applying c) of Lemma C.4.4 we obtain

0 ≤ kQxk2 ≤ ( Qx | x ) ≤ ( Ax | x ) for all x ∈ D(A). Now if there is x ∈ D(A) such that kxk = 1 and ( Ax | x ) = 0 it follows that 0 = Qx = A(A + 1)−1 x. But this implies 0 6= (A + 1)−1 x ∈ N(A), whence 0 ∈ P σ(A). Similarly, if xn ∈ D(A) such that kxn k = 1 and ( Axn | xn ) → 0, then Qxn → 0. Hence (xn )n is a generalized eigenvector for Q and 0 ∈ σ(Q). But this immediately implies 0 ∈ σ(A).

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Appendix C. Operator Theory on Hilbert Spaces

Corollary C.4.6. Let A be a bounded and selfadjoint operator on A. Then sup W (A), inf W (A) ∈ σ(A). In particular, kAk = r(A), where r(A) denotes the spectral radius of the operator A. Proof. Let α := inf W (A) and β := sup W (A). Then 0 = inf W (A − α) = inf W (β − A), whence 0 ∈ σ(A − α) ∩ σ(β − A) by Proposition C.4.5. From b) of Lemma C.4.4 we know that r(A) ≤ kAk = max |α| , |β| ≤ r(A).

Note that positive operators are special cases of m-accretive operators (see Section C.7 below). Proposition C.4.7. Let S, T ∈ L(H) be selfadjoint, and let 0 ≤ S ≤ T in the sense that ( Sx | x) ≤ ( T x | x) for all x ∈ H. If S is invertible then T is and T −1 ≤ S −1 .

Proof. By Corollary C.4.6, if S is invertible, there is 0 < α ∈ R such that α ≤ S. Hence α ≤ T , whence T is invertible. Let x ∈ H and define y := T −1 x, z := S −1 x. Then ˛ ´` ˛ ´ ` −1 ˛ ´2 ` T x ˛ x = ( y | Sz )2 ≤ ( y | Sy ) ( z | Sz ) ≤ ( y | T y ) ( z | Sz ) = T −1 x ˛ x S −1 x ˛ x ˛ ´ ` ˛ ´ ` by Cauchy-Schwarz. Consequently, T −1 x ˛ x ≤ S −1 x ˛ x .

C.5 Equivalent Scalar Products and the Lax-Milgram Theorem Let H be a Hilbert space. We denote by H ∗ the space of continuous conjugatelinear functionals on H, endowed with the norm kϕkH ∗ := sup{|ϕ(x)| | x ∈ H, kxk = 1} for ϕ ∈ H ∗ . One sometimes writes hϕ, xi instead of ϕ(x) , where x ∈ H, ϕ ∈ H ∗ . Let a be a continuous sesquilinear form on H. Then we have an induced linear mapping La := (u 7−→ a(u, ·)) : H −→ H ∗ (C.4) which is continuous. The Riesz-Fr´echet Theorem [113, Theorem 4.12] implies that for each u ∈ H there is a unique Qu ∈ H such that a(u, ·) = La (u) = ( Qu | ·) Obviously, Q is a linear and bounded operator (kQk = kLa k) and a is uniquely determined by Q. On the other hand, given Q ∈ L(H), the form aQ defined by aQ (u, v) := ( Qu | v ) is sesquilinear and continuous. Hence the mapping (Q 7−→ aQ ) : L(H) −→ {continuous, sesquilinear forms on H} is an isomorphism. We have aQ = aQ∗ , whence the form a is symmetric if and only if Q is selfadjoint, and the form is positive if and only if Q ≥ 0.

C.5. Equivalent Scalar Products and the Lax-Milgram Theorem

235

Proposition C.5.1. Let Q ∈ L(H). Then the form aQ is a scalar product on H if and only if Q is positive and injective. The norm induced by this scalar product is equivalent to the original one if and only if Q is invertible. Proof. The form aQ is positive semi-definite if and only if Q ≥ 0. This is clear from the definitions. If Q is not injective, then obviously aQ is not definite. Let Q be injective and positive. By Proposition C.4.5 we conclude that 0 ∈ / W (Q). But this means that aQ is definite. Since aQ is continuous, equivalence of the induced norm is the same as the existence of a δ > 0 such that δ ≤ Q. But this is equivalent to Q being invertible by Proposition C.4.5.

A scalar product on H which induces a norm equivalent to the original one is simply called an equivalent scalar product. Lemma C.5.2. Let A be a m.v. operator on H and let ( · | ·)◦ := aQ be an equivalent scalar product on H. Denote by A◦ the adjoint of A with respect to the new scalar product. Then (x, y) ∈ A◦ ⇐⇒ (Qx, Qy) ∈ A∗

for all x, y ∈ H. In particular, A◦ = Q−1 A∗ Q if A∗ is single-valued.

Proof. Fix x, y ∈ H. Since Q is positive, (x, y) ∈ A◦ ⇔ ( v | x )◦ = ( u | y )◦ ∀(u, v) ∈ A ⇔ ( v | Qx ) = ( u | Qy ) ∀(u, v) ∈ A ⇔ (Qx, Qy) ∈ A∗ . The rest is straightforward.

The next theorem gives a sufficient condition for the mapping La to be an isomorphism. Theorem C.5.3. (Lax-Milgram) Let a be a continuous sesquilinear form on H. Assume that there is a δ > 0 such that 2 Re a(u) ≥ δ kuk (C.5)

for all u ∈ H. Then the mapping La : H −→ H ∗ defined by (C.4) is an isomorphism.

The inequality (C.5) is called a coercivity condition and a form which satisfies (C.5) for some δ > 0 is called coercive. Proof. Take u ∈ H with kuk = 1. Then kLa (u)k ≥ |La (u)(u)| = |a(u)| ≥ Re a(u) ≥ δ. Hence δ kuk ≤ kLa (u)k for all u, whence La is injective and has a closed range. So it remains to show that E := R(La ) is dense in H ∗ . The Riesz-Fr´ echet Theorem implies that Φ := (x 7−→ ( x | · )) : H −→ H ∗

is an isomorphism. Hence E is dense in H ∗ if and only if Φ−1 (E) is dense in H if and only if Φ−1 (E)⊥ = 0. Now y ∈ Φ−1 (E)⊥ ⇔ ( x | y ) = 0 ∀ x ∈ Φ−1 (E) ⇔ ϕ(y) = 0 ∀ ϕ ∈ E ⇔ a(x, y) = 0 ∀ x ∈ H

for each y ∈ H. In particular we have y ∈ Φ−1 (E)⊥ ⇒ a(y, y) = 0, but this implies y = 0 by coercivity. Thus, the proposition is proved.

Remark C.5.4. The coercivity condition in the Lax-Milgram Theorem can be 2 weakened to |a(u)| ≥ δ kuk for all u ∈ H. This is easily seen from the proof.

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Appendix C. Operator Theory on Hilbert Spaces

C.6 Weak Integration Let (Ω, µ) be a measure space. A measurable map F : Ω −→ L(H) is called weakly integrable, if Z |( F (ω)x | y )| µ(dω) < ∞ Ω

for all x, y ∈ H. (By the Polarization Identity it suffices to have it for all x = y ∈ H.) Lemma C.6.1. Let F : Ω −→ L(H) be weakly integrable. Then there is a unique operator Q ∈ L(H) such that ( Qx | y ) = One usually writes

R

Ω

Z

Ω

( F (ω)x | y ) µ(dω)

F (ω) µ(dω) := Q.

Proof. Fix x ∈ H. Then the mapping y 7−→ ( F (·)x | y ) is linear from H to L1 (Ω, µ) by hypothesis. The closed graph theorem shows that it must be continuous. The Riesz-Fr´ echet Theorem [113, Theorem 4.12] shows that there is a unique Qx ∈ H such that ( Qx | y ) = R ( F (ω)x | y ) µ(dω) for all y ∈ H. Obviously Q : H −→ H is linear. The same argument as before shows that if y is fixed, the mapping (x 7−→ ( F (·)x | y )) : H −→ L1 (Ω, µ) is continuous. Therefore, Q is weakly continuous, whence continuous.

The following lemma is almost trivial. Lemma C.6.2. Let F : Ω −→ L(H) be weakly integrable. Then also F ∗ := (ω 7−→ F (ω)∗ ) is weakly integrable with Z

∗ Z F (ω) µ(dω) = F (ω)∗ µ(dω).

In particular, if F (ω) is selfadjoint for µ-almost all ω then also R joint. If F (ω) ≥ 0 for µ-almost all ω then F dµ ≥ 0.

R

F dµ is selfad-

The next result is of fundamental importance in the theory of functional calculus on Hilbert spaces.

Proposition C.6.3. Let (Ω, µ) be a measure space and S : Ω −→ L(H) measurable such that the mapping (ω 7−→ S(ω)∗ S(ω)) is weakly integrable. Then for each F ∈ L∞ (Ω, µ; L(H)) the mapping (ω 7−→ S(ω)∗ F (ω)S(ω)) is weakly integrable and

Z

Z

S(ω)∗ F (ω)S(ω)µ(dω) ≤ kF k S(ω)∗ S(ω) µ(dω) ∞

C.7. Accretive Operators

237

R R Proof. Let x, y ∈ H and define T := S ∗ S dµ and R := S ∗ F S dµ. Then Z Z |( S(ω)∗ F (ω)S(ω)x | y )| µ(dω) = |( F (ω)S(ω)x | S(ω)y )| µ(dω) Ω Z ≤ kF k∞ kS(ω)xk kS(ω)yk µ(dω) ≤ kF k∞

„Z

= kF k∞

„Z

« 1 „Z «1 2 2 kS(ω)yk2 µ(dω) kS(ω)xk2 µ(dω)

« 1 „Z «1 2 2 ( S(ω)∗ S(ω)x | x ) µ(dω) ( S(ω)∗ S(ω)y | y ) µ(dω) 1

1

≤ kF k∞ ( T x | x ) 2 ( T y | y ) 2 ≤ kF k∞ kT k kxk kyk

Corollary C.6.4. Let T : Ω −→ L(H) be weakly integrable and assume T (ω) ≥ 0 for µ-almost all ω. Then

Z

Z

f (ω)T (ω) µ(dω) ≤ kf k T (ω) µ(dω) ∞

for all f ∈ L∞ (Ω, µ).

1

Proof. Apply Proposition C.6.3 withR F := f and S(ω) := T (ω) 2 . There is also a direct proof not using square roots: Define Q := T dµ and compute Z Z |( f (ω)T (ω)x | y )| µ(dω) ≤ kf k∞ |( T (ω)x | y )| µ(dω) Z 1 1 ≤ kf k∞ ( T (ω)x | x ) 2 ( T (ω)y | y ) 2 µ(dω) ≤ kf k∞

for all x, y ∈ H.

„Z

« 1 „Z «1 2 2 ( T (ω)x | x ) µ(dω) ( T (ω)y | y ) µ(dω) 1

1

= kf k∞ ( Qx | x ) 2 ( Qy | y ) 2 ≤ kf k∞ kQk kxk kyk

C.7 Accretive Operators Here are the defining properties of accretive operators. Lemma C.7.1. Let A be an operator on the Hilbert space H and let µ > 0. The following assertions are equivalent. (i) Re ( Au | u ) ≥ 0 for all u ∈ D(A), i.e., W (A) ⊂ {Re z ≥ 0}.

(ii) k(A + µ)uk ≥ k(A − µ)uk for all u ∈ D(A).

(iii) k(A + λ)uk ≥ (Re λ) kuk for all u ∈ D(A), Re λ ≥ 0. (iv) k(A + λ)uk ≥ λ kuk for all u ∈ D(A) and all λ ≥ 0.

An operator A which satisfies the equivalent conditions (i) − (iv) is called accretive. An operator A is called dissipative if −A is accretive.

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Appendix C. Operator Theory on Hilbert Spaces

Proof. We have k(A + µ)uk2 − k(A − µ)uk2 = 4 Re ( Au | u ) for all u ∈ D(A). This gives (i) ⇔ (ii). For λ > 0 we have k(A + λ)uk2 − λ2 kuk2 = kAuk2 + 2λ ( Au | u ) for all u ∈ D(A). This shows (i) ⇒ (iv); dividing by λ and letting λ → ∞ gives the reverse implication. The implication (iii) ⇒ (iv) is obvious. Assume (i) and let Re λ ≥ 0. Let α := Im λ. Then (i) holds with A replaced by A + iα. Since we have already proved (i) ⇒ (iv), we know that k((A + iα) + Re λ)uk ≥ (Re λ) kuk for all u ∈ D(A). But this is (iii).

Note that an operator A is symmetric if and only if ±iA both are accretive. An operator A is called m-accretive if A is accretive and closed and R(A + 1) is dense in H. Proposition C.7.2. Let A be an operator on H, α ∈ R and λ > 0. The following assertions are equivalent. (i) A is m-accretive. (ii) A + iα is m-accretive. (iii) A + ε is m-accretive for all ε > 0.

(iv) −λ ∈ ̺(A) and (A − λ)(A + λ)−1 ≤ 1. (v) {Re λ < 0} ⊂ ̺(A) and

kR(λ, A)k ≤

1 |Re λ|

(Re λ < 0).

(vi) (−∞, 0) ⊂ ̺(A) and supt>0 t(t + A)−1 ≤ 1.

(vii) A is closed and densely defined, and A∗ is m-accretive. The operator (A − 1)(A + 1)−1 is called the Cayley transform of A.

Proof. Let B be a closed accretive operator on H. By Proposition C.3.1 and its corollary, if R(B + µ) is dense for some µ with Re µ > 0 then {Re µ < 0} ⊂ ̺(B). This consideration is of fundamental importance and is used several times in the sequel. In particular, it shows (i) ⇔ (ii) ⇔ (iii). The equivalence (iv) ⇔ (i) holds by (ii) of Lemma C.7.1. Similarly, (v) ⇔ (i) and (vi) ⇔ (i) hold by (iii) and (iv) of Lemma C.7.1. Part (vi) shows in particular, that an m-accretive operator is sectorial (see §2.1 of Chapter2), and as such is densely defined, since H is reflexive (see h) of Proposition 2.1.1). Moreover (vi) implies (vi) with A replaced by A∗ , whence (vi) ⇒ (vii) follows. Finally, assume (vii). Since we already have proved (i) ⇒ (vii), we conclude that A = A = A∗∗ is m-accretive.

Theorem C.7.3. (Lumer-Phillips) An operator A is m-accretive if and only if −A generates a strongly continuous contraction semigroup. Proof. If A is m-accretive, parts (vi) and (vii) of Proposition C.7.2 show that the Hille-Yosida Theorem A.8.6 (with ω = 0 and M = 1) is applicable to the operator −A. Conversely, suppose that −A generates the C0 -semigroup T such that kT (t)k ≤ 1 for all t ≥ 0. Take x ∈ D(A). Then the function t 7−→ T (t)x is differentiable with derivative t 7−→ −AT (t)x. Thus, t −→ kT (t)xk2 is differentiable with d |t=0 kT (t)xk2 = ( T (0)x | − AT (0)x ) = ( −AT (0)x | T (0)x ) = −2 Re ( Ax | x ) . dt

C.8. The Theorems of Plancherel and Gearhart

239

Since T is a contraction semigroup, the mapping t 7−→ kT (t)xk2 is decreasing. This implies d/dt|t=0 kT (t)xk2 ≤ 0. Hence, Re ( Ax | x ) ≥ 0, i.e., A is accretive. Since 1 ∈ ̺(−A) we conclude that A is in fact m-accretive.

Theorem C.7.4. (Stone) An operator B on the Hilbert space H generates a C0 − group (U (t))t∈R of unitary operators if and only if B = iA for some selfadjoint operator A. Proof. If B = iA then B and −B are both m-accretive. By the Lumer-Phillips Theorem both operators generate C0 -contraction semigroups. Hence B generates a unitary C0 -group. This proof works also in the reverse direction.

C.8 The Theorems of Plancherel and Gearhart We state without proof two theorems which are “responsible” for the fact that life is so much more comfortable in Hilbert spaces. Theorem C.8.1. (Plancherel) Let f ∈ L1 (R, H) ∩ L2 (R, H) and define F (f ) : R −→ H by Z F (f )(t) := f (s)e−ist ds (t ∈ R) R

Then F (f ) ∈ C0 (R, H) ∩ L2 (R, H) with kF (f )kL2 =

√ 2π kf kL2 .

A proof can bee found in [6, page 46]. Let X be a Banach space and T a C0 -semigroup on X with generator A. From Proposition A.8.1 we know that {Re z ≥ ω} ⊂ ̺(A)

and

sup{kR(z, A)k | Re z ≥ ω}

(C.6)

for every ω > ω0 (T ). (Recall that ω0 (T ) is the growth bound of T , see Appendix A, Section A.8.) When we define s0 (A) := inf{ω ∈ R | (C.6) holds},

(C.7)

this is equivalent to the statement “s0 (A) ≤ ω0 (T )”. The number s0 (A) is called the abszissa of uniform boundedness of the generator A. The next theorem states that there is in fact equality if X = H is a Hilbert space. Theorem C.8.2. (Gearhart) Let A be the generator of a C0 -semigroup on a Hilbert space H. Then s0 (A) = ω0 (T ). One of the many proofs of this theorem uses the Plancherel Theorem and Datko’s theorem which states that ω0 (T ) < 0 if and only if T (·)x ∈ L2 (R+ , H) for every x ∈ H. It can be found in [6, page 347].

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Appendix C. Operator Theory on Hilbert Spaces

References The following books underly our presentation: [65] and [115, Chapter XII] for sesquilinear forms and numerical range, [22, Chapter III] for adjoints, [20, Chapter X, §2] and [111, Chapter VIII] for symmetric and selfadjoint operators, [120, Chapter 2] for accretive operators and the Lax-Milgram theorem, and [40, Chapter II.3.b] for the Lumer-Phillips Theorem and Stone’s theorem. The idea how to achieve part c) of Lemma C.4.4 came from [77, Chapter XVIII, §4].

Appendix D

The Spectral Theorem The spectral theorem for normal operators appears in many versions. Basically one can distinguish the “spectral measure approach” and the “multiplicator approach”. Following Halmos’s article [56] (and R. Nagel’s preaching), we will give a consequent “multiplicator” account of the subject matter. In this form, the spectral theorem essentially says that a given selfadjoint operator on a Hilbert space acts “like” the multiplication of a real function on an L2 -space. In contrast to usual expositions we will stress that the underlying measure space can be chosen locally compact and the real function can be chosen continuous.

D.1 Multiplication Operators A Radon measure space is a pair (Ω, µ) where Ω is a locally compact Hausdorff space and µ is a positive functional on Cc (Ω). By the Riesz Representation Theorem [113, Theorem 2.14] we can identify µ with a σ-regular Borel measure on Ω. If the measure µ has the property Z ϕ ∈ Cc (Ω), 0 ≤ ϕ 6= 0 ⇒ ϕ dµ > 0, (D.1) then the Radon measure space is called a standard measure space. Property D.1 is equivalent to the fact that a nonempty open subset of Ω has positive µ-measure. As a consequence of this we obtain that the natural mapping C(Ω) −→ L1loc (Ω, µ) is injective. Let f ∈ C(Ω) be a continuous function on Ω. The multiplication operator Mf on L2 (Ω, µ) is defined by D(Mf ) := {g ∈ L2 (Ω, µ) | gf ∈ L2 (Ω, µ)}

and Mf g := f g (f ∈ D(Mf )).

The following proposition summarizes the properties of this operator.

242

Appendix D. The Spectral Theorem

Proposition D.1.1. Let f ∈ C(Ω) where (Ω, µ) is a standard measure space. The following assertions hold. a) The operator (Mf , D(Mf )) is closed. b) The space Cc (Ω) is a core for Mf . c) One has (Mf )∗ = Mf . d) One has Mf ∈ L(L2 ) if and only if f is bounded. In this case kMf kL(L2 ) = kf k∞ . e) The operator Mf is injective if and only if µ({f = 0} ∩ K) = 0 for every K ⊂ Ω compact, i.e., {f = 0} is locally µ-null. f ) One has σ(Mf ) = f (Ω). Furthermore, R(λ, Mf ) = M(λ−f )−1 for λ ∈ ̺(Mf ).

g) If Mf ⊂ 0 then f = 0.

h) Let also g ∈ C(Ω). Then Mf Mg ⊂ Mf g . One has equality if g is bounded or if Mf is invertible. Proof. Ad a). Assume gn ∈ D(Mf ) such that gn → g and f gn → h in L2 . For ϕ ∈ Cc (Ω) we have ϕgn → ϕg and ϕf gn → ϕh in L2 , whence ϕf g = ϕh. Since ϕ was arbitrary it follows that f g = h. S Ad b). Let g ∈ D(Mf ). Since g ∈ L2 , it vanishes outside a set L = n Kn , where Kn ⊂ Kn+1 and each Kn is compact. Then 1Kn g → g and 1Kn f g → f g. Hence we can assume without restriction that g vanishes outside a compact set K. By Urysohn’s lemma [113, Lemma 2.12] one an find a function ψ ∈ Cc (Ω) such that 1K ≤ ψ ≤ 1. Since Cc (Ω) is dense in L2 , there is a sequence ϕn ∈ Cc (Ω) such that ϕn → g in L2 . Then ϕn ψ → ψg = g and f ϕn ψg → ψf g = f g. Ad c). Let x, y ∈ L2 . Then Z Z (x, y) ∈ (Mf )∗ ⇔ uy dµ = uf x dµ ∀ u ∈ D(Mf ). Ω

Ω

Since Cc (Ω) ⊂ D(Mf ), this is the case if and only if f x ∈ L2 . b 2 2 Ad ‚ d). ‚ Let f ∈ C (Ω). Then kf gk2 ≤ kf k∞ kgk2 for each g ∈ L . Hence Mf ∈ L(L ) and ‚Mf ‚ ≤ kf k . Assume that Mf is bounded and choose ω ∈ Ω. For every neighborhood U of ω ∞ 1

we define ϕU := µ(U )− 2 1U . (Note that µ(U ) 6= 0 since (Ω, µ) is standard.) Then ˛ ˛ Z ˛ ˛ ˛ ˛ ‚ ‚ ‚ ‚ ˛µ(U )−1 ˛ = ˛( f ϕU | ϕU ) 2 ˛ ≤ ‚Mf ‚ kϕU k2 = ‚Mf ‚ . f dµ L 2 ˛ ˛ U

R Since f is µ(U )−1 U f dµ → f (ω) if U shrinks to {ω}. Thus, f is bounded and ‚ continuous, ‚ kf k∞ ≤ ‚Mf ‚. Ad e). Suppose there is K such that µ(A) > 0 where A := {f = 0} ∩ K. Then 0 6= 1A ∈ L2 and Mf (1A ) = f 1A = 0. Hence Mf is not injective. Now assume that µ({f = 0} ∩ K) = 0 holds for every compact K. Let g ∈ L2 such that f g = 0 and choose K compact. Then f g1K = 0, whence K = ({f = 0} ∩ K) ∪ ({g = 0} ∩ K). From the hypothesis it follows that µ({g = 0} ∩ K) = µ(K). Since K was arbitrary and g ∈ L2 , we conclude that g = 0 µ-a.e.. Ad f ). Obviously, we have λ − Mf = Mλ−f for every λ ∈ C. Thus it suffices to consider the case λ = 0. If 0 ∈ / f (Ω) then f −1 ∈ Cb (Ω) and it is easy to see that in this case (Mf )−1 = Mf −1 . Let 0 ∈ f (Ω), say ϕU defined as in the proof of d). Then ‚ f (ω)‚= 0. Consider the functions R kϕU k2 = 1 and ‚Mf ϕU ‚2 = kf ϕU k = µ(U )−1 U |f |2 → |f (ω)|2 = 0 if U shrinks to {ω}. Hence (ϕU )U is an approximate eigenvector for 0. Ad g). Assume Mf ⊂ 0. Since Cc (Ω) is a core for Mf , we conclude that Mf = 0. Hence we

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have f = 0 by d). Ad h). It is straightforward to prove Mf Mg ⊂ Mf g . Assume that g is bounded. Then Mf Mg is closed by Lemma A.1.3. Since obviously Cc (Ω) ⊂ D(Mf Mg ) and Cc (Ω) is a core for Mf g we obtain Mf g ⊂ Mf Mg . If Mf is invertible then f is injective and f −1 is bounded (by f )). If ψ ∈ D(Mf g ), i.e., f gψ ∈ L2 , we have also gψ = f −1 f gψ ∈ L2 , whence ψ ∈ D(Mf Mg ).

Corollary D.1.2. a) Mf is symmetric if and only if Mf is selfadjoint if and only if f is real valued. b) Mf is accretive if and only if Mf is m-accretive if and only if Re f ≥ 0.

c) Mf is positive if and only if f (Ω) ⊂ [0, ∞).

Proof. Just apply the definitions and Proposition D.1.1.

In this section we have considered only a very special situation of multiplication operators, namely those who are induced by continuous functions on a standard measure space. Of course one can define Mf on L2 (Ω, µ) also if f is just a locally integrable measurable function and if the measure µ is not standard. Proposition D.1.1 and Corollary D.1.2 remain true after some small changes. This will be of some interest in Section D.6.

D.2 Commutative C ∗-Algebras. The Cyclic Case In this section we start with a Hilbert space H and a commutative sub-C ∗ -algebra with unit A of L(H). We assume the reader to be familiar with the basic notions and results of Gelfand theory, as can be found for example in [114, Chapter 11] or [38, Chapters 3 and 4]. Let H ′ be another Hilbert space and A′ be a commutative sub-C ∗ -algebra with unit of L(H ′ ). We say that (A, H) and (A′ , H ′ ) are unitarily equivalent, if there is an unitary isomorphism U : H −→ H ′ such that the mapping (T 7−→ U T U −1) : A −→ A′ is bijective. Suppose there is a cyclic vector v, i.e., we have {T v | T ∈ A} = H. This implies readily that the mapping (T 7−→ T v) : A −→ H is injective with dense range A v := {T v | T ∈ A}.

Proof. Injectivity is seen as follows. Let T v = 0 for some T ∈ A. Then, T Sv = ST v = S0 = 0 for every S ∈ A. Hence, T = 0 on the dense subspace A v.

Let Ω denote the spectrum (Gelfand space) of A . Then Ω is compact. We want to find a Radon measure µ on Ω which turns (Ω, µ) into a standard measure space, and a sub-C ∗ -algebra B of C(Ω) such that (A, H) is unitarily equivalent to (B, L2 (Ω, µ)). By the Gelfand-Naimark Theorem [114, Theorem 11.18], the canonical embedding Φ : A −→ C(Ω) is an isomorphism of C ∗ -algebras. Define the functional µ on C(Ω) by Z µ(f ) = f dµ := Φ−1 (f )v v H (f ∈ C(Ω)).

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If f ≥ 0, there is a g such that g ∗ g = f . This implies µ(f ) = µ(g ∗ g) =

2 Φ−1 (g)v Φ−1 (g)v H = Φ−1 (g)v ≥ 0. Hence µ is positive. If in addition µ(f ) = 0, we must have Φ−1 (g)v = 0. But v is cyclic, so Φ−1 (g) = 0. This implies g = 0, whence f = 0. Thus we have shown that (Ω, µ) is a standard measure space. We now construct the unitary operator U : H −→ L2 (Ω, µ) as follows. For w = T v ∈ A v we define U (w) = U (T v) := Φ(T ) ∈ C(Ω) ⊂ L2 (Ω, µ) Because v is a cyclic vector, U is well-defined and of course linear. The computation Z Z ∗ ∗ ( T v | Sv )H = ( S T v | v ) = Φ(S T ) dµ = Φ(S)Φ(T ) dµ = ( U (T v) | U (Sv))L2 (Ω,µ) ,

where T, S ∈ A, shows that U is isometric. Note that the range of U is all of C(Ω) which is a dense subspace of L2 (Ω, µ). Hence, U has a unique extension to an isometric isomorphism from H to L2 (Ω, µ). We are left to show that in fact U induces an unitary equivalence of (A, H) and (C(Ω), L2 (Ω, µ)). Let T ∈ A. It suffices to show that U T U −1 = MΦ(T ) . To prove this it is enough to check the action of both operators on the dense subspace C(Ω) of L2 (Ω, µ). If f ∈ C(Ω), there is a unique S ∈ A such that Φ(S) = f . Hence we have U T U −1(f ) = U T U −1 Φ(S) = U (T Sv) = Φ(T S) = Φ(T )Φ(S) = MΦ(T ) (f ). We thus have proved the following theorem. Proposition D.2.1. Let H be a Hilbert space and A a commutative sub-C ∗ -algebra with unit of L(H). Suppose that H has a cyclic vector with respect to A. Let Ω be the Gelfand space of A. Then there is a standard Radon measure µ on Ω such that (A, H) and (C(Ω), L2 (Ω, µ)) are unitarily equivalent. Remark D.2.2. If H is separable and A is a maximal commutative sub-W ∗ -algebra of L(H), then there is a cyclic vector, see [38, Theorem 4.65]. Of course, by an application of Zorn’s lemma one can show that each commutative selfadjoint subalgebra of L(H) is contained in a maximal commutative one (which a forteriori must be a W ∗ -algebra). Hence a bounded normal operator on a separable Hilbert space is unitarily equivalent to multiplication by a continuous function on an L2 -space over a compact space. This is one version of the spectral theorem (cf. Corollary D.3.3).

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D.3 Commutative C ∗-Algebras. The General Case Suppose we are given a Hilbert space H and a commutative sub-C ∗ -algebra with unit A of L(H), but there is no cyclic vector. We choose any vector 0 6= v ∈ H Then Hv := A v is a closed subspace of H that reduces A, i.e., it is A-invariant (clear) and even its orthogonal complement Hv⊥ is A-invariant. Proof. Let w ⊥ Hv , S ∈ T , and x ∈ Hv . Then ( Sw | x ) = ( w | S ∗ x ) = 0, because S ∗ x ∈ Hv again.

If we restrict the operators from A to the space Hv we obtain a selfadjoint subalgebra with unit Av of L(Hv ). Moreover, v is a cyclic vector with respect to A. Clearly, the whole procedure can be repeated on the Hilbert space Hv⊥ . Therefore, a standard application of Zorn’s lemma yields the following lemma. ∗ Lemma D.3.1. Let H be a Hilbert space, and A ⊂ L(H) L a commutative sub-C algebra with unit. Then there is a decomposition H = α∈I Hα as a Hilbert space direct sum such that each Hα is A-invariant and has a cyclic vector with respect to A.

Note that if H is separable, the decomposition in Lemma D.3.1 is actually countable. Theorem D.3.2. Let H be a Hilbert space and A ⊂ L(H) a commutative selfadjoint subalgebra. Then there is a standard measure space (Ω, µ) and a subalgebra B of Cb (Ω) such that (A, H) is unitarily equivalent to (B, L2 (Ω, µ)). ∗ Proof. Without restriction we can assume L that A is a sub-C -algebra with unit of L(H). By Lemma D.3.1 we can decompose H = α∈I Hα , where the Hα are A-invariant and have cyclic vectors, say, vα . We let Aα := A |Hα ⊂ L(Hα ), and define Ωα to be the spectrum of Aα . Proposition D.2.1 (cyclic case) yields an unitary isomorphism

Uα : (Aα , Hα ) −→ (C(Ωα ), L2 (Ωα , µα )) where µα is a standard measure on Ωα . In fact, Uα (Tα vα ) = Φα (Tα ) for each Tα ∈ Aα , where Φα : Aα −→ C(ΩαS) is the Gelfand isomorphism (cf. the proof of Proposition D.2.1). We now let Ω := Ωα the (disjoint) topological direct sum of the Ωα . Clearly, Ω is a locally compact Hausdorff space. Furthermore, each Ωα is an open subset of Ω. If f ∈ C(Ωα0 ) for some particular α0 , we can extend f continuously to the whole of Ω by letting f |Xα = 0 for every other α. We can therefore identify the continuous functions on Ωα0 with the continuous functions on Ω having support contained in Ωα0 . Note that for each ϕ ∈ Cc (Ω) there are only finitely many α such that ϕ|Ωα 6= 0. Hence, by Z X µ(ϕ) = ϕ dµ := µα (ϕ|Ωα ) Ω

α

a positive functional on Cc (Ω) is defined. Obviously, (Ω, µ) is a standard measure space (each (Ωα , µα ) is!). We now construct the unitary operator U . The subspace H0 := span{Tα vα | α ∈ I, Tα ∈ Aα } is dense in H. We define X X U := ( Tα vα ) 7−→ Φα (Tα ) : H0 −→ Cc (Ω) α∈F

α∈F

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where F ⊂ I is a finite subset, and Tα ∈ Aα for all α. Note that Φα (Tα ) is a continuous function on Ωα , hence can be viewed as a continuous function on Ω. It is clear that U is linear and surjective. The computation ˛ 0 1 ˛X X X ˛ ˛ ( x | y )H = @ Tα vα ˛ Sβ vβ A = ( Tα Vα | Sα Vα ) ˛ β∈G α∈F α∈F ∩G 0 10 1 Z X Z X X @ = Φ(Tα )Φ(Sα ) dµα = Φα (Tα )A @ Φβ (Sβ )A dµ α∈F ∩G Xα

Ω

α∈F

β∈G

= ( U x | U y )L2 (Ω,µ) , P P where x = α∈F Tα vα , y = β∈G Sβ vβ ∈ H0 , shows that U is even isometric. Hence, U extends to a unitary isomorphism U : H −→ L2 (Ω, µ). To conclude the proof we show U T U −1 = Mf

(T ∈ A), where f ∈ C(Ω) is defined by f |Xα = Φα (T |Hα ) ∈ C(Ωα ).

Note that, if this is true, it follows that f ∈ Cb (Ω) because the operator Mf is bounded on L2 (Ω, µ) (cf. part d) of Proposition D.1.1). To show the claim we only have to check the action of both operators on the dense subspace Cc (Ω) of L2 (Ω, µ). Therefore, let ϕ ∈ Cc (Ω) and define P ϕα := ϕ|Ωα . Then, ϕ = P ϕ where the sum is actually finite. We let Sα := Φ−1 (ϕα ). Then α α we have ϕ = U x with x = α Sα vα ∈ H0 . Now, X X U T U −1 (ϕ) = U T x = U T ( Sα vα ) = U ( Tα Sα vα ) α

=

X

Φα (Tα Sα ) =

α

α

X

Φα (Tα )Φα (Sα ) =

α

X = f( Φα (Sα )) = f ϕ = Mf (ϕ).

X α

f |Xα Φα (Sα )

α

This completes the proof.

Corollary D.3.3. (Spectral Theorem I) Let H be a Hilbert space and (Tj )j∈J a family of commuting bounded normal operators on H. Then there is a standard measure space (Ω, µ) and a family of bounded continuous functions (fj )j∈J on Ω such that ((Tj )j∈J , H) is unitarily equivalent to ((fj )j∈J , L2 (Ω, µ)). Proof. By Fugledge’s theorem [20, Chapter IX, Theorem 6.7] Tk∗ Tj = Tj Tk∗ for all j, k ∈ J. Hence the ∗-algebra A generated by (Tj )j∈J is commutative. Now we can apply Theorem D.3.2.

Remark D.3.4. One should note that in the case of a single operator T , Fugledge’s theorem is not needed for the proof of Corollary D.3.3 (see also Section D.4 below. On the other hand one can deduce Fugledge’s theorem from the spectral theorem for a single operator, as Halmos in [56] points out.

D.4 The Spectral Theorem: Bounded Normal Operators Let H be a Hilbert space and T ∈ L(H) a bounded normal operator on H. Denote by A the sub-C ∗ -algebra of L(H) which is generated by T . Because T is normal, A

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is commutative. We have σ(A) = σ(T ) and Tˆ (the Gelfand transform of T , being a continuous function on σ(A)) is the coordinate function (z 7−→ z). Moreover, R(λ, T ) ∈ A for every λ ∈ ̺(T ). Proof. From elementary Gelfand theory it follows that Tˆ : σ(A) −→ σ(T ) is surjective. But it is also injective, since T generates A. Hence Tˆ is a homeomorphism, identifying σ(A) and σ(T ). With this identification Tˆ becomes just the coordinate function (z 7−→ z). If λ ∈ ̺(T ), rλ := (λ−z)−1 is a continuous function on σ(T ), and the Gelfand-Naimark Theorem implies that there is an operator Rλ ∈ A which corresponds to rλ . But obviously we have Rλ = R(λ, T ).

In the following we will review the construction of the proof of Theorem D.3.2. For this we need a lemma whose proof is straightforward. Lemma D.4.1. Let H0 be a closed subspace of H, with orthogonal projection P : H −→ H0 . The subspace H0 is A-invariant, if and only if T P = P T . In this case one has σ(T |H0 ) ⊂ σ(T ) and R(λ, T |H0 ) = R(λ, T )|H0 for each λ ∈ ̺(T ). The C ∗ -closure A0 of {S|H0 | S ∈ A} is generated by T |H0 and σ(A0 ) = σ(T |H0 ). Proceeding along L the lines of the proof of Theorem D.3.2, we decompose the Hilbert space H = α∈I Hα into A-invariant and cyclic subspaces Hα . Let Tα := T |Hα and Ωα := σ(Aα ) = σ(Tα ). Each Ωα carries a standard measure µα such that (Tα , Hα ) is unitarily equivalent to (Mz , L2 (Ωα , µα )), where Mz is multiplication by the coordinate function (see above). The locally compact Hausdorff space Ω is the disjoint union of the σ(Tα ), and the standard measure µ on Ω is defined by µ|Ωα = µα . Finally, (T, H) and (Mf , L2 (Ω, µ)) are unitarily equivalent, where f ∈ Cb (Ω) is the coordinate function on each σ(Tα ). By Lemma D.4.1, σ(Tα ) is a closed subset of σ(T ) for each α. Therefore, Ω can be viewed as a closed subset of σ(T ) × I. The measure µ on Ω extends canonically to a measure on σ(T ) × I with support Ω which we denote again by µ. Note that this extension in general is not a standard measure any more. But obviously, there is a unitary equivalence of (Mf , L2 (Ω, µ)) and (Mh , L2 (σ(T ) × I, µ)), where f : σ(T ) × I −→ C is defined by h(z, α) = z for (z, α) ∈ σ(T ) × I. Thus, we have proved the following theorem. Theorem D.4.2. (Spectral Theorem II) Let H be a Hilbert space and T ∈ L(H) a bounded normal operator on H. Then there is a discrete set I and a (not necessarily standard) Radon measure µ on σ(T ) × I such that (T, H) is unitarily equivalent to (Mh , L2 (σ(T ) × I, µ)), where h is defined by h = ((z, α) 7−→ z) : σ(T ) × I −→ C. If H is separable, I = N.

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D.5 The Spectral Theorem: Unbounded Selfadjoint Operators Let H be a Hilbert space and A a (not necessarily bounded) selfadjoint operator on H. Then σ(A) ⊂ R. The spectral theorem for A states that one can find a standard measure space (Ω, µ) and a real-valued continuous function f on Ω such that (A, H) and (Mf , L2 (Ω, µ)) are unitarily equivalent. This means that there is a unitary isomorphism U : H −→ L2 (Ω, µ) which takes the graph of A bijectively onto the graph of Mf , i.e., it holds (x, y) ∈ A if and only if (U x, U y) ∈ Mf . The situation can be reduced to the bounded case by taking resolvents. Let T := (i − A)−1 . Then clearly T is a bounded normal operator on H. By Corollary D.3.3 we find a standard measure space (Ω1 , µ1 ) and a bounded continuous function g1 on Ω such that (T, H) and (Mg1 , L2 (Ω1 , µ1 )) are unitarily equivalent. Now T is injective, hence (g1 = 0) is a closed subset of Ω1 which is locally µ1 -null (see part e) of Proposition D.1.1). If we set Ω := Ω1 \ (g1 = 0) and µ := µ1 |Ω , we obtain a standard measure space (Ω, µ). It is easy to see that (Mg1 , L2 (Ω1 , µ1 )) is unitarily equivalent to (Mg , L2 (Ω, µ)), where g := g1 |Ω . But g does not vanish, hence f := i − g −1 defines a continuous function on Ω. Clearly, (A, H) is unitarily equivalent to (Mf , L2 (Ω, µ)). If we use Theorem D.4.2 instead of Corollary D.3.3, we obtain a discrete set I and a Radon measure ν on σ(T ) × I such that (T, H) is unitarily equivalent to (h, L2 (σ(T ) × I, ν)), where h : σ(T ) × I −→ C is the projection onto the first coordinate. Thanks to the spectral mapping theorem for resolvents (Proposition A.3.1) we have σ(T ) = ϕ(σ(A)), with ϕ(w) = (i − w)−1 . We define the Radon measure µ on σ(A) × I by Z Z f (w, α) dµ(w, α) := f (ϕ−1 (z), α) dν(z, α) σ(A)×I

σ(T )×I

for f ∈ Cc (σ(A) × I). Then it is immediate that (A, H) is unitarily equivalent to (Mf , L2 (σ(A) × I, µ)) where f : σ(A) × I −→ R is the projection onto the first coordinate. We summarize our considerations in the next theorem. Theorem D.5.1. (Spectral Theorem III) Let H be a Hilbert space and A a selfadjoint operator on H. a) There is a standard measure space (Ω, µ) and a function f ∈ C(Ω, R) such that (A, H) is unitarily equivalent to (Mf , L2 (Ω, µ)). b) There is a discrete set I and a positive Radon measure µ on σ(A) × I such that (A, H) is unitarily equivalent to (Mf , L2 (σ(A)×I, µ)), where f is given by f = ((z, α) 7−→ z) : σ(A) × I −→ R.

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249

If H is separable, I = N. Remark D.5.2. We focused our considerations on selfadjoint operators but with the same proofs can obtain a similar result for unbounded, normal operators.

D.6 The Functional Calculus The spectral theorem(s) allow(s) us to define a functional calculus for a normal operator on a Hilbert space. Let Ω be a locally compact space, µ a positive Radon measure on Ω, and f ∈ C(Ω) a continuous function on Ω. We let X := f (Ω) ⊂ C. Denote by B(X) the bounded Borel measurable functions on X. If g ∈ B(X), g ◦ f ∈ B(Ω), hence Mg◦f is a bounded operator on L2 (Ω, µ) satisfying kMg◦f kL(L2 ) ≤ kg ◦ f k∞ ≤ kgk∞ . Obviously, the mapping (g 7−→ g(Mf ) := Mg◦f ) : B(X) −→ L(L2 (Ω, µ)) is a homomorphism of C ∗ -algebras. Moreover, if gn is unifomly bounded sequence in B(X) converging pointwise to g ∈ B(X), then Lebesgue’s Dominated Convercence theorem yields that gn (Mf ) → g(Mf ) strongly. If we put this together with the spectral theorem(s), we obtain the following result. Theorem D.6.1. Let A be a selfadjoint operator on a Hilbert space H. Then there exists a unique mapping Ψ : B(σ(A)) −→ L(H) with the following properties. 1) Ψ is a ∗-homomorphism. 2) Ψ((λ − z)−1 ) = R(λ, A) for all λ ∈ / R. 3) If (gn )n ⊂ B(σ(A)) is uniformly bounded and gn → g pointwise, then Ψ(gn ) → Ψ(g) strongly. One usually writes g(A) instead of Ψ(g). Proof. Existence is clear from the remarks at the beginning of the section and the spectral theorem D.5.1. We show uniqueness. Observe that the (selfadjoint) algebra which is generated by the set {(λ − z)−1 | λ ∈ / R} is uniformly dense in C0 (R) by the Stone-Weierstrass theorem. The sequence of functions gn (z) := in(in − z)−1 is uniformly bounded on R and converges pointwise to the constant 1. On the other hand is is clear that inR(in, A) → I strongly. Hence Ψ(1) = I. Moreover, we see from this that Ψ is determined on Cb (R). By the Tietze Extension theorem we know that each bounded continuous function on σ(A) is the restriction of a bounded continuous function on R. Hence Ψ is determined on Cb (σ(A)). From this follows that Ψ is determined on the smallest class M of functions which contains the bounded continuous ones and is closed under bounded and pointwise convergence. Now, σ(A) is a metric, separable, locally compact space, whence by Urysohn’s lemma [113, Lemma 2.12] the characteristic functions of compact sets are contained in M. The class A := {M ⊂ σ(A) | 1M ∈ M} is easily seen to be a σ-algebra on σ(A) which contains the compact subsets. Hence A = B(σ(A)), the Borel σ-algebra on σ(A). But a standard result from measure theory says that each bounded Borel measurable function can be approximated uniformly by a sequence of Borel simple functions. Altogether this implies B(σ(A)) = M.

Remark D.6.2. The spectral theorem allows to define g(A) for a Borel measurable function g : σ(A) −→ C which is not bounded any more. This is done in the same

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way as for bounded functions g. For example, if A ≥ 0, i.e., σ(A) ⊂ [0, ∞) one can define the fractional powers Aα as (z α )(A) in this way.

References A highly readable account of the spectral theorem including also the “spectral measure version” is [111, Chaper VII and Section VIII.3]. One may also profit from [114, p.321 and p. 368] and [20, Chapter IX and ChapterX], where the stress is on the spectral measures. We have based our exposition mainly on the book [116], especially its Chapter IX, from which we learned about standard measure spaces and the “continuous multiplicator version” of the spectral theorem. Theorem D.4.2 in the separable case is [30, Theorem 2.5.1], but it is proved with entirely different methods. For some historical remarks on the spectral theorem see [112].

Appendix E

Approximation by Rational Functions In this appendix we provide some results from approximation theory. The objective is to approximate a given continuous function f on a compact subset K of the Riemann sphere C∞ by rational functions in some sense. Unfortunately, there is not enough room to develop the necessary complex function theory. Hence we have to refer to the literature. However, we could not find any account of the topic which served our purposes perfectly. Therefore, we will take two results from the book [45] of Gamelin as a starting point and modify them according to our needs. Note that a subset K ⊂ C∞ is called finitely connected if C∞ \ K has a finite number of connected components. If K ⊂ C∞ is compact, we consider ˚ A(K) := {f ∈ C(K) | f is holomorphic on K}. The set of all rational functions is denoted by C(z). We view a rational function r ∈ C(z) as a continuous (or holomorphic) function from C∞ to C∞ . A point λ ∈ C∞ is called a pole of r, if r(λ) = ∞. Given any subset K ⊂ C∞ we define R(K) := {r ∈ C(z) | r(K) ⊂ C} to be the set of rational functions with poles lying outside of M . If K is compact, we denote by R(K) the closure of R(K) in C(K). Then, A(K) is a closed subalgebra of C(K) with R(K) ⊂ A(K). Proposition E.0.3. [45, Chapter II, Theorem 10.4] Let K ⊂ C be compact and finitely connected. Then A(K) = R(K), i.e. each ˚ can be approximated uniformly on function f ∈ C(K) which is holomorphic on K K by rational functions rn which have poles outside of K. The other result we will need is concerned with pointwise bounded approximation. We say that a sequence fn of functions on a set Ω ⊂ C∞ converges pointwise boundedly on Ω to a function f , if supn supz∈Ω |fn (z)| < ∞ and fn (z) → f (z) for all z ∈ Ω. For Ω ⊂ C∞ open we let H ∞ (Ω) := {f : Ω → C | f is bounded and holomorophic}

252

Appendix E. Approximation by Rational Functions

be the Banach algebra of bounded holomorphic functions on Ω. We will occasionally write kf kΩ = kf k∞,Ω to denote the supremum norm of f ∈ H ∞ (Ω). Proposition E.0.4. [45, Chapter VI, Theorem 5.3] ˚ there is Let K ⊂ C be compact and finitely connected. Then, for every f ∈ H ∞ (K) a sequence of rational functions rn with poles outside K such that krn kK ≤ kf kK ˚ ˚ In particular, rn → f pointwise boundedly. and rn → f pointwise on K. Propositions E.0.3 and E.0.4 refer only to subsets K of the plane C. But one can easily extend these results to (strict) subsets K ⊂ C∞ of the Riemann sphere by a rational change of coordinates, a so-called M¨obius transformation. These are the mappings az + b : C∞ −→ C∞ m(a, b, c, d) := z 7−→ cz + d with complex numbers a, b, c, d ∈ C, ad−bc 6= 0. It is well known that each M¨obius transformation is invertible, its inverse being again a M¨obius tranformation. In particular, they are homeomorphisms of C∞ . Let K ⊂ C∞ be compact. If ∞ ∈ K ⊂ C∞ and K is not the whole sphere, there is some p ∈ C \ K. The transformation ϕ := m(0, 1, 1, −p) has the property ˚ = that ϕ(K) is a compact subset of C. Because ϕ is a homeomorphism, ϕ(K) ◦ ϕ(K) . Moreover, K is finitely connected if and only if ϕ(K) is. If r is a rational function with poles outside of ϕ(K), r◦ϕ is a rational function with poles outside of K, and one has krkϕ(K) = kr ◦ ϕkK . Finally, if f : K → C, then f ∈ A(K) if and ˚ if and only if f ◦ ϕ−1 ∈ H ∞ (ϕ(K)◦ ). only if f ◦ ϕ−1 ∈ A(ϕ(K)) and f ∈ H ∞ (K) These considerations show that Propositions E.0.3 and E.0.4 remain true for subsets K ⊂ C∞ with K 6= C∞ . We now deal with some special sets K. Let Ω ⊂ C be open. We denote by K the closure of Ω in C∞ , while we keep the notation Ω for the closure of Ω in C. We assume K 6= C∞ ,

∞ ∈ K,

˚ Ω = K,

and

K is finitely connected.

For example, Ω can be anything from the list • Sω = {z | z 6= 0, |arg z| < ω} the sector of angle 2ω symmetric about the positive real line, where ω < π. • Hω = {z | |Im z| < ω} the horizontal strip of height 2ω, symmetric about the real line, where ω > 0 is arbitrary. • Σω = Sω ∪ −Sω a double sector, where ω <

π 2.

• Πω = {z | (Im z)2 < 4ω 2 Re z} a horizontal parabola, where ω > 0 is arbitrary.

253 ∞ We define R∞ (Ω) := R(Ω) ∩ H ∞ (Ω) and R∞ 0 (Ω) := {r ∈ R (Ω) | r(∞) = 0}. ∞ Then it is clear that R(K) = R (Ω) and ∞ ∞ R∞ 0 (Ω) = R (Ω) ∩ C0 (Ω) ⊂ R (Ω)

⊂ A(K) = {f ∈ H ∞ (Ω) ∩ C(Ω) | lim f (z) ex.} z→∞

⊂ {f ∈ C(Ω) | lim f (z) ex.} = C(K). z→∞

Proposition E.0.5. We have R∞ (Ω) = {

p | p, q ∈ C[z], (q = 0) ∩ Ω = ∅, deg(p) ≤ deg(q)} q

(E.1)

R∞ 0 (Ω) = {

p | p, q ∈ C[z], (q = 0) ∩ Ω = ∅, deg(p) < deg(q)}. q

(E.2)

and

−1 The algebra R∞ (λ ∈ / Ω). 0 (Ω) is generated by the elementary rationals (λ − z) The algebra R∞ (Ω) is generated by the elementary rationals (λ − z)−1 (λ ∈ / Ω) together with the constant 1 function. The closure of R∞ (Ω) with respect to k·kΩ 0 is H ∞ (Ω) ∩ C0 (Ω). The closure of R∞ (Ω) with respect to k·kΩ is A(K).

Proof. Let r = p/q ∈ C(z) be a rational function. Then r is bounded on Ω if and only if it is bounded on K (view r as a continuous function from C∞ to itself). So its poles lie outside of K and r(∞) ∈ C. This implies deg p ≤ deg q. If r(∞) = 0 it follows that deg p < deg q. The other inclusions are clear. Obviously, every elementary rational (λ − z)−1 with λ ∈ / Ω is contained in R∞ 0 (Ω). Since we can write µ−z µ−λ α = α( + 1) (α, µ, λ ∈ C, λ ∈ / Ω) λ−z λ−z it follows from (E.2) and the Fundamental Theorem of Algebra that the elementary rationals ∞ ∞ generate R∞ 0 (Ω). From (E.1) it is clear that R (Ω) = R0 (Ω) ⊕ 1. From Proposition E.0.3 we know that R(K) = A(K). Let f ∈ A(K) suchthat f (∞) = 0. We can find rn ∈ R∞ (Ω) suchthat krn − f kΩ → 0. Since ∞ is in the closure of Ω in C∞ , this implies rn (∞) → f (∞) = 0. Hence k(rn − rn (∞)) − f kΩ → 0 and rn − rn (∞) ∈ R∞ 0 (Ω).

Proposition E.0.6. Let Ω and K be as above and f ∈ H ∞ (Ω). Then there is a sequence of rational functions rn ∈ R∞ (Ω) such that krn kΩ ≤ kf kΩ for all n and rn → f pointwise on Ω.

Proof. The statement is just a reformulation of Proposition E.0.4 combined with the remarks immediately after it.

254

Appendix E. Approximation by Rational Functions

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