The truncated Hamburger moment problems with gaps in the index set
aa r X i v : . [ m a t h . F A ] J u l THE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET
ALJA ˇZ ZALARA
BSTRACT . In this article we solve four special cases of the truncated Hamburger moment problem (THMP) of degree k with one or two missing moments in the sequence. As corollaries we obtain, by using appropriate substitutions, the solutionsto bivariate truncated moment problems of degree k for special curves. Namely, for the curves y = x (first solved byFialkow [Fia11]), y = x , y = x where a certain moment of degree k + 1 is known and y = x with a certain momentgiven. The main technique is the completion of the partial positive semidefinite matrix (ppsd) such that the conditions ofCurto and Fialkow’s solution of the THMP are satisfied. The main tools are the use of the properties of positive semidefiniteHankel matrices and a result on all completions of a ppsd matrix with one unknown entry, proved by the use of the Schurcomplements for × and × block matrices.
1. I
NTRODUCTION
For x = ( x , . . . , x d ) ∈ R d and i = ( i , . . . , i d ) ∈ Z d + , we set | i | = i + . . . + i d and x i = x i · · · x i d d . Given a real d -dimensional multisequence β = β (2 k ) = { β i } i ∈ Z d + , | i |≤ k of degree k and a closed subset K of R d , the truncatedmoment problem (TMP) supported on K for β asks to characterize the existence of a positive Borel measure µ on R with support in K , such that(1.1) β i = Z K x i dµ ( x ) for i ∈ Z d + , | i | ≤ k. If such measure exists, we say that β has a representing measure supported on K and µ is its K - representing measure .We denote by M ( k ) = M ( k )( β ) = ( β i,j ) ki,j =0 the moment matrix associated with β , where the rows and columnsare indexed by X i , | i | ≤ k , in degree-lexicographic order. Let R [ x ] k := { p ∈ R [ x ] : deg p ≤ k } stand for theset of polynomials in d variables of degree at most k . To every p := P i ∈ Z d + , | i |≤ k a i x i ∈ R [ x ] k , we denote by p ( X ) = P i ∈ Z d + , | i |≤ k a i X i the vector from the column space C ( M ( k )) of the matrix M ( k ) . Recall from [CF96], that β has a representing measure µ with the support supp µ being a subset of Z p := { x ∈ R d : p ( x ) = 0 } if and only if p ( X ) = 0 . We say that the matrix M ( k ) is recursively generated (rg) if for p, q, pq ∈ R [ x ] k such that p ( X ) = 0 , itfollows that ( pq )( X ) = 0 . The full moment problem (MP), where β i is given for every i ∈ Z d + , being the classical question in analysis and alsodue to its relation with real algebraic geometry via the duality with positive polynomials given by Haviland’s theorem[Hav35], has been widely studied, see e.g., [Akh65, AhK62, KN77, Las09, Lau05, Lau09, Mar08, PS06, PS08, Put93,PV99, Sch91, Sch03, Sch17]. The TMP, which is more general than the full MP [Sto01], has been intensively studied ina series of papers by Curto and Fialkow [CF91, CF96, CF98a, CF98b, CF02, CF04, CF05, CF08] with the celeberatedflat extension theorem they established being the main tool of their research. Recently, Fialkow’s core variety [Fia17]approach led to many new results on the TMP; see also [BF+, DS18]. A concrete solution to the TMP is a setof necessary and sufficient conditions for the existence of a K -representing measure. Among necessary conditions, M ( k ) must be psd and rg [CF91, CF98b], which also suffice in some cases. Concrete solutions to the TMP are knownin the following cases:(1) (Truncated Hamburger moment problem (THMP)) d = 1 and K = R . See [AhK62, Theorem I.3] or [Ioh82,Theorem A.II.1] for the special case of even k with an invertible moment matrix and [CF91, Section 3] for thegeneral case. F ACULTY OF C OMPUTER AND I NFORMATION S CIENCE , U
NIVERSITY OF L JUBLJANA , V E ˇ CNA POT
JUBLJANA , S
LOVENIA
E-mail address : [email protected] . Date : July 21, 2020.2010
Mathematics Subject Classification.
Primary 47A57, 47A20, 44A60; Secondary 15A04, 47N40.
Key words and phrases.
Hamburger moment problem, truncated moment problems, representing measure, moment matrix.Supported by the Slovenian Research Agency grant P1-0288. (2) (Truncated Hausdorff moment problem) d = 1 and K = [0 , ∞ ) . See [KN77, p. 175] for the special case of aninvertible moment matrix and [CF91, Section 5] for the general case.(3) (Truncated Stieltjes moment problem) d = 1 and K = [ a, b ] , a < b . See [KN77, Theorems III.2.4 and II.2.3]and [CF91, Section 4] for the general case.(4) d = 2 and K is a curve p ( x, y ) = 0 with deg p ≤ . See [CF02, CF04, CF05, FN10, Fia14, CS16].(5) d = 2 and K is a curve y = x . See [Fia11].(6) d = 2 and the moment matrix has a special feature called recursive determinateness . See [CF13] for details.(7) (Extremal case) The rank of the moment matrix is the same as the cardinality of the corresponding variety; see[CFM08].(8) Some special cases are solved in [CS15, Fia17, Ble15, BF+].In (5), β must satisfy certain numerical conditions, which are equivalent to the conditions from Corollary 3.3 below.The proof is by separating the nonsingular case from the singular one. In the nonsingular case the existence of a flatextension is established by a detailed and technically demanding analysis, while the singular case is done by the use ofadditional features of the moment matrix such as recursive determinateness and known results for such matrices.In this article we present concrete solutions to the four cases of the THMP of degree k with some unknownmoments β i , . . . , β i j , ≤ i ≤ · · · ≤ i j ≤ k − , in the sequence, which we call the THMP with gaps ( β i , . . . , β i j ) .Namely, we solve the THMP with gaps ( β k − ) , ( β k − , β k − ) , ( β ) and ( β , β ) . The motivation to solve this casesof the THMP with gaps is to obtain the solutions to the special cases of the 2-dimensional TMP. Namely, the solutionof the THMP with gaps:(1) ( β k − ) gives an alternative solution to the TMP with d = 2 and K being the curve y = x (see (5) above).The advantage of our approach is that the proof is short and we also do not need to separate three subcases,i.e., k = 1 , k = 2 and k ≥ .(2) ( β k − , β k − ) solves the TMP with d = 2 , K being the curve y = x and in addition the moment β , k − ofdegree k + 1 is known. To solve the TMP for the curve y = x without this additional moment, one needs tosolve the THMP with gaps ( β k − , β k − , β k − ) which is a possible topic of future research.(3) ( β ) solves the TMP with d = 2 and K being the curve y = x .(4) ( β , β ) solves the TMP with d = 2 , K being the curve y = x and known β , . By β , we mean themoment of x , i.e., R K x dµ . To solve the TMP for the curve y = x without this additional information,one needs to solve the THMP with gaps ( β , β , β ) , which is another open question for future research.1.1. Readers Guide.
The paper is organized as follows. In Section 2 we present the tools used in the proofs of ourmain results: • Generalized Schur complements and verification of positive semidefiniteness of block matrices (Subsection2.1). • Properties of psd Hankel matrices (Subsection 2.2). • The solution to the THMP (Subsection 2.3). • A result about psd completions of partial psd matrices with one unknown entry (Subsetion 2.4). • An extension principle for psd matrices (Subsection 2.5). • A result about subsequences of moment sequences (Subsection 2.6).In Section 3 we solve the THMP of degree k with gaps ( β k − ) (see Theorem 3.1) and ( β k − , β k − ) (see Theorem3.5). Corollary 3.3, being a special case of the ( β k − ) -case, is the solution to the TMP with d = 2 and the curve y = x as K , while Corollary 3.6, being a special case of the ( β k − , β k − ) -case, is the solution to the TMP with d = 2 , the curve y = x as K and an additional moment β , k − known.In Section 4 we solve the THMP of degree k with gaps ( β ) (see Theorem 4.1) and ( β , β ) (see Theorem 4.5).Corollary 4.4, being a special case of the ( β ) -case, is the solution to the TMP with d = 2 and the curve y = x as K ,while Corollary 4.7, being a special case of the ( β , β ) -case, is the solution to the TMP with d = 2 , the curve y = x as K and an additional moment β , known. 2. P RELIMINARIES
In this section we present some tools which will be needed in the proofs of our main results in Sections 3 and 4.
HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 3
We write M n,m (resp. M n ) for the set of n × m (resp. n × n ) real matrices. For a matrix M we denote by C ( M ) itscolumn space. The set of real symmetric matrices of size n will be denoted by S n . For a matrix A ∈ S n the notation A ≻ (resp. A (cid:23) ) means A is positive definite (pd) (resp. positive semidefinite (psd)).2.1. Generalized Schur complements.
Let(2.1) M = (cid:18) A BC D (cid:19) ∈ S n + m be a real matrix where A ∈ M n , B ∈ M n,m , C ∈ M m,n and D ∈ M m . The generalized Schur complement [Zha05]of A (resp. D ) in M is defined by M/A = D − CA + B ( resp. M/D = A − BD + C ) , where A + (resp. D + ) stands for the Moore-Penrose inverse of A (resp. D ). Remark . (1) If A (resp. D ) is invertible, then M/A (resp.
M/D ) is the usual Schur complement of A (resp. D ) in M .(2) Note that M/A = (cid:18) D CB A (cid:19) /A .The following theorem gives conditions for verifying positive semidefiniteness of a block matrix of size 2. Theorem 2.2. [Alb69]
Let (2.2) M = (cid:18) A BB T C (cid:19) ∈ S n + m be a real symmetric matrix where A ∈ S n , B ∈ M n,m and C ∈ S m . Then the following conditions are equivalent:(1) M (cid:23) .(2) C (cid:23) , C ( B T ) ⊆ C ( C ) and M/C (cid:23) .(3) A (cid:23) , C ( B ) ⊆ C ( A ) and M/A (cid:23) . If m = 1 in (2.2), then rank M ∈ { rank A, rank A + 1 } . The following proposition characterizes w.r.t. the valueof M/A when each of the possibilities occurs in the case M is psd. Proposition 2.3.
Let M = (cid:18) A bb T c (cid:19) ∈ S n +1 be a real symmetric matrix where A ∈ S n , b ∈ R n and c ∈ R . Then rank M = rank A if and only if M/A = 0 .Otherwise rank M = rank A + 1 .Proof. By Theorem 2.2, the psd assumption implies that b ∈ C ( A ) . By the properties of the Moore-Penrose inverse { A + b + w : w ∈ ker A } is the set of solutions z of the system Az = b . Therefore,(2.3) C ( M ) = C (cid:0) (cid:18) A b T c − b T ( A + b + w ) (cid:19) (cid:1) = C (cid:0) (cid:18) A b T M/A (cid:19) (cid:1) , where the second equality follows from the fact that A is symmetric, b ∈ C ( A ) and w ∈ ker A . Now, the statement ofthe proposition follows from (2.3). (cid:3) The following proposition gives an explicit formula, called the quotient formula [CH69], for expressing the Schurcomplement of a × upper left-hand or a × lower right-hand block in a × block matrix using × blocksubmatrices. Proposition 2.4.
Let K = A B DB T C ED T E T F = M DED T E T F = A B DB T D T N ∈ S n + n + n be a × block real matrix, where A ∈ S n , C ∈ S n , F ∈ S n are real symmetric matrices and B ∈ M n ,n , D n ,n , E n ,n are rectangular matrices. If M and A are nonsingular, then (2.4) K/M = (cid:18) A DD T F (cid:19) /A − (cid:20)(cid:18) A BD T E T (cid:19) . A (cid:21) ( M/A ) − (cid:20)(cid:18) A DB T E (cid:19) . A (cid:21) . ALJAˇZ ZALAR If N and C are nonsingular, then (2.5) K/N = (cid:18) C B T B A (cid:19) . C − (cid:20)(cid:18) C EB D (cid:19) . C (cid:21) ( N/C ) − (cid:20)(cid:18) C B T E T D T (cid:19) . C (cid:21) . Proof.
By an easy calculation we have that
K/A = M/A (cid:18)
A DB T E (cid:19) /A (cid:18) A BD T E T (cid:19) /A (cid:18) A DD T F (cid:19) /A . Now the quotient formula [CH69]
K/M = (
K/A ) / ( M/A ) yields (2.4).By Remark 2.1 (2), it is true that K/N = L/N where L = N B T D T B D A . Now (2.5) follows from (2.4). (cid:3)
Hankel matrices.
Let k ∈ N . For β = ( β , . . . , β k ) ∈ R k +1 , we denote by A β := ( β i + j ) ki,j =0 = β β β · · · β k β β . . . . . . β k +1 β . . . . . . . . . ...... . . . . . . . . . β k − β k β k +1 · · · β k − β k ∈ S k +1 the corresponding Hankel matrix. We denote by v j := ( β j + ℓ ) kℓ =0 the ( j + 1) -th column of A β , ≤ j ≤ k , i.e., A β = (cid:0) v · · · v k (cid:1) . As in [CF91], the rank of β , denoted by rank β , is defined by rank β = (cid:26) k + 1 , if A β is nonsingular , min { i : v i ∈ span { v , . . . , v i − }} , if A β is singular . We denote the upper left-hand corner of A β of size m + 1 by A β ( m ) = ( β i + j ) mi,j =0 ∈ S m +1 . The following proposition is the alternative description of rank β if A β is singular. Proposition 2.5. [CF91, Proposition 2.2]
Let k ∈ N , β = ( β , . . . , β k ) , and assume that A is positive semidefiniteand singular. Then rank β = min { j : 0 ≤ j ≤ k such that A β ( j ) is singular } . Important property of psd Hankel matrices is the following rank principle.
Theorem 2.6. [CF91, Corollary 2.5]
Let k ∈ N , β = ( β , . . . , β k ) , e β = ( β , . . . , β k − ) , A β (cid:23) and r = rank e β .Then:(1) rank A e β = r .(2) r ≤ rank A β ≤ r + 1 .(3) rank A β = r + 1 if and only if β k > ϕ β k − r + . . . + ϕ r − β k − , where ( ϕ , . . . , ϕ r − ) := A β ( r − − ( β r , . . . , β r − ) T We will use the following corollary of Proposition 2.5 and Theorem 2.6 in the sequel.
HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 5
Corollary 2.7.
In the notation of Theorem 2.6, under the assumptions A β (cid:23) , A β is singular, and r = rank e β , then r = rank β = rank A β ( r −
1) = rank A β ( r ) = . . . = rank A β ( k −
1) = rank A e β . We denote the lower right-hand corner of A β of size m + 1 by A β [ m ] = ( β i + j ) ki,j = m − k = β k − m ) β k − m )+1 β k − m +1) · · · β k − m β k − m )+1 β k − m +1) . . . . . . β k − m +1 β k − m +1) . . . . . . . . . ...... . . . . . . . . . β k − β k − m β .k − m +1 · · · β k − β k ∈ S m +1 Let β (rev) := ( β k , β k − , . . . , β ) be the sequence obtained from β by reversing the order of numbers. Using Corollary 2.7 for a reversed sequenceimplies the following corollary. Corollary 2.8.
In the notation of Theorem 2.6, under the assumption A β (cid:23) , A β is singular and r = rank e β (rev) ,where e β (rev) := ( β k , . . . , β ) , it holds that r = rank β (rev) = rank A β [ r −
1] = rank A β [ r ] = . . . = rank A β [ k −
1] = rank A e β (rev) . Proof.
Corollary 2.7 used for β (rev) implies that(2.6) r = rank β (rev) = rank A β (rev) ( r −
1) = rank A β (rev) ( r ) = . . . = A β (rev) ( k −
1) = rank A e β (rev) . For ℓ = 0 , . . . , k define the permutation matrices P ℓ : R ℓ +1 → R ℓ +1 by e ( ℓ ) i e ( ℓ ) ℓ +2 − i , i = 1 , . . . , ℓ + 1 , where e ( ℓ )1 , . . . , e ( ℓ ) ℓ +1 is the standard basis for R ℓ +1 . Note that A β (rev) ( ℓ ) = P Tℓ A β [ ℓ ] P ℓ and hence rank A β (rev) ( ℓ ) = rank A β [ ℓ ] ,which together with (2.6) implies the statement of the corollary. (cid:3) A sequence β = ( β , . . . , β k ) with r := rank β is positively recursively generated if A β ( r − ≻ and denoting ( ϕ , . . . , ϕ r − ) := A β ( r − − ( β r , . . . , β r − ) T , it is true that(2.7) β j = ϕ β j − r + · · · + ϕ r − β j − for j = r, . . . , k. Note that (2.7) is equivalent to(2.8) v j = ϕ v j − r + · · · + ϕ r − v j − for j = r, . . . , k. Solution of the truncated Hamburger moment problem.Theorem 2.9. [CF91, Theorem 3.9]
For k ∈ N and β = ( β , . . . , β k ) with β > , the following statements areequivalent:(1) There exists a representing measure for β supported on K = R .(2) There exists a (rank β ) -atomic representing measure for β .(3) β is positively recursively generated.(4) A β (cid:23) and rank A β = rank β . A straightforward corollary of Theorem 2.9 and Corollary 2.7 is the following.
Corollary 2.10.
Let k ∈ N and β = ( β , . . . , β k ) with β > . Suppose that A β is singular. The following statementsare equivalent:(1) There exists a representing measure for β supported on K = R .(2) There exists a (rank β ) -atomic representing measure for β .(3) β is positively recursively generated.(4) A β (cid:23) and rank A β = rank A β ( k − . ALJAˇZ ZALAR
Partially positive semidefinite matrices and their completions. A partial matrix A = ( a ij ) ni,j =1 is a matrixof real numbers a ij ∈ R , where some of the entries are not specified.A partial symmetric matrix A = ( a ij ) ni,j =1 is partially positive semidefinite (ppsd) (resp. partially positivedefinite (ppd) ) if the following two conditions hold:(1) a ij is specified if and only if a ji is specified and a ij = a ji .(2) All fully specified principal minors of A are psd (resp. pd).It is well-known that a ppsd matrix A ( x ) of the form as in Lemma 2.11 below admits a psd completion. (Thisfollows from the fact that the corresponding graph is chordal, see e.g. [GJSW84, Dan92, BW11].) In the notation ofLemma 2.11, if A ( x ) , x ∈ R , is a psd Hankel matrix, then Corollary 2.7 implies that (2.9) below holds. Since wewill need an additional information about the rank of the completion A ( x ) and the explicit interval of all possible x for our results, we give a proof of Lemma 2.11 based on the use of generalized Schur complements assuming (2.9)holds. Lemma 2.11.
Let A ( x ) := A a ba T α xb T x β ∈ S n be a partially positive semidefinite symmetric matrix, where A ∈ S n − , a, b ∈ R n − , α, β ∈ R and x is a variable.Let A := (cid:18) A aa T α (cid:19) ∈ S n − , A := (cid:18) A bb T β (cid:19) ∈ S n − , and x ± := b T A +1 a ± p ( A /A )( A /A ) ∈ R . Suppose the following holds: (2.9) A is invertible or rank A = rank A . Then:(1) A ( x ) is positive semidefinite if and only if x ∈ [ x − , x + ] .(2) rank A ( x ) = (cid:26) max (cid:8) rank A , rank A (cid:9) , for x ∈ { x − , x + } , max (cid:8) rank A , rank A (cid:9) + 1 , for x ∈ ( x − , x + ) . (3) If A ( x ) is partially positive definite, then A ( x ′ ) is positive definite for x ′ ∈ ( x − , x + ) .Proof. By Theorem 2.2, A ( x ) (cid:23) if and only if(2.10) A (cid:23) , (cid:18) bx (cid:19) ∈ C ( A ) and f ( x ) := A ( x ) /A ≥ , The first condition of (2.10) is true by the ppsd assumption.Since A (cid:23) , it follows by Theorem 2.2 that a ∈ C ( A ) and hence by the properties of the Moore-Penrose inversewe have that A ( A +1 a ) = a . Thus,(2.11) C ( A ) = C (cid:0) (cid:18) A a T α − a T A +1 a (cid:19) (cid:1) = C (cid:0) (cid:18) A a T A /A (cid:19) (cid:1) . Now we separate two cases according to A /A . Case 1: A /A > .(2.11) and the assumption of Case 1 imply that C ( A ) = C ( A ⊕ . Since A (cid:23) , it follows by Theorem 2.2 that b ∈ C ( A ) . Therefore (cid:0) b x (cid:1) T ∈ C ( A ⊕ for every x ∈ R . Thus the second condition of (2.10) is true for every x ∈ R .Note that the assumption of Case 1 and Proposition 2.3 imply that rank A > rank A and hence the assumption(2.9) implies invertibility of A and A . By Proposition 2.4, used for A ( x ) as K , A as M and A as A , we have that(2.12) f ( x ) = A /A − ( A /A ) − ( x − b T A +1 a ) . HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 7
Therefore f ( x ) ≥ if and only if x ∈ [ x − , x + ] , which is the third condition of (2.10). Now by Proposition 2.3 weknow that rank A ( x ) > rank A if and only if f ( x ) > , which establishes (1),(2) in the case A /A > . Case 2: A /A = 0 .(2.11) and the assumption of Case 2 imply that(2.13) C ( A ) = C (cid:0) (cid:18) A a T (cid:19) (cid:1) . Therefore, using (2.13), it is true that(2.14) (cid:18) bx (cid:19) ∈ C ( A ) ⇔ (cid:18) bx (cid:19) = (cid:18) A a T (cid:19) z = (cid:18) A za T z (cid:19) for some z ∈ R n − . Since A (cid:23) , it follows by Theorem 2.2 that b ∈ C ( A ) and hence by the properties of the Moore-Penrose inverse { A +1 b + w : w ∈ ker A } is the set of all solutions z of the system A z = b . Therefore, using (2.14), it follows that (cid:18) bx (cid:19) ∈ C ( A ) ⇔ x ∈ { a T A +1 b + a T w : w ∈ ker A } = { a T A +1 b } , where we used the fact that A is symmetric, a ∈ C ( A ) and w ∈ ker A for the last equality. So only x = a T A +1 b satisfies the second condition of (2.10).Now by definition of the generalized Schur complement, we have f ( x ) = β − (cid:0) b T x (cid:1) A +2 (cid:18) bx (cid:19) . By the properties of the Moore-Penrose inverse A +2 (cid:18) bx (cid:19) = (cid:18) A +1 b (cid:19) + v for some v ∈ ker A . Hence, f ( x ) = β − (cid:0) b T x (cid:1) (cid:16) (cid:18) A +1 b (cid:19) + v (cid:17) = β − b T A +1 b = A /A ≥ , where the second equality follows from the fact that A is symmetric, (cid:0) b T x (cid:1) T ∈ C ( A ) and v ∈ ker A , andthe last inequality follows by the ppsd assumption. Note that x = x + = x − and by Proposition 2.3, rank A ( x ) =rank A if and only if A /A = 0 , in which case also rank A = rank A . Otherwise we have f ( x ) = A /A > ,which implies by Proposition 2.3 that rank A ( x ) = rank A = rank A + 1 . Thus (1),(2) are true in the case A /A = 0 .(3) follows from (2) by noticing that A /A > , A /A > and rank A = rank A = n − . (cid:3) Extension principle.
The extension principle for psd matrices is the following.
Lemma 2.12.
Let A ∈ S n be a positive semidefinite matrix, Q ⊆ { , . . . , n } a subset and A Q be the restriction of A to rows and columns from the set Q . If v ∈ ker A Q is a nonzero vector from the kernel of A Q , then the vector b v withthe only nonzero entries in rows from Q and such that the restriction b v | Q to the rows from Q equals to v , belongs to ker A .Proof. By permuting rows and columns we may assume that A is of the form A = (cid:18) A Q BB T C (cid:19) . We have to prove that(2.15) A (cid:18) v (cid:19) = 0 . Since A is psd, for every w := (cid:0) v T u T (cid:1) ∈ R n we have that(2.16) ≤ wAw T = 2 u T B T v + u T Cu. If B T v = 0 , then we define u := − αB T v where α > is an arbitrary positive real number, and plug into (2.16) to get(2.17) ≤ − α (cid:13)(cid:13) B T v (cid:13)(cid:13) + α v T BCB T v = α ( αv T BCB T v − (cid:13)(cid:13) B T v (cid:13)(cid:13) ) =: αS ( α ) . Since lim α → S ( α ) = − (cid:13)(cid:13) B T v (cid:13)(cid:13) < , (2.17) cannot be true for α small enough. Hence B T v = 0 , which proves(2.15). (cid:3) ALJAˇZ ZALAR
Subsequences of one-dimensional moment sequences.Proposition 2.13.
Let k ∈ N and β = ( β , . . . , β k ) with β > be a sequence which admits a representing measuresupported on K = R . Then for every i, j ∈ N , where ≤ i ≤ j ≤ k , a subsequence β ( i,j ) := ( β i , . . . , β j ) alsoadmits a representing measure supported on K = R .Proof. Note that A β is of the form A β = A β (0 ,i − ∗ ∗∗ A β ( i,j ) ∗∗ ∗ A β ( j +1 ,k ) . By Theorem 2.9, A β (cid:23) and hence A β ( i,j ) (cid:23) . For i = j the statement is clear, i.e., the representing atom is β i withdensity 1. Assume that i < j . We separate two cases according to the invertibility of A β ( i,j ) .(1) If A β ( i,j ) ≻ , then rank A β ( i,j ) = rank β ( i,j ) = j − i + 1 and by Theorem 2.9, β ( i,j ) admits a measure.(2) Else A β ( i,j ) = (cid:18) A β ( i,j − v T v β j (cid:19) is singular, where v = (cid:0) β j · · · β j − (cid:1) . We separate two cases according to the invertibility of A β ( i,j − . • If A β ( i,j − is invertible, then rank A β ( i,j − = rank A β ( i,j ) . • Else A β ( i,j − is singular and by Corollary 2.7 used for β ( i,j ) as β , we get rank A β ( i,j − = rank A β ( i,j − .This implies that the last column of A β ( i,j − is in the span of the other columns of A β ( i,j − . By Lemma2.12, the j -th column of A β is in the span of the columns i + 1 , . . . , j − . Since β is positively recursivelygenerated, the ( j + 1) -th column of A β is in the span of the columns i + 2 , . . . , j and in particular the lastcolumn of A β ( i,j ) is in the span of the other columns of A β ( i,j ) . Hence rank A β ( i,j − = rank A β ( i,j ) .In both subcases of (2), rank A β ( i,j − = rank A β ( i,j ) and Corollary 2.10 implies that β ( i,j ) admits a measure. (cid:3)
3. T
RUNCATED H AMBURGER MOMENT PROBLEM OF DEGREE k WITH GAP ( β k − ) AND ( β k − , β k − ) In this section we solve the THMP of degree k with gaps ( β k − ) (see Theorem 3.1) and ( β k − , β k − ) (seeTheorem 3.5). As a corollary of Theorem 3.1 we obtain the solution to the TMP for the curve y = x (see Corollary3.3), while as a corollary of Theorem 3.5 we get the solution to the TMP for the curve y = x and an additionalmoment β , k − given (see Corollary 3.6).3.1. Truncated Hamburger moment problem of degree k with gap ( β k − ) .Theorem 3.1. Let k ∈ N and β ( x ) := ( β , β , . . . , β k − , x, β k ) be a sequence where each β i is a real number, β > and x is a variable. Let b β := ( β , . . . , β k − ) and e β := ( β , . . . , β k − ) be subsequences of β ( x ) , v := ( β k · · · β k − ) a vector and e A := (cid:18) A b β v T v β k (cid:19) a matrix. Then the following statements are equivalent:(1) There exists x ∈ R and a representing measure for β ( x ) supported on K = R .(2) There exists x ∈ R and a (rank e β ) -atomic representing measure for β ( x ) .(3) A β ( x ) is partially positive semidefinite and one of the following conditions is true:(a) k = 1 .(b) k > and one of the following conditions is true:(i) A e β ≻ .(ii) rank A b β = rank A e β = rank e A. HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 9
Proof.
First we prove the implication (1) ⇒ (3). By Theorem 2.9, A β ( x ) (cid:23) and rank A β ( x ) = rank β ( x ) . A β ( x ) (cid:23) in particular implies that A β ( x ) is ppsd. If k = 1 , then (3a) holds. Otherwise k > . If A e β ≻ , then(3(b)i) holds. Else A e β is singular and hence(3.1) rank A b β = rank A e β = rank β ( x ) = A β ( x ) , where the first two equalities follow by Corollary 2.7 used for β ( x ) as β and the last by Theorem 2.9. A b β being aprincipal submatrix of e A and e A being a principal submatrix of A β ( x ) = A b β u T v T u β k − x v x β k , where u = ( β k − · · · β k − ) , imply together with (3.1) that (3(b)ii) holds and concludes the proof of theimplication (1) ⇒ (3).Second we prove the implication (3) ⇒ (2). We separate two cases according to k . • k = 1 . We have that A β ( x ) = (cid:18) β xx β (cid:19) . For x = √ β β , A β ( x ) is of rank 1 and the second columnis the multiple of the first. Hence, by Corollary 2.10, a 1-atomic measure exists, proving the implication(3) ⇒ (2) in this case. • k > . Notice that A β ( x ) is of the same form as A ( x ) from Lemma 2.11, where A b β , A e β , e A correspond to A , A , A , respectively. Since both cases (3(b)i) and (3(b)ii) satisfy the assumption (2.9), it follows by Lemma2.11 that there exists x such that A β ( x ) (cid:23) and(3.2) rank A β ( x ) = max n rank A e β , rank e A o . Since in the case (3(b)i), it holds that rank e A ≤ rank A e β , while in the case (3(b)ii), rank e A = rank A e β , weobtain from (3.2) that rank A β ( x ) = rank A e β . By Corollary 2.10, (rank e β ) -representing measure for β ( x ) exists, which proves (2).The implication (2) ⇒ (1) is trivial. (cid:3) Example 3.2.
For k = 9 , let β (1) ( x ) = (1 , , , , , , , , , , , , , , , , , x, ,β (2) ( x ) = (cid:16) , , , − , , − , , − , , − , , − , , − , , − , , x, (cid:17) ,β (3) ( x ) = (8 , , , , , , , , , , , , , , , , ,x, . Let e A ( i ) , i = 1 , , , denote e A from Theorem 3.1 corresponding to β ( i ) ( x ) . Using Mathematica [Wol] one can checkthat: • e A ( i ) (cid:23) for i = 1 , , . • A e β (1) ≻ , A e β (2) , A e β (3) (cid:23) and dim (cid:16) ker A e β (3) (cid:17) = 1 . • rank A b β (3) = rank e A (3) = rank A e β (3) = 8 .Therefore: • A β (1) ( x ) is ppsd and e β (1) satisfies (3(b)i) of Theorem 3.1, implying that a 9-atomic measure for β (1) ( x ) exists. • A β (2) ( x ) is not ppsd and by Theorem 3.1, there is no representing measure for β (2) ( x ) . • A β (3) ( x ) is ppsd and e β (3) satisfies (3(b)ii) of Theorem 3.1, implying that an 8-atomic measure for β (3) ( x ) exists.The following corollary is a consequence of Theorem 3.1 and is an alternative solution of the bivariate TMP for thecurve y = x , first solved by Fialkow in [Fia11]. Corollary 3.3.
Let k ∈ N and β = ( β i,j ) i,j ∈ Z ,i + j ≤ k be a 2-dimensional real multisequence of degree k . Suppose M ( k ) is positive semidefinite and recursively generated. Let u ( i ) := ( β ,i , β ,i , β ,i ) for i = 0 , . . . , k − , b β := ( u (0) , . . . , u (2 k − ) and e β := ( u (0) , . . . , u (2 k − , β , k − , β , k − ) be subsequences of β . Then β has a representing measure supported on y = x if and only if the following statementshold:(1) One of the following holds: • If k ≥ , then Y = X is a column relation of M ( k ) . • If k = 2 , then the equalities β , = β , , β , = β , , β , = β , hold.(2) One of the following holds:(a) A e β ≻ .(b) A e β (cid:23) and rank A b β = rank A e β = rank M ( k ) .Moreover, if the representing measure exists, then: • If A e β is nonsingular, there exists a k -atomic measure. • If A e β is singular, then the measure is (rank M ( k )) -atomic.Proof. For m ∈ { , . . . , k − , k } we define the numbers e β m by the following rule e β m := β m (mod 3) , ⌊ m ⌋ . Claim 1.
Every number e β m is well-defined.We have to prove that m (mod 3) + ⌊ m ⌋ ≤ k. We separate three cases according to m . • m ≤ k − : ⌊ m ⌋ + m (mod 3) ≤ (2 k −
2) + 2 = 2 k . • m ∈ { k − , k − } : ⌊ m ⌋ + m (mod 3) ≤ (2 k −
1) + 1 = 2 k . • m = 6 k : ⌊ m ⌋ + m (mod 3) = 2 k + 0 = 2 k. Claim 2.
Let t ∈ N . The atoms ( x , x ) , . . . ( x t , x t ) with densities λ , . . . , λ t are the ( y − x ) -representingmeasure for β if and only if the atoms x , . . . , x t with densities λ , . . . , λ t are the R -representing measure for e β ( x ) = ( e β , . . . , e β k − , x, e β k ) .The if part follows from the following calculation: e β m = β m (mod 3) , ⌊ m ⌋ = t X ℓ =1 λ ℓ x m (mod 3) ℓ x ⌊ m ⌋ ℓ = t X ℓ =1 λ ℓ x m (mod 3)+3 ⌊ m ⌋ ℓ = t X ℓ =1 λ ℓ x mℓ , where m = 0 , , . . . , k − , k .The only if part follows from the following calculation: β i,j = β i − ,j +1 = · · · = β i (mod 3) ,j + ⌊ i ⌋ = e β i (mod 3)+3( j + ⌊ i ⌋ ) = t X ℓ =1 λ ℓ x i (mod 3)+3( j + ⌊ i ⌋ ) ℓ = t X ℓ =1 λ ℓ x i (mod 3)+3 ⌊ i ⌋ ℓ x jℓ = t X ℓ =1 λ ℓ x iℓ ( x ℓ ) j , where the equalities in the first line follow by M ( k ) being rg.Using Claim 2 and a theorem of Bayer and Teichmann [BT06], implying that if a finite sequence has a K -representing measure, then it has a finitely atomic K -representing measure, the statement of the Corollary followsby Theorem 3.1. (cid:3) Remark . (1) Corollary 3.3 in case k = 1 is an improvement of [Fia11, Proposition 5.6.ii)] by decreasing thenumber of atoms from 6 to 3.(2) For M (1) ≻ and A e β , (2) of Corollary 3.3 is not satisfied and hence the measure does not exist. Sincethis is the case under the assumptions of [Fia11, Proposition 5.6.iii)], the additional conditions in [Fia11,Proposition 5.6.iii)] are never satisfied. HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 11 (3) Examples in the Example 3.2 above are derived from [Fia11, Example 5.2], [Fia08, Example 4.18], [Fia08,Example 3.3], which demonstrate the solution of the moment problem for the curve y = x .3.2. Truncated Hamburger moment problem of degree k with gaps ( β k − , β k − ) .Theorem 3.5. Let k ∈ N , k > , and β ( x, y ) := ( β , β , . . . , β k − , y, x, β k ) be a sequence, where each β i is a real number, β > and x, y are variables. Let b β := ( β , . . . , β k − ) and e β := ( β , . . . , β k − ) be subsequences of β ( x, y ) , u := (cid:0) β k · · · β k − (cid:1) , s := (cid:0) β k − · · · β k − (cid:1) and w := (cid:0) β k − · · · β k − (cid:1) vectors and e A := (cid:18) A b β u T u β k (cid:19) a matrix. Then the following statements are equivalent:(1) There exist x , y ∈ R and a representing measure for β ( x , y ) supported on K = R .(2) There exist x , y ∈ R and a (rank e β ) or (rank e β + 1) -atomic representing measure for β ( x , y ) .(3) A β ( x,y ) is partially positive semidefinite and one of the following conditions holds:(a) k = 2 and β β ≤ √ β β .(b) k > , the inequality (3.3) sA + e β s T ≤ uA + b β w T + q ( A e β /A b β )( e A/A b β ) . holds and one of the following conditions is true:(i) A e β ≻ .(ii) rank A b β = rank A e β = rank (cid:0) A e β s T (cid:1) = rank e A .Moreover, if the representing measure for β exists, then: • If k = 2 , then there is a -atomic measure if β β = √ β β . Otherwise there is a 2-atomic measure. • If k > , there exists a (rank e β ) -atomic if and only if one of the equalities (3.4) sA + e β s T = uA + b β w T − q ( A e β /A b β )( e A/A b β ) or sA + e β s T = uA + b β w T + q ( A e β /A b β )( e A/A b β ) holds.Proof. Note that β ( x, y ) admits a measure if and only if there exist y ∈ R such that β ( x, y ) admits a measure.Theorem 3.1 implies that the following claim holds. Claim 1. β ( x, y ) admits a measure if and only if the following conditions hold:(1) A β ( x,y ) is ppsd.(2) One of the following is true:(a) A ( e β,β k − ,y ) ≻ , where A ( e β,β k − ,y ) = (cid:18) β β β y (cid:19) , if k = 2 , (cid:18) A e β s T s y (cid:19) = A b β w T s T w β k − β k − s β k − y where s T = β k − ...β k − , otherwise . (b) rank A e β = rank b A ( y ) , where b A ( y ) := (cid:18) β yy β (cid:19) , if k = 2 , (cid:18) A e β u ( y ) T u ( y ) β k (cid:19) = A b β w T u T w β k − yu y β k and u ( y ) := (cid:0) u y (cid:1) , otherwise . Claim 2.
Let k > . Assume A b β ≻ or rank A b β = rank A e β . Then b A ( y ) (cid:23) if and only if(3.5) b A ( y ) is ppsd and y ∈ h uA + b β w T − q ( A e β /A b β )( e A/A b β ) , uA + b β w T + q ( A e β /A b β )( e A/A b β ) i =: [ y − , y + ] . Moreover,(3.6) rank b A ( y ) = ( max (cid:8) rank A e β , rank e A (cid:9) , for y ∈ { y − , y + } , max (cid:8) rank A e β , rank e A (cid:9) + 1 , for y ∈ ( y − , y + ) . The assumption (2.9) of Lemma 2.11 used for b A ( y ) , A b β , A e β , e A as A ( x ) , A , A , A , respectively, are by the as-sumption of Claim 2 satisfied and hence Claim 2 follows by Lemma 2.11. Claim 3.
Let k > . Assume A b β ≻ or rank A b β = rank A e β . Then A β ( x,y ) is ppsd for some y ∈ R if and only if A β ( x,y ) is ppsd, s T ∈ C ( A e β ) and (3.3) holds.Note that A β ( x,y ) is ppsd if and only if A ( e β,β k − ,y ) (cid:23) and b A ( y ) (cid:23) . By Theorem 2.2, A ( e β,β k − ,y ) (cid:23) ifand only if(3.7) A e β (cid:23) , s T ∈ C ( A e β ) and A ( e β,β k − ,y ) /A e β = y − sA + e β s T ≥ , By Claim 2, b A ( y ) is psd if and only if (3.5) holds. Now note that the first condition of (3.5) (which also includes thefirst condition of (3.7)) is equivalent to A β ( x,y ) being ppsd and that y satisfying the third condition of (3.7) and thesecond condition of (3.5) exists if and only if (3.3) holds. This proves Claim 3.First we prove the implication (1) ⇒ (3). By Claim 1, in particular A β ( x,y ) is ppsd.If k = 2 , then A ( e β,β ,y ) (cid:23) , which implies that y ≥ β β , and b A ( y ) (cid:23) , which implies that y ≤ √ β β . Hence, β β ≤ √ β β , which is (3a). Since A β ( x,y ) being ppsd implies that also A β ( x,y ) is ppsd, this proves the implication(1) ⇒ (3) in this case.It remains to prove (1) ⇒ (3) in the case k > . We separate two cases according to the invertibility of A e β . • A e β ≻ : Using Claim 3, A β ( x,y ) is ppsd, (3.3) and (3(b)i) holds, which proves the implication (1) ⇒ (3) inthis case. • A e β : It follows that A ( e β,β k − ,y ) and hence (2b) of Claim 1 must hold. Corollary 2.7 used for ( e β, β k − , y ) as β implies that(3.8) rank A b β = rank A e β . By Proposition 2.13, ( e β, β k − , y ) also admits a measure and Corollary 2.10 used for ( e β, β k − , y ) as β implies that(3.9) rank A e β = rank A ( e β,β k − ,y ) . (2b) of Claim 1 together with (3.8) implies that all the inequalities in the estimate rank A b β ≤ rank e A ≤ rank b A ( y ) are equalities and in particular,(3.10) rank A b β = rank e A. (3.8), (3.9), (3.10) and Claim 3 imply that A β ( x,y ) is ppsd, (3.3) and (3(b)ii) holds, which proves the implication(1) ⇒ (3) in this case. HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 13
Second we prove the implication (3) ⇒ (1). We separate two cases according to k .If k = 2 , then we are in the case (3a). For y = √ β β , β ( x, y ) is ppsd and satisfies (2a) of Claim 1 if β β < √ β β and (2b) if β β = √ β β . In both cases Claim 1 implies the implication (3) ⇒ (1) is true in this case.Else k > . If (3(b)i) holds, then in particular A b β ≻ . Otherwise (3(b)ii) holds and in particular rank A b β =rank A e β . In both cases the assumptions of Claims 2 and 3 are fulfilled. By Claim 3, the matrix A β ( x,y + ) is ppsd andby (3.6) of Claim 2, rank b A ( y + ) = max { rank A e β , rank e A } . If (3(b)i) holds, then rank b A ( y + ) = rank A e β = k − .Else (3(b)ii) holds and rank b A ( y + ) = rank A e β = rank e A . In both cases, β ( x, y + ) satisfies (1) and (2b) of Claim 1above and thus the measure exists which proves the implication (3) ⇒ (1).The implication (2) ⇒ (1) is trivial.Now we prove the implication (1) ⇒ (2). If β ( x, y ) has a representing measure, then: • By Theorem 3.1 it has a (rank( e β, β k − , y )) -atomic representing measure. • By Proposition 2.13, e β and ( e β, β k − , y ) also have measures and hence by Theorem 2.9, rank A e β = rank e β and rank( e β, β k − , y ) = rank A ( e β,β k − ,y ) . Since rank A ( e β,β k − ,y ) ∈ { rank A e β , rank A e β + 1 } , the implication (1) ⇒ (2) is true.It remains to prove the moreover part. We separate two cases according to k . • If k = 2 , then rank A e β = rank( β ) = 1 . So 1-atomic measure exists if and only if rank A ( β ,β ,y ) =rank b A ( y ) = 1 for some y . But from the form of A ( β ,β ,y ) and b A ( y ) this is possible only if y = β β = √ β β . Otherwise there is a 2-atomic measure. • Else k > . By Proposition 2.3 and (3.7) above, rank A ( e β,β k − ,y ) = rank A e β if and only if y = sA + e β s T .In the proof of the implication (3) ⇒ (1) we see that rank A e β ≥ rank e A . Using this in (3.6) above, it followsthat sA + e β s T must be equal to y − or y + , which is exactly (3.4).This concludes the proof of the theorem. (cid:3) The following corollary is a consequence of Theorem 3.5 and solves the bivariate TMP for the curve y = x wherealso β , k − is given. Corollary 3.6.
Let β = ( β i,j ) i,j ∈ Z ,i + j ≤ k be a 2-dimensional real multisequence of degree k and let β , k − bealso given. Suppose M ( k ) is positive semidefinite and recursively generated. Let u ( i ) = ( β ,i , β ,i , β ,i , β ,i ) for i = 0 , . . . , k − , b β := ( u (0) , . . . , u (2 k − , β , k − , β , k − , β , k − ) and e β := ( b β, β , k − , β , k − ) be subsequences of β , u := (cid:0) u ( k ) · · · u (2 k − β , k − (cid:1) , s := (cid:0) β ,k − u ( k ) · · · u (2 k − β , k − (cid:1) ,w := (cid:0) β ,k − β ,k − u ( k ) · · · u (2 k − β , k − β , k − β , k − (cid:1) vectors and e A := (cid:18) A b β u T u β , k (cid:19) a matrix. Then β has a representing measure supported on y = x if and only if sA + e β s T ≤ uA + b β w T + q ( A e β /A b β )( e A/A b β ) . one of the following statements hold:(1) One of the following holds: • If k ≥ , then Y = X is a column relation of M ( k ) . • If k = 3 , then the equalities β , = β , , β , = β , , β , = β , hold. • If k = 2 , then the equality β , = β , holds.(2) One of the following conditions holds:(a) A e β ≻ .(b) A e β (cid:23) and rank A b β = rank A e β = rank (cid:0) A e β s T (cid:1) = rank e A . Moreover, if the representing measure exists, then there is a (rank e β ) -atomic measure if sA + e β s T ∈ (cid:26) uA + b β w T − q ( A e β /A b β )( e A/A b β ) , uA + b β w T + q ( A e β /A b β )( e A/A b β ) (cid:27) . and (rank e β + 1) -atomic otherwise.Proof. For m ∈ { , . . . , k − , k } we define the numbers e β m by the following rule e β m := β m (mod 4) , ⌊ m ⌋ . Claim 1.
Every number e β m is well-defined.We will prove that m (mod 4) + ⌊ m ⌋ ≤ k if m = 8 k − , while for m = 8 k − we have e β k − = β , k − . Weseparate three cases according to m . • m < k − : ⌊ m ⌋ + m (mod 4) ≤ (2 k −
3) + 3 = 2 k . • m ∈ { k − , k − , k − } : ⌊ m ⌋ + m (mod 4) ≤ (2 k −
2) + 2 = 2 k . • m ∈ { k − , k − } : ⌊ m ⌋ + m (mod 4) ≤ (2 k −
1) + 1 = 2 k . • m = 8 k : ⌊ m ⌋ + m (mod 3) = 2 k + 0 = 2 k. Claim 2.
Let t ∈ N . The atoms ( x , x ) , . . . ( x t , x t ) with densities λ , . . . , λ t are the ( y − x ) -representing measurefor β and β , k − if and only if the atoms x , . . . , x t with densities λ , . . . , λ t are the R -representing measure for e β ( x, y ) = ( e β , . . . , e β k − , y, x, e β k ) .The if part follows from the following calculation: e β m = β m (mod 4) , ⌊ m ⌋ = t X ℓ =1 λ ℓ x m (mod 4) ℓ x ⌊ m ⌋ ℓ = t X ℓ =1 λ ℓ x m (mod 4)+4 ⌊ m ⌋ ℓ = t X ℓ =1 λ ℓ x mℓ , where m = 0 , . . . , k − , k .The only if part follows from the following calculation for i + j ≤ k : β i,j = β i − ,j +1 = · · · = β i (mod 4) ,j + ⌊ i ⌋ = e β i (mod 4)+4( j + ⌊ i ⌋ ) = t X ℓ =1 λ ℓ x i (mod 4)+4( j + ⌊ i ⌋ ) ℓ = t X ℓ =1 λ ℓ x i (mod 4)+4 ⌊ i ⌋ ℓ x jℓ = t X ℓ =1 λ ℓ x iℓ ( x ℓ ) j , where the equalities in the first line follow by M ( k ) being rg, and β , k − = e β k − = t X ℓ =1 λ ℓ x k − ℓ = t X ℓ =1 λ ℓ x ℓ ( x ℓ ) k − . Using Claim 2 and a theorem of Bayer and Teichmann [BT06], implying that if a finite sequence has a K -representing measure, then it has a finitely atomic K -representing measure, the statement of the Corollary followsby Theorem 3.5. (cid:3)
4. T
RUNCATED H AMBURGER MOMENT PROBLEM OF DEGREE k WITH GAP ( S ) ( β ) , ( β , β ) In this section we solve the THMP of degree k with gaps ( β ) (see Theorem 4.1) and ( β , β ) (see Theorem 4.5).As a corollary of Theorem 4.1 we obtain the solution to the TMP for the curve y = x (see Corollary 4.4), while as acorollary of Theorem 4.5 we get the solution to the TMP for the curve y = x and an additional moment β , given(see Corollary 4.7).4.1. Truncated Hamburger moment problem of degree k with gaps ( β ) .Theorem 4.1. Let k ∈ N , k > , and β ( x ) := ( β , x, β , . . . , β k ) be a sequence where each β i is a real number, β > and x is a variable. Let b β := ( β , . . . , β k − ) , e β := ( β , . . . , β k ) , β := ( β , . . . , β k − ) and ( β := ( β , . . . , β k ) HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 15 be subsequences of β ( x ) , v := (cid:0) β · · · β k − (cid:1) and u := (cid:0) β . . . β k (cid:1) vectors, and e A := (cid:18) β vv T A β (cid:19) and b A := (cid:18) β uu T A ( β (cid:19) matrices. Then the following statements are equivalent:(1) There exists x ∈ R and a representing measure for β ( x ) supported on K = R .(2) There exists x ∈ R and a (rank e β ) or a (rank e β + 1) -atomic representing measure for β ( x ) .(3) A β ( x ) is partially positive semidefinite and one of the following conditions is true:(a) (i) k = 2 and A e β ≻ .(ii) k > , A e β ≻ and e A ≻ .(b) rank A b β = rank A e β = rank A ˘ β . Moreover, if the representing measure exists, then there does not exist a (rank e β ) -atomic measure if and only if (3b) holds and rank A b β < rank b A .Proof. First we prove the implication (1) ⇒ (3). By Theorem 2.9, A β ( x ) (cid:23) and rank A β ( x ) = rank β ( x ) . Thecondition A β ( x ) (cid:23) implies that A β ( x ) is ppsd. We separate two cases according to the invertibility of A e β . • A e β ≻ : Since A e β is a principal submatrix of A β ( x ) , we conclude that rank A β ( x ) ≥ rank A e β = k ,and hence A β ( x ) is either invertible or rank A β ( x ) is singular and by Corollary 2.10 used for β ( x ) as β , rank A β ( x ) = rank A ( β ,x , b β ) . In both cases A ( β ,x , b β ) = (cid:18) β x x β (cid:19) , if k = 2 , β x vx β v v T v T A β where v = (cid:0) β · · · β k (cid:1) , if k > , , is invertible. If k > , e A is a principal submatrix of A ( β ,x , b β ) and it follows that e A ≻ . Hence, (3a) holds.Together with A β ( x ) being ppsd, proves the implication (1) ⇒ (3) in this case. • A e β is singular: Since e β is a subsequence of β ( x ) of the form from Proposition 2.13 with i = 1 , j = k , itadmits a measure. By Corollary 2.10 used for e β as β , it follows that(4.1) rank A e β = rank A b β . By Corollary 2.8 used for β ( x ) as β , it follows that(4.2) rank A e β = rank A ˘ β . Hence, (4.1) and (4.2) imply that (3b) holds. Together with A β ( x ) being ppsd, proves the implication (1) ⇒ (3)in this case.Second we prove the implication (3) ⇒ (2). Let P : R k +1 → R k +1 be the following permutation matrix P = I k − , where stands for the row of k − zeros and I k − is the identity matrix of size k − . Then P T A β ( x ) P is of the form P T A β ( x ) P = A ( β w T u T w β xu x β , where w = (cid:0) β · · · β k +1 (cid:1) is a vector. Claim. A β ( x ) is psd if and only if x ∈ h uA +˘ β w T − q ( A e β /A ( β )( b A/A ( β ) , uA +˘ β w T + q ( A e β /A ( β )( b A/A ( β ) i =: [ x − , x + ] . Moreover,(4.3) rank A β ( x ) := ( max { rank A e β , rank b A } , if x ∈ { x − , x + } , max { rank A e β , rank b A } + 1 , if x ∈ ( x − , x + ) . Denoting the matrices A := (cid:18) A ˘ β w T w β (cid:19) and B := (cid:18) A ˘ β u T u β (cid:19) , and the permutation matrix P : R k → R k by P = (cid:18) I k − (cid:19) , where stands for the row of k − zeroes and I k − the identity matrix of size k − , we have that A = P T A e β P and B = P T b AP . In particular,(4.4) rank A = rank A e β and rank B = rank b A. If (3a) holds, then A e β ≻ implies that A ˘ β ≻ . If (3b) holds, then in particular rank A ˘ β = rank A e β = rank A .Hence, the assumption (2.9) of Lemma 2.11 used for P T A β ( x ) P , A ˘ β , A , B as A ( x ) , A , A , A , respectively, is sat-isfied and using also A /A ˘ β = A e β /A ˘ β and B /A ˘ β = b A/A ˘ β , Claim follows.First assume that (3a) holds. We separate two cases according to the inverbility of b A . • b A ≻ : From A e β ≻ and b A ≻ it follows, using Proposition 2.3, that A e β /A ( β > and b A/A ( β > . Henceby the definition of x ± , we have x − < x + and by Claim, A β ( x ) ≻ for x ∈ ( x − , x + ) . By Theorem 2.9, (rank β ( x )) = (rank e β + 1) -atomic representing measure for β ( x ) exists, which proves the implication(3) ⇒ (2) in this case. • b A is singular: From A e β ≻ it follows that A ˘ β ≻ . Since b A is singular, Proposition 2.3 implies that b A/A ( β =0 , and hence by the definition of x ± , we have x − = x + . By Claim, A β ( x ± ) (cid:23) with rank A β ( x ± ) = rank A e β .We separate two cases according to k . – k = 2 : Since b A = (cid:18) β β β β (cid:19) and β > , it follows that the second (also the last) column of b A is inthe span of the first (also the others) one. – k > : By assumptions e A ≻ and b A = (cid:18) e A u T u β k (cid:19) being singular, where the u is equal to u = (cid:0) β k β k +2 · · · β k − (cid:1) , it follows that the last column of b A is in the span of the others.By Lemma 2.12, the last column of A β ( x ± ) is also in the span of the others and by Corollary 2.10, we havethat (rank β ( x ± )) = (rank e β ) -atomic representing measure for β ( x ± ) exists, which proves the implication(3) ⇒ (2) in this case.Otherwise (3b) holds. Proposition 2.3 implies that A e β /A b β = 0 , and hence by the definition of x ± , we have x − = x + . By Claim, A β ( x ± ) (cid:23) . The assumption rank A e β = rank A b β , also implies that the last column of A e β = (cid:18) A b β u T u β k (cid:19) , where u = (cid:0) β k · · · β k − (cid:1) , is in the span of the others. By Lemma 2.12, the lastcolumn of A β ( x ± ) is in the span of the others. Hence, by Corollary 2.10, (rank β ( x ± )) -atomic measure for β ( x ± ) exists. Since e β is a subsequence of β ( x ) of the form from Proposition 2.13 with i = 1 , j = k , it admits a measure andhence Theorem 2.9 implies that rank A e β = rank e β . From (4.3), it follows that: • If rank b A ≤ rank A e β , then rank β ( x ± ) = rank A e β = rank e β . • Else rank b A = rank A e β + 1 and rank β ( x ± ) = rank b A = rank e β + 1 . HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 17
This proves the implication (3) ⇒ (2) in this case.The implication (2) ⇒ (1) is trivial.It remains to prove the moreover part. Observe that in the proof of the implication (3) ⇒ (2), (rank e β ) -atomicmeasure might not exist if (3a) holds with b A ≻ and does not exist if (3b) holds with rank A e β < rank b A . We willprove that in the first case there always exists a (rank e β ) -atomic measure. Assume that A e β ≻ and b A ≻ . We willprove that one of A β ( x ± ) or A β ( x + ) satisfies(4.5) rank A β ( x ± ) = rank A β ( x ± ) ( k − , and hence by Corollary 2.10, a (rank β ( x ± )) = (rank e β ) -atomic measure exists. Using Proposition 2.4 with for A β ( x ) , A e β , A b β as K, N, C , respectively, and denoting u := A e β /A b β , we have that f ( x ) := A β ( x ) /A e β = (cid:16) β − e ( x ) A − b β e ( x ) T (cid:17) − u (cid:16) β k − e ( x ) A − b β z T (cid:17) =: g ( x ) − u h ( x ) , where e ( x ) := (cid:0) x β · · · β k − (cid:1) and z := (cid:0) β k +1 · · · β k − (cid:1) . From the proof of the implication (3) ⇒ (2), we know that x − < x + and(4.6) f ( x − ) = f ( x + ) = 0 . Note that g ( x ) = A ( β ,x, b β ) /A b β . If(4.7) g ( x − ) = g ( x + ) = 0 , then h ( x − ) = h ( x + ) = 0 . But h ( x ) is a linear function in x , so this is possible only if h ( x ) = 0 for every x ∈ R . Thisis possible only if(4.8) A − b β z T = (cid:0) b · · · b k − (cid:1) T for some b , . . . , b k ∈ R and β k = k − X i =2 β i b i . We write ( A β ( x ) ) | S ,S for the restriction of A β ( x ) to rows from S and columns from S . Since A β ( x ) is a Hankelmatrix, we have ( A β ( x ) ) { ,...,X k − } , { X,...,X k } = ( A β ( x ) ) { X,...,X k } , { ,...,X k − } , which is equal to (cid:18) e ( x ) β k A b β z T (cid:19) = (cid:18) e ( x ) T A b β β k z (cid:19) . (4.8) implies that the last column of ( A β ( x ) ) { ,...,X k − } , { X,...,X k } is in the span of the columns , . . . , k − . From ( A β ( x ) ) { X,...,X k } , { ,...,X k − } this in particular implies that the last column of A b β is in the span of the others and A b β issingular, which is a contradiction with the assumption A e β ≻ . Therefore (4.7) cannot be true and one of g ( x − ) and g ( x + ) is positive. By Proposition 2.3, this means that A ( β ,x + , b β ) ≻ or A ( β ,x − , b β ) ≻ and hence rank A ( β ,x − , b β ) = k or rank A ( β ,x + , b β ) = k . By Proposition 2.3 and (4.6), rank A β ( x − ) = rank A β ( x + ) = rank A e β = k . Therefore rank A β ( x − ) = rank A ( β ,x − , b β ) or rank A β ( x + ) = rank A ( β ,x + , b β ) . Noticing that A ( β ,x ± , b β ) = A β ( x ± ) ( k − , itfollows that one of x ± satisfies (4.5). This concludes the proof of the moreover part. (cid:3) Remark . For k = 1 , the THMP with gaps ( β ) coincides with the THMP with gaps ( β k − ) and hence the case k = 1 is already covered by Theorem 3.1. Example 4.3.
For k = 9 , let β (1) ( x ) = (cid:16) , x, , , , , , , , , , , , , , , , , (cid:17) ,β (2) ( x ) = (cid:16) , x, , , , , , , , , , , , , , , , , (cid:17) ,β (3) ( x ) = (cid:16) , x, , , , , , , , , , , , , , , , , (cid:17) ,β (4) ( x ) = 19 (9 , x, , − , , − , , − , , − , , − , , − , , − , , − , , Let e A ( i ) and b A ( i ) , i = 1 , , , denote e A , b A , respectively, from Theorem 4.1 corresponding to β ( i ) ( x ) . Using Mathe-matica [Wol] one can check that: • b A (1) ≻ , while for i = 2 , , it holds that b A ( i ) (cid:23) and dim (cid:16) ker b A ( i ) (cid:17) = 1 . • For i = 1 , we have e A ( i ) ≻ for i = 1 , , while for i = 2 , it holds that e A ( i ) (cid:23) and dim (cid:16) ker e A ( i ) (cid:17) = 1 . • A e β ( i ) ≻ for i = 1 , , , A e β (3) (cid:23) and dim (cid:16) ker A e β (3) (cid:17) = 1 . • A b β (3) ≻ and A ˘ β (3) ≻ .Therefore: • A β (1) ( x ) is ppsd and (3a) of Theorem 4.1 is true, implying that a 9-atomic measure for β (1) ( x ) exists. • β (2) ( x ) does not satisfy (3a) neither (3b) of Theorem 4.1, implying there is no representing measure for β (2) ( x ) . • A β (3) ( x ) is ppsd and e β (3) satisfies (3b) of Theorem 4.1 together with rank A b β (3) = rank b A (3) , implying thatan 8-atomic measure for β (3) ( x ) exists. • A β (4) ( x ) is ppsd and e β (4) satisfies (3a) of Theorem 4.1, implying that a 9-atomic measure for β (4) ( x ) exists.The following corollary is a consequence of Theorem 4.1 and gives the solution of the bivariate TMP for the curve y = x . Corollary 4.4.
Let β = ( β i,j ) i,j ∈ Z ,i + j ≤ k be a 2-dimensional real multisequence of degree k . Suppose M ( k ) ispositive semidefinite and recursively generated. Let u ( i ) := ( β ,i , β ,i +1 , β ,i ) for i = 0 , . . . , k − , b β := ( u (0) , . . . , u (2 k − ) , e β := ( b β, β , k − , β , k ) , β := ( β , , u (1) , . . . , u (2 k − ) and ( β := ( β, β , k − , β , k ) be subsequences of β , v := (cid:0) u (0) · · · u ( k − β ,k − (cid:1) a vector and e A := (cid:18) β vv T A β (cid:19) a matrix. Then β has a representing measure supported on y = x if and only if the following statements hold:(1) One of the following holds: • If k ≥ , then Y = X is a column relation of M ( k ) . • If k = 2 , then the equalities β , = β , , β , = β , , β , = β , hold.(2) One of the following holds:(a) A e β ≻ and e A ≻ . (b) A e β (cid:23) and rank A b β = rank A e β = rank A ˘ β . HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 19
Moreover, if the representing measure exists, then there exists a (rank e β ) -atomic measure if (2a) is true or (2b) holdswith rank A b β = rank M ( k ) . Otherwise there is a (rank e β + 1) -atomic measure.Proof. For m ∈ { , . . . , k } we define the numbers e β m by the following rule e β m := β , m , if m (mod 3) = 0 ,β , ⌊ m ⌋− , if m (mod 3) = 1 ,β , ⌊ m ⌋ , if m (mod 3) = 2 . Claim 1.
Every number e β m is well-defined.We have to prove that i + j ≤ k , where i, j are indices of β i,j used in the definition of e β m . We separate three casesaccording to m : • m (mod 3) = 0 : m ≤ k . • m (mod 3) = 1 : ⌊ m ⌋ − ≤ k −
2) = 2 k. • m (mod 3) = 2 : ⌊ m ⌋ ≤ k −
1) = 2 k. Claim 2.
Let t ∈ N . The atoms ( x , x ) , . . . ( x t , x t ) with densities λ , . . . , λ t are the ( y − x ) -representingmeasure for β if and only if the atoms x , . . . , x t with densities λ , . . . , λ t are the R -representing measure for e β ( x ) = ( e β , x, e β , . . . , e β k ) .The if part follows from the following calculation: e β m = β , m , if m (mod 3) = 0 ,β , ⌊ m ⌋− , if m (mod 3) = 1 ,β , ⌊ m ⌋ , if m (mod 3) = 2 , = P tℓ =1 λ ℓ ( x ℓ ) m , if m (mod 3) = 0 , P tℓ =1 λ ℓ ( x ℓ ) ( x ℓ ) ⌊ m ⌋− , if m (mod 3) = 1 , P tℓ =1 λ ℓ x ℓ ( x ℓ ) ⌊ m ⌋ , if m (mod 3) = 2 , = t X ℓ =1 λ ℓ x mℓ , where m = 0 , , . . . , k .The only if part follows from the following calculation: β i,j = β i − ,j +2 = · · · = β i (mod 3) ,j +2 ⌊ i ⌋ = e β i (mod 3))+3( j +2 ⌊ i ⌋ ) = t X ℓ =1 λ ℓ x i (mod 3))+3( j +2 ⌊ i ⌋ ) ℓ = t X ℓ =1 λ ℓ x i (mod 3)+3 ⌊ i ⌋ ) ℓ x jℓ = t X ℓ =1 λ ℓ ( x ℓ ) i ( x ℓ ) j , where the first three equalities in the first line follow by M ( k ) being rg.Using Claim 2 and a theorem of Bayer and Teichmann [BT06], implying that if a finite sequence has a K -representing measure, then it has a finitely atomic K -representing measure, the statement of the Corollary followsby Theorem 4.1. (cid:3) Truncated Hamburger moment problem of degree k with gaps ( β , β ) .Theorem 4.5. Let k ∈ N , k > , and β ( x, y ) := ( β , x, y, β , . . . , β k ) be a sequence, where each β i is a real number, β > and x, y are variables. Let e β := ( β , . . . , β k − ) , β := ( β , . . . , β k ) , ˘ β := ( β , . . . , β k − ) , β := ( β , . . . , β k ) be subseqeunces of β ( x, y ) , v := (cid:0) β . . . β k − (cid:1) , u := (cid:0) β · · · β k (cid:1) , s := (cid:0) β · · · β k +1 (cid:1) , w := (cid:0) β · · · β k +2 (cid:1) , vectors, and A := (cid:18) β vv T A ˘ β (cid:19) and e A := (cid:18) β uu T A β (cid:19) matrices. Then the following statements are equivalent:(1) There exist x , y ∈ R and a representing measure for β ( x , y ) supported on K = R .(2) There exist x , y ∈ R and a (rank β ) or (rank β + 1) -atomic representing measure for β ( x , y ) . (3) A β ( x,y ) is partially positive semidefinite, (4.9) sA + β s T ≤ uA + β w T + q ( A β /A β )( e A/A β ) and one of the following statements is true:(a) A β ≻ and one of the following holds:(i) (A) k = 3 and the inequality in (4.9) is strict..(B) k > , A ≻ and the inequality in (4.9) is strict.(ii) The following inequalities holds: uA + e β u T < sA + β s T and uA + β w T − q ( A β /A β )( e A/A β ) ≤ sA + β s T . (b) rank A e β = rank A β = rank (cid:0) s T A β (cid:1) .Moreover, if the representing measure exists, then there is a (rank β ) -atomic if and only if (3(a)ii) or (3b) holds.Proof. Note that β ( x, y ) admits a measure if and only if there exist y ∈ R such that β ( x, y ) admits a measure.Theorem 4.1 implies the following claim holds. Claim 1. β ( x, y ) admits a measure if and only if the following conditions hold:(1) A β ( x,y ) is ppsd.(2) Denoting e β ( y ) := ( y , β , . . . , β k − ) and β ( y ) := ( y , β , . . . , β k ) , one of the following is true:(a) A β ( y ) = (cid:18) y ss T A β (cid:19) ≻ and A ( y ) ≻ , where A ( y ) := β y β yβ A e β , if k = 3 , (cid:18) β v ( y ) v ( y ) T A e β (cid:19) = β y vy β w v T w T A ˘ β , v ( y ) T = (cid:18) yv (cid:19) and w T = β ...β k +2 , otherwise . (b) rank A e β ( y ) = rank A β ( y ) = rank A β . We denote by e A ( y ) := (cid:18) β u ( y ) u ( y ) T A β (cid:19) where u ( y ) = (cid:0) y u (cid:1) . Claim 2.
Assume A β ≻ or rank A β = rank A β . Then e A ( y ) (cid:23) if and only if(4.10) e A ( y ) is ppsd and y ∈ h uA + β w T − q ( A β /A β )( e A/A β ) , uA + β w T + q ( A β /A β )( e A/A β ) i =: [ y − , y + ] . Moreover,(4.11) rank e A ( y ) = ( max { rank A β , rank e A } , y ∈ { y − , y + } , max { rank A β , rank e A } + 1 , y ∈ ( y − , y + ) . Let P be the permutation matrix as in the proof of Theorem 4.1. We have that P T e A ( y ) P is of the form(4.12) P T e A ( y ) P = A β w T u T w β yu y β , and denoting the matrices A := (cid:18) A β w T w β (cid:19) and B := (cid:18) A β u T u β (cid:19) , HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 21 and the permuation matrix P : R k − → R k − by P = (cid:18) I k − (cid:19) , where stands for the row of k − zeros and I k − is the identity matrix of size k − , we have that(4.13) A = P T A β P and B = P T e AP . By the assumptions in Claim 2 and (4.13), A β ≻ or rank A β = rank A . Hence, the assumption (2.9) of Lemma2.11 used for P T e A ( y ) P , A β , A , B as A ( x ) , A , A , A , respectively, is satisfied and using also A /A β = A β /A β , B /A β = e A/A β , Claim 2 follows.Theorem 2.2 implies the following claim. Claim 3.
It is true that:(1) A β ( y ) (cid:23) if and only if(4.14) A β (cid:23) , s T ∈ C ( A β ) and A β ( y ) /A β = y − sA + β s T ≥ . (2) A e β ( y ) (cid:23) if and only if(4.15) A e β (cid:23) , u T ∈ C ( A e β ) and A e β ( y ) /A e β = y − uA + e β u T ≥ . Claim 4.
Assume A β ≻ or rank A β = rank A β . Then A β ( x,y ) is ppsd for some y ∈ R if and only if A β ( x,y ) isppsd, s T ∈ C ( A β ) and (4.9) holds.Note that A β ( x,y ) is ppsd if and only if A β ( y ) (cid:23) and e A ( y ) (cid:23) . The first condition of (4.10) (which alsoincludes the first condition of (4.14)) is equivalent to A β ( x,y ) being ppsd. Further on, y satisfying the third conditionof (4.14) and the second condition of (4.10) exists if and only if (4.9) holds. This proves Claim 4.First we prove the implication (1) ⇒ (3). By Claim 1, in particular A β ( x,y ) (and hence also A β ( x,y ) ) is ppsd. Since e β ( y ) also admits a measure by Proposition 2.13, we either have A e β ( y ) ≻ and in particular A β ≻ , or A e β ( y ) issingular and it follows by Corollary 2.8 that A β ≻ or rank A β = rank A β . If (2a) of Claim 1 holds, then in particular A β ≻ and if k > also A ≻ . Since A β ( y ) ≻ , it follows usingProposition 2.3 that A β ( y ) /A β > or equivalently y > sA β + s T . Since by Claim 2, y ∈ [ y − , y + ] , this implies that sA β + s T < y + which means that the inequality in (4.9) is strict. Hence, A β ( x,y ) is ppsd, A β ≻ and (3(a)i) holds.This proves the implication (1) ⇒ (3) in this case.Assume now that (2b) of Claim 1 holds. There are two cases to consider: • A β ≻ : It follows that rank A e β ( y ) = rank A β ( y ) = k − , which implies that: – A e β ( y ) ≻ since A e β ( y ) is of size k − . – By Proposition 2.3, y = sA + β s T since A β ( y ) = (cid:18) y ss T A β (cid:19) is singular. – k − ≤ rank A β ( x ,y ) ≤ k for some x ∈ R such that A β ( x ,y ) (cid:23) , since A β ( x ,y ) = (cid:18) β u ( x , y ) u ( x , y ) T A β ( y ) (cid:19) where u ( x , y ) = (cid:0) x y u (cid:1) . From A e β ( y ) ≻ and y = sA + β s T , it follows by Proposition 2.3 and (4.15) that uA + e β u T < sA + β s T . Furtheron, y − ≤ sA + β s T since by Claim 2, e A ( y ) (cid:23) implies that y ∈ [ y − , y + ] . Hence, A β ( x,y ) is ppsd, A β ≻ and (3(a)ii) holds. This proves the implication (1) ⇒ (3) in this case. • A β : By Lemma 2.13, β also admits a measure and hence by Corollary 2.10 used for β as β , rank A e β =rank A β . Together with the second condition in (4.14), this implies that (3b) holds. Since A β ( x,y ) is ppsd and(4.9) holds, this proves the implication (1) ⇒ (3) in this case. Second we prove the implication (3) ⇒ (1). If (3a) holds, then A β ≻ and in particular A β ≻ . Else (3b) holdsand in particular A β is singular. By Claim 3, A β ( sA + β s T ) (cid:23) and hence by Corollary 2.8 used for β ( sA + β s T ) as β weconclude that rank A β = rank A β . Hence the assumption of Claims 2 and 4 is satisfied and A β ( x,y ) is ppsd for every y from the interval [max { y − , sA + β s T } , y + ] . We separate cases three cases according to the assumptions: • Case (3(a)i): We separate two cases according to the invertibility of e A . – e A ≻ : Since A β ≻ and e A ≻ , it follows that A β /A β > and e A/A β > . By the form of y ± given in Claim 2, we have that y − < y + . Since by assumption also the inequality (4.9) is strict,the interval (max { y − , sA + β s T } , y + ) is not empty and hence for every y ∈ (max { y − , sA + β s T } , y + ) , A β ( x,y ) satisfies (2a) above by Claims 2 and 3. This proves the implication (3) ⇒ (1) in this case. – e A in singular: First we show that the last column of e A is in the span of others. We separate two casesaccording to k . ∗ k = 3 : Since e A = (cid:18) β β β β (cid:19) and β > , it follows that the second (also the last) column of e A is a multiple of the first (also it the span of the others). ∗ k > : Since A ≻ , the last column of e A = (cid:18) A r T r β k (cid:19) is in the span of the others, where r = (cid:0) β k β k +2 · · · β k − (cid:1) .Since A β ≻ , it follows that A β ≻ and e A/A β = 0 . By the form of y ± given in Claim 2, we havethat y − = y + . By Claim 2, e A ( y + ) = (cid:18) A ( y + ) r T r β k (cid:19) (cid:23) , where r = (cid:0) β k · · · β k − (cid:1) , and rank e A ( y + ) = rank A β = k − . By Lemma 2.12, the last column of e A ( y + ) is in the span of the othersand hence A ( y + ) ≻ . Since by assumption also the inequality (4.9) is strict, A β ( y + ) ≻ by Claim 3.Hence, (2a) of Claim 1 holds for y = y + , which proves the implication (3) ⇒ (1) in this case. • Case (3(a)ii): β ( x, sA + β s T ) is ppsd. Since A β ( sA + β s T ) = A e β ( sA + β s T ) r T r β β k ! , where r = (cid:0) β k +1 · · · β k − (cid:1) , is singular, the assumption uA + e β u T < sA + β s T and Claim 3 implythat A e β ( sA + β s T ) ≻ , hence rank A e β ( sA + β s T ) = rank A β ( sA + β s T ) = rank A β . Hence, (2b) of Claim 1 for y = sA + β s T holds, which proves the implication (3) ⇒ (1) in this case. • Case (3b): By assumption rank A e β = rank A β , it follows that the last column of A β is in the span of theothers. There exists x ∈ R such that A β ( x ,y + ) is psd and by Lemma 2.12, the last column of A β ( y + ) is in thespan of the others and hence rank A e β ( y + ) = rank A β ( y + ) . Since A β ( y + ) is singular, using Corollary 2.8 with β equal to β ( x , y + ) , we get rank A β ( y + ) = rank A β , which in particular implies that y + = sA + β s T . Hence, rank A e β ( y + ) = rank A β ( y + ) = rank A β , which is (2b) of Claim 1. This proves the implication (3) ⇒ (1) inthis case.It remains to prove the implication (1) ⇒ (2). By Theorem 4.1, if β ( x, y ) has a representing measure, then thereis a (rank β ( y )) or (rank β ( y ) + 1) -atomic representing measure. By Corollary 2.8, rank β ( y ) = rank A β ( y ) =rank A β if A β ( y ) is singular and rank β ( y ) = rank A β + 1 = rank β + 1 otherwise.For the moreover part, note from the previous paragraph that (rank β ) -atomic measure exists if and only if A β ( y ) =rank A β for some y such the β ( x, y ) admits a measure. The only y ∈ R satisfying rank A β ( y ) = rank A β is sA + β s T and hence a (rank β ) -atomic measure exists if and only if β ( x, sA + β s T ) admits a measure. From the proof ofthe implication (3) ⇒ (1) we see that this is true in the cases (3(a)ii) and (3b). Finally, if (3(a)i) holds, then we see that: • If e A ≻ , then we must have y − ≤ sA + β s T and uA + e β u T < sA + β s T (see the proof of (3(a)ii)), which meansthat (3(a)ii) holds. • If e A is singular, then sA + β s T < y − = y + and β ( x, sA + β s T ) does not admit a (rank β ) -atomic measure.This establishes the proof of the moreover part. (cid:3) HE TRUNCATED HAMBURGER MOMENT PROBLEMS WITH GAPS IN THE INDEX SET 23
Remark . For k = 2 , the THMP with gaps ( β , β ) coincides with the THMP with gaps ( β k − , β k − ) and hencethe case k = 2 is already covered by Theorem 3.5 .The following corollary is a consequence of Theorem 4.5 and solves the bivariate TMP for the curve y = x wherealso β , is given. Here β , stands for the integral of x w.r.t. µ , i.e., R K x dµ . Corollary 4.7.
Let β = ( β i,j ) i,j ∈ Z ,i + j ≤ k β be a 2-dimensional real multisequence of degree k and let β , be alsogiven. Suppose M ( k ) is positive semidefinite and recursively generated. Let u (1) = ( β , , β , , β , , β , ) , u ( i ) = ( β ,i , β ,i − , β ,i − , β ,i ) for i = 2 , . . . , k − , e β := ( u (1) , . . . , u (2 k − , β , k − , β , k − , β , k − ) , β := ( e β, β , k − , β , k ) , ˘ β := ( b β, β , k − , β , k − ) and β := ( ˘ β, β , k − , β , k ) be subsequences of β , v := (cid:0) β , u (1) · · · u ( k − β ,k − β ,k − β ,k − β ,k − (cid:1) , u := (cid:0) v β ,k (cid:1) ,s := (cid:0) u β ,k − (cid:1) , w := (cid:16) β , β , β , u (2) · · · u ( k − β ,k β ,k − β ,k − (cid:17) vectors and A := (cid:18) β vv T A ˘ β (cid:19) and e A := (cid:18) β uu T A β (cid:19) matrices. Then β has a representing measure supported on y = x if and only if (4.16) sA + β s T ≤ uA + β w T + q ( A β /A β )( e A/A β ) one of the following statements hold:(1) One of the following holds: • If k ≥ , then Y = X is a column relation of M ( k ) . • If k = 3 , then the equalities β , = β , , β , = β , , β , = β , , β , = β , , β , = β , . • If k = 2 , then the equality β , = β , holds.(2) One of the following holds:(a) A β ≻ , A ≻ and the inequality in (4.16) is strict.(b) A β ≻ and the following inequalities holds: uA + e β u T < sA + β s T and uA + β w T − q ( A β /A β )( e A/A β ) ≤ sA + β s T . (c) A β (cid:23) and rank A e β = rank A β = rank (cid:0) s T A β (cid:1) .Moreover, if the representing measure exists, then there exists a (rank β ) -atomic measure if and only if (2b) or (2c) holds. Otherwise there is a (rank β + 1) -atomic measureProof. For { , , , , . . . , k } we define the numbers e β m by the following rule e β m := β , m , if m (mod 4) = 0 ,β , ⌊ m ⌋− , if m (mod 4) = 1 ,β , ⌊ m ⌋− , if m (mod 4) = 2 ,β , ⌊ m ⌋ , if m (mod 4) = 3 . Claim 1.
Every number e β m is well-defined.We have to prove that i + j ≤ k , where i, j are indices of β i,j used in the definition of e β m . We separate four casesaccording to m : • m (mod 4) = 0 : m ≤ k . • m (mod 4) = 1 : ⌊ m ⌋ − ≤ (2 k −
1) + 1 = 2 k. • m (mod 4) = 2 : ⌊ m ⌋ − ≤ (2 k −
1) + 1 = 2 k. • m (mod 4) = 3 : ⌊ m ⌋ + 1 ≤ (2 k −
1) + 1 = 2 k. We also define e β := β , . Claim 2.
Let t ∈ N . The atoms ( x , x ) , . . . ( x t , x t ) with densities λ , . . . , λ t are the ( y − x ) -representing measurefor β with β , known if and only if the atoms x , . . . , x t with densities λ , . . . , λ t are the R -representing measure for e β ( x, y ) = ( e β , x, y, e β , . . . , e β k ) .The if part follows from the following calculation: e β m = β , m , if m (mod 4) = 0 ,β , ⌊ m ⌋− , if m (mod 4) = 1 ,β , ⌊ m ⌋− , if m (mod 4) = 2 ,β , ⌊ m ⌋ , if m (mod 4) = 3 , = P tℓ =1 λ ℓ ( x ℓ ) m , if m (mod 4) = 0 , P tℓ =1 λ ℓ ( x ℓ ) ( x ℓ ) ⌊ m ⌋− , if m (mod 4) = 1 , P tℓ =1 λ ℓ ( x ℓ ) ( x ℓ ) ⌊ m ⌋− , if m (mod 4) = 2 , P tℓ =1 λ ℓ x ℓ ( x ℓ ) ⌊ m ⌋ , if m (mod 4) = 3 , = t X ℓ =1 λ ℓ x mℓ , where m = 0 , , , , . . . , k and e β = β , = t X ℓ =1 λ ℓ ( x ℓ ) = t X ℓ =1 λ ℓ x ℓ . The only if part follows from the following calculation: β i,j = β i − ,j +3 = · · · = β i (mod 4) ,j +3 ⌊ i ⌋ = e β i (mod 4))+4( j +3 ⌊ i ⌋ ) = t X ℓ =1 λ ℓ x i (mod 4))+4( j +3 ⌊ i ⌋ ) ℓ = t X ℓ =1 λ ℓ x i (mod 4)+4 ⌊ i ⌋ ) ℓ x jℓ = t X ℓ =1 λ ℓ ( x ℓ ) i ( x ℓ ) j , where the first three equalities in the first line follow by M ( k ) being rg and β , = e β = t X ℓ =1 λ ℓ x ℓ = t X ℓ =1 λ ℓ ( x ℓ ) . Using Claim 2 and a theorem of Bayer and Teichmann [BT06], implying that if a finite sequence has a K -representing measure, then it has a finitely atomic K -representing measure, the statement of the Corollary followsby Theorem 4.5. (cid:3) R EFERENCES[Akh65] N. I. Akhiezer,
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