HOUSTON JOURNAL OF MATHEMATICS
Volume 17, No. 4, 1991
RECURSIVENESS, POSITIVITY, AND TRUNCATED MOMENT PROBLEMS
I•A•L E. CURTO LAWRENCE A. FIALKOW
Abstract. Using elementary techniques from linear algebra, we de- scribe a recursire model for singular positive Hankel matrices. We then use this model to obtain necessary and sufficient conditions for existence or uniqueness of positive Borel measures which solve the truncated moment problems of Hamburger, Hausdorff and Stieltjes. We also present analogous results concerning Toeplitz matrices and the truncated trigonometric moment problem.
Dedicated to the memory of Domingo A. Herrero
1. Introduction. Given an infinite sequenceof complexnumbers 7 - {70,71,--' } and a subsetK C_C, the K-powermoment problem with data 7 entails finding a positiveBorel measurey on C suchthat (1.1) / tJdp(t)=7j (J •- O) and supp p C_K. The classicaltheorems of Stieltjes,Toeplitz, Hamburger and Hausdorffpro- vide necessaxyand sufficientconditions for the solubilityof (1.1) in case K = [0,+o•),K ---T := (t e C: Itl- 1},K - • andK - [a,bl(a,b e [•), respectively([AhKI, [Akhl, [KrNI, [Lanl, [Sat], [ShT]). For example, when K = •, Hamburger'sTheorem implies that (1.1) is solubleif and
Both authorswere partially supportedby grantsfrom NSF. Rafil Curto was alsopartially supportedby a Universityof Iowa faculty scholaraward.
603 604 R. E. CURTO AND L. A. FIALKOW onlyif theHankel matrices A(n):= (7i+j)in,j=o(n = 0,1,... ) arepositive semidefinite[ShT, Theorem1.1.2]. The classicaltheory also providesuniqueness theorems and parame- terizationsof the setsof solutions.Parts of the theory havebeen extendedto covermore general support sets for/• ([SEMI, [Gas],[Sch]), and the "mul- tidimensionalmoment problem" has also been studied extensively (lAtz], [Ber], [Fug],[Put]). For 0 _• m < •x•,let ? = (7o,..-,7m) E Cre+l, and considerthe truncated K-power moment problem (1.2) / tJdl•(t)=7j (0 _< j <_ m). and supp/• C_K. For example, the "even" caseof the truncated Hamburgerproblem corre- spondsto m = 2k for somek _>0 and K = It. On the basisof Hamburger's Theorem,one might surmise that the criterionfor solubilityof (1.2) in this caseis A(k) _>O. As wewill seein the sequel,this positivity condition is al- waysnecessary, and is sufficientwhen A(k) is nonsingular[ABE, Theorem 1.3];however, when A(k) is singular,the positivity of A(k) is not sufficient. In particular, the classicaltheory for the "full" moment problem does not include the theory of the truncated momentproblem as a specialcase. On the other hand, it is knownthat the theoryof the truncatedmoment prob- lem can be used to solve the full moment.problem and to recover such resultsas Hamburger'sTheorem [Lan, p. 5], [Akh]. Observethat the truncatedStieltjes moment problem differs slightly fromthe classicaltruncated Stieltjes moment problem for 7 = (70,..., 7m), defined by: øøtJdl•(t)= 7j (O_< j _ (1.3) tmd[2(t)_• •m. For purposesof parameterizingthe solutionspaces, the classicalproblem has provedmore amenablethan the problemwe consider.However, for cer- tain basicinterpolation problems in operatortheory (discussed below), it is TRUNCATED MOMENT PROBLEMS 605 necessaryto considerthe Stieltjesmoment problem of (1.2), i.e., wherethe inequalityin (1.3) is replacedby equality.Despite the importanceof trun- cated moment problems,their treatment in the literature seemssomewhat fragmentary. Most sourcesignore the "odd" caseentirely, although for us this caseis fundamental.The usualapproach (e.g. [Ioh, TheoremA.II.1], [ShT, p. 28 if.I) is to treat the "nonsingulareven" case of Hamburger's problem and then reducethe "singulareven" caseto the nonsingularcase by means of quasi-orthogonalpolynomials and Gaussianquadrature. In the present note we present a comprehensiveand unified treat- ment of necessaryand sufficientconditions for the solubilityof (1.2) in case K - R, [a,b], [0,-t-•x•) or T. Our approachrevolves around the notionof "recursiveness"for positiveHankel and Toeplitz matrices(Theorems 2.4 and 6.4). This key ingredient,apparently neglected in the literature,allows us to recover many classicalresults and to obtain many new ones. Our basic result, Theorem 3.1, providessimple operator-theoreticnecessary and sufficientconditions for solubility of the "odd" caseof the truncated Ham- burger moment problem. Our solution is based on elementary techniques concerningnonnegative polynomials, Lagrange interpolation, and linear al- gebra (Proposition3.3); for the "singualrodd" case,the solutionrests on a descriptionof positive singularHankel matrices. Theorem 2.4 showsthat eachentry of sucha matrix (exceptperhaps the lowerright-hand entry!) is recursivelydetermined by the entriesin the largestnonsingular upper left-hand"corner" of the matrix. Theorem2.4 immediatelyallows us to re- ducesingular Hamburger moment problems to equivalentnonsingular ones without recourseto quasi-orthogonalpolynomials or Gaussianquadrature (Corollary3.4). The "even"cases of Hamburger'sproblem are then treated as consequencesof the "odd"cases (Theorem 3.9). Theorems3.8 and 3.10 contain our uniquenesscriteria for the truncatedHamburger problem. So- lutions of the Hausdorffand Stieltjesmoment problems are corollariesof this approachand are presentedin Sections4 and 5, respectively.In Sec- tion 6 we treat the trigonometricmoment problem (K - T) througha corresponding(but independent)analysis of positiveTocplitz matrices. The presentwork supplements[CuF], where,as part of a detailed study of k-hyponormalityand quadratic hyponormalityfor unilateral weightedshifts, we solvedthe following"subnormal completion problem" of J. G. Stampfii[Sta]: When may a finite sequenceof positivenumbers, 70,... ,7m, be "completed"to a sequence? t%)n=0 whichis the mo- 606 R. E. CURTO AND L. A. FIALKOW ment sequenceof a subnormalunilateral weighted shift Wv ? In [CF] the subnormalityof Wv was establishedusing Berger's Theorem [Con]' We essentiallysolved the truncated Stieltjes moment problem with a measure producedfrom the spectral measureof a normal operatorßOur motivation for the present work was the desireto developa more transparent method of solving such truncated moment problems,without recourseto such an- alytic tools as the spectral theorem or quasi-orthogonalpolynomials: our new treatment satisfiesthis requirement. In particular, Section 3 concludes with a simplecomputational algorithm for solvingthe truncated Hamburger problem. 2. Recursire Models for Positive Hankel Matrices. Our goal in this section is to prove a structure theorem for positive semi-definiteHankel matrices.If A is a Hankelmatrix of sizek+ 1, let 7 = (70,..., 72k)E R2k+• representthe entriesof A, in the sensethat A = Av := (Ti+5)i,j=O'k The j-th columnof A will be denotedbyvj := (7j+e)e=o,k 0 _ Proof: (ii) The fact that A(r- 1) is invertibleis containedin [CF, Propo- sition 2.12]ß To provethe identityfor (I'(7), let (I,' •-- (9,•,..., •v--1)' :-- TRUNCATED MOMENT PROBLEMS 607 A(r- 1)-Xv(r,r - 1). We employthe minimality condition in the definition of r - rank V'•(V)- (T0,...,Tr-x) E Rr is the uniquesolution of In particular,•0?j + '" + •?-x?j+?-x = ?j+?,0 _ • -- f["{r+SJr,s=O •u+x= (v(k -•- A1, k)* v(k+"{2k+21,k)) ' The followingelementary, but veryuseful, result will be neededin the proof of Theorem2.4; an operatorversion of this resultis implicit in the work of Smul'jan[Smu] (cf. [CuF]). Lemma 2.3. (cf. /CuF, Proposition2.3]) Let A • Mn(C),b • Cn, c • C, and let •:= b* c ' (i) If • > O,then A _>0, b = Aw forsome w • C'•, and c > w*Aw. 608 R. E. CURTO AND L. A. FIALKOW (ii) LetA> -- Oandb=Awforsomew6C n. Then fi>__ 0if and only if c _>w* Aw. (iii) IœA >_0 andb = Aw, thenrank fi = rankA if andonly if c = w*Aw. We now presentour structure theorem for positive Hankel matrices. Theorem2.4. Letff = (fro,...,ff2k), ff2k+l, ff2k+2, fix be as in (2.1),assume that 2i >_O, and let r := rank7. Then (i) A(r - 1) is positiveand invertible; (ii) • = (To,...,Tr-1):= A(r- 1)-lv(r,r- 1) satisfies (2.2) 7r+j = •O?jq-''' q-•r--l?r+j--1 (0 • ?k+p+ r+ 1 -- CO'r+p q- ''' q- Ck?k+r+p = CO(lt907pq-''' q-•r--l"•p+r--1) q-''' q-Ck(•OTp+k q-''' q-lt9r-17p+k+r--1) = •O(C07pq-''' q-Ckfp+k) q-''' q-•r-l(COfp+r-1 q-''' q-Ckfp+k+r-1) -- //907p+k+1 q- ... q- lt9r-17p+k+r; TRUNCATED MOMENT PROBLEMS 609 thus (ii) followsby induction. For the converse,staxting with A(r- 1) _>0, we use(2.2) to prove inductivelythat A(j) _>O, r - 1 _ A(j+1) = (A(j) W•-k1 '72j+2Wj+l ) , where wj+•= ' . '72j+• Let uj = (0,...,0,(/90,...,(/9r_1)• Rj+l. Then (2.2) impliesthat A(j + 1)uj = Wj+l andthat '72j+2 = ujwj+l; thusLemma 2.3 implies that A(j + 1) _>0. By induction,we haveA(k) _>O, and nowa similar argumentusing (iii) showsthat .• _>0. [] In thecase when .• issingular, Theorem 2.4 (ii) showsthat all of the elementsof .• (exceptperhaps '72t•+2) axe determined recursively by the elementsof A(r - 1). The proofof the "converse"direction of Theorem2.4 ~ thenshows that if A _>0, thenrank A(j q-1) = rankA(j),r - 1 _ '72k+2> •0'72k+2-v q- ''' q- •v-l'72k+l. The following"extension principle" for positiveHankel matrices will proveuseful in developingour resultson moments. Theorem 2.6. Let A := A(k) andlet r := rank7. If A _>O, thenthe followingare equivalent: (i) A hasa positiveHankel extension; 610 R. E. CURTO AND L. A. FIALKOW (ii) rankA=r; (iii) There exist To,...,T•-i • H such that 7j = •07j--r '4-''' '4-•r--17j--1, T __ v(k + 1, k) := ' ½ Ran A. 72k+1 If A is invertible,v(k + 1, k) ½ Ran A for every72k+1 ½ I•. We may thus assume1 < r < k. Since r < k, Proposition2.2 imphes that r = rank7 = rank (7o,...,Ta•-a). Theorem2.4(ii) (apphedwith A(k) as 2i and.4(k- l) as.4) impliesthat there are unique scalars •o,... ,•-1 suchthat 7•+j = •o7i+'"+•-1%+•-1(0 < j < 2k-l-r). Thus, A(k- 1)i, = v(k,k- •), wherei, := (0,...,0,;•o,...,;•-1) • a•; also, • = rank.4(k): rankA(k- •). Lemma2.3 (iii) nowimphes that 7a• = •90•Y2k_vq- ''' q- •v_l•Y2k_l . Let 72kq-1:= •072k-i-1-r q-''' q- •r-172k; then the recursiverelations imply that ¾k+l • span{vo,...,v•_•} = Ran A. Moreover,if/• y• 72k+1has the property;hat (7•+1,-..,72k,/•) ½ Ran .4, then (0,0,..., •) • span{Vo,...,v•-l}, so there existscalars Co,..., c•_1, not all zero,such that Coy(0, r- 1)+..-+C•-lV(r- 1,r- 1) = 0, a contradictionto the invertibilityof A(r - 1). [] Remark 2.7. Assumethat A(k) hasa positiveHankel extension (as just described).Examination of the precedingargument shows that when TRUNCATED MOMENT PROBLEMS 611 A is invertible, ?2k+l can be chosenarbitrarily, while if A is singular,?2k+l is uniquelydetermined. In either case,once 72k+1 is prescribed,72k+2 can be chosenas any valuesatisfying Theorem 2.4 (iii). Let 7 = (7o,.-., ?2k)and let r := rank 7- In the sequelwe saythat 7 is positively recursivelygenerated if A(r- 1) >_ 0 and ?j = (1907j_vq-... q-(/9v_x?j_x(r_< j _<2k); equivalently,if A(k) > 0 andthe con- ditionsof Theorem2.6 hold. We saythat • := (7o,..., 72k+1)is positively recursivelygenerated if A(r- 1) > 0 and?j = •o?j-• q-"' q-99r-17j-x( r _< j <_2k q- 1); equivalently,A(k) >_0 and v(k q- 1,k) E Ran A(k) (Theorem 2.4). 3. The Truncated Hamburger Moment Problem. Our aim in this section is to present an elementaryalgebraic treatment of the truncated Hamburgermoment problem. For k > 0, let ? - (7o,...,72•) E R2•+1, 7o > 0. For 72•+1 • R, let • = (7o,... ,72•+1). As in the previoussection, we defineA(m) = (?i+j)o Vr -- •oVo q-''' q- •r--lVr--1- Notethat • = A(r- 1)-lv(r,r - 1)and that if r _ g•(t) := t• - (•o +'" + •v-1tv-1) (t e a) as the generating.function of •. We begin with an existencetheorem for the "odd case" of the Hamburgerpower moment problem (3.1) ?j= /tJdl•(t), O_ (ii) There exists a compactly supported representingmeasure for•, (iii) There existsa finitely atomic representingmeasure for •, (iv) There exists a rank '-atomic representingmeasure for •, whosesupport consists of the rootsof g•; A :- A(k) >_O andvk+• 6 Ran A; (vi) A(k + >_0 eorsome choice oe ß R (i.e., hasa positiveHankel extension using '/•+• ); (vii) • is positively recursivelygenerated. Theorem3.1 ((i) •=• (v)) reducesthe questionof existenceof a rep- resenting measure to two standard problems in finite dimensionallinear algebra and operator theory: the problemof determiningwhether v•+• ß Ran A(/•), andthe problemof verifyingthat the eigenvaluesof A(/•) are nonnegative;at the end of this section,we providea simplealgorithm which checksboth conditionssimultaneously. We defer the proof of Theorem 3.1 in favor of somepreliminaries. For Y ß R, 5u denotesthe probabilitymeasure on • suchthat 5u({y}) - 1. Thus, a finitely atomic positive measureon • with atoms Y0,. ß., y• may be expressedas•u - •']•j-0PJSyj, where each pj >_O. Thenext result relates rank '/to the numberof atomsin any finitely atomicrepresenting measure of if. Lemma 3.2. ff• has a finitely atomicrepresenting measure with m atoms, then m >_ rank ra--1 Proof:Let • = •j-o PJSyjbe a representingmeasure for • withdistinct atomsYo,...,Y,•-•. Let g(t):= (t-yo)...(t-y,•_•)= c0+.--+c,•_•t "•-• +t "•. In L•(•), wehave tJg = 0(0 _ XO Xm Vx:= . . ; m Vx is invertibleif and only if the x'is axedistinct. Proposition 3.3. Assumethat A = A(k) is positiveand invertible.Then g• has k + I distinct real roots, x0,...,xk. Thus Vx is invertible, and if p -- (po,...,pk) := V•Xv0, thenpj > 0(0 _ s(p) = s p&t = = (At,,q). i i,j=O Thusif p • 0 andA is positiveand invertible, then 0 < (A•, •) = $(Ip 2). Now supposedeg r _<2k, r • 0, and r(t) _>O(t e 1•). Then degr = 2m for somem, 0 _ (3.2) { Adeg_> r0 Proof of Proposition 3.3: The result is trivial for k = 0, so we assume k > 0. SinceA is invertible,rank 7 = k + 1, andfrom the definition of •, we have ?k+j+•= •0?j +"' + •k?j+• (0 <_j _ k 0< (A•(J), •©) - S([f ©2) __ f [f(j)[2dl a_ y•pi[f(j)(xi)]2i=0 _pj. We axenow prepared to proveTheorem 3.1 (v) =, (iv). Corollary 3.4. Assumethat A = A(k) >_0 and thatv(k+ 1, k) • Ran A. Then thereexists a rank7-atomic positive measure/•, with suppIa = Z(g• ), the zero set ot'g•, suchthat 7j= / tJdlu (O_ Proof.' If A is invertible, the result follows from Proposition 3.3. Let r :- rank 7; if A is singular,then r <_ k,A(r - 1) is positiveand in- vertible(Theorem 2.4 (i)), and• = (•0,...,•-1) is theunique solution of A(r- 1)i, = (•,...,•2•-•). Wenow apply Proposition 3.3 to these- quence7(r) := (70,..-,72•-2), so that the role of A in Proposition3.3 is playedby A(r - 1) and the role of • in Proposition3.3 is playedby 7(r)~ = (70,... ,72•-•). Note,moreover, that g,r(•)~ = g•. Thusrroposi- •'--1 tion 3.3 yieldsa positivemeasure •u := Y•-i=0Pit•xi (where x0,..., X•_l are the distinctreal rootsof g•) suchthat (3.4) •j= j tjdlu (O_< j_<2r-1). The hypothesesand Lemma2.3 (ii) imply that A admitsa positiveHankel extension.• = A(k+ 1) with72k+• from •. Since.• >_0, Theorem2.4 (ii) implies that Startingwith (3.4),we use (3.5) to proveinductively that 7j= J tJ dl• (2r <_j_< 2k+l ). 616 R. E. CURTO AND L. A. FIALKOW Supposethat 2r - 1 •_j _•2k and7i = f tidy for0 _•i _•j. Then tJ+ldy=/tJ+l-rtrdy =/ tJ+l-•(q50+... + qS•_lt•-l)dy (sinceg• • 0 on suppy) ----(•0')'j+l-r -[- ''' -[- •r-17j ----')'j+l (by 0.5)). [] Remark 3.5. The precedingproof showsthat for any •,g• has rank 7 distinctreal roots, whichwe denotein the sequelby Z(g•) :- Lemma 3.6. If rank 7 •- k and y is a solutionof (3.1), then supp y C_ Proof: Let r := rank 7; then7j = q507j-•q-'-'q-•r-17j--1(T •__j <_r+k). Thus,by (3.1),f tJgsd•= 0, 0 •_j _•k. Now, -•jtJg•(t)dy-O(O•_j•_r-1) and Fg•(t)dy=0 since r < k. Thus / g•(t)2dy=f - (•o+... + •r-ltr-1))g•(t)dy =O, whenceg• = 0 in L2 (y). Sincey isa positiveBorel measure, it follows that suppy C_Z(g•). [] Proof of Theorem 3.1: (i)•(vi) Let y be a representingmeasure for if, let p(t) -- Po+ '" + Pktk(t • R) bea polynomialwith complex coefficients, and let •) - (p0,... ,pk). Then as in the discussionpreceding the proofof Proposition3.3, (A•),•))= S(Ip2), and(i)implies S(]p[ 2) = f p]2dy_> 0; thus A >_0. If A is invertible,then rank A = rank 7 = k + 1, so Theorem 2.6 ((ii)=• (i)) impliesthat A hasa positiveHankel extension of the form A(k + 1). If A is singular,then rank 7 <- k, so Lemma3.6 impliesthat TRUNCATED MOMENT PROBLEMS 617 supp/t C_Z(g•). Let •2•+2 := ft2•+2d/t< +0% andlet A(k + 1) := A(k + 1)(/•). We maynow extend the functional$ to polynomialsof degree upto 2k+ 2 by S(tj) := '/j(0 _ = i=0 heresupp • = {/0,...,/•-1} = Z(g•) andp := (p0,...,p•)is givenby p = V•-lvo,where x := (Xo,... ,xr-1). 618 R. E. CURTO AND L. A. FIALKOW (ii) If r = k + 1, then (3.1) hasinfinitely many solutions: Let (I>:= A(k)-lvk+l andlet c := T*Vk+l. If p is a representingmeasure for if, then 72k+2(P) _>c; moreover,there is a uniquerepresenting measure • with ft2•+2d•(t) = c, and• hask+l atoms.For each 7•+• > cand any 7•+3 E R,•/ := (70,... ,7•+3) hasa representingmeasure, and any such measurep satisfiescard (suppp) _>k + 2. Proof: (i) The proofof Corollary3.4 showsthat (3.6) is a representing measurefor if. Supposev is also a representingmeasure. Sincerank ? <_ k, Lemma 3.6 impliesthat supp v C_Z(g•); thus y is of the form y = Y•i=0r--1 eiSx•. Since 7j = f tJdy(0 _ vxe = ß = vxp, whencee = p and •: p. (ii) Suppose,is a representingmeasure of•. Weclaim that f t2k+•dp(t) k c. We may assumef t•+•dp(t) < q-•, and thuswe may define A(k q- 1) :: A(k q-1)(,) (usingff•+2 :: ft•+•dp(t)). Justas in the proofof Theorem3.1 ((i) =• (vi)), A(k + 1) _>0, soLemma 2.3 (ii) implies f t•+•dp(t)>(I)*vk-{-1 =C. Wenext show that there is a uniquerepresenting measure fi with f t2•+2dfi(t) = c. Let )•2•+2:=/t2•+2dp(t): c= •*V•+l =(/907k+1 q-''' q- Let Wo,..., Wk+l denotethe successivecolumns of the correspondingHan- kel matrix A(k + 1), so that Wk+ 1 -- •0W0 q- ... q- •kWk . Let72•+3 '---- 99o7k+2+ ... + 99•72•+2and let w•+2:= (7j}j=k+2'•2•+3 Finally, let½ :: (7o,..., 72k+3).Lemma 2.3 implies A(k + 1) >_0, andclearly w•+2 TRUNCATED MOMENT PROBLEMS 619 ß Ran A(k + 1). SinceA(k + 1) is singular,Theorem 3.1 andpart (i) above implythat hasa uniquerepresenting measure • and that card (supp •) = k + 1;in particular,f tak+ad•(t)= c. Lett• beany representing measure for (7o,.-. ,72k+1,c). SinceA(k + 1) is sin•lar, Le•a 3.7 impliesthat suppv • Z(g•). Thusf t2•+•d•(t) < +• andw•+2(v) ½ • A(k + 1) (A(k + 2)(v) is positive).Since w&+2 and w&+2(v) •e in R• A(k + 1), if f t2k+3d•(t)• 72k+3,then R• A(k + 1) containsthe fin• b•is vector ek+l := (0,...,0,1) ofRk+2. Since Wk+l • span(w0,...,w&), thereexist scalarsco,..., c&,not aH zero,such that e&+l = c0w0+-.. + c&w&,whence 0 = c0v0+ ... + c&v&,contradicting ra• 7 = k + 1. Thusf t2&+adv(t)= 72&+a,and v is a representingme•ure of •, whence(by (i) above)v = •. Thus• isthe unique representing me•ure of • forwhich f t2&+2d•= c, and c•d (supp•)= k + 1. Finally,it is cle• that if 72&+2> c, then A(k + 1) is positiveand invertible.Theorem 3.1 thus implies that foreach 72&+a • •, (70,-.., 72&+a) h• a representingme•ure; moreover,Le•a 3.2 imphesthat for each representingmeasure •, card(supp/•)_> rank (7o,..-,72&+2)= k + 2. [] We nowturn our attentionto the "evencase" of the Hamburgertrun- cated moment problem: (3.7) ?j= j tJdl•(t)(0<_ j _< 2/½). Theorem 3.9. (Existence,Even Case) For k > O,let 7 = (7o,..., 70 > 0. The followingare equivalent: (i) There existsa positiveBorel measure/• on R satisfying (ii) There existsa compactlysupported representing measure for ?; (iii) Thereexists a tinitelyatomic representing measure for 7; (iv) Thereexists a (rank7)-atomic representing measure for 7; (v) t(k) > 0 andrank A(k) = ,; (vi) A(k) hasa positiveHankel extension; (vii) ? is positivelyrecursively generated. 62O R. E. CURTO AND L. A. FIALKOW Versionsof this theoremappear in the literature. Shohatand Tamarkin establisha correspondencebetween solutions of the "even"moment problem and certain analyticfunctions with prescribedrational part [ShT, Theo- rem 2.2]. The casewhen A(k) is nonsingularis treatedin [AhK, Theorem 1.3] and [Ioh, TheoremA.II.1] usingmethods from the theory of quasi- orthogonalpolynomials. Additionally, [Ioh, Remark,p. 205] treats the casewhen rank A(k) = k. Proof: (i) =• (vi) As in the proofof Theorem3.1 (i) =• (vi), the exis- tenceof a representingmeasure p impliesA(k) •_ O. We considertwo cases, dependingon the value of r :- rank 7- If r •_ k, the existenceof a posi- tive Hankelextension follows from Theorem 2.6 (sincerank A - rank 7)- If r ( k, let ff :- (70,--.,72•-1); sincep interpolatesif, by Theorem 3.8 (i), p is the uniqueinterpolating measure, and thuscoincides with the finitelyatomic measure produced in (3.6). Thus,7•+• :- f t2•+idPand 7•+• :- f t2•+2dPare finite and, as in theproof of Theorem 3.1 ((i) =•(vi)), A(k + 1)(p) is a positiveSankel extension of A(k). (vi)**(v) Theorem2.6. (vi)•(iv) SupposeA(• + x) is a positiveHankel extension of A(k). Then Lemma 2.3 impliesA(k) •_ 0 and v(k -[- 1,k) 6 Ran A(k). Let r :- rank 7- Theorem3.1 ((v)=y(iv)) impliesthat ff :- (70,---, '•2k-I-1)has an r-atomic representingmeasure, and this is clearly a representingmeasure for 7. (iv)=• (iii)=•(ii) =•(i) Trivial. (v)•(vii) Remark2.7. [] Theorem $.10. (Uniqueness,Even Case)Let 7 = (%,...,72k),7o > 0. Suppose7 has a representingmeasure, i.e., A(k) >_0 and rank A(k) = rank 3/. (i) Supposer := rank7 <- k. Let •I, := •I,(7) andlet 72k+• := P072k+•-r + "' + Pr-•72k. Then V := has the uniquerepresenting measure or (3.6), whichis alsothe uniquerepresenting measure of 7. (ii) Supposer = k + 1. For each72k+• E R, let • = (7o,--.,72•+•)- Then • has infinitelymany representingmeasures (described by Theorem 3.8 (ii)) and eachis a representingmeasure oœ 7- TRUNCATED MOMENT PROBLEMS 621 Proof: (i) Theorem3.9 impliesthat A(k) hasa positiveHankel extension, and Theorem 2.6 implies that in any such extension, 72k+x is given recur- sivelyvia •(7). If it is a representingmeasure for 7, then by Lemma3.7, suppit is finite,whence 72k+•(it) := f t2•+•dit< c•. SinceA(k + 1)(it)is a positiveHankel extension of A(k), we musthave 72k+•(it) = 72k+•;thus it is a representingmeasure for the "singularodd" system•. By Theorem 3.8(i), it must coincidewith the measurein (3.6). (ii) Apply Remark2.7 and Theorem3.8(ii). [] We conclude this section with our elementary algorithm for solv- ing the truncated Hamburger problem. Indeed, Theorems 3.1 and 3.9 lead to a simple iterative procedure for determining whether or not 7 :- (70,..., 7-•), 70 > 0, admitsan interpolatingmeasure. Let dn := det A(n), 0 _ 4. The Truncated Hausdorff Moment Problem. Let a < b; given m >_0 and 70,. ß., %• E [•, 70 > 0, we seeknecessary a•nd sufficient condi- tions for the existenceof a positive Borel measure it such that 622 R. E. CURTO AND L. A. FIALKOW (4.1) ?j=f tJdl•(t),(O_ (4.2) supp•u c_ [a, b]. We first considerthe casem = 2k + 1 for somek ) 0. Let B(m) := (?i+j+•)i,"•=o,0 _< m _ Remark 4.2. In [KrN, TheoremIII.2.4], Kreinand Nudel'man prove that (i) is equivalentto (•) •(•) _>•(•) _>a•(•). In casea = 0, b - 1, the sameresult is givenin [Akh, p. 74]. We know that (v)is equivalentto (iv) (becauseeach condition is equivalentto (i)); it is easyto showdirectly that (v)=• A(k) >_O, but we do not knowa direct proofthat (v)=• v(k + 1,k) E Ran A(k). Indeed,we have never seen "range conditions"of this kind in the literature of truncated momentproblems. Proof of Theorem 4.1: (i)•(iv) Suppose/•is a positiveBorel measure satisfying(4.1) and (4.2). Theorem3.1 impliesthat A :- A(k) >_0 and v(kq- 1,k) E RanA. For•) = (P0,'"• pk)• ck-{-1, let p(t) = •j=oPJ• t j (t ½ a). We have k k i,j=O i,j=O TRUNCATED MOMENT PROBLEMS 623 dl•- b f p2dl• _•f t[p[2dl• k -- E 7i+j+lPi•--(B•),•)); i,j=o thus bA > B. A similar calculation shows that B > aA. (iv)•(iii) The conditionsA _>0 and v(k + 1, k) • Ran A, together with Theorem 3.1, imply that there exists a representingmeasure p for • suchthat supp p = Z(g•) and card (suppp) = rank 3'. Let p(t) = Po+ "' + pktk, with i5• Ck+l. SincebA - B >_O, it followsas above that f(b--t)lP12dl•= ((bA-B)•,•) >_O. Supposethat thereexists to E suppp suchthat to > b. Thencard (supp •u ffl (-oo, b]) <_( rank 7)- 1 <_k. Thus, by Lagrangeinterpolation, there is a polynomialP0 with degP0 _ (iii) There existsan r-atomic representingmeasure p /or 7 with suppp _C[a, b]; (iv) A(k) >_ 0 and there exists72k+, e R suchthat v(k + 1, k) e va() > > Remark 4.4. Kreinand Nudel'man proved in [KrN, Theorem11.2.3] that (i) is equivalentto (v) A(k) > 0 and (a + b)B(k - 1) > abA(k- 1) + C, whereC := (')'i+j)i,j=l'Of course,this condition ismore concrete than our condition(iv), but the proofdepends on the Maxkov-Luk•csRepresenta- tion Theoremfor polynomialsthat are nonnegativeon [a,b]; our proofof (iv)=•(i) doesnot usethis theorem.The casewhen a = 0, b = 1, is treated in [Akh, p. 74]. ProofofTheorem4.3: (i) =• (iv)Since supp p C_[a, b], 3'2kq. 1 := f t2&+l dp is finite, and p is a representingmeasure for • := (70,..-, 7•&+1);(iv) now followsfrom Theorem4.1 ((i) • (iv)). (iv)=•(iii) Theorem4.1 imphesthat • hasa representingmeasure it whichis r-atomicand satisfiessupp p C_[a, b]; clearly, p is alsoa represent- ing measurefor 7. (iii)=l•(ii) =• (i) Trivial. [] Remark 4.5. (Uniquenessin the TruncatedHausdorff Problem) For the singular case, if there is a solution, it is unique. Indeed, if p and •, interpolate3' = (70,--., 3'm),have supports inside [a, b], and if rank 3' < k, then p and v are measureson (-0% +o•) whichinterpolate 3', so from Theorem3.8 or Theorem3.10, p = v. For the nonsingularcase, we do not know when uniquenessholds. The issueseems to be whether the matrix inequalitiesof Theorem4.1 (iv) and Theorem4.3 (iv) can be extendedto A(k + 1) and B(k + 1) for somechoices of 3'2•+2and 3'2k+3.We havenot seenuniqueness for the truncatedHausdorff problem treated in the standard references. The resultsof this section(and their proofs)show that whenthe Hausdorffmoment problemis solvable,i.e., the conditionsof Theorem 4.1 (iv) or Theorem4.3 (iv) aresatisfied, the algorithmat the endof Section3 can be used to producea solution. 5. The Truncated Stieltjes Moment Problem. We considerthe trun- TRUNCATED MOMENT PROBLEMS 625 cated Stieltjes moment problem (5.1) 3'j=J tJdla (0_• j _• m) and (•.e) supp• c_[o, +•). Our first resultfollows from Theorem4.1 (and its proof) by letting a = 0 and b -• +oc; we omit the details. Theorem 5.1. (Existence,Odd Case)Let q = (3'0,.-., 3'2•+1),3'0 > 0, and let r := rank 3'. The œo//ow/ngare equivalent: (i) Thereexists a positiveBorel measure/• satisfying (5.1) and (5.2); (ii) Thereexists a finiteIFatomic representing measure I• for • satis- fying (5.2); (iii) Thereexists an r-atomicrepresenting measure/• for • satisfying (5.2); (i•) •(•) > 0,•(•) > 0 • •(• + 1,•) e • •(•). Krein and Nudel'man treat the classical truncated Stieltjes moment problem(1.3) [KrN, p. 175]. Moreover,they showthat the "odd" case of the truncatedStieltjes problem (1.2) has a solutionif A(k) and B(k) are positiveand invertible;of coursethis is a specialcase of (iv)•(iii) in Theorem 5.1. Theorem 5.2. (Uniqueness,Odd Case)Let • = (3'o,..., 3'2k+1),3'0 > 0, and assumethat •/ has a representingmeasure for the Stieltjes problem, i.e., A(k) > 0, B(k) > 0, •na v(• + •,•) e Ran A(•). •:he• uniquerepresenting measure if andon/y if eitherA(k) is singuJar,or A(k) is nonsingularand B(k) is singular. Proof.' Let/• be a representingmeasure for •. Supposefirst that A(k) is singular;since any representingmeasure for • is a solutionof the "singular odd" Hamburgerproblem for •, then Theorem3.8 (i) showsthat/• is the uniquesolution of this Hamburgerproblem, whence/• is alsounique for the Stieltjesproblem. Suppose next that bothA(k) andB(k) areboth positive and nonsingular.By Proposition2.3(ii), for suitablylarge choices of and 3'2•+3,A(k+ 1) and B(k + 1) are positiveand nonsingular.Thus, by 626 R. E. CURTO AND L. A. FIALKOW Theorem5.1, the Stieltjesproblem with data (70,.-., 72•+a)has a solution; since72•+2 is essentiallyarbitrary, we obtain infinitely many solutions. Finally,we seekto showthat if A(k) is nonsingularand B(k) is sin- gular,then it is the uniquesolution. Since A(k) is nonsingularand B(k) is singular,rank B(k) = k, and, moreover,rank (7•,..-,7•+•) = k. In par- ticular, there existsunique scalars •,..., •, suchthat •v(1, k) +.-. + ;•v(•,•) = v(• + 1,•). Theorem2.6 ((ii)• (i)) showsthat B(•) has a positiveHankel extension, and in any suchextension B(k + 1), 7•+• is uniquelydetermined by 7•+2 = •7•+• + "' + •72•+• (Remark2.7). It follows that v(k + 1,k + 1)= 99xv(1,k + 1) + .- - + •p•v(k,k + 1), whenceA(k + 1) is singular.We claimthat it has momentsof all orders. Without lossof generality,assume that it y• 6o, and let dy(t) := tdit(t); clearly,y is a solutionof the Stieltjesproblem with data (7i,-.-,7a•+l)- SinceB(k) is singular,Lemma 3.7 imphesthat y is finitelyatomic, hence it hasmoments of all orders.Thus, in B(k + 1)(it), wemust have 7a•+a(it) := f teA+edit= 72•+•- Thus it solvesthe "singular even" Hamburger problem with data (70,.--, %•+•). By Theorem3.10(i), we concludethat it is the unique solution of the Stieltjes problem. [] Theorem 5.3. (Existence,Even Case)Let 7 = (7o,..., 72•), 70 > 0, and let r := rank 7. The followingare equivalent: (i) Thereexists a positiveBore1 measure it satisfying(5.1) and (5.2); (ii) Thereexists a iinitelyatomic representing measure it for 7 satis- fying(5.2); (iii) Thereexists an r-atom/crepresenting measure it/'or 7 satisfying (5.2); (iv) A(k) _>0, B(k- 1) _>0, andv(k + 1,k- 1) e Ran B(k - 1). Krein and Nudel'man proved that the "even" case of the truncated Stieltjesproblem has a solutionwhen A(k) and B(k- 1) axepositive and nonsingular[KrN, p. 175];this is a specialcase of (iv)=•(iii) of Theorem 5.3. Proof: (i)=• (iv) As in the proofof Theorem3.1 ((i)=• (vi)), the existence of a representingmeasure it imphesA(k) >_O. Also,if p(t) = Po+"' + p•_•t•-•, then (B(k- 1))t), t))= f tp(t)12dit >0; TRUNCATED MOMENT PROBLEMS 627 thus B(k- 1) _>0. We next showthat B(k- 1) has a positiveHenkel extension.Without lossof generality,assume that/• is not the point mass at the origin.Define a positiveBorel measure • by d•(t) := tdl•(t).• solves the "odd" Hamburgerproblem with data (3`•,... ,3`2k). By Theorem3.1 ((i)=•(vi)), B(k- 1) hasa positiveHenkel extension using 3`2k. By Propo- sition2.3(i) we concludethat v(k q- 1,k- 1) ß Ran B(k- 1). (iv)=•(iii) SinceB(k- 1) >_0 and v(k + 1,k- 1) ß Ran B(k- 1), we may define72•+1 recursivelyso that B(k) > 0 and v(k + 1, k) ß span{Vl,...,v•} C_ Ran A(k). Theorem5.1 nowimplies that ff has an r-atomic representingmeasure tt with supptt C_[0,+oo), and tt also represents 3'. The other implicationsare trivial. [] Theorem 5.4. (Uniqueness,Even Case)Let -/ = (70,..-,%u),% > 0. Assume that 3' has a representingmeasure for the Stieltjes problem, i.e., A(k) k O, B(k -1) k 0, v(k+l,k-1) ß RanB(k-1). Then7 hasa uniquerepresenting measure if and onlyif A(k) is singular. Proof.' Let •ube a representingmeasure for 7. If A(k) is singular,•u is a solutionof a "singulareven" Hamburgerproblem, so uniquenessfollows fromTheorem 3.10(i). SupposeA(k) is nonsingular.The hypothesesimply that B(k- 1) has a positiveHenkel extension B(k), and sinceA(k) is nonsingular,the Stieltjesproblem with data (%,... ,72•+1) has a solution (Theorem5.1). Now -/2k+•is essentiallyarbitrary, so -/has infinitelymany representingmeasures. [] The resultsof this section(and their proofs)show that when the Stieltjes moment problem is solvable,i.e., the conditionsin Theorem 5.1 (iv) or Theorem5.3(iv) are satisfied,the algorithmat the endof Section3 can be used to produce a solution. 6. Positive Toeplitz Matrices. In this sectionwe givea simpledescrip- tion of positiveToeplitz matrices,along the linesof the descriptiongiven in Section2 for positiveHenkel matrices. Let "/= (7-•, ßß ß, 7-1,70,71,..-, begiven, assume that ?_j -•j(j - 1,..., k) andthat 70 > 0, andconsider 628 R. E. CURTO AND L. A. FIALKOW the Toeplitz matrix ')'•--•. ')'•- ß ß ß ')'k ')'r- 2 7r- 1 ß ß ß ')'k- •. := 7o ')'1 ß ' ß ')'k+•.-r 7-1 70 "' 7k-r 7-k 3/-k+r-1 7-k+v ' ' ' 70 By analogywith the definition of rank for Hankel matrices, we define the (Toeplitz)rank of 7 asfollows: If T(k) is invertible,then rank 7 = k + 1. If T(k) singular,then rank 7 is the smallestinteger r, 1 < r, suchthat w• = w(r,k) := (7•,..-,7•-k) ½ span{w0,...,w•_l}. Then {Wo,...,w•_•} is linearlyindependent, so there exists a unique E C • suchthat w• = ;o0w0+ ... + ;or_•w•_•, i.e., 7j=qo07j-•+"'+qo•-•7j-• (r-k The results of Section 2 show that the Hankel rank of a finite collection of real numbersis intimately related to the rank of its associatedHankel matrix. In particular, for 7 = (70,..., 72k)E !t2k+l , 70 =fi 0, r := rank7 and A _..A,• _..A(k) ;'--(7i+j )ik, j=O, Lemma2.1(ii) impliesthat A(r-1) is alwaysinvertible. By wayof contrast, the next exampleshows that the analogousresult doesnot hold for Toeplitz matricesß Example 6.1. Let 7 = (x, 2, 1,2, 1, 2, 1, 2, x), x =fi1, so that 1 2 1 2 x 2 1 2 1 2 T := T(4)= 1 2 1 2 1 2 1 2 1 2 x 2 1 2 1 TRUNCATED MOMENT PROBLEMS 629 Sincex • 1, 2 I 2 • 0, so •w0,wl,w•) are linearlyindependent. Since x21121[ w3-w•,itfollowsthatrank7=3. On the other hand,T(2)- 2•2 , 121 which is singular. Note that T(1) hasnegative determinant, so T is not positivesemi- definite.We will showin the sequelthat if T _>0, then T(r - 1) is positive emd invertible. First we establishthe recursire structure of a positive sin- gular Toeplitz matrix. Our next propositionalso shows that in the context of Toeplitz matricesthere is a "remkprinciple" similar to the one studied in Section 2 for Hemkelmatrices, but this time we require positivity. Proposition 6.2. Let 7 := (?-k,...,70,...,?k) and T(k) be as above, assumethat T(k) is singularand positive,let r :- rank 7, and let ß := •(7). Then œorr - k _• j _• k, (6.1) 7j = •0o7j-,-+ .-. + •o,--•7j-•; moreover,rank T (k) - r. Proof: Write T(k)=(T(k-1) w(k,k- 1)* w(k,k- 70 1)) ' wherew(i,m):-(7i_j)jm__o . Since T(k) •_O, there exist scalars Co,... ,c•_• such that ?m --- CO?m--k-•- ''' -•-Ck-l•m-1,1 •__m •_ k (by Lemma2.3 (i)). By definitionof r, (6.1) holdsfor r- k <_j <_r. Supposethat (6.1) holdsfor r - k •_j •_p wherer _•p _• k - 1. Then --' C07p4-1-k-• ''' -• Ck-17p = co[•o7p+1-•-•+-.. + •-•Tp-•] +--' + c•-1[•o7p-• +.--+ •-17p-•] = •oo[coTp+•_k_•+-.. + c•_•Tp_•]+... + •o•_1[co7p_•+... + c•_•%_•] -- •Offp-r+l • "' • •r-lffp. Thus, using• inductive•gument, (6.1) holdsfor r - k • j • k. It now followsby anotherinductive argument that wj • sp• {Wo,..., w•_• ) for • • j • •, wne.ce •=k T(•) • •. •.t si.ce {wo,..., W•_l} is U.e•ly independent,we obtainrank T(k) = r. • The next resultis • •o•e of Le•a 2.1 (ii). 630 R.E. CURTO AND L. A. FIALKOW Proposition 6.3. Let %T(k), and r be as before,and assumethat T(k) > O. ThenT(r - 1) is nonsingular. Proof: We mayassume r < k. We seekto provethat (w(0,r- 1),..., w(r- 1,r- 1)} islineaxly independent. Suppose that cow(O, r- 1)+--. + Cr_lW(r -- 1, r - 1) = 0. We claimthat CoW(0,k) -+-...-+-Cv_lW(r- 1,k) -0. We prove this by induction. Supposethat (6.2) c0w(0,j) +---+ C•_lw(r - 1,j) = 0 for somej, r - I < j < k. We'll showthat c0w(0,j + 1)+-..+ Cv_lw(r- 1,j + 1)= 0; for this, it sut•ces to prove that CO•--j-1•- ''' -•-C•-1•-j-1+•-1 -- 0. By Proposition 6.2, we know that •m = •0•m--v -•-'' '-•- •--1•m--1, r -- k _• m < k, and we alsohave V-m = Vm, 0 < m < k. Thus CO•-j-1 •- '' ' •- C•-1•-j-1+•-1 -- C0•--j+1 -+--..-+-C•_i•j+i_•+ 1 -- CO[•OVj-]-l--v -]-'''-]- •v--lVj]-]- ' ' ' -+-Cv--l[•0•+2--2v -•-'''-+- •v--l•j--v+l] -- •0 [C0•j+l--v-•-'''-•- Cv--l•j+2--2v]-•-' ' ' -•-•v_l [C0•j-•-'''-•- Cv_l•j_v+l] ----•0 [C0•_j_l+v-•-'-'-•- Cv_l•_j_l+v+(v_l)] -•-'''-•-•v_l[C0•_j -•-'''-•-Cv_l•_j+v_l] -- 0 by (6.2). Thusc0w(0, j + 1)+... + C,lw(r- 1,j + 1)= 0. By induction, COW(0,k) -•-'''-•- Cv_lW(r-- 1,k) = 0, whence co = -.- = c,1 = 0 by the definition of rank "/. Thus {w(0, r - 1),...,w(r - 1,r- 1)) is lineaxlyindependent, and therefore T(r - 1) is nonsingular. [] We next presentour structuretheorem for positiveToeplitz matrices. TRUNCATED MOMENT PROBLEMS 631 Theorem 6.4. Let '7, T(k), r be as beœore,and assumethat T(k) is sin- gular and positive. Then (i) T(r - 1) is positiveand invertible,and rank T(k) - r; (ii) • := •(?) satisfies ?j=•o0?j_•+...+•o•-x?•-l, (r-k_•j_•k). Conversely,if there exist r, 1 _• r _• k, and scalars•oo,..., •o•-1, such that T(r - 1) •_ 0 and (ii) holds,then T(k) is positive. Proof.- Apply Propositions 6.2 and 6.3. For the converse,apply Lemma 2.3. [] Remark 6.5. Notice that in the abovemodel, every?j is uniquely determined;in the Hankelcase, the entry72k+2 was free (cf. Theorem2.4). Becauseof this, the (linearalgebraic) rank of a Toeplitzmatrix equalsour (recursive)rank. In the Hankelcase, we onlyhad r _• rankA(k) _• r + 1. We now focus on extensionsof positive Toeplitz matrices. In the sequel,to emphasizethe dependenceon the data ?, we denoteT(k) by Tv. Corollary 6.6. Let ? and Tv be as before,assume that Tv _• 0 and that Tv is singulax.Then T v admitsa uniquepositive Toeplitz extension. Proof.' Firstwe establish uniqueness. For •k+l • C, let • = (•+1, ?-•,-.-, 70,... ,?•,•k+l), and assumethat T• is positive.By Proposition6.3 and Lemma2.1 (i), rank• - min {p' T• (p) is singular) and,since T v is singu- lax and positiveit followsthat rank • - rank ?. Let r - rank ?, 1 _• r _• k. Thus,T•(r- 1) - Tv(r-1 ) is invertible,and so •(•) - •(?). SinceT• _• 0, Proposition 6.2 implies that (6.3) Thus •+1 is uniquelydetermined. ASfor existence,let •k+l be definedby (6.3), where(qo0,..., qO•-l) := ß (?). Then Theorem6.4 (ii), appliedto •,T(k + 1) and r, showsthat w(k+ 1,k) satisfiesw(k+ 1,k) = T(k)•I,, where•I, = (0,..., 0, •o0,..., •o•_1) E Ck+l. Moreover,Theorem 6.4 (ii) alsoimplies that •I,*w(k + 1,k) = •o?• + '"+ •-171 = (•07-• + '" + •,-17-1)- = 7o = 70, 632 R. E. CURTO AND L. A. FIALKOW so Lemma2.3 impliesat oncethat T(k + 1) _• 0. [] Remark 6.7. An alternative proof of Corollary 6.6 can be based on Theorem6.4 (i) and [Ioh, Theorem13.2] (observe that p in [Ioh] is the usualrank). Supposenow that T• is positiveand invertible. Let let and set Proposition 6.8. Assumethat T• is positiveand invertible,and let • be definedas above.Then T• is positiveand invertible. Proof: SinceT• is positiveand invertible,Lemma 2.3 impliesthat 70 > (I,*w(k, k- 1)= •o% + "' + •k71 (= •O•k -• '''-• •k--1•1)--). Let u = (0,•o0,...,•Ok_l),so that Tvu = •r(k + 1,k)= (•k+l,7k,. -. ,71). Then u*T•u = •*w(k, k - 1) < 70: (T•)k+2,u+2. SinceT• is positiveand invertible, it nowfollows from Lemma 2.3 that is positive and invertible. Corollary 6.9. Assumethat T• is positiveand invertible. Then T• admits infinitely many positive and invertible Toeplitz extensions. Proof:For A E C, let • = (•,%•,...,70,...,7•,A). SinceT• is posi- tive and invertible,there exists5 > 0 suchthat if [A- %+•[ < 5, then det TX > 0, whencethe nesteddeterminants criterion implies that TX is positive and invertible. [] The problemof producinga singularpositive Toeplitz extensionof a nonsingularpositive Toeplitz matrix is surprisinglydifficult, and the solu- tion entails Sylvester'sformula for the determinant of a bordered matrix [Ioh, p. 98]. We nowproceed to applythe modelfor positiveToeplitz matri- cesto theproblem of interpolating Toeplitz data. Let 7 - (7-k,..., 70,..., 7•) E C•k+x,70 > 0,be given, where 7-j - 7'-jfor all j, andlet us seek a repre- sentingmeasure for 7, that is, a positiveBorel measure/z with supp/z C_T andsatisfying 7j = f tJdl•0 _ Proposition 6.10. Assumethat ? admits a representingmeasure. Then T,>O. Proof.-For a = (no,...,ak), let a(z) = ao+... + akzk. Then la(z)l• - k ' ' •]]i,j=oaia7z'-3, z &T. If p isa representingmeasure for 7, then k k i,j=0 i,j=O For the next result,we let r := rank 7 <- k, • := •(7), and we define g,(•) := z•-(•o+...+•_•-•). Corollary 6.11. Assumethat 7 admitsa representingmeasure p and that rank 7 -< k. Then suppp C_Z(g,). Proof: We definethe hnear functionalR as follows:For a trigonometric polynomialh(z) = a_kz-k +... + a0+... + akzk, let R(h) := a-k7-k + ß-. + ao•o+... + a•. •f a(z) = ao+... + a•z•, •(•) = •o+... + •z •, a := (no,...,ak), and f, := (0o,...,0k), we then have Also, R(g,) = 7r - (990+'" + •9r--l•r--1): 0, and, fromthe definitionsof rank 7 and R(g,7j) = O (r - k _ Thus,gv is R-orthogonalto zk-r,..., z, 1,z-•,..., z-k, whichimphes that S(gvy-•,= 0, andthis in turn gives (T(k)•v, •) = 0. Therefore,f Igvl2dp = 0, or g.•= 0 a.e. [p],so that supp• C_2(g.•). [] We are nowready to solvethe truncatedtrigonometric moment prob- lem. Thefollowing result is yerysimilar to [AhK, TheoremI.I.12] and to [Ioh, p. 211].Unlike these authors, we avoid Gaussian quadrature in treat- ing the singularcase, and instead we use the recursivemodel of Theorem 6.4. 634 R. E. CURTO AND L. A. FIALKOW Theorem6.12. Let'= ('-k,... ,'o,... ,'k) • C2k+1, 70 > 0,7-j -- •j, be given. Then there existsa representingmeasure/• for ? if and only if T• _• O. In this case,p canbe chosento haver atoms,where r := rank 7- Proof. Case 1: (r _• k) If 7 has a representingmeasure, the conclusion followsfrom Proposition6.10 and Corollary 6.11. Conversely,if T• •_ 0, we apply Theorem6.4, [Ioh, p. 210], and the methodof the proof of Proposition3.3 and Corollary 3.4 to producea representingmeasure with r atoms. Case 2. (r = k + 1) As above,if ? has a representingmeasure, then Tv _• 0. Assumingthat Tv is positiveand invertible,we ca• apply [Ioh, Theorem13.1 and Remark 1] to obtaina singularHermitian extension T(k + 1) of Tv. By Lemma2.3, we see at oncethat T(k + 1) •_ 0. Moreover, T(k + 1) is singular,so we ca• apply Case I to producea representing measure with r atoms. [] Remark 6.13. As in Section 3, it is not difficult to show that in the nonsingularcase there are infinitely many solutions. For the singular case, we claim that there exists a unique solution. Let /• a•d y be two interpolating measures. Observefirst that /• and y have momentsof all orders,and we can then defineT,(j) and T,(j) for all j _>0. However, it followsfrom Corollary6.6 that T,(j) -- T,(j) for all j •_ 0. Thus, f zJdl•= f zJdyfor all j _• 0. The F. andM. RieszTheorem implies that Borel measureson T are determined by their trigonometric power moments,so we must have/• - y. (Forthe classof finitelyatomic measures, uniquenessis provedin [AhK, Theorem1.1.12].) REFERENCES [AhK] N. I. Ahiezerand M. Krein, SomeQuestions in the Theoryof Moments,Transl. Math. Monographs2 (1962), American Math. Soc., Providence. [Akh] N. I. Akhiezer, The ClassicalMoment Problem,Hafner Publ. Co. (1965), New York. latz] A. Atzmon, A momentproblem for positivemeasures on the unit disc,Pacific J. Math. 59 (1975), 317-325. [Ber] C. Berg, The multidimensionalmoment problem and semigroups,Proc. Sym- posia Appl. Math. 37 (1987), 110-124. [BeM] C. Berg and P. H. 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Curto Department of Mathematics The University of Iowa Iowa City, Iowa 52242 [email protected] Lawrence A. Fialkow Dept. of Math. and Comp. Science SUNY at New Paltz New Paltz, NY 12561 fia•kowœ@snynewvm.bitnet ReceivedAugust 5, 1991