Kyungpook Mathematical Journal 2018; 58(4): 711-732
Multivariable Recursively Generated Weighted Shifts and the 2-variable Subnormal Completion Problem
Kaissar Idrissi, and El Hassan Zerouali*
Centre of Mathematical research in Rabat (CeReMaR), Mohamed V University, Rabat, Morocco
e-mail : kaissar.idrissi@gmail.com and zerouali@fsr.ac.ma
*Corresponding Author.
Received: January 6, 2018; Revised: October 16, 2018; Accepted: October 24, 2018; Published online: December 23, 2018.

This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License (http://creativecommons.org/licenses/by-nc/3.0/) which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract

In this paper, we give a new approach to solving the 2-variable subnormal completion problem (SCP for short). To this aim, we extend the notion of recursively generated weighted shifts, introduced by R. Curto and L. Fialkow, to 2-variable case. We next provide ”concrete” necessary and sufficient conditions for the existence of solutions to the 2-variable SCP with minimal Berger measure. Furthermore, a short alternative proof to the propagation phenomena, for the subnormal weighted shifts in 2-variable, is given.

Keywords: subnormal completion problem, multivariable weighted shift, generalized Fibonacci sequence, moment problems, propagation phenomena.
1. Introduction and Results

The notion of recursively generated weighted shifts, largely studied in the literature, is employed to solve various questions in operator theory. R. Curto and L. Fialkow [3, 4] have used this concept to solve the Subnormal Completion Problem (SCP for short) in one variable. In this paper, we extend this notion to 2-variables, and we use it to provide a new approach to solve the open problem of 2-variable SCP. A concrete solution to the minimal 2-variable SCP (see Section 4) is given as well as an alternative proof for the propagation phenomena for subnormal 2-variable weighted shifts.

First we recall some definitions and notations. A bounded linear operator T ∈ ℬ(ℋ) on a complex Hilbert space ℋ is normal if TT* = T*T, subnormal if T = N |, where N is normal and N(ℋ) ⊆ ℋ, and hyponormal if T*TTT* ≥ 0. The n-tuple T ≡ (T1, …, Tn) is said to be normal if T is commuting and each Ti is normal, and T is subnormal if T is the restriction of a normal n-tuple to a common invariant subspace.

For $α≡{αn}n=0∞$ a bounded sequence of positive real numbers (called weights), let Wα : l2(ℝ+) → l2(ℝ+) be the associated unilateral weighted shift, defined by Wαen := αnen+1, where ${en}n=0∞$ is the canonical orthonormal basis in l2(ℝ+). Given k ∈ ℝ+, the moments of α of order k is given by

$γk≡γk(α):=‖wαke0‖2={1if k=0,α02…αk-12if k≥1.$

It is easy to see that Wα is never normal. In the case where Wα is subnormal, Stampfli showed in [13] that a propagation phenomenon occurs which forces the flatness of Wα, that is, if αk = αk+1 for some k ≥ 0, then αn = αn+1 for every n ≥ 0.

Consider double-indexed positive bounded sequences $α≡{αk}k∈ℤ+2$ and $β≡{βk}k∈ℤ+2$. We define in a similar way the 2-variable weighted shift T ≡ (T1, T2) acting on the Hilbert space $l2(ℤ+2)$, associated with, $α≡{αk}k∈ℤ+2$ and $β≡{βk}k∈ℤ+2$ by

$T1ek:=αkek+ɛ1 and T2ek:=βkek+ɛ2,$

where ε1 := (1, 0) and ε2 := (0, 1). We recall here that, the vector space $l2(ℤ+2)$ is canonically isometrically isomorphic to l2(ℝ+)⊗l2(ℝ+), equipped with its canonical orthonormal basis ${ek}k∈ℤ+2$.

Clearly,

$T1T2=T2T1⇔βk+ɛ1αk=αk+ɛ2βk(for all k∈ℤ+2).$

Given $k≡(k1,k2)∈ℤ+2$, the moment of (α, β) of order k is

$γk:=‖T1k1T2k2e0‖2={1if k=0,α(0,0)2…α(k1-1,0)2if k1≥1 and k2=0,β(0,0)2…β(0,k2-1)2if k1=0 and k2≥1,α(0,0)2…α(k1-1,0)2·β(k1,0)2⋯β(k1,k2-1)2if k1≥1 and k2≥1.$

Conversely, one can recover the weights from the moments, by using the following relations:

$αk=γk+ɛ1γk and βk=γk+ɛ2γk.$

The presence of consecutive equal weights, of a 2-variable subnormal weighted shift, leads to horizontal or vertical flatness (see Definition 3.4). Explicitly, if, for some k1, k2 ≥ 1, α(k1, k2) = α(k1+1, k2) (resp. β(k1, k2) = β(k1, k2+1)), then α(k1, k2) = α(1,1) (resp. β(k1, k2) = β(1,1)) for all k1, k2 ≥ 1. This result is known as propagation phenomena. We give new short proof of this fact in Theorem 3.4.

A characterization of subnormality for multivariable weighted shifts is given in [10]. More precisely, T admits a commuting normal extension if and only if there is a probability measure μ, which we call the Berger measure of T, defined on the 2-dimensional rectangle R = [0, ||T1||2] × [0, ||T2||2] such that

$γk=∫Rxk1yk2dμ(x,y), for all k≡(k1,k2)∈ℤ+2.$

A measure μ satisfies (1.4) is also known as representing measure for ${γk}k∈ℤ+2$.

With a given bi-sequence $γ(2n)≡{γi}i∈ℤ+2,∣i∣≤2n≡{γij}i+j≤2n$, we associate the moment matrix M(n) ≡ M(n)(γ2n), introduced by R. Curto and L. Fialkow [5, 6], build as follows.

$M(n)=(M[0,0]M[0,1]…M[0,n]M[1,0]M[1,1]…M[1,n]⋮⋮⋱⋮M[n,0]M[n,1]…M[n,n]),$

where

$M(i,j)=(γi+j,0γi+j-1,1…γi,jγi+j-1,1γi+j-2,2…γi-1,j-1⋮⋮⋱⋮γi,jγi-1,j+1…γ0,i+j).$

Considering the following lexicographic order, to denote rows and columns of the moment matrix M(n),

$1,X,Y,X2,XY,Y2,…,Xn,Xn-1Y,…,XYn-1,Yn.$

The matrix M(n) detects the positivity of the Riesz functional

$Λγ(2n):p(x,y)≡∑0≤i+j≤2naijxiyj→∑0≤i+j≤2naijγij$

on the cone generated by {p2 : p ∈ ℝn[x, y]}, the sum of squares of polynomials (sometimes abbreviated as SOS), where ℝn[x, y] is the vector space of polynomials in two variables with real coefficients and total degree less than or equal to n.

For reason of simplicity, we identify a polynomial p(x, y) ≡ ∑aijxiyj with its coefficient vector p = (aij) with respect to the basis of monomials of ℝn[x, y] in degree-lexicographic order (see (1.7)). Clearly M(n) acts on these coefficient vectors as follows:

$=qTM(n)p=Λγ(2n)(pq), p,q∈ℝn[x,y].$

Furthermore, let g ∈ ℝ[x, y] be with coefficient vector {gβ} and let g * γ denote the shifted vector in $ℝℤ+2$ whose α-th entry is $(g*γ)α:=∑βgβγβ+α$. The moment matrix $M(n) (g*γ)=Mg(n+[1+deg g2])$ is called the localizing matrix with respect to γ and g. For example, given γ(4) ≡ {γij}i+j≤4, then

$M(1)=(γ00γ10γ01γ10γ20γ11γ01γ11γ02),M(2)=(γ00γ10γ01γ20γ11γ02γ10γ20γ11γ30γ21γ12γ01γ11γ02γ21γ12γ03γ20γ30γ21γ40γ31γ22γ11γ21γ12γ31γ22γ13γ02γ12γ03γ22γ13γ04),Mx(2)=(γ10γ20γ11γ20γ30γ21γ11γ21γ12) and My(2)=(γ01γ11γ02γ11γ21γ12γ02γ12γ03).$

In the one variable case, the SCP was stated and solved abstractly by J. Stampfli in [13]:

### Problem 1.(One-Variable Subnormal Completion Problem)

Given m ≥ 0 and a finite collection of positive numbers$Ωm={αk}k=0m$, find necessary and sufficient conditions on Ωm to guarantee the existence of a subnormal weighted shift whose initial weights are given by Ωm.

When m = 0 or m = 1 the solution can be given by the canonical completion α0, α0, α0, … and α0 < α1, α1, …, with Berger measure $μ:=δα02$ and $μ:=α02α12δα12$, respectively. In [13], J. Stampfli showed that given $α:a,b,c$ with a < b < c, there always exists a subnormal completion of α (this solves the case m = 2), but for $α:a,b,c,d$, with a < b < c < d, such a subnormal completion may not exist. The complete solution of the SCP in one variable was given by R. curto and L. Fialkow [3], the explicit calculation requires recursively generated weighted shifts (such shifts have finite atomic Berger measures).

We now state the 2-variable SCP:

### Problem 2.(2-Variable Subnormal Completion Problem)

Given m ≥ 0 and a finite collection of pairs of positive numbers Ωm = {(αk; βk)}|k|≤m satisfying (1.1) for all | k |≤ m (where | k |:= k1 + k2), find necessary and sufficient conditions to guarantee the existence of a subnormal 2-variable weighted shift whose initial weights are given by Ωm.

In [8], R. Curto, S. H. Lee and J. Yoon have introduced an approach to the 2-variable SCP based on positivity and rank-preserving extension of the moment matrix, developed in [5, 6, 7]. Although this lead to an explicit criterion for SCP with quadratic moment data (i.e., m = 1), the 2-Variable Subnormal Completion Problem remains open. Recently, S. H. Lee and J. Yoon [11] found, by using the recursiveness, a necessary and sufficient conditions for the existence of 2-Variable subnormal completion with minimal Berger measure for the case m = 2 ( i.e., rank M(1)-atomic Berger measure). We will provide, in Theorem 4.2, a complete (and concrete) solution for the 2-Variable SCP with minimal Berger measure.

In the present paper, we will show that if a given, finite, collection of weights Ωm admits a subnormal completion in 2-variable, then there exists a 2-variable subnormal weighted shift (solution of the SCP for Ωm) with moment sequence obey to some recurrence relations, we shall refer to such shifts as 2-variable recursively generated weighted shifts (see Definition 2.5). In Theorem 3.1, we will provide necessary and sufficient conditions for the existence of 2-variable recursively generated weighted shift completion, and thus a solution to the SCP in 2-variable. As application, we provide a generalization of a recent result of S. H. Lee and J. Yoon [11, Theorem 2.2]. More precisely, in Theorem 4.2 we give a concrete necessary and sufficient conditions for the existence of a subnormal completion with minimal Berger measure.

This paper is organized as follows. In Section 2, we introduce the recursively generated weighted shifts and we exhibit some useful results. We devote Section 3 to provide a solution to the 2-Variable Subnormal Completion Problem (Theorem 3.1 ) and the phenomena propagation for subnormal weighted shifts. In section 4, we solve the minimal SCP in 2-variable.

2. The 2-variable Recursively Weighted Shifts

We introduce below the notion of 2-variable weighted shifts, which will play a central role in this paper, and we give some useful properties.

### Definition 2.1

Let T ≡ (T1, T2) be a 2-variable weighted shift and let γ ≡ {γij}i,j∈ℤ+ be its associated moment sequence. A polynomial $p(x,y)=∑i,jpijxiyj∈ℝ[x,y]$ is said to be characteristic polynomial associated with T, or with γ, if

$∑i,jpi,jγi+n,j+m=0, for all n,m∈ℤ+.$

### Remark 2.2

If p is a characteristic polynomial associated with γ, then, for every q ∈ ℝ[x, y], the polynomial pq is also a characteristic polynomial. In particular, the set of all characteristic polynomials associated with γ is an ideal in ℝ[x, y].

The following proposition is an immediate consequence of relations (1.8) and (2.1).

### Proposition 2.3

Let γ ≡ {γij}i,j∈ℤ+be a bi-indexed sequence and let p(x, y) ∈ ℝ[x, y]. Then p is a characteristic polynomial of γ if and only if M(∞)(γ)p = 0.

### Definition 2.4

A sequence γ ≡ {γij}i,j∈ℤ+ is said to be recursive double indexed sequence (RDIS in short) if it has two characteristic polynomials p1, p2 ∈ ℝ[x, y],

with

${p1(x,y)=xr+1-∑i+j≤raijxiyjp2(x,y)=ys+1-∑i+j≤s+1,j≠s+1bi,jxiyj,$

or in the symetric form

${p1(x,y)=xr+1-∑i+j≤r+1,i≠r+1aijxiyjp2(x,y)=ys+1-∑i+j≤sbi,jxiyj.$

Without loss of generality, throughout this paper, we assume that the pair of characteristic polynomials of the RDIS γ ≡ {γij}i,j∈ℤ+ is given in the form (2.2).

### Definition 2.5

A 2-variable weighted shift is said to be recursively generated if its associated moment sequence is a RDIS.

Let us show that RDIS are well defined. Consider a RDIS γ ≡ {γij}i,j≥0 associated with a pair of characteristic polynomials as in (2.2), that is, for all n, m, e ∈ ℝ+ with nr and ms,

$γn+1,e=∑i+j≤raijγn-r+i,e+j and γe,m+1∑l+k≤sblkγe+l,n-s+k.$

A direct computation shows that

$γn+1,m+1=∑i+j≤raijγn-r+i,m+1+j (=∑l+k≤sblkγn+1+l,m-s+k)=∑i+j≤r,l+k≤saijblkγn-r+i+l,m-s+j+k.$

Equation (2.5) gives the compatibility condition of the two relations in (2.4). Hence the sequence γ ≡ {γij}i,j≥0 is well defined. The other case (i.e., when the characteristic polynomial are defined as in (2.3)) is treated similarly.

Different pair of characteristic polynomials can be associated with the same RDIS, as shown in the following example.

### Example 1

Let us consider the bi-sequence γ ≡ {γij}i,j∈ℤ+ defined by γn,m = a2n +b3m, where a and b are real numbers. The polynomials $x2-4x-12y+92$ and y + 2x − 5 are two characteristic polynomials of γ. Indeed,

$γn+2,m-4γn+1,m-12γn,m+1+92γn,m=a2n+2+b3m-4a2n+1-4b3m-12a2n-12b3m+1+92a2n+92b3m=0,$

and

$γn,m+1+2γn+1,m-5γm,n=a2n+b3m+1+2a2n+1+2b3m-5a2n-5b3m=0.$

On the other hand, we have

$γn+2,m-3γn+1,m+2γn,m =a2n+2+b3m-3a2n+1-3b3m+2a2n+2b3m =0,γn,m+2-4γn,m+1+3γn,m =a2n+b3m+2-4a2n-4b3m+1+3a2n+3b3m =0.$

Hence (x2−3x+2; y2−4y+3) is, also, a pair of characteristic polynomials associated with γ. (The last characteristic polynomials are analytic, i.e., (x2 − 3x + 2; y2 − 4y + 3) ∈ ℝ[X] × ℝ[Y ]).

Let γ ≡ {γij}i,j∈ℤ+ be a RDIS, we denote by ℘γ (⊆ ℝ[x, y] × ℝ[x, y]) the set of pair of characteristic polynomials associated with γ and we denote by (⊆ ℝ[X]×ℝ[Y ]) the family of analytic, monic, characteristic polynomials associated with γ.

### Remark 2.6

The pair of characteristic polynomials (p1, p2) ∈ ℘γ, together with the initial conditions ${γij}0≤i≤deg p1-1,0≤i≤deg p2-1.$, are said to define the sequence γ.

We use structural properties of moment matrices to get the following interesting lemma.

### Lemma 2.7

Let γ ≡ {γij}i,j∈ℤ+be a bi-indexed sequence and let f, g, h ∈ ℝ[x, y], then

$fTM(∞)(γ)(gh)=(fg)TM(∞)(γ)h.$
Proof

We write $f=∑i1,12f(i1,i2)xi1yi2,g=∑j1,j2g(j1,j2)xj1yj2$ and $h=∑k1,k2h(k1,k2)xk1yk2$. As the entry, of the infinite moment matrix M(∞)(γ), corresponding to the column XnY m and the line XlY k is γn+l,m+k, we obtain

$fTM(γ)(gh)=(∑i1,i2f(i1,i2)xi1yi2)TM(∞)(γ)(∑j1,j2,k1,k2g(j1,j2)h(k1,k2)xj1+k1yj2+k2)=∑i1,i2,j1,j2,k1,k2f(i1,i2)g(ji,j2)h(k1,k2)γi1+j1+k1,i2+j2+k2=(fg)TM(γ)h,$

which is the required result.

It follows that

### Proposition 2.8

Let γ ≡ {γij}i,j≥0be a bi-indexed sequence and let M(∞)(γ) be the associated infinite moment matrix. If M(∞)(γ) ≥ 0, Then, for any polynomial p ∈ ℝ[x, y] and any integer n ≥ 1, we have

$M(∞)(γ)pn=0⇒M(∞)(γ)p=0.$
Proof

If M(∞)(γ)p2 = 0, then 0 = M(∞)(γ)p2 = 1TM(∞)(γ)p2 = pTM(∞)(γ)p, from (2.6); since M(∞)(γ) ≥ 0, we obtain M(∞)(γ)p = 0 and hence (2.7) holds for n = 2. By induction, (2.7) remains valid for any power of 2. Now, if M(∞)(γ)pn = 0 we choose r in such a way that r + k is a power of 2 to ensure that

$M(∞)(γ)pn+r=(pr)TM(∞)(γ)pn=0.$

Which gives M(∞)(γ)p = 0.

Before continuing our investigations on RDIS, we recall the next result on weighted r-generalized Fibonacci sequences ${yA(n)}n=0+∞$ where the initial conditions $A≡{αn}n=0r-1⊂ℂ$ are given and the sequence is associated with the characteristic polynomial p(x) = xra1xr−1–…–ar ∈ ℂ[x]. It is also is the sequence generated by the difference equation with initial values:

${yA(n)=αn; n=0,1,…,r-1,yA(n+r)=a1yA(n+r-1)+…+aryA(n); n=r,r+1,....$

We have

### Theorem 2.9.([9, Theorem 1])

Let${yA(n)}n=0+∞$be a RDIS and p(x) be the associated polynomial as above, with$p(x)=∏i=1k(x-λi)mi$, (m1 + …+ mk = r). Then the difference equation (2.8) has r independent solution$njλln$(j = 0, …, ml−1; l = 1, …, k). Moreover, any solution of (2.8) is of the form

$yA(n)=∑l=1k∑j=0ml-1el,jnjλln,$

where el,j are solutions of the following system of r-equations

$∑l=1k∑j=0ml-1el,jnjλln=yA(n), n=0,…,r-1.$

As observed in [2, Proposition 2.1], among all characteristic polynomials defining S, there exists a unique monic characteristic polynomial p0 of minimal degree, called the minimal characteristic polynomial, and which, moreover, divides every characteristic polynomial. The next proposition gives a generalization of this result to the 2-variable case.

### Proposition 2.10

For every RDIS γ ≡ {γij}i,j≥0, with, there exists a unique pair of characteristic polynomials$(p1γ,p2γ)∈Aγ$with minimal degree. Moreover, for every, the polynomials$p1γ$and$p2γ$divide Q1and Q2, respectively.

Proof

Let , with Q1(x) = xr+1a0xr − …− ar and Q2(y) = ys+1b0ys − …− bs.

Given an integer j ∈ ℝ+. Since, for all i ∈ ℝ+,

$γi+r+1,j=a0γi+r,j+…+arγi,j,$

then Q1 is a characteristic polynomial of the Fibonacci sequence γj ≡ {γij}i∈ℤ+ : iγij, hence there exists a minimal characteristic polynomial $Q1(j)$, which divides Q1. Thus $Q1γ=∧j≥0Q1(j)$, the smallest common multiple of {$Q1(j)$; j ∈ ℝ+}, divides Q1 and is a characteristic polynomial of γ.

Similarly, Given an integer i ≥ 0. Q2 is a characteristic polynomial for the Fibonacci sequence γi ≡ {γij}j∈ℤ+ : jγij, then there exists a minimal characteristic polynomial $Q2(i)$ of γi, and hence $Q2γ=∧i≥0Q2(i)$ is a characteristic polynomial of γ, which divides Q2. We conclude that the pair of analytic characteristic polynomials ($Q1γ,Q2γ$) provides a positive answer to the proposition.

In the following proposition, as well as in the remainder of this paper, we associate every RDIS γ (with ) with its pair of minimal polynomials. The next theorem, of independent interest, is a crucial point in our approach.

### Theorem 2.11

Let γ ≡ {γij}i,j∈ℤ+be a RDIS and let$(p1,p2)=(p1γ,p2γ)∈Aγ$. If M(∞)(γ) ≥ 0, then the polynomials p1and p2have distinct roots.

Proof

Let l = 1,2 and let $pl(x)=∏i=1m(x-λi)di$, where λi ∈ ℂ. We notice that since pl(x) ∈ ℝ[x, y], then if λi ∉ ℝ, for some i, hence there exists ji such that $λi=λj¯$ and di = dj. Setting $M=maxi=1mdi$. Since pl is a characteristic polynomial of γ, then, from Proposition 2.3, $M(∞)(γ)∏i=1m(x-λi)di=0$ and hence $M(∞)(γ)∏i=1m(x-λi)M=0$. By using Proposition 2.8, it follows that $M(∞)(γ)∏i=1m(x-λi)=0$. Hence, via Proposition 2.3, the polynomial $∏i=1m(x-λi)$ is a characteristic polynomial of γ, which divides pl. As pl is minimal, we deduce that $pl(x)=∏i=1m(x-λi)$, as desired.

In the next, we give an extension of Theorem 2.9 to 2-variables.

### Lemma 2.12

Let γ ≡ {γij}i,j∈ℤ+be a RDIS and let, with$P1(x)=∏l=1k1(x-λl)ml$and$P2(y)=∏l=1k2(y-βl)nl$, l, βl ∈ ℂ). Then

$γij=∑a=1k1∑b=0ma-1∑c=1k2∑d=0nc-1ea,b,c,dibjdλaiβcj,$

where ea,b,c,d are determined by using the initial condition {γij}0≤ir−1,0≤js−1.

Proof

Given an integer j ∈ ℝ+, the single sequence γj : iγij is a general Fibonacci sequence associated with P1, setting r := deg P1. This implies, by virtue of Theorem 2.9, that

$γij=∑a=1k1∑b=0ma-1ea,b(j)ibλai,$

where ea,b are determined uniquely by the initial conditions {γij}0≤ir−1.

Since, for every i, the sequence jγij is a general Fibonacci sequence associated with P2(x) = xsp1xs−1 − …− ps, then

$γi,j+s-p1γi,j+s-1-…-psγi,j=0.$

Hence, using (2.10), we obtain

$∑a=1k1∑b=0ma-1(ea,b(j+s)-p1ea,b(j+s-1)-…-psea,b(j))ibλai=0.$

As $i→ibλai$ (with b = 0, …, ma − 1; a = 1, …, k1) are linearly independent, see Theorem 2.9, then

$ea,b(j+s)-p1ea,b(j+s-1)-…-psea,b(j)=0.$

Since j is arbitrary, the sequences jea,b(j) (b = 0, …, ma − 1; a = 1, …, k1) are general Fibonacci sequences, associated with the characteristic polynomial P2. Hence, for all b = 0, …, ma − 1 and a = 1, …, k1, we obtain

$ea,b(j)=∑c=1k2∑d=0nc-1ea,b,c,djdβcj,$

where ea,b,c,d are determined by the initial conditions {ea,b(j)}0≤js−1. We conclude, from (2.11) and (2.12), that

$γij=∑a=1k1∑b=0ma-1∑c=1k2∑d=0nc-1ea,b,c,dibjdλaiβcj.$

By virtue of Theorem 2.11 and Lemma 2.12, we have the next corollary.

### Corollarly 2.13

Let γ ≡ {γij}i,j∈ℤ+be a RDIS, with M(∞)(γ) ≥ 0, and let. Then

$γij=∑a=1r+1∑c=1s+1ea,cλaiβcj,$

where Z(P1) := {z ∈ ℂ such that P(z, z̄) = 0} = {λ1, …, λr+1} and Z(P2) = {β1, …, βs+1}.

The next lemma establishes, for a RDIS γ, a link between the positivity of infinite moment matrix and that of the finite one.

### Lemma 2.14

Let γ ≡ {γij}i,j∈ℤ+be a RDIS and let (p1, p2) ∈ ℘γ. Then

$M(∞)(γ)≥0⇔M(deg p1+deg p2-2)(γ)≥0.$

Moreover, rankM(∞)(γ) = rankM(deg p1 +deg p2 − 2)(γ).

Proof

Let (p1, p2) be as in (2.2) and let H ∈ ℝv+1[x, y], with v = r + s. First, we will show that, for every e = 0, …, v+1, there exist some real numbers {$αlk(e)$; l, k ∈ ℝ+ with l + kv} such that

$Λγ(xeyv+1-eH)=∑l+k≤vαlk(e)Λγ(xlykH).$

To this end we distinguish two cases.

• When er + 1. We have, from Remark 2.2, Λγ(p1xer−1yv+1−eH) = 0, then $Λγ(xeyv+1-eH-∑i+j≤raijxi+e-r-1yj+v+1-eH)=0$.

Hence $Λγ(xeyv+1-eH)=∑i+j≤raijΛγ(xi+e-r-1yj+v+1-eH)=∑l+k≤vαlk(e)Λγ(xlykH),$

with l = i + er − 1, k = j + v + 1 − e and $al-e+r+1,k-v-1+e=αlk(e)$.

• When er. It follows, from Remark 2.2, that Λ(p2xeyvseH) = 0. Since $p2=ys+1-∑f=1s+1bf,s+1-fxfys+1-f-∑i+j≤sbijxiyj$, then

$Λγ(xeyv+1-eH-∑f=1s+1bf,s+1-fxf+eyv+1-e-fH-∑i+j≤sbijxi+eyj+v+1-eH)=0.$

Hence,

$Λγ(xeyv+1-eH)=Λγ(∑f=1s+1bf,s+1-fxf+eyv+1-e-fH)+Λγ(∑i+j≤sbijxi+eyj+v+1-eH).$

In the first term of the right-hand expression of equation (2.14), the Riesz functional Λγ is applied to a sum of monomials of total degree v+1 times H. With the aim of lowering the degree of the associated polynomial, we employ the method used in the case (i), for the monomials with power of x greater than r + 1 (i.e., f +er +1). When f +er +1, we reapply the technic of the case (ii) in order to increase strictly the power of x. It follows that each time when applying case (ii) the minimum power of x, of all monomials, increases strictly. Since we can decrease the total degree of each monomial with power in x greater than or equal r + 1, by applying the case (i), we write

$Λγ(xeyv+1-eH)=∑l+k≤vαlk(e)Λγ(xlykH).$

Now we construct a m(v) × v-matrix Wv with successive rows defined by the relation,

$Pxeyv+1-e=∑l+k≤vαlk(e)e(l,k),$

where e = 0, …, v + 1 and {e(l,k); k + lv} the canonical basis of ℝm(n).

Therefore, it is easy to show that:

$M(v+1)(γ)=(M(v)(γ)BB*C)$

with B = M(v)(γ)Wv and C = B*Wv. Since M(v)(γ) ≥ 0, then, by using Smul’jan’s theorem [12] (see also Section 4), M(v + 1)(γ) ≥ 0 and rankM(v) = rankM(v + 1). In the same way one can show that M(v + 2)(γ) ≥ 0 and rank(v + 2) = rankM(v + 1) = rankM(v). And thus, by induction, we conclude that M(∞)(γ) ≥ 0 and rankM(v)(γ) = rankM(∞)(γ), as desired.

3. The 2-variable SCP and the propagation phenomena

In this section we involve the 2-variable recursively generated weighted shifts to obtain a solution to the SCP in 2-variable. Also, a simple and new proof to the propagation phenomena, for the 2-variable subnormal weighted shift, is given.

### Theorem 3.1

Let Ωn := {(α(k1, k2), β(k1, k2)); k1 + k2 ≤ 2n} be a given collection of weights obeying (1.1) and γ(2n+1)be the associated moment sequence, given by (1.2). The following statements are equivalent.

• Ωn admits a 2-variable subnormal completion,

• Ωn admits a recursively generated 2-variable subnormal completion,

• There exists a RDIS γ ≡ {γij}i,j≥0such that γ(2n+1)γ and the matrices M(deg p1+deg p2−2)(γ), Mx(deg p1+deg p2−1)(γ) and My(deg p1+deg p2− 1)(γ) are positive, where (p1, p2) ∈ ℘γ.

Proof

First, let us show that i) ⇒ ii). If Ωn admits a 2-variable subnormal completion, then there is a Berger’s measure ν, supported in $ℝ+2$, such that

$γij=∫xiyjdν, i+j≤2n+1.$

A result of C. Bayer and J. Teichmann in [1] states that if a finite bi-sequence of positive numbers {γij}0≤i+j≤2n has a probability measure verifies (3.1), supported in $ℝ+2$, then it has a finitely atomic positive measure μ verify the same relation and supported, also, in $ℝ+2$. Write suppμ ⊂ {λ1, λ2, …, λr} × {β1, β2, …, βs}, where λ1, λ2, …, λr and β1, β2, …, βs are real numbers. It is easy to see that $∫xi+1-ryj∏k=1r(x-λk)dμ=∫xiyj+1-s∏k=1s(x-βk)dμ=0$, for all ir−1 and js − 1. Then μ is a representing measure (that is, μ satisfies the relation (1.4)) for the RDIS γ ≡ {γij}i,j≥0, defined by the initial conditions ${γij}0≤i≤r-1,0≤i≤s-1.$ and by the following linear recurrence relations:

$γi+1,j=∑n=0r-1anγi-n,j and γi,j+1=∑m=0s-1bmγi,j-m,$

for all ir − 1 and js − 1, where

${ak-1=(-1)k-1∑1≤i1

Remark that $xr-∑n=0r-1anxr-1-n=∏k=1r(x-λk)$ and $xs-∑n=0s-1bnxs-1-n=∏k=1s(x-βk)$. Hence γ is a RDIS admitting a Berger’s measure, and thus the recursively generated 2-variable subnormal completion associated with γ gives a positive answer to the SCP associated with Ωn.

To show that ii) ⇒ iii), it suffices to remark that if Ωn admits a recursively generated 2-variable subnormal completion, then there exists a finite Berger’s measure μ and a RDIS γ ≡ {γij} such that γ(2n)γ and ∫ xiyj = γij, for all i, j ∈ ℝ+. Hence, for every Q ∈ ℝ[x, y] with degQ ≤ deg p1 +deg p2 − 2, we have

$QTM(p1+deg p2-2)(γ)Q=∫Q2dμ≥0.$

Therefore M(p1 + degp2 − 2)(γ) ≥ 0. Similarly, let H ∈ ℝ[x, y] be with degH ≤ deg p1 + degp2 − 2, since μ is a positive measure supported in $ℝ+2$, we get

$HTMx(p1+deg p2-1)(γ)H=∫xH2dμ≥0$

and

$HTMy(p1+deg p2-1)(γ)H=∫yH2dμ≥0.$

Hence Mx(p1 + deg p2 − 1)(γ) and My(p1 + degp2 − 1)(γ) are positive.

It remains to prove the implication iii) ⇒ i). From Lemma 2.14 it follows that M(∞)(γ) has finite rank, then there exists an integer n (resp., an integer m) such that, in the matrix M(∞)(γ), the column Xn+1 (resp., the column Y m+1) is a linearly combination of the columns 1, X, …, Xn (resp., the columns 1, Y, …, Y m). So we write

$Xn+1=α0Xn+…+an1 and Ym+1=β0Ym+…+bm.$

Hence, for every i, j ∈ ℝ+, we have

${(XiYj)TM(∞)(γ)Xn+1=(XiYj)M(∞)(γ)(a0Xn+…+an1)(XiYj)TM(∞)(γ)Xm+1=(XiYj)M(∞)(γ)(b0Ym+…+bm1),$

that is,

${γi+n+1,j=a0γi+n,j+…+anγi,jγi,j+m+1=b0γi,j+m+…+bmγi,j.$

Setting Q1(x) = a0xn+…+an and Q2(y) = b0ym+…+bm, we have . Since M(deg p1 + degp2 − 2)(γ) ≥ 0, then, via Lemma 2.14, M(∞)(γ) ≥ 0 and thus there exits, by using Theorem 2.11, a pair of minimal analytic characteristic polynomials with distinct roots, writing $H1(x)=∏l=1r+1(x-λl)$ and $H2(x)=∏l=1s+1(y-βl)$, where λl, βl ∈ ℂ.

Hence, via Corollary 2.13,

$γij=∑a=1r+1∑c=1s+1ea,cλaiβcj, for all i,j∈ℤ+,$

and thus the measure $μ=∑a=1r+1∑c=1s+1ea,cδ(λa,βc)$ (Here, δ(λac) is the Dirac measure at (λa, βc) ∈ ℝ2, having mass (λa, βc) at x and mass 0 elsewhere) verifies

$γij=∫xiyjdμ, for all i,j∈ℤ+.$

We will show that μ is a Berger measure, that is, μ is a positive measure supported in $ℝ+2$, which implies that γ is a moment sequence of some 2-variable subnormal weighted shifts, so that Ωn admits a 2-variable subnormal completion.

Let Iμ = {(a, b) such that ea,c ≠ 0} and let

$L(λi,βj)(x,y)=∏0≤l≤r+1l≠i(x-λiλl-λi)∏0≤k≤s+1k≠j(y-βjλk-λj)$

Clearly

$L(λi,βj)(x,y)={1if (i,j)=(l,k),0if (i,j)≠(l,k).$

Thus, for every (a, b) ∈ Iμ, we have

$ea,c=∫∣L(λa,βc)∣2 dμ=L(λa,βc)TM(∞)(γ)L(λa,βc)≥0.$

and then ea,c > 0, because ea,cIμ. Therefore, μ is a positive measure.

It remains to show that $(λa,βb)∈ℝ+2$, for all (a, b) ∈ Iμ. To this aim, let (a, b) ∈ Iμ, we have

$ea,cλa=∫∣L(λa,βc)∣2xdμ=L(λa,βc)TMx(∞)(γ)L(λa,βc)≥0$

and

$ea,cβc=∫∣L(λa,βc)∣2ydμ=L(λa,βc)TMy(∞)(γ)L(λa,βc)≥0.$

Since ea,c > 0, we deduce that $(λa,βc)∈ℝ+2$, as desired.

The following corollary, of Theorem 3.1, characterizes subnormal recursively generated 2-variables weighted shifts.

### Crollary 3.2

Let T ≡ (T1, T2) be a recursively generated 2-variable weighted shift and let γ ≡ {γij}i,j≥0be the associated moment sequence, with. Then T is subnormal if and only if M(deg(P1) + deg(P2) − 2)(γ) ≥ 0.

In particular, we have

### Crollary 3.3

Every collection of positive numbers α(0,0), β(0,0), α(0,1), β(1,0), such that β(1,0)α(0,0) = α(0,1)β(0,0), admits a subnormal completion.

Proof

Let {γij}0≤i,j≤1 be the collection of moments associated with ω ≡ {α(0,0), β(0,0), α(0,1), β(1,0)}, given by (1.2), and let γ ≡ {γij}i,j∈ℤ+ be the RDIS defined by the initial conditions {γij}0≤i,j≤1 and by the pair of, minimal, analytic characteristic polynomials (P1, P2). We set P1(X) = (Xλ0)(Xλ1) and P2(X) = (Xβ0)(Xβ1), with 0 ≤ λ0λ1 and 0 ≤ β0β1.

The measure μ = C00δ(λ0, β0) + C10δ(λ1, β0) + C01δ(λ0, β1) + C11δ(λ1, β1) is a representing measure for γ if and only if (Cij)0≤i,j≤1 are nonnegative and satisfy the following linear system of 4 equations

${C00+C10+C01+C11=γ00,C00β0+C10β0+C01β1+C11β1=γ01,C00λ0+C10λ1+C01λ0+C11λ1=γ10,C00β0λ0+C10β0λ1+C01β1λ0+C11β1λ1=γ11.$

Since the determinant of the preceding system is

$|1111β0β0β1β1λ0λ1λ0λ1β0λ0β0λ1β1λ0β1λ1|=-((λ1-λ0)(β1-β0))2≠0,$

we obtain the existence of {C00, C10, C01, C11}. Thus for the existence of T ≡ (T1, T2) a subnormal completion of {γij}0≤i,j≤1 it suffices show that {Cij}0≤i,j≤1 can be positive.

The numbers C00, C10, C01 and C11 are given by

${C00=γ11-λ1γ01-β1γ10+λ1β1(λ1-λ0)(β1-β0),C01=-γ11+λ1γ01+β0γ10-λ1β0(λ1-λ0)(β1-β0),C10=-γ11+λ0γ01+β1γ10-λ0β1(λ1-λ0)(β1-β0),C11=γ11-λ0γ01-β0γ10+λ0β0(λ1-λ0)(β1-β0).$

Direct computations show that C00, C10, C01 and C11 are positive numbers precisely when,

${max{2γ10,2γ11γ01}≤λ1,max{2γ01,2γ11γ10}≤β1,0≤λ0≤min{γ112γ01,γ102},0≤β0≤min{γ012,γ112γ10}.$

Therefore it suffices to choose the roots of the polynomials p1 and p2 obeying (6).

We employ Berger’s Theorem and Equality (1.3) to give a simple proof to the propagation phenomena for 2-variable subnormal weighted shifts.

### Definition 3.4

A 2-variable weighted shift T ≡ (T1, T2) is horizontally flat ( resp. vertically flat) if α(k1, k2) = α(1,1), for all k1, k2 ≥ 1 (resp. β(k1, k2) = β(1,1)).

### Theorem 3.5

LetT ≡ (T1, T2) be a subnormal 2-variable weighted shift associated with the weight sequences${αk}k∈ℤ+2$and${βk}k∈ℤ+2$. If α(k1, k2) = α(k1+1, k2)for some k1, k2 ≥ 1 (resp. β(k1, k2) = β(k1, k2+1)), thenTis horizontally flat (resp. vertically flat).

Proof

Let γ ≡ {γij}i,j≥0 be the moment sequence of T, see (1.2). Given an arbitrary number n0 ≥ max(k1 + 1, k2), let γ(n0) ≡ {γij}i,jn0 be a truncated subsequence of γ. Since γ admits a Berger measure, then, via Tchakaloff’s Theorem [1], there exists a finite supported Berger measure μ of γ(n0), write $μ=∑0≤i≤p,0≤j≤q.ρi,jδ(λi,ϑj)$. Notice that {ϑj; 0 ≤ jq} ≠ {0} because the moments of the subnormal weighted shift are strictly positive.

Since α(k1, k2) = α(k1+1, k2), then it follows from (1.3) that

$γ(k1+1,k2)γ(k1,k2)=γ(k1+2,k2)γ(k1+1,k2),$

hence

$(∑0≤i≤p,0≤j≤q.ρijλik1+1ϑjk2)2=(∑0≤i≤p,0≤j≤q.ρijλik1ϑjk2)(∑0≤i≤p,0≤j≤q.ρijλik1+2ϑjk2),$

that is,

$∑0≤i

We deduce that λi = λk whenever λik ≠ 0, and thus

$γnl=λ1n(∑0≤j≤qρjlϑjl), for all n,l≤n0.$

Since n0 is arbitrary, then

$α(n,m)=γ(n+1,m)γ(n,m)=γ(2,1)γ(1,1)=α(1,1), for all n,m∈ℕ*.$

The vertically flatness can be proved analogously.

4. The Minimal 2-variable SCP

Given an integer n ≥ 0, let Ωn := {(α(k1, k2), β(k1, k2)); k1 + k2 ≤ 2n} be a given collection of weight obeying (1.1) and let γ(2n+1) ≡ {γij}i+j≤2n+1 be the corresponding moment sequence given by (1.2). We say that Ωn admits a minimal 2-variable subnormal completion if it has a subnormal completion with rankM(n)-atomic Berger measure.

For n = 1, a characterization of weights admitting minimal 2-variable subnormal completion is given by S. H. Lee and J. Yoon [11]. In this section we give the complete solution of this problem.

Recall that, an extension of M(n)(γ(2n)) is a block matrix of the form

$M=(M(n)(γ(2n))BB*C).$

A theorem of Smul’jan [12] shows that M(n)(γ(2n)) admits a positive extension in the form of moment matrix if, and only if,

• M(n) ≥ 0,

• there exists a matrix W such that B = AW,

• CB*W,

• C is a Hankel (n + 2) × (n + 2) matrix.

Moreover, M(n)(γ(2n)) admits a flat extension, or M is flat, if, and only if, C = B*W. Notice that the condition iv) serves only to ensure that M is a moment matrix, see (1.5) and (1.6). If the conditions i)–iv) are satisfied, we set MM(n + 1)( γ̂(2n+2)), where γ(2n)γ̂(2n+2), i.e., γij = γ̂ij for all i + j ≤ 2n. To simplify our notations, we set γ(2n+2) = γ̂(2n+2).

Observe that if M(n+1)(γ(2n+2)) ≡ M(n+1) is flat (that is, rankM(n+1) = rankM(n)), then every column of M(n + 1), indexed by a monomial of degree n+1, is a linear combination of columns indexed by monomials of degree less than or equal to n. Explicitly, the columns in M(n + 1) satisfies

$Xn+1-eYe=∑i+j≤naij(e)XiYj with aij(e)∈ℝ and e=0,…,n+1.$

Hence

$(XlYk)TM(n+1)Xn+1-eYe=∑i+j≤naij(e)(XlYk)TM(n+1)XiYj,$

for (l + kn + 1), that is

$γn+1-e+l,e+k=∑i+j≤naij(e)γi+l,j+k (e=0,…,n+1 and l+k≤n+1).$

Setting $PXn+1-eYe(x,y)=∑i+j≤naij(e)xiyj$.

### Lemma 4.1

Let γ(2n+2) ≡ {γij}i+j≤2n+2be a finite collection of non negative numbers such that rankM(n + 1) = rankM(n) and let PXn+1−eYebe as above. Then, for all e ∈ {0, …, n+1}, the polynomial xn+1−eyePXn+1−eYeis a characteristic one of the RDIS defined by the initial conditions {γij}i,jn and by the pair of characteristic polynomials (xn+1PXn+1, yn+1PYn+1).

Proof

Let γ ≡ {γij}i,j∈ℤ+ be a RDIS defined by the initial conditions {γij}i,jn and by the pair of characteristic polynomials (xn+1PXn+1, yn+1PYn+1). Corresponding to the sequence γ, the Riesz functional Λ : ℝ[x, y] → ℝ defined by $Λ(∑i,jpijxiyj)=∑i,jpijγij$. Then, for all a, b ∈ ℝ+, we have

$Λ(xayb(xn+1-PXn+1))=Λ(xayb(yn+1-PYn+1))=0.$

Let e ∈ {0, …, n + 1} be a fixed integer, we will show that xn+1−eyePXn+1−eYe is a characteristic polynomial of γ, that is, for all a, b ∈ ℝ+,

$Λ(xayb(xn+1-eye-PXn+1-eYe))=0.$

We distinguish two cases;

• a+bn+1, then (4.3) implies that Λ(xa+n+1−eyb+e) = Λ(xaybPXn+1−eYe) and hence (4.5) is verified.

• a + b = n + 2, we show first that

$Λ(xfygQk)=0 for all f+g≤n+1,$

where Qk = yPXn+1−kYkxPXnkYk+1 and k ∈ {0, …, n + 1}.

For f + gn, we have

$Λ(xfygQk)=Λ(xfyg(yPXn+1-kYk-xPXn-kYk+1))=Λ(xfyg+1PXn+1-kYk-xf+1ygPXn-kYk+1))=Λ(xf+n+1-kyg+1+k-xf+1+n-kyg+k+1), due to (4.3),=0.$

For f + g = n + 1, since degQkn + 1, and according to (4.3), we derive that Λ(xf ygQk) = Λ(PXfYgQk). As deg PXfYgn, then (4.7) implies that Λ(PXfYgQk) = 0, and hence Λ(xf ygQk) = 0 (for all f + gn + 1).

Now we consider the case a + b = n + 2, which we split in two subcases.

i) If ae: according to (4.6), we have

$Λ(x1-eyb+ePXn+1)=Λ(xa-e+1yb+e-1PXnY)=…=Λ(xaybPXn+1-eYe)$

Thus

$0=Λ(xa-eyb+ePXn+1)-Λ(xaybPXn+1-eYe)=Λ(xa-e+n+1yb+e)-Λ(xaybPXn+1-eYe), by applying (4.4),=Λ(xa+n+1-eyb+e-xaybPXn+1-eYe)=Λ(xayb(xn+1-eye-PXn+1-eYe)).$

ii) If ae: since a+b = n+2, then b+en+1 and b+en+1. According to (4.7), we have

$Λ(xaybPXn+1-eYe)=Λ(xa+1yb-1PXn-eYe+1) =…=Λ(xa+n+1-eyb-n-1+ePYn+1).$

Hence

$0=Λ(xa+n+1-eyb-n-1+ePYn+1)-Λ(xaybPXn+1-eYe)=Λ(xa+n+1-eyb+e)-Λ(xaybPXn+1-eYe), due to (4.4),=Λ(xa+n+1-eyb+e-xaybPXn+1-eYe)=Λ(xayb(xn+1-eye-PXn+1-eYe)).$

Therefore, we conclude that

$Λ(xayb(xn+1-eye-PXn+1-eYe))=0, for all a+b≤n+2.$

By induction, we obtain Λ(xayb(xn+1−eyePXn+1−eYe)) = 0, for all a, b ∈ ℝ+.

Now we are able to give a concrete solution to the minimal 2-variable SCP. To this aim, let $γ(2n)≡{γi}i∈ℤ+2,∣i∣≤2n≡{γij}i+j≤2n$ be a given bi-sequence and let M[i, j] (i, j = 0, 1, …, n) be as in (1.6). In the next theorem, we denote by B the following matrix

$B≡B(n+1):=(M[0,n+1]⋮M[n-1,n+1]M[n,n+1]).$

### Theorem 4.2

Let Ωn := {(α(k1, k2), β(k1, k2)); k1 + k2 ≤ 2n} be a given collection of weights obeying (1.1) and let γ(2n+1), M(n), Mx(n), My(n), B and Pxn+1−iyi(i = 0, …, n + 1) be as above, with RangBRangM(n) and M(n), Mx(n) and My(n) are positive. Then Ωn admits a minimal subnormal completion if and only if

$Pxn+1-jyjTM(n)Pxn+1-iyi=Pxn-jyj+1TM(n)Pxn+2-iyi-1$

for all integers i and j with 0 ≤ jn − 1 and 2 + jin + 1.

Proof

Let mm(n) denote the number of rows (or columns) in M(n). Adopting the notation in (4.1), with B is a m × (n + 1) matrix. Observe that, the sequence γ(2n+1) fills only the matrices M(n) and B. We are looking for new numbers (entries) for the matrix C in order that M be a flat moment matrix.

Let the rows Xn+1−jY j, columns Xn+1−iY i (i, j = 0, …, n + 1) entry of the matrix M be equal to $Pxn+1-jyjTM(n)Pxn+1-iyi$ (i.e., C = (M(n)W)*W). With this completion, M becomes a flat completion of M(n).

Let us show that M is a moment matrix, that is, C is Hankel. The relation (4.11) implies that the upper left triangular part of the matrix C is Hankel type. Since M(n) is symmetric, then C is also symmetric, and thus C = (M(n)W)*W is a Hankel matrix. Setting M = M(n + 1)(γ(2n+2)). It follows, from Lemma 4.1, that there exists a DIRS γ such that rankM(n) = rankM(∞)(γ) (where M(n) = M(n)(γ(2n)) = M(n)(γ)).

We show now that rankMx(n + 1)(γ) = rankMx(∞)(γ) and rankMy(n + 1)(γ) = rankMy(∞)(γ). We prove first that the columns in Mx(n + 2) verify the relation $XeYn+1-e=∑i+j≤naij(e)XiYj(=Pxeyn+1-e(X,Y))$. We have, for all a + bn + 1,

$(XaYb)TMx(n+2)(γ)(XeYn+1-e-Pxeyn+1-e(X,Y))= , see (1.8),= =0, since xeyn+1-e-Pxeyn+1-e is a characteristic polynomial of γ.$

Hence Mx(n + 2)(γ)(XeY n+1−e) = Mx(n + 2)(γ)(Pxeyn+1−e (X, Y )). Since deg Pxeyn+1−en, then rankMx(n + 1)(γ) = rankMx(n + 2)(γ), and thus we obtain, by induction, rankMx(n + 1)(γ) = rankMx(∞)(γ). Similarly, one shows that rankMy(n+1)(γ) = rankMy(∞)(γ). Since Mx(n+1)(γ) and My(n+1)(γ) are positive, then, via Schmul’jan’s theorem, Mx(∞)(γ) and My(∞)(γ) are positive.

Let . We conclude, from above, that M(deg p1 + degp2 − 2)(γ), My(deg p1 + degp2 − 1)(γ) and My(deg p1 + degp2 − 1)(γ) are positive. Now, by applying Theorem 3.1, Ωn admits a subnormal completion with finite Berger measure, say $μ=∑(a,b)∈Iμe(a,b)dδ(λa,βb)$ (with e(a,b) ≠ 0, for all (a, b) ∈ Iμ).

It remain to show that card suppμ = rankM(n)(γ). Let $ζ(λa,βc):=(1,λa,βc,λa2,λaβc,βc2,…)=(λaiβcj)(i,j)∈ℤ+2∈ℝℤ+2$. By using Corollary 2.13, the infinite moment matrix can be formulated as follows $μ(∞)(γ)=∑(a,c)∈Iμe(a,c)ζ(λa,βc)ζ(λa,βc)T$, then

$rankM(∞)(γ)≤card Iμ=card suppμ.$

On the other hand, Let L(λa, βc) be as in (3.5) and let

$M(∞)(γ)∑(a,c)∈Iμκ(a,c)1e(a,c)L(λa,βc)=0,$

where {κ(a,c)}(a,c)∈Iμ are real numbers (not all zero).

Since $1e(i,j)L(λi,βj)TM(∞)(γ)1e(n,m)L(λn,βm)$ is 1 if (i, j) = (n, m) and equal to 0 if (i, j) ≠ (n, m), then

$0=(∑(a,c)∈Iμκ(a,c)1e(a,c)L(λa,βc))TM(∞)(γ)∑(a,c)∈Iμκ(a,c)1e(a,c)L(λa,βc)=∑(a,c)∈Iμκ(a,c)2,$

a contradiction. Hence card suppμrankM(∞)(γ), and thus card suppμ = rankM(n)(γ), as desired.

References
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