A rational moment problem on the unit circle�A. Bultheely, P. Gonz�alez-Veraz, E. Hendriksenx, O. Nj�astad{AbstractLet f�kg1k=1 be a sequence of not necessarily distinct points on the complex unitcircle. We consider the moment problem where it is to �nd a positive measure on[��; �] such that for !0 = 1 and !n(z) = (z � �1) � � �(z � �n), n = 1; 2; : : : we haveZ ��� d�(�) = 1; Z ��� d�(�)!n(ei�) = �n; n = 1; 2; : : :for a given sequence of moments f�ng1n=0. This paper gives results which to someextend generalise the limit point - limit circle situation of classical moment problems.Keywords: general moment problem, orthogonal rational functions, nested disks.AMS Classi�cation: 30E05, 42C10, 42A99, 47A571 IntroductionLet T = fz 2 C : jzj = 1g be the complex unit circle and let f�kg1k=1 be a sequence ofnot necessarily distinct points on Tn f1g. Introduction of the \forbidden" point 1 is nota severe restriction because there is only a countable number of �k's so that there alwaysexists such a point on T, which by a simple rotation can be brought to the position 1. De�ne!n(z) = Qnk=1(z � �k) for n � 1 and set !0 = 1. By �n we denote the set of polynomials ofdegree at most n. We consider the spacesLn = ( pn(z)!n(z) : pn 2 �n) ; L = [1k=0Lk:For any complex function f , let f�(z) = f(1=z). It is obvious that Ln� = ff : f� 2 Lng = Lnand similarly L� = L. Finally, we set Rn = Ln � Ln and R = L � L. Let M be a linearfunctional de�ned on L � L which is real and positive, i.e., which satis�esMff�g =Mffg; f 2 L � L and Mfff�g > 0; 0 6= f 2 L:This functional de�nes an inner product byhf; gi =Mffg�g:�This research was performed as part of the European project rolls under contract CHRX-CT93-0416.yDepartment of Computer Science, K.U.Leuven, BelgiumzDepartment An�alisis Math., Univ. La Laguna, Tenerife, SpainxDepartment of Math., Univ. of Amsterdam, The Netherlands{Department of Math. Sciences, Norwegian Univ. of Science and Technology, Trondheim, Norway1

An example of such a linear functional is given byMffg = Z ��� f(ei�)d�(�); hf; gi = Z ��� f(ei�)g(ei�)d�(�)where � is a positive measure on [��; �]. The subject of this paper, just as in Section 8 of[3], is to solve the following moment problem: Given the moments f�ng1n=0, de�ned byMf1g = �0 and M � 1!n� = �n; n = 1; 2; : : :(without loss of generality we may assume that �0 = 1), does there exist a �nite and positivemeasure � on [��; �] such thatM � 1!n� = �n = Z ��� d�(�)!n(ei�) ; n = 0; 1; : : :It was shown in [3] that this moment problem always has at least one solution. This existenceresult was proved in a constructive way. Such a solution was obtained as the limit of aconverging subsequence of quadrature formulas. Since there could be di�erent subsequences,converging to di�erent limits, there might be (in�nitely) many solutions. Here we will addressthe problem whether the moment problem is determinate or indeterminate, i.e., whether asolution � is unique or not. We shall, to a certain extend, generalise the classical limitingpoint versus limiting disk situation for a sequence of nested disks. This technique wasknown to Weyl [23] and used by Akhiezer [1] and Shohat and Tamarkin [20] when theystudied the classical Hamburger moment problem to characterize the situation where themoment problem has a unique solution. See also the work of Stone [22].For more information on moment problems and nested disks see [10, 11, 12, 21, 14].Uniqueness criteria for the strong Hamburger moment problem can be found in [17]. Anextended Hamburger moment problem was discussed in [15]. In the Hamburger problem, theunit circle is replaced by the real line. Similar results about unique solvability were obtainedin [16] for the extended Hamburger moment problem where a �nite number of points �kon the real line are cyclically repeated. Multipoint matrix versions of the Hamburger andStieltjes moment problems are found in e.g., [19, 18]. In the case where the points �k arelocated inside the unit disk, results similar to the ones of this paper were given in [7] and [4].Here we treat the case of the unit circle and at the same time we consider a general sequenceof �k's which need not be cyclically repeated. We emphasize that the theory where all thepoints �k lie inside the open unit disk is substantially di�erent from the theory where theyare on its boundary. In the boundary case, nontrivial new problems arise e.g. due to the factthat functions in L are not continuous on the unit circle anymore. Analog results for thereal line can be obtained in a similar way. However, there the problem is that the supportof the measure is not compact and special attention has to be paid to the point at in�nitywhich causes some trouble. For simplicity, we discuss here only the case of the unit circle.An essential role will be played by the orthogonal rational functions. They play the role oforthogonal polynomials in the Hamburger case or the orthogonal Laurent polynomials [13]in the case of the strong Hamburger moment problem. For the case of points inside theunit disk, such orthogonal rational functions were �rst studied in [8, 2, 9]. For points on theboundary, they appear in [16, 3, 6, 5] 2

2 Orthogonal rational functions and recurrence rela-tionFirst we observe that Ln = spanfb0; b1; : : : ; bngwith b0 = 1; bn = Z1Z2 � � �Zn; n � 1where Zk(z) = i(z � 1)(�k � 1)z � �k ; k � 1:We use this notation also for k = 0, in which case we set �0 = �1. ThusZ0(z) = 2i1� z1 + z :Note that the basis functions bk satisfy bk� = bk . By a Gram-Schmidt procedure, these basisfunctions are orthogonalised to give the orthonormal functions �n, n = 0; 1; : : :. Let�n(z) = �n(1) + � � �+ �0nbn�1(z) + �nbn(z): (2.1)The orthonormal functions can be �xed uniquely by requiring �n > 0. This is what willalways be assumed when we refer to the orthonormal functions. Note that then, because ofMf1g = 1 we �nd �0 = �0 = 1. Furthermore�n = "�n(z)bn(z) #z=�n and �n + �0nZn(�n�1) = "�n(z)bn(z) #z=�n�1 ; n � 1: (2.2)If we set �n = pn=!n, with pn 2 �n, then we say that �n (and also its index n) is singularif pn(�n�1) = 0. Otherwise, they are called regular. The system f�ng is called regular if allthe indices are regular.With this normalization we haveLemma 2.1 The orthonormal functions �n have real coe�cients with respect to the basis bkand �n� = �n.Proof. Because bk� = bk, it is obvious that if the coe�cients are real, also �n� = �n. Theproof of real coe�cients follows easily by induction. The result is true for n = 0. Suppose itis true for i � n � 1, then by the Gram-Schmidt procedure�n = �n=k�nk; with �n = bn � n�1Xi=0 i�i; i = hbn; �ii :Using Mff�g = Mffg, hf; gi = Mffg�g, bn� = bn and �i� = �i for i < n, it follows thatthe coe�cients i = hbn; �ii = Mfbn�i�g = Mfbn��ig = Mfbn�i�g = i are real. Since�i has real coe�cients with respect to the basis bk, also �n and thus also �n will have realcoe�cients with respect to the basis bk. 2The following is a slight adaptation of Theorem 4.1 in [3].3

Theorem 2.2 Suppose the system f�ng is regular. Then the following recurrence holds�n(z) = AnZn(z) +Bn Zn(z)Zn�2(z)!�n�1(z) + Cn Zn(z)Zn�2(z)�n�2(z); n = 2; 3; : : : (2.3)with constants An; Bn; Cn satisfying the conditionsEn = An +Bn=Zn�2(�n�1) 6= 0; k = 2; 3; : : : (2.4)Cn 6= 0; k = 2; 3; : : : (2.5)Let us introduce the real numbersDn = 1Zn�2(z) � 1Zn�1(z) = �i �n�1 � �n�2(1� �n�1)(1 � �n�2) 2 R: (2.6)Since this is a constant not depending on z and because 1=Zk(�k) = 0 for all k, we can useDn = 1Zn�2(�n�1) = � 1Zn�1(�n�2) 2 Rfor any n � 2 when it will be convenient.Lemma 2.3 Let �n and �0n be as in (2.1). De�ne for n � 1En = 1�n�1 "�n + �0nZn(�n�1)# : (2.7)Then En 2 R.If the recurrence relation holds, and hence An and Bn are de�ned, then the En of (2.7)coincide with the En of (2.4), i.e., En = An + BnDn. The latter also holds for n = 1 if weset by de�nition A1 = �1 and B1 = �01.Proof. Because the coe�cients of the orthonormal functions are real and Dn is real, itfollows that also En is real. To show that En = An + BnDn, we divide the recurrencerelation (2.3) by bn(z) and set z = �n�1. With (2.2) and 1=Zk(�k) = 0, we get the result.2Note that it easily follows from this de�nition of En that En = 0 if and only if n is a singularindex.Lemma 2.4 Suppose the recurrence relation for the orthonormal functions �n holds withcoe�cients An, Bn and Cn. Let Dn be as in (2.6) and En = An +BnDn. ThenEn = �CnEn�1; n � 2: (2.8)Proof. Because bn�1=Zn 2 Ln�1, it is orthogonal to �n. Using the recurrence relation for�n, we get 0 = h�n; bn�1=Zni= An hbn�1; �n�1i +Bn * bn�1Zn�2 ; �n�1+ + Cn * bn�1Zn�2 ; �n�2+ :4

Because hbn�2; �n�1i = 0 and hbi; �ii = 1=�i, we have0 = An�n�1 + BnDn�n�1 + Cn�n�2 + CnDn hbn�1; �n�2i : (2.9)The remaining inner product can be evaluated when we usebn�1(z) = �n�1(z)�n�1 � �0n�1�n�1 bn�2(z) + � � �so that hbn�1; �n�2i = *�n�1�n�1 ; �n�2+� �0n�1�n�1 hbn�2; �n�2i = � �0n�1�n�1�n�2 : (2.10)When we combine (2.9) and (2.10), we get0 = An�n�1 + BnDn�n�1 + Cn�n�2 � CnDn�0n�1�n�1�n�2= An +BnDn�n�1 + Cn�n�2 "1 �Dn�0n�1�n�1 #= En�n�1 + Cn�n�1 � 1�n�2 "�n�1 + �0n�1Zn�1(�n�2)#= 1�n�1 [En + CnEn�1];which gives us the expression we wanted. 2For solutions of the recurrence relation, we can prove a general summation theorem. Toformulate this formula, we de�ne for n � 1H(z;w) = 1Zn�1(z) � 1Zn�1(w) = �i z � w(w � 1)(z � 1) : (2.11)Note that this expression does not depend on n. Furthermore we setHn(z;w) = 1Zn�2(w)Zn�1(z) � 1Zn�2(z)Zn�1(w)= " 1Zn�2(w)Zn�1(z) � 1Zn�2(w)Zn�2(z)#+ " 1Zn�2(w)Zn�2(z) � 1Zn�2(z)Zn�1(w)#= �DnZn�2(w) + DnZn�2(z) = DnH(z;w): (2.12)Theorem 2.5 Let xn(z) and yn(z) be two solutions of the recurrence relation (2.3) andde�ne Fn(z;w) = xn(w)yn�1(z)Zn(w)Zn�1(z) � yn(z)xn�1(w)Zn(z)Zn�1(w) :Then, with H(z;w) as in (2.11) and En as in (2.7)Fn(z;w) = yn�1(z)xn�1(w)H(z;w)En �CnFn�1(z;w)= "n�1Xk=1 yk(z)xk(w)#H(z;w)En + (�1)nCnCn�1 � � �C2F1(z;w):5

Proof. We use the recurrence relation for xn and yn in the de�nition of Fn(z;w), whichgives Fn(z;w) = Anxn�1(w)yn�1(z) " 1Zn�1(z) � 1Zn�1(w)#+Bnxn�1(w)yn�1(z) " 1Zn�1(z)Zn�2(w) � 1Zn�1(w)Zn�2(z)#�CnFn�1(z;w):Using the expressions (2.11) and (2.12), we �ndFn(z;w) = xn�1(w)yn�1(z)H(z;w)[An +BnDn]� CnFn�1(z;w)= xn�1(w)yn�1(z)H(z;w)En �CnFn�1(z;w):An induction argument leads to the result. 2It is possible to derive from this formula the Christo�el-Darboux type formulas which aregiven below. However, this would require that the system �n is regular, since it is basedon the existence of the recurrence relation. It is possible however to prove the Christo�el-Darboux formulas without using the recurrence relation and only relying on the orthogonalityproperties of the �n. This is what we shall do here.3 Functions of the second kindWith the orthonormal functions �n, we associate functions of the second kind n de�ned by 0 = �1� z1 + z = �Z0(z)Z0(0) n = MtfD(t; z)[�n(t)� �n(z)]g; n � 1;where D(t; z) is the Riesz-Herglotz kernelD(t; z) = t+ zt� z :We also introduced the notationMt to indicate thatM operates on its argument as a functionof t. We prove that these functions of the second kind are also solutions of the recurrence(2.3).Theorem 3.1 Suppose that the system of orthogonal rational functions �n is regular and let n be the associated functions of the second kind. Then these n satisfy the same recurrencerelation (2.3) as the �n.Proof. We use the recurrence relation for �n(t) and �n(z) in the de�nition of n. This givesfor n � 2 n(z) = AnMt fD(t; z) [Zn(t)�n�1(t)� Zn(z)�n�1(z)]g+BnMt (D(t; z) " Zn(t)Zn�2(t)�n�1(t)� Zn(z)Zn�2(z)�n�1(z)#)6

+CnMt (D(t; z) " Zn(t)Zn�2(t)�n�2(t)� Zn(z)Zn�2(z)�n�2(z)#)= AnZn(z) n�1(z) +Bn Zn(z)Zn�2(z) n�1(z) + Cn Zn(z)Zn�2(z) n�2(z)+Mt fD(t; z)fn(t; z)g+ �n2Z2(z)Z0(0)C2 (3.1)withfn(t; z) = An[Zn(t)� Zn(z)]�n�1(t) + " Zn(t)Zn�2(t) � Zn(z)Zn�2(z)# [Bn�n�1(t) + Cn�n�2(t)]We note that Zn(t)� Zn(z) = �i (t� z)(�n � 1)2(t� �n)(z � �n)and Zn(t)Zn�2(t) � Zn(z)Zn�2(z) = (t� z)(1� �n)(�n�2 � �n)(t� �n)(z � �n)(1� �n�2) :Thereforefn(t; z) = (t� z)(1� �n)(t� �n)(z � �n) "Ani(�n � 1)�n�1(t) + �n�2 � �n1� �n�2 [Bn�n�1(t) + Cn�n�2(t)]#Next we split D(t; z) as D1(t; z) +D2(t; z):D(t; z) = t� �nt� z + z + �nt� z :Thus in the argument of Mt in (3.1), the factor t� z in the numerator of fn(t; z) cancels thedenominator of D(t; z).Using the orthogonality of the �k, we �ndMtfD1(t; z)fn(t; z)g = 0; for n � 3= C2 (�2 � 1)(1 + �2)2(z � �2) ; for n = 2.For the second term with D2(t; z), we use again the recurrence relation to write fn(t; z) asfn(t; z) = t� zz � �n "��n(t)� i(1� �n)An�n�1(t) + 1� �n1� �n�2 �Bn�n�1(t) + Cn�n�2(t)�# :Again, by the orthogonality of the �k, we getMtfD2(t; z)fn(t; z)g = 0; for n � 3= C2 (z + �2)(1� �2)2(z � �2) ; for n = 2.This proves the recurrence relation for n � 3 directly. For n = 2, we can put together allthe terms involved and we �nd that then also the recurrence relation is satis�ed becauseZ2(z)Z0(0)C2 � C2 (1� �2)(1 + �2)2(z � �2) + C2 (z + �2)(1 � �2)2(z � �2) = 0:7

2It is possible to build some redundancy in the de�nition of the n. For an analogous proofin the case where the points �k are inside the unit disk, see [7].Lemma 3.2 Let �n be the orthonormal system and n the associated functions of the secondkind. For n > 0 and for any f such that (as a function of t) D(t; z)[f(t)� f(z)] 2 Ln�1 wehave n(z)f(z) =MtfD(t; z) [�n(t)f(t)� �n(z)f(z)]g:Proof. SinceMtfD(t; z)[f(t)�n(t)� f(z)�n(z)]g = f(z) n(z) +MtfD(t; z)[f(t)� f(z)]�n(t)g;the result follows by the orthogonality of the �n. 2Note that in particular we could take f 2 Ln�1 or f(t) = g(t)(t� �n)=(t+ z) with g 2 Ln,We shall now derive a Liouville-Ostrogradskii type determinant formula.Theorem 3.3 (determinant formula) Let �n be the orthonormal functions and n thefunctions of the second kind, then for n � 1 we have with En as in (2.7)�n�1(z) n(z)� �n(z) n�1(z) = 2iz(1 � z)2EnZn�1(z)Zn(z)= �2iz(�n � 1)(�n�1 � 1)En(z � �n�1)(z � �n) :Proof. We note that�n�1(z)[�n(t)� �n(z)]� �n(z)[�n�1(t)� �n�1(z)]= �n�1(t)[�n(t)� �n(z)]� �n(t)[�n�1(t)� �n�1(z)]:Multiply with D(t; z) and apply Mt to get for the left-hand side�n�1(z) n(z)� �n(z) n�1(z)while for the right-hand side we �ndMtf�n�1(t)D(t; z)[�n(t)� �n(z)]g �Mtf�n(t)D(t; z)[�n�1(t)� �n�1(z)]g:Note that in the second term D(t; z)[�n�1(t)� �n�1(z)] 2 Ln�1 so that this term is zero bythe orthogonality of �n. To compute the �rst term, we de�neh(t) = D(t; z)[�n(t)� �n(z)] = nbn(t) + 0nbn�1(t) + � � �where n = " h(t)bn(t)#t=�n = D(�n; z)�nand n + 0nZn(�n�1) = " h(t)bn(t)#t=�n�1 = D(�n�1; z) "�n + �0nZn(�n�1)# = D(�n�1; z)�n�1En:8

Thus, by the orthogonality of �n�1 = �(n�1)�Mf�n�1hg = nMf�n�1bng+ 0nMf�n�1bn�1g:Because �k = �kbk + �0kbk�1 + � � �, it follows by orthogonality thatMf�n�1bng = � �0n�n�n�1 and Mf�n�1bn�1g = 1�n�1 :So we obtainMf�n�1hg = � n �0n�n�n�1 + 0n�n�1= � n�0n�n�n�1 + [D(�n�1; z)En � n�n�1 ]Zn(�n�1)= �D(�n; z) �0n�n�1 +D(�n�1; z)EnZn(�n�1)�D(�n; z)�nZn(�n�1)= �D(�n; z)EnZn(�n�1) +D(�n�1; z)EnZn(�n�1)= EnZn(�n�1)[D(�n�1; z)�D(�n; z)]:Working this out gives the result. 24 Christo�el-Darboux relationsWe now prove some Chisto�el-Darboux type relations.Theorem 4.1 (Christo�el-Darboux relation) Suppose �n are the orthonormal functionsand let H(z;w) and En be de�ned by (2.11) and (2.7). Then�n(w)�n�1(z)Zn(w)Zn�1(z) � �n(z)�n�1(w)Zn(z)Zn�1(w) = H(z;w)En n�1Xk=0 �k(z)�k(w):Proof. De�neg(z;w) = (w � �n)(z � �n�1)�n(w)�n�1(z) and G(w) = g(z;w)� g(w; z): (4.1)Then the Christo�el-Darboux relation which has to be shown is equivalent withF (w) = G(w)z �w = �i(�n � 1)(�n�1 � 1)En n�1Xk=0 �k(z)�k(w): (4.2)Observe that F (w) 2 Ln�1 so that it can be written asF (w) = n�1Xk=0 k(z)�k(w); k(z) =MfF�k�g =MfF�kg:We have k(z) =MfF�kg =MwfF (w)[�k(w) � �k(z)]g+ �k(z)MfFg:9

The �rst term is zero. To see this, we write it out asMwfF (w)[�k(w)� �k(z)]g = (z � �n�1)�n�1(z)Mwf�n(w)w � �nz � w [�k(w)� �k(z)]g�(z � �n)�n(z)Mwf�n�1(w)w � �n�1z � w [�k(w)� �k(z)]g:Because �n ? Ln�1, the �rst term is zero and because �n�1 ? Ln�2, the second term is zero.Thus it remains that k(z) = �k(z)MfFg. We note thatMfFg = (z � �n�1)�n�1(z)fn(z)� (z � �n)�n(z)fn�1(z); fi(z) =Mwfw � �iz � w �i(w)g:By adding and subtracting (z��n�1)(z��n)�n�1(z)�n(z)D(z;w)=(2z), we can rewrite thisas MfFg = (z � �n�1)�n�1(z)gn(z)� (z � �n)�n(z)gn�1(z)with for i = n; n� 1 > 0gi(z) =Mw �D(z;w) �w � �iz + w �i(w)� z � �i2z �i(z)�� = z � �i2z i(z) (4.3)(the last equality is by Lemma 3.2). For i = 0, one has to use �0 = 1, �0 = �1 and 0(z) = (z � 1)=(z + 1) and it then follows that the previous relation also holds for i = 0.Thus, using the determinant relation of Theorem 3.3, this givesMfFg = (z � �n)(z � �n�1)2z [�n�1(z) n(z)� �n(z) n�1(z)]= �i(�n � 1)(�n�1 � 1)Enso that eventually F (w) = n�1Xk=0 k(z)�k(w) =MfFg n�1Xk=0 �k(z)�k(w)yields (4.2). 2We also have the following generalization of the determinant formula.Theorem 4.2 Let f�ng be the orthonormal functions and n the associated functions of thesecond kind. De�ne Fn(z;w) = n(w)�n�1(z)Zn(w)Zn�1(z) � �n(z) n�1(w)Zn(z)Zn�1(w) :Let H(z;w) be as in (2.11), En as in (2.7) and D(z;w) the Riesz-Herglotz kernel. Then, forw 6= z, Fn(z;w) = "n�1Xk=1 �k(z) k(w)�D(z;w)#H(z;w)En:For w = z, this reduces to the determinant formula.10

Proof. De�ne g(z;w) as in (4.1) and setG(z;w) = g(z;w)� g(w; z):Then, by the Chisto�el-Darboux formulaG(z;w) = �i(z � w)(�n � 1)(�n�1 � 1)En n�1Xk=0 �k(z)�k(w):Apply Mt on the expression D(t; w)t+ w G(t; z)� D(t; w)2w G(w; z)and this gives with (4.3)(z � �n)(w � �n�1)�n(z) n�1(w)� (w � �n)(z � �n�1) n(w)�n�1(z)= �i(�n � 1)(�n�1 � 1)En "n�1Xk=1 �k(z)(w � z) k(w) + (z + w)#= i(�n � 1)(�n�1 � 1)(z �w)En "n�1Xk=1 �k(z) k(w)�D(z;w)#This can be rewritten in the required form. 2An analog of the Christo�el-Darboux relations for the functions of the second kind may alsobe obtained.Theorem 4.3 Let n be the functions of the second kind. Then, in analogy with theChristo�el-Darboux relation we have n(w) n�1(z)Zn(w)Zn�1(z) � n(z) n�1(w)Zn(z)Zn�1(w) = H(z;w)En "n�1Xk=1 k(z) k(w)� 1#with H(z;w) as in (2.11) and En as in (2.7).Proof. We setG(z;w) = (w � �n)(z � �n�1)�n(w) n�1(z)� (z � �n)(w � �n�1)�n�1(w) n(z):Then by the previous theoremG(z;w) = i(�n � 1)(�n�1 � 1)(z � w)En "n�1Xk=1 �k(w) k(z)�D(z;w)# :By applying Mt on D(t; w)t+ w G(z; t)� D(t; w)2w G(z;w)we get by using (4.3)(w � �n)(z � �n�1) n(w) n�1(z)� (z � �n)(w � �n�1) n�1(w) n(z)= �i(�n � 1)(�n�1 � 1)(z �w)En "n�1Xk=1 k(z) k(w) � 1#and this is equivalent with the required formula. 2Finally, we give a combination of the previous summation formulas.11

Theorem 4.4 Let �n be the orthonormal functions, n the functions of the second kind andde�ne for an arbitrary complex parameter s�n(z; s) = n(z) + s�n(z):Then �n(w; t)�n�1(z; s)Zn(w)Zn�1(z) � �n(z; s)�n�1(w; t)Zn(z)Zn�1(w)= H(z;w)En "n�1Xk=1 �k(z; s)�k(w; t) + [st� 1 +D(z;w)(t� s)]#with H(z;w) as in (2.11) and En as in (2.7).Proof. This is directly obtained by working out the left-hand side and using the threeprevious theorems. 25 Green's formulaWe now give complex forms of the previous Christo�el-Darboux type formulas. Thereforewe introduce ~H(z;w) = 1Zn(z) � 1Zn(w) = � i(1� zw)(1� z)(1� w) ; n � 0: (5.1)Note that ~H(z;w) = H�(z;w) where the substar is with respect to z. For w = z, we have~H(z; z) = �iP (1; z) with P (t; z) = 1 � jzj2jt� zj2the Poisson kernel. Furthermore we de�ne for n � 2~Hn(z;w) = 1Zn�2(w)Zn�1(z) � 1Zn�1(w)Zn�2(z) = Dn ~H(z;w): (5.2)We now give without proof the following complex analog of Theorem 2.5. A possible proofwould be to take substar conjugates of the relations from the previous section. Note thatwe then need the fact that the numbers En and Cn are real, which they are by Lemma 2.3and Lemma 2.4.Theorem 5.1 (Green's formula) Let xn(z) and yn(z) be both solutions of the recurrencerelation (2.3) and de�neGn(z;w) = xn(w)Zn(w)! yn�1(z)Zn�1(z)!� yn(z)Zn(z)! xn�1(w)Zn�1(w)!:Then, with ~H(z;w) as in (5.1) and En as in (2.7)Gn(z;w) = yn�1(z)xn�1(w) ~H(z;w)En �CnGn�1(z;w)= "n�1Xk=1 yk(z)xk(w)# ~H(z;w)En + (�1)nCnCn�1 � � �C2G1(z;w):12

One can use this theorem and take xk = yk = �k or xk = yk = k or xk = �k andyk = k. One then obtains the formulas in the theorem below which are then proved underthe assumpsion that they satisfy the recurrence relation, in particular that the �k form aregular system. However, it is equally simple to follow the way of reasoning used in theprevious section to arrive at the Christo�el-Darboux type formulas where orthogonality wasthe only assumption needed, and again one will arrive at the results of the theorem below.Thus the results hold without assumimg that there is a recurrence relation, i.e., without thesystem �n being regular. We just give the result without proof.Theorem 5.2 Let �n be the orthonormal functions and n the functions of the second kind.Let ~H(z;w) be as de�ned in (5.1), En as in (2.7) and D(z;w) the Riesz-Herglotz kernel.Then �n(w)Zn(w)! �n�1(z)Zn�1(z)!� �n(z)Zn(z)! �n�1(w)Zn�1(w)! = "n�1Xk=0 �k(z)�k(w)# ~H(z;w)En n(w)Zn(w)! n�1(z)Zn�1(z)!� n(z)Zn(z)! n�1(w)Zn�1(w)! = "n�1Xk=1 k(z) k(w) + 1# ~H(z;w)En n(w)Zn(w)! �n�1(z)Zn�1(z)!� �n(z)Zn(z)! n�1(w)Zn�1(w)! = "n�1Xk=1 �k(z) k(w)�D�(z;w)# ~H(z;w)En;where in the last equation, the substar is with respect to z.Note that this also holds for w = z. In that caseD�(z; z) = 1 + jzj21 � jzj2 :The previous relations can be combined to give the followingTheorem 5.3 Let �n be the orthonormal functions and n the functions of the second kind.For an arbitrary complex s, we set�n(z; s) = n(z) + s�n(z):Then �n(w; t)Zn(w) ! �n�1(z; s)Zn�1(z) !� �n(z; s)Zn(z) ! �n�1(w; t)Zn�1(w) != "n�1Xk=1 �k(z; s)�k(w; t) + 1 + st� (s+ t)D�(z;w)# ~H(z;w)En: (5.3)In particular for z = w and t = s, we obtain �n(z; s)Zn(z) ! �n�1(z; s)Zn�1(z) !� �n(z; s)Zn(z) ! �n�1(z; s)Zn�1(z) != "n�1Xk=1 j�k(z; s)j2 + j1 � sj2# ~H(z; z)En + (s+ s)En 2ijzj2jz � 1j2 : (5.4)13

Proof. By the previous theorem, we can evaluate the left-hand side and obtain the formula(5.3). For z = w and t = s, this equals"n�1Xk=1 j�k(z; s)j2# ~H(z; z)En + EnY (z; s)with Y (z; s) = ~H(z; z)[1� (s+ s)D�(z; z) + jsj2]= ~H(z; z)j1� sj2 + (s+ s) ~H(z; z)[1�D�(z; z)]:Because ~H(z; z)[1�D�(z; z)] = i1 � jzj2j1� zj2 "1 + jzj21 � jzj2 � 1# = i 2jzj2j1� zj2 :we get the result (5.4). 26 Quasi-orthogonal functionsWe de�ne quasi-orthogonal functions asQn(z; � ) = �n(z) + � Zn(z)Zn�1(z)�n�1(z); � 2 R= R[ f1g; n � 1: (6.1)For � =1, we set Qn(z;1) = Zn(z)Zn�1(z)�n�1(z):It was proved in [3] that if �n is regular, then there always exists in�nitely many so called reg-ular values � = �n 2 R such that Qn(z; �n) has n simple zeros, all lying on Tnf�1; �2; : : : ; �ng.In fact all, except �nitely many � 2 R are regular values. Suppose we have chosen a regularvalue �n and suppose that the zeros of Qn(z; �n) are given by �k = �nk(� ), k = 1; : : : ; n. Theycan be used to construct quadrature formulas. These formulas use the �k for their abscissasand their weights �nk(� ) are given by�nk(� ) =MfLnkg =MfLnk(z; � )Lnk�(z; � )g > 0with Lnk(z; � ) = (z � �1) � � � (z � �k�1)(z � �k+1) � � � (z � �n)!n�1(�k)(�k � �1) � � � (�k � �k�1)(�k � �k+1) � � � (�k � �n)!n�1(z) :It was also proved that these quadrature formulas are exact in Rn�1 = Ln�1 � Ln�1.Theorem 6.1 With the notation just introduced, we have thatMffg = nXk=1 �nk(� )f(�nk(� ))for all f 2 Rn�1 = Ln�1 � Ln�1. 14

Alternative expressions for the weights can be obtained as follows. First introduce quasi-orthogonal functions of the second kind byPn(z) = Pn(z; � ) = n(z) + � Zn(z)Zn�1(z) n�1(z); � 2 R; n � 1:We then haveTheorem 6.2 The weights of the quadrature formula are given by�nk = 12�k Pn(�k)Q0n(�k)where the prime means derivative.Proof. From the partial fraction decomposition we see thatRn(z) = �Pn(z)Qn(z) = nXj=1 �njD(�j ; z)with D(t; z) the Riesz-Herglotz kernel. Therefore(z � �k)Rn(z) =Xj 6=k �njD(�j ; z)(z � �k) + �nkD(�k; z)(z � �k):Taking the limit for z ! �k cancels the sum and we get�Pn(�k)Q0n(�k) = �nk limz!�kD(�k; z)(z � �k)and this gives the result. 2Another expression, which clearly shows the positivity of the weights isTheorem 6.3 The weights of the quadrature formula are given by�nk = 24n�1Xj=0 j�j(�k)j235�1 :Proof. Since Qn(�k) = 0, we get from the �rst formula in Theorem 5.2 with w = �k andz = t �n�1(�k)Zn�1(�k)Zn(t) [Qn(�k)�Qn(t)] = 24n�1Xj=0 �j(t)�j(�k)35 ~H(t; �k)En:Dividing by �k � t and letting t tend to �k we get�n�1(�k)Zn�1(�k)Zn(�k)Q0n(�k) = kn�1(�k; �k)En limt!�k ~H(t; �k)�k � twhere kn�1(z; z) = Pn�1j=0 j�j(z)j2 and the limit in the right-hand side is i=(1 � �k)2, providedthat �k 6= 1. 15

Similarly, using the third formula in Theorem 5.2 with z = w = �k and taking into accountthat Qn(�k) = 0, we get �n�1(�k)Zn�1(�k)Zn(�k)Pn(�k) = �En limt!�kD�(t; �k) ~H(t; �k)where the limit is (again for �k 6= 1) �i 2j1� �kj2 :We now plug the values of Q0n(�k) and Pn(�k) into the expression of �nk of the previoustheorem and we get the result for �k 6= 1.For �k = 1, we use a perturbation technique. First we choose a continuous perturbation� (�) 2 R of the parameter � in Qn. Assume that lim�!0 � (�) = � . The coe�cients of Qndepend continuously on � and hence also its zero �k. Therefore we can always choose � 6= 0such that �k(�) 6= 1. Then the relation we want does holds true by our previous arguments.Since �k(�)! 1 as �! 0 and since both Pn and Q0n as well as their ratio depend continuouslyon � in the neighbourhood of � = 0, we �nally obtain the result for �k = 1 since1kn�1(1; 1) = lim�!0 1kn�1(�k(�); �k(�)) = lim�!0 12�k(�) Pn(�k(�))Q0n(�k(�)) = 12 Pn(1)Q0n(1) = �nk:This concludes the proof. 2The discrete measure which takes weights �nk(� ) in the points �nk(� ), k = 1; : : : ; n will bedenoted by �n = �n(�; � ). The set of regular � values for which such a measure exists isdense in R.7 Nested disksIn this section we use the notations = sn(z) = Rn(z; � ) = �Pn(z; � )Qn(z; � )where Qn are the quasi-orthogonal functions and Pn are the quasi-orthogonal functions ofthe second kind. Obviously when n is a regular index, then � ! sn(z) maps (for a �xedz 2 C nTwith C = C [f1g) the extended real line R onto a circleKn(z). Furthermore, theupper half plane is mapped onto its interior. This follows for instance from Theorem 7.1(2)below. The closed disk with boundary Kn(z) is denoted as �n(z).For z 2 T, the circle Kn(z) degenerates to a line (the imaginary axis).When n is a singular index, then the transformation is degenerate. In that case, thewhole plane is mapped to a point. Indeed, since for a singular n we have En = 0, it followsfrom the Christo�el-Darboux relation that�n(w)Zn(w) �n�1(z)Zn�1(z) = �n(z)Zn(z) �n�1(w)Zn�1(w)16

for any z and w. Choose w such that �n�1(w)=Zn�1(w) 6= 0, then we see that there shouldbe a constant cn such that �n(z) = cn Zn(z)Zn�1(z)�n�1(z):It follows similarly from Theorem 4.3 that the same holds for the functions of the secondkind n, i.e., we may replace � by in the previous relation and the constant cn is the samefor � and as follows from the determinant fromula of Theorem 3.3. Hence we getRn(z; � ) = �Pn(z; � )Qn(z; � ) = �(cnZn=Zn�1 + � ) n�1(z)(cnZn=Zn�1 + � )�n�1(z) = � n�1(z)�n�1(z) :This is independent of � for any � 2 C (not only the real ones) and thus the whole Riemannsphere is mapped to a point, which we denote as Kn(z).We have the following theorem.Theorem 7.1 Suppose that the index n is regular. Then for z 2 C nT1. The equation of the circle Kn(z) is given byn�1Xk=1 j k(z) + s�k(z)j2 + j1� sj2 = (s+ s) 2jzj21 � jzj2 :2. The (closed) disk �n(z) is obtained by replacing the equality sign by �.3. The center cn and the radius rn of the disk arecn = � n�n�1 � n�1�n�n�n�1 � �n�1�n = 1kn�1(z; z) 1 + jzj21 � jzj2rn = ����� n�n�1 � n�1�n�n�n�1 � �n�1�n ����� = 1kn�1(z; z) ����� z1 � jzj2 �����where kn�1(z; z) =Pn�1k=0 j�k(z)j2.4. The circle Kn(z) is in the right (left) half plane i� jzj < 1 (jzj > 1) .5. If m is also a regular index and m > n, then �m(z) � �n(z).6. The circles Kn(z) and Kn�1(z) touch.Proof. We can solve the relation s = Rn(z; � ) for � and we then get� = �Zn�1(z)Zn(z) n(z) + s�n(z) n�1(z) + s�n�1(z)= � jZn�1j2ZnZn�1 [ n + s�n][ n�1 + s�n�1]j n�1 + s�n�1j2 :Taking the imaginary part gives=� = 12i jZn�1j2j n�1 + s�n�1j2 gn(z; s)17

with gn(z; s) = ( n + s�n)( n�1 + s�n�1)ZnZn�1 � ( n + s�n)( n�1 + s�n�1)ZnZn�1 :Now suppose that the index n is regular. Then we can write the equation of the circleKn(z),which corresponds to =� = 0 as gn(z; s) = 0. With the help of Green's formula (5.4) (sincen is a regular index and hence En 6= 0) we may also write the equation for the circle asn�1Xk=1 j k(z) + s�k(z)j2 + j1� sj2 = (s+ s) X(z)~H(z; z) :with X(z) = 2ijzj2=jz � 1j2. This is (1) of the theorem.Since the denominator 1� jzj2 is positive, negative or zero, i� jzj < 1, jzj = 1 or jzj > 1,this means that the circle will be in the right or left half plane depending on jzj being lessthan 1 or greater than 1 respectively. This is (4).The disk �n(z) with boundary Kn(z) is given by putting a � sign instead of equality.Since (s+ s) 2jzj21� jzj2 � n�1Xk=1 j k(z) + s�k(z)j2 � j1� sj2 = Ajsj2 +Bs+Bs+Dwith A = �Pn�1k=0 j�kj2 < 0, it follows that this will become negative for jsj su�ciently large.Thus this expression is negative outside the disk. Therefore, the closed disk is described byn�1Xk=1 j k(z) + s�k(z)j2 + j1� sj2 � (s+ s) 2jzj21� jzj2 : (7.1)This is (2).Since the sum in the left-hand side of (7.1) is non-decreasing with n, it follows thatwe have nested disks for regular indices, i.e., if m > n and m and n regular indices, then�m(z) � �n(z). This is (5). We have to exclude here the singular indices because when nis singular, then �n collapses to a point and a subsequent disk with positive radius can notbe a subset of �n.Since Rn(z;1) = Rn�1(z; 0), the circles will touch, even if the index n is singular. In thelatter case, Kn(z) is a point from the circle Kn�1(z).The expressions for center and radius for a general linear fractional transform (� 2 R)s = �a� �bc� �d = �ad� bccd� dc + ad� bccd� dc � c� �dc� �dare obviously c = �ad� bccd� dc and r = �����ad� bccd � dc ����� :Using the Green and Christo�el-Darboux formulas, the expressions for cn and rn as in (3)will follow. 2Corollary 7.2 For z 2 f0;1g, all the circles Kn(z) n = 1; 2; : : : reduce to the same points(z) � 1. 18

Proof. When n is regular, it follows from the expression for the radius that for z 2 f0;1git is zero. From the equation of the circle, it follows that s = 1.Since successive circles touch also for singular indices, we get the same point for all n. 2Assume from now on that there are in�nitely many regular indices. Because the disks arenested, it follows that when n(�) is the sequence of regular indices, then �1 = \n(�)�n(�) isa disk with radius r(z) = lim�!1 rn(�)(z)which may reduce to a point when this radius is zero. We have the followingLemma 7.3 Suppose z 2 C 0 = C n (T[ f0g).If �1(z) is a disk (with positive radius), then1Xk=0 j�k(z)j2 <1 and 1Xk=0 j k(z)j2 <1:If �1(z) is a point, then1Xk=0 j�k(z)j2 =1 and 1Xk=0 j k(z)j2 =1:Proof. It follows from the expression for the radii that r(z) is positive (zero) i� k1(z; z) is�nite (in�nite), i.e., i� P1k=0 j�k(z)j2 is �nite (in�nite).Let s be a point from the disk �1(z). Then it follows from (7.1) that in any case (diskor point) 1Xk=1 j k(z) + s(z)�k(z)j2 <1:Thus P1k=1 j�k(z)j2 <1 i� P1k=1 j k(z)j2 <1. 2Before we can prove an invariance theorem, we also need the followingLemma 7.4 If �n are the orthonormal functions and n the functions of the second kind,then for any w 2 C 0 = C n (T[ f0g)�n(z) = �n(w) + z � w�n � z "A0n + n�1Xk=1 �k(z)Akn(w)# n(z) = n(w) + z � w�n � z "B0n + n�1Xk=1 k(z)Akn(w)#where A0n(w) = Vn(w)�n(w)� Yn(w) n(w)B0n(w) = Vn(w) n(w)� Yn(w)�n(w);Akn(w) = Yn(w)akn(w); akn(w) = k(w)�n(w)� �k(w) n(w); k = 1; : : : ; n� 1;Yn(w) = w � �n2w and Vn(w) = w + �n2w :19

Proof. From the Christo�el-Darboux formula in Theorem 4.1 and the mixed formula inTheorem 4.2, we get�n(z)�n�1(w)Zn(z)Zn�1(w) � �n(w)�n�1(z)Zn(w)Zn�1(z) = �H(z;w)En n�1Xk=0 �k(z)�k(w)�n(z) n�1(w)Zn(z)Zn�1(w) � n(w)�n�1(z)Zn(w)Zn�1(z) = �H(z;w)En "n�1Xk=1 �k(z) k(w) �D(z;w)# :Elimination of �n�1(z) gives�n(z)Zn(z)Zn�1(w)an�1;n(w) = �H(z;w)En "n�1Xk=1 �k(z)ak;n(w)� [ n(w) +D(z;w)�n(w)]# :From the determinant formula of Theorem 4.2, we �ndan�1;n(w) = �EnZn�1(w)Zn(w) 2iw(1 �w)2 :Hence �H(z;w)EnZn(z)Zn�1(w)an�1;n(w) = z � w�n � z �w � �n2w � = z �w�n � zYn(w):Thus �n(z) = z � w�n � zYn(w) "n�1Xk=1 �k(z)akn(w)� [ n(w) +D(z;w)�n(w)]# :Next we compute� z � w�n � zYn(w)D(z;w) = (z + w)(�n � w)2w(�n � z) = 1 + (z � w)(w + �n)(�n � z)2w :Thus �n(z) = �n(w) + z � w�n � z [Vn(w)�n(w) � Yn(w) n(w)]+ z � w�n � z "n�1Xk=1 �k(z)[Yn(w)akn(w)]# :This is the �rst formula required.The second formula is proved similarly. 2We can now prove the following invariance theorem.Theorem 7.5 (invariance) Suppose w 2 C 0 = C n (T[f0g) and suppose that �1(w) is adisk with positive radius. Then �1(z) is a disk with positive radius for every z 2 C 0 andnXk=0 j�k(z)j2 and nXk=0 j k(z)j2converge locally uniformly in C 0 as n!1. 20

Proof. From Lemma 7.4, we know that with xn = �n and yn = n or visa versa, we have,with the notation as in Lemma 7.4xn(z) = xn(w) + z � w�n � zYn(w) n�1Xk=1 xk(z)akn(w)+ z � w�n � z [Vn(w)xn(w)� Yn(w)yn(w)]:From the de�nition of akn(w), it follows1Xn=1 n�1Xk=1 jakn(w)j2 � 1Xn=1 n�1Xk=1 �j k(w)j2j�n(w)j2 + j�k(w)j2j n(w)j2�� 2 1Xn=1 j n(w)j2! 1Xn=1 j�n(w)j2! <1by Lemma 7.3. Now suppose thet C is a compact subset of C 0 . Then for arbitrary z 2 Cand w 2 C 0 �xed z � w�n � zYn(w) and z � w�n � zVn(w)are uniformly bounded, say���� z � w�n � zYn(w)���� � R1 and ���� z � w�n � zVn(w)���� � R2:So NXn=m jxn(z)j2!1=2 � NXn=m jxn(w)j2!1=2 +R1 NXn=m jyn(w)j2!1=2+R2 NXn=m jxn(w)j2!1=2 +R10@ NXn=m �����n�1Xk=1 akn(w)xk(z)�����21A1=2 :For any 0 < � < 1, choose 1 < m = m(�;R1; R2) such that 1Xn=m jyn(w)j2!1=2 < �R1 ; 1Xn=m jxn(w)j2!1=2 < �1 +R2 ;and 1Xn=m n�1Xk=1 jakn(w)j2!1=2 < �R1 :Then NXn=m jxn(z)j2!1=2 � �+ �+R1 " NXn=m n�1Xk=1 jakn(w)j2! n�1Xk=1 jxk(z)j2!#1=2� 2�+R1 NXk=1 jxk(z)j2!1=2 1Xn=m n�1Xk=1 jakn(w)j2!1=2� 2�+ � NXk=1 jxk(z)j2!1=2� 2�+ � NXk=m jxk(z)j2!1=2 + � m�1Xk=1 jxk(z)j2!1=2 :21

Hence (1� �) NXn=m jxn(z)j2!1=2 � 2�+ � m�1Xk=1 jxk(z)j2!1=2 :Because Pm�1k=1 jxk(z)j2 is a continuous function in C, there is an M > 0 such that m�1Xk=1 jxk(z)j2!1=2 �Mholds uniformly in C. Thus NXk=1 jxk(z)j2!1=2 � (2 +M)�1 � � ; N � m:This proves the uniform convergence of P1n=0 jxk(z)j2 in C. 2We can now also prove the analyticity theorem.Theorem 7.6 (analyticity) Let �1(z) be a point. Then the limits(z) = limn!1Rn(z; � ); z 2 C nTis an analytic function of z not depending on � . Moreover<s(z)1� jzj2 > 0; z 2 C nT:Proof. By Theorem 7.1 (6), it follows that if �1(z) is a point thenlimn!1Rn(z; � ) = limn!1 sn(z) = s(z)exists and is independent of � .Obviously sn(z) = Rn(z; � ) is analytic for z 2 C nT. Thus the analyticity of s(z) willfollow if the functions sn(z) are uniformly bounded in compact subsets of C n T. This isshown as follows. We know thatj1 � sn(z)j2 + n�1Xk=1 j k(z) + sn(z)�n(z)j2 = (sn(z) + sn(z)) 2jzj21� jzj2 :Thus 1 + jsnj2 + n�1Xk=1 j k + sn�nj2 = (sn + sn)1 + jzj21� jzj2or, using jsn + snj � 2jsnj,jsnj2 < 1 + jsnj2 + n�1Xk=1 j k + sn�nj2 � 2jsnj �����1 + jzj21 � jzj2 ����� :Therefore jsn(z)j � �����1 + jzj21 � jzj2 ����� :Since the right-hand side is uniformly bounded in compact subsets of C nT, the analyticityof s(z) follows.The last inequality is a direct consequence of Theorem 7.1 (4). 222

8 The moment problemWe are now ready to border the moment problem as it was described in the introduction.Thus given the linear functionalM , real and positive on R, can we �nd the measure �, �niteand positive on [��; �], such thatM � 1!n� = �n = Z ��� d�(�)!n(ei�) ; n = 0; 1; : : :Note that if � is a solution of this moment problem, then it is easily seen that Mffg =R ��� f(ei�)d�(�) for all f 2 L. Thus � de�nes the linear functional M in the space L. Thegiven functional M was however de�ned on the larger space R = L � L. Thus strictlyspeaking, to solve the moment problem, it would have been su�cient to de�neM on L only.As we have seen however, our approach so far relied heavily on orthogonality properties, i.e.,on the fact that we have a real and positive inner product, and that is why we needed thefunctional M to be de�ned on R.Another, more general moment problem which could have been considered, is to writethe functional M as an integral with a positive measure � for all functions in R = L � L(and not just for functions in L). Thus Mffg = R ��� f(ei�)d�(�) for all f 2 R. Then weshould consider the problem of �nding a positive measure � such that the following momentrelations M � 1!n!m�� = �nm = Z ��� d�(�)!n(ei�)!m�(ei�) ; n;m = 0; 1; : : : (8.1)hold. This moment problem is indeed equivalent to �nding a positive measure � such thathf; gi =Mffg�g = Z ��� f(ei�)g(ei�)d�(�); 8f; g 2 L: (8.2)We shall call this the moment problem in R = L � L, while the previously explained simplerform is called the moment problem in L. Note that the moment problem in R prescribesalso the moments �n0 = �n, which implies that a solution of the moment problem in R willautomatically be a solution of the moment problem in L. It is not clear whether the converseis true in general.It was shown in [3] that the moment problem in L has at least one solution. Since theproof of [3] is rather compact, we include it here in more detail. The arguments which aremade explicit in this proof also show why the same arguments do not lead to a solution ofthe moment problem in R.Theorem 8.1 Let M be a real positive functional de�ned on R = L � L and assume thatthe sequence of orthogonal rational functions �n has in�nitely many regular indices. Thenthere exists at least one measure � on Twhich solves the moment problem in L, i.e., whichsatis�es (8.1).Proof. The proof is based on the fact that the discrete measures �k representing thequadrature formulas kXi=1 �ki(� )f(�ki(� )) = Z ��� f(ei�)d�k(�)which are exact in Rk�1, hence also in Lk�1, are uniformly bounded by Mf1g = 1 andhence by Helly's selection principle, the sequence f�kg will have a convergent subsequence23

�k(j) ! �. Next we prove that such a � solves the moment problem in L, i.e., thatM � 1!n� = Z ��� d�(�)!n(ei�) ; n = 0; 1; : : : :For n = 0, we can apply Helly's convergence theorem since 1 is a continuous function in theinterval [��; �]. Thus it remains to prove this for a �xed n > 0. We denote the elementsof the subsequence k(j) by k for simplicity. Note that for n > 0 we can not simply applyHelly's convergence theorem because the !�1n are not continuous on the interval I = [��; �].However, consider any subset J � I which does not contain the �1; : : : ; �n, then !�1n isindeed continuous on J and this means that we can �nd a k = k(j) large enough such thatfor a given � > 0, �����ZJ d�(�)� d�k(�)!n(ei�) ����� < �3 : (8.3)On the other hand, for k = k(j) > n,�����ZI�J d�k(�)!n(ei�) ����� = �����ZI�J !n�(ei�)d�k(�)j!n(ei�)j2 ������ maxI�J j!n�(ei�)j ZI�J d�k(�)j!n(ei�)j2 � maxI�J j!n�(ei�)jM � 1!n!n�� :We can always choose J large enough such that the maximum of j!n�j in I � J is arbitrarysmall. Because Mf[!n!n�]�1g is �nite, it follows that for any � > 0, we can make J largeenough to satisfy �����ZI�J d�k(�)!n(ei�) ����� < �3 : (8.4)Note that this holds for any k > n, i.e., J is independent of k. Next we want to show thatJ can also be made large enough to satisfy�����ZI�J d�(�)!n(ei�) ����� < �3 : (8.5)To obtain this, we consider sets Jp none of which contain �1; : : : ; �n and such that Jq � Jpfor p > q and [pJp = I. Note that for k = k(j) > n and for any pZJp d�(�)!n(ei�) = ZJp d�(�) � d�k(�)!n(ei�) + ZJp d�k(�)!n(ei�) :Thus if p > q and k > n�����ZJp�Jq d�(�)!n(ei�) ����� � �����ZJp�Jq d�(�) � d�k(�)!n(ei�) �����+ �����ZJp�Jq d�k(�)!n(ei�) ����� : (8.6)By (8.4), there is always a p and q large enough such that for any � > 0�����ZJp�Jq d�k(�)!n(ei�) ����� < �2 (8.7)24

for all large k. By (8.3), we can make k so large that for any � > 0�����ZJp�Jq d�(�) � d�k(�)!n(ei�) ����� < �2 : (8.8)Combining (8.6), (8.7) and (8.8) shows that it is possible to make�����ZJp d�(�)!n(ei�) � ZJq d�(�)!n(ei�) ����� < �for any � > 0. This means that (ZJp d�(�)!(ei�))pis a Cauchy sequence so that its limit for p !1 (which is the integral over I) exists, thusthat for any � > 0 there exists a p large enough such that�����ZJp d�(�)!n(ei�) � ZI d�(�)!n(ei�) ����� < �3which proves (8.5). Finally, by (8.4), (8.5) and (8.3),�����ZI d�k(�)� d�(�)!n(ei�) ����� � �����ZI�J d�k(�)!n(ei�) �����+ �����ZI�J d�(�)!n(ei�) �����+ �����ZJ d�k(�)� d�(�)!n(ei�) ����� < �because each of the terms in the right-hand side can be bounded by �=3. Thuslimk(j)!1 ZI d�k(j)(�)!n(ei�) =M � 1!n� = ZI d�(�)!n(ei�) :This proves the theorem. 2We now use our framework of nested disks to obtain information about when the solutionis unique. Let us denote by ML the set of solutions of the moment problem in L and MRthe set of solutions of the moment problem in R. Then we haveTheorem 8.2 Assume that the sequence of orthonormal functions �n has in�nitely manyregular indices. Hence ML 6= ?. Fix z 2 C nT and de�ne for � 2 ML its Riesz-Herglotztransform �(z) = Z ��� D(ei�; z)d�(�):Then f�(z) : � 2 MRg � �1(z) � f�(z) : � 2 MLg:Proof. Let s = �(z) for some � 2 MR. Note that the system f�ng is then orthonormalwith respect to the inner product de�ned by the measure �. Let f(z) = D(t; z), t 2 T.Writing the generalized Fourier series of f(z) asf(z) � 1Xk=0 k�k(z); k = hf; �ki ;25

then using k = Z ��� D(t; z)�k(t)d�(�);it turns out that 0 = s and k = k(z) + s�k(z) for k � 1 because k = Z ��� D(t; z)�k(t)d�(�) = Z ��� D(t; z)[�k(t)� �k(z)]d�(�) + �k(z)�(z); t = ei�:On the other hand, it can be shown that for t 2 T�1 + 1 + jzj21� jzj2 [D(t; z) +D(t; z)] = jt+ zj2jt� zj2 = jD(t; z)j2:Using Bessel's inequality, it then follows that1Xk=0 j kj2 = jsj2 + 1Xk=1 j k + s�kj2 � �1 + 1 + jzj21� jzj2 (s + s);which can be rearranged asj1 � sj2 + 1Xk=1 j k + s�kj2 � 2jzj21� jzj2 (s+ s):This means that s 2 �1(z).Next it is shown that if s 2 �1(z), then it is the Riesz-Herglotz transform of some� 2 ML. This is readily shown by using the quadrature formulas we have discussed. Weconsider the limiting point and the limiting disk case separately.If �1(z) is a point, then since s 2 �1(z), there must exist sn 2 Kn(z) such that sn ! s.Since there is for each n some �n such thatsn = Rn(z; �n) = Z ��� D(ei�; z)d�n(�):Helly's selection criterion then yields that there is a subsequence �n(j) ! �. By the proof ofthe previous theorem, � 2 ML and by Helly's convergence theoremlimj!1 Z ��� D(t; z)d�n(j)(�) = Z ��� D(t; z)d�(�) = s:Thus s is the Riesz-Herglotz transform of a � 2 ML.If �1(z) is a disk, let s be a point on the boundary K1(z). Recall that for a �xed n,we can, except for �nitely many values of � , associate a quadrature formula with Rn(z; � ).Let us denote the discrete measure that is associated with this quadrature by �n(�; � ). Itdepends on n but also on the choice of � . We can then, for every regular index n choose ansn 2 Kn(z) such that these sn tend to s and such that sn = �n(z), where �n = �n(�; �n)and where �n is chosen such that sn = Rn(z; �n). By Helly's theorems and the proof of theprevious theorem, there exists a � 2 ML such that �(z) = s.Thus every s on the boundary K1(z) is of the form �(z) with � 2 ML. Now let s be aninterior point of �1(z). Then it can be found as a convex combination s = �s1 + (1� �)s2(0 < � < 1) of points s1; s2 on the boundary K1(z). Thus there exist �1; �2 2 ML suchthat sj = Z ��� D(t; z)d�j (�); t = ei�:Thus � = ��1 + (1� �)�2 2 ML and s = �(z). 2Now the following corollary is obvious. 26

Corollary 8.3 In the case of a limiting disk, for each s 2 �1(z), z 2 C 0 = C n (T[ f0g),there are in�nitely many � 2 ML such that s = �(z). The moment problem in L hasin�nitely many solutions.In the case of a limiting point, a solution of the moment problem in R is unique.References[1] N.I. Akhiezer. The classical moment problem. Oliver and Boyd, Edinburgh, 1969.Originally published Moscow, 1961.[2] A. Bultheel and P. Dewilde. Orthogonal functions related to the Nevanlinna-Pick prob-lem. In P. Dewilde, editor, Proc. 4th Int. Conf. on Math. Theory of Networks andSystems at Delft, pages 207{212, North-Hollywood, 1979. Western Periodicals.[3] A. Bultheel, P. Gonz�alez-Vera, E. Hendriksen, and O. Nj�astad. Orthogonal rationalfunctions with poles on the unit circle. J. Math. Anal. Appl., 182:221{243, 1994.[4] A. Bultheel, P. Gonz�alez-Vera, E. Hendriksen, and O. Nj�astad. Orthogonal rationalfunctions and nested disks. J. Approx. Theory, 1994. Accepted.[5] A. Bultheel, P. Gonz�alez-Vera, E. Hendriksen, and O. Nj�astad. Orthogonality andboundary interpolation. In A.M. Cuyt, editor, Nonlinear Numerical Methods and Ra-tional Approximation II, pages 37{48. Kluwer, 1994.[6] A. Bultheel, P. Gonz�alez-Vera, E. Hendriksen, and O. Nj�astad. Recurrence relationsfor orthogonal functions. In S.C. Cooper and W.J. Thron, editors, Continued Fractionsand Orthogonal Functions, volume 154 of Lecture Notes in Pure and Appl. Math., pages24{46. Marcel Dekker, 1994.[7] A. Bultheel, P. Gonz�alez-Vera, E. Hendriksen, and O. Nj�astad. Orthogonal rationalfunctions and modi�ed approximants. Numer. Algorithms, 11:57{69, 1996.[8] P. Dewilde, A. Viera, and T. Kailath. On a generalized szeg}o-Levinson realizationalgorithm for optimal linear predictors based on a network synthesis approach. IEEETrans. Circuits and Systems, CAS-25:663{675, 1978.[9] M.M. Djrbashian. A survey on the theory of orthogonal systems and some open prob-lems. In P. Nevai, editor, Orthogonal polynomials: Theory and practice, volume 294 ofSeries C: Mathematical and Physical Sciences, pages 135{146, Boston, 1990. NATO-ASI, Kluwer Academic Publishers.[10] H. Hamburger. Ueber eine Erweiterung des Stieltjesschen Moment Problems I. Math.Ann., 81:235{319, 1920.[11] H. Hamburger. Ueber eine Erweiterung des Stieltjesschen Moment Problems II. Math.Ann., 82:120{164, 1921.[12] H. Hamburger. Ueber eine Erweiterung des Stieltjesschen Moment Problems II. Math.Ann., 82:168{187, 1921. 27

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