arX

iv:1

203.

5859

v2 [

mat

h.C

A]

13 A

pr 2

012

SPARSE MOMENT SEQUENCES

SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

Abstract. The well-known theorems of Stieltjes, Hamburger and Hausdorff establish con-ditions on infinite sequences of real numbers to be moment sequences. Further, works byCaratheodory, Schur and Nevanlinna connect moment problems to problems in functiontheory and functions belonging to various spaces. In many problems associated with re-alization of a signal or an image, data may be corrupt or missing. Reconstruction of afunction from moment sequences with missing terms is an interesting problem leading toadvances in image and/or signal reconstruction. It is easy to show that a subsequence of amoment sequence may not be a moment sequence. Conditions areobtained to show howrigid the space of sub-moment sequences is and necessary andsufficient conditions for asequence to be a sub-moment sequence are established. A deepconnection between thesub-moment measures and the moment measures is derived and the determinacy of the mo-ment and sub-moment problems are related. This problem is further related to completionof positive Hankel matrices.

1. Introduction

While numerous applications of moment problems with a complete set of moments havebeen been identified, they are mostly limited to theoreticalobservations. For practical im-plementation of moment problems, it is vital to be able to deal with missing moment datasince data obtained from physical experiments and phenomena are often corrupt or incom-plete. We call a positive subsequence of a moment sequence a sub-moment sequence andthe moment problem of sub-moment sequence a sub-moment problem. In this paper wegive a complete characterization of the sub-moment problemand develop a deep connec-tion between the measures arising from the sub-moment problem and the original momentproblem. Further, we relate this to completion of positive Hankel matrices and arrive at anecessary and sufficient condition for the completion of such matrices. Recentadvancesin image reconstruction from sparse MRI data [6], partial data transmission and sparsesignals, make this problem timely and of broad interest.

In his memoir ”Recherches sur les Fractions Continues” [30]from 1894-95, Stieltjesintroduced the ”Problem of Moments” on [0,∞): find a bounded non-decreasing functionσ(u) in the interval [0,∞) such that its ”moments”,sn, are given as

∫ ∞

0undσ(u), n ∈N0 = {0,1,2, · · · }

Stieltjes took the terminology ”Problem of Moments” from Mechanics. He often usedmechanical concepts of mass, density, stability, etc., in solving analytical problems. We

can considerdσ(u) to be mass distributed over the interval [u,u+ du] so that∫ u

0dσ(t)

gives the mass distributed over [0,u]. Accordingly,∫ ∞

0udσ(u) and

∫ ∞

0u2dσ(u) respec-

tively represent the first moment and the second moment, alsocalled moment of inertia,

2010Mathematics Subject Classification.44A60.

1

2 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

with respect to 0 of the total mass∫ ∞

0dσ(u) distributed over the interval [0,∞). Stieltjes

generalizes this concept to call∫ ∞

0undσ(u) then-th moment of the given mass distribution

characterized by the functionσ(u). Stieltjes proves several results concerning solubilityofthe problem of moments and uniqueness of the solution. He shows that for a sequence{sn}the moment problem is solvable if and only if the determinants of the matrices

Dn =

s0 s1 · · · sn

s1 s2 · · · sn+1...

.... . .

...

sn sn+1 · · · s2n

, D(1)n =

s1 s2 · · · sn+1

s2 s3 · · · sn+2...

.... . .

...

sn+1 sn+2 · · · s2n+1

, n ∈N0

are non-negative. There have been several sufficiency conditions established for unique-ness of the solution to the moment problem. While Stieltjes had mechanics in mind, Cheby-shev and Markov had explicitly probability theory as aims ofmoment problems.

Hamburger continued the work of Stieltjes in his series of papers ”Uber eine Erweiterungdes Stieltjes Momentenproblems [16]”, from 1920-21. He treated the moment problemsas a theory of its own. He extended the Stieltjes’ moment problem to the whole real axis.The Hamburger Moment Problem is stated as: find a non-decreasing functionσ(u) in theinterval (−∞,∞) such that its moments,sn, are given by

∫

R

undσ(u), n ∈N0.

He established that a necessary and sufficient condition for the moment solution to existis the positivity of moment sequence. Later, mathematicians including R. Nevanlinna, M.Riesz, T. Carleman, F. Hausdorff, M. Stone, and C. Caratheodory further studied severalmoment problems.

A moment problem that is not well-posed may not be solvable. Criteria for existence ofsolution is studied to a great extent for every type of momentproblem. Once the solubilitycriteria are determined, then comes the question of uniqueness of the solution since it isoften the case that there is more than one solution. Two solutions of a moment problem arenot considered to be distinct if their difference is a constant at all the points of continuityof the difference. Such a moment problem with no more than one distinct solution is calleddeterminate, andindeterminateotherwise.

Hausdorff’s moment problem deals with finding a functionσ(u) with support on theclosed unit interval [0,1] such that a given sequence{sn} satisfies

sn =

∫ 1

0unσ(u), n ∈N0.

Hausdorff establishes that there is a unique functionσ(u) in [0,1] if and only if the se-quence{sk} is completely monotonic, that is, it satisfies

(−1)n∆nsk ≥ 0, for all n,k≥ 0,

where∆ is the difference operator defined as

∆nsk =

n∑

i=0

(

ni

)

(−1)n−i si+k.

This condition for existence and uniqueness holds for the moment problem in any finiteinterval. In fact, the Hausdorff moment problem is often stated on the interval [−1,1] justfor the sake of simplicity.

SPARSE MOMENT SEQUENCES 3

The problem of moments stands as a very important problem in analysis up to thepresent day. Results on moment problems have numerous applications in the areas ofextremal problems, optimization theory and limit theoremsin probability theory. Ratherthan the usual infinite moment sequence problem, the truncated moment problem is moreapplicable in mathematical and physical sciences. Further, multivariable moment problemis more applicable than a single variable one. Multivariateextension of moment problemis also largely studied (see [11] and [26] for instance).

Moments of a geometric object conveys a lot of its topological information such as howround it is, where it is located, what direction it is tapered, where its mass is centered,etc. For simple objects like an ellipsoid, just a few terms inthe moment sequence areenough to identify it. However, only infinitely many terms can uniquely identify morecomplicated shapes like a polygon or a quadrature domain. The authors in [25] give anexample of identifying a polygon from its given moments. They prove that the verticesof a nondegenerate, simply connected n-gon in the plane are uniquely determined by itsmoments up to order 2n− 3. Theory of moments can also be applied to some modelingproblems. An example of speech modeling can be found in [9]. Additional examples ofdata reconstruction with moments can be found in [10], [14],[15], [19] and [29].

Various generalizations and extensions of the problems of moments have been studiedto deal with several applications. Some generalizations include replacing the sequence offunctions{un} with a more general sequencefn(u) such as{einu}, and replacing integralswith more general functionals in abstract spaces. Lasser [23], for example, studies momentproblem by replacing{un} with polynomials. However, we will not consider these gener-alizations in this paper. The important variation of the moment problem we look at is themoment problem with some of the moments missing. Moment dataobtained from variousexperiments are often corrupt and incomplete. Such reconstruction of moments can be thenapplied in problems associated with realization of a signalor an image. Reconstruction ofa function from moment sequences with missing terms is an interesting problem leading toseveral advances in image and/or signal reconstruction. In this paper we will work towardsfinding techniques for reconstructing the missing moment data.

Several authors have studied this problem by replacing the missing moments with anappropriate value, leading to a perturbation of the moment sequences followed by thequestion of sensitivity of the corresponding orthogonal polynomials. Because of necessaryrequirements for a sequence to be moments, this approach turns out to be very restrictive inchoosing and replacing the missing moments. Thus, we look for ideas to extract a positivesubsequence of a positive sequence and look for its moment solution.

We primarily have two questions to answer:

Q1. How do we characterize positive subsequences of a positive sequence?Q2. How do we characterize the corresponding non-decreasing measures arising from

the sub-moment data and what, if any, is the relationship of these measures to theoriginal measures?

In this paper we establish several methods for extracting positive subsequences and lookat solutions to the sub-moment problems. Unless stated otherwise, bymoment problemweshall refer to theHamburger moment problemthroughout the paper. In the first sectionof this paper, we will thoroughly discuss some classical results and properties of positivesequences and corresponding polynomials. Then we will survey some of the prominentworks in the field of moment problem in the next couple of sections. The results describedhere, mostly without proof and with reference to literature, are those that will play an es-sential role in the sub-moment problem. The second part of this paper is dedicated to the

4 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

study of the sub-moment sequences, sub-moment problems andpositive Hankel matrixcompletion problem. We shall establish necessary and sufficient conditions for a subse-quence of a moment sequence to be positive. We will also find conditions for determinacyof sub-moment problems. These results give a complete answer to the problem of Hankelmatrix completion.

2. Positive Sequences

Positive sequences play an important role in the theory of orthogonal polynomials, pos-itive functionals and moment problems. This section is concerned with an overview ofsome characteristics of positive sequences. In this section, we will describe many of theclassical theorems and results that we will need in the ensuing sections of this paper. Inmost cases the proofs are not given and the reader is referredto one of the several paper orclassical monographs on the theory. In a few instances, for later reference and for readers’convenience, brief proofs are provided.

Definition 2.1. An infinite sequence{sk} is positive if the quadratic formm

∑

i, j=0

xi x j si+ j

is positive for any m and x0, x1, · · · , xm ∈R. This is equivalent to saying that the determi-nant

(1) Dm=

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

s0 s1 · · · sm

s1 s2 · · · sm+1...

.... . .

...

sm sm+1 · · · s2m

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

is non-negative for m∈N0.

Proposition 2.2. Term-wise sum and product of two positive sequences is positive.

Proof. The claim that the term-wise sum is positive follows readilyfrom the definition.The claim that the term-wise product is positive follows from the Schur product theorem[18, Theorem 7.5.3]. �

Given a sequence{sk}, define a linear functionalΛ such that

Λ(P(u)) = p0s0+ p1s1+ · · ·+ pnsn

where

P(u) = p0+ p1u+ · · ·+ pnun.

Definition 2.3. A linear functionalΛ is called positive if for any polynomial P: R→ R,P(u) ≥ 0 and P(u) . 0, it follows thatΛ(P(u)) > 0.

Theorem 2.4. For any sequence{sk}, the functionalΛ defined above is positive if and onlyif {sk} is positive.

Proof. Suppose{sk} is positive. Any polynomialP(u) = p0+ p1u+ · · ·+ pnun ≥ 0 andP(u) . 0 can be written in the form

P(u) = [A(u)]2+ [B(u)]2

SPARSE MOMENT SEQUENCES 5

whereA(u),B(u) are polynomials with real coefficients. Then we have

Λ(P(u)) =m

∑

i, j=0

xi x j si+ j +

m∑

i, j=0

yiy j si+ j ≥ 0.

Now supposeΛ is positive. Thenm

∑

i, j=0

xi x j si+ j = Λ(

[P(u)]2)

,

whereP(u) = x0+ x1u+ · · ·+ xmum. Hence{sk} is positive. �

Given a positive sequence{sk}, we can construct a set of polynomials{Pn(u)} such thatPn(u) is of degreen with positive leading coefficient and they are orthonormal with respectto the functionalΛ. These polynomials are explicitly given by the formula

(2) Pn(u) =1√

Dn−1Dn

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

s0 s1 · · · sn−1 sn

s1 s2 · · · sn sn+1...

.... . .

......

sn−1 sn · · · s2n s2n−1

1 u · · · un−1 un

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

,

where

Dn =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

s0 s1 · · · sn

s1 s2 · · · sn+1...

.... . .

...

sn sn+1 · · · s2n

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

, andD−1 = 1.

It is well known that this orthonormal set of polynomials{Pn(u)} satisfy the three termrecurrence relation

(3) uPn(u) = bnPn+1(u)+anPn(u)+bn−1Pn−1(u), n= 1,2,3, · · · ,where

an = Λ(

u[Pn(u)]2)

, andbn =

√Dk−1Dk+1

Dk

and

P0(u) = 1, andP1(u) =u−a0

b0.

Together with the orthonormal polynomials{Pn(u)}, the polynomials of the second kind{Qn(u)} are also generated from the same three-term recurrence relation (3) with initial

conditionsQ0(u) = 0,Q1(u) =1b0

. {Pn(u)} and{Qn(u)} are linearly independent solutions

to (3). In fact,Qn(u) is a polynomial of degreen−1 and can be expressed explicitly as

(4) Qn(u) = Λ

(

Pn(u)−Pn(v)u−v

)

.

Orthonormal sets of polynomials{Pn(u)} and{Qn(u)} play an important role in the the-ory of moment problems. See [13] for further details on the polynomials and their recur-rence relation. Several conditions on these polynomials have been established to determinethe existence and uniqueness of solutions of moment problems. We will elaborate more onthe concepts and usage of these polynomials in the later sections.

6 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

3. Classical Moment Problems & Determinacy

In this section we review some of the key results in the Hamburger’s moment prob-lem, also called power moment problem, which is stated as: given an infinite sequence ofnumbers{sk}, find a non-decreasing functionσ(u) such that

(5) sk =

∫

R

ukσ(u).

This problem was formulated and solved by Hamburger. Later works by mathematiciansincluding Fischer, Akheizer and Krein establish conditions on the spectrum of the momentsolutions. Some of these results are summarized in the theorems below. The reader isreferred [1] and [28] for an extensive account of the principle results.

Theorem 3.1(Hamburger). There is a non-decreasing functionσ(u) with an infinite num-ber of points of increase such that equation(5) holds if and only if{sk} is positive.

Theorem 3.2. For a non-decreasing functionσ(u) with precisely k+ 1 distinct points ofincrease such that equation(5) holds, it is necessary and sufficient that

D0,D1, · · · ,Dk > 0, Dk+1 = Dk+2 = · · · = 0,

where Di is the determinant of the Hankel matrix defined in(1).

For a proof of the Hamburger Theorem, refer to [1]. The proof uses the Helly’s Theoremextensively. In some instances, such as in [24] and [31], theproperties of the determinantsDm have been studied to describe the moment solution.

The orthonormal set of polynomials{Pn(u)} and{Qn(u)} are used to establish the con-ditions on determinacy of the moment problems. So before moving on to results on deter-minacy we will define the circleKn(λ) as follows.

Theorem 3.3(Hellinger). Letλ ∈ C be fixed withℑλ > 0 (ℑλ < 0) and letτ ∈R. Then

(6) wn(λ,τ) = −Qn(λ)− τQn−1(λ)Pn(λ)− τPn−1(λ)

= −Qn(λ,τ)Pn(λ,τ)

describes a circular contour Kn(λ) in the open half planeℑw> 0 (ℑw< 0). The center ofthis circle is at the point

(7) − Qn(λ)Pn−1(λ)−Qn−1(λ)Pn(λ)

Pn(λ)Pn−1(λ)−Pn−1(λ)Pn(λ)with radius

(8)1

∣

∣

∣λ−λ∣

∣

∣

n−1∑

k=0

|Pk(λ)|2.

Furthermore, the equation of the circle can be written in theform

(9)w−w

λ−λ−

n−1∑

k=0

|wPk(λ)+Qk(λ)|2 = 0

Proof. Refer to [17]. �

Note thatKn+1(λ) ⊆ Kn(λ). We call the limiting circleK∞(λ). It can be shown that thequantity

(10) w(λ) =∫

R

dσ(u)u−λ

SPARSE MOMENT SEQUENCES 7

falls on the circumference ofK∞(λ), [1, p. 30-34].Now we look at some results concerning determinacy of the Hamburger moment prob-

lem. Most of the determinacy theorems we talk about here dealwith the nature of thelimiting circle K∞(λ). It is useful to present the following theorem and its proofon deter-minacy since we will be using it for the case of sub-moment problem in a later section.

Theorem 3.4. If {sk} is a positive sequence such that its corresponding limitingcirclesK∞(λ) is a point, then the Hamburger moment problem of sequence{sk} is determinate.

Proof. Suppose the moment problem has two solutionsσ1(u) andσ1(u). Consider thefunctions

(11) f1(u) =∫

R

dσ1(u)u−λ , and f2(u) =

∫

R

dσ2(u)u−λ ,

for someλ ∈ C such thatℑλ , 0. Now for a fixedλ, each off1(λ) and f2(λ) are inK∞(λ)by Theorem 2.2.4 in [1]. But ifK∞(λ) is a point, thenf1(λ) and f2(λ) coincide. Putting

α(u) = σ1(u)−σ2(u),

we have∫

R

dα(u)u−λ = 0.

By the Stieltjes and Perron inversion formula,

α(u−0)+α(u+0)2

= c for some constantc.

Thereforeσ1(u) andσ2(u) represent the same solution of the moment problem. �

The converse of this theorem is also true. That is, if the limiting circle K∞(λ) is not apoint, then the Hamburger moment problem is indeterminate.Carleman[7] proved severalstriking results regarding determinacy of the Hamburger moment problem. We mention acouple of them here.

Theorem 3.5(Carleman). If for a positive sequence{sk},∞∑

n=1

1bn=∞,

where bn are constants as given in the three term recurrence relation(3), then the momentproblem is determinate.

Theorem 3.6(Carleman). If for a positive sequence{sk},∞∑

n=1

s− 1

2n2n =∞,

then the moment problem is determinate.

4. Function Theory and theMoment Problems

For a measureσ(u) which arises as a moment solution to a positive sequence, wecandefine theLp space in the usual sense:

(12) Lpσ =

{

f : R→ C : f isσ−measurable and∫

R

| f (u)|pdσ(u) <∞}

.

8 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

The space of functionsLpσ with the norm

(13) ‖ f (u)‖Lpσ=

(∫

R

| f (u)|pdσ(u)

)1p

is complete and is a Hilbert Space whenp= 2 where inner product defined as

(14) (f (u),g(u))σ =∫

R

f (u)g(u)dσ(u)

for any functionsf ,g ∈ Lpσ.

Use of the spaceL2σ to study the properties such as determinacy of the solution(s)σ(u)

for the moment problem is very prevalent. This is done mostlyin connection to the poly-nomials generated by the corresponding sequence{sk}. Several theorems by M. Rieszcharacterize solutions of a moment problem and its determinacy [27]. Below is a veryuseful consequence of these theorems.

Corollary 4.1. If σ is the solution of a determinate moment problem, then the setof all thepolynomials is dense in L2

σ.

The Nevanlinna formula relates all the solutions of an indeterminate moment problem tothe functions inN−class. This formula establishes a one-to-one correspondence betweenthe aggregate of all the solutionsσ of an indeterminate moment problem and the aggregateof all the functionsφ in class N, which we will define below, augmented by the constantat∞. Applying the Stieltjes-Perron inversion formula to the Nevanlinna formula one candetermineσ in terms ofφ.

Definition 4.2. Nevanlinna matrix is any matrix of the form

[

a(z) b(z)c(z) d(z)

]

where the elements a(z),b(z),c(z) are entire transcendental functions, and the followingtwo conditions are satisfied:

(1) a(z)d(z)−b(z)c(z)= 1(2) For any fixed real numberτ ∈R the function

w(z) =τa(z)−c(z)τb(z)−d(z)

is regular in each open half plane and satisfies

ℑw(z)ℑz

> 0.

SPARSE MOMENT SEQUENCES 9

Now, let us consider the four families of polynomials given as

An(z) =zn−1∑

k=0

Qk(0)Qk(z)

Bn(z) =−1+zn−1∑

k=0

Qk(0)Pk(z)

Cn(z) =1+zn−1∑

k=0

Pk(0)Qk(z)

Dn(z) =zn−1∑

k=0

Pk(0)Pk(z).

If the moment problem is indeterminate, then the sums

∞∑

k=0

|Pk(z)|2 ,∞∑

k=0

|Qk(z)|2

are bounded and converge uniformly to respective limits on compact subsets ofC. Bythe Cauchy-Bunyakovskii inequality, the polynomialsAn(z),Bn(z),Cn(z),Dn(z) convergeuniformly to the entire transcendental functions

A(z) =z∞∑

k=0

Qk(0)Qk(z)

B(z) =−1+z∞∑

k=0

Qk(0)Pk(z)

C(z) =1+z∞∑

k=0

Pk(0)Qk(z)

D(z) =z∞∑

k=0

Pk(0)Pk(z)

on compact subsets ofC.It has already been shown that the limit functionsA(z),B(z),C(z),D(z) are entire tran-

scendental. It can also be shown that the functionsA(z),B(z),C(z),D(z) satisfy the require-ments for elements of a Nevanlinna matrix. Thus the matrix

[

A(z) B(z)D(z) D(z)

]

is a Nevanlinna matrix.Before we state the Nevanlinna formula, we define the Nevanlinna class, N-class, of

functions.

Definition 4.3 (N-class). The Nevanlinna class (Class N) consists of all holomorphic func-tions w= f (z) on the upper half plane such thatℑw > 0. By the Riesz-Herglotz integralrepresentation formula, a function f∈ N may be expressed as

f (z) = µz+ ν+∫

1+uzu−z

dτ(u)

whereµ,ν ∈R,µ ≥ 0 andτ(u) is a non-decreasing function with bounded variation.

10 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

Theorem 4.4. If ℑz> 0, for an indeterminate moment problem,∫

dσ(u)u−z

= −A(z)φ(z)−C(z)B(z)φ(z)−D(z)

,

where A(z),B(z),C(z),D(z) form a Nevanlinna matrix,σ is a solution to the concernedmoment problem, andφ ∈ N.

This theorem leads to the question of relation between the moment solutions of subse-quences to the functions in the N-class. It is worthwhile to look for the moment solutionsof subsequences in the same N-class.

5. Incomplete and Sparse Moment Sequence

There are two main methods for dealing with missing moments:(i) Perturbation/modfication,and (ii) Subsequences. Perturbation or modification methodis done by replacing the miss-ing data with an appropriate value, whereas the method of subsequences is based on simplyremoving the missing moments. In some applications, peoplehave tried replacing the lostmoments with zero as well. Because of Hamburger’s theorem, in both cases the modifieddata has to be positive to guarantee a moment solution. Let uslook at an example of apositive sequence and some of its perturbations and subsequences.

The sequence given bysk =1

k+1is a positive sequence since, for everyk, determinant

of Hankel matrix is simply given as

Dk =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

112

13· · · 1

k+112

13

14· · · 1

k+2

13

14

15· · ·

...

......

.... . .

...

1k+1

1k+2

· · · · · · 12k+1

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

=

k∏

i=0

(2i +1)

(

2ii

)2

,

which is positive. So the sequence{sk} is positive and has a moment solution.By settings3 = 0, the second Hankel matrix has determinant

D2 =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

112

12

0

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

< 0.

This shows how easily we lose positivity of the sequence if perturbations of modifica-tions is not done properly. Thus, perturbation is a very restrictive method because of therequirement of positivity. Due to this restriction and involvement of extensive matrix com-putations, the study of sensitivity of corresponding polynomials and moment solutionsbecomes computationally expensive, if not intractable. Inthis chapter we will mostly dealwith the method of subsequences.

SPARSE MOMENT SEQUENCES 11

The subsequence given byski =1ki !∈ {sk} is not a positive sequence since the determi-

nant of its second Hankel matrix is

D2 =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

112

12

16

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

< 0

and hence has no solution to its moment problem. This exampleshows that an arbitrarysubsequence of a positive sequence is not necessarily positive. Thus we need to look forideas to appropriately remove moments while preserving positivity of subsequences.

5.1. Perturbation of a Positive Sequence.In this section we will make a few notes re-garding perturbations of terms in a positive sequence. Modification of a term in a mo-ment sequence very significantly affects its moment solution and orthogonal polynomial itgenerates. Several people have studied the sensitivity of orthogonal polynomials to mod-ifications of the moments. Gautschi [12], for example, studies sensitivity of orthogonalpolynomials to perturbations in moments. Reader is advisedto refer to [4] and [22] forsome more examples on modified moments.

The fact thats0 > |si |, i = 1,2,3, · · · , suggests that we can have a wide range of choicesfor s0 keeping the rest of terms constant and maintaining positivity. However, we are notmuch interested in modifyings0 since we can always scale this term. The theorem belowshows how rigid positive sequences are with respect to a single termsn, n, 0.

Theorem 5.1. Let {sk}k≥0 be a positive sequence. If

sk

{

sk if k, nτ if k= n

is positive andsk ≤ sk, thenτ = sn.

Proof. Consider an arbitrary system of numberx0, x1, x2, · · · , xm. Then we can write thesum

(15)m

∑

i, j=0

xi x j si+ j =

m∑

i, j=0i+ j=n

xi x j sn+

m∑

i, j=0i+ j,n

xi x j si+ j ≥ 0.

Similarly, the positivity of{sk}k≥0 gives

(16)m

∑

i, j=0

xi x j si+ j =

m∑

i, j=0i+ j=n

xi x jτ+

m∑

i, j=0i+ j,n

xi x j si+ j ≥ 0.

Combining (15) and (16), and using ˜sk ≤ sk,m

∑

i, j=0i+ j=n

xi x jτ−m

∑

i, j=0i+ j=n

xi x j sn+

m∑

i, j=0

xi x j si+ j ≥ 0,

which implies

(17)m

∑

i, j=0

xi x j si+ j ≥ [sn− τ]m

∑

i, j=0i+ j=n

xi x j .

12 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

Since {sk}k≥0 is a positive sequence the left side of (17) should be positive for anysystem of numbersx0, x1, x2, · · · , xm. Now, letsn > τ and choosex0, x1, x2, · · · , xm= 1. Then

m∑

i, j=0i+ j=n

xi x j = (n+1). Then (17) givesm

∑

i, j=0

xi x j si+ j ≥ (n+1)(sn− τ) > 0. But this is not true

since, scaling the sequence{sk} to have the the first term ass0 =(n+1)(sn− τ)

2(m+1)2and using

the fact thatsi ≤ s0 for all i, we havem

∑

i, j=0

xi x j si+ j ≤m

∑

i, j=0

xi x j s0 = (m+1)2s0 =(n+1)(sn− τ)

2< (n+1)(sn− τ).

Thereforesn ≤ τ.

Similarly using (15) and (16) again we can obtain

(18)m

∑

i, j=0

xi x j si+ j ≥ [τ− sn]m

∑

i, j=0i+ j=n

xi x j .

Then using the same argument as above we see thatsn ≥ τ. This completes the proof. �

It is noteworthy that if we perturb more than one term then we have more flexibility. Themoment sequence tends to get less rigid as we replace more terms. If the momentss1 ands2 are replaced byx1 andx2 respectively, does imposing positivity implys1 = x1, s2 = x2?Supposespi , i = 1,2, ...,n are replaced byτpi . Then we can write the sum

(19)m

∑

i, j=0

xi x j si+ j =

n∑

q=1

m∑

i, j=0i+ j=pq

xi x j spq +

m∑

i, j=0i+ j,p1,...,pn

xi x j si+ j ≥ 0.

Similarly, the positivity of{sk}k≥0 gives

(20)m

∑

i, j=0

xi x j si+ j =

n∑

q=1

m∑

i, j=0i+ j=pq

xi x jτpq +

m∑

i, j=0i+ j,p1,...,pn

xi x j si+ j ≥ 0.

Combining (19) and (20), and using ˜sk ≤ sk gives

(21)n

∑

q=1

ξpq

m∑

i, j=0i+ j=pq

xi x j

≤m

∑

i, j=0

xi x j si+ j

whereξpq = spq − τpq. Now settingxi = 1 for i = 0,1, ...,m from (21) we have

(22)n

∑

q=1

[

ξpq(pq+1)]

≤ (m+1)2s0.

Although very restrictive and not much revealing, perturbation can be useful in somecases. An example of an application of modified moments to harmonic solids is givenin [5]. An important application of modification ins0 is that it can lead to a determinatesolution. It was proved in Stieltjes’ 1894-Memoir [30] thata determinate moment solutioncan be obtained from an indeterminate one with a modificationin s0

SPARSE MOMENT SEQUENCES 13

The moment sequencesn = q−(n+1)2

2 gives the Stieltjes-Wigert polynomialsPn(u;q),which are orthogonal in a log-normal distribution, known tobe indeterminate. The modi-fied sequence{sk}k≥0 defined as

s0 = s0−1

∑∞n=0[Pn(0;q)]2

andsn = sn for all n≥ 1 has a determinate moment solution ˜σ(u) given by

σ(u) =∑

u∈Ucuδu,

whereU is the zero set of the reproducing kernel

K(0,w) =∞∑

n=0

Pn(0)Pn(w)

and

cu =1

∑∞k=0[Pk(u;q)]2

,u ∈ U.

5.2. Positive Subsequences.As was seen in an above example, an arbitrary subsequenceof a positive sequence is not necessarily positive. We will develop some methods to extractpositive subsequences from a positive sequence.

First we give a simple construction of a positive sequence bythe following theorem andproposition, which can then be used to generate positive subsequences.

Theorem 5.2. Let fk : R→R be a sequence of functions such that

(23) fi (u) f j(u) = fi+ j (u).

Then the sequence{sk} given as

sk =

∫

fk(u)dσ(u)

is positive for any non-decreasing functionσ(u) with an infinite number of points of in-crease.

Proof. For any finite set of numbersx0, x1, · · · , xm, we have

m∑

i, j=0

xi x j si+ j =

m∑

i, j=0

xi x j

∫

fi+ j (u)dσ(u)

=

m∑

i, j=0

xi x j

∫

fi (u) f j(u)dσ(u)

=

∫ m∑

i, j=0

xi x j fi (u) f j(u)dσ(u)

=

∫

m∑

i=0

xi fi (u)

2

dσ(u) ≥ 0.

14 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

Note that the equality sign holds if the functionm

∑

i=0

xi fi (u) is zero at every point where

σ(u) increases. Sinceσ(u) increases at infinitely many points then the functionm

∑

i=0

xi fi (u)

is identically zero. Therefore we have the equality sign only if xi = 0 for all i. �

Proposition 5.3. The sequence fk(u) satisfies the condition fi (u) f j(u)= fi+ j (u) for i, j ∈N0

if and only if

fk(u) = ( f1(u))k.

Proof. Clearly the conclusion holds fork = 1. Assumefn(u) = ( f1(u))n holds. Thenfn+1(u) = fn(u) f1(u) = ( f1(u))n f1(u) = ( f1(u))n+1, inductively. Proof of the converse is triv-ial. �

Note that Theorem 5.2 and Proposition 5.3 allow many constructions of positive se-quences. For example, lettingfk(u) = uk we obtain precisely the classical power momentsequence for any non-decreasing measureσ(u). Similarly, letting fk(u) to beaku for someconstanta or (φ(u))k for anyσ−measurable functionφ gives us more positive sequencescorresponding to a non-decreasing measureσ(u).

Now we are interested in extracting positive subsequences from a moment sequence.Using Theorem 5.2 and Proposition 5.3, we can construct several positive subsequencesfrom a moment sequence. For example, let{sk} be a moment sequence constructed from anon-decreasing measureσ(u) defined as

sk =

∫

(φ(u))kdσ(u)

for some functionφ(u). Then for a fixedℓ ∈N0 the sequence

skℓ :=∫

(

(φ(u))ℓ)k

dσ(u) =∫

(φ(u))kℓdσ(u) = skℓ

is a positive subsequence of{sk}. But how can we characterize all the positive subsequenceswe can extract from a moment sequence? Given a positive sequence{sk}, the problem ofidentifying all the positive subsequences requires findingall the sequences{ℓk} ⊆N0 suchthat the sequence given as

sk = sk+ℓk

is positive.

Theorem 5.4. Let {sk} be a positive sequence. Then the sequencesk = sk+ℓk is a positivesequence if{uℓk} is a positive sequence for all u∈R.

Proof. By Hamburger’s theorem{sk} is a positive sequence is equivalent to the fact thatthere is a non-decreasing functionσ(u) such that

sk =

∫

R

ukdσ(u), k ∈N0.

SPARSE MOMENT SEQUENCES 15

For x0, x1, · · · , xm ∈R we havem

∑

i, j=0

xi x j si+ j =

m∑

i, j=0

xi x j si+ j+ℓi+ j

=

m∑

i, j=0

xi x j

∫

R

ui+ j+ℓi+ j dσ(u)

=

∫

R

m∑

i, j=0

xi x jui+ j+ℓi+ j

dσ(u)

=

∫

R

m∑

i, j=0

uℓi+ j (xiui)(x ju

j)

dσ(u).(24)

But the integrand in (24) can be written as

(25)m

∑

i, j=0

uℓi+ j (xiui)(x ju

j) =[

x0 x1u · · · xmum]

uℓ0 uℓ1 · · · uℓm

uℓ1 uℓ2 · · · uℓm+1

......

. . ....

uℓm uℓm+1 · · · uℓ2m

x0

x1u...

xmum

.

Thus, the integral (24) is positive if the Hankel matrix for the sequence{uℓk} is positive forall u ∈R. This is equivalent to the sequence{uℓk} being positive for allu ∈R. �

However, the following example due to David Kimsey (privatecorrespondence) showsthat the converse of this theorem does not hold in general. Wegratefully acknowledge thiscontribution.

Consider the sequence{sk} where sk =1

2k. It is easy to verify that this sequence is

positive. Notice that{sk}, where

sk = sk+1 =1

2k+1=

(

12

)

1

2k,

is a positive sequence, but{uℓk}, whereℓk = 1, k ∈N0, is not a positive sequence in generalfor u ∈R . Indeed, if we putu= −1 then we arrive at a negative sequence.

It is easy to see that the difficulty pointed out by this example is thatu can take nega-tive values. Restrictingu to nonnegative values inR (i.e. the Stieltjes moment problem)would have made the converse of the above theorem to also hold. Therefore, it should notbe surprising that for even powers ofu stronger results hold. Now we give the follow-ing corollaries which give further insights on the characteristics of the sequence{ℓk} forextracting a positive subsequence.

Corollary 5.5. Let {sk} be a positive sequence. Then the sequence{sk} defined bysk = sk+ℓ

for anyℓ ∈ 2N0 is a positive sequence.

Proof. Note that the constant sequence{uℓ} for ℓ ∈ 2N0 is positive for anyu ∈ R. ApplyTheorem 5.4. �

Corollary 5.6. If the sequence{uℓk} is positive then

(1) ℓi ∈ 2N0 for every i∈ 2N0, and

(2) for every odd positive integer i,ℓi =ℓi−1+ ℓi+1

2.

16 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

Proof. By Corollary 5.5, since every tail of the sequence{uℓk}with ℓk starting from an eveninteger is positive,

uℓi ≥ 0 for everyu ∈R.Henceℓi ∈ 2N0 for everyi ∈ 2N0.

Again using Corollary 5.5, for every (i −1)∈ 2N0, the two by two matrix of the form[

uℓi−1 uℓi

uℓi uℓi+1

]

must have nonnegative determinant. Then

(26) uℓi−1+ℓi+1 −u2ℓi ≥ 0.

Since the inequality (26) is true for everyu ∈R, the statement (2) follows. �

The converse of this corollary does not hold in general. To see this, consider the se-quence{uℓk}with {ℓk}= {0,1,2,4,6, ...}. The determinant of the corresponding 3 x 3 Hankelis −u6(u−1)2, which is a negative function.

Theorem 5.7. Let {sk} be a positive sequence. The subsequence{sk} given assk = sk+ℓk ispositive ifℓk = kd+ ℓ0 for any d∈N0, andℓ0 ∈ 2N0.

Proof. Supposeℓk = kd+ ℓ0 whered ∈ N0, and ℓ0 ∈ 2N0. Then for all real numbersx0, x1, x2, · · · , xm, we have

m∑

i, j=0

xi x j si+ j =

m∑

i, j=0

xi x j si+ j+(i+ j)d+ℓ0

=

m∑

i, j=0

xi x j

∫

R

u(1+d)i+(1+d) j+ℓ0dσ(u)

=

∫

R

uℓ02

m∑

i=0

xiu(1+d)i

2

dσ(u) ≥ 0.

�

The above results show that any subsequence of a positive sequence is positive if theyare extracted in a certain periodic manner. A converse to these results will be provedin a later section using matrix completion method. Specifically, we will establish thatextraction of a subsequence from a positive sequence must bein a periodic manner for thesubsequence to be positive. Here we provide an example. Consider the sequence given by

sk =1

(k+1)ek+1.

This sequence is positive since, for every m, the determinant of the Hankel matrix givenby

Dm=

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

1e

1

2e2· · · 1

(m+1)em+1

1

2e2

1

3e3· · · 1

(m+2)em+2

......

. . ....

1

(m+1)em+1

1

(m+2)em+2· · · 1

(2m+1)e2m+1

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

=1

e(m+1)2

m∏

i=0

(2i +1)

(

2ii

)2

,

SPARSE MOMENT SEQUENCES 17

is positive. But the subsequence given as

sk =1

(k+1)2e(k+1)2

is not positive since the terms are not taken in a periodic manner as defined by the abovetheorem. We can easily verify this by looking at the determinant of the second Hankelmatrix:

D2 =

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

1e

1

2e2

1

2e2

1

4e4

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

< 0.

6. Sub-Moment Problems

Given any two positive sequences{sk} and{sk} with moment solutionsσ(u) andσ(u)respectively,σ(u) = σ(u) if and only if sk = sk for all k. Therefore the moment solutions toany two distinct positive sequences can never be equal. If{sk} is a positive subsequence of{sk} with corresponding moment solutions ˜σ(u) andσ(u) respectively, then these solutionscan not be the same. In this section, we develop techniques tocompareσ(u) andσ(u).

Consider a positive sequence{sk} with its moment solutionσ(u). If some of the termsfrom {sk} are missing then we obtain a subsequence which, if positive,gives another mo-ment solution. For an appropriately chosen{ℓk} ⊆N0, assume that{sk} = {sk+ℓk} is positiveand has the moment solution ˜σ(u).Write

(27)∫

R

ukdσ(u) =∫

R

uk+ℓkdσ(u).

Then for any polynomialP(u) =n

∑

k=0

akuk,we have

(28)∫

R

P(u)dσ(u) =∫

R

Pℓk(u)dσ(u),

wherePℓk(u) =n

∑

k=0

akuk+ℓk.

Here, we present some results concerning moment solution toa shifted sequence ˜sk =

sk+ℓ, ℓ ∈ 2N0, which has already been proved to be a positive subsequence of a positivesequencesk.

Theorem 6.1. Let {sk} be a positive sequence and{sk} be its subsequence given assk =

sk+ℓ for a fixedℓ ∈ 2N0. Letσ(u) and σ(u) respectively be the moment solutions of thesequences{sk} and{sk}. If one of the moment problems is determinate, then

(29)∫

R

f (u)uℓdσ(u) =∫

R

f (u)dσ(u), f ∈ L2σ∩L2

σ.

Proof. To prove the first assertion, define the following two linear functionals:

Φ1( f (u)) =∫

R

f (u)uℓdσ(u), Φ2(g(u)) =∫

R

g(u)dσ(u)

for f ∈ L2σ andg ∈ L2

σ. To seeΦ1 andΦ2 are bounded, we have

Φ1( f (u)) ≤∫

R

∣

∣

∣ f (u)uℓ∣

∣

∣dσ(u) ≤ ‖ f (u)‖L2σ

√s2ℓ <∞

18 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

and

Φ2( f (u)) ≤∫

R

|g(u)|dσ(u) ≤ ‖g(u)‖L2σ

√s0 <∞.

Now for any functionf ∈ L2σ∩L2

σ,

Φ( f (u)) = Φ1( f (u))−Φ2( f (u))

is a bounded linear functional.Observe that for any polynomialP(u) = amum+ · · ·+a1u+a0,

Φ(P(u)) =∫

R

P(u)uℓdσ(u)−∫

R

P(u)dσ(u)

=

∫

R

[

amum+ℓ + · · ·+a1u1+ℓ +a0uℓ]

dσ(u)−∫

R

[

amum+ · · ·+a1u+a0]

dσ(u)

=am(sm+ℓ − sm)+ · · ·+a1(s1+ℓ − s1)+a0(sℓ − s0)

=0.

By Corollary 4.1 and due to the determinacy condition of the moment problems, the setof polynomials is dense inL2

σ or L2σ. Hence the set of polynomials is dense inL2

σ ∩ L2σ.

Therefore, by the Hahn-Banach Theorem,Φ is identically zero. Thus equation (29) holds.�

The above theorem is particularly useful when the first finitely many moments are miss-ing. It says that we can find the moment solution with the remaining data and reconstructthe missing moments by using formula (29). Whether the moment solution of a subse-quence is determinate or not is to be studied more carefully in comparison with the deter-minacy of the moment solution of the original sequence. It is, however, known that the firstterm in the subsequence{skℓ} can be modified to make the moment problem determinate,but the solution will not remain the same.

As a remark toL2 space of sub-moment solutions, the following theorem givesus a rela-tion betweenL2 spaces related to a moment sequence and its sub-moment sequence. Herewe require the evenness assumption onℓ to guarantee thatν is a nondecreasing measure.

Corollary 6.2. Consider a positive sequence{sk} and its subsequence{sk} = {sk+ℓ} forℓ ∈ 2N0. Letσ(u) andσ(u) be the moment solutions of the sequence and the subsequencerespectively. Then there is a measureν(u) absolutely continuous with respect toσ(u) suchthat

(30)(

L2σ∩L2

σ

)

⊆ L2ν .

Proof. Let f ∈ L2σ∩L2

σ. By equation (29) we have∫

R

f (u)uℓdσ(u) =∫

R

f (u)dσ(u), f ∈ L2σ∩L2

σ.

Setdν = uℓdσ. Thenν << σ and

(31)∫

R

f (u)dν(u) =∫

R

f (u)dσ(u).

Hencef ∈ L2ν . �

Now consider a general sub-moment sequence{sk} given as ˜sk = sk+ℓk for an appropriate{ℓk} ⊆N0. It was shown in the previous section that for anyk,

ℓk = kd+ ℓ0, whered ∈N0 andℓ0, ℓ1 ∈ 2N0.

SPARSE MOMENT SEQUENCES 19

Modifying the functionalΦ1 in the proof of Theorem 6.1 gives the following result.

Theorem 6.3. Let {sk} be a positive sequence and{sk} be its subsequence given bysk =

skd+ℓ0. Letσ(u) and σ(u) respectively be the moment solutions of the sequences{sk} and{sk}. If one of the moment problems is determinate, and f(ud) ∈ L2

σ for any f ∈ L2σ, then

(32)∫

R

f (ud)uℓ0dσ(u) =∫

R

f (u)dσ(u), f ∈ L2σ∩L2

σ.

A precise relation betweenL2 spaces of moment solutions to the original sequence andthat of one of its positive subsequences can be useful in characterizing the sub-momentsolutions and hence approximating and/or reconstructing the missing data.

Exploring equation (32) a little further, forλ ∈ C,y= ℑλ , 0, define

f (u) =1

u−λ.

Then we have∫

R

| f (u)|2dσ(u) =∫

R

dσ(u)

|u−λ|2 ≤∫

R

dσ(u)

y2=

s0

y2<∞.

since|u−λ| ≥ |y|. Therefore,f ∈ L2σ. By the same argument,f ∈ L2

σ. Similarly, it can beshown thatf (ud) ∈ L2

σ. Then by Theorem 6.3,

(33)∫

R

uℓ0dσ(u)

ud−λ =∫

R

dσ(u)u−λ .

Applying Theorem 4.4 to the equation (33) yields the following result which connects amoment problem with the polynomials corresponding to its sub-moment problem.

Theorem 6.4. Let {sk} be a positive sequence and{sk} be its subsequence given assk =

skd+ℓ0. Letσ(u) and σ(u) respectively be the moment solutions of the sequences{sk} and{sk}. If the moment problem of{sk} is indeterminate, then

(34)∫

R

uℓ0dσ(u)

ud−λ = −A(λ)φ(λ)− C(λ)

B(λ)φ(λ)− D(λ),

whereA(λ), B(λ),C(λ), D(λ) form a Nevanlinna matrix of the sub-moment problem, andφ ∈ N.

Next, for the shifted sub-sequence we discussed in Theorem 6.1, we want to investigatedeterminacy of its moment problem of a special sub-moment sequence in relation to deter-minacy of the original moment problem. Recall that the limitcirclesK∞(λ),λ ∈ C,ℑλ , 0defined in the Section 2 provide a fundamental concept for studying determinacy of mo-ment problems.

Theorem 6.5. For the same moment sequences and conditions as in Theorem 6.1, forλ ∈ C,ℑλ , 0 the following holds.

(35)∫

R

dσ(u)u−λ =C+λℓ

∫

R

dσ(u)u−λ .

where C is a constant depending onλ.

Proof. It is already shown that, for

f (u) =1

u−λ, λ ∈ C,ℑλ , 0,

20 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

f ∈ L2σ∩L2

σ. Then by Theorem 6.1,∫

R

uℓdσ(u)u−λ =

∫

R

dσ(u)u−λ .

Butuℓ

u−λ = R(u)+λℓ

u−λ , whereR(u) =ℓ−1∑

i=0

λiuℓ−1−i . Therefore,

∫

R

dσ(u)u−z

=

∫

R

R(u)dσ(u)+λℓ∫

R

dσ(u)u−λ .

Hence∫

R

dσ(u)u−λ =C+λℓ

∫

R

dσ(u)u−λ ,

whereC =ℓ−1∑

i=0

λi sℓ−1−i . �

Let

wσ(λ) =∫

R

dσ(u)u−λ andwσ(λ) =

∫

R

dσ(u)u−λ .

Then Theorem 6.5 states that

(36) wσ(λ) =C+λℓwσ(λ).

for sk andsk specified in that theorem.Recall that pointswσ andwσ lie on the circumferences of the circlesK∞(λ) andK∞(λ)

corresponding to the moment problems of{sk} and {sk} respectively. Equation (36) pre-cisely describes the distortion of the circle corresponding to the original moment sequencein relation to its tail.

7. Completion of Positive Hankel Matrix

Recall from Definition 2.1 that the positivity of an infinite sequence{sk} is equivalent tothe positivity of Hankel matrices of the form

Hm=

s0 s1 · · · sm

s1 s2 · · · sm+1...

.... . .

...

sm sm+1 · · · s2m

.

Determining whether a subsequence of a positive sequence{sk} is positive is then equiv-alent to a matrix completion problem. A partial Hankel matrix is a partial matrix that isHankel to the extent to which it is specified. All the specifiedentries lie along certain skew-diagonals (positive sloping diagonals), and all the entries along a specified skew-diagonalhave the same value. For example, for a moment sequence{sk} with every (2i+1)-th entrymissing we have a corresponding Hankel matrix

H5 =

s0 ∗ s2 ∗ s4 ∗∗ s2 ∗ s4 ∗ s6

s2 ∗ s4 ∗ s6 ∗∗ s4 ∗ s6 ∗ s8

s4 ∗ s6 ∗ s7 ∗∗ s6 ∗ s8 ∗ s10

.

SPARSE MOMENT SEQUENCES 21

Essentially, we want to determine whether a particular partial Hankel matrix can be com-pleted to a positive Hankel matrix. Note that here we are interested in completing thepartial Hankel matrices corresponding to a sparse moment sequence.

The problem of matrix completion has been studied in great details (see [2], [3], [6],[20], for example). Most common matrices that are of interest in matrix completion arepositive definite matrices, positive semi-definite matrices, co-positive matrices, totally pos-itive matrices and Toeplitz matrices. Matrix completion properties have several applica-tions in sparse moment problem. Some results on multidimensional trigonometric momentproblem have been proved using moment completion in [2].

Most of the results in matrix completion deal with the case when the main diagonalentries are specified. This creates a problem in completing apartial Hankel matrix. When-ever a (2i)-th term of a moment sequence is missing, the correspondingHankel matrix hasa diagonal entry missing.

Definition 7.1. Given an n× n matrix A, let w,v be subsets ofN0. A submatrix A[w,v]consists of the entries in rows w and columns v. The principlesubmatrix is A[w] = A[w,w].

For example,

A=

1 2 3 45 6 7 89 10 11 1213 14 15 16

, A[{2,4}] = A[{2,4}, {2,4}] =[

6 814 16

]

.

Matrix completion encounters an important inheritance structure. IfA is a matrix withpropertyX, then in most cases of matrix completion of interest, the fully specified principlesubmatrices ofA inherit the propertyX. Positive (semi) definite matrix completions havethis inheritance structure. But the converse is not true fora general positive (semi) definitematrix. That is, a partial matrix with all of its fully specified principle submatrices positive(semi) definite does not necessarily have a positive (semi) definite completion, [20]. Simi-larly, if a partial matrixA admits a completion of rank≤ k, then every specified submatrixof A has rank≤ k.

With the inheritance structure of matrix completion, the matrix completion problem forpositive Hankel matrices becomes equivalent to finding all the positive subsequences of amoment sequence. That is, the principle sub-Hankel matrices of a positive Hankel matrixare those corresponding to positive subsequences of a moment sequence.

Given a Hankel matrixH corresponding to a moment sequence{sk} consider a fullyspecified principle submatrixH[α], whereα = {α0,α1,α2, . . . ,αn} ⊆ N0. For H[α] to bealso a Hankel matrix we require that

αi1 +α j1 = αi2 +α j j wheneveri1+ i2 = j1+ j2.

Then for anyk, we have

αk =αk−1+αk+1

2.

Also note that every principal submatrixH[α] has (1,1) entry s2i for somei. With thisobservation and Theorem 5.7, we have proved the following result:

Theorem 7.2. Let {sk} be a positive sequence. A subsequence{sk} given assk = sk+ℓk , ispositive if and only ifℓk = kd+ ℓ0 for some d∈N0 andℓ0 ∈ 2N0.

In terms of matrix completion the above theorem gives a necessary and sufficient condi-tion for completion of a partial Hankel matrix. Interestingly, this result for Hankel matrices

22 SAROJ ARYAL, FARHAD JAFARI, AND MIHAI PUTINAR

seems to be similar with the result for Toeplitz matrices proved in [21]. The reader is re-ferred to [3] for more on matrix completion problem for Hankel and Toeplitz matrices.

8. Remarks

The problem of identifying positive subsequences of momentsequences and the cor-responding matrix completion problem is the forward problem associated with sparse se-quences and sparse matrices. The corresponding inverse problem of extension of positivesequences as sparse subsets of moment sequences is closely related. The necessary andsufficient conditions obtained suggests that positive sparse sequences can be completedby extending the measures associated with these sequences by Theorem 6.4 and calculat-ing the arising moments. This suggests that the moment sequence can be recovered to agreat degree from its positive sparse subsequences. The entropy concept introduced by J.Chover [8] for extensions of positive definite functions wasused to find a unique extensionof positive definite functions which maximizes the entropy.In parallel to this classical andwell-known theory, we anticipate that for each positive sequence arising from the Nevan-linna class there is a unique extension in the Nevanlinna class that maximizes the entropy.The corresponding phrasing of this problem for positive Hankel matrices gives a norm, in-duced by the entropy, which allows recovery of these structured matrices from their sparsepositive submatrices. Relationship of this approach to finding the matrix with minimumnuclear norm that fits the data in Candes and Recht’s [6] is not clear to us at the presenttime, but is a subject of further investigation.

References

1. N. I. Akhiezer,The Classical Moment Problem and Some Related Questions in Analysis, Translated by N.Kemmer, Hafner Publishing Co., New York, 1965.

2. Mihaly Bakonyi and Geir Naevdal,On the matrix completion method for multidimensional moment problems,Acta Sci. Math. (Szeged)64 (1998), no. 3-4, 547–558.

3. Mihaly Bakonyi and Hugo J. Woerdman,Matrix Completions, Moments, and Sums of Hermitian Squares,Princeton University Press, Princeton, NJ, 2011.

4. Bernhard Beckermann and Emmanuel Bourreau,How to choose modified moments?, J. Comput. Appl. Math.98 (1998), no. 1, 81–98.

5. Carl Blumstein and John C. Wheeler,Modified-moments method: Applications to harmonic solids, Phys.Rev. B.8 (1973), no. 4, 1764–1776.

6. Emmanuel Candes and Benjamin Recht,Exact matrix completion via convex optimization, Found. Comput.Math.9 (2009), no. 6, 717–772.

7. P.L. Carleman,Les Fonctions Quasi-Analytiques. Collection Borel, Gauthier-Villars, Paris, 1926.8. J. Chover,On normalized entropy and the extensions of a positive-definite function, J. Math. Mech.10(1961),

927–945.9. Ph. Delsarte, Y. Genin, Y. Kamp, and P. Van Dooren,Speech modelling and the trigonometric moment

problem, Philips J. Res.37 (1982), no. 5-6, 277–292.10. Ky Fan,On positive definite sequences, Ann. of Math. (2)47 (1946), 593–607.11. Bent Fuglede,Multidimensional moment problem, Exposition. Math.1 (1983), no. 1, 47–65.12. Walter Gautschi,On the sensitivity of orthogonal polynomials to perturbations in the moments, Numer. Math.

48 (1986), no. 4, 369–382.13. C. D. Godsil,Algebraic combinatorics, Chapman and Hall Mathematics Series, Chapman & Hall, New York,

1993.14. Rao S. Govindaraju and Bhabani S. Das,Moment Analysis for Subsurface Hydrologic Applications, Springer,

Dordrecht, The Netherlands, 2007.15. Bjorn Gustafsson, Mihai Putinar, Edward B. Saff, and Nikos Stylianopoulos,Bergman polynomials on an

archipelago: estimates, zeros and shape reconstruction, Adv. Math.222(2009), no. 4, 1405–1460.16. Hans Hamburger,Uber eine Erweiterung des Stieltjesschen Momentenproblems, Math. Ann.82(1920, 1921),

120–164, 168–187.17. Ernst Hellinger,Zur Stieltjesschen Kettenbruchtheorie, Math. Ann.86 (1922), no. 1-2, 18–29.

SPARSE MOMENT SEQUENCES 23

18. Roger A. Horn and Charles R. Johnson,Matrix analysis, Cambridge University Press, Cambridge, 1985.19. Ales Jaklic and Franc Solina,Moments of superellipsoids and their application to range image registration,

IEEE Transactions on Systems, Man, and Cybernetics33 (2003), no. 4, 648–657.20. Charles R. Johnson,Matrix completion problems: a survey, Matrix theory and applications (Phoenix, AZ,

1989), Proc. Sympos. Appl. Math., vol. 40, Amer. Math. Soc.,Providence, RI, 1990, pp. 171–198.21. Charles R. Johnson, Michael Lundquist, and Geir Nævdal,Positive definite Toeplitz completions, J. London

Math. Soc. (2)59 (1999), no. 2, 507–520.22. Sun-Mi Kim and Lothar Reichel,Sensitivity analysis for Szego polynomials, Numer. Math.113(2009), no. 2,

265–279.23. Rupert Lasser,On the problem of modified moments, Proc. Amer. Math. Soc.90 (1984), no. 3, 360–362.24. Bruce G. Lindsay,On the determinants of moment matrices, Ann. Statist.17 (1989), no. 2, 711–721.25. Peyman Milanfar, Mihai Putinar, James Varah, Bjorn Gustafsson, and Gene Golub,Shape reconstruction

from moments: Theory, algorithms, and applications, Proceedings of SPIE Conference 4116, AdvancedSignal Processing Algorithms, Architectures, and Implementations4116(2000), 406–416.

26. Mihai Putinar and Florian-Horia Vasilescu,Solving moment problems by dimensional extension, Ann. ofMath. (2)149(1999), no. 3, 1087–1107.

27. M. Riesz,Sur le probleme des moments et le theoreme de parseval correspondant, Acta Litt. ac., Azeged1(1922), 209–225.

28. J. A. Shohat and J. D. Tamarkin,The Problem of Moments, American Mathematical Society Mathematicalsurveys, vol. II, American Mathematical Society, New York,1943.

29. Barry Simon,The classical moment problem as a self-adjoint finite difference operator, Adv. Math. 137(1998), no. 1, 82–203.

30. Thomas Jan Stieltjes,Œuvres completes/Collected papers. Vol. I, II, Springer-Verlag, Berlin, 1993, Reprintof the 1914–1918 edition.

31. Harold Widom,Hankel matrices, Trans. Amer. Math. Soc.121(1966), 1–35.

Department of Mathematics, University ofWyoming, Laramie, WY 82070E-mail address: saryal@uwyo.edu

Department of Mathematics, University ofWyoming, Laramie, WY 82070E-mail address: fjafari@uwyo.edu

Department of Mathematics, University of California at Santa Barbara, Santa Barbara, CA 93106E-mail address: mputinar@math.ucsb.edu