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The truncated moment problem on reducible cubic curves I: Parabolic and Circular type relations

Seonguk Yoo Department of Mathematics Education and RINS, Gyeongsang National University, Jinju, 52828, Korea. [email protected]  and  Aljaž Zalar2,Q Aljaž Zalar, Faculty of Computer and Information Science, University of Ljubljana & Faculty of Mathematics and Physics, University of Ljubljana & Institute of Mathematics, Physics and Mechanics, Ljubljana, Slovenia. [email protected]
Abstract.

In this article we study the bivariate truncated moment problem (TMP) of degree 2k2k on reducible cubic curves. First we show that every such TMP is equivalent after applying an affine linear transformation to one of 8 canonical forms of the curve. The case of the union of three parallel lines was solved in [Zal22a], while the degree 6 cases in [Yoo17b]. Second we characterize in terms of concrete numerical conditions the existence of the solution to the TMP on two of the remaining cases concretely, i.e., a union of a line and a circle y(ay+x2+y2)=0,a{0}y(ay+x^{2}+y^{2})=0,a\in\mathbb{R}\setminus\{0\}, and a union of a line and a parabola y(xy2)=0y(x-y^{2})=0. In both cases we also determine the number of atoms in a minimal representing measure.

Key words and phrases:
Truncated moment problems; KK–moment problems; KK–representing measure; Minimal measure; Moment matrix extensions
2020 Mathematics Subject Classification:
Primary 44A60, 47A57, 47A20; Secondary 15A04, 47N40.
2Supported by the Slovenian Research Agency program P1-0288 and grants J1-50002, J1-2453, J1-3004.
QThis work was performed within the project COMPUTE, funded within the QuantERA II Programme that has received funding from the EU’s H2020 research and innovation programme under the GA No 101017733 \euflag

1. Introduction

Given a real 22–dimensional sequence

ββ(2k)={β0,0,β1,0,β0,1,,β2k,0,β2k1,1,,β1,2k1,β0,2k}\beta\equiv\beta^{(2k)}=\{\beta_{0,0},\beta_{1,0},\beta_{0,1},\ldots,\beta_{2k,0},\beta_{2k-1,1},\ldots,\beta_{1,2k-1},\beta_{0,2k}\}

of degree 2k2k and a closed subset KK of 2\mathbb{R}^{2}, the truncated moment problem (KK–TMP) supported on KK for β(2k)\beta^{(2k)} asks to characterize the existence of a positive Borel measure μ\mu on 2\mathbb{R}^{2} with support in KK, such that

(1.1) βi,j=Kxiyj𝑑μfori,j+, 0i+j2k.\beta_{i,j}=\int_{K}x^{i}y^{j}d\mu\quad\text{for}\quad i,j\in\mathbb{Z}_{+},\;0\leq i+j\leq 2k.

If such a measure exists, we say that β(2k)\beta^{(2k)} has a representing measure supported on KK and μ\mu is its KKrepresenting measure (KK–rm).

In the degree-lexicographic order

1,X,Y,X2,XY,Y2,,Xk,Xk1Y,,Yk\mathit{1},X,Y,X^{2},XY,Y^{2},\ldots,X^{k},X^{k-1}Y,\ldots,Y^{k}

of rows and columns, the corresponding moment matrix to β\beta is equal to

(1.2) (k)(k;β):=([0,0](β)[0,1](β)[0,k](β)[1,0](β)[1,1](β)[1,k](β)[k,0](β)[k,1](β)[k,k](β)),\mathcal{M}(k)\equiv\mathcal{M}(k;\beta):=\left(\begin{array}[]{cccc}\mathcal{M}[0,0](\beta)&\mathcal{M}[0,1](\beta)&\cdots&\mathcal{M}[0,k](\beta)\\ \mathcal{M}[1,0](\beta)&\mathcal{M}[1,1](\beta)&\cdots&\mathcal{M}[1,k](\beta)\\ \vdots&\vdots&\ddots&\vdots\\ \mathcal{M}[k,0](\beta)&\mathcal{M}[k,1](\beta)&\cdots&\mathcal{M}[k,k](\beta)\end{array}\right),

where

[i,j](β):=(βi+j,0βi+j1,1βi+j2,2βi,jβi+j1,1βi+j2,2βi+j3,3βi1,j+1βi+j2,2βi+j3,3βi+j4,4βi2,j+2βj,iβj1,i+1βj2,i+2β0,i+j).\mathcal{M}[i,j](\beta):=\left(\begin{array}[]{ccccc}\beta_{i+j,0}&\beta_{i+j-1,1}&\beta_{i+j-2,2}&\cdots&\beta_{i,j}\\ \beta_{i+j-1,1}&\beta_{i+j-2,2}&\beta_{i+j-3,3}&\cdots&\beta_{i-1,j+1}\\ \beta_{i+j-2,2}&\beta_{i+j-3,3}&\beta_{i+j-4,4}&\cdots&\beta_{i-2,j+2}\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ \beta_{j,i}&\beta_{j-1,i+1}&\beta_{j-2,i+2}&\cdots&\beta_{0,i+j}\\ \end{array}\right).

Let [x,y]k:={p[x,y]:degpk}\mathbb{R}[x,y]_{\leq k}:=\{p\in\mathbb{R}[x,y]\colon\deg p\leq k\} stand for the set of real polynomials in variables x,yx,y of total degree at most kk. For every p(x,y)=i,jai,jxiyj[x,y]kp(x,y)=\sum_{i,j}a_{i,j}x^{i}y^{j}\in\mathbb{R}[x,y]_{\leq k} we define its evaluation p(X,Y)p(X,Y) on the columns of the matrix (k)\mathcal{M}(k) by replacing each capitalized monomial XiYjX^{i}Y^{j} in p(X,Y)=i,jai,jXiYjp(X,Y)=\sum_{i,j}a_{i,j}X^{i}Y^{j} by the column of (k)\mathcal{M}(k), indexed by this monomial. Then p(X,Y)p(X,Y) is a vector from the linear span of the columns of (k)\mathcal{M}(k). If this vector is the zero one, i.e., all coordinates are equal to 0, then we say pp is a column relation of (k)\mathcal{M}(k). A column relation pp is nontrivial, if p0p\not\equiv 0. We denote by 𝒵(p):={(x,y)2:p(x,y)=0}\mathcal{Z}(p):=\{(x,y)\in\mathbb{R}^{2}\colon p(x,y)=0\}, the zero set of pp. We say that the matrix (k)\mathcal{M}(k) is recursively generated (rg) if for p,q,pq[x,y]kp,q,pq\in\mathbb{R}[x,y]_{\leq k} such that pp is a column relation of (k)\mathcal{M}(k), it follows that pqpq is also a column relation of (k)\mathcal{M}(k). The matrix (k)\mathcal{M}(k) is pp–pure, if the only column relations of (k)\mathcal{M}(k) are those determined recursively by pp. We call a sequence β\beta pp–pure, if (k)\mathcal{M}(k) is pp–pure.

A concrete solution to the TMP is a set of necessary and sufficient conditions for the existence of a KK–representing measure μ\mu, that can be tested in numerical examples. Among necessary conditions, (k)\mathcal{M}(k) must be positive semidefinite (psd) and rg [CF04, Fia95], and by [CF96], if the support supp(μ)\mathrm{supp}(\mu) of μ\mu is a subset of 𝒵(p)\mathcal{Z}(p) for a polynomial p[x,y]kp\in\mathbb{R}[x,y]_{\leq k}, then pp is a column relation of (k)\mathcal{M}(k). The bivariate KK–TMP is concretely solved in the following cases:

  1. (1)

    K=𝒵(p)K=\mathcal{Z}(p) for a polynomial pp with 1degp21\leq\deg p\leq 2.

    Assume that degp=2\deg p=2. By applying an affine linear transformation it suffices to consider one of the canonical cases: x2+y2=1x^{2}+y^{2}=1, y=x2y=x^{2}, xy=1xy=1, xy=0xy=0, y2=yy^{2}=y. The case x2+y2=1x^{2}+y^{2}=1 is equivalent to the univariate trigonometric moment problem, solved in [CF02]. The other four cases were tackled in [CF02, CF04, CF05, Fia15] by applying the far-reaching flat extension theorem (FET) [CF96, Theorem 7.10] (see also [CF05b, Theorem 2.19] and [Lau05] for an alternative proof), which states that β(2k)\beta^{(2k)} admits a (rank(k))(\operatorname{rank}\mathcal{M}(k))–atomic rm if and only if (k)\mathcal{M}(k) is psd and admits a rank–preserving extension to a moment matrix (k+1)\mathcal{M}(k+1). For an alternative approach with shorter proofs compared to the original ones by reducing the problem to the univariate setting see [BZ21, Section 6] (for xy=0xy=0), [Zal22a] (for y2=yy^{2}=y), [Zal22b] (for xy=1xy=1) and [Zal23] (for y=x2y=x^{2}).

    For degp=1\deg p=1 the solution is [CF08, Proposition 3.11] and uses the FET, but can be also derived in the univariate setting (see [Zal23, Remark 3.3.(4)])

  2. (2)

    K=2K=\mathbb{R}^{2}, k=2k=2 and (2)\mathcal{M}(2) is invertible. This case was first solved nonconstructively using convex geometry techniques in [FN10] and later on constructively in [CY16] by a novel rank reduction technique.

  3. (3)

    KK is one of 𝒵(yx3)\mathcal{Z}(y-x^{3}) [Fia11, Zal21], 𝒵(y2x3)\mathcal{Z}(y^{2}-x^{3}) [Zal21], 𝒵(y(ya)(yb))\mathcal{Z}(y(y-a)(y-b)) [Yoo17a, Zal22a], a,b{0}a,b\in\mathbb{R}\setminus\{0\}, aba\neq b, or 𝒵(xy21)\mathcal{Z}(xy^{2}-1) [Zal22b]. The main technique in [Fia11] is the FET, while in [Zal21, Zal22a, Zal22b] the reduction to the univariate TMP is applied.

  4. (4)

    (k)\mathcal{M}(k) has a special feature called recursive determinateness [CF13] or extremality [CFM08].

  5. (5)

    (3)\mathcal{M}(3) satisfies symmetric cubic column relations which can only cause extremal moment problems. In order to satisfy the variety condition, another symmetric column relation must exist, and the solution was obtained by checking consistency [CY14].

  6. (6)

    Non-extremal sextic TMPs with rank(3)8\operatorname{rank}\mathcal{M}(3)\leq 8 and with finite or infinite algebraic varieties [CY15].

  7. (7)

    (3)\mathcal{M}(3) with reducible cubic column relations [Yoo17b].

The solutions to the KK–TMP, which are not concrete in the sense of definition from the previous paragraph, are known in the cases K=𝒵(yq(x))K=\mathcal{Z}(y-q(x)) and K=𝒵(yq(x)1)K=\mathcal{Z}(yq(x)-1), where q[x]q\in\mathbb{R}[x]. [Fia11, Section 6] gives a solution in terms of the bound on the degree mm for which the existence of a positive extension (k+m)\mathcal{M}(k+m) of (k)\mathcal{M}(k) is equivalent to the existence of a rm. In [Zal23] the bound on mm is improved to m=degq1m=\deg q-1 for curves of the form y=q(x)y=q(x), degq3\deg q\geq 3, and to m=+1m=\ell+1 for curves of the form yx=1yx^{\ell}=1, {1}\ell\in\mathbb{N}\setminus\{1\}.

References to some classical work on the TMP are monographs [Akh65, AK62, KN77], while for a recent development in the area we refer a reader to [Sch17]. Special cases of the TMP have also been considered in [Kim14, Ble15, Fia17, DS18, BF20, Kim21], while [Nie14] considers subspaces of the polynomial algebra and [CGIK+] the TMP for commutative \mathbb{R}–algebras.

The motivation for this paper was to solve the TMP concretely on some reducible cubic curves, other than the case of three parallel lines solved in [Zal22a]. Applying an affine linear transformation we show that every such TMP is equivalent to the TMP on one of 8 canonical cases of reducible cubics of the form yc(x,y)=0yc(x,y)=0, where c[x,y]c\in\mathbb{R}[x,y], degc=2\deg c=2. In this article we solve the TMP for the cases c(x,y)=ay+x2+y2c(x,y)=ay+x^{2}+y^{2}, a{0}a\in\mathbb{R}\setminus\{0\}, and c(x,y)=xy2c(x,y)=x-y^{2}, which we call the circular and the parabolic type, respectively. The main idea is to characterize the existence of a decomposition of β\beta into the sum β()+β(c)\beta^{(\ell)}+\beta^{(c)}, where β()={βi,j()}i,j+, 0i+j2k\beta^{(\ell)}=\{\beta_{i,j}^{(\ell)}\}_{i,j\in\mathbb{Z}_{+},\;0\leq i+j\leq 2k} and β(c)={βi,j(c)}i,j+, 0i+j2k\beta^{(c)}=\{\beta_{i,j}^{(c)}\}_{i,j\in\mathbb{Z}_{+},\;0\leq i+j\leq 2k} admit a \mathbb{R}–rm and a 𝒵(c)\mathcal{Z}(c)–rm, respectively. Due to the form of the cubic yc(x,y)=0yc(x,y)=0, it turns out that all but two moments of β()\beta^{(\ell)} and β(c)\beta^{(c)} are not already fixed by the original sequence, i.e., β0,0()\beta_{0,0}^{(\ell)}, β1,0()\beta_{1,0}^{(\ell)}, β0,0(c)\beta_{0,0}^{(c)}, β1,0(c)\beta_{1,0}^{(c)} in the circular type case and β0,0()\beta_{0,0}^{(\ell)}, β2k,0()\beta_{2k,0}^{(\ell)}, β0,0(c)\beta_{0,0}^{(c)}, β2k,0(c)\beta_{2k,0}^{(c)} in the parabolic type case. Then, by an involved analysis, the characterization of the existence of a decomposition β=β()+β(c)\beta=\beta^{(\ell)}+\beta^{(c)} can be done in both cases. We also characterize the number of atoms in a minimal representing measure, i.e., a measure with the minimal number of atoms in the support.

1.1. Readers Guide

The paper is organized as follows. In Section 2 we present some preliminary results needed to establish the main results of the paper. In Section 3 we show that to solve the TMP on every reducible cubic curve it is enough to consider 8 canonical type relations (see Proposition 3.1). In Section 4 we present the general procedure for solving the TMP on all but one of the canonical types and prove some results that apply to them. Then in Sections 5 and 6 we specialize to the circular and the parabolic type relations and solve them concretely (see Theorems 5.1 and 6.1). In both cases we show, by numerical examples, that there are pure sequences β(6)\beta^{(6)} with a psd (3)\mathcal{M}(3) but without a rm (see Examples 5.3 and 6.3).

2. Preliminaries

We write n×m\mathbb{R}^{n\times m} for the set of n×mn\times m real matrices. For a matrix MM we call the linear span of its columns a column space and denote it by 𝒞(M)\mathcal{C}(M). The set of real symmetric matrices of size nn will be denoted by SnS_{n}. For a matrix ASnA\in S_{n} the notation A0A\succ 0 (resp. A0A\succeq 0) means AA is positive definite (pd) (resp. positive semidefinite (psd)). We write 𝟎t1,t2\mathbf{0}_{t_{1},t_{2}} for a t1×t2t_{1}\times t_{2} matrix with only zero entries and 𝟎t=𝟎t,t\mathbf{0}_{t}=\mathbf{0}_{t,t} for short, where t1,t2,tt_{1},t_{2},t\in\mathbb{N}. The notation Ei,j()E^{(\ell)}_{i,j}, \ell\in\mathbb{N}, stands for the usual ×\ell\times\ell coordinate matrix with the only nonzero entry at the position (i,j)(i,j), which is equal to 1.

In the rest of this section let kk\in\mathbb{N} and ββ(2k)={βi,j}i,j+, 0i+j2k\beta\equiv\beta^{(2k)}=\{\beta_{i,j}\}_{i,j\in\mathbb{Z}_{+},\;0\leq i+j\leq 2k} be a bivariate sequence of degree 2k2k.

2.1. Moment matrix

Let (k)\mathcal{M}(k) be the moment matrix of β\beta (see (1.2)). Let Q1,Q2Q_{1},Q_{2} be subsets of the set {XiYj:i,j+, 0i+jk}\{X^{i}Y^{j}\colon i,j\in\mathbb{Z}_{+},\;0\leq i+j\leq k\}. We denote by ((k))Q1,Q2(\mathcal{M}(k))_{Q_{1},Q_{2}} the submatrix of (k)\mathcal{M}(k) consisting of the rows indexed by the elements of Q1Q_{1} and the columns indexed by the elements of Q2Q_{2}. In case Q:=Q1=Q2Q:=Q_{1}=Q_{2}, we write ((k))Q:=((k))Q,Q(\mathcal{M}(k))_{Q}:=(\mathcal{M}(k))_{Q,Q} for short.

2.2. Affine linear transformations

The existence of representing measures is invariant under invertible affine linear transformations of the form

(2.1) ϕ(x,y)=(ϕ1(x,y),ϕ2(x,y)):=(a+bx+cy,d+ex+fy),(x,y)2,\phi(x,y)=(\phi_{1}(x,y),\phi_{2}(x,y)):=(a+bx+cy,d+ex+fy),\;(x,y)\in\mathbb{R}^{2},

a,b,c,d,e,fa,b,c,d,e,f\in\mathbb{R} with bfce0bf-ce\neq 0. Namely, let Lβ:[x,y]2kL_{\beta}:\mathbb{R}[x,y]_{\leq 2k}\to\mathbb{R} be a Riesz functional of the sequence β\beta defined by

Lβ(p):=i,j+,0i+j2kai,jβi,j,wherep=i,j+,0i+j2kai,jxiyj.L_{\beta}(p):=\sum_{\begin{subarray}{c}i,j\in\mathbb{Z}_{+},\\ 0\leq i+j\leq 2k\end{subarray}}a_{i,j}\beta_{i,j},\qquad\text{where}\quad p=\sum_{\begin{subarray}{c}i,j\in\mathbb{Z}_{+},\\ 0\leq i+j\leq 2k\end{subarray}}a_{i,j}x^{i}y^{j}.

We define β~={β~i,j}i,j+, 0i+j2k\widetilde{\beta}=\{\widetilde{\beta}_{i,j}\}_{i,j\in\mathbb{Z}_{+},\;0\leq i+j\leq 2k} by

β~i,j=Lβ(ϕ1(x,y)iϕ2(x,y)j).\widetilde{\beta}_{i,j}=L_{\beta}(\phi_{1}(x,y)^{i}\cdot\phi_{2}(x,y)^{j}).

By [CF04, Proposition 1.9], β\beta admits a (rr–atomic) KK–rm if and only if β~\widetilde{\beta} admits a (rr–atomic) ϕ(K)\phi(K)–rm. We write β~=ϕ(β)\widetilde{\beta}=\phi(\beta) and (k;β~)=ϕ((k;β))\mathcal{M}(k;\widetilde{\beta})=\phi(\mathcal{M}(k;\beta)).

2.3. Generalized Schur complements

Let

M=(ABCD)(n+m)×(n+m)M=\left(\begin{array}[]{cc}A&B\\ C&D\end{array}\right)\in\mathbb{R}^{(n+m)\times(n+m)}

be a real matrix where An×nA\in\mathbb{R}^{n\times n}, Bn×mB\in\mathbb{R}^{n\times m}, Cm×nC\in\mathbb{R}^{m\times n} and Dm×mD\in\mathbb{R}^{m\times m}. The generalized Schur complement [Zha05] of AA (resp. DD) in MM is defined by

M/A=DCAB(resp.M/D=ABDC),M/A=D-CA^{\dagger}B\quad(\text{resp.}\;M/D=A-BD^{\dagger}C),

where AA^{\dagger} (resp. DD^{\dagger}) stands for the Moore–Penrose inverse of AA (resp. DD).

The following lemma will be frequently used in the proofs.

Lemma 2.1.

Let n,mn,m\in\mathbb{N} and

M=(ABBTC)Sn+m,M=\left(\begin{array}[]{cc}A&B\\ B^{T}&C\end{array}\right)\in S_{n+m},

where ASnA\in S_{n}, Bn×mB\in\mathbb{R}^{n\times m} and CSmC\in S_{m}. If rankM=rankA\operatorname{rank}M=\operatorname{rank}A, then the matrix equation

(2.2) (ABT)W=(BC),\begin{pmatrix}A\\ B^{T}\end{pmatrix}W=\begin{pmatrix}B\\ C\end{pmatrix},

where Wn×mW\in\mathbb{R}^{n\times m}, is solvable and the solutions are precisely the solutions of the matrix equation AW=BAW=B. In particular, W=ABW=A^{\dagger}B satisfies (2.2).

Proof.

The assumption rankM=rankA\operatorname{rank}M=\operatorname{rank}A implies that

(2.3) (ABT)W=(AWBTW)=(BC)\begin{pmatrix}A\\ B^{T}\end{pmatrix}W=\begin{pmatrix}AW\\ B^{T}W\end{pmatrix}=\begin{pmatrix}B\\ C\end{pmatrix}

for some Wn×mW\in\mathbb{R}^{n\times m}. So the equation (2.2) is solvable. In particular, AW=BAW=B. It remains to prove that any solution WW to AW=BAW=B is also a solution to (2.3)\eqref{140722-1056}. Note that all the solutions of the equation AW~=BA\widetilde{W}=B are

(2.4) W~=AB+Z,\widetilde{W}=A^{\dagger}B+Z,

where each column of Zn×mZ\in\mathbb{R}^{n\times m} is an arbitrary vector from kerA\ker A. So WW satisfiying (2.3)\eqref{140722-1056} is also of the form AB+Z0A^{\dagger}B+Z_{0} for some Z0n×mZ_{0}\in\mathbb{R}^{n\times m} with columns belonging to kerA\ker A. We have that

(2.5) C=BTW=BT(AB+Z0)=BTAB+BTZ0=BTAB,C=B^{T}W=B^{T}(A^{\dagger}B+Z_{0})=B^{T}A^{\dagger}B+B^{T}Z_{0}=B^{T}A^{\dagger}B,

where we used the fact that each column of BB belongs to 𝒞(A)\mathcal{C}(A) and ker(A)=𝒞(A).\ker(A)^{\perp}=\mathcal{C}(A). Replacing WW with any W~\widetilde{W} of the form (2.4) in the calculation (2.5) gives the same result, which proves the statement of the proposition. ∎

The following theorem is a characterization of psd 2×22\times 2 block matrices.

Theorem 2.2 ([Alb69]).

Let

M=(ABBTC)Sn+mM=\left(\begin{array}[]{cc}A&B\\ B^{T}&C\end{array}\right)\in S_{n+m}

be a real symmetric matrix where ASnA\in S_{n}, Bn×mB\in\mathbb{R}^{n\times m} and CSmC\in S_{m}. Then:

  1. (1)

    The following conditions are equivalent:

    1. (a)

      M0M\succeq 0.

    2. (b)

      C0C\succeq 0, 𝒞(BT)𝒞(C)\mathcal{C}(B^{T})\subseteq\mathcal{C}(C) and M/C0M/C\succeq 0.

    3. (c)

      A0A\succeq 0, 𝒞(B)𝒞(A)\mathcal{C}(B)\subseteq\mathcal{C}(A) and M/A0M/A\succeq 0.

  2. (2)

    If M0M\succeq 0, then

    rankM=rankA+rankM/A=rankC+rankM/C.\operatorname{rank}M=\operatorname{rank}A+\operatorname{rank}M/A=\operatorname{rank}C+\operatorname{rank}M/C.

2.4. Extension principle

Proposition 2.3.

Let 𝒜Sn\mathcal{A}\in S_{n} be positive semidefinite, QQ a subset of the set {1,,n}\{1,\ldots,n\} and 𝒜|Q\mathcal{A}|_{Q} the restriction of 𝒜\mathcal{A} to the rows and columns from the set QQ. If 𝒜|Qv=0\mathcal{A}|_{Q}v=0 for a nonzero vector vv, then 𝒜v^=0\mathcal{A}\widehat{v}=0, where v^\widehat{v} is a vector with the only nonzero entries in the rows from QQ and such that the restriction v^|Q\widehat{v}|_{Q} to the rows from QQ equals to vv.

Proof.

See [Fia95, Proposition 2.4] or [Zal22a, Lemma 2.4] for an alternative proof. ∎

2.5. Partially positive semidefinite matrices and their completions

A partial matrix A=(ai,j)i,j=1nA=(a_{i,j})_{i,j=1}^{n} is a matrix of real numbers ai,ja_{i,j}\in\mathbb{R}, where some of the entries are not specified.

A partial symmetric matrix A=(ai,j)i,j=1nA=(a_{i,j})_{i,j=1}^{n} is partially positive semidefinite (ppsd) (resp. partially positive definite (ppd)) if the following two conditions hold:

  1. (1)

    ai,ja_{i,j} is specified if and only if aj,ia_{j,i} is specified and ai,j=aj,ia_{i,j}=a_{j,i}.

  2. (2)

    All fully specified principal minors of AA are psd (resp. pd).

For nn\in\mathbb{N} write [n]:={1,2,,n}[n]:=\{1,2,\ldots,n\}. We denote by AQ1,Q2A_{Q_{1},Q_{2}} the submatrix of An×nA\in\mathbb{R}^{n\times n} consisting of the rows indexed by the elements of Q1[n]Q_{1}\subseteq[n] and the columns indexed by the elements of Q2[n]Q_{2}\subseteq[n]. In case Q:=Q1=Q2Q:=Q_{1}=Q_{2}, we write AQ:=AQ,QA_{Q}:=A_{Q,Q} for short.

It is well-known that a ppsd matrix A(𝐱)A(\mathbf{x}) of the form as in Lemma 2.4 below admits a psd completion (This follows from the fact that the corresponding graph is chordal, see e.g., [GJSW84, Dan92, BW11]). Since we will need an additional information about the rank of the completion A(x0)A(x_{0}) and the explicit interval of all possible x0x_{0} for our results, we give a proof of Lemma 2.4 based on the use of generalized Schur complements.

Lemma 2.4.

Let A(𝐱)A(\mathbf{x}) be a partially positive semidefinite symmetric matrix of size n×nn\times n with the missing entries in the positions (i,j)(i,j) and (j,i)(j,i), 1i<jn1\leq i<j\leq n. Let

A1\displaystyle A_{1} =(A(𝐱))[n]{i,j},a=(A(𝐱))[n]{i,j},{i},b=(A(𝐱))[n]{i,j},{j},α=(A(𝐱))i,i,γ=(A(𝐱))j,j.\displaystyle=(A(\mathbf{x}))_{[n]\setminus\{i,j\}},\;a=(A(\mathbf{x}))_{[n]\setminus\{i,j\},\{i\}},\;b=(A(\mathbf{x}))_{[n]\setminus\{i,j\},\{j\}},\;\alpha=(A(\mathbf{x}))_{i,i},\;\gamma=(A(\mathbf{x}))_{j,j}.

Let

A2=(A(𝐱))[n]{j}=(A1aaTα)Sn1,A3=(A(𝐱))[n]{i}=(A1bbTγ)Sn1,A_{2}=(A(\mathbf{x}))_{[n]\setminus\{j\}}=\begin{pmatrix}A_{1}&a\\ a^{T}&\alpha\end{pmatrix}\in S_{n-1},\qquad A_{3}=(A(\mathbf{x}))_{[n]\setminus\{i\}}=\begin{pmatrix}A_{1}&b\\ b^{T}&\gamma\end{pmatrix}\in S_{n-1},

and

x±:=bTA1a±(A2/A1)(A3/A1).x_{\pm}:=b^{T}A_{1}^{\dagger}a\pm\sqrt{(A_{2}/A_{1})(A_{3}/A_{1})}\in\mathbb{R}.

Then:

  1. (i)

    A(x0)A(x_{0}) is positive semidefinite if and only if x0[x,x+]x_{0}\in[x_{-},x_{+}].

  2. (ii)
    rankA(x0)={max{rankA2,rankA3},forx0{x,x+},max{rankA2,rankA3}+1,forx0(x,x+).\operatorname{rank}A(x_{0})=\left\{\begin{array}[]{rl}\max\big{\{}\operatorname{rank}A_{2},\operatorname{rank}A_{3}\big{\}},&\text{for}\;x_{0}\in\{x_{-},x_{+}\},\\[5.0pt] \max\big{\{}\operatorname{rank}A_{2},\operatorname{rank}A_{3}\big{\}}+1,&\text{for}\;x_{0}\in(x_{-},x_{+}).\end{array}\right.
  3. (iii)

    The following statements are equivalent:

    1. (a)

      x=x+x_{-}=x_{+}.

    2. (b)

      A2/A1=0A_{2}/A_{1}=0 or A3/A1=0A_{3}/A_{1}=0.

    3. (c)

      rankA2=rankA1\operatorname{rank}A_{2}=\operatorname{rank}A_{1} or rankA3=rankA1\operatorname{rank}A_{3}=\operatorname{rank}A_{1}.

Proof.

We write

A(𝐱)\displaystyle A(\mathbf{x}) =(A11a12A13a14A15(a12)Tα(a23)T𝐱(a25)T(A13)Ta23A33a34a35(a14)T𝐱(a34)Tγ(a45)T(A15)Ta25(A35)Ta45A55)\displaystyle=\begin{pmatrix}A_{11}&a_{12}&A_{13}&a_{14}&A_{15}\\[1.99997pt] (a_{12})^{T}&\alpha&(a_{23})^{T}&\mathbf{x}&(a_{25})^{T}\\[1.99997pt] (A_{13})^{T}&a_{23}&A_{33}&a_{34}&a_{35}\\[1.99997pt] (a_{14})^{T}&\mathbf{x}&(a_{34})^{T}&\gamma&(a_{45})^{T}\\[1.99997pt] (A_{15})^{T}&a_{25}&(A_{35})^{T}&a_{45}&A_{55}\end{pmatrix}
(Si1(i1)×1(i1)×(ji1)(i1)×1(i1)×(nj)1×(i1)1×(ji1)1×(nj)(j11)×(i1)(ji1)×1Sji1(ji1)×1(ji1)×(nj)1×(i1)1×(ji1)1×(nj)(nj)×(i1)(nj)×1(nj)×(ji1)(nj)×1Snj)\displaystyle\in\begin{pmatrix}S_{i-1}&\mathbb{R}^{(i-1)\times 1}&\mathbb{R}^{(i-1)\times(j-i-1)}&\mathbb{R}^{(i-1)\times 1}&\mathbb{R}^{(i-1)\times(n-j)}\\ \mathbb{R}^{1\times(i-1)}&\mathbb{R}&\mathbb{R}^{1\times(j-i-1)}&\mathbb{R}&\mathbb{R}^{1\times(n-j)}\\ \mathbb{R}^{(j-1-1)\times(i-1)}&\mathbb{R}^{(j-i-1)\times 1}&S_{j-i-1}&\mathbb{R}^{(j-i-1)\times 1}&\mathbb{R}^{(j-i-1)\times(n-j)}\\ \mathbb{R}^{1\times(i-1)}&\mathbb{R}&\mathbb{R}^{1\times(j-i-1)}&\mathbb{R}&\mathbb{R}^{1\times(n-j)}\\ \mathbb{R}^{(n-j)\times(i-1)}&\mathbb{R}^{(n-j)\times 1}&\mathbb{R}^{(n-j)\times(j-i-1)}&\mathbb{R}^{(n-j)\times 1}&S_{n-j}\end{pmatrix}

Let PP be a permutation matrix, which changes the order of columns to

1,2,,i1,i+1,,j1,j+1,,n,i,j.1,2,\ldots,i-1,i+1,\ldots,j-1,j+1,\ldots,n,i,j.

Then

PTA(𝐱)P=(A11A13A15a12a14(A13)TA33A35a23a34(A15)T(A35)TA55a25a45(a12)T(a23)T(a25)Tα𝐱(a14)T(a34)T(a45)T𝐱γ)P^{T}A(\mathbf{x})P=\begin{pmatrix}A_{11}&A_{13}&A_{15}&a_{12}&a_{14}\\[1.99997pt] (A_{13})^{T}&A_{33}&A_{35}&a_{23}&a_{34}\\[1.99997pt] (A_{15})^{T}&(A_{35})^{T}&A_{55}&a_{25}&a_{45}\\[1.99997pt] (a_{12})^{T}&(a_{23})^{T}&(a_{25})^{T}&\alpha&\mathbf{x}\\[1.99997pt] (a_{14})^{T}&(a_{34})^{T}&(a_{45})^{T}&\mathbf{x}&\gamma\end{pmatrix}

Note that

(2.6) PTA(𝐱)P=(A1abaTα𝐱bT𝐱γ)andPTA(𝐱)P0A(𝐱)0.P^{T}A(\mathbf{x})P=\begin{pmatrix}A_{1}&a&b\\[1.99997pt] a^{T}&\alpha&\mathbf{x}\\[1.99997pt] b^{T}&\mathbf{x}&\gamma\end{pmatrix}\qquad\text{and}\qquad P^{T}A(\mathbf{x})P\succeq 0\;\Leftrightarrow\;A(\mathbf{x})\succeq 0.

Lemma 2.4 with the missing entries in the positions (n1,n)(n-1,n) and (n,n1)(n,n-1) was proved in [Zal21, Lemma 2.11] using computations with generalized Schur complements under one additional assumption:

(2.7) A1is invertibleorrankA1=rankA2.A_{1}\;\text{is invertible}\quad\text{or}\quad\operatorname{rank}A_{1}=\operatorname{rank}A_{2}.

Here we explain why the assumption (2.7) can be removed from [Zal21, Lemma 2.11]. The proof of [Zal21, Lemma 2.11] is separated into two cases: A2/A1>0A_{2}/A_{1}>0 and A2/A1=0A_{2}/A_{1}=0. The case A2/A1=0A_{2}/A_{1}=0 does not use (2.7). Assume now that A2/A1>0A_{2}/A_{1}>0 or equivalently rankA2>rankA1\operatorname{rank}A_{2}>\operatorname{rank}A_{1}. Invertibility of A1A_{1} (and by A2/A1>0A_{2}/A_{1}>0 also A2A_{2} is invertible) is used in the proof of [Zal21, Lemma 2.11] for the application of the quotient formula ([CH69])

(2.8) (A(x)/A2)=(A(x)/A1)/(A2/A1),(A(x)/A_{2})=\big{(}A(x)/A_{1}\big{)}\big{/}\big{(}A_{2}/A_{1}\big{)},

where

A(x)/A1=(A2/A1(A1baTx)/A1(A1abTx)/A1A3/A1)A(x)/A_{1}=\begin{pmatrix}A_{2}/A_{1}&\begin{pmatrix}A_{1}&b\\ a^{T}&x\end{pmatrix}\big{/}A_{1}\\ \begin{pmatrix}A_{1}&a\\ b^{T}&x\end{pmatrix}\big{/}A_{1}&A_{3}/A_{1}\end{pmatrix}

However, the formula (2.8) has been generalized [CHM74, Theorem 4] to noninvertible A1A_{1}, A2A_{2}, where all Schur complements are the generalized ones, under the conditions:

(2.9) (bx)T𝒞(A2)anda𝒞(A1).\begin{pmatrix}b&x\end{pmatrix}^{T}\in\mathcal{C}(A_{2})\qquad\text{and}\qquad a\in\mathcal{C}(A_{1}).

So if we show that the conditions (2.9) hold, the same proof as in [Zal21, Lemma 2.11] can be applied in the case A1A_{1} is singular. From A2A_{2} (resp. A3A_{3}) being psd, a𝒞(A1)a\in\mathcal{C}(A_{1}) (resp. b𝒞(A1)b\in\mathcal{C}(A_{1})) follows by Theorem 2.2, used for (M,A):=(A2,A1)(M,A):=(A_{2},A_{1}) (resp. (M,A):=(A3,A1)(M,A):=(A_{3},A_{1})). The assumption A2/A1>0A_{2}/A_{1}>0 implies that (aα)T𝒞((A1aT)T)\begin{pmatrix}a&\alpha\end{pmatrix}^{T}\notin\mathcal{C}(\begin{pmatrix}A_{1}&a^{T}\end{pmatrix}^{T}). Since a𝒞(A1)a\in\mathcal{C}(A_{1}), it follows that (01)T𝒞(A2)\begin{pmatrix}0&1\end{pmatrix}^{T}\in\mathcal{C}(A_{2}). Hence, (bx)T𝒞(A2)\begin{pmatrix}b&x\end{pmatrix}^{T}\in\mathcal{C}(A_{2}) for every xx\in\mathbb{R}, which concludes the proof of (2.9). ∎

2.6. Hamburger TMP

Let kk\in\mathbb{N}. For v=(v0,,v2k)2k+1v=(v_{0},\ldots,v_{2k})\in\mathbb{R}^{2k+1} we define the corresponding Hankel matrix as

(2.10) Av:=(vi+j)i,j=0k=(v0v1v2vkv1v2vk+1v2v2k1vkvk+1v2k1v2k)Sk+1.A_{v}:=\left(v_{i+j}\right)_{i,j=0}^{k}=\left(\begin{array}[]{ccccc}v_{0}&v_{1}&v_{2}&\cdots&v_{k}\\ v_{1}&v_{2}&\iddots&\iddots&v_{k+1}\\ v_{2}&\iddots&\iddots&\iddots&\vdots\\ \vdots&\iddots&\iddots&\iddots&v_{2k-1}\\ v_{k}&v_{k+1}&\cdots&v_{2k-1}&v_{2k}\end{array}\right)\in S_{k+1}.

We denote by 𝐯𝐣:=(vj+)=0k\mathbf{v_{j}}:=\left(v_{j+\ell}\right)_{\ell=0}^{k} the (j+1)(j+1)–th column of AvA_{v}, 0jk0\leq j\leq k, i.e.,

Av=(𝐯𝟎𝐯𝐤).A_{v}=\left(\begin{array}[]{ccc}\mathbf{v_{0}}&\cdots&\mathbf{v_{k}}\end{array}\right).

As in [CF91], the rank of vv, denoted by rankv\operatorname{rank}v, is defined by

rankv={k+1,if Av is nonsingular,min{𝐢:𝐯𝐢span{𝐯𝟎,,𝐯𝐢𝟏}},if Av is singular.\operatorname{rank}v=\left\{\begin{array}[]{rl}k+1,&\text{if }A_{v}\text{ is nonsingular},\\[1.99997pt] \min\left\{\bf{i}\colon\bf{v_{i}}\in\operatorname{span}\{\bf{v_{0}},\ldots,\bf{v_{i-1}}\}\right\},&\text{if }A_{v}\text{ is singular}.\end{array}\right.

For mkm\leq k we denote the upper left–hand corner (vi+j)i,j=0mSm+1\left(v_{i+j}\right)_{i,j=0}^{m}\in S_{m+1} of AvA_{v} of size m+1m+1 by Av(m)A_{v}(m). A sequence vv is called positively recursively generated (prg) if for r=rankvr=\operatorname{rank}v the following two conditions hold:

  • Av(r1)0A_{v}(r-1)\succ 0.

  • If r<k+1r<k+1, denoting

    (2.11) (φ0,,φr1):=Av(r1)1(vr,,v2r1)T,(\varphi_{0},\ldots,\varphi_{r-1}):=A_{v}(r-1)^{-1}(v_{r},\ldots,v_{2r-1})^{T},

    the equality

    (2.12) vj=φ0vjr++φr1vj1v_{j}=\varphi_{0}v_{j-r}+\cdots+\varphi_{r-1}v_{j-1}

    holds for j=r,,2kj=r,\ldots,2k.

The solution to the \mathbb{R}–TMP is the following.

Theorem 2.5 ([CF91, Theorems 3.9–3.10]).

For kk\in\mathbb{N} and v=(v0,,v2k)2k+1v=(v_{0},\ldots,v_{2k})\in\mathbb{R}^{2k+1} with v0>0v_{0}>0, the following statements are equivalent:

  1. (1)

    There exists a \mathbb{R}–representing measure for β\beta.

  2. (2)

    There exists a (rankAv)(\operatorname{rank}A_{v})–atomic \mathbb{R}–representing measure for β\beta.

  3. (3)

    AvA_{v} is positive semidefinite and one of the following holds:

    1. (a)

      Av(k1)A_{v}(k-1) is positive definite.

    2. (b)

      rankAv(k1)=rankAv\operatorname{rank}A_{v}(k-1)=\operatorname{rank}A_{v}.

  4. (4)

    vv is positively recursively generated.

2.7. TMP on the unit circle

The solution to the 𝒵(x2+y21)\mathcal{Z}(x^{2}+y^{2}-1)–TMP is the following.

Theorem 2.6 ([CF02, Theorem 2.1]).

Let p(x,y)=x2+y21p(x,y)=x^{2}+y^{2}-1 and β:=β(2k)=(βi,j)i,j+,i+j2k\beta:=\beta^{(2k)}=(\beta_{i,j})_{i,j\in\mathbb{Z}_{+},i+j\leq 2k}, where k2k\geq 2. Then the following statements are equivalent:

  1. (1)

    β\beta has a 𝒵(p)\mathcal{Z}(p)–representing measure.

  2. (2)

    β\beta has a (rank(k))(\operatorname{rank}\mathcal{M}(k))–atomic 𝒵(p)\mathcal{Z}(p)–representing measure.

  3. (3)

    (k)\mathcal{M}(k) is positive semidefinite and the relations β2+i,j+βi,2+j=βi,j\beta_{2+i,j}+\beta_{i,2+j}=\beta_{i,j} hold for every i,j+i,j\in\mathbb{Z}_{+} with i+j2k2i+j\leq 2k-2.

2.8. Parabolic TMP

We will need the following solution to the parabolic TMP (see [Zal23, Theorem 3.7]).

Theorem 2.7.

Let p(x,y)=xy2p(x,y)=x-y^{2} and β:=β(2k)=(βi,j)i,j+,i+j2k\beta:=\beta^{(2k)}=(\beta_{i,j})_{i,j\in\mathbb{Z}_{+},i+j\leq 2k}, where k2k\geq 2. Let

={1,Y,X,XY,X2,X2Y,,Xi,XiY,,Xk1,Xk1Y,Xk}.\mathcal{B}=\{\mathit{1},Y,X,XY,X^{2},X^{2}Y,\ldots,X^{i},X^{i}Y,\ldots,X^{k-1},X^{k-1}Y,X^{k}\}.

Then the following statements are equivalent:

  1. (1)

    β\beta has a 𝒵(p)\mathcal{Z}(p)–representing measure.

  2. (2)

    β\beta has a (rank(k))(\operatorname{rank}\mathcal{M}(k))–atomic 𝒵(p)\mathcal{Z}(p)–representing measure.

  3. (3)

    (k)\mathcal{M}(k) is positive semidefinite, the relations β1+i,j=βi,2+j\beta_{1+i,j}=\beta_{i,2+j} hold for every i,j+i,j\in\mathbb{Z}_{+} with i+j2k2i+j\leq 2k-2 and one of the following statements holds:

    1. (a)

      ((k)){Xk}\big{(}\mathcal{M}(k)\big{)}_{\mathcal{B}\setminus\{X^{k}\}} is positive definite.

    2. (b)

      rank((k)){Xk}=rank(k).\operatorname{rank}\big{(}\mathcal{M}(k)\big{)}_{\mathcal{B}\setminus\{X^{k}\}}=\operatorname{rank}\mathcal{M}(k).

  4. (4)

    The relations β1+i,j=βi,2+j\beta_{1+i,j}=\beta_{i,2+j} hold for every i,j+i,j\in\mathbb{Z}_{+} with i+j2k2i+j\leq 2k-2 and γ=(γ0,,γ4k)\gamma=(\gamma_{0},\ldots,\gamma_{4k}), defined by γi=βi2,imod 2\gamma_{i}=\beta_{\lfloor\frac{i}{2}\rfloor,i\;\mathrm{mod}\;2}, admits a \mathbb{R}–representing measure.

Remark 2.8.

The equivalence (3)(4)\eqref{parabolic--pt3}\Leftrightarrow\eqref{131023-1346-pt4} is part of the proof of [Zal23, Theorem 3.7].

3. TMP on reducible cubics - case reduction

In this section we show that to solve the TMP on reducible cubic curves it suffices, after applying an affine linear transformation, to solve the TMP on 8 canonical forms of curves.

Proposition 3.1.

Let kk\in\mathbb{R} and β:=β(2k)=(βi,j)i,j+,i+j2k\beta:=\beta^{(2k)}=(\beta_{i,j})_{i,j\in\mathbb{Z}_{+},i+j\leq 2k}. Assume (k;β)\mathcal{M}(k;\beta) does not satisfy any nontrivial column relation between columns indexed by monomials of degree at most 2, but it satisfies a column relation p(X,Y)=𝟎p(X,Y)=\bf{0}, where p[x,y]p\in\mathbb{R}[x,y] is a reducible polynomial with degp=3\deg p=3. If β\beta admits a representing measure, then there exists an invertible affine linear transformation ϕ\phi of the form (2.1) such that the moment matrix ϕ((k;β))\phi\big{(}\mathcal{M}(k;\beta)\big{)} satisfies a column relation q(x,y)=0q(x,y)=0, where qq has one of the following forms:

Parallel lines type:

q(x,y)=y(a+y)(b+y)q(x,y)=y(a+y)(b+y), a,b{0}a,b\in\mathbb{R}\setminus\{0\}, aba\neq b.

Circular type:

q(x,y)=y(ay+x2+y2)q(x,y)=y(ay+x^{2}+y^{2}), a{0}a\in\mathbb{R}\setminus\{0\}.

Parabolic type:

q(x,y)=y(xy2).q(x,y)=y(x-y^{2}).

Hyperbolic type 1:

q(x,y)=y(1xy)q(x,y)=y(1-xy).

Hyperbolic type 2:

q(x,y)=y(x+y+axy)q(x,y)=y(x+y+axy), a{0}a\in\mathbb{R}\setminus\{0\}.

Hyperbolic type 3:

q(x,y)=y(ay+x2y2)q(x,y)=y(ay+x^{2}-y^{2}), aa\in\mathbb{R}.

Intersecting lines type:

q(x,y)=yx(y+1)q(x,y)=yx(y+1),

Mixed type:

q(x,y)=y(1+ay+bx2+cy2)q(x,y)=y(1+ay+bx^{2}+cy^{2}), a,b,ca,b,c\in\mathbb{R}, b0b\neq 0.

Remark 3.2.

The name of the types of the form qq in Proposition 3.1 comes from the type of the conic q(x,y)y=0\frac{q(x,y)}{y}=0. The conic x+y+axy=0x+y+axy=0, a{0}a\in\mathbb{R}\setminus\{0\}, is a hyperbola, since the discriminant a2a^{2} is positive. Similarly, the conic ay+x2y2=0ay+x^{2}-y^{2}=0, aa\in\mathbb{R}, is a hyperbola, since its discriminant is equal to 4. Clearly, the conic ay+x2+y2=0ay+x^{2}+y^{2}=0, aa\in\mathbb{R}, is a circle with the center (0,a2)(0,-\frac{a}{2}) and radius a2\frac{a}{2}.

Now we prove Proposition 3.1.

Proof of Proposition 3.1.

Since p(x,y)p(x,y) is reducible, it is of the form p=p1p2p=p_{1}p_{2}, where

p1(x,y)\displaystyle p_{1}(x,y) =a0+a1x+a2ywith ai,(a1,a2)(0,0),\displaystyle=a_{0}+a_{1}x+a_{2}y\quad\text{with }a_{i}\in\mathbb{R},\;(a_{1},a_{2})\neq(0,0),
p2(x,y)\displaystyle p_{2}(x,y) =b0+b1x+b2y+b3x2+b4xy+b5y2with bi,(b3,b4,b5)(0,0,0).\displaystyle=b_{0}+b_{1}x+b_{2}y+b_{3}x^{2}+b_{4}xy+b_{5}y^{2}\quad\text{with }b_{i}\in\mathbb{R},\;(b_{3},b_{4},b_{5})\neq(0,0,0).

Without loss of generality we can assume that a20a_{2}\neq 0, since otherwise we apply the alt (x,y)(y,x)(x,y)\mapsto(y,x) to exchange the roles of xx and yy. Since a20a_{2}\neq 0, the alt

ϕ1(x,y)=(x,a0+a1x+a2y)\phi_{1}(x,y)=(x,a_{0}+a_{1}x+a_{2}y)

is invertible and hence:

(3.1) A sequence ϕ1(β) has a moment matrix ϕ1((k;β)) satisfying the column relationc0Y+c1X+c2Y2+c3X2Y+c4XY2+c5Y3=𝟎 with ci,(c3,c4,c5)(0,0,0).\displaystyle\begin{split}&\text{A sequence }\phi_{1}(\beta)\text{ has a moment matrix }\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying the column relation}\\ &c_{0}Y+c_{1}X+c_{2}Y^{2}+c_{3}X^{2}Y+c_{4}XY^{2}+c_{5}Y^{3}=\mathbf{0}\text{ with }c_{i}\in\mathbb{R},\;(c_{3},c_{4},c_{5})\neq(0,0,0).\end{split}

We separate two cases according to the value of c3c_{3}.

Case 1: c3=0c_{3}=0. In this case (3.1) is equal to

(3.2) A sequence ϕ1(β) has a moment matrix ϕ1((k;β)) satisfying the column relationc0Y+c1XY+c2Y2+c4XY2+c5Y3=𝟎 with ci,(c4,c5)(0,0).\displaystyle\begin{split}&\text{A sequence }\phi_{1}(\beta)\text{ has a moment matrix }\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying the column relation}\\ &c_{0}Y+c_{1}XY+c_{2}Y^{2}+c_{4}XY^{2}+c_{5}Y^{3}=\mathbf{0}\text{ with }c_{i}\in\mathbb{R},\;(c_{4},c_{5})\neq(0,0).\end{split}

If c0=c1=c2=0c_{0}=c_{1}=c_{2}=0, then (3.2) is equal to c4XY2+c5Y3=𝟎c_{4}XY^{2}+c_{5}Y^{3}=\mathbf{0}. Since by assumption β\beta and hence ϕ1(β)\phi_{1}(\beta) admit a rm, supported on

𝒵(y2(c4x+c5y))=𝒵(y(c4x+c5y)),\mathcal{Z}(y^{2}(c_{4}x+c_{5}y))=\mathcal{Z}(y(c_{4}x+c_{5}y)),

it follows by [CF96] that c4XY+c5Y2=𝟎c_{4}XY+c_{5}Y^{2}=\mathbf{0} is a nontrivial column relation in ϕ1((k;β))\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}. Hence, also (k;β)\mathcal{M}(k;\beta) satisfies a nontrivial column relation between columns indexed by monomials of degree at most 2, which is a contradiction with the assumption of the proposition. Therefore (c0,c1,c2)(0,0,0).(c_{0},c_{1},c_{2})\neq(0,0,0).

Case 1.1: c00c_{0}\neq 0. Dividing the relation in (3.2) by c0c_{0}, we get:

(3.3) A sequence ϕ1(β) has a moment matrix ϕ1((k;β)) satisfying the column relationY+c~1XY+c~2Y2+c~4XY2+c~5Y3=𝟎 with c~i,(c~4,c~5)(0,0).\displaystyle\begin{split}&\text{A sequence }\phi_{1}(\beta)\text{ has a moment matrix }\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying the column relation}\\ &Y+\widetilde{c}_{1}XY+\widetilde{c}_{2}Y^{2}+\widetilde{c}_{4}XY^{2}+\widetilde{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widetilde{c}_{i}\in\mathbb{R},\;(\widetilde{c}_{4},\widetilde{c}_{5})\neq(0,0).\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.1: c~1=0\widetilde{c}_{1}=0. In this case (3.3) is equivalent to:

(3.4) A sequence ϕ1(β) has a moment matrix ϕ1((k;β)) satisfying the column relationY+c~2Y2+c~4XY2+c~5Y3=𝟎 with c~i,(c~4,c~5)(0,0).\displaystyle\begin{split}&\text{A sequence }\phi_{1}(\beta)\text{ has a moment matrix }\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying the column relation}\\ &Y+\widetilde{c}_{2}Y^{2}+\widetilde{c}_{4}XY^{2}+\widetilde{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widetilde{c}_{i}\in\mathbb{R},\;(\widetilde{c}_{4},\widetilde{c}_{5})\neq(0,0).\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.1.1: c~4=0\widetilde{c}_{4}=0. In this case (3.4) is equivalent to

(3.5) A sequence ϕ1(β) has a moment matrix ϕ1((k;β)) satisfying the column relationY+c~2Y2+c~5Y3=𝟎 with c~2,c~5{0}.\displaystyle\begin{split}&\text{A sequence }\phi_{1}(\beta)\text{ has a moment matrix }\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying the column relation}\\ &Y+\widetilde{c}_{2}Y^{2}+\widetilde{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widetilde{c}_{2}\in\mathbb{R},\widetilde{c}_{5}\in\mathbb{R}\setminus\{0\}.\end{split}

The quadratic equation 1+c~2y+c~5y2=01+\widetilde{c}_{2}y+\widetilde{c}_{5}y^{2}=0 must have two different real nonzero solutions, otherwise 𝒵(y(1+c~2x+c~5y))\mathcal{Z}(y(1+\widetilde{c}_{2}x+\widetilde{c}_{5}y)) is a union of two parallel lines. Then it follows by [CF96] that there is a nontrivial column relation in (k;β)\mathcal{M}(k;\beta) between columns indexed by monomials of degree at most 2, which is a contradiction with the assumption of the proposition. So we have the parallel lines type relation from the proposition.

Case 1.1.1.2: c~40\widetilde{c}_{4}\neq 0. In this case the alt

ϕ2(x,y)=(c~2c~4xc~5y,y)\phi_{2}(x,y)=(-\widetilde{c}_{2}-\widetilde{c}_{4}x-\widetilde{c}_{5}y,y\Big{)}

is invertible and applying it to ϕ1(β)\phi_{1}(\beta), we obtain:

A sequence (ϕ2ϕ1)(β) has a moment matrix (ϕ2ϕ1)((k;β)) satisfyingthe hyperbolic type 1 relation from the proposition.
\displaystyle\begin{split}&\text{A sequence }(\phi_{2}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{2}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the hyperbolic type 1 relation from the proposition.}\vskip 12.0pt plus 4.0pt minus 4.0pt\end{split}

Case 1.1.2: c~10\widetilde{c}_{1}\neq 0. We apply the alt

ϕ3(x,y)=(1+c~1x,y)\phi_{3}(x,y)=(1+\widetilde{c}_{1}x,y)

to ϕ1(β)\phi_{1}(\beta) and obtain:

(3.6) A sequence (ϕ3ϕ1)(β) has a moment matrix (ϕ3ϕ1)((k;β)) satisfyingthe column relation XY+c^2Y2+c^4XY2+c^5Y3=𝟎 with c^i,(c^4,c^5)(0,0).\displaystyle\begin{split}&\text{A sequence }(\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }XY+\widehat{c}_{2}Y^{2}+\widehat{c}_{4}XY^{2}+\widehat{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widehat{c}_{i}\in\mathbb{R},\;(\widehat{c}_{4},\widehat{c}_{5})\neq(0,0).\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.2.1: c^40\widehat{c}_{4}\neq 0. We apply the alt

ϕ4(x,y)=(xc^5c^4y,y)\phi_{4}(x,y)=\Big{(}x-\frac{\widehat{c}_{5}}{\widehat{c}_{4}}y,y\Big{)}

to (ϕ3ϕ1)(β)(\phi_{3}\circ\phi_{1})(\beta) and obtain:

(3.7) A sequence (ϕ4ϕ3ϕ1)(β) has a moment matrix (ϕ4ϕ3ϕ1)((k;β)) satisfyingthe column relation XY+c˘2Y2+c^4XY2=𝟎 with c˘2,c^4,c^40.\displaystyle\begin{split}&\text{A sequence }(\phi_{4}\circ\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{4}\circ\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }XY+\breve{c}_{2}Y^{2}+\widehat{c}_{4}XY^{2}=\mathbf{0}\text{ with }\breve{c}_{2},\widehat{c}_{4}\in\mathbb{R},\;\widehat{c}_{4}\neq 0.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.2.1.1: c˘2=0\breve{c}_{2}=0. In this case the relation in (3.7) is of the form

XY+c^4XY2=𝟎with c^4{0}.XY+\widehat{c}_{4}XY^{2}=\mathbf{0}\quad\text{with }\widehat{c}_{4}\in\mathbb{R}\setminus\{0\}.

Applying the alt

ϕ5(x,y)=(x,c^4y)\phi_{5}(x,y)=(x,\widehat{c}_{4}y)

to (ϕ4ϕ3ϕ1)(β)(\phi_{4}\circ\phi_{3}\circ\phi_{1})(\beta) we obtain:

A sequence (ϕ5ϕ4ϕ3ϕ1)(β) has a moment matrix (ϕ5ϕ4ϕ3ϕ1)((k;β)) satisfyingthe intersecting lines type relation from the proposition.\displaystyle\begin{split}&\text{A sequence }(\phi_{5}\circ\phi_{4}\circ\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{5}\circ\phi_{4}\circ\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the intersecting lines type relation from the proposition.}\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.2.1.2: c˘20\breve{c}_{2}\neq 0. We apply the alt

ϕ6(x,y)=(x,c˘2y)\phi_{6}(x,y)=(x,\breve{c}_{2}y)

to (ϕ4ϕ3ϕ1)(β)(\phi_{4}\circ\phi_{3}\circ\phi_{1})(\beta) and obtain:

A sequence (ϕ6ϕ4ϕ3ϕ1)(β) has a moment matrix (ϕ6ϕ4ϕ3ϕ1)((k;β)) satisfyingthe hyperbolic type 2 relation in the proposition.\displaystyle\begin{split}&\text{A sequence }(\phi_{6}\circ\phi_{4}\circ\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{6}\circ\phi_{4}\circ\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the hyperbolic type 2 relation in the proposition.}\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.2.2: c^4=0\widehat{c}_{4}=0. In this case (3.6) is equivalent to:

(3.8) A sequence (ϕ3ϕ1)(β) has a moment matrix (ϕ3ϕ1)((k;β)) satisfyingthe column relation XY+c^2Y2+c^5Y3=𝟎 with c^2,c^5,c^50.\displaystyle\begin{split}&\text{A sequence }(\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }XY+\widehat{c}_{2}Y^{2}+\widehat{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widehat{c}_{2},\widehat{c}_{5}\in\mathbb{R},\;\widehat{c}_{5}\neq 0.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.2.2.1: c~2=0\widetilde{c}_{2}=0. Applying the alt

ϕ7(x,y)=(x,c^5y),\phi_{7}(x,y)=(x,-\widehat{c}_{5}y),

to (ϕ3ϕ1)(β)(\phi_{3}\circ\phi_{1})(\beta) we obtain:

A sequence (ϕ7ϕ3ϕ1)(β) has a moment matrix (ϕ7ϕ3ϕ1)((k;β)) satisfyingthe parabolic type relation in the proposition.\displaystyle\begin{split}&\text{A sequence }(\phi_{7}\circ\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{7}\circ\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the parabolic type relation in the proposition.}\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 1.1.2.2.2: c~20\widetilde{c}_{2}\neq 0. Applying the alt

ϕ8(x,y)=(x,c^2y)\phi_{8}(x,y)=(x,\widehat{c}_{2}y)

to (ϕ3ϕ1)(β)(\phi_{3}\circ\phi_{1})(\beta) and obtain:

(3.9) A sequence (ϕ8ϕ3ϕ1)(β) has a moment matrix (ϕ8ϕ3ϕ1)((k;β)) satisfyingthe column relation XY+Y2+c˘5Y3=𝟎 with c˘5,(c50.\displaystyle\begin{split}&\text{A sequence }(\phi_{8}\circ\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{8}\circ\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }XY+Y^{2}+\breve{c}_{5}Y^{3}=\mathbf{0}\text{ with }\breve{c}_{5}\in\mathbb{R},\;\mathchoice{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{2.42343pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{1.73102pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle c\hss$\crcr}}}\limits}_{5}\neq 0.\end{split}

Further on, the relation in (3.9) is equivalent to

(3.10) (

(

c
5
)
1
(XY+Y2)
+Y3
=𝟎
with 

(

c
5
,

(

c
5
0
.
(\mathchoice{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{2.42343pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{1.73102pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle c\hss$\crcr}}}\limits}_{5})^{-1}(XY+Y^{2})+Y^{3}=\mathbf{0}\quad\text{with }\mathchoice{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{2.42343pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{1.73102pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle c\hss$\crcr}}}\limits}_{5}\in\mathbb{R},\;\mathchoice{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{2.42343pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{1.73102pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle c\hss$\crcr}}}\limits}_{5}\neq 0.

Finally, applying the alt

ϕ9(x,y)=((

(

c
5
)
1
(x+y)
,y)
\phi_{9}(x,y)=\big{(}(-\mathchoice{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\displaystyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{3.46205pt}{3.8889pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\textstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{2.42343pt}{2.72223pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptstyle c\hss$\crcr}}}\limits}{\mathop{\vbox{\halign{#\cr\kern 0.80002pt$\hss\leavevmode\resizebox{1.73102pt}{1.94444pt}{\rotatebox[origin={c}]{90.0}{(}}\hss$\crcr\nointerlineskip\cr$\hss\scriptscriptstyle c\hss$\crcr}}}\limits}_{5})^{-1}(x+y),y\big{)}

to (ϕ8ϕ3ϕ1)(β)(\phi_{8}\circ\phi_{3}\circ\phi_{1})(\beta), we obtain:

A sequence (ϕ9ϕ8ϕ3ϕ1)(β) has a moment matrix (ϕ9ϕ8ϕ3ϕ1)((k;β))satisfying the parabolic type relation in the proposition.
\displaystyle\begin{split}&\text{A sequence }(\phi_{9}\circ\phi_{8}\circ\phi_{3}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{9}\circ\phi_{8}\circ\phi_{3}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\\ &\text{satisfying the parabolic type relation in the proposition.}\vskip 12.0pt plus 4.0pt minus 4.0pt\end{split}

Case 1.2: c0=0c_{0}=0. In this case (3.2) is equivalent to:

(3.11) A sequence ϕ1(β) has a moment matrix ϕ1((k;β)) satisfying the column relationc1XY+c2Y2+c4XY2+c5Y3=𝟎 with ci,(c4,c5)(0,0).\displaystyle\begin{split}&\text{A sequence }\phi_{1}(\beta)\text{ has a moment matrix }\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying the column relation}\\ &c_{1}XY+c_{2}Y^{2}+c_{4}XY^{2}+c_{5}Y^{3}=\mathbf{0}\text{ with }c_{i}\in\mathbb{R},\;(c_{4},c_{5})\neq(0,0).\end{split}

Assume that c1=0c_{1}=0. Since by assumption β\beta and hence ϕ1(β)\phi_{1}(\beta) admits a rm, supported on

𝒵(y2(c2+c4x+c5y))=𝒵(y(c2+c4x+c5y)),\mathcal{Z}(y^{2}(c_{2}+c_{4}x+c_{5}y))=\mathcal{Z}(y(c_{2}+c_{4}x+c_{5}y)),

it follows by [CF96] that c2Y+c4XY+c5Y2=𝟎c_{2}Y+c_{4}XY+c_{5}Y^{2}=\mathbf{0} is a nontrivial column relation in ϕ1((k;β))\phi_{1}\big{(}\mathcal{M}(k;\beta)\big{)}. Hence, also (k;β)\mathcal{M}(k;\beta) satisfies a nontrivial column relation between columns indexed by monomials of degree at most 2, which is a contradiction with the assumption of the proposition. Hence, c10.c_{1}\neq 0. Applying the alt (x,y)(c1x,y)(x,y)\mapsto(c_{1}x,y) to ϕ1(β)\phi_{1}(\beta), we obtain a sequence with the moment matrix satisfying the column relation of the form (3.6) and we can proceed as in the Case 1.1.2 above.

Case 2: c30c_{3}\neq 0. Applying the alt

ϕ10(x,y)=(|c3|x,y)\phi_{10}(x,y)=\big{(}\sqrt{|c_{3}|}x,y\big{)}

to ϕ1(β)\phi_{1}(\beta), we obtain:

(3.12) A sequence (ϕ10ϕ1)(β) has a moment matrix (ϕ10ϕ1)((k;β)) satisfyingthe column relation c0Y+c~1XY+c2Y2+|c3|c3X2Y+c~4XY2+c5Y3=𝟎 with ci,c~i.\displaystyle\begin{split}&\text{A sequence }(\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }c_{0}Y+\widetilde{c}_{1}XY+c_{2}Y^{2}+\frac{|c_{3}|}{c_{3}}X^{2}Y+\widetilde{c}_{4}XY^{2}+c_{5}Y^{3}=\mathbf{0}\text{ with }c_{i},\widetilde{c}_{i}\in\mathbb{R}.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.1: c~1=0\widetilde{c}_{1}=0. In this case (3.12) is equivalent to:

(3.13) A sequence (ϕ10ϕ1)(β) has a moment matrix (ϕ10ϕ1)((k;β)) satisfyingthe column relation c0Y+c2Y2+|c3|c3X2Y+c~4XY2+c5Y3=𝟎 with ci,c~i.\displaystyle\begin{split}&\text{A sequence }(\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }c_{0}Y+c_{2}Y^{2}+\frac{|c_{3}|}{c_{3}}X^{2}Y+\widetilde{c}_{4}XY^{2}+c_{5}Y^{3}=\mathbf{0}\text{ with }c_{i},\widetilde{c}_{i}\in\mathbb{R}.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.1.1: c0=0c_{0}=0. Dividing the relation in (3.13) with |c3|c3\frac{|c_{3}|}{c_{3}}, (3.13) is equivalent to:

(3.14) A sequence (ϕ10ϕ1)(β) has a moment matrix (ϕ10ϕ1)((k;β)) satisfyingthe column relation c~2Y2+X2Y+c^4XY2+c~5Y3=𝟎 with c~2,c^4,c~5.\displaystyle\begin{split}&\text{A sequence }(\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }\widetilde{c}_{2}Y^{2}+X^{2}Y+\widehat{c}_{4}XY^{2}+\widetilde{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widetilde{c}_{2},\widehat{c}_{4},\widetilde{c}_{5}\in\mathbb{R}.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Applying the alt

ϕ11(x,y)=(x+c^42y,y)\phi_{11}(x,y)=\Big{(}x+\frac{\widehat{c}_{4}}{2}y,y\Big{)}

to (ϕ10ϕ1)(β)(\phi_{10}\circ\phi_{1})(\beta), we obtain:

(3.15) A sequence (ϕ11ϕ10ϕ1)(β) has a moment matrix (ϕ11ϕ10ϕ1)((k;β))satisfying the column relation c˘2Y2+X2Y+c˘5Y3=𝟎 with c˘2,c˘5.\displaystyle\begin{split}&\text{A sequence }(\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{11}\circ\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\\ &\text{satisfying the column relation }\breve{c}_{2}Y^{2}+X^{2}Y+\breve{c}_{5}Y^{3}=\mathbf{0}\text{ with }\breve{c}_{2},\breve{c}_{5}\in\mathbb{R}.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.1.1.1: c˘5=0\breve{c}_{5}=0. Since by assumption of the proposition, (ϕ11ϕ10ϕ1)(β)(\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta) admits a rm, supported on 𝒵(y(c˘2y+x2))\mathcal{Z}(y(\breve{c}_{2}y+x^{2})), c˘2\breve{c}_{2} in (3.15) cannot be equal to 0. Indeed, c˘2=0\breve{c}_{2}=0 would imply that 𝒵(y(c˘2y+x2))=𝒵(yx2)=𝒵(yx)\mathcal{Z}(y(\breve{c}_{2}y+x^{2}))=\mathcal{Z}(yx^{2})=\mathcal{Z}(yx) and by [CF96], XY=𝟎XY=\mathbf{0} would be a nontrivial column relation in (ϕ11ϕ10ϕ1)((k;β))(\phi_{11}\circ\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}. Hence, also (k;β)\mathcal{M}(k;\beta) would satisfy a nontrivial column relation between columns indexed by monomials of degree at most 2, which is a contradiction with the assumption of the proposition. Since c˘20\breve{c}_{2}\neq 0, after applying the alt

ϕ12(x,y)=(x,c˘2y)\phi_{12}(x,y)=(x,-\breve{c}_{2}y)

to (ϕ11ϕ10ϕ1)(β)(\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta), we obtain:

A sequence (ϕ12ϕ11ϕ10ϕ1)(β) has a moment matrix (ϕ12ϕ11ϕ10ϕ1)((k;β))satisfying the parabolic type relation in the proposition.\displaystyle\begin{split}&\text{A sequence }(\phi_{12}\circ\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{12}\circ\phi_{11}\circ\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\\ &\text{satisfying the parabolic type relation in the proposition.}\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.1.1.2: c˘5>0\breve{c}_{5}>0. Applying the alt

ϕ13(x,y)=(x,c˘5y)\phi_{13}(x,y)=(x,\sqrt{\breve{c}_{5}}y)

to (ϕ11ϕ10ϕ1)(β)(\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta) we obtain:

A sequence (ϕ13ϕ11ϕ10ϕ1)(β) has a moment matrix (ϕ13ϕ11ϕ10ϕ1)((k;β))satisfying the circular type relation in the proposition.\displaystyle\begin{split}&\text{A sequence }(\phi_{13}\circ\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{13}\circ\phi_{11}\circ\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\\ &\text{satisfying the circular type relation in the proposition.}\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.1.1.3: c˘5<0\breve{c}_{5}<0. Applying the alt

ϕ14(x,y)=(x,c˘5y)\phi_{14}(x,y)=(x,\sqrt{-\breve{c}_{5}}y)

to (ϕ11ϕ10ϕ1)(β)(\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta), we obtain:

A sequence (ϕ14ϕ11ϕ10ϕ1)(β) has a moment matrix (ϕ14ϕ11ϕ10ϕ1)((k;β))satisfying the hyperbolic type 3 relation in the proposition.\displaystyle\begin{split}&\text{A sequence }(\phi_{14}\circ\phi_{11}\circ\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{14}\circ\phi_{11}\circ\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\\ &\text{satisfying the hyperbolic type 3 relation in the proposition.}\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.1.2: c00c_{0}\neq 0. Dividing the relation in (3.13) with c0c_{0}, (3.13) is equivalent to:

(3.16) A sequence (ϕ10ϕ1)(β) has a moment matrix (ϕ10ϕ1)((k;β)) satisfyingthe column relation Y+c~2Y2+c~3X2Y+c^4XY2+c~5Y3=𝟎 with c~i,c^4,c~30.\displaystyle\begin{split}&\text{A sequence }(\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying}\\ &\text{the column relation }Y+\widetilde{c}_{2}Y^{2}+\widetilde{c}_{3}X^{2}Y+\widehat{c}_{4}XY^{2}+\widetilde{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widetilde{c}_{i},\widehat{c}_{4}\in\mathbb{R},\widetilde{c}_{3}\neq 0.\end{split}

Applying the alt

ϕ15(x,y)=(x+c^42c~3,y)\phi_{15}(x,y)=\Big{(}x+\frac{\widehat{c}_{4}}{2\widetilde{c}_{3}},y\Big{)}

to (ϕ10ϕ1)(β),(\phi_{10}\circ\phi_{1})(\beta), we obtain:

A sequence (ϕ15ϕ10ϕ1)(β) has a moment matrix (ϕ15ϕ10ϕ1)((k;β))satisfying the mixed type relation in the proposition.\displaystyle\begin{split}&\text{A sequence }(\phi_{15}\circ\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{15}\circ\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\\ &\text{satisfying the mixed type relation in the proposition.}\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.2: c~10\widetilde{c}_{1}\neq 0. Dividing the relation in (3.12) with |c3|c3\frac{|c_{3}|}{c_{3}}, (3.12) is equivalent to:

(3.17) A sequence (ϕ10ϕ1)(β) has a moment matrix (ϕ10ϕ1)((k;β)) satisfying thecolumn relation c^0Y+c^1XY+c^2Y2+X2Y+c^4XY2+c^5Y3=𝟎 with c^i,c^10.\displaystyle\begin{split}&\text{A sequence }(\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\text{ satisfying the}\\ &\text{column relation }\widehat{c}_{0}Y+\widehat{c}_{1}XY+\widehat{c}_{2}Y^{2}+X^{2}Y+\widehat{c}_{4}XY^{2}+\widehat{c}_{5}Y^{3}=\mathbf{0}\text{ with }\widehat{c}_{i}\in\mathbb{R},\widehat{c}_{1}\neq 0.\end{split}

Now we apply the alt

ϕ16(x,y)=(x+c^12,y)\phi_{16}(x,y)=\Big{(}x+\frac{\widehat{c}_{1}}{2},y\Big{)}

to (ϕ10ϕ1)(β)(\phi_{10}\circ\phi_{1})(\beta) and obtain:

(3.18) A sequence (ϕ16ϕ10ϕ1)(β) has a moment matrix (ϕ16ϕ10ϕ1)((k;β))satisfying the column relation c˘0Y+c˘2Y2+X2Y+c˘4XY2+c˘5Y3=𝟎 with c˘i.\displaystyle\begin{split}&\text{A sequence }(\phi_{16}\circ\phi_{10}\circ\phi_{1})(\beta)\text{ has a moment matrix }(\phi_{16}\circ\phi_{10}\circ\phi_{1})\big{(}\mathcal{M}(k;\beta)\big{)}\\ &\text{satisfying the column relation }\breve{c}_{0}Y+\breve{c}_{2}Y^{2}+X^{2}Y+\breve{c}_{4}XY^{2}+\breve{c}_{5}Y^{3}=\mathbf{0}\text{ with }\breve{c}_{i}\in\mathbb{R}.\vskip 6.0pt plus 2.0pt minus 2.0pt\end{split}

Case 2.2.1: c˘0=0\breve{c}_{0}=0. In this case the relation in (3.18) becomes equal to the relation in (3.14) from the Case 2.1.1, so we can proceed as above.

Case 2.2.2: c˘00\breve{c}_{0}\neq 0. Dividing the relation in (3.18) with c˘0\breve{c}_{0}, it becomes equal to the relation in (3.16) from the Case 2.1.2, so we can proceed as above. ∎

4. Solving the TMP on canonical reducible cubic curves

Let β={βi}i+2,|i|2k\beta=\{\beta_{i}\}_{i\in\mathbb{Z}_{+}^{2},|i|\leq 2k} be a sequence of degree 2k2k, kk\in\mathbb{N}, and

(4.1) 𝒞={1,X,Y,X2,XY,Y2,,Xk,Xk1Y,,Yk}\mathcal{C}=\{\mathit{1},X,Y,X^{2},XY,Y^{2},\ldots,X^{k},X^{k-1}Y,\ldots,Y^{k}\}

the set of rows and columns of the moment matrix (k;β)\mathcal{M}(k;\beta) in the degree-lexicographic order. Let

(4.2) p(x,y)=yc(x,y)[x,y]3p(x,y)=y\cdot c(x,y)\in\mathbb{R}[x,y]_{\leq 3}

be a polynomial of degree 3 in one of the canonical forms from Proposition 3.1. Hence, c(x,y)c(x,y) a polynomial of degree 2. β\beta will have a 𝒵(p)\mathcal{Z}(p)–rm if and only if it can be decomposed as

(4.3) β=β()+β(c),\beta=\beta^{(\ell)}+\beta^{(c)},

where

β()\displaystyle\beta^{(\ell)} :={βi()}i+2,|i|2khas a representing measure on y=0,\displaystyle:=\{\beta_{i}^{(\ell)}\}_{i\in\mathbb{Z}_{+}^{2},|i|\leq 2k}\quad\text{has a representing measure on }y=0,
β(c)\displaystyle\beta^{(c)} :={βi(c)}i+2,|i|2khas a representing measure on the conic c(x,y)=0,\displaystyle:=\{\beta_{i}^{(c)}\}_{i\in\mathbb{Z}_{+}^{2},|i|\leq 2k}\quad\text{has a representing measure on the conic }c(x,y)=0,

and the sum in (4.3) is a component-wise sum. On the level of moment matrices, (4.3) is equivalent to

(4.4) (k;β)=(k;β())+(k;β(c)).\mathcal{M}(k;\beta)=\mathcal{M}(k;\beta^{(\ell)})+\mathcal{M}(k;\beta^{(c)}).

Note that if β\beta has a 𝒵(p)\mathcal{Z}(p)–rm, then the matrix (k;β)\mathcal{M}(k;\beta) satisfies the relation p(X,Y)=𝟎p(X,Y)=\mathbf{0} and it must be rg, i.e.,

(4.5) XiYjp(X,Y)=𝟎for i,j=0,,k3 such that i+jk3.X^{i}Y^{j}p(X,Y)=\mathbf{0}\quad\text{for }i,j=0,\ldots,k-3\text{ such that }i+j\leq k-3.

We write X(0,k):=(1,X,,Xk)\vec{X}^{(0,k)}:=(\mathit{1},X,\ldots,X^{k}). Let 𝒯𝒞\mathcal{T}\subseteq\mathcal{C} be a subset, such that the columns from 𝒯\mathcal{T} span the column space 𝒞((k;β))\mathcal{C}(\mathcal{M}(k;\beta)) and

(4.6) P is a permutation matrix such that moment matrix ~(k;β):=P(k;β)PThas rows and columns indexed in the order X(0,k),𝒯X(0,k),𝒞(X(0,k)𝒯).\displaystyle\begin{split}&P\text{ is a permutation matrix such that moment matrix }\widetilde{\mathcal{M}}(k;\beta):=P\mathcal{M}(k;\beta)P^{T}\\ &\text{has rows and columns indexed in the order }\vec{X}^{(0,k)},\mathcal{T}\setminus\vec{X}^{(0,k)},\mathcal{C}\setminus(\vec{X}^{(0,k)}\cup\mathcal{T}).\end{split}

In this new order of rows and columns, (4.4) becomes equivalent to

(4.7) ~(k;β)=~(k;β())+~(k;β(c)).\widetilde{\mathcal{M}}(k;\beta)=\widetilde{\mathcal{M}}(k;\beta^{(\ell)})+\widetilde{\mathcal{M}}(k;\beta^{(c)}).

We write

(4.9) ~(k;β)\displaystyle\widetilde{\mathcal{M}}(k;\beta) = (\@arstrutX(0,k)TX(0,k)C(X(0,k)T)\\(X(0,k))TA11A12A13\\[0.3em](TX(0,k))T(A12)TA22A23\\[0.3em](C(X(0,k)T))T(A13)T(A23)TA33\\) .\displaystyle=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(0,k)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{T}\setminus\vec{X}^{(0,k)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{C}\setminus(\vec{X}^{(0,k)}\cup\mathcal{T})\\(\vec{X}^{(0,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{11}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{13}\\[0.3em](\mathcal{T}\setminus\vec{X}^{(0,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{12})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{23}\\[0.3em](\mathcal{C}\setminus(\vec{X}^{(0,k)}\cup\mathcal{T}))^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{13})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{33}\\$\hfil\kern 5.0pt\crcr}}}}\right)$}}.

By the form of the atoms, we know that ~(k;β())\widetilde{\mathcal{M}}(k;\beta^{(\ell)}) and ~(k;β(c))\widetilde{\mathcal{M}}(k;\beta^{(c)}) will be of the forms

(4.10) ~(k;β(c))= ( \@arstrutX(0,k)TX(0,k)C(X(0,k)T)\\(X(0,k))TAA12A13\\[0.3em](TX(0,k))T(A12)TA22A23\\[0.3em](C(X(0,k)T))T(A13)T(A23)TA33\\) ,\\[0.5em]~(k;β())= (\@arstrutX(0,k)TX(0,k)C(X(0,k)T)\\(X(0,k))TA11-A00\\[0.3em](TX(0,k))T000\\[0.3em](C(X(0,k)T))T000\\) \displaystyle\begin{split}\widetilde{\mathcal{M}}(k;\beta^{(c)})&=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(0,k)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{T}\setminus\vec{X}^{(0,k)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{C}\setminus(\vec{X}^{(0,k)}\cup\mathcal{T})\\(\vec{X}^{(0,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{13}\\[0.3em](\mathcal{T}\setminus\vec{X}^{(0,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{12})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{23}\\[0.3em](\mathcal{C}\setminus(\vec{X}^{(0,k)}\cup\mathcal{T}))^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{13})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{33}\\$\hfil\kern 5.0pt\crcr}}}}\right)$}},\\[0.5em]\widetilde{\mathcal{M}}(k;\beta^{(\ell)})&=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(0,k)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{T}\setminus\vec{X}^{(0,k)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{C}\setminus(\vec{X}^{(0,k)}\cup\mathcal{T})\\(\vec{X}^{(0,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle A_{11}-A$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}\\[0.3em](\mathcal{T}\setminus\vec{X}^{(0,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}\\[0.3em](\mathcal{C}\setminus(\vec{X}^{(0,k)}\cup\mathcal{T}))^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathbf{0}\\$\hfil\kern 5.0pt\crcr}}}}\right)$}}\end{split}

for some Hankel matrix ASk+1.A\in S_{k+1}. Define matrix functions :Sk+1S(k+1)(k+2)2\mathcal{F}:S_{k+1}\to S_{\frac{(k+1)(k+2)}{2}} and :Sk+1Sk+1\mathcal{H}:S_{k+1}\to S_{k+1} by

(4.11) (𝐀)\displaystyle\mathcal{F}(\mathbf{A}) =(𝐀A12A13(A12)TA22A23(A13)T(A23)TA33)and(𝐀)=A11𝐀.\displaystyle=\begin{pmatrix}\mathbf{A}&A_{12}&A_{13}\\[3.00003pt] (A_{12})^{T}&A_{22}&A_{23}\\[3.00003pt] (A_{13})^{T}&(A_{23})^{T}&A_{33}\end{pmatrix}\quad\text{and}\quad\mathcal{H}(\mathbf{A})=A_{11}-\mathbf{A}.

Using (4.10), (4.7) becomes equivalent to

(4.12) ~(k;β)=(A)+(A)𝟎k(k+1)2\widetilde{\mathcal{M}}(k;\beta)=\mathcal{F}(A)+\mathcal{H}(A)\oplus\mathbf{0}_{\frac{k(k+1)}{2}}

for some Hankel matrix ASk+1A\in S_{k+1}.

Lemma 4.1.

Assume the notation above. The sequence β={βi}i+2,|i|2k\beta=\{\beta_{i}\}_{i\in\mathbb{Z}_{+}^{2},|i|\leq 2k}, where k3k\geq 3, has a 𝒵(p)\mathcal{Z}(p)–representing measure if and only if there exist a Hankel matrix ASk+1A\in S_{k+1}, such that:

  1. (1)

    The sequence with the moment matrix (A)\mathcal{F}(A) has a 𝒵(c)\mathcal{Z}(c)–representing measure.

  2. (2)

    The sequence with the moment matrix (A)\mathcal{H}(A) has a \mathbb{R}–representing measure.

Proof.

First we prove the implication ()(\Rightarrow). If β\beta has a 𝒵(p)\mathcal{Z}(p)–rm μ\mu, then μ\mu is supported on the union of the line y=0y=0 and the conic c(x,y)=0c(x,y)=0. Since the moment matrix, generated by the measure supported on y=0y=0, can be nonzero only when restricted to the columns and rows indexed by X(0,k){\vec{X}}^{(0,k)}, it follows that the moment matrix generated by the restriction μ|{c=0}\mu|_{\{c=0\}} (resp. μ|{y=0}\mu|_{\{y=0\}}) of the measure μ\mu to the conic c(x,y)=0c(x,y)=0 (resp. line y=0y=0), is of the form (A)\mathcal{F}(A) (resp. (A)𝟎k(k+1)2\mathcal{H}(A)\oplus\mathbf{0}_{\frac{k(k+1)}{2}}) for some Hankel matrix ASk+1A\in S_{k+1}.

It remains to establish the implication ()(\Leftarrow). Let (c)(k)\mathcal{M}^{(c)}(k) (resp. ()(k)\mathcal{M}^{(\ell)}(k)) be the moment matrix generated by the measure μ1\mu_{1} (resp. μ2\mu_{2}) supported on 𝒵(c)\mathcal{Z}(c) (resp. y=0y=0) such that

(4.13) P(c)(k)PT\displaystyle P\mathcal{M}^{(c)}(k)P^{T} =(A),P()(k)PT=(A)𝟎k(k+1)2,\displaystyle=\mathcal{F}(A),\quad P\mathcal{M}^{(\ell)}(k)P^{T}=\mathcal{H}(A)\oplus\mathbf{0}_{\frac{k(k+1)}{2}},

respectively, where PP is as in (4.6). The equalities (4.13) imply that (k;β)=(c)(k)+()(k;β)\mathcal{M}(k;\beta)=\mathcal{M}^{(c)}(k)+\mathcal{M}^{(\ell)}(k;\beta). Since the measure μ1+μ2\mu_{1}+\mu_{2} is supported on the curve 𝒵(c){y=0}=𝒵(p)\mathcal{Z}(c)\cup\{y=0\}=\mathcal{Z}(p), the implication ()(\Leftarrow) holds. ∎

Lemma 4.2.

Assume the notation above and let the sequence β={βi}i+2,|i|2k\beta=\{\beta_{i}\}_{i\in\mathbb{Z}_{+}^{2},|i|\leq 2k}, where k3k\geq 3, admit a 𝒵(p)\mathcal{Z}(p)–representing measure. Let A:=A(β0,0(c),β1,0(c),,β2k,0(c))Sk+1A:=A_{\big{(}\beta_{0,0}^{(c)},\beta_{1,0}^{(c)},\ldots,\beta_{2k,0}^{(c)}\big{)}}\in S_{k+1} be a Hankel matrix such that (A)\mathcal{F}(A) admits a 𝒵(c)\mathcal{Z}(c)–representing measure and (A)\mathcal{H}(A) admits a \mathbb{R}–representing measure. Let c(x,y)c(x,y) be of the form

(4.14) c(x,y)=a00+a10x+a20x2+a01y+a02y2+a11xywith aijand exactly one of the coefficients a00,a10,a20 is nonzero.\displaystyle\begin{split}&c(x,y)=a_{00}+a_{10}x+a_{20}x^{2}+a_{01}y+a_{02}y^{2}+a_{11}xy\quad\text{with }a_{ij}\in\mathbb{R}\\ &\text{and exactly one of the coefficients }a_{00},a_{10},a_{20}\text{ is nonzero}.\end{split}

If:

  1. (1)

    a000a_{00}\neq 0, then

    βi,0(c)=1a00(a01βi,1+a02βi,2+a11βi+1,1)for i=0,,2k2.\beta_{i,0}^{(c)}=-\frac{1}{a_{00}}(a_{01}\beta_{i,1}+a_{02}\beta_{i,2}+a_{11}\beta_{i+1,1})\quad\text{for }i=0,\ldots,2k-2.
  2. (2)

    a100a_{10}\neq 0, then

    βi,0(c)=1a10(a01βi,1+a02βi,2+a11βi+1,1)for i=1,,2k1.\beta_{i,0}^{(c)}=-\frac{1}{a_{10}}(a_{01}\beta_{i,1}+a_{02}\beta_{i,2}+a_{11}\beta_{i+1,1})\quad\text{for }i=1,\ldots,2k-1.
  3. (3)

    a200a_{20}\neq 0, then

    βi,0(c)=1a20(a01βi,1+a02βi,2+a11βi+1,1)for i=2,,2k.\beta_{i,0}^{(c)}=-\frac{1}{a_{20}}(a_{01}\beta_{i,1}+a_{02}\beta_{i,2}+a_{11}\beta_{i+1,1})\quad\text{for }i=2,\ldots,2k.
Proof.

By Lemma 4.1, (A)\mathcal{F}(A) has a 𝒵(c)\mathcal{Z}(c)–rm for some Hankel matrix ASk+1A\in S_{k+1}. Hence, (A)\mathcal{F}(A) satisfies the rg relations XiYjc(X,Y)=𝟎X^{i}Y^{j}c(X,Y)=\mathbf{0} for i,j+i,j\in\mathbb{Z}_{+}, i+jk2i+j\leq k-2. Let us assume that a000a_{00}\neq 0 and a10=a20=0a_{10}=a_{20}=0. In particular, (A)\mathcal{F}(A) satisfies the relations

(4.15) a001+a01Y+a02Y2+a11XY=𝟎,a00Xk2+a01Xk2Y+a02Xk2Y2+a11Xk1Y=𝟎.\displaystyle\begin{split}a_{00}\mathit{1}+a_{01}Y+a_{02}Y^{2}+a_{11}XY&=\mathbf{0},\\ a_{00}X^{k-2}+a_{01}X^{k-2}Y+a_{02}X^{k-2}Y^{2}+a_{11}X^{k-1}Y&=\mathbf{0}.\end{split}

Observing the rows 1,X,,Xk\mathit{1},X,\ldots,X^{k} of (A)\mathcal{F}(A), the relations (4.15) imply that

(4.16) βi,0(c)=1a00(a01βi,1(c)+a02βi,2(c)+a11βi+1,1(c))for i=0,,2k2.\beta_{i,0}^{(c)}=-\frac{1}{a_{00}}\big{(}a_{01}\beta_{i,1}^{(c)}+a_{02}\beta_{i,2}^{(c)}+a_{11}\beta_{i+1,1}^{(c)}\big{)}\quad\text{for }i=0,\ldots,2k-2.

Using the forms of ~(k;β)\widetilde{\mathcal{M}}(k;\beta) and (A)\mathcal{F}(A) (see (4.9) and (4.11)), it follows that βi,1(c)=βi,1\beta_{i,1}^{(c)}=\beta_{i,1} and βj,2(c)=βj,2\beta_{j,2}^{(c)}=\beta_{j,2} for each i,ji,j. Using this in (4.16) proves the statement (1) of the lemma. The proofs of the statements (2) and (3) are analogous. ∎

Lemma 4.2 states that for all canonical relations from Proposition 3.1 except for the mixed type relation, all but two entries of the Hankel matrix AA from Lemma 4.1 are uniquely determined by β\beta. The following lemma gives the smallest candidate for AA in Lemma 4.1 with respect to the usual Loewner order of matrices.

Lemma 4.3.

Assume the notation above and let β={βi}i+2,|i|2k\beta=\{\beta_{i}\}_{i\in\mathbb{Z}_{+}^{2},|i|\leq 2k}, where k3k\geq 3, be a sequence of degree 2k2k. Assume that ~(k;β)\widetilde{\mathcal{M}}(k;\beta) is positive semidefinite and satisfies the column relations (4.5). Then:

  1. (1)

    (A)0\mathcal{F}(A)\succeq 0 for some ASk+1A\in S_{k+1} if and only if AA12(A22)(A12)TA\succeq A_{12}(A_{22})^{\dagger}(A_{12})^{T}.

  2. (2)

    (A12(A22)(A12)T)0\mathcal{F}\big{(}A_{12}(A_{22})^{\dagger}(A_{12})^{T}\big{)}\succeq 0 and (A12(A22)(A12)T)0\mathcal{H}\big{(}A_{12}(A_{22})^{\dagger}(A_{12})^{T}\big{)}\succeq 0.

  3. (3)

    (A12(A22)(A12)T)\mathcal{F}\big{(}A_{12}(A_{22})^{\dagger}(A_{12})^{T}\big{)} satisfies the column relations XiYjc(X,Y)=0X^{i}Y^{j}c(X,Y)=0 for i,j+i,j\in\mathbb{Z}_{+} such that i+jk2i+j\leq k-2.

  4. (4)

    We have that

    rank~(k;β)\displaystyle\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta) =rankA22+rank(A11A12(A22)(A12)T)\displaystyle=\operatorname{rank}A_{22}+\operatorname{rank}\big{(}A_{11}-A_{12}(A_{22})^{\dagger}(A_{12})^{T}\big{)}
    =rank(A12(A22)(A12)T)+rank(A12(A22)(A12)T).\displaystyle=\operatorname{rank}\mathcal{F}\big{(}A_{12}(A_{22})^{\dagger}(A_{12})^{T}\big{)}+\operatorname{rank}\mathcal{H}\big{(}A_{12}(A_{22})^{\dagger}(A_{12})^{T}\big{)}.
Proof.

By the equivalence between (1a) and (1b) of Theorem 2.2 used for (M,A)=(~(k;β),A11)(M,A)=(\widetilde{\mathcal{M}}(k;\beta),A_{11}) and (M,A)=((~(k;β))X(0,k)𝒯,A11)(M,A)=(\big{(}\widetilde{\mathcal{M}}(k;\beta)\big{)}_{\vec{X}^{(0,k)}\cup\mathcal{T}},A_{11}), it follows in particular that

(4.17) 𝒞(((A12)T(A13)T))𝒞((A22A23(A23)TA33)),𝒞(A12T)𝒞(A22).\displaystyle\begin{split}\mathcal{C}\big{(}\begin{pmatrix}(A_{12})^{T}\\ (A_{13})^{T}\end{pmatrix}\big{)}&\subseteq\mathcal{C}\big{(}\begin{pmatrix}A_{22}&A_{23}\\[3.00003pt] (A_{23})^{T}&A_{33}\end{pmatrix}\big{)},\\[3.00003pt] \mathcal{C}(A_{12}^{T})&\subseteq\mathcal{C}(A_{22}).\end{split}

and

(4.18) (Amin)0,\mathcal{H}(A_{\min})\succeq 0,

where

Amin:=(A12A13)(A22A23(A23)TA33)((A12)T(A13)T).A_{\min}:=\begin{pmatrix}A_{12}&A_{13}\end{pmatrix}\begin{pmatrix}A_{22}&A_{23}\\[3.00003pt] (A_{23})^{T}&A_{33}\end{pmatrix}^{\dagger}\begin{pmatrix}(A_{12})^{T}\\ (A_{13})^{T}\end{pmatrix}.

Using the equivalence between (1a) and (1b) of Theorem 2.2 again for the pairs (M,A)=((A),A)(M,A)=(\mathcal{F}(A),A) and (M,A)=(((A))X(0,k)𝒯,A)(M,A)=(\big{(}\mathcal{F}(A)\big{)}_{\vec{X}^{(0,k)}\cup\mathcal{T}},A), it follows that

(4.19) (A)0AAmin,((A))X(0,k)𝒯0AA12(A22)(A12)T=:A~min.\displaystyle\begin{split}\mathcal{F}(A)\succeq 0\quad&\Leftrightarrow\quad A\succeq A_{\min},\\[5.0pt] \big{(}\mathcal{F}(A)\big{)}_{\vec{X}^{(0,k)}\cup\mathcal{T}}\succeq 0\quad&\Leftrightarrow\quad A\succeq A_{12}(A_{22})^{\dagger}(A_{12})^{T}=:\widetilde{A}_{\min}.\end{split}

Since (A)0\mathcal{F}(A)\succeq 0 implies, in particular, that ((A))X(0,k)𝒯0\big{(}\mathcal{F}(A)\big{)}_{\vec{X}^{(0,k)}\cup\mathcal{T}}\succeq 0, (4.19) implies that

(4.20) AminA~min.A_{\min}\succeq\widetilde{A}_{\min}.\vskip 3.0pt plus 1.0pt minus 1.0pt

Claim. Amin=A~minA_{\min}=\widetilde{A}_{\min}.\\

Proof of Claim. By (4.19) and (4.20), it suffices to prove that (A~min)0\mathcal{F}(\widetilde{A}_{\min})\succeq 0. By definition of 𝒯\mathcal{T} and the relations XiYjp(X,Y)=XiYj+1c(X,Y)=𝟎X^{i}Y^{j}p(X,Y)=X^{i}Y^{j+1}c(X,Y)=\mathbf{0}, i,j+,i+jk3i,j\in\mathbb{Z}_{+},i+j\leq k-3, which hold in ~(k;β)\widetilde{\mathcal{M}}(k;\beta), it follows, in particular, that

(4.21) 𝒞((A23A33))𝒞((A22(A23)T))\mathcal{C}\big{(}\begin{pmatrix}A_{23}\\ A_{33}\end{pmatrix}\big{)}\subseteq\mathcal{C}\big{(}\begin{pmatrix}A_{22}\\[3.00003pt] (A_{23})^{T}\end{pmatrix}\big{)}

(4.17) and (4.21) together imply that

(4.22) 𝒞(((A12)T(A13)T))𝒞((A22(A23)T)).\mathcal{C}\big{(}\begin{pmatrix}(A_{12})^{T}\\ (A_{13})^{T}\end{pmatrix}\big{)}\subseteq\mathcal{C}\big{(}\begin{pmatrix}A_{22}\\[3.00003pt] (A_{23})^{T}\end{pmatrix}\big{)}.

(4.17) and (4.22) can be equivalently expressed as

(4.23) (A22(A23)T)W=(A23A33)for some matrix W,(A22(A23)T)X=((A12)T(A13)T)for some matrix X.\displaystyle\begin{split}\begin{pmatrix}A_{22}\\[3.00003pt] (A_{23})^{T}\end{pmatrix}W&=\begin{pmatrix}A_{23}\\ A_{33}\end{pmatrix}\;\text{for some matrix }W,\\[3.00003pt] \begin{pmatrix}A_{22}\\[3.00003pt] (A_{23})^{T}\end{pmatrix}X&=\begin{pmatrix}(A_{12})^{T}\\ (A_{13})^{T}\end{pmatrix}\;\text{for some matrix }X.\end{split}

We have that

0\displaystyle 0 (XTIWT)A22(XIW)\displaystyle\preceq\begin{pmatrix}X^{T}\\ I\\ W^{T}\end{pmatrix}A_{22}\begin{pmatrix}X&I&W\end{pmatrix}
=(XTA22XXTA22XTA22WA22XA22A22WWTA22XWTA22WTA22W)\displaystyle=\begin{pmatrix}X^{T}A_{22}X&X^{T}A_{22}&X^{T}A_{22}W\\[3.00003pt] A_{22}X&A_{22}&A_{22}W\\[3.00003pt] W^{T}A_{22}X&W^{T}A_{22}&W^{T}A_{22}W\end{pmatrix}
=(A12(A22)(A12)TA12A13(A12)TA22A23(A13)T(A23)TA33)=(A~min)\displaystyle=\begin{pmatrix}A_{12}(A_{22})^{\dagger}(A_{12})^{T}&A_{12}&A_{13}\\[3.00003pt] (A_{12})^{T}&A_{22}&A_{23}\\[3.00003pt] (A_{13})^{T}&(A_{23})^{T}&A_{33}\end{pmatrix}=\mathcal{F}(\widetilde{A}_{\min})

where II is the identity matrix of the same size as A22A_{22} and we used (4.23) in the second equality. This proves the Claim. \blacksquare\\

Using (4.18), (4.19) and Claim, the statements (1) and (2) follow. By Theorem 2.2.(2), used for (M,A)=(~(k;β),A11)(M,A)=(\widetilde{\mathcal{M}}(k;\beta),A_{11}), we have that

(4.24) rank~(k;β)=rank(A22A23(A23)TA33)+rank(Amin)=rank(Amin)+rank(Amin).\displaystyle\begin{split}\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)&=\operatorname{rank}\begin{pmatrix}A_{22}&A_{23}\\[3.00003pt] (A_{23})^{T}&A_{33}\end{pmatrix}+\operatorname{rank}\mathcal{H}(A_{\min})\\[3.00003pt] &=\operatorname{rank}\mathcal{F}(A_{\min})+\operatorname{rank}\mathcal{H}(A_{\min}).\end{split}

By (4.21) and

B:=(A22A23(A23)TA33)0,B:=\begin{pmatrix}A_{22}&A_{23}\\[3.00003pt] (A_{23})^{T}&A_{33}\end{pmatrix}\succeq 0,

it follows by Theorem 2.2, used for (M,A)=(B,A22)(M,A)=(B,A_{22}), that rankB=rankA22\operatorname{rank}B=\operatorname{rank}A_{22}. Using this and the Claim, (4.24) implies the statement (4).

Since ~(k;β)\widetilde{\mathcal{M}}(k;\beta) satisfies the relations (4.5), it follows that the restriction ((A~min))𝒞X(0,k),𝒞\big{(}\mathcal{F}(\widetilde{A}_{\min})\big{)}_{\mathcal{C}\setminus\vec{X}^{(0,k)},\mathcal{C}} satisfies the column relations XiYjc(X,Y)=𝟎X^{i}Y^{j}c(X,Y)=\mathbf{0} for i,j+i,j\in\mathbb{Z}_{+} such that i+jk2i+j\leq k-2. By Proposition 2.3, these relations extend to (A~min)\mathcal{F}(\widetilde{A}_{\min}), which proves (3). ∎

Remark 4.4.

By Lemmas 4.14.3, solving the 𝒵(p)\mathcal{Z}(p)–TMP for the sequence β={βi}i+2,|i|2k\beta=\{\beta_{i}\}_{i\in\mathbb{Z}_{+}^{2},|i|\leq 2k}, where k3k\geq 3, with pp being any but the mixed type relation from Proposition 3.1, the natural procedure is the following:

  1. (1)

    First compute Amin:=A12(A22)A12A_{\min}:=A_{12}(A_{22})^{\dagger}A_{12}. By Lemma 4.3.(3), there is one entry of AminA_{\min}, which might need to be changed to obtain a Hankel structure. Namely, in the notation (4.14), if:

    1. (a)

      a000a_{00}\neq 0, then the value of (Amin)k,k(A_{\min})_{k,k} must be made equal to (Amin)k1,k+1(A_{\min})_{k-1,k+1}.

    2. (b)

      a100a_{10}\neq 0, then the value of (Amin)1,k+1(A_{\min})_{1,k+1} must be made equal to (Amin)2,k(A_{\min})_{2,k}.

    3. (c)

      a200a_{20}\neq 0, then the value of (Amin)2,2(A_{\min})_{2,2} must be made equal to (Amin)3,1(A_{\min})_{3,1}.

    Let A^min\widehat{A}_{\min} be the matrix obtained from AminA_{\min} after performing the changes described above.

  2. (2)

    Study if (A^min)\mathcal{F}(\widehat{A}_{\min}) and (A^min)\mathcal{H}(\widehat{A}_{\min}) admit a 𝒵(c)\mathcal{Z}(c)–rm and a \mathbb{R}–rm, respectively. If the answer is yes, β\beta admits a 𝒵(p)\mathcal{Z}(p)–rm. Otherwise by Lemma 4.2, there are two antidiagonals of the Hankel matrix A^min\widehat{A}_{\min}, which can by varied so that the matrices (A^min)\mathcal{F}(\widehat{A}_{\min}) and (A^min)\mathcal{H}(\widehat{A}_{\min}) will admit the corresponding measures. Namely, in the notation (4.14), if:

    1. (a)

      a000a_{00}\neq 0, then the last two antidiagonals of A^min\widehat{A}_{\min} can be changed.

    2. (b)

      a100a_{10}\neq 0, then the left–upper and the right–lower corner of A^min\widehat{A}_{\min} can be changed.

    3. (c)

      a200a_{20}\neq 0, then the first two antidiagonals of A^min\widehat{A}_{\min} can be changed.

    To solve the 𝒵(p)\mathcal{Z}(p)–TMP for β\beta one needs to characterize, when it is possible to change these antidiagonals in such a way to obtain a matrix A˘min\breve{A}_{\min}, such that (A˘min)\mathcal{F}(\breve{A}_{\min}) and (A˘min)\mathcal{H}(\breve{A}_{\min}) admit a 𝒵(c)\mathcal{Z}(c)–rm and a \mathbb{R}–rm, respectively.

In Sections 5 and 6 we solve concretely the TMP on reducible cubic curves in the circular and parabolic type form (see the classification from Proposition 3.1). The parallel lines type form was solved in [Zal22a], while the hyperbolic type forms will be solved in the forthcoming work [YZ+].

5. Circular type relation: p(x,y)=y(ay+x2+y2)p(x,y)=y(ay+x^{2}+y^{2}), a{0}a\notin\mathbb{R}\setminus\{0\}.

In this section we solve the 𝒵(p)\mathcal{Z}(p)–TMP for the sequence β={βi,j}i,j+,i+j2k\beta=\{\beta_{i,j}\}_{i,j\in\mathbb{Z}_{+},i+j\leq 2k} of degree 2k2k, k3k\geq 3, where p(x,y)=y(ay+x2+y2)p(x,y)=y(ay+x^{2}+y^{2}), a{0}a\in\mathbb{R}\setminus\{0\}. Assume the notation from Section 4. If β\beta admits a 𝒵(p)\mathcal{Z}(p)–TMP, then (k;β)\mathcal{M}(k;\beta) must satisfy the relations

(5.1) aY2+jXi+Y1+jX2+i=Y3+jXifor i,j+ such that i+jk3.aY^{2+j}X^{i}+Y^{1+j}X^{2+i}=-Y^{3+j}X^{i}\quad\text{for }i,j\in\mathbb{Z}_{+}\text{ such that }i+j\leq k-3.

In the presence of all column relations (5.1), the column space 𝒞((k;β))\mathcal{C}(\mathcal{M}(k;\beta)) is spanned by the columns in the set

(5.2) 𝒯=X(0,k)YX(0,k1)Y2X(0,k2),\mathcal{T}=\vec{X}^{(0,k)}\cup Y\vec{X}^{(0,k-1)}\cup Y^{2}\vec{X}^{(0,k-2)},

where

YiX(j,):=(YiXj,YiXj+1,,YiX)with i,j,+,j,i+k.Y^{i}\vec{X}^{(j,\ell)}:=(Y^{i}X^{j},Y^{i}X^{j+1},\ldots,Y^{i}X^{\ell})\quad\text{with }i,j,\ell\in\mathbb{Z}_{+},\;j\leq\ell,\;i+\ell\leq k.

Let ~(k;β)\widetilde{\mathcal{M}}(k;\beta) be as in (4.10). Let

(5.3) Amin:=A12(A22)(A12)T.A_{\min}:=A_{12}(A_{22})^{\dagger}(A_{12})^{T}.

As described in Remark 4.4, AminA_{\min} might need to be changed to

A^min=Amin+ηE2,2(k+1),\widehat{A}_{\min}=A_{\min}+\eta E_{2,2}^{(k+1)},

where

η:=(Amin)1,3(Amin)2,2.\eta:=(A_{\min})_{1,3}-(A_{\min})_{2,2}.

Let (𝐀)\mathcal{F}(\mathbf{A}) and (𝐀)\mathcal{H}(\mathbf{A}) be as in (4.11). Write

(5.4) (A^min):= (\@arstrut1XX(2,k)\\1β0,0-(Amin)1,1β1,0-(Amin)1,2(h12(1))T\\[0.2em]Xβ1,0-(Amin)1,2β2,0-(Amin)1,3(h12(2))T\\[0.2em](X(2,k))Th12(1)h12(2)H22) ,\\[0.5em]H1:=((A^min)){1}X(2,k)= (\@arstrut1X(2,k)\\[0.2em]1β0,0-(Amin)1,1(h12(1))T\\[0.2em](X(2,k))Th12(1)H22) ,\\[0.5em]H2:=((A^min))X(1,k)= ( \@arstrutXX(2,k)\\[0.2em]Xβ2,0-(Amin)1,3(h12(2))T\\[0.2em](X(2,k))Th12(2)H22) .\displaystyle\begin{split}\mathcal{H}(\widehat{A}_{\min})&:=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(2,k)}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{1,0}-(A_{\min})_{1,2}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12}^{(1)})^{T}\\[0.2em]X$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{1,0}-(A_{\min})_{1,2}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2,0}-(A_{\min})_{1,3}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12}^{(2)})^{T}\\[0.2em](\vec{X}^{(2,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}^{(1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}^{(2)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt\crcr}}}}\right)$}},\\[0.5em]H_{1}&:=(\mathcal{H}(\widehat{A}_{\min}))_{\{1\}\cup\vec{X}^{(2,k)}}=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(2,k)}\\[0.2em]\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12}^{(1)})^{T}\\[0.2em](\vec{X}^{(2,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}^{(1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt\crcr}}}}\right)$}},\\[0.5em]H_{2}&:=(\mathcal{H}(\widehat{A}_{\min}))_{\vec{X}^{(1,k)}}=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(2,k)}\\[0.2em]X$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2,0}-(A_{\min})_{1,3}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12}^{(2)})^{T}\\[0.2em](\vec{X}^{(2,k)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}^{(2)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt\crcr}}}}\right)$}}.\end{split}

Define also the matrix function

(5.5) 𝒢:2Sk+1,𝒢(𝐭,𝐮)=A^min+𝐭E1,1(k+1)+𝐮(E1,2(k+1)+E2,1(k+1)).\mathcal{G}:\mathbb{R}^{2}\to S_{k+1},\qquad\mathcal{G}(\mathbf{t},\mathbf{u})=\widehat{A}_{\min}+\mathbf{t}E_{1,1}^{(k+1)}+\mathbf{u}\big{(}E_{1,2}^{(k+1)}+E_{2,1}^{(k+1)}\big{)}.

The solution to the cubic circular type relation TMP is the following.

Theorem 5.1.

Let p(x,y)=y(ay+x2+y2)p(x,y)=y(ay+x^{2}+y^{2}), a{0}a\in\mathbb{R}\setminus\{0\}, and β=(βi,j)i,j+,i+j2k\beta=(\beta_{i,j})_{i,j\in\mathbb{Z}_{+},i+j\leq 2k}, where k3k\geq 3. Assume also the notation above. Then the following statements are equivalent:

  1. (1)

    β\beta has a 𝒵(p)\mathcal{Z}(p)–representing measure.

  2. (2)

    ~(k;β)\widetilde{\mathcal{M}}(k;\beta) is positive semidefinite, the relations

    (5.6) aβi,2+j+β2+i,1+j=βi,3+jhold for every i,j+ with i+j2k3a\beta_{i,2+j}+\beta_{2+i,1+j}=-\beta_{i,3+j}\quad\text{hold for every }i,j\in\mathbb{Z}_{+}\text{ with }i+j\leq 2k-3

    and one of the following statements holds:

    1. (a)

      η=0\eta=0 and one of the following holds:

      1. (i)

        rank((Amin))X(0,k1)=k\operatorname{rank}(\mathcal{H}(A_{\min}))_{\vec{X}^{(0,k-1)}}=k.

      2. (ii)

        rank(H2)X(1,k1)=rankH2\operatorname{rank}(H_{2})_{\vec{X}^{(1,k-1)}}=\operatorname{rank}H_{2}.

    2. (b)

      η>0\eta>0, H2H_{2} is positive semidefinite and defining a real number

      (5.7) u0=β1,0(Amin)1,2(h12(1))T(H22)h12(2),\displaystyle\begin{split}u_{0}&=\beta_{1,0}-(A_{\min})_{1,2}-(h_{12}^{(1)})^{T}(H_{22})^{\dagger}h_{12}^{(2)},\end{split}

      a function

      (5.8) h(𝐭)=(H1/H22𝐭)(H2/H22)h(\mathbf{t})=\sqrt{(H_{1}/H_{22}-\mathbf{t})(H_{2}/H_{22})}

      and a set

      (5.9) ={(t,ηt)+×+:ηt=u0+h(t)},{(t,ηt)+×:ηt=u0h(t)},{(t,ηt)+×+:ηt=u0+h(t)},{(t,ηt)+×:ηt=u0h(t)},\displaystyle\begin{split}\mathcal{I}&=\big{\{}(t,\sqrt{\eta t})\in\mathbb{R}_{+}\times\mathbb{R}_{+}\colon\sqrt{\eta t}=u_{0}+h(t)\},\\ &\hskip 14.22636pt\cup\big{\{}(t,\sqrt{\eta t})\in\mathbb{R}_{+}\times\mathbb{R}_{-}\colon\sqrt{\eta t}=u_{0}-h(t)\},\\ &\hskip 14.22636pt\cup\big{\{}(t,-\sqrt{\eta t})\in\mathbb{R}_{+}\times\mathbb{R}_{+}\colon-\sqrt{\eta t}=u_{0}+h(t)\},\\ &\hskip 14.22636pt\cup\big{\{}(t,-\sqrt{\eta t})\in\mathbb{R}_{+}\times\mathbb{R}_{-}\colon-\sqrt{\eta t}=u_{0}-h(t)\},\\ \end{split}

      one of the following holds:

      1. (i)

        The set \mathcal{I} has two elements and H2H_{2} is positive definite.

      2. (ii)

        ={(t~,u~)}\mathcal{I}=\{(\tilde{t},\tilde{u})\} and

        (5.10) rank(((𝒢(t~,u~)))X(0,k1))=rank(𝒢(t~,u~)).\operatorname{rank}\big{(}\big{(}\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u}))\big{)}_{\vec{X}^{(0,k-1)}}\big{)}=\operatorname{rank}\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})).

Moreover, if a 𝒵(p)\mathcal{Z}(p)–representing measure for β\beta exists, then:

  • There exists at most (rank~(k;β)+1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1)–atomic 𝒵(p)\mathcal{Z}(p)–representing measure.

  • There exists a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–representing measure if and only if any of the following holds:

    • η=0\eta=0.

    • η>0\eta>0 and (Amin)\mathcal{H}(A_{\min}) is positive definite.

In particular, a pp–pure sequence β\beta with a 𝒵(p)\mathcal{Z}(p)–representing measure admits a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–representing measure.

Remark 5.2.

In this remark we explain the idea of the proof of Theorem 5.1 and the meaning of the conditions in the statement of the theorem.

By Lemmas 4.14.2, the existence of a 𝒵(p)\mathcal{Z}(p)–rm for β\beta is equivalent to the existence of t,ut,u\in\mathbb{R} such that (𝒢(t,u))\mathcal{F}(\mathcal{G}(t,u)) admits a 𝒵(ay+x2+y2)\mathcal{Z}(ay+x^{2}+y^{2})–rm and (𝒢(t,u))\mathcal{H}(\mathcal{G}(t,u)) admits a \mathbb{R}–rm. Let

1\displaystyle\mathcal{R}_{1} ={(t,u)2:(𝒢(t,u))0}and2={(t,u)2:(𝒢(t,u))0}.\displaystyle=\big{\{}(t,u)\in\mathbb{R}^{2}\colon\mathcal{F}(\mathcal{G}(t,u))\succeq 0\big{\}}\quad\text{and}\quad\mathcal{R}_{2}=\big{\{}(t,u)\in\mathbb{R}^{2}\colon\mathcal{H}(\mathcal{G}(t,u))\succeq 0\big{\}}.

We denote by Ri\partial R_{i} and R̊i\mathring{R}_{i} the topological boundary and the interior of the set RiR_{i}, respectively. By the necessary conditions for the existence of a 𝒵(p)\mathcal{Z}(p)–rm [CF04, Fia95, CF96], ~(k;β)\widetilde{\mathcal{M}}(k;\beta) must be psd and the relations (5.6) must hold. Using also Theorem 2.6, Theorem 5.1.(1) is equivalent to

(5.11) ~(k;β)0, the relations (5.6) hold and (t0,u0)12:(𝒢(t0,u0)) admits a –rm.\displaystyle\begin{split}&\widetilde{\mathcal{M}}(k;\beta)\succeq 0,\text{ the relations }\eqref{221023-1857-equation}\text{ hold and }\\ &\exists(t_{0},u_{0})\in\mathcal{R}_{1}\cap\mathcal{R}_{2}:\mathcal{H}(\mathcal{G}(t_{0},u_{0}))\text{ admits a }\mathbb{R}\text{--rm}.\end{split}

In the proof of Theorem 5.1 we show that (5.11) is equivalent to Theorem 5.1.(2):

  1. (1)

    First we establish (see Claims 1 and 2 below) that the form of:

    • 1\mathcal{R}_{1} is one of the following:

      [Uncaptioned image] [Uncaptioned image]

      where the left case occurs if η>0\eta>0 and the right if η=0\eta=0. The case η<0\eta<0 cannot occur.

    • 2\mathcal{R}_{2} is one of the following:

      [Uncaptioned image] [Uncaptioned image]

      where the left case occurs if H2/H22>0H_{2}/H_{22}>0 and the right if H2/H22=0H_{2}/H_{22}=0.

  2. (2)

    If η=0\eta=0, then we show that (5.11) is equivalent to

    ~(k;β)0, the relations (5.6) hold and (𝒢(0,0)) admits a –rm.\displaystyle\begin{split}&\widetilde{\mathcal{M}}(k;\beta)\succeq 0,\text{ the relations }\eqref{221023-1857-equation}\text{ hold and }\mathcal{H}(\mathcal{G}(0,0))\text{ admits a }\mathbb{R}\text{--rm}.\end{split}

    The latter statement is further equivalent to Theorem 5.1.(2a).

  3. (3)

    If η>0\eta>0, then by the forms of 1\mathcal{R}_{1} and 2\mathcal{R}_{2}, =12\mathcal{I}=\partial\mathcal{R}_{1}\cap\partial\mathcal{R}_{2} is one of the following: (i) \emptyset, (ii) a one-element set, (iii) a two-element set. In the case:

    • (i), a 𝒵(p)\mathcal{Z}(p)–rm for β\beta clearly cannot exist.

    • (ii), then denoting ={(t~,u~)}\mathcal{I}=\{(\tilde{t},\tilde{u})\}, (5.11) is equivalent to

      ~(k;β)0, the relations (5.6) hold and (𝒢(t~,u~)) admits a –rm.\displaystyle\begin{split}&\widetilde{\mathcal{M}}(k;\beta)\succeq 0,\text{ the relations }\eqref{221023-1857-equation}\text{ hold and }\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u}))\text{ admits a }\mathbb{R}\text{--rm}.\end{split}

      The latter statement is equivalent to Theorem 5.1.(2(b)ii).

    • (iii), (5.11) is equivalent to H2H_{2} being positive definite, which is Theorem 5.1.(2(b)i). Moreover, in this case for at least one of the points (t,u)(t,u)\in\mathcal{I}, a 𝒵(ay+x2+y2)\mathcal{Z}(ay+x^{2}+y^{2})–rm and a \mathbb{R}–rm exist for (𝒢(t,u))\mathcal{F}(\mathcal{G}(t,u)) and (𝒢(t,u))\mathcal{H}(\mathcal{G}(t,u)), respectively.

Proof of Theorem 5.1.

Let 1,2\mathcal{R}_{1},\mathcal{R}_{2} be as in Remark 5.2. As explained in Remark 5.2, Theorem 5.1.(1) is equivalent to (5.11), thus it remains to prove that (5.11) is equivalent to Theorem 5.1.(2).\\

First we establish a few claims needed in the proof. Claim 1 (resp. 2) describes 1\mathcal{R}_{1} (resp. 2\mathcal{R}_{2}) concretely.\\

Claim 1. Assume that ~(k;β)0\widetilde{\mathcal{M}}(k;\beta)\succeq 0. Then

(5.12) 1={{(t,u)2:t0,u[ηt,ηt]},if η0,,if η<0.\mathcal{R}_{1}=\left\{\begin{array}[]{rl}\big{\{}(t,u)\in\mathbb{R}^{2}\colon t\geq 0,u\in\left[-\sqrt{\eta t},\sqrt{\eta t}\right]\big{\}},&\text{if }\eta\geq 0,\\[3.00003pt] \emptyset,&\text{if }\eta<0.\end{array}\right.

If η0\eta\geq 0, we have

(5.16) rank(𝒢(t,u))={rank(Amin),if t=0,η=0,rank(Amin)+1,if (t>0 or η>0) and u{ηt,ηt},rank(Amin)+2,if t>0,η>0,u(ηt,ηt),\displaystyle\operatorname{rank}\mathcal{F}(\mathcal{G}(t,u))=\left\{\begin{array}[]{rl}\operatorname{rank}\mathcal{F}(A_{\min}),&\text{if }t=0,\eta=0,\\[3.00003pt] \operatorname{rank}\mathcal{F}(A_{\min})+1,&\text{if }(t>0\text{ or }\eta>0)\text{ and }u\in\{-\sqrt{\eta t},\sqrt{\eta t}\},\\[3.00003pt] \operatorname{rank}\mathcal{F}(A_{\min})+2,&\text{if }t>0,\eta>0,u\in\left(-\sqrt{\eta t},\sqrt{\eta t}\right),\end{array}\right.

where AminA_{\min} is an in (5.3).\\

Proof of Claim 1. Note that

(5.17) 𝒢(𝐭,𝐮)=Amin+ηE2,2(k+1)+𝐭E1,1(k+1)+𝐮(E1,2(k+1)+E2,1(k+1))=Amin+(𝐭𝐮𝐮η)𝟎k1.\displaystyle\begin{split}\mathcal{G}(\mathbf{t},\mathbf{u})&=A_{\min}+\eta E_{2,2}^{(k+1)}+\mathbf{t}E_{1,1}^{(k+1)}+\mathbf{u}\big{(}E_{1,2}^{(k+1)}+E_{2,1}^{(k+1)}\big{)}\\ &=A_{\min}+\begin{pmatrix}\mathbf{t}&\mathbf{u}\\ \mathbf{u}&\eta\end{pmatrix}\oplus\mathbf{0}_{k-1}.\end{split}

By Lemma 4.3, we have that

(5.18) (𝒢(t,u))0𝒢(t,u)Amin\mathcal{F}(\mathcal{G}(t,u))\succeq 0\quad\Leftrightarrow\quad\mathcal{G}(t,u)\succeq A_{\min}

Using (5.17), (5.18) and the definition of 1\mathcal{R}_{1}, we have that

(5.19) (t,u)1\displaystyle(t,u)\in\mathcal{R}_{1}\quad (tuuη)0t0,η0,tηu2,\displaystyle\Leftrightarrow\quad\begin{pmatrix}t&u\\ u&\eta\end{pmatrix}\succeq 0\quad\Leftrightarrow\quad t\geq 0,\eta\geq 0,t\eta\geq u^{2},

which proves (5.12).

To prove (5.16) first note that by construction of (Amin)\mathcal{F}(A_{\min}), the columns 11 and XX are in the span of the columns indexed by 𝒞X(0,k)\mathcal{C}\setminus\vec{X}^{(0,k)}. Hence, there are vectors

(5.20) v1,v2ker(Amin)v_{1},v_{2}\in\ker\mathcal{F}(A_{\min})

of the forms

v1=(1𝟎1,k(v~1)T)T(k+1)(k+2)2andv2=(01𝟎1,k2(v~2)T)T(k+1)(k+2)2.v_{1}=\begin{pmatrix}1&\mathbf{0}_{1,k}&(\tilde{v}_{1})^{T}\end{pmatrix}^{T}\in\mathbb{R}^{\frac{(k+1)(k+2)}{2}}\quad\text{and}\quad v_{2}=\begin{pmatrix}0&1&\mathbf{0}_{1,k-2}&(\tilde{v}_{2})^{T}\end{pmatrix}^{T}\in\mathbb{R}^{\frac{(k+1)(k+2)}{2}}.

Let r:=rank(tuuη)r:=\operatorname{rank}\begin{pmatrix}t&u\\ u&\eta\end{pmatrix}. Clearly,

(5.21) rank(𝒢(t,u))rank(Amin)+r.\operatorname{rank}\mathcal{F}(\mathcal{G}(t,u))\leq\operatorname{rank}\mathcal{F}(A_{\min})+r.

We separate three cases according to rr. \\

Case 1: r=0r=0. In this case t=u=η=0t=u=\eta=0 and 𝒢(0,0)=Amin\mathcal{G}(0,0)=A_{\min}. In this case (5.16) clearly holds.\\

Case 2: r=1r=1. In this case tη=u2t\eta=u^{2}. Together with (5.19), this is equivalent to (t>0 or η>0) and u{ηt,ηt}(t>0\text{ or }\eta>0)\text{ and }u\in\{-\sqrt{\eta t},\sqrt{\eta t}\}. By (5.21) and (𝒢(t,u))(Amin)\mathcal{F}(\mathcal{G}(t,u))\succeq\mathcal{F}(A_{\min}) to prove (5.16), it suffices to find vker(Amin)v\in\ker\mathcal{F}(A_{\min}) and vker(𝒢(t,u))v\notin\ker\mathcal{F}(\mathcal{G}(t,u)). Note that at least one of v1,v2v_{1},v_{2} from (5.20) is such a vector, since

(v1)T(𝒢(t,u))v1=tand(v2)T(𝒢(t,u))v2=η.(v_{1})^{T}\mathcal{F}(\mathcal{G}(t,u))v_{1}=t\quad\text{and}\quad(v_{2})^{T}\mathcal{F}(\mathcal{G}(t,u))v_{2}=\eta.\vskip 3.0pt plus 1.0pt minus 1.0pt

Case 3: r=2r=2. In this case tη>u2t\eta>u^{2}. Together with (5.19), this is equivalent to t>0,η>0,u(ηt,ηt)t>0,\eta>0,u\in(-\sqrt{\eta t},\sqrt{\eta t}). Note that

(5.22) (𝒢(t,u))=(𝒢(u2η,u))+(tu2η)𝟎(k+1)(k+2)21(𝒢(u2η,u)).\mathcal{F}(\mathcal{G}(t,u))=\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{u^{2}}{\eta},u\Big{)}\Big{)}+\begin{pmatrix}t-\frac{u^{2}}{\eta}\end{pmatrix}\oplus\mathbf{0}_{\frac{(k+1)(k+2)}{2}-1}\succeq\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{u^{2}}{\eta},u\Big{)}\Big{)}.

By Case 2, we have rank(𝒢(u2η,u))=rank(Amin)+1\operatorname{rank}\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{u^{2}}{\eta},u\Big{)}\Big{)}=\operatorname{rank}\mathcal{F}(A_{\min})+1. By (5.21) and (5.22), to prove (5.16), it suffices to find vker(𝒢(u2η,u))v\in\ker\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{u^{2}}{\eta},u\Big{)}\Big{)} and vker(𝒢(t,u))v\notin\ker\mathcal{F}(\mathcal{G}(t,u)). We will check below, that v3v_{3}, defined by

v3=v1uηv2=(1uη(v~3)T)T(k+1)(k+2)2,v_{3}=v_{1}-\frac{u}{\eta}v_{2}=\begin{pmatrix}1&-\frac{u}{\eta}&(\tilde{v}_{3})^{T}\end{pmatrix}^{T}\in\mathbb{R}^{\frac{(k+1)(k+2)}{2}},

is such a vector. This follows by

(𝒢(u2η,u))v3=(Amin)v3+((u2ηuuη)𝟎(k+1)(k+2)12)v3=𝟎(k+1)(k+2)2,1\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{u^{2}}{\eta},u\Big{)}\Big{)}v_{3}=\mathcal{F}(A_{\min})v_{3}+\left(\begin{pmatrix}\frac{u^{2}}{\eta}&u\\ u&\eta\end{pmatrix}\oplus\mathbf{0}_{\frac{(k+1)(k+2)-1}{2}}\right)v_{3}=\mathbf{0}_{\frac{(k+1)(k+2)}{2},1}

and

(v3)T(𝒢(t,u))v3=tu2η>0.(v_{3})^{T}\mathcal{F}(\mathcal{G}(t,u))v_{3}=t-\frac{u^{2}}{\eta}>0.

This concludes the proof of Claim 1. \blacksquare\\

Claim 2. Assume that ~(k;β)0\widetilde{\mathcal{M}}(k;\beta)\succeq 0. Let u0u_{0}, h(𝐭)h(\mathbf{t}) be as in (5.7),(5.8) and

t0=β0,0(Amin)1,1(h12(1))T(H22)h12(1).t_{0}=\beta_{0,0}-(A_{\min})_{1,1}-(h_{12}^{(1)})^{T}(H_{22})^{\dagger}h_{12}^{(1)}.

Then

(5.23) 2={{(t,u)2:tt0,u[u0h(t),u0+h(t)]},if H20,,if H20.\mathcal{R}_{2}=\left\{\begin{array}[]{rl}\big{\{}(t,u)\in\mathbb{R}^{2}\colon t\leq t_{0},u\in[u_{0}-h(t),u_{0}+h(t)]\big{\}},&\text{if }H_{2}\succeq 0,\\[3.00003pt] \emptyset,&\text{if }H_{2}\not\succeq 0.\end{array}\right.

If H20H_{2}\succeq 0, we have that

(5.27) rank(𝒢(t,u))={rankH2,for t=t0,u=u0,rankH22+1,for t<t0,u{u0h(t),u0+h(t)},rankH22+2,for t<t0,u(u0h(t),u0+h(t)).\displaystyle\operatorname{rank}\mathcal{H}(\mathcal{G}(t,u))=\left\{\begin{array}[]{rl}\operatorname{rank}H_{2},&\text{for }t=t_{0},u=u_{0},\\[1.99997pt] \operatorname{rank}H_{22}+1,&\text{for }t<t_{0},u\in\{u_{0}-h(t),u_{0}+h(t)\},\\[1.99997pt] \operatorname{rank}H_{22}+2,&\text{for }t<t_{0},u\in(u_{0}-h(t),u_{0}+h(t)).\end{array}\right.

Proof of Claim 2. Write

H(𝐭)\displaystyle H(\mathbf{t}) :=((𝒢(𝐭,𝐮))1X(2,k)= (\@arstrut1X(2,k)\\1β0,0-(Amin)1,1-t(h12(1))T\\[0.3em]X(2,k)h12(1)H22) \displaystyle:=\big{(}\mathcal{H}(\mathcal{G}(\mathbf{t},\mathbf{u})\big{)}_{1\cup\vec{X}^{(2,k)}}=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(2,k)}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}-\mathbf{t}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12}^{(1)})^{T}\\[0.3em]\vec{X}^{(2,k)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}^{(1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt\crcr}}}}\right)$}}

Note that H(0)=((Amin)){1}X(2,k)H(0)=(\mathcal{H}(A_{\min}))_{\{\mathit{1}\}\cup\vec{X}^{(2,k)}}. By Lemma 4.3.(2), (Amin)0\mathcal{H}(A_{\min})\succeq 0 and hence, H(0)0H(0)\succeq 0. By Theorem 2.2, used for (M,C)=(H(0),H22)(M,C)=(H(0),H_{22}), it follows that H20H_{2}\succeq 0 and h12(1)𝒞(H22).h_{12}^{(1)}\in\mathcal{C}(H_{22}). Again, by Theorem 2.2, used for (M,C)=(H(t),H22)(M,C)=(H(t),H_{22}), it follows that H(t)0H(t)\succeq 0 iff tt0t\leq t_{0}. For a fixed tt satisfying tt0t\leq t_{0}, Lemma 2.4, used for A(𝐱)=(𝒢(t,𝐱))A(\mathbf{x})=\mathcal{H}(\mathcal{G}(t,\mathbf{x})), together with H(t)/H22=H1/H22tH(t)/H_{22}=H_{1}/H_{22}-t, implies (5.23)–(5.27) and proves Claim 2. \blacksquare\\

Claim 3. If η=0\eta=0, then (0,0)12(0,0)\in\partial\mathcal{R}_{1}\cap\mathcal{R}_{2}.\\

Proof of Claim 3. By Claim 1, η=0\eta=0 implies that (0,0)1(0,0)\in\partial\mathcal{R}_{1}. By (5.17) and η=0\eta=0, (Amin)=(𝒢(0,0))\mathcal{H}(A_{\min})=\mathcal{H}(\mathcal{G}(0,0)). By Lemma 4.3.(2), (Amin)0\mathcal{H}(A_{\min})\succeq 0. Hence, (0,0)2(0,0)\in\mathcal{R}_{2}, which proves Claim 3. \blacksquare\\

Claim 4. If η>0\eta>0, then:

  • The set \mathcal{I} (see (5.9)) has at most 2 elements.

  • 12\mathcal{R}_{1}\cap\mathcal{R}_{2}\neq\emptyset if and only if .\mathcal{I}\neq\emptyset.

  • If \mathcal{I} has two elements, then H2/H22>0H_{2}/H_{22}>0.

  • If \mathcal{I} has one element, which we denote by (t~,u~)(\tilde{t},\tilde{u}), then one of the following holds:

    • 12=\mathcal{R}_{1}\cap\mathcal{R}_{2}=\mathcal{I}.

    • 2=2={(t,u0):tt0}\partial\mathcal{R}_{2}=\mathcal{R}_{2}=\{(t,u_{0})\colon t\leq t_{0}\} and 12={(t,u0):t~tt0}\mathcal{I}\subsetneq\mathcal{R}_{1}\cap\mathcal{R}_{2}=\{(t,u_{0})\colon\tilde{t}\leq t\leq t_{0}\}.

Proof of Claim 4. Note that the set \mathcal{I} is equal to 12\partial{\mathcal{R}}_{1}\cap\partial{\mathcal{R}}_{2} (see (5.12) and (5.23)). Further on, 1\partial{\mathcal{R}}_{1} is the union of the square root functions ±η𝐭\pm\sqrt{\eta\mathbf{t}}, defined for 𝐭[0,)\mathbf{t}\in[0,\infty). Similarly, 2\partial{\mathcal{R}}_{2} is the union of the square root functions u0±(H1/H22𝐭)(H2/H22)u_{0}\pm\sqrt{(H_{1}/H_{22}-\mathbf{t})(H_{2}/H_{22})}, defined for 𝐭(,t0]\mathbf{t}\in(-\infty,t_{0}]. If H2/H22=0H_{2}/H_{22}=0, then the latter could be a half-line {(t,u0):tt0}\{(t,u_{0})\colon t\leq t_{0}\}. If 12\mathcal{R}_{1}\cap\mathcal{R}_{2}\neq\emptyset, then geometrically it is clear that \mathcal{I} contains one or two elements. Assume that \mathcal{I} contains only one element, denoted by (t~,u~)(\tilde{t},\tilde{u}). Clearly, 12\mathcal{I}\subseteq\mathcal{R}_{1}\cap\mathcal{R}_{2}. Further on, we either have =12\mathcal{I}=\mathcal{R}_{1}\cap\mathcal{R}_{2} or 12\mathcal{I}\subsetneq\mathcal{R}_{1}\cap\mathcal{R}_{2}. By the forms of 1\partial\mathcal{R}_{1} and 2\partial\mathcal{R}_{2}, the latter case occurs if H2/H22=0H_{2}/H_{22}=0 or equivalently 2=2={(t,u0):tt0}\partial\mathcal{R}_{2}=\mathcal{R}_{2}=\{(t,u_{0})\colon t\leq t_{0}\}. But then the whole line segment {(t,u0:t~tt0}\{(t,u_{0}\colon\tilde{t}\leq t\leq t_{0}\} lies in 1\mathcal{R}_{1}, which proves Claim 4. \blacksquare\\

Claim 5. Let H2H_{2} (see (5.4)) be positive definite, (t1,u1)2,(t2,u2)2(t_{1},u_{1})\in\partial\mathcal{R}_{2},(t_{2},u_{2})\in\partial\mathcal{R}_{2} and u1u2u_{1}\neq u_{2}. Then at least one of (𝒢(t1,u1))\mathcal{H}(\mathcal{G}(t_{1},u_{1})) and (𝒢(t2,u2))\mathcal{H}(\mathcal{G}(t_{2},u_{2})) admits a \mathbb{R}–rm.\\

Proof of Claim 5. Note that (𝒢(ti,ui))\mathcal{H}(\mathcal{G}(t_{i},u_{i})), i=1,2i=1,2, is of the form

(𝒢(ti,ui))\displaystyle\mathcal{H}(\mathcal{G}(t_{i},u_{i})) = (\@arstrut1XX(2,-k1)Xk\\1β0,0-(Amin)1,1-tiβ1,0-(Amin)1,2-ui(^h12(1))T~βk,0\\[0.3em]Xβ1,0-(Amin)1,2-uiβ2,0-(Amin)1,3(^h12(2))T~β+k1,0\\[0.3em](X(2,-k1))T^h12(1)^h12(2)^H2^h3\\[0.3em]Xk~βk,0~β+k1,0(^h3)T~β2k,0) .\displaystyle=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(2,k-1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}-t_{i}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{1,0}-(A_{\min})_{1,2}-u_{i}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(\widehat{h}_{12}^{(1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widetilde{\beta}_{k,0}\\[0.3em]X$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{1,0}-(A_{\min})_{1,2}-u_{i}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2,0}-(A_{\min})_{1,3}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(\widehat{h}_{12}^{(2)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widetilde{\beta}_{k+1,0}\\[0.3em](\vec{X}^{(2,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widehat{h}_{12}^{(1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widehat{h}_{12}^{(2)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widehat{H}_{2}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widehat{h}_{3}\\[0.3em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widetilde{\beta}_{k,0}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widetilde{\beta}_{k+1,0}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(\widehat{h}_{3})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widetilde{\beta}_{2k,0}$\hfil\kern 5.0pt\crcr}}}}\right)$}}.

Assume on the contrary that none of (𝒢(t1,u1))\mathcal{H}(\mathcal{G}(t_{1},u_{1})) and (𝒢(t2,u2))\mathcal{H}(\mathcal{G}(t_{2},u_{2})) admits a \mathbb{R}–rm. Theorem 2.5 implies that the column XkX^{k} of (𝒢(ti,ui))\mathcal{H}(\mathcal{G}(t_{i},u_{i})), i=1,2i=1,2, is not in the span of the other columns. Using this fact, the facts that (𝒢(ti,ui))\mathcal{H}(\mathcal{G}(t_{i},u_{i})), i=1,2i=1,2, are not pd (by (ti,ui)2(t_{i},u_{i})\in\partial\mathcal{R}_{2}, i=1,2i=1,2) and H2H_{2} is pd, it follows that there is a column relation 1=j=1k1αj(i)Xj,\mathit{1}=\sum_{j=1}^{k-1}\alpha^{(i)}_{j}X^{j}, αj(i)\alpha_{j}^{(i)}\in\mathbb{R}, in (𝒢(ti,ui))\mathcal{H}(\mathcal{G}(t_{i},u_{i})), i=1,2i=1,2. Since (𝒢(ti,ui))0\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\succeq 0, i=1,2i=1,2, it follows in particular by Theorem 2.2, used for (M,A)=((𝒢(ti,ui)),((𝒢(ti,ui)))X(0,k1))(M,A)=(\mathcal{H}(\mathcal{G}(t_{i},u_{i})),(\mathcal{H}(\mathcal{G}(t_{i},u_{i})))_{\vec{X}^{(0,k-1)}}), i=1,2i=1,2, that

(5.28) (β~k,0β~k+1,0(h^3)T)T\displaystyle\begin{pmatrix}\widetilde{\beta}_{k,0}&\widetilde{\beta}_{k+1,0}&(\widehat{h}_{3})^{T}\end{pmatrix}^{T} 𝒞(((𝒢(ti,ui)))X(0,k1)),i=1,2.\displaystyle\in\mathcal{C}\Big{(}\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(0,k-1)}}\Big{)},\quad i=1,2.

Since the first column of (𝒢(ti,ui))0\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\succeq 0, i=1,2i=1,2, is in the span of the others, (5.28) is equivalent to

(5.29) (β~k,0β~k+1,0(h^3)T)T\displaystyle\begin{pmatrix}\widetilde{\beta}_{k,0}&\widetilde{\beta}_{k+1,0}&(\widehat{h}_{3})^{T}\end{pmatrix}^{T} 𝒞(((𝒢(ti,ui)))X(0,k1),X(1,k1)),i=1,2.\displaystyle\in\mathcal{C}\Big{(}\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(0,k-1)},\vec{X}^{(1,k-1)}}\Big{)},\quad i=1,2.

Since

H~2:=((𝒢(ti,ui)))X(1,k1),i=1,2,\widetilde{H}_{2}:=\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(1,k-1)}},\quad i=1,2,

is invertible as a principal submatrix of H2H_{2}, it follows that

(5.30) (β~k,0β~k+1,0(h^3)T)T\displaystyle\begin{pmatrix}\widetilde{\beta}_{k,0}&\widetilde{\beta}_{k+1,0}&(\widehat{h}_{3})^{T}\end{pmatrix}^{T} =(((𝒢(ti,ui)))X(0,k1),X(1,k1))v,i=1,2.\displaystyle=\Big{(}\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(0,k-1)},\vec{X}^{(1,k-1)}}\Big{)}v,\quad i=1,2.

with

v=H~21(β~k+1,0h^3)T=(v1v2vk1)T.v=\widetilde{H}_{2}^{-1}\begin{pmatrix}\widetilde{\beta}_{k+1,0}&\widehat{h}_{3}\end{pmatrix}^{T}=\begin{pmatrix}v_{1}&v_{2}&\cdots&v_{k-1}\end{pmatrix}^{T}.

If v10v_{1}\neq 0, this contradicts to (5.30) since u1u2u_{1}\neq u_{2}. Hence, v1=0v_{1}=0. By the Hankel structure of (𝒢(ti,ui))\mathcal{H}(\mathcal{G}(t_{i},u_{i})), i=1,2i=1,2, we have that

((𝒢(ti,ui)))X(0,k2),X(2,k)=((𝒢(ti,ui)))X(1,k1),X(1,k1),i=1,2.\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(0,k-2)},\vec{X}^{(2,k)}}=\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(1,k-1)},\vec{X}^{(1,k-1)}},\quad i=1,2.

Then (5.30) and v1=0v_{1}=0 imply that

(5.31) (((𝒢(ti,ui)))X(0,k2),X(2,k))v~=(((𝒢(ti,ui)))X(1,k1),X(1,k1))v~=𝟎k+1,1,\Big{(}\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(0,k-2)},\vec{X}^{(2,k)}}\Big{)}\widetilde{v}=\Big{(}\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(1,k-1)},\vec{X}^{(1,k-1)}}\Big{)}\widetilde{v}=\mathbf{0}_{k+1,1},

where v~=(v2vk11).\widetilde{v}=\begin{pmatrix}v_{2}&\cdots&v_{k-1}&-1\end{pmatrix}. Since ((𝒢(ti,ui)))X(1,k1),X(1,k1),\big{(}\mathcal{H}(\mathcal{G}(t_{i},u_{i}))\big{)}_{\vec{X}^{(1,k-1)},\vec{X}^{(1,k-1)}}, i=1,2i=1,2, is a principal submatrix of H2H_{2}, (5.31) contradicts to H2H_{2} being pd. This proves Claim 5. \blacksquare\\

Now we prove the implication (5.11)Theorem 5.1.(2)\eqref{251023-1603-v2}\Rightarrow\text{Theorem }\ref{221023-1854}.\eqref{221023-1854-pt3}. Since (t0,u0)1(t_{0},u_{0})\in\mathcal{R}_{1}, it follows that 1.\mathcal{R}_{1}\neq\emptyset. By (5.12), η0\eta\geq 0. We separate two cases according to the value of η\eta.\\

Case 1: η=0\eta=0. We separate two cases according to the invertibility of H2H_{2}.\\

Case 1.1: H2H_{2} is not pd. Since H2H_{2} is not pd, then by Theorem 2.5, the last column of (𝒢(t0,u0))\mathcal{H}(\mathcal{G}(t_{0},u_{0})) is in the span of the previous ones. But then by rg, the last column of H2H_{2} is in the span of the previous ones. This is the case Theorem 5.1.(2(a)ii).\\

Case 1.2: H2H_{2} is pd. We separate two cases according to the invertibility of ((Amin))X(0,k1)(\mathcal{H}(A_{\min}))_{\vec{X}^{(0,k-1)}}.\\

Case 1.2.1: rank((Amin)X(0,k1))=k\operatorname{rank}(\mathcal{H}(A_{\min})_{\vec{X}^{(0,k-1)}})=k. This is the case Theorem 5.1.(2(a)i).\\

Case 1.2.2: rank((Amin)X(0,k1))<k\operatorname{rank}(\mathcal{H}(A_{\min})_{\vec{X}^{(0,k-1)}})<k. We will prove that this case cannot occur. It follows from the assumption in this case that rank(Amin)=rankH2=k\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{2}=k. Further on, the last column of (Amin)\mathcal{H}(A_{\min}) cannot be in the span of the previous ones (otherwise rank(Amin)<k\operatorname{rank}\mathcal{H}(A_{\min})<k). Hence, by Theorem 2.5, (Amin)=(𝒢(0,0))\mathcal{H}(A_{\min})=\mathcal{H}(\mathcal{G}(0,0)) does not admit a \mathbb{R}–rm. Using this fact and Claim 3, (0,0)2(0,0)\in\partial\mathcal{R}_{2}. If t0=0t_{0}=0, then 12={(0,0)}\mathcal{R}_{1}\cap\mathcal{R}_{2}=\{(0,0)\}, which contradicts to the third condition in (5.11). So 0<t00<t_{0} must hold. Since η=0\eta=0, Claim 1 implies that 1={(t,0):t0}\mathcal{R}_{1}=\{(t,0)\colon t\geq 0\} is a horizontal half-line. By the form of 2\partial\mathcal{R}_{2}, which is the union of the graphs of two square root functions on the interval (,t0](-\infty,t_{0}], intersecting in the point (t0,u0)(t_{0},u_{0}) and such that (t0,u0)R2(t_{0},u_{0})\in\partial R_{2}, it follows that 12={(0,0)}\mathcal{R}_{1}\cap\mathcal{R}_{2}=\{(0,0)\}. Note that by H20H_{2}\succ 0, we have H2/H22>0H_{2}/H_{22}>0 and hence h(t)0h(t)\not\equiv 0 (see (5.8)), which implies that the square root functions are indeed not just a horizontal half-line. As above this contradicts to the third condition in (5.11). Hence, Case 1.2.2 cannot occur.\\

Case 2: η>0\eta>0. By assumptions, (t0,u0)12(t_{0},u_{0})\in\mathcal{R}_{1}\cap\mathcal{R}_{2}. By Claim 4, \mathcal{I}\neq\emptyset and \mathcal{I} has one or two elements. We separate two cases according to the number of elements in \mathcal{I}.\\

Case 2.1: \mathcal{I} has two elements. By Claim 4, H2/H22>0H_{2}/H_{22}>0. If H2H_{2} is not pd, then the fact that (𝒢(t0,u0))\mathcal{H}(\mathcal{G}(t_{0},u_{0})) has a \mathbb{R}–rm, implies that H2/H22=0H_{2}/H_{22}=0, which is a contradiction. Indeed, if H2/H22>0H_{2}/H_{22}>0 and H2H_{2} is not pd, then there is a nontrivial column relation among columns X2,,XkX^{2},\ldots,X^{k} in H2H_{2}. By Proposition 2.3, the same holds for (𝒢(t0,u0))\mathcal{H}(\mathcal{G}(t_{0},u_{0})). Let i=0k2ciXi+2=𝟎\sum_{i=0}^{k-2}c_{i}X^{i+2}=\mathbf{0} be the nontrivial column relation in (𝒢(t0,u0))\mathcal{H}(\mathcal{G}(t_{0},u_{0})). But then 𝒵(x2i=0k2cixi)=𝒵(xi=0k2cixi)\mathcal{Z}(x^{2}\sum_{i=0}^{k-2}c_{i}x^{i})=\mathcal{Z}(x\sum_{i=0}^{k-2}c_{i}x^{i}) and it follows by [CF96] that i=0k2ciXi+1=𝟎\sum_{i=0}^{k-2}c_{i}X^{i+1}=\mathbf{0} is also a nontrivial column relation in (𝒢(t0,u0))\mathcal{H}(\mathcal{G}(t_{0},u_{0})). In particular, H2/H22=0H_{2}/H_{22}=0. Hence, H2H_{2} is pd. This is the case Theorem 5.1.(2(b)i). \\

Case 2.2: \mathcal{I} has one element. Let us denote this element by (t~,u~)(\tilde{t},\tilde{u}). By Claim 4, =12\mathcal{I}=\mathcal{R}_{1}\cap\mathcal{R}_{2} or 2=2={(t,u0):tt0}\partial\mathcal{R}_{2}=\mathcal{R}_{2}=\{(t,u_{0})\colon t\leq t_{0}\} and 12={(t,u0):t~tt0}\mathcal{I}\subsetneq\mathcal{R}_{1}\cap\mathcal{R}_{2}=\{(t,u_{0})\colon\tilde{t}\leq t\leq t_{0}\} . We separate two cases according to these two possibilities.\\

Case 2.2.1: =12\mathcal{I}=\mathcal{R}_{1}\cap\mathcal{R}_{2}. In this case (t0,u0)=(t~,u~)(t_{0},u_{0})=(\tilde{t},\tilde{u}) and hence (𝒢(t~,u~)\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u}) admits a \mathbb{R}–rm. Since (t~,u~)1(\tilde{t},\tilde{u})\in\partial\mathcal{R}_{1}, (𝒢(t~,u~))\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})) is not pd. Hence, by Theorem 2.5, the statement Theorem 5.1.(2(b)ii) holds.\\

Case 2.2.2: 2=2={(t,u0):tt0}\partial\mathcal{R}_{2}=\mathcal{R}_{2}=\{(t,u_{0})\colon t\leq t_{0}\} and 12={(t,u0):t~tt0}\mathcal{I}\subsetneq\mathcal{R}_{1}\cap\mathcal{R}_{2}=\{(t,u_{0})\colon\tilde{t}\leq t\leq t_{0}\}. By (5.23), it follows that H2/H22=0H_{2}/H_{22}=0 (see the definition (5.8) of h(𝐭)h(\mathbf{t})). Since H2H_{2} is not pd, Theorem 2.5 used for (𝒢(t0,u0))\mathcal{H}(\mathcal{G}(t_{0},u_{0})), implies that the last column of H2H_{2} is in the span of the others. Hence, the same holds by Proposition 2.3 for (𝒢(t~,u~))\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})) and (𝒢(t~,u~))\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})) admits a \mathbb{R}–rm by Theorem 2.5. Since (𝒢(t~,u~))\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})) is not pd, it in particular satisfies (5.10). Hence, we are in the case Theorem 5.1.(2(b)ii). \\

This concludes the proof of the implication (5.11)Theorem 5.1.(2)\eqref{251023-1603-v2}\Rightarrow\text{Theorem }\ref{221023-1854}.\eqref{221023-1854-pt3}.\\

Next we prove the implication Theorem 5.1.(2)(5.11)\text{Theorem }\ref{221023-1854}.\eqref{221023-1854-pt3}\Rightarrow\eqref{251023-1603-v2}. We separate four cases according to the assumptions in Theorem 5.1.(2)\text{Theorem }\ref{221023-1854}.\eqref{221023-1854-pt3}.\\

Case 1: Theorem 5.1.(2(a)i) holds. By Claim 3, (0,0)12(0,0)\in\mathcal{R}_{1}\cap\mathcal{R}_{2}. This and the assumption rank((Amin))X(0,k1)=k\operatorname{rank}(\mathcal{H}(A_{\min}))_{\vec{X}^{(0,k-1)}}=k, imply by Theorem 2.5, that (𝒢(0,0))=(Amin)\mathcal{H}(\mathcal{G}(0,0))=\mathcal{H}(A_{\min}) admits a \mathbb{R}–rm. This proves (5.11) in case of Theorem 5.1.(2(a)i).\\

Case 2: Theorem 5.1.(2(a)ii) holds. By Claim 3, (0,0)12(0,0)\in\mathcal{R}_{1}\cap\mathcal{R}_{2}. Since the last column of H2H_{2} is by assumption in the span of the previous ones, the same holds for (𝒢(0,0))\mathcal{H}(\mathcal{G}(0,0)) by Proposition 2.3. By Theorem 2.5, (𝒢(0,0))\mathcal{H}(\mathcal{G}(0,0)) admits a \mathbb{R}–rm. This proves (5.11) in case of Theorem 5.1.(2(a)ii).\\

Case 3: Theorem 5.1.(2(b)i) holds. By assumption, =12={(t1,u1),(t2,u2)}\mathcal{I}=\partial\mathcal{R}_{1}\cap\partial\mathcal{R}_{2}=\{(t_{1},u_{1}),(t_{2},u_{2})\}. Since H2H_{2} is pd, 2\partial\mathcal{R}_{2} is not a half-line and hence u1u2u_{1}\neq u_{2}. By Claim 5, at least one of (𝒢(t1,u1))\mathcal{H}(\mathcal{G}(t_{1},u_{1})) and (𝒢(t2,u2))\mathcal{H}(\mathcal{G}(t_{2},u_{2})) admits a \mathbb{R}–rm. This proves (5.11) in case of Theorem 5.1.(2(b)i). \\

Case 4: Theorem 5.1.(2(b)ii) holds. The assumptions imply by Theorem 2.5, that (𝒢(t~,u~))\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})) admits a \mathbb{R}–rm. This proves (5.11) in case of Theorem 5.1.(2(b)ii).\\

This concludes the proof of the implication Theorem 5.1.(2)(5.11).\text{Theorem }\ref{221023-1854}.\eqref{221023-1854-pt3}\Rightarrow\eqref{251023-1603-v2}.\\

Up to now we established the equivalence (1)(2)\eqref{221023-1854-pt1}\Leftrightarrow\eqref{221023-1854-pt3} in Theorem 5.1. It remains to prove the moreover part. We observe again the proof of the implication (2)(5.11)\eqref{221023-1854-pt3}\Rightarrow\eqref{251023-1603-v2}. By Lemma 4.3.(4),

(5.32) rank~(k;β)=rank(Amin)+rank(Amin).\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)=\operatorname{rank}\mathcal{F}(A_{\min})+\operatorname{rank}\mathcal{H}(A_{\min}).

In the proof of the implications Theorem 5.1.(2(a)i)(5.11)\ref{221023-1854}.\eqref{221023-1857-pt3.1.1}\Rightarrow\eqref{251023-1603-v2} and Theorem 5.1.(2(a)ii)(5.11)\ref{221023-1854}.\eqref{221023-1857-pt3.1.2}\Rightarrow\eqref{251023-1603-v2} we established that (𝒢(0,0))\mathcal{H}(\mathcal{G}(0,0)) has a \mathbb{R}–rm. By Theorem 2.5, there also exists a (rank(𝒢(0,0)))(\operatorname{rank}\mathcal{H}(\mathcal{G}(0,0)))–atomic one. By Theorem 2.6, the sequence with the moment matrix (𝒢(0,0))\mathcal{F}(\mathcal{G}(0,0)) can be represented by a (rank(𝒢(0,0)))(\operatorname{rank}\mathcal{F}(\mathcal{G}(0,0)))–atomic 𝒵(ay+x2+y2)\mathcal{Z}(ay+x^{2}+y^{2})–rm. By (5.32) and 𝒢(0,0)=Amin\mathcal{G}(0,0)=A_{\min} if η=0\eta=0, in these two cases β\beta has a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm.

In the proof of the implication Theorem 5.1.(2(b)i)(5.11)\ref{221023-1854}.\eqref{221023-1857-pt3.2.1}\Rightarrow\eqref{251023-1603-v2} we established that (𝒢(t,u))\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime})) has a \mathbb{R}–rm for some (t,u)(t^{\prime},u^{\prime})\in\mathcal{I}. Analogously as for the point (0,0)(0,0) in the previous paragraph, it follows that β\beta has a (rank(𝒢(t,u))+rank(𝒢(t,u)))(\operatorname{rank}\mathcal{F}(\mathcal{G}(t^{\prime},u^{\prime}))+\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime})))–atomic 𝒵(p)\mathcal{Z}(p)–rm. Using (5.16), (5.27) and rankH2=rankH22+1\operatorname{rank}H_{2}=\operatorname{rank}H_{22}+1 (by H2H_{2} being pd), it follows that

(5.33) rank(𝒢(t,u))+rank(𝒢(t,u))=rank(Amin)+rankH2+1.\operatorname{rank}\mathcal{F}(\mathcal{G}(t^{\prime},u^{\prime}))+\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime}))=\operatorname{rank}\mathcal{F}(A_{\min})+\operatorname{rank}H_{2}+1.

We separate two cases:

  • If (Amin)\mathcal{H}(A_{\min}) is pd, then rank(Amin)=rankH2+1\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{2}+1. This, (5.32) and (5.33) imply that β\beta admits a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm.

  • If (Amin)\mathcal{H}(A_{\min}) is not pd, then we must have rank(Amin)=rankH2\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{2}, Otherwise we have ((Amin))X(1,k)/H22=0(\mathcal{H}(A_{\min}))_{\vec{X}^{(1,k)}}/H_{22}=0 and hence ((AminηE2,2(k+1)))X(1,k)/H22<0(\mathcal{H}(A_{\min}-\eta E_{2,2}^{(k+1)}))_{\vec{X}^{(1,k)}}/H_{22}<0, which contradicts to (AminηE2,2(k+1))\mathcal{H}(A_{\min}-\eta E_{2,2}^{(k+1)}) being psd. Hence, in this case β\beta has a (rank~(k;β)+1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1)–atomic 𝒵(p)\mathcal{Z}(p)–rm. Moreover, there cannot exist a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm. Indeed, since η>0\eta>0, at least rank(Amin)+1\operatorname{rank}\mathcal{F}(A_{\min})+1 (resp. rankH2\operatorname{rank}H_{2}) atoms are needed to represent (𝒢(t′′,u′′))\mathcal{F}(\mathcal{G}(t^{\prime\prime},u^{\prime\prime})) (resp. (𝒢(t′′,u′′))\mathcal{H}(\mathcal{G}(t^{\prime\prime},u^{\prime\prime}))) for any (t′′,u′′)12(t^{\prime\prime},u^{\prime\prime})\in\mathcal{R}_{1}\cap\mathcal{R}_{2} (see (5.16) and (5.27)). Hence, at least rank(Amin)+rankH2+1\operatorname{rank}\mathcal{F}(A_{\min})+\operatorname{rank}H_{2}+1 atoms are needed in a 𝒵(p)\mathcal{Z}(p)–rm for any (t′′,u′′)12(t^{\prime\prime},u^{\prime\prime})\in\mathcal{R}_{1}\cap\mathcal{R}_{2}.

In the proof of the implication Theorem 5.1.(2(b)ii)(5.11)\ref{221023-1854}.\eqref{221023-1857-pt3.2.2}\Rightarrow\eqref{251023-1603-v2} we established that (𝒢(t~,u~))\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})) has a \mathbb{R}–rm. Analogously as for the point (0,0)(0,0) in two paragraphs above, it follows that β\beta has a (rank(𝒢(t~,u~))+rank(𝒢(t~,u~)))(\operatorname{rank}\mathcal{F}(\mathcal{G}(\tilde{t},\tilde{u}))+\operatorname{rank}\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u})))–atomic 𝒵(p)\mathcal{Z}(p)–rm. By (5.16) and (5.27), this measure is (rank(Amin)+rankH22+2)(\operatorname{rank}\mathcal{F}(A_{\min})+\operatorname{rank}H_{22}+2)–atomic.

  • If (Amin)\mathcal{H}(A_{\min}) is pd, then rank(Amin)=rankH22+2\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{22}+2. This and (5.32) imply that β\beta admits a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm.

  • If (Amin)\mathcal{H}(A_{\min}) is not pd, then we have rank(Amin)=rankH22+1\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{22}+1, since otherwise the equality ((Amin))X(1,k)/H22=0(\mathcal{H}(A_{\min}))_{\vec{X}^{(1,k)}}/H_{22}=0 implies ((AminηE2,2(k+1)))X(1,k)/H22<0(\mathcal{H}(A_{\min}-\eta E_{2,2}^{(k+1)}))_{\vec{X}^{(1,k)}}/H_{22}<0, which contradicts to (AminηE2,2(k+1))\mathcal{H}(A_{\min}-\eta E_{2,2}^{(k+1)}) being psd. Hence, in this case β\beta has a (rank~(k;β)+1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1)–atomic 𝒵(p)\mathcal{Z}(p)–rm. Moreover, there cannot exist a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm in this case. Indeed,

    (12)=(1̊2)(̊12)(̊1̊2).(\mathcal{R}_{1}\cap\mathcal{R}_{2})\setminus\mathcal{I}=(\partial\mathcal{R}_{1}\cap\mathring{\mathcal{R}}_{2})\cup(\mathring{\mathcal{R}}_{1}\cap\partial\mathcal{R}_{2})\cup(\mathring{\mathcal{R}}_{1}\cap\mathring{\mathcal{R}}_{2}).

    Using (5.16) and (5.27), in every point from (12)(\mathcal{R}_{1}\cap\mathcal{R}_{2})\setminus\mathcal{I} at least rank(Amin)+rankH22+2\operatorname{rank}\mathcal{F}(A_{\min})+\operatorname{rank}H_{22}+2 atoms are needed in a 𝒵(p)\mathcal{Z}(p)–rm.

This concludes the proof of the moreover part.

Since for a pp–pure sequence with ~(k;β))0\widetilde{\mathcal{M}}(k;\beta))\succeq 0, (5.32) implies that (Amin)\mathcal{H}(A_{\min}) is pd, it follows by the moreover part that the existence of a 𝒵(p)\mathcal{Z}(p)–rm implies the existence of a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm. ∎

The following example demonstrates the use of Theorem 5.1 to show that there exists a bivariate y(2y+x2+y2)y(-2y+x^{2}+y^{2})–pure sequence β\beta of degree 6 with a positive semidefinite (3)\mathcal{M}(3) and without a 𝒵(y(2y+x2+y2))\mathcal{Z}(y(-2y+x^{2}+y^{2}))–rm.

Example 5.3.

Let β\beta be a bivariate degree 6 sequence given by

β00\displaystyle\beta_{00} =10,\displaystyle=10, β10\displaystyle\beta_{10} =385,\displaystyle=\frac{38}{5}, β01\displaystyle\beta_{01} =395,\displaystyle=\frac{39}{5},
β20\displaystyle\beta_{20} =60225,\displaystyle=\frac{602}{25}, β11\displaystyle\beta_{11} =325,\displaystyle=\frac{3}{25}, β02\displaystyle\beta_{02} =31325,\displaystyle=\frac{313}{25},
β30\displaystyle\beta_{30} =9152125,\displaystyle=\frac{9152}{125}, β21\displaystyle\beta_{21} =421125,\displaystyle=\frac{421}{125}, β12\displaystyle\beta_{12} =3125,\displaystyle=\frac{3}{125},
β03\displaystyle\beta_{03} =2709125,\displaystyle=\frac{2709}{125}, β40\displaystyle\beta_{40} =172118625,\displaystyle=\frac{172118}{625}, β31\displaystyle\beta_{31} =27625,\displaystyle=\frac{27}{625},
β22\displaystyle\beta_{22} =2717625,\displaystyle=\frac{2717}{625}, β13\displaystyle\beta_{13} =3625,\displaystyle=\frac{3}{625}, β04\displaystyle\beta_{04} =24373625,\displaystyle=\frac{24373}{625},
β50\displaystyle\beta_{50} =33033683125,\displaystyle=\frac{3303368}{3125}, β41\displaystyle\beta_{41} =77893125,\displaystyle=\frac{7789}{3125}, β32\displaystyle\beta_{32} =273125,\displaystyle=\frac{27}{3125},
β23\displaystyle\beta_{23} =193813125,\displaystyle=\frac{19381}{3125}, β14\displaystyle\beta_{14} =33125,\displaystyle=\frac{3}{3125}, β05\displaystyle\beta_{05} =2243493125,\displaystyle=\frac{224349}{3125},
β60\displaystyle\beta_{60} =4156,\displaystyle=4156, β51\displaystyle\beta_{51} =24315625,\displaystyle=\frac{243}{15625}, β42\displaystyle\beta_{42} =4445315625,\displaystyle=\frac{44453}{15625},
β33\displaystyle\beta_{33} =2715625,\displaystyle=\frac{27}{15625}, β24\displaystyle\beta_{24} =14935715625,\displaystyle=\frac{149357}{15625}, β15\displaystyle\beta_{15} =315625,\displaystyle=\frac{3}{15625},
β06\displaystyle\beta_{06} =209413315625.\displaystyle=\frac{2094133}{15625}.

Assume the notation as in Theorem 5.1. ~(3)\widetilde{\mathcal{M}}(3) is psd with the eigenvalues 4445\approx 4445, 189.2\approx 189.2, 16.6\approx 16.6, 11.9\approx 11.9, 3.2\approx 3.2, 1.22\approx 1.22, 0.57\approx 0.57, 0.022\approx 0.022, 0.0030\approx 0.0030, 0 and the column relation

2Y2+X2Y+Y3=0.-2Y^{2}+X^{2}Y+Y^{3}=0.

We have that

Amin=(32433055873132789278915772527125132789278915418009113945752712514936257725271251493625243312527125149362524331253343715625)A_{\min}=\begin{pmatrix}\frac{324330}{55873}&\frac{132789}{278915}&\frac{77}{25}&\frac{27}{125}\\[5.0pt] \frac{132789}{278915}&\frac{4180091}{1394575}&\frac{27}{125}&\frac{1493}{625}\\[5.0pt] \frac{77}{25}&\frac{27}{125}&\frac{1493}{625}&\frac{243}{3125}\\[5.0pt] \frac{27}{125}&\frac{1493}{625}&\frac{243}{3125}&\frac{33437}{15625}\end{pmatrix}

and so

η=772541800911394575=460855783.\eta=\frac{77}{25}-\frac{4180091}{1394575}=\frac{4608}{55783}.

The matrix H2H_{2} is equal to:

H2\displaystyle H_{2} =(217327373273105727310576490406315625).\displaystyle=\begin{pmatrix}21&73&273\\ 73&273&1057\\ 273&1057&\frac{64904063}{15625}\end{pmatrix}.

The eigenvalues of H2H_{2} are 4441.1\approx 4441.1, 6.74\approx 6.74, 0.019\approx-0.019 and hence H2H_{2} is not psd. By Theorem 5.1, β\beta does not have a 𝒵(y(2y+x2+y2))\mathcal{Z}(y(-2y+x^{2}+y^{2}))–rm, since by (2b) of Theorem 5.1, H2H_{2} should be psd.

6. Parabolic type relation: p(x,y)=y(xy2)p(x,y)=y(x-y^{2}).

In this section we solve the 𝒵(p)\mathcal{Z}(p)–TMP for the sequence β={βi}i,j+,i+j2k\beta=\{\beta_{i}\}_{i,j\in\mathbb{Z}_{+},i+j\leq 2k} of degree 2k2k, k3k\geq 3, where p(x,y)=y(xy2)p(x,y)=y(x-y^{2}). Assume the notation from Section 4. If β\beta admits a 𝒵(p)\mathcal{Z}(p)–TMP, then (k;β)\mathcal{M}(k;\beta) must satisfy the relations

(6.1) Y3+jXi=Y1+jXi+1for i,j+ such that i+jk3.Y^{3+j}X^{i}=Y^{1+j}X^{i+1}\quad\text{for }i,j\in\mathbb{Z}_{+}\text{ such that }i+j\leq k-3.

In the presence of all column relations (6.1), the column space 𝒞((k;β))\mathcal{C}(\mathcal{M}(k;\beta)) is spanned by the columns in the set

(6.2) 𝒯=X(0,k)YX(0,k1)Y2X(0,k2),\mathcal{T}=\vec{X}^{(0,k)}\cup Y\vec{X}^{(0,k-1)}\cup Y^{2}\vec{X}^{(0,k-2)},

where

YiX(j,):=(YiXj,YiXj+1,,YiX)with i,j,+,j,i+k.Y^{i}\vec{X}^{(j,\ell)}:=(Y^{i}X^{j},Y^{i}X^{j+1},\ldots,Y^{i}X^{\ell})\quad\text{with }i,j,\ell\in\mathbb{Z}_{+},\;j\leq\ell,\;i+\ell\leq k.

Let ~(k;β)\widetilde{\mathcal{M}}(k;\beta) be as in (4.9). Let

(6.3) Amin:=A12(A22)(A12)T.A_{\min}:=A_{12}(A_{22})^{\dagger}(A_{12})^{T}.

As described in Remark 4.4, AminA_{\min} might need to be changed to

A^min=Amin+η(E1,k+1(k+1)+Ek+1,1(k+1)),\widehat{A}_{\min}=A_{\min}+\eta\big{(}E_{1,k+1}^{(k+1)}+E_{k+1,1}^{(k+1)}\big{)},

where

η:=(Amin)2,k(Amin)1,k+1.\eta:=(A_{\min})_{2,k}-(A_{\min})_{1,k+1}.

Let (𝐀)\mathcal{F}(\mathbf{A}) and (𝐀)\mathcal{H}(\mathbf{A}) be as in (4.11). Define also the matrix function

(6.4) 𝒢:2Sk+1,𝒢(𝐭,𝐮)=A^min+𝐭E1,1(k+1)+𝐮Ek+1,k+1(k+1).\mathcal{G}:\mathbb{R}^{2}\to S_{k+1},\qquad\mathcal{G}(\mathbf{t},\mathbf{u})=\widehat{A}_{\min}+\mathbf{t}E_{1,1}^{(k+1)}+\mathbf{u}E_{k+1,k+1}^{(k+1)}.

Write

(6.5) (A^min)= (\@arstrut1X(1,-k1)Xk\\1β0,0-(Amin)1,1(h12)Tβk,0-(Amin)2,k\\[0.2em](X(1,-k1))Th12H22h23\\[0.2em]Xkβk,0-(Amin)2,k(h23)Tβ2k,0-(Amin)+k1,+k1) ,\\[0.5em]H1:=((A^min))X(0,k1)= (\@arstrut1X(1,-k1)\\1β0,0-(Amin)1,1(h12)T\\[0.2em](X(1,-k1))Th12H22) ,\\[0.5em]H2:=((A^min))X(1,k)= (\@arstrutX(1,-k1)Xk\\(X(1,-k1))TH22h23\\[0.2em]Xk(h23)Tβ2k,0-(Amin)+k1,+k1) .\displaystyle\begin{split}\mathcal{H}(\widehat{A}_{\min})&=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(1,k-1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{2,k}\\[0.2em](\vec{X}^{(1,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{23}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{2,k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2k,0}-(A_{\min})_{k+1,k+1}$\hfil\kern 5.0pt\crcr}}}}\right)$}},\\[0.5em]H_{1}&:=(\mathcal{H}(\widehat{A}_{\min}))_{\vec{X}^{(0,k-1)}}=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(1,k-1)}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12})^{T}\\[0.2em](\vec{X}^{(1,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt\crcr}}}}\right)$}},\\[0.5em]H_{2}&:=(\mathcal{H}(\widehat{A}_{\min}))_{\vec{X}^{(1,k)}}=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(1,k-1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}\\(\vec{X}^{(1,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{23}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2k,0}-(A_{\min})_{k+1,k+1}$\hfil\kern 5.0pt\crcr}}}}\right)$}}.\end{split}

Let us define the matrix

K\displaystyle K :=(A^min)/H22\displaystyle:=\mathcal{H}(\widehat{A}_{\min})/H_{22}
=(β0,0(Amin)1,1βk,0(Amin)2,kβk,0(Amin)2,kβ2k,0(Amin)k+1,k+1.)((h12)T(h23)T)(H22)(h12h23)\displaystyle=\begin{pmatrix}\beta_{0,0}-(A_{\min})_{1,1}&\beta_{k,0}-(A_{\min})_{2,k}\\[1.99997pt] \beta_{k,0}-(A_{\min})_{2,k}&\beta_{2k,0}-(A_{\min})_{k+1,k+1}.\end{pmatrix}-\begin{pmatrix}(h_{12})^{T}\\ (h_{23})^{T}\end{pmatrix}(H_{22})^{\dagger}\begin{pmatrix}h_{12}&h_{23}\end{pmatrix}
:=(β0,0(Amin)1,1(h12)T(H22)h12βk,0(Amin)2,k(h12)T(H22)h23βk,0(Amin)2,k(h23)T(H22)h12β2k,0(Amin)k+1,k+1(h12)T(H22)h12)\displaystyle:=\begin{pmatrix}\beta_{0,0}-(A_{\min})_{1,1}-(h_{12})^{T}(H_{22})^{\dagger}h_{12}&\beta_{k,0}-(A_{\min})_{2,k}-(h_{12})^{T}(H_{22})^{\dagger}h_{23}\\[3.00003pt] \beta_{k,0}-(A_{\min})_{2,k}-(h_{23})^{T}(H_{22})^{\dagger}h_{12}&\beta_{2k,0}-(A_{\min})_{k+1,k+1}-(h_{12})^{T}(H_{22})^{\dagger}h_{12}\end{pmatrix}
:=(k11k12k12k22).\displaystyle:=\begin{pmatrix}k_{11}&k_{12}\\ k_{12}&k_{22}\end{pmatrix}.

Let

𝒯^={1,Y,X,XY,X2,X2Y,,Xi,XiY,,Xk1,Xk1Y,Xk},\widehat{\mathcal{T}}=\{\mathit{1},Y,X,XY,X^{2},X^{2}Y,\ldots,X^{i},X^{i}Y,\ldots,X^{k-1},X^{k-1}Y,X^{k}\},

and

(6.6) P^ be a permutation matrix such that moment matrix ^(k;β):=P^(k;β)(P^)Thas rows and columns indexed in the order 𝒯^,𝒞𝒯^.\displaystyle\begin{split}&\widehat{P}\text{ be a permutation matrix such that moment matrix }\widehat{\mathcal{M}}(k;\beta):=\widehat{P}\mathcal{M}(k;\beta)(\widehat{P})^{T}\\ &\text{has rows and columns indexed in the order }\widehat{\mathcal{T}},\mathcal{C}\setminus\widehat{\mathcal{T}}.\end{split}

Write

(6.7) ^(𝐭,𝐮)=P^(𝒢(𝐭,𝐮))(P^)T= (\@arstrut1^T{1,Xk}XkC^T\\[0.2em]1(Amin)1,1+t(f12)T(Amin)2,k(f14)T\\[0.2em](^T{1,Xk})Tf12F22f23F24\\[0.2em]Xk(Amin)2,k(f23)T(Amin)+k1,+k1+u(f34)T\\[0.2em]C^Tf14(F24)Tf34F44) .\displaystyle\begin{split}&\widehat{\mathcal{F}}(\mathbf{t},\mathbf{u})=\widehat{P}\mathcal{F}(\mathcal{G}(\mathbf{t},\mathbf{u}))(\widehat{P})^{T}\\ &=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widehat{\mathcal{T}}\setminus\{\mathit{1},X^{k}\}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{C}\setminus\widehat{\mathcal{T}}\\[0.2em]\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{1,1}+\mathbf{t}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{12})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{2,k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{14})^{T}\\[0.2em](\widehat{\mathcal{T}}\setminus\{\mathit{1},X^{k}\})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle F_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{23}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle F_{24}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{2,k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{k+1,k+1}+\mathbf{u}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{34})^{T}\\[0.2em]\mathcal{C}\setminus\widehat{\mathcal{T}}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{14}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(F_{24})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{34}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle F_{44}$\hfil\kern 5.0pt\crcr}}}}\right)$}}.\end{split}

The solution to the cubic parabolic type relation TMP is the following.

Theorem 6.1.

Let p(x,y)=y(xy2)p(x,y)=y(x-y^{2}) and β:=β(2k)=(βi,j)i,j+,i+j2k\beta:=\beta^{(2k)}=(\beta_{i,j})_{i,j\in\mathbb{Z}_{+},i+j\leq 2k}, where k3k\geq 3. Assume also the notation above. Then the following statements are equivalent:

  1. (1)

    β\beta has a 𝒵(p)\mathcal{Z}(p)–representing measure.

  2. (2)

    ~(k;β)\widetilde{\mathcal{M}}(k;\beta) is positive semidefinite, the relations

    (6.8) βi,j+3=βi+1,j+1hold for every i,j+ with i+j2k3,\beta_{i,j+3}=\beta_{i+1,j+1}\quad\text{hold for every }i,j\in\mathbb{Z}_{+}\text{ with }i+j\leq 2k-3,

    (A^min)\mathcal{H}(\widehat{A}_{\min}) is positive semidefinite, defining real numbers

    (6.9) t1=H1/H22=β0,0(Amin)1,1(h12)T(H22)h12,u1=H2/H22=β2k,0(Amin)k+1,k+1(h23)T(H22)h23,\displaystyle\begin{split}t_{1}&=H_{1}/H_{22}=\beta_{0,0}-(A_{\min})_{1,1}-(h_{12})^{T}(H_{22})^{\dagger}h_{12},\\[1.99997pt] u_{1}&=H_{2}/H_{22}=\beta_{2k,0}-(A_{\min})_{k+1,k+1}-(h_{23})^{T}(H_{22})^{\dagger}h_{23},\end{split}

    and the property

    (6.10) ((A^min))X(0,k1)0orrank((A^min))X(0,k1)=rank(A^min),\displaystyle(\mathcal{H}(\widehat{A}_{\min}))_{\vec{X}^{(0,k-1)}}\succ 0\quad\text{or}\quad\operatorname{rank}(\mathcal{H}(\widehat{A}_{\min}))_{\vec{X}^{(0,k-1)}}=\operatorname{rank}\mathcal{H}(\widehat{A}_{\min}),

    one of the following statements holds:

    1. (a)

      F22F_{22} is not positive definite, η=0\eta=0 and (6.10) holds.

    2. (b)

      F22F_{22} is positive definite, H22H_{22} is not positive definite and one of the following holds:

      1. (i)

        u1=η=0u_{1}=\eta=0.

      2. (ii)

        u1>0u_{1}>0, t1>0t_{1}>0, t1u1η2t_{1}u_{1}\geq\eta^{2} and βk,0(Amin)2,k=(h12)T(H22)h23.\beta_{k,0}-(A_{\min})_{2,k}=(h_{12})^{T}(H_{22})^{\dagger}h_{23}.

    3. (c)

      F22,H22F_{22},H_{22} are positive definite and one of the following holds:

      1. (i)

        η=0\eta=0 and (6.10) holds.

      2. (ii)

        η0\eta\neq 0 and

        (6.11) (k11k22sign(k12)k12)2η2,(\sqrt{k_{11}k_{22}}-\operatorname{sign}(k_{12})k_{12})^{2}\geq\eta^{2},

        where sign\operatorname{sign} is the sign function and sign(0)=0\operatorname{sign}(0)=0.

Moreover, if a 𝒵(p)\mathcal{Z}(p)–representing measure for β\beta exists, then:

  • There exists at most (rank~(k;β)+1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1)–atomic 𝒵(p)\mathcal{Z}(p)–representing measure.

  • There exists a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–representing measure if and only if any of the following holds:

    • η=0\eta=0.

    • rank(Amin)=rankH22+2.\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{22}+2.

    • rank(Amin)=rankH22+1\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{22}+1 and one of the following holds:

      • *

        H22H_{22} is not positive definite and t1u1=η2t_{1}u_{1}=\eta^{2}.

      • *

        H22H_{22} is positive definite, k12=0k_{12}=0 and k11k22=η2k_{11}k_{22}=\eta^{2}.

In particular, a pp–pure sequence β\beta with a 𝒵(p)\mathcal{Z}(p)–representing measure admits a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–representing measure.

Remark 6.2.

In this remark we explain the idea of the proof of Theorem 6.1 and the meaning of conditions in the statement of the theorem.

By Lemmas 4.14.2, the existence of a 𝒵(p)\mathcal{Z}(p)–rm for β\beta is equivalent to the existence of t,ut,u\in\mathbb{R} such that (𝒢(t,u))\mathcal{F}(\mathcal{G}(t,u)) admits a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm and (𝒢(t,u))\mathcal{H}(\mathcal{G}(t,u)) admits a \mathbb{R}–rm. Let

1\displaystyle\mathcal{R}_{1} ={(t,u)2:(𝒢(t,u))0}and2={(t,u)2:(𝒢(t,u))0}.\displaystyle=\big{\{}(t,u)\in\mathbb{R}^{2}\colon\mathcal{F}(\mathcal{G}(t,u))\succeq 0\big{\}}\quad\text{and}\quad\mathcal{R}_{2}=\big{\{}(t,u)\in\mathbb{R}^{2}\colon\mathcal{H}(\mathcal{G}(t,u))\succeq 0\big{\}}.

We denote by Ri\partial R_{i} and R̊i\mathring{R}_{i} the topological boundary and the interior of the set RiR_{i}, respectively. By the necessary conditions for the existence of a 𝒵(p)\mathcal{Z}(p)–rm [CF04, Fia95, CF96], ~(k;β)\widetilde{\mathcal{M}}(k;\beta) must be psd and the relations (6.8) must hold. Then Theorem 6.1.(1) is equivalent to

(6.12) ~(k;β)0, the relations (6.8) hold and (t0,u0)12:(𝒢(t0,u0)) and (𝒢(t0,u0)) admit 𝒵(xy2)–rm and a –rm, respectively.\displaystyle\begin{split}&\widetilde{\mathcal{M}}(k;\beta)\succeq 0,\text{ the relations }\eqref{131023-0847-equation}\text{ hold and }\\ &\exists(t_{0},u_{0})\in\mathcal{R}_{1}\cap\mathcal{R}_{2}:\mathcal{F}(\mathcal{G}(t_{0},u_{0}))\text{ and }\mathcal{H}(\mathcal{G}(t_{0},u_{0}))\text{ admit }\\ &\hskip 113.81102pt\text{a }\mathcal{Z}(x-y^{2})\text{--rm and a }\mathbb{R}\text{--rm, respectively.}\end{split}

In the proof of Theorem 6.1 we show that (6.12) is equivalent to Theorem 6.1.(2):

  1. (1)

    First we establish (see Claims 1 and 2 below) that the form of:

    • 1\mathcal{R}_{1} is one of the following:

      [Uncaptioned image] [Uncaptioned image]

      where the left case occurs if η0\eta\neq 0 and the right if η=0\eta=0.

    • 2\mathcal{R}_{2} is one of the following:

      [Uncaptioned image] [Uncaptioned image]

      where the left case occurs if k120k_{12}\neq 0 and the right if k12=0k_{12}=0.

  2. (2)

    If F22F_{22} is only positive semidefinite but not definite, then we show that (6.12) is equivalent to

    (6.13) ~(k;β)0, the relations (6.8) hold,η=0 and (𝒢(0,0)) admits a –rm.\displaystyle\begin{split}&\widetilde{\mathcal{M}}(k;\beta)\succeq 0,\text{ the relations }\eqref{131023-0847-equation}\text{ hold},\eta=0\text{ and }\mathcal{H}(\mathcal{G}(0,0))\text{ admits a }\mathbb{R}\text{--rm}.\end{split}

    The latter statement is further equivalent to Theorem 6.1.(2a).

  3. (3)

    Assume that F22F_{22} is positive definite and H22H_{22} is only positive semidefinite but not definite. If:

    • u1=0u_{1}=0, then we show that (6.12) is equivalent to (6.13). The latter statement is further equivalent to Theorem 6.1.(2(b)i).

    • u1>0u_{1}>0, then we show that (6.12) is equivalent to

      ~(k;β)0, the relations (6.8) hold, (𝒢(t1,u1)) and (𝒢(t1,u1)) admit a 𝒵(xy2)–rm and a –rm, respectively.\displaystyle\begin{split}&\widetilde{\mathcal{M}}(k;\beta)\succeq 0,\text{ the relations }\eqref{131023-0847-equation}\text{ hold, }\mathcal{F}(\mathcal{G}(t_{1},u_{1}))\text{ and }\\ &\mathcal{H}(\mathcal{G}(t_{1},u_{1}))\text{ admit a }\mathcal{Z}(x-y^{2})\text{--rm and a }\mathbb{R}\text{--rm, respectively.}\end{split}

      The latter statement is further equivalent to Theorem 6.1.(2(b)ii).

    • u1<0u_{1}<0, then (6.12) cannot hold.

  4. (4)

    Assume that F22F_{22} and H22H_{22} are positive definite. If:

    • η=0\eta=0, then we show that (6.12) is equivalent to (6.13). The latter statement is further equivalent to Theorem 6.1.(2(c)i).

    • η0\eta\neq 0, then we show that (6.12) is equivalent to 12\mathcal{R}_{1}\cap\mathcal{R}_{2}\neq\emptyset. The latter statement is further equivalent to Theorem 6.1.(2(c)ii).

Proof of Theorem 6.1.

Let 1,2\mathcal{R}_{1},\mathcal{R}_{2} be as in Remark 6.2. As explained in Remark 6.2, Theorem 6.1.(1) is equivalent to (6.12), thus it remains to prove that (6.12) is equivalent to Theorem 6.1.(2).\\

First we establish a few claims needed in the proof. Claim 1 (resp. 2) describes 1\mathcal{R}_{1} (resp. 2\mathcal{R}_{2}) concretely.\\

Claim 1. Assume that ~(k;β)0\widetilde{\mathcal{M}}(k;\beta)\succeq 0. Then

(6.14) 1={(t,u)2:t0,u0,tuη2}.\mathcal{R}_{1}=\big{\{}(t,u)\in\mathbb{R}^{2}\colon t\geq 0,u\geq 0,tu\geq\eta^{2}\big{\}}.

If (t,u)1(t,u)\in\mathcal{R}_{1}, we have

(6.19) rank(𝒢(t,u))={rank(Amin),if η=t=u=0,rank(Amin)+1,if (η=t=0,u>0) or (η=u=0,t>0) or (η0,tu=η2),rank(Amin)+2,if tu>η2.\displaystyle\operatorname{rank}\mathcal{F}(\mathcal{G}(t,u))=\left\{\begin{array}[]{rl}\operatorname{rank}\mathcal{F}(A_{\min}),&\text{if }\eta=t=u=0,\\[3.00003pt] \operatorname{rank}\mathcal{F}(A_{\min})+1,&\text{if }(\eta=t=0,u>0)\text{ or }(\eta=u=0,t>0)\\[1.99997pt] &\text{ or }(\eta\neq 0,tu=\eta^{2}),\\[3.00003pt] \operatorname{rank}\mathcal{F}(A_{\min})+2,&\text{if }tu>\eta^{2}.\end{array}\right.

where AminA_{\min} is as in (6.3).\\

Proof of Claim 1. Note that

(6.20) 𝒢(𝐭,𝐮)=Amin+η(E1,k+1(k+1)+Ek+1,1(k+1))+𝐭E1,1(k+1)+𝐮Ek+1,k+1(k+1)=Amin+(𝐭𝟎1,k1η𝟎k1,1𝟎k1𝟎k1,1η𝟎1,k1𝐮).\displaystyle\begin{split}\mathcal{G}(\mathbf{t},\mathbf{u})&=A_{\min}+\eta\big{(}E_{1,k+1}^{(k+1)}+E_{k+1,1}^{(k+1)}\big{)}+\mathbf{t}E_{1,1}^{(k+1)}+\mathbf{u}E_{k+1,k+1}^{(k+1)}\\[3.00003pt] &=A_{\min}+\begin{pmatrix}\mathbf{t}&\mathbf{0}_{1,k-1}&\eta\\ \mathbf{0}_{k-1,1}&\mathbf{0}_{k-1}&\mathbf{0}_{k-1,1}\\ \eta&\mathbf{0}_{1,k-1}&\mathbf{u}\end{pmatrix}.\end{split}

By Lemma 4.3, we have that

(6.21) (𝒢(t,u))0𝒢(t,u)Amin\mathcal{F}(\mathcal{G}(t,u))\succeq 0\quad\Leftrightarrow\quad\mathcal{G}(t,u)\succeq A_{\min}

Using (6.20), (6.21) and the definition of 1\mathcal{R}_{1}, we have that

(6.22) (t,u)1\displaystyle(t,u)\in\mathcal{R}_{1}\quad (tηηu)0t0,u0,tuη2,\displaystyle\Leftrightarrow\quad\begin{pmatrix}t&\eta\\ \eta&u\end{pmatrix}\succeq 0\quad\Leftrightarrow\quad t\geq 0,u\geq 0,tu\geq\eta^{2},

which proves (6.14).

To prove (6.19) first note that by construction of (Amin)\mathcal{F}(A_{\min}), the columns 11 and XkX^{k} are in the span of the columns indexed by 𝒞X(0,k)\mathcal{C}\setminus\vec{X}^{(0,k)}. Hence, there are vectors

(6.23) v1,v2ker(Amin)v_{1},v_{2}\in\ker\mathcal{F}(A_{\min})

of the forms

(6.24) v1=(1𝟎1,k(v~1)T)T(k+1)(k+2)2andv2=(𝟎1,k1(v~2)T)T(k+1)(k+2)2.v_{1}=\begin{pmatrix}1&\mathbf{0}_{1,k}&(\tilde{v}_{1})^{T}\end{pmatrix}^{T}\in\mathbb{R}^{\frac{(k+1)(k+2)}{2}}\quad\text{and}\quad v_{2}=\begin{pmatrix}\mathbf{0}_{1,k}&1&(\tilde{v}_{2})^{T}\end{pmatrix}^{T}\in\mathbb{R}^{\frac{(k+1)(k+2)}{2}}.

Let r:=rank(tηηu)r:=\operatorname{rank}\begin{pmatrix}t&\eta\\ \eta&u\end{pmatrix}. Clearly,

(6.25) rank(𝒢(t,u))rank(Amin)+r.\operatorname{rank}\mathcal{F}(\mathcal{G}(t,u))\leq\operatorname{rank}\mathcal{F}(A_{\min})+r.

We separate three cases according to rr. \\

Case 1: r=0r=0. In this case t=u=η=0t=u=\eta=0 and 𝒢(0,0)=Amin\mathcal{G}(0,0)=A_{\min}. In this case (6.19) clearly holds.\\

Case 2: r=1r=1. In this case tu=η2tu=\eta^{2}. Together with (6.22), this is equivalent to (η=t=0,u>0)(\eta=t=0,u>0) or (η=u=0,t>0)(\eta=u=0,t>0) or (η0,tu=η2).(\eta\neq 0,tu=\eta^{2}). By (6.25) and (𝒢(t,u))(Amin)\mathcal{F}(\mathcal{G}(t,u))\succeq\mathcal{F}(A_{\min}) to prove (6.19), it suffices to find vker(Amin)v\in\ker\mathcal{F}(A_{\min}) and vker(𝒢(t,u))v\notin\ker\mathcal{F}(\mathcal{G}(t,u)). Note that at least one of v1,v2v_{1},v_{2} from (6.24) is such a vector, since

(v1)T(𝒢(t,u))v1=tand(v2)T(𝒢(t,u))v2=u.(v_{1})^{T}\mathcal{F}(\mathcal{G}(t,u))v_{1}=t\quad\text{and}\quad(v_{2})^{T}\mathcal{F}(\mathcal{G}(t,u))v_{2}=u.\vskip 3.0pt plus 1.0pt minus 1.0pt

Case 3: r=2r=2. In this case tu>η2tu>\eta^{2}. Note that

(6.26) (𝒢(t,u))=(𝒢(η2u,u))+(tη2u)𝟎(k+1)(k+2)21(𝒢(η2u,u)).\mathcal{F}(\mathcal{G}(t,u))=\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{\eta^{2}}{u},u\Big{)}\Big{)}+\begin{pmatrix}t-\frac{\eta^{2}}{u}\end{pmatrix}\oplus\mathbf{0}_{\frac{(k+1)(k+2)}{2}-1}\succeq\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{\eta^{2}}{u},u\Big{)}\Big{)}.

By Case 2, we have rank(𝒢(η2u,u))=rank(Amin)+1\operatorname{rank}\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{\eta^{2}}{u},u\Big{)}\Big{)}=\operatorname{rank}\mathcal{F}(A_{\min})+1. By (6.25) and (6.26), to prove (6.19), it suffices to find vker(𝒢(η2u,u))v\in\ker\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{\eta^{2}}{u},u\Big{)}\Big{)} and vker(𝒢(t,u))v\notin\ker\mathcal{F}(\mathcal{G}(t,u)). We will check below, that v3v_{3}, defined by

v3=v1ηuv2=(1𝟎1,k1ηu(v~3)T)T(k+1)(k+2)2,v_{3}=v_{1}-\frac{\eta}{u}v_{2}=\begin{pmatrix}1&\mathbf{0}_{1,k-1}&-\frac{\eta}{u}&(\tilde{v}_{3})^{T}\end{pmatrix}^{T}\in\mathbb{R}^{\frac{(k+1)(k+2)}{2}},

is such a vector. This follows by

(𝒢(η2u,u))v3=𝟎(k+1)(k+2)2,1\mathcal{F}\Big{(}\mathcal{G}\Big{(}\frac{\eta^{2}}{u},u\Big{)}\Big{)}v_{3}=\mathbf{0}_{\frac{(k+1)(k+2)}{2},1}

and

(v3)T(𝒢(t,u))v3=tη2u>0.(v_{3})^{T}\mathcal{F}(\mathcal{G}(t,u))v_{3}=t-\frac{\eta^{2}}{u}>0.

This concludes the proof of Claim 1. \blacksquare\\

Note that

(𝒢(𝐭,𝐮))\displaystyle\mathcal{H}(\mathcal{G}(\mathbf{t},\mathbf{u})) = (\@arstrut1X(1,-k1)Xk\\1β0,0-(Amin)1,1-t(h12)Tβk,0-(Amin)2,k\\[0.2em](X(1,-k1))Th12H22h23\\[0.2em]Xkβk,0-(Amin)2,k(h23)Tβ2k,0-(Amin)+k1,+k1-u) .\displaystyle=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(1,k-1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}-\mathbf{t}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{2,k}\\[0.2em](\vec{X}^{(1,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{23}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{2,k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2k,0}-(A_{\min})_{k+1,k+1}-\mathbf{u}$\hfil\kern 5.0pt\crcr}}}}\right)$}}.

Define the matrix function

(6.27) 𝒦(𝐭,𝐮)=(𝒢(𝐭,𝐮))/H22=(A^min)/H22(𝐭00𝐮)=K(𝐭00𝐮)=(k11𝐭k12k12k22𝐮).\displaystyle\begin{split}\mathcal{K}(\mathbf{t},\mathbf{u})=\mathcal{H}(\mathcal{G}(\mathbf{t},\mathbf{u}))\big{/}H_{22}&=\mathcal{H}(\widehat{A}_{\min})\big{/}H_{22}-\begin{pmatrix}\mathbf{t}&0\\ 0&\mathbf{u}\end{pmatrix}\\ &=K-\begin{pmatrix}\mathbf{t}&0\\ 0&\mathbf{u}\end{pmatrix}=\begin{pmatrix}k_{11}-\mathbf{t}&k_{12}\\ k_{12}&k_{22}-\mathbf{u}\end{pmatrix}.\end{split}

Claim 2. Assume that ~(k;β)0\widetilde{\mathcal{M}}(k;\beta)\succeq 0. Then

(6.28) 2={(t,u)2:𝒦(t,u)0}={(t,u)2:tk11,uk22,(k11t)(k22u)k122}.\displaystyle\begin{split}\mathcal{R}_{2}&=\big{\{}(t,u)\in\mathbb{R}^{2}\colon\mathcal{K}(t,u)\succeq 0\big{\}}\\ &=\big{\{}(t,u)\in\mathbb{R}^{2}\colon t\leq k_{11},u\leq k_{22},(k_{11}-t)(k_{22}-u)\geq k_{12}^{2}\big{\}}.\end{split}

If (t,u)2(t,u)\in\mathcal{R}_{2}, we have

(6.32) rank(𝒢(t,u))={rankH22,if k12=0,t=k11,u=k22,rankH22+1,if (k11t)(k22u)=k122,(tk11 or uk22),rankH22+2,if (k11t)(k22u)>k122.\displaystyle\operatorname{rank}\mathcal{H}(\mathcal{G}(t,u))=\left\{\begin{array}[]{rl}\operatorname{rank}H_{22},&\text{if }k_{12}=0,t=k_{11},u=k_{22},\\[3.00003pt] \operatorname{rank}H_{22}+1,&\text{if }(k_{11}-t)(k_{22}-u)=k_{12}^{2},(t\neq k_{11}\text{ or }u\neq k_{22}),\\[3.00003pt] \operatorname{rank}H_{22}+2,&\text{if }(k_{11}-t)(k_{22}-u)>k_{12}^{2}.\end{array}\right.

where AminA_{\min} is as in (6.3).\\

Proof of Claim 2. Permuting rows and columns of (𝒢(𝐭,𝐮))\mathcal{H}(\mathcal{G}(\mathbf{t},\mathbf{u})) we define

~(𝒢(𝐭,𝐮))\displaystyle\widetilde{\mathcal{H}}(\mathcal{G}(\mathbf{t},\mathbf{u})) = (\@arstrut1XkX(1,-k1)\\1β0,0-(Amin)1,1-tβk,0-(Amin)2,k(h12)T\\[0.2em]Xkβk,0-(Amin)2,kβ2k,0-(Amin)+k1,+k1-u(h23)T\\[0.2em](X(1,-k1))Th12h23H22) .\displaystyle=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(1,k-1)}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}-\mathbf{t}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{2,k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12})^{T}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{2,k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2k,0}-(A_{\min})_{k+1,k+1}-\mathbf{u}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{23})^{T}\\[0.2em](\vec{X}^{(1,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{23}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt\crcr}}}}\right)$}}.

Note that

(𝒢(t,u))0~(𝒢(t,u))0\mathcal{H}(\mathcal{G}(t,u))\succeq 0\quad\Leftrightarrow\quad\widetilde{\mathcal{H}}(\mathcal{G}(t,u))\succeq 0

and

(6.34) (Amin)\displaystyle\mathcal{H}(A_{\min}) = (\@arstrut1X(1,-k1)Xk\\1β0,0-(Amin)1,1(h12)Tβk,0-(Amin)1,+k1\\[0.2em](X(1,-k1))Th12H22h23\\[0.2em]Xkβk,0-(Amin)1,+k1(h23)Tβ2k,0-(Amin)+k1,+k1) .\displaystyle=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(1,k-1)}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{1,k+1}\\[0.2em](\vec{X}^{(1,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{23}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{1,k+1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2k,0}-(A_{\min})_{k+1,k+1}$\hfil\kern 5.0pt\crcr}}}}\right)$}}.

By Lemma 4.3.(2), (Amin)0\mathcal{H}(A_{\min})\succeq 0. Permuting rows and columns, this implies that

~(Amin)\displaystyle\widetilde{\mathcal{H}}(A_{\min}) = (\@arstrut1XkX(1,-k1)\\1β0,0-(Amin)1,1βk,0-(Amin)1,+k1(h12)T\\[0.2em]Xkβk,0-(Amin)1,+k1β2k,0-(Amin)+k1,+k1(h23)T\\[0.2em](X(1,-k1))Th12h23H22) 0.\displaystyle=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\vec{X}^{(1,k-1)}\\\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{0,0}-(A_{\min})_{1,1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{1,k+1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{12})^{T}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{k,0}-(A_{\min})_{1,k+1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\beta_{2k,0}-(A_{\min})_{k+1,k+1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(h_{23})^{T}\\[0.2em](\vec{X}^{(1,k-1)})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle h_{23}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle H_{22}$\hfil\kern 5.0pt\crcr}}}}\right)$}}\succeq 0.

By Theorem 2.2, used for (M,C)=(~(Amin),H22)(M,C)=(\widetilde{\mathcal{H}}(A_{\min}),H_{22}), it follows that H220H_{22}\succeq 0 and h12,h23𝒞(H22)h_{12},h_{23}\in\mathcal{C}(H_{22}). Let

:S2Sk+1,(𝐀)=(𝐀((h12)T(h23)T)(h12h23)H22).\mathcal{L}:S_{2}\to S_{k+1},\quad\mathcal{L}(\mathbf{A})=\begin{pmatrix}\mathbf{A}&\left(\begin{array}[]{c}(h_{12})^{T}\\ (h_{23})^{T}\end{array}\right)\\ \left(\begin{array}[]{cc}h_{12}&h_{23}\end{array}\right)&H_{22}\end{pmatrix}.

be a matrix function. Using Theorem 2.2 again for (M,C)=((A),H22)(M,C)=(\mathcal{L}(A),H_{22}), it follows that

(6.35) (A)0A((h12)T(h23)T)(H22)(h12h23)\mathcal{L}(A)\succeq 0\quad\Leftrightarrow\quad A\succeq\begin{pmatrix}(h_{12})^{T}\\ (h_{23})^{T}\end{pmatrix}(H_{22})^{\dagger}\begin{pmatrix}h_{12}&h_{23}\end{pmatrix}

and

(6.36) rank(A)=rankH22+rank(A((h12)T(h23)T)(H22)(h12h23))\operatorname{rank}\mathcal{L}(A)=\operatorname{rank}H_{22}+\operatorname{rank}\Big{(}A-\begin{pmatrix}(h_{12})^{T}\\ (h_{23})^{T}\end{pmatrix}(H_{22})^{\dagger}\begin{pmatrix}h_{12}&h_{23}\end{pmatrix}\Big{)}

Further, (6.35) implies that

~(𝒢(t,u))0\displaystyle\widetilde{\mathcal{H}}(\mathcal{G}(t,u))\succeq 0\quad\Leftrightarrow
(β0,0(Amin)1,1tβk,0(Amin)2,kβk,0(Amin)2,kβ2k,0(Amin)k+1,k+1u)((h12)T(h23)T)(H22)(h12h23)0\displaystyle\Leftrightarrow\quad\begin{pmatrix}\beta_{0,0}-(A_{\min})_{1,1}-t&\beta_{k,0}-(A_{\min})_{2,k}\\ \beta_{k,0}-(A_{\min})_{2,k}&\beta_{2k,0}-(A_{\min})_{k+1,k+1}-u\end{pmatrix}-\begin{pmatrix}(h_{12})^{T}\\ (h_{23})^{T}\end{pmatrix}(H_{22})^{\dagger}\begin{pmatrix}h_{12}&h_{23}\end{pmatrix}\succeq 0
𝒦(t,u)0,\displaystyle\Leftrightarrow\quad\mathcal{K}(t,u)\succeq 0,

where we use the definition (6.27) of 𝒦(t,u)\mathcal{K}(t,u) in the last equivalence. Moreover, rank~(𝒢(t,u))=rankH22+rank𝒦(t,u)\operatorname{rank}\widetilde{\mathcal{H}}(\mathcal{G}(t,u))=\operatorname{rank}H_{22}+\operatorname{rank}\mathcal{K}(t,u). This proves (6.28) and (6.32). \blacksquare\\

Claim 3. If (t,u)2(+)2(t,u)\in\mathcal{R}_{2}\cap(\mathbb{R}_{+})^{2}, then

tu(k11k22sign(k12)k12)2=:pmax.tu\leq(\sqrt{k_{11}k_{22}}-\operatorname{sign}(k_{12})k_{12})^{2}=:p_{\max}.

The equality is achieved if:

  • k12=0k_{12}=0, in the point (t,u)=(k11,k22)(t,u)=(k_{11},k_{22}).

  • k12>0k_{12}>0, in the point (t,u)=(k11k12k11k22,k22+k12k22k11)(t_{-},u_{-})=(k_{11}-\frac{k_{12}\sqrt{k_{11}}}{\sqrt{k_{22}}},k_{22}+\frac{k_{12}\sqrt{k_{22}}}{\sqrt{k_{11}}}).

  • k12<0k_{12}<0, in the point (t+,u+)=(k11+k12k11k22,k22k12k22k11)(t_{+},u_{+})=(k_{11}+\frac{k_{12}\sqrt{k_{11}}}{\sqrt{k_{22}}},k_{22}-\frac{k_{12}\sqrt{k_{22}}}{\sqrt{k_{11}}}).

Moreover, if k120k_{12}\neq 0, then for every p[0,pmax]p\in[0,p_{\max}] there exists a point (t~,u~)2(+)2(\tilde{t},\tilde{u})\in\mathcal{R}_{2}\cap(\mathbb{R}_{+})^{2} such that t~u~=p\tilde{t}\tilde{u}=p and (k11t~)(k22u~)=k122(k_{11}-\tilde{t})(k_{22}-\tilde{u})=k_{12}^{2}.

Proof of Claim 3. If k12=0k_{12}=0, then (t,u)2(+)2=[0,k11]×[0,k22](t,u)\in\mathcal{R}_{2}\cap(\mathbb{R}_{+})^{2}=[0,k_{11}]\times[0,k_{22}] and Claim 3 is clear.

Assume that k120k_{12}\neq 0. Then clearly tutu is maximized in some point satisfying (k11t)(k22u)=k122(k_{11}-t)(k_{22}-u)=k_{12}^{2}. Let f(t):=t(k22k122k11t).f(t):=t\big{(}k_{22}-\frac{k_{12}^{2}}{k_{11}-t}\big{)}. We are searching for the maximum of f(t)f(t) on the interval [0,k11][0,k_{11}]. The stationary points of ff are t±=k11±k12k11k22.t_{\pm}=k_{11}\pm\frac{k_{12}\sqrt{k_{11}}}{\sqrt{k_{22}}}. Then u±=k22k12k22k11u_{\pm}=k_{22}\mp\frac{k_{12}\sqrt{k_{22}}}{\sqrt{k_{11}}}. If k12>0k_{12}>0, then t[0,k11]t_{-}\in[0,k_{11}] (note that k11k22k122k_{11}k_{22}\geq k_{12}^{2} if 2(+)2\mathcal{R}_{2}\cap(\mathbb{R}_{+})^{2}\neq\emptyset). Further on, tu=(k11k22k12)2t_{-}u_{-}=(\sqrt{k_{11}k_{22}}-k_{12})^{2}. Similarly, if k12<0k_{12}<0, then t+[0,k11]t_{+}\in[0,k_{11}] and t+u+=(k11k22+k12)2t_{+}u_{+}=(\sqrt{k_{11}k_{22}}+k_{12})^{2}. The moreover part follows by noticing that f(0)=0f(0)=0 and hence on the interval [0,t±][0,t_{\pm}], ff attains all values between 0 and pmaxp_{\max}. \blacksquare\\

In the proof of Theorem 6.1 we will need a few further observations:

  • Observe that

    (6.37) ((𝒢(t,u)))X(0,k1)=H1tE1,1(k),((𝒢(t,u)))X(1,k1)=H22,((𝒢(t,u)))X(1,k)=H2uEk,k(k).\begin{split}(\mathcal{H}(\mathcal{G}(t,u)))_{\vec{X}^{(0,k-1)}}&=H_{1}-tE_{1,1}^{(k)},\\ (\mathcal{H}(\mathcal{G}(t,u)))_{\vec{X}^{(1,k-1)}}&=H_{22},\\ (\mathcal{H}(\mathcal{G}(t,u)))_{\vec{X}^{(1,k)}}&=H_{2}-uE_{k,k}^{(k)}.\end{split}
  • We have

    (6.38) ((𝒢(t,u)))X(0,k1)/((𝒢(t,u)))X(1,k1)=H1/H22t=t1t,(\mathcal{H}(\mathcal{G}(t,u)))_{\vec{X}^{(0,k-1)}}\big{/}(\mathcal{H}(\mathcal{G}(t,u)))_{\vec{X}^{(1,k-1)}}=H_{1}/H_{22}-t=t_{1}-t,

    where in the first equality we used (6.37) and in the second the definition of t1t_{1} (see (6.9)).

  • We have

    (6.39) ((𝒢(t,u)))X(1,k)/((𝒢(t,u)))X(1,k1)=H2/H22u=u1u,(\mathcal{H}(\mathcal{G}(t,u)))_{\vec{X}^{(1,k)}}\big{/}(\mathcal{H}(\mathcal{G}(t,u)))_{\vec{X}^{(1,k-1)}}=H_{2}/H_{22}-u=u_{1}-u,

    where in the first equality we used (6.37) and in the second the definition of u1u_{1} (see (6.9)).

First we prove the implication (6.12)Theorem 6.1.(2)\eqref{151123-0822}\Rightarrow\text{Theorem }\ref{131023-0847}.\eqref{131023-0847-pt3}. By the necessary conditions for the existence of a 𝒵(p)\mathcal{Z}(p)–rm [CF04, Fia95, CF96], ~(k;β)\widetilde{\mathcal{M}}(k;\beta) must be psd and the relations (6.8) must hold. By Lemma 4.3.(2), (Amin)0\mathcal{F}(A_{\min})\succeq 0. Hence,

(6.40) F^=P^(Amin)(P^)T= (\@arstrut1^T{1,Xk}XkC^T\\[0.2em]1(Amin)1,1(f12)T(Amin)1,+k1(f14)T\\[0.2em](^T{1,Xk})Tf12F22f23F24\\[0.2em]Xk(Amin)1,+k1(f23)T(Amin)+k1,+k1(f34)T\\[0.2em]C^Tf14(F24)Tf34F44) 0.,\displaystyle\begin{split}&\widehat{F}=\widehat{P}\mathcal{F}(A_{\min})(\widehat{P})^{T}\\ &=\hbox{}\vbox{\kern 0.86108pt\hbox{$\kern 0.0pt\kern 2.5pt\kern-5.0pt\left(\kern 0.0pt\kern-2.5pt\kern-6.66669pt\vbox{\kern-0.86108pt\vbox{\vbox{ \halign{\kern\arraycolsep\hfil\@arstrut$\kbcolstyle#$\hfil\kern\arraycolsep& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep&& \kern\arraycolsep\hfil$\@kbrowstyle#$\ifkbalignright\relax\else\hfil\fi\kern\arraycolsep\cr 5.0pt\hfil\@arstrut$\displaystyle$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\widehat{\mathcal{T}}\setminus\{\mathit{1},X^{k}\}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle\mathcal{C}\setminus\widehat{\mathcal{T}}\\[0.2em]\mathit{1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{1,1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{12})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{1,k+1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{14})^{T}\\[0.2em](\widehat{\mathcal{T}}\setminus\{\mathit{1},X^{k}\})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{12}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle F_{22}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{23}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle F_{24}\\[0.2em]X^{k}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{1,k+1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{23})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(A_{\min})_{k+1,k+1}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(f_{34})^{T}\\[0.2em]\mathcal{C}\setminus\widehat{\mathcal{T}}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{14}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle(F_{24})^{T}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle f_{34}$\hfil\kern 5.0pt&5.0pt\hfil$\displaystyle F_{44}$\hfil\kern 5.0pt\crcr}}}}\right)$}}\succeq 0.,\end{split}

where P^\widehat{P} is as in (6.6). In particular, F220F_{22}\succeq 0. We separate two cases according to the invertibility of F22F_{22}.\\

Case 1: F22F_{22} is not pd. Let β(c)\beta^{(c)} be a sequence corresponding to the moment matrix (𝒢(t0,u0))\mathcal{F}(\mathcal{G}(t_{0},u_{0})). Let γ=(γ0,,γ4k)\gamma=(\gamma_{0},\ldots,\gamma_{4k}) be a sequence defined by γi=βi2,imod 2(c)\gamma_{i}=\beta^{(c)}_{\lfloor\frac{i}{2}\rfloor,i\;\text{mod}\;2}. Note that

(^(𝒢(t0,u0)))𝒯^{1,Xk}=(F^)𝒯^{1,Xk}=F22=Aγ^,\big{(}\widehat{\mathcal{F}}(\mathcal{G}(t_{0},u_{0}))\big{)}_{\widehat{\mathcal{T}}\setminus\{\mathit{1},X^{k}\}}=(\widehat{F})_{\widehat{\mathcal{T}}\setminus\{\mathit{1},X^{k}\}}=F_{22}=A_{\widehat{\gamma}},

where γ^=(γ2,,γ4k2)\widehat{\gamma}=(\gamma_{2},\ldots,\gamma_{4k-2}). Since F22F_{22} is not pd, it follows that there is a non-trivial column relation in F22F_{22}, which is also a column relation in AγA_{\gamma} by Proposition 2.3. By Theorem 2.7, γ\gamma has a \mathbb{R}–rm, which implies by Theorem 2.5, that AγA_{\gamma} is rg. Hence, the last column of Aγ=^(𝒢(t0,u0))A_{\gamma}=\widehat{\mathcal{F}}(\mathcal{G}(t_{0},u_{0})) is in the span of the columns in 𝒯^{1,Xk}\widehat{\mathcal{T}}\setminus\{1,X^{k}\}. It follows that

(6.41) ((f12)TF22(f23)T)(F22)f23=((Amin)2,kf23(Amin)k+1,k+1+u0).\begin{pmatrix}(f_{12})^{T}\\ F_{22}\\ (f_{23})^{T}\end{pmatrix}(F_{22})^{\dagger}f_{23}=\begin{pmatrix}(A_{\min})_{2,k}\\[1.99997pt] f_{23}\\[1.99997pt] (A_{\min})_{k+1,k+1}+u_{0}\end{pmatrix}.

On the other hand, by construction of F^\widehat{F}, the column XkX^{k} is also in the span of the columns in 𝒯^{1,Xk}\widehat{\mathcal{T}}\setminus\{1,X^{k}\}. Hence,

(6.42) ((f12)TF22(f23)T)(F22)f23=((Amin)1,k+1f23(Amin)k+1,k+1).\begin{pmatrix}(f_{12})^{T}\\ F_{22}\\ (f_{23})^{T}\end{pmatrix}(F_{22})^{\dagger}f_{23}=\begin{pmatrix}(A_{\min})_{1,k+1}\\[1.99997pt] f_{23}\\[1.99997pt] (A_{\min})_{k+1,k+1}\end{pmatrix}.

By (6.41) and (6.42), it follows that (Amin)2,k=(Amin)1,k+1(A_{\min})_{2,k}=(A_{\min})_{1,k+1} or equivalently η=0\eta=0, and u0=0u_{0}=0. Note that

(6.43) ^(𝒢(t0,u0))=^(𝒢(t0,0))^(𝒢(0,0))=(Amin),(𝒢(t0,u0))=(𝒢(t0,0))(𝒢(0,0))=(Amin).\displaystyle\begin{split}\widehat{\mathcal{F}}(\mathcal{G}(t_{0},u_{0}))=\widehat{\mathcal{F}}(\mathcal{G}(t_{0},0))&\succeq\widehat{\mathcal{F}}(\mathcal{G}(0,0))=\mathcal{F}(A_{\min}),\\ \mathcal{H}(\mathcal{G}(t_{0},u_{0}))=\mathcal{H}(\mathcal{G}(t_{0},0))&\preceq\mathcal{H}(\mathcal{G}(0,0))=\mathcal{H}(A_{\min}).\end{split}

Further on, ^(Amin)\widehat{\mathcal{F}}(A_{\min}) has a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm by Theorem 2.7 and (Amin)\mathcal{H}(A_{\min}) by Theorem 2.5. Indeed, the column XkX^{k} of ^(Amin)\widehat{\mathcal{F}}(A_{\min}) is in the span of the others and since (𝒢(t0,0))\mathcal{H}(\mathcal{G}(t_{0},0)) satisfies the conditions in Theorem 2.5, the same holds for (Amin)\mathcal{H}(A_{\min}). But then the property (6.10) holds (note that η=0\eta=0). This is the case Theorem 6.1.(2a). \\

Case 2: F22F_{22} is pd. By Lemma 4.3.(2), (Amin)0\mathcal{H}(A_{\min})\succeq 0 (see (6.34)). In particular, H220H_{22}\succeq 0. We separate two cases according to the invertibility of H22H_{22}.\\

Case 2.1: H22H_{22} is not pd. By (6.39) and Theorem 2.5, it follows that

(6.44) u1=u0.u_{1}=u_{0}.

By (6.14),

(6.45) u00.u_{0}\geq 0.

We separate two cases according to the value of u1u_{1}. \\

Case 2.1.1: u1=0u_{1}=0. By (6.44), it follows that u0=0u_{0}=0. Note that

(6.46) (^(𝒢(t0,u0)))𝒯^{1}=(^(𝒢(t0,0)))𝒯^{1}=(F^)𝒯^{1}.\big{(}\widehat{\mathcal{F}}(\mathcal{G}(t_{0},u_{0}))\big{)}_{\widehat{\mathcal{T}}\setminus\{\mathit{1}\}}=\big{(}\widehat{\mathcal{F}}(\mathcal{G}(t_{0},0))\big{)}_{\widehat{\mathcal{T}}\setminus\{\mathit{1}\}}=\big{(}\widehat{F}\big{)}_{\widehat{\mathcal{T}}\setminus\{\mathit{1}\}}.

Since in F^\widehat{F} we have the column relation (6.42) by construction, (6.46) and Proposition 2.3 imply that

(^(𝒢(t0,0)))𝒯^,𝒯^{1,Xk}(F22)f23=(^(𝒢(t0,0)))𝒯^,{Xk},\big{(}\widehat{\mathcal{F}}(\mathcal{G}(t_{0},0))\big{)}_{\widehat{\mathcal{T}},\widehat{\mathcal{T}}\setminus\{\mathit{1},X^{k}\}}(F_{22})^{\dagger}f_{23}=\big{(}\widehat{\mathcal{F}}(\mathcal{G}(t_{0},0))\big{)}_{\widehat{\mathcal{T}},\{X^{k}\}},

or equivalently (6.41) with u0=0u_{0}=0. By (6.41) and (6.42), it follows that (Amin)2,k=(Amin)1,k+1(A_{\min})_{2,k}=(A_{\min})_{1,k+1} or equivalently η=0\eta=0. This is the case Theorem 6.1.(2(b)i). \\

Case 2.1.2: u1>0u_{1}>0. Since the column XkX^{k} of (𝒢(t0,u1))\mathcal{H}(\mathcal{G}(t_{0},u_{1})) is in the span of the columns in X(1,k1)\vec{X}^{(1,k-1)}, it first follows by observing the first row of (𝒢(t0,u1))\mathcal{H}(\mathcal{G}(t_{0},u_{1})) that

(6.47) βk,0(Amin)2,k=(h12)T(H22)h23.\beta_{k,0}-(A_{\min})_{2,k}=(h_{12})^{T}(H_{22})^{\dagger}h_{23}.

Further on,

(6.48) (𝒢(t,u1))/((𝒢(t,u1)))X(1,k)=((𝒢(t,u1)))X(0,k1)/((𝒢(t,u1)))X(1,k1)=t1t,\mathcal{H}(\mathcal{G}(t,u_{1}))\big{/}(\mathcal{H}(\mathcal{G}(t,u_{1})))_{\vec{X}^{(1,k)}}=(\mathcal{H}(\mathcal{G}(t,u_{1})))_{\vec{X}^{(0,k-1)}}\big{/}(\mathcal{H}(\mathcal{G}(t,u_{1})))_{\vec{X}^{(1,k-1)}}=t_{1}-t,

where we used (6.38) in the second equality. By (6.48) and Theorem 2.2 used for (M,C)=((𝒢(t,u1)),((𝒢(t,u1)))X(1,k))(M,C)=(\mathcal{H}(\mathcal{G}(t,u_{1})),(\mathcal{H}(\mathcal{G}(t,u_{1})))_{\vec{X}^{(1,k)}}), it follows that (𝒢(t1,u1))0\mathcal{H}(\mathcal{G}(t_{1},u_{1}))\succeq 0. By Theorem 2.5, (𝒢(t1,u1))\mathcal{H}(\mathcal{G}(t_{1},u_{1})) admits a \mathbb{R}–rm. Note that

(6.49) ^(𝒢(t0,u0))=^(𝒢(t0,u1))^(𝒢(t1,u1)),\displaystyle\begin{split}\widehat{\mathcal{F}}(\mathcal{G}(t_{0},u_{0}))=\widehat{\mathcal{F}}(\mathcal{G}(t_{0},u_{1}))&\preceq\widehat{\mathcal{F}}(\mathcal{G}(t_{1},u_{1})),\end{split}

where we used that t0t1t_{0}\leq t_{1} by (6.48). By Theorem 2.7, (^(𝒢(t1,u1)))𝒯^{Xk}(\widehat{\mathcal{F}}(\mathcal{G}(t_{1},u_{1})))_{\widehat{\mathcal{T}}\setminus\{X^{k}\}} must be pd. (Here we used that since u1>0u_{1}>0 and F220F_{22}\succ 0, it follows that (^(𝒢(t1,u1)))𝒯^{1}0.(\widehat{\mathcal{F}}(\mathcal{G}(t_{1},u_{1})))_{\widehat{\mathcal{T}}\setminus\{1\}}\succ 0.) Therefore Claim 1 implies that t1>0t_{1}>0 and t1u1η2t_{1}u_{1}\geq\eta^{2}. Together with (6.47), this is the case Theorem 6.1.(2(b)ii). \\

Case 2.2: H22H_{22} is pd. We separate two cases according to the value of η.\eta.\\

Case 2.2.1: η=0\eta=0. By Lemma 4.3.(2), (Amin)0\mathcal{H}(A_{\min})\succeq 0 (see (6.34)).

If (Amin)\mathcal{H}(A_{\min}) does not admit a \mathbb{R}–rm, it follows by Theorem 2.5, that ((Amin))X(0,k1)(\mathcal{H}(A_{\min}))_{\vec{X}^{(0,k-1)}} is not pd and u1>0u_{1}>0. Equivalently,

t1=((Amin))X(0,k1)/H22=0,t_{1}=(\mathcal{H}(A_{\min}))_{\vec{X}^{(0,k-1)}}\big{/}H_{22}=0,

which by (6.38) implies that t0=0t_{0}=0. By Theorem 2.7, since ^(𝒢(t0,u0))=^(𝒢(0,u0))\widehat{\mathcal{F}}(\mathcal{G}(t_{0},u_{0}))=\widehat{\mathcal{F}}(\mathcal{G}(0,u_{0})) admits a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm, F220F_{22}\succ 0 and (^(𝒢(0,u0)))𝒯^{Xk}(\widehat{\mathcal{F}}(\mathcal{G}(0,u_{0})))_{\widehat{\mathcal{T}}\setminus\{X^{k}\}} is not pd, it follows that u0=0u_{0}=0. But then (𝒢(t0,u0))=(𝒢(0,0))=(Amin)\mathcal{H}(\mathcal{G}(t_{0},u_{0}))=\mathcal{H}(\mathcal{G}(0,0))=\mathcal{H}(A_{\min}) does not admit a \mathbb{R}–rm, which is a contradiction.

Hence, (Amin)\mathcal{H}(A_{\min}) admits a \mathbb{R}–rm, which is equivalent to (6.10) (using η=0\eta=0). This is the case Theorem 6.1.(2(c)i). \\

Case 2.2.2: η0\eta\neq 0. By (6.19) it follows that t0u0η2.t_{0}u_{0}\geq\eta^{2}. This fact and Claim 3 imply the second condition in the case Theorem 6.1.(2(c)ii). \\

This concludes the proof of the implication (6.12)Theorem 6.1.(2)\eqref{151123-0822}\Rightarrow\text{Theorem }\ref{131023-0847}.\eqref{131023-0847-pt3}.\\

Next we prove the implication Theorem 6.1.(2)(6.12)\text{Theorem }\ref{131023-0847}.\eqref{131023-0847-pt3}\Rightarrow\eqref{151123-0822}. We separate five cases according to the assumptions in Theorem 6.1.(2)\text{Theorem }\ref{131023-0847}.\eqref{131023-0847-pt3}.\\

Case 1: Theorem 6.1.(2a)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt1} holds. By Lemma 4.3.(2), (Amin)0\mathcal{F}(A_{\min})\succeq 0 and (Amin)0\mathcal{H}(A_{\min})\succeq 0. Since η=0\eta=0, both matrices have a moment structure. Since by construction, the column XkX^{k} of ^(Amin)\widehat{\mathcal{F}}(A_{\min}) is in the span of the others, it has a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm by Theorem 2.7. Since (Amin)\mathcal{H}(A_{\min}) satisfies (6.10) (using η=0\eta=0), it admits a \mathbb{R}–rm by Theorem 2.5. This proves (6.12) in this case. \\

Case 2: Theorem 6.1.(2(b)i)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt2-1} holds. By the same reasoning as in the Case 1 above, ^(Amin)\widehat{\mathcal{F}}(A_{\min}) has a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm. Since u1=0u_{1}=0, the column XkX^{k} of (Amin)\mathcal{H}(A_{\min}) is in the span of the other columns. By Theorem 2.5, (Amin)\mathcal{H}(A_{\min}) admits a \mathbb{R}–rm. This proves (6.12) in this case. \\

Case 3: Theorem 6.1.(2(b)ii)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt2-2} holds. By (6.38), (6.39) and the fourth assumption of (2(b)ii), it follows that (𝒢(t1,u1))\mathcal{H}(\mathcal{G}(t_{1},u_{1})) is psd and the columns 1,Xk1,X^{k} are in the span of the columns in X(1,k1)\vec{X}^{(1,k-1)}. By Theorem 2.5, (𝒢(t1,u1))\mathcal{H}(\mathcal{G}(t_{1},u_{1})) admits a \mathbb{R}–rm. Since (t1,u1)1(t_{1},u_{1})\in\mathcal{R}_{1} by (6.14) and the assumptions in (2(b)ii), it follows that F^(𝒢(t1,u1))\widehat{F}(\mathcal{G}(t_{1},u_{1})) is psd and by construction, (F^(𝒢(t1,u1)))𝒯^{Xk}\big{(}\widehat{F}(\mathcal{G}(t_{1},u_{1}))\big{)}_{\widehat{\mathcal{T}}\setminus\{X^{k}\}} is pd. By Theorem 2.7, it has a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm. This proves (6.12) in this case. \\

Case 4: Theorem 6.1.(2(c)i)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt3-1} holds. ^(Amin)\widehat{\mathcal{F}}(A_{\min}) has a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm and (Amin)\mathcal{H}(A_{\min}) has a \mathbb{R}–rm by the same reasoning as in the Case 1 above. This proves (6.12) in this case.\\

Case 5: Theorem 6.1.(2(c)ii)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt3-2} holds. We separate three cases according to the sign of k12k_{12}.

  • If k12=0k_{12}=0, then by Claim 2, (𝒢(k11,k22))\mathcal{H}(\mathcal{G}(k_{11},k_{22})) is psd and the column XkX^{k} is in the span of the previous ones. Since (𝒢(0,0))=(A^min)\mathcal{H}(\mathcal{G}(0,0))=\mathcal{H}(\widehat{A}_{\min}) is psd by assumption, it follows that k110k_{11}\geq 0 and k220k_{22}\geq 0. Since η0\eta\neq 0 and k11k22η2k_{11}k_{22}\geq\eta^{2} by (6.11), it follows that k11>0k_{11}>0 and k22>0k_{22}>0. By Claim 1, ^(𝒢(k11,k22))0\widehat{\mathcal{F}}(\mathcal{G}(k_{11},k_{22}))\succ 0. By Theorem 2.7, it has a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm. This proves (6.12) in this case.

  • If k12>0k_{12}>0, then by Claim 3, (𝒢(t,u))\mathcal{H}(\mathcal{G}(t_{-},u_{-})) is psd and tuη2t_{-}u_{-}\geq\eta^{2}. By construction, rank(𝒢(t,u))=k\operatorname{rank}\mathcal{H}(\mathcal{G}(t_{-},u_{-}))=k and since t<k11t_{-}<k_{11}, it follows that ((𝒢(t,u)))X(0,k1)(\mathcal{H}(\mathcal{G}(t_{-},u_{-})))_{\vec{X}^{(0,k-1)}} is pd. Hence, the column XkX^{k} of (𝒢(t,u))\mathcal{H}(\mathcal{G}(t_{-},u_{-})) is in the span of the others. By Theorem 2.5, (𝒢(t,u))\mathcal{H}(\mathcal{G}(t_{-},u_{-})) admits a \mathbb{R}–rm. By Claim 1 and tuη2t_{-}u_{-}\geq\eta^{2}, it follows that ^(𝒢(t,u))0\widehat{\mathcal{F}}(\mathcal{G}(t_{-},u_{-}))\succeq 0. Since t>0t_{-}>0, it follows that (F^(𝒢(t,u)))𝒯^{Xk}\big{(}\widehat{F}(\mathcal{G}(t_{-},u_{-}))\big{)}_{\widehat{\mathcal{T}}\setminus\{X^{k}\}} is pd. By Theorem 2.7, it has a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm. This proves (6.12) in this case.

  • If k12<0k_{12}<0, then the proof of (6.12) is analogous to the case k12>0k_{12}>0 by replacing (t,u)(t_{-},u_{-}) with (t+,u+)(t_{+},u_{+}).\\

This concludes the proof of the implication Theorem 6.1.(2)(6.12).\text{Theorem }\ref{131023-0847}.\eqref{131023-0847-pt3}\Rightarrow\eqref{151123-0822}.\\

By now we established the equivalence (1)(2)\eqref{131023-0847-pt1}\Leftrightarrow\eqref{131023-0847-pt3} in Theorem 6.1. It remains to prove the moreover part. We observe again the proof of the implication (2)(6.12).\eqref{131023-0847-pt3}\Rightarrow\eqref{151123-0822}. By Lemma 4.3.(4),

(6.50) rank~(k;β)=rank^(Amin)+rank(Amin).\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)=\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+\operatorname{rank}\mathcal{H}(A_{\min}).

In the proofs of the implications Theorem 6.1.(2a)(6.12)\ref{131023-0847}.\eqref{parabolic-pt1}\Rightarrow\eqref{151123-0822}, Theorem 6.1.(2(b)i)(6.12)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt2-1}\Rightarrow\eqref{151123-0822} and Theorem 6.1.(2(c)i)(6.12)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt3-1}\Rightarrow\eqref{151123-0822}, we established that ^(Amin)\widehat{\mathcal{F}}(A_{\min}) and (Amin)\mathcal{H}(A_{\min}) admit a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm and a \mathbb{R}–rm, respectively. By Theorems 2.5 and 2.7, there also exist a (rank^(Amin))(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min}))–atomic and a (rank(Amin))(\operatorname{rank}\mathcal{H}(A_{\min}))–atomic rms. By (6.50), β\beta has a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm.

Assume that Theorem 6.1.(2(b)ii)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt2-2} holds. We separate two cases according to the value of η\eta:

  • η=0\eta=0. We separate two cases according to the existence of a \mathbb{R}–rm of (Amin)\mathcal{H}(A_{\min}):

    • The last column of (Amin)\mathcal{H}(A_{\min}) is in the span of the previous ones. Then as in the previous paragraph, ^(Amin)\widehat{\mathcal{F}}(A_{\min}) and (Amin)\mathcal{H}(A_{\min}) admit a (rank^(Amin))(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min}))–atomic 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm and a (rank(Amin))(\operatorname{rank}\mathcal{H}(A_{\min}))–atomic \mathbb{R}–rm, respectively. Hence, β\beta has a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm.

    • The last column of (Amin)\mathcal{H}(A_{\min}) is not in the span of the previous ones. Since also t1>0t_{1}>0, it follows that rank(Amin)=rankH22+2\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{22}+2. But then rank(𝒢(t1,u1))=rankH22\operatorname{rank}\mathcal{H}(\mathcal{G}(t_{1},u_{1}))=\operatorname{rank}H_{22} and rank^(𝒢(t1,u1))=rank^(Amin)+2\operatorname{rank}\widehat{\mathcal{F}}(\mathcal{G}(t_{1},u_{1}))=\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+2 (see (6.19)). This implies that ~(β;k)\widetilde{\mathcal{M}}(\beta;k) admits a (rank~(k);β)(\operatorname{rank}\widetilde{\mathcal{M}}(k);\beta)–atomic 𝒵(p)\mathcal{Z}(p)–rm.

  • η0\eta\neq 0. We separate two cases according to rank(Amin)\operatorname{rank}\mathcal{H}(A_{\min}), which can be either rankH22+2\operatorname{rank}H_{22}+2 or rankH22+1\operatorname{rank}H_{22}+1 (since t1>0t_{1}>0).

    • rank(Amin)=rankH22+2\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{22}+2. Then as in the second Case of the case η=0\eta=0 above, in the point (t1,u1)(t_{1},u_{1}) there is a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm for β\beta. (Note that t1u1t_{1}u_{1} is automatically strictly larger than η2\eta^{2}, otherwise the measure was (rank~(k;β)1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)-1)–atomic, which is not possible.)

    • rank(Amin)=rankH22+1\operatorname{rank}\mathcal{H}(A_{\min})=\operatorname{rank}H_{22}+1. In this case we have

      rank(𝒢(t1,u1))+rank^(𝒢(t1,u1))\displaystyle\operatorname{rank}\mathcal{H}(\mathcal{G}(t_{1},u_{1}))+\operatorname{rank}\widehat{\mathcal{F}}(\mathcal{G}(t_{1},u_{1})) =rankH22+rank^(𝒢(t1,u1))\displaystyle=\operatorname{rank}H_{22}+\operatorname{rank}\widehat{\mathcal{F}}(\mathcal{G}(t_{1},u_{1}))
      ={rankH22+rank^(Amin)+1,if t1u1=η2,rankH22+rank^(Amin)+2,if t1u1>η2,\displaystyle=\left\{\begin{array}[]{rl}\operatorname{rank}H_{22}+\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1,&\text{if }t_{1}u_{1}=\eta^{2},\\[3.00003pt] \operatorname{rank}H_{22}+\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+2,&\text{if }t_{1}u_{1}>\eta^{2},\end{array}\right.
      ={rank~(k;β),if t1u1=η2,rank~(k;β)+1,if t1u1>η2,\displaystyle=\left\{\begin{array}[]{rl}\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta),&\text{if }t_{1}u_{1}=\eta^{2},\\[3.00003pt] \operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1,&\text{if }t_{1}u_{1}>\eta^{2},\end{array}\right.

      where we used (6.19) in the second and (6.50) in the third equality. Hence, β\beta has a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic rm if t1u1=η2t_{1}u_{1}=\eta^{2} and (rank~(k;β)+1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1)–atomic rm if t1u1>η2t_{1}u_{1}>\eta^{2}. It remains to show that in the case t1u1>η2t_{1}u_{1}>\eta^{2}, there does not exist a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic rm. Since H22H_{22} is not pd and u1>0u_{1}>0, if (𝒢(t,u))\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime})) has a \mathbb{R}–rm, then u=u1u^{\prime}=u_{1}. Since η0\eta\neq 0, then ^(𝒢(t,u1))\widehat{\mathcal{F}}(\mathcal{G}(t^{\prime},u_{1})) with a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm is at least (rank^(Amin)+1)(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1)–atomic (see (6.19)). If tt1t^{\prime}\neq t_{1}, then rank(𝒢(t,u1))=rankH22+1\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u_{1}))=\operatorname{rank}H_{22}+1. Hence,

      rank(𝒢(t,u1))+rank^(𝒢(t,u1))(rankH22+1)+(rank^(Amin)+1)=rank~(k;β)+1,\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u_{1}))+\operatorname{rank}\widehat{\mathcal{F}}(\mathcal{G}(t^{\prime},u_{1}))\geq(\operatorname{rank}H_{22}+1)+(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1)=\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1,

      where we used (6.50) in the last equality.

  • Assume that Theorem 6.1.(2(c)ii)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt3-2} holds. We separate two cases according to the value of k12k_{12}.

    • k12=0k_{12}=0. We separate two cases according to rank(Amin)\operatorname{rank}\mathcal{H}(A_{\min}), i.e., rank(Amin){k,k+1}\operatorname{rank}\mathcal{H}(A_{\min})\in\{k,k+1\}. Note that rank(Amin)\operatorname{rank}\mathcal{H}(A_{\min}) cannot be k1k-1, since η0\eta\neq 0 and k12=0k_{12}=0 imply that ((Amin)/H22)120\big{(}\mathcal{H}(A_{\min})/H_{22}\big{)}_{12}\neq 0.

      • *

        rank(Amin)=k+1\operatorname{rank}\mathcal{H}(A_{\min})=k+1. Then as in the second case of the case η=0\eta=0 of Theorem 6.1.(2(b)ii)\text{Theorem }\ref{131023-0847}.\eqref{parabolic-pt2-2} above, in the point (t1,u1)(t_{1},u_{1}) there is a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm for β\beta. (Note that t1u1t_{1}u_{1} is automatically strictly larger than η2\eta^{2}, otherwise the measure was (rank~(k;β)1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)-1)–atomic, which is not possible.)

      • *

        rank(Amin)=k\operatorname{rank}\mathcal{H}(A_{\min})=k. In this case we have

        rank(𝒢(k11,k22))+rank(𝒢(k11,k22))\displaystyle\operatorname{rank}\mathcal{H}(\mathcal{G}(k_{11},k_{22}))+\operatorname{rank}\mathcal{F}(\mathcal{G}(k_{11},k_{22})) ={rankH22+rank(Amin)+1,if k11k22=η2,rankH22+rank(Amin)+2,if k11k22>η2,\displaystyle=\left\{\begin{array}[]{rl}\operatorname{rank}H_{22}+\operatorname{rank}\mathcal{F}(A_{\min})+1,&\text{if }k_{11}k_{22}=\eta^{2},\\[3.00003pt] \operatorname{rank}H_{22}+\operatorname{rank}\mathcal{F}(A_{\min})+2,&\text{if }k_{11}k_{22}>\eta^{2},\end{array}\right.
        ={rank~(k;β),if k11k22=η2,rank~(k;β)+1,if k11k22>η2,\displaystyle=\left\{\begin{array}[]{rl}\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta),&\text{if }k_{11}k_{22}=\eta^{2},\\[3.00003pt] \operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1,&\text{if }k_{11}k_{22}>\eta^{2},\end{array}\right.

        where we used (6.19) in the first and (6.50) in the second equality. Hence, β\beta has a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic rm if k11k22=η2k_{11}k_{22}=\eta^{2} and (rank~(k;β)+1)(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1)–atomic rm if k11k22>η2k_{11}k_{22}>\eta^{2}. It remains to show that in the case k11k22>η2k_{11}k_{22}>\eta^{2}, there does not exist a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic rm. Since η0\eta\neq 0, if (𝒢(t,u))\mathcal{F}(\mathcal{G}(t^{\prime},u^{\prime})) is psd, it follows that tuη2t^{\prime}u^{\prime}\geq\eta^{2} by (6.14). But then if ^(𝒢(t,u))\widehat{\mathcal{F}}(\mathcal{G}(t^{\prime},u^{\prime})) also admits a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm, this rm is at least (rank^(Amin)+1)(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1)–atomic (see (6.19)). If t<k11t^{\prime}<k_{11} or u<k22u^{\prime}<k_{22}, then rank(𝒢(t,u))rankH22+1\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime}))\geq\operatorname{rank}H_{22}+1. Hence,

        rank(𝒢(t,u))+rank^(𝒢(t,u))(rankH22+1)+(rank^(Amin)+1)=rank~(k;β)+1,\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime}))+\operatorname{rank}\widehat{\mathcal{F}}(\mathcal{G}(t^{\prime},u^{\prime}))\geq(\operatorname{rank}H_{22}+1)+(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1)=\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1,

        where we used (6.50) in the last equality.

    • k120k_{12}\neq 0. We separate two cases according to rank(Amin)\operatorname{rank}\mathcal{H}(A_{\min}), i.e. rank(Amin){k,k+1}\operatorname{rank}\mathcal{H}(A_{\min})\in\{k,k+1\}. Note that rank(Amin)\operatorname{rank}\mathcal{H}(A_{\min}) cannot be k1k-1, since otherwise (A^min)/H22=(0ηη0)\mathcal{H}(\widehat{A}_{\min})/H_{22}=\begin{pmatrix}0&\eta\\ \eta&0\end{pmatrix}, which cannot be psd by η0\eta\neq 0. By Claim 3, there is a point (t~,u~)2(+)2(\tilde{t},\tilde{u})\in\mathcal{R}_{2}\cap(\mathbb{R}_{+})^{2}, such that t~u~=η2\tilde{t}\tilde{u}=\eta^{2} and (k11t~)(k22u~)=k122(k_{11}-\tilde{t})(k_{22}-\tilde{u})=k_{12}^{2}. By (6.19) and (6.32) we have

      rank(𝒢(t~,u~))+rank^(𝒢(t~,u~))\displaystyle\operatorname{rank}\mathcal{H}(\mathcal{G}(\tilde{t},\tilde{u}))+\operatorname{rank}\widehat{\mathcal{F}}(\mathcal{G}(\tilde{t},\tilde{u})) =(rankH22+1)+(rank^(Amin)+1)\displaystyle=(\operatorname{rank}H_{22}+1)+(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1)
      ={rank~(k;β),if rank(Amin)=k+1,rank~(k;β)+1,if rank(Amin)=k,\displaystyle=\left\{\begin{array}[]{rl}\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta),&\text{if }\operatorname{rank}\mathcal{H}(A_{\min})=k+1,\\[3.00003pt] \operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1,&\text{if }\operatorname{rank}\mathcal{H}(A_{\min})=k,\end{array}\right.

      where we used (6.50) in the second equality. It remains to show that in the case rank(Amin)=k\operatorname{rank}\mathcal{H}(A_{\min})=k, there does not exist a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic rm. Since η0\eta\neq 0, if ^(𝒢(t,u))\widehat{\mathcal{F}}(\mathcal{G}(t^{\prime},u^{\prime})) is psd, it follows that tuη2t^{\prime}u^{\prime}\geq\eta^{2} by (6.14). But then if ^(𝒢(t,u))\widehat{\mathcal{F}}(\mathcal{G}(t^{\prime},u^{\prime})) also admits a 𝒵(xy2)\mathcal{Z}(x-y^{2})–rm, this rm is at least (rank^(Amin)+1)(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1)–atomic (see (6.19)). Since k120k_{12}\neq 0, rank(𝒢(t,u))rankH22+1\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime}))\geq\operatorname{rank}H_{22}+1 by (6.32). Hence,

      rank(𝒢(t,u))+rank^(𝒢(t,u))(rankH22+1)+(rank^(Amin)+1)=rank~(k;β)+1,\operatorname{rank}\mathcal{H}(\mathcal{G}(t^{\prime},u^{\prime}))+\operatorname{rank}\widehat{\mathcal{F}}(\mathcal{G}(t^{\prime},u^{\prime}))\geq(\operatorname{rank}H_{22}+1)+(\operatorname{rank}\widehat{\mathcal{F}}(A_{\min})+1)=\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta)+1,

      where we used (6.50) in the last equality.

This concludes the proof of the moreover part.

Since for a pp–pure sequence with ~(k;β))0\widetilde{\mathcal{M}}(k;\beta))\succeq 0, (6.50) implies that (Amin)\mathcal{H}(A_{\min}) is pd, it follows by the moreover part that the existence of a 𝒵(p)\mathcal{Z}(p)–rm implies the existence of a (rank~(k;β))(\operatorname{rank}\widetilde{\mathcal{M}}(k;\beta))–atomic 𝒵(p)\mathcal{Z}(p)–rm. ∎

The following example demonstrates the use of Theorem 6.1 to show that there exists a bivariate y(xy2)y(x-y^{2})–pure sequence β\beta of degree 6 with a positive semidefinite (3)\mathcal{M}(3) and without a 𝒵(y(xy2))\mathcal{Z}(y(x-y^{2}))–rm.

Example 6.3.

Let β\beta be a bivariate degree 6 sequence given by

β00\displaystyle\beta_{00} =12281531372615,\displaystyle=\frac{1228153}{1372615}, β10\displaystyle\beta_{10} =9710,\displaystyle=\frac{97}{10}, β01\displaystyle\beta_{01} =2110,\displaystyle=\frac{21}{10},
β20\displaystyle\beta_{20} =228910,\displaystyle=\frac{2289}{10}, β11\displaystyle\beta_{11} =44110,\displaystyle=\frac{441}{10}, β02\displaystyle\beta_{02} =9110,\displaystyle=\frac{91}{10},
β30\displaystyle\beta_{30} =6720710,\displaystyle=\frac{67207}{10}, β21\displaystyle\beta_{21} =1220110,\displaystyle=\frac{12201}{10}, β12\displaystyle\beta_{12} =4552,\displaystyle=\frac{455}{2},
β03\displaystyle\beta_{03} =44110,\displaystyle=\frac{441}{10}, β40\displaystyle\beta_{40} =214269310,\displaystyle=\frac{2142693}{10}, β31\displaystyle\beta_{31} =37676110,\displaystyle=\frac{376761}{10},
β22\displaystyle\beta_{22} =6717110,\displaystyle=\frac{67171}{10}, β13\displaystyle\beta_{13} =1220110,\displaystyle=\frac{12201}{10}, β04\displaystyle\beta_{04} =4552,\displaystyle=\frac{455}{2},
β50\displaystyle\beta_{50} =7134072710,\displaystyle=\frac{71340727}{10}, β41\displaystyle\beta_{41} =1231316110,\displaystyle=\frac{12313161}{10}, β32\displaystyle\beta_{32} =4285192,\displaystyle=\frac{428519}{2},
β23\displaystyle\beta_{23} =37676110,\displaystyle=\frac{376761}{10}, β14\displaystyle\beta_{14} =6717110,\displaystyle=\frac{67171}{10}, β05\displaystyle\beta_{05} =1220110,\displaystyle=\frac{12201}{10},
β60\displaystyle\beta_{60} =243823650910,\displaystyle=\frac{2438236509}{10}, β51\displaystyle\beta_{51} =41599868110,\displaystyle=\frac{415998681}{10}, β42\displaystyle\beta_{42} =7134045110,\displaystyle=\frac{71340451}{10},
β33\displaystyle\beta_{33} =1231316110,\displaystyle=\frac{12313161}{10}, β24\displaystyle\beta_{24} =4285192,\displaystyle=\frac{428519}{2}, β15\displaystyle\beta_{15} =37676110,\displaystyle=\frac{376761}{10},
β06\displaystyle\beta_{06} =6717110.\displaystyle=\frac{67171}{10}.

Assume the notation as in Theorem 6.1. ~(3)\widetilde{\mathcal{M}}(3) is psd with the eigenvalues 2.51108\approx 2.51\cdot 10^{8}, 47179\approx 47179, 112.1\approx 112.1, 7.4\approx 7.4, 1.11\approx 1.11, 0.1\approx 0.1, 0.03\approx 0.03, 0.0005\approx 0.0005, 4.9106\approx 4.9\cdot 10^{-6}, 0, and the column relation Y3=YXY^{3}=YX. We have that

Amin=(553792309110455261999553923091104552671711042851924552671711042851927134045110619995539230428519271340451104500982093091846)A_{\min}=\begin{pmatrix}\frac{5537}{9230}&\frac{91}{10}&\frac{455}{2}&\frac{61999553}{9230}\\[5.0pt] \frac{91}{10}&\frac{455}{2}&\frac{67171}{10}&\frac{428519}{2}\\[5.0pt] \frac{455}{2}&\frac{67171}{10}&\frac{428519}{2}&\frac{71340451}{10}\\[5.0pt] \frac{61999553}{9230}&\frac{428519}{2}&\frac{71340451}{10}&\frac{450098209309}{1846}\end{pmatrix}

and so

η=6717110619995539230=72923.\eta=\frac{67171}{10}-\frac{61999553}{9230}=-\frac{72}{923}.

The matrices F22F_{22} and H22H_{22} are equal to:

F22\displaystyle F_{22} =(9110441104552122011067171104411045521220110671711037676110455212201106717110376761104285192122011067171103767611042851921231316110671711037676110428519212313161107134045110),H22=(75185185495).\displaystyle=\begin{pmatrix}\frac{91}{10}&\frac{441}{10}&\frac{455}{2}&\frac{12201}{10}&\frac{67171}{10}\\[5.0pt] \frac{441}{10}&\frac{455}{2}&\frac{12201}{10}&\frac{67171}{10}&\frac{376761}{10}\\[5.0pt] \frac{455}{2}&\frac{12201}{10}&\frac{67171}{10}&\frac{376761}{10}&\frac{428519}{2}\\[5.0pt] \frac{12201}{10}&\frac{67171}{10}&\frac{376761}{10}&\frac{428519}{2}&\frac{12313161}{10}\\[5.0pt] \frac{67171}{10}&\frac{376761}{10}&\frac{428519}{2}&\frac{12313161}{10}&\frac{71340451}{10}\end{pmatrix},\qquad H_{22}=\begin{pmatrix}\frac{7}{5}&\frac{18}{5}\\[5.0pt] \frac{18}{5}&\frac{49}{5}\end{pmatrix}.

They are both pd with the eigenvalues 7.3106\approx 7.3\cdot 10^{6}, 1987.6\approx 1987.6, 5.6\approx 5.6, 0.099\approx 0.099, 0.0013\approx 0.0013 and 11.1\approx 11.1, 0.068\approx 0.068, respectively. The matrix KK is equal to

K=(k11k12k12k22)=(605032948143098510395395494141487685)K=\begin{pmatrix}k_{11}&k_{12}\\ k_{12}&k_{22}\end{pmatrix}=\begin{pmatrix}\frac{6050329}{48143098510}&\frac{3}{95}\\[1.99997pt] \frac{3}{95}&\frac{4941414}{87685}\end{pmatrix}

and thus

(6.51) (k11k12k12)2η2=0.0033<0.(\sqrt{k_{11}k_{12}}-k_{12})^{2}-\eta^{2}=-0.0033<0.

By Theorem 6.1, β\beta does not have a 𝒵(y(xy2))\mathcal{Z}(y(x-y^{2}))–rm, since by (2(c)ii) of Theorem 6.1, (6.51) should be positive.

References

  • [Alb69] A. Albert, Conditions for positive and nonnegative definiteness in terms of pseudoinverses, SIAM J. Appl. Math. 17 (1969), 434–440.
  • [Akh65] N.I. Akhiezer. The classical moment problem and some related questions in analysis, New York: Hafner Publishing Co., 1965.
  • [AK62] N.I. Akhiezer, M. Krein. Some questions in the theory of moments, Transl. Math. Monographs 2. Providence: American Math. Soc., 1962.
  • [BZ21] A. Bhardwaj, A. Zalar. The tracial moment problem on quadratic varieties, J. Math. Anal. Appl.  498 (2021). Available from: https://doi.org/10.1016/j.jmaa.2021.124936.
  • [BW11] M. Bakonyi, H.J. Woerdeman, Matrix Completions, Moments, and Sums of Hermitian Squares, Princeton University Press, Princeton, 2011.
  • [Ble15] G. Blekherman. Positive Gorenstein ideals, Proc. Amer. Math. Soc. 143 (2015) 69–86. Available from: https://doi.org/10.1090/S0002-9939-2014-12253-2.
  • [BF20] G. Blekherman, L. Fialkow. The core variety and representing measures in the truncated moment problem, Journal of Operator Theory 84 (2020) 185–209.
  • [CGIK+] R. Curto, M. Ghasemi, M. Infusino, S. Kuhlmann. The truncated moment problems for unital commutative \mathbb{R}-algebras, Journal of Operathor Theory, to appear. Available from: https://arxiv.org/pdf/2009.05115.pdf.
  • [CHM74] D. Carlson, E. Haynsworth, T. Markham, A generalization of the Schur complement by means of the Moore-Penrose inverse, SIAM J. Appl. Math.. 26(1) (1974) 169–175.
  • [CH69] D. Crabtree, E. Haynsworth, An identity for the Schur complement of a matrix. Proc. Am. Math. Soc. 22 (1969) 364–366.
  • [CF91] R. Curto, L. Fialkow, Recursiveness, positivity, and truncated moment problems, Houston J. Math. 17 (1991) 603–635.
  • [CF96] R. Curto, L. Fialkow, Solution of the truncated complex moment problem for flat data, Mem. Amer. Math. Soc. 119 (1996).
  • [CF02] R. Curto, L. Fialkow, Solution of the singular quartic moment problem, J. Operator Theory 48 (2002) 315–354.
  • [CF04] R. Curto, L. Fialkow, Solution of the truncated parabolic moment problem, Integral Equations Operator Theory 50 (2004), 169–196.
  • [CF05] R. Curto, L. Fialkow, Solution of the truncated hyperbolic moment problem, Integral Equations Operator Theory 52 (2005) 181–218.
  • [CF05b] R. Curto, L. Fialkow, Truncated KK-moment problems in several variables, J. Operator Theory 54 (2005) 189–226.
  • [CF08] R. Curto, L. Fialkow, An analogue of the Riesz-Haviland theorem for the truncated moment problem, J. Funct. Anal. 225 (2008) 2709–2731.
  • [CF13] R. Curto, L. Fialkow, Recursively determined representing measures for bivariate truncated moment sequences, J. Operator Theory 70(2) (2013) 401–436.
  • [CFM08] R. Curto, L. Fialkow, H. M. Möller, The extremal truncated moment problem, Integral Equations Operator Theory 60(2) (2008) 177-200.
  • [CY14] R. Curto, S. Yoo, Cubic column relations in the truncated moment problems, J. Funct. Anal. 266(3) (2014) 1611–1626.
  • [CY15] R. Curto, S. Yoo, Non-extremal sextic moment problems, J. Funct. Anal. 269(3) (2015) 758–780.
  • [CY16] R. Curto, S. Yoo, Concrete solution to the nonsingular quartic binary moment problem, Proc. Amer. Math. Soc. 144 (2016) 249–258.
  • [Dan92] J. Dancis, Positive semidefinite completions of partial hermitian matrices, Linear Algebra Appl. 175 (1992) 97–114.
  • [DS18] P.J. di Dio, K. Schmüdgen. The multidimensional truncated Moment Problem:Atoms, Determinacy, and Core Variety, J. Funct. Anal. 274 (2018) 3124–3148. Available from: https://doi.org/10.1016/j.jfa.2017.11.013.
  • [Fia95] L. Fialkow, Positivity, extensions and the truncated complex moment problem, Contemporary Math. 185 (1995), 133–150.
  • [Fia11] L. Fialkow, Solution of the truncated moment problem with variety y=x3y=x^{3}, Trans. Amer. Math. Soc. 363 (2011) 3133–3165.
  • [Fia15] L. Fialkow, The truncated moment problem on parallel lines, In: Theta Foundation International Book Series of Mathematical Texts 20 (2015), 99–116.
  • [Fia17] L. Fialkow. The core variety of a multisequence in the truncated moment problem, J. Math. Anal. Appl. 456 (2017) 946–969. Available from: https://doi.org/10.1016/j.jmaa.2017.07.041.
  • [FN10] L. Fialkow, J. Nie, Positivity of Riesz functionals and solutions of quadratic and quartic moment problems, J. Funct. An. 258 (2010), 328–356.
  • [GJSW84] R. Grone, C. R. Johnson, E. M. Sá, H. Wolkowicz, Positive definite completions of partial hermitian matrices, Linear Algebra Appl. 58 (1984), 109–124.
  • [Kim14] D.P. Kimsey. The cubic complex moment problem, Integral Equations Operator Theory 80 (2014) 353-–378. Available from: https://doi.org/10.1007/s00020-014-2183-4.
  • [Kim21] D.P. Kimsey. On a minimal solution for the indefinite truncated multidimensional moment problem, J. Math. Anal. Appl. 500 (2021) Available from: https://doi.org/10.1016/j.jmaa.2021.125091.
  • [KN77] K.G. Krein, A.A. Nudelman. The Markov moment problem and extremal problems, Translations of Mathematical Monographs, Amer. Math. Soc.; 1977.
  • [Lau05] M. Laurent, Revising two theorems of Curto and Fialkow on moment matrices, Proc. Amer. Math. Soc. 133 (2005), 2965–2976.
  • [Nie14] J. Nie. The 𝒜\mathcal{A}-truncated KK-moment problem, Found. Comput. Math. 14 (2014) 1243–1276.
  • [Sch17] K. Schmüdgen. The moment problem, Graduate Texts in Mathematics 277. Cham: Springer; 2017.
  • [Wol] Wolfram Research, Inc., Mathematica, Version 12.0, Wolfram Research, Inc., Champaign, IL, 2020.
  • [Yoo17a] S. Yoo, Sextic moment problems on 3 parallel lines, Bull. Korean Math. Soc. 54 (2017), 299–318.
  • [Yoo17b] S. Yoo, Sextic moment problems with a reducible cubic column relation, Integral Equations Operator Theory 88 (2017), 45–63.
  • [YZ+] S. Yoo, A. Zalar, The truncated moment problem on reducible cubic curves II: Hyperbolic type relations, in preparation.
  • [Zal21] A. Zalar. The truncated Hamburger moment problem with gaps in the index set, Integral Equations Operator Theory 93 (2021) 36 pp. Available from: https://doi.org/10.1007/s00020-021-02628-6.
  • [Zal22a] A. Zalar. The truncated moment problem on the union of parallel lines, Linear Algebra and its Applications 649 (2022) 186–239. Available from: https://doi.org/10.1016/j.laa.2022.05.008.
  • [Zal22b] A. Zalar. The strong truncated Hamburger moment problem with and without gaps, J. Math. Anal. Appl. 516 (2022) 21pp. Available from: https://doi.org/10.1016/j.jmaa.2022.126563.
  • [Zal23] A. Zalar. The truncated moment problem on curves y=q(x)y=q(x) and yx=1yx^{\ell}=1, Linear and Multilinear Algebra (2023) 45pp. Available from: https://doi.org/10.1080/03081087.2023.2212316.
  • [Zha05] F. Zhang, The Schur Complement and Its Applications, Springer-Verlag, New York, 2005.