JOURNAL OF MATHEMATICAL ANALYSIS AND APPLICATIONS 197, 227᎐248Ž. 1996 ARTICLE NO. 0017
Solutions of the Strong Hamburger Moment Problem
Olav Njastad*˚
Department of Mathematical Sciences, Uni¨ersity of TrondheimᎏNTH, N-7034 Trondheim, Norway
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Received December 27, 1993 provided by Elsevier - Publisher Connector
The strong Hamburger moment problem for a bi-infinite sequence Äcn: n s 0, " 1, " 2, . . .4Ž. can be described as follows: 1 Find conditions for the existence of a Ž. Ž.ϱ ϱ HϱnŽ. Ž. positive measure on y , such that cn s yϱ td tfor all n. 2 When there is a solution, find conditions for uniqueness of the solution.Ž. 3 When there is more than one solution, describe the family of all solutions. In this paper a theory concerning questionŽ. 3 is developed. In particular, an analog to the Nevanlinna parametrization describing the solutions of the classical Hamburger moment problem is given. ᮊ 1996 Academic Press, Inc.
1. INTRODUCTION
The classical Hamburger moment problem can be defined as follows: A Ä4 Ž. sequence cn: n s 0, 1, 2, . . . of real numbers is given. 1 Find conditions Ž. Ž. for there to exist a positive measure on yϱ, ϱ such that cn s Hϱ n Ž. Ž. yϱ td tfor n s 0, 1, 2, . . . . 2 When there is a solution, i.e., when at least one such measure exists, find conditions for uniqueness of the solution.Ž. 3 When there is more than one solution, describe the family of all solutions. The problem is called determinate when there exists exactly one solution, indeterminate when there exists more than one solution. The problem was first discussed by Stieltjeswx 28 for the case that has support inw 0, ϱ.Žthe classical Stieltjes moment problem . , and then by Hamburger wx9 for the general case that the support of is only required to be contained in Ž.yϱ, ϱ . This initial work was followed by an extensive development of a theory of moment problems, where the connection with the theory of orthogonal polynomials plays a central role. We refer the reader to the papers and monographswx 1, 6, 9, 13, 17, 19, 20, 26᎐30 .
U Research supported by the Norwegian Research Council.
227
0022-247Xr96 $12.00 Copyright ᮊ 1996 by Academic Press, Inc. All rights of reproduction in any form reserved. 228 OLAV NJASTAD˚
The strong Hamburger moment problemŽ and strong Stieltjes moment problem. can be formulated in the same way as the classical problem, Ä4 except that here bi-infinite sequences cn: n s 0, " 1, " 2, . . . are in- volved. This problem was introduced by Jones and Thron about 1980 wx14᎐16Ž and for the Stieltjes case by Jones et al.wx 17. A theory of these problems and their connection with orthogonal Laurent polynomials was developed in the ensuing years as far as questionsŽ. 1 and Ž. 2 are con- cerned. Seewxwx 7, 10᎐12, 23᎐25 . Also in 2 the strong moment problem was briefly discussed. In this paper we continue these investigations to develop a theory concerning questionŽ. 3 , and present an analog of the Nevanlinna parametrization in the classical case. The classical and strong moment problems are special cases of a more general theory, where moments corresponding to an arbitrary countable sequence Ä4␣n of points are involvedŽ in the classical and strong cases the points are Ä4ϱ repeated and Äϱ, 0 4 cyclically repeated, respectively. , and where orthogonal rational functions play the role of orthogonal polynomi- als and Laurent polynomials. For an introduction to orthogonal rational functions and discussions of questionŽ.Ž 1 and question Ž. 2 in the case of a finite number of points cyclically repeated. , seewx 3᎐5, 8, 21, 22 .
2. PRELIMINARIES
The basic theory of orthogonal Laurent polynomials and strong moment problems including the results sketched in this section can be found in wx7, 10᎐12, 14᎐17, 23᎐25 . Ž. For any pair p, q of integers with p F q, let H p, q denote the complex j linear space spanned by the functions z , j s p,...,q. We shall write H 2 m s Hym, m and H 2 mq1 s HyŽmq1., m for m s 0, 1, 2, . . . , and H s ϱ D ns0 H n. An element of H is called a Laurent polynomial. Let M be a linear functional on H, and assume that M is positive on Ž.Žyϱ,ϱ i.e., MLwx)0 for all L g H where LzŽ.k0, Lz Ž.G0 when Ž.. n zgyϱ,ϱ. The moments cnnare defined for all n g ޚ by c s Mzwx. The strong Hamburger moment problem Ž.SHMP consists of finding all Ž.ϱ ϱ Hϱ Ž. Ž. positive Borel measures on y , such that MLwxs yϱ L d n Hϱ n Ž. for all L g H, or equivalently such that Mzw xs yϱ d for all n g ޚ. A necessary and sufficient condition for there to exist at least one such measure is that M is positive on Ž.Žyϱ, ϱ as defined above. . The problem is called determinate if there is only one solution, indeterminate otherwise. ²: An inner product , is defined on Hޒޒ= H by
²:P,QsMPzŽ.иQz Ž..2.1 Ž . STRONG HAMBURGER MOMENT PROBLEM 229
Ž j Here Hޒ denotes the real space spanned by z , j s 0, " 1, " 2, . . . . By orthonormalization of the baseÄ 1, zy1, z, zy22, z ,..., zynn, z ,...4 , or-
thonormal Laurent polynomials Ä4n are obtained. They have the form q 2 m , ym m 2mŽ.z sqиии qqz2m,m ,q2m,m)02.2 Ž . zm q 2mq1,yŽmq1. m 2m 12Ž.z sqиии qqzm1, m , q zmq1 q
q2mq1,yŽmq1.)02.3Ž.
for m s 0, 1, 2, . . . . We shall in the following assume that M gi¨es rise to a regular system, which means that q2 m, ym / 0, q2 mq1, m / 0 for all m. The associated orthogonal Laurent polynomials Ä4n are defined by
nnŽ.y Ž.z nŽ.zsM .2.4 Ž . yz Ž.The functional is applied to its argument as a function of . We note that 02' 0, mg Hym,my12, mq1g HyŽmq1.,my1. We further define quasi-orthogonal Laurent polynomials nŽ.z, and their associated Ž. Ž . Laurent polynomials nz, of order nns2mand n s 2m q 1by 2mŽz, .s 2m Ž.zy z2my1 Ž.z Ž2.5 . 2m 12Žz, .sm12 Ž.zy m Ž.z Ž2.6 . q qz 2mŽz, .s2m Ž.zyz2my1 Ž.z Ž2.7 . 2m 12Žz, .sm12 Ž.zy m Ž.z Ž2.8 . q qz
Here g ޒˆs ޒ j Ä4ϱ . For n s 1, 2, . . . we may also write
nnŽ.,y Ž.z, nŽ.z,sM .2.9 Ž. yz
The quasi-approximants RnŽ. z, are defined by
nŽ.z, RznŽ.,s . Ž. 2.10 nŽ.z,
For s 0 they are simply called approximants. Ž. All zeros of n z, are real and simple, and for g ޒ there are n of Ž.Žn.Ž. Žn.Ž. them while for s ϱ there are n y 1 . Let 1 ,...,n denote 230 OLAV NJASTAD˚ these zeros. Then the following quadrature formulas are valid, where Žn. Žn. 1 Ž.,...,n Ž.are positive weights: n Žn.Žn. MFwxsÝkkŽ.FŽ. Ž. Ž2.11 . ks1 valid for F g Hy2 m,2my2 when n s 2m, for F g Hy2m,2m when n s 2mq1. In particular, 2m p p Ž2 m.Ž2m. MzwxsÝkkŽ. Ž. for p sy2m,...,2my2 ks1 Ž.2.12
2m1 q p p Ž2mq1.Ž2mq1. Mzwxs ÝkkŽ. Ž. for p sy2m,...,2m. ks1 Ž.2.13 Ž. Ž Ž . Ž ..Ž . The function f s nn, y z, r yzbelongs to Hy2m,2my2 for n s 2m, and to Hy2 m,2m for n s 2m q 1. Therefore by the quadra- ŽŽŽn.Ž. . . ture formulas taking into account that nk ,s0 we may write n Žn. k Ž. RznŽ.,sy Ý . Ž. 2.14 Žn.Ž. z ks1 k y It follows fromŽ.Ž. 2.12 ᎐ 2.14 that
2my1 k 2 m Rz2mŽ.,sy ÝczyŽkq1.qOz Ž.at 0, Ž. 2.15 ks0 2my1 1 yk Rz2mkŽ.,sÝcz1qO at ϱ, Ž. 2.16 yž/z2m ks1 2my1 k 2m Rz2mq1Ž.,sy ÝczyŽkq1.qOz Ž.at 0, Ž. 2.17 ks0 2mq1 1 yk Rz2m1Ž.,sÝczk1 qO at ϱ. Ž. 2.18 q y ž/z2mq2 ks1 The following determinant formulas are valid:
q2 m , m y z 2mŽ.z 2my12 Ž.z y z my12 Ž.z m Ž.z s Ž2.19 . qmy1,ym
q2m 1, m q z 2mŽ.z 2mq12 Ž.z yz mq12 Ž.z m Ž.z s . Ž 2.20 . q2m,m STRONG HAMBURGER MOMENT PROBLEM 231
The following general Christoffel᎐Darboux formulas are valid for arbitrary complex coefficients a, b, c, d, and for z, g ރ y Ä40: и za 2my12Ž.z qb my12 Ž.z c m Ž .qd2m Ž . и yc2my12Ž.qdmy12 Ž.am Ž.zqb2m Ž.z
2my1 q2m,ym sŽ.Ž.ac y bd q z y ÝŽaj Ž.z q 2my1,ym js0
qbjjŽ.zc.Ž Ž.qd j Ž. . Ž2.21 .
и za 2mq12Ž.z qb mq12 Ž.z c m Ž .qd2m Ž . и yc2mq12Ž.qdmq12 Ž.am Ž.zqb2m Ž.z 2m q2mq1, m s Ž.Ž.Ž.ac y bd q z y ÝŽajz q 2m,m js0
qbjjŽ.zc.Ž Ž.qd j Ž. . . Ž 2.22 .
Let U denote the open upper half-plane: U s Ä4 g ރ:Im)0 . For each z g U and each n the mapping ª w is defined by
w s wnnsyRzŽ.,. Ž. 2.23 This linear fractional transformation maps ޒˆonto a circle bounding a disk contained in U. We use the notation ⌬ nnŽ.z for the open disk, Ѩ⌬ Ž.z for Ž. Ž. Ž. the boundary circle, and ⌬ nnnz for the closed disk ⌬ z j Ѩ⌬ z .By solvingŽ. 2.23 with respect to we get
2mŽ.zw2mq2m Ž.z , n 2m,Ž. 2.24 s s z2my12Ž.zwmq 2my1 Ž.z
z2m 12Ž.zwm12qm1 Ž.z q q q ,n2m1.Ž. 2.25 s sq 2mŽ.zw2mq12q m Ž.z Ž. Ž The circle Ѩ⌬ n z is given by Im s 0, from which follows by use of Ž.Ž..2.15 ᎐ 2.22 that
ny1 2 w y w wg⌬njjŽ.zÝ<< Ž.zqw Ž.zF. Ž 2.26 . m zz js0 y 232 OLAV NJASTAD˚
This shows that ⌬ ⌬ nq1Ž.z ; n Ž.z . Ž 2.27 . ⌬ Ž.ϱ ⌬ Ž. It follows that the intersection ϱ z s F ns1 n z is either a single point or a closed disk. We write ⌬ϱϱŽ.z for the interior and Ѩ⌬ Ž.z for the boundary of ⌬ϱŽ.z . The radius nn Ž.z of ⌬ Ž.z is given by
1 ny1 y 2 njŽ.zs<<< ϱ y1 2 ϱŽ.zs<<< ϱ dŽ. FzŽ.sH . Ž 2.30 . yϱ yz The quadrature formulas described earlier give rise to discrete measures Žn. Žn. Žn. Ž.,having masses of magnitude kkŽ.at the points Ž..It follows fromŽ.Ž. 2.12 , 2.13 that Ž2m.Ž.,solves the truncated moment problem ϱ p H dŽ.scp for p sy2m,...,2my2, Ž 2.31 . yϱ and Ž2mq1.Ž.,solves the truncated moment problem ϱ p H dŽ.scp for p sy2m,...,2m. Ž 2.32 . yϱ FormulaŽ. 2.14 shows that Žn. FzŽ.syRzn Ž, .when Ž.s Ž, .. Ž 2.33 . Ž. Ž. Ž. Žn.Ž. This means that Fz gѨ⌬n zwhen s,, and every point in Ѩ⌬ nŽ.z is the value Fz Ž.of the Stieltjes transform of a measure Žn. Ž.sŽ.,. STRONG HAMBURGER MOMENT PROBLEM 233 We use the following notations for sets of values of Stieltjes transforms for solutions of moment problems: Ž.z Fz Ž.:solution of the truncated moment problem Ý s ½ q,r ϱ pdŽ.c for p q,...,r Ž2.34 . H s p s 5 yϱ Ž.z Fz Ž.:solution of the moment problem Ý s½ ϱ ϱ pdŽ.c for p 0, 1, 2,... . Ž 2.35 . H s p s " " 5 yϱ It follows by compactness argumentsŽ. Helly’s theorems that ϱ ÝŽ.z s F Ý Ž.z . Ž 2.36 . ϱ yq,rs1q,r From the remarks above it follows that Ѩ⌬ Ѩ⌬ 2mŽ.z ; ÝÝ Ž.z and 2mq1 Ž.z ; Ž.z . y2m,2my2 y2m,2m ⌬ Ž.Ý Ž.⌬ Ž.Ý Ž.Ý Ž. Then also 2 m z ; y2 m,2my22z , mq1z ; y2m,2mpz since ,qz is a convex set. Bessel’s inequality for the function fŽ. s 1r Ž y z .and Ž.Ý Ž. the use of 2.26 show that if w g y2 m,2my2 z , respectively, w g Ý Ž.⌬ Ž.⌬ Ž. y2 m,2m z, then also w g 2m z , respectively, w g 2mq1 z . Thus ⌬ ⌬ 2mŽ.z s ÝÝ Ž.z , 2mq1 Ž.z s Ž.z . Ž 2.37 . y2m,2my2 y2m,2m It follows that ⌬ϱŽ.z s Ý Ž.z . Ž 2.38 . ϱ Thus ⌬ϱŽ.z consists of exactly the values of Stieltjes transforms of solu- tions of the moment problem. Since a measure is uniquely determined by its Stieltjes transform, it follows that the moment problem is determinate if and only if ⌬ϱŽ.z reduces to a single point Ž for one z, or for all z .. It is seen by the compactness argumentsŽ. Helly’s theorems applied ŽnŽk.. above that every subsequence ÄŽ.,nŽk. 4contains a subsequence converging to a measure Ž.which is a solution of the moment problem. 234 OLAV NJASTAD˚ Solutions that can be obtained in this way shall be called quasi-natural solutions. Solutions obtained from the orthogonal Laurent polynomials Ž. nŽk. s0 for all k are the natural solutions. 3. SEPARATE CONVERGENCE AND N-EXTREMAL SOLUTIONS Ä4 Let x0 be an arbitrary, fixed point on ޒ y 0 . We define functions ␣nnnnŽ.z, Ž.z,␥ Ž.z,␦ Ž.Žz depending on x0 .by ny1 ␣nŽ.zs Žzyx0 .Ýj Žx0 .j Ž.z Ž3.1 . js1 ny1 nŽ.zsy1q Žzyx0 .Ýj Žx0 .j Ž.z Ž3.2 . js1 ny1 ␥nŽ.zs1q Žzyx0 .Ýj Žx0 .j Ž.z Ž3.3 . js1 ny1 ␦nŽ.zs Žzyx0 .Ýj Žx0 .j Ž.z.3.4 Ž . js0 Since the coefficients in jjŽ.z and Ž.z are real, it follows that ␣nnnnŽ.z, Ž.z,␥ Ž.z,␦ Ž.zare real for real z. The definition of ␣nnnnŽ.z ,  Ž.z , ␥ Ž.z , ␦ Ž.z together with the Christoffel᎐Darboux-type formulasŽ.Ž. 2.21 , 2.22 gives ␣ 2mŽ.z s Kz2m 2my12 Ž.zm Žx002 .yxm Ž.z2my10 Žx . Ž3.5 . ␣ 2mq12Ž.zsKzmq12mq12 Ž.zm Žx002 .yxm Ž.z2mq10 Žx . Ž3.6 .  2mŽ.zsKz2m 2m Ž.x02my102 Ž.zyxmy10 Ž.x 2m Ž.z Ž3.7 .  2mq12Ž.zsKzmq12m Ž.x02mq102 Ž.zyxmq10 Ž.x 2m Ž.Žz3.8 . ␥ 2mŽ.zsKz2m 2my12 Ž.zm Žx002 .yxmy10 Žx . 2m Ž.z Ž3.9 . ␥ 2mq12Ž.zsKzmq12mq12 Ž.zm Ž.x002yxmq10 Ž.x 2m Ž.Žz3.10 . ␦ 2mŽ.zsKz2m 2my12 Ž.zm Žx002 .yxm Ž.z2my10 Žx . Ž3.11 . ␦ 2mq12Ž.zsKzmq12mq12 Ž.zm Žx002 .yxm Ž.z2mq10 Žx ., Ž 3.12 . where qq2my1,ym 2m, m K2ms , K2mq1s .Ž. 3.13 qq2m,ym 2mq1, m STRONG HAMBURGER MOMENT PROBLEM 235 We note that 2 mŽ.z , ␦2 m Ž.z are quasi-orthogonal Laurent polynomials of order 2m and ␣2 mŽ.z , ␥ 2 m Ž.z are associated quasi-orthogonal Laurent y1 Ž. y1␦ Ž. polynomials of order 2m, while z 2 mq12z , z mq1z are quasi- orthogonal Laurent polynomials of order 2m q 1 and y1␣ y1␥ z 2mq12Ž.z , z mq1 Ž.z are associated quasi-orthogonal Laurent polynomials of order 2m q 1. By elimination in the formulasŽ.Ž. 3.5 ᎐ 3.12 we can express n Ž.z and nŽ.zas follows: 2mŽ.z s 2m Žx02 .␦ m Ž.z y 2m Žx02 . m Ž.z Ž3.14 . ␦  2mŽ.zs2m Žx02 .mq12 Ž.zym Žx02 .mq1 Ž.z Ž3.15 . x0 2m 12Ž.zsm10 Žx .␦ 2m12 Ž.zym10 Žx . 2m1 Ž.z Ž3.16 . q z qq q q x0 2m 12Ž.zs m10 Žx .␦ 2m Ž.zy2m10 Žx . 2m Ž.z Ž3.17 . y z y y 2mŽ.zs2m Žx02 .␥m Ž.zy2m Žx02 .␣m Ž.z Ž3.18 . ␥ ␣ 2mŽ.zs2m Žx02 .mq12 Ž.zym Žx02 .mq1 Ž.z Ž3.19 . x0 2m 12Ž.zsm10 Žx .␥ 2m12 Ž.zym10 Žx .␣ 2m1 Ž.z Ž3.20 . q z qq q q x0 2m 12Ž.zs m10 Žx .␥ 2m Ž.zy2m10 Žx .␣ 2m Ž.z. Ž 3.21 . y z y y It follows that all quasi-orthogonal Laurent polynomials of order 2m, Ž. Ž. respectively 2m q 1, are linear combinations of 2 m z , ␦2 m z , resp. y1 Ž. y1␦ Ž. z 2mq12z,z mq1z. By substituting from the expressionsŽ.Ž. 3.14 ᎐ 3.21 in the determinant formulasŽ.Ž. 2.19 , 2.20 at the point x0 we get ␣2mŽ.z ␦2m Ž.z y 2m Ž.z ␥ 2m Ž.z s 1 Ž 3.22 . ␣ ␦  ␥ 2mq12Ž.z mq12 Ž.zy mq12 Ž.z mq1 Ž.zs1. Ž 3.23 . It follows fromŽ.Ž. 3.22 , 3.23 that for an arbitrary complex constant t, Ž. Ž. Ž. Ž. ␣nnzty␥ z and  nnzty␦ z have no common zeros. Substituting from the formulasŽ.Ž. 3.14 ᎐ 3.21 into the expressions Ž.Ž.2.5 ᎐ 2.8 for nn Žz, .and Žz, .we obtain ␣2mŽ.zt2m Ž. y␥2m Ž.z Rz2mŽ.,s Ž.3.24 2mŽ.zt2m Ž. y␦2m Ž.z ␣2m 12Ž.ztm12 Ž. y␥m1 Ž.z RzŽ., q q q Ž.3.25 2mq1 s ␦ 2mq12Ž.ztmq12 Ž.y mq1 Ž.z 236 OLAV NJASTAD˚ where 2mŽ.x002y x m10 Ž.x tŽ. y Ž3.26 . 2m s 2mŽ.x002y x my10 Ž.x x02m10Ž.xy 2m Ž.x0 t Ž. q . Ž 3.27 . 2mq1 s x02mq10Ž.x y 2m Ž.x0 Ž. ˆ We note that the linear fractional transformation ª t s tn maps ޒ bi-uniquely onto ޒˆ. We set ␣nnŽ.zty␥ Ž.z TznŽ.,ts . Ž 3.28 . nnŽ.zty␦ Ž.z We may then write RznnŽ.,sTz Ž.,t Ž.3.29 where t is obtained from by the transformationsŽ.Ž. 3.26 , 3.27 . Žn. We shall denote by t Ž.the discrete measure determined by the Ž. Ž. Žn.Ž. quadrature formula associated with nnzty␦ z. Then ts Žn.Ž. Ž. Ž.Ž.Ž.Ž. ,, where t s tn . It follows by 2.14 , 2.33 , 3.28 , and 3.29 that ␣ Ž.zt ␥ Ž.z nny Žn. FzŽ.sy when s t . Ž 3.30 . nnŽ.zty␦ Ž.z THEOREM 3.1. Assume that the moment problem is indeterminate. Then Ž. Ž. Ž. Ž. Ä4 the functions ␣nnnnz ,  z , ␥ z , ␦ z con¨erge locally uniformly in ރ y 0 to analytic functions ␣Ž.z ,  Ž.z , ␥ Ž.z , ␦ Ž.zgi¨en by ϱ ␣ Ž.z s Žz y x0 .Ýj Žx0 .j Ž.z Ž3.31 . js1 ϱ Ž.zsy1q Žzyx0 .Ýj Žx0 .j Ž.z Ž3.32 . js1 ϱ ␥Ž.zs1q Žzyx0 .Ýj Žx0 .j Ž.z Ž3.33 . js1 ϱ ␦Ž.zs Žzyx0 .Ýj Žx0 .j Ž.z. Ž 3.34 . js0 The functions ␣Ž.z ,  Ž.z , ␥ Ž.z , ␦ Ž.z satisfy the equation ␣ Ž.Ž.z ␦ z y  Ž.Ž.z ␥ z s 1. Ž 3.35 . STRONG HAMBURGER MOMENT PROBLEM 237 Žn.Ž. Ž. For each t g ޒˆ the discrete measures ttcon¨erge to a solution of the SHMP, and ϱ dtŽ. ␣ Ž.zty␥ Ž.z H sy .Ž. 3.36 yϱ yz Ž.zty␦ Ž.z Proof. The locally uniform convergence of ␣nnnnŽ.z ,  Ž.z , ␥ Ž.z , ␦ Ž.z and the form of the limit functions ␣Ž.z ,  Ž.z , ␥ Ž.z , ␦ Ž.z follow from the Ýϱ < Ž.<2 Ýϱ < Ž.<2 locally uniform convergence of the series js0 jjz and s1jz ŽŽ..see Section 2, after formula 2.29 together with Schwarz’ inequality. It Ž. HϱŽŽn.Ž.Ž .. follows from 3.30 that yϱ d t ryzconverge locally uniformly ŽŽ.Ž. to an analytic function in ރ y R since all the zeros of nnzty␦ z are real.Ž. . A standard type of argument involving Helly’s theorems then shows that Žn.Ž.converges to a measure Ž., that Fz Ž. ŽŽ.␣ zt tttsy y Ž..Ž Ž. Ž.. Ž. ␦zrzty␦ z , and that t is a solution of the moment prob- lem. FormulaŽ. 3.35 follows from Ž.Ž. 3.22 , 3.23 . Set w syŽ␣ Žzt .y␥ Žz ..r Ž Žzt .y␦ Žz ... It follows by the use of formulaŽ. 3.35 that dw 1 s 2.Ž. 3.37 dt  Ž.zty␦ Ž.z Hence for real z, dwrdt ) 0. Thus in this situation w increases along ޒ as tincreases along ޒ, and consequently the mapping t ª yŽŽ.␣ zty ␥Ž..Žzr Ž.zty␦ Ž..t maps U onto U. A continuity argument then shows that for z g U the mapping t ª yŽ␣ Žzt .y␥ Žz ..r Ž Žzt .y␦ Žz .. maps ޒonto Ѩ⌬ϱϱŽ.z and U onto ⌬ Ž.z . See the argument given in the classical situation inwx 1, p. 98 . THEOREM 3.2. The mapping t ª t establishes a one-to-one correspon- dence between ޒˆ and the set of all quasi-natural solutions of the SHMP. Proof. Different values of t give different functions yŽŽ.␣ zty Ž..Ž Ž. Ž.. ˆ ␥zrzty␦ z , hence the mapping t ª t is one-to-one from ޒ onto all solutions of the form t. Let be a quasi-natural solution. By ŽnŽk.. definition there exists a sequence ÄŽ.,nŽk. 4Žcf. Section 2, after formulaŽ.. 2.30 converging to . For every k there is a tk such that ŽnŽk..Ž., ŽnŽk..Ž.. Since nŽk. stk ŽnŽk.. ϱ d Ž. ␣ Ž.zt ␥ Ž.z tk nŽk. kny Žk. H sy , yϱ yz nŽk.Ž.ztkny␦Žk. Ž.z since ␣nŽk.Ž.z , nŽk. Ž.z , ␥nŽk. Ž.z , ␦nŽk. Ž.z converge to ␣ Ž.z ,  Ž.z , ␥ Ž.z , ␦ Ž.z , and since Hϱ ŽdŽnŽk..Ž.Žz ..converges, it follows that Ä4t con- yϱ tkkry 238 OLAV NJASTAD˚ Ž . Ž Ž . Ž .. Ž Ž . Ž .. verges to a value t g ޒˆ. Thus Fz sy ␣ zty␥ z rzty␦ z , hence s t. It follows from the way the quasi-natural solutions t are obtained that FzŽ.␦⌬ Ž.zfor all z U. A solution which has this property t g ϱ g Ž. Ž. Fz gѨ⌬ϱ zfor all z g U shall be called a Ne¨anlinna extremal,or N-extremal solution, as in the classical case. Thus all quasi-natural solu- tions are N-extremal solutions. We shall later show that they are the only N-extremal solutions.Ž In fact, all other solutions have Stieltjes trans- Ž. Ž. . forms F with values Fz in the open disk ⌬ϱ z for all z g U. We close this section with a result concerning separate convergence of subsequences of the orthogonal Laurent polynomials nŽ.z and their associated Laurent polynomials nŽ.z , and of the discrete measures deter- mined by nŽ.z . Ä4 THEOREM 3.3. Assume that for some x0 g ޒ y 0 a subsequence Ž. Ž. Ä4 nŽk. x0 rnŽk. x00 con¨erges to a ¨alue t g ޒˆ y 0. Then the Laurent ÄŽ. Ž .4ÄŽ. Ž .4 polynomials nŽk. z rnŽk. x0 and nŽk. z rnŽ k. x0 con¨erge locally uniformly to analytic functions Ž.z , Ž.z respecti¨ely in ރ y Ä40, and the measures ŽnŽk..Ž.,0 con erge to the solution . ¨ t0 Proof. Let ␣nnnnŽ.z ,  Ž.z , ␥ Ž.z , ␦ Ž.z be defined in terms of the point x0 Ž.Ž. Ž. in the assumption. It follows from 3.26 , 3.27 that tnŽk. 0 s Ž. Ž. Ž. nŽk. x0 rnŽk. x0 , hence by assumption tnŽk. 0 converges to t0. It follows Ž. Ž. Ž. Ž. Ž. that ␣nŽk. ztnŽk. 0 y␥nŽk. z converges to ␣ zt0 y␥ z and Ž. Ž. Ž. Ž. Ž. nŽk. ztnŽk. 0y␦nŽk. z converges to  zt0 y␦ z. From the formulas Ž.Ž.Ž.Ž. Ä Ž. Ž .4 3.14 , 3.16 , 3.18 , and 3.20 we then conclude that nŽ k. z rnŽ k. x0 Ä Ž. Ž .4 Ž.Ä4 and nŽk. z rnŽk. x0 converge locally uniformly in ރ y 0 . Further- more, ŽnŽk..Ž.,0 ŽnŽk.. , hence ÄŽnŽk..Ž., 04 converge to . s tnŽ k .Ž0. t0 4. NEVANLINNA PARAMETRIZATION In the following let N denote the class of Nevanlinna functions, i.e., analytic functions in U mapping U into U y Ä4ϱ Žthe closed finite upper half-plane. , and let N* denote the extended class of Nevanlinna functions: Ä4 Ž. N*sNjϱ. We recall that the Stieltjes transform Fz s Hϱ ŽŽ.Ž .. yϱ d r yz of a finite positive measure belongs to N. Note that a function in N either maps U into U,orŽ.z 'r, where r is a real constant. Thus N* consists of the analytic functions mapping U into U and the constants in ޒˆ. In the whole of this section we assume that the SHMP is indeterminate. STRONG HAMBURGER MOMENT PROBLEM 239 PROPOSITION 4.1. Let be an arbitrary solution of the moment problem. Then there exists a g N* such that ϱ dŽ. ␣ Ž.z Ž.z y␥ Ž.z H sy .4.1Ž. yϱ yz Ž.z Ž.zy␦ Ž.z Proof. Define the function by the formula FzŽ.Ž.␦zq␥ Ž.z Ž.zs .4.2 Ž . FzŽ. Ž.zq␣ Ž.z This function satisfiesŽ. 4.1 . We shall show that g N*. Ž. Ž Ž. Ž.. Ž . Ž. Assume first that Fz sy ␣ z r z . Then by 4.2 , z ' ϱ, which is consistent withŽ. 4.1 . Ž. Ž Ž. Ž.. Ž . Next assume that Fz ky␣zrz. Then as defined by 4.2 is Ž. Ž. analytic in U, except possibly for poles. We know that Fz g⌬ϱ z.We also know that t g U if and only if yŽ␣ Žzt .y␥ Žz ..r Ž Žzt .y␦ Žz .. g Ž. Ž. Ž Ž. ⌬ϱ z. Let z g U. For the value t s z we have y ␣ zty Ž..Ž Ž. Ž.. Ž. Ž. Ž. ␥zrzty␦ z sFz g⌬ϱ z, hence t s z g U. It follows that Ž.z g U for z g U. This mapping property excludes the possibility of poles in U. Thus is an analytic function mapping U into U. We recall the formulas 2my1 2mŽ.z sy1q Žzyx0 .Ýk Žx0 .k Ž.z Ž4.3 . ks1 2m  2mq10Ž.zsy1q Žzyx .Ý k Žx0 .k Ž.z .4.4 Ž . ks1 We may write b 2 m , ym m 2mŽ.z sqиии qbz2m,m Ž4.5 . zm b 2mq1,yŽmq1. m 2m 12Ž.z sqиии qbzm1, m .4.6 Ž . q zmq1 q  Ž. We shall call n z regular if b2m,ym/ 0 and b2m,m/ 0 for n s 2m,if b2 mq1, yŽmq1. /0 and b2mq1, m / 0 for n s 2m q 1. ROPOSITION  Ž. Ž. P 4.2. For each m, either 2 m zor 2 mq1 z is regular. Ž. Ž. Ž. Proof. It follows from 4.3 and 4.5 that if 2 my10x / 0, then Ž. Ž. b2 m, ym /0 and b2 m, m / 0. Similarly, it follows from 4.4 and 4.6 that if Ž. Ž 2 m x02/0, then b mq1, yŽmq1. / 0 and b2mq1, m / 0. Recall that we 240 OLAV NJASTAD˚ have assumed that the orthogonal system Ä4n is regular.. Thus the coefficients q2 m, m, q2 mq1, yŽmq1., q2m, ym, q2mq1, m are all different from Ž.Ž. Ž. Ž. zero. See 2.2 , 2.3 and the following remark. Since n x0 and nq10x ŽŽ.Ž.. cannot both be zero cf., e.g., 2.19 , 2.20 we conclude that b2 m, ym / 0, b2 m, m / 0, or b2 mq1, yŽmq1. / 0, b2mq1, m / 0. Let be an arbitrary function in N*. For every natural number n we define ␣nnŽ.z Ž.z y ␥ Ž.z wznŽ.sy .4.7 Ž . nnŽ.z Ž.zy␦ Ž.z Ž. Ž. Ž. First let z k ϱ. Then for z g U we have t s z g U, hence wzn g Ž. ⌬nnz;U. It follows that w is analytic and maps U into U. Next let Ž. Ž. Ž Ž. Ž.. z'ϱ. Then wznnnsy ␣ z r z gU. Thus in all cases, w ng N. An arbitrary function in N can be written in the form ϱ 1 q uz Ž.z s Az q B q H d Ž.u ,4.8 Ž . yϱ uyz where A G 0, B g R, is a finite positive measure.Ž See, e.g.,w 1, p. 92; 27, p. 23x ..Ž. It follows from this and Julia᎐Caratheodory’s´ lemma that z can be written in the form D Ž.z s Cz qq⌽ Ž.z Ž4.9 . z where C G 0, D F 0, and ⌽ is a bounded function in N. By taking into accountŽ.Ž. 2.22 , 2.23 we find that ␣nŽ.z 1 wznŽ.qsy . Ž 4.10 . nnnnŽ.z  Ž.z  Ž.z Ž.zy␦ Ž.z LEMMA 4.3. Assume that C / 0, D / 0 in Ž.4.9 . A. If 2 mŽ.z is regular, then ␣2mŽ.z 1 wz2mŽ.qsO for z ϱ Ž4.11 . Ž.z ž/z2mq1 ª 2m ␣2mŽ.z 2mq1 wz2mŽ.qsOz Ž .for z ª 0. Ž 4.12 . 2mŽ.z STRONG HAMBURGER MOMENT PROBLEM 241  Ž. B. If 2 mq1 z is regular, then ␣2m 1Ž.z wzŽ.q OŽ.1z2mq3 for z ϱ Ž4.13 . 2mq1qs r ª 2mq1Ž.z ␣2m 1Ž.z wzŽ.q Oz Ž2mq1 .for z 0. Ž 4.14 . 2mq1 qs ª 2mq1Ž.z Proof. Under the stated assumptions, y1 1 Ž.2mŽ.z 2m Ž.z Ž.z q ␦2m Ž.z s Ž4.15 . zzzO2m2mŽ.1 m y1 z Ž.2mŽ.z 2m Ž.z Ž.z q␦2m Ž.z s Ž4.16 . zzOymy1Ž.1 y1 1 Ž.2m12Ž.z m12 Ž.z Ž.z q␦m1 Ž.z s Ž4.17 . qq qzzzOmq1mq1Ž.1 m y1 z Ž.2m12Ž.z m12 Ž.z Ž.z q␦m1 Ž.z s .4.18 Ž . qq qzOyŽmq1.Ž.1 The results then follow byŽ. 4.10 . LEMMA 4.4. The following formulas hold: 2my1 ␣2mŽ.z yk y2m ysyÝczk1 qOzŽ.for z ϱ Ž.4.19 Ž.z y ª 2m ks1 ␣ Ž.z 2my1 2m k 2m ysÝczŽk1.qOzŽ.for z 0 Ž. 4.20 Ž.z yq ª 2m ks0 2mq1 ␣2m 1Ž.z q yk yw2mq2x ysyÝczk1qOzŽ.for z ϱ Ž.4.21 Ž.z y ª 2mq1 ks1 ␣ Ž.z 2my1 2mq1 k 2m ysÝczŽk1.qOzŽ.for z 0. Ž. 4.22  Ž.z yq ª 2mq1 ks1 Ž. Ž. Proof. The functions ␣nnz r z are quasi-approximants, so the re- sults follow fromŽ.Ž. 2.15 ᎐ 2.18 . 242 OLAV NJASTAD˚ PROPOSITION 4.5. Assume that C / 0, D / 0 in Ž.4.9 . A. If 2 mŽ.z is regular, then 2my1 yk y2 m ϱ wz2mkŽ.qÝczy1sO Ž z . for z ª Ž4.23 . ks1 2my1 k2m wz2mŽ.yÝczyŽkq1. sO Ž z . for z ª 0 Ž 4.24 . ks0  Ž. B. If 2 mq1 z is regular, then 2mq1 yk yw2mq2x ϱ wz2mq1Ž.qÝczky1 sO Ž z . for z ª Ž4.25 . ks1 2my1 k 2m wz2mq1Ž.yÝczyŽkq1. sO Ž z . for z ª 0. Ž 4.26 . ks0 Proof. The results follow by combining Lemmas 4.3 and 4.4. PROPOSITION 4.6. Assume that C / 0, D / 0 in Ž.4.9 . Ž. A. If 2 m z is regular, then there exists a positi¨e measure 2 m such that ϱ d2mŽ. wz2mŽ.sH Ž4.27 . yϱ yz and ϱ k cks H d2mŽ., ksy2m,...,2my2. Ž 4.28 . yϱ  Ž. B. If 2 mq1 z is regular, then there exists a positi¨e measure 2 mq1 such that ϱ d 2mq1Ž. wz2mq1Ž.sH Ž4.29 . yϱ yz and ϱ k cks H d 2mq1Ž.,ksy2m,...,2m. Ž 4.30 . yϱ Proof. It follows fromŽ.Ž. 4.23 , 4.25 that sup < Ž.Ž. Ž. zsxqiy when nnz is regular. Since wzbelongs to N there exists a measure n such that ϱ dnŽ. wznŽ.sH . Ž 4.32 . yϱ yz Žsee, e.g.,wx 1, p. 93; 27, pp. 24᎐25 .. From the asymptotic expansions ϱϱdŽ.ϱ d Ž. nnk HH;Ýzk1 Ž.4.33 ϱz ϱq yyks0y ϱ ϱϱdnŽ. ykky1 HH;yÝzdnŽ. Ž4.34 . ϱz ϱ yyks1y together withŽ.Ž. 4.23 ᎐ 4.24 , respectively Ž.Ž. 4.25 ᎐ 4.26 , we conclude that Ž.4.28 , respectively Ž. 4.29 , holds under the conditions stated. PROPOSITION 4.7. Assume that C / 0, D / 0 in Ž.4.9 . Then there exists a measure which sol¨es the SHMP ϱ k cks H dŽ., ks0," 1," 2,..., Ž 4.35 . yϱ and such that ϱ dŽ. ␣ Ž.z Ž.z y␦ Ž.z H sy .Ž. 4.36 yϱ yz Ž.z Ž.zy␦ Ž.z Proof. For each n such that nŽ.z is regular there exists by Proposition 4.6 a measure n which solves the truncated moment problemŽ. 4.28 , respectivelyŽ. 4.30 , and such that ϱ dnnŽ. ␣ Ž.z Ž.zy␥ n Ž.z H sy .Ž. 4.37 yϱ yz nnŽ.z Ž.zy␦ Ž.z ŽŽ.Ž.cf. 4.7 , 4.27 , and Ž 4.19 .. . It follows from Proposition 4.2 that there are infinitely many indices such that nŽ.z is regular. By using Helly’s theo- rems and the convergence of ␣nnnnŽ.z ,  Ž.z , ␥ Ž.z , ␦ Ž.z we then establish the existence of a measure satisfyingŽ. 4.35 and Ž. 4.36 . THEOREM 4.8. Assume that the functional M gi¨es rise to a regular system. Then there exists a one-to-one correspondence between the functions in the extended Ne¨anlinna class N* and the measures in the class M of solutions of the SHMP. The correspondence is gi¨en by ϱ dŽ. ␣ Ž.z Ž.z y␥ Ž.z H sy .Ž. 4.38 yϱ yz Ž.z Ž.zy␦ Ž.z 244 OLAV NJASTAD˚ Proof. According to Proposition 4.1 there exists for every in M a function in N* satisfyingŽ. 4.38 . Now let be a function in N. It follows from Proposition 4.7 that there exists a solution of the moment problem such thatŽ. 4.38 is satisfied if C / 0, D / 0 in Ž. 4.9 . If this is not the case, we consider a Nevanlinna function B A,BŽ.z s Az qq Ž.z Ž4.39 . z of the desired form. Then according to Proposition 4.7 a solution A, B of the moment problem exists such that ϱ dA,BAŽ.z ␣ Ž.z ,B Ž.z y ␥ Ž.z H sy .Ž. 4.40 yϱ yz Ž.zA,B Ž.zy␦ Ž.z Ž. Ž. Letting A, B ª 0, hence A, B z ª z , and again using Helly’s theo- rems, we obtain a solution of the SHMP such thatŽ. 4.38 holds. Finally, Ž. Ž. Ž Ž. Ž.. if z ' ϱ, then wznnnsy ␣ z r z . Then the existence of mea- sures n as in Proposition 4.6 follows directly from Lemma 4.4. Hence the existence of a solution with the desired properties follows as in the proof of Proposition 4.7. That the correspondence is one-to-one follows directly from formulas Ž.Ž.4.1 , 4.2 together with the fact that a measure is determined by its Ž.Hϱ Ž Ž.Ž .. Stieltjes transform Fz syϱ d r yz . 5. CANONICAL SOLUTIONS A solution of the SHMP is called a canonical solution if the Nevan- linna function inŽ 4.38 . is of the form Ž.z s Pz Ž.rQz Ž., where P and Q are polynomials with real coefficients.Ž This is analogous to the defini- tion of canonical solution in the classical situation.. Note that all real constants and the constant ϱ are among these functions. The canonical solution is said to be of order r if maxŽ. degree P, degree Q s r, where P and Q have no common factors. A canonical solution of order 0 is then a solution where ϱ dŽ. ␣ Ž.zty␥ Ž.z H sy , tgޒˆ.5.1Ž. yϱ yz Ž.zty␦ Ž.z According to Theorem 3.2 these solutions are exactly the quasi-natural solutions. It follows from Theorem 4.8 that these solutions are also exactly the N-extremal solutions. We note that if P and Q are polynomials, then the functions PzŽ.␣ Ž.zyQz Ž.Ž.␥ z and Pz Ž. Ž.zyQz Ž.Ž.␦z STRONG HAMBURGER MOMENT PROBLEM 245 are analytic in ރ y Ä40 . The zeros of these functions thus form discrete sets in ރ y Ä40 , with 0 and ϱ as the only possible limit points. We also note that if P and Q have no common zeros, then ␣Ž.zPz Ž.y␥ Ž.zQz Ž.and Ž.zPz Ž.y␦ Ž.zQz Ž.have no common zeros. This follows from the fact Ž.Ž. Ž.Ž. that z00000being a common zero would imply ␣ z ␦ z y  z ␥ z s 0, which is not the caseŽ cf. Ž 3.35 .. . Thus the function y Ž␣ ŽzPz . Ž . y ␥Ž.zQz Ž..Žr Ž.zPz Ž.y␦ Ž.zQz Ž..has singularities exactly at the zeros of Ž.zPz Ž.y␦ Ž.zQz Ž., and possibly at the origin. The zeros of Ž.zPz Ž.y␦ Ž.zQz Ž.are real and simple since the Ž . Ž Ž . Ž . Ž . Ž .. Ž Ž . Ž . Ž . Ž .. function Fz s␣zPzy␥zQz r zPzy␦zQz is an- alytic in U and maps U into U. Thus all singularities of FzŽ.are simple poles on the real axis, except for 0 which is a limit point for the poles if it is a singularity. We note that cannot have a mass point at the origin, since the negative moments cyn of are assumed to exist. To obtain a description of the measure we shall use the Stieltjes᎐ Perron inversion formulaŽ see, e.g.,wx 1, pp. 124᎐125. . It follows from this that Ž.b qq Ž.by Ž.aqq Ž.ay y 22 1b slimH Im FŽ.Ž. q i d 5.2 ª0a for arbitrary points a, b on ޒ. We note that if ŽŽ x qq . Žxyr ..2is constant on an open interval, then Ž.x is also constant on that interval. ŽŽ.We have here used x for the distribution function which determines the measure as well as for the measure itself.. THEOREM 5.1. Let be a canonical solution of the SHMP, with Ž.z s PzŽ.rQz Ž.,where P and Q are polynomials with no common factors. Then the spectrum of consists of a discrete subset of ޒ y Ä40, namely the set Ä4Ž. Ž. Ž. Ž. zk :ks1, 2, . . . of zeros of  zPzy␦zQz, and in addition the origin. The measure has a mass of magnitude kkkksy at z , where is the residue of FŽ. z at zk . The origin is a point of continuity. Proof. FormulaŽ. 4.38 may be written as ϱ dŽ. ␣ Ž.zPz Ž.y␥ Ž.zQz Ž. H sy Ž.5.3 yϱ yz Ž.zPz Ž.y␦ Ž.zQz Ž. in this case. It is known that the integral to the left is analytic for z not in the spectrum of . It then follows from the foregoing discussion that all the zeros of Ž.zPz Ž.y␦ Ž.zQz Ž.belong to the spectrum, and the origin ifŽ. and only if it is a limit point for the zeros. We shall show that the spectrum contains no other points. 246 OLAV NJASTAD˚ Ž. Ž. Ž. Ž. Let 12and be two zeros of  zPzy␦zQz such that there are no zeros between, and such that both are positive or both are negative. Assume 12- , and let 1- a - b - 2. Let ⌫ denote the contour in the complex plane consisting of the line segments ⌫0 from Ž.Ž.b,0 to a,0 , ⌫a from Ž.Ž.a,0 to a,i , ⌫ from Ž.Ž.a, i to b, i , and ⌫b from Ž.b, i to Ž.b, 0 . By Cauchy’s integral theorem H Fzdz Ž. 0. Clearly, H Fdzand ⌫ s ⌫a H FŽ.zdztend to 0 when tends to 0. It follows that ⌫b lim HHFzdzŽ.q Fzdz Ž.s0, Ž 5.4 . ª0⌫⌫0 i.e., a b lim HHFŽ. d q F Ž q i .d s 0, Ž 5.5 . ª0ba hence also a b limHH Im FŽ. d q Im F Ž q i .d s 0. Ž 5.6 . ª0ba Since FzŽ.is real for real z, this implies b limH Im FŽ. q i d s 0. Ž. 5.7 ª0a So fromŽ. 5.2 we conclude that Ž.x is constant on Ž12, .. This shows that the spectrum of contains no other points than the zeros of Ž.zPz Ž.y␦ Ž.zQz Ž., and possibly the origin, and since the zeros are isolated, has point masses there. It remains to determine the magnitude of the masses. Ž. Ž. Ž. Ž. Let zk be one of the zeros of  zPzy␦zQz, and let k denote the residue of FŽ.z sy Ž␣ Ž.zPz Ž.y␥ Ž.zQz Ž..Žr Ž.zPz Ž.y ␦Ž.zQz Ž..at zkk. Let a - z - b, let there be no other zeros in Ža, b ., and 1i let ⌬ denote the contour consisting of the half circle S: z y zk s 2 e , 1 gwx0, , and the line segments ⌬ akfrom z y 2 to a; ⌫ afrom a to 0 aqi;⌫ from a q i to b q i; ⌫bbfrom b q i to b , and ⌬ from b 1 Ž. to zk q 2 . By Cauchy’s integral theorem H⌬ Fzdz s0. Clearly H FzdzŽ.,H Fzdz Ž. tend to zero when tends to zero. Hence ⌫ab ⌫ lim HHHHFzdzŽ.q Fzdz Ž.q Fzdz Ž.q Fzdz Ž. s0. ϱ ª⌬⌬abS⌫ Ž.5.8 Since FzŽ.is real for real z, this implies b lim Im HHFzdzŽ.q Im F Ž q i .d s 0. Ž 5.9 . ª0 Sa STRONG HAMBURGER MOMENT PROBLEM 247 Since zk is a simple pole of FzŽ., we have lim HFzdzŽ.sik, Ž 5.10 . ª0 S hence lim Im HFzdzŽ.sk. Ž 5.11 . ª0 S It follows fromŽ.Ž.Ž. 5.2 , 5.9 , 5.10 , and the fact that Ž.x is constant on Ž.a,zkkand Ž.z , b that Ž.by Ž.asyk. Ž 5.12 . Ž. 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