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[不等式] 正定(psd)多项式表示为平方和(sos)

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hbghlyj 发表于 2023-3-23 22:57 |阅读模式
本帖最后由 hbghlyj 于 2023-5-14 13:17 编辑 $type 不 等 式 研 究 通 讯2003.5z.pdf (376.84 KB, 下载次数: 3)
不等式研究通讯 2003.5
杨路教师 2003 年 9 月 21 日在论坛讲了一段话,现摘录如下:
过去有人向我提过一个问题: 多项式 $x^2 y^2 (x^2+y^2-1)+1$ 能否表成实系数多项式的平方和?

要么表示出来,要么证明不能表. 这个问题难度怎么样? 大家不妨试试. 这显然是一个正定的多项式.
或者,谁愿意举出任何一个这样的多项式,它是正定的,但却不能表成多项式的平方和?
针对此,浙江的吴裕东先生给出了一个分拆式:
$x^2 y^2 (x^2+y^2-1)+1=\left[\frac{1}{\sqrt{2}}\left(\sqrt{\sqrt[3]{x^{4} y^{2}}+\sqrt[3]{x^{2} y^{4}}+1}\right)\left(\sqrt[3]{x^{4} y^{2}}-\sqrt[3]{x^{2} y^{4}}\right)\right]^{2}$
$+\left[\frac{1}{\sqrt{2}}\left(\sqrt{\sqrt[3]{x^{4} y^{2}}+\sqrt[3]{x^{2} y^{4}}+1}\right)\left(\sqrt[3]{x^{4} y^{2}}-1\right)\right]^{2}+\left[\frac{1}{\sqrt{2}}\left(\sqrt{\sqrt[3]{x^{4} y^{2}}+\sqrt[3]{x^{2} y^{4}}+1}\right)\left(\sqrt[3]{x^{2} y^{4}}-1\right)\right]^{2}+(\sqrt{2} x y)^{2}$
但杨路老师认为吴裕东先生的解答虽然“很有技巧,但不大合要求.因为把它表成了根式的平方和,而要求是表成多项式的平方和”.
姚勇发现, $x^2 y^2 (x^2+y^2-1)$ 可拆分为\begin{array}{c}x^{2} y^{2}\left(x^{2}+y^{2}-1\right)+1=r 1^{2}+r 2^{2}+r 3^{2}+r 4^{2} \\ r 1=\frac{x^{3} y^{2}+x^{3} y-x}{x^{2}+1} \quad r 2=\frac{y^{2} x^{2}-x^{4} y-1}{x^{2}+1}, \\ r 3=\frac{y^{2} x^{2}-\sqrt{3} x y-x^{2}}{x^{2}+1} \quad r 4=\frac{\sqrt{3} x^{2} y+x y^{2}-x}{x^{2}+1}\end{array}并且说不能表示为多项式的平方和.

2~5#内容提要

$x^2 y^2 (x^2+y^2-3)+1$就是Motzkin多项式$p(x)=x^{4} y^{2}+x^{2} y^{4}-3 x^{2} y^{2}+1$
证明“它不能表示为多项式的平方和”见4#
用4#的方法为什么能完全判定“不能表示为多项式”见2#
如何得出吴裕东先生的分拆式见5#
按5#可以把$x^6+y^6+z^6-3x^2y^2z^2$分拆成9项平方和
但$x^6+y^6+z^6-3x^2y^2z^2$最少能拆成4项, 见3#
这相当于$3$元的AM-GM不等式的sos证明.
关于$n$元的AM-GM不等式的sos证明, 见3#所引用的Hurwitz的德语论文

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 楼主| hbghlyj 发表于 2023-3-24 00:46


A note on mediated simplices | Arxiv
Abstract. Many homogeneous polynomials that arise in the study of sums of squares and Hilbert’s 17th problem are those formed by monomial substitutions into the arithmetic-geometric inequality. In 1989, Reznick [Math. Ann. 283, 431–464] gave a necessary and sufficient condition for such a form to have a representation as a sum of squares of forms, involving the arrangement of lattice points in the simplex whose vertices were the $n$-tuples of the exponents used in the substitution. Further, a claim was made, and not proven, that sufficiently large dilations of any such simplex will also satisfy this condition. The aim of this short note is to prove the claim, and provide further context for the result, both in the study of Hilbert’s 17th Problem and the study of lattice point simplices.
1. Introduction
In 1989, the second author considered [14] a class of homogeneous polynomials (forms) which had arisen in the study of Hilbert’s 17th Problem as monomial substitutions into the arithmetic-geometric inequality. The goal was to determine when such a form, which must be positive semidefinite, had a representation as a sum of squares of forms. The answer was a necessary and sufficient condition involving the arrangement of lattice points in the simplex whose vertices were the $n$-tuples of the exponents used in the substitution. Further, a claim was made in [14], and not proven, that sufficiently large dilations of any such simplex will also satisfy this condition. The aim of this short note is to prove the claim, and provide further context for the result, both in the study of Hilbert’s 17th Problem and the study of lattice point simplices. The second author is happy to acknowledge that the return to this claim was triggered by two nearly simultaneous events: an invitation to speak at the 2019 SIAM Conference on Applied Algebraic Geometry, and a request from Jie Wang for a copy of [15], which was announced in [14] but never written.
2. Preliminaries
We work with homogeneous polynomials (forms) in $\mathbb{R}[x]=\mathbb{R}\left[x_{1}, \ldots, x_{n}\right]$, the ring of real polynomials in $n$ variables. Write the monomial $x_{1}^{\alpha_{1}} \cdots x_{n}^{\alpha_{n}}$ as $x^{\alpha}$, for $\alpha=$ $\left(\alpha_{1}, \ldots, \alpha_{n}\right) \in \mathbb{Z}^{n}$. For $S \subset \mathbb{Z}^{n},\operatorname{cvx}(S)$ denotes the convex hull of $S$. For $p(x)=\sum_{\alpha} c(\alpha) x^{\alpha} \in \mathbb{R}[x]$, let $\operatorname{supp}(p)=\{\alpha \mid c(\alpha) \neq 0\}$, write $\operatorname{New}(p)$ for the Newton polytope of $p$, that is, $\operatorname{New}(p)=\operatorname{cvx}(\operatorname{supp}(p))$, and let $C(p)=\operatorname{New}(p) \cap \mathbb{Z}^{n}$.

A form $p \in \mathbb{R}[x]$ is positive semidefinite or psd if $p(x) \geq 0$ for all $x \in \mathbb{R}^{n}$. It is a sum of squares or sos if $p=\sum_{j} h_{j}^{2}$ for forms $h_{j} \in \mathbb{R}[x]$. Clearly, every sos form is psd. In 1888, D. Hilbert [8] proved that there exist psd forms which are not sos. The arithmetic-geometric inequality (or AGI) states that if $t_{i} \geq 0, \lambda_{i} \geq 0$ and $\sum_{i=1}^{n} \lambda_{i}=1$, then \[ \lambda_{1} t_{1}+\cdots+\lambda_{n} t_{n} \geq t_{1}^{\lambda_{1}} \cdots t_{n}^{\lambda_{n}}, \] with equality only if the $t_{i}$ 's are equal. In 1891, A. Hurwitz [9] gave a proof of the AGI, in which the key step was setting $\lambda_{i}=a_{i} / N$ where $a_{i} \in \mathbb{Z}^{n}$ with $\sum a_{i}=N$ for even $N$, and $t_{i}=x_{i}^{N}$. Under this substitution and a scaling, one obtains the form \[ a_{1} x_{1}^{N}+\cdots+a_{n} x_{n}^{N}-N x_{1}^{a_{1}} \cdots x_{n}^{a_{n}} . \] Hurwitz then proves that each such form is sos (in fact, a sum of squares of binomials), and hence psd. (He cites [8] to observe that this is not automatic.) For example, after a scaling and relabeling of the $x_{i}$ 's as $x, y, z$, we have \[ \begin{gathered} H(x, y, z):=x^{6}+y^{6}+z^{6}-3 x^{2} y^{2} z^{2} \\ =\frac{3}{2}\left(x^{2} y-y z^{2}\right)^{2}+\left(x^{3}-x y^{2}\right)^{2}+\frac{1}{2}\left(x^{2} y-y^{3}\right)^{2}+\left(z^{3}-y^{2} z\right)^{2}+\frac{1}{2}\left(y z^{2}-y^{3}\right)^{2} . \end{gathered} \] For more on Hurwitz' proof, see [13], where Eq. (3.5) gives a representation of $H$ as a sum of four squares, one of which is the square of a trinomial.

The first explicit example of a psd form which is not sos was presented in 1967 by T. Motzkin [11]. It, too, arises as a substitution into the AGI: let $t_{1}=x^{4} y^{2}, t_{2}=$ $x^{2} y^{4}, t_{3}=z^{6}, \lambda_{i}=\frac{1}{3}$ and scale: \[ M(x, y, z):=x^{4} y^{2}+x^{2} y^{4}+z^{6}-3 x^{2} y^{2} z^{2} \] The proof that $M$ is not sos was based on a preliminary argument that if $M=\sum h_{j}^{2}$, then $h_{j}(x, y, z)=c_{1 j} x^{2} y+c_{2 j} x y^{2}+c_{3 j} z^{3}+c_{4 j} x y z$ : the coefficient of $x^{2} y^{2} z^{2}$ in $\sum h_{j}^{2}$ is then $\sum c_{4 j}^{2} \neq-3$. The argument of Motzkin's proof was formalized in [12], where it is shown that, in general, $p=\sum h_{j}^{2}$ implies that $C\left(h_{j}\right) \subseteq \frac{1}{2} C(p)$. The following machinery was developed in [12,14] to analyze such forms. Consider a set $\mathcal{U}=\left\{u_{1}, \ldots, u_{m}\right\}$ with $u_{i} \in\left(2 \mathbb{Z}_{\geq 0}\right)^{n}$ and $\sum_{j=1}^{m} u_{i j}=2 d$. We further assume that $\operatorname{cvx}(\mathcal{U})$ is a simplex, and $w \in \operatorname{cvx}(\mathcal{U}) \cap \mathbb{Z}^{n}$ has the barycentric representation $w=\sum \lambda_{i} u_{i}, \lambda_{i} \geq 0$ and $\sum_{i=1}^{m} \lambda_{i}=1$. In this way, the substitution $\left\{t_{i}=x^{u_{i}}\right\}$ into the AGI yields a psd form of degree $2 d$ \[ p(x)=\lambda_{1} x^{u_{1}}+\cdots+\lambda_{m} x^{u_{m}}-x^{w} . \] This was called an agiform in [14] and the set $\mathcal{U}$ was called a trellis. Observe that $C(p)=\operatorname{cvx}(\mathcal{U}) \cap \mathbb{Z}^{n}$. More generally, a polynomial $p(x)=\sum_{i=1}^{m} a_{i} x^{u_{i}}-b x^{w}$ with $a_{i}>0$ for $i=1, \ldots, m$ is called a circuit polynomial. Circuit polynomials have recently been studied by M. Dressler, J. Forsgård, S. Iliman, T. de Wolff, and J. Wang; see for example $[10],[2],[3],[16]$. Interest in circuit polynomials is in part due to their use in finding efficiently-computable certificates of positivity based on the AGI, which are then independent of sos representations.

There is a geometric criterion which determines whether an agiform is sos.
Definition. Suppose $\mathcal{U}$ is given as above, and let $S \subset \operatorname{cvx}(\mathcal{U}) \cap \mathbb{Z}^{n}$ be a set of lattice points containing the $u_{i}$ 's. Then $S$ is $\mathcal{U}$-mediated if for every $y \in S$, either $y=u_{i}$ for some $i$, or there exist $z_{1} \neq z_{2} \in S \cap(2 \mathbb{Z})^{n}$ so that $y=\frac{1}{2}\left(z_{1}+z_{2}\right)$. In other words, $S$ is $\mathcal{U}$-mediated if every point in $S$ is either a vertex of $\mathcal{U}$ or an average of two different even points in $S$.
Theorem 2.1. [14, Cor. 4.9] With $\mathcal{U}, \lambda_{i}$ as above, the agiform $\lambda_{1} x^{u_{1}}+\cdots+\lambda_{m} x^{u_{m}}-x^{w}$ is sos if and only if there is a $\mathcal{U}$-mediated set containing $w$.

Let $\mathcal{U}_{1}=\{(4,2,0),(2,4,0),(0,0,6)\}$ and $\mathcal{U}_{2}=(\{(6,0,0),(0,6,0),(0,0,6)\} .$ Up to scaling, both $H$ and $M$ are agiforms, since $w=(2,2,2)$ is the centroid to both $\operatorname{cvx}\left(\mathcal{U}_{1}\right)$ and $\operatorname{cvx}\left(\mathcal{U}_{2}\right)$. By Theorem 2.1, $M$ is not sos because $\operatorname{cvx}\left(\mathcal{U}_{1}\right) \cap(2 \mathbb{Z})^{3}=\left\{u_{1}, u_{2}, u_{3}, w\right\}$ and it is impossible to write $w$ as an average of two different members of this set. However, it is easy to check that the set \[ S=\{(6,0,0),(0,6,0),(0,0,6),(2,2,2),(4,2,0),(2,4,0),(0,2,4),(0,4,2)\} \] is $\mathcal{U}_{2}$-mediated, providing an independent proof that $H$ is sos. We refer the reader to [14] for the separate proofs of the necessity and sufficiency in Theorem 2.1.
3. MAIN THEOREM
The following theorem was asserted in [14, Prop.2.7].
Theorem 3.1. For every integer $k \geq \max \{2, m-2\}, \operatorname{cvx}(k \mathcal{U}) \cap \mathbb{Z}^{n}$ is $(k \mathcal{U})$-mediated.
Corollary 3.2. Any agiform $p \in \mathbb{R}\left[x_{1}, \ldots, x_{n}\right]$ can be written as a sum of squares of forms in the variables $x_{i}^{1 / k}$ for $k \geq \max \{2, m-2\}$.

To prove Theorem 3.1, we show that any non-vertex $w \in \operatorname{cvx}(k \mathcal{U}) \cap \mathbb{Z}^{n}$ is the average of two different points in $\operatorname{cvx}(k \mathcal{U}) \cap(2 \mathbb{Z})^{n}$. The proof for $k \geq m-1$ is easy, while the proof for the remaining case $(m \geq 4$ and $k=m-2)$ is more delicate. We defer the discussion of Corollary $3.2$ to the next section.

We start with some notation and lemmas. First, recall that $t \in \mathbb{R}$ may be written as $t=\lfloor t\rfloor+\{t\}$, where $\lfloor t\rfloor \in \mathbb{Z}$ and $\{t\} \in[0,1)$. If $v=\sum a_{i} u_{i} \in \operatorname{cvx}(k \mathcal{U})$ with $a_{i} \in \mathbb{Z}_{\geq 0}, \sum a_{i}=k$, then we say that $v$ is a bead. Observe that beads are always even.

Lemma 3.3. Suppose $k>1$ and $v \in \operatorname{cvx}(k \mathcal{U})$ is a non-vertex bead. Then $v$ is an average of two different beads in $\operatorname{cvx}(k \mathcal{U})$.

Proof. Suppose that $v=\sum a_{i} u_{i}$ is a non-vertex bead. At least two of the $a_{i}$ 's must be positive; without loss of generality, suppose $a_{1}, a_{2} \geq 1$. Then $v$ is the average of the beads $v \pm\left(u_{1}-u_{2}\right)$ in $\operatorname{cvx}(k \mathcal{U})$

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 楼主| hbghlyj 发表于 2023-3-24 01:35
A quantitative version of Hurwitz' theorem on the arithmetic-geometric inequality, Bruce Reznick
式(2.9)\begin{aligned} 2 u^{6}+2 v^{6}+2 w^{6}-6 u^{2} v^{2} w^{2}= & 2\left(u^{3}-u v^{2}\right)^{2}+2\left(w^{3}-v^{2} w\right)^{2} \\ & +\left(u^{2} v-v^{3}\right)^{2}+\left(v w^{2}-v^{3}\right)^{2}+3\left(u^{2} v-v w^{2}\right)^{2}\end{aligned} Screenshot 2023-03-23 at 17-35-28 Journal for Pure and Applied Mathematics - GDZ.png
式(3.5)\begin{array}{l}4\left\{u^{6}+v^{6}+w^{6}-3 u^{2} v^{2} w^{2}\right\} \\ \quad=4\left(u^{2} v-v^{3}\right)^{2}+4\left(u^{2} w-w^{3}\right)^{2}+7\left(u v^{2}-u w^{2}\right)^{2}+\left(u v^{2}+u w^{2}-2 u^{3}\right)^{2}\end{array}
Screenshot 2023-03-23 at 17-33-37 Journal for Pure and Applied Mathematics - GDZ.png

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 楼主| hbghlyj 发表于 2023-3-24 01:41
SOS Proofs and the Motzkin Polynomial

Let $P(x)$ be a polynomial with real coefficients such that $P(x) \ge 0$ for every real $x$. Prove that$$p(x)=f_1(x)^2+f_2(x)^2+.......+f_n(x)^2$$
Proof: Let $p(x)=r(x)q(x)$, where $r(x)$ has all real solutions and $q(x)$ has all non-real solutions. Since $p(x)\geq0$ for all $x$, in can never cross the $x$-axis and thus all real roots must have even multiplicity, so $r(x)=s(x)^2$ for some polynomial $s(x)$. Then, since the polynomial has real coefficients, every non-real root must also have its complex conjugate as a solution. Expanding these factors we get $(x-a-bi)(x-a+bi)=(x-a)^2+b^2$. Expanding out $q(x)$ fully this way should show it is the sum of the squares of polynomials, so then $q(x)r(x)=q(x)s(x)^2=p(x)$ also has this property.$\square$
(实际上,可以不经过因式分解,p减去p的最小值是有偶重根的多项式,除去这根的多项式记为q,q减去q的最小值,如此下去...)

One can put this result in the context of Hilbert's 17th Problem:
The same result is false if we consider polynomials in more than one variable, a counter-example being
\[1-3x^2y^2+x^4y^2+x^2y^4.\]
However, amazingly, the result becomes true again if we allow rational functions instead of polynomials (and now in any number of variables). This is the content of Hilbert's 17th Problem which was solved by Artin.
In the example above, the "solution" would be
\[1-3x^2y^2+x^4y^2+x^2y^4=\left(\frac{1+x^2-2x^2y^2}{1+x^2}\right)^2+\left(\frac{x(1-x^2)y^2}{1+x^2}\right)^2+\left(\frac{x(1-x^2)y}{1+x^2}\right)^2+\left(\frac{x^2(1-x^2)y}{1+x^2}\right)^2.\]

Motzkin Polynomial Non-negativity
Motzkin Polynomial:
$$
p(x, y)=x^{4} y^{2}+x^{2} y^{4}-3 x^{2} y^{2}+1
$$
Applying AM-GM with $x^{4} y^{2}, y^{2} x^{4}, 1$,
$x^{2} y^{2}=\sqrt[3]{\left(x^{4} y^{2}\right) \cdot\left(y^{2} x^{4}\right) \cdot 1} \leq \frac{x^{4} y^{2}+y^{2} x^{4}+1}{3}$
Multiplying this by 3,  $p(x, y) \geq 0$

Newton Polytope
Given a polynomial, assign a point to each monomial based on the degree of each variable. Examples:
1. $x^{2} y$ is assigned the point $(2,1)$
2. $y^{5}$ is assigned the point $(0,5)$
3. $x y^{2} z^{3}$ is assigned the point $(1,2,3)$
The Newton polytope of a polynomial is the convex hull of the points assigned to each monomial.

Newton Polytope of a Sum of Squares
Let $f$ be a sum of squares, i.e. $f=\sum_{j} g_{j}^{2}$
Claim: The Newton polytope of $f$ is $2 X$ where $X$ is the convex hull of all the points corresponding to some monomial in some $g_{j}$
Proposition: If $p, q$ are monomials with corresponding points $a, b$ then $p q$ corresponds to the point $a+b$
One direction: Let $X_{j}$ be the Newton polytope of $g_{j}$. The Newton polytope of $g_{j}^{2} \subseteq 2 X_{j} \subseteq 2 X$. Thus, the Newton polytope of $f \subseteq 2 X$.
Other direction: If $p, q, r$ are monomials where $p r=q^{2}$ and $a, b, c$ are the corresponding points, $a+c=2 b$
Corollary: If $b$ is a vertex of $X$ corresponding to a monomial $q$ then if
1. $p, r$ are monomials appearing in some $g_{j}$ (and thus their corresponding points $a, c$ are in $X$ )
2. $p r=q^{2}$
then $p=r=q$ as otherwise $b$ would be between $a$ and $c$ and thus not a vertex of $X$
Corollary: If $b$ is a vertex of $X$ corresponding to a monomial $q$ then $q^{2}$ appears with positive coefficient in $f=\sum_{j} g_{j}^{2}$.
This implies that $2 X \subseteq$ the Newton polytope of $f$
Putting everthing together, the Newton polytope of $f$ is $2 X$.

Motzkin Polynomial Newton Polytope
Motzkin polynomial:$p(x)=x^{4} y^{2}+x^{2} y^{4}-3 x^{2} y^{2}+1$

If $p(x)$ were a sum of squares of polynomials, their corresponding points would have to be inside the following polytope.


Motzkin is not a Sum of Squares
If $p(x)=x^{4} y^{2}+x^{2} y^{4}-3 x^{2} y^{2}+1$ were a sum of squares of polynomials, it would have to be a sum of terms of the form
$$
\left(a x^{2} y+b x y^{2}+c x y+d\right)^{2}
$$
However, no such term has a negative coefficient of $x^{2} y^{2}$. Contradiction.

jump.dev/SumOfSquares.jl/stable/generated/Getting%20started/motzkin/
en.wikipedia.org/wiki/Positive_polynomial
people.kth.se/~apote/SOSseminar/
math.tamu.edu/~sottile/conferences/CIMPA17/Notes/Powers.pdf

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 楼主| hbghlyj 发表于 2023-3-24 01:45
1#吴裕东的分拆式可以写成
\[x^4 y^2+x^2y^4+1-3x^2y^2=\frac12\left(\sqrt[3]{x^{4} y^{2}}+\sqrt[3]{x^{2} y^{4}}+1\right)\left[\left(\sqrt[3]{x^{4} y^{2}}-\sqrt[3]{x^{2} y^{4}}\right)^{2}
+\left(\sqrt[3]{x^{4} y^{2}}-1\right)^{2}+\left(\sqrt[3]{x^{2} y^{4}}-1\right)^{2}\right]\]
把$x^2,y^2$换成$x,y$得
\[x^2y+xy^2+1-3xy=\frac12\left(\sqrt[3]{x^2y}+\sqrt[3]{xy^2}+1\right)\left[\left(\sqrt[3]{x^2y}-\sqrt[3]{xy^2}\right)^{2}
+\left(\sqrt[3]{x^2y}-1\right)^{2}+\left(\sqrt[3]{xy^2}-1\right)^{2}\right]\]
把$x,y$换成$x^3,y^3$得
\[x^6y^3+x^3y^6+1-3x^3y^3=\frac12\left(x^2y+xy^2+1\right)\left[\left(x^2y-xy^2\right)^{2}
+\left(x^2y-1\right)^{2}+\left(xy^2-1\right)^{2}\right]\]把$x^2y,xy^2$换成$x,y$得\[x^3+y^3+1-3xy=\frac12\left(x+y+1\right)\left[\left(x-y\right)^{2}
+\left(x-1\right)^{2}+\left(y-1\right)^{2}\right]\]
齐次化
\[x^3+y^3+z^3-3xyz=\frac12\left(x+y+z\right)\left[\left(x-y\right)^{2}
+\left(x-z\right)^{2}+\left(y-z\right)^{2}\right]\]

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