Analysis of Five Diagonal Reproducing Kernels

Transcription

1 Bucknell University Bucknell Digital Commons Honors Theses Student Theses 2015 Analysis of Five Diagonal Reproducing Kernels Cody Bullett Stockdale Follow this and additional works at: Recommended Citation Stockdale, Cody Bullett, "Analysis of Five Diagonal Reproducing Kernels" 2015) Honors Theses This Honors Thesis is brought to you for free and open access by the Student Theses at Bucknell Digital Commons It has been accepted for inclusion in Honors Theses by an authorized administrator of Bucknell Digital Commons For more information, please contact

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3 ii Acknowledgments Thanks to the faculty of the Bucknell mathematics department for their continual support I would particularly like to thank my advisors, Gregory Adams and Paul McGuire, for their dedication to the successful completion of this work

4 iii Contents Abstract iv 1 Introduction 1 2 Operators on RKHS 12 3 Tridiagonal Reproducing Kernels 16 4 Five Diagonal Reproducing Kernels 20 5 Summative Conclusion 34 6 Open Questions 36 7 Appendix 38

5 iv Abstract The focus of this work is on a specific class of reproducing kernel Hilbert spaces In Adams and McGuire [2], the tridiagonal reproducing kernels were introduced, and in [3], a specific example of a tridiagonal reproducing kernel Hilbert space was investigated In particular, a careful functional comparison was made between this tridiagonal space and the well-known Hardy space This tridiagonal example is studied further in this thesis via the determination of the spectrum of the multiplication by z operator The main results of this thesis generalize this example to the five diagonal case A general framework is developed for functionally comparing different bandwidth spaces, and this framework is applied to outline the relationship between the generalized five diagonal example and the Hardy space

6 1 Chapter 1 Introduction The work in this thesis concerns a specific class of reproducing kernel Hilbert spaces This section begins by providing a definition for a Hilbert space and giving some examples Next, a reproducing kernel is defined and the relationship between a reproducing kernel and a corresponding Hilbert space is discussed Some important types of reproducing kernels that foreshadow the main examples studied in this thesis are highlighted Before defining a Hilbert space, some necessary terms are defined Throughout we will assume that all vector spaces are over the field C of complex numbers The unit disk {z : z < 1} will be denoted by D and its boundary, the unit circle, by D Definition 11 Inner Product) Let V be a vector space An inner product,, : V V C, is a function satisfying the following conditions for all x, y, z V and scalars α C: x, x 0 with x, x = 0 if and only if x = 0; x, y = y, x ; x + y, z = x, z + y, z ; and αx, y = α x, y A vector space equipped with an inner product is called an inner product space An inner product provides additional structure to a vector space For example, an

7 CHAPTER 1 INTRODUCTION 2 inner product gives a way to quantify the interaction between two elements, or vectors, of an inner product space This gives a sense of angle between two vectors in the space It also allows one to assign lengths to vectors of a vector space via an induced norm In order to discuss the norm induced by an inner product, first recall the definition of a norm Definition 12 Norm) Let V be a vector space A norm, : V R +, is a function satisfying the following conditions for all x, y V and scalars α C: x 0 and x = 0 if and only if x = 0; αx = α x ; and x + y x + y An inner product naturally induces a norm on an inner product space via x = x, x From this point on, we will assume that this is the norm associated to any given inner product space An inner product space is complete if every Cauchy sequence of elements in the space converges to an element in the space Recall a sequence {x n } is Cauchy if for each ε > 0, there is an N such that x n x m < ε whenever n, m > N Definition 13 Hilbert Space) A Hilbert space is a complete inner product space A collection of vectors {f n } in a Hilbert space is orthonormal if { 1 if n = m f n, f m = 0 if n m It is well known that every Hilbert space has a basis of orthonormal vectors Two Hilbert spaces, H 1 and H 2, are isomorphic if there is a one-to-one and onto mapping between the spaces that preserves the inner product We will denote such an isomorphism by H 1 = H2 Often we will look at Hilbert spaces whose vectors are functions defined on a domain D When two such Hilbert spaces H 1 and H 2 contain the same functions, but are not necessarily isomorphic, we will denote that relationship by H s 1 = H 2 Similarly, if all of the functions in H 1 are in H 2, we will write H 1 s H2

8 CHAPTER 1 INTRODUCTION 3 Examples of some well known Hilbert spaces, each with an accompanying orthonormal basis, are listed Example 14 1 Let C n = {x 1,, x n ) : x i C} with inner product defined by x, y = n x i y i i=1 Note that C n is a finite dimensional Hilbert space with orthonormal basis {e i } n i=1 where e i is the n-tuple of complex numbers whose entries are 1 in the i th slot and 0 elsewhere 2 Let { } L 2 D) = f : D C : fs) 2 ds < D with inner product defined by f, g = 1 2π D fs)gs) ds where ds denotes arc length measure on D Note that L 2 D) is an infinite dimensional Hilbert space with orthonormal basis {f n } n= where f n e iθ ) = e inθ Elements of L 2 D) can be thought of as equivalence classes of square integrable functions on D where f g if and only if f g = 0 almost everywhere 3 Let { } l 2 Z) = {x i } i Z : x i C and i Z x i 2 < with inner product defined by x, y = i Z x i y i Note that l 2 Z) is an infinite dimensional Hilbert space with orthonormal basis {e n } n Z where e n is the sequence indexed by Z whose entries are 1 in the n th slot and 0 elsewhere Notice that l 2 Z) = L 2 D) via the isomorphism φ : l 2 Z) L 2 D) defined by φ{a n }) = n Z a ne inθ which identifies the orthonormal basis {e n } of l 2 Z) with the orthonormal basis {f n } of L 2 D) It is often convenient to use this identification to interchange back and forth between the two spaces

9 CHAPTER 1 INTRODUCTION 4 4 Let { l 2 +Z) = {x i } i=0 : x i C and with inner product defined by } x i 2 < i=0 x, y = x i y i i=0 Note that l 2 +Z) is a subspace of l 2 Z) with orthonormal basis {e n } 5 Let H 2 D) be the space of analytic functions fz) = a nz n on D such that a n 2 < If fz) = a nz n and gz) = b nz n are in H 2 D), then the inner product of f and g is defined by f, g = a n b n Note that H 2 D) is an infinite dimensional Hilbert space with orthonormal basis {f n } where f n z) = z n Also, note that l 2 +Z) = H 2 D) via the isomorphism φ : l 2 +Z) H 2 D) defined by φ{a n }) = a nz n As with l 2 Z) and L 2 D), we will frequently use this identification to interchange between the two spaces l 2 +Z) and H 2 D) This interchange is well known and both spaces are referred to in the literature as the Hardy space The focus of this thesis will be on separable Hilbert spaces, which are Hilbert spaces with countable bases Some elementary and well known facts about separable Hilbert spaces and the operators on such spaces are included below for completeness These facts will be used later in this section when defining the reproducing kernel Proposition 15 If H is a separable Hilbert space with orthonormal basis {f n }, then for each g H, g = a n f n = g, f n f n Moreover, g 2 = a n 2 = g, f n 2

10 CHAPTER 1 INTRODUCTION 5 Proof Since {f n } is a basis, g = a nf n for some sequence {a n } of complex numbers Since {f n } is an orthonormal basis, the proposition follows on noting that g, f n = lim N N j=0 a j f j, f n = lim N N j=0 a j f j, f n = lim N N a n = a n j=0 Definition 16 Let H and K be Hilbert spaces 1 A linear operator is a function L : H K such that for all f, g H and α C Lαf + g) = αlf) + Lg) 2 A linear operator L is bounded if there exists a nonnegative real number M such that Lh M h for all h H The smallest such constant M is denoted L and is called the norm of L 3 A linear operator L : H C is called a linear functional The following proposition shows that boundedness is equivalent to continuity for a linear operator Proposition 17 If H and K are Hilbert spaces and L : H K is a linear operator, then L is bounded if and only if L is continuous Proof First assume that L is bounded Let M be a constant such that Lh M h for all h H Then Lf Lg = Lf g) M f g Therefore L is uniformly) continuous Now assume that L is continuous Choose an element f 0 H Since L is continuous at f 0, there exists a δ > 0 such that Lf f 0 ) = Lf Lf 0 < 1 whenever f f 0 < δ If f H is arbitrary and nonzero, then δ f + f 2 f 0) f 0 = δ < δ 2 Hence L δ f ) < 1, and so Lf 2 f Notice that Lf 2 f holds for 2 f δ δ f = 0 Since f was arbitrary, L is bounded by the constant 2 δ

11 CHAPTER 1 INTRODUCTION 6 The next result asserts that all bounded linear functionals on a Hilbert space have a particularly nice representation Theorem 18 Riesz Representation Theorem) If L : H C is a bounded linear functional on a Hilbert space H, then there exists a unique g H such that Lf) = f, g for all f H Proof If L = 0, then g = 0 is a vector in H is such that Lf) = f, g for all f H Suppose L 0 and let M = kerl) Note that since L 0, M H So M {0} and there is a vector g 0 M such that Lg 0 ) = 1 If f H and α = Lf), then Lf αg 0 ) = 0 So f αg 0 M Therefore 0 = f Lf)g 0, g 0 = f, g 0 α g 0 2 g since g 0 M Hence Lf) = α = f, 0 So g = g 0 is such that Lf) = f, g for g 2 g 2 all f H It remains to show that g is unique If h g is a vector in H such that Lf) = f, h for all f H, then f, h g = Lf) Lf) = 0 for all f H Thus h g = 0, and we have established uniqueness The Hilbert spaces studied in this thesis are separable Hilbert spaces whose elements are complex-valued functions on a domain D In this case, it makes sense to define for each w D, an evaluation map given by E w f) = fw) If E w is bounded, then the Riesz representation theorem implies the existence of a unique function k w H such that E w f) = fw) = f, k w for all f H This property is called the reproducing property Definition 19 Reproducing Kernel Hilbert Space) A Hilbert space H of functions on D is a Reproducing Kernel Hilbert space if E w is bounded for each w D and H is the closed linear span of {k w : w D} The fact that the Riesz representation theorem guarantees the uniqueness of k w for each w D allows for the following definition Definition 110 Reproducing Kernel) The reproducing kernel of a reproducing kernel Hilbert space H is the unique function K : D D C defined by Kz, w) = k w z)

12 CHAPTER 1 INTRODUCTION 7 Since k w H, it is true that Kz, w) = k w, k z Therefore the reproducing kernel satisfies Kz, w) = k w, k z = k z, k w = Kw, z) Also, the reproducing kernel satisfies the following property which is called positive definiteness: n α i α j Kw i, w j ) > 0 i,j=1 for any finite set {α 1, α 2,, α n } C \ {0}, and {w 1, w 2, w n } D To see this, notice n n n α i α j Kw i, w j ) = α i α j k wj, k wi i,j=1 n = α j k wj, j=1 i=1 i=1 j=1 n n 2 α i k wi = α i k wi > 0 Since Kz, w) = Kw, z), it follows that if Kz, w) is analytic in z, then K is coanalytic in w It is also well known that { n } α i k wi : {α 1, α 2,, α n } C and {w 1, w 2, w n } D is dense in H j=1 Conversely, if K : D D C is positive definite, then there is a unique Hilbert space HK) such that K is the reproducing kernel for HK) Namely, HK) is the closed linear span of K Notice that the existence of the reproducing kernel guarantees that point evaluation is a bounded linear functional for each w D Therefore, HK) is a reproducing kernel Hilbert space We could equivalently define a reproducing kernel Hilbert space to be a Hilbert space that contains a reproducing kernel The following proposition appeals to Proposition 15 in order to obtain a formula for Kz, w) Proposition 111 If {f n } is an orthonormal basis for a reproducing kernel Hilbert space H, then Kz, w) = f nw)f n z) i=1

13 CHAPTER 1 INTRODUCTION 8 Proof Let f H and w D Since f = f, f n f n, we have fz) = f, k z = f, f n f n, k z = f, f n f n, k z = f, f n f n z) This holds true for all f H, in particular it holds true for k w Thus Kz, w) = k w z) = k w, f n f n z) = f n w)f n z) Example 112 Consider the Hardy space H 2 D) We will show that H 2 D) is a reproducing kernel Hilbert space with kernel Kz, w) = 1 Recall that if f 1 wz H 2 D), then fz) = a nz n where a n 2 < Also note that if w D, then {w n } l 2 +Z) and w n 2 2 = w 2n 1 = 1 w 2 Hence E w f) = fw) = a n w n = {a n }, {w n } l 2 + By the Cauchy-Schwarz inequality, E w f) {a n } 2 w n 1 2 = f 2 1 w 2 Therefore, E w is a bounded linear functional on H 2 D) To see that H 2 D) is the closed linear span of {k w : w D}, note that if f H 2 D) is orthogonal to k w for all w, then fw) = f, k w = 0 for all w D Hence f = 0 This shows that H 2 D) is a reproducing kernel Hilbert space By Proposition 111, the reproducing kernel for H 2 D) is given by Kz, w) = w n z n = wz) n = 1 1 wz We now focus on spaces HK) whose elements are analytic functions on a connected domain D C In this case, Kz, w) = k w z) is analytic in z on D and

14 CHAPTER 1 INTRODUCTION 9 coanalytic in w Since K : D D C is analytic in the first variable z and coanalytic in the second variable w, K has a convergent Taylor series expansion, Kz, w) = i,j=0 a ijz z 0 ) i w w 0 ) j, about each point z 0, w 0 ) D D By a translation and a rescaling, we may assume that z 0, w 0 ) = 0, 0) and consider only kernels of the form Kz, w) = i,j=0 a ijz i w j where D contains the open unit disk D The kernel K can be written more compactly by writing Kz, w) = z Aw where z denotes the column vector whose transpose is 1, z, z 2, ) and A = [a i,j ] i,j=0 Recall, a complex n n matrix M is positive semi-definite positive definite) if Mf, f 0 for all f C n if Mf, f > 0 for all nonzero f C n ) Similarly, the infinite matrix A = [a i,j ] i,j=0 is positive semi-definite if Af, f 0 for each nonzero f l 2 + If Af, f is strictly positive, then A is positive definite Since n n ) n ) α i α j Kw j, w i ) = A α i w i, α j w j i,j=0 and {w : w D} has dense span in l 2 +, it is clear that K is positive semi-definite as a function on D D if and only if A is a positive semi-definite matrix on l 2 + Therefore K is positive definite if and only if A is positive definite, which is true if and only if kera) = {0} Henceforth, A will be assumed to have trivial kernel and HA) will denote the space HK) where Kz, w) = z Aw Recall that A is positive definite if and only if A = BB for some matrix B Also recall that if B is a bounded operator, then the range space of B is the Hilbert space RB) = {Bx : x l 2 +} with inner product Bx, By RB) = x, y l 2 + Note RB) is a reproducing kernel Hilbert space with i=0 Kz, w) = z BB )w = z Aw Since reproducing kernels give rise to unique reproducing kernel Hilbert spaces, RB) = HA) = HK) where Bx is identified with the function Bx, z l 2 + With this identification, Corollary 32 of [1] implies that { Be n, z l 2 + } is an orthonormal basis for HA) Therefore, the columns of B are the coefficient vectors of a basis for the space HA) Since A is positive definite, we can use the well known Cholesky algorithm to write A = LL where L is a lower triangular matrix This decomposition makes the identification of a particularly useful orthonormal basis an easy task, as illustrated in the following examples j=0

15 CHAPTER 1 INTRODUCTION 10 Example Let HK) be the reproducing kernel Hilbert space with Kz, w) = a n wz) n = a n z n ) a n w n ) The associated coefficient matrix A is a a 1 0 A = 0 a a 3 0 a0 0 a0 0 0 a1 0 0 a1 0 = 0 a2 0 0 a2 0 0 a3 0 0 a3 0 Notice that HK) has { a n z n } as an orthonormal basis Spaces HK) of this form are called diagonal spaces 2 Let HK) be the reproducing kernel Hilbert space with Kz, w) = f n z)f n w) where f n z) = a n,0 + + a n,j z J )z n The associated matrix is a 0,0 0 a 0,1 a 1,0 0 a 1,1 a 0,0 a 0,1 a 0,J 0 A = a 0,J 0 a 1,0 a 1,1 a 1,J a 1,J 0 The main results of this thesis concern spaces of this form with J = 2 These spaces are called five diagonal reproducing kernel Hilbert spaces because the matrix A is nonzero only on the five central diagonals

16 CHAPTER 1 INTRODUCTION 11 In previous work, these spaces were studied and in some cases, HK) was specifically described In particular, when J = 1, p > 0, and HK) has orthonormal basis ) p f n z) = 1 z) z n, HK) was explicitly described in Admas and McGuire [3] In section 3, we will consider this space and determine the spectrum of the operator of multiplication by z Section 2 consists of some well known preliminary results regarding operators on reproducing kernel Hilbert spaces that are needed for sections 3 and 4 In section 4, the work in [2] is naturally extended to the five diagonal reproducing kernel Hilbert space with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n Section 5 provides a summative conclusion of the work in this thesis, and section 6 provides a collection of open questions Section 7 is an appendix of background results in topology and analysis on which the results in section 2 depend

17 12 Chapter 2 Operators on reproducing kernel Hilbert spaces The results in this section are well known in the literature and most can be found in Conway [5] The primary focus will be on multiplication operators, which will be defined and motivated shortly Theorem 21 Closed Graph Theorem) If X and Y are Banach spaces and T : X Y is a bounded linear transformation such that the graph of T, is closed, then T is continuous GT ) = {x T x X Y : x X} Proof Since the direct sum X Y is a Banach space and GT ) is closed, GT ) is a Banach space Define P : GT ) X by P x T x) = x It is straightforward to check that P is bounded and bijective By the Inverse Mapping Theorem see section 7), P 1 : X GT ) is continuous Thus T : X Y is the composition of the continuous map P 1 : x GT ) and the continuous map of GT ) Y defined by x T x T x Therefore T is continuous Definition 22 Multiplier) A function φ defined on a set D is a multiplier of a reproducing kernel Hilbert space H of functions defined on D if φf is in H whenever f is in H Note that if H contains the constant functions, then any multiplier φ of H must be in H

18 CHAPTER 2 OPERATORS ON RKHS 13 For φ H, define the operator of multiplication by φ by M φ where M φ f = φf for f H Note that φ is a multiplier of H if M φ H) H Multiplication operators are a very well studied and natural class of operators to consider on any function or L 2 space In linear algebra one shows that a linear transformation T operator) is diagonalizable if and only if there exists a basis {v 1,, v n } of eigenvectors associated with the eigenvalues {λ 1,, λ n } In this case T can be written as T = E λ1 E λn where E λi denotes the projection onto the eigenvector v i of the eigenvalue λ i If the eigenvalues are distinct and m i denotes the unit point mass measure at λ i, then the Hilbert space C n can be viewed as L 2 m) where m = m m n In this case the domain of the functions in L 2 m) is D = {λ 1,, λ n }, the operator T is identifiable with M z on L 2 m), and M φ is identifiable with the operator φt ) While the situation is much more complicated for general L 2 and function spaces, the motivations to study the multiplication operators are sufficiently strong that identifying the multipliers and understanding M z is of fundamental importance The following result is a consequence of the Closed Graph Theorem and shows that the operator M φ is in fact continuous whenever φ is a multiplier and M φ is well defined Proposition 23 If HK) is a reproducing kernel Hilbert space of analytic functions defined on D and φ HK), then φ is a multiplier of HK) if and only if the multiplication operator M φ is bounded on HK) Proof Suppose that φ is a multiplier of an analytic reproducing kernel Hilbert space HK) of functions defined on a domain D The Closed Graph Theorem will be used to show that M φ is bounded Let f n M φ f n f g in HK) HK) Hence f n f and M φ f n g The graph of M φ will be shown to be closed by proving M φ f = g Note that M φ f is well defined since φ is assumed to be a multiplier of HK) Since f n f, for each w D, < f n f, k w > 0 Thus f n w) fw) for all w D Since ) < φf n φf, k w >= φw)f n w) φw)fw) = φw) f n w) fw) 0, and {k w : w G} is a dense subset of HK), M φ f n M φ f and g = M φ f as desired Now if M φ is bounded on HK), then there is an M such that M φ f M f < Therefore M φ f HK) and φ is a multiplier Example 24 The function φz) = z is easily seen to be a multiplier of H 2 D) Recall fz) = a nz n H 2 D) if and only if a n 2 < Hence M z f)z) =

19 CHAPTER 2 OPERATORS ON RKHS 14 a nz n+1 is in H 2 D) whenever fz) = a nz n is in H 2 D) Moreover, if φ is any bounded analytic function with φz) M, then φf 2 ds M 2 f 2 ds M 2 f 2 2 D which shows that φf is in H 2 D) considered as a subspaces of L 2 D) D Definition 25 Spectrum) The spectrum of a bounded linear operator T on a Banach space X over C, denoted σt ), is the set {λ C : T λi is not invertible} where I is the identity operator Lemma 26 If T is a bounded linear operator on a Banach Space X over C and T I < 1, then T is invertible Proof Let B = I T Note that B n B n r n < 1 for some r < 1 Therefore Bn < and Bn converges to a continuous linear operator on X, call it A If A N = N Bn, then A N I B) = I B N+1 Since B N+1 r N+1, B N+1 0 as n Therefore AI B) = lim N A NI B) = lim N I BN+1 = I Therefore, A is a left inverse of I B A similar argument shows I B)A = I So A = I B) 1 Since B = I T, T = I B, and so T is invertible Proposition 27 The spectrum of a bounded linear operator T on a Banach space X over C is a closed and bounded subset of C Proof Let G be the set of invertible bounded linear operators on X and let L G Let S be a bounded linear operator on X such that S L < 1 L Then SL 1 I = S L)L 1 S L L 1 < 1 L 1 L 1 = 1 By Lemma 26, SL 1 is invertible Therefore, SL 1 )T = S is invertible since it is the composition of invertible bounded linear operators Hence, the set G of invertible bounded linear operators on X is an open set Let φ be the map from C to the set of bounded linear operators on X defined by φλ) = λi T Since λ n λ as n implies λ n I T λi T as n, φ is continuous Thus, φ 1 G) = C \ σt ) is open in C Therefore σt ) is closed

20 CHAPTER 2 OPERATORS ON RKHS 15 Let λ C with λ > T Then I 1 T ) I = 1 T < 1 By Lemma 26, λ λ I 1 T is invertible So λi 1 T ) = T λi is invertible Thus σt ) is contained λ λ in the closed disk of radius T and σt ) is bounded Definition 28 Spectral Radius) The spectral radius of a bounded linear operator T is rt ) = max{ λ : λ σt )} The multiplication operator M z is an operator of particular interest The spectrum of M z is determined for a specific example in Theorem 33 When studying a reproducing kernel Hilbert space, HK), it is useful to know how it compares to other reproducing kernel Hilbert spaces, say HK 1 ) Since the space HK) = HA) is isomorphic to RB) where A = BB and HK 1 ) = HA 1 ) is isomorphic to RB 1 ) where A 1 = B 1 B 1, we can determine the comparison between HK) and HK 1 ) by determining the comparison between RB) and RB 1 ) The following result of Douglas [6] allows for one to make this determination Theorem 29 Douglas Lemma) If T and L are bounded linear operators on a Hilbert space H, then RangeT ) RangeL) if and only if there exists a bounded linear operator C on H such that T = LC In Section 4, a specific reproducing kernel Hilbert space is studied Douglas s Lemma is a fundamental tool used to compare that reproducing kernel Hilbert space to H 2 D)

21 16 Chapter 3 Tridiagonal Reproducing Kernels A particularly interesting class of reproducing kernel Hilbert spaces is the class of finite diagonal reproducing kernel Hilbert spaces It is well known, see Aronszajn [4], that if {f n z)} is an orthonormal basis for a reproducing kernel Hilbert space of functions on E, then Kz, w) = f n z)f n w) for all z, w in E Moreover if the largest common domain E of the functions {f n z)} is larger than E, then the largest natural domain of HK) is given by DomK) = {z E : f n z) 2 < } In the very well studied diagonal case where f n z) = a n z n see Shields [7]), DomK) is always a disk In Adams and McGuire [2] the domain of functions in the reproducing kernel Hilbert space with orthonormal basis {f n } where f n z) = a n,0 + + a n,j z J )z n is shown to be either an open or closed disk about the origin together with at most J points not in the disk This result not only illustrates a key distinction between the finite bandwidth reproducing kernel Hilbert spaces and the diagonal spaces, but also motivates interest in the role those additional domain points play in the properties of the spaces The space HK p ) with orthonormal basis {f n } where f n z) = 1 n+1 n+2 )p z)z n is the main focus of study in Adams and McGuire [3] A straightforward argument verifies that the reproducing kernel K p has domain D {1} Since the additional point in the domain of K p is on D, this space inherits some interesting structure The next result is proved in [3]

22 CHAPTER 3 TRIDIAGONAL REPRODUCING KERNELS 17 Theorem 31 Adams-McGuire) If HK p ) is the tridiagonal reproducing kernel Hilbert space with orthonormal basis {f n } where f n z) = 1 n+1 n+2 )p z)z n, then the following hold 1 if p > 1 2, then M z is bounded on HK p ); 2 if p > 1 2, then HK p) contains the polynomials; 3 the space HK p ) s H 2 D) for all p > 0; and 4 The space HK p ) decomposes as follows a) If p > 1 2, then HK p) s = 1 z)h 2 D) + CK p z, 1) b) If p = 1 2, then HK p) s = 1 z)a p +CK p z, 1) where A p is dense in H 2 D), but not equal to H 2 D) c) If 0 < p < 1, then HK 2 p) = s 1 z)a p + CK p z, 1) where A p is the orthogonal complement in H 2 D) of the function g p z) = 1 ) p ) ) p z n This section adds to the above result in Adams and McGuire [3] by showing that the spectrum of M z is the closed unit disk D We make use of the following result in Adams and McGuire [2] regarding the spectral radius of M z on a space with kernel Kz, w) = f n z)f n w) where f n z) = a n + b n z)z n Theorem 32 Adams-McGuire) If M z is bounded with spectral radius ρm z ), then ρm z ) α where α = lim sup k sup n 0 a n+1 a n+k + 2 sup n 0 c n,k c n+k 1,1 ) 1 k and c n,k = b n a n+k+1 a n a n+k b n+k a n+k+1

23 CHAPTER 3 TRIDIAGONAL REPRODUCING KERNELS 18 Theorem 33 If HK p ) is the tridiagonal reproducing kernel Hilbert space with orthonormal basis {f n } where f n z) = 1 n+1 n+2 )p z)z n, then the spectrum of M z is the closed unit disk Proof If λ D, then Mz λi)k λ = 0 which implies D σmz ) Hence σm z ), which is the complex conjugate of σmz ), contains the closed unit disk Apply theorem 32 with a n = 1, b n = n+1 n+2 )p, and c n,k = n+1 n+2 )p n+k+1 n+k+2 )p to obtain an upper bound for the spectral radius of M z which we compute as α = lim k = lim k = lim k = lim k sup n sup n sup n sup n 0 ) n+k+1 n+k+2 )p n+1 1 n+2 )p k n+k+1 n+k+2 )p n+k n+k+1 )p ) 1 n+1 n+2 )p n+k+2 1 n+k+1 )p k 1 n+k n+k+1 )p n+k+2 n+k+1 )p n+2 )p n+k+1 )p n+k+1 )p n+k+1 )p )) k n+2)n+k+1) )p k ) n+k+1) p 2 ) 1 k Let Rn, k) = ) 1 1 k n+2)n+k+1) )p ) n+k+1) p 2 Notice that the numerator and denominator of Rn, k) both involve terms of the form fx) = 1 x) p where 0 x 1/2 The first order Taylor approximation for f at 0 gives the bounds ) ) 1 px 1 2 where 0 t x 1/2 sup f t x ) 0 t x 1/2 x 2 fx) 1 px sup f t x ) 0 t x 1/2 Since f x) is continuous when 0 x 1/2, sup 0 t x 1/2 f t x ) M for some M constant M Let k be large enough so that 2n+k+1) 2 p 2 Applying the above x 2

24 CHAPTER 3 TRIDIAGONAL REPRODUCING KERNELS 19 bounds to Rn, k) gives Rn, k) = pk n+2)n+k+1) + M 2 p n+k+1) 2 M 2 pkn+k+1) n+2) + Mk2 2n+2) 2 p M 2n+k+1) 2 pkn+k+1) n+2) + Mk2 2n+2) 2 p 2 2kn + k + 1) ) 2 k n+2)n+k+1) ) 4 1 n+k+1) + Mk2 p) 2 Substituting this bound for Rn, k), we obtain )) 1 2kn + k + 1) α lim sup + Mk2 k k n 0 p) 2 Since for each k > 1, 2kn+k+1) and Mk2 are both non-increasing in n, the supremum n+2 pn+2) 2 occurs when n = 0 Take the logarithm of both sides to compute the limit in k Thus α is bounded above by lim k ) kk + 1) + Mk2 k = 1 2p Since D σm z ), α 1 Hence α = 1 and σm z ) is the closed unit disk

25 20 Chapter 4 Five Diagonal Reproducing Kernels This section is focused on the study of the five diagonal or two bandwidth) reproducing kernel Hilbert space HLL ) with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n We will often write where a n = 1, b n = f n z) = a n + b n z + c n z 2) z n ) 1 + e iθ 0 ), and cn = ) 2 e iθ 0 This is a generalization of the example in section 3 to the five diagonal case Only the parameter p = 1 is considered in order for the computations to be manageable The next proposition shows that the domain of the functions in HLL ) is D {1, e iθ 0 } Proposition 41 If HLL ) is the reproducing kernel Hilbert space with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n, then the largest natural domain of HLL ) is D {1, e iθ 0 }

26 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 21 Proof Let DomLL ) be the largest natural domain of HLL ) By Aronszajn [4], DomLL ) = {z C : f n z) 2 < } If z < 1, then f n z) 2 = 1 ) ) z 1 Also, since 1 n+1 n+2 eiθ 2 and 1 n+1 n+2 = 1 n+2, f n 1) 2 = f n e iθ 0 ) 2 = 1 ) 1 eiθ 0 1 e iθ 0 Hence D {1, e iθ 0 } DomLL ) If z D {1, e iθ 0 }, then z 1 and ) ) ) ) 1 z 1 e iθ 0 z ) ) 2 e iθ 0 z z n < 16 ) 2 < 4 1 ) 1 ) 2 < 4 z 2n < z) 1 e iθ 0 z ) 0 <, and 2 < Hence f n z) 2 = 1 and DomLL ) = D {1, e iθ 0 } ) ) z 1 ) ) 2 e iθ 0 z z n = The next result shows that the functions in HLL ) are contained in H 2 D) for general θ 0 [0, 2π) Theorem 42 If HLL ) is the reproducing kernel Hilbert space with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n, then HLL ) s H 2 D) for each θ 0 [0, 2π)

27 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 22 Proof Let f HLL ) Then f = α n 1 ) ) z 1 for some {α n } l 2 + Expand this expression to write f = α 0 + α 1 α )) e iθ 0 z+ 2 Since n < 1 and {α n+1 n} l 2 +, α n n=2 ) ) e iθ 0 z z n n ) α n e iθ 0 + { } n 1 + e iθ 0 ) αn 1 l 2 + By the same reasoning, { n ) 2 1 α n 2e 0} iθ l 2 n + Thus { α n and hence f H 2 D) n α n 1 ) 1 + e iθ 0 ) 2 n 1 + α n 2e 0} iθ l 2 + n ) n 1 2 α n 2e 0) iθ z n n Before focusing further on this example, a few general observations about five diagonal reproducing kernel Hilbert spaces are in order We begin by considering the relationship between HLL ) and H L L ) with a 0 0 b 0 a 1 0 c 0 b 1 a 2 0 L = 0 c 1 b 2 a c 2 b 3 a 4 0

28 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 23 and â 0 0 b0 â 1 0 ĉ L = 0 b1 â ĉ 1 b2 â ĉ 2 b3 â 4 0 where the diagonal entries of L and L are assumed to be nonzero Suppose that the matrix C = c j,k ) j,k=0 satisfies L = LC Since L and L have finite bandwidth, are lower triangular, and have nonzero diagonal entries, C is also a well defined lower triangular matrix with nonzero diagonal entries Remark Range Inclusion) In order to study the relationship between HLL ) and H L L ), we study properties of C For example, applying Douglas Lemma allows one to see that C is bounded if and only if H L L ) s HLL ); C is invertible if and only if HLL ) and H L L ) contain the same functions; and the columns of C are bounded if and only if the columns of L are the power series coefficients of functions in HLL ) Note that if L is the identity matrix, then H L L ) = HI) is the Hardy space H 2 D) In particular, the existence of a bounded matrix C such that I = L = LC is equivalent to the Hardy space H 2 D) being contained in HLL ) It is possible that I = L = LC where C is not bounded In that case we can still observe that the orthonormal basis vectors {e n } are in the range space of L if and only if the columns of C are vectors in l 2 + Thus the polynomials are in HLL ) if and only if there is a C whose columns are vectors in l 2 + such that I = L = LC The entries of C are obtained from L and L in a way that is similar to the three bandwidth case in Adams and McGuire [2] However what is a scalar argument in [2] requires a matrix argument in the five bandwidth setting We first need to introduce some notation and ultimately define a new matrix C e, which is related to C, that will help us set up the matrix argument we require Suppose first that C is a matrix satisfying L = LC where L = [ l j,k ] and L = [l j,k ] Since L and L are lower triangular, it follows that C is lower triangular If 0 k j, then l j,k = l j,s c s,k s=0

29 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 24 Since L is lower triangular, l j,s = 0 if s > j Therefore l j,k = j l j,s c s,k s=0 Since L has finite bandwidth, l j,s = 0 if s < j 2 So l j,k = j s=j 2 l j,s c s,k Using the specific form of L and being careful to note throughout the difference between the entries c j, ĉ j in the matrices L and L and the entries c j,k in the matrix C, we obtain the equation l j,k = c j 2 c j 2,k + b j 1 c j 1,k + a j c j,k where c x,y = 0 if x happens to be negative Depending on the respective cases j = k, j = k + 1, j = k + 2, or j > k + 2, the equation j l j,k = l j,s c s,k leads to the equations s=j 2 â j = l j,j = a j c j,j, 41) bj 1 = l j,j 1 = b j 1 c j 1,j 1 + a j c j,j 1 42) ĉ j 2 = l j,j 2 = c j 2 c j 2,j 2 + b j 1 c j 1,j 2 + a j c j,j 2 43) 0 = l j,k = c j 2 c j 2,k + b j 1 c j 1,k + a j c j,k 44) Notice that equation 4)) can be conveniently expressed using vector notation To cj 1,k this end, let v j,k = where again c 1,k = 0 and let c j,k 0 1 M n = c n 2 a n Equation 4) is now encoded by the equation b n 1 a n v j,k = M j v j 1,k )

30 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 25 where j > k + 2, since v j,k = cj 1,k c j,k ) = 0 1 = c n 2 a n Moreover, if j > k + 2, then b n 1 a n c j 2 a j ) cj 2,k c j 1,k c j 2,k b j 1 a j c j 1,k c j 1,k ) = M j v j 1,k ) v j,k = M j v j 1,k = M j M j 1 v j 2,k = = M j M j 1 M k+3 v k+2,k The recursion suggests we introduce the expanded matrix C e of C defined by 0 0 c 0,0 0 c 0,0 0 v 0,0 0 0 c 1,0 c 1,1 0 v 1,0 v 1,1 0 C e = c 1,0 c 1,1 0 = v 2,0 v 2,1 v 2,2 c 2,0 c 2,1 c 2,2 0 c 2,0 c 2,1 c 2,2 0 With the above recursion, notice v 0,0 0 v 1,0 v 1,1 0 v 2,0 v 2,1 v 2,2 0 C e = M 3 v 2,0 v 3,1 v 3,2 v 3,3 0 M 4 M 3 v 2,0 M 4 v 3,1 v 4,2 v 4,3 v 4,4 0 M 5 M 4 M 3 v 2,0 M 5 M 4 v 3,1 M 5 v 4,2 v 5,3 v 5,4 v 5,5 0 Notice that by the Range Inclusion Remark 4, C is bounded if and only if C e is bounded, and the columns of C are elements of l 2 + if and only if the columns of C e are elements of l 2 + The matrix C e allows for the easy observation of the recursion present in the entries of C Therefore, it is convenient to consider C e in lieu of C

31 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 26 We now return to consideration of the space HLL ) with orthonormal basis {f n } where f n z) = 1 n+1 n+1 )z)1 n+2 n+2 )eiθ 0 z)z n We will show that when θ 0 = π and θ 0 = 2π, the polynomials are contained in 3 HLL ) Additionally, we will show that when θ 0 = π, not all of the functions in H 2 D) are contained in HLL ) We first treat the case where θ 0 = π Our immediate goal is to show that HLL ) contains the polynomials In order to establish this result, we will let L be the identity matrix and show that the columns of C e are vectors in l 2 + ) 0 1 Notice for this example, M n = n 1 ) 2 In order to simplify this argument, 0 n let W n = M n M n 1 M 3 Note that if W n 2 <, then M n M n 1 M k 2 < n=3 for each k 3 To see this, observe that M j is invertible for each j 3 and M n M n 1 M k = W n M3 1 M4 1 M 1 k 1 Thus Notice n=k M n M n 1 M k 2 M 1 3 M 1 4 M 1 k 1 2 v j,k = M j M j 1 M k+3 v k+2,k n=k W n 2 < = M j M j 1 M k+3 M k+2 M k+1 M k M 1 k M 1 k+1 M 1 k+2 v k+2,k for j k + 3 So if M n M n 1 M k 2 <, then j=k n=k v j,k 2 M 1 k M 1 k+1 M 1 k+2 v k+2,k 2 n=k M j M j 1 M k 2 < j=k Hence the k th column of C e is in l 2 + Therefore, it suffices to show Lemma 43 gives W n explicitly W n 2 < n=3 Lemma 43 Let θ 0 = π and x n = n 1 n )2 If HLL ) is the reproducing kernel Hilbert space with orthonormal basis {f n } where f n z) = 1 ) z ) 1 ) ) e iθ 0 z z n, then

32 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 27 W n = n 1 2 k=1 0 n 1 2 k=2 x 2k+1 0 x 2k n 2 1 x 2k+1 0 k=1 for odd n and W n = n 2 0 k=2 x 2k for even n Proof We will use induction on n and will first consider the case where n is odd The base case holds true since ) 0 1 W 3 = M 3 = x 3 0 Suppose the claim holds true for n Then W n+2 = M n+2 M n+1 W n = ) ) x n+2 0 x n+1 0 n k=2 = n+1 2 x 2k+1 0 k=1 x 2k n 1 2 Now consider the n even case The base case holds true since ) ) ) x3 0 W 4 = M 4 M 3 = = x 4 0 x x 4 k=1 0 n 1 2 k=2 x 2k+1 0 Suppose the claim holds true for n Then n 2 1 ) ) x 2k k=1 W n+2 = M n+2 M n+1 W n = n x n+2 0 x n k=2 x 2k x 2k

33 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 28 n 2 x 2k+1 0 k=1 = n k=2 x 2k Theorem 44 Let θ 0 = π and x n = n 1 n )2 If HLL ) is the reproducing kernel Hilbert space with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n, then HLL ) contains the polynomials Proof Recall from the Range Inclusion Remark 4 that HLL ) contains the polynomials if and only if the columns of C e are in l 2 + Therefore if W n 2 <, then HLL ) contains the polynomials For all n, n 2 1 n 2 W n max x 2k+1, x 2k, Observe that n 2 1 x 2k+1 = k=1 n 2 x 2k = k=2 n 1 2 k=2 n 1 2 k=1 x 2k = x 2k+1 = 2 3 ) 2 ) k=1 k=2 ) 2 n 2 n 1 ) 2 ) 2 ) n n 3 4 ) 2 ) ) 2 n 2 n 1 ) 2 ) 2 ) n n Thus W n 3 for any n So W n n 2 the polynomials n= n 1 2 k=2 x 2k, ) ) 3 4 ) ) 4 5 ) ) 4 5 ) ) 3 4 n=1 n 1 2 k=1 n=3 x 2k+1 n 2 n 1 n 1 n n 2 n 1 n 1 n ) ) n 1 = 2 n n, ) ) n = 3, ) ) n 1 = 3 n n, ) ) n = 2 ) 2 3 < and HLL ) contains n

34 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 29 We will now treat the case where θ 0 = 2π/3 Our goal with this case is to show that HLL ) contains the polynomials In order to establish this result, we will let L be the identity matrix and show the columns of C e are bounded Unfortunately, in this case, the product M n M n 1 M 3 does not take such a nice form as the θ 0 = π case We therefore employ a different approach to show the columns of C e are bounded Specifically, we show that, for each k, the sequence { v j,k } is square summable in j Notice v j,k 2 = v 3j,k 2 + v 3j+1,k 2 + v 3j+2,k 2 j=0 j=0 j=0 We will show { v j,k } is square summable in j by proving the three subsequences { v 3j,k }, { v 3j+1,k }, and{ v 3j+2,k } are square summable in j Theorem 45 Let θ 0 = 2π If 3 HLL ) is the reproducing kernel Hilbert space with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n, then HLL ) contains the polynomials j=0 Proof We will prove that { v 3j,k }, { v 3j+1,k }, and{ v 3j+2,k } are square summable in j Notice that ) 0 1 M n = n 1 n )2 e 2π 3 i n 1 + e 2π 3 i ) n+1 Compute to see Let = M n+1 M n M n 1 = 1 + e 2π 3 i ) 1 3 ) ) e 2π 3 i ) r n = 1 + e 2π 3 i ) nn 2) 2 e 2π n+1)n 1) 2 3 i n 1 n 2 n+1) nn 2) 2 n 5 +n 4 3n 3 n 2 +n+1 n+2)n 2 1) 2 n 2 n+1) 2 n+2) 2 n n 1) 2 nn 2) 2 n+1)n+2)n 1) 2 2 n n 1) 2 nn 2) 2 n+1)n+2)n 1) 2 e 2π 3 i n 1 n 2 n 3 +3n 2 +n+1 ) n 2 n+1)n+2) ) e 2π 3 i n 1 n 2 n 3 +3n 2 +n+1 ) n 2 n+1)n+2) ) Notice lim r n = 0, so there exists an N 0 > 0 such that if n > N 0, r n 1 Let n N = max{n 0, k + 2} So if n > N, M n+1 M n M n rn ) 1 2 ) n+1 n+1 n+1 )

35 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 30 Let W 0,j = M N+3j M N+3j 1 M N+3j 2, W 1,j = M N+3j+1 M N+3j M N+3j 1, W 2,j = M N+3j+2 M N+3j+1 M N+3j Recall N is chosen so that N k + 2 The application of recursion ) relating to the columns of C e yields v N+1,k = M N+1 v N,k Therefore, we have v N+3j,k = W 0,j W 0,j 1 W 0,1 v N,k, v N+3j+1,k = W 1,j W 1,j 2 W 1,1 v N+1,k, and v N+3j+2,k = W 2,j W 2,j 1 W 2,1 v N+2,k Thus, to show { v 3j,k }, { v 3j+1,k }, { v 3j+2,k } are square summable in j, it suffices to show W 0,j W 0,j 1 W 0,1 2 <, j=1 W 1,j W 1,j 1 W 1,1 2 <, j=1 and and W 2,j W 2,j 1 W 2,1 2 < respectively j=1 Since M n+1 M n M n ) for all n > N, n+1 W 0,j 1 2 N + 3j ), W 2 1,j 1 N + 3j + 1 ), and W 2 2,j 1 N + 3j + 2 ) Thus W 0,j, W 1,j, W 2,j 1 2 j ) 2 1 < j=1 m=1 Let 2 N + 3m + 2 y = log j m=1 N+3j+2 ) 2 1 = N + 3m + 2 Since log1 x) x for x, 1), y j m=1 ) We will complete the proof by showing j m=1 2 N + 3m + 2 ) 2 log 1 N + 3m + 2

36 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 31 2 Let fx) = Since f decreases as x increases, N+3x+2 j j+1 Therefore, So Thus m=1 y 2 N + 3m + 2 j m=1 1 ) 2/3 2 N + 5 dx = log N + 3x + 2 N + 3j j+1 ) 2/3 N + 3m N + 5 dx = log 1 N + 3x + 2 N + 3j + 5 j m=1 j j=1 m=1 ) ) 2/3 2 N = e y N + 3m + 2 N + 3j ) ) 2 2 N + 3m + 2 j=1 ) 4/3 N + 5 < N + 3j + 5 The next result shows that when θ 0 = π, not all of the functions in H 2 D) are contained in HLL ) As we will see, this argument generalizes to show that H 2 D) s HLL ) whenever θ 0 0, 2π), Theorem 46 Let θ 0 = π If HLL ) is the reproducing kernel Hilbert space with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n, then H 2 D) s HLL ) Proof By the Range Inclusion Remark 4, H 2 D) s HLL ) if and only if I = LC for some bounded C Notice that such a C would be a right inverse of L, so to prove H 2 D) s HLL ), we show no such bounded right inverse exists Notice that L =

37 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 32 If LC = I, we see that C = {c j,k } j,k=0 with 0 if j < k or j + k 1 mod 2 c j,k = 1 if j = k j 1 ) 2 c j j 2,k otherwise Explicitly, C = ) ) )2 3 4 ) )2 0 Let K Z + be given Choose j Z + such that j+1 j+2 )2K > 1 Notice that the first 2 K nonzero terms of Ce j the j th column of C) are 1, j+1 j+2 )2, j+1 j+2 )2 j+3 j+4 )2,, and j+1 j+2 )2 j+3 j+4 )2 j+2k 3 j+2k 2 )2 Since each of the first K nonzero terms of Ce j are greater than j+1 j+2 )2K, our assumption on j guarantees each of the first K nonzero terms of Ce j are greater than 1 2 Therefore, Ce j 2 > 1 4 is not bounded We conclude that H 2 D) s HLL ) K Hence sup j Z + Ce j =, and so C The argument above which shows that H 2 D) is not contained in HLL ) for θ 0 = π generalizes for all θ 0 0, 2π) In that case, we effectively wrote M n = M + 1 n E n where ) 0 1 M = e iθ e iθ and sup E 0 n < ) 0 1 When θ 0 = π, M = The argument above relied on the fact that, for each 1 0 K Z +, there exists an N such that if n N and k K, then v n+k,n = M n+k M n+k 1 M n+3 v n+2,n M k v n+2,n For θ 0 = π, M has eigenvalues 1 and 1, and M k v n+2,n = v n+2,n For general θ 0 0, 2π), M has eigenvalues 1 and e iθ 0, so this must be slightly modified and we find a constant δ such that M k v n+2,n δ v n+2,n

38 CHAPTER 4 FIVE DIAGONAL REPRODUCING KERNELS 33 for 0 k K and n N This shows that the l 2 + norm of the n th column of C e exceeds Kδ v n+2,n A computation based on the four recursion equations from this section shows that for large n, v n+2,n 1 + e iθ e iθ 0 + e 2iθ 0 Thus the l 2 + norm of the n th column of C e exceeds d Kδ where d = ) ) 1 + e iθ e iθ 0 + e 2iθ 0 In particular, the l 2 + norms of the columns of C e are not bounded This leads to the following theorem Theorem 47 Let θ 0 0, 2π) If HLL ) is the reproducing kernel Hilbert space with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n, then H 2 D) s HLL )

39 34 Chapter 5 Summative Conclusion The motivation for the work in this thesis was to generalize the study of diagonal reproducing kernel Hilbert spaces to the bandwidth J case This was first done by Adams and McGuire in [2] One particularly interesting result about the bandwidth J spaces is that the natural domain of the functions in such a space is an open or closed disk together with at most J additional points The tridiagonal space HK p ) with orthonormal basis {f n } where f n z) = 1 ) p z) z n was studied in [2] and [3] In this space, an additional domain point is added on the boundary of the unit disk, leading to distinctly different behavior from the Hardy space or any diagonal space Theorem 33 furthers the study of this space HK p ) by proving the spectrum of M z is the closed unit disk The main goal of this thesis is to understand the structure of higher bandwidth spaces and their multiplication operators, in order to generalize the results of the tridiagonal case Section 4 uses Douglas Lemma to provide a framework for analyzing these higher bandwidth spaces In particular, we can study the relationship between two spaces H L L ) and HLL ) by examining the matrix C where L = LC If H L L ) and HLL ) are five diagonal spaces, some essential properties of the matrix C hold if and only if those same properties hold for an expanded matrix C e In this case, C e takes a nice form where the columns are given by a recursion involving 2 by 2 matrices

40 CHAPTER 5 SUMMATIVE CONCLUSION 35 The five diagonal space HLL ) with orthonormal basis {f n } where ) ) ) ) f n z) = 1 z 1 e iθ 0 z z n generalizes the tridiagonal example of [2] and [3] so that two additional domain points are now included on the boundary of the unit disk The framework of section 4 is applied to this example to show that HLL ) properly contains the polynomials and is properly contained in H 2 D) Furthermore, this framework lays the foundation for future work in determining the multiplier algebra, answering the open questions in section 6, and investigating the general structure of higher bandwidth spaces

41 36 Chapter 6 Open Questions The following questions pertain to the space HK) studied in section 4 Question 61 It is reasonable to expect that HK) behaves somewhat analogously to the three-diagonal space studied in Example 31 motivating the following question Are the functions in HK) all of the form [ z 1)z e iθ )g + αk 1 z) + β K e iθz) K1, ] eiθ ) K1, 1) K 1z) where α, β C and g H 2 D)? Note that the last two terms simply describe the span of the two vectors K 1 and K e iθ Question 62 Does HK) contain the polynomials for all θ 0 [0, 2π)? It is expected that the methods used in Theorem 45 generalize to answer this question affirmatively where θ 0 is a rational multiple of π The answer to this question for θ 0 not of this form will likely require a somewhat different approach The following questions relating to multiplication operators on HK) were all motivating questions for this thesis and time did not permit there consideration It is hoped that the framework developed in this thesis will be helpful in answering them Question 63 What are the multipliers of HK)? Equivalently, which multiplication operators M φ are bounded?

42 CHAPTER 6 OPEN QUESTIONS 37 Question 64 Is z a multiplier of HK)? Equivalently, is M z bounded on HK)? Question 65 Assuming that M z is bounded on HK) 1 What is the spectrum of M z? 2 What is the spectral radius of M z? 3 What is the norm of M z? 4 Is M z similar to a rank two perturbation of the unilateral shift S on H 2 D)? Question 66 How does the example in section 4 generalize if the orthonormal basis {f n } is changed to f n z) = J m=1 1 ) p e z) iθm z n? What role does p > 0 play? Are there intervals of p which separate the behavior of HK) as in Example 31? One could first consider the simplest case p = 1 in which case the orthonormal basis {f n } is given by f n z) = J m=1 1 ) ) e iθm z z n

43 38 Chapter 7 Appendix The following are basic results that can be found in any functional analysis text see [5]) Theorem 71 Cantor Intersection Theorem) If X is a complete metric space, {F n } is a sequence of nonempty closed sets such that F n+1 F n for all n, and d n 0 where d n is the diameter of F n, then F n is nonempty Proof For each n, let x n F n Since d n 0 and F n+1 F n for all n, {x n } is a Cauchy sequence Since X is complete, {x n } converges to some x X Since {x k } k n F n for all n N, x is a limit point of each F n Since each F n is closed, x F n for each n Thus x F n Definition 72 Baire Space) A Baire space X is a topological space such that whenever {O n } is a countable collection of open dense sets in X, the intersection On is a dense set Theorem 73 Every complete metric space X is a Baire space Proof Let {O n } be a sequence of dense open sets in X and let O X be an arbitrary open set Since O 1 is dense and open, O 1 O has nonempty interior Let x 1 into 1 O) Hence there is a closed disk Bx 1, r 1 ) of radius r 1 about x 1 entirely contained in O 1 O For the same reason, O 2 Bx 1, r 1 ) contains a closed disc Bx 2, r 2 ) where x 2 into 2 Bx 1, r 1 )) and 0 < r 2 < 1r 2 1 Continuing in this way,

44 CHAPTER 7 APPENDIX 39 we obtain a nested sequence {Bx n, r n )} of closed disks whose diameters r n 0 By the Cantor Intersection Theorem, ) ) O O n = O O 1 ) O n Bx n, r n ) n=1 n=2 Thus O n=1 O n, and so n=1 O n is a dense set n=1 Definition 74 Let S be a subset of a topological space X 1 The set S is nowhere dense in X if the closure of S has empty interior 2 The set S is of first category in X if S is the countable union of nowhere dense sets in X 3 The set S is of second category in X if it is not of first category in X Example 75 The set of rational numbers Q is of first category in R since it is the countable union of singleton sets, each of which is nowhere dense in R Theorem 76 Baire Category Theorem) A topological space X is a Baire space if and only each nonempty open set is of second category in X Proof Suppose X is a Baire space and O is a nonempty open set Let {S n } be a sequence of sets which is nowhere dense in X and let O n = X \ S n Each O n is a dense open set of X and, since X is a Baire space, O n is dense In particular O intersects O n which means that O contains a point not in X \ O n ) = S n This means O contains a point not in S n, so O S n and so O is of second category in X Conversely, if X is not a Baire space, then there is a sequence {O n } of dense open sets whose intersection is not dense Hence, there exists a nonempty open set O such that O does not intersect O n Let S n = O \ O n and note that each S n is nowhere dense since S n X \ O n Also, O = O \ O n = O \ O n ) = S n which shows O is of first category in X Definition 77 Banach Space) A space X is a Banach Space if it a complete normed vector space Theorem 78 Open Mapping Theorem) Let X and Y be Banach spaces If T : X Y is a continuous linear surjection, then T is an open map