 # A sufficient condition for local nonnegativity

A real polynomial f is called local nonnegative at a point p, if it is nonnegative in a neighbourhood of p. In this paper, a sufficient condition for determining this property is constructed. Newton's principal part of f (denoted as f_N) plays a key role in this process. We proved that if every F-face, (f_N)_F, of f_N is strictly positive over (ℝ∖ 0)^n, then f is local nonnegative at the origin O.

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## 1 Introduction

Before begin, several notations will be given. Let denote the set of all natural numbers, real numbers, positive real numbers and nonnegative real numbers, respectively. Let us start with a question.

What is local nonnegative? Here is an explicit definition.

Let be a real polynomial. Consider the point . If there is a ’s neighbourhood such that

 ∀x∈B(ϵ,p), f(x)≥0,

then is called local nonnegative at .

There is a variety of mathematical problems related to local nonnegativity. For instance, the problem of determining the local minimum of a polynomial, finding the isolated singular point of a real algebraic variety, constructing Lyapunov functions, and proving algebra inequalities.

So, what kind of function is local nonnegative at a point?

This problem will be discussed in two cases. Pick a point , and let be a real function. Case 1: . It is a trivial situation. Since a familiar result said that if is continuous in an open set , and there is a point such that , then, there exists a neighbourhood of such that is still positive in the neighbourhood. Thus, it is obvious that is local nonnegative at the point . Case 2: . It becomes a tricky problem. To facilitate discussion, we will assume to be a polynomial, and denote as the origin. By coordinate transformation, the problem that whether is local nonnegative at an arbitrary point can be reduced to determining whether it is local nonnegative at the origin. Hence, throughout the paper, the local nonnegativity of means that it is local nonnegative at the origin. Then, there is a renowned result about Hessian matrix in the standard textbooks (Apo ,Zor ,CS ), which may come in handy.

###### Theorem 1.

Let be an open set. Assume that all second-order partial derivatives of exist and are continuous at the origin , and assume further that

 f(O)=0, ∂f∂xi(O)=0, 1≤i≤n.

If Hessian matrix

 Hf(O)=(∂2f∂xi∂xj(O))1≤i,j≤n

is positive definite, then is local nonnegative (and has no other zeros neighbouring ).

Unfortunately, this law will failed if Hessian matrix is not positive definite. Here is an example that Hessian matrix is a singular matrix (i.e., the determinant of Hessian matrix is ), whereas the polynomial is local nonnegative.

###### Example 1.

For an arbitrary real number , the polynomial

 f=x2+y4+z6−sxy2z3, f∈R[x,y,z].

is local nonnegative.

Roughly speaking, if Hessian matrix is singular, then it is tough to determine local nonnegativity.

However, it is easy to find a necessary condition of local nonnegativity:

Let be arbitrary polynomials in a variable. Then the polynomial is local nonnegative only if, , or the degree of the leading term (i.e. the term with lowest degree ) of is even and its leading coefficient (i.e. the coefficient of leading term ) is positive.

This result can be used to prove that a polynomial is not local nonnegative. For instance, if there exist polynomials , such that the leading coefficient of  is negative, then is not local nonnegative. Here is an example.

###### Example 2.

Consider the polynomial

 f=x16+y18−x7y3+x12y15+x4y2−2x3y3+y4x2.

Pick , the above polynomial can written as

 f(t,t)=t27+t18+t16−t10=t10(−1+t6+t8+t17).

It is obvious that the coefficient of leading term is . Then, when . Thus, is not local nonnegative.

In contrast, it is a hard problem to obtain a sufficient condition of local nonnegativity. We hardly get any other result about the sufficient condition except the Theorem 1. The aim of this paper is to construct an available sufficient condition for determining the local nonnegativity.

## 2 Preliminaries

Let with . Then, a polynomial can be written as

 f=∑αaαxα.

The support of is defined as , and let

 Sup+(f)={α∈Nn| aα>0},Sup−(f)={α∈Nn| aα<0}.

Here and stand for the positive and negative support of respectively. Denote

 f+=∑α∈Sup+(f)aαxα, f−=∑α∈Sup−(f)aαxα.

Let be the Newton polytope of . It is the convex hull of , that is,

 N(f)=Conv{α∈Sup(f)}.

Let be a set of all the nonempty faces of . For , the -face of the polynomial is

 fF=∑α∈(F⋂Sup(f))aαxα.

Noticed that when .

###### Definition 2.1.

Let denote the set of vertices of . Define

 fV=∑υ∈Vxυ.

is called the characteristic polynomial of vertices of .

To study the local nonnegativity of polynomials, we need to present the definition of Newton’s principal part. For a polynomial , define set of points

 Nf=Conv(⋃α∈Sup(f)(α+Rn≥0)). (1)

The compact face of is denoted as , which is called the Newton’s diagram of . Then, Newton’s principal part is defined as follows.

###### Definition 2.2.

The polynomial

 fN=∑α∈(Γf⋂Sup(f))aαxα,

is called Newton’s principal part of .

Thus, an arbitrary polynomial can be written as

 f=fN+f¯¯¯¯N,

where is the Newton’s principal part of , and is the remainder (it may be zero polynomial). Especially for the polynomial in one variable, is the term of with lowest degree.

The definition of Newton’s principal part of a polynomial is a key to this paper. It is easy to see that for a polynomial , the condition that Hessian matrix is positive definite in Theorem 1 is exactly the requirement that the Newton’s principal part is a positive definite quadratic form. Hence, a normal idea is to generalize this result and hope that a polynomial is local nonnegative based on the premise that its Newton’s principal part is strictly positive on . Unfortunately, this conjecture is proved wrong. The counterexample is Example 2. Given

 f=x16+y18−x7y3+x12y15+x4y2−2x3y3+y4x2,

its Newton’s principal part is

 fN = x16+x4y2−2x3y3+y4x2+y18 = x16+y18+(x2y−y2x)2.

is strictly positive on , whereas is not local nonnegative. Moreover, we noticed that one -face, , of is not strictly positive definite on . Because it vanishes when . The Newton’s diagram of is a polygonal line in Fig. 1.

It is time to present the main result of this paper.

###### Theorem 2.

Let be the Newton’s principal part of a polynomial (see Definition 2.2). If every -face, , of satisfies

 ∀x∈(R∖0)n, (fN)F>0,

then is local nonnegative.

## 3 Proof of the Main Result

In this section, we will present the proof of the main result (Theorem 2). Firstly, several lemmas are developed.

###### Lemma 3.1.

Given , can be written as a convex combination of  , i.e., there exists nonnegative real numbers satisfying

 β=λ1α(1)+⋯+λmα(m), λ1+⋯+λm=1.

Then, there exists a positive constant such that

 ∀x∈Rn≥0, xα(1)+⋯+xα(m)≥kβxβ.
###### Proof.

By generalized mean inequality HLP ,

 ∀x∈Rn≥0, λ1xα(1)+⋯+λmxα(m)≥xλ1α(1)+⋯+λmα(m)=xβ.

Let . Then,

 xα(1)+⋯+xα(m)≥kβxβ

holds. ∎

###### Lemma 3.2 (HandelmanHan ; At ).

Let be a nonzero polynomial. Then, there exists a positive integer such that all the coefficients of are nonnegative real numbers if and only if,

 ∀F∈F(h), ∀x∈Rn>0, hF>0.
###### Remark 1.

All of the coefficients of are nonnegative real numbers.

###### Lemma 3.3.

Let be a nonzero polynomial. Then, there exists a constant satisfying

 ∀x∈Rn>0, h≥τhV

if and only if,

 ∀F∈F(h), ∀x∈Rn>0, hF>0,

where is a characteristic polynomial of vertices of .

###### Proof.

: Let be the set of vertices of , where is the Newton polytope of . Then, the characteristic polynomial of vertices of is

 hV=d∑i=1xv(i).

Moreover, is also the set of vertices of because and have the same Newton polytope. For the polynomial and , the Newton polytope of their multiplication corresponds to the Minkowski sum of their polytopes[7, 8], that is,

 N(pq)=N(p)+N(q).

Hence,

 N(h(h+−h−)m)=N(h)+N(h)+⋯+N(h)m+1=N((h+−h−)m+1).

Note that are vertices of both and . Thus we have

 (m+1)v(1),⋯,(m+1)v(d)∈Sup(h(h+−h−)m).

Applying Handelman’s theorem (Lemma 3.2) to , then there exists a positive integer such that all coefficients of are nonnegative real numbers. There exists a positive constant , therefore, satisfying

 ∀x∈Rn>0, h(h+−h−)m≥k1d∑i=1x(m+1)v(i). (2)

By Hölder’s inequality, it holds that

 dmd∑i=1x(m+1)v(i)≥(d∑i=1xv(i))m+1=(hV)m+1. (3)

(2), together with (3) indicate that

 h≥k1(hV)m+1dm(h+−h−)m. (4)

It is well known that all the points on a convex polyhedron can be presented by convex combinations of the elements in the set of its vertices. Hence, each term of can leads to an inequality by using Lemma 3.1. Add these inequalities, and then, there exists a positive constant satisfying

 ∀x∈Rn>0, (h+−h−)≤khV(x). (5)

Pick . Combining (4) and (5) imply that

 h≥τhV.

: Conversely, assume that there exists a -face  and a point  such that  . On one hand, by Definition 2.1, it is clearly,

 x′∈Rn>0, (hF)V(x′)>0.

On the other hand, choose a vector

such that the dot product arrives at the maximum on the face , which is denoted as . Note that the dot product   arrives at the same on .

Let . Then,

 h(x′1eν1t,…,x′neνnt)=∑α∈Sup(h)aα(x′)αe(α⋅ν)t,

and

 hV(x′1eν1t,…,x′neνnt)=∑α∈Vaα(x′)αe(α⋅ν)t.

Hence, we have

 limt→∞e−Mth(x′1eν1t,…,x′neνnt)=hF(x′), (6)

and

 limt→∞e−MthV(x′1eν1t,…,x′neνnt)=(hF)V(x′). (7)

If there is a positive constant  satisfying

 ∀x∈Rn>0, h≥τhV. (8)

Then, (6) together with(7) and (8), indicate that

 hF(x′)≥τ(hF)V(x′), (9)

contradicting the assumption that  while . ∎

Before proving the main result, we continue discuss the polynomial of Example 1, which will illuminate the proof of Theorem 2.

Consider

 g=x2+y4+z6−sxy2z3,

where is a given real number. If write as a polynomial in , then its lowest degree is with coefficient . Namely, the coefficient of the term in with lowest degree is indefinite. However,

is still local nonnegative. In fact, by inequality of arithmetic and geometric mean, it holds that

 x2+y4+z6−3x23y43z2≥0.

Thus can be written as

 g=(x2+y4+z6−3x23y43z2)+(x13y23z)2(3−sx13y23z).

Noticed that

 lim(x,y,z)→(0,0,0)sx13y23z=0.

Hence, is local nonnegative.

###### Theorem 2.

Let be the Newton’s principal part of a polynomial (see Definition 2.2). If every -face, , of satisfies

 ∀x∈(R∖0)n, (fN)F>0,

then is local nonnegative.

###### Proof.

The result obviously holds if since is global nonnegative, and so, in particular it is local nonnegative.

Next, if , then can be written as

 f=fN+f¯¯¯¯N=fN+∑β∈Sup(f¯¯¯¯N)aβxβ. (10)

Let T be the number of elements in the set , i.e. . By hypothesis, is nonnegative over , this means that for all , and, hence, components of each vertex must be even ( possibly be zero ). Thus, it reduces to discussing whether the result holds in . Consequently, assume that .

Firstly, by Lemma 3.3, there exists a positive constant such that

 fN−τ(fN)V≥0. (11)

Secondly, consider every term of . By (1), there exists a monomial such that

 ^β∈N(fN), β=^β+δ, δ∈Rn≥0.

Note that By Lemma 3.1, there exists a positive constant satisfying

 (fN)V≥k^βx^β. (12)

Moreover,

 (13)

This, together with (10) and (11), yields

 f = fN+f¯¯¯¯N = (fN−τ(fN)V)+(τ(fN)V+f¯¯¯¯N) = (fN−τ(fN)V)+∑β∈Sup(f¯¯¯¯N)τT((fN)V−k^βx^β)+∑β∈Sup(f¯¯¯¯N)x^β(τk^βT+aβxδ).

It is easy to see that

 lim∥x∥→0(τkβT+aβxδ)=τkβT>0. (14)

This, together with (11) and (12), implies that is local nonnegative. Finally we proved the desired result. ∎

Two immediate consequences of Theorem 2 are as follows.

###### Corollary 1.

Let be the Newton’s principal part of a polynomial . If is homogenous and, moreover, positive definite ( i.e., is strictly positive over ), then is local nonnegative.

###### Corollary 2.

Let be the Newton’s principal part of a polynomial , and let is the characteristic polynomial of vertices of . If every -face, , of satisfies

 ∀x∈(R∖0)n, (fN)F>0 (15)

and, moreover, there is a term such that

 x2dii∈(fN)V, ∀xi∈Rn,i=1,…,n. (16)

Then is an isolated singular point of the variety .

For an example of Corollary 2, consider the polynomial in Example 1.

 f=x2+y4+z6−sxy2z3

It is easy to see that for every -face of , (15)holds and so are (16). Then is an isolated singular point of the variety .

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