Katholieke Universiteit Leuven Department of Computer Science

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1 Growth properties of Nevanlinna matrices for rational moment problems Adhemar Bultheel, Pablo González-Vera, Erik Hendriksen, Olav Njåstad eport TW 56, February 200 Katholieke Universiteit Leuven Department of Computer Science Celestijnenlaan 200A B-300 Heverlee (Belgium)

2 Growth properties of Nevanlinna matrices for rational moment problems Adhemar Bultheel, Pablo González-Vera, Erik Hendriksen, Olav Njåstad eport TW 56, February 200 Department of Computer Science, K.U.Leuven Abstract We consider rational moment problems on the real line with their associated orthogonal rational functions. There exists a Nevanlinna type parameterization relating to the problem, with associated Nevanlinna matrices of functions having singularities in the closure of the set of poles of the rational functions belonging to the problem. We prove results related to the growth at the singularities of the functions in a Nevanlinna matrix, and in particular provide bounds on the growth analogous to the corresponding result in the classical polynomial case, when the number of singularities is finite. Keywords : rational moment problems, orthogonal rational functions, Nevanlinna parameterization, iesz type criterion, growth estimates. MSC : 30D5, 30E05, 42C05, 44A60,

3 Growth properties of Nevanlinna matrices for rational moment problems A. Bultheel,5,6 P. González-Vera 2,5 E. Hendriksen 3 O. Njåstad 4 Abstract We consider rational moment problems on the real line with their associated orthogonal rational functions. There exists a Nevanlinna type parameterization relating to the problem, with associated Nevanlinna matrices of functions having singularities in the closure of the set of poles of the rational functions belonging to the problem. We prove results related to the growth at the singularities of the functions in a Nevanlinna matrix, and in particular provide bounds on the growth analogous to the corresponding result in the classical polynomial case, when the number of singularities is finite. Key words: rational moment problems, orthogonal rational functions, Nevanlinna parameterization, iesz type criterion, growth estimates 2000 MSC: 30D5, 30E05, 42C05, 44A60 Introduction We use the following notations. C denotes the complex plane, Ĉ the one point compactification of C (the extended complex plane), the real line, ˆ the closure of in Ĉ, U the open upper half-plane, Û the closure of U in C. Department of Computer Science, K.U.Leuven, Belgium. Corresponding author: adhemar.bultheel@cs.kuleuven.be 2 Department of Mathematical Analysis, La Laguna University, Tenerife, Spain. pglez@ull.es 3 Beerstratenlaan 23, 242 GN, Nieuwkoop, The Netherlands. erik.hendriksen@planet.nl 4 Department of Math. Sc., Norwegian Univ. of Science and Technology, Trondheim, Norway. njastad@math.ntnu.no 5 This work is partially supported by the Direción General de Investigación, Ministerio de Educación y Ciencia, under grant MTM The work of this author is partially supported by the Fund for Scientific esearch (FWO), projects AM: ational modelling: optimal conditioning and stable algorithms, grant #G and the Belgian Programme on Interuniversity Poles of Attraction, initiated by the Belgian State, Prime Minister s Office for Science, Technology and Culture. The scientific responsibility rests with the author. Preprint 7 February 200

4 A function f is called a Pick function if it is holomorphic in U and maps U into Û. A Pick function is either a constant in ˆ or maps U into U. Let µ be a finite positive measure on. The Stieltjes transform S µ of µ is defined as S µ (z) = C(t, z) dµ(t), C(t, z) = t z. The Herglotz-iesz-Nevanlinna transform Ω µ of µ is defined as Ω µ (z) = D(t, z) dµ(t), D(t, z) = + tz t z. Both of these functions are Pick functions. Furthermore Ω µ (z) = ( + z 2 )S µ (z) + dµ(t). Thus for fixed z there is a one-to-one correspondence between Ω µ and S µ as functions of µ. Let M be a Hermitian, positive definite linear functional on the space P of polynomials, and define its moments c n by c n = M[z n ], n = 0,, 2,.... A solution of the Hamburger moment problem for {c n } (or M) is a measure µ on which satisfies tn dµ(t) = c n for all n. (Such measures exist.) A moment problem is called determinate if it has exactly one solution, indeterminate if it has more than one solutions. There is a one-to-one correspondence between all Pick functions f and all solutions µ of an indeterminate problem given by A(z)f(z) C(z) S µ (z) = B(z)f(z) D(z) (Nevanlinna parameterization of the solutions). Here A, B, C, D are entire transcendent functions where the growth is restricted as follows: Let F be any of the functions A, B, C, D. Then for every positive ε, there exists a constant M(ε) such that F (z) M(ε) exp{ε z }. (Thus the function is of at most minimal type of order.) For detailed treatments of important aspects of the Hamburger moment problem, see e.g. [,3 5, 3,6,22 26]. 2

5 The strong Hamburger moment problem is analogous to the classical problem, with the space of polynomials replaced by the space of Laurent polynomials (linear combinations of z k, k = 0, ±, ±2,....) A similar parameterization of the set of solutions of an indeterminate problem holds, with the appropriate functions A, B, C, D holomorphic in C \ {0}. When F is any of the functions A, B, C, D, there exist for every positive ε, constants M (ε) and M 0 (ε) such that F (z) M (ε) exp(ε z ) and F (z) M 0 (ε) exp(ε/ z ). For detailed treatments on the theory of strong Hamburger moment problems, see e.g., [4,7 2]. In this paper, we treat a rational moment problem, where polynomials are replaced by rational functions with prescribed poles in ˆ. A Nevanlinna parameterization for solutions of an indeterminate problem in terms of Ω µ and Pick functions was proved by A. Almendral in [2]. The classical Hamburger moment problem is a special case of the rational problem under consideration. Thus in this case there is an alternative parameterization in terms of Ω µ. Our aim in this paper is to establish growth conditions at the singularities of the functions A, B, C, D appearing in the parameterization formula. In Section 2 we introduce the rational spaces on which the rational moment problems are defined, and sketch the theory of orthogonal rational functions and their use in the theory of rational moment problems, including the Nevanlinna parameterization of the solutions of indeterminate problems. Section 3 is devoted to establishing a iesz type criterion for such indeterminate problems when the number of singularities is finite. This criterion is crucial for the further development of the growth properties. (For the classical iesz criterion, see e.g., [],[22 24].) Finally in Section 4 we prove our result on the restriction on the growth of the functions A, B, C, D at the singularities. The organization and presentation of the material in Sections 3 and 4 is strongly influenced by Akhiezer s work [] on the classical moment problem. Other very instructive treatments of the classical problem can be found in the treatises by M. iesz [22 24] and by Shohat and Tamarkin [25] and Stone [26]. This classical approach has to be modified in a number of ways, but the final results are of basically the same structure. emark. A parameterization result for rational moment problems associated with poles outside the closed unit disk and measures on the unit circle T was proved in [0]. Here Ω µ is replaced by the Herglotz-iesz transform T t+z dµ(t) and Pick functions are replaced by t z Carathéodory functions (holomorphic in the open unit disk and mapping this disk to the closed right half-plane). All the isolated singularities of the relevant functions are poles in this case. 3

6 2 Orthogonal rational functions and rational moment problems Let {α k } be a sequence of arbitrary points (interpolation points or singularities) in ˆ \ {0}, α 0 =. We denote by G the set of points α in ˆ \ {0} for which there is at least one k such that α k = α. For α G we denote by Γ α the subsequence of {α k } consisting of those α k for which α k = α. Set π 0 =, π n (z) = n ( z ), n =, 2,..., b n (z) = zn, n = 0,, 2,.... αk π n (z) The set {b 0, b,..., b n } is a basis for the space L n = { } p(z) π n (z) : p P n where P n denotes the space of polynomials of degree at most n. We set L = n=0l n. We shall also consider the space = L L consisting of products of two functions in L. Note that if Γ α is infinite for all α G, then = L. emark 2. The space of Laurent polynomials is not formally included in this setting. The exclusion of the origin as interpolation point is for technical reasons. A discussion of basic properties in the general case when also the origin is included among the possible interpolation points can be found in [9]. Let M be a Hermitian, positive definite linear functional on. Thus M[f] = M[f] for f and M[g g] > 0 for g L, g 0. For convenience we assume M normalized such that M[] =. The moments µ m,n of M are defined as µ m,n = M[b m b n ]. (Note that b n = b n.) A measure µ on is said to solve the rational moment problem on L if b m is integrable with respect to µ and b m (t) dµ(t) = µ m,0 for m = 0,, 2,.... Equivalently g(t) dµ(t) = M[g] for g L. 4

7 A measure µ on is said to solve the rational moment problem on if b m b n is integrable with respect to µ and b m (t)b n (t) dµ(t) = µ m,n for m, n = 0,, 2,.... Equivalently f(t) dµ(t) = M[f] for f. A solvable rational moment problem is said to be determinate if it has exactly one solution, indeterminate if it has more than one solution. We denote by M(L ) the set of solutions of the problem on L, and by M( ) the set of solutions of the problem on. Let {ϕ n } n=0 be the sequence of functions obtained by orthonormalization (with respect to M) of the sequence {b n } n=0. We fix them uniquely by multiplying with a unimodular constant, so that the coefficient of b n in the expansion of ϕ n with respect to the basis {b n } is positive. The function ϕ n has the form ϕ n (z) = pn(z) π n(z), p n P n. Note that by our normilization, the coefficients are real, hence ϕ n (x) is real for x. The functions ψ n of the second kind are defined by ψ 0 (z) = z, ψ n (z) = M t [D(t, z){ϕ n (t) ϕ n (z)}], n =, 2,.... We shall also consider the rational functions σ n given by (M t refers to M applied to t-variable) σ n (z) = M t [C(t, z){ϕ n (t) ϕ n (z)}], n = 0,, 2,.... We observe that both ψ n and ϕ n belong to L n, and that both functions are real for real z. Furthermore we find that σ n (z) = + z 2 [zϕ n(z) + ψ n (z)], n = 0,, 2,.... (2.) The sequences {ϕ n }, {ψ n }, and {σ n } satisfy a three-term recurrence relation of the form σ n (z) { } σ n (z) z z/α n 2 ψ n (z) = E n + B n z/α n z/α n ψ n (z) ϕ n (z) ϕ n (z) + C n z/α n 2 z/α n σ n 2 (z) ψ n 2 (z) ϕ n 2 (z) 5

8 with initial conditions σ 0 (z) 0 ψ 0 (z) = z. ϕ 0 (z) Here B n, C n, E n are real numbers satisfying E n = C n E n for n = 2, 3,.... See [8, Sections.,.2,.9]. Note that σ has the form σ (z) = κ/( z/α ), where κ is a constant. We define χ n (z) = κ ( z/α )σ n+ (z), n = 0,, 2,.... (2.2) Note that χ n (x) is real for real x. The sequence {χ n } satisfies the recurrence relation χ n (z) = { E n+ } z z/α n z/α n + B n+ χ n (z) + C n χ n 2 (z) z/α n+ z/α n+ z/α n+ for n = 2, 3,..., with χ 0 =. Set π 0 =, π n (z) = n+ k=2 ( z ), n =, 2,... and, bn (z) = zn, for n = 0,, 2,.... αk π n (z) Let L n denote the space spanned by { b 0, b,..., b n }, and set L = n=0 L n, = L L. We then have χ n L n. According to the Favard type theorem for orthogonal rational functions (see [8, Section.9]), it follows that there is a positive functional M on such that the sequence {χ n } is orthonormal with respect to M. We can then consider moment problems on L and for the functional M. We shall call these moment problems associated moment problems. Since M is positive, the moment problem on L is always solvable. We shall use the notation ω n (z) = + n ϕ k (z) 2, Ω n (z) = + n ψ k (z) 2, ω n (z) = + n χ k (z) 2. 6

9 We also set ω α,n (z) = n α k Γα ϕ k (z) 2. Note that ω n (z) = + α G ω α,n (z). Let x 0 be a point in \[Ĝ {0}], where Ĝ denotes the closure in Ĉ of the set G of interpolation points. For technical reasons, x 0 is chosen such that ψ n (x 0 ) 0 and q n (α k, x 0 ) 0 for k =, 2,..., n, for all n, where q n (z, τ) is the numerator polynomial of the rational function ϕ n (z)+ τ z/α n z/α n ϕ n (z). Such choice is always possible, see [8, Lemma.5.4]. In the following x 0 shall be kept fixed, and will not be included in the notation for A n, B n, C n, D n below. We set H(z, x 0 ) = x 0 z x 0 and define z A n (z) = H(z, x 0 ) B n (z) = H(z, x 0 ) C n (z) = H(z, x 0 ) D n (z) = H(z, x 0 ) [ [ [ [ + n D(z, x 0 ) D(z, x 0 ) + + n ψ k (x 0 )ψ k (z) n n ϕ k (x 0 )ϕ k (z) ] ψ k (x 0 )ϕ k (z) ϕ k (x 0 )ψ k (z) (Note that the definitions differ from those used in [2] by a real constant factor E n.) We set C G = Ĉ \ [ Ĝ { i, i} ]. For z C G and t ˆ we define ] ] ] T n (z, t) = A n(z)t C n (z) B n (z)t D n (z) (which means A n (z)/b n (z) when t = ). The functions A n, B n, C n, D n can also be expressed in the following way: A n (z) = E n x 0 z [f n(x 0, z)ψ n (x 0 )ψ n (z) f n (z, x 0 )ψ n (x 0 )ψ n (z)] B n (z) = E n x 0 z [f n(x 0, z)ψ n (x 0 )ϕ n (z) f n (z, x 0 )ψ n (x 0 )ϕ n (z)] C n (z) = E n x 0 z [f n(x 0, z)ϕ n (x 0 )ψ n (z) f n (z, x 0 )ϕ n (x 0 )ψ n (z)] D n (z) = E n x 0 z [f n(x 0, z)ϕ n (x 0 )ϕ n (z) f n (z, x 0 )ϕ n (x 0 )ϕ n (z)] 7

10 where f n (z, w) = ( ) ( ) z α n w α n. It follows by a simple argument from [8, Corollary.5.6] that the functions B n (z)t D n (z), t ˆ, have all their zeros on. According to [8, Lemma.0.6], the function z T n (z, t) for t ˆ is a Pick function, hence all the zeros of A n (z)t C n (z) are also real. The index n (or the function ϕ n ) is said to be regular if p n (α n ) 0 (p n ( ) 0 means that p n has degree exactly like n). For z fixed, the linear fractional transformation t T n (z, t) maps for a regular index n the closed lower half-plane onto a proper closed disk n (z) in the open right half-plane. When m > n, we have m (z) n (z). Let Λ denote the sequence of regular indices, and set (z) = n Λ n (z). The (z) is a proper, closed disk or a single point, independent of z in C G. Furthermore (z) is a proper disk if and only if the series k=0 ϕ k (z) 2 converges locally uniformly in the domain C G. This is the case if and only if the series ψ k (z) 2 converges. We shall in the following assume that the set Λ is infinite. For simplicity of notation we let without loss of generality Λ consist of the natural numbers. We shall use the notation ω(z) = + ϕ k (z) 2, Ω(z) = + ψ k (z) 2, ω(z) = + χ k (z) 2. The following inclusions hold: {Ω µ (z) : µ M( )} (z) {Ω µ (z) : µ M(L )}. It follows that if the moment problem on is indeterminate, then the series ϕ k (z) 2 and ψ k (z) 2 converge. Furthermore, if the series ϕ k (z) 2 converges, then the moment problem on L is indeterminate. Now assume that the moment problem for M on is indeterminate. Then ψ k (z) 2 converges, hence also χ k (z) 2 converges. Thus the associated moment problem for M on L is indeterminate. Because of the closely related recursion formulas, it is reasonable to expect that the moment problem for M on is indeterminate when the problem for M on is indeterminate. We have no proof of this, but we shall make this assumption in the proof of (3.9) and Proposition 4.2 for F equal to A or C. However, when all the sets Γ α are infinite, then = L and the moment problem on L and coincide. In this case = L = L and thus the assumption above is automatically satisfied. Note also emark 4.5, where the assumption is not needed. Thus our main result Theorem 4.4 does not depend on this assumption. The theory of orthogonal rational functions with poles on the extended real line is equivalent to a theory of orthogonal rational functions with poles on the unit circle. See especially [8] and 8

11 [6]. For more details on the properties of orthogonal rational functions and rational moment problems that we have discussed so far, we refer to [2],[6],[7], [8, Chap ], [9]. The convergence results and the parameterization results below were obtained by A. Almendral in [2]. Assume that (z) is a proper disk (the limit circle case in contrast to the limit point case). Then the functions A n, B n, C n, D n converge locally uniformly in C G to holomorphic functions A, B, C, D. We may then write [ ] A(z) = H(z, x 0 ) + ψ k (x 0 )ψ k (z) [ ] B(z) = H(z, x 0 ) D(z, x 0 ) ψ k (x 0 )ϕ k (z) [ ] C(z) = H(z, x 0 ) D(z, x 0 ) + ϕ k (x 0 )ψ k (z) [ ] D(z) = H(z, x 0 ) + ϕ k (x 0 )ϕ k (z). The collection {A, B, C, D} is called a Nevanlinna matrix for the problem. The functions A, B, C, D appear in the following Nevanlinna type parameterization for an indeterminate rational moment problem. Theorem 2.2 Assume that the moment problem on is indeterminate, and consider the formula A(z)f(z) C(z) Ω µ (z) = B(z)f(z) D(z). (2.3) Then (i) For every Pick function f, there exists a µ M(L ) such that (2.3) is satisfied. (ii) For every µ M( ), there exists a Pick function f such that (2.3) is satisfied. emark 2.3 The correspondence between µ and Ω µ is one-to-one. When Γ α is infinite for all α G, we have {Ω µ (z) : µ M(L )} = (z). Hence in this situation (2.3) establishes a one-to-one correspondence between Pick functions and solutions of the moment problem (on L or ). 9

12 3 A iesz type criterion Let µ and µ 2 be two distinct solutions of the moment problem on. The function Ω µ (z) Ω µ2 (z) is holomorphic in C \ [ { i, i}], hence the zeros are isolated. It follows that there exist positive γ, γ such that Ω µ (β + iγ) Ω µ2 (β + iγ) for all β. Note that we then also have S µ (β + iγ) S µ2 (β + iγ) for all β. We choose a fixed γ with this property, and use the notation ζ β = β + iγ. Let us start by stating a general Poisson formula first. Lemma 3. Suppose ξ is in the lower half plane and ζ β = β+γi, β 0, γ > 0, and α \{0}, then γ ln t ξ dt = ln π t ζ β 2 ζ β ξ, (3.) and γ π ln t α dt = ln t ζ β 2 ζ β α. (3.2) POOF. This can be proved by standard complex analysis arguments. emark 3.2 Note that (3.) also follows from [, p.53] where in a footnote it is remarked that π ln s c s 2 + ds = ln i, (Im(c) < 0). c Change of variables s = (t β)/γ and c = (ξ β)/γ also yields (3.). In the following positive function shall always mean strictly positive function. Proposition 3.3 Let be a function in which is positive on. Then there exists a function L L such that L(x) x ζ β = (x) x ζ β exp γ 2π ln (t) t ζ β dt 2 for all x in. 0

13 POOF. By dividing out possible common factors in the numerator and denominator of we may write (z) = P (z) ( ) 2 ( ) 2 z α k z α kp where P is a polynomial of degree n, P (x) positive for x. The polynomial P then has the form ( P (z) = A 2 z ) ) ) ( ( ( zξn zξ z ) ξ ξ n for a suitable constant A and ξ,..., ξ n in the lower half-plane. We define Q(z) = A ( ) ( z ξ z ( ) ( ). z α k z α kp ξ n ) Then Q(z) 2 = A 2 ( ) ( ) ( z ξ z ξ n z z 2 α k z ξ ) α kp 2 ( ) z ξ n and for x : Q(x) 2 = A ( ) ( ) ( ) ( ) 2 x ξ x ξ n x x ξ ξ n x 2 α k x 2 = (x). α kp We further define L(z) = Q(z) Q(ζ β ) z ζ β. Note that L L. We have z ζ β L(z) = Q(z) (z ζ β )Q(ζ β ),

14 hence for x : L(x) z ζ β = Q(x) x ζ β Q(ζ β ) = (x) x ζ β Q(ζ β ). (3.3) Using the identity dt t ζ 2 dt = π γ, and Lemma 3. we get n ln Q(ζ β ) = ln A + ln ζ β p j= ξ j ln ζ β j= α kj = ln A γ dt n π t ζ β + ln 2 ζ β p j= ξ j ln ζ β j= α kj = γ n π ln A + ln ζ β p j= ξ j ln ζ β dt j= α kj t ζ β 2 and hence ln Q(ζ β ) = γ π ln Q(t) t ζ β dt = γ 2 2π ln (t) dt. (3.4) t ζ β 2 It follows from (3.3) and (3.4) that L(x) x ζ β = (x) x ζ β exp γ 2π ln (t) dt t ζ β 2. which concludes the proof. Corollary 3.4 For each non-negative integer n and each α G there exists an L n L such that for x : + ωα,n L n (x) x ζ β = (x) exp x ζ β γ ln[ + ω α,n (t)] dt 2π t ζ β 2. POOF. The function +ω α,n (x) is the restriction to of the function + n ;α k Γ α ϕ k (z) 2, which belongs to and is positive on. Consequently the result follows from Proposition

15 Proposition 3.5 There exists a finite constant K such that for every in which is positive on we have γ exp 2π ln (t) dt t ζ β 2 K sup µ M( ) (x) dµ(x) x ζ β, where K is independent of. POOF. ecall that S µ (ζ β ) S µ2 (ζ β ), where µ, µ 2 are two different measures in M( ), cf. the introduction to this section. Set K 0 = S µ (ζ) S µ2 (ζ) 0. Since L(x) dµ (x) = L(x) dµ 2(x) for all L L we may write ( ) ( ) K 0 = L(x) dµ (x) L(x) dµ 2 (x) x ζ β x ζ β hence K 0 L(x) x ζ β dµ (x) + L(x) x ζ β dµ 2(x), and consequently K 0 2 sup L(x) x ζ β dµ(x), where the supremum is taken over all µ M( ). Let be an arbitrary function in which is strictly positive on. Then we conclude from Proposition 3.3 that K 0 2 exp γ ln (t) dt 2π t ζ β 2 sup (x) x ζ β dµ(x), hence γ exp 2π ln (t) dt t ζ β 2 K sup (t) x ζ β dµ(x) where K = 2/ K 0. 3

16 Corollary 3.6 There exists a constant K independent of the index n such that for every α G, γ ln[ + ω α,n (t)] dt exp 2π t ζ β 2 K sup + ωα,n (t) dµ(x) x ζ β where the supremum is taken over all µ M( ). POOF. The function +ω α,n (x) is the restriction to of the function + n ;α k Γ α ϕ k (z) 2, which belongs to and is positive on. Thus the conditions of Proposition 3.5 are satisfied, and so the result follows from this proposition. Lemma 3.7 Let α G, α, β. Then the following inequality holds for α k = α, µ M( ): (x α)(x β)ϕ k (x) 2 dµ(x) x ζ β 2 α ζ [ β ϕk (ζ γ 2 β )ψ k (ζ β ) + Ω µ (ζ β )ϕ k (ζ β ) 2 ]. (3.5) POOF. For any β we may write ( x )ϕ α k(x) 2 = ( x x ζ β α )ϕ k(x) ϕ k (ζ β ) ϕ k (x) ζ β + α ϕ x ζ β + ζβ 2 k (ζ β )ϕ k (x) + + ζ β [ α D(x, ζβ ){ϕ + ζβ 2 k (x) ϕ k (ζ β )}ϕ k (ζ β ) + D(x, ζ β )ϕ k (ζ β ) 2]. We observe that ( x ) ϕ k(x) ϕ k (ζ β ) α x ζ β belongs to L k. Thus the integral of the first term to the right vanishes by orthogonality. Also the integral of the second term vanishes by orthogonality. We then get ( x α )ϕ k(x) 2 x ζ β dµ(x) = ζβ [ α ϕk (ζ + ζβ 2 β )ψ k (ζ β ) + ϕ k (ζ β ) 2 Ω µ (ζ β ) ]. Hence by taking the real part of the equation we get (x α)(x β)ϕ k (x) 2 dµ(x) x ζ β 2 α ζ [ β + ζβ 2 ϕk (ζ β )ψ k (ζ β ) + ϕ k (ζ β ) 2 Ω µ (ζ β ) ]. We find that + ζ 2 β 2 = ( + β 2 γ 2 ) 2 + 4β 2 γ 2 ( γ 2 ) 2, from which (3.5) now follows. 4

17 Lemma 3.8 Let µ be a positive measure in and let f be a non-negative function on. Let [a, b] be a bounded interval and let β. Then there exist positive numbers m(β) and M(β) such that m(β) f(x) dµ(x) x ζ β 2 f(x) dµ(x) M(β) x ζ α 2 f(x) dµ(x) x ζ β 2 for all α [a, b]. POOF. The function φ(x) = x ζ β 2 / x ζ α 2 has a positive minimum m(α, β) and a finite maximum M(α, β) since φ(x) as x ±. The values m(α, β) and M(α, β) are continuous functions of α, hence there exist a positive m(β) and a finite M(β) such that m(β) m(α, β), M(α, β) M(β) for all α [a, b]. We may write f(x) dµ(x) = x ζ α 2 f(x) x ζ β 2 x ζ β 2 x ζ α 2 dµ(x), hence m(β) f(x) dµ(x) x ζ β 2 f(x) dµ(x) M(β) x ζ α 2 f(x) dµ(x) x ζ β 2 for all α [a, b]. Proposition 3.9 Assume that G is bounded, α G, β. Then there exists a constant K 2 (α, β) independent of the index n and the measure µ M( ) such that + ωα,n (x) dµ(x) K 2 (α, β). (3.6) x ζ β POOF. We know that the series ϕ k (ζ β ) 2 and ψ k (ζ β ) 2 converge, and by Schwarz inequality then also the series ϕ k (ζ β )ψ k (ζ β ) converges. Furthermore, Ω µ (ζ β ) (ζ β ), which implies that Ω µ (ζ β ) is bounded independently of µ M( ). It follows from Lemma 3.7 that (x α) 2 ω α,n (x) x ζ α 2 dµ(x) γ γ 2 n α k Γ α { ϕk (ζ α )ψ k (ζ α ) + Ω µ (ζ α )ϕ k (ζ α ) 2 }. Taking into account Lemma 3.8 we then get 5

18 (x α) 2 ω α,n (x) x ζ β 2 dµ(x) γ m(β) γ 2 n α k Γ α { ϕk (ζ α )ψ k (ζ α ) + Ω µ (ζ α )ϕ k (ζ α ) 2 }. Thus there exists a constant K 3 (α, β) independent of n and µ M( ) such that (x α) 2 [ + ω α,n (x)] x ζ β 2 dµ(x) K 3 (α, β). (3.7) We may write + ωα,n (x) x ζ β = x α + ω α,n (x) x ζ β x α. Hence by Schwarz inequality we get + ωα,n (x) dµ(x) x ζ β (x α) 2 [ + ω α,n (x)] dµ(x) x ζ β 2 /2 dµ(x) (x α) 2 /2. (3.8) dµ(x) (x α) 2 The factor /(x α) 2 belongs to, hence equals a finite constant K 4 (independent of µ). Setting K 2 = K 3 (α, β)k 4, we obtain (3.6) from (3.7) and (3.8). Theorem 3.0 (iesz type criterion) Assume that the moment problem on is indeterminate. Assume that G is finite and let β. Then ln ω(t) dt t ζ β 2 < and ln Ω(t) dt <. (3.9) t ζ β 2 POOF. According to Proposition 3.9 we have sup µ M( ) + ωα,n (x) dµ(x) K 2 (α, β) x ζ β where K 2 (α, β) is independent of n. Consequently there is a constant K 2 (β) such that sup α G µ M( ) + ωα,n (x) dµ(x) K 2 (β). x ζ β 6

19 It then follows from Corollary 3.6 that there is a constant K(β) such that α G ln[ + ω α,n (t)] t ζ β 2 dt K(β) for all n. For any non-negative numbers t, t 2,..., t N we have ( ) N N ln + t k ln( + t k ). Consequently we may conclude from the fact that ω n (t) = + α G ω α,n (t): ln ω n (t) t ζ β 2 dt { } ln[ + ω t ζ β 2 α,n (t)] dt K(β) α G for all n, from which the first inequality in (3.9) follows. Similarly, since {χ n } are the orthonormal functions associated with the indeterminate moment problem on, we find ln ω n (t) t ζ β 2 dt Then from (2.) and (2.2) we infer that also the second inequality in (3.9) is satisfied. emark 3. By considering the imaginary part in (3.7) when G = { }, we get K 3 γ and hence by Schwartz inequality ω n (x) dµ(x) x ζ β ω n (x) dµ(x) x ζ β 2 /2 K3 γ. ω n(x) dµ(x) x ζ β 2 It follows from Corollary 3.6 that in this case ln ω(t) dt t ζ β <. 2 4 Growth estimates in the finite case We continue to assume that the moment problem on is indeterminate. Let α be a fixed point in G. For the sake of simplicity we formulate the results and carry out the arguments only for the case α. By adapting the arguments given in this section, estimates in appropriate 7

20 form can be proved also in the case α =. In the following β shall denote an arbitrary point in G. We set m(z) = max {, D(z, x 0 )}, p(z) = max{, H(z, x 0 )}, q = max { [ ϕ k (x 0 ) 2 ] /2, [ ψ k (x 0 ) 2 ] /2}, and define [ { }] [ { }] Φ(t) = ln p(t) m(t) + q ω(t), Ψ(t) = ln p(t) m(t) + q Ω(t). It follows from Theorem 3.0 (the iesz type criterion) that Φ(t) dt t ζ β 2 < and Ψ(t) dt < for any β. (4.) t ζ β 2 Note that Φ(t) 0 and Ψ(t) 0 for all t. Let η (0, π/2). We introduce the notation (α, η) = {z C : η arg(z α) π η}. As usual we set z = x + yi. f(t) dt t ζ α 2 Lemma 4. Assume that f is a non-negative function on satisfying some α. Then for every ε > 0 there exists a disk U α with center at α such that y π f(t) dt t z < ε 2 z α < for (4.2) for z U α (α, ε). POOF. We have y2 f(t) f(t) a.e., and t z 2 t α γ f(t) dt < since t z 2 is bounded for t α γ. Hence it follows by Lebesgue s dominated convergence theorem, that y 2 π t α γ f(t) dt 0, hence y2 t z 2 y 0 π t α γ f(t) dt t z < ε sin η (4.3) 2 2 for y sufficiently small. For z (α, η) we have t z 2 t α 2 sin 2 η. For t x γ this implies t x 2 8

21 t 2 z 2 sin 2 η. Hence y 2 π t α γ f(t) dt t z 2y2 2 π sin 2 η t α γ f(t) dt t z 2. Since t α γ f(t) dt t z 2 y 2 π < we find that t α γ f(t) dt 0, hence y2 t z 2 y 0 π t α γ f(t) dt t z < ε sin η (4.4) 2 2 for y sufficiently small. We have z α sin η < y when z (α, η), and so (4.2) follows from (4.3) and (4.4). Proposition 4.2 Assume that G is finite, and let α G. Let V α be a disk with center at α and let F denote any of the functions A, B, C, D. Then there exists for every ε > 0 a constant M (ε, η) such that F (z) M (ε, η) exp { ε } z α (4.5) for z V α (α, η). POOF. Let H n denote any of the functions B n, D n. It follows by Schwartz inequality that H n (t) p(t)[m(t) + q ω(t)], hence ln H n (t) Φ(t) for t. ecall that all the zeros and poles of H n are real. From Poisson s formula and Lemma 3., applied to the rational function H n (t) with all poles on, we find ln H n (z) = y π ln H n (t) dt t z 2, for z and hence ln H n (z) y π Φ(t) dt t z 2. 9

22 It now follows from Lemma 4. that ln H n (z) ε for z U z α α (α, η), where U α is sufficiently small. In (V α \ U α ) (α, η) we have ln H n (z) M (ε, η) where the constant M (ε, η) is independent of n, since H n is uniformly convergent in (V α \ U α ) (α, η). Thus ln H n (z) M (ε, η) + ε z α in V α (ε, η) for all n. Simiarly let G n denote any of the functions A n, C n. Then ln G n (t) Ψ(t) for all t, and all the zeros and poles of G n are real. It follows from (4.) by the same kind of reasoning as above that there exists a constant M (ε, η) such that ln G n (z) M ε (ε, η) + z α for z V α (α, η). Setting M (ε, η) = max{exp[m (ε, η)], exp[m (ε, η)]}, we obtain (4.5). V α (α,η) η α ρ α η t S (α,η) (α,η) Fig.. Elements appearing in the proof of Proposition 4.3 Proposition 4.3 Assume that G is finite, G. Let α G and let V α be a disk with center at α containing no other point in G. Then for every ε > 0 there exists a constant M 2 (ε, η) such that F (z) M 2 (ε, η) exp { ε } z α for z V α [C \ (α, η)] where F is any of the functions A, B, C, D. POOF. Let ρ α denote the radius of V α and let S(α, η) denote the sector of V α [C \ (α, η)] lying to the right of α. According to Proposition 4.2 there is for every ε > 0 a constant 20

23 M (ε cos η, η) such that F (z) M (ε cos η, η) exp { } ε cos η z α for z Vα (α, η), hence a constant M2 (ε cos η, η) such that F n (z) M 2 (ε cos η, η) exp { } ε cos η z α (4.6) for z V α (α, η), where F n is any of the functions A n, B n, C n, D n. We now consider z in the closure S(α, η) of the sector S(α, η). The function { Q n (z) = F n (z) exp ε } z α is holomorphic in S(α, η) \ {α}. We have { Q n (z) = F n (z) exp ε e }, (4.7) z α hence by (4.6) Q n (z) M 2 (ε cos η, η) exp { } { ε cos η exp ε x α }. (4.8) z α z α 2 Let z be a point on one of the line segments of the boundary S(α, η). Then x α = z α cos η, hence Q n (z) M 2 (ε cos η, η). (4.9) Next let z be a point on the circular arc of S(α, η). Then we have { } { } ε cos η Q n (z) M2 ε(x α) (ε cos η, η) exp exp = M 2 (ε cos η, η) exp ρ α { ε ρ 2 α ρ 2 α } [ρ α cos η (x α)], hence { } ε cos η Q n (z) M2 (ε cos η, η) exp. ρ α Thus there is a constant M 3 (ε cos η, η) such that Q n (z) M 3 (ε cos η, η) for z S(α, η) \ {α}. 2

24 ecall that F n is a rational function. Therefore there exists for every ε > 0 a constant k n (ε) such that F n (z) k n (ε) exp { } ε cos η z α for z Vα. For z S(α, η)\{α} we have x α z α cos η. Hence { } { ε cos η Q n (z) k n (ε) exp exp ε x α } k z α z α 2 n (ε) for z S(α, η) \ {α}. It follows that lim sup Q n (z) k n (ε) <. z α z S(α,η)\{α} Then, according to a version of the maximum principle (see for example [5, Part II, p.208]) we have Q n (z) M 3 (ε cos η, η) for z S(α, η), and hence, according to (4.7) F n (z) M 3 (ε cos η, η) exp { } ε, for z S(α, η). (4.0) z α In the same way we find an estimate { } ε F n (z) M3 (ε cos η, η) exp, (4.) z α for z in the sector of V α [C \ (α, η)] to the left of α. Letting n tend to infinity in (4.0)-(4.) and combining the resulting inequalities, the proof is completed. Theorem 4.4 Assume that G is finite, G. Let α G and let V α be a disk with center at α containing no other point of G. Then for every ε > 0 there exists a constant M(ε) such that F (z) M(ε) exp { ε } z α for all z V α, where F is any of the functions A, B, C, D. POOF. Choose a fixed η = η 0, and define M(ε) = max{m (ε, η 0 ), M 2 (ε, η 0 )}. The result then follows from Proposition

25 emark 4.5 For fixed t ˆ, the rational function T n (z, t) = A n(z)t C n (z) B n (z)t D n (z) has a partial fraction decomposition of the form T n (z, t) = n λ n,k (t) + ξ n,k(t)z ξ n,k (t) z, with ξ n,k and λ n,k > 0 for k =,..., n, and n λ n,k (t) =. (See [2, Sections 0-], [8, Sction.0].) Since ξ n,k (t) z y z α sin η for z (α, η), we find T n (z, t) + z 2 z α + z. In particular A n (z) B n (z) + z 2 z α + z, C n (z) D n (z) + z 2 z α + z, for z (α, η). Since m exp{ ε } for arbitrary z α z α ε > 0 and suitable m, we conclude that if B and D satisfy an estimate of the form F (z) M exp{ ε } for z (α, η) for some z α M, then also A and C do. Hence the result of Proposition 4.2 can be obtained without making use of the iesz type criterion ln Ω(t) t ζ β dt < (see (3.9)). The result in Theorem 4.4 can 2 thus be established without use of this criterion. emark 4.6 When G consists of the only point (i.e. when α k = for all k), the functions F n are polynomials with all zeros in. Hence F n (z) is an increasing function of y, and an estimate of the form F n (z) M(ε, η) exp{ε z } in {z C : η arg z π η} can easily be extended to an estimate of the same kind in the whole plane. See e.g. [, Chap. 2]. An argument of this kind is not possible in the general case. emark 4.7 If α is an isolated point in G and there is only a finite number m of elements α k in some Γ α, then F has a pole of order m at α. Thus in a neighborhood V α we have in this case the stronger estimate F (z) M(ε) z α m. eferences [] N.I. Akhiezer. The classical moment problem and some related questions in analysis. Hafner, New York, 965. [2] A. Almendral Vazquez. Nevanlinna parametrization of the solutions of some rational moment problems. Analysis, 23:07 24, [3] C. Berg. Indeterminate moment problems and the theory of entire functions. J. Comput. Appl. Math., 65:27 55,

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