%~Mouliné par MaN_auto v.0.42.3 (a17591c0) 2026-08-31 15:26:05 \documentclass[OJMO,Unicode]{cedram} \usepackage{amssymb} \usepackage{mathtools} \usepackage{bm} \usepackage{enumitem} \DeclareMathOperator*{\Res}{Res} \DeclareMathOperator*{\tr}{tr} \DeclareMathOperator*{\sign}{sign} \DeclareMathOperator*{\supp}{spt} \newcommand{\R}{\mathbb{R}} \newcommand{\N}{\mathbb{N}} \newcommand{\C}{\mathbb{C}} \newcommand{\abs}[1]{\lvert #1\rvert} \newcommand{\ceq}{:=} \newcommand{\dd}{\mathrm{d}} \newcommand{\dx}{\dd x} \newcommand{\dz}{\dd z} \newcommand{\dv}{\dd v} \newcommand{\dmu}{\dd \mu} \newcommand{\dtheta}{\dd \theta} \let\originalleft\left \let\originalright\right \renewcommand{\left}{\mathopen{}\mathclose\bgroup\originalleft} \renewcommand{\right}{\aftergroup\egroup\originalright} \newcommand{\rout}{\mathrm{out}} \newcommand{\meqref}[1]{(\ref{#1})} \graphicspath{{./figures/}} \newcommand*{\mk}{\mkern -1mu} \newcommand*{\Mk}{\mkern -2mu} \newcommand*{\mK}{\mkern 1mu} \newcommand*{\MK}{\mkern 2mu} \hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red} \newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}} \newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}} \newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}} \newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}} \let\oldforall\forall \renewcommand*{\forall}{\mathrel{\oldforall}} \newcommand{\gge}{\mathrel{\,\ge\,}} \newcommand{\lle}{\mathrel{\,\le\,}} \newcommand{\eeq}{\mathrel{\,=\,}} \title[Solving Stengle’s Example in Rational Arithmetic]{Solving Stengle’s Example in Rational Arithmetic: Exact Values of the Moment-SOS Relaxations} \author{\firstname{Didier} \lastname{Henrion}} \address{LAAS-CNRS, University of Toulouse, France and Faculty of Electrical Engineering, Czech Technical University in Prague, Czechia} \email{henrion@laas.fr} \begin{abstract} We revisit Stengle's classical univariate polynomial optimization example $\min 1-x^2$ s.t. $(1-x^2)^3\ge 0$ whose constraint description is degenerate at the minimizers. We prove that the moment-SOS hierarchy of relaxation order $r\ge 3$ has the exact value { $-1/(r(r-2))$}. For this we construct in rational arithmetic a dual polynomial sum-of-squares (SOS) certificate and a primal moment sequence representing a finitely atomic measure. The key ingredients are elementary trigonometric properties of Chebyshev and Gegenbauer polynomials, and a Christoffel--Darboux kernel argument. \end{abstract} \begin{document} \maketitle \section{Introduction} Given a polynomial optimization problem (POP), the moment-SOS hierarchy (aka the Lasserre hierarchy~\cite{Lasserre2001}) builds a sequence of semidefinite relaxations of increasing size indexed by a relaxation order $r$. On the primal side, one optimizes a linear functional over truncated moment sequences subject to semidefinite constraints on the moment matrix and localizing matrices. On the dual side, one searches for a sum-of-squares (SOS) decomposition of a shifted polynomial. See~\cite{Henrion2020,Nie2023,Theobald2024} for recent general overviews. Recent work has developed \emph{worst-case upper bounds} on the relaxation error: for broad classes of instances one can guarantee that the gap after $r$ steps is at most $O(r^{-k})$, with the exponent $k$ depending on the geometry of the domain and regularity assumptions; see~\cite{Gribling2025, Laurent2024, Schlosser2026, Slot2022} and references therein. Such results are \emph{sufficiency} statements: they provide a safety guarantee that the hierarchy cannot converge slower than a stated rate, even on the hardest instances in the class. In contrast, \emph{lower bounds} on the relaxation error show that certain upper bounds are essentially sharp: one constructs explicit instances where the hierarchy converges no faster than a given rate. These hardness results are \emph{necessity} statements: they quantify intrinsic limitations of the method (or of the representation class), and rules out uniformly better guarantees without changing assumptions or algorithms. The earliest quantitative lower bound predates the modern formulation of the moment-SOS hierarchy and is due to Stengle~\cite{Stengle1996}; see the discussions in~\cite[Ch.~17]{Schmuedgen2017} and~\cite[\S 3.5.1]{Nie2023}. Stengle provided a univariate example showing a lower bound of $\Omega(1/r^2)$ and an upper bound of $O((\log r)^2/r^2)$. This example is characterized by a degenerate description of the domain. Recently, the paper~\cite{Baldi2024} gives lower bounds for domains described by non-degenerate inequalities, connecting these with quantitative non-stability phenomena~\cite{Scheiderer2005}. Lower bounds were also constructed for a specific parametric univariate POP in~\cite{Henrion2025}, but in this case the moment-SOS hierarchy has always finite (i.e. non-asymptotic) convergence, contrary to the Stengle example. \paragraph*{Contribution.} The Stengle example appears as Example~1 in~\cite{Kocvara2025}, where high-precision semidefinite programming experiments suggest a $\Theta(1/r^2)$ behavior. We solve Stengle's example analytically, in rational arithmetic, and prove that the relaxation error is \emph{exactly} {$-1/(r(r-2))$}. The proof is based on convex duality and basic properties of orthogonal polynomials (Chebyshev and Gegenbauer polynomials, and the Christoffel--Darboux kernel) and elementary trigonometric identities, mirroring the role of orthogonal polynomials in upper-bound analyses~\cite{Laurent2024, Schlosser2026, Slot2022}. \section{Stengle's example and its moment-SOS relaxations} \subsection{The POP and its degeneracy} Consider the univariate constrained polynomial optimization problem (POP) \begin{equation}\label{eq:POP} \tag{POP} \min_{x\in\R}\ f(x)\ceq 1-x^2 \qquad\text{s.t.}\qquad g(x)\ceq (1-x^2)^3\ \ge 0. \end{equation} Since $g(x)\ge 0$ holds if and only if $x\in[-1,1]$, the problem reduces to \[ f^* \ceq \min_{|x|\le 1} (1-x^2)=0, \] attained at the boundary points $x^\star=\pm 1$. The POP is ill-posed because the feasible set is described by a single inequality $g(x)\ge 0$ whose gradient vanishes at the minimizers: \[ g'(x) = -6x(1-x^2)^2,\qquad g'(\pm 1)=0. \] Thus the active constraint at $x^\star=\pm 1$ is \emph{degenerate}: the linearization carries no first-order information. The standard linear independence constraint qualifications of nonlinear programming are violated. For illustration, let us consider a classical interior-point method consisting of minimizing, for a given barrier parameter $\mu>0$, a log-barrier function \[ h_\mu(x)\ceq f(x) - \mu \log g(x) \;=\; 1-x^2 - 3\mu \log(1-x^2), \qquad x\in(-1,1). \] A stationary point satisfies \[ { h'_\mu(x)} = -2x + \frac{6\mu x}{1-x^2}=0, \] hence either $x=0$ or $1-x(\mu)^2 = 3\mu$ which yields \[ x(\mu)=\sqrt{1-3\mu}\ \xrightarrow[\mu\downarrow 0]{}\ 1. \] The Lagrange multiplier induced by the log-barrier stationarity is \[ { f'(x(\mu)) - \lambda(\mu)g'(x(\mu))}=0 \quad\text{with}\quad \lambda(\mu)=\frac{\mu}{g(x(\mu))}=\frac{\mu}{(3\mu)^3}=\frac{1}{27\mu^2}\ \xrightarrow[\mu\downarrow 0]{}\ +\infty. \] Moreover, the curvature of the barrier objective is \[ {h''_\mu(x)}= -2 + 6\mu\frac{1+x^2}{(1-x^2)^2}, \] and substituting $x(\mu)^2=1-3\mu$ gives \[ {h''_\mu(x(\mu))} = -2 + 6\mu\frac{2-3\mu}{(3\mu)^2} = \frac{4}{3\mu}-4\ \xrightarrow[\mu\downarrow 0]{}\ +\infty. \] Thus the formulation~\eqref{eq:POP} is a clean toy model for ill-conditioning of barrier methods: the primal iterates converge to the boundary while the dual multiplier and the barrier function Hessian blow up. { As we will see next, the moment-SOS hierarchy is affected by the same degeneracy. Since $f'(\pm 1)=\mp 2\neq 0$ while $g'(\pm 1)=0$, there is no Lagrange multiplier $\lambda\ge 0$ with $f'(x^\star)=\lambda\,g'(x^\star)$, so the linear independence constraint qualification condition of~\cite[\S 5.1]{Nie2023} fails at both minimizers, and with it the sufficient conditions for finite convergence of the hierarchy~\cite[Thm.~5.4.1]{Nie2023}. Below we will quantify exactly the resulting loss of accuracy for the moment-SOS hierarchy.} \subsection{Moment and SOS relaxations} Fix an integer $r\ge 3$. Let $\R[x]_r$ denote the vector space of polynomials of the scalar indeterminate $x \in \R$ of degree at most $r$. In the monomial basis $b(x):=(1,x,\dots,x^r)^\top$ we identify a polynomial $p(x)$ with its coefficient vector $p \in \R^{r+1}$, i.e. $p(x) = p^\top b(x)$. A truncated moment sequence is a vector $y=(y_0,y_1,\dots,y_{2r})$. The associated Riesz functional $\ell_y$ acts linearly on $\R[x]_{2r}$ by $\ell_y(x^k)=y_k$ and extension by linearity. The moment matrix $M_r(y)$ of order $r$ is the Hankel matrix indexed by monomials $b(x)$: \begin{equation}\label{eq:mommat} p^\top M_r(y)p = \ell_y(p(x)^2)\qquad\forall p\in\R[x]_r. \end{equation} The localizing matrix for $g$ of order $r-3$ is \begin{equation}\label{eq:locmat} q^\top M_{r-3}(g\,y)\,q = \ell_y(g(x)\,q(x)^2)\qquad\forall q\in\R[x]_{r-3}. \end{equation} The order-$r$ moment relaxation of~\eqref{eq:POP} is \begin{equation}\label{eq:MOM} \tag{MOM$_r$} \begin{aligned} &\inf_y \ell_y(f)\\ &\,\text{s.t. } \ell_y(1)=1,\quad M_r(y)\succeq 0,\quad M_{r-3}(g\,y)\succeq 0 \end{aligned} \end{equation} and the order-$r$ SOS relaxation of~\eqref{eq:POP} is \begin{equation}\label{eq:SOS} \tag{SOS$_r$} \begin{aligned} &\sup_{\varepsilon,\,p,\,q} \varepsilon\\ &\,\text{s.t. } f(x)-\varepsilon = p(x) + g(x)\,q(x), \quad p\in\Sigma[x]_{2r},\quad q\in\Sigma[x]_{2(r-3)} \end{aligned} \end{equation} where the convex cone $\Sigma[x]_{2d} \subset \R[x]_{2d}$ consists of SOS polynomials of degree at most $2d$. The moment-SOS hierarchy for POP was introduced in this exact form in~\cite{Lasserre2001}, see e.g.~\cite{Henrion2020,Nie2023,Theobald2024} for recent overviews. \begin{lemm}[Weak duality]\label{prop:weak-duality} For every $r\ge 3$, any feasible $y$ in~\eqref{eq:MOM} and any feasible $(\varepsilon,p,q)$ in~\eqref{eq:SOS} satisfy $\ell_y(f)\ge \varepsilon$. \end{lemm} \begin{proof} If $f-\varepsilon=p+g q$ with $p,q$ SOS, then for any feasible $y$, \[ \ell_y(f)-\varepsilon = \ell_y(p)+\ell_y(g q) \ge 0, \] because $\ell_y(p)\ge 0$ follows from $M_r(y)\succeq 0$ and $\ell_y(g q)\ge 0$ follows from $M_{r-3}(g y)\succeq 0$. \end{proof} \subsection{Strong duality and non-attainment at zero} The quadratic module generated by $g$ is the set of polynomials $p+gq$ for $p$ and $q$ SOS. We say that a quadratic module is Archimedean when it contains $R^2-x^2$ for some $R>0$. Note that this implies compactness of the set described by the inequality $g(x) \geq 0$. The convergence of the moment-SOS hierarchy, as originally proved in~\cite{Lasserre2001}, relies on Putinar's Positivstellensatz (Psatz) \cite{Putinar1993}, which in turn relies on the Archimedean property of the quadratic module. The quadratic module generated by $g(x)=(1-x^2)^3$ is Archimedean since \begin{equation}\label{eq:archi} 2-x^2 = \underbrace{(x^3-2x)^2 + (x^2-1)^2}_{p(x)} + (1-x^2)^3 \underbrace{(1)}_{q(x)}. \end{equation} Therefore Putinar's Psatz applies and the moment-SOS hierarchy converges, i.e. the values of the problems~\eqref{eq:MOM} and~\eqref{eq:SOS} tend to $f^* =0$ when $r$ tends to infinity, as proved in~\cite{Lasserre2001}. Note however that for our POP the quadratic module and the preordering are the same, since we have only one generator $g$. Convergence of the moment-SOS hierarchy then also follows from Schmüdgen's Psatz~\cite{Schmudgen1991} which applies without resorting to the Archimedean property, because the feasibility domain $[-1,1]$ is compact. \begin{lemm}[Strong duality and attainment]\label{prop:slater} For every $r\ge 3$, strong duality, primal attainment and dual attainment hold, i.e. \[ \varepsilon_r^\star := \inf\eqref{eq:MOM} = \min\eqref{eq:MOM} = \sup\eqref{eq:SOS} = \max\eqref{eq:SOS}. \] \end{lemm} \begin{proof} Let $\mu$ be the normalized Lebesgue measure on $[-1,1]$ and let $y$ be its moment sequence ($y_k=\int_{-1}^1 x^k\,\dmu$). Then for any nonzero $p\in\R[x]_{r}$, $\int p^2\,\dmu>0$, hence $M_r(y)\succ 0$. Similarly, $g(x)=(1-x^2)^3>0$ for all $x\in(-1,1)$, so for nonzero $q\in\R[x]_{r-3}$, $\int g q^2\,\dmu>0$ and $M_{r-3}(g y)\succ 0$. Thus~\eqref{eq:MOM} is strictly feasible and conic duality yields strong duality and dual attainment. To prove primal attainment, we verify Slater's condition for the dual~\eqref{eq:SOS}. Fix $r\ge 3$ and choose {$q(x):=\sum_{k=0}^{r-3} x^{2k}$ in the interior of $\Sigma[x]_{2(r-3)}$}. Define the polynomial $h(x):=f(x)-g(x)\,q(x)$ which has degree~$2r$ and strictly positive leading coefficient, hence it is coercive and it attains a global minimum $m:=\min_{x\in\mathbb R} h(x)$. Let $\varepsilon:=m-1$ and $p(x):=h(x)-\varepsilon=h(x)-m+1$. Then $p(x)\ge 1$ for all $x\in\mathbb R$ and $\deg p=2r$, so $p(x)$ is in the interior of $\Sigma[x]_{2r}$ and we have the identity $f(x)-\varepsilon=p(x)+g(x)q(x)$. Therefore $(\varepsilon,p,q)$ is strictly feasible for~\eqref{eq:SOS}. Conic duality implies strong duality and primal attainment. \end{proof} \begin{lemm}\label{lem:eps-neg} Any dual-feasible triple $(\varepsilon,p,q)$ in~\eqref{eq:SOS} satisfies $\varepsilon\le 0$. Moreover, $\varepsilon=0$ is \emph{not} feasible in~\eqref{eq:SOS} for any finite $r$. In particular, $\varepsilon_r^\star<0$ for all $r\ge 3$. \end{lemm} \begin{proof} Evaluating $f(x)-\varepsilon=p(x)+g(x)q(x)$ at $x=1$ gives $-\varepsilon=p(1) \ge 0$. Assume by contradiction that $\varepsilon=0$ is feasible, i.e. $1-x^2 = p(x) + (1-x^2)^3 q(x)$ for $p, q$ SOS. At $x=1$, the right-hand side equals $p(1)$, so $p(1)=0$. Since $p$ is SOS, every real root has even multiplicity; therefore $p$ vanishes at $x=1$ with multiplicity at least $2$. The term $(1-x^2)^3 q(x)$ vanishes at $x=1$ with multiplicity at least $3$. Hence the right-hand side vanishes at $x=1$ with multiplicity at least $2$, whereas the left-hand side $1-x^2$ has a simple zero at $x=1$, a contradiction. \end{proof} \subsection{Main theorem and proof strategy} \begin{theo}[Exact relaxation value]\label{thm:main} For every integer $r\ge 3$, the order-$r$ moment--SOS relaxation value is \[ \varepsilon_r^\star = -\frac{1}{r(r-2)}. \] \end{theo} To prove this result, we will construct analytically: \begin{itemize}[leftmargin=2em] \item in Section~\ref{sec:SOS}, a dual-feasible SOS certificate with $\varepsilon=-1/(r(r-2))$; \item in Section~\ref{sec:Moments}, a primal-feasible moment sequence $y$ with $\ell_y(1-x^2)=-1/(r(r-2))$. \end{itemize} The proof of Theorem~\ref{thm:main} is then given in Section~\ref{sec:proof}. \section{The SOS side: a pure-square identity}\label{sec:SOS} Our numerical experiments when solving~\eqref{eq:SOS} suggest that optimal SOS polynomials are perfect squares. Motivated by this observation, we seek an identity of the form \begin{equation}\label{eq:identity} r(r-2)(1-x^2)+1 \;=\; A_r(x)^2 + 4\,(1-x^2)^3\,B_r(x)^2, \end{equation} for some $A_r \in \R[x]_r$, $B_r \in \R[x]_{r-3}$, which would yield the order-$r$ certificate \begin{equation}\label{eq:certificate} 1-x^2+\frac{1}{r(r-2)}=\underbrace{\frac{A_r(x)^2}{r(r-2)}}_{p(x)} + (1-x^2)^3\underbrace{\frac{4B_r(x)^2}{r(r-2)}}_{q(x)}, \end{equation} hence $\varepsilon=-1/(r(r-2))$ would be feasible in~\eqref{eq:SOS}. \subsection{Chebyshev and Gegenbauer polynomials} Let $T_k$ denote the Chebyshev polynomial of the first kind, defined by the three-term recurrence \[ T_{k+1}(x)=2xT_k(x)-T_{k-1}(x),\qquad T_0(x)=1,\quad T_1(x)=x. \] See~\cite[Ch.~22, \S22.7, Table~22.7.4]{AbramowitzStegun1964} for the recurrence and~\cite[Ch.~22, \S22.4, Eq.~(22.4.4)]{AbramowitzStegun1964} for the initial values. Let $P_k^{(2)}$ denote the Gegenbauer polynomial of parameter $2$, defined by the recurrence \[ (k+1)P_{k+1}^{(2)}(x)=2(k+2)x\,P_k^{(2)}(x)-(k+3)P_{k-1}^{(2)}(x),\qquad P_0^{(2)}(x)=1,\quad P_1^{(2)}(x)=4x. \] Gegenbauer polynomials are also called Jacobi's ultraspherical polynomials~\cite[\S 4.7]{Szego1975}, see~\cite[(4.7.17)]{Szego1975} for the recurrence. See also~\cite[Ch.~22, \S22.7, Eq.~(22.7.3)]{AbramowitzStegun1964} for the recurrence and~\cite[Ch.~22, \S22.4, Eq.~(22.4.2)]{AbramowitzStegun1964} for the initial values. Define, for $r\ge 3$, \begin{equation}\label{eq:def-Ar-Br} A_r(x)\ceq \frac{r-2}{2}T_r(x)-\frac{r}{2}T_{r-2}(x), \qquad B_r(x)\ceq P_{r-3}^{(2)}(x). \end{equation} \subsection{Proof of the identity via a Pythagorean trigonometric argument} \begin{lemm}\label{lem:sos} For every $r\ge 3$, the polynomials $A_r,B_r$ defined in~\eqref{eq:def-Ar-Br} satisfy the identity~\eqref{eq:identity}. \end{lemm} \begin{proof} Fix $\theta\in\R$ and $x=\cos\theta$. Recall the classical trigonometric representations~\cite[Ch.~22, \S22.3, Eqs.~(22.3.15)--(22.3.16)]{AbramowitzStegun1964}, \cite[(1.12.3)]{Szego1975}: \begin{equation}\label{eq:trigo} T_k(\cos\theta)=\cos(k\theta), \qquad U_k(\cos\theta)=\frac{\sin(k+1)\theta}{\sin\theta} \end{equation} where $U_k$ denote the Chebyshev polynomial of the second kind, defined by the recurrence \[ U_{k+1}(x)=2x\,U_k(x)-U_{k-1}(x),\qquad U_0(x)=1,\quad U_1(x)=2x. \] See~\cite[Ch.~22, \S22.7, Eq.~(22.7.5)]{AbramowitzStegun1964} for the recurrence and~\cite[Ch.~22, \S22.4, Eq.~(22.4.5)]{AbramowitzStegun1964} for the initial values. Moreover, the Chebyshev polynomial of the second kind is a special Gegenbauer polynomial: \[ U_k(x)=P_k^{(1)}(x) \] see~\cite[Ch.~22, \S22.5, Eq.~(22.5.34)]{AbramowitzStegun1964}) or~\cite[(4.7.2)]{Szego1975}. Using the Gegenbauer derivative identity~\cite[(4.7.14)]{Szego1975}: \[ \frac{\dd}{\dx}P_{k+1}^{(1)}(x)=2\,P_k^{(2)}(x), \] we obtain \[ P_k^{(2)}(x)=\tfrac12\,\frac{\dd}{\dx}U_{k+1}(x). \] Combining this with~\eqref{eq:trigo} and the chain rule $\frac{\dd}{\dx}=\frac{\dtheta}{\dx}\frac{\dd}{\dtheta}=-\frac{1}{\sin\theta}\frac{\dd}{\dtheta}$ yields the trigonometric form \begin{equation}\label{eq:crcos} P_{k}^{(2)}(\cos\theta) =\frac{\sin(k+2)\theta\ \cos\theta-(k+2)\ \cos(k+2)\theta\ \sin\theta}{2\sin^3\theta}. \end{equation} Let $\phi\ceq (r-1)\theta$. From~\eqref{eq:def-Ar-Br} and~\eqref{eq:trigo}, we obtain \begin{equation}\label{eq:Ar-cos-expand} A_r(\cos\theta) =\frac{r-2}{2}\cos r\theta-\frac{r}{2}\cos (r-2)\theta. \end{equation} Recall the addition formulas \begin{gather*} \cos r\theta =\cos(\phi+\theta)=\cos\phi\ \cos\theta-\sin\phi\ \sin\theta, \\ \cos (r-2)\theta =\cos(\phi-\theta)=\cos\phi\ \cos\theta+\sin\phi\ \sin\theta. \end{gather*} Substitute these into~\eqref{eq:Ar-cos-expand}: \begin{align} A_r(\cos\theta) &=\frac{r-2}{2}\bigl(\cos\phi\ \cos\theta-\sin\phi\ \sin\theta\bigr) -\frac{r}{2}\bigl(\cos\phi\ \cos\theta+\sin\phi\ \sin\theta\bigr) \nonumber \\[0.3em] &=\Bigl(\frac{r-2}{2}-\frac{r}{2}\Bigr)\cos\phi\ \cos\theta +\Bigl(-\frac{r-2}{2}-\frac{r}{2}\Bigr)\sin\phi\ \sin\theta \nonumber \\[0.3em] &=-\cos\phi\ \cos\theta-(r-1)\sin\phi\ \sin\theta.\label{eq:arcos} \end{align} Define \[ z\ceq \bigl(\cos\theta-i(r-1)\ \sin\theta\bigr)\,e^{i\phi}. \] Then \[ |z|^2 = \bigl|\cos\theta-i(r-1)\ \sin\theta\bigr|^2 = \cos^2\theta+(r-1)^2\ \sin^2\theta = 1+r(r-2)\ \sin^2\theta. \] Moreover, expanding $z$ reveals its real and imaginary parts \[ \Re z=\cos\theta\ \cos\phi+(r-1)\ \sin\theta\ \sin\phi, \qquad \Im z=\cos\theta\ \sin\phi-(r-1)\ \sin\theta\ \cos\phi. \] By~\eqref{eq:def-Ar-Br}, \eqref{eq:crcos}, \eqref{eq:arcos} we get \[ A_r(\cos\theta)^2 = (\Re z)^2, \qquad 4\sin^6\theta\,B_r(\cos\theta)^2 = (\Im z)^2. \] Therefore \[ A_r(\cos\theta)^2 + 4(1-\cos^2\theta)^3\ B_r(\cos\theta)^2 = (\Re z)^2+(\Im z)^2 = |z|^2 = 1+r(r-2)\ \sin^2\theta. \] Since $\sin^2\theta=1-\cos^2\theta=1-x^2$, we obtain~\eqref{eq:identity} at $x=\cos\theta$. Both sides are polynomials in $x$ and agree on infinitely many $x\in[-1,1]$, hence they agree identically on $\R$. \end{proof} \subsection{Expressions for low relaxation orders} \begin{figure}[h!] \centering \includegraphics[width=\textwidth]{stengle_sos.png} \caption{For each relaxation order $r\in\{3,\dots,8\}$ we represent, over $x\in[-1,1]$, the SOS identity~\eqref{eq:certificate}. The dashed gray curve is the shifted objective {$1-x^2+1/(r(r-2))$}, the plain gray curve is the term $A_r(x)^2/r(r-2)$, and the black curve is the term $4(1-x^2)^3B_r(x)^2/r(r-2)$. The value {$-\varepsilon_r = 1/(r(r-2))$} can be read from the gray curves at $x=\pm 1$.}\label{fig:stengle-sos-identity} \end{figure} For illustration, the polynomials $(A_r,B_r)$ in the monomial basis are: \begin{align*} A_3(x)&= 2x^3-3x,&B_3(x)&=1,\\ A_4(x)&=8x^4-12x^2+3,&B_4(x)&=4x,\\ A_5(x)&=24x^5-40x^3+15x,&B_5(x)&=12x^2-2,\\ A_6(x)&=64x^6-120x^4+60x^2-5,&B_6(x)&=32x^3-12x,\\ A_7(x)&=160x^7-336x^5+210x^3-35x,&B_7(x)&=80x^4-48x^2+3,\\ A_8(x)&=384x^8-896x^6+672x^4-168x^2+7,&B_8(x)&=192x^5-160x^3+24x. \end{align*} The corresponding SOS identities~\eqref{eq:certificate} are visualized on Figure~\ref{fig:stengle-sos-identity}. \section{The moment side: atoms, weights, and positivity}\label{sec:Moments} In this section we construct a primal-feasible moment sequence $y$ attaining the value $\ell_y(1-x^2)=-1/(r(r-2))$. \subsection{Complementarity suggests atomic support on the roots of \texorpdfstring{$A_r$}{A\_r}} The identity~\eqref{eq:identity} implies that if $(\varepsilon^\star_r,s_0^\star,s_1^\star)$ was an optimal certificate then \[ \varepsilon_r^\star = -\frac{1}{r(r-2)}, \qquad s_0^\star(x)=\frac{1}{r(r-2)}A_r(x)^2, \qquad s_1^\star(x)=\frac{4}{r(r-2)}B_r(x)^2. \] Complementarity with a primal optimal $y$ would read \[ \ell_y(s_0^\star)=0, \qquad \ell_y(g s_1^\star)=0. \] Since $s_0^\star$ and $g s_1^\star$ are nonnegative polynomials, these equalities force $s_0^\star=0$ and $g s_1^\star=0$ on the support of any representing measure of $y$. In particular, $\supp\mu\subseteq\{x \in \R :A_r(x)=0\}$ is a natural candidate. \subsection{Roots of \texorpdfstring{$A_r$}{A\_r} and their location} A key structural fact is that $A_r$ is (up to scaling) a Jacobi polynomial $P^{(\alpha,\beta)}_r$ with parameters $\alpha=\beta=-\tfrac32$. See e.g.~\cite[\S 2.4]{Szego1975} for the definition of Jacobi polynomials, and~\cite[Ch.~4]{Szego1975} for their properties. \begin{lemm}[Jacobi representation]\label{lem:Ar-Jacobi} For every $r\ge 3$, \[ A_r(x)=\frac{2\sqrt{\pi}\,\Gamma(r+1)}{\Gamma(r-\tfrac12)}\,P_r^{(-3/2,-3/2)}(x) \] where $\Gamma$ is the Gamma function satisfying $\Gamma(\tfrac12)=\sqrt{\pi}$, $\Gamma(1)=1$ and $\Gamma(k+1)=k!$ for integer $k$. \end{lemm} \begin{proof} Fix $\theta\in\R$ and $x=\cos\theta$. First let us establish a derivative identity for $A_r$. Using the classical trigonometric representations~\eqref{eq:trigo}, differentiate $T_r(\cos\theta)=\cos(r\theta)$ with respect to $\theta$ and use $\frac{\dx}{\dtheta}=-\sin\theta$ to obtain the standard derivative formula \begin{equation}\label{eq:chebder} \frac{\dd}{\dx}T_r(x)=r\,U_{r-1}(x). \end{equation} Using~\eqref{eq:def-Ar-Br}, a direct differentiation and a short simplification yield \begin{equation}\label{eq:Arprime} \frac{\dd}{\dx}A_r(x)=r(r-2)\,T_{r-1}(x). \end{equation} Chebyshev polynomials have the following Jacobi representation: \begin{equation}\label{eq:T-as-Jacobi} T_k(x) = \frac{k!\sqrt{\pi}}{\Gamma(k+\tfrac12)}\,P_k^{(-1/2,-1/2)}(x), \end{equation} see~\cite[Ch.~22, \S22.5, Eq.~(22.5.31)]{AbramowitzStegun1964} and~\cite[Ch.~4]{Szego1975}. Applying~\eqref{eq:T-as-Jacobi} with $k=r-1$ in~\eqref{eq:Arprime} gives \begin{equation}\label{eq:Arprime-Jacobi} \frac{\dd}{\dx}A_r(x) = \frac{r(r-2)\,\Gamma(r)\sqrt{\pi}}{\Gamma(r-\tfrac12)}\, P_{r-1}^{(-1/2,-1/2)}(x). \end{equation} For Jacobi polynomials one has the derivative relation \[ \frac{\dd}{\dx}P_r^{(\alpha,\beta)}(x) = \frac12\,(r+\alpha+\beta+1)\,P_{r-1}^{(\alpha+1,\beta+1)}(x), \] see~\cite[Eq.~(4.5.5)]{Szego1975}. With $\alpha=\beta=-\tfrac32$, it becomes \begin{equation}\label{eq:Jacobi-derivative-special} \frac{\dd}{\dx}P_r^{(-3/2,-3/2)}(x) = \frac12\,(r-2)\,P_{r-1}^{(-1/2,-1/2)}(x). \end{equation} Comparing~\eqref{eq:Arprime-Jacobi} with~\eqref{eq:Jacobi-derivative-special}, we see that \[ \frac{\dd}{\dx} A_r(x) = \Biggl(\frac{2\sqrt{\pi}\,\Gamma(r+1)}{\Gamma(r-\tfrac12)}\Biggr)\; \frac{\dd}{\dx}P_r^{(-3/2,-3/2)}(x). \] Therefore there exists a constant $c_r\in\R$ such that \begin{equation}\label{eq:Ar-affine} A_r(x)=\frac{2\sqrt{\pi}\,\Gamma(r+1)}{\Gamma(r-\tfrac12)}\,P_r^{(-3/2,-3/2)}(x)+c_r. \end{equation} Recall the standard normalization \[ P_r^{(-3/2,-3/2)}(1) = \frac{\Gamma(r-\tfrac12)}{\Gamma(-\tfrac12)\,\Gamma(r+1)} \] see~\cite[Ch.~22, \S22.2, Table~22.2.1]{AbramowitzStegun1964}. Evaluating~\eqref{eq:Ar-affine} at $x=1$ gives \[ A_r(1) = \frac{2\sqrt{\pi}\,\Gamma(r+1)}{\Gamma(r-\tfrac12)}\, \frac{\Gamma(r-\tfrac12)}{\Gamma(-\tfrac12)\,\Gamma(r+1)} +c_r = \frac{2\sqrt{\pi}}{\Gamma(-\tfrac12)}+c_r. \] Using $\Gamma(-\tfrac12)=-2\sqrt{\pi}$, we get $A_r(1)=-1+c_r$. From~\eqref{eq:def-Ar-Br}, it holds $A_r(1)=-1$ and hence $c_r=0$, and~\eqref{eq:Ar-affine} reduces to the desired proportionality. \end{proof} \begin{lemm}[Root distribution]\label{lem:roots} The roots of $A_r$ are symmetric with respect to the origin, and zero is a root if $r$ is odd. The roots are all real and simple. Moreover, $r-2$ roots lie inside $(-1,1)$ and 2 roots lie outside $[-1,1]$. \end{lemm} \begin{proof} Since $T_k(-x)=(-1)^k T_k(x)$, the definition~\eqref{eq:def-Ar-Br} gives \[ A_r(-x)=\frac{r-2}{2}(-1)^rT_r(x)-\frac{r}{2}(-1)^{r-2}T_{r-2}(x) =(-1)^rA_r(x) \] which shows that the roots are symmetric with respect to the origin, and also that $A_r(0)=0$ if $r$ is odd. By Lemma~\ref{lem:Ar-Jacobi}, $A_r$ and $P_r^{(-3/2,-3/2)}$ have the same roots. Then we apply~\cite[Thm.~6.72]{Szego1975} with the notations $\alpha=\beta=-\tfrac32$ and $n:=r$. First, the excluded cases (6.72.1)--(6.72.3) do not occur: $\alpha,\beta$ are not negative integers, and $n+\alpha+\beta=n-3\ge 0$ for $n\ge 3$. Hence the roots are different from $\pm1$ and $\infty$, and (by the discussion following (6.72.3)) are distinct. Next, compute the integers $X,Y,Z$ from (6.72.5) using Klein's symbol $E(\cdot)$ in (6.72.4). Here $2n+\alpha+\beta+1=2n-2>0$ and $|\alpha|=|\beta|=\tfrac32$, so \[ X=E\left(\frac12\bigl((2n-2)-\tfrac32-\tfrac32+1\bigr)\right)=E(n-2)=n-3, \qquad Y=Z=E\left(\frac12(-2n+3)\right)=0. \] Moreover, \[ \binom{n+\alpha}{n}=\binom{n-\tfrac32}{n} =\frac{\Gamma(n-\tfrac12)}{\Gamma(n+1)\Gamma(-\tfrac12)}<0, \qquad \binom{2n+\alpha+\beta}{n}=\binom{2n-3}{n}>0. \] Therefore, in (6.72.6) we obtain $N_1=n-2$ (both parity cases give the same result), while in (6.72.7)--(6.72.8) we are in the strictly negative branches and get $N_2=N_3=1$. Thus $P_n^{(-3/2,-3/2)}$ has exactly $n-2$ roots in $(-1,1)$, one in $(-\infty,-1)$ and one in $(1,\infty)$, and they are all simple. \end{proof} Based on Lemma~\ref{lem:roots}, denoting by $\{x_i\}_{i=1}^r$ the roots of $A_r$, their pattern is as follows: \[ \begin{aligned} &r\ \text{even:}&& 00$ large enough so that all poles lie inside. By the residue formula~\cite[Ch.~3, \S2]{Stein2003}, \[ \int_{\Gamma_R}F(z)\,\dz = 2\pi i \sum_{a \in \mathcal P}\Res_{z=a}F(z). \] As $z\to\infty$, $G(z)=O(1/z)$ and $(1-z^2)^{-2}=O(1/z^4)$, hence $F(z)=O(1/z^5)$. Thus $\int_{\Gamma_R}F(z)\,\dz\to 0$ as $R\to\infty$, and we obtain \begin{equation}\label{eq:sumresFzero-SS-short} \sum_{a \in \mathcal P}\Res_{z=a}F(z)=0. \end{equation} By Lemma~\ref{lem:roots}, $x_i$ is a simple zero of $A_r$ and $x_i\neq\pm1$. By~\cite[Ch.~3, Thm.~1.1]{Stein2003} we may write $A_r(z)=(z-x_i)h(z)$ near $x_i$ with $h$ holomorphic and non-vanishing, and $A_r'(x_i)=h(x_i)$. Hence $F$ has a simple pole at $x_i$ and, by~\eqref{eq:res}, \[ \Res_{z=x_i}F(z)=\lim_{z\to x_i}\frac{A_r'(z)}{h(z)(1-z^2)^2}=\frac{1}{(1-x_i^2)^2}. \] At $z=1$, $F$ has a pole of order $2$, hence from~\eqref{eq:resn}: \[ \Res_{z=1}F(z)=\lim_{z\to 1}\frac{\dd}{\dz}\Bigl((z-1)^2F(z)\Bigr) = \left.\frac{\dd}{\dz}\Bigl(\frac{G(z)}{(1+z)^2}\Bigr)\right|_{z=1} =\frac{G'(1)-G(1)}{4}. \] Similarly, \[ \Res_{z=-1}F(z)=\lim_{z\to -1}\frac{\dd}{\dz}\Bigl((z+1)^2F(z)\Bigr) =\left.\frac{\dd}{\dz}\Bigl(\frac{G(z)}{(1-z)^2}\Bigr)\right|_{z=-1} =\frac{G'(-1)+G(-1)}{4}. \] Now $A_r(1)=-1$, $A_r(-1)=-(-1)^r$ and from~\eqref{eq:Arprime} we have $A_r'(1)=r(r-2)$, $A_r'(-1)=r(r-2)(-1)^{r-1}$ which implies \[ G(1)=-r(r-2), \qquad G(-1)=r(r-2). \] Moreover $G'(z)=A_r''(z)/A_r(z)-(A_r'(z)/A_r(z))^2$. Using~\eqref{eq:chebder} and~\eqref{eq:Arprime} we have $A_r'(x)=r(r-2)T_{r-1}(x)$, $T_{r-1}'(x)=(r-1)U_{r-2}(x)$, and $U_{r-2}(1)=r-1$, $U_{r-2}(-1)=(-1)^{r-2}(r-1)$. So we get $A_r''(1)=r(r-2)(r-1)^2$, $A_r''(-1)=r(r-2)(-1)^r(r-1)^2$, hence \[ G'(1)= -r(r-2)(r-1)^2-r^2(r-2)^2,\qquad G'(-1)= -r(r-2)(r-1)^2-r^2(r-2)^2. \] Therefore $G'(1)-G(1)=-2r^2(r-2)^2$ and $G'(-1)+G(-1)=-2r^2(r-2)^2$, so \[ \Res_{z=1}F(z)=\Res_{z=-1}F(z)=-\frac{r^2(r-2)^2}{2}. \] Substituting the residues into~\eqref{eq:sumresFzero-SS-short} yields \[ \sum_{a \in \mathcal P}\Res_{z=a}F(z) = \sum_{i=1}^r\frac{1}{(1-x_i^2)^2} -\frac{r^2(r-2)^2}{2}-\frac{r^2(r-2)^2}{2}=0, \] which is~\eqref{eq:sum-1over1mx2sq}. \let\qed\relax \end{proof} \begin{proof}[Proof of~\meqref{eq:sum-1over1mx2}] Define \[ H(z):=\frac{A_r'(z)}{A_r(z)}\frac{1}{1-z^2}=G(z)\frac{1}{(1-z)(1+z)}. \] The poles are simple at $x_i$ and at $\pm1$. As $z\to\infty$, $H(z)=O(1/z^3)$, so the same contour argument gives \[ \sum_{a\in\mathcal P}\Res_{z=a}H(z)=0. \] {As above, with the same factorization $A_r(z)=(z-x_i)h(z)$ from~\cite[Ch.~3, Thm.~1.1]{Stein2003}, \eqref{eq:res} gives \[ \Res_{z=x_i}H(z)=\lim_{z\to x_i}\frac{A_r'(z)}{h(z)(1-z^2)}=\frac{1}{1-x_i^2}. \] } Also, \[ \Res_{z=1}H(z)=\lim_{z\to 1}(z-1)H(z)=\lim_{z\to 1}G(z)\frac{z-1}{(1-z)(1+z)} =-\frac{G(1)}{2}=\frac{r(r-2)}{2}, \] and \[ \Res_{z=-1}H(z)=\lim_{z\to -1}(z+1)H(z)=\lim_{z\to -1}G(z)\frac{z+1}{(1-z)(1+z)} =\frac{G(-1)}{2}=\frac{r(r-2)}{2}. \] Thus \[ \sum_{a \in \mathcal P}\Res_{z=a}H(z) = \sum_{i=1}^r\frac{1}{1-x_i^2}+\frac{r(r-2)}{2}+\frac{r(r-2)}{2}=0, \] which is~\eqref{eq:sum-1over1mx2}. \end{proof} \let\qed\relax \end{proof} \begin{lemm}[Probability and objective value]\label{lem:prob-and-obj} The measure $\mu$ in~\eqref{eq:mu} satisfies \[ y_0 = \int \dmu(x) = 1, \qquad \ell_y(1-x^2) = \int (1-x^2)\,\dmu(x) = -\frac{1}{r(r-2)}. \] \end{lemm} \begin{proof} The mass is \[ \int \dmu(x) = \frac{1}{(r(r-2))^2}\sum_{i=1}^r \frac{1}{(1-x_i^2)^2}=1 \] by~\eqref{eq:sum-1over1mx2sq}. Next, \[ \int (1-x^2)\,\dmu(x) =\frac{1}{(r(r-2))^2}\sum_{i=1}^r \frac{1-x_i^2}{(1-x_i^2)^2} =\frac{1}{(r(r-2))^2}\sum_{i=1}^r \frac{1}{1-x_i^2} = -\frac{1}{r(r-2)} \] by~\eqref{eq:sum-1over1mx2}. \end{proof} \subsection{Positivity of the moment matrix} \begin{lemm}\label{lem:moment-matrix-psd} Let $\mu$ be the measure~\eqref{eq:mu} with moment vector $y$. Then its moment matrix of order $r$ is positive semidefinite: \[ M_r(y)\succeq 0. \] \end{lemm} \begin{proof} Given any polynomial $p \in \R[x]_r$, using the definition~\eqref{eq:mommat} of the moment matrix: \[ p^\top M_r(y)\,p = \ell_y(p^2) = \int p(x)^2\,\dmu(x). \] Since $\mu$ is a nonnegative measure, it holds \[ \int p(x)^2\,\dmu(x)\ge 0, \] which implies $M_r(y)\succeq 0$. \end{proof} \subsection{Positivity of the localizing matrix} \begin{lemm}\label{lem:psd-localizing} Let $\mu$ be the measure~\eqref{eq:mu} with moment vector $y$. Then its localizing moment matrix of order $r-3$ is positive semidefinite: \[ M_{r-3}(g y)\succeq 0. \] \end{lemm} \begin{proof} Set $n:=r-3$ and write $b(x):=(1,x,\dots,x^n)^\top$. By~\eqref{eq:locmat}, $M_n(g\,y)\succeq0$ is equivalent to \[ \int (1-x^2)^3 p(x)^2\,\dmu(x)\ge 0\qquad\forall\,p\in\mathbb R[x]_n. \] Using~\eqref{eq:mu} it holds \[ \int (1-x^2)^3 p(x)^2\,\dmu(x) =\sum_{i=1}^r \frac{1-x_i^2}{(r(r-2))^2}\,p(x_i)^2. \] {From} Lemma~\ref{lem:roots} we know that among the $r$ roots of $A_r$, exactly $r-2$ satisfy $|x_i|<1$ and two are $\pm x_{\rout}$ with $x_{\rout}>1$. Define the positive weights \[ w_i:=\frac{1-x_i^2}{(r(r-2))^2}>0\quad(|x_i|<1),\qquad w_{\rout}:=\frac{x_{\rout}^2-1}{(r(r-2))^2}>0, \] and the matrices \[ H:=\sum_{|x_i|<1} w_i\,b(x_i)b(x_i)^\top,\qquad V:=\sqrt{w_{\rout}}\,[\,b(x_{\rout})\quad b(-x_{\rout})\,]. \] Then the localizing matrix decomposes as \begin{equation}\label{eq:MHV} M_n(g\,y)=H-VV^\top. \end{equation} Since the inner measure $\sum_{|x_i|<1}w_i\delta_{x_i}$ has $n+1$ distinct nodes in $(-1,1)$ with positive weights, its moment matrix is positive definite, i.e. $H\succ0$. Therefore, by a rank-two Schur complement, we have the equivalence \begin{equation}\label{eq:SchurG} M_n(gy) = H-VV^\top\succeq 0 \quad\Longleftrightarrow\quad \begin{pmatrix}H & V \\ V^\top & I_2 \end{pmatrix} \succeq 0 \quad\Longleftrightarrow\quad G\ceq V^\top H^{-1}V \preceq I_2. \end{equation} Define the Christoffel--Darboux kernel of the inner measure \[ K_n(x,y):=b(x)^\top H^{-1}b(y), \] see e.g.~\cite[Ch.~2]{Lasserre2022}. Then \[ G =w_{\rout} \begin{pmatrix} K_n(x_{\rout},x_{\rout}) & K_n(x_{\rout},-x_{\rout})\\ K_n(x_{\rout},-x_{\rout}) & K_n(x_{\rout},x_{\rout}) \end{pmatrix} = \begin{pmatrix}\alpha&\beta\\ \beta&\alpha \end{pmatrix}, \] with $\alpha:=w_{\rout}K_n(x_{\rout},x_{\rout})$ and $\beta:=w_{\rout}K_n(x_{\rout},-x_{\rout})$. Hence \[ \lambda_+=\alpha+\beta\ \text{with eigenvector}\ u_+=(1,1),\qquad \lambda_-=\alpha-\beta\ \text{with eigenvector}\ u_-=(1,-1). \] Let $q(x):=B_r(x)=P_n^{(2)}(x)$ and let $q$ be its coefficient vector in the basis $b$, i.e. $q(x)=q^\top b(x)$. We claim that \begin{equation}\label{eq:Mq0-short} M_n(g\,y)\,q=0. \end{equation} Indeed, it suffices to show $\ell_y(g\,x^k q)=0$ for $k=0,\dots,n$. Using~\eqref{eq:mu}, \[ \ell_y(g\,x^k q)=\frac{1}{(r(r-2))^2}\sum_{i=1}^r (1-x_i^2)\,x_i^k\,q(x_i), \] so we have to prove the discrete orthogonality relation \begin{equation}\label{eq:disc-orth-short} \sum_{i=1}^r (1-x_i^2)\,x_i^k\,B_r(x_i)=0,\qquad k=0,\dots,n. \end{equation} Recall the two polynomial identities derived from the trigonometric formulas~\eqref{eq:trigo}, \eqref{eq:crcos} and~\eqref{eq:arcos}: \begin{align} A_r(x) &= -\Bigl(x\,T_{r-1}(x) + (r-1)(1-x^2)\,U_{r-2}(x)\Bigr), \label{eq:Ar-TU-recall} \\ 2(1-x^2)\,B_r(x) &= x\,U_{r-2}(x) - (r-1)\,T_{r-1}(x). \label{eq:Br-TU-recall} \end{align} Let $x_i$ be such that $A_r(x_i)=0$. Since $x_i\neq \pm1$, \eqref{eq:Ar-TU-recall} gives \[ x_i\,T_{r-1}(x_i)=-(r-1)(1-x_i^2)\,U_{r-2}(x_i), \qquad\text{hence}\qquad U_{r-2}(x_i)=-\frac{x_i}{(r-1)(1-x_i^2)}\,T_{r-1}(x_i). \] Substitute this into~\eqref{eq:Br-TU-recall} at $x=x_i$: \begin{align*} 2(1-x_i^2)\,B_r(x_i) &=x_i\,U_{r-2}(x_i)-(r-1)T_{r-1}(x_i) =-\left(\frac{x_i^2}{(r-1)(1-x_i^2)}+(r-1)\right)T_{r-1}(x_i)\\ &=-\frac{x_i^2+(r-1)^2(1-x_i^2)}{(r-1)(1-x_i^2)}\,T_{r-1}(x_i) =-\frac{1+r(r-2)(1-x_i^2)}{(r-1)(1-x_i^2)}\,T_{r-1}(x_i), \end{align*} where we used the algebraic identity \(x^2+(r-1)^2(1-x^2)=1+r(r-2)(1-x^2). \) Multiplying by $(r-1)(1-x_i^2)$ yields \begin{equation}\label{eq:pre-pointwise} 2(r-1)(1-x_i^2)^2\,B_r(x_i)=-\bigl(1+r(r-2)(1-x_i^2)\bigr)\,T_{r-1}(x_i). \end{equation} Now evaluate the SOS identity~\eqref{eq:identity} at $x=x_i$. Since $A_r(x_i)=0$, it reduces to \begin{equation}\label{eq:identity-at-root} 1+r(r-2)(1-x_i^2)=4(1-x_i^2)^3\,B_r(x_i)^2. \end{equation} Insert~\eqref{eq:identity-at-root} into~\eqref{eq:pre-pointwise}: \[ 2(r-1)(1-x_i^2)^2\,B_r(x_i) =-4(1-x_i^2)^3\,B_r(x_i)^2\,T_{r-1}(x_i). \] We now justify division by $(1-x_i^2)^2B_r(x_i)$. First, $x_i\neq\pm1$ so $(1-x_i^2)^2\neq0$. Second, the Gegenbauer polynomial $B_r=P^{(2)}_{r-3}$ has all its real zeros in $(-1,1)$, and if $|x_i|<1$ then~\eqref{eq:identity-at-root} gives $1+r(r-2)(1-x_i^2)>1$, so $B_r(x_i)\neq0$ as well. Hence $B_r(x_i)\neq0$ for every root $x_i$ of $A_r$. Dividing by $(1-x_i^2)^2B_r(x_i)$ gives \[ (1-x_i^2)\,T_{r-1}(x_i)\,B_r(x_i)=-\frac{r-1}{2}. \] Using the derivative identity~\eqref{eq:Arprime}: \[ A_r'(x)=r(r-2)\,T_{r-1}(x), \] one gets the pointwise relation \begin{equation}\label{eq:BoverAprime-short} (1-x_i^2)\,B_r(x_i)=-\frac{r(r-1)(r-2)}{2}\,\frac{1}{A_r'(x_i)}. \end{equation} Next, for each $k\le n=r-3$, consider the meromorphic function $F_k(z):=z^k/A_r(z)$. { Its only finite poles are the simple zeros $x_i$ of $A_r$. Factoring $A_r(z)=(z-x_i)h(z)$ as in the proof of Lemma~\ref{lem:residue-sums}, with $h$ holomorphic and non-vanishing near $x_i$ and $h(x_i)=A_r'(x_i)$~\cite[Ch.~3, Thm.~1.1]{Stein2003}, \eqref{eq:res} gives \[ \Res_{z=x_i}F_k(z)=\lim_{z\to x_i}\frac{z^k}{h(z)}=\frac{x_i^k}{A_r'(x_i)}. \] } Since $\deg A_r=r$ and $k\le r-3$, we have $F_k(z)=O(1/z^3)$ as $z\to\infty$; hence the contour integral over $|z|=R$ vanishes as $R\to\infty$. By the residue theorem~\cite[Ch.~3, \S 2]{Stein2003}, \[ 0=\sum_{i=1}^r \Res_{z=x_i}F_k(z)=\sum_{i=1}^r \frac{x_i^k}{A_r'(x_i)}. \] Multiplying by $-\frac{r(r-1)(r-2)}{2}$ and using~\eqref{eq:BoverAprime-short} yields~\eqref{eq:disc-orth-short}, hence~\eqref{eq:Mq0-short} holds. Now combine~\eqref{eq:MHV} and~\eqref{eq:Mq0-short}: \[ (H-VV^\top) q=0\quad\Longrightarrow\quad H q=VV^\top q. \] Left-multiplying by $V^\top H^{-1}$ gives \[ V^\top q=(V^\top H^{-1}V)(V^\top q)=G\,(V^\top q), \] so $1$ is an eigenvalue of $G$ with eigenvector $V^\top q\neq 0$. Moreover, since $q(-x)=(-1)^n q(x)$, \[ V^\top q =\sqrt{w_{\rout}}\,(q(x_{\rout}),q(-x_{\rout}))^\top =\sqrt{w_{\rout}}\,q(x_{\rout})\,(1,(-1)^n)^\top, \] so the eigenvector is precisely $u_{(-1)^n}=(1,(-1)^n)$ and therefore \begin{equation}\label{eq:eig1} \lambda_{(-1)^n}=1. \end{equation} It remains to show $\max(\lambda_+,\lambda_-)=1$. For this we use only the sign of $\beta$. Let $\{\pi_k\}$ be the monic orthogonal polynomials for the inner product $\langle f,g\rangle_{\mathrm{inn}}:=\sum_{|x_i|<1}w_i f(x_i)g(x_i)$ and $h_k:=\langle\pi_k,\pi_k\rangle_{\mathrm{inn}}>0$. The Christoffel--Darboux formula~\cite[\S 3.1.1]{Lasserre2022} gives, for $x\neq y$, \[ K_n(x,y)=\frac{1}{h_n}\,\frac{\pi_{n+1}(x)\pi_n(y)-\pi_n(x)\pi_{n+1}(y)}{x-y}. \] Since {the inner product is defined on symmetric atoms}, $\pi_k(-x)=(-1)^k\pi_k(x)$, so with $(x,y)=(x_{\rout},-x_{\rout})$, \[ K_n(x_{\rout},-x_{\rout}) =\frac{(-1)^n}{x_{\rout}h_n}\,\pi_n(x_{\rout})\pi_{n+1}(x_{\rout}). \] All zeros of $\pi_k$ lie in $(-1,1)$, hence $\pi_k(x_{\rout})>0$ for $x_{\rout}>1$, so $K_n(x_{\rout},-x_{\rout})\neq 0$ and $\sign(K_n(x_{\rout},-x_{\rout}))=(-1)^n$. As $w_{\rout}>0$, we obtain \[ \sign(\beta)=(-1)^n, \] so the eigenvalue associated with $u_{(-1)^n}$ is exactly $\alpha+|\beta|=\max(\lambda_+,\lambda_-)=1$. Together with~\eqref{eq:eig1}, this yields $\max(\lambda_+,\lambda_-)=1$ hence $G\preceq I_2$. Finally, \eqref{eq:SchurG} implies $M_n(g\,y)= M_{r-3}(g\,y)\succeq 0$. \end{proof} \subsection{Further properties of the moments} { The moment vector $y$ is feasible for the relaxation~\eqref{eq:MOM}, but it is not the moment vector of any probability measure supported on the feasible set $[-1,1]$: such a measure would satisfy $y_{2k}\le 1$ for every $k$, whereas $\ell_y(1-x^2)<0$ forces $y_2>1$. The quantity $y_{2k}-1>0$ therefore measures how far the optimal solution of the relaxation is from solving~\eqref{eq:POP} itself, and the two results below locate the largest such violation. } \begin{lemm}\label{lem:moments-basic} For the moment sequence $y$ of the measure $\mu$ in~\eqref{eq:mu}, it holds $y_k=0$ for odd $k$ and {$1+\tfrac{k}{r(r-2)} \leq y_{2k}\le y_{2(k+1)}$} for all $k\ge 0$. \end{lemm} \begin{proof} The roots of $A_r$ are symmetric: if $x_i$ is a root then so is $-x_i$. The weights in~\eqref{eq:mu} depend only on $x_i^2$, hence the atom at $x_i$ and the atom at $-x_i$ have the same weight. Therefore, for any odd integer $k$, \[ y_k=\int x^k\,\dmu(x) =\int (-x)^k\,\dmu(x) =-\int x^k\,\dmu(x), \] hence $y_k=0$. Fix an integer $k\ge 1$ and set $u(x):=x^2\ge 0$. Then $y_{2k}=\int u(x)^k\,\dmu(x)$. By convexity of $u\mapsto u^k$ on $[0,\infty)$, its tangent inequality at $u=1$ reads \[ {u(x)^k} \ge 1 + k(u(x)-1)\qquad \forall x. \] Integrating this inequality with respect to $\mu$ gives \[ y_{2k}=\int u(x)^k\,\dmu(x) \ge \int \bigl(1+k(u(x)-1)\bigr)\,\dmu(x) = 1+k\Bigl(\int u(x)\,\dmu(x)-\int 1\,\dmu(x)\Bigr) = 1+k(y_2-1), \] where we used Lemma~\ref{lem:prob-and-obj} and $\int \dmu=1$. The same Lemma~\ref{lem:prob-and-obj} yields \[ y_2=\int x^2\,\dmu(x)=1-\int(1-x^2)\,\dmu(x)=1+\frac{1}{r(r-2)}>1, \] {so that $y_{2k}\ge 1+k/(r(r-2))$, and in particular $y_{2k}>1$ for all $k\ge 1$.} To prove that the even moments are nondecreasing, define two nondecreasing functions on $[0,\infty)$: \[ f(u):=u^{k-1},\qquad g(u):=u-1. \] Consider the product measure $\mu\otimes\mu$ on $\R\times\R$. Since $f$ and $g$ are nondecreasing, for all $x,x'\in\R$ we have the pointwise inequality \[ \bigl(f(u(x))-f(u(x'))\bigr)\bigl(g(u(x))-g(u(x'))\bigr)\, \ge\, 0. \] Integrating with respect to $\mu(x)\mu(x')$ gives \begin{align*} 0 &\le \iint \bigl(f(u(x))-f(u(x'))\bigr)\bigl(g(u(x))-g(u(x'))\bigr)\,\dmu(x)\,\dmu(x')\\ &= 2\int f(u(x))g(u(x))\,\dmu(x) -2\biggl(\int f(u(x))\,\dmu(x)\biggr)\biggl(\int g(u(x))\,\dmu(x)\biggr). \end{align*} Hence \begin{equation}\label{eq:chebyshev-mu} \int f(u(x))g(u(x))\,\dmu(x) \ \ge\ \biggl(\int f(u(x))\,\dmu(x)\biggr)\biggl(\int g(u(x))\,\dmu(x)\biggr). \end{equation} Substituting $f(u)=u^{k-1}$ and $g(u)=u-1$ in~\eqref{eq:chebyshev-mu} yields \[ \int u(x)^{k-1}(u(x)-1)\,\dmu(x) \ \ge\ \biggl(\int u(x)^{k-1}\,\dmu(x)\biggr)\biggl(\int (u(x)-1)\,\dmu(x)\biggr). \] Now \[ \int u(x)^{k-1}(u(x)-1)\,\dmu(x) =\int u(x)^k\,\dmu(x)-\int u(x)^{k-1}\,\dmu(x) =y_{2k}-y_{2(k-1)}, \] and \[ \int (u(x)-1)\,\dmu(x)=\int x^2\,\dmu(x)-\int 1\,\dmu(x)=y_2-1. \] Therefore we obtain the moment gap bound: \[ y_{2k}-y_{2(k-1)} \gge y_{2(k-1)}(y_2-1). \] Since $y_{2(k-1)}\ge 0$ and $y_2-1=\frac{1}{r(r-2)}>0$, it follows that $y_{2k}\ge y_{2(k-1)}$ for all $k\ge 1$. \end{proof} \begin{lemm}\label{prop:y2r} For the moment sequence $y$ of the measure $\mu$ in~\eqref{eq:mu}, it holds $\lim_{r\to\infty} y_{2r} = 1$. \end{lemm} \begin{proof} By Lemma~\ref{lem:moments-basic}, $y_{2r}\ge 1$. For the upper bound, note that $\mu$ is supported on $\abs{x}\le x_{\rout}$, hence \[ y_{2r}=\int x^{2r}\,\dmu \le x_{\rout}^{2r}. \] {Let $\kappa=1.1997\ldots<\tfrac54$ denote the unique positive root of $\kappa\tanh\kappa=1$; uniqueness holds because $t\mapsto t\tanh t$ increases from $0$ to $+\infty$ on $[0,\infty)$. We claim that \begin{equation}\label{eq:tr-bracket} \frac{\kappa}{r}\lle t_r\lle\frac{\kappa}{r-2}, \qquad\text{where}\qquad x_{\rout}=\cosh t_r,\quad t_r>0. \end{equation} Such a $t_r$ exists and is unique since $\cosh$ maps $(0,\infty)$ bijectively onto $(1,\infty)$ and, by Lemma~\ref{lem:roots}, $A_r$ has a single root in $(1,\infty)$. Using $T_k(\cosh t)=\cosh(kt)$, the equation $A_r(x_{\rout})=0$ reads \[ (r-2)\cosh(r t_r) = r\cosh((r-2)t_r), \qquad\text{that is}\qquad \Phi(t_r)=\log\frac{r}{r-2}, \] where $\Phi(t):=\log\cosh(rt)-\log\cosh((r-2)t)$. Since $\frac{\partial}{\partial v}\log\cosh(vt)=t\tanh(vt)$, we have the integral representation \[ \Phi(t)=\int_{r-2}^{r} t\,\tanh(vt)\,\dv, \] and, $\tanh$ being increasing, for every $t>0$ \begin{equation}\label{eq:Phi-sandwich} 2t\,\tanh\bigl((r-2)t\bigr)\lle\Phi(t)\lle2t\,\tanh(rt). \end{equation} Moreover, applying $\log z\le z-1$ with $z=\frac{r}{r-2}$ and $\log z \ge 1-1/z$ gives \[ \frac{2}{r}\lle\log\frac{r}{r-2}\lle\frac{2}{r-2}. \] Combining the left inequality of~\eqref{eq:Phi-sandwich} with the upper bound on $\log\frac{r}{r-2}$ yields $(r-2)t_r\tanh\bigl((r-2)t_r\bigr)\le 1$, hence $(r-2)t_r\le\kappa$. Combining the right inequality of~\eqref{eq:Phi-sandwich} with the lower bound yields $r t_r \tanh(r t_r)\ge 1$, hence $r t_r\ge \kappa$. This proves~\eqref{eq:tr-bracket}. Finally, from $\cosh z\le e^{z^2/2}$, valid for all $z\in\R$ because $(2k)!\ge 2^k k!$ for every $k\in\N$, the upper bound in~\eqref{eq:tr-bracket} gives \[ 1\lle y_{2r}\lle x_{\rout}^{2r}=\cosh^{2r}(t_r) \lle\exp\left(\frac{r\,\kappa^2}{(r-2)^2}\right) \eeq 1+O(1/r), \] so $y_{2r}\to 1$.} \end{proof} \begin{coro}\label{cor:top-moment} For every $r\ge 3$ the even moments are nondecreasing, so that $y_{2r}=\max_{0\le k\le r} y_{2k}$, and \[ 1+\frac{1}{r-2}\lle y_{2r}\lle\exp\left(\frac{r\,\kappa^2}{(r-2)^2}\right), \] where $\kappa=1.1997\ldots$ is the unique positive root of $\kappa\tanh\kappa=1$. In particular $y_{2r}-1=\Theta(1/r)$, whereas every probability measure supported on $[-1,1]$ satisfies $y_{2r}\le 1$. \end{coro} \begin{proof} Monotonicity is Lemma~\ref{lem:moments-basic}, which taken at $k=r$ also gives $y_{2r}\ge 1+r/(r(r-2))=1+1/(r-2)$. The upper bound is the estimate established in the proof of Lemma~\ref{prop:y2r}, and equals $1+O(1/r)$. The last assertion holds since $x^{2r}\le 1$ on $[-1,1]$. \end{proof} \subsection{Rational expressions of the moments} The optimal moments are actual \emph{rational} numbers. To see why, define a polynomial $P_r(u)\in\mathbb Q[u]$ by extracting the even part of $A_r$: \[ A_r(x)= \begin{cases} P_r(x^2),& r\ \text{even},\\ x\,P_r(x^2),& r\ \text{odd}. \end{cases} \] Then the nonzero squared nodes $u_i>0$ are precisely the roots of $P_r(u)$. The crucial point is that the moment sum \[ y_{2p}:=\int x^{2p}\,\dmu_r(x)=w_0 0^p + 2\sum_{u_i> 0}w_i {u_i^p},\qquad p\ge 0 \] is a \emph{symmetric rational function} of the roots of an explicit polynomial with rational coefficients. Let $C_r\in\mathbb Q^{m\times m}$ denote the companion matrix of the monic polynomial proportional to $P_r$ (here $m=\deg P_r=\lfloor r/2\rfloor$), so that the eigenvalues of $C_r$ are exactly the roots $(u_i)$ of $P_r$. Since $u=1$ is not a root of $P_r$, the matrix $I-C_r$ is invertible, and the spectral mapping theorem yields \[ \tr\left(C_r^p\,(I-C_r)^{-2}\right) =\sum_{u_i>0}\frac{u_i^p}{(1-u_i)^2}. \] Therefore, for every $p\ge 1$, \begin{equation}\label{eq:trace-moment} y_{2p}= \frac{2}{r^2(r-2)^2}\tr\left(C_r^p\,(I-C_r)^{-2}\right). \end{equation} \subsection{Moment vectors for low relaxation orders}\label{subsec:explicit-moments} Using the trace formula~\eqref{eq:trace-moment} we can compute the first moment vectors. \paragraph*{Order $r=3$.} \[ (y_0,y_2,y_4,y_6)=\left(1,\,\frac{4}{3},\,2,\,3\right). \] \paragraph*{Order $r=4$.} \[ (y_0,y_2,y_4,y_6,y_8) =\left(1,\,\frac{9}{8},\,\frac{21}{16},\,\frac{99}{64},\,\frac{117}{64}\right). \] \paragraph*{Order $r=5$.} \[ (y_0,y_2,y_4,y_6,y_8,y_{10}) =\left(1,\,\frac{16}{15},\,\frac{52}{45},\,\frac{34}{27},\,\frac{223}{162},\,\frac{1465}{972}\right). \] \paragraph*{Order $r=6$.} \[ (y_0,y_2,y_4,y_6,y_8,y_{10},y_{12}) =\left(1,\,\frac{25}{24},\,\frac{35}{32},\,\frac{295}{256},\,\frac{7475}{6144},\,\frac{21075}{16384},\,\frac{178425}{131072} \right). \] \paragraph*{Order $r=7$.} \[ (y_0,y_2,y_4,y_6,y_8,y_{10},y_{12},y_{14}) =\left(1,\,\frac{36}{35},\,\frac{186}{175},\,\frac{963}{875},\,\frac{2853}{2500},\,\frac{29613}{25000},\,\frac{1230417}{1000000},\,\frac{12788307}{10000000} \right). \] \paragraph*{Order $r=8$.} \[ (y_0,y_2,y_4,y_6,y_8,y_{10},y_{12},y_{14},y_{16}) =\left(1,\,\frac{49}{48},\,\frac{301}{288},\,\frac{3703}{3456},\,\frac{22799}{20736},\,\frac{561883}{497664},\,\frac{3463859}{2985984},\,\frac{42726971}{35831808},\,\frac{65904559}{53747712} \right). \] \section{Proof of Theorem~\ref{thm:main}}\label{sec:proof} From Lemma~\ref{lem:sos}, for every $r\ge 3$, the polynomials $A_r,B_r$ defined in~\eqref{eq:def-Ar-Br} satisfy the identity~\eqref{eq:identity}. Thus the triple \[ \varepsilon:=-\frac{1}{r(r-2)},\qquad p(x):=\frac{A_r(x)^2}{r(r-2)},\qquad q(x):=\frac{4B_r(x)^2}{r(r-2)} \] is feasible for~\eqref{eq:SOS}. Therefore \begin{equation}\label{eq:lower-bound-epsr} \sup\eqref{eq:SOS}\gge -\frac{1}{r(r-2)}. \end{equation} Let $\mu$ be the atomic measure~\eqref{eq:mu} supported on the roots of $A_r$ and let $y$ be its moment vector. By Lemma~\ref{lem:prob-and-obj}, $\ell_y(1)=y_0=1$. By Lemma~\ref{lem:moment-matrix-psd}, $M_r(y)\succeq 0$. By Lemma~\ref{lem:psd-localizing}, $M_{r-3}(g\,y)\succeq 0$. Hence $y$ is feasible for~\eqref{eq:MOM}. Again by Lemma~\ref{lem:prob-and-obj}, \[ \ell_y(f)=-\frac{1}{r(r-2)}, \] so \begin{equation}\label{eq:upper-bound-epsr} \inf\eqref{eq:MOM}\lle-\frac{1}{r(r-2)}. \end{equation} By the weak duality {Lemma~\ref{prop:weak-duality}}, we have \[ \inf\eqref{eq:MOM}\gge \sup\eqref{eq:SOS}. \] Combining this inequality with~\eqref{eq:lower-bound-epsr} and~\eqref{eq:upper-bound-epsr} yields \[ \inf\eqref{eq:MOM}=\sup\eqref{eq:SOS}=-\frac{1}{r(r-2)}. \] By Lemma~\ref{prop:slater}, strong duality and attainment hold on both sides, so this common value is precisely the relaxation optimum $\varepsilon_r^\star$. This proves Theorem~\ref{thm:main}. \qed \section{Conclusion} In this paper we show that trigonometric properties of orthogonal polynomials can be exploited to construct in rational arithmetic an analytic solution of the semidefinite relaxations of Stengle's example~\cite{Stengle1996}, thereby settling the question of the exact convergence rate of the moment-SOS hierarchy. It would be interesting to investigate whether similar techniques could be used to solve analytically the other low-dimensional POP examples used as challenging benchmarks for high-precision semidefinite solvers~\cite{Kocvara2025}. Key to the analytic solution of problem~\eqref{eq:SOS} is the pure square form of the polynomials $p$ and $q$, expressed as a quadratic algebraic equation~\eqref{eq:identity} in polynomials $A_r$ and $B_r$. The rank-one structure of the Gram matrices of the SOS polynomials was also exploited in~\cite{Henrion2025} to derive tight upper and lower bounds on the values of the moment-SOS hierarchy for a parametric POP problem. It is currently not well understood for which {classes} of POP does optimality of a {moment-SOS} relaxation {implies} that the Gram matrices of the SOS dual multipliers are rank-one. \section*{Acknowledgments} The author is grateful to Michal Ko\v cvara for sharing the slides of his talk~\cite{Kocvara2025} and for insightful discussions. This paper benefited from the feedback by an anonymous reviewer. The author acknowledges the use of AI for assistance with brainstorming ideas, mathematical development, coding and drafting the manuscript. The final content, analysis and conclusions remain the sole responsibility of the author. \bibliographystyle{plain+eid} \bibliography{OJMO_Henrion_233} \end{document}