Orthogonal Polynomials: Interaction Between Orthogonality in L -Spaces

Technische Universit¨at Munc¨ hen Zentrum Mathematik

Orthogonal Polynomials: Interaction between Orthogonality in L2-spaces and Orthogonality in Reproducing Kernel Spaces

Frank Hofmaier

Vollständiger Abdruck der von der Fakultät fur¨ Mathematik der Technischen Universität Munc¨ hen zur Erlangung des akademischen Grades eines Doktors der Naturwissenschaften (Dr. rer. nat.) genehmigten Dissertation.

Vorsitzender: Univ.-Prof. Dr. Gregor Kemper Prufer¨ der Dissertation: 1. apl. Prof. Dr. Domenico Castrigiano 2. Univ.-Prof. Dr. Rupert Lasser 3. Prof. Dr. Christian Berg, Universit¨at Kopenhagen/D¨anemark (schriftliche Beurteilung)

Die Dissertation wurde am 12. März 2007 bei der Technischen Universität eingereicht und durch die Fakultät fur¨ Mathematik am 9. Juli 2007 angenommen.

About this Work

Given a sequence (Pn)n 0 of polynomials in one complex variable where P0 1 and ≥ ≡ deg(Pn) = n for all n, we look for a Borel mesaure µ supported on some subset of C such that Pn Pm dµ = δn,m which we then will refer to as an orthonormalizing measure (om) for (Pn)n. If there exists an om µ, we are also interested in the questions whether (Pn)n R 2 is an orthonormal basis (onb) in Lµ and whether µ is uniquely determined.

The classical approach to this problem uses the multiplication operator D defined by (Dp)(z) := z p(z) for p C[z]. One constructs an abstract Hilbert space containing ∈ H C[z] as a dense linear subspace such that the given polynomials (Pn)n form an orthonormal basis and D becomes a densely defined (not necessarily bounded) linear operator in . 2 H If there exists an om then we can isometrically embed into Lµ and multiplication by z 2 H in Lµ is a normal operator extending D. In particular, there exists an om if and only if D is subnormal, i.e. has a normal extension which, however, might only exist in a larger space containing as a closed subspace. Unfortunately, in general, it is comparatively K H difficult to determine whether an operator is subnormal or not.

In addition to the abstract space , we will construct a Reproducing Kernel Hilbert Space H C 2 (RKHS) consisting of complex-valued functions defined on E := z : Pn(z) n ` with kernel ∈ ∈ K : E E C , K(z, w) = Pn(z) Pn(w) ; × → n 0 ≥ P denote that space by (K). We point out that E might be finite or even empty and if H E is infinite, it still can happen that (Pn)n is not an orthogonal system in (K). Yet, H 2 in many cases there exists an om µ and (K) can be isometrically embedded into Lµ. H 2 This is of particular interest because the elements of Lµ, in general, only are classes of functions while the members of an RKHS are functions defined pointwisely on E. Making use of this fact, we will be able to deduce particular qualities of om; for example, if the 2 operator D is essentially normal then there exists a unique om µ, (Pn)n is an onb in Lµ, E = z C : µ( z ) > 0 , and (P ) is an onb in the kernel space (K) if and only if µ is ∈ { } n n H discrete. It can also be shown that if there exists an om µ such that µ(E) = 1 then (P ) n n is an onb in the associated RKHS. Another remarkable fact – to name just a few – is that, if the interior of E is non-empty then there exists an open set G that is dense in E and where all the members of (K) are holomorphic. If, in addition, there exists an om µ such H2 C that (Pn)n is an onb in Lµ then G must be a µ-nullset and Λµ := z : µ( z ) > 0 cannot have a limit point in G. It remains an open question whether∈it is possible{ } that the largest such set G is a proper subset of the interior of E. ii

As convergence in the (K)-norm implies pointwise convergence on E, we will also be able H to make statements about the limits of pointwisely – on a subset of C – convergent series of polynomials, concerning the domain where that limit is holomorphic. In connection with orthogonal polynomials on the unit circle ∂D we can, given any compact uncountable Lebesgue1-nullset K ∂D, construct an om µ such that µ(K) > 0 and a sequence of ⊂ polynomials (qn)n which is convergent pointwisely in the open unit disk D and Lebesgue- almost everywhere in ∂D with limit 0 while µ-almost everywhere on K the limit is 1.

Orthogonal polynomials, especially in the real case, have been studied intensely for the past 70 years. The textbooks by Szeg˝o [Sze] and Chihara [Chi] may serve as standard references. A short historical note on orthogonal polynomials can be found in [As, pp. 12- 13]; for an overview concerning applications to random matrices, discrete Schr¨odinger operators, and birth-and-death processes, see [As, Ch. 6], for instance. Polynomials orthogonal with respect to measures supported on the unit circle are very well understood, too. A two-volume monograph by Simon [Si] presents a vast treatise on this topic. Polynomial sequences for which one tries to ﬁnd orthonormalizing measures may arise from numerical problems such as interpolation or quadrature, for example. As mentioned before, existence of orthonormalizing measures is equivalent to subnormality of the multiplication operator D. For the theory of bounded subnormal operators, we refer to [Con1]; unbonded subnormal operators have been observed in [StSz1], [StSz2], and [StSz3]; see also [CaKl2] and [Kl2].

Our work is organized as follows. The ﬁrst chapter is a short introduction stating the main problem and presenting the ideas how to handle it.

In chapter 2 we construct reproducing kernel spaces (K) associated to a given sequence (P ) . We will gather some statements about the elemenH ts of (K) concerning (amongst n n H other things) analyticity and we also realize some RKHS as a subspace of an L2-space.

Chapter 3 is the heart of this work presenting all the theoretical aspects on orthonormalizing measures, the multiplication operator D, its normal extensions, and all the remarkable connections between L2 and (K) mentioned above. µ H Finally, in chapter 4 we study some special cases; in particular, orthogonal polynomials on the real line and on the unit circle – both of which have been studied intensely in the past. However, we see how they ﬁt into the theory as special cases of the more general problem. Afterwards, we conclude the work with a variety of examples.

1Here we refer to one-dimensional Lebesgue measure on the unit circle. iii

Acknowledgments

I am very indebted to Professor Konrad Königsberger who encouraged me to stay at Technische Universität Munc¨ hen after receiving my Diplom in summer 2001. I would like to thank Professor Gregor Kemper as well as Professor Christian Berg for their support and Professor Rupert Lasser for a very pleasant period of work and many helpful com- ments. I would also like to thank Professor Gero Friesecke, whose chair I was working at for several teaching terms, for fruitful discussions and very useful hints. Especially, I would like to thank Professor Domenico Castrigiano for introducing me to the field of orthogonal polynomials. His excellent guidance made a great contribution to this work and he always provided a very comfortable atmosphere which undoubtedly supported the development of this work in a crucial way. iv CONTENTS

Contents

1 Introduction 1 1.1 Orthonormalizing Measures ...... 1 1.2 Hessenberg Operators ...... 3 1.3 Subnormal Operators ...... 4 1.4 Reproducing Kernel Hilbert Spaces (RKHS) ...... 8

2 Orthogonality in RKHS 11 2.1 Series Expansions of Kernel Functions ...... 11 2.2 Analytic Kernels ...... 17 2 2.3 Analytic Functions Forming Subspaces of Lµ ...... 24

2 3 Relations between the Spaces , (K), and Lµ 29 3.1 Preliminaries ...... H. . H...... 29 3.2 Subnormal Hessenberg Operators ...... 33 3.3 Spectral Properties of the Multiplication Operator ...... 40 C 2 3.4 Denseness of [z] in Lµ ...... 46 3.5 Open Subsets of E ...... 51 3.6 Point Evaluation on C[z] ...... 56 3.7 Orthonormal Bases in and (K) ...... 61 3.8 Concluding Remarks .H. . . .H...... 68

4 Applications and Examples 70 4.1 Orthogonal Polynomials on the Real Line ...... 70 4.2 Orthogonal Polynomials on the Unit Circle ...... 76 4.3 Orthogonal Polynomials on α-sets ...... 88 4.4 Weighted Shifts ...... 89 4.5 Various Examples ...... 93

Appendix 106 A.1 The Spectral Theorem for Normal Operators ...... 106 A.2 Normal Extensions of Hessenberg Operators ...... 109 A.3 Reproducing Kernel Hilbert Spaces ...... 111

References 116 1

1 Introduction

Throughout this work, for any complex Hilbert space , we will denote its inner product and norm by , and , respectively. Note thatHthe inner product is linear in the H second argumenh ·t.· iWe mayk·omitkH the index if it is clearly understood. B C C 2 Furthermore, for a measure µ on the σ-algebra ( ) of Borel subsets of , let Lµ be the Hilbert space of classes of (complex valued) functions square integrable with respect to µ, with inner product f , g = f g dµ. h i · Finally, let N denote the positive integers and N := N 0 . R 0 ∪ { }

1.1 Orthonormalizing Measures

1.1.1 Deﬁnition. Let (Pn)n 0 be a sequence in C[z] such that P0 1 and deg(Pn) = n ≥ ≡ for all n. A measure µ on B(C) is called an orthonormalizing measure (om) for (Pn)n, if all p C[z] are square integrable with respect to µ and ∈

Pn Pm dµ = δn,m , R 2 C i.e., if (Pn)n is an orthonormal system in the Hilbert space Lµ. We regard [z] as a linear 2 subspace of Lµ via the canonical embedding.

Note that, at this point, we do not require that (Pn)n be an orthonormal basis (in Hilbert 2 C 2 space sense); we will write Pµ for the closure of [z] in Lµ. 2 Furthermore, as 1 = P0 dµ = 1 dµ = µ(C), an om always is a probability measure. Note also that, in general,| | supp(µ) = C. In particular, µ with supp(µ) R will play a R R 6 ⊂ special role and have been widely studied. The textbooks by Szeg˝o [Sze], Chihara [Chi], and van Assche [As] may be stated as references here; in section 4.1 we will discuss some general results about polynomials orthogonal w.r.t. a measure supported on the real line.

Polynomials with om supported on (possibly subsets of) the unit circle z C : z = 1 have also been examined in depth. Here the recently published two-volume{ b∈ook b|y |Barry} Simon [Si] deﬁnitely has to be mentioned. 2 1 INTRODUCTION

On the other hand, if µ is a probability measure on B(C) such that every p C[z] is ∈ square integrable with respect to µ, then µ obviously is an om for such a sequence of 2 2 polynomials: One only has to apply the Gram-Schmidt algorithm in Lµ to 1, z, z , . . ., say. For more detailed information on measure theoretical issues, we refer to [Bau], [El], or also [LL, Ch. 1 and 2].

The Newton Polynomials,

n n z 1 ( 1) P (z) := ( 1) − = − (z 1)(z 2) (z n) , n − n n! − − · · · − 2 may serve as a ﬁrst example. They form an orthonormal set in the space Lµ where µ is given by 2 ∞ n+1 1 Γ( 2 +iy) dµ(x, y) = 2π | Γ(n+1) | dy (z = x + iy) , n=0 see 4.5.7 for details. P In section 4.5 we will give many more examples of polynomials and their orthonormalizing measures as well as sequences of polynomials which do not have an om.

We now turn to our central question: Which sequences of polynomials admit an om?

1.1.2 Main Problem. Let (Pn)n 0 be a sequence of polynomials in C[z] satisfying ≥ P 1 and deg(P ) = n for all n. 0 ≡ n Is there an om for (P ) ? • n n If there exists an om, is it unique ? • If µ is an om for (P ) , is P 2 a proper subspace of L2 ? • n n µ µ

Without further assumptions on the given polynomials, these questions can not be answered. It is our aim to state necessary and/or suﬃcient criteria for the existence or uniqueness of orthonormalizing measures as well as properties of these measures which can be deduced from certain qualities of the Pn.

The following is closely related to 1.1.2.

1.1.3 The Complex Moment Problem. Let (sij)i,j N0 be a doubly indexed sequence ∈ i j of complex numbers. Is there a measure µ on B(C) such that z z dµ = sij for all i, j ?

If such a measure exists, (sij)i,j N0 is called a complex moment sequence. ∈ R This generalizes the Hamburger moment problem where, given a sequence (mn)n 0 of n ≥ real numbers, one asks for a measure µ supported on a subset of R such that x dµ = mn for all n. R 3

For further reference on the complex moment problem we mention two papers by Kilpi [Ki1] and [Ki2] as well as [StSz2]. Treatises on the Hamburger moment problem can be found in [Ak] and [ShTa]; see also [BCh] or [BD], for example.

1.1.4 Proposition. Let (Pn)n be a sequence of polynomials as in 1.1.1. Then there are n n unique coeﬃcients aij C such that z = aknPk for all n N0. Furthermore, set ∈ k=0 ∈ P i sij := aki akj . k=0 P Then (sij)i,j is a complex moment sequence associated to µ if and only if µ is an om for (Pn)n.

The proof is a simple calculation and can be found in [Kl2, 1.2.2], for instance.

1.2 Hessenberg Operators

In the following, (Pn)n 0 will always be a sequence of polynomials as in 1.1.1. These ≥ polynomials form a vector space basis of C[z] and there is a uniquely deﬁned inner product on C[z] such that P , P = δ . h n m i n,m Let now denote the abstract completion of C[z] with respect to the norm induced H by this inner product. Then, by construction, (Pn)n 0 is an orthonormal basis (onb) of ≥ the Hilbert space and if an om µ exists then is isometrically isomorphic to P 2 via H H µ P P . n 7→ n The operator D deﬁned below has become a standard tool, see e.g. [Sz4] or [Kl2], to deal with the question on exsistence of orthonormalizing measures for a given sequence of polynomials.

1.2.1 Definition. Let D : C[z] C[z] be the multiplication operator defined by (Dp)(z) := z p(z). → We point out, that D is a densely defined linear operator in with invariant domain Hn dom(D) = C[z]. Moreover, D is cyclic, i.e. the linear span of D P0 : n N0 (which is equal to C[z], of course), is dense in . { ∈ } H As, for all n N0, DPn is a polynomial of degree n + 1, there exist d0n, . . . , dn+1,n C such that ∈ ∈ n+1 DPn = din Pi (1.1) i=0 P 4 1 INTRODUCTION

and, with respect to the onb (P ) of , D is represented by the matrix n n H d d d 00 01 02 · · · d d d  10 11 12 · · ·  0 d21 d22 (dij)i,j 0 =  · · ·  ≥  0 0 d32   · · ·   0 0 0     . . . .   . . . ..      where d = 0 for all n. n+1,n 6

Note that dij = 0 if i > j + 1. A matrix with this property is called (upper) Hessenberg matrix. Accordingly, we say that D is a Hessenberg operator. Note also that, in terms of this matrix, the elements of C[z] correspond to the vectors having only ﬁnitely many non-zero entries. However, the formal multiplication with this matrix might have a larger domain. In 3.2.7 we will examine the matrix representation of D more closely.

1.2.2 Example (Unilateral Shift). Let P (z) := zn for n N . Then, n ∈ 0 π π 1 it it 1 i(m n)t 2π Pn(e ) Pm(e ) dt = 2π e − dt = δn,m . π π − − 1 R R Hence, µ := 2π λ, where λ is the one-dimensional Lebesgue measure on the unit circle, is an om for (Pn)n. The Polynomials (Pn)n, however, do not form an orthonormal basis of 2 n 2 Lµ. An onb is given, for example, by (z )n Z. This shows that Pµ is a proper subspace 2 ∈ of Lµ. Here we have DP = P for all n, i.e. d = 1 and d = 0 if i = j + 1. Thus D is n n+1 n+1,n ij 6 the unilateral shift on C[z] with respect to the basis (Pn)n. As D is continuous, it has a 2 continuous extension on Pµ and, as we shall see, a uniquely determined normal extension 2 2 on the entire space Lµ. Clearly, this extension is the multiplication by z on Lµ. We will return to this example in 1.3.8.

Note that, in general, D may be unbounded.

1.3 Subnormal Operators

Concerning subnormality we start with some deﬁnitions. For further reference on linear operators in Hilbert space see e.g. [AG], [EE], and [W], or textbooks on functional analysis such as [Ru2].

1.3.1 Deﬁnition. Let T be a (not necessarily bounded) densely deﬁned linear operator in a Hilbert space and T ∗ its adjoint. H 5

We say that the operator T is hyponormal if dom(T ) dom(T ∗) and T ∗x T x ⊂ k k ≤ k k for all x dom(T ). ∈ A hyponormal operator T is formally normal if T ∗x = T x for all x dom(T ). k k k k ∈ An operator T is normal if it is formally normal and dom(T ) = dom(T ∗). Furthermore, T is subnormal if there exist a Hilbert space containing as a linear subspace and a normal operator N in such that dom(T ) Kdom(N) and HT x = Nx for all x dom(T ). The operator N is thenK called a normal extension⊂ of T . ∈ Finally, T is essentially normal if T is closable and its closure T is normal.

1.3.2 Proposition. Let (Ω, , µ) be a measure space and ϕ : Ω C measurable. Set 2 2 A 2 → B := f Lµ : ϕ f Lµ and deﬁne Mϕ : B Lµ, Mf := ϕ f. { ∈ · ∈ } 2 → · Then Mϕ is a normal operator in Lµ with adjoint Mϕ∗ = Mϕ (the bar denoting complex conjugation).

Proof: Following [W, 4.1 Beispiel 1], we show that Mϕ is densely deﬁned to ensure that N Mϕ∗ exists. For n set Xn := x Ω : ϕ(x) n . Then, for all n, Xn , ∈ { ∈ | 2 | ≤ } ∈ A X X , and ∞ X = Ω. Now take f L and deﬁne f := 1 f for n N, n ⊂ n+1 n=1 n ∈ µ n Xn ∈ where 1X denotes the indicator function of the set X. Then fn dom(Mϕ) = B for all n 2 S ∈ and fn f in Lµ as n . → → ∞ 2 As f was arbitrarily chosen, we see that dom(Mϕ) is dense in Lµ. Obviously, dom(M ) = dom(M ) and M f = M f for all f dom(M ), as well as ϕ ϕ k ϕ k k ϕ k ∈ ϕ f ϕg dµ = ϕf g dµ for all f, g dom(M ), hence M is normal and M ∗ = M . ∈ ϕ ϕ ϕ ϕ R R B C C 2 If µ is a probability measure on ( ) and [z] Lµ, we write Mµ for the operator Mϕ 2 ⊂ in Lµ where ϕ(z) := z. Note that dom(Mµ) for such µ always contains C[z]. Moreover, C[z] is an invariant subspace of M , i.e. M p C[z] for all p C[z]. µ µ ∈ ∈

Let now (Pn)n 0 be a sequence of polynomials as in 1.1.1. If there exists an om µ, we ≥ 2 regard D as an operator in Lµ. Then Mµ is a normal extension of D. In other words, we have just proved the following.

1.3.3 Corollary. If there exists an om µ for (Pn)n 0, then D is subnormal. ≥

To prove that the contrary is also true we will make use of the spectral theorem for normal operators, see A.1.3 in the appendix. The correspondence between the existence of orthonormalizing measures on the one hand and subnormality of the multiplication operator D on the other hand is also the main topic of [Kl2], [CaKl1], and [CaKl2]. The theory of subnormal operators and their normal extensions has been vastly treated in a series of three papers by Stochel and Szafraniec [StSz1], [StSz2], [StSz3] and a famous textbook by Conway [Con1] presents the theory of bounded subnormal operators. A proof of the next theorem can be found e.g. in [Kl2, 1.3.3]. For bounded operators, see also [Con1, II. 5]. We will present a proof in A.2.1 in the appendix. § 6 1 INTRODUCTION

1.3.4 Theorem. For (Pn)n 0 as in 1.1.1 deﬁne and D as before. Assume that there ≥ H exists a normal extension N of D in a Hilbert space and denote by σ(N) the spectrum of N and by E the spectral measure of N. K ⊃ H

Then the measure µ on B(C) given by µ(∆) := P0 , E(∆)P0 is an om and supp(µ) 2 h i ⊂ σ(N). Furthermore, Lµ can be embedded isometrically into such that D Mµ N 2 K ⊂ ⊂ and Mµ is a normal extension of D (in Lµ) with supp(µ) = σ(Mµ).

After introducing two more deﬁnitions we can proceed to the main statement of this section.

1.3.5 Deﬁnition. Let T be an operator in a Hilbert space and a closed subspace of K H . Denote by P the projection from onto . We say that reduces T if P T T P , i.e.K for all x dom(T ), we have P x Kdom(TH) and P T x = T PHx. ⊂ ∈ ∈

1.3.6 Deﬁnition. Let T be a subnormal operator in a Hilbert space and N a normal H extension of T in a Hilbert space . The operator N is called a minimal normal extension of T , if the only closed Ksubspace of containing and reducing N is the K H space itself. K

Remark. Stochel and Szafraniec [StSz3] call this minimal normal extension of spectral type. In [StSz3, Proposition 1] they show that N is a minimal normal extension of spectral type of T if and only if E(∆)x : x , ∆ B(C) is dense in where E { ∈ H ∈ } K denotes the spectral measure of N. This justiﬁes their name for this property. As, for our purposes, we do not need another kind of minimality, we will keep with the shorter.

Now we can state the main result concerning the existence of orthonormalizing measures in terms of the theory of subnormal operators: There is a bijective correspondence between the orthonormalizing measures for (Pn)n and – up to unitary equivalence – the minimal normal extensions of the Hessenberg operator D. More precisely, the following holds.

1.3.7 Theorem. If, in the situation of 1.3.4, N is a minimal normal extension of D 2 C then there exists an isomorphism β : Lµ such that β(p) = p for all p [z] and 1 → K ∈ β− Nβ = Mµ.

If, on the other hand, µ is an om for (Pn)n then Mµ is a minimal normal extension of D. In particular, if D is subnormal then there always exists a minimal normal extension. Furthermore, if D is essentially normal then its only minimal normal extension is D, 2 there exists a unique om µ, and (Pn)n is an orthonormal basis of Lµ.

For a proof, see A.2.4 in the appendix.

We point out that there need not be a unique minimal extension. It may happen that there exist “diﬀerent” normal extensions N and N of D in spaces and , respectively, K L K L 7

such that there is no isomorphism β : which is the identity on satisfying 1 K → L H β− N β = N . K L Unfortunately, in general, one does not know if a linear operator is subnormal or not while, as mentioned before, in the case of a symmetric operator, we can give an answer. For the theory of self-adjoint extensions of symmetric operators, see [AG, Kap. VIII], [Ru2, 13.20], or [W, Kap. 8], for example. Suppose now that D is symmetric. Then the matrix representation of D is tri-diagonal with non-vanishing conjugate entries in the upper and lower diagonal and reals in the main diagonal; such operators are commonly called Jacobi operators. If such D is subnormal then its normal extensions must be self-adjoint, hence their spectral measures and, therefore, also the corresponding orthonormalizing measures are supported in R. It is a well known fact, that D has a self-adjoint extension in if and only if the deﬁciency H indices dim(ran(D i id)⊥) are equal. In 3.1.3 we will be able to compute these deﬁciency · indices using the given polynomials (Pn)n. The theory of real orthogonal ploynomials, like the theory of self-adjoint extensions, has been widely examined. As standard references see the textbooks by Szeg˝o [Sze] and Chihara [Chi]. We will deal with this situation in section 4.1. Many properties of real orthogonal polynomials, however, do not even have an equivalent in our more general problem. In particular, formally normal or hyponormal operators will not be an adequate counterpart to symmetric operators in the real case. As a matter of fact, every subnormal operator is hyponormal, see [J, I. Example 4] or [OSch]2, while there exist hyponormal and even formally normal operators that are not subnormal. Examples to the latter can be found in a short note by Schmudgen¨ [Sch] and in [Cod]. We now proceed to a subnormal operator which is not formally normal.

1.3.8 Unilateral Shift (continued). The best-known example of a subnormal operator which does not have a normal extension in the same space, is the unilateral shift in `2(N ). Let e , e , . . . be the standard basis of `2(N ) and define S : `2(N ) `2(N ) 0 { 0 1 } 0 0 → 0 by Sei := ei+1 for i N0. It is not difficult to see that its adjoint is given by S∗ei = ei 1 ∈ − for i 1, S∗e = 0, and hence S is not normal and not formally normal, either. ≥ 0 2 2 However, S is subnormal. To see that, we understand ` (N0) as a linear subspace of ` (Z) 2 via the canonical embedding. Denote the standard basis of ` (Z) by ei : i Z and define T : `2(Z) `2(Z) by T e := e for i Z. { ∈ } → i i+1 ∈ One can easily see that T ∗ is given by T ∗ei = ei 1 for i Z and that dom(T ) = dom(T ∗) = 2 2 − ∈ ` (Z) with T x = T ∗x for all x ` (Z). Thus T is normal and extends S. k k k k ∈ Note that T ∗ is not an extension of S∗. Let us now return to 1.2.2, where P (z) = zn for n N . If we identify P 2 with `2(N ) via n ∈ 0 µ 0 Pn en, then S and also T extend the Hessenberg operator D. Hence D is subnormal. Moreo7→ ver, we can identify L2 and `2(Z) via zn e , n Z, and see that T = M is a µ 7→ n ∈ µ minimal normal extension of D.

2Otaˆ and Schmudgen¨ [OSch] use the terms hyponormal for what we call subnormal and formally hyponormal for what we call hyponormal. 8 1 INTRODUCTION

1.4 Reproducing Kernel Hilbert Spaces (RKHS)

Assume that µ is an om for a sequence of polynomials (Pn)n 0 as in 1.1.1 and that there 2 ≥ exists z C such that Pn(z) < . ∈ n 0 | | ∞ ≥ P 2 Then kz := Pn(z) Pn is well-defined and kz , Pm Lµ = Pm(z) for all m. By linearity, n 0 h i ≥ P k , p 2 = p(z) for all p C[z] . h z iLµ ∈ 2 C 2 2 Recall that Pµ denotes the closure of [z] in Lµ. For any f Pµ , there exists a sequence (q ) in C[z] such that q f with respect to the L2 -norm∈and for every such sequence, n n n → µ 2 2 qn(z) = kz , qn Lµ kz , f as n . Therefore, to f Pµ we can assign a function h i → h i → ∞ 2 ∈ pointwisely defined on the set E := z C : Pn(z) < in a canonical way. ∈ n 0 | | ∞ ≥ This approach leads to a Reproducing KernelPHilbert Space; we start with some definitions.

1.4.1 Deﬁnition. Let E be an arbitrary non-empty set and (E) be the vector space of all functions f : E C. Furthermore, let (K) be a linearFsubspace of (E) which → H F is a Hilbert space with the following property: For every z E there exists K (K) ∈ z ∈ H such that f(z) = Kz , f for all f (K). Then (K) is calledh a Reproi ducing∈ HKernel Hilbert Space (RKHS) with domain E H and K : E E C, K(z, w) := Kz(w) = Kw , Kz is the reproducing kernel or just kernel of ×(K).→The functions K are alsoh sometimesi referred to as kernel functions. H z Note that L2-spaces cannot match this deﬁniton when their elements are not functions but equivalence classes of functions.

Note also that Kz is unique: Take h (K) such that f(z) = h , f for all f (K), then K h , f = 0 for all f (K∈)Hand hence K h = 0. h i ∈ H h z − i ∈ H z − Obviously, K(z, z) = K , K = K 2 0 and K(z, w) = K(w, z) for all z, w E. h z z i k zk ≥ ∈ Without loss of generality we can assume K(z, z) > 0 for all z E because if K(z0, z0) = 0 for some z E then K = 0 in (K) and 0 = K , f =∈f(z) for all z E. In this 0 ∈ z0 H h z0 i ∈ case we can simply restrict to E z0 , see A.3.6 in the appendix for a more detailed discussion. \ { }

Moreover, Kz , f = 0 for all z E is equivalent to f = 0; hence Kz : z E is total in (K). h i ∈ { ∈ } H

1.4.2 Example. For any non-empty set E, the space `2(E) is an RKHS with kernel 2 K(z, w) = δz,w because, for α = (αλ)λ E ` (E) and z E, we have ∈ ∈ ∈

αz = δz,λ αλ = Kz , α . λ E h i ∈ P In particular, as dim `2(E) = E , an RKHS need not be separable. | | 9

A fundamental work concerning RKHS is the survey by Aronszajn [Ar]. An introduction to this topic can also be found in [Me] as well as in [Do, Chapter X]. We will state the most important properties of RKHS here, too, including some very straightforward proofs.

1.4.3 Lemma. A Hilbert space (E) is an RKHS if and only if point evaluation f f(z) is a continuous linear functionalH ⊂ F for every z E. 7→ ∈ Proof: Obviously, if is an RKHS with kernel K and domain E then, for ﬁxed z E, H ∈ the mapping f f(z) = K , f is continuous and linear. 7→ h z i Conversely, let now f f(z) be continuous. According to the Riesz representation 7→ theorem, there exists K such that K , f = f(z) for all f . z ∈ H h z i ∈ H

Another important property of RKHS is that convergence in the norm always implies pointwise convergence.

1.4.4 Lemma. Let (f ) be a sequence in (K) which is weakly convergent to f, say, n n H i.e. lim h , fn = h , f for every h (K). Then lim fn(z) = f(z) for all z E. n n →∞h i h i ∈ H →∞ ∈ Moreover, if f f (in the norm) as n then (f ) is uniformly convergent in every n → → ∞ n n subset of E where K(z, z) is bounded.

Proof: The ﬁrst assertion follows immediately from g(z) = K , g for all g (K). h z i ∈ H Let now fn f 0 as n and c > 0, A E such that K(z, z) c for all z A. Cauchy-Sck hw−arz kyields→ → ∞ ⊂ ≤ ∈

1 f (z) f(z) = K , f f K f f c 2 f f | n − | |h z n − i| ≤ k zk k n − k ≤ k n − k for all z A. Hence f f uniformly on A. ∈ n →

In the setting of the above deﬁnition, for arbitrary ﬁnite subsets z1, . . . , zn E and c , . . . , c C, we have { } ⊂ 1 n ∈ n n n n n n

0 ci Kzi , ci Kzi = cj ck Kzj , Kzk = ck cj K(zk, zj) . (1.2) ≤ i=1 i=1 k=1 j=1 h i k=1 j=1 D P P E P P P P The converse is also true:

1.4.5 Proposition. Let E be an arbitrary set and K : E E C a map such that × → n n ci cj K(zi, zj) 0 i=1 j=1 ≥ P P for all ﬁnite sets z1, . . . , zn E and c1, . . . , cn C. Then K is the kernel{ function} ⊂ of an RKHS. Mor∈ e precisely, there is a unique linear subspace V of (E) such that (K) = V . F H 10 1 INTRODUCTION

We will provide a proof in the appendix, see A.3.1.

Remark. In other words, the premises of this theorem read for any z , . . . , z E { 1 n} ⊂ the matrix (kij)1 i,j n with kij := K(zi, zj) is positive semi-deﬁnite. Therefore, K is ≤ ≤ sometimes, as in [Do] or [Me, Kap. V], referred to as positive matrix.

1.4.6 Lemma. Let (K) be an RKHS with domain E. If (hι)ι I is an onb in (K) H ∈ H then K(z, w) = hι(z)hι(w) for all z, w E. ι I ∈ ∈ P Proof: Using the Parseval equation, we see

K(z, w) = Kw , Kz = Kw , hι hι , Kz = hι(w) hι(z) . h i ι Ih i h i ι I ∈ ∈ P P

For (Pn)n 0 as in 1.1.1 we aim to construct an RKHS such that (Pn)n is an onb. As we ≥ have just seen, its kernel must be of the form K(z, w) = Pn(z)Pn(w) for z, w E with n 0 ∈ suitable E C. ≥ ⊂ P We will deal with this topic in the next chapter.

As for now, we conclude this chapter with a necessary and suﬃcient criterion for a function to be a member of a given RKHS. For a proof see A.3.4 in the appendix and also [Sz1] or [Sz2].

1.4.7 RKHS-Test. Let (K) be an RKHS with domain E and f a complex-valued H function on E. Then f (K) if and only if there exists r > 0 such that ∈ H n n n 2 cicj K(zi, zj) r ci f(zi) i=1 j=1 ≥ i=1

P P P for all ﬁnite sets z , . . . , z E and c , . . . , c C. { 1 n} ⊂ 1 n ∈

Note that if this holds for some r0 > 0 then it holds for all 0 < r < r0. 11

2 Orthogonality in RKHS

Given a sequence of polynomials (Pn)n 0 as in 1.1.1, we will in this chapter state necessary ≥ and suﬃcient conditions for the existence of an RKHS such that (Pn)n is an orthonormal 2 basis (onb) in this space. For E := z C : Pn(z) < deﬁne ∈ n 0 | | ∞ ≥ P K : E E C , K(z, w) := Pn(z)Pn(w) . × → n 0 ≥ P As we shall see, K is the kernel of some RKHS but (Pn)n need not be an onb of this space.

2.1 Series Expansions of Kernel Functions

We now construct an RKHS from a (pointwisely on a set E) convergent series of functions. The following theorem contains [Sz3, Proposition 3.1].

2.1.1 Theorem. Let (hι)ι I , I an index set, be a family of complex-valued functions ∈ 2 deﬁned on a set E such that hι(z) < for all z E. Then ι I | | ∞ ∈ ∈ P K : E E C , K(z, w) := hι(z)hι(w) × → ι I ∈ P is the kernel of a reproducing kernel Hilbert space (K). For any α = (α ) `2(I), H ι ι ∈

β(α) := αι hι ι I ∈ P is summable in (K) and deﬁnes a continuous linear map β : `2(I) (K) with H → H β 1. Moreover, h : ι I is total in (K). k k ≤ { ι ∈ } H

Proof: For z E set h(z) := (hι(z))ι I . Now take z1, . . . , zn E and c1, . . . , cn C. ∈ ∈ { } ⊂ ∈ Then n n n n n 2 cicj K(zi, zj) = cicj h(zi) , h(zj) `2 I = ci h(zi) 0 (2.1) ( ) 2 i=1 j=1 i=1 j=1 h i i=1 ` (I) ≥

P P P P P and, according to 1.4.5, (K) exists. H 12 2 ORTHOGONALITY IN RKHS

2 For α = (αι)ι ` (I), deﬁne f : E C, f(z) := αιhι(z) = α , h(z) `2(I). ∈ → ι I h i ∈ In order to show f (K), we use Cauchy-SchwParz and (2.1) to obtain ∈ H n 2 n 2 ci f(zi) = α , ci h(zi) `2(I) i=1 i=1 n 2 n n P 2 P 2 α 2 ci h(zi) = α 2 cicj K(zi, zj) . ` (I) 2 ` (I) ≤ k k · i=1 ` (I) k k i=1 j=1

P P P Thus the RKHS-test 1.4.7 implies f (K). ∈ H We next show that αιhι is summable in (K) with sum f. ι I H ∈ P n Let 0 := lin Kz : z E and u = biKzi , see also (A.3) in the appendix. Then H { ∈ } i=1 P n 2 n n 2 2 2 2 u , f (K) = bi f(zi) α `2(I) bibj K(zi, zj) = α `2(I) u (K) . h iH i=1 ≤ k k i=1 j=1 k k · k kH

P P P 2 2 2 According to 1.4.1, 0 is dense in (K); this yields g , f (K) α `2(I) g (K) H H |h iH | ≤ k k · k kH for all g (K). ∈ H 4 2 2 In particular, f (K) α `2(I) f (K). Thus f (K) α `2(I). k kH ≤ k k · k kH k kH ≤ k k Therefore, the linear map

2 β : ` (I) (K) , β(α)(z) := αι hι(z) → H ι I ∈ P (pointwise limit for z E) is well-defined and continuous with β 1. ∈ k k ≤ Let now ε > 0. There exists a finite set Jε I such that, for every finite J I with J J, the sequence α `2(I) given by α =⊂α for ι J, and α = 0 otherwise,⊂satisfies ε ⊂ ∈ ι ι ∈ ι α α `2(I) < ε. Using continuity of β, we get f αι hι (K) < β ε ε. Thus k − k k − ι J kH k k · ≤ f = αι hι in (K).e e ∈ e ι I H P ∈e P In particular, hι(z) hι = Kz for z E and ι I ∈ ∈ P f(z) = Kz , f (K) = hι(z) hι , f (K) = hι(z) hι , f (K) (2.2) h iH ι I H ι I h iH ∈ ∈ P P for all f (K). ∈ H For the last assertion take f h : ι I ⊥. Then, by (2.2), f(z) = 0 for all z E, i.e. ∈ { ι ∈ } ∈ f = 0, which proves that h : ι I is total in (K). { ι ∈ } H

Note that, as β need not be one-to-one, in general αι = hι , f (K). 6 h iH In 1.4.6 we have seen that the kernel always has a series expansion with respect to an onb. We will now turn to the question whether (hι)ι I is an onb of (K). ∈ H 13

2.1.2 Lemma. In the situation of 2.1.1, (hι)ι I is an onb of (K) if and only if β is ∈ H an isomorphism.

2 Proof: Let (eκ)κ I be the standard onb of ` (I), i.e. [eκ]ι = δικ. Then β(eκ) = hκ for all ∈ κ I which shows that (hκ)κ I is an onb if and only if β is an isomorphism. ∈ ∈

Recall the deﬁnition := lin K : z E , see also (A.3). Set W := lin h(z) : z E , H0 { z ∈ } 0 { ∈ } where h(z) = (hι(z))ι I and the bar denotes complex conjugation. Then W0 is a linear 2 ∈ subspace of ` (I) and β h(z) = hι(z) hι = Kz for all z E. Thus β(W0) = 0. ι I ∈ H ∈ 2 2 Furthermore, let W be the closurePof W in ` (I). Then ` (I) = W W ⊥. 0 ⊕

2.1.3 Lemma. Using the above notation, β : `2(I) (K) is a partial isometry with → H (β) = W ⊥. N Proof: For z , . . . , z E and c , . . . , c C we have { 1 n} ⊂ 1 n ∈ n 2 n 2 β cih(zi) = ci β h(zi) i=1 (K) i=1 (K) H H P Pn 2 n n n 2 (1.2) (2.1) = ci Kz = ci cj K(zi, zj) = ci h(zi) . i 2 i=1 (K) i=1 j=1 i=1 ` (I) H P P P P

Hence β(α) (K) = α `2(I) for all α W0. k kH k k ∈ As β is continuous, it isometrically maps W onto the closure of in (K), which is H0 H (K) itself. H Finally, if α W ⊥ then ∈

0 = h(z) , α `2(I) = αι hι(z) = β(α)(z) for all z E . h i ι I ∈ ∈ P Thus β(α) = 0.

Combining 2.1.2 and 2.1.3, we have the following necessary and suﬃcient criterion for (hι)ι I being an onb of (K). ∈ H

2.1.4 Corollary. In the situation of 2.1.1, (hι)ι I is an onb of (K) if and only if 2 ∈ H V := hι(z) ι I : z E is total in ` (I). ∈ ∈ Proof: According to 2.1.2, (hι)ι I is an onb of (K) if and only if β is an isomorphism. ∈ The partial isometry β is an isomorphism if Hand only if its null space is trivial, i.e. 2 W ⊥ = 0 . The latter is equivalent to the totality of V in ` (I) because W is the closure { } of the linear span of V . 14 2 ORTHOGONALITY IN RKHS

Recall that, for every κ I, we have β(e ) = h , where e : ι I denotes the standard ∈ κ κ { ι ∈ } unit vectors of `2(I). If (h ) is an onb in (K) then β is an isomorphism `2(I) (K). ι ι H → H If, on the other hand, hκ , hι (K) = 0 for some ι = κ I, then at least one of the h iH 6 6 ∈ vectors eι, eκ must not belong to W . Analogously, if hι (K) = 1 for some ι then β is k kH 6 not an isomorphism and W = `2(I). 6 (ι) (ι) (ι) For ι I there exists a unique b W such that β(b ) = hι and, as β(b eι) = ∈ (ι) ∈ − h h = 0, we have b e W ⊥. ι − ι − ι ∈ Now, for arbitrary f (K), let a = (a ) W such that β(a) = f. Then ∈ H ι ι ∈ (ι) (ι) hι , f (K) = b , a `2(I) = b eι , a `2(I) + eι , a `2(I) . h iH h i h − i h i (ι) Due to b eι W ⊥, the ﬁrst summand vanishes, leaving hι , f (K) = eι , a `2(I) = aι − ∈ h iH h i (κ) as well as β [ hι , f (K)]ι = f. In particular, for f = hκ, we get hι , hκ (K) = bι . h iH h iH We have just proved the follo wing lemma.

2.1.5 Lemma. Let f (K). Then hι , f (K) ι W and f = hι , f (K) hι ∈ H h iH ∈ ι Ih iH in (K). ∈ H P Moreover, if f = cιhι with (cι)ι W then (cι)ι = hι , f (K) ι. ι I ∈ h iH ∈ P Let B be the orthogonal projection in `2 onto W . Then β = β B and we have the following characterization of B. ◦

2 (κ) 2.1.6 Proposition. For x = (xι)ι ` (I), let Bx := bι xι . ∈ ι I κ ∈ Then x Bx is the orthogonal projection in `2(I) ontoPW . 7→ (κ) (κ) (κ) Proof: We start with x W ⊥. As b W , we have bι xι = b , x `2(I) = 0 for ∈ ∈ ι I h i all κ I. Hence Bx = 0. ∈ ∈ P On the other hand, for x W , deﬁne f := β(x) (K). Then xι = hι , f (K) for H ι I and ∈ ∈ H h i ∈ (κ) (κ) bι xι = b , x `2(I) = hκ , f (K) = xκ . ι I h i h iH ∈ Thus Bx = x. P

Remark. Although (h ) need not be an onb in (K), 2.1.5 shows that any f (K) is ι ι H ∈ H represented by a Fourier-like series with respect to the sequence (hι)ι. Such a sequence is commonly called a Parseval frame. For details on the theory of frames in Hilbert spaces we refer to [HL] or [Chr]. In view of this theory, our next theorem is a special case of [Chr, 5.4.7].

2.1.7 Theorem. Let κ I. Then hκ (K) 1 and ∈ k kH ≤

hκ (K) = 1 hκ , hι (K) = 0 for all ι = κ hκ / lin hι : ι = κ . k kH ⇐⇒ h iH 6 ⇐⇒ ∈ { 6 } 15

Proof: As β is a partial isometry, hκ (K) = β(eκ) (K) eκ `2(I) = 1. k kH k kH ≤ k k 2 (κ) 2 (κ) 2 2 Moreover, hκ , hκ (K) = hκ (K) = b `2(I) = bι = hι , hκ (K) . h iH k kH k k ι I ι I h iH ∈ ∈ P P 2 Equivalently, hκ , hκ (K) 1 hκ , hκ (K) = hι , h κ (K) . h iH − h iH ι=κ h iH 6 P This shows hκ , hκ (K) = 1 hι , hκ (K) = 0 for all ι = κ . h iH ⇐⇒ h iH 6 Clearly, hι , hκ (K) = 0 for all ι = κ is equivalent to hκ lin hι : ι = κ , implying h iH 6 ⊥ { 6 } h / lin h : ι = κ . κ ∈ { ι 6 } For the converse, suppose hκ , hκ (K) = 1. h iH 6 (κ) (κ) Then hκ = β b = bι hι = hι , hκ (K) hι implies ι I ι Ih iH ∈ ∈ P P 1 hκ = hι , hκ (K) hι . 1 hκ ,hκ H (K) ι=κh iH −h i 6 P Hence h lin h : ι = κ . κ ∈ { ι 6 }

2.1.8 Notations. We now apply the previous results to a given sequence of polynomials. 2 For (Pn)n 0 as in 1.1.1, set E := z C : Pn(z) < , ≥ ∈ n 0 | | ∞ ≥ P K : E E C , K(z, w) := Pn(z)Pn(w) , × → n 0 ≥ P and denote the associated RKHS by (K). Then, for all f (K) and all z E, H ∈ H ∈

f(z) = Kz , f (K) = Pn(z) Pn , f (K) = Pn , f (K)Pn(z) . (2.3) h iH n 0 H n 0h iH ≥ ≥ P P 2 2 2 In the following, when we write ` , we always mean ` (N0), i.e. the vectors in ` start with the 0th entry.

Deﬁne P (E) := Pn(z) n 0 : z E ; in accordance to the previous results, let W0 ≥ ∈ 2 be the linear span of P (E) and let W be the closure of W0 in ` . Then, see 2.1.3, 2 β : ` (K) , (αn)n 0 αn Pn , is a partial isometry with initial space W . → H ≥ 7→ n 0 ≥ Denote by α the canonical isomorphismP from the abstract space deﬁned in section 1.2 H onto `2. Then β α maps p C[z] to p (K) and ◦ ∈ ⊂ H ∈ H C p (K) p for all p [z] . (2.4) k kH ≤ k kH ∈

Furthermore, note that the map z Pn(z) n is one-to-one, as Pn(z) n = Pn(w) n implies p(z) = p(w) for all p C[z]7→and hence z = w. ∈

2 2.1.9 Corollary. (P ) is an onb of (K) if and only if P (E) is total in ` , i.e. W ⊥ = 0 . n n H { }

This was just the adaption of 2.1.4 to the situation above. 16 2 ORTHOGONALITY IN RKHS

Obviously, if E = then the set P : n N is linearly independent in (K) and, | | ∞ { n ∈ 0} H due to the last assertion in 2.1.1, C[z] is dense in (K). H Therefore, we can apply Gram-Schmidt orthonormalization to (P ) in (K) obtaining n n H an onb (Qn)n 0 of (K) where Qn is a polynomial of degree n and ≥ H

K(z, w) = Pn(z) Pn(w) = Qn(z) Qn(w) for all z, w E . n 0 n 0 ∈ ≥ ≥ P P 1 1 Note that Q0(z) = P P0(z) = P ; hence (Qn)n need not match 1.1.1. k 0kH(K) k 0kH(K)

2 2.1.10 Series indexed by N0. Let α = (αn)n 0 ` and f := αn Pn (K). Then ≥ ∈ n 0 ∈ H ≥ P ∞ f(z) = αn Pn(z) = αn Pn(z) for z E . n 0 n=0 ∈ ≥ P P We will use the latter notation especially when we regard this as a series of functions deﬁned on some subset of C without using the Hilbert space structure of (K); it may also represent a convergent but not absolutely convergent series for some αH / `2(N ) or ∈ 0 z / E, when the former does not make sense. ∈

2.1.11 Example (Weighted Shift). For n N , let P (z) := b zn where b C 0 ∈ 0 n n n ∈ \ { } and b0 = 1. Then ∞ 2 2n E = z C : bn z < ∈ n=0 | | | | ∞ n o is an either open or closed disk centered atP0 (including the cases E = 0 and E = C). { } We now consider the case E = 0 . For m N , (2.2) yields 6 { } ∈ 0

m ∞ ∞ n bmz = Pm(z) = Pn , Pm (K) Pn(z) = Pn , Pm (K) bnz for all z E . n=0h iH n=0h iH ∈ P P Now uniqueness of power series implies Pn , Pm (K) = δnm. Thus (Pn)n is an ortho- h iH normal system in (K). Moreover, according to 2.1.1, this sytem is total in (K) and, consequently, an onHb. H Here the multiplication operator D, deﬁned as in 1.2.1, is a weighted shift: In its matrix bn representation, dn+1,n = for all n N0 and dij = 0 if i = j + 1. bn+1 ∈ 6 We will return to this situation in 2.2.4 concerning analycity of the functions in (K) and in section 4.4 regarding subnormality of weighted shifts. H Now we turn to the special case b = 1 for all n N . Here E = z C : z < 1 and n ∈ 0 { ∈ | | } ∞ n 1 K(z, w) = (z w) = 1 z w for z, w E. n=0 − ∈ n 2 Recall 1.2.2Pand 1.3.8 where we have seen that (z )n Z is an onb in Lµ where µ is the ∈ normalized one-dimensional Lebesgue measure on the unit circle and we can identify (K) C 2 H with the closure of [z] in Lµ. 17

1 n 1 2.1.12 Example. Let P : 1 and P (z) := z − (z n) for n 1. Then 0 ≡ n √n! − ≥

∞ 2 ∞ 1 2n 2 2 ∞ 1 2n 2 2 2 Pn(z) = 1 + n! z − z n 1 + n! z − 2 z + n n=0 | | n=1 | | | − | ≤ n=1 | | · | | P P 2 P ∞ 1 2n ∞ n 2n 2 C = 1 + 2 n! z + 2 n! z − < for all z . n=1 | | n=1 | | ∞ ∈ P P Hence E = C and

∞ ∞ 1 n 1 K(z, w) = Pn(z) Pn(w) = 1 + n! (z w) − (z n)(w n) n=0 n=1 − −

P ∞ 1 n P ∞ 1 n 1 ∞ n n 1 = 1 + n! (z w) (z + w) (n 1)! (z w) − + (n 1)! (z w) − n=1 − n=1 − n=1 − P P P = exp(z w) (z + w) exp(z w) + (z w + 1) exp(z w) − = 1 + (1 z)(1 w) exp(z w) for z, w C . − − ∈ We next show that (P ) is not an on b in (K). n n H 1 2 For n N0 set xn := . Then x := (xn)n 0 ` and ∈ √n! ≥ ∈

∞ ∞ 1 n 1 xn Pn(z) = 1 + n! z − (z n) n=0 n=1 − P P ∞ 1 n ∞ 1 n 1 C = 1 + n! z (n 1)! z − = exp(z) exp(z) = 0 for all z . n=1 − n=1 − − ∈ P P Thus 0 = x W ⊥ and 2.1.9 implies that (P ) is not an onb in (K). Moreover, in 4.5.2 6 ∈ n n H we will be able to conclude that there exists no om for (Pn)n.

These two examples show that, for (Pn)n 0 as in 1.1.1, both cases – being an onb in ≥ (K) or not – can occur. If, in particular, E is ﬁnite then (K) is ﬁnite-dimensional H H and (Pn)n 0 cannot be an onb in (K). It may also happen that E = ∅. ≥ H

2.2 Analytic Kernels

If the polynomials (P ) form an onb in (K) then this space is isometrically isomorphic n n H to the abstract space defined in section 1.2 via Pn Pn, and if there exists an om µ H2 7→ for (Pn)n then also to Pµ . We observe that, on the one hand, point evaluation in (K) is a continuous linear func- H 2 tional while, on the other hand, point evaluation in general is not even defined in Lµ. In this section we will state some sufficient conditions for the functions of (K) being analytic – at least in some subset G E – in order to characterize (K) Has a proper ⊂ 2 H (because in general not every element of Lµ has an analytic representative) subspace of 18 2 ORTHOGONALITY IN RKHS

2 Lµ. In this situation, D is sometimes said to have an analytic model, see e.g. [StSz3]. At this point, however, we do not know any connections between the domain E of (K) and the support of the om µ yet; they may be disjoint, which actually can happen,Hrecall n 1.2.2 and 1.3.8 where Pn(z) = z , supp(µ) is the unit circle, and E is the open unit disk; see also section 4.2.

n 2.2.1 Notations. For A C set A∗ := z : z A and for p(z) = a0 + a1z + + anz , ⊂ { ∈ n } · · · deﬁne p∗ C[z] by p∗(z) := a + a z + + a z . ∈ 0 1 · · · n Let now (Pn)n 0 be as in 1.1.1; we deﬁne (K) as in 2.1.8 and regard C[z] as a dense ≥ H (due to the last assertion in 2.1.1) linear subspace of (K). H Furthermore, set κ : E C, κ(z) := K(z, z) and → C H : E∗ E , H(z, w) := K(z, w) = Pn∗(z) Pn(w) . × → n 0 ≥ P Note that κ(z) 1 for all z E and that κ need not be a member of (K). ≥ ∈ H

2.2.2 Lemma. Let G be an open non-empty subset of E such that all f (K) are ∈ H holomorphic in G. Then H is holomorphic in G∗ G. In particular, κ is continuous in G. ×

Proof: For fixed z G∗, the function H(z, ) = K (K) is holomorphic in G. ∈ · z ∈ H Now fix w G and define h : G∗ C, h (z) := H(z, w) = K(z, w) = K(w, z) = K (z). ∈ w → w w Clearly, hw is holomorphic in G∗, as Kw is holomorphic in G. Recall that functions of several complex variables are holomorphic if and only if they are holomorphic in each variable seperately, see e.g. [GR, I.A.2].

In particular, H is continuous in G∗ G which implies that z κ(z) = H(z, z) is continuous in G. × 7→

Remark. Note that we did not have to require that (K) contains polynomials here. H

2.2.3 Lemma. Let G be an open non-empty subset of E and suppose that for every compact K G there exists cK > 0 such that κ(z) cK for all z K. Then all f ⊂ (K) are holomorphic in G. In particular,≤ K is∈holomorphic in G for ∈ H z every z E. ∈ Moreover, H is holomorphic in G∗ G with ×

∂zH(z, w) = Pn∗0(z) Pn(w) , ∂wH(z, w) = Pn∗(z) Pn0 (w) . n 0 n 0 ≥ ≥ P P Proof: As C[z] is dense in (K), for arbitrary f (K), there exists a sequence (qn)n N C H ∈ H ∈ in [z] such that f qn (K) 0 as n . Then qn f uniformly on K (see 1.4.4). k − kH → → ∞ → According to the Weierstrass convergence theorem (see e.g. [Re, 8. 4]), the limit of a pointwisely convergent sequence of holomorphic functions G C §which is uniformly → convergent on every compact K G, is holomorphic in G and q0 f 0 uniformly on ⊂ n → every compact subset of G. 19

n Now ﬁx w E. For f = H( , w) and qn = Pk(w) Pk∗, we obtain qn0 f 0 on G∗. Hence ∈ · k=0 → P d dz H(z, w) = Pn∗0(z) Pn(w) . n 0 ≥ P Note that, for ﬁxed z E∗, we get an analogous result concerning the derivative of H with respect to w on G∈. Therefore, H is holomorphic in G∗ G with partial derivatives as asserted. ×

For (z , w ) G∗ G, there exist coeﬃcients a C and r > 0 such that 0 0 ∈ × jk ∈

∞ ∞ j k K(z, w) = H(z, w) = ajk(z z0) (w w0) for z z0 < r , w w0 < r . j=0 k=0 − − | − | | − | P P If, for example, G = E is an open disk centered at z0 then we can write

∞ ∞ j k K(z, w) = ajk(z z0) (w z0) for all z, w E j=0 k=0 − − ∈ P P with some a C. Here, K is oftene referred to as an analytic kernel. jk ∈ e 2.2.4 Weighted Shift (continued). For n N let P (z) := b zn, where b = 1 and ∈ 0 n n 0 bn C 0 for n > 0, see also 2.1.11. Assume E = 0 ; then E is an open or closed disk cen∈tered\ {at}0 or E = C and (b zn) `2 for all z 6 E{ .}Now, for x = (x ) `2 we set n n ∈ ∈ n n ∈

∞ n fx : E C , fx(z) := xnbnz → n=0 P and := f : x `2 . Then β(x) := f deﬁnes linear map β : `2 which is F { x ∈ } x → F one-to-one, as fx = 0, i.e. fx(z) = 0 for all z E, implies xnbn = 0 for all n and hence x = 0. ∈ As β is onto by construction, β is a bijective map `2 . We deﬁne an inner product 2 → F n on by fx , fy := x , y `2 . Then = β(` ) and, in particular, bnz : n N0 is an F onbFin h. i h i F { ∈ } F For ζ E, set k(ζ) := ( b ζn ) . This yields ∈ n n

∞ n fk(ζ) , fy = k(ζ) , y `2 = bnζ yn = fy(ζ) for all fy and all ζ E . h iF h i n=0 · ∈ F ∈ P Therefore, is an RKHS with kernel F

∞ n n ∞ K(z, w) = fk(w) , fk(z) = bnw bnz = Pn(z) Pn(w) h iF n=0 n=0 P P and thus coincides with (K) when deﬁned as in 2.1.8. H Furthermore, every f (K) = is analytic. ∈ H F 20 2 ORTHOGONALITY IN RKHS

2.2.5 Lemma. Let (K) be an RKHS with domain E such that all f (K) are H ∈ H holomorphic in G E. ⊂ Then, for z G, there exists L (K) such that L , f = f 0(z) for all f (K). ∈ z ∈ H h z i ∈ H Proof: Fix z G and choose a sequence (h ) in C 0 such that z + h G for all n ∈ n n \ { } n ∈ and h 0 as n . Then, for all f (K), n → → ∞ ∈ H 1 f(z+hn) f(z) Kz+hn Kz , f = − f 0(z) as n . hn − hn → → ∞ A well-known corollary of theBanac h-Steinhaus theorem, see e.g. [W, 4.24 d], implies 1 that there exists Lz (K) such that Lz , f = lim Kz+hn Kz , f for all n hn ∈ H h i →∞ − f (K). This yields L , f = f 0(z). ∈ H h z i

Suppose now that (K) is deﬁned via polynomials as in 2.1.8 and G E exists such that all f (K) areH holomorphic in G. Let L be as in 2.2.5. ⊂ ∈ H z

2 2.2.6 Theorem. Let z G. Then Lz = Pn0 (z) Pn. In particular, Pn0 (z) n ` . ∈ n 0 ∈ ≥ P Proof: According to 2.2.5, L , f = f 0(z) for all f (K). h z i ∈ H 2 Using 2.1.5, we get ( Pn , Lz )n ` and Lz = Pn , Lz Pn = Pn0 (z) Pn. h i ∈ n 0h i n 0 ≥ ≥ P P (n) (n) Remark. Analogously, one can deﬁne higher-order derivatives Lz satisfying Lz , f = f (n)(z). h i

If the interior E◦ of E is non-empty, the question arises whether all f (K) are ∈ H holomorphic there. Recall that f can be approximated (in (K) and therefore also pointwisely) by a sequence of polynomials. According to OsgoHod’s theorem, see [Osg, Theorem I] and also [BM], the limit of a sequence of holomorphic functions which is pointwisely convergent in an open set U, is holomorphic in an open set that is dense in U. In the following, we will be able to chracterize a dense open subset of E◦ in our particular situation where all f (K) are holomorphic; some intermediate results will be useful later on, too. ∈ H

2.2.7 Lemma. For R > 0, the set A := z E : κ(z) R is closed in C. R { ∈ ≤ }

Proof: Let (zm)m N be a convergent sequence in C such that zm AR for all m N and ∈ ∈ ∈ set z0 := lim zm. Fix N N; then m →∞ ∈ N 2 ∞ 2 Pn(zm) Pn(zm) = κ(zm) R for all m N n=0 | | ≤ n=0 | | ≤ ∈ P P N 2 and continuity implies Pn(z0) R. Now N yields κ(z0) R. n=0 | | ≤ → ∞ ≤ Thus z A and A isPa closed subset of C. 0 ∈ R R 21

Note that Ar AR whenever 0 < r < R and E = Ak is an Fσ-set. k N ⊂ ∪∈ Suppose that κ is bounded. In other words, there is R > 0 such that κ(z) R for all z E. Then E = A is a closed subset of C. ≤ ∈ R

2.2.8 Corollary. If E is not closed then κ is unbounded. More generally, if (zn)n N is ∈ a sequence in E such that z z / E as n then κ(z ) . n → 0 ∈ → ∞ n → ∞

Proof: Assume that κ(zn) 9 as n . Then there exists a subsequence (zn )k N ∞ → ∞ k ∈ and R > 0 such that κ(znk ) R for all k. Thus znk AR for all k. As AR is closed, this implies z A E and yields≤ a contradiction. ∈ 0 ∈ R ⊂ Remark. If E is closed, κ still may be unbounded. In 4.1.3 and 4.1.4 we will be able to construct examples where E is a countable compact set and κ is unbounded.

2.2.9 Proposition. Suppose E◦ = ∅ and set G := Ak◦ E◦. k N 6 ∪∈ ⊂ Then G is dense in E◦ (in particular, G = ∅) and all f (K) are holomorphic in G. 6 ∈ H

Proof: Take z E◦ and let F be a closed disk centered at z and contained in E. Then 0 ∈ 0 F = F Ak = (F Ak). k N k N ∩ ∪∈ ∪∈ ∩ According to Baire category theory, applied to the complete metric space F , there exists k N such that F A◦ = ∅ and, in particular, F G = ∅. ∈ ∩ k 6 ∩ 6 As z0 and F were chosen arbitrarily, this proves that G is dense in E◦.

If Ak◦ = ∅ for some k then, by 2.2.3, all f (K) are holomorphic in Ak◦. Therefore, all 6 ∈ H f (K) are holomorphic in Ak◦ E◦. k N ∈ H ∪∈ ⊂

It remains an open question whether the case G = E◦ can occur. We will show that κ 6 obeys a maximum principle in E◦ and that if z E◦ G exists then κ must be unbounded 0 ∈ \ in every neighborhood of z0.

2.2.10 Corollary. In the situation of 2.2.9, suppose that E◦ G = ∅. Then, for every \ 6 z0 E◦ G, there exists a sequence (zn)n N in G such that zn z0 and κ(zn) as ∈ n ∈ . \In particular, κ is not continuous at z . → → ∞ → ∞ 0

Proof: Choose a sequence (εn)n N such that εn > 0 for all n, εn 0 as n , and ∈ → → ∞ U := z C : z z < ε E◦ for all n. n { ∈ | − 0| n} ⊂ Now, for ﬁxed n, assume that U A . This implies U A◦ G in contradiction to n ⊂ n n ⊂ n ⊂ z0 / G. Thus Un (C An) = ∅. As this is an open set contained in E◦, it must have non-empt∈ y intersection∩ \with the6 dense set G. Hence there exists z U (C A ) G for every n N. n ∈ n ∩ \ n ∩ ∈ By construction, the sequence (zn)n N has the asserted properties, the discontinuity of κ ∈ at z0 is an immediate consequence. 22 2 ORTHOGONALITY IN RKHS

2.2.11 Lemma. For R > 0, we have κ(z) < R whenever z A◦ . ∈ R

Proof: Assume A◦ = ∅ and fix z0 A◦ . R 6 ∈ R Let U be an open disk centered at z such that U A . Then K is continuous in 0 ⊂ R z0 U and holomorphic in U. Using the maximum principle we find z1 ∂U such that K (z) < K (z ) for all z U and hence κ(z ) = K (z ) < K (z ) .∈ | z0 | | z0 1 | ∈ 0 z0 0 | z0 1 | Analogously, we now apply the maximum principle to Kz1 and find z2 in ∂U such that K (z ) < K (z ) . | z1 0 | | z1 2 | Finally,

κ(z ) < K (z ) = K (z ) < K (z ) 0 | z0 1 | | z1 0 | | z1 2 | 1 1 = K , K K K = κ(z ) 2 κ(z ) 2 R . |h z2 z1 i| ≤ k z2 k · k z1 k 1 2 ≤

Thus κ(z) < R for all z A◦ . ∈ R

2.2.12 Maximum Principle for κ.

In E◦, κ does not attain a local maximum.

Proof: Assume that there exists a local maximum M := κ(z ) at z E◦, i.e. there 0 0 ∈ exists an open disk U E◦ centered at z such that κ(z) M for all z U which implies ⊂ 0 ≤ ∈ U A and thus z A◦ . Now 2.2.11 implies κ(z ) < M which is a contradiction. ⊂ M 0 ∈ M 0

2.2.13 Lemma. Let f (K) and R > 0. The restriction f A is continuous on A . ∈ H | R R

Proof: In analogy to the proof of 2.2.3, we can ﬁnd a sequence (qn)n in C[z] such that qn f in (K) as n and qn AR f AR uniformly on AR. As the uniform limit of a sequence→ Hof continuous→ ∞functions|is con→tin|uous, the proof is complete.

Remark. Note that we cannot conclude that f is continuous on E. In 4.2.10 we will see that (K) may contain functions which are not continuous on E. H In 3.5.6 we will be able to show that G defined in 2.2.9 is precisely the largest open set contained in E◦ where all f (K) are holomorphic. We will then give characterizations of that set by other means as∈ Hwell. As for now, we conclude this section by a different approach to finding subsets of E where all f (K) are holomorphic. ∈ H

2.2.14 Theorem. Let γ be a simply closed rectiﬁable curve in AR for some R > 0 and denote by V the bounded component of C γ. \ Then V E and all f (K) are holomorphic in V and continuous in V . ⊂ ∈ H 23

Proof: Fix z V . Then there exists d > 0 such that z z d for all z γ. 0 ∈ | − 0| ≥ ∈ Now let a = (a ) `2. Using the Cauchy integral formula, for m > n we have k k ∈ m m m 1 1 1 1 akPk(z0) = akPk(z) dz akPk(z) dz 2πi z z0 2π d k=n γ − k=n ≤ γ k=n | | P P P R m 1 m R1 m 1 1 2 2 2 2 2 2 √R 2πd ak Pk(z) dz ak 2πd l(γ) ≤ k=n | | k=n | | | | ≤ k=n | | · γ R P P P ∞ where l(γ) denotes Euclidean length of γ. This shows that akPk(z0) exists, and k=0 P ∞ √R akPk(z0) a `2 2πd l(γ) k=0 ≤ k k · ·

P ∞ 2 implies that a akPk(z0) is a continuous linear functional; thus Pk(z0) k ` . 7→ k=0 ∈ In other words, z P E. Therefore, V = V γ E. 0 ∈ ∪ ⊂ For f (K), according to 2.1.5, f = Pk , f (K)Pk. Set ∈ H k 0h iH ≥ nP fn := Pk , f (K)Pk k=0h iH P for n N. The sequence (f ) is convergent in (K), hence pointwisely in V and uni- ∈ n n H formly, as γ is contained in AR (see 1.4.4), on ∂V = γ. It is a well known fact (see [Re, 8.5.4], for instance) that then the limit is holomorphic in V and continuous in V .

The previous theorem can also be found in [StSz3, Theorem 8]. If we do not require κ to be bounded but at least integrable on γ, we can prove a similar result:

2.2.15 Theorem. Let γ be a simply closed rectiﬁable curve in E such that κ(z) dz < . Denote by V the bounded component of C γ. γ | | ∞ \ Then V E and all f (K) are holomorphic in V . R ⊂ ∈ H Proof: For any compact set K V there exists d > 0 such that z w d for all z K and all w γ. Then ⊂ | − | ≥ ∈ ∈ l(γ) 1 dζ for all z K ζ z 2 d2 γ | − | | | ≤ ∈ where l(γ) denotes EuclideanR length of γ. Now, for n N and z K, we have ∈ ∈ 2 2 2 1 Pn(ζ) 1 Pn(ζ) Pn(z) = 2πi ζ z dζ 4π2 ζ z dζ | | − ≤ − | | γ γ R R l(γ) 1 P (ζ) 2 dζ 1 dζ P (ζ) 2 dζ . 4π2 n ζ z 2 4π2d2 n ≤ γ | | | | · γ | − | | | ≤ γ | | | | R R R 24 2 ORTHOGONALITY IN RKHS

Using dominated convergence, we obtain

2 l(γ) 2 l(γ) κ(z) = Pn(z) 4π2d2 Pn(ζ) dζ = 4π2d2 κ(ζ) dζ =: cK n 0 | | ≤ n 0 | | | | | | ≥ γ ≥ γ P R P R for all z K. Thus K E. Moreover, by 2.2.3, all f (K) are holomorphic in V . ∈ ⊂ ∈ H Note that in 2.2.15 we do not have to require γ E. There might exist a nullset (with 2 ⊂ respect to line measure) where Pn(z) = . n 0 | | ∞ ≥ P

2 2.3 Analytic Functions Forming Subspaces of Lµ

In this section we construct RKHS whose elements are analytic functions which are square integrable with respect to some suitable measure µ. If all polynomials in C[z] are square integrable with respect to µ and µ(C) = 1 then we can use Gram-Schmidt orthonormalization to obtain a system of polynomials as in 1.1.1 such that µ is an om. Hence, in this case, we will be able to embed (K), deﬁned as in 2.1.8, into L2 . H µ

2 B C 2.3.1 The Space Aµ. Let µ be a measure on ( ), absolutely continuous with respect to Lebesgue measure, µ = αλ, and U be an open subset of C. 1 Furthermore, assume that α dλ < for all compact B U. B ∞ ⊂ Now deﬁne R A2 := f holomorphic in U : f(z) 2 dµ < . µ | | ∞ n U o 2 R 2 Obviously, Aµ is a vector space and inherits an inner product from Lµ. We will show 2 that Aµ is an RKHS which is a well-known fact but sometimes proved under stronger conditions, see [Hall, Theorem 2.2], for example, where α is supposed to be continuous. See also [Ga, Chapter I] or [Con2, 29] where U is bounded and µ is normalized Lebesgue 2 § measure on U. In this case, Aµ is often referred to as Bergman space. For a generalization of Bergman spaces to unbounded domains of ﬁnite Lebesgue measure, see [CJK].

In the following, for z C and r > 0, set B (z) := ζ C : ζ z r . ∈ r { ∈ | − | ≤ }

2.3.2 Lemma. For every z U, there exists c > 0 such that ∈ z 2 f(z) cz f 2 for all f Aµ . | | ≤ k kLµ ∈ 25

Proof: Let z U. Then there exists r > 0 such that B := B (z) U. ∈ r ⊂ Now we have 1 1 2 1 2 > α dλ = 1B α α dλ = 1B α dµ ∞ B U · · U · R R R which implies 1 1 L2 . Using the mean v alue theorem, for f A2 , we get B · α ∈ µ ∈ µ

1 1 2 1B , f 2 = 1B f α dλ = f dλ = r π f(z) α Lµ α · U · · · B R R 1 1 and Cauchy-Schwarz yields f(z) 2 1B 2 f L2 . | | ≤ r π · α Lµ · k k µ 1 1 Therefore, set cz := 2 1B 2 to conclude the proof. r π · α Lµ

2.3.3 Lemma. If Br(z0) U then there exists c > 0 such that f(z) c f 2 for all ⊂ | | ≤ k kLµ f A2 and all z B (z ). ∈ µ ∈ r 0

Proof: There exists R > r such that Br(z0) BR(z0) U. Set ρ := R r, then ⊂ ⊂ − r Bρ(z) BR(z0) for all z Br(z0) . ⊂ ∈ z0 R Furthermore, 1 1 L2 , and, for arbitrary z B (z ), U Br(z0) · α ∈ µ ∈ r 0 we have

1 1 Figure 1. 1B (z)(w) 1B (z0)(w) for all w U ρ · α(w) ≤ R · α(w) ∈ On the proof of 2.3.3 implying 1 1 1Bρ(z) 2 1BR(z0) 2 for all z Br(z0) . · α Lµ ≤ · α Lµ ∈

In analogy to 2.3.2, we get

1 1 1 1 f(z) 2 1Bρ(z) 2 f L2 2 1BR(z0) 2 f L2 | | ≤ ρ π · α Lµ · k k µ ≤ ρ π · α Lµ · k k µ =: c 2 for all f A and all z Br(z0). ∈ µ ∈ | {z }

2 2.3.4 Proposition. The space Aµ is an RKHS.

2 2 Proof: According to 2.3.2, point evaluation in Aµ is continuous. Regarding Aµ as a 2 2 2 subspace of Lµ, it only remains to show that Aµ is closed in Lµ. 2 Let (fn)n N be a Cauchy sequence in Aµ. ∈ For any closed disk B U, due to 2.3.3, there exists c > 0 such that ⊂

fn(z) fm(z) c fn fm 2 for all z B and m, n N . | − | ≤ k − kLµ ∈ ∈ 26 2 ORTHOGONALITY IN RKHS

Hence (fn)n is uniformly convergent on every closed disk contained in U. Denote the limit by f and note that f : U C is holomorphic. → 2 2 Therefore, Aµ is a Hilbert space. Using 1.4.3, we see that Aµ is an RKHS.

2 2.3.5 Orthogonal Polynomials in Aµ. Suppose now that µ(U) = 1 and that all p C[z] are square integrable with respect to µ. Then we can regard C[z] as a subspace of∈A2 . Furthermore, Gram-Schmidt orthonormalization applied to 1, z, z2, . . . yields an µ { } orthonormal system (Pn)n 0 of polynomials with P0 = 1 and deg Pn = n for n 1. ≥ 2 ≥ There exists an onb (ek)k I of Aµ, I a suitable index set, that extends (Pn)n 0, i.e. for N ∈ ≥ every n 0, there exists kn I such that Pn = ekn . ∈ 2 ∈ Let K be the kernel of Aµ, then K(z, w) = ek(z) ek(w) by 1.4.6. In particular, k I ∈ P e 2 e 2 Pn(z) ek(z) < for all z U . n 0 | | ≤ k I | | ∞ ∈ ≥ ∈ P P Therefore, U E. ⊂ Remark. For bounded U and continuous α a proof of the last statement, U E, is also ⊂ given in [GTV, Proposition 2].

2 2 2 Note that Aµ is a proper subspace of Lµ, as Lµ contains elements which do not have a 2 holomorphic representative. In particular, (Pn)n is not an onb in Lµ.

2.3.6 Lemma. In the situation of 2.3.5, (P ) is an onb in (K). n n H

Proof: Let m N0. According to 2.1.5, Pm = Pn , Pm (K)Pn holds in (K) and, ∈ n 0h iH H by (2.3), ≥ P Pm(z) = Pn , Pm (K)Pn(z) for all z E . n 0h iH ∈ ≥ P 2 2 In particular, Pn , Pm (K) n ` ; thus f := Pn , Pm (K)Pn Aµ exists. h iH ∈ n 0h iH ∈ ≥ 2 P As point evaluation in Aµ is continuous, f(z) = Pn , Pm (K)Pn(z) holds for all z U. n 0h iH ∈ ≥ P Using U E, we obtain f(z) = P (z) for all z U; hence f = P in A2 . ⊂ m ∈ m µ 2 As (Pn)n is an onb in Aµ, this implies Pn , Pm (K) = δmn, completing the proof. h iH

2.3.7 Example (Bergman space on the unit disk). Denote by D the open unit disk and let µ be normalized Lebesgue measure on D. n A short calculation shows that the polynomials Pn(z) := √n + 1 z , n N0, form an orthonormal system in L2 . Clearly, C[z] A2 . It is well-known that ∈here P 2 = A2 µ ⊂ µ µ µ holds, see [Halm, Problem 25], for instance. According to 2.3.6, (K) is isometrically isomorphic to A2 via P P . Moreover, H µ n 7→ n n 1 K(z, w) = (n + 1)( zw) = (1 zw)2 . n 0 ≥ − P 27

In the following, we will turn to a similar situation where µ is not supported on an open set but on its boundary.

2.3.8 A2 -spaces on a Curve. In the remainder of this section, suppose that γ : [0, 1] µ → C is a simply closed piecewise continuously diﬀerentiable curve. For abbreviation, we denote γ([0, 1]) by γ again. Let µ be absolutely continuous w.r.t. line measure on γ with density α and assume 1 α(z) dz < . γ | | ∞ 2 R Now < 1 dz = 1 dµ yields 1 L2 . ∞ α(z) | | α α ∈ µ γ

Moreover, letR V be the boundedR component of C γ. \

2.3.9 Lemma. For any compact L V , there exists c > 0 such that ⊂

f(z) c f 2 ≤ · Lµ for all z L and all functions f which are holomorphic in an open set containing V ∈ and satisfy f 2 dµ < . | | ∞ Proof: ClearlyR , d := inf z w : z L, w γ > 0. | − | ∈ ∈ Let now z L. Then, using the Cauchy integral formula, we get ∈ 1 1 1 f(γ(t)) 1 1 f(z) = γ0(t) dt f γ(t) γ0(t) dt = f(ζ) dζ 2πi γ(t) z ≤ 2πd · 2πd | | 0 − 0 γ 1 R 1 R 1 1 1 1 R = f(ζ) α(ζ ) dζ = f , 2 f 2 2πd α(ζ) 2πd α 2 2πd α Lµ Lµ γ | | | | Lµ ≤ D E R 1 1 for all z L. In particular, the constant c := 2 is independent of f. ∈ 2πd α Lµ

2.3.10 Lemma. Let U V be open and (fn)n N be a sequence of holomorphic functions ∈ f : U C such that ⊃ n → f 2 dµ < for all n | n| ∞ R 2 and (fn)n is convergent w.r.t. the Lµ-norm.

Then f : V C, f(z) := lim fn(z), exists and is holomorphic in V . n → →∞ Proof: For any compact set L V , according to 2.3.9, there exists c > 0 such that ⊂

fn(z) fm(z) c fn fm 2 for all z L . − ≤ − Lµ ∈ Hence f f uniformly in every compact subset of V which implies that f is holomorphic n → in V . 28 2 ORTHOGONALITY IN RKHS

Note that if, in particular, µ(γ) = 1 then every p C[z] is square integrable w.r.t. µ and, 2 ∈ applying Gram-Schmidt in Lµ to the monomials, we obtain an orthonormal system (Pn)n such that P 1 and deg(P ) = n for all n. 0 ≡ n

2.3.11 Theorem. In the situation of 2.3.8, assume that µ is an om for (Pn)n 0. Then 2 2 ≥ Pn(z) < for all z V and, for any a = (an)n 0 ` , n 0 | | ∞ ∈ ≥ ∈ ≥ P q(z) := anPn(z) n 0 ≥ P deﬁnes a holomorphic function q : V C. → m Proof: Clearly, qm : C C, qm(z) := anPn(z) is an entire function for every m N. → n=0 ∈ 2 2 Moreover, as (Pn)n is an orthonormal Psystem in L , we have qm q in L as m µ → µ → ∞ and, according to 2.3.10, q : V C, q(z) := anPn(z), is holomorphic in V . → n 0 ≥ P Now ﬁx z V . By 2.3.9, there exists c > 0 such that qm(z) c qm 2 for all m. This ∈ | | ≤ k kLµ yields

anPn(z) c anPn = c a 2 2 ` n 0 ≤ n 0 Lµ k k ≥ ≥ P P which shows that a anPn(z) is a con tinuous linear functional. 7→ n 0 ≥ P 2 Therefore, by the Riesz representation theorem, we obtain Pn(z) n ` to complete the proof. ∈

Remark. For continuous α, this result can be found in [Sze, Theorem 16.3] and is also part of [GTV, Proposition 2].

2 Note that, by 2.3.11, every f Pµ can in a natural way be deﬁned pointwisely in V and represents a holomorphic function∈ there. One can show that this gives rise to an RKHS whose domain contains V . We will do this in a more general situation in section 3.6; in 2 2 particular, see 3.6.8. See also 3.3.7 where we can already conclude that Pµ $ Lµ and section 4.2 for the special case that γ is the unit circle. 29

3 Relations between the Spaces , (K), and L2 H H µ

Throughout this chapter, (Pn)n 0 always denotes a sequence of polynomials as in 1.1.1; ≥ the Hilbert space and the multiplication operator D are defined as in 1.2.1. H 2 Furthermore, recall E = z C : Pn(z) < and define (K) as in 2.1.8. ∈ n 0 | | ∞ H ≥ P By construction, (Pn)n is an onb of . Moreover, these polynomials are total in (K) but need not be orthogonal there, seeH section 2.1. If they are orthogonal then Hand H (K) are isometrically isomorphic via the identity map on C[z]. H 2 C 2 2 Finally, if an om µ exists then let Pµ be the closure of [z] in Lµ and Pµ is isometrically isomorphic to via P P , too. Note that in this case D is subnormal and the H n 7→ n multiplication operator Mµ is a minimal normal extension of D, see 1.3.7. As mentioned in 1.4.3, a Hilbert space consisting of functions E C is an RKHS if and → only if point evaluation is a bounded linear functional for every z E; therefore, if (Pn)n 2 ∈ is an onb in (K) – as well as in Pµ , here by definition – the elements of E are commonly H 2 referred to as the bounded point evaluations of Lµ, see e.g. [Con1] or [Tr].

3.1 Preliminaries

We start with some deﬁnitions concerning linear operators in Hilbert space. For more details, see chapter A.1 in the appendix.

3.1.1 Deﬁnitions. Let A be a densely deﬁned linear operator in a Hilbert space. We denote by dom(A), ran(A), and (A) its domain, range, and null-space, respectively. Let N σ(A) be the spectrum of A, and ρ(A) its resolvent set, ρ(A) = C σ(A). If A is closable then let A denote its closure. \ A point z C is called a regular value of A, if there exists a constant c > 0 such that (A z id)∈f c f for all f dom(A). The set reg(A) of all regular points of A is the k − k ≥ k k ∈ domain of regularity of A.

The space ran(A z id)⊥ is called deﬁciency space of A in z, its dimension is the − deﬁciency index of A in z. 30 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

The spectrum σ(A) can be decomposed into three disjoint subsets,

(i) the point spectrum σ (A) := z C : (A z id) is not one-to-one. , p { ∈ − } (ii) the continuous spectrum

1 σ (A) := z C : (A z id)− is densely defined and not continuous , c { ∈ − } 1 (iii) the residual spectrum σ (A) := z C : (A z id)− is not densely defined ; r { ∈ − } 1 in (ii) and (iii) the operator (A z id) is supposed to be one-to-one, i.e. (A z id)− is −1 − well-defined with dom(A z id)− = ran(A z id). − −

Note that at every z σr(A) the deﬁciency index is diﬀerent from 0 and that at every z ρ(A) σ (A) it is ∈0. ∈ ∪ c The following properties are also well-known. A proof can be found in any textbook concerning Hilbert space theory, such as [AG], [EE], or [W], for example.

3.1.2 Proposition. Let A be a densely deﬁned linear operator in a Hilbert space. Then its adjoint A∗ exists and ran(A z id)⊥ = (A∗ z id). − N − The domain of regularity is an open subset of C and ρ(A) reg(A) ρ(A) σ (A). In ⊂ ⊂ ∪ r particular, if A is normal then σr(A) = ∅ and ρ(A) = reg(A). If A is closable then reg(A) = reg(A). Moreover, in any connected subset of reg(A), the deﬁciency index is constant.

For z reg(A), denote by Pz the orthogonal projection onto ran(A z id)⊥. If (zn)n N is ∈ − ∈ a convergent sequence in reg(A) with limit z reg(A) then Pzn Pz∗ 0 as n . ∗ ∈ − → → ∞ Let now A be symmetric. Then C R ρ(A) and the deﬁciency indic es in the upper and lower half-planes are constant. Thus\ ⊂one can restrict to the deﬁciency indices at i. The operator A has a self-adjoint extension in if and only if these are equal and is essentially self-adjoint if and only if they are 0. H

Concerning the following theorem, see also [Kl2, 2.1.5] or [StSz3, Proposition 6]. However, we will state a proof here, too, as it provides an essential connection between the operator D in and the domain E of (K). H H For z E, we can regard ∈ Pn(z) Pn n 0 ≥ as K (K) or as an element of the abstractP space . We will denote the latter by k . z ∈ H H z

3.1.3 Theorem. For z C, the deﬁciency index of D in z is either 0 or 1. In particular, ∈ 0 for z / E , ran(D z id)⊥ = (D∗ z id) = { } ∈ − N − ( C kz for z E . · ∈ 31

Proof: The ﬁrst equality is due to 3.1.2.

Now ﬁx z C and let f (D∗ z id). Using (1.1), for any n N , we get the recursion ∈ ∈ N − ∈ 0 formula

n+1 z f , Pn = z f , Pn = D∗f , Pn = f , DPn = din f , Pi . h i h i h i h i i=0 h i P Starting with c := f , P , again by (1.1), we recursively obtain f , P = c P (z) for h 0 i h n i n all n N . Now the Parseval equation yields ∈ 0

f = Pn , f Pn = c Pn(z) Pn . n 0h i n 0 ≥ ≥ P P Hence c P (z) `2. For z / E, this implies c = 0 and, therefore, f = 0. n n ∈ ∈ C If z E then f= c Pn(z) Pn = c kz. Hence (D∗ z id) = kz. ∈ n 0 N − · ≥ P

3.1.4 Corollary. Let E := E reg(D). This is an open subset of C. reg ∩

Proof: The case Ereg = ∅ is trivial. Otherwise, consider z0 Ereg. As reg(D) is open, there exists an open disk U containing z such that U reg(D∈). According to 3.1.2, the 0 ⊂ deﬁciency index of D is constant in any connected subset of reg(D). Note that, by 3.1.3, the deﬁciency index in z is equal to 1 if and only if z E. This implies U E; hence U E . Thus E is open. ∈ ⊂ ⊂ reg reg

Note that kz , Pn = Pn(z) for all n N0. By linearity, we get the following analogon to the reproh ducingikernel property in ∈ (K). H

3.1.5 Corollary. Let z E. Then kz , p = p(z) for all p C[z]. ∈ h iH ∈

Note that Kz , f (K) = f(z) for all f (K) while in 3.1.5 we can only speak of h iH ∈ H polynomials as, in general, elements of need not be functions. However, to any f K H K ∈ H we can assign a function f : E C satisfying f (z) = kz , f . We will take care of H this construction in section 3.6. → h i

3.1.6 Theorem. If D is bounded then E is bounded. More precisely, if D is bounded then E z C : z D . ⊂ { ∈ | | ≤ k k}

Proof: It is well-known that if D is bounded then D∗ is bounded, too, and D = D∗ , see [W, Satz 4.14], for instance. k k k k

For z E, using 3.1.3, we get z k = z k = D∗k D∗ k implying ∈ | | k zk k zk k zk ≤ k k k zk D∗ z for all z E. Thus z D for all z E. k k ≥ | | ∈ | | ≤ k k ∈ Remark. In the case of the Hermite polynomials, see 4.5.11, D is unbounded and E = ∅. 32 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

Suppose now that µ is an om for (Pn)n 0 and deﬁne Λµ := z C : µ( z ) > 0 . ≥ ∈ { } 2 Note that Λµ is precisely the set of eigenvalues of the multiplication operator M µ in Lµ and Mµ1 z = z 1 z for all z Λµ. In particular, Λµ = σp(Mµ). { } { } ∈ Furthermore, for z E, deﬁne κ(z) := K(z, z). ∈

3.1.7 Lemma. Let µ be an om for (Pn)n 0. Then ≥ 1 µ( z ) κ(z)− for all z C (3.1) { } ≤ ∈ 1 where κ(z)− := 0 if z / E. In particular, Λ E. ∈ µ ⊂ 2 2 In (3.1) equality holds if and only if 1 z Pµ . Moreover, if (Pn)n is an onb in Lµ then equality holds for all z Λ . { } ∈ ∈ µ Proof: Obviously, (3.1) holds for z / Λ . ∈ µ 2 Let now z Λµ. Then 1 z Lµ 0 . Since f , 1 z L2 = f(z) µ( z ), point evaluation ∈ { } ∈ \{ } h { } i µ · { } at z is well-deﬁned for all f L2 . ∈ µ 2 N N Choose an onb (ei)i 0 of Lµ such that for every n 0 there exists in 0 with Pn = ein . ≥ For k N, set ∈ ∈ ∈ k k sk := ei , 1 z ei = µ( z ) ei(z) ei . i=0h { } i { } i=0 P P Then sk 1 z as k . Thus there exists a subsequence (sk )l N such that sk 1 z { } l ∈ l { } µ-almost→everywhere →as ∞l . In particular, → → ∞

kl lim µ( z ) ei(z) ei(w) = 1 z (w) for all w Λµ . l { } →∞ { } i=0 ∈ P With w = z we get

kl 1 1 2 2 2 µ( z ) 1 z (z) = µ( z ) = lim ei(z) = ei(z) Pn(z) = κ(z) (3.2) { } l i=0 | | i 0 | | ≥ n 0 | | { } { } →∞ ≥ ≥ P P P 2 2 where equality holds if and only if 1 z Pµ . If, in particular, (Pn)n is an onb in Lµ then { } ∈ equality holds in (3.2) for every z Λ . ∈ µ To complete the proof, note that, for z Λµ, in (3.1) equality holds if and only if equality holds in (3.2). ∈

n In general, however, Λµ $ E. Let Pn(z) := z for n N0. Recall 2.1.11 where we have found E = z C : z < 1 and 1.2.2 where we have∈seen that, up to a constant factor, { ∈ | | } Lebesgue measure on the unit circle is an om, hence Λµ = ∅.

3.1.8 Lemma. Let µ be an om for (Pn)n 0 and denote by P the orthogonal projection 2 2 ≥ in Lµ onto Pµ . Then P 1 z = µ( z ) kz for all z Λµ. { } { } · ∈ 33

2 Proof: As (Pn)n is an onb of Pµ , we obtain

P 1 z = Pn , 1 z Pn = µ( z ) Pn(z) Pn = µ( z ) kz . { } n 0h { } i n 0 { } { } · ≥ ≥ P P

2 2 R Let us now have a short look at the case that Pµ = Lµ and µ is supported in . Then Mµ is self-adjoint, D is symmetric, and its deﬁciency indices in the upper and lower half- planes are equal, see also section 4.1 for more details. Combining this with 3.1.3, we get the following.

R 2 2 3.1.9 Proposition. Suppose µ is an om for (Pn)n 0 with supp(µ) and Pµ = Lµ. ≥ ⊂ Then either C R E or (C R) E = ∅. The latter is the case if and only if D is \ ⊂ \ ∩ essentially self-adjoint.

In 4.1.6 we will show that if D is symmetric and not essentially self-adjoint then E = C and there exists an om µ such that P 2 = L2 as well as an om ν such that P 2 = L2 . µ µ ν 6 ν

3.2 Subnormal Hessenberg Operators

As already mentioned in 1.3.7, there exists an om for a given sequence (Pn)n 0 of polyno- ≥ mials if and only if the Hessenberg operator D is subnormal. In general, however, there is no canonical method to construct normal extensions of a linear operator, it might even be diﬃcult to determine whether there exist any normal extensions at all.

3.2.1 Lemma. Let N be a normal operator in a Hilbert space which extends a densely L deﬁned operator A, i.e. dom(A) dom(N) and Ax = Nx for all x dom(A). Then A is formally normal. ⊂ ∈

Proof: Fix x dom(N ∗). Then, for all y dom(A), we have x , Ay = x , Ny = ∈ ∈ h i h i N ∗x , y showing x dom(A∗) and A∗x = N ∗x. h i ∈ Therefore, dom(A) dom(N) = dom(N ∗) dom(A∗) and Ax = Nx = N ∗x = ⊂ ⊂ k k k k k k A∗x for all x dom(A). Hence A is formally normal. k k ∈

Note that, if A is subnormal but has no normal extension in the space but in a larger L space then A need not be formally normal and A∗ need not extend N ∗. An example K ⊃ L for a subnormal but not formally normal operator is the unilateral shift, see 1.3.8 for more details.

2 2 3.2.2 Corollary. Let µ be an om for (Pn)n 0. If Pµ = Lµ then D is formally normal. ≥ 34 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

2 2 Proof: As (Pn)n is an onb in Lµ, we can regard D as a densely deﬁned operator in Lµ. Moreover, the multiplication operator Mµ is a normal extension of D, see 1.3.4. Now 3.2.1 completes the proof.

Remark. Although we know that there exist formally normal operators which are not subnormal, see [Sch] or [Cod], we do not know whether this is possible in the case of Hessenberg operators.

3.2.3 Lemma. Let A be a formally normal operator in a Hilbert space . L (i) A is closable and its closure A is formally normal, too.

(ii) If A is bounded then A is essentially normal.

Proof: Due to dom(A) dom(A∗), the adjoint operator A∗ is densely defined in . It is ⊂ L well-known, see e.g. [W, Satz 5.3], that a densely defined operator is closable if and only if its adjoint is densely defined; moreover, then A∗∗ = A and A∗ = A∗. To see that A is formally normal, we follow [Kl1, Satz 1.1.3]. Let x dom(A). Then there exists a sequence (x ) in dom(A) such that x x and Ax = Ax∈ Ax as n . n n n → n n → → ∞ Moreover, as A is formally normal, A∗(x x ) = A(x x ) for all n, m N. This k n − m k k n − m k ∈ shows that (A∗x ) is a Cauchy sequence and, as A∗ is closed, this yields x dom(A∗) as n n ∈ well as A∗x A∗x. We now obtain n →

Ax = lim Axn = lim A∗xn = A∗x ; n n k k →∞ k k →∞ k k k k hence A is formally normal. Let now A be bounded. Then dom(A) = dom(A∗) = . As we have just seen, A is formally normal. Hence here A is normal and A is essentiallyL normal.

3.2.4 Corollary. Let (Pn)n 0 be a sequence of polynomials such that the Hessenberg ≥ operator D is continuous and formally normal. Then there exists a unique om µ, supp(µ) 2 is compact, and (Pn)n is an onb in Lµ.

Proof: According to 3.2.3, D is essentially normal. Now 1.3.7 yields existence and 2 uniqueness of µ as well as (Pn)n being an onb in Lµ. Moreover, D = Mµ is continuous, too, which implies σ(Mµ) = supp(µ) is compact.

We will now have a closer look at the matrix representation of the Hessenberg operator D to determine a necessary condtition for D being subnormal as well as a necessary and sufficient condition for D being formally normal. According to 1.2.1, D is a densely defined operator in with dom(D) = C[z] where is H H the abstract completion of C[z] with respect to the inner product defined by Pn , Pm := h iH δn,m. Note that (1.1) yields dij = Pi , DPj for i, j N0. Recall that dij = 0 whenever h iH ∈ i > j + 1. 35

th 2 2 For i N0, let ei be the i standard unit vector of ` and denote by β : ` the ∈ 2 → H1 isometric isomorphism given by β(ei) := Pi. We deﬁne an operator in ` by D2 := β− Dβ with dom(D2) = (xn)n 0 : xi = 0 for ﬁnitely many i =: `0. { ≥ 6 } N Clearly, ei , D2ej `2 = dij for all i, j 0 and (D2x)i = dijxj. ∈ j i 1 ≥ − P

3.2.5 Lemma. The domain of D2∗ is given by

2 ∞ 2 dom(D2∗) = (yn)n ` : dkj yk ` . ∈ k=0 j ∈ n P o j+1 ∞ Moreover, D2∗y j = ej , D2∗y `2 = dkj yk = dkj yk for all y dom(D2∗). k=0 k=0 ∈ P P

Proof: Fix y dom(D∗) and set z := D∗y. Then z = z , e 2 = y , D e 2 and ∈ 2 2 j h j i` 2 j `

∞ ∞ ∞ y , D2ej 2 = ykek , D2ej = yk ek , D2ej 2 = dkj yk . (3.3) ` 2 ` k=0 ` k=0 k=0 D P E P P ∞ ∞ 2 Therefore, D2∗y j = zj = dkj yk and dkj yk ` for all y dom(D2∗). k=0 k=0 j ∈ ∈ P P 2 ∞ 2 To show the converse inclusion, let y ` such that z := dkj yk ` . For ﬁxed j, we can use (3.3) to obtain ∈ k=0 j ∈ P ∞ dkj yk = D2ej , y `2 . k=0 P ∞ Now e , z 2 = z = d y yields e , z 2 = D e , y for all j and, by h j i` j kj k h j i` 2 j `2 linearity, k=0 P x , z 2 = D x , y for all x ` . h i` 2 `2 ∈ 0 Thus y dom(D∗) and z = D∗y. ∈ 2 2 The last equality in the assertion is due to dkj = 0 for k > j + 1.

∞ 2 3.2.6 Theorem. For ﬁxed i N0, Pi dom(D∗) if and only if dij < . ∈ ∈ j=0 | | ∞ ∞ 2 Furthermore, if D is formally normal then dij < for all iP N0. j=0 | | ∞ ∈ P Proof: Clearly, P dom(D∗) if and only if e dom(D∗) and, according to 3.2.5, i ∈ i ∈ 2

∞ 2 2 ∞ 2 ei dom(D2∗) dkj δik ` dij j ` dij < . ∈ ⇐⇒ k=0 j ∈ ⇐⇒ ∈ ⇐⇒ j=0 | | ∞ P P If D is formally normal then, by deﬁnition, P dom(D∗) for all i. Thus the second i ∈ assertion is an immediate consequence. 36 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

3.2.7 Matrix Representation of D2 and D. Let d be the infinite matrix d := (dij)i,j 0 ≥ and d∗ its Hermitian d∗ = (dij∗ )i,j 0 where dij∗ := dji. ≥ For i N , denote by c := (d ) the ith column of the matrix d which contains only ∈ 0 i ki k finitely many entries different from 0, i.e. c ` . Thus the formal matrix product d∗d is i ∈ 0 well-defined and

min i,j +1 ∞ ∞ { } (d∗d)ij = dik∗ dkj = dki dkj = dki dkj = ci , cj `2 . k=0 k=0 k=0 h i P P P Moreover, for i, j N , ∈ 0 ∞ ∞ DPi , DPj = D2ei , D2ej `2 = D2ei , ek `2 ek , D2ej `2 = dki dkj = (d∗d)ij H k=0 k=0 P P and if P , P dom(D∗) then i j ∈ ∞ D∗Pi , D∗Pj = D2∗ei , D2∗ej `2 = D2∗ei , ek `2 ek , D2∗ej `2 H k=0 P ∞ ∞ ∞ = ei , D2ek `2 D2ek , ej `2 = dik djk = dik dk∗j . k=0 k=0 k=0 P P P If, in particular, C[z] dom(D∗) then the matrix product dd∗ is well-defined, too. ⊂ This leads to a necessary and sufficient criterion for formal normality of the Hessenberg operator D in terms of the coefficients dij; we just prove a short lemma in advance.

3.2.8 Lemma. Let V and W be complex vector spaces and X V such that linX = V . Furthermore, let , be an inner product in W and A, B line⊂ ar mappings V W . Then h · · i →

Av = Bv for all v V Ax , Ay = Bx , By for all x, y X . k k k k ∈ ⇐⇒ h i h i ∈ where is the norm on W induced by the inner product. k·k Proof: Using the polarisation identity, one can easily see that Av = Bv for all v V k k k k ∈ implies Ax , Ay = Bx , By for all x, y V . h i h i ∈ To prove the converse implication, assume Ax , Ay = Bx , By for all x, y X. For arbitrary v V , there exist n N, a , . . . , ah C, andi xh, . . . , x i X such that∈ ∈ ∈ 1 n ∈ 1 n ∈ n v = ak xk k=1

n n P n n 2 2 and we obtain Av = ai aj Axi , Axj = ai aj Bxi , Bxj = Bv . k k i=1 j=1 h i i=1 j=1 h i k k P P P P

∞ 2 3.2.9 Theorem. The operator D is formally normal if and only if dij < for j=0 | | ∞ all i N0 and d∗d = dd∗. ∈ P 37

∞ 2 Proof: Let D be formally normal. Then C[z] dom(D∗) and 3.2.6 yields dij < ⊂ j=0 | | ∞ for all i N0. ∈ P Now use 3.2.8 with V = C[z], X = P : n 0 , W = , A = D, and B = D∗ to { n ≥ } H conclude DPi , DPj = D∗Pi , D∗Pj for all i, j, and the calculations in 3.2.7 show h iH h iH d∗d = dd∗. ∞ 2 As to the converse implication, note that, by 3.2.6, dij < implies Pi dom(D∗). j=0 | | ∞ ∈ Therefore C[z] dom(D∗). ⊂ P Now 3.2.7 shows DPi , DPj = D∗Pi , D∗Pj for all i, j. Use 3.2.8 again to conclude h iH h iH that Dp = D∗p for all p C[z]; hence D is formally normal. k k k k ∈

Note the following consequence. If D is formally normal then

j+1 ∞ 2 2 djk = dij for all j N0 . (3.4) k=0 | | i=0 | | ∈ P P For bounded D, we have the following stronger result concerning formal normality.

2 3.2.10 Theorem. Let µ ba a compactly supported om for (Pn)n 0 and deﬁne j Lµ by ≥ ∈ j(z) := z. The following properties are equivalent.

2 2 (i) Pµ = Lµ, (ii) j P 2, ∈ µ

∞ 2 2 (iii) d0j = d10 , j=1 | | | | P (iv) D is formally normal.

(v) D is essentially normal.

Moreover, µ is the only om for (Pn)n.

Proof: Note that, as supp(µ) = σ(Mµ) is compact, Mµ and hence D is bounded. Clearly, j L2 and (i) (ii) is obvious. ∈ µ ⇒ 2 2 2 Now assume j Pµ . As D is continuous, dom(D) = Pµ and Pµ is an invariant subspace ∈ 2 of Mµ. Therefore, (Pµ )⊥ is an invariant subspace of Mµ∗. Moreover, p j = p(Mµ)j = 2 2 · p(D)j P for all p C[z]. Let f (P )⊥. Then ∈ µ ∈ ∈ µ

M f , p = f(z) z p(z) dµ(z) = f , p(D)j = 0 for all p C[z] . h µ i h i ∈

2 R 2 Thus Mµf (Pµ )⊥ and hence (Pµ )⊥ is an invariant subspace of Mµ as well. It is a well- ∈ 2 known fact, see e.g. [W, Aufgabe 5.39] that, as Mµ is bounded and (Pµ )⊥ is an invariant 2 2 subspace of Mµ as well as of Mµ∗, the space (Pµ )⊥ and also Pµ itself reduce Mµ. According to 1.3.7, M is a minimal normal extension of D, implying P 2 = L2 . Thus (ii) (i). µ µ µ ⇒ 38 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

We next show (ii) (iii). ⇐⇒ 2 2 2 By the parseval equation, we have j Pµ j = j , Pn . Note that ∈ ⇐⇒ k k n 0 |h i| ≥ P j 2 = j 2 = DP 2 = d 2 + d 2 . k k k k k 0k | 00| | 10| n+1 Furthermore, j , Pn = z Pn(z) dµ(z) = P0 , DPn = P0 , dinPi = d0n. Hence h i h i i=0 D E R P 2 2 ∞ 2 2 j = j , Pn d0j = d10 . k k n 0 |h i| ⇐⇒ j=1 | | | | ≥ P P The implication (iv) (iii) is an immediate consequence of (3.4) and (i) (iv) is due to 3.2.2. ⇒ ⇒ For (iv) (v), see 3.2.3, and (v) (i) is due to 1.3.7. ⇒ ⇒ Finally, uniqueness of µ has already been proved in 3.2.4.

For more equivalences, see also 3.5.9. Furthermore, if supp(µ) is compact then µ is unique even if P 2 = L2 , see 3.2.14. µ 6 µ

A weaker condition than (3.4) holds if D is subnormal. Remember that there exist formally normal operators which are not subnormal as well as subnormal operators which are not formally normal.

3.2.11 Theorem. If D is subnormal then

j+1 ∞ 2 2 djk dij for all j N0 . (3.5) k=0 | | ≤ i=0 | | ∈ P P Proof: As D is subnormal, there exist a Hilbert space and a normal operator N K ⊃ H in such that Dp = Np for all p C[z]. K ∈ Fix j N0. Then djk = Pj , DPk = Pj , NPk = N ∗Pj , Pk for all k N0 and ∈ h iH h iK h iK ∈ ∞ 2 ∞ 2 djk = N ∗Pj , Pk < , k=0 | | k=0 h iK ∞ P P as (P ) is an orthonormal system in . Moreo ver, using the Bessel inequality, we obtain n n K j+1 2 j+1 ∞ 2 2 2 2 2 N ∗Pj , Pk N ∗Pj = NPj = DPj = dij Pi = dij k=0 h iK ≤ k kK k kK k kH i=0 i=0 | | H P P P to complete the proof.

We point out that (3.4) and (3.5) are necessary but not suﬃcient for formal normality or subnormality of the Hessenberg operator D. However, they will come in handy, see 4.4.5 where we will deduce a necessary condition for the existence of an om when D is a weighted shift, and 4.5.3 or 4.5.7 in order to prove that D is not formally normal or not subnormal, respectively. 39

3.2.12 Formally Normal Extension of D. Let us now canonically extend the multiplication operator D to C[z, z] := z q(z, z) : q C[z, w] . 7→ ∈ Given (Pn)n 0 as in 1.1.1, we can also extend the inner product on C[z] – with respect to ≥ which (P ) is an orthonormal sequence – to a sesquilinear form , on C[z, z] via n n h · · i uv , w = u , vw for all u, v, w C[z, z] (3.6) h i h i ∈ where u is defined by u(z) := u(z). It is the matter of a lengthy but not difficult calculation, see [Kl2, 1.1.4], that this is indeed well-defined and leads to a uniquely determined sesquilinear extension of the inner product on C[z]. Yet, it may happen that u , u < 0 for some u C[z, z], see [Kl2, 1.1.5]. h i ∈ In the case that u , u 0 for all u C[z, z], let := u C[z, z] : u , u = 0 . Now , becomesh ani inner≥ product ∈on the quotienNt space{ C∈[z, z]/ whoseh abstracti } h · · i N completion, in analogy to the completion of C[z] defined in section 1.2, we will denote by . Regard as a closed subspaceH of . Then the canonical extension of the multiplicationK operatorH D to C[z, z] leads to aKwell-defined operator F in with K dom(F ) = C[z, z]/ , see [Kl2, 1.3.12], for instance. N Assume that there exists an om µ for (Pn)n. Clearly, every u C[z, z] is square integrable with respect to µ. As shown in [Kl2, 1.1.6], when we regard∈u C[z, z] as a member of ∈ L2 , the inner product on L2 extends , constructed in (3.6) and µ µ h · · i = u C[z, z] : u 2 dµ = 0 . N ∈ | | 2 R 2 Now can be canonically embedded into Lµ and the operator Mµ in Lµ is a minimal normalK extension not only of D but also of F . In particular, every normal extension of D is a normal extension of F as well. Note that, if F is essentially normal then the unique minimal normal extension of F is its closure F which must also be the unique minimal normal extension of D. However, in this situation D need not be essentially normal.

Finally, we point out that F n is formally normal for all n N. To see that, set u(z) := zn ∈ for ﬁxed n; now (3.6) yields v , F nw = v , uw = uv , w for all v, w C[z, z] , h i h i h i ∈ n n n hence C[z, z] dom (F )∗ and (F )∗v (z) = (uv)(z) = z v(z); using (3.6) once again, ⊂ we obtain n 2 n 2 F v = uv , uv = uv , uv = (F )∗v . k kK h i h i k kK

3.2.13 Example. Let µ be an om for (Pn)n 0 such that supp(µ) z C : z = 1 ; ≥ ⊂ { ∈ | | } set q(z, w) := 1 zw, and deﬁne u C[z, z] by u(z) := q(z, z) = 1 z 2. − ∈ − | | Then u(z) = 0 whenever z supp(µ) and hence u . ∈ ∈ N

The operator F is studied in depth in [Kl2]; we have only mentioned here some of its most important properties in order to give a short proof of the following fundamental fact concerning uniqueness of compactly supported om. 40 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

3.2.14 Proposition. If there exists a compactly supported om µ for (Pn)n 0 then µ is ≥ the only om at all.

Proof: As supp(µ) = σ(Mµ) is compact, Mµ is bounded. Thus the operator F deﬁned in 3.2.12 is bounded, too, and 3.2.3 implies that F is essentially normal. Therefore, F has a unique minimal extension F . As we have seen in 3.2.12, every normal extension of D is a normal extension of F . Thus Mµ is the unique minimal normal extension of D, and 1.3.7 implies uniqueness of the orthonormalizing measure.

3.2.15 Deﬁnition. Let S be a (not necessarily densely deﬁned) linear operator in a Hilbert space with invariant domain, i.e. ran(S) dom(S). Then S is said to obey H ⊂ the Halmos condition (see [Kl2, 1.3.20], [Br, Introduction], or [StSz2, p.156 (H)], for instance), if n n j i S qi , S qj 0 i=0 j=0h iH ≥ P P for all n N and all sequences (q ) in dom(S). ∈ 0 i i

3.2.16 Proposition. If D is bounded then the following properties are equivalent.

(i) There exists an om, (ii) there exists a unique om, (iii) there exists a compactly supported om, (iv) u , u 0 for all u C[z, z] with , deﬁned as in 3.2.12, h i ≥ ∈ h · · i (v) D obeys the Halmos condition.

For a proof we refer to [Kl2, 1.3.21 and 1.5.4]. The main eﬀort to prove this, is to see that F is bounded whenever D is bounded and obeys the Halmos condition which is, for example, shown in [Kl2, 1.5.1] following an essay by Bram [Br].

3.3 Spectral Properties of the Multiplication Operator

In this section, we will prove more detailed relations between the set E, the spectrum of D, and – if an om µ exists – the support of µ. Recall that in 3.1.3 we have seen that E is precisely the set of numbers where the deﬁciency index of D is equal to 1 and, according to 3.1.7, the discrete mass points of an om must belong to E. Some statements in the following theorem can be found in [StSz3, Th. 1 and Cor. 12] and (v) extends [Halm, Problem 158]. 41

3.3.1 Theorem. Let µ be an om for (Pn)n 0. Regard D as a densely deﬁned operator 2 ≥ 2 in Pµ and Mµ as a minimal normal extension of D acting in Lµ (see also 1.3.7). Then

(i) σ(M ) σ(D), µ ⊂ (ii) σ (D) σ (M ) = Λ E and σ (D) E E , p ⊂ p µ µ ⊂ p ⊂ \ reg (iii) E = σ (D) σ (D) and σ (D) = σ(D) E, p ∪ r c \ (iv) σ (D) σ (M ), c ⊂ c µ (v) if A ρ(M ) is connected then either A ρ(D) or A E , ⊂ µ ⊂ ⊂ reg (vi) σ(M ) E = σ(D) E = σ (D), µ \ \ c (vii) E σ(M ) = σ(D). ∪ µ

Proof: First note that D is subnormal and Mµ is a minimal normal extension of D. In particular, D is closable.

(i) Equivalently, we show ρ(D) ρ(Mµ). The following is taken from [StSz3, Th. 1 - 9◦]. ⊂1 Let λ ρ(D). Then (D λ id)− is a continuous linear operator on . Now choose ε > 0 ∈ −1 C H 2 such that ε (D λ id)− < 1; deﬁne ∆ := z : z λ < ε and h := 1∆ Lµ. k − kn { ∈ | − | } ∈ Then h dom(M λ id)∗ and ∈ µ −

n n µ(∆) = h P dµ = h , (D λ id) (D λ id)− P · 0 h − − 0 i

n n n n = R h , (Mµ λ id) (D λ id)− P0 = (Mµ λ id)∗ h , (D λ id)− P0 h − − i h − − 1 i 2 n 1 n 2n 1 n (M λ id)∗ h (D λ id)− P = z λ dµ (D λ id )− ≤ k µ − k · k − k · k 0k − k − k ∆ 1 1 2n 2 1 n R 1 n ε µ(∆) (D λ id)− = µ(∆) 2 ε (D λ id)− ≤ k − k k − k for all n N. This implies µ(∆) = 0. Thus ∆ is an open µ-nullset and, therefore, ∈ ∆ C supp(µ) = ρ(M ). In particular, λ ρ(M ). ⊂ \ µ ∈ µ (ii) σ (D) σ (M ) is obvious. For σ (M ) = Λ E see 3.1.7 and, by deﬁnition, an p ⊂ p µ p µ µ ⊂ eigenvalue of D cannot be a regular value of D. Thus σ (D) E E . p ⊂ \ reg (iii) Due to 3.1.3, E = λ C : ran(D λid) is not dense in P 2 . Note that ran(D λid) { ∈ − µ } − is not dense in P 2 if and only if ran(D λ id) is not dense in P 2. Hence E σ(D). µ − µ ⊂ Furthermore, by deﬁnition of the residual and continuous spectrum, we have σ (D) E r ⊂ and σc(D) E = ∅. ∩ Finally, σ (D) E, according to (ii). p ⊂ As σ(D) is the union of the disjoint sets σ (D) σ (D) σ (D), we get E = σ (D) σ (D) r ∪ c ∪ p p ∪ r and σ (D) = σ(D) E. c \ 42 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

1 2 (iv) If λ σc(D) then (D λ id)− is densely deﬁned in Pµ and not continuous. Suppose ∈ − 1 2 1 λ ρ(M ). Then (M λid)− is a continuous linear operator in L extending (D λid)− ∈ µ µ − µ − which is a contradiction. Hence σ (D) σ(M ). c ⊂ µ Moreover, σc(D) E = ∅ according to (iii). Now (ii) implies σc(D) σp(Mµ) = ∅. As ∩ ∩ Mµ is normal, σr(Mµ) = ∅. Thus σc(D) σc(Mµ). ⊂

(v) Obviously, if λ is not a regular value of D then it is not a regular value of Mµ, either. As M is normal, reg(M ) = ρ(M ). Thus ρ(M ) reg(D). µ µ µ µ ⊂ Let now A ρ(M ) be connected. According to 3.1.2, the deﬁciency index is constant in ⊂ µ every connected subset of reg(D) = reg(D). Hence either A E or A E = ∅. ⊂ ∩ In the ﬁrst case we obtain A E . ⊂ reg Consider now A C E. By (iii) we have C E = C σ (D) σ (D) = ρ(D) σ (D). ⊂ \ \ \ p ∪ r ∪ c Moreover, σc(D) σc(Mµ), see (iv). Using A ρ(Mµ) this yields A σc(D) = ∅. ⊂ ⊂ ∩ Hence only A ρ(D) remains. ⊂ (vi) Using the previous results, we get (i) (iii) (iv) σ(M ) E σ(D) E = σ (D) = σ (D) E σ(M ) E , µ \ ⊂ \ c c \ ⊂ µ \ thus σ(M ) E = σ(D) E = σ (D). µ \ \ c (vii) Using (iii) and (vi), we see σ(M ) E = σ(M ) E E = σ (D) σ (D) σ (D) = µ ∪ µ \ ∪ c ∪ p ∪ r σ(D).

Ereg (open)

Λµ\ ρ(Mµ) \ Ereg ∩ σc(Mµ) \ ρ(D) σp(D) Ereg ∩ Λµ σc(D) (Ereg ∪ σp(D)) ρ(D) σc(Mµ) (Ereg ∪ σc(D))

(open) (open)

σp(Mµ) = Λµ σc(Mµ)

σr(D)

Figure 2. This graphically summarizes the assertions of 3.3.1.

We have ρ(D) ρ(Mµ) and σp(D) σp(Mµ) as well as σc(D) σc(Mµ), ⊂ ⊂ ⊂ while σr(D) can contain parts of σp(Mµ), σc(Mµ), and also of ρ(Mµ).

Note that ρ(Mµ) and hence all its connected components are open. As shown in 3.3.1(v), every connected component of ρ(Mµ) is contained in either ρ(D) or Ereg. Therefore, ρ(Mµ) ρ(D) Ereg is open. \ ⊂ Recall that, according to 3.1.4, Ereg is open, too. 43

If D is essentially normal, i.e. D is normal, we obtain the following.

3.3.2 Corollary. If D is essentially normal then there exists a unique om µ, D = Mµ, and E = Λ = z C : µ( z ) > 0 . µ { ∈ { } }

Proof: According to 1.3.7, there exists a unique om µ and D = Mµ is the only minimal normal extension of D. As the residual spectrum of a normal operator is empty, 3.3.1(ii) and (iii) yield σp(Mµ) = E = Λµ.

Note that E can be ﬁnite or even empty. Note also that if, in 3.3.1, D is bounded then σ(D) and hence also E is bounded. In 3.1.6, however, we have already proved that D bounded implies E bounded, even if there exists no om.

Remark. If Mµ is bounded then D is bounded, too. On the other hand, if D is bounded and there exists an om µ then 3.3.1(i) shows that D bounded implies Mµ bounded. Hence D is bounded if and only if Mµ is bounded, provided that there exists an om µ at all.

The following is an immediate consequence of 3.3.2 in connection with 3.2.10.

3.3.3 Proposition. Let µ be an om for (Pn)n 0 and supp(µ) be compact. If (Pn)n is 2 2 2 ≥ an onb in Lµ (in other words, Pµ = Lµ) then D is essentially normal, E = Λµ, and µ is unique.

Remark. Recall that in 3.2.14 we have seen that if there exists a compactly supported om then it is uniquely determined – even if P 2 = L2 . µ 6 µ To prove uniqueness in this case, we could also use the fact that, according to [Con1, II.2.7], a bounded subnormal operator has a unique – up to unitary equivalence – minimal normal extension.

Let us now have a closer look at the boundary of E,

∂E := z C : U E = ∅ and U (C E) = ∅ for all open sets U containing z . { ∈ ∩ 6 ∩ \ 6 } We start with a short lemma.

3.3.4 Lemma. If there exists an om µ for (Pn)n 0 then C supp(µ) reg(D). ≥ \ ⊂ Proof: As M is normal, ρ(M ) = reg(M ). Clearly, z / reg(D) implies z / reg(M ). µ µ µ ∈ ∈ µ Now C supp(µ) = ρ(M ) completes the proof. \ µ 44 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

3.3.5 Theorem. If there exists an om µ for (Pn)n 0 then ∂E supp(µ). ≥ ⊂

Proof: In the cases ∂E = ∅ or supp(µ) = C, the assertion is trivial. Now assume that there exists z ∂E C supp(µ) . As C supp(µ) = ρ(M ) is ∈ ∩ \ \ µ open, we can ﬁnd an open disk U containing z such that U ρ(M ). By 3.3.1(v), either µ U ρ(D) or U E . Moreover, ρ(D) (C E), see 3.3.1(iii),⊂ and E E. Thus ⊂ ⊂ reg ⊂ \ reg ⊂ either U (C E) or U E which contradicts z ∂E. Hence ∂E C supp(µ) = ∅ ⊂ \ ⊂ ∈ ∩ \ as asserted.

3.3.6 Corollary. Let G be a simply connected bounded open subset of C and µ an om for (Pn)n such that supp(µ) ∂G. If E supp(µ) = ∅ then ⊂ \ 6

(i) G E G ∂G and G = E◦ = E , ⊂ ⊂ ∪ reg (ii) supp(µ) = ∂G,

(iii) P 2 = L2 and D is not essentially normal. µ 6 µ

Proof: (i) We ﬁrst show E G. ⊂ Let U := C G and assume that there exists z E U. Note that, as G and hence also \ ∈ ∩ G is simply connected and bounded, U is connected. By assumption, U supp(µ) = ∅, thus U ρ(M ) and 3.3.1(v) implies U E . In particular, E is not bounded.∩ ⊂ µ ⊂ reg On the other hand, as supp(µ) is bounded, the multiplication operator Mµ and, consequently, D is bounded. Now, by 3.1.6, E is bounded and we have a contradiction. Therefore, E U = ∅, in other words, E G. ∩ ⊂ Now we show G E. ⊂ Note that ∂E supp(µ) and E supp(µ) = ∅ imply E◦ = ∅. Hence E◦ G = ∅ and 3.3.1(v) implies⊂G E . \ 6 6 ∩ 6 ⊂ reg Now E◦ = G E is an immediate consequence. According to 3.1.4, E is open; hence ⊂ reg reg E E◦ and, ﬁnally, we obtain G = E◦ = E . reg ⊂ reg (ii) Furthermore, G E G ∂G implies ∂E = ∂G and, again using ∂E supp(µ), we get ∂G supp(µ). Giv⊂ en⊂supp∪(µ) ∂G, we now have ∂G = supp(µ). ⊂ ⊂ ⊂ 2 2 (iii) If D is essentially normal or Pµ = Lµ then 3.3.2 and 3.3.3, respectively, imply E = Λµ which contradicts E supp(µ) = ∅. \ 6

Remark. If supp(µ) is a proper subset of ∂G then 3.3.6(ii) implies E supp(µ) and supp(µ) is an α-set in accordance to section 4.3. In that case, 4.3.4 sho⊂ws that D is essentially normal. However, it is also possible that D is essentially normal if supp(µ) = ∂G, as we will see in section 4.2, where G is the open unit disk. 45

3.3.7 Example. Let γ : [0, 1] C be a simply closed piecewise continuously diﬀeren- → tiable curve and µ be absolutely continuous with respect to line measure on γ, having density α, such that 1 α(z) dz < γ | | ∞ R (see 2.3.8 for the details).

Moreover, assume µ(γ) = 1, hence µ is an om for some (Pn)n, and denote by G the bounded component of C γ. \ Then, according to 2.3.11, G E. Thus 3.3.6 is applicable to this situation. ⊂

Remark. If supp(µ) contains an open set, a similar integrability condition concerning the absolutely continuous (with respect to two-dimensional Lebesgue mesaure) part of µ implies P 2 = L2 , see 3.6.5. µ 6 µ

3.3.8 Theorem. Let µ be an om for (Pn)n 0 and denote by P the orthogonal projection 2 2 ≥ 1 in L onto P . For a E and b E supp(µ), deﬁne f := (M a)∗[(M b)− ]∗k . µ µ ∈ ∈ \ a,b µ − µ − a Then P f = f , P k . a,b h a,b 0 i b 2 Moreover, fa,b = 0 if and only if a Λµ and 1 a Pµ . ∈ { } ∈

1 2 Proof: As b / supp(µ), (Mµ b)− is a bounded linear operator in Lµ. Furthermore, z a ∈ − ϕ(z) := z−b defines a bounded function on supp(µ). − 1 2 2 Hence (M a)∗[(M b)− ]∗f = ϕ f L for all f L and f is well-defined. µ − µ − · ∈ µ ∈ µ a,b For arbitrary p C[z], define q C[z] by q(z) := p(z) p(b). Note that kb , q = q(b) = 0 and k , P =∈P (b) = 1. This∈yields − h i h b 0 i 0 f f , P k , p a,b − h a,b 0 i b = f , q + f , p(b)P f , P k , q f , P k , p(b)P h a,b i h a,b 0 i − h a,b 0 ih b i − h a,b 0 ih b 0 i 1 = f , q = k , (M b)− (M a)q = 0 , h a,b i h a µ − µ − i

1 as (M b)− (M a)q C[z] has a zero at a. µ − µ − ∈ 2 2 Thus f f , P k (P )⊥ and, as k P , we obtain P f = f , P k . a,b − h a,b 0 i b ∈ µ b ∈ µ a,b h a,b 0 i b Note that ϕ f = 0 if and only if f = c1 a for some c C. According to 3.1.8, if a Λµ · 2 { } ∈ ∈ and 1 a Pµ then 1 a = µ( a ) ka; hence fa,b = 0. As to the converse implication, { } ∈ { } { } 2 0 = fa,b = ϕ ka implies ka = c1 a showing a Λµ and 1 a Pµ . · { } ∈ { } ∈ 46 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

C 2 3.4 Denseness of [z] in Lµ

C 2 Let µ be an om for (Pn)n 0. We now consider the case that [z] is dense in Lµ. In other 2 2 ≥ words, Pµ = Lµ. Assume that these polynomials form a basis of (K), too. If E◦ = ∅ then there exists H 6 an open set G which is dense in E◦ and all f (K) are holomorphic in G, see 2.2.9. 2 ∈ H On the other hand, in general, Lµ contains elements which do not have a holomorphic representative. 2 2 Thus, if Pµ = Lµ, one can suppose that G – except for some discrete points – is a µ-nullset. We will prove that, in this case, µ(E Λµ) = 0 even if the (Pn)n are not orthogonal in (K). \ H As for now, we start with a more general situation.

3.4.1 Definitions. Let (Ω, A, µ) be a σ-finite measure space and (hn)n 0 be an onb in 2 ≥ Lµ. Define 1 2 2 π : Ω [0, ] , π(ω) := hn(ω) → ∞ n 0 ≥ P and Eπ := ω Ω : π(ω) < . Then π is µ-measurable and Eπ A (note that π and { ∈ ∞} ∈ Eπ depend on the particular choice of the representatives of hn).

Moreover, deﬁne K : Eπ Eπ C, K(ω0, ω) := hn(ω0) hn(ω). × → n 0 ≥ A set B A is called a block (or atom) if µ(B)P> 0 and ∈ µ(C) = 0 or µ(B C) = 0 for all C A , C B . \ ∈ ⊂ Note that if B is a block and N a nullset then B N and B N are blocks, too. ∪ \ A (possibly ﬁnite and) at most countable family (Aj)j of mutually disjoint sets in A such that Aj = A is called a partition of A. ∪j

3.4.2 Lemma. If A A, A Eπ, µ(A) < , and π dµ < , then ∈ ⊂ ∞ A ∞ R µ(A) = K d µ µ . A A ⊗ ×R N Proof: For N N, hn(ω0) hn(ω) π(ω0) π(ω), and ∈ n=0 ≤ P N N N 2 1A , hn = hn dµ hn dµ = hn hn d µ µ n=0 |h i| n=0 A A n=0 A A × ⊗ P P P × RN R R = hn hn d µ µ . A A n=0 × ⊗ ×R P 47

As π π is integrable on A A, dominated convergence yields × ×

2 ∞ 2 µ(A) = 1A = 1A , hn = K d µ µ . k k n=0 |h i| A A ⊗ P ×R

Now denote by Z(A) the class of all partitions of A; for P = (A ) Z(A) deﬁne s(P) := j j ∈ sup µ(Aj) and set sA := inf s(P). j P Z(A) ∈

3.4.3 Lemma. Let A be as in 3.4.2. If, in addition, µ(A) > 0 then sA > 0.

N P n Z Proof: Assume sA = 0. Then, for n , there exists n = (Aj )j (A) such that P 1 ∈ ∈ s( n) < n µ(A) . Without restriction we can assume P P for all n, i.e. for every B P there n+1 ⊂ n ∈ n+1 exists An P such that B An. j ∈ n ⊂ j n n For abbreviation, set Cn := Aj Aj . We have ∪j ×

n n n 2 1 n 1 µ µ(Cn) = µ µ(Aj Aj ) = µ(Aj ) n µ(A) µ(Aj ) = n ⊗ j ⊗ × j ≤ j P P P n and, using 3.4.2, we obtain µ(A) = µ(Aj ) = K d µ µ = K d µ µ. j j n n ⊗ ⊗ Aj Aj Cn P P × 1 R R Due to µ µ(Cn) = n and Cn+1 Cn for all n, the integral on the right hand side tends to 0 as n⊗ . This implies µ(A⊂) = 0 which is a contradiction. Hence s > 0. → ∞ A

3.4.4 Lemma. Let A be as in 3.4.2. If, in addition, 0 < µ(A) < then there exists a block B A. ∞ ⊂ Proof: According to 3.4.3, s := s > 0. Note that s < , as µ(A) < . Thus we can A ∞ ∞ choose a partition P Z(A) with s(P) < 6 s. Now we can ﬁnd A P such that ∈ 5 1 ∈ µ(C) 4 s or µ(A C) 4 s for all C A , C A ≥ 5 1 \ ≥ 5 ⊂ 1 ∈ 4 because otherwise we could construct P0 Z(A) with s(P0) s. ∈ ≤ 5 Note that µ(A C) 4 s implies µ(C) < 2 s, as µ(A ) < 6 s. Thus 1 \ ≥ 5 5 1 5 µ(C) 4 s or µ(C) < 2 s for all C A , C A . (3.7) ≥ 5 5 ⊂ 1 ∈ A 4 2 Let Ci , Ci A1, and µ(Ci) 5 s for i 1, 2 . Assume µ(C1 C2) < 5 s. Then µ(C C∈ ) = µ(C⊂)+µ(C ) µ(C ≥ C ) 8 s∈ {2 s =} 6 s in contradiction∩ to µ(A ) < 6 s. 1 ∪ 2 1 2 − 1 ∩ 2 ≥ 5 − 5 5 1 5 Thus (3.7) yields µ(C C ) 4 s. ( ) 1 ∩ 2 ≥ 5 ∗ 4 A N Now set s1 := inf µ(C) : C , C A1 , µ(C) 5 s and choose a sequence (Cn)n { ∈ ⊂4 ≥ } ∈ in A such that Cn A1 and µ(Cn) s for all n N such that lim µ(Cn) = s1. 5 n ⊂ ≥ ∈ →∞ 48 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

n For n N, let Cn := Ci. ∈ i∩=1 Note that µ(Cn)eis monotonically decreasing as n and, in analogy to ( ), inductively, 4 → ∞ ∗ we obtain µ(Cn) 5 s while, by construction, µ(Cn) µ(Cn) for all n. Hence e ≥ ≤ 4 e s lim µ(Cn) lim eµ(Cn) = s1 . (3.8) 5 n n ≤ →∞ ≤ →∞ e Define B := Cn. n N ∩∈ Then B A, B A , and (3.8) reads 4 s µ(B) s . On the other hand, µ(B) 4 s e 1 5 1 5 implies µ∈(B) s⊂, simply by definition of s≤. Thus≤only µ(B) = s remains. ≥ ≥ 1 1 1 Finally, we show that B is a block. Assume that there exists C A, C B such ∈ ⊂ 4 that 0 < µ(C) < µ(B) = s1. Recall the definition of s1 again to see that µ(C) 5 s 4 ≥ or µ(B C) 5 s would imply µ(C) s1 or µ(B C) s1, respectively, and yield a \ ≥ ≥ 2 \ ≥ 2 contradiction. Now (3.7) shows that µ(C) < 5 s and also µ(B C) < 5 s in contradiction to µ(B) 4 s, completing the proof. \ ≥ 5

3.4.5 Lemma. Let A be as in 3.4.2. Then A is the union of at most countably many blocks A = Bn and µ(Bi Bj) = 0 for i = j. ∪n ∩ 6

Proof: If B1, B2 A are blocks such that µ(B1 B2) > 0 then µ(B1) = µ(B1 B2) = µ(B ) and hence ⊂µ (B B ) (B B ) = 0. ∩In this case B and B are said∩ to be 2 1 ∪ 2 \ 1 ∩ 2 1 2 equivalent (note that µ(B B ) > 0 and µ(B B ) > 0 for blocks B , B , B A implies 1 2 2 3 1 2 3 µ(B B ) > 0). ∩ ∩ ⊂ 1 ∩ 3 Assume that there exist n N and a sequence (Bk)k N of mutually non-equivalent blocks ∈ in A such that µ(B ) > 1 ∈for all k N. k n ∈ As µ(Bi Bj) = 0 for i = j, this implies µ Bk = µ(Bk) = in contradiction to N ∩ 6 k∪ k N ∞ µ(A) < . ∈ ∈ ∞ P Hence, for every n N, there exist only ﬁnitely many mutually non-equivalent blocks ∈ 1 whose measure exceeds n . Therefore, there exist at most countably many classes of equivalence.

Now let (Bi)i be a sequence of blocks in A which contains exactly one element of each equivalence class and set B := Bi. Then A B does not contain a block and, by 3.4.4, i µ(A B) = 0. ∪ \ \ Clearly, B1 (A B) is a block, too, equivalent to B1. Thus A = Bi B1 (A B) ∪ \ i=1 ∪ ∪ \ completes the proof. ∪6

3.4.6 Theorem. In the situation of 3.4.1, Eπ is the union of at most countably many blocks.

Proof: As µ is σ-ﬁnite, there exists a partition (A ) of Ω such that µ(A ) < for all n. n n n ∞ Now, for m N set ∈ A := A ω Ω : π(ω) < m . nm n ∩ { ∈ } 49

If µ(Anm) > 0 then, by construction, Anm fulﬁls the premises of 3.4.5 and hence is the union of at most countably many blocks. As the union af a block and a nullset is a block, too, Eπ is the union of at most countably many blocks as well.

We now have a closer look at what blocks look like in Rd with respect to a σ-ﬁnite measure deﬁned on the Borel sets.

3.4.7 Theorem. Denote by B(Rd) the Borel σ-algebra in Rd and let µ be a σ-ﬁnite measure on B(Rd). If B B(Rd) is a block with respect to µ then there exists b B such that µ( b ) > 0 ∈ ∈ { } and µ(B b ) = 0. \ { } N Rd m m 1 Proof: For m , let = Cn where each Cn is a closed ball with radius . ∈ n N m ∪∈ 1 1 For an arbitrary block B, in particular, µ(B Cn) = 0 or µ(B Cn) = 0. Assume µ(B C1) = 0 for all n. Then µ(B) = 0 whic∩h is a contradiction.\ Thus there exists ∩ n n N such that µ(B C1 ) = 0. 1 ∈ \ n1 1 N Set B1 := B Cn1 . Then µ(B1) = µ(B). In analogy to the above, we can ﬁnd n2 such that µ(B∩ C2 ) = 0 and with B := B C2 we get µ(B ) = µ(B ). Repeating∈ 1 \ n2 2 1 ∩ n2 2 1 this, we obtain a sequence (Bj)j N satisfying Bj+1 Bj and µ(Bj) = µ(B) for all j N. ∈ ⊂ 1 ∈ Furthermore, each Bj is contained in a ball with radius . Therefore, B := Bj j ∗ j N contains at most one point. ∩∈ Moreover, µ(B ) = lim µ(Bj) = µ(B) > 0, hence B = ∅. ∗ j ∗ →∞ 6 Thus B = b for some b Rd and µ( b ) = µ(B) > 0, µ(B b ) = 0. ∗ { } ∈ { } \ { }

3.4.8 Corollary. Let µ be an om such that P 2 = L2 . Then µ(E Λ ) = 0. µ µ \ µ Proof: Λ E is due to 3.1.7 and, by 3.4.6, E is the union of at most countably many µ ⊂ blocks, E = Bn, say. ∪n According to 3.4.7, for each of these blocks there exists b B such that µ( b ) > 0 n ∈ n { n} and µ(Bn bn ) = 0. Hence bn Λµ and Bn bn Λµ = ∅ for all n. \ { } ∈ \ { } ∩ Therefore, µ(E Λµ) = µ (Bn bn ) = 0. \ ∪n \ { } Remark. We will return to this situation in 3.5.7, where we will be able to prove that Ereg does not contain a limit point of Λµ.

If D is essentially normal then E = Λµ, see 3.3.2. Thus, in that case, 3.4.8 is trivial. 2 2 Note that, if D is continuous and has a normal extension in (i.e. Pµ = Lµ) then D is 2 H essentially normal, as D is deﬁned on the whole space Lµ.

2 2 Now consider the case that Pµ = Lµ and D is not essentially normal, hence Mµ is a proper extension of D in the same space. In particular, then D is not continuous. 50 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

3.4.9 Lemma. Let S and T be closed linear operators in a Hilbert space such that H dom(S) dom(T ) and T f = Sf for all f dom(S). If ρ(S) ρ(T ) = ∅ then S = T . ⊂ ∈ ∩ 6 1 1 Proof: Let z ρ(S) ρ(T ). Then (S z id)− and (T z id)− are continuous linear operators in ∈and for∩arbitrary f ,− − H ∈ H 1 1 1 (T z id)(T z id)− f = f = (S z id)(S z id)− f = (T z id)(S z id)− f . − − − − − − 1 1 1 As (T z id) is one-to-one, this implies (T z id)− f = (S z id)− f. Hence (T z id)− = − 1 − − − (S z id)− . − 1 1 Let now g dom(T ). Then g = (T z id)− (T z id)g = (S z id)− (T z id)g dom(S). Thus dom(∈T ) dom(S) which completes− the −proof. − − ∈ ⊂

3.4.10 Theorem. Let (Pn)n 0 be as in 1.1.1. Assume that there exists an om µ such 2 2 ≥ that Pµ = Lµ and the multiplication operator D is not essentially normal. Then

(i) σ(D) = C,

(ii) reg(D) = Ereg,

(iii) ρ(M ) E and E supp(µ) = C, µ ⊂ reg reg ∪

(iv) E◦ Λ = ρ(M ). \ µ µ

Proof: (i) According to 3.3.1(i), we have ρ(D) ρ(Mµ). By 3.4.9, ρ(D) ρ(Mµ) = ∅ ⊂ ∩ 6 would imply D = Mµ in contradiction to D being not essentially normal. Hence only ρ(D) = ∅ remains.

(ii) Recall reg(D) = reg(D) ρ(D) σr(D), see 3.1.2, and σr(D) E, see 3.3.1(iii). Thus here reg(D) E and, therefore,⊂ ∪reg(D) = reg(D) E = E . ⊂ ⊂ ∩ reg (iii) As ρ(D) = ∅, the inclusion ρ(Mµ) Ereg is an immediate consequence of 3.3.1(v). ⊂ Now C = ρ(M ) σ(M ) E σ(M ) implies E σ(M ) = C. µ ∪ µ ⊂ reg ∪ µ reg ∪ µ (iv) E◦ Λ is open and, by 3.4.8, a µ-nullset. Thus E◦ Λ C supp(µ) = ρ(M ). \ µ \ µ ⊂ \ µ For the converse inclusion note that Λµ = σp(Mµ), hence ρ(Mµ) Λµ = ∅. By (iii), ∩ ρ(M ) E. Thus ρ(M ) E Λ . As ρ(M ) is open, this yields ρ(M ) E◦ Λ . µ ⊂ µ ⊂ \ µ µ µ ⊂ \ µ

3.4.11 Corollary. If, in the situation of 3.4.10, E◦ = ∅ then supp(µ) = C.

Proof: According to 3.1.4, Ereg is open; therefore, E◦ = ∅ implies Ereg = ∅. Now, using 3.4.10(iii), we obtain supp(µ) = C as asserted.

Remark. In 3.6.5 we will see that the density of the absolutely continuous (w.r.t. Lebesgue measure) part of µ must be of rather peculiar appearance (or vanish) in order to have 2 2 Pµ = Lµ. See there for details. 51

3.5 Open Subsets of E

If E contains a non-void open set then there exists an open set G which is dense in E◦ such that all f (K) are holomorphic in G, see 2.2.9. ∈ H As we shall prove in this section, if there exists an om then Ereg is the largest open set contained in E where all functions of (K) are holomorphic. H Deﬁne G := G E : all f (K) are holomorphic in G . K ⊂ ∈ H As the propertSy “holomorphic” is only deﬁned in open sets, GK is the union of open sets and hence open. Moreover, all f (K) are holomorphic in G and there exists no ∈ H K proper superset G0 % GK such that all f (K) are holomorphic in G0. ∈ H

3.5.1 Lemma. The map z K(z, z) is continuous on E . 7→ reg

kz 1 Proof: For z E, deﬁne hz := = kz. kz ∈ k kH √K(z,z) Moreover, for z E let Q be the orthogonal projection in onto ran(D z id)⊥. ∈ reg z H − Taking into account that ran(D z id)⊥ = C hz (see 3.1.3) and hz = 1, we get − · k kH

Qzf = hz , f hz for all f . h iH ∈ H

Let now (zn)n N be a convergent sequence in reg(D) such that lim zn =: z reg(D). ∈ n ∗ →∞ ∈ Recall that, by 3.1.2, Q Q ∗ 0 as n . Thus k zn − z k → → ∞

(Qzn Qz∗ )hz∗ = hzn , hz∗ hzn hz∗ , hz∗ hz∗ = hzn , hz∗ hzn hz∗ (3.9) − h iH − h iH h iH − implies hzn , hz∗ hzn hz∗ as n . Due to hzn = 1 for all n, we obtain h iH → → ∞ k kH hzn , hz∗ 1. Moreover, (3.9) yields |h iH| →

(Qzn Qz∗ )hz∗ , P0 = hzn , hz∗ hzn , P0 hz∗ , P0 − h iH h iH − h iH H D E 1 1 = hzn , hz∗ . h iH √K(zn,zn) − √K(z∗,z∗)

1 1 Using hzn , hz∗ 1, this implies and, ﬁnally, |h iH| → √K(zn,zn) → √K(z∗,z∗)

K(zn, zn) K(z , z ) as n , → ∗ ∗ → ∞ showing continuity of z K(z, z). 7→

3.5.2 Theorem. If Ereg = ∅ then all f (K) are holomorphic in Ereg. 6 ∈ H Proof: According to 3.5.1, the map κ : E [1, ), κ(z) := K(z, z) is continuous in E . Thus, for every compact K E there→ exists∞ c > 0 such that κ(z) c for all reg ⊂ reg K ≤ K z K. Now 2.2.3 yields the assertion. ∈ 52 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

The following corollary, given there exists an om µ, characterizes E supp(µ) as an open \ set in which all f (K) are holomorphic. Note that we do not require supp(µ) E. ∈ H ⊂

3.5.3 Corollary. Let µ be an om for (Pn)n 0. Then E supp(µ) is open. Moreover, if ≥ \ E supp(µ) = ∅ then all f (K) are holomorphic in E supp(µ). \ 6 ∈ H \

Proof: According to 3.3.4, C supp(µ) reg(D). This implies E supp(µ) Ereg. Now ∂E supp(µ), see 3.3.5, yields\ that ⊂E supp(µ) is open and, as\all f (⊂K) are ⊂ \ ∈ H holomorphic in Ereg, see 3.5.2, the proof is complete.

In 3.5.2 we have seen that Ereg GK. To prove the converse inclusion, we use the following lemma. ⊂

3.5.4 Lemma. Suppose (f ) is a sequence in (K), M > 0, and λ E such that n n H ∈ fn M for all n and lim fn(z) exists for all z E λ . (K) n k kH ≤ →∞ ∈ \ { } Then there exist a subsequence (f ) and f (K) such that nk k ∈ H

lim fn , h (K) = f , h (K) for all h (K) k k H H →∞h i h i ∈ H and, in particular, lim fn (z) = f(z) for all z E. k k →∞ ∈

Proof: As fn(λ) = Kλ , fn (K) Kλ (K) fn (K) M Kλ (K) for all n, we h iH ≤ k kH k kH ≤ k kH can ﬁnd a subsequence (fn )k such that lim fn (λ) exists. k k k →∞ As (fnk )k is a bounded and pointwisely convergent sequence in an RKHS, it is weakly convergent, see A.3.2; in other words, lim fnk , h (K) = f , h (K) for all h (K) k H H →∞h i h i ∈ H where f(z) := lim fn (z) and f (K). k k →∞ ∈ H

3.5.5 Theorem. If there exists an om µ for (Pn)n 0 then GK Ereg. ≥ ⊂

Proof: The case GK = ∅ is trivial. Let now λ G . According to 2.2.2, z κ(z) := K(z, z) is continuous in G . Hence ∈ K 7→ K there exist c > 0 and r > 0 such that κ(z) c for all z U := z C : z λ < r . ≤ ∈ { ∈ | − | } Assume λ / reg(D). Then there exists a sequence (qn)n in C[z] such that qn = 1 for ∈ k kH C all n and (D λ id)qn 0 as n . By (2.4), p (K) p for all p [z]; k − kH → → ∞ k kH ≤ k kH ∈ hence (D λ id)qn (K) 0 and qn (K) 1 for all n. k − kH → k kH ≤ Now, by 1.4.4, (z λ)q (z) 0 for all z E which implies q (z) 0 for all z E λ − n → ∈ n → ∈ \ { } and we can apply 3.5.4 to obtain a subsequence weakly convergent in (K). We tacitly H assume that (qn)n is already such a subsequence and denote the weak limit by q. As the weak limit is also the pointwise limit and q (K) is holomorphic in an open ∈ H neighborhood of λ, we get q(λ) = 0. 53

As µ is an om,

2 2 2 z λ 2 2 1 = qn = qn 2 = | − |2 qn(z) dµ + qn(z) dµ Lµ z λ H C U | − | U \ R R 1 2 2 2 r2 z λ qn(z) dµ + qn(z) dµ ≤ C U | − | U \ R R 1 2 2 2 (D λ id)qn 2 + qn(z) dµ for all n . (3.10) r Lµ ≤ − U

R Furthermore, for z U, we have qn(z) κ(z) qn (K) √c. ∈ ≤ k kH ≤ 2 Using dominated convergence and lim qn(zp) = 0, we obtain lim qn(z) dµ = 0. n n →∞ →∞ U

Together with (D λ id)qn 0, this yields R k − kH → 1 2 2 lim 2 (D λ id)qn + qn(z) dµ = 0 n r →∞ − H U in contradiction to (3.10). R Therefore, we get λ reg(D) to complete the proof. ∈

We are now able to characterize the set GK by varoius means. For R > 0, recall AR = z E : κ(z) R , see 2.2.7. { ∈ ≤ }

3.5.6 Theorem. The set GK can be characterized as follows.

G = z E◦ : κ is continuous in a neighborhood of z K ∈ = z E◦ : κ is bounded in a neighborhood of z ∈ 2 = z E◦ : Pn(z) is uniformly convergent in a neighborhood of z ∈ n 0 ≥ P = Ak◦ . k N ∪∈ If, in addition, there exists an om for (Pn)n then GK = Ereg and Ereg = ∅ E◦ = ∅. 6 ⇐⇒ 6

Proof: Recall that, by construction, G is open and G E◦. K K ⊂ (a) Let z G . By 2.2.2, κ is continuous in G . In particular, κ is continuous in a 0 ∈ K K neighborhood of z0.

(b) Let now z E◦ such that κ is continuous in a neighborhood of z . Then clearly there 0 ∈ 0 exists a neighborhood of z0 where κ is bounded.

(c) Let z0 E◦ such that κ is bounded in a neighborhood of z0. Then there exist an open disk U∈ centered at z and m N such that κ(z) m for all z U, in other words, 0 ∈ ≤ ∈ U Am. As U is open, this implies z0 Am◦ Ak◦. k N ⊂ ∈ ⊂ ∪∈ (d) The inclusion Ak◦ GK is due to 2.2.9. k N ∪∈ ⊂ 54 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

2 (e) We show GK = z E◦ : Pn(z) is uniformly convergent in a neighborhood of z . ∈ n 0 ≥ P Due to (a) and (b), for z G , we can ﬁnd m N such that z A◦ and a compact 0 ∈ K ∈ 0 ∈ m disc C Am◦ centered at z0 such that κ is continuous on C. Now Dini’s theorem (see ⊂ 2 [K¨o, 15.7], for example) yields uniform convergence of Pn(z) on C. n 0 ≥ For the converse inclusion, use that the limit of a uniformlyP con vergent sequence of con- 2 tinuous functions is continuous. Suppose that Pn(z) is uniformly convergent in a n 0 neighborhood U of z . ≥ 0 P This implies that κ is continuous in U; using (b), (c), and (d), we see z G . 0 ∈ K (f) If there exists an om for (Pn)n 0 then GK = Ereg is an immediate consequence of 3.5.2 ≥ and 3.5.5. Moreover, by 2.2.9, GK is dense in E◦; hence GK = ∅ E◦ = ∅. 6 ⇐⇒ 6

Remark. In the case that there exists a compactly supported om, the characterization

G = z E◦ : κ is continuous in a neighborhood of z K ∈ is proved in [Con1, II.7.6]. In the more special case that supp(µ) is contained in the closed unit disk, the equality GK = Ereg is the subject of [Tr, Theorem 1.1].

3.5.7 Theorem. Let µ be an om such that P 2 = L2 . Then µ(E Λ ) = 0 and E Λ µ µ \ µ reg ∩ µ does not have a limit point within E . In particular, E Λ is a discrete set. reg reg ∩ µ Proof: In 3.4.8, we have already shown µ(E Λ ) = 0. \ µ Assume now that there exists a convergent sequence (zn)n of mutually diﬀerent points in Ereg Λµ such that z := lim zn Ereg. ∗ n ∩ →∞ ∈ 1 Recall that, by 3.1.7, µ( zn ) = for all n and note that µ( zn ) 0 as n . { } κ(zn) { } → → ∞ According to 3.5.6, κ is continuous on Ereg which yields a contradiction. Therefore, a limit point of E Λ cannot belong to E . reg ∩ µ reg

Remark. Note that we did not have to require z Λµ. ∗ ∈ Furthermore, recall that if D is essentially normal then E = Λµ, see 3.3.2; hence in that case we have Ereg = ∅ and 3.5.7 is trivial. Moreover, note that if supp(µ) is compact and 2 2 Pµ = Lµ then D is essentially normal, see 3.2.10.

3.5.8 Open Questions.

(i) Is G = E◦ possible ? K 6 2 2 (ii) Given an om µ such that P = L , is E◦ always non-empty ? µ 6 µ According to [Con1, II.7.11], (i) is an open question even if there exists a compactly supported om. Regarding orthonormalizing measures supported on the closed unit disk, 55

this question also appears – and remains unanswered – in [Tr, Remark after Cor. 1.3]. In 4.5.4 we will give an example where E = z C : z < 1 1 and κ is not continuous { ∈ | | } ∪ { } at z = 1 but still continuous in E◦.

Assume now that there exists an om µ for (P ) such that G = E◦. As shown in 3.5.6, n n K 6 G = E , hence λ E◦ G is not a regular value of D and, therefore, not a regular value K reg ∈ \ K of M , either. As M is normal, reg(M ) = ρ(M ), implying E◦ G σ(M ) = supp(µ). µ µ µ µ \ K ⊂ µ Moreover, GK does not contain “holes” in the sense that the interior of a domain, whose boundary is a simply connected rectiﬁable curve in GK , is completely contained in GK, see 2.2.14; furthermore, as shown in 2.2.9, GK is dense in E◦. For any f (K), there exists a sequence (q ) of polynomials convergent to f w.r.t. the ∈ H n n norm in (K), hence pointwisely on E and, in particular, uniformly on every compact H subset of GK. Thus, if E◦ GK = ∅ then there exists a sequence of polynomials con- \ 6 vergent pointwisely on E and uniformly on any compact subset of GK whose limit is not holomorphic in (at least some subset of) E◦ G . \ K Given a pointwisely, on some open set A C, convergent sequence of polynomials, the question what those domains look like where⊂ the limit function is not holomorphic, was already discussed in a paper by Hartogs and Rosenthal [HR1] published in 1928. The answer is rather complicated; more recently, that topic appears in a paper by Davie [Da]. Yet, we still cannot give an answer to question (i) above. It is worth mentioning that Beardon and Minda [BM] show how to construct a sequence (q ) of polynomials such that q (z) 1 whenever z is a non-negative real number while, n n n → otherwise, qn(z) 0. They also present a sequence of polynomials convergent to 0 on the whole complex→plane except for one point where the limit is 1.

Concerning question (ii), Conway proves [Con1, VIII.4.3] that if there exists a compactly 2 2 supported om µ such that Pµ = Lµ then E◦ = ∅, using an idea by Thomson [Th]. In a later paper [CY, sec. 4] he calls6 that method “excellen6 t but complicated”, supposing that there might exist a somewhat simpler proof. However, we add the following.

3.5.9 Theorem. Let µ be a compactly supported om for (Pn)n 0. The following proper- ≥ ties are equivalent.

2 2 (i) Pµ = Lµ, (ii) D is formally normal, (iii) D is essentially normal, (iv) (z z) P 2, 7→ ∈ µ ∞ 2 2 (v) d0j = d10 , j=1 | | | | P (vi) E = Λµ,

(vii) E◦ = ∅.

(viii) Ereg = ∅. 56 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

Proof: We have already proved equivalence of the first five properties in 3.2.10. The implication (iii) (vi) is due to 3.3.2 and (vi) (vii) is clear. By 3.5.6, we have (vii) (viii), and, ⇒finally, (vii) implies (i) by [Con1,⇒VIII.4.3]. ⇐⇒

Remark. The implications (iii) (i), (iii) (vi), and (i) (ii) as well as the equivalence (vii) (viii) hold even if supp⇒(µ) is un⇒bounded, but w⇒e still do not know whether, in ⇐⇒ 2 2 the unbounded case, E◦ = ∅ implies Pµ = Lµ.

3.6 Point Evaluation on C[z]

While in an RKHS point evaluation always is well deﬁned and continuous, point evaluation 2 C in Lµ in general is not deﬁned. However, point evaluation exists on [z] and if µ is an om 2 for (Pn)n 0 then we can establish relations between the members of Pµ and the members ≥ 2 of (K). In particular, we will see that every element in Pµ has a representative that poinHtwisely on E coincides with a function in (K). H

3.6.1 Lemma. Let µ be a measure on B(C) such that µ(C) = 1, and C[z] L2 . Fix ⊂ µ a C and deﬁne ηa := inf p 2 : p C[z] , p(a) = 1 . ∈ k kLµ ∈

There exists a constant ca > 0 such that

p(a) ca p 2 for all p C[z] (3.11) ≤ k kLµ ∈ if and only if ηa > 0. Moreover, if η > 0 then c = 1 is the smallest constant satisfying (3.11). a a ηa

Proof: Suppose that there exists ca > 0 such that p(a) ca p 2 for all p C[z] Lµ ≤C k k ∈ and assume ηa = 0. Then there exists a sequence ( pn)n in [z] such that pn(a) = 1 for all n and pn 2 0 as n . On the other hand, 1 = pn(a) ca pn 2 for all n k kLµ → → ∞ ≤ k kLµ which is a contradiction. Hence ηa > 0.

For the converse, suppose that ηa > 0. Now choose arbitrary p C[z] with p(a) = 0, and set p(z) := p(z) . Then p(a) = 1 and ∈ 6 p(a)

2 2 2 2 e 2 2 2 e 2 p 2 = p(z) dµ = p(a) p(z) dµ = p(a) p 2 ηa p(a) . k kLµ k kLµ ≥

R R e e 57

1 Thus p(a) p 2 . Note that this is also valid if p(a) = 0. ≤ ηa k kLµ 1 It remains to show that ca = is the smallest constant satisfying (3.11). ηa By deﬁnition of ηa, there exists a sequence (pn)n in C[z] such that pn(a) = 1 for all n and pn 2 ηa as n . Let now c > 0 satisfy (3.11). Then 1 = pn(a) c pn 2 cηa k kLµ → → ∞ ≤ k kLµ → as n , which implies 1 c and the proof is complete. ηa → ∞ ≤

C 2 3.6.2 Theorem. Let µ be as in 3.6.1 and set A := a : ηa > 0 . For f Pµ , there C 2 { ∈ } ∈ exists a sequence (pn)n in [z], convergent in Lµ with limit f. Moreover, the function

K K f : A C , f (z) := lim pn(z) , n → →∞ is well-deﬁned, i.e. the limit exists and does not depend on the particular choice of (pn)n.

C 2 Proof: Let (pn)n and (qn)n be sequences in [z], both convergent to f in Lµ (clearly, 1 such sequences exist). Fix a A. Now 3.6.1 yields pn(a) pm(a) pn pm 2 for ∈ − ≤ ηa k − kLµ all m, n N. Thus there exists fa := lim pn(a). Analogously, ga := lim qn(a) exists. n n ∈ →∞ →∞ 2 The sequence p1, q1, p2, q2, . . . converges to f in Lµ, too, and the inequality above yields fa = ga.

3.6.3 Corollary. In the situation of 3.6.2,

K 1 2 f (a) f 2 for all f Pµ and all a A . (3.12) ≤ ηa k kLµ ∈ ∈

2 K Moreover, if (fn)n is a conver gent sequence in Pµ with limit f, then fn is pointwisely on A convergent with limit f K.

K Proof: According to 3.6.2, f (a) = lim pn(a) where (pn)n is a sequence of polynomials n 2 →∞ convergent to f in Lµ. 1 Due to 3.6.1, we have pn(a) pn 2 . Letting n , we obtain (3.12). ≤ ηa k kLµ → ∞ Let now (f ) be a sequence in P 2, f f in L2 . Then f P 2 and (3.12) yields n n µ n → µ ∈ µ K K 1 N fn (a) f (a) fn f 2 for all n − ≤ ηa − Lµ ∈ and, with n , the assertion follo ws. → ∞

We can now apply this to orthonormalizing measures.

3.6.4 Lemma. Let µ be an om for (Pn)n 0 and deﬁne ηa as in 3.6.1. ≥ 1 Then ηa > 0 if and only if a E. Furthermore, if a E then ηa = . ∈ ∈ √K(a,a) 58 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

2 Proof: We start with ηa > 0. Fix b = (bn)n ` . For ε > 0, we can ﬁnd N N such m 1 ∈ ∈ 2 2 that bk < ε for all n, m N with m > n > N and 3.6.1 yields k=n | | ∈ P m m 1 1 bkPk(a) bkPk < ε , ηa 2 ηa k=n ≤ k=n Lµ

P P ∞ showing that bkPk(a) exists. Using 3.6.1 again, we obtain k=0 P M M 1 1 bnPn(a) bnPn b 2 for all M N . ηa 2 ηa ` n=0 ≤ n=0 Lµ ≤ k k ∈

P P ∞ 2 Therefore, b bn Pn(a) is a continuous linear functional on ` . 7→ n=0 2 Now the Riesz represenP tation theorem implies Pn(a) ` ; in other words, a E. n ∈ ∈ N For the converse, suppose a E. Let p C[z]. Then p = an Pn where N = deg p ∈ ∈ n=0 and a , . . . , aN C. Moreover, 0 ∈ P N N 1 N 1 2 2 2 2 p(a) = an Pn(a) an Pn(a) p 2 K(a, a) . Lµ n=0 ≤ n=0 | | n=0 ≤ k k p P P P Now 3.6.1 implies η > 0 and 1 K(a, a). It remains to show that 1 K(a, a). a ηa ηa 1 ≤ ≥ By 3.6.1, p(a) p 2 for all p C[z]. Thus ≤ ηa k kLµ p∈ p

M M M M 1 2 1 1 2 2 Pn(a) = Pn(a) Pn (a) Pn(a) Pn = Pn(a) ηa 2 ηa n=0 n=0 ≤ n=0 Lµ n=0 P 1 P P P K(a, a) for all M N . ≤ ηa ∈ p For M this yields K(a, a) 1 K(a, a), hence 1 K(a, a) as asserted. → ∞ ≤ ηa ηa ≥ p p As we have just seen, E = a C : η > 0 is another characterization of the set E if an { ∈ a } om exists.

3.6.5 Theorem. Let µ = αλ + µ be an om with αλ and µ its absolutely continuous ⊥ ⊥ and singular parts w.r.t. 2-dimensional Lebesgue measure, respectively. 1 2 2 If there exists an open set U such that α dλ < then U E and Pµ = Lµ. U ∞ ⊂ 6 R 2 C 2 Proof: For ν := αλ U, deﬁne Aν as in 2.3.1. Clearly, [z] Aν and p 2 p 2 for | ⊂ k kAν ≤ k kLµ all p C[z]. Let a U. Using 2.3.2, we ﬁnd ca > 0 such that p(a) ca p 2 ca p 2 ∈ ∈ | | ≤ k kAν ≤ k kLµ for all p C[z] and, according to 3.6.1, η > 0 which, by 3.6.4, is equivalent to a E. ∈ a ∈ Thus U E. ⊂ 59

Assume P 2 = L2 . Then 3.4.8 implies µ(E Λ ) = 0 and hence µ(U Λ ) = 0. µ µ \ µ \ µ 1 Thus α vanishes λ-almost everywhere on U in contradiction to α dλ < . U ∞ Therefore, P 2 = L2 . µ 6 µ R

2 Recall, for a E, ka = Pn(a) Pn Pµ . The following lemma extends 3.1.5. ∈ n 0 ∈ ≥ P

K 3.6.6 Lemma. Let µ be an om for (Pn)n 0 and E = ∅. Then ka , f L2 = f (a) for ≥ 6 h i µ all a E and all f P 2. ∈ ∈ µ If µ( z ) > 0 for some z then z E and f K (z) = f(z) for all f P 2. { } ∈ ∈ µ

N 2 2 Proof: Let f P . Note that P , f 2 P f in L as N . For a E, ∈ µ h n iLµ n → µ → ∞ ∈ 3.6.2 yields n=0 P K ∞ f (a) = Pn(a) Pn , f L2 = Pn(a) Pn , f = ka , f L2 . µ 2 µ n=0 h i n 0 Lµ h i ≥ P D P E If µ( z ) > 0 then, according to 3.1.7, z E. Moreover, the expression f(z) is well- deﬁned{ }for all f L2 . ∈ ∈ µ 2 2 By 3.1.8, P 1 z = µ( z ) kz, where P is the orthogonal projection in Lµ onto Pµ . This { } yields { } · 1 K 2 2 f(z) = µ( z ) 1 z , f Lµ = kz , f Lµ = f (z) . { } h { } i h i

2 3.6.7 Proposition. Let µ be an om for (Pn)n 0 and E = ∅. For f Pµ , deﬁne ≥ α(f) := f K. Then α is a partial isometry P 2 (K). 6 ∈ µ → H

2 2 2 Proof: Let ι : Pµ ` be the isomorphism given by ι(f) := Pn , f Lµ n 0. → h i ≥ Now let β : `2 (K) as in 2.1.8, → H 2 β(b) = bnPn where b = (bn)n ` n 0 ∈ ≥ P 2 and recall that β is a partial isometry. Therefore, β ι is a partial isometry Pµ (K) and ◦ → H K 2 β ι(f) = Pn , f Lµ Pn = f = α(f) . ◦ n 0h i ≥ P Hence α : P 2 (K), α(f) := f K, is a partial isometry and, in particular, f K (K) µ → H ∈ H for all f P 2. ∈ µ

Note that α maps P P 2 to P (K). n ∈ µ n ∈ H 60 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

3.6.8 Example. Let γ : [0, 1] C be a simply closed piecewise continuously diﬀeren- → tiable curve and µ absolutely continuous w.r.t. line measure on γ, having density α, such that 1 α(z) dz < , γ | | ∞ see 2.3.8 for the details; see also 3.3.7.R Moreover, assume µ(γ) = 1, hence µ is an om for some (Pn)n, and let V be the bounded component of C γ. \ 2 Recall 2.3.11, where we have shown V E and also, to f Pµ , have assigned a holomorphic function V C which, as we kno⊂w now, coincides with∈ f K V . → | Using 3.3.6 and 3.3.7, we also see E V = V ∂V = V γ; in general, we do not know whether E and γ are disjoint or not.⊂ ∪ ∪

3.6.9 Corollary. Let µ be an om for (Pn)n 0 and E = ∅. The following properties are ≥ equivalent. 6 (i) (P ) is not an onb in (K). n n H (ii) There exists f P 2, f = 0, such that f K = 0. ∈ µ 6 2 (iii) There exists f P , f = 0, such that k , f 2 = 0 for all a E. ∈ µ 6 h a iLµ ∈ 2 (iv) There exists b = (bn)n 0 ` , b = 0, such that bn Pn(z) = 0 for all z E. ≥ ∈ 6 n 0 ∈ ≥ P Proof: The equivalence (ii) (iii) is an immediate consequence of 3.6.6; the equivalence (i) (iv) is due to ⇐2.1.9.⇒ As to (iii) (iv), note that, for z E, we have ⇐⇒ ⇐⇒ ∈ 2 2 kz , f Lµ = bn Pn(z) whenever f = bn Pn Pµ . h i n 0 n 0 ∈ ≥ ≥ P P If we do not know whether there exists an om then, for f , we can deﬁne ∈ H K K f : E C , f (z) := kz , f = Pn(z) Pn , f → h iH n 0 h iH ≥ P K and Cauchy-Schwarz yields f (z) kz f = K(z, z) f ; matching (3.12) if an om exists. Using this deﬁnition≤fork fkKHand· k replacingkH L2 and· kP 2kHby in the proofs p µ µ H of 3.6.7 and 3.6.9, we get the following proposition.

K 3.6.10 Proposition. Let (Pn)n 0 such that E = ∅. For f deﬁne α(f) := f . ≥ Then α is a partial isometry (K). 6 ∈ H H → H Furthermore, the following properties are equivalent. (i) (P ) is not an onb in (K). n n H (ii) There exists f , f = 0, such that f K = 0. ∈ H 6 (iii) There exists f , f = 0, such that kz , f = 0 for all z E. ∈ H 6 h iH ∈ 2 (iv) There exists b = (bn)n 0 ` , b = 0, such that bn Pn(z) = 0 for all z E. ≥ ∈ 6 n 0 ∈ ≥ P 61

3.6.11 Summary. Let W be the closure of the linear span of Pn(z) n 0 : z E 2 2 ≥ ∈ in ` , see also 2.1.8, and note that the canonical isomorphism ` maps W ⊥ onto → H f : f K = 0 = (α). { ∈ H } N Taking into account that ran(D a id)⊥ = C k for all a E, see 3.1.3, which implies − · a ∈ k ⊥ = ran(D a id) in , we obtain { a} − H K f : f = 0 = (α) = f : ka , f = 0 for all a E = ran(D a id) . H a E { ∈ H } N ∈ H h i ∈ ∩∈ −

2 2 ` W ⊥ W ` W ⊥ W

(K) (K) H H

(α) (α)⊥ (α) N N 2 N H Lµ

2 (Pµ )⊥ Figure 3. The arrows denote isometric isomorphisms. Figure 4. The canonical isomorphism `2 given → H If there exists an om for (Pn)n 0 then we by ≥ can embed isometrically into L2 . (bn)n 0 bn Pn ( ) H µ ≥ 7→ n 0 ∗ ≥ P maps W and W onto (α) and (α), ⊥ N ⊥ N respectively. We can also regard ( ) as the ∗ map β : `2 (K) deﬁned in 2.1.8 which → H is a partial isometry with initial space W ; hence its restricton to W is an isometric isomorphism W (K). → H

3.7 Orthonormal Bases in and (K) H H

In this section, we will study conditions for (Pn)n being an onb in (K). If (Pn)n is an onb in (K) then and (K) are isometrically isomorphic via the Hidentity map on C[z]. H H H 2 Recall that (Pn)n is an onb in (K) if and only if (Pn(z))n : z E is total in ` or, H { 2 ∈ } using the notations from 2.1.8 and 3.6.11, W ⊥ = 0 , i.e. W = ` . See also 3.6.9. { }

2 3.7.1 Lemma. Let µ be an om for (Pn)n 0. If there exist b = (bn)n 0 ` , b = 0, and ≥ ≥ M B(C) such that ∈ 6 ∈ ∞ bn Pn(z) = 0 for all z M n=0 ∈ then µ(C M) > 0. P \ 62 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

2 Proof: For f := bn Pn Lµ 0 , there exists a sequence (fk)k N, n 0 ∈ \ { } ∈ ≥ P nk fk(z) := bj Pj(z) , j=0 P which is µ-a.e. convergent to a representative f of f where (nk)k is a monotonically increasing sequence of integers.

Define A := z M : f(z) = lim fk(z) . Hence M A is a µ-nullset. k ∈ →∞ \ Note that lim fk(z) = 0 for all z M. Therefore, f(z) = 0 for all z A. Moreover, k →∞ ∈ ∈ 2 2 2 2 0 = f 2 = f dµ + f dµ + f dµ . Lµ 6 k k A | | M A | | C M | | R R\ \R The first integral vanishes, as f(z) = 0 for all z A; the second integral vanishes, too, ∈ as M A is a µ-nullset. Thus the last integral must be different from 0 which implies µ(C \M) > 0. \ 2 Remark. Note that we did not have to require Pn(z) n 0 ` for z M here. ≥ ∈ ∈

Recall that E is an Fσ-set (see 2.2.7) and, therefore, measurable.

3.7.2 Theorem. If there exists an om µ for (Pn)n 0 such that µ(E) = 1 then (Pn)n is ≥ an onb in (K). H Proof: Note that µ(C) = 1; hence here C E is a µ-nullset. \ We have to show W ⊥ = 0 . Choose b = (b ) W ⊥. In other words, { } n n ∈

bn Pn(z) = 0 for all z E . n 0 ∈ ≥ P If b = 0 then 3.7.1 implies µ(C E) > 0 which is a contradiction. Thus b = 0 and 6 \ W ⊥ = 0 as asserted. { }

The converse is not true. Our most famous example, see also 1.2.2, 1.3.8, and 2.1.11, shows that even µ(E) = 0 and E open is possible while (P ) is an onb in (K): n n H

n 3.7.3 Example. Let Pn(z) := z for n N0. Then one-dimensional Lebesgue measure on the unit circle is, up to a constant factor,∈ an om. For the moment denote this measure by µ. Here, E is the interior of the unit disk and hence µ(E) = 0. However, (Pn)n is an onb in (K). Note that, according to 3.2.14, µ is the only om for (P ) . H n n

3.7.4 Theorem. Let µ be a compactly supported om for (Pn)n 0 and E = ∅. If (Pn)n ≥ 6 is not an onb in (K) then dim(W ⊥) = . H ∞ 63

Proof: As (P ) is not an onb in (K), there exists h P 2, h = 0, such that hK = 0, n n H ∈ µ 6 see 3.6.9.

As supp(µ) is compact, the multiplication operator Mµ is continuous; hence the Hessen- berg operator D is continuous, too, and its closure is deﬁned on the whole space which we identify with P 2. Moreover, 0 = (M )jh P 2 for all j N. In particular, H µ 6 µ ∈ µ ∈

j j K 3.6.6 j j Pn , (Mµ) h L2 Pn(z) = (Mµ) h (z) = kz , (Mµ) h L2 = kz , D h n 0 µ µ H ≥ P j 3.1.3 j = (D∗) kz , h = λ kz , h H H j j 3.6.6 j K = λ kz , h = λ kz , h P 2 = λ h (z) = 0 h iH h i µ

(j) (j) j for all z E and all j N. Thus α := (αn ) := P , (M ) h 2 W ⊥. ∈ ∈ n h n µ iLµ n ∈ (j) Assume now that dim(W ⊥) = k N. Then α : j = 0, . . . , k cannot be linearly ∈ { } independent in W ⊥. k (j) 2 Hence there exist c0, . . . , ck C, ci = 0 for some i, such that cj α = 0 in ` which means ∈ 6 j=0 k k P j j 0 = cj Pn , (Mµ) h 2 = Pn , cj(Mµ) h for all n . Lµ 2 j=0 j=0 Lµ P D P E k k 2 j 2 j As (Pn)n is an onb in Pµ and cj(Mµ) h Pµ , we get cj(Mµ) h = 0; in other words, j=0 ∈ j=0 P P k 2 j 2 0 = cj(Mµ) h = ph dµ 2 j=0 Lµ | |

P R where p(z) := c + c z + + c zk. Thus ph = 0 µ-almost everywhere. 0 1 · · · k If µ( z ) > 0 for some z then z E, see 3.1.7, and, by 3.6.6, h(z ) = hK(z ) = 0. { 0} 0 0 ∈ 0 0 As p has only ﬁnitely many zeros, we ﬁnally obtain h = 0 µ-almost everywhere in contra- 2 diction to h = 0 in P . Thus dim(W ⊥) = . 6 µ ∞

Due to 2.1.9, the following is an immediate consequence.

3.7.5 Corollary. If, for a sequence of polynomials (Pn)n 0, the Hessenberg operator D ≥ is continuous and 0 < dim(W ⊥) < then there exists no om. ∞

n For a Ereg, deﬁne Va := ran ( D a id) . Clearly, Va . n N ∈ ∩∈ − ⊂ H

3.7.6 Lemma. For (Pn)n 0 such that Ereg = ∅ and ﬁxed a Ereg, denote by Z the ≥ 6 ∈ connected component of E containing a. Then V = f : f K Z = 0 . reg a ∈ H | 64 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

Proof: If f V then, for every n N, there exists f dom(D) such that ∈ a ∈ n ∈ ⊂ H f = ( D a id)nf . According to 3.6.10, for z E, we obtain − n ∈ K n n f (z) = kz , f = kz , ( D a id) fn = (D∗ a id) kz , f h iH − H − H n n K = (z a) kz , fn = (z a) fn (z) − h iH − K K where f and fn are members of (K) which, see 3.5.2, are holomorphic in Ereg. Thus, for every n N, f K has a zero ofHorder n at a. As f K is holomorphic in Z, and Z is ∈ ≥ connected by deﬁnition, this yields f K(z) = 0 for all z Z; in other words, f K Z = 0. ∈ | K To prove the converse inclusion, let f such that f Z = 0. Choose a sequence (pn)n C ∈ H | K K in [z] such that pn f w.r.t. as n . Thus pn f in (K) which yields p (z) f K(z) for all→z E; in particulark·kH p→(a∞) 0. Deﬁne→ H n → ∈ n → C C pn(z) pn(a) qn : , qn(z) := z−a for z = a , qn(a) := pn0 (a) . → − 6

Clearly, qn C[z] and ( D a id)qn = pn pn(a)P0 f in . Now a reg(D) = reg(D) implies that∈(q ) is convergen− t in to −f , say. As→( D Ha id) is a closed∈ operator, we n n H 1 − obtain ( D a id)qn ( D a id)f1, hence f = ( D a id)f1 ran( D a id). − → − −K ∈ K− Moreover, an analogous calculation as above yields f (z) = (z a)f1 (z) for all z E and, as f K Z is holomorphic, f K Z = 0. − ∈ 1 | 1 | Starting with f1 instead of f, we can see that f1 ran( D a id) and obtain f2 such K ∈ − ∈ H n that ( D a id)f2 = f1 and f2 Z = 0 and, by induction, ﬁnally f ran ( D a id) − | ∈ n N − as asserted. ∩∈

Now let A Ereg such that A contains exactly one point of each connected component ⊂ K of Ereg and deﬁne V := Va = f : f Ereg = 0 . ∞ a A ∩∈ { ∈ H | } As we have just seen, V does not depend on the particular choice of A. Moreover, if ∞ K E = Ereg then V = f : f = 0 . ∞ { ∈ H }

3.7.7 Corollary. If Ereg = ∅ and V = 0 then (Pn)n 0 is an onb in (K). 6 ∞ { } ≥ H Proof: Assume that (P ) is not an onb in (K). According to 3.6.10, there exists n n H f , f = 0, such that f K = 0. This implies f V ; hence V = 0 in contradiction ∞ ∞ to∈theH premises.6 Thus (P ) is an onb in (K). ∈ 6 { } n n H

Recall that if an om exists then Ereg = GK is dense in E◦ and Ereg = ∅ E◦ = ∅, see section 3.5. 6 ⇐⇒ 6

3.7.8 Corollary. Let µ be a compactly supported om for (Pn)n 0 such that E◦ = ∅. If ≥ 6 dim(V ) < then (Pn)n is an onb in (K). ∞ ∞ H

Proof: If (Pn)n is not an onb in (K) then 3.7.4 implies dim(W ⊥) = . Furthermore, H 2 ∞ K recall that the canonical isomorphism ` maps W ⊥ onto f : f = 0 , see → H { ∈ H } 3.6.11, and f : f K = 0 V . This yields dim(V ) = . { ∈ H } ⊂ ∞ ∞ ∞ 65

Note that if E◦ = ∅ then V remains undeﬁned. If, for example, D is essentially normal ∞ then E = Λµ = z C : µ( z ) > 0 where µ is the unique om for (Pn)n 0, see 3.3.2; in ∈ { } } ≥ this case E◦ = ∅. 2 2 Note also that if µ is an om such that Pµ = Lµ but D is not essentially normal then E supp(µ) = C, see 3.4.10(iii). If, in particular, supp(µ) = C then E◦ = ∅. ∪ 6 6

In section 4.2, concerning om supported on the unit circle, we will encounter the space V from another point of view. ∞

3.7.9 Theorem. Let µ be an om for (Pn)n 0 such that E◦ = ∅ and assume that there ≥ 6 exists λ C satisfying Df = λf for some f dom(D), f = 0. Then f = c kλ V ∞ with some∈ c C. ∈ 6 · ∈ ∈

2 Proof: Identify with Pµ and note that Mµf = λf. Hence µ( λ ) > 0 and f = c0 1 λ H 2 { } · { } for some c0 = 0. By assumption, f Pµ , and 3.1.8 yields 1 λ = µ( λ )kλ; in particular, { } λ E. Hence6 f = c k for some c∈ C. { } ∈ · λ ∈

Clearly, λ / reg(D). Therefore, λ a = 0 for a Ereg and (D a id)kλ = (λ a)kλ. This ∈ n − 6 ∈ − − shows that kλ ran(D a id) for all n N and all a Ereg. Hence kλ V . ∈ − ∈ ∈ ∈ ∞

K Remark. According to 3.7.6, V = f : f Ereg = 0 . Therefore, in the situation of ∞ { ∈ H | } 3.7.9, k , k 2 = 0 for all a E . Moreover, h a λ iLµ ∈ reg

2 0 = ka , kλ Lµ = Pn(λ) Pn(a) = Ka , Kλ (K) for all a Ereg h i n 0 h iH ∈ ≥ P regardless of (P ) being an onb in (K) or not. n n H If, in particular, λ is a boundary point of E then (K) contains a function which is reg H not continuous at λ, namely Kλ. In 4.2.10 we will see that this can actually occur.

3.7.10 Discrete om. Suppose now that there exists a discrete om µ for (Pn)n 0, i.e. ≥ there are mutually diﬀerent complex numbers z , i N, and i ∈

µ = giδzi i N ∈ P where δzi denotes Dirac measure at zi and gi are positive numbers such that their sum is equal to 1. Now consider Λµ := z C : µ( z ) > 0 = z1, z2, . . . . By construction, µ(Λ ) = 1 and, according to 3.1.7, Λ∈ E. {This} implies µ{(E) = 1 and,} using 3.7.2, we µ µ ⊂ conclude that (Pn)n 0 is an onb in (K). ≥ H 66 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

3.7.11 Proposition. Let µ be a discrete om for (Pn)n 0 such that Λµ is closed. Then ≥ the following properties hold.

(i) Either E = C or E = Λµ.

(ii) If, in addition, Λµ is bounded then E = Λµ and µ is the unique om for (Pn)n.

Proof: As Λ is closed, Λ = supp(µ) = σ(M ). As C Λ is connected, 3.3.1(v) implies µ µ µ \ µ either C Λµ E or (C Λµ) E = ∅. Combining this with Λµ E, see 3.1.7, we get \ ⊂ \ ∩ ⊂ either E = C or E = Λµ.

If, in addition, Λµ is bounded then Mµ, and hence D, is bounded. Now 3.1.6 implies that E is bounded. Thus only the case E = Λµ remains. Moreover, according to 3.2.14, µ is the unique om for (Pn)n.

We now turn to the situation that (P ) is an onb in (K) as well as in L2 . n n H µ First consider the case that D is essentially normal, i.e. its closure D is the multiplication operator Mµ.

3.7.12 Theorem. Suppose that D is essentially normal. Then there exists a unique 2 2 om µ, Pµ = Lµ, and E = Λµ. Moreover, (Pn)n is an onb in (K) if and only if µ is discrete. H

2 2 Proof: Due to 1.3.7, there exists a unique om µ and Pµ = Lµ; by 3.3.2, E = Λµ. According to 3.1.8, 1 z = µ( z ) kz for all z Λµ. { } { } · ∈ Let now (Pn)n be an onb in (K) and assume that µ is not discrete, i.e. there exists 2 H 2 0 = f Lµ such that 1 z , f L2 = 0 for all z Λµ. As (Pn)n is an onb in Lµ, 6 ∈ h { } i µ ∈ too, there exists an isometry ι : L2 (K) which is the identity on C[z]. Therefore, µ → H Kz , ι(f) (K) = 0 for all z Λµ = E. Furthermore, by 2.1.4, Kz : z E is total in h (K) implyingiH ι(f) = 0 whic∈h is a contradiction. Hence µ is discrete.{ ∈ } H As to the converse implication, note that if there exists a discrete om then (Pn)n is an onb in (K), see 3.7.10. H

Now suppose that D is not essentially normal but has a minimal normal extension in 2 2 2 the same space Pµ , i.e. there exists an om µ such that Pµ = Lµ and the multiplication operator Mµ is a proper extension of D.

2 2 3.7.13 Theorem. Let µ be an om for (Pn)n 0 such that Pµ = Lµ. If µ(E) = 1 then µ ≥ is discrete. Furthermore, (P ) is an onb in (K), too. n n H Proof: According to 3.7.2, µ(E) = 1 implies that (P ) is an onb in (K). n n H Moreover, due to 3.4.8, µ(E Λ ) = 0 which implies µ(Λ ) = µ(E) = 1. Hence µ is \ µ µ discrete. 67

3.7.14 Deﬁnitions. Let (K) be an RKHS with domain E. H A set ∆ E is called domain of determinacy for (K) if f(z) = 0 for all z ∆ implies f ⊂= 0. H ∈ If, for example, (K) is an RKHS with a connected domain E C such that all f (K) are holomorphicH in E then every subset of E having a limit⊂ point is a domain of∈ Hdeterminacy for (K). H A set ∆ E is called diagonal if K(z, w) = 0 for all z, w ∆, z = w. ⊂ ∈ 6 Note that, for (K) as in 2.1.8, K(z, w) = 0 if and only if P (z) P (w) in `2. H n n⊥ n n Thus ∆ E is diagonal if and only if P (∆) := P (z) : z ∆ is an orthogonal ⊂ n n ∈ system in `2.

3.7.15 Theorem. Let (Pn)n 0 be a sequence of polynomials as in 1.1.1, deﬁne (K) as ≥ in 2.1.8, and let ∆ E. The following properties are equivalent. H ⊂ 2 1 (i) (Pn)n is an onb in Lµ where µ = K(z,z) δz with δz the Dirac measure at z, z ∆ ∈ P (ii) P (∆) is an orthogonal basis of `2,

(iii) ∆ is a diagonal domain of determinacy for (K) and (P ) is an onb in (K). H n n H

Proof: We start with (i) (ii). ⇒ According to 3.1.8, 1 z = µ( z ) kz for all z ∆ = Λµ E. { } { } ∈ ⊂ In 3.7.10 we have seen that if a discrete om exists then (Pn)n is an onb in (K). Moreover, 2 H here Lµ and (K) are isometrically isomorphic via Pn Pn and for z, w ∆, z = w, we obtain H 7→ ∈ 6

1 1 0 = µ( w ) 1 w , µ( z )1 z L2 = Kw , Kz (K) = K(z, w) = Pn(z) Pn(w) . { } { } µ h iH n 0 { } { } ≥ P 2 2 Hence P (∆) is an orthogonal system in ` . Now choose a = (an)n ` such that 2 ∈ an Pn(z) = 0 for all z ∆ and set f := an Pn Lµ. Then n 0 ∈ n 0 ∈ ≥ ≥ P P 1 0 = a P (z) = f , k 2 = f , 1 for all z ∆ n n z Lµ µ( z ) z L2 n 0 h i { } µ ∈ ≥ { } P which implies f = 0 and hence a = 0. Thus P (∆) is an orthogonal basis in `2.

Next we prove (ii) (iii). ⇒ Obviously, if P (∆) is an orthogonal system in `2 then ∆ is diagonal. As here P (∆) is 2 total in ` , 2.1.9 implies that (Pn)n is an onb in (K). H 2 Let now f := an Pn (K) with a = (an)n 0 ` such that f(z) = 0 for all z ∆. n 0 ∈ H ≥ ∈ ∈ ≥ P 68 3 RELATIONS BETWEEN THE SPACES , (K), AND L2 H H µ

This means 0 = f(z) = Kz , f (K) = an Pn(z) for all z ∆ h iH n 0 ∈ ≥ P and, again because P (∆) is total in `2, we get a = 0 and hence f = 0. Thus ∆ is a domain of determinacy for (K). H It remains to show that (iii) implies (i).

As ∆ is diagonal, Kw , Kz (K) = K(z, w) = 0 for z, w ∆, z = w. Thus Kz : z ∆ H is an orthogonal systemh in i (K). Furthermore, since ∆ is∈ a domain6 of determinacy{ ∈then} H 1 0 = f(z) = Kz , f (K) for all z ∆ implies f = 0. Hence Kz − Kz : z ∆ is an h iH 1 ∈ 1 {k k ∈ } onb in (K). Recall that Kz − = . H k k √K(z,z) Clearly, ∆ is countable. For z ∆, let δ denote Dirac measure at z and define ∈ z 1 µ := K(z,z) δz . z ∆ ∈ P 2 By construction, K(z, z) 1 z : z ∆ is an onb in Lµ. { } ∈ 2 Thus we can definepan isometric isomorphism β : (K) Lµ by β(Kz) := K(z, z) 1 z 2 H → { } for z ∆. Then β(Pn) n is an onb in Lµ since (Pn)n is an onb in (K). The members of L2 ∈are defined pointwisely for z ∆ and H µ ∈ 1 β(Pn)(z) = K z,z 1 z , β(Pn) 2 = Kz , Pn (K) = Pn(z) for all z ∆ . ( ) { } Lµ h iH ∈

Thus β maps P (K) to P L2 and we are done. n ∈ H n ∈ µ

3.8 Concluding Remarks

Before we turn to applications of the theory developed so far, we graphically summarize some remarkable properties and implications in the following ﬁgure. 2 2 If µ is an om for (Pn)n 0 such that Pµ = Lµ, we will, for abbreviation, call µ an obm. ≥

Figure 5. (opposite page) The diagram shows some of the implications proved so far. Note, for example, that if µ is a compactly supported (hence D is continuous) discrete obm for (Pn)n then all of the shown statements apply. 69

there exists om µ (Pn)n is ⊥ W = {0} s.t. µ(E) = 1 an onb in H(K)

there exists a discrete om

there exists a unique om which is and a discrete obm there exists om µ

s.t. E = Λµ there exists a discrete obm

D is essentially normal

∃∆ ⊆ E s.t. {(Pn(z))n : z ∈ ∆} is an onb in `2

there exists an obm and

there exists om µ D is

s.t. µ(E \ Λµ) = 0 formally normal

E is bounded D is continuous

if om exists

there exists a unique om 70 4 APPLICATIONS AND EXAMPLES

4 Applications and Examples

Given a sequence of polynomials (Pn)n 0 as in 1.1.1, it seems, in general, be rather difficult ≥ to determine whether there exists an om, and it is even more difficult to find an om. In some cases, however, we can quite easily conclude that there exists no om, see 4.5.2 or 4.5.4, for instance. Before we turn to a collection of various examples, we will have a look at orthogonal polynomials on (possibly subsets of) the real line or the unit circle. In the former case, as already mentioned in 3.1.9, we can restrict to the theory of self-adjoint extensions. In the latter case, the Hessenberg operator D must be isometric because here the multiplication operator M in L2 is isometric, as z = 1 for all z supp(µ). µ µ | | ∈

4.1 Orthogonal Polynomials on the Real Line

Recall that the support of an om for (Pn)n 0 is precisely the spectrum of a minimal ≥ normal extension of the Hessenberg operator D. Thus, if there exists an om µ such that supp(µ) R then D must have a normal extension with spectrum in R, i.e. a self-adjoint extension.⊂ In this case, D is symmetric. Thus its matrix representation is of the form a c 0 0 0 0 · · · c a c 0  0 1 1 · · ·  0 c1 a2 c2    ..   0 0 c2 a3 .   . . .   . .. ..   0 .    where ci C 0 and ai R for all i. A matrix of this form is called a Jacobi matrix. Accordingly∈ , D\ {is}called Jacobi∈ operator and (1.1) can be written as a 3-term recurrence,

z Pn(z) = cn 1Pn 1(z) + anPn(z) + cnPn+1(z) (4.1) − − where c 1 := 0, P 1 : 0. − − ≡ Note that the orthogonal polynomials associated to a particular om µ are uniquely determined up to a factor of absolute value 1. For given µ we can obtain these by the 71

Gram-Schmidt algorithm applied to 1, z, z2, . . . . Performing Gram-Schmidt in L2 where { } µ supp(µ) R, we can always achieve that these polynomials have real coeﬃcients. On ⊂ the other hand, given (Pn)n 0 such that D is symmetric, we can explicitly calculate poly- ≥ nomials (Qn)n 0 such that Qn R[z] having positive leading coeﬃcient and Qn = αnPn ≥ ∈ with α = 1 for all n N , as we shall see now. | n| ∈ 0

4.1.1 Lemma. Let (Pn)n 0 be a sequence of polynomials as in 1.1.1 such that D is ≥ symmetric. Using the notations in (4.1), for n N deﬁne ∈ n 1 − ck αn := ck k=0 | | Q and Q := α P , Q := P 1. Then α = 1 and Q R[z] with positive leading n n n 0 0 ≡ | n| n ∈ coeﬃcient for all n N . ∈ 0

Proof: Clearly, αn = 1 for all n. With α 1 := α0 := 1 and Q 1 : 0, (4.1) yields | | − − ≡

αn+1 Qn+1(z) = αn+1Pn+1(z) = (z an)Pn(z) cn 1Pn 1(z) cn − − − − αn+1 1 αn+1 cn−1 = (z an) Qn(z) Qn 1(z) αn · cn − − αn−1 · cn − for all n 0. Moreover, by ≥

αn+1 1 c 1 1 αn+1 cn−1 cncn−1 cn−1 cn−1 = n = and = = | | αn cn cn cn cn αn− cn cncn− cn cn · | | · | | 1 · | 1| · | | we can inductively conclude the assertions.

Note that we did not have to require that an om for (Pn)n exists. Note also that, if p R[z] then p(z) = p(z). Therefore, if D is symmetric then z E if and only if z E. ∈As the deficiency indices of the symmetric operator D in the ∈upper and lower half-planes∈ z C : Im (z) > 0 and z C : Im (z) < 0 , respectively, are { ∈ } { ∈ } constant, this yields either C R E or E R. It is well-known that a symmetric operator in a Hilbert space \is closable,⊂ having⊂ self-adjoint extensions in if and only H H if the deficiency indices in the upper and lower half-planes are equal, see [Ru2, 13.20], for example. Thus the question whether there exists an om for (Pn)n 0 with support ≥ in R can be answered with comparatively little effort. The theory of real orthogonal polynomials has been studied intensely for many years; as standard references we mention the textbooks by Chihara [Chi] and Szeg˝o [Sze] as well as Borwein and Erdélyi [BE]. The assertions in this section are already well-known. However, we will consider the situation of real orthogonal polynomials as a special case of complex polynomials. We point out that if all the coefficients of the polynomials (Pn)n (and hence all entries in the matrix representation of D) are real and D is not symmetric, there still might exist an om µ with supp(µ) (C R) = ∅ which, for example, is the case for the Newton Polynomials, ∩ \ 6 see 4.5.7, and the “modified” Hermite polynomials in 4.5.12. 72 4 APPLICATIONS AND EXAMPLES

4.1.2 Theorem. Let (Pn)n 0 be a sequence of polynomials as in 1.1.1 such that D is ≥ symmetric. The operator D is essentially self-adjoint if and only if ran(D i id)⊥ = 0 . · { } Moreover, if D is essentially self-adjoint then (i) there exists a unique om µ with supp(µ) R and P 2 = L2 , ⊂ µ µ (ii) E = Λµ, (iii) (P ) is an onb in (K) if and only if µ is discrete. n n H Proof: It is a well-known fact that any densely deﬁned symmetric operator S in a Hilbert space is essentially self-adjoint if and only if ran(S i id)⊥ = ran(S + i id)⊥ = 0 , see also [Ru2, 13.20]. − · · { } Taking into account that the spectrum of a self-adjoint operator is contained in the real line, and that a self-adjoint operator is normal, the remainder of this proposition is just a special case of 3.7.12.

∞ 4.1.3 Example. We start with a sequence (gn)n 0 of positive reals such that gn = 1 ≥ and an inﬁnite countable set z , z , z , . . . [0, 1]. Then n=0 { 0 1 2 } ⊂ P ∞ µ := gnδzn , n=0 P where δz denotes Dirac measure at z, deﬁnes a discrete om for some (Pn)n. The operator D is bounded and symmetric, hence essentially self-adjoint. Due to 4.1.2, µ is the unique om for (P ) , E = Λ = z , z , z , . . . , and (P ) is an onb in (K). n n µ { 0 1 2 } n n H

Remark. If, in addition, Λ = z , z , z , . . . is closed (i.e. contains all its limit points) µ { 0 1 2 } then we could also use 3.7.11 to conclude uniqueness of µ and E = Λµ.

∞ 4.1.4 Example. Again, choose a sequence (gn)n 0 of positive reals such that gn = 1 ≥ and an inﬁnite countable set z , z , z , . . . [0, 1]. n=0 { 0 1 2 } ⊂ P Let λ denote Lebesgue measure on [0, 1] and δz be Dirac measure at the point z. Then

1 ∞ µ := 2 λ + gnδzn n=0 P is an om for some (Pn)n with supp(µ) = [0, 1]. As in the previous example, the operator D is bounded and symmetric and hence essentially self-adjoint. Now 4.1.2 shows that µ is the unique om for (P ) , E = Λ = z , z , z , . . . , and (P ) is not an onb in (K). n n µ { 0 1 2 } n n H

Remark. As usual, for z Λµ, we define κ(z) := K(z, z). According to 3.1.7, µ( zn ) = 1 ∈ { } κ(zn)− for all z Λµ. For arbitrary ε > 0, in 4.1.3 and 4.1.4 we have gn < ε for all but ∈ 1 at most finitely many n. This implies κ(zn) > ε for all but at most finitely many n. In particular, κ is unbounded. Recall that, by 2.2.8, if κ is bounded then E is closed. The converse is not true: In 4.1.3 or 4.1.4, E = Λ = z , z , z , . . . can be chosen to be closed while κ is unbounded. µ { 0 1 2 } 73

We now turn to the case that D is symmetric but not essentially self-adjoint. As the proof of the following lemma requires some lengthy calculations rather than operator theoretical means, we will not give a proof here but refer to [Ak, Theorem 1.3.2], or [Bay, Satz 2.2.12 and Lemma 2.2.1], for instance.

4.1.5 Lemma. Let (Pn)n 0 be a sequence of polynomials in R[z] satisfying a symmetric ≥ 2 3-term recurrence (4.1). If there exists z C R such that Pn(z) < then ∈ \ n 0 | | ∞ ≥ 2 P Pn(z) < for all z C . n 0 | | ∞ ∈ ≥ P Moreover, the series is uniformly convergent in any compact subset of C.

4.1.6 Theorem. Let (Pn)n 0 be a sequence of polynomials as in 1.1.1 such that D is ≥ symmetric. If D is not essentially self-adjoint then

(i) E = Ereg = C, (ii) (P ) is an onb in (K) and all members of (K) are entire functions, n n H H 2 2 2 (iii) there exists an om µ such that (Pn)n is an onb in Lµ (i.e. Pµ = Lµ), (iv) there exists an om ν such that P 2 = L2, ν 6 ν (v) supp(µ) R for all om µ. ⊂ Proof: As D is symmetric and not essentially self-adjoint, there exists λ C R such ∈ \ that ran(D λ id)⊥ = 0 , in other words, see 3.1.3, λ E. − 6 { } ∈ Without restriction, see 4.1.1, we can assume Pn R[z]; 4.1.5 yields E = C and uniform 2 ∈ convergence of the series κ(z) = Pn(z) in any compact subset of C. n 0 | | ≥ P According to 3.5.6, all f (K) are entire functions and, in particular, Ereg = C, provided there exists an om.∈ H If µ is an om then µ(E) = µ(C) = 1 and 3.7.2 shows that (P ) is an onb in (K). n n H Moreover, E = C implies dim(ran(D i id)⊥) = dim(ran(D + i id)⊥) = 1; hence − · · there exists a self-adjoint extension of D in which we can regard as the multiplication 2 H 2 operator Mµ in Lµ for some om µ. In particular, (Pn)n is an onb in Lµ. Every symmetric operator which is not essentially self-adjoint also has self-adjoint extensions in a larger space, see [AG, IX.111, Satz 1]. These correspond to om ν such that 2 2 2 (Pn)n is not an onb in Lν, hence Pν = Lν. 6 2 Let now µ be an om and regard D as an operator in Pµ . Note that P0(z) = 1, DP0(z) = z, 2 2 and D P0(z) = z . Therefore,

2 2 2 2 2 2 2 z z dµ = 2 z z z dµ = 2 DP0 2 P0 , D P0 L2 D P0 , P0 L2 | − | | | − − k kLµ − h i µ − h i µ 2 R = 2R DP0 2 DP0 , DP0 L2 DP0 , DP0 L2 = 0 k kLµ − h i µ − h i µ because D is symmetric. This implies that C R is a µ-nullset. As C R is open, we \ \ obtain supp(µ) R. ⊂ 74 4 APPLICATIONS AND EXAMPLES

Remark. The last step of the proof, in particular, shows that a symmetric Hessenberg operator cannot have a minimal normal extension which is not self-adjoint.

Combining the preceding results, we see that in the symmetric case there always exists an om.

4.1.7 Corollary. Let (Pn)n 0 be a sequence of polynomials as in 1.1.1. There exists an ≥ om µ with supp(µ) R if and only if D is symmetric. ⊂ Proof: Clearly, if there exists an om supported in R then D is symmetric. On the other hand, if D is symmetric then either 4.1.2 or 4.1.6 is applicable and in both cases existence of an om supported on the real line follows.

4.1.8 N-Extremal Measures. We have just seen that if D is symmetric then there 2 2 always exists an om. In particular, there always exists an om µ such that Pµ = Lµ. These measures are called Nevanlinna-extremal or N-extremal for short, see [BCh] and also [Ak, Theorem 2.3.3]. It is known that N-extremal measures corresponding to a not essentially self-adjoint Jacobi operator are discrete. We can prove that, too, using results from chapter 3. Before we do so, note the following proposition concerning self-adjoint extensions with particularly desired eigenvalues.

4.1.9 Proposition. Let S be a symmetric operator in a Hilbert space with equal K deﬁciency indices dim(ran(S i id)⊥) = dim(ran(S + i id)⊥) = m where 0 < m < . For any λ R reg(S), ther−e exists· a self-adjoint extension· T of S within the space∞ ∈ ∩ K having λ as an eigenvalue of multiplicity m.

For a proof, see [AG, VIII.105, Satz 3].

4.1.10 Theorem. Assume that D is symmetric but not essentially self-adjoint. 2 2 If µ is an om for (Pn)n 0 satisfying Pµ = Lµ then µ is discrete and every bounded interval ≥ contains at most ﬁnitely many points of supp(µ). For arbitrary x R, there exists a unique discrete om ν such that ν( x ) > 0 and ∈ { } L2 = P 2. Moreover, then K(x, w) = 0 w supp(ν) x . ν ν ⇐⇒ ∈ \ { } Proof: By 4.1.6, we have E = C which implies µ(E) = 1 and, according to 3.7.13, µ is discrete. In particular, with Λµ = z C : µ( z ) > 0 and δz the Dirac measure at z, using 3.1.7, we obtain ∈ { } 1 µ = K(z,z) δz . z Λµ ∈ P By 3.7.15, P (Λ ) is an onb in `2, implying K(z, w) = 0 for all z, w Λ , z = w. Assume µ ∈ µ 6 that there exists a bounded interval containing inﬁnitely many points of Λµ. Thus there exists a limit point of Λ . Let z Λ and recall that K is a member of (K) and, by µ ∈ µ z H 4.1.6(ii), an entire function. On the other hand, K(z, w) = 0 for w Λ z implies ∈ µ \ { } 75

that the zeros of Kz have a limit point and therefore Kz must vanish identically which is a contradiction. Hence every bounded interval contains at most ﬁnitely many points of Λµ. In particular, Λµ is closed and supp(µ) = Λµ.

Now ﬁx x R and use that E = C, see 4.1.6 again. According to 4.1.9, there exists a ∈ reg self-adjoint extension T of D in such that x is an eigenvalue of T . By identifying with 2 2 H H Lν = Pν as usual, T corresponds to the multiplication operator Mν whose eigenvalues are exactly the elements of Λ ; hence ν( x ) > 0. ν { } Assume that there exists b C supp(ν) such that K(x, b) = 0. Hence k , k 2 = 0. ∈ \ h b x iLν 2 1 x a Recall that ν is discrete and ν( x ) kx = 1 x in Lν . Thus (Mν b)− (Mν a)kx = x−b kx. { } { } − − − C 2 According to 3.3.8, for arbitrary a , we have fa,b fa,b , P0 Lν kb = 0 where fa,b = 1 ∈ − h i (M a)∗[(M b)− ]∗k . This yields ν − ν − a 1 x a 0 = f , k 2 = k , (M b)− (M a)k 2 = K(x, a) a,b x Lν a ν ν x Lν x−b h i h − − i − for all a C. Hence the entire function Kx vanishes on C a implying Kx = 0 which is a contradiction.∈ Thus K(x, b) = 0 for all b C supp(ν).\ { } 6 ∈ \ As we have already shown, K(x, b) = 0 whenever b supp(ν) x . Therefore, supp(ν) = C 1∈ \{ } Λν = x b : K(x, b) = 0 and ν = K(z,z) δz is uniquely determined. { } ∪ { ∈ } z Λν ∈ P Note that we could also use 3.5.7 to prove that every bounded interval contains only ﬁnitely many points of Λµ.

4.1.11 Corollary. Let D be a Jacobi operator in a Hilbert space which is not essen- H tially self-adjoint. Then there exist self-adjoint extensions of D in the space as well as H in a larger space % . The spectrum of any self-adjoint extension in is discrete. K H H Moreover, for every x R, there exists a self-adjoint extension T in such that x σ(T ) which is uniquely determine∈ d up to unitary equivalence. H ∈

This was just a re-formulation of the previous theorem.

Remark. In terms of the Hamburger moment problem, cf. 1.1.3, the case where there exists a unique om is referred to as the determinate case while otherwise one speaks of the indeterminate case. The property E = C in the indeterminate case, see 4.1.6(i), can be found in [ShTa, Cor. 2.7], for example. For more details, see also [BCh].

4.1.12 Orthogonal polynomials on an arbitrary line. It is not diﬃcult to generalize the case of ortonormalizing measures supported in R to om with support contained in 1 an arbitrary straight line given by a− (t b) : t R where a, b C and a = 1. Replacing D by aD + b id, one can pro{ ve analogous− results∈ } to the above.∈ In particular,| | D is still tri-diagonal. More generally, a formally normal Hessenberg operator whose matrix representation only has ﬁnitely many non-vanishing entries in every row, is tri-diagonal. For details, we refer to [Kl2, Kap. 3.1] and [CaKl1], for example. 76 4 APPLICATIONS AND EXAMPLES

4.2 Orthogonal Polynomials on the Unit Circle

Let D denote the open unit disk, D := z C : z < 1 . In this section we will deal with { ∈ | | } om µ such that supp(µ) ∂D = z C : z = 1 . Note that then, for f dom(M ), ⊂ { ∈ | | } ∈ µ 2 2 2 2 Mµf 2 = z f(z) dµ(z) = f(z) dµ(z) = f L2 . Lµ k k µ

R R 1 Hence Mµ is isometric. Therefore, D must be an isometry, too. Note that z = z− for 1 2 z ∂D and M ∗ is multiplication by z− in L . This shows M M ∗ = id 2 = M ∗M , thus ∈ µ µ µ µ Lµ µ µ Mµ is unitary. We will apply some of our results to this situation. However, orthogonal polynomials on the unit circle have been widely studied in the past few years. A two-volume textbook by Simon [Si] presents a vast treatise on this topic. For a proof of the following proposition, see also [Kl2, Satz 3.3.2], for example.

4.2.1 Proposition. If D is isometric then there exists a unique om µ. Moreover, supp(µ) ∂D. ⊂

The monomials 1, z, z2, . . . may serve as a ﬁrst example. They form an orthonormal 2 { } D system in Lµ where µ is the normalized Lebesgue measure on ∂ , see 1.2.2, 1.3.8, and also 2.1.11.

Recall that, whenever there exists an om µ for (Pn)n 0, the boundary of the set E is ≥ contained in supp(µ). Moreover, for supp(µ) ∂D, the following holds. ⊂

4.2.2 Lemma. Let µ be an om for (Pn)n 0 such that supp(µ) ∂D. The following ≥ properties are equivalent. ⊂

2 2 (i) Pµ = Lµ, (ii) D is essentially normal,

(iii) E = Λµ, (iv) E supp(µ), ⊂ 2 (v) Pn(0) = , n 0 | | ∞ ≥ P Moreover, given these properties are satisﬁed, (Pn)n is an onb in (K) if and only if µ is discrete. H

Proof: For (i) (ii) (iii) see 3.5.9; (iii) (iv) (v) is obvious. ⇐⇒ ⇐⇒ ⇒ ⇒ According to 3.1.3, (v) implies ran(D)⊥ = 0 . As D is isometric, ran(D) is a closed subspace of . Therefore, ran(D) = , hence{ D} is unitary and D is essentially normal. H H For the last assertion, see 3.7.12. 77

4.2.3 Lemma. Let µ be an om for (Pn)n 0 such that supp(µ) ∂D. The following ≥ ⊂ properties are equivalent.

(i) P 2 = L2 , µ 6 µ (ii) E supp(µ) = ∅, \ 6 (iii) D E D ∂D, ⊂ ⊂ ∪ 2 (iv) Pn(0) < . n 0 | | ∞ ≥ P Moreover, if these properties are satisﬁed then Ereg = E◦ = D and supp(µ) = ∂D.

Proof: The implication (ii) (iii) is due to 3.3.6, the implications (iii) (iv) and (iv) (ii) are obvious, and, by 4.2.2, (i)⇒ (iv). ⇒ ⇒ ⇐⇒ For the last assertion, see 3.3.6 again.

Remark. We could also use E z C : z D , see 3.1.6, and D = 1 to prove ⊂ ∈ | | ≤ k k k k E D ∂D. ⊂ ∪

4.2.4 Example. For n Λ := z C : z2 = 1 , n N ∪∈ { ∈ } deﬁne a measure µ such that µ(Λ) = 1 and µ( λ ) > 0 for all λ Λ. Thus Λ = Λµ and D C 2 { } ∈ supp(µ) = ∂ . Clearly, [z] Lµ, µ is an om for some (Pn)n. Moreover, Λ E. ⊂ 2n ⊂ For n N, deﬁne qn C[z] by qn(z) := z . Note that lim qn(z) = 1 for all z Λ. n ∈ ∈ →∞ ∈ By dominated convergence, lim qn P0 2 = 0; (2.4) implies lim qn P0 = 0. n Lµ n (K) →∞ k − k →∞ k − kH Recall that convergence in the (K)-norm implies pointwise convergence on E. Therefore, H lim qn(z) = 1 for all z E. In particular, 0 / E. n →∞ ∈ ∈ 2 2 Now 4.2.2 shows that E = Λ and Pµ = Lµ. Hence here D is a unitary operator with pure point spectrum dense in ∂D.

C 2 We will now have a closer look at the case that [z] is not total in Lµ.

D 2 2 4.2.5 Lemma. Let µ be an om for (Pn)n 0 supported on ∂ and suppose that Pµ = Lµ. 2 ≥ 6 Then κ := Pn(0) < . Moreover, n 0 | | ∞ ≥ P 1 2 h := √κ Pn(0) Pn Pµ n 0 ∈ ≥ P and (M )nh : n Z is an orthonormal system in L2 . { µ ∈ } µ Finally, let λ denote normalized one-dimensional Lebesgue measure on ∂D. 2 C Then λ( h = 0 ) = 0 and λ( f = 0 ) = 0 for all f Pµ (h [z])⊥. { b } { 6 } ∈ ∩ · b b 78 4 APPLICATIONS AND EXAMPLES

2 2 Proof: According to 4.2.2, κ = is equivalent to Pµ = Lµ. Hence here κ < and h is ∞ 2 ∞ well-deﬁned. By construction, h Pµ , h 2 = 1. ∈ k kLµ 2 2 Regard D and D as bounded operators in Pµ . Clearly, dom(D) = Pµ and Mµ is a normal 2 extension of D as well as D, acting in the larger space Lµ. m n 2 Z m n − 2 Note that h Pµ and, for m, n , m > n, we have (Mµ) − h = D h Pµ . As Mµ ∈ n ∈ ∈ is unitary, (Mµ) h 2 = 1 and Mµ∗Mµ = idL2 . Now we get Lµ µ n m n n m n m n (Mµ) h , (Mµ) h 2 = (Mµ∗) (Mµ) h , (Mµ) − h 2 = h , (Mµ) − h 2 Lµ Lµ Lµ m n − m n = h , D h 2 = (D∗) − h , h 2 . Pµ Pµ n m n Z By 3.1.3, D∗h = 0, hence (Mµ) h , (Mµ) h 2 = 0. Therefore, (Mµ) h : n is an Lµ { ∈ } orthonormal system in L2 . µ In particular, for m, n N , ∈ 0 n m n m 2 δmn = (Mµ) h , (Mµ) h 2 = z z h(z) dµ(z) Lµ

2 R shows that h µ is an om for the monomials which, according to 3.2.14, is uniquely | | determined. Note that λ is an om for (zn) ; thus h 2µ = λ and we obtain λ( h = 0 ) = 0. n | | { } 2 C 2 N Let now f Pµ (h [z])⊥, i.e. f Pµ and, for n 0, ∈ ∩ · b ∈ ∈ b b 2 0 = f(z) znh(z) dµ(z) = f(z)/h(z) zn h(z) dµ(z) = f(z)/h(z) zn dλ(z) . Furthermore,R R R b

n n n n f(z) z− h(z) dµ(z) = f , (Mµ∗) h 2 = (Mµ) f , h 2 = D f , h 2 Lµ Lµ Pµ n N R = f , (D∗) h 2 = f , 0 P 2 = 0 for all n . Pµ h i µ ∈

Hence 0 = f(z) znh(z) dµ(z) = f(z)/h(z) zn dλ(z) holds for all n Z. ∈ n 2 2 As z : n Z is an onb in Lb, we obtain f/h= 0 in Lb and λ( f = 0 ) = 0. { ∈R } λ R b λ { 6 }

4.2.6 Lebesgue Decomposition. Let µ be a Borel measureb on ∂D. We denote its absolutely continuous, discrete, and singularly continuous parts by µac, µd, and µsc, respectively. Moreover, let λ denote Lebesgue measure on ∂D. Then µac = αλ with measurable non-negative α. For abbreviation, set µ := µd + µsc. ⊥ Note that µac, µd, and µsc are mutually orthogonal, i.e. there exist mutually disjoint D subsets Aac, Ad, and Asc of ∂ such that

µsc(Aac) = 0 = µd(Aac) , µac(Ad) = 0 = µsc(Ad) , µac(Asc) = 0 = µd(Asc) , and µ(Aac) = µac(Aac) > 0, µ(Ad) = µd(Ad) > 0, µ(Asc) = µsc(Asc) > 0. The set Ad is at most countable while Aac and Asc are uncountable (or empty). For more details, we refer to [El, VII. 4], for example. § 79

4.2.7 Proposition. In the situation of 4.2.5, let µ = αλ + µ . 1 ⊥ Then α = 2 holds λ-almost everywhere and µ ( h = 0 ) = 0. h ⊥ | | { 6 }b b 2 2 2 Proof: Recall that λ = h µ = h µ + h µac. | | | | ⊥ | | 2 1 1 Therefore, h µ = 0 and µac = 2 λ; hence µ ( h = 0 ) = 0 and α = 2 . ⊥ b h ⊥ h | | | | { 6 } | | b 2 K K For f P , let f be as in 3.6.2 and recall that f (z) = k , f 2 for all z E, see ∈ µ h z iLµ ∈ 3.6.6.

4.2.8 Theorem. In the situation of 4.2.5,

n 2 K D 2 V = ran D = f Pµ : f = 0 = f Pµ : λ( f = 0 ) = 0 . ∞ n N ∩∈ ∈ | ∈ { 6 } b Proof: First note that, due to 4.2.3, we have Ereg = E◦ = D which is a connected set n 2 K D and 3.7.6 yields V = ran D = f Pµ : f = 0 . ∞ n N ∩∈ ∈ | n 2 Let now f ran D Pµ . ∈ n∩N ⊂ ∈ m Then, for every m N, there exists f P 2 such that f = D f = (M )mf . Thus ∈ m ∈ µ m µ m n n n+1 n n (Mµ) h , f 2 = (Mµ) h , (Mµ) fn+1 2 = (Mµ∗) (Mµ) h , Mµfn+1 2 Lµ Lµ Lµ

= h , Mµfn+1 2 = h , Dfn+1 2 = D∗h , fn+1 2 = 0 Lµ Pµ Pµ for all n N and 4.2.5 yields λ( f = 0 ) = 0. ∈ 0 { 6 } As to the converse inclusion, let f P 2 such that λ( f = 0 ) = 0. Then b ∈ µ { 6 } b h , f L2 = h f dµ = h f dµ . h i µ ⊥ R R Note that, by 4.2.7, µ ( h = 0 ) = 0. Hence the above integral vanishes and we obtain ⊥ { 6 } 2 h , f Lµ = 0. h i 2 As D is an isometry, ran(D) is a closed subspace of Pµ ; 3.1.3 implies ran(D) = h ⊥. 2 { } Therefore, f ran(D) and there exists f1 Pµ such that f = Df1 = Mµf1. This shows 1 ∈ ∈ f = (M )− f = M ∗f. As f = 0 only on a null-set w.r.t. Lebesgue measure, so is f . 1 µ µ 6 1 Now we can apply this result to f1 instead of f. Repeating that n times, we see that n f ran(D ) for arbitrary n. ∈

4.2.9 Theorem. Let µ be an om for (Pn)n 0 such that supp(µ) ∂D. For z E, recall 2 ≥ ⊂ ∈ kz = Pn(z) Pn Pµ . n 0 ∈ ≥ P D 2 We have E ∂ = Λµ and 1 z = µ( z ) kz Pµ for all z Λµ. ∩ { } { } ∈ ∈ 80 4 APPLICATIONS AND EXAMPLES

2 Proof: In the case that (Pn)n is an onb in Lµ, 4.2.2 shows E = Λµ. The other assertion follows easily using 3.1.8. n Otherwise, 4.2.3 shows D E D ∂D and, according to 4.2.5, D h : n N0 = n N ⊂ ⊂ ∪ 2 { ∈ } Mµ h : n 0 is an orthonormal system in Pµ . { ∈ } n 2 2 This yields f , D h 2 ` for all f P . h iPµ n ∈ ∈ µ D 2 Now let a E ∂ . Then ka := Pn(a) Pn Pµ and D∗ka = aka. This implies ∈ ∩ n 0 ∈ ≥ P n n n k , D h 2 = (D∗) k , h 2 = a k , h for all n N h a iPµ h a iPµ h a i ∈ 2 and, as a = 1, we obtain an ` -sequence only if k , h 2 = 0. | | h a iPµ Once again we use that ran(D) = h ⊥. Hence ka ran(D) and, as DD∗ is the orthogonal 2 { } ∈ projection in Pµ onto ran(D), we ﬁnally see a ka = DD∗a ka = Da a ka = Dka = Mµka.

As the eigenvalues of the multiplication operator Mµ are precisely the indicator functions 2 D 1 z for z Λµ, we obtain 1 a = c ka Pµ with some c = 0. Thus E ∂ Λµ. The { } ∈ { } · ∈ 6 ∩ ⊂ converse inclusion is clear, as Λ E. µ ⊂ 2 Now, according to 3.1.8, 1 z = µ( z ) kz Pµ for all z Λµ. { } { } ∈ ∈ Remark. This theorem is not new. Our proof, however, mainly uses the theory of subnormal Hessenberg operators. The more general statement 3.1.8, consequently, cannot be found in works concerning om supported on the unit circle only. For other ways proving 4.2.9, see e.g. [Si, 2.7.3 and 4.3.15(a)].

So far, in this section we have not yet had a look at the reproducing kernel space (K). H In the case that (P ) is an onb in L2 , we know that E = Λ and (K) is isometrically n n µ µ H isomorphic to the closed linear span of 1 z : z Λµ via 1 z µ( z ) Kz. Otherwise, { { } ∈ } { } 7→ { } we have D E; here the connections between L2 and (K) are much more interesting. ⊂ µ H

D 2 2 4.2.10 Theorem. Let µ be an om for (Pn)n 0 supported on ∂ such that Pµ = Lµ. The ≥ following properties hold. 6

(i) E = D Λ , ∪ µ (ii) all f (K) are holomorphic in D, ∈ H (iii) kλ , ka L2 = Kλ , Ka (K) = 0 for all λ Λµ and all a D, h i µ h iH ∈ ∈ (iv) kλ V for all λ Λµ, ∈ ∞ ∈ (v) if (zn)n N is a sequence in D such that lim zn =: z ∂D then ∈ n ∗ →∞ ∈ κ(z ) as n n → ∞ → ∞ 2 where κ(z) = K(z, z) = Pn(z) for z E. n 0 | | ∈ ≥ P 81

Proof: Assertion (i) is an immediate consequence of 4.2.9 in connection with 4.2.3(iii).

As to (ii), according to 4.2.3, we have Ereg = D and, by 3.5.2, all members of (K) are holomorphic in D. H

Concerning (iii) and (iv), let Λµ = ∅. For any λ Λµ, we have Dkλ = λkλ, see 4.2.9, 6 K D ∈ which implies kλ V . Now 4.2.8 yields kλ = 0, in other words, kλ , ka = 0 for all ∞ H a D. Isometrically∈ embedding into L2 as| usual, we obtain h i ∈ H µ

2 D 0 = kλ , ka Lµ = Pn(λ) Pn(a) = Kλ , Ka (K) for all λ Λµ and all a . h i n 0 h iH ∈ ∈ ≥ P For the proof of (v), assume that there exists R > 0 such that κ(z ) R for all n N, n ≤ ∈ i.e. z : n N A = z E : κ(z) R . As A is closed, see 2.2.7, we { n ∈ } ⊂ R { ∈ ≤ } R obtain z AR and 2.2.13 yields that Kz∗ AR is continuous. Note that Kz∗ (z ) = 0 and ∗ ∈ |N ∗ 6 Kz∗ (zn) = Kzn , Kz∗ (K) = 0 for all n by (iii). This is a contradiction; therefore, H there existshno such Ri and κ(z ) . ∈ n → ∞ Note that in (ii) and (iii) we do not require (P ) be an onb in (K), see also the remark n n H following 3.7.9. If Λµ = ∅, here (K) contains functions which are not continuous. 6 H

In order to determine a necessary and suﬃcient condition for (Pn)n being an onb in (K), we will now have a closer look at the space V . H ∞

4.2.11 Von Neumann-Wold Decomposition. Let S be an isometric operator in a n Hilbert space , 0 := ran(S)⊥, and V := ran(S ). ∞ n N H H ∩∈ Then V is a closed subspace of which reduces S and the restriction of S to V is ∞ H ∞ unitary. Moreover, V ⊥ is the orthogonal sum ∞

V ⊥ = n ∞ n N0 H ∈⊕ of spaces n where S n 1 = n for all n N. H H − H ∈ For a proof we refer to [Con1, I.3.6].

Note that S is unitary if and only if V ⊥ = 0 . Furthermore, recall that in the case of ∞ { } 2 2 an om µ supported on the unit circle, we have Pµ = Lµ if and only if D is unitary. Thus 2 2 2 2 either V = Pµ = Lµ or V $ Pµ $ Lµ. ∞ ∞ 2 In the latter case, to avoid confusion with orthogonal complements taken in Lµ, denote 2 2 the orthogonal complement of V in Pµ by V+, hence we have Pµ = V V+. Moreover, ∞ ∞ ⊕ the orthogonal complement of ran(D) in P 2 is given by = C k where µ H0 · 0

k0 = Pn(0) Pn , n 0 ≥ P 82 4 APPLICATIONS AND EXAMPLES

k0 see also 3.1.3; normalizing this by h := k , we obtain k 0k n n n V+ = D 0 = lin D h : n N0 = lin (Mµ) h : n N0 n N0 H { ∈ } { ∈ } ∈⊕ n and D h : n N0 is an orthonormal basis of V+. { ∈ } n Note that we already know that D h : n N is an onb of V from 4.2.5. { ∈ 0} +

D 2 2 4.2.12 Lemma. Let µ be an om for (Pn)n 0 supported on ∂ such that Pµ = Lµ. Then ≥ 6 kλ V for all λ Λµ and ka V+ for all a D. ∈ ∞ ∈ ∈ ∈ Proof: The ﬁrst assertion is due to 4.2.10(iv). Denote by P the orthogonal projection in P 2 onto V and, for a D, set j := P k . Then µ + ∈ a a n n n n n n ja = D h , ka D h = h , (D∗) ka D h = h , ka a D h n 0 n 0 n 0h i ≥ ≥ ≥ P P P and, using D∗h = 0, D∗Dh = h, we obtain

n n n n 1 n+1 n D∗ja = h , ka a D∗D h = h , ka a D − h = h , ka a D h = a ja . n 0h i n 1h i n 0h i ≥ ≥ ≥ P P P Hence j (D∗ a id) and, therefore, j = c k with some c C. As j = P k a ∈ N − a a a a ∈ a a with P an orthogonal projection, only ca = 0 or ca = 1 is possible. Note that ca = 0 K D means ja = P ka = 0 and, therefore, ka V which implies ka = 0 in contradiction to ∞ kK D (a) = k , k = 0. ∈ | a | h a a i 6 Thus j = P k = k and k V . a a a a ∈ +

In order to show that V reduces not only D but also Mµ we need the following. ∞

4.2.13 Lemma. Let U be a unitary operator in a Hilbert space and be a closed H L subspace of . Then reduces U if and only if U = . H L L L Proof: Let P be the orthogonal projection in onto . H L Assume that reduces U. Then P U = UP and, for any f , we have Uf = UP f = L ∈ L P Uf ; hence U . Moreover, f = P UU ∗f = UP U ∗f shows U . ∈ L L ⊂ L 1 L ⊃ L 1 As to the converse implication, from U ∗ = U − and U = we obtain U ∗ = U − = . L L L L L Then UP = P UP and U ∗P = P U ∗P . The latter implies P U = P UP and altogether we have UP = P U showing that reduces U. L

4.2.14 Corollary. Let µ be an om for (Pn)n 0 supported on the unit circle. Then V ≥ ∞ reduces Mµ.

Proof: According to 4.2.11, the restriction of D to V is unitary. This implies DV = V ∞ ∞ ∞ and, as Mµ extends D, also MµV = V . Now we can apply 4.2.13 to conclude that V ∞ ∞ ∞ reduces Mµ. 83

Recall that, according to 3.6.9, (P ) is not an onb in (K) if and only if there exists n n H f P 2, f = 0, such that f K = 0. Therefore, we deﬁne W := f P 2 : f K = 0 = ∈ µ 6 K { ∈ µ } f P 2 : k , f = 0 for all a E and obtain that (P ) is an onb in (K) if and only { ∈ µ h a i ∈ } n n H 2 K D if WK = 0 . Note that WK f Pµ : f = 0 = V , see also 4.2.8, and if we { } ⊂ { ∈ | } ∞ deﬁne W := lin k : a E , we get P 2 = W W . { a ∈ } µ K ⊕ Finally, let V := lin k : a Λ . d { a ∈ µ}

4.2.15 Lemma. Using the notations above, we have

V = Vd WK and V+ = lin ka : a D . ∞ ⊕ { ∈ }

Moreover, Vd and WK reduce Mµ.

Proof: Let X := lin k : a D . Then W = V X. { a ∈ } d ∪ 2 According to 4.2.12, Vd V and X V+. In connection with Pµ = V V+, this yields ∞ ∞ W = V X. Using P 2⊂= W W , ⊂we get P 2 = W V X. ⊕ d ⊕ µ K ⊕ µ K ⊕ d ⊕ With WK Vd V and X V+, we obtain Vd WK = V and X = V+. ⊕ ⊂ ∞ ⊂ ⊕ ∞ Now we show that both Vd and WK are invariant subspaces for Mµ. As Vd is the closure of the linear span of the eigenvectors of Mµ, clearly Mµf Vd for all f Vd. Let now f W . Then, for all a E, ∈ ∈ ∈ K ∈

0 = a k , f = a k , f = D∗k , f = k , Df h a i a a a and, therefore, M f = Df W . µ ∈ K By 4.2.13, MµV = V . As V = Vd WK , we obtain MµVd = Vd as well as MµWK = WK ∞ ∞ ∞ ⊕ and can use 4.2.13 to conclude that Vd and WK reduce Mµ.

Remark. We can also deduce MµVd = Vd from the fact that Vd is the closure of the linear span of the eigenvectors of Mµ and 0 is not an eigenvalue.

D 2 2 4.2.16 Lemma. Let µ be an om for (Pn)n 0 supported on ∂ such that Pµ = Lµ and ≥ 6 n N 2 ⊥ deﬁne V := lin (Mµ)− h : n . Then V = Pµ and − { ∈ } − 2 Lµ = V V+ V . ∞ ⊕ ⊕ − n Z 2 Moreover, Vs := V+ V = lin (Mµ)− h : n = 1 h=0 f : f Lµ and V = 2 ⊕ − { ∈ } { { 6 } ∈ } ∞ 1 h=0 f : f Lµ . { { } ∈ } n n Proof: Using 4.2.5, we see that (Mµ) h : n N0 and (Mµ)− h : n N are { ∈ } {N 2 ∈ } orthonormal bases of V+ and V , respectively. For arbitrary n and p Pµ , we obtain − ∈ ∈ n n n n (Mµ)− h , p 2 = h , (Mµ) p 2 = h , D p 2 = (D∗) h , p 2 = 0 . Lµ Lµ Pµ Pµ

84 4 APPLICATIONS AND EXAMPLES

2 ⊥ This yields V Pµ and, in particular, V V . As V+ was deﬁned as the orthogonal − ⊂ 2 ∞ ⊥ − complement of V in Pµ , we now have V Vs. ∞ ∞ ⊥ Clearly, MµVs = Vs and, according to 4.2.13, Vs reduces Mµ. Thus V Vs also reduces 2 ∞ ⊕ Mµ. Furthermore, Pµ V Vs and, as Mµ is a minimal normal extension of D, the ⊂ ∞ ⊕ 2 2 only reducing subspace of Mµ containing Pµ is the whole space Lµ.

2 2 ⊥ Therefore, Lµ = V Vs and V = Pµ . ∞ ⊕ − 2 2 2 Now we use that Lµ = 1 h=0 f : f Lµ 1 h=0 f : f Lµ . { { 6 } ∈ } ⊕ { { } ∈ } 2 Assume g V 1 h=0 f : f Lµ . According to 4.2.8, λ( g = 0 ) = 0 and, according ∞ { 6 } to 4.2.7, µ∈( h ∩={0 ) = 0 whic∈ h implies} µ ( g = 0 ) = 0.{ Altogether,6 } we now have ⊥ { 6 } ⊥ { 6 } µ( g = 0 ) = 0 and hence g = 0. b { 6 } 2 Thus 1 h=0 f : f Lµ V . { { 6 } ∈ } ⊥ ∞ On the other hand, for any f L2 and n Z, ∈ µ ∈

n n 1 h=0 f , (Mµ) h = f(z) z h(z) dµ(z) = 0 { } h=0 { R } 2 showing 1 h=0 f : f Lµ Vs. { { } ∈ } ⊥ 2 2 Now we ﬁnally can conclude V = 1 h=0 f : f Lµ and Vs = 1 h=0 f : f Lµ . ∞ { { } ∈ } { { 6 } ∈ }

D 4.2.17 Theorem. Let µ = αλ + µd + µsc be an om for (Pn)n 0 supported on ∂ such ≥ that P 2 = L2 . Then µ 6 µ b (i) µ ( h = 0 ) = 0 = µsc( h = 0 ), d { 6 } { 6 }

(ii) λ(Λµ) = 0 = µsc(Λµ),

(iii) λ( h = 0 Λ ) = 0 = µ ( h = 0 Λ ), b { } \ µ d { } \ µ and L2 = V V W where µ b s ⊕ d ⊕ K 2 Vs = 1 h=0 f : f Lµ , • { { 6 } ∈ } V = 1 f : f L2 , • d { Λµ ∈ µ} 2 WK = 1 h=0 Λµ f : f Lµ . • { { }\ ∈ } Furthermore, (K) is isometrically isomorphic to V V via P P and (P ) is an H + ⊕ d n 7→ n n n onb in (K) if and only if µsc = 0. H Moreover, if µsc = 0 then Pn / Vd V+ for all n. 6 ∈ ⊕ Proof: (i) is due to 4.2.7; (ii) is clear because Λ is countable; Λ = z C : µ( z ) > 0 µ µ { ∈ { } } also proves the second equality in (iii). Finally, according to 4.2.5, λ( h = 0 Λ ) = 0. { } \ µ 2 The decomposition Lµ = Vs Vd WK was the subject of 4.2.16 where we have already ⊕2 ⊕ b proved Vs = 1 h=0 f : f Lµ . { { 6 } ∈ } 85

2 From 4.2.16 we also know that V = 1 h=0 f : f Lµ and, by 4.2.15, V = Vd WK. ∞ { { } ∈ } ∞ ⊕ As Vd is spanned by the eigenvectors of Mµ which are the 1 λ for λ Λµ, we see 2 2 { } ∈ Vd = 1Λµ f : f Lµ . Now WK = 1 h=0 Λµ f : f Lµ is clear, too. { ∈ } { { 6 }\ ∈ } Note that W is the null-space of the partial isometry α : P 2 (K), f f K, deﬁned K µ → H 7→ in 3.6.7; hence α isometrically maps Vd V+ onto (K). Moreover, (Pn)n is an onb in (K) if and only if W = 0 . By the⊕characterizationsH above, this is the case if and H K { } only if µsc = 0.

2 Concerning the last assertion, assume Pn Vd V+ for some n. Hence Pn dµsc = 0. ∈ ⊕ | | This implies Pn(z) = 0 for µsc-almost all z in contradiction to Pn having only ﬁnitely many zeros. R

2 V = WK Vd Vs = V+ (Pµ )⊥ ∞ ⊕ ⊕ −n {(Mµ) h : n ∈ N} is onb

2 WK Vd (Pµ )⊥ V+ 2 K 2 K = {f ∈ Pµ : f = 0} = {f ∈ Pµ : f |D = 0} = lin{ka : a ∈ D} = lin{ka : a ∈ Λµ}

n {(Mµ) h : n ∈ N0} is onb

{h = 0} \ Λµ Λµ {h 6= 0}

Figure 6. D 2 2 2 In the case of an om µ supported on ∂ such that Pµ = Lµ, the space Lµ 2 2 6 can be decomposed into L = WK Vd V (P ) . µ ⊕ ⊕ + ⊕ µ ⊥ 2 The spaces WK , Vd, and Vs = V (P ) reduce Mµ and can be character- + ⊕ µ ⊥ ized by the sets h = 0 Λµ, Λµ, and h = 0 as well as by the singularly { } \ { 6 } continuous, discrete, and absolutely continuous parts of µ, respectively.

As mentioned before, om supported on ∂D have been widely studied. In the remainder of this section we will give a short overview on other remarkable aspects concerning orthogonal polynomials on the unit circle.

D 4.2.18 Deﬁnition. A Borel measure µ = αλ + µd + µsc on ∂ is said to obey the Szeg˝o condition if ln α dλ > . −∞ R Note that µ cannot obey the Szeg˝o condition if supp(µ) $ ∂D.

Let µ be an om for (Pn)n 0. It is well-known, see [Sze, Chapter XI] and also [Con1, III.12.9] ≥ or [Si, 2.7.15], for example, that µ obeys the Szeg˝o condition if and only if P 2 = L2 . µ 6 µ 86 4 APPLICATIONS AND EXAMPLES

4.2.19 Example. Let µ = αλ be an om supported on ∂D such that

b 1 dλ < . α ∞ R b D 2 2 Recall 2.3.11 which implies that in this case E. By 4.2.3, then Pµ = Lµ; hence µ must obey the Szeg˝o condition. We will verify that⊂ by a short calculation.6 First note that the zeros of α form a Lebesgue null-set because otherwise the integral above would not be ﬁnite. 1 1 Clearly, 0 ln(x) < x for x 1 and, for 0 < x < 1, we have ln(x) = ln x x 1 = 1 x 1 ≤ ≥ ≤ − − . Thus, for A := 0 < α < 1 and B := α 1 , the integrals x ≤ x { } { ≥ }

1 ln(α) dλ α dλ and ln(α) dλ < α dλ A − ≤ A B B R R R R are ﬁnite and, consequently, ln α dλ is ﬁnite as well. One may ask if in 2.3.11 the condition 1 L1 can be weakened, too. R b α ∈

4.2.20 Corollary. Let µ be an om for (Pn)n 0 such that supp(µ) ∂D. The following ≥ properties are equivalent. ⊂

(i) P 2 = L2 , µ 6 µ 2 (ii) Pn(0) < , n 0 | | ∞ ≥ P 2 2 2 ⊥ (iii) Pn(0) < and Lµ = WK Vd V+ Pµ , n 0 | | ∞ ⊕ ⊕ ⊕ ≥ P (iv) E = D Λ , ∪ µ (v) µ obeys the Szeg˝o condition, (vi) p / V for all p C[z] 0 . ∈ ∞ ∈ \ { } Proof: Combining 4.2.3, 4.2.10, and 4.2.17, we obtain (i) (ii) (iii) (iv). ⇐⇒ ⇐⇒ ⇐⇒ Furthermore, we have already mentioned (i) (v) in 4.2.18. ⇐⇒ 2 2 2 According to 4.2.11, Pµ = Lµ is equivalent to V = Pµ . Hence (vi) implies (i). ∞ 2 2 C For the converse, assume that Pµ = Lµ and there exists 0 = p [z] V . Then, 6 6 ∈ ∩ ∞ by 4.2.17, p 2 α dλ = 0, implying p(z)α(z) = 0 Lebesgue-almost everywhere and | | hence α(z) = 0 for λ-almost all z. Therefore, α does not obey the Szeg˝o condition, R2 2 b implying Pµ = Lµ which is a contradiction. b We can now conclude that, given an om µ supported on the unit circle such that P 2 = L2 µ 6 µ and all its parts µac, µd, and µsc are non-trivial, there must exist an uncountable Lebesgue- nullset B, which then can be chosen B = h = 0 Λ , having positive measure. { } \ µ 87

4.2.21 Corollary. Let µ be a measure on ∂D with µ(∂D) = 1 obeying the Szeg˝o condition and having non-trivial singularly continuous part. Then there exists a sequence (qn)n in C[z] with the following properties.

(i) lim qn(z) = 0 for all z D, n →∞ ∈ (ii) lim qn(z) = 0 for Lebesgue-almost all z ∂D, n →∞ ∈ (iii) lim qn(z) = 1 for uncountably many z ∂D. n →∞ ∈ (again, we refer to one-dimensional Lebesgue measure on ∂D).

2 Proof: Let (Pn)n 0 be orthonormal polynomials in Lµ satisfying P0 1 and deg(Pn) = n ≥ ≡ for all n. Then, according to 4.2.20, P 2 = L2 and h exists. µ 6 µ With B := h = 0 Λµ, 4.2.17 yields λ(B) = 0 = µ (B) and µ(B) = µsc(B) > 0. Hence { } \ d B is an uncountable Lebesgue-nullset.

Now deﬁne the sequence (pk)k by b

k pk(z) := 1B , Pi Pi(z) , i=0h i P then there exists a subsequence (qn)n of (pk)k which is µ-a.e. convergent to 1B. Moreover, as 1 W , we obtain q (z) = k , q k , 1 = 0 for z D. B ∈ K n h z n i → h z B i ∈ Hence (qn)n has the asserted properties.

Remark. The result of [Si, Th. 2.5.1] is similar to 4.2.21 but does not specify the set where the limit is diﬀerent from 0. In 4.5.13 and 4.5.16 we will construct such sets using a Cantor-like method. In particular, we will see how to construct a singularly continuous measure supported on a given uncountable compact set.

We conclude this section with another characterization of isometric Hessenberg operators which Simon [Si] calls the GGT representation.

4.2.22 Proposition. Let G = (gij)i,j 0 be an isometric Hessenberg matrix. Then there ≥ 2 2 exist sequences (αn)n 0 and (βn)n 0 of complex numbers satsifying αn + βn = 1 for ≥ ≥ all n and | | | | j 1 − αj αi 1 βj for 0 i j , − − k=i ≤ ≤ gij =   βj Q for i = j + 1 ,   0 for i > j + 1 , that is,   α α β α β β α β β β 0 1 0 2 0 1 3 0 1 2 · · · β α α α α β α α β β  0 − 1 0 − 2 0 1 − 3 0 1 2 · · ·  G = 0 β1 α2 α1 α3 α1 β2 .  − − · · ·   0 0 β2 α3 α2   − · · ·       · · · · · · · · · · · · · · ·  88 4 APPLICATIONS AND EXAMPLES

For a proof see [Si, Prop. 4.1.2].

Note that multiplication with a complex number of absolute value 1 does not aﬀect or- 2 thogonality in Lµ; therefore, if G represents the multiplication operator D, we can assume 2 0 < βn = 1 αn for all n. − | | p D 2 2 4.2.23 Proposition. Let µ be an om for (Pn)n 0 such that supp(µ) ∂ and Pµ = Lµ. N n 2 ≥ ⊂ 26 For n deﬁne n := (Mµ∗) Pµ and let Q n be the orthogonal projection in Lµ onto ∈ H− 2 − n, Q0 the orthogonal projection onto Pµ . H− n ϕ−n 2 Then with ϕ n := Q1 n(Mµ∗) P0 and P n := , we obtain an onb of Lµ given by − − − ϕ−n L2 P : n Z . k k µ { n ∈ } In terms of this basis, Mµ has the matrix representation M = (mij)i,j Z where ∈ g for i, j 0 , ij ≥ δ for i, j 1 ,  i+1,j ≤ − mij =  ∞  αi 1 βk for i 0, j = 1 ,  − − k=i ≥ − 0 Q otherwise ,   with gij as in 4.2.22.  This construction is discussed in detail in [Si, Section 4.1], see there for a proof.

Remark. The coefficients αn in 4.2.22 and 4.2.23 also appear in the following context. For m any polynomial p C[z], p(z) = b0 + b1z + . . . = bmz , we define the reversed polynomial ∈ m by p∗(z) := bm + bm 1z + . . . + b0z . Furthermore, for n N, let Φn = cnPn with cn C − ∈ ∈ such that the leading coefficient of Φn is 1. Then, according to [Si, Theorem 1.5.2],

Φ (z) = z Φ (z) + α Φ∗ (z) and Φ∗ (z) = Φ∗ (z) z α Φ (z) . n+1 n n n n+1 n − n n Simon [Si] calls these αn the Verblunsky Coeﬃcients and, amongst many other things, also shows [Si, Theorem 2.7.15] that P 2 = L2 if and only if (α ) `2. µ 6 µ n n ∈ For details on isometric Hessenberg operators, we also refer to [Kl2, 3.3].

4.3 Orthogonal Polynomials on α-sets

Given a probability measure µ supported on some subset of C, without further assumptions on supp(µ) it might be rather intricate to calculate an orthonormal sequence of 2 2 2 polynomials in Lµ via the Gram-Schmidt algorithm, or to check whether Pµ = Lµ or not. The following deﬁnition is due to [HR1, 6]. §

4.3.1 Deﬁnition. A set K C is called an α-set if, for any continuous function ⊂ f : K C, there exists a sequence (qn)n N in C[z] such that qn f uniformly on K as → ∈ → n . → ∞ 89

Note that, according to the Weierstrass approximation theorem, every compact subset of R is an α-set.

4.3.2 Proposition. A set K C is an α-set if and only if K is compact, its interior is ⊂ empty, and its complement in C is connected.

This characterization is the subject of [HR2], see there for a proof.

4.3.3 Lemma. Let µ be a Borel measure such that K := supp(µ) is an α-set and µ(K) < . Then P 2 = L2 . ∞ µ µ Proof: It is a well-known fact (see [Ru1, 3.14], for instance) that, as µ(A) < for every ∞ compact A C, the space Cc(C) of all compactly supported (complex-valued) functions C ⊂ 2 C on is dense in Lµ. Clearly, f K C(K) whenever f Cc( ) and C(K), in a canonical 2| ∈ ∈ way, is a dense subspace of Lµ as well. Moreover, as K is an α-set, any f C(K) can uniformly on K be approximated by a sequence of polynomials. Note that, in∈connection with µ(K) < , uniform convergence on K implies convergence with respect to the 2 ∞C Lµ-norm. Hence [z] is a dense (w.r.t. 2 ) subspace of C(K). k·kLµ C 2 2 2 Thus [z] is a dense subspace of Lµ, too. In other words, Pµ = Lµ.

4.3.4 Corollary. Let µ be an om for (Pn)n 0 such that supp(µ) is an α-set. Then (Pn)n 2 ≥ is an onb in Lµ, the multiplication operator D is essentially normal, and E = Λµ.

2 2 2 Proof: According to 4.3.3, Pµ = Lµ, hence (Pn)n is an onb in Lµ. By the characterization 4.3.2, supp(µ) is compact; now 3.3.3 shows the remaining assertions.

Remark. Using 4.3.2, we see that, in particular, every compact proper subset of the unit circle, is an α-set. More generally, if G C is open, bounded, simply connected, and ⊂ supp(µ) $ ∂G then supp(µ) is an α-set. In that case, 4.3.4 gives rise to another proof of 3.3.6(ii).

4.4 Weighted Shifts

n In the case that Pn(z) = bnz the Hessenberg operator D is a weighted shift with respect to the onb (Pn)n, see also 2.1.11 and 2.2.4.

4.4.1 Deﬁnition. A linear operator S in a Hilbert space is called a weighted shift H if there exists an onb (en)n 0 such that en dom(S) and there exist an C satisfying ≥ ∈ ∈ en+1 = an+1Sen for all n. Clearly, an = 0 for all n and it is easy to see that S is bounded if and only if sup 1 < and then this6 supremum equals S . an n N ∞ k k ∈

90 4 APPLICATIONS AND EXAMPLES

Subormality of weighted shifts is quite well-understood and is the subject of a variety of papers as [Sta], [BCZ], and [StSz3, pp. 132ﬀ]; see also [Con1, II. 6]. We will give a brief survey here. §

4.4.2 Stieltjes Moment Problem. Let (mn)n 0 be a sequence of positive real numbers. ≥ We look for a Borel measure τ supported on [0, ) such that ∞

∞ n t dτ(t) = mn 0 R for all n N . If such τ exists then (m ) is called a Stieltjes moment sequence. ∈ 0 n n

n 4.4.3 Proposition. Let Pn(z) = bnz where b0 = 1 and bn = 0 for all n N. 2 6 ∈ − There exists an om µ if and only if bn n 0 is a Stieltjes moment sequence. | | ≥ This is a well-known fact, see [Kl2, 4.2.2] or [StSz2, Theorem 4] for a proof.

Note that (Pn)n is an orthogonal system with respect to λ (normalized Lebesgue measure on the unit circle). The following is part of [Kl2, 4.2.2], too. See also [BT]. b

n 4.4.4 Proposition. Let µ be a rotation invariant om for (Pn)n 0. Then Pn(z) = bnz ≥ where b = 1 and b = 0 for all n N. 0 n 6 ∈ 2 − Conversely, assume that, in the situation of 4.4.3, bn n 0 is a Stieltjes moment | | ≥ sequence and let τ be a solution of the associated moment problem. Then µ := τ λ is an om for (P ) . ⊗ n n b 4.4.5 Lemma. Let µ be an om for (Pn)n 0 as in 4.4.3. Then ≥ 2 bn 1bn+1 bn for all n N . (4.2) | − | ≤ | | ∈ Moreover, P 2 = L2 . µ 6 µ Proof: As DP = bn P , the matrix representation of D is of the form d = bn n bn+1 n+1 n+1,n bn+1 and d = 0 whenever i = j + 1. Now (3.5) yields ij 6 2 j+1 2 bj−1 2 ∞ 2 2 2 bj = dj,j = djk dij = dj ,j = (4.3) bj 1 +1 bj+1 | − | k=0 | | ≤ i=0 | | | |

P P for all j N and (4.2) is an immediate consequence. ∈ 2 2 Now assume Pµ = Lµ. Then, according to 3.2.2, D is formally normal. However, D cannot satisfy (3.4), as in the 0-th row of its matrix representation all entries vanish while the 0-th column contains one element diﬀerent from 0. Hence we have a contradiction and, therefore, P 2 = L2 . µ 6 µ 91

Note that, according to (4.3), we can deﬁne a monotonically decreasing sequence by 1 a := bn+1 and obtain P (z) 2 = b 2 z2 n < if z < lim a − . n bn n n n n 0 | | n 0 | | | | ∞ | | n ≥ ≥ →∞ P P

4.4.6 Lemma. If, in the situation of 4.4.3, an om exists then E is an open disk centered at the origin (including the case E = C). More precisely, with a := lim bn+1 , n bn →∞

C 1 z : z < a if a > 0 , E = ∈ | | ( C if a = 0 .

Proof: It only remains to show that z / E if a > 0 and z = 1 . ∈ | | a As (an)n is monotonically decreasing, we observe that

2 1 2 1 1 2 1 N bn 2 n bn 2 2 n−1 = bn 1 2 n−1 for all n | | (a ) ≥ | | an−1 (a ) | − | (a ) ∈ 2 1 n 2 1 implying bn a2 = and hence Pn(z) = whenever z = a . n 0 | | ∞ n 0 | | ∞ | | ≥ ≥ P P

n 4.4.7 Summary on Weighted Shifts. Let Pn(z) = bnz where b0 = 1 and bn C 0 2 ∈ \{ } for all n N. If there is an om µ for (Pn)n then bn 1bn+1 bn for all n and − a := lim ∈bn+1 exists. | | ≤ | | n bn →∞ Note that D is bounded with D = 1 if and only if a > 0. Recall that, according to a 3.2.16, if D is bounded and therek existsk an om then it is unique. C 1 C Moreover, E = z : z < a or E = if a > 0 or a = 0, respectively, and, see 2.1.11 and 2.2.4, (P∈) is |an| onb in the reproducing kernel space (K), consisting of n n holomorphic functions deﬁned on E, whose kernel is given by H

2 n K : E E C , K(z, w) = Pn(z) Pn(w) = bn (z w) . × → n 0 n 0 | | ≥ ≥ P P 2 Hence (K) is, via Pn Pn, isometrically isomorphic to Pµ which is a proper subspace 2 H 7→ of Lµ. Note that, by 3.5.6, E = Ereg.

n Our favorite example, Pn(z) := z , see 1.2.2, 1.3.8, and 2.1.11, ﬁts into this situation. C 1 Here E = z : z < 1 and (Pn)n is an onb in (K) where K(z, w) = 1 zw and λ is the unique{ ∈om. | | } H − b

4.4.8 Corollary. Whenever, for (Pn)n 0 as in 4.4.7, E is not an open disk (and = C) ≥ then there exists no om. 6

This is an immediate consequence of the above.

In the following two examples, D is a non-subnormal weighted shift. 92 4 APPLICATIONS AND EXAMPLES

n 1 4.4.9 Example. Let Pn(z) := bnz where b0 = 1 and b2n 1 := b2n := for n 1. − √(n 1)! ≥ − 1 1 2 For n > 2 we have b2n 2 b2n = > = b2n 1 and, according to 4.4.5, | − | √n 1 (n 2)! (n 1)! | − | there exists no om. − · − − However,

2 ∞ 1 2(2k 1) 2 2k 2 2 4 Pn(z) = 1 + (k 1)! z − + z · = 1 + z 1 + z exp z < n 0 | | k=1 | | | | | | | | | | ∞ ≥ − P P for all z C. Hence, see 2.2.4, (Pn)n is an onb in (K) which consists of entire functions and an analogous∈ calculation to the above yields KH(z, w) = 1 + z w 1 + z w exp (z w)2 .

n 2 4.4.10 Example. Let Pn(z) := n! z for n N0. Here Pn(z) < z = 0. ∈ n 0 | | ∞ ⇐⇒ ≥ Hence E = 0 is not an open disk and, according to 4.4.8,P there exists no om. { } Note that, as E only consists of a single point, the question whether (Pn)n is an onb in (K), is easily answered. H

It is worth mentioning that for a subnormal weighted shift the case E = C can occur, as we shall see now.

4.4.11 Gauss Measure. It is the matter of an easy calculation that the polynomials 1 n (Pn)n 0, given by Pn(z) := z , satisfy ≥ √n!

1 2 2 Pn(x + iy) Pm(x + iy) π exp (x + y ) dx dy = δnm (z = x + iy) . C − R Hence here we have an om µ such that supp(µ) = C. Moreover,

2 ∞ 1 2 n 2 Pn(z) = n! z = exp z n 0 | | n=0 | | | | ≥ P P and therefore, E = C. Clearly, the kernel of (K) is K(z, w) = exp(zw). H Recall the spectral properties given in 3.3.1 as well as 3.1.3. Accordingly, this is an example for an unbounded subnormal operator D such that σ(D) = σr(D) = C = σ(D∗) = σp(D∗) which has a minimal normal extension Mµ acting in a larger space with σ(Mµ) = σc(Mµ) = C.

Remark. A characterization of the spectra of unbounded subnormal weighted shift operators can also be found in [StSz3, p. 136]. 93

4.5 Various Examples

In addition to the examples already discussed in the previous sections and chapters, we will now present some more sequences of polynomials and associated orthonormalizing measures as well as such sequences which do not have an om at all. The following theorem provides a nice tool for proving the non-existence of orthonormalizing measures.

4.5.1 Theorem. Let (Pn)n 0 be a sequence of polynomials as in 1.1.1 such that E = C. 2 ≥ If there exists (an)n 0 ` 0 satisfying ≥ ∈ \ { } ∞ anPn(z) = 0 for all z C n=0 ∈ P then there exists no om for (Pn)n.

Proof: According to 2.1.9, (P ) is not an onb in (K). n n H Now assume that there exists an om µ. Then µ(E) = µ(C) = 1 and 3.7.2 yields that (P ) is an onb in (K); thus we have a contradiction. n n H Hence there exists no om.

1 n 1 4.5.2 Example (Continuation of 2.1.12). Let P 1 and P (z) = z − (z n) 0 ≡ n √n! − for n 1. ≥ 2 ∞ 1 C In 2.1.12 we have already seen that Pn(z) < and √ Pn(z) = 0 for all z . n 0 | | ∞ n=0 n! ∈ ≥ Therefore, by 4.5.1, there exists no omP for (Pn)n. P

An apparently slight modiﬁcation of the above leads to the following example which looks very similar but where 4.5.1 is not applicable.

1 n 1 4.5.3 Example. Let P0 : 1 and Pn(z) := z − (z 1) for n 1. ≡ √(n 1)! − ≥ − 2 2 2 It is not diﬃcult to see that Pn(z) = 1 + z 1 exp z for all z C, hence n 0 | | | − | | | ∈ E = C. ≥ P 2 Now let x = (xn)n 0 ` such that ≥ ∈

∞ ∞ xn n 1 0 = xnPn(z) = x0 + (z 1) z − for all z C . n=0 − n=1 √(n 1)! ∈ − P P For z < 1 we obtain | | ∞ xn n 1 x0 ∞ n 1 z − = 1 z = x0 z − n=1 √(n 1)! n=1 − − P P and uniqueness of power series implies x = (n 1)! x for all n N. As x `2, only n − · 0 ∈ ∈ p 94 4 APPLICATIONS AND EXAMPLES

x = 0 remains and 2.1.9 implies that (P ) is an onb in (K). Furthermore, n n H

K(z, w) = Pn(z) Pn(w) = 1 + (z 1)(w 1) exp z w . n 0 − − ≥ P In particular, (Pn)n does not satisfy the premises of 4.5.1.

Moreover, we do not know whether there exists an om for (Pn)n at all. Yet we see that the multiplication operator D is unbounded by 3.1.6 since E = C; a simple calculation shows that the matrix representation of D is given by

1 0 0 0 · · · 1 0 0 0  · · ·  0 1 0 0  · · ·  (dij)i,j 0 =  .  ≥  . 0 √2 0   . · · ·   . √   . 0 3   . ·.· ·   . ..      which does not obey (3.4). Hence D is not formally normal and therefore, if there exists an om µ then, according to 3.2.2, P 2 = L2 . µ 6 µ Note that the matrix coeﬃcients dij obey (3.5) which, however, is just a necessary criterion for existence of an om; it remains an open question if there exists an om for these (Pn)n.

The following example looks similar to the previous one but leads to a situation where we can use (3.5) to conclude that there exists no orthonormalizing measure. Moreover, we will see that it is possible that E is connected and (K) contains functions which are not continuous on all of E. However, this will not giveHan answer to the question whether all f (K) have to be holomorphic on E◦ in general. ∈ H

n 1 4.5.4 Example. Let P : 1 and P (z) := n z − (z 1) for n 1. Now 0 ≡ n − ≥ N N 2 2 2 2 n 1 Pn(z) = 1 + z 1 n z − n=0 | | | − | n=1 | | P P 2 shows that Pn(z) < z < 1 or z = 1. n 0 | | ∞ ⇐⇒ | | ≥ We can constructP the reproducing kernel space (K) via the kernel K : E E C, H × → 1 if z = 1 or w = 1 and

K(z, w) = Pn(z) Pn(w) = ∞ 2 n 1 n 0  1 + (z 1)(w 1) n (z w) − otherwise , ≥ − − P  n=1 P where E = z C : z < 1 1 . { ∈ | | } ∪ { } Note that K(z, z) is bounded on every compact subset of E◦ = z C : z < 1 and { ∈ | | } hence, by 2.2.3, all f (K) are holomorphic in E◦. ∈ H 95

2 ∞ Now let x = (xn)n ` such that xnPn(z) = 0 for all z E. ∈ n=0 ∈

For z = 1, we immediately obtain Px0 = 0. Then

∞ ∞ n 1 ∞ n 0 = xnPn(z) = (z 1) xn n z − = x1 + n xn (n + 1) xn+1 z for all z E n=1 − n=1 − n=1 − ∈ P P P implies x = 0 and x = n x for all n 1. This yields x = 0 for all n and, 1 n+1 n+1 n ≥ n according to 2.1.9, (P ) is an onb in (K). n n H We next show that (K) contains a function which is not continuous at z = 1. H 1 N 2 Set a0 := 1 and an := n for n . Clearly, a := (an)n ` and f := anPn (K) ∈ ∈ n 0 ∈ H ≥ is well-deﬁned. In particular, P

N N N 1 n 1 n n 1 f(z) = lim an Pn(z) = lim 1 + n z − (z 1) = lim 1 + z z − N N n N →∞ n=0 →∞ n=1 − →∞ n=1 − P P P 1 for z = 1 , = lim zN = N ( 0 for z E 1 . →∞ ∈ \ { } Thus we have found a member of (K) which is not continuous at z = 1. However, all H f (K) are holomorphic in E◦. ∈ H Having in mind 4.2.10, one could expect to ﬁnd an om supported on the unit circle having positive point mass at z = 1. To see that this is not the case here, let us examine the Hessenberg operator D more closely. We have zP0(z) = z 1 + 1 = P0(z) + P1(z) and n − zP (z) = P (z) for n 1. A look at the associated Hessenberg matrix n n+1 n+1 ≥ 1 0 0 0 · · · 1 0 0 0  · · ·  0 1 0 0  2 · · ·  (dij)i,j 0 =  . 2  ≥  . 0 0   3 · · ·   . 3   . 0 4   . ·.· ·   . ..      2 ∞ 2 1 2 shows that d1k = 1 > 4 = di1 and, due to 3.2.11, there exists no om. k=0 | | i=0 | | P P n Finally, note that, for arbitrary p = ck Pk C[z], k=0 ∈ P n n 2 k 2 2 2 2 Dp = c0(P0 + P1) + ck k+1 Pk+1 2 c0 + ck 2 p , k k k=1 ≤ | | k=1 | | ≤ k k P P showing that D is continuous. 96 4 APPLICATIONS AND EXAMPLES

4.5.5 Polynomial Sequences of Finite Bandwidth. In [AM] Adams and McGuire k deﬁne a sequence (fk)k 0 in C[z] to be of bandwidth j if fk(z) = z pk(z) where every pk is ≥ of degree j. Moreover, an RKHS (KA) with domain E C is said to be of bandwidth j H ⊂ 2 if it has an onb (fk)k of polynomials of bandwitdth j and E = z C : fk(z) < . ∈ k 0 | | ∞ ≥ According to 1.4.6, the kernel of (KA) is then given by P H A K (z, w) = fk(z) fk(w) . k 0 ≥ P If (fk)k 0 is of bandwidth j then we can construct a sequence (Pn)n 0 as in 1.1.1 by ≥ ≥ deﬁning Pn := fn j for n j and arbitrarily adding P0, . . . , Pj 1. Clearly, the spaces − − (K) and (KA) have the≥same domain and H H j 1 − A K(z, w) = Pn(z) Pn(w) = Pn(z) Pn(w) + K (z, w) . n 0 n=0 ≥ P P We point out that, for (Pn)n 0 such that (Pn+j)n is of bandwidth j, it is possible that A ≥ (Pn+j)n is an onb in (K ) while (Pn)n is not an onb in (K). As an example, recall H 1 n 1 H 2.1.12 where P 1 and P (z) = z − (z n) for n 1. We know that the kernel of 0 ≡ n √n! − ≥ (K) is given by H K : C C C , K(z, w) = 1 + (1 z)(1 w) exp(z w) × → − − and that (P ) is not an onb in (K). n n H Leaving out P , we obtain a sequence of bandwith j = 1 and (KA) with kernel 0 H

A ∞ K (z, w) = Pn(z) Pn(w) = K(z, w) 1 . n=1 − P ∞ Let a = (an)n 1 such that an Pn(z) = 0 for all z C. Then ≥ n=1 ∈ P ∞ 1 n 1 ∞ 1 n ∞ n n 1 0 = √ anz − (z n) = √ anz √ anz − n=1 n! − n=1 n! − n=1 n! P P P ∞ 1 n ∞ √n+1 n C N yields √ anz = √ an+1z for all z and we obtain an = 0 for all n . n=1 n! n=0 n! ∈ ∈ Now 2.1.4P shows that P(P ) is an onb in (KA). n+1 n H

Remark. According to [AM, Theorem 1], the domain of any RKHS of bandwidth j is the union of an open disk and at most j points oﬀ that disk.

Note that, when we leave out P0 in the examples 4.5.3 or 4.5.4, we obtain a sequence of bandwidth j = 1. Moreover, the following lemma shows that they are orthonormal bases in the corresponding (KA)-spaces. H 97

4.5.6 Lemma. Let (Pn)n 0 be a sequence as in 1.1.1 and define E and (K) as in 2.1.8. ≥ H Fix j N, then ∈ A A K : E E C , K (z, w) := Pn(z) Pn(w) × → n j ≥ P A is the kernel of an RKHS (K ) and if (Pn)n is an onb in (K) then (Pn+j)n is an onb in (KA). H H H Proof: Clearly, KA is well defined on E E; the existence of (KA) is due to 2.1.1. × H 2 ∞ Now choose a = (an)n 0 ` such that an Pn+j(z) = 0 for all z E. Define b = (bn)n 0 ≥ ∈ ∈ ≥ by n=0 P an j for n j , − bn := ≥ ( 0 otherwise .

∞ Then bn Pn(z) = 0 for all z E. As (Pn)n is an onb in (K), 2.1.4 implies b = 0. n=0 ∈ H A Thus aP= 0 and, using 2.1.4 again, we see that (Pn+j)n is an onb in (K ). H

4.5.7 Newton Polynomials. For n N , let ∈ 0 n n z 1 ( 1) P (z) := ( 1) − = − (z 1)(z 2) (z n) . n − n n! − − · · · − Due to z P (z) = (z n 1)P (z) + (n + 1)P (z) = (n + 1)P (z) + (n + 1)P (z), the n − − n n − n+1 n matrix representation of D is given by

1 0 0 · · · 1 2 0  − · · ·  (dij)i,j 0 = 0 2 3 ≥  − · · ·   .   . 0 3   . − ·.· ·   . ..      which, as one can easily see, does not obey (3.4) but obeys (3.5). Hence D is not formally normal but may be subnormal. Indeed, it is subnormal, as we shall see below. If there exists an om µ then 3.2.2 implies P 2 = L2 . µ 6 µ Clearly, P (z) `2 if z is a positive integer, as in these cases only ﬁnitely many entries n n ∈ are diﬀerent from 0. Thus N E. More precisely, if k N then Pn(k) = 0 for all n k ⊂ ∈ 2 ≥ and Pk 1(k) = 0 which shows that Pn(k) : k N is total in ` and, by 2.1.9, (Pn)n − n is an onb in 6 (K). ∈ H The space (K) is known explicitly. According to [Kst, Theorem 1(ii)], here H C 2 C 1 E = z : Pn(z) < = z : Re (z) > 2 , ∈ n 0 | | ∞ ∈ ≥ P z 1 w 1 Γ(z+w 1) K(z, w) = = − , and all f (K) are holomorphic in E. −n n− Γ(z)Γ(w) n 0 ∈ H ≥ P 98 4 APPLICATIONS AND EXAMPLES

Moreover, see [Kst, Theorem 1(iv)], there exists an om µ which is given by

2 1 Γ(x+iy) dµ(x, y) = 2π | Γ(2x) | dy dγ(x) (z = x + iy) ,

n+1 where γ is the discrete measure having unit mass at the points x = 2 , n = 0, 1, 2, . . .

Note that here we have an example for a sequence (Pn)n 0 with the remarkable properties ≥ (i) D is an unbounded subnormal operator which is not formally normal,

(ii) ∂E is a proper subset of supp(µ),

(iii) E supp(µ) = ∅, E C supp(µ) = ∅, and (C E) supp(µ) = ∅, ∩ 6 ∩ \ 6 \ ∩ 6 (iv) all f (K) are holomorphic in E, ∈ H

(v) all Pn have real coeﬃcients but supp(µ) is not contained in the real line.

More generally, we will now examine sequences (Pn)n where the zeros of any Pn are zeros of Pn+1, too.

4.5.8 Example. Let a , a , . . . be mutually diﬀerent complex numbers and c , c , . . . 1 2 1 2 ∈ C 0 . Deﬁne (Pn)n 0 by P0 : 1 and \ { } ≥ ≡ P (z) := c (z a )(z a ) (z a ) n n − 1 − 2 · · · − n for n 1. One can easily check that A := a1, a2, . . . E and that Pn(a) n : a A is total≥ in `2. Hence 2.1.9 imples that (P ){ is an onb}in⊂ (K). ∈ n n H cn For n N, we obtain z Pn(z) = (z an+1)Pn(z) + an+1Pn(z) = Pn+1(z) + an+1Pn(z) ∈ − cn+1 and z P (z) = 1 P (z) + a P (z). Thus the matrix representation of D is given by 0 c1 1 1 0

a 0 0 0 1 · · · 1 a2 0 0  c1 · · ·  c1 (dij)i,j 0 = 0 c a3 0 ≥  2 · · ·   c2   0 0 a4   c3   . . . .   ......      which, as a 2 + 1 2 = a 2, does not obey (3.4) and hence D is not formally normal. 1 c1 1 | | 6 | | 2 2 Therefore, if an om µ exists then Pµ = Lµ by 3.2.2 and, furthermore, (3.5) implies 2 2 6 cn−1 2 2 cn 2 + an+1 an+1 + which ﬁnally yields cn 1cn+1 cn for all n 1. cn | | ≤ | | cn+1 | − | ≤ | | ≥ Note that we have an analogous necessary condition (4.2) for existence of an om in the weighted shift case. Note also that the Newton polynomials ﬁt into this situation and, in contrary to the weighted shift case, here E need not be a disk. 99

While a point which is a zero of all but ﬁnitely many Pn must belong to E, it is possible that there exists z C which is a zero of inﬁnitely many Pn and yet does not belong to E, as we will see in∈the following.

4.5.9 Example. Choose a, b C and deﬁne ∈ (z a)n for even n , Pn(z) := − ( 1 (z b)n for odd n . n! − 2 2 As P2n(z) < z a < 1 and P2n+1(z) < for all z C, we obtain n 0 | | ∞ ⇐⇒ | − | n 0 | | ∞ ∈ ≥ ≥ P P E = z C : z a < 1 . { ∈ | − | } If, in particular, b a 1 then b / E. | − | ≥ ∈

As we have already (and not just once) pointed out that polynomials orthogonal with respect to a measure supported on the real line are quite well understood, we should not go without mentioning the probably best-known class of orthogonal polynomials.

(α,β) 4.5.10 Jacobi Polynomials. For α, β > 1, let (Pn )n 0 be (real) polynomials such − ≥ that deg(Pn) = n, satisfying the orthogonality relation 1 (α,β) (α,β) α β Pn (x)Pm (x) (1 x) (1 + x) dx = 0 whenever n = m . (4.4) 1 − 6 − R (α,β) In other words, (Pn )n is a sequence of polynomials having an om µ, deﬁned via the above integral, where supp(µ) is the interval [ 1, 1]. It is well known that the Jacobi polynomials, −

n n (α,β) ( 1) α β d α β 2 n P (z) = − (1 z)− (1 + z)− (1 z) (1 + z) (1 z ) , n 2n n! − dzn − − (α,β) satisfy (4.4); a short calculation shows that indeed each Pn is a polynomial of degree n and square integrable with respect to µ. See [Sze, Chapter IV] for more details; see also [Chi, V.2(A)] or [BE, 2.3 Exercise 5]. Note that, to match 1.1.1, we still have to normalize the Jacobi polynomials. According to [Sze, (4.3.4)], 1 2n+α+β+1 Γ(n+1)Γ(n+α+β+1) 2 (α,β) Pn := 2α+β+1 Γ(n+α+1)Γ(n+β+1) Pn , 2 yields an orthonormal system in Lµ. Note also that, as supp(µ) is contained in the real line, the multiplication operator D is symmetric and the Jacobi polynomials obey a 3-term recurrence (4.1) which is explicitly known, see [Sze, (4.5.1)] or [BE, 2.3 Exercise 5], for example. 2 2 Now we have several means to show Pµ = Lµ, for example, using the Stone-Weierstrass theorem. Another way goes via the operator D which is symmetric and bounded and hence essentially self-adjoint. According to 1.3.7 and 4.1.2, we obtain 100 4 APPLICATIONS AND EXAMPLES

2 2 2 (i) Pµ = Lµ, i.e. the (normalized) Jacobi Polynomials form an onb in Lµ, (ii) µ is the unique om,

2 (iii) Pn(z) = for all z C; in other words, E = ∅. n 0 | | ∞ ∈ ≥ P Remark. The Jacobi polynomials in the case α = β are called ultraspherical polynomials 1 and sometimes referred to as Gegenbauer polynomials, too. In particular, for α = β = 2 1 − or α = β = 2 , we obtain the so called Chebyshev polynomials or Chebyshev polynomials of the second kind, respectively. The simplest case is α = β = 0 where we get the Legendre polyonomials with normalized Lebesgue measure on [ 1, 1] as (the unique) om. −

In the following example, D is an unbounded essentially self-adjoint operator.

4.5.11 Hermite Polynomials. One can easily verify that, for n N , ∈ 0 n H (z) := ( 1)n exp(z2) d exp( z2) n − dzn − is a polynomial of degree en and, due to n+1 n n−1 d 2 d 2 d 2 exp( z ) = 2z exp( z ) 2n − exp( z ) , dzn+1 − − dzn − − dzn 1 − we obtain the 3-term recurrence Hn+1(z) = 2z Hn(z) 2n Hn 1(z) for n 1. − − ≥ 1 Re-normalizing this sequence by Hn := √ n Hn yields e 2 n! e e 1 1 1 1 n+1 2 n 2 1e 2 n 2 2 Hn+1(z) + 2 Hn 1(z) = 2n+2 n! Hn+1(z) + 2n (n 1)! Hn 1(z) − − − 1 1 2 e e = 2n n! z Hn(z) = z Hn(z) . (4.5) Note that z H (z) = z = 1 H (z) and D is symmetric. 0 √2 1 e 1 n 2 As (4.5) implies Hn+1(0) = Hn 1(0), we inductively obtain n+1 −

1 1 3 (2k 1) 2 N H2k(0) = · 2···4 2−k and H2k+1(0) = 0 for all k 0 . · ··· ∈ 2 In particular, Hn(0) = , i.e. 0 / E. n 0 | | ∞ ∈ ≥ Moreover, D isPessentially self-adjoint because otherwise 4.1.6 would imply E = C in contradiction to the above. Now 4.1.2 shows that there exists a unique om µ supported on the real line satisfying P 2 = L2 . In particular, 3.1.7 yields µ( 0 ) = 0. µ µ { } Note that one can conclude these facts without knowledge of the measure itself. Indeed, the om for (Hn)n is explicitly known. As shown in [BE, 2.3 Exercise 6], for instance, the Hermite polynomials satisfy

∞ exp( x2) Hn(x)Hm(x) √−π dx = δnm . −∞R 101

A treatise on the Hermite polynomials can also be found in [HS, (16.25)]. The fact that 2 2 2 they are a basis in Lµ, i.e. Pµ = Lµ, can be proved using knowledge on analytic functions; see [HS, (21.64)] for a sketch of proof which is far from being trivial. Finally, using 3.3.2, we see E = ∅.

The following example is taken from [EM].

4.5.12 Modified Hermite Polynomials. Keeping with the notations in 4.5.11 (which slightly differ from those in [EM]), for 0 < A < 1, define

n A 1 A 2 Hn := 1+−A Hn .

A According to [EM, Theorem 3.2], the set of functions (Ψn )n 0 deﬁned by ≥ 1 A 1 A 2 1 2 A Ψ (z) := − exp z H (z) n π√A − 2 n is an onb in a Hilbert space XA consisting precisely of all entire functions ϕ satisfying

ϕ(x + iy) 2 exp Ax2 1 y2 dx dy < | | − A ∞ R with inner product ϕ , ψ = ϕ(x + iy) ψ(x + iy) exp Ax2 1 y2 dx dy. h iXA − A Now R A 1 A 2 1 2 HA(x + iy) H (x + iy) − exp (1 A)x 1 y dx dy n m π√A − − − A − R = ΨA(x + iy) ΨA (x + iy) exp Ax2 1 y2 dx dy = δ n m − A mn R A A is the matter of a simple calculation; hence we have found an om µ for (Hn )n with supp(µ) = C.

Moreover, as shown in [EM, Corollary 3.3], XA is an RKHS with kernel

2 2 2 ∞ A A 1 A 1+A 2 2 1 A KXA (z, w) = Ψn (z) Ψn (w) = 2−πA exp 4A (z + w ) + −2A zw n=0 − P A 2 C which shows that Hn (z) < for all z . n 0 | | ∞ ∈ ≥ Thus we can now deﬁneP an RKHS (KA) with domain E = C and kernel H A ∞ A A π√A 1 2 2 ∞ A A K (z, w) = Hn (z) Hn (w) = 1 A exp 2 z + w Ψn (z) Ψn (w) n=0 − n=0 P P 2 2 1+A (1 A) 2 2 1 A = exp − (z + w ) + − zw 2√A − 4A 2A 1+A 1 A 2 2 = exp − A(z + w) (z w) . 2√A 4A − − 102 4 APPLICATIONS AND EXAMPLES

Clearly, KA(z, z) is bounded on every compact subset of C and, see 2.2.3, all members A A A A of (K ) are entire functions. Note that (Hn )n have an om µ such that µ (E) = 1. H A A 2 Thus, by 3.7.2, (Hn )n is an onb in the space (K ) which we now can identify with PµA 2 2 H and 3.7.13 yields P A = L A . Note that (4.5) yields µ 6 µ 1 1 1 1 A n+1 2 1+A 2 A n 2 1 A 2 A z Hn (z) = 2 1 A Hn+1(z) + 2 1+−A Hn 1(z) , − − showing that the multiplication operator D in thiscase is notsymmetric but still tri- diagonal and we obtain (4.5) when letting A 0. → Moreover, D is not formally normal, as (3.4) is not satisﬁed here, and 3.2.2 implies that A 2 2 there exists no om ν for (Hn )n such that Pν = Lν. A A However, we do not know whether µ is the only om for (Hn )n.

4.5.13 Measures on Fractal Sets. The method described below is a standard tool, see [F, Prop. 1.7], for example, to construct measures on Cantor-like fractal sets.

We start with a sequence (Ek)k N0 of subsets of a metric space (X, d) such that each Ek, ∈ k > 0, is the union of ﬁnitely many non-empty disjoint sets

mk Ek = Uk,i i∪=1 where every Uk,j contains some Uk+1,i and is contained in one Uk 1,j. Hence Ek Ek 1 − ⊂ − for all k N. ∈

The sets Uk,i will be referred to as basic sets; for ﬁxed k, the sets Uk,1, . . . , Uk,mk are called the basic sets of level k. Let U be the collection of all basic sets and deﬁne F := Ek. k ∩N0 Note that, in general, F may be empty. ∈

U1,1 U1,2

m2 E0 E1 = U1,1 U1,2 E2 = U2,i ∪ i∪=1 Figure 7. An example to illustrate the construction of Cantor-like sets.

Furthermore, we deﬁne the maximum diameter dk of Ek by

dk := max sup d(x, y) : x, y Uk,i i=1,...,mk ∈ and assume d 0 as k . We assign a mass distribution ρ to the sets U such that k → → ∞ k,i 0 < ρ(Uk,i) < and ρ(Uk,i) = ρ(Uk+1,j) for k > 0 , ∞ j P the sum to be taken over all j satisfying U U . k+1,j ⊂ k,i 103

Now, for A X, deﬁne ⊂

µ(A) := inf ρ(Uj) : (A F ) Uj where Uj U for all j . (4.6) j ∩ ⊂ ∪j ∈ nP o It is the matter of an almost straightforward calculation that µ(A B) = µ(A) + µ(B) ∪ whenever A, B X with d(A, B) > 0. Therefore, by Carath´eodory’s lemma, the restriction of µ to the⊂Borel subsets of X is a measure, see [Ma, 4.1 and 4.2] or [El, 9.2 and 9.3] for details.

4.5.14 Theorem. In the situation of 4.5.13, let X as well as all basic sets be compact and ρ(E ) = 1. Then µ(U) = ρ(U) for all U U and µ(F ) = 1. 0 ∈

Proof: Compactness of the basic sets yields F = ∅ and V F = V Ei = ∅ for 6 ∩ i N ∩ 6 all V U. ∩∈ ∈ Next we show that V F is relatively open w.r.t. F for any V U. To see that, let V be a basic set of level k. ∩By construction, ∈

O := X U : U is basic set of level k , U = V \ 6 is open in X and O F = V[ F . Hence V is relatively open as asserted. ∩ ∩ Let now U be a basic set of level p and V a family of basic sets satisfying (U F ) V . ∩ ⊂ V∪V In order to see that for A = U in (4.6) the inﬁmum is attained when we simply ∈cover U F with U, we have to show ρ(U) ρ(V ). ∩ ≤ V V ∈ First we note that (U F ) V yieldsP(U F ) (V F ). V V V V ∩ ⊂ ∪∈ ∩ ⊂ ∪∈ ∩ As U F is compact and all V F are relatively open, there exist ﬁnitely many V1, . . . , Vn such that∩ ∩ U F (V V ) F V V . (4.7) ∩ ⊂ 1 ∪ · · · ∪ n ∩ ⊂ 1 ∪ · · · ∪ n It is no restriction to assume that V , . . . , V all are of level q with some q > p. 1 n ≤ Now, for i 1, . . . , n , let W , . . . , W be te basic sets of level q contained in V and ∈ { } i,1 i,ki i let U1, . . . , Uk0 be the basic sets of level q contained in U. Then

ki k0 ρ(Vi) = ρ(Wi,j) and ρ(U) = ρ(Ul) . j=1 l=1 P P For ﬁxed l 1, . . . , k0 , choose x Ul∗ F . By (4.7), there exist i and j such that ∗ ∈ { } ∈ ∩ ∗ ∗ x W ∗ ∗ . As two basic sets of the same level either have empty intersection or coincide, ∈ i ,j this yields Ul∗ = Wi∗,j∗ . Therefore,

k0 n ki n ρ(U) = ρ(Ul) ρ(Wi,j) = ρ(Vi) ρ(V ) l=1 ≤ i=1 j=1 i=1 ≤ V V ∈ P P P P P and hence µ(U) = ρ(U) for all basic sets U.

Finally, note that ρ(Ek) = µ(Ek) = 1 for all k and hence µ(F ) = lim µ(Ek) = 1. k →∞ 104 4 APPLICATIONS AND EXAMPLES

4.5.15 Example. Let now K be a compact uncountable Lebesgue-nullset3 contained in [0, 1] and define µ(K) := 1. Then there exists x 1 , 3 such that x / K. Set ∈ 4 4 ∈ U1,1 := [0, x] K, U1,2 := [x, 1] K, and E1 := U1,1 U1,2. Note that these sets are disjoint ∩ ∩ ∪ 1 and compact. If both U1,1 and U1,2 are uncountable, then let µ(U1,1) := µ(U1,2) := 2 . If U1,1 is at most countable then U1,2 must be uncountable and we define µ(U1,1) := 0 and µ(U1,2) := 1. Otherwise, we set µ(U1,1) := 1 and µ(U1,2) := 0. As K is a Lebesgue-nullset, any interval [a, b] with a < b intersects the complement of K. Therefore, repeating the technique above, we can recursively define E2, E3, . . . such that dk 0 as k . On the other hand, Ek = K for all k, and hence K = Ek. k N0 → → ∞ ∈∩ Note that some of the Uk,i defined in this way may be empty which, however, does not affect the soundness of this construction; we can simply leave them out. According to 4.5.14, µ can be extended to a Borel measure with µ(K) = 1.

Clearly, one can analogously construct a Borel measure supported on an arbitrary uncountable compact set K ∂D which is s nullset w.r.t. one-dimensional Lebesgue measure on the unit circle. This giv⊂es rise to the following remarkable example.

4.5.16 Example. Let λ denote normalized Lebesgue measure on the unit circle ∂D and K ∂D be an uncountable compact set such that λ(K) = 0. Now construct a Borel ⊂ measure ν with and ν(Kb) = 1 in analogy to 4.5.15. 1 b Then the measure µ := 2 (λ + ν) is an om for a sequence (Pn)n 0 of polynomials as usual, 1 ≥ satisfying the Szeg˝o condtition 4.2.18, and 2 ν is the singularly continuous part of µ. 2 b 2 D According to 4.2.20, Pµ = Lµ and E = . Moreover, K, up to a µ-nullset, coincides with 6 2 the set B in the proof of 4.2.21 and 1K Pµ . Hence there exists a sequence (qn)n of polynomials satisfying ∈

(i) lim qn(z) = 0 for all z D, n →∞ ∈ (ii) lim qn(z) = 0 for Lebesgue-almost all z ∂D, n →∞ ∈ (iii) lim qn(z) = 1 for µ-almost all and hence uncountably many z K. n →∞ ∈

Remark. According to [HS, (10.55)], every uncountable borel set in a complete metric space contains a non-void perfect set which, see [HS, (6.61)] for the deﬁnition, is uncountable and compact.

4.5.17 Summary. The table on the following page presents a survey on (almost) all of the examples mentioned so far. For abbreviation, we will call µ an obm if it is an om such that P 2 = L2 . Moreover, set H := z C : Re (z) > 1 . µ µ r { ∈ 2 } Recall that, according to 3.5.6, if there exists an om µ for a given sequence (Pn)n of polynomials then E is precisely the largest subset of E where all members of (K) are reg H holomorphic. Note that we do not know an example where E = E◦. reg 6 3As an example, one may think of the middle-third Cantor set. 105

(P ) where n n om is E Ereg onb in om obm to ﬁnd ∃ ∃ unique (K) H 1 n 1 2.1.12 Pn(z) = z (z n) C C no no no n/a √n! − − 4.5.2 1 n 1 Pn(z) = z − (z 1) 4.5.3 C C yes ? no ? √(n 1)! − − n 1 Pn(z) = n z (z 1) 4.5.4 D 1 D yes no no n/a − − ∪ { } n Pn(z) = bnz where 1 4.4.9 C C yes no no n/a b2n 1 = b2n = − √(n 1)! − 1.2.2 1.3.8 P (z) = zn D D yes yes no yes n 2.1.11 4.4.7 2.3.7 P (z) = √n + 1 zn D D yes yes no yes n 4.4.7 D symmetric but not 4.1.6 C C yes yes yes no essentially self-adjoint Jacobi polynomials 4.5.10 ∅ ∅ n/a yes yes yes Hermite polynomials n ( 1) 2 P (z) = − exp(z ) n √2nn! 4.5.11 ∅ ∅ n/a yes yes yes n d [exp( z2)] · dzn − modiﬁed 4.5.12 C C yes yes no ? Hermite polynomials Newton polynomials n ( 1)n 4.5.7 Hr Hr yes yes no ? P z z k n( ) = −n! ( ) k=1 − discrete om µ Q 3.7.11 Λµ ∅ yes yes yes yes where Λµ is compact om µ supported on [0, 1]; not discrete but having 4.1.4 Λµ ∅ no yes yes yes some discrete mass points om µ supported on ∂D; not discrete but having 4.2.2 some discrete mass points Λµ ∅ no yes yes yes 4.2.18 and not obeying the Szeg˝o condition om µ supported on ∂D; 4.2.17 obeying the Szeg˝o condition D Λµ D yes yes no yes 4.2.20 ∪ and µsc = 0 4.2.17 om µ supported on ∂D; 4.2.20 obeying the Szeg˝o condition D Λµ D no yes no yes 4.2.21 ∪ and µsc = 0 6 4.5.16 106 APPENDIX

Appendix

A.1 The Spectral Theorem for Normal Operators

In a complex Hilbert space we denote the inner product and norm by , and H , respectively. We will omitH the index if it is clearly understood. Note thath · the· i inner kpro·kHduct is linear in the second argument. Moreover, let B( ) be the set of bounded H linear operators in . H We will denote the spectrum of a linear operator A by σ(A). For standard deﬁnitions and theorems concerning (possibly unbounded) linear operators in a Hilbert space we refer to [AG], [W], or [Ru2, Ch. 13], for instance. 2 In particular, the spaces Lµ, where µ is an om for a given sequence of polynomials, are of special interest here. For mesaure theoretical details see e.g. [Bau] or [El]. In this section we will state the spectral theorem for normal operators which plays a central role in the theory of orthogonal polynomials, as the orthonormalizing measures for a given sequence of polynomials correspond to the spectral measures of certain normal operators, see 1.3.7. Furthermore, we will quote some facts from the theory of subnormal operators and we will also have a look at fundamental properties of Reproducing Kernel Hilbert Spaces (RKHS) including proofs of the most important statements cited in section 1.4. A densely deﬁned linear operator N in a Hilbert space is normal, if dom(N) = H dom(N ∗) and Nx = N ∗x for all x dom(N). In particular, every self-adjoint k k k k ∈ operator is normal.

A.1.1 Proposition. Let N be a densely deﬁned closed operator in a Hilbert space . H The following properties are equivalent.

(i) N is normal.

(ii) N ∗ is normal.

(iii) N ∗N = NN ∗.

For a proof see e.g. [W, 5.6, Folgerung 2]. 107

A.1.2 Resolutions of the Identity. The following deﬁnitions are standard tools of spectral theory of linear operators in a Hilbert space. Let be a Hilbert space and a σ-algebra on a set Ω. A map E : B( ) is called a resolutionH of the identity orA projection valued measure if itAsatisﬁes→ Hthe following properties. (i) E(∆) is an orthogonal projection for every ∆ . ∈ A (ii) E(Ω) = id.

(iii) If (∆n)n N is a sequence of mutually disjoint sets in then ∈ A ∞ ∞ E ∆n x = E(∆n)x for all x . n∪=1 n=1 ∈ H P As an almost immediate consequence, we get E(∅) = 0 and E(∆1 ∆2) = E(∆1)E(∆2) ∩ for all ∆1, ∆2 . ∈ A 2 For x we define Ex : R, Ex(∆) := x , E(∆)x = E(∆)x . Then Ex is a measure∈ onH and E (Ω) =A x→2. h i k k A x k k 1 For an elementary function u : Ω C, i.e. the image u(Ω) is finite and u− ( a ) for → { } ∈ A all a u(Ω), we define the integral ∈ 1 u dE := a E u− ( a ) a u(Ω) { } ∈P which clearly is a bounded linearR operator in . H 1 Let now f : Ω C be measurable, i.e. f − (B) for all B B(C). Then one can define a linear op→erator by ∈ A ∈

2 dom f dE = x : f dEx < , f dE x := lim ux,n dE x ∈ H | | ∞ n →∞ R R R R where (u ) is a sequence of elementary functions such that u f 2 dE 0 x,n n | x,n − | x → as n . → ∞ R It is well known that Ψ(f) := f dE is a densely deﬁned linear operator in and H R 2 Ψ(f)x = f 2 dE for all x dom Ψ(f) . | | x ∈ R A.1.3 Spectral Theorem for Normal Operators. Let N be a normal operator in . Then there exists a unique resolution of the identity E on B(C) such that H

N = idC dE . R In particular, E(∆) = 0 whenever ∆ σ(N) = ∅. ∩ Moreover, E(∆)S = SE(∆) for all ∆ B(C) and all S B( ) satisfying SN NS. ∈ ∈ H ⊂ 108 APPENDIX

A proof of the spectral theorem can be found in [Ru2, 13.33] or [W, Satz 7.32], for instance. Note that E σ(N) = id and E(∆) = E σ(N) ∆ for all ∆ B(C). Therefore, E is H also referred to as the spectral decomposition∩or the spectral∈ measure of N.

It is not diﬃcult to see that, for a σ-ﬁnite measure µ on B(C), the spectral decomposition 2 of the multiplication operator Mµ in Lµ (which is normal, see 1.3.2) is given by E(∆) = 2 M∆ where M∆ denotes multiplication with 1∆ in Lµ.

A.1.4 Lemma. Let (Ω, A, µ) be a finite measure space and T a bounded linear operator 2 in Lµ that commutes with multiplication by 1∆ for every ∆ , i.e. T (1∆ f) = 1∆ T f 2 ∈ A · · for all f L . Then T acts as multiplication with an L∞-function. ∈ µ µ Proof: As µ is finite, we have 1 L2 for all ∆ A. ∆ ∈ µ ∈ Set g := T 1 and note that g L2 implies g L1 for finite µ. Define γ L2 by Ω ∈ µ ∈ µ ∈ µ

g(z) | | if g(z) = 0 , γ(z) := g(z) 6 ( 1 otherwise .

We obtain

g dµ = γ g dµ γ 1∆ , T 1Ω γ 1∆ T 1Ω = µ(∆) T 1Ω ∆ | | ∆ · ≤ h · i ≤ k · k k k k k k k k k

R R and, therefore, g(z) T 1 for µ-almost all z. This shows g L∞. Moreover, | | ≤ k k k Ωk ∈ µ T 1 = T (1 1 ) = 1 T 1 = 1 g ∆ ∆ · Ω ∆ · Ω ∆ · for all ∆ A. As 1 : ∆ A is total in L2 , this yields T f = f g for all f L2 . ∈ { ∆ ∈ } µ · ∈ µ

A.1.5 Symbolic Calculus for Normal Operators. Let N be a normal operator in a Hilbert space and E its spectral decomposition. For measurable f : σ(N) C, we H → deﬁne Ψ(f) := f dE R as above. This is well-deﬁned as the integral only depends on the values of f on σ(N). For arbitrary p C[z], one can show that Ψ(p) = p(N) where dom Ψ(f) is to be chosen as the canonical∈domain of p(N). For more details, see e.g. [Ru2, 12.24]. In particular,

p(N)x 2 = p 2 dE for all x dom p(N) . (A.1) | | x ∈ R 109

A.2 Normal Extensions of Hessenberg Operators

A densely deﬁned linear operator F in a Hilbert space is formally normal, if dom(F ) H ⊂ dom(F ∗) and F x = F ∗x for all x dom(F ). Consequently, every normal operator is k k k k ∈ formally normal. A densely deﬁned linear operator S in is subnormal, if there exist a Hilbert space H K and a normal operator N in such that can be embedded isometrically into and F x = Nx for all x dom(F ).KIn particular,Hevery symmetric operator is subnormal,K as if S is a symmetric op∈erator in then there always exists a self-adjoint operator extending H S acting in a possibly larger space , see [AG, IX. 111. Satz 1], for example. K The textbook by Conway [Con1] provides extensive information on the theory of bounded subnormal operators. Concerning the theory of unbounded subnormal operators, we will in particular refer to a trilogy of papers by Stochel and Szafraniec, [StSz1], [StSz2], and [StSz3]. As mentioned before, see 1.3.4 and 1.3.7 (for both of which we will present a proof here), subnormal Hessenberg operators play an important role in the theory of orthogonal polynomials. Therefore, subnormality is one of the main topics of any work dealing with questions about orthonormalizing measures, as [CaKl1], [CaKl2], or [Kl2], to name just a few. The following can be found in [CaKl2, Theorem 1(iii)], for instance.

A.2.1 (Proof of 1.3.4). Let N be a normal extension of D in a Hilbert space K ⊃ H and let E denote the spectral measure of N. For ∆ B(C) deﬁne µ(∆) := E (∆) = ∈ P0 P0 , E(∆)P0 . By construction, supp(µ) σ(N). K ⊂ 2 The assignmen t 1∆ E(∆)P0 gives rise to a linear map β : Lµ which is an isometry since 7→ → K

0 0 1∆ , 1∆ L2 = 1∆ 1∆ dµ = µ(∆ ∆0) = P0 , E(∆ ∆0)P0 µ ∩ ∩ K

= RP0 , E(∆)E(∆0)P0 = E(∆)P0 , E(∆0)P0 K K 2 for ∆, ∆0 B(C). Consequently, := β(L ) is the closure of the linear span of E(∆)P : ∈ L µ { 0 ∆ B(C) in and β : L2 is an isomorphism. If ∈ } K µ → L k

u = αj1∆j j=1 P is an elementary function where α , . . . , α C and, without loss of generality, ∆ , . . . , ∆ 1 k ∈ 1 k ∈ B(C) are mutually disjoint then

k βu = αjE(∆j)P0 = u dE P0 . j=1 P R Considering arbitrary p C[z] as a member of , we have p = p(N)P = p(D)P and ∈ K 0 0 (A.1) yields 2 2 2 p(N)P0 = p dEP0 = p dµ K | | | |

R R 110 APPENDIX

showing that p is square integrable with respect to µ. We now show that β maps p L2 ∈ µ to p . ∈ H There exists a sequence (un)n of elementary functions such that un p 2 0 which k − kLµ → implies βun βp 0 as n . On the other hand, k − kK → → ∞

βu = u dE P p dE P n n 0 → 0 R R with respect to . Using A.1.5, we obtain k·kK

p dE P = Ψ(p)P = p(N)P = p(D)P = p . 0 0 0 0 ∈ H R Therefore, βp = p for all p C[z]; in particular, (P ) is an orthonormal system in L2 . ∈ n n µ Hence µ is an om for (Pn)n.

Denote by F the spectral measure of Mµ and recall that F (∆) is multiplication with 1∆ for ∆ B(C). Thus ∈

β F (∆)1∆0 = β1∆ ∆0 = E(∆ ∆0)P0 = E(∆)E(∆0)P0 = E(∆)β1∆0 ∩ ∩ for all ∆, ∆0 B(C) which shows β(Mµf) = N(βf) for f dom(Mµ). Therefore, 1 ∈ ∈ βMµβ− = N dom(N) and D N dom(N) N. 2 |L ∩ ⊂ |L ∩ ⊂ 2 Let P be the completion of C[z] w.r.t. 2 . Then β maps P onto and we can regard µ Lµ µ k·k 2 H D as a densely deﬁned subnormal operator in Pµ with normal extension Mµ acting in the 2 possibly larger space Lµ. Clearly, supp(µ) = σ(Mµ).

A.2.2 Lemma. Let N be a normal operator in a Hilbert space and be a closed subspace of . Then reduces N if and only if the restriction of NKto Ldom(N) is a normal operKator in .L L ∩ L For self-adjoint operators this result is well-known. The generalization to normal operators uses that the real and imaginary parts of a normal operator are self-adjoint, see [Kl2, Satz 1.3.9] or [CaKl2, Theorem 17] for a detailed proof.

It is also well-known, see e.g. [W, 5.6 - Aufgabe 5.39], that reduces N if and only if L dom(N) + ⊥ dom(N) = dom(N) and both as well as ⊥ are invariant under N, L ∩ L ∩ L L i.e. N maps dom(N) and ⊥ dom(N) into and ⊥, respectively. L ∩ L ∩ L L

A.2.3 Theorem. Let µ be a ﬁnite measure on the Borel σ-algebra in C. A closed subspace L2 reduces M if and only if = 1 f : f L2 with some Borel set ∆. L ⊂ µ µ L { ∆ ∈ µ} 2 Proof: For an arbitrary Borel set ∆, deﬁne = 1∆f : f Lµ . 2 L { ∈ } 2 Clearly, ⊥ = 1C ∆f : f Lµ and as well as ⊥ are closed subspaces of Lµ, invariant L { \ ∈ } L L under M . Hence (and also ⊥) reduce M . µ L L µ 111

Now let reduce M , i.e. P M M P where P denotes the orthogonal projection in M µ µ ⊂ µ L2 onto . Now the spectral theorem A.1.3 shows P E(∆) = E(∆)P for all ∆ B(C) µ M ∈ where E is the spectral measure of Mµ. As E(∆) is multiplication with 1∆, we can apply 2 A.1.4 and see that P acts as multiplication with an Lµ∞-function g. Note that P = P which implies that g can only attain the values 0 or 1. Hence g = 1 0 for some ∆0 B(C) ∆ ∈ and P f = 1 0 f for all f L2 . Therefore, = 1 0 f : f L2 as asserted. ∆ ∈ µ M { ∆ ∈ µ}

A.2.4 (Proof of 1.3.7). Let N be a minimal normal extension of D acting in a Hilbert 2 space . In analogy to A.2.1 we can deﬁne an om µ and embed Lµ isometrically K ⊃ H 1 into . Then βMµβ− is normal in and A.2.2 implies that reduces N; by deﬁnition K L L 1 of minimality we obtain = . Hence β is an isomorphism and β− Nβ = M . L K µ As to the converse, let µ be an om. Then Mµ is a normal extension of D; we have to show that it is minimal.

Denote by E be the spectral measure of Mµ. Assume that is a closed subspace of 2 M B C Lµ reducing Mµ and P0 . According to A.2.3, there exists ∆ ( ) such that 2 ∈ M ∈ = 1 f : f L while on the other hand, P = 1C. Therefore, only ∆ = C remains. M { ∆ ∈ µ} 0 Hence = L2 and, therefore, M is minimal. M µ µ For the last assertion, let D be essentially normal, i.e. the closure D of D is a normal operator in . Then any normal extension of D also extends D and as an immediate consequenceHof A.2.2, D is the only minimal extension of D. Hence there exists a unique om µ and we can identify D with M . In particular, is isometrically isomorphic to L2 µ H µ via P P and (P ) is an onb in L2 . n 7→ n n n µ

A.3 Reproducing Kernel Hilbert Spaces

The theory of Reproducing Kernel Hilbert Spaces (RKHS) is treated in a very detailed way by Aronszajn whose article [Ar] may be given as standard reference on this topic. Other references are a textbook by Meschkowski [Me] and also [Do, Chapter X]. We will here construct an RKHS such that a given positive matrix (in accordance to 1.4.5) becomes its kernel and also give a proof of the RKHS-test 1.4.7 as well as provide some examples of RKHS.

The following is well-known, see [Ar, I.4], for example.

A.3.1 (Proof of 1.4.5). Let E be an arbitrary non-empty set and K : E E C a × → map satisfying n n ci cj K(zi, zj) 0 (A.2) i=1 j=1 ≥ for all ﬁnite sets z , . . . , z EPandP c , . . . , c C. { 1 n} ⊂ 1 n ∈ We immediately see that K(z, z) 0 for all z E and a short calculation yields K(z, w) = ≥ ∈ K(w, z) for all z, w E. ∈ 112 APPENDIX

For z E we set K : E C, K (w) := K(z, w). ∈ z → z Let now be the linear span of K : z E , i.e. every u can be written as a H0 { z ∈ } ∈ H0 ﬁnite(!) sum

u = ak Kzk (A.3) k P where a C and z E. We deﬁne a map , : C as follows. For u as k ∈ k ∈ h i H0 × H0 → above and v = bj Kwj set u , v := ak bj K(wj, zk). Taking into account that j h i k j P P P

u(wj) bj = u , v = ak v(zk) , j h i k P P we observe that u , v is well-deﬁned although the representations of u and v as in (A.3) h i may not be unique. In particular, for u = K we get K , v = v(z). z h z i Obviously, u , v = v , u and , is linear in the second argument. Moreover, (A.2) yields u , uh 0.i Therefore,h i we hcani use Cauchy-Schwarz to show h i ≥ u(z) 2 = K , u 2 u , u K , K . | | |h z i| ≤ h i · h z z i

Hence u , u = 0 implies u(z) = 0 for all z E, i.e. u = 0 in 0. Thus , is a positive definiteh sesquilineari form on . ∈ H h i H0 1 Let denote the abstract completion of with respect to the norm u := u , u 2 . H H0 k k h i We now define a map ι : (E) as follows (here (E) denotes the set of all maps e H → F F E C). For f there exists a Cauchy sequence (u ) in with u f as n . → ∈ H n n H0 n → → ∞ Set f : E C, f(z) := lime Kz , un and ι(f) := f. → n h i e e →∞ e Note that ι is well-defined and its restrictione to 0 is the canonical embedding into (E). As is a dense subspace of H, the range ran(H ι) is a Hilbert space isometrically F H0 isomorphic to . H e Thus (K) := ran(ι) is a completion of 0 and f(z) = lim Kz , un e = Kz , f e = H e H n h iH h iH Kz , f (K). →∞ h iH e According to 1.4.1, (K) is an RKHS with kernel K(z, w) := K (w) = K , K . H z h w z i

From pointwise convergence, in general, we can not conclude norm convergence in (K). The following lemma shows that pointwise convergence of a bounded sequence impliesH weak convergence in (K). H

A.3.2 Lemma. Let (f ) be a sequence in (K) and M > 0 such that f M for n n H k nk ≤ all n and f(z) := lim fn(z) exists for all z E. Then n →∞ ∈

f (K) and lim fn , h = f , h for all h (K) . n ∈ H →∞h i h i ∈ H 113

Proof: Choose arbitrary F (K) and deﬁne 0 as in A.3.1. ∈ H k H

For ε > 0 there exists g 0, g = αjKzj say, such that F g < ε. Then ∈ H j=1 k − k P k k

g , fn = αjKzj , fn = αj fn(zj) h i j=1 j=1 D P E P and, in particular, lim g , fn exists. n →∞h i Let now N N such that g , f g , f < ε for all m, n N. We obtain ∈ h n i − h m i ≥

F , fn F , fm F g , fn fm + g , fn fm h i − h i ≤ h − − i h − i F g f f + g , f g , f ε 2M + ε ≤ k − k k n − mk h n i − h m i ≤ · for all m, n N. Hence F , fn is convergent. ≥ h i n Therefore, L : (K) C, Lh :=lim fn , h , is a well-deﬁned linear functional. More- H → n h i over, →∞ f , h f h M h for all n N and all h (K) . h n i ≤ k nk k k ≤ k k ∈ ∈ H Hence L is bounded and the Riesz representation theorem provides f (K) such that ∈ H Lh = f , h . Furthermore, h i e ef(z) = Kz , f = L Kz = lim fn , Kz = lim fn(z) for all z E . n n h i →∞h i →∞ ∈ e e Thus f = f (K) and lim fn , h = Lh = f , h as asserted. n ∈ H →∞h i h i e Given an RKHS with domain E, we would like to be able to decide whether a function f : E C belongs to this space. →

A.3.3 Lemma. Let (K) be an RKHS with domain E. If g is a complex-valued function deﬁned on a non-emptyH set F E such that ⊂ n n n 2 cicj K(zi, zj) ci g(zi) (A.4) i=1 j=1 ≥ i=1

P P P for all ﬁnite sets z , . . . , z F and c , . . . , c C. { 1 n} ⊂ 1 n ∈ Then there exists f (K) satisfying f(z) = g(z) for all z F and (A.4) remains valid for f instead of g and∈ Hall z , . . . , z E. ∈ { 1 n} ⊂ Proof: Let (F ) be the linear span of K : z F , in analogy to (A.3). H0 { z ∈ } For z F set ϕ(K ) := g(z). In order to show that ϕ can be extended to a linear form ∈ z on 0(F ), suppose H m

ak Kvk = 0 . k=1 P 114 APPENDIX

Then (A.4) yields

m m m m

0 = ak Kvk , ak Kvk = aiaj Kvi , Kvj k=1 k=1 i=1 j=1 h i D E mPm P m P P 2 = aiaj K(vj, vi) ai g(vi) . i=1 j=1 ≥ i=1

P P P n Thus for h = ci Kzi 0(F ) the expression i=1 ∈ H P n ϕ(h) := ci g(zi) i=1 P is well-deﬁned. Using (A.4) once more, we get

n 2 n n n 2 2 2 ϕ(h) = ci g(zi) ci cj K(zi, zj) = ck Kzk = h . | | i=1 ≤ i=1 j=1 k=1 k k

P P P P According to the Hahn-Banac h theorem, ϕ can be extended to a con tinuous linear functional on (K) with ϕ(h) h for all h (K). By the Riesz representation theorem there existsH f (K|) such| ≤thatk kϕ(h) = f∈, hH for all h (K) and f 1. ∈ H h i ∈ H k k ≤ For z F then f(z) = Kz , f = ϕ(Kz) = g(z) and for arbitrary z1, . . . , zn E and c , . . . ,∈c C, ﬁnally, h i { } ⊂ 1 n ∈ n 2 n 2 n 2

ci f(zi) = ci Kzi , f = ci Kzi , f i=1 i=1 h i i=1

P Pn 2 Pn n 2 ci Kzi f cicj K(zi, zj) ≤ i=1 · k k ≤ i=1 j=1

P P P

A.3.4 (Proof of 1.4.7). Let (K) be an RKHS with domain E and f : E C. H → (i) Suppose there exists r > 0 such that

n n n 2 cicj K(zi, zj) r ci f(zi) i=1 j=1 ≥ i=1

P P P for all ﬁnite sets z , . . . , z E and c , . . . , c C. Then g := r f obeys (A.4) with { 1 n} ⊂ 1 n ∈ · F = E. Thus g (K) and, obviously, f (K). ∈ H ∈ H (ii) Suppose now f (K) 0 . Take z , . . . , z E; using Cauchy-Schwarz, we get ∈ H \ { } { 1 n} ⊂ n 2 n 2 n 2 n n 2 2 ci f(zi) = ci Kzi , f ci Kzi f f cicj K(zi, zj) . i=1 i=1 ≤ i=1 · k k ≤ k k · i=1 j=1

P P P P P 1 Setting r := f 2 completes the proof. k k

The following theorem is also well-known, see [Ar, I.5], for example. 115

A.3.5 Theorem. Let (K) be an RKHS with domain E und suppose F is a non-empty H subset of E. Then L := K F F is the kernel of an RKHS which we will denote by (L). | × H In particular, (K) = ⊥, where := f (K) : f(z) = 0 for all z F , and H F ⊕ F F { ∈ H ∈ } ⊥ is isometrically isomorphic to (L) via f f F . F H 7→ | Proof: Note that is a closed linear subspace of (K), as convergence in (K)-norm F H H implies pointwise convergence on E. Consequently, (K) = ⊥. H F ⊕ F Deﬁne a linear map β from (K) onto a set L say, such that β(f) := f F , i.e. L is a subset of the set of all functionsH F C. Then β(f) = 0 if and only if f | . → ∈ F In order to show that ϕ := β ⊥ is one-to-one, take f1, f2 ⊥ such that ϕ(f1) = ϕ(f2). Then, on the one hand, 0 =|Fϕ(f f ) = β(f f ) whic∈h Fimplies f f but, on 1 − 2 1 − 2 1 − 2 ∈ F the other hand, f f ⊥. Thus f f = 0. 1 − 2 ∈ F 1 − 2 1 1 Now deﬁne an inner product on L by f , g L := ϕ− (f) , ϕ− (g) (K). Then L, with h i h iH respect to this inner product, is a Hilbert space isometrically isomorphic to ⊥. F For z F let L := ϕ(K ), then ∈ z z 1 Lz , f L = Kz , ϕ− (f) (K) = f(z) for allf L and all z F . h i h iH ∈ ∈ Hence L is an RKHS with domain F and kernel L := K F F . | ×

A.3.6 Positivity of K(z, z). Let (K) be an RKHS with domain E and suppose there is E E such that K(z, z) = 0 forHall z E . 0 ⊂ ∈ 0 Then f(z) = K , f K f = K(z, z) f = 0 for all z E . Thus, if | | |h z i| ≤ k zk · k k · k k ∈ 0 f (K) vanishes on F := E E it is the zero function. Using the notation of A.3.5, ∈ H \ 0 we have ⊥ = (K). F H Therefore, without loss of generality, we always assume K(z, z) > 0 for all z E. ∈ 116

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