On the comparison of powers of differential operators (polynomials)

Abstract

Necessary and sufficient conditions are obtained for the generalized-homogeneous or common polynomial P to be more powerful then the generalized-homogeneous polynomial (in particular monomial) Q (denoted as $P>Q$ or $Q<P).$ Conditions are formulated in terms of orders of generalized-homogeneity of these polynomials and the multiplicities of their roots.

1 Notation, statement of the problem and preliminary results

Let ${\mathbb {E}}^n$ and ${\mathbb {R}}^n$ be n-dimensional Euclidean spaces of points (vectors) $x=(x_1,\ldots ,x_n)$ and $\xi = (\xi _1,\ldots ,\xi _n)$ respectively, ${\mathbb {R}}^{n,+}: = \{ \xi \in {\mathbb {R}}^n: \xi _{j} \ge 0 (j = 1,\ldots , n) \},$ ${\mathbb {R}}^{n, 0}: = \{ \xi \in {\mathbb {R}}^n: \xi _{1} \ldots \xi _{n} \ne 0 \},$ ${\mathbb {C}}^{n}:= {\mathbb {R}}^n \times i {\mathbb {R}}^n,$ ${\mathbb {N}}$ denotes the set of all natural numbers, ${\mathbb {N}}_0:= {\mathbb {N}} \cup \{0\}$, ${\mathbb {N}}_{0}^{n}:= {\mathbb {N}}_0\times \ldots \times {\mathbb {N}}_0 = \{ \alpha = (\alpha _{1},\ldots , \alpha _{n}): \alpha _{i} \in {\mathbb {N}}_0 (i = 1,\ldots ,n) \}$ is the set of all n-dimensional multi-indices, i.e. the set of all points with non-negative integer coordinates.

For $\, t> 0, \, \, \xi \in {\mathbb {R}}^n, \, \alpha \in {\mathbb {N}}_{0}^{n}, \,$ $\lambda \in {\mathbb {R}}^{n, +} \cap {\mathbb {R}}^{n, 0} \,$ and $\, \nu \in {\mathbb {R}}^{n,+} \,$ we denote $\, |\xi |:= \sqrt{\xi ^{2}_1+\cdots +\xi ^{2}_n},$ $\, |\xi , \lambda |:= \sqrt{|\xi _1|^{2/\lambda _{1}}+\cdots + |\xi _n|^{2/\lambda _{n}}},$ $| \nu |:=\nu _1+\cdots +\nu _n,$ $\, \xi ^{\alpha }:= \xi _{1} ^{\alpha _{1}} \ldots \, \xi _{n} ^{\alpha _{n}},$ $\, | \xi ^{\nu }|: = | \xi _{1} |^{\nu _{1}} \ldots | \xi _{n} |^{\nu _{n}}, \,$ $t^{\lambda }:= (t^{\lambda _{1}},\ldots , t^{\lambda _{n}} ),\,$ $\, t^{\lambda } \, \xi := (t^{\lambda _{1}} \,\xi _{1},\ldots , t^{\lambda _{n}} \,\xi _{n}).$

For $\alpha \in {\mathbb {N}}_{0}^{n} \,$ we denote $D^\alpha := D_{1}^{\alpha _1}\ldots D_{n}^{\alpha _n},$ where $D_j = {1\over i} \partial /\partial x_j \,\,$ or $D_{j} = \partial /\partial \xi _j \,\,$ $(j = 1,\ldots n). \,$ Let ${\mathcal {A}} = \{\nu ^{j} = (\nu _{1}^{j},\ldots , \nu _{n}^{j}) \, \}_{j = 1}^{M}$ be a finite set of points $\, \nu ^{j} \in {\mathbb {R}}^{n,+}.$ We call the Newton polyhedron of the set ${\mathcal {A}}$ the least convex hull (which is a polyhedron) $\Re = \Re ({\mathcal {A}})$ in ${\mathbb {R}}^{ n},$ containing all points of ${\mathcal {A}}.$

A polyhedron $\Re \,$ with vertices in $\, {\mathbb {R}}^{n, +} \,$ is said to be complete if $\Re \,$ has a vertex at the origin of coordinate axis and vertices on each coordinate axis of $\, {\mathbb {R}}^{n, +}, \,$ other than the origin.

Everywhere below, except when otherwise specified, we will assume that all polyhedrons are complete.

By $\, \Re _{i}^{k} \,\,\, (i = 1,\ldots , M_{k}, k = 0,1,\ldots , n-1 )$ we denote the $\ k -$dimensional faces of a polyhedron $\, \Re \,$. The faces of the Newton polyhedron will be considered (by definition) closed sets. It will be convenient for us to denote the set of vertices of the polyhedron $\, \Re \,$ by $\, \Re ^{0}, \,$ so $\, \Re ^{0} = \{ \Re _{i}^{0} \,\, (i = 1,\ldots , M_{0}) \}.$

The unit outward normal to a supporting hyper-plane of a polyhedron $\, \Re \,$ containing the face $\, \Re _{i}^{k} \,$ and not containing any other face of dimension greater then $\, k \,$ will be simply called an outward normal $(\Re$ - normal) of the face $\, \Re _{i}^{k}. \,$ Thus a given unit vector $\, \lambda \,$ can serve as an outward normal to only one face of $\, \Re . \,$ We denote by $\, \Lambda _{i}^{k} \,$ the set of all outward normals of the face $\, \Re _{i}^{k} \,$ $\, \, (i = 1,\ldots , M_{k}, k = 0,1,\ldots , n-1 ).$ Note that either the set $\, \Lambda _{i}^{k} \,$ consists of one vector (when $\, k = n - 1 ),$ or it is an open set (when $\,0 \le k < n - 1 ).$

For any $\, \lambda \in \Lambda _{i}^{k} \,$ $\, \, ( 1 \le i \le M_{k}, 0 \le k \le n-1 )$ there exists a number $d = d_{i,k} = d_{i,k}(\lambda ) \ge 0 \,$ such that $\, (\lambda , \alpha ) = d \,$ for all $\, \alpha \in \Re _{i}^{k} \,$ and $\, (\lambda , \alpha ) < d \,$ for all $\, \alpha \in \Re {\setminus } \Re _{i}^{k}. \,$ Moreover, the $\, \Re -$normal of the $\, (n - 1) -$dimensional face $\Re _{i}^{n - 1} \,$ $\,\, (1 \le i \le M_{n - 1}) \,$ of the polyhedron $\Re \,$ (hence, the number $d_{i,n - 1}(\lambda ) )$ is determined uniquely.

Definition 1.1

(see [32]) A face $\, \Re _{i}^{k} \,$ of a polyhedron $\, \Re \,$ is said to be principal if one of the following (obviously equivalent) conditions is satisfied. 1) $\, \Re _{i}^{k} \,$ does not go through the origin 2) among the $\, \Re$-normals of this face there is one with at least one positive component. We say that a point $\alpha \in \Re$ is principal if $\alpha$ lies on some principal face of polyhedron $\Re$. Obviously, all subfaces of the principal face are principal.

Definition $1.1'$ (see [18]) A principal face $\, \Re _{i}^{k} \,$ $\, \, ( 1 \le i \le M_{k}, 0 \le k \le n-1 ) \,$ of the complete polyhedron $\, \Re \,$ is said to be completely regular if among the $\, \Re$-normals of this face there is one whose components are all positive. A complete polyhedron $\, \Re \,$ is called completely regular if all of its principal faces are completely regular.

Let $P(D) = P (D_{1},\ldots , D_{n}) = \sum _{\beta } \gamma _{\beta } \, D^{\beta }$ be a linear differential operator with constant coefficients and $P(\xi ) = \sum _{\beta } \gamma _{\beta } \, \xi ^{\beta }$ be its complete symbol (the characteristic polynomial). Here the sum goes over a finite set of multi-indices $(P):= \{ \beta \in {\mathbb {N}}_{0}^{n}: \gamma _{\beta } \ne 0 \}.$

The Newton polyhedron of the set $\,\, (P) \cup \{0 \}$, denoted by $\Re (P)$, is called the Newton polyhedron of the operator P(D) ( polynomial $P(\xi )$). Note that $\Re (P) \,$ may have a dimension less than $\, n.$ The Newton polyhedron of any operator $\, P(D) \,$ (polynomial $P(\xi )) \,$ is actually constructed as the Newton polyhedron of the operator $I+P(D) \,\,$ (polynomial $1 +P(\xi )),$ where $\, I \,$ is the identity operator.

Let $\, \Re (P) \,$ be the Newton polyhedron of the polynomial $P(\xi ) \,$ with $\, \Re _{i}^{k} \,\, ( i = 1,\ldots , M_{k}; k = 0,1,\ldots , n - 1 ) \,$ being its faces. The polynomial

$$\begin{aligned} P^{i,k} (\xi ): = \sum \limits _{\beta \in \Re _{i}^{k} } \gamma _{\beta } \, \xi ^{\beta } \,\,\,\,\, (1 \le i \le M_{k}; 0 \le k \le n) \end{aligned}$$

will be called the sub-polynomial of polynomial $\, P(\xi ), \,$ corresponding to the face $\, \Re _{i}^{k}. \,$

Definition 1.2

([14], XI.1 or [32]) Let $\lambda = (\lambda _{1},\ldots , \lambda _{n}) \,$ be a vector with rational components. A polynomial $R(\xi )$ is called $\lambda -$homogeneous (generalized homogeneous) of $\lambda -$order $\, d = d(\lambda ) \,$ (which is also a rational number), if $\, R(t^{\lambda } \, \xi ): =$ $R(t^{\lambda _{1}} \, \xi _{1},\ldots , t^{\lambda _{n}} \, \xi _{n})$ = $t^{d} \,\, R(\xi )$ for all $\xi \in {\mathbb {R}}^{n} \,\,$ and for any $t > 0.$

Note that whenever $\, \lambda _{1} = \lambda _{2} = \ldots = \lambda _{n} \, ( = 1), \,$ $R(\xi )$ is an ordinary homogeneous polynomial.

Lemma 1.1

[32] Let $\Re = \Re (P)$ be the Newton polyhedron of the polynomial $\, P \,$ and $\lambda$ be any $\Re -$normal to the face $\, \Re _{i}^{k} \,$ $(\lambda \in \Lambda _{i}^{k} )$ $(1 \le i \le M_{k}; 0 \le k \le n - 1)$ of the polyhedron $\, \Re . \,$ Then subpolynomial $\, P^{i, k} \,$ is $\lambda -$homogeneous.

Proof

Let the $\, k -$dimensional face $\, \Re _{i}^{k} \,$ is obtained by the intersection of $\, (n - 1) -$dimensional faces $\Re _{i_{1}}^{n - 1},\ldots , \Re _{i_{p}}^{n - 1}$ with outward normals $\, \lambda ^{1},\ldots , \lambda ^{p}$ respectively. Then, obviously, the vector $\lambda$ can be represented as any linear combination of vectors $\, \lambda ^{1},\ldots , \lambda ^{p}: \,\,$ $\lambda = a_{1} \lambda ^{1} +\cdots + a_{p} \lambda ^{p}.\,\,$ Since the equation of the $(n - 1) -$dimensional hyperplane passing through the face $\Re _{i_{j}}^{n - 1}$ can be written in the form $(\lambda ^{j}, \alpha ) = b_{j} \,\,$ $(j = 1,2,\ldots , p); \,\,$ for any $\, t>0 \,$ and $\, \xi \in {\mathbb {R}}^{n} \,$ we have $P^{i,k} (t^{\lambda } \, \xi ) = t^{d} \, P^{i,k} (\xi ), \,$ where $\, d = \sum _{j = 1}^{p} a_{j} \, b_{j}. \,$ This meanes that the polynomial $\, P \,$ is $\lambda -$homogeneous of $\lambda -$order $\, d.$ $\square$

Remark 1.1

(1) Note that if $\, \lambda \,$ is a unit vector   ($|\lambda | = 1$) and $\, P (\xi ) = \sum _{\alpha \in (P)} \gamma _{\alpha } \xi ^{\alpha } \,$ is a polynomial, then, clearly, there exist (uniquely defined) numbers $d_{j} (\lambda ) \,$ and $\lambda -$homogeneous polynomials $\, P_{j}: = P_{d_{j} (\lambda )} \,\,$ $(j = 0,1,\ldots ,M(\lambda ), \,\,$ $d_{0}(\lambda )> d_{1}(\lambda )> \ldots > d_{M(\lambda )}(\lambda ))$ such that polynomial $P(\xi )$ can be represented in the form

$$\begin{aligned} P(\xi ) = \sum \limits _{j=0}^{M(\lambda )} P_{j} (\xi ) = \sum \limits _{j=0}^{M(\lambda )} P_{d_{j} (\lambda )} (\xi ) = \sum \limits _{j=0}^{M(\lambda )} \sum \limits _{(\lambda , \alpha ) = d_{j} (\lambda )} \gamma _{\alpha } \, \xi ^{\alpha }, \end{aligned}$$

(1.1)

where the set of numbers $\, \{d_{j} = d_{j} (\lambda ) \} \,$ coincides with the finite set of values $\{(\lambda , \alpha ):\alpha \in \Re (P) \}.$

In particular, if $\, \lambda _{1} = \lambda _{2} = \cdots = \lambda _{n}$ then the polynomial $P(\xi )$ is represented as the sum of homogeneous polynomials and the representation (1.1) takes the form

Wherein, (1) if $\, \Re _{i}^{k} \,$ is some principal face of $\Re (P) \,$ and $\lambda \in \Lambda (\Re _{i}^{k})$, then $\, (\lambda , \alpha ) = d_{0}(\lambda )$ is the equation of the $\, (n - 1) -$dimensional support hyperplane to $\, \Re (P)$ with the outward (with respect to $\, \Re (P) ) \,\,$ normal $\, \lambda , \,$ containing this face, and $P_{d_{0} (\lambda )} (\xi ) \equiv P^{i,k} (\xi ).$

(2) It follows from lemma 1.1 that the sub-polynomial $\, P^{i,k} \,\, (1\le i \le M_{k}, 0 \le k \le n - 1)$ of polynomial $\, P \,$ is $\lambda -$homogeneous for any $\, \lambda \in \Lambda _{i}^{k} (\Re (P)), \,$ i.e. there exists a number $\, d_{i, k} = d_{i, k} (\lambda ) \ge 0 \,$ such that the sub-polynomial $\, P^{i,k} \,$ can be represented in form

$$\begin{aligned} P^{i,k} (\xi ) = \sum \limits _{(\lambda , \beta ) = d_{i, k}} \gamma _{\beta }\, \xi ^{\beta }. \end{aligned}$$

We draw the reader’s attention to the fact that in all the above notations (e.g., $d_{j}(\lambda )$, $P_{d_{j} (\lambda )} (\xi )$) the dependence on the face $\Re _{i}^{k}$ is not clearly visible. This is because each vector $\, \lambda \,$ uniquely determines the face of the polyhedron, the outward normal of which is $\, \lambda . \,$ Therefore, the dependence of these quantities on $\Re _{i}^{k}$ is expressed through their dependence on $\, \lambda . \,$

A face $\, \Re _{i}^{k} \,\, (1 \le i \le M_{k}, 0 \le k \le n - 1 )$ of the polyhedron $\, \Re (R) \,$ of a polynomial $\, R(\xi ) \,$ is said to be non-degenerate. (see [32] ) if $\, R^{i, k} (\xi ) \ne 0 \,$ for $\, \xi \in {\mathbb {R}}^{n, 0}. \,$ If there exists a point $\, \eta \in {\mathbb {R}}^{n, 0}, \,$ such that $\, P^{i, k} (\eta ) = 0, \,$ then the face $\, \Re _{i}^{k} \,$ is said to be degenerate.

A polynomial $\, P(\xi ) \,$ with Newton polyhedron $\, \Re (P) \,$ is said to be non-degenerate if all its principal faces are non-degenerate.

An operator P(D) (a polynomial $P(\xi )$) is called hypoelliptic (see [14], Definition 11.1.2 and Theorem 11.1.1) if the following equivalent conditions hold:

• (1) For any $\Omega \subset {\mathbb {E}}^{n}$ and $f \in C^{\infty } ( \Omega )$ all the solutions of equation $P(D)u = f$, $u \in D'(\Omega ) \,\,$ are infinitely differentiable i.e. belong to $C^{\infty }(\Omega )$.

• (2) $D^{\alpha } P(\xi )/P(\xi ) \rightarrow 0 \,\,$ when $| \xi | \rightarrow \infty ,$ for any $\alpha \in {\mathbb {N}}_{0}^{n}$, $\alpha \ne 0$.

Definition 1.3

• (1) (see [34] or [18]) We say that the polynomial P is more powerful then the polynomial Q (the polynomial Q is less powerful then the polynomial P) and write $P > Q$ ( $Q < P$), if there exists a constant $\, c > 0 \,$ such that

  $$\begin{aligned} | Q (\xi ) | \le c \left( | P (\xi ) | + 1 \right) \forall \xi \in {\mathbb {R}}^{n}. \end{aligned}$$

  (1.2)

  If $Q< P < Q,$ then we say that the polynomials P and Q have the same power.

• (2) (see [14], Definition 10.3.4) We say that the polynomial P is stronger (by L. H$\ddot{o }$rmander) then the polynomial $\, Q \,$ $\, ( Q \,$ is weaker than $\, P) \,$ and write $P \succ Q$ $\, ( Q \prec P),$ if there exists a constant $\, c > 0$ such that

  $$\begin{aligned} \tilde{Q} (\xi ) \le c \tilde{P} (\xi ) \forall \xi \in {\mathbb {R}}^{n}, \end{aligned}$$

  (1.3)

  where, according to L.H$\ddot{o }$rmander, $\tilde{R}$ for the polynomial $\, R \,$ is defined by the formula

  $$\begin{aligned} \tilde{R} (\xi ) = \left( \sum \limits _{| \alpha | \ge 0} | D^{\alpha } R (\xi ) |^{2} \right) ^{1/2}, \xi \in {\mathbb {R}}^{n}. \end{aligned}$$

If $Q \prec P \prec Q,$ then we say that polynomials $\, P \,$ and $\, Q \,$ have the same strength (by L. H$\ddot{o }$rmander).

Obviously, if $Q < P \,$ then $Q \prec P \,$ and when polynomial $\,P \,$ is hypoelliptic then (see Lemma 10.4.2 from [14]) the relations $Q < P$ and $Q \prec P \,$ are equivalent.

Our goal in this paper is to find algebraic conditions under which the polynomial $\,\, Q(\xi ) = Q(\xi _{1},\ldots , \xi _{n}) \,\,$ is less powerful than the polynomial $\, P(\xi ) = P(\xi _{1},\ldots , \xi _{n})$, or in another formulation describe the set of polynomials $\, \{ Q \}$ that are less powerful then the given polynomial $\,\, P.$

In another paper, we intend to describe the set of polynomials $\, \{Q\} \,$ weaker then $\, P \,$ for a given polynomial P.

We bring several reasons to confirm the relevance of posing such questions:

• (1) Estimates of monomials in terms of a given polynomial play an important role in the general theory of differential equations. In particular, they are used to obtain coercive estimates for derivatives of functions in terms of differential operators applied to these functions (see [3, 4]),

• (2) if P(D) and Q(D) have the same strength and P(D) is hypoelliptic, it follows that Q(D) is also hypoelliptic (see Theorem 11.1.9 and Corollary 10.4.8 from [14]),

• (3) Let $\, P(\xi ) \,$ and $\, Q(\xi ) \,$ be polynomials with real coefficients of orders $\, d_{P}\,$ and $\, d_{Q}\,$ respectively ($\, d_{P} > d_{Q}$). If for any real number $\,\, a \,\,$ the polynomial $\, P + a \, Q \,$ is hypoelliptic then $\, P \,$ dominates $\, Q \,\,$ (they write $\, Q \prec \prec P ), \,$ (see [31]) i.e. (see definition 10.4.4 from [14])

  $$\begin{aligned} \sup \limits _{\xi } \tilde{Q} (\xi , t) / \tilde{P} (\xi , t) \rightarrow 0 \,\,\,\, for \,\,\, t \rightarrow \infty , \end{aligned}$$

• (4) Another interesting application of the operator comparison is the question of adding lower order terms to a given (elliptic, hypoelliptic, almost hypoelliptic, hyperbolic, etc.) operator that does not violate its character, i.e. preserves its ellipticity, hypoellipticity, etc. (see, for example, [5,6,7, 10, 12]). For example, in [19] (see also [22]) the following result is proved: Let, for some $\lambda \in \Lambda (\Re (P)), \,$ the hypoelliptic polynomial P is represented in the form (1.1), where $M =1.$ Let R be a $\lambda -$homogeneous polynomial of $\lambda -$order $d(R): d_{1}< d(R) < d_{0}$ and $R < P_{0},$ then $P + R$ is also hypoelliptic,

• (5) If a polynomial P with complex coefficients is hypoelliptic and $Q < P,$ then there exists a number $\varepsilon > 0$ such that for any complex number $a: | a | < \varepsilon \,\,$ the polynomial $P + a \, Q \,$ is also hypoelliptic (see Theorem 2 from [21]),

• (6) Let $\, P_{m} \,$ be a homogeneous polynomial of order m hyperbolic with respect to the vector $\, N \in {\mathbb {E}}^{n}$ and $\, Q \,$ be a polynomial such that $\, ord(Q) < m. \,$ Then polynomial $\, P_{m} + Q \,$ is hyperbolic (with respect to the $N) \,$ if and only if $\, Q \prec P_{m}.$ (see [10, 38, 40], as well as [25, 26]).

These examples, the list of which can be continued, show the relevance of the problem of finding polynomials $\, \{Q \} \,\,$ (in particular, monomials $\, \{\xi ^{\nu } \})\,\,$ which have less power then given (in particular, a generalized-homogeneous) polynomial $\, P. \,$ The present work is devoted to this topic.

We note that many significant results have been obtained in this direction. Let’s list some of them.

In paper [34] (see also [35, 36]) by B. Pini conditions for the hypoellipticity of an operator with constant coefficients were found in terms of a comparison of differential operators.

In [2] Barozzi gives explicit conditions under which for a given set $\,\, \{ \alpha ^{k} \}^{N}_{ k = 1} \,\,$ of multi-indices there exists a polynomial $\, P(\xi ) = \sum _{k = 1}^{N} p_{k} \, \xi ^{\alpha ^{k}} \,$ stronger (by L.H$\ddot{o}$rmander) than any polynomial $\, Q(\xi ) = \sum _{k = 1}^{N} q_{k} \, \xi ^{\alpha ^{k}}. \,$

In paper [9] of Friberg the concept of polynomial with characteristics of constant multiplicity is introduced: let $\, P(\xi ) = P(\xi _{1},\ldots , \xi _{n})$ is a generalized homogeneous polynomial. We say that $\, P \,$ is a polynomial with characteristics of constant multiplicity, if for any $\eta \in {\mathbb {R}}^{n} \,\,$ such that $\eta \ne 0$ and $\, P(\eta ) = 0, \,\,$ there exists a neighborhood $\, O(\eta ) \subset {\mathbb {R}}^{n}, \,\,$ sufficiently smooth generalized homogeneous functions $\, q(\xi ) = q_{\eta }(\xi ), \, \,$ $r(\xi ) = r_{\eta }(\xi )$ and natural number $\, m = m (\eta ), \,$ such that $\, q (\eta ) = 0, \,$ $\, r(\eta ) \ne 0 \,$ and $\, P(\xi ) = [ q(\xi )]^{m} \cdot r(\xi )$ $\,\,\, \forall \xi \in O(\eta ).$

In [6] Cattabriga obtained a certain class of non-elliptic polynomials that are hypoelliptic and in [7] he first drew attention to the close relationship between the smoothness of solutions of general linear differential equations and the geometric properties of Newton polyhedrons corresponding to these equations. In this regard, he proved that the Newton polyhedron of the hypoelliptic (by H$\ddot{o}$rmander) operator (polynomial) is completely regular.

The monograph [39] of Rodino is devoted to weakly hyperbolic operators. There, in particular, it is proved that the condition of weak hyperbolisity is necessary, and the condition of strong hyperbolicity is sufficient for equation $\, P(D) u = 0 \,$ to have a solution $\, u \in C^{\infty }.$

It was proved in paper [41] of Zanghirati and in paper [8] of Corli that for the s-hyperbolic operator P,  the equation $P(D) u = f \,\,$ (where $f \in {\dot{G}}^{^{ \rho } } (\Omega ), \rho < s)$ has a solution $u \in G^{s}$ such that $\, supp \, (u) \subset \Omega ,$ where $\, G^{s}$ is an isotropic Gevrey space.

It was proved in paper [5] of D. Calvo that the Cauchy problem for the equation $P(D) u = 0 \,\,$ has a solution $u \in G^{\delta \cdot \Re } ({\mathbb {E}}^{n}), \,$ where $\, G \,$ is some weighted multi-anisotropic Gevrey space (see also [25]).

The following simple statement gives us the right to consider only non-negative polynomials $\, P \,$ and $\, Q \,$ with real coefficients, when finding conditions for fulfilling the relation $\, Q < P. \,$

Lemma 1.2

(see [18]) Let $\, \Re = \Re (P) \,$ be the complete Newton polyhedron of the polynomial $\, P \,$ and ${\mathfrak {M}} = {\mathfrak {M}} (|P|^{2})$ be Newton polyhedron of the polynomial $\, | P |^{2}. \,$ Then $\, \Re \,$ is similar to $\, {\mathfrak {M}} \,$ with similarity ratio 2 and center of similarity at the origin. Further, if similar faces are denoted by the same indices $\, (i,k)\,$ then $\, [|P |^{2} ]^{i,k} (\xi ) \equiv | P^{i,k} (\xi ) |^{2} \,$. This means, in particular, that if a face $\, \Re _{i}^{k}\,$ of $\, \Re \,$ is principal (degenerate, non -degenerate) then the face ${\mathfrak {M}}^{k}_{i} \,$ of ${\mathfrak {M}} \,$ is also principal (degenerate, non -degenerate) and vice versa.

Remark 1.2

Based on the above lemma, when comparing two polynomials (or a monomial with a polynomial), we assume that a) the polynomial $\, P(\xi ) \,$ (with which monomials or other polynomials are compared) has real coefficients b) polynomial $\, P(\xi ) \,$ and all its sub-polynomials $\, P^{i,k} (\xi ) \,$ are non-negative for all $\, \xi \in {\mathbb {R}}^{n}.$ In particular, this meanes that all the principal vertices of the polyhedron $\, \Re \,$ have even coordinates and the coefficients at these vertices are positive.

We also note the following important factor for further discussion: let $P(\xi ) \,$ be a polynomial with complete Newton polyhedron $\Re (P) \,$ and $\,\Re ({\mathcal {M}}) \,$ be Newton polyhedron of the set of multi-indices ${\mathcal {M}}$ $= \{\nu : \nu \in \Re (P), \xi ^{\nu } < P \}$. Since $\, {\mathcal {M}} \subset \Re (P), \,$ then it is obvious that for any polynomial Q with Newton polyhedron $\Re (Q) \subset \Re ({\mathcal {M}}) \,$ the relation $Q < P \,$ holds. Therefore, it is natural to describe first the set of multi-indices $\{\nu : \nu \in {\mathbb {N}}^{n}_{0} \}$ such that $\xi ^{\nu } < P$. We will deal with this in the next paragraph.

2 Estimates of monomials through a given polynomial

In this paragraph, for a given polynomial $\, P(\xi ) = P(\xi _{1},\ldots , \xi _{n}),$ we describe the set of monomials $\{ \xi ^{\nu } \}$, for which the inequality

$$\begin{aligned} | \xi ^{\nu } | \le c \, [ | P(\xi ) | + 1] \,\,\, \forall \xi \in {\mathbb {R}}^{n} \end{aligned}$$

(2.1)

holds with corresponding constants $\, c = c(\nu ) = c(\nu , P) > 0.$

We denote by $\, I_{n} \,$ the set of polynomials $\, P(\xi ) = P(\xi _{1},\ldots ,\xi _{n}) \,\,$ of $\, n \,$ variables with generally speaking complex coefficients such that $\, | P(\xi ) | \rightarrow \infty \,$ as $\, | \xi | \rightarrow \infty . \,$ Note that a polynomial $\, P \in I_{n} \,$ if and only if $\, | P |^{2} \in I_{n}. \,$ Therefore, without loss of generality, we will further assume that the polynomials from $\, I_{n} \,$ have real coefficients and are non-negative.

Note that the definition of $\, I_{n} \,$ naturally arises when describing hypoelliptic operators $\, P(D) \,$ ($\, |P(\xi )|^2 \,$ belongs to $\, I_{n})$, solutions to which are infinitely differentable functions. Although they can also have better smoothness properties, for example, they can belong to certain Gevrey class $\, \Gamma ^{(\sigma )}, \, (\sigma \ge 1) \,$ (see [11, 33, 39]). As it is known, the Gevrey class $\, \Gamma ^{(\sigma )} \,$ is intermediate between the class of infinitely differentiable and the class of real-analitic functions. Moreover, if for the hypoelliptic differential operator $\, P(D) \,$ there exist positive constants $\, c \,$ and $\, k \,$ such that $1+ | P(\xi )| \ge c \, (1 + |\xi |^{k}) \,$ for all $\, \xi \in {\mathbb {R}}^{n}, \,$ then the value of $\sigma$ directly depends on the value of $\, k \,$ ([5, 10, 28,29,30, 33]). Therefore, the need to describe the set of multi-indices $\, {\mathbb {B}} = {\mathbb {B}} (P)\,$ for which the estimate $1+ | P(\xi )| \ge c \, \sum _{\beta \in {\mathbb {B}} } |\xi ^{\beta } | \,\,\, \forall \xi \in {\mathbb {R}}^{n}$ is valid with some constant $\, c > 0$ naturally arises.

Mikhailov in [32] described a class of non-degenerate polynomials P with a complete Newton polyhedron, for which the set $\, {\mathbb {B}}$ coincides with the set $\Re (P), \,$ i.e. an “optimal” result (in a certain sense) was obtained. Similar result for an incomplete Newton polyhedron was obtained by Gindikin in [13]. The class of polynomials considered by these authors are certainly different from the class of elliptic ones, but they are close in character to an elliptic operators in the sense that they are non-degenerate. The case when the polynomial $\, P(\xi ) \,$ is degenerate was first considered in work [20]. Here we prove a similar proposition for one wider class of degenerate polynomials $\, \{ P: P \in I_{n} \} \,$ (see Theorem 2.1 below).

The following easily verifiable properties of polynomials from $\, I_{n} \,$ will greatly help us further.

Lemma 2.1

Let $\, \Re = \Re (P) \,$ be the Newton polyhedron of the polynomial $\, P \in I_{n} \,$ and $\,\, \Re _{i}^{k} \,\, ( i = 1,2,\ldots , M_{k}, \, k = 0,1,\ldots , n - 1)~-$ the principal faces of $\, \Re , \,$ then

• (a) polyhedron $\, \Re = \Re (P) \,$ is complete,

• (b) $P^{i,k} (\xi ) \ge 0 \,$ for any $\, \xi \in {\mathbb {R}}^{n} \,\,$ $( i = 1,2,\ldots , M_{k}, \, k = 0,1,\ldots , n - 1)$,

• (c) suppose that the pair $\, (i,k) \,\, (1 \le i \le M_{k}, \,0 \le k \le n - 1), \,$ the vector $\, \lambda \in \Lambda ( \Re _{i}^{k}) \,$ and the point $\, \eta \in {\tilde{\Sigma }} (P^{i,k}): = \{ \xi \in {\mathbb {R}}^{n}, | \xi | = 1, P^{i,k} (\xi ) = 0 \} \,$ are fixed and (see representation (1.1)) $\, P_{j} (\eta ) = 0 \,\,$ $\, (j = 0,1,\ldots , l -1), \,\,$ $\,\, P_{l} (\eta ) \ne 0 \,\,$ $(1\le l \le M ). \,\,$ Then $\, P_{l} (\eta ) > 0.$

Proof

Property (a) is obvious. In both cases (b) and (c), assuming that at some point $\, \eta \in {\tilde{\Sigma }} (P^{i,k}) \,$ $\,\, P_{0} (\eta ):= P^{i,k} (\eta ) <0 \,\,$ (respectively, $\, P_{l} (\eta ) < 0), \,$ we get that on the sequence $\, \{ \xi ^{s}: = s^{\lambda } \, \eta \}_{s = 1}^{\infty } \,$ $\, P(\xi ^{s}) \rightarrow - \infty \,$ as $\, s \rightarrow \infty , \,$ which contradicts our assumption $\, P(\xi ) \ge 0 \,$ for all $\, \xi \in {\mathbb {R}}^{n}.$ $\square$

In addition to this lemma, we will prove one simple proposition that we will use frequently.

Lemma 2.2

Let a point $\, \alpha \in {\mathbb {R}}^{n,+}$ (not necessarily an integer) belong to the convex hull $\Re \,$ of the points $\alpha ^{1},\ldots , \alpha ^{N}, \,$ i.e. $\, \alpha = \sum _{j = 1}^{N} \, \sigma _{j}\, \alpha ^{j}$ for some $\sigma _{j} \ge 0, \,\,$ $(j = 1,\ldots , N), \,\,\,$ $\sum _{j = 1}^{N} \, \sigma _{j} = 1, \,$ $\, N \ge n. \,$ Then for all $\xi \in {\mathbb {R}}^{n}$ we have $\, | \xi ^{\alpha } | \le$ $h(\xi ): = |\xi ^{\alpha ^{1}} | +\cdots + | \xi ^{\alpha ^{N}} |.$ Moreover, if $\, \alpha \,$ is the interior point of the $\Re , \,$ then $| \xi ^{\alpha } | / [ h(\xi ) ] \rightarrow 0 \,$ as $\, h(\xi ) \rightarrow \infty .$

Proof

By Young’s inequality, we have

$$\begin{aligned} |\xi ^{\alpha } | = | \xi ^{\sum \limits _{j=1}^{N} \sigma _{j} \, \alpha ^{j}} | = | \xi ^{{\alpha }^{1}} |^{\sigma _{1}} \ldots | \xi ^{{\alpha }^{N}} |^{\sigma _{N}} \le \sum \limits _{j=1}^{N} \sigma _{j} \, | \xi ^{{\alpha }^{j}} | \le \sum \limits _{j=1}^{N} \, | \xi ^{{\alpha }^{j}} |. \end{aligned}$$

To prove the second part of the lemma, note that if $\, \alpha \,$ is an interior point of $\Re , \,$ then for some $\, t>1 \,$ the point $\,t \, \alpha \,$ is also an interior point of $\Re , \,$ hence $|\xi ^{t \, \alpha } | =$ $|\xi ^{ \alpha } |^{t} \le$ $c \, h(\xi ),\,$ i.e. $|\xi ^{ \alpha } | \le$ $c^{1 / t} \, h(\xi )^{1 / t}.\,$ Therefore $\, |\xi ^{\alpha } | / h(\xi )$ $\le c^{1 / t} \, h(\xi )^{ (1 / t) - 1}$ $\rightarrow 0 \,$ for $h(\xi ) \rightarrow \infty .$

To describe the set of monomials that are estimated through a given polynomial $\, P \in I_{n}, \,$ we first consider the simplest case when only one principal face of the polyhedron $\, \Re (P) \,$ of the polynomial $\, P \,$ is degenerate, in particular this is a $\, (n - 1) -$dimensional face. Obviously, the fact that we consider the case when only one face is degenerate does not violate the generality. Considering the presence of only $\, (n - 1)-$dimensional degenerate faces is justified by the fact that, first, in the two-dimensional case, this is the only possible case, and secondary, that a more general case is considered below (see Theorem 5.3).

So let $\, \Re = \Re (P) \,$ be the complete Newton polyhedron of the polynomial $\, P \in I_{n}. \,$ Let all of the principal faces of $\, \Re \,$ with the exception of possibly one $\, (n - 1) -$dimensional face $\, \Gamma :=\Re _{i_{0}}^{n - 1} \,\,$ (with the outward normal $\, \mu \,$ which in this case is uniquely determined) are non degenerate.

If the face $\, \Gamma \,$ is degenerate, then with respect to the vektor $\, \mu \,$ we represent the polynomial $\, P \,$ as a sum of $\, \mu -$homogeneous polynomials in the form (1.1) and consider the following two cases

• (1) $\mu _{j} > 0 \,$ $(j= 1,\ldots , n). \,$ In this case we introduce the set $\, \Sigma (P_{0}) \,$ $\, = \Sigma (P_{0}, \mu ): =$ $\{\xi : \xi \in {\mathbb {R}}^{n}, | \xi , \mu | = 1,$ $P_{0} (\xi ) = 0\} \,$ (see representation 1.1 for $P_0$),

• (2) some of the coordinates of the vector $\, \mu \,$ are zero or negative. In this case we denote $\Sigma ' (P_{0}): =$ $\, \Sigma ' (P_{0}, \mu ) \, =$ $\{\xi : \xi \in {\mathbb {R}}^{n, 0},$ $\, P_{0} (\xi ) = 0\}.$ Note that in this case the set $\Sigma ' (P_{0}, \mu ) \,$ may not be a compact set.

$\square$

2.1 The case of degeneracy of the $(n - 1) -$dimensional face

2.1.1 The case $\mu _{j} > 0 \, (j= 1,\ldots , n) \,$

Theorem 2.1

Let $\, \Gamma :=\Re _{i_{0}}^{n - 1} \,\,$ (with the outward normal $\mu$ for which $\mu _{j} > 0, \,j= 1,\ldots , n$) be the only degenerate face of the complete Newton polyhedron $\, \Re = \Re (P) \,$ of the polynomial $\, P \in I_{n}, \,$ the polynomial $\, P \,$ is represented in the form (1.1), $\, \Sigma (P_{0}, \mu )\cap \Sigma (P_{1}, \mu ) = \emptyset$ and $\Re ^{*}: =\{ \beta \in \Re : (\mu ,\beta ) \le d_{1}\}. \,$ Then inequality (2.1) holds for $\, \nu \in {\mathbb {R}}^{n,+} \,$ if and only if $\, \nu \in \Re ^{*}.$

Proof

We will prove the Necessity and the Sufficiency of the condition.

Necessity. Let us prove that inequality (2.1) cannot hold for any $\nu \notin \Re ^{*}.$ Without loss of generality we may assume that $\, \nu \in \Re \setminus \Re ^{*}.$ Let $\, \nu \in \Re {\setminus } \Re ^{\star }, \,$ $\eta \in \Sigma (P_{0}, \mu ) \cap {\mathbb {R}}^{n,0}.$ Note that the existence of $\eta$ follows from the definition of the degeneracy of $\, P.$ Further, taking into account that $\, d_{1} < (\mu , \nu ) \le d_{0}\,$ and denoting $\tau : = | \eta _{1}^{\nu _{1}} | \ldots | \eta _{n}^{\nu _{n}} | \ne 0$, $\, \xi (t): = t^{\mu } \, \eta \,$, we have $|\xi (t)^{\nu } | = t^{(\mu , \nu )} \tau ,$ $\,\,| P(\xi (t)) | = t^{d_{1}} \, | P_{1} (\eta ) | (1 + o(1)) \,\,$ as $\, t \rightarrow \infty \,$ i.e. $\, |\xi (t)^{\nu } | / [1 + | P(\xi (t)) | ]$ $\rightarrow \infty \,$ which means that inequality (2.1) doesn’t hold for $\nu$. The necessity of the condition is proved.

Sufficiency. Suppose the opposite, that there exists a point $\nu \in {\mathbb {R}}^{n,+} \cap \Re ^{\star } \,$ and a sequence $\, \{ \xi ^{s} \}_{s = 1}^{\infty } \,$ such that $\, |\xi ^{s}, \mu | \rightarrow \infty \,$ (which is equivalent to $\, | \xi ^{s} | \rightarrow \infty \,$ ) when $s \rightarrow \infty \,\,$ and

$$\begin{aligned} | (\xi ^{s})^{\nu } | / [| P ( \xi ^{s}) | + 1] \rightarrow \infty . \end{aligned}$$

(2.2)

After denoting $\, \tau ^{s}: = \xi ^{s} / |\xi ^{s}, \mu |^{\mu } \,\, (s = 1,2, \ldots ) \,$ we get $| \tau ^{s}, \mu | = 1 \,\, (s = 1,2, \ldots )$. By the Bolzano–Weierstrass theorem there exits a subsequence of $\, \{ \xi ^{s} \} \,$ (which we also denote by $\, \{ \xi ^{s} \} ) \,$ and a point $|\, \tau |$ such that $|\tau , \mu | = 1, \,$ for which $\, | \tau ^{s} - \tau , \lambda | \rightarrow 0 \,$ as $s \rightarrow \infty . \,\,$ From (2.2) we have

when $s \rightarrow \infty$. Since $\, (\mu ,\beta ) \le d_{1}, \,$ it follows from the relation (2.2’) (see Lemma 2.1) that $\tau \in \Sigma (P_{0}) \,$ and, therefore, by virtue of the condition $\, P \in I_{n} \,$ and due to Lemma 2.1, we have that $\, P_{d_{0}} (\xi ) \ge 0 \,$ and $\, P_{d_{1}} (\eta ) > 0. \,$ Therefore (recall that $d_{0}> d_{1}> \ldots > d_{M} ), \,$ for sufficiently large $\, s \,$ we have

$$\begin{aligned} | P(\xi ^{s}) |= & {} | \sum \limits _{j = 0}^{M} P_{d_{j}} (\xi ^{s}) | = | \sum \limits _{j = 0}^{M} | \xi ^{s}, \mu |^{d_{j}} \, P_{d_{j}} (\tau ^{s}) |\\\ge & {} |\, | \xi ^{s}, \mu |^{d_{0}} \, P_{d_{0}} (\tau ^{s}) + | \xi ^{s}, \mu |^{d_{1}} \, P_{d_{1}} (\tau ^{s}) | - \sum \limits _{j = 2}^{M} \, | \xi ^{s}, \mu |^{d_{j}} | P_{d_{j}} (\tau ^{s}) |\\\ge & {} \, | \xi ^{s}, \mu |^{d_{0}} \, | P_{d_{0}} (\tau ^{s}) | + | \xi ^{s}, \mu |^{d_{1}} \, | P_{d_{1}} (\tau ^{s}) | - \sum \limits _{j = 2}^{M} \, | \xi ^{s}, \mu |^{d_{j}} | P_{d_{j}} (\tau ^{s}) |\\\ge & {} | \xi ^{s}, \mu |^{d_{1}} \, | P_{d_{1}} (\tau ) | (1 + o(1)). \end{aligned}$$

Hence, by virtue of estimate (2.$2'$)

$$\begin{aligned} | (\tau ^{s})^{\nu } | \, |\xi ^{s}, \mu |^{(\mu ,\nu )} / | \xi ^{s}, \mu |^{d_{1}} \, | P_{d_{1}} (\tau ^{s}) | (1 + o(1)) \rightarrow \infty . \end{aligned}$$

when $\, s \rightarrow \infty \,$. This contradicts the condition $\, (\mu ,\nu ) \le d_{1} \,$and proves the theorem. $\square$

2.1.2 The case when some coordinates of the vector $\, \mu \,$ are non positive

Theorem 2.1$'$Let $\, \Re = \Re (P) \,$ be the complete Newton polyhedron of the polynomial $\, P \in I_{n}. \,$ Let all of the principal faces of $\, \Re \,$ with the exception of possibly one $\, (n - 1) -$dimensional face $\, \Gamma :=\Re _{i_{0}}^{n - 1} \,\,$ be non degenerate (with the outward normal $\, \mu , \,$ some coordinates of which may be non positive). Then

• (1) If the face $\, \Gamma \,$ is non-degenerate, then for all multi-indices $\, \nu \in \Re \,$ inequality (2.1) holds

• (2) If the face $\, \Gamma \,$ is degenerate, then with respect to the vector $\, \mu \,$ we represent the polynomial $\, P \,$ as a sum of $\, \mu -$homogeneous polynomials in the form (1.1). If $\, P_{1} (\eta ) \ne 0 \,$ for all $\, \eta \in \Sigma ' (P_{0}): = \{ \xi \in {\mathbb {R}}^{n,0}, P_{0} (\xi ) = 0 \}, \,$ then inequality (2.1) holds for $\, \nu \in {\mathbb {R}}^{n,+} \,$ if and only if $\, \nu \in \Re ^{*}: =$ $\{ \beta \in \Re , (\mu ,\beta ) \le d_{1}\}.$

Proof

We carry out the proof simultaneously for both cases 1) and 2). Suppose, on the contrary, that there exist a multiindex $\, \nu \in \Re \,$ and a sequence $\, \{\xi ^{s} \}_{s = 1}^{\infty } \,$ such that the relation (2.2) is fulfilled.

It can be assumed without loss of generality that all coordinates of $\, \xi ^{s} \,$ are positive. Let

$$\begin{aligned} \rho _{s}: = \exp \left( \sqrt{ \sum \limits _{k = 1}^{n} ( \ln \xi _{k}^{s})^{2}}\right) , \,\,\, \lambda _{i}^{s}: = \frac{\ln \xi _{i}^{s}}{\ln \rho _{s}} \,\,\, (i = 1,\ldots , n), \end{aligned}$$

(2.3)

then $\,\, \lambda ^{s} = ( \lambda _{1}^{s},\ldots , \lambda _{n}^{s}) \,\,$ is a unite vector and

It is obvious that $\, \rho _{s} \rightarrow \infty \,$, if $\, | \xi ^{s} | \rightarrow \infty \,\,$ or when one of the coordinates of $\, \xi ^{s} \,$ tends to zero.

Since the vectors $\, \lambda ^{s} \,$ are on the unite sphere, the sequence $\, \{ \lambda ^{s} \} \,$ has a limit point $\, \lambda ^{\infty }. \,\,$ It can be assumed that $\, \lambda ^{s} \rightarrow \lambda ^{\infty },\,$ $\,\,| \lambda ^{\infty }| =1.$ From the convexity of the polyhedron $\, \Re (P) \,$ it follows that $\, \lambda ^{\infty } \,$ is an outward normal to one and only one face of $\, \Re (P). \,$

Denote $\, \lambda ^{\infty } \,$ by $e^{1,1},$ and choose $\, n~-$ dimensional vectors $\, (e^{1,1}, e^{1,2},\ldots , e^{1, n}) \,$ in such a way that they formed an orthonormal basis in $\, {\mathbb {R}}^{n}.$ Then $\, \lambda ^{s} = \sum _{i = 1}^{n} \lambda _{1,i}^{s} \, e^{1,i} \,\,$ $(s = 1,2,\ldots ). \,\,$ Since $\, \lambda ^{s} \rightarrow \lambda ^{\infty } = e^{1,1} \,$ when $\, s \rightarrow \infty ,$ therefore $\, \lambda _{1,1}^{s} \rightarrow 1, \,$ $\, \lambda _{1,i}^{s} = o (\lambda _{1,1}^{s}) \,$ for $\, i= 2,3,\ldots , n.$

If $\, \sum _{j =2}^{n} \lambda _{1,j}^{s} e^{1,j} =0 \,$ for all sufficiently large $\, s, \,$ we denote the basis $\, ( e^{1,2},\ldots , e^{1, n}) \,$ by $\, e^{1},\ldots ,e^{n}. \,$ Otherwise, by an appropriate choice of a subsequence we may assume that $\, \sum _{j =2}^{n} \lambda _{1,j}^{s} e^{1,j} \ne 0 \,$ for all $s = 1,2,\ldots$, and as $s \rightarrow \infty$

$$\begin{aligned} \sum \limits _{i = 2}^{n} \lambda _{1,i}^{s} e^{1,i} / \, \left| \sum \limits _{i = 2}^{n} \lambda _{1,i}^{s} e^{1,i} \right| \, \rightarrow e^{2,2}. \end{aligned}$$

In the subspace spanned by $\, (e^{1,2}, e^{1,3},\ldots , e^{1, n}) \,$ we pass to a new orthonormal basis $\, (e^{2,2}, e^{2,3},\ldots , e^{2, n}) \,$ with the vector $e^{2,2}$ defined above. Then, if $n \ge 3 \,$

$$\begin{aligned} \lambda ^{s} = \lambda _{1,1}^{s} \, e^{1,1} + \lambda _{2,2}^{s} \, e^{2,2} + \sum \limits _{i = 3}^{n} \lambda _{2,i}^{s} \, e^{2,i}, \,\, (s=1,2,\ldots ), \end{aligned}$$

in which clearly $\, \lambda _{1,1}^{s} \rightarrow 1,\,$ $\, \lambda _{2,2}^{s} = o ( \lambda _{1,1}^{s}), \,\,\$ $\lambda _{2,i}^{s} = o ( \lambda _{2,2}^{s}), \,\,$ $i = 3,\ldots , n \,$ as $\, s \rightarrow \infty .$

Reasoning analogously in the subspace with basis $\, (e^{2,3},\ldots , e^{2,n})$ etc., we finally obtain (after modifying the notation) that $\lambda ^{s} = \sum _{i = 1}^{n} \lambda _{i}^{s} \, e^{i}, \,\,$ where $(e^{1},\ldots , e^{n}) \,$ is an orthonormal basis, and $\lambda _{1}^{s} \rightarrow 1, \lambda _{i +1}^{s} = o (\lambda _{i }^{s}), \,\, i = 1,\ldots , n-1 \,$ as $s \rightarrow \infty .$

Moreover, there exist numbers $\, s_{0}$ and $\, m: \, 1 \le m \le n \,$ such that for all $\, s \ge s_{0} \,$ we have $\lambda _{i}^{s} > 0 \,\,$ for $(i = 1,\ldots ,m) \,\,$ and $\lambda _{i}^{s} = 0 \,\,$ $(i = m + 1,\ldots , n). \,\,$ By choosing a subsequence, we may assume that $\, s_{0}=1, \,\,$ $\lambda _{i}^{s} > 0 \,\,$ for all $(i = 1,\ldots ,m) \,\,$ and $s \in {\mathbb {N}}.$

Now we associate the constructed basis with the polyhedron $\, \Re .$ We select the faces $\,\, \Re _{i_{1}}^{k_{1}}, \Re _{i_{2}}^{k_{2}},\ldots , \Re _{i_{m}}^{k_{m}} \,\,$ as follows (see [1]): denote by $\,\, \Re _{i_{1}}^{k_{1}} \,$ the face of $\, \Re (P)\,$ which lies in the supporting hyper-plane of $\, \Re (P)\,$ with outward normal $\, e^{1}, \,$ then each face $\,\Re _{i_{j}}^{k_{j}} \,$ $\, (j = 2,\ldots ,n)\,\,$ either coincides with the previous one or is a sub face of it, and in either cases lies in the hyper-plane, for whose points $\, \alpha \,$ the quantity $\, (e^{j}, \alpha ) \,$ is largest possible.

From the construction of the faces $\,\, \Re _{i_{1}}^{k_{1}}, \Re _{i_{2}}^{k_{2}},\ldots , \Re _{i_{m}}^{k_{m}} \,\,$ it is obvious that their dimensions satisfy the relation: $\, k_{1} \ge k_{2} \ge \ldots \ge k_{m} \,$, and (see (2.3), (2.3$'$))

$$\begin{aligned} \xi ^{s} = \rho _{s}^{\sum \limits _{ i=1}^{n} { \lambda _{i}^{s} e^{i} }} (s=1,2,\ldots ). \end{aligned}$$

it can be assumed that $\rho _{s} \rightarrow \infty$ as $s \rightarrow \infty$ and for some $r \,\, (1 \le r \le m)$

$$\begin{aligned} \rho _{s}^{\lambda _{j}^{s}} \rightarrow \infty \,\, (j=1,\ldots ,r), \,\,\, \rho _{s}^{\lambda _{r+1}^{s}} \rightarrow b \ge 1. \end{aligned}$$

When $r = m =n$ we assume by definition that $\lambda _{n+1}^{s} =0 \,\, (s=1,2,\ldots ),$ and $e^{n+1} \,$ is an arbitrary unit vector.

As earlier, let $\, P^{i_{j}, k_{j}} (\xi ) \,$ be the sub-polynomial (see Lemma 1.1) of $\,P(\xi )\,$ corresponding to the face $\,\, \Re _{i_{j}}^{k_{j}},$ i.e. $P^{i_{j}, k_{j}} (\xi ):= \sum _{\beta \in \Re _{i_j}^{k_j} } \gamma _{\beta } \, \xi ^{\beta }.$ We will study the behavior of the polynomial $P(\xi )$ and the monomial $\xi ^{\nu }$ on the sequence $\xi ^{s} = \rho _{s}^{ \lambda _{1}^{s} e^{1} + \lambda _{2}^{s} e^{2} +\cdots + \lambda _{n}^{s} e^{n} }$, when $\rho _{s} \rightarrow \infty$.

Further for the simplicity we omit index $\, s \,$ in the notation, where this does not cause misunderstanding.

From the $e^{j} -$homogeneity of the polynomials $\, P^{i_{j}, k_{j}} (\xi ) \,$ and from the convexity of $\Re (P)$ and its faces there exist positive $\sigma _{1},\ldots , \sigma _{r}$ such that for an arbitrary multi-index $\, \alpha \,$ belonging to all $\,\, \Re _{i_{j}}^{k_{j}} \,\, (j = 1,\ldots , r),$ i.e. $\alpha \in \Re _{i_{r}}^{k_{r}} \,\,$, we have

$$\begin{aligned} P(\xi )= & {} \rho ^{(\alpha , \lambda _{1} \, e^{1}) } [ P^{i_{1}, k_{1}} (\rho ^{\sum \limits _{j= 2}^{n+1} \lambda _{j} e^{j}} ) + o(\rho ^{-\sigma _{1}\, \lambda _{1}}) ]\nonumber \\= & {} \rho ^{(\alpha , \lambda _{1} \, e^{1} +\lambda _{2} \, e^{2}) } [ P^{i_{2}, k_{2}}(\rho ^{\sum \limits _{j= 3}^{n+1} \lambda _{j} e^{j}}) + o(\rho ^{-\sigma _{2}\, \lambda _{2}}) ] =\nonumber \\{} & {} \quad \quad \quad \quad \quad \ldots ..\nonumber \\= & {} \rho ^{(\alpha , \sum \limits _{j=1}^{r}\lambda _{j} \, e^{j})} [ P^{i_{r}, k_{r}} (\rho ^{\sum \limits _{j= r+1}^{n+1} \lambda _{j} e^{j}}) + o(\rho ^{-\sigma _{r}\, \lambda _{r})} ]. \end{aligned}$$

(2.4)

Since $\, \rho ^{\lambda _{r+1}}\rightarrow b \ge 1$ then $\rho ^{(\alpha , \sum _{r+1}^{n+1}\lambda _{j} \, e^{j})} \rightarrow b^{e^{r+1}}=:\eta .$ Clearly, $\, 0< \eta _{i} < \infty$ for all $\, i = 1,\ldots ,n \,$ (in accordance with the definition of the $\eta _{i}$ as finite powers of a positive number).

We consider two cases: a) $(e^{1}, \alpha )>0$ and b) $(e^{1}, \alpha )=0. \,$ The case $(e^{1}, \alpha )<0 \,$ is impossible, which is due to the fact that the equation for the supporting hyperplane with outward normal $\lambda \,$ of the complete polyhedron $\Re \,$ can be written in the form $\, (\lambda , \alpha ) = d, \,$ where $\, d \ge 0 \,$ is the distance from the origin to the given hyperplane and $\, \alpha \,$ is a point on the hyperplane (see, for example, [1, 24]).

Case a.1) Suppose first, that $P^{i_{r},k_{r}} (\eta ) \ne 0.$ This is only possible in case 1) of our theorem. Since $(e^{1}, \alpha )>0$ and $\lambda _{i} = o(\lambda _{1}) \,$ for $i = 2,\ldots ,n$, we eventually have for sufficiently large $\, s \,$ that $(\alpha , \sum _{1}^{r} \lambda _{j} \, e^{j}) >0. \,$ Therefore from (2.4) we imply

$$\begin{aligned} P(\xi ) = \rho ^{(\alpha , \sum \limits _{j=1}^{r}\lambda _{j} \, e^{j}) } [ P^{i_{r}, k_{r}} (\eta ) + o(1) ]. \end{aligned}$$

(2.5)

For the monomial $\, \xi ^{\nu } \,$ we have

$$\begin{aligned} \xi ^{\nu } = \rho ^{(\nu , \sum \limits _{j=1}^{r}\lambda _{j} \, e^{j}) } [\eta ^{\nu } + o(1) ]. \end{aligned}$$

(2.6)

We are going to show, that

$$\begin{aligned} \left( \nu , \sum \limits _{j=1}^{r}\lambda _{j} \, e^{j}\right) \le \left( \alpha ,\sum \limits _{j=1}^{r}\lambda _{j} \, e^{j}\right) .\, \end{aligned}$$

(2.7)

Since $\, \nu \in \Re ,$ $\, \alpha \in \Re _{i_{1}} ^{k_{1}}, \,$ and $\, e^{1} \,$ is the normal of the face $\, \Re _{i_{1} } ^{k_{1}},$ hence $\,(\nu , e^{1}) \le (\alpha ,e^{1}).$ If $\,(\nu , e^{1}) < (\alpha ,e^{1}),$ then the inequality (2.7) follows from the fact that $\, \lambda _{1} \rightarrow 1 \,$ and $\lambda _{j+1} = o(\lambda _{j}) \,$ for $\, j = 1,2,\ldots ,r -1.$ If $\,(\nu , e^{1}) = (\alpha ,e^{1}),$ then the points $\, \nu$ and $\, \alpha \,$ belong to the same face $\Re _{i_{1}}^{k_{1}}. \,$ Since $\, \nu \in \Re , \,$ hence $\, (\nu ,e^{2}) \le (\alpha ,e^{2}). \,$ If $\, (\nu ,e^{2}) < (\alpha ,e^{2}),\,$ then the inequality (2.7) follows from the same fact regarding numbers $\,{ \lambda _{j}}. \,$ If $\, (\nu ,e^{2}) = (\alpha ,e^{2}),\,$ then this means that the points $\, \nu$ and $\, \alpha \,$ belong to the same face $\Re _{i_{2}}^{k_{2}} \,$ and then $\, (\nu ,e^{3}) \le (\alpha ,e^{3}) \,$ and so on.

Continuing this process, after a finite number of steps, we either arrive at relation $\, (\nu ,e^{j}) = (\alpha , e^{j}) \,$ $\, j = 1,2,\ldots ,q -1,\,$ $\, (\nu ,e^{j}) < (\alpha , e^{j}) \,$ for some $q < r,$ or $\, (\nu ,e^{j}) = (\alpha , e^{j}) \,$ for all $\, j = 1,\ldots ,r.$ In both cases the inequality (2.7) is obvious.

This also proves that (for sufficiently large values of $\, s \,$) the vectors $h^{s}:= \sum _{j = 1}^{r} \lambda _{j}^{s} \, e^{j} \,$ are normals to the face $\, \Re _{i_{r} } ^{k_{r}}.$ Moreover, since $\, (\alpha , h^{s}) > 0,\,$ the face $\, \Re _{i_{r} } ^{k_{r}} \,$ is the principal one, and therefore, in the case under consideration, it is non-degenerate, i.e. $\,P^{i_{r}, k_{r}} (\eta )\ne 0.$ Then relations (2.5)-(2.7) together contradict our assumption (2.2) and complete the proof of the sub-case a.1).

Case a.2) $P^{i_{r},k_{r}} (\eta ) = 0.$ This case corresponds to the case 2) of our theorem, when $P^{i_{0}, n-1} (\eta ) = 0.$ In this case, the face $\Re _{i_{r}}^{k_{r}} \,$ coincides with the $\, (n-1)$ dimensional degenerate face $\, \Gamma :=\Re _{i_{0}}^{n - 1} \,\,$ (with the outward normal $\, \mu ) \,$ and $r =m =1, \,$ $k_{r}=k_{1}= n-1, \,$ $e^{1} = \mu ,$ $\eta \in \Sigma ' (\Gamma ).$

With respect to the vector $e^{1} = \mu ,$ we represent the polynomial $P(\xi )$ in the form (1.5)

$$\begin{aligned} P(\xi ) = \sum \limits _{j=0}^{M} P_{j} (\xi ): = \sum \limits _{j=0}^{M} \sum \limits _{(e^{1}, \beta ) = d_{j}} \gamma _{\beta } \, \xi ^{\beta }, \end{aligned}$$

(2.8)

and denote $q(\xi ):= P(\xi ) - [P_{0}(\xi )+P_{1}(\xi )].$ Then

$$\begin{aligned} P(\xi ) = P_{0}(\xi )+P_{1}(\xi ) +q(\xi ). \end{aligned}$$

(2.9)

Substituting the value

$$\begin{aligned} \xi (= \xi ^{s} ) = \rho ^{\sum \limits ^{n+1}_{j = 1} \lambda _{j} e^{j}} = \rho _{s}^{\sum \limits ^{n+1}_{j = 1} \lambda _{j}^{s} e^{j} } \end{aligned}$$

into (2.9) and using the $e^{1} = \mu ~-$ homogeneity of polynomials $P_{0}(\xi )$ and $P_{1}(\xi ),$ we get

$$\begin{aligned} P(\xi ) = \rho ^{\lambda _{1} \, d_{0}} P_{0} \left( \rho ^{\sum \limits ^{n+1}_{j = 2} \lambda _{j} e^{j} }\right) + \rho ^{\lambda _{1} \, d_{1}} P_{1} \left( \rho ^{\sum \limits ^{n+1}_{j = 2} \lambda _{j} e^{j} }\right) + q(\xi ). \end{aligned}$$

(2.10)

Let $\xi \,\, ( = \xi ^{s}) \rightarrow \infty ,$ i.e. $\rho \,\, (= \rho _{s}) \rightarrow \infty .$

Then $(\alpha , \lambda _{1} e^{1}) \rightarrow d_{0}$ for $\, \alpha \in (P_{0}), \,$ $\, (\beta , \lambda _{1} e^{1}) \rightarrow d_{1}$ for $\, \beta \in (P_{1}), \,$ $\rho ^{\sum \limits ^{n+1}_{j = 2} \lambda _{j} e^{j} } \rightarrow \eta , \,$ and by Lemma 1.1$P_{0} (\eta ) \ge 0,\,$ $P_{1} (\eta )> 0.$ On the other hand, since ord(q) $\le d_{2} < d_{1},$ then by Lemma 2.2$| q(\xi ) | = o(\rho ^{d_{1}}),$ consequently from (2.10) (for sufficiently large value of $\, s) \,$ we have with a constant $C_1 > 0$

$$\begin{aligned} |P(\xi ) | \ge C_{1} \rho ^{\lambda _{1} \, d_{1}}. \end{aligned}$$

(2.11)

As for monomial $\xi ^{\nu },$ then $\nu \in \Re ^{*},$ therefore $(\nu , \lambda _{1} \, e_{1}) \rightarrow a \le d_{1}$ and consequently (for sufficiently large values of $\, s$) we have with a constant $C_{2} > 0$

Relations (2.11), (2.11$'$) together contradict our assumption (2.2) and complete the proof of the sub-case a.2).

Case b) $(e^{1}, \alpha ) = 0.$ First note, that if the Newton polyhedron $\Re \,$ of the polynomial $P(\xi ) = P(\xi _{1},\ldots ,\xi _{n})$ is complete, then the Newton polyhedron of polynomial $P(\xi )|_{\xi _{j} =0 } \,$ for $j = 1, 2,\ldots , n$ is also complete in the corresponding $(n - 1)-$dimensional subspace. Secondly, in the case b) the face, whose outward normal is $\, e^{1} \,$ clearly passes through the origin and hence is not a principal face of $\Re$, consequently $e_{i}^{1} \le 0 \,$ $\, (i=1,\ldots ,n).$ Due to this, if the non principal face with outward normal $\, e^{1} \,$ has dimension $\, l \le n-1, \,$ then $\, l \,$ numbers among $e_{i}^{1} \, \, ( 1\le i \le n )$ are equal to zero with the remaining numbers being negative. It can be assumed without loss of generality that $e_{1}^{1} = \ldots = e_{l}^{1} =0,^{\,\,}$ $e_{l+1}^{1}< 0,\ldots , e_{n}^{1} < 0.$

Since

$$\begin{aligned} e_{j}^{1} = \lim \limits _{\xi \rightarrow \infty } \left[ \ln \xi _{j}/\left( \sum \limits _{k = 1}^{n} (\ln \xi _{k})^{2}\right) ^{1/2}\right] \,\, < 0 \,\, (j = l + 1,\ldots , n), \, \end{aligned}$$

starting from some number $\, s_{0} \,$ (we assume that$\, s_{0} = 1 )$ $\xi _{j}^{s} <1 \,\, (j = l + 1,\ldots , n) \,\, (s = 1,2,\ldots ).$ On the other hand, since $\, | \xi ^{s} | \rightarrow \infty \,\,$ as $s \rightarrow \infty , \,$ we have $\xi _{i}^{s} \rightarrow \infty \,$ for certain $\, i \in [1,l].$ Since $e_{i}^{1} = 0 \,$ for such $\,i, \,$ $\, \xi _{j}^{s} \rightarrow 0 \,$ as $s \rightarrow \infty \,$ for at list one $\, j \in (l,n ]$ (for some sub-sequence of the sequence $\, {\xi ^{s} } \,$)

Suppose that (after a possible renumbering) $\, \xi _{j}^{s} \rightarrow \infty \,\,$ $j = 1,\ldots ,l_{0}$ $(l_{0} \le l) \,$ and $\, \xi _{j}^{s} \rightarrow 0$ $j = l + 1,\ldots , l + l_{1}$ $( l + l_{1} \le n) \,$ as $s \rightarrow \infty . \,$

Let $\psi (\xi ):= \max _{1 \le j \le l_{0}} \xi _{j}, \,$ then it is obvious that as $s \rightarrow \infty$

$$\begin{aligned} \ln \psi (\xi ^{s}) / \left[ \sum \limits _{k = 1} ^{n}(\ln \xi ^{s}_{k})^{2}\right] ^{1/2} \rightarrow 0. \end{aligned}$$

(2.12)

On the other hand, there exist positive constants $\, C_{3}, C_{4} \,$ such that

$$\begin{aligned} C_{3} \le \sum \limits _{k = 1}^{l_{0}} (\ln \xi _{k}^{s})^{2} / (\ln \psi (\xi ^{s}))^{2} \le C_{4} \,\,\,\,\, (s = 1,2,\ldots ). \end{aligned}$$

(2.13)

It follows from (2.12)–(2.13) that

$$\begin{aligned} \sum \limits _{k = l_{0} + 1}^{n} (\ln \xi _{k}^{s})^{2} / (\ln \psi (\xi ^{s}))^{2} \rightarrow \infty \,\,\,\,\, as \,\,\,\, s \rightarrow \infty . \end{aligned}$$

(2.14)

From this result, by going over a sub-sequence, if necessary, we get that for some $j \in [l + 1, n ]$

$$\begin{aligned} |\ln \xi _{j}^{s} | / \ln \psi (\xi ^{s}) \rightarrow \infty \,\,\,\,\, as \,\,\,\, s \rightarrow \infty , \end{aligned}$$

(2.15)

i.e. $|\ln \xi _{j}^{s} | \rightarrow \infty \,$ “faster” than $\ln \psi (\xi ^{s}) \rightarrow \infty .$ Hence $\xi _{j}^{s} = o([\psi (\xi ^{s})]^{- \sigma })\,$ for any $\sigma > 0.$ Or, bearing in mind that $\, - \sigma \,\alpha _{1} + \alpha _{2} < 0 \,\,$ for sufficiently large $\sigma > 0, \,$ equivalently,

$$\begin{aligned} ( \xi _{j}^{s})^{\alpha _{1}}. [\psi (\xi ^{s})]^{\alpha _{2}} \rightarrow 0 \,\,\,\,\,\,\, as \,\,\,\,\,\, |\xi | \rightarrow \infty \end{aligned}$$

(2.16)

for $\alpha _{1} > 0$ and $\alpha _{2} \ge 0, \,$ $\, - \sigma \,\alpha _{1} + \alpha _{2} < 0 \,$ for sufficiently large $\sigma > 0$.

Let $\breve{\xi } = ( \breve{\xi _{1}},\ldots , \breve{\xi _{n}}), \,$ where $\breve{\xi _{j} } = 0 \,$, if $\, j \,$ satisfies condition (2.15), and $\breve{\xi } = \xi _{j} \,$ otherwise.

Taking into consideration (2.16), it follows from (2.2) that (with the possibility of going over a sub-sequence of $\{\xi ^{s} \}$ multiple times).

$$\begin{aligned} | (\breve{\xi ^{s}})^{\nu } | / (1 + | P(\breve{\xi ^{s}}) |) \rightarrow \infty \,\,\,\,\,\,\, as \,\,\,\,\,\, s \rightarrow \infty . \end{aligned}$$

(2.17)

As a result, from the polynomial $\, P(\xi ) = P(\xi _{1},\ldots , \xi _{n}) \,$ we get the polynomial $\, \breve{ P} (\xi ):= P(\breve{\xi })\,$ from less than $\, n \,$ variables. Consequently, the dimension of the polyhedron $\breve{ \Re } (P): = \Re (\breve{ P})$ is less than the dimension of the polyhedron $\, \Re (P), \,$ while the non-degenerate faces of $\Re \,$ correspond to the non-degenerate faces of $\, \breve{ \Re } \,$ and vice versa.

Thus the relation (2.2) leads either to a contradiction or to relation (2.17), which is analogous to (2.2) but corresponds to a space of dimension less then or equal to $\, n -1.$

Repeating the arguments presented above in the proof with respect to the polynomial $\,\breve{ P}, \,$ and so on, after a finite number of steps we clearly arrive at either a contradiction or relation (2.17) for polynomials of one variable.

For polynomials of one variable the case 2) of Theorem 1,1 is excluded, and the polyhedron $\ \Re \,$ has the shape of segment, any point of which is estimated through the endpoints of this segment.

It remains to prove the part of statement 2) of the theorem, relating to necessity. Namely, we prove that for any point $\, \nu \notin \Re ^{*}$ the inequality (2.1) cannot hold. Obviously, it is enough to consider the case $\, \nu \in \Re {\setminus } \Re ^{*}.$ In this case $(\mu ,\nu ) > d_{1}, \,$ where $\mu -$ is the outward normal of the degenerate face $\, \Gamma :=\Re _{i_{0}}^{n - 1} \,\,$ and $\Re ^{*}: =\{ \beta \in \Re , (\mu ,\beta ) \le d_{1}\}. \,$

Let $\eta \in \Sigma ' (P_{0}): = \Sigma ' (P^{i_{0}, n - 1})$ and $\xi ^{s}: = s^{\mu } \, \eta = (s^{\mu _{1}} \eta _{1} \,,\ldots ,s^{\mu _{n}} \, \eta _{n} ) \,\, \,\, (s = 1,2,\ldots ). \,$ With respect to the vector $e^{1} = \mu ,$ we represent the polynomial $P(\xi )$ in the form (2.9).

Because of $\, \mu -$homogeneity of polynomials $P_{0}$ and $P_{1}$ and due to the fact that $P_{0} (\eta ) = 0,$ it follows from the definition of the number $d_{1}$ and the set $\Re ^{*}$, that $\,\, P(\xi ^{s}) = s^{d_{0}} \, P_{0} (\eta )$ $+ s^{d_{1}} \, P_{1} (\eta )$ $+ o(s^{d_{1}}) \,\,$ when $s \rightarrow \infty$, i.e. $| P(\xi ^{s}) | = | P_{1} (\eta ) | s^{d_{1}}$ $\, (1 + o(1)).$ Similarly, $| (\xi ^{s})^{\nu } | = | \eta ^{\nu } |\, s^{(\mu , \nu )} \,$ for all $\, s = 1,2,\ldots \cdot$ Since $\, \eta \in {\mathbb {R}}^{n,0}, \,$ it follows that $| \eta ^{\nu } | \ne 0,$ moreover, $(\mu , \nu ) > d_{1}: \,$ therefore, it follows from the last two representations that $| (\xi ^{s})^{\mu } |/[1 + | P(\xi ^{s}) |]$ $\rightarrow \infty$ as $s \rightarrow \infty . \,$

Thus, statement 2) of the theorem is also completely proved. $\square$

Corollary 2.1

Let $\, \Re = \Re (P) \,$ be the complete Newton polyhedron of the polynomial $\, P \in I_{n}. \,$ Let all of the principal faces of $\, \Re \,$ with the exception of possibly one $\, (n - 1) -$dimensional face $\, \Gamma :=\Re _{i_{0}}^{n - 1} \,\,$ (with the outward normal $\, \mu ) \,$ are non-degenerate. Then

• (1) If the face $\, \Gamma \,$ is non-degenerate, then $\, P^{i,k} < P \,$ for all $i = 1,\ldots , M_{k} \,$ and $\, k = 0,1,\ldots , n-1.$

• (2) If the face $\, \Gamma = \Re _{i_{0}}^{n - 1}\,$ is degenerate, and with respect to the vector $\, \mu \,$ we represent the polynomial $\, P \,$ in the form (1.1). Then $\, P_{0} < P, \,\,$ $P_{1} < P.$

• (3) $\, ( | P_{2}(\xi ) | +\cdots + | P_{M}(\xi ) |) / (|P(\xi ) | + 1) \rightarrow 0 \,$ for $\, | \xi | \rightarrow \infty .$

Proof

Before the proof recall, that without loss of generality we can assume that $\, P(\xi ) \ge 0 \,\,$ for all $\, \xi \in {\mathbb {R}}^{n}\,$ and for any polynomial $\, P \in {I}_{n}$.

If the face $\Gamma$ is non-degenerate, then the statement 1) of the corollary directly follows from Theorem 2.1. Let the face $\,\Gamma$ is degenerate. The relation $\, P_{1} < P\,$ follows directly from Theorem $2.1'$, if we keep in mind that $(P_{1}) \subset \Re ^{*}.$ Let us prove that $\, P_{0} < P.$ Since $(P) \cap \{\Re {\setminus } \Re ^{*}\} = (P_{0}) \,$ and $\, Q < P \,$ for any polynomial $\, Q(\xi ) = \sum _{\alpha \in \Re ^{*} } q_{\alpha } \, \xi ^{\alpha }, \,$ then $\, P - P_{0} < P, \,$ hence $\, P_{0} = P - (P - P_{0} ) < P.$

The third point of the corollary follows from Lemma 2.2, if we bear in mind that $\, d_{j } < d_{1} \,$ for $\, j = 2,\ldots , M.$

Corollary 2.1 is proved. $\square$

Remark 2.1

The condition $\, P \in {I}_{n} \,$ in Theorem 2.1 may seem too strict. However, this is not the case for many reasons: (1) these are hypoelliptic (hence elliptic) polynomials (2) if we trace the proof of this theorem, then it is easy to see, that during the proof we use the condition $\, P \in {I}_{n} \,$ basically in the fact that in a neighborhood of the zeros of the polynomial $\, P_{0}:= P^{i_{0}, n - 1}, \,$ the polynomials $\, P_{0} \,$ and $\, P_{1}, \,$ have the same sign. Therefore, without involving the condition $\, P \in {I}_{n}, \,$ the theorem can be rephrased as follows.

Theorem $2.1''$. Let either

• I) all the principal faces of the complete Newton polyhedron $\Re = \Re (P) \,$ of the polynomial $\, P \,$ be non-degenerate, or

• II) $\Gamma = \Re _{i_{0}}^{n - 1}$ is its only principal degenerate face. Then

  • 1) in case I) for all $\, \nu \in \Re$ inequality (2.1) holds,

  • 2) in case II), inequality (2.1)cannot hold for any point $\, \nu \in {\mathbb {R}}^{n,+} {\setminus } \Re ^{*}$,

  • 3) in order for inequality (2.1) to be satisfied for all $\, \nu \in \Re ^{*}$ it is necessary and sufficient, that $\, P_{1} (\eta ) \ne 0 \,$ for all $\, \eta \in \Sigma ' (P_{0}), \,$ and for any point $\ \eta \in \Sigma ' (P_{0})$ there exists a neighborhood $\, U(\eta ) \,$ of this point, such that $P_{0} (\xi ) \, P_{1} (\xi ) \ge 0$ for all $\, \xi \in U(\eta ).$

Proof

Obviously, we need to prove only the part of the theorem, that relates to the necessity of the condition $P_{0} (\xi ) \, P_{1} (\xi ) \ge 0$ for all $\, \xi \in U(\eta ) \,$ and for all $\, \eta \in \Sigma (P_{0}). \,$ First note, that instead of comparing $\, \xi ^{\nu }\,$ and $P \,$ we can compere $\,| \xi ^{\nu } |^{2} \,$ and $\, | P |^{2}. \,$ So we can assume that $\ P(\xi ) \ge 0 \,$ for all $\, \xi \in {\mathbb {R}}^{n}, \,$ from where, it can be obtained, that $\ P_{0}(\xi ) \ge 0 \,$ for all $\, \xi \in {\mathbb {R}}^{n}. \,$ Thus, our condition is equivalent to the condition $\, P_{1} (\eta ) > 0 \,$ for all $\, \eta \in \Sigma ' (P_{0}). \,$

Assume that $\, P_{1} (\eta ^{0}) <0 \,$ for some $\, \eta ^{0} \in \Sigma ' (P_{0}). \,$ Then on the sequence $\xi ^{s} = s^{\mu } \, \eta ^{0} \,$ we have

$$\begin{aligned} P(\xi ^{s}) = s^{d_{1}} \, P_{1} (\eta ^{0}) + o (s^{d_{1}}) \rightarrow - \infty \,\,\,\,\, as \,\,\,\, s \rightarrow \infty , \end{aligned}$$

which contradicts our assumption, that $\, P (\xi ) \ge 0. \,$ $\square$

Now we prove a statement, which we will strengthen the Theorem 2.1.

Proposition 2.1

Let $\, \lambda = (\lambda _{1},\ldots , \lambda _{n}), \,\,$ $\lambda _{j} > 0 \,\, (j = 1,\ldots , n) \,$ and $\, P \in {I}_{n} \,$ is a polynomial with a complete polyhedron $\, \Re ,\,$ all principal faces of which are non degenerate except for the $\, (n - 1) -$dimensional face $\Gamma = \Re _{i_{0}}^{ n - 1} \,$ with outward normal $\, \lambda . \,$ Let $\, \{\xi ^{s} \} \,$ be a sequence, such that $\, \xi ^{s} \rightarrow \infty \,$ as $\, s \rightarrow \infty \,$, and there exists a point $\, \eta \in {\mathbb {R}}^{n} \,$, such that $\,\xi ^{s} / | \lambda , \xi ^{s} |^{\lambda } \rightarrow \eta , \,$ where $\, \eta _{1} \cdot \ldots \cdot \eta _{n} = 0. \,$ Then there is a constant $c>0 \,$, such that for any $\nu \in \Re$ the inequality

$$\begin{aligned} | (\xi ^{s})^{\nu } | \le c \, [| P(\xi ^{s}) | + 1] \end{aligned}$$

(2.17)

holds.

Proof

Suppose the opposite, that for some sequence $\, \{\xi ^{s} \} \,$ and for some multi-index $\nu \in \Re , \,$ satisfying the conditions of the theorem

$$\begin{aligned} | (\xi ^{s})^{\nu } | / [| P(\xi ^{s}) | + 1] \rightarrow \infty \end{aligned}$$

(2.18)

as $\, s \rightarrow \infty \,$. Carrying out arguments, similar to those in the proof of Theorem $\, 2.1' \,$, we obtain the existence of a number $\, r \in {\mathbb {N}}, r \le n, \,$ vectors $\, e^{1},\ldots , e^{n}, e^{n + 1}, \,$ sequences $\, \{ \rho _{s}\}_{s = 1}^{\infty }, \,\,$ $\, \{ \lambda _{j}^{s} \ge 0 \}_{s = 1}^{\infty }, \,\,$ $j = 1,\ldots ,n,n+1, \,\,$ faces $\, \Gamma _{1} \supset \Gamma _{2} \supset \ldots \supset \Gamma _{r} \,$ of the polyhedron $\Re \,$, such that $\rho _{s} \rightarrow \infty , \,$ $\lambda _{1}^{s} \rightarrow 1, \,\,$ $\lambda _{j +1}^{s} = o(\lambda _{j}^{s})$, $j = 1,\ldots , n - 1, \,\,$ $\lambda _{n + 1}^{s} = 0 \,\,\, (s = 1,2, \ldots )$, $\rho _{s}^{\lambda _{r}^{s}} \rightarrow \infty , \,\,$ $\rho _{s}^{\lambda _{r + 1}^{s}} \rightarrow b \, (\ne 0) \,\,$ as $\, s \rightarrow \infty \,$ and vectors $\, \mu ^{s}: = \sum _{j = 1}^{r} \lambda _{j}^{s} \, e^{j}\,\, (s = 1,2, \ldots ) \,$ are normals of the face $\Gamma _{r}. \,$

Representation (2.4) implies the existence of a number $\sigma _{r} > 0, \,$ such that for all $\, \alpha \in \Gamma _{r} \,$ we have

$$\begin{aligned} P(\xi ^{s}) =\rho _{s}^{(\alpha ,\mu ^{s})} \left[ P_{\Gamma _{r}} (\rho _{s}^{ \sum _{j = r + 1}^{n + 1} (\lambda _{j}^{s} \, e^{j}\,)}) + o(\rho _{s}^{- \sigma _{r} \,\lambda _{r}^{s} })\right] , \end{aligned}$$

where $\, P_{\Gamma _{r}} \,$ is the subpolynomial of $\, P \,$ corresponding to $\, \Gamma _{r}. \,$ Therefore, for $\, s \rightarrow \infty \,$ we have

$$\begin{aligned} P(\xi ^{s}) =\rho _{s}^{(\alpha ,\mu ^{s})} [P_{\Gamma _{r}} (b^{e^{r + 1}}) + o (1) ]. \end{aligned}$$

(2.19)

Consider the following possible cases: a) $r > 1$ and b) $r = 1$.

In case a) the dimension of $\, \Gamma _{r} \,$ is less then $\, n - 1 \,$ and $\,b^{e^{r + 1}} \in {\mathbb {R}}^{n,0}, \,$ hence $\, P_{\Gamma _{r}} (b^{e^{r + 1}}) =: c_{1} \ne 0.$

On the other hand, for $\, s \rightarrow \infty \,$ we have

$$\begin{aligned} | (\xi ^{s})^{\nu } | = \rho _{s}^{(\nu ,\mu ^{s})} \, \rho _{s}^{ \sum _{j = r + 1}^{n + 1} \lambda _{j}^{s} \, (e^{j}, \nu ) \,)} = \rho _{s}^{(\nu ,\mu ^{s})} \, [b^{(e^{r + 1}, \nu )} + o(1)]. \end{aligned}$$

(2.20)

Since $\rho _{s} \rightarrow \infty \,$ for $\, s \rightarrow \infty \,$ and $\nu \in \Re , \,$ we have $\, (\nu , \mu ^{s}) \le (\alpha , \mu ^{s}) \,\,$ $\, (s = 1,2, \ldots ), \,$ then, due to the fact that $\, c_{1} >0, \,$ the relations (2.19)–(2.20) contradict the relation (2.18).

In the case b) consider the following sub-cases: b.1) $\, e^{1} \ne \lambda \,$ and b.2) $\, e^{1} = \lambda . \,$ Since in sub-case b.1) $\, \Gamma _{1} \ne \Gamma , \,$ then by the hypothesis of the lemma $\, c_{1} >0, \,$ and again the relations (2.19)–(2.20) contradict the relation (2.18).

Consider the sub-case b.2). In this case for any $\, l = 1,\ldots ,n \,$ we have

$$\begin{aligned} \xi ^{s}_{l} = \rho _{s}^{ \sum _{j = 1}^{n + 1} \lambda _{j}^{s} \, e^{j}_{l}} = \rho _{s}^{\lambda _{1}^{s} \, \lambda _{l}} \, \rho _{s}^{ \sum _{j = 2 }^{n } \lambda _{j}^{s} \, e^{j}_{l}} = \rho _{s}^{\lambda _{1}^{s} \, \lambda _{l}} [b^{e_{l}^{2}} + o(1) ]. \end{aligned}$$

From where by the hypothesis of the lemma, we obtain

$$\begin{aligned} \eta _{l}= & {} \lim \limits _{ s \rightarrow \infty } \xi ^{s}_{l} / |\xi _{l}^{s}, \lambda |^{\lambda _{l}}\\= & {} \lim \limits _{ s \rightarrow \infty } \frac{ \rho _{s}^{\lambda _{1}^{s} \, \lambda _{l}} \left[ b^{e_{l}^{2}} + o(1) \right] }{ \left\{ \sum _{j = 1}^{n } \left[ \rho _{s}^{\lambda _{1}^{s} \, \lambda _{j}} \, (b^{e_{l}^{2}} + o(1))\right] ^{2 / \lambda _{j}} \right\} ^{\lambda _{l} / 2} }\\= & {} \lim \limits _{ s \rightarrow \infty } \frac{ \rho _{s}^{\lambda _{1}^{s} \, \lambda _{l}} \left[ b^{e_{l}^{2}} + o(1) \right] }{\rho _{s}^{\lambda _{1}^{s} \, \lambda _{l}} \left[ \sum _{j = 1}^{n } (b^{e_{j}^{2}} + o(1))^{2 /\lambda _{j}} \right] ^{\lambda _{l} / 2} }\\= & {} \frac{b^{e_{l}^{2}}}{ \left[ \sum _{j = 1}^{n} (b^{e_{l}^{2}} )^{2 /\lambda _{j}} \right] ^{\lambda _{l} / 2}} \ne 0 \,\,\, (j = 1,\ldots , n). \end{aligned}$$

This contradicts the condition $\,\eta _{1}\, \eta _{2} \ldots \eta _{n} = 0 \,$ of the lemma and proves the theorem. $\square$

We are now going to strengthen the Theorem 2.1. Namely, the following is true

Proposition 2.2

Let the polynomial $\, P \,$ satisfies all the conditions of the Theorem 2.1, except for the condition $\, \Sigma (P_{0}, \mu )\cap \Sigma (P_{1}, \mu ) = \emptyset , \,$ which is replaced by the condition $\, \Sigma (P_{0}, \mu )\cap \Sigma (P_{1}, \mu ) \cap {\mathbb {R}}^{n,0} = \emptyset . \,$ Then inequality (2.1) holds for $\, \nu \in {\mathbb {R}}^{n,+} \,$, if and only if $\, \nu \in \Re ^{*}.$

Proof

Necessity is proved similarly to the corresponding part of Theorem 2.1. To prove sufficiency, assume the converse, that there exists a multi-index $\nu \in \Re ^{*}, (\lambda , \nu ) \le d_{1} \,$ and a sequence $\{\xi ^{s}\}$, such that $\xi ^{s} \rightarrow \infty$ for $\, s \rightarrow \infty \,$ and

This relation implies the existence of a sub-sequence (which we also denote by $\{\xi ^{s}\}$) and a point $\, \eta \in {\mathbb {R}}^{n}, |\lambda , \eta | = 1 \,$ such that $\, \xi ^{s} / | \lambda , \xi ^{s} |^{\lambda } \rightarrow \eta \,$ for $\, s \rightarrow \infty . \,$ Obviously, it is sufficient to consider the case when $\, \eta \in \Sigma (P_{0}).$

Consider the following possible cases: I) $\, \eta \in {\mathbb {R}}^{n,0}, \,$ II) $\, \eta \notin {\mathbb {R}}^{n,0}. \,$

In case I) a contradiction with (2.1$'$) is obtained by carrying out similar arguments as in the proof of sufficiency in Theorem 2.1. A contradiction with (2.1$'$) is directly obtained by applying Proposition 2.1. The resulting contradictions prove the part of the proposition related to sufficiency.

Proposition 2.1 is proved. $\square$

2.2 The case of degeneracy of a k-dimensional face ($0< k < n - 1)$

In this section we study estimates of the form (2.1), when the Newton polyhedron of the polynomial in $n > 2$ variables can have a degenerate face of dimension $k < n - 1.$

Note, that this case is possible only when $\, n > 2 \,$. An example of such a polynomial for $\, n = 3 \,$ is $R(\xi ) = (\xi _{1} - \xi _{2})^{2} + \xi _{3}^{2},$ with one-dimensional degenerate face $\, (2,0,0) - (0,2,0).$

Let $\Re = \Re (P)$ be the complete Newton polyhedron of the polynomial $P, \,\,$ $\Re _{i}^{k} \,\, (i = 1,\ldots , M_{k}; k = 0, 1,\ldots , n -1) \,$ be its principal faces. Select some principal face $\Gamma : = \Re _{i_{0}}^{k_{0}} \,\,$ $(1 \le i_{0} \le M_{k}; 1 \le k \le n - 1) \,$ and let $\lambda \in \Lambda (\Gamma ).$We represent polynomial $P, \,$ in the form (1.5) with respect to the vector $\lambda$ and denote

$$\begin{aligned} \Re _{\lambda }^{*}: = \{ \alpha \in \Re , (\lambda , \alpha ) \le d_{1}(\lambda ) \}; \,\,\, \Re ^{*}: = \cap _{\lambda \in \Lambda (\Gamma )} \Re _{\lambda }^{*}. \end{aligned}$$

Theorem 2.2

Let $\Re \,$ be the complete Newton polyhedron of the polynomial $P \in {I}_{n}.$ Let all the principal faces $\Re _{i}^{k} \,\,$ $(i = 1,\ldots , M_{k}, \,\, k=0,1,\ldots ,n - 1)$ of the polyhedron $\, \Re , \,$ except from one face $\, \Gamma := \Re _{i_{0}}^{ k_{0}} \,\,$ $(1 \le i_{0} \le M_{k_{0}}: 1\le k_{0} < n - 1 ) \,$ are non-degenerate, and the face $\, \Gamma$ is degenerate. Let $\lambda \in \Lambda (\Gamma ) \,$ and, according to the vector $\lambda ,$ the polynomial $\, P \,$ is represented in the form (1.1), with $P_{d_{1} (\lambda )} \ne 0 \,$ for all $\eta \in \Sigma ' (\Gamma ).$ Then inequality (2.1) holds for $\nu \in \Re ^{*} \,$ if and only if $\, P_{d_{1} (\lambda )} (\eta ) > 0 \,$ for all $\lambda \in \Lambda (\Gamma ) \,$ and for all $\, \eta \in \Sigma ' (\Gamma ).$

Proof

Repeating the arguments used in the proof of Theorem $2.1'$, we obtain the relations (2.5)–(2.7), with $e^{r} \in \Lambda (\Re _{i_{r}, k_{r}}).$ Obviously, it suffices to consider the case $(e^{r}, \alpha ) > 0.$

If $P^{i_{r}, k_{r}} (\eta ) \ne 0,$ then these relations contradict (2.2). If $P^{i_{r}, k_{r}} (\eta ) = 0,$ then the face $\Re _{i_{r}}^{ k_{r}} \,$ coincides with the (unique) degenerate principal face $\, \Gamma := \Re _{i_{0}} ^{ k_{0}}, \,\,$ and $h^{1,s}: = \sum _{j = 1}^{r} \lambda _{j}^{s} \, e^{j} \in \Lambda (\Gamma ) \,$ for all $\, s = 1,2,\ldots \cdot$

Note, that due to the finite number of multi-indices of the set $\Re ^{*},$ to an infinite set of vectors $\, \{ h^{1,s} \}$ corresponds only finitely many different polynomials $\, \{P_{1, h^{1,s}}\}.$ Therefore, by choosing a subsequence, we can assume that the sequence $\, \{ h^{1,s} \}$ corresponds to the same polynomial $P_{1}(\xi ): = P_{1, h^{1,s} }$ and hence the same polynomial $r(\xi ): = P(\xi ) - P_{0} (\xi ) - P_{1} (\xi ), \,$ where $\, P_{0} (\xi ) \equiv P^{i_{0}, k_{0}} (\xi ).$

Thus, the polynomial P can be represented as

$$\begin{aligned} P(\xi ^{s}) = P_{0} (\xi ^{s}) + P_{1} (\xi ^{s}) + r(\xi ^{s}), \end{aligned}$$

(2.21)

where $P_{0}$ and $P_{1}$ are $h^{1,s} -$homogeneous polynomials for all $s=1,2,\ldots \cdot$ Moreover, for any point $\alpha \in (P_{0})$ and $\beta \in (P_{1})$ we have $(\alpha , h^{1,s}) =:d_{0,s} \rightarrow d_{0}$ and $(\beta , h^{1,s}) =:d_{1,s} \rightarrow d_{1}$ as $s \rightarrow \infty .$

Then, due to the $h^{1,s} -$homogeneity of these polynomials and from the representation (2.21), for all $s=1,2,\ldots$ we have

$$\begin{aligned} P(\xi ^{s}) = \rho _{s}^{d_{0,s}} \, P_{0} (\rho _{s}^{h^{1,s}}) + \rho _{s}^{d_{1,s}} \, P_{1} (\rho _{s}^{h^{1,s}}) + r(\xi ^{s}). \end{aligned}$$

(2.22)

Since $\rho _{s}^{h^{1,s}} \rightarrow \eta \in \Sigma ' (\Gamma )$ for $s \rightarrow \infty$ and by the hypothesis of the theorem $P_{1}(\eta ) \ne 0,$ then $P_{1} (\rho _{s}^{h^{1,s}}) \ne 0$ for sufficiently large s. Without loss of generality, we can assume that $P_{1} (\rho _{s}^{h^{1,s}}) \ne 0$ for all $s \in {\mathbb {N}}.$ On the other hand, since $P \in {I}_{n},$ by Lemma 2.1$P_{0} (\rho _{s}^{h^{1,s}}) \ge 0$ for all $s=1,2,\ldots ,$ and $P_{1} (\eta ) >0.$ As a result in the case under consideration $P_{0} (\rho _{s}^{h^{1,s}}) \ge 0$ and $P_{1} (\rho _{s}^{h^{1,s}}) \ge 0$ for all $s=1,2,\ldots \cdot$

From geometric considerations (see [22], also Lemma 2.2) it follows that

$$\begin{aligned} |r(\xi ^{s}) | = o(\rho _{s}^{d_{1,s}}) \end{aligned}$$

(2.23)

when $s \rightarrow \infty$. Therefore, from (2.22) - (2.23) it follows, that for $s \rightarrow \infty$ we have

$$\begin{aligned} P(\xi ^{s}) \ge P_{1} (\eta ) \rho _{s}^{d^{1,s}} (1+ o(1)). \end{aligned}$$

(2.24)

Since $\, \nu \in \Re ^{*},$ with some constant $c>0$ we have

$$\begin{aligned} | (\xi ^{s})^{\nu } | \le c \, \rho _{s}^{d^{1,s}} \,\, s= 1,2,\ldots \cdot \end{aligned}$$

(2.25)

Relations (2.24)–(2.25) together contradict (2.2) and prove Theorem 2.2. $\square$

Corollary 2.2

A polynomial $\, P \,$ with a complete Newton polyhedron $\, \Re \,$ satisfies the relation

$$\begin{aligned} \sum \limits _{\nu \in \Re } |\xi ^{\nu } | \le c \, ( | P(\xi ) | + 1) \,\,\, \forall \xi \in {\mathbb {R}}^{n} \end{aligned}$$

(2.26)

with some constant $c > 0$, if and only if $\, P \,$ is non-degenerate.

Proof

The fact, that inequality (2.26) follows from the non-degeneracy of $\, P \,$, is proved in Theorem 2.1. Let us prove the converse, that (2.26) implies the non-degeneracy of polynomial $\, P. \,$ Let the opposite, some principal face $\Gamma := \Re _{i}^{k} \,$, $\,\, 1 \le i \le M_{k}, 1 \le k \le n - 1 \,\,$ is degenerate. Let $\, P^{i,k} \,$ be a sub-polynomial of polynomial $\, P \,$ corresponding to this face, $\lambda \in \Lambda (\Gamma ), \,$ $\eta \in \Sigma ' (\Gamma ) \,$ and $\, (\lambda , \alpha ) = d_{0} \,$ is the equation of $(n - 1) -$dimensional supporting hyperplane of $\, \Re \,$ passing through the face $\Gamma .$

We represent the polynomial $\, P \,$ in the form (1.1) with respect to the vector $\, \lambda \,$ and put $\, \xi ^{s}:= s^{\lambda } \, \eta \,\, (s=1,2,\ldots ).$ Then $\, P^{i,k}(\eta ) = 0 \,$ and for all $\, s=1,2,\ldots$ we have

$$\begin{aligned} P(\xi ^{s})= & {} P^{i,k}(s^{\lambda } \, \eta ) + \sum \limits _{j = 1}^{M} P_{j} (s^{\lambda } \, \eta )\\= & {} s^{d_{0}} \,P^{i,k}(\eta ) + \sum \limits _{j = 1}^{M} s^{d_{j}} \, P_{j} (\eta ) = \sum \limits _{j = 1}^{M} s^{d_{j}} \, P_{j} (\eta ), \end{aligned}$$

moreover, $|(\xi ^{s})^{\nu } | = s^{d_{0}} \, |\eta _{1}^{\nu _{1}} \ldots \eta _{n}^{\nu _{n}} | \,$ for any point $\nu \in \Gamma .$

Since $\eta \in {\mathbb {R}}^{n,0}$, hence $\,| \eta _{1} |^{\nu _{1}} \ldots |\eta _{n} |^{\nu _{n}} \ne 0.$ On the other hand $d_{j} < d_{0}$ for all $\, j = 1,\ldots , M. \,$ Therefore, from these relations it follows that $|P(\xi ^{s}) | = o(s^{d_{o}}) \,$ for sufficiently large $\, s, \,$ while $\, | (\xi ^{s})^{\nu } | / s^{d_{o}} = | \eta _{1}^{\nu _{1}} \ldots \eta _{n}^{\nu _{n}} | > 0 \,$ $(s = 1,2,\ldots ).$ This contradicts (2.26) and proves the Corollary 2.2. $\square$

3 Comparison of the power of generalized homogeneous polynomials

3.1 General (n-dimensional) case

Let $\lambda = (\lambda _{1},\ldots , \lambda _{n} ) \in {\mathbb {E}}^{n} \,$ be a vector with positive rational coordinates and

$$\begin{aligned} R (\xi ) = R (\xi _{1},\ldots ,\xi _{n}) = \sum \limits _{(\lambda , \alpha ) = d_{R} } \gamma _{\alpha }^{R} \xi ^{\alpha } \end{aligned}$$

be a $\lambda -$homogeneous polynomia. As usual, we denote by (R) the set of multi-indices $\{\alpha \}$ for which $\gamma _{\alpha }^{R} \ne 0 \,$ and by $\Re (R)$ we denote the Newton polyhedron of the set $(R)\cup {0}.$

Further, we will assume that the polyhedron $\Re (R)$ is n-dimensional. We denote $\Sigma (R): = \{ \xi : \xi \in {\mathbb {R}}^{n,0}, | \xi , \lambda | = 1, R (\xi ) = 0 \}$. For the points $\eta \in \Sigma (R)$ we denote

$$\begin{aligned} {\mathfrak {A}} (\eta , R)&: =&\{ \nu : \nu \in {\mathbb {N}}_{0}^{n}, D^{\nu }R (\eta ) \ne 0 \}\nonumber \\ \Delta (\eta , R)&: =&\min \limits _{\nu \in {\mathfrak {A}} (\eta , R)} (\lambda ,\nu ). \end{aligned}$$

(3.1)

The following proposition was proved in [18], Theorem 1]

Theorem 3.1

Let $P$ and $Q$ be $\lambda -$homogeneous polynomials of $\lambda -$orders $d_{P}$ and $d_{Q}$respectively, $\, \lambda _{j} > 0 \,\, (j = 1,2,\ldots , n)$. Let all the principal faces $\, \Re (P) \,$ of the polynomial $P$ be non-degenerate, except possibly the $(n - 1) -$dimensional face $\, \Gamma$ containing the set $\, (P).$ Then

• (I) If   face $\, \Gamma$ is non-degenerate, then $Q < P$ for any polynomial Q such that $\, \Re (Q) \subset \Re (P)$.

• (II) If   face $\, \Gamma$ is degenerate, then $P > Q$, if and only if the following conditions are simultaneously satisfied

  • (1) $d_{Q} \le d_{P} \,$

  • (2) $\Sigma (Q) \supset \Sigma (P) \,$

  • (3) $\Re (Q) \subset \Re (P) \,\,$

  • (4) (see notation (3.1))

    $$\begin{aligned} \frac{d_{Q}}{d_{P}} \le \frac{\Delta (\eta , Q)}{\Delta (\eta , P)} \quad \forall \eta \in \Sigma (P). \end{aligned}$$

    (3.2)

In order to simplify the proof of the theorem, we make the following remarks

Remark 3.1

Along points a) and b) of Remark 1.2 we add the following: c) it is geometrically obvious that sub-polynomials corresponding to the faces $\, \Re _{i}^{k} \,$ of the polyhedron $\Re (P) = \Re ((P)\bigcup \{0\})$ have the form $P^{i,k} (\xi )$ or $P^{i,k} (\xi ) + 1.$ Therefore, point b) implies, that only the faces, that don’t include the point zero, can be degenerate, because the remaining faces correspond to polynomials of the form $P^{i,k} (\xi ) + 1.$

Remark 3.2

Since the numbers $\lambda _{1},\ldots , \lambda _{n}$ are positive and rational, and any $\lambda -$homogeneous polynomial R is also $(k \, \lambda ) -$homogeneous, then choosing a natural number $\, k \,$ in an appropriate way (which does not affect the $\, d_{Q} / d_{P} \,$ ratios), we can assume that the numbers $d_{Q}$ and $d_{P}$ are natural, hence the functions $P^{d_{Q}}$ and $Q^{d_{P}}$ are also polynomials. Therefore, we can compare their power. Moreover, the following proposition holds

Lemma 3.1

Let $P$ and $Q$ be $\lambda -$homogeneous polynomials of $\lambda -$orders $d_{P}$ and $d_{Q}$respectively, where $d_{P} \ge d_{Q}.$ Then a) $P > Q$ if and only if $Q^{d_{P}} < P ^{d_{Q}},$ (which is the same as $\, Q < P^{\, d_{Q}/ d_{P}})$ i.e there is a number $c > 0$, such that

$$\begin{aligned} | Q (\xi ) | ^{d_{P}} \le c \, \left[ 1 + | P (\xi ) |^{\, d_{Q} } \right] \forall \xi \in {\mathbb {R}}^{n}. \end{aligned}$$

(3.3)

b) If $d_{Q} < d_{P},$ then $\, |Q(\xi ) | / | P(\xi ) | \rightarrow 0$, when $\, | P(\xi ) | \rightarrow \infty .$ c) Let $P > Q,$ $d_{P} \ge d_{Q},$ $\delta < d_{Q} / d_{P}, \,\,$ $\eta \in \Sigma (P) \,$ and $\eta ^{s} \rightarrow \eta \,$, as $\, s \rightarrow \infty . \,$ Then $| Q (\eta ^{s}) | / | P (\eta ^{s}) |^{\delta } \rightarrow 0$, as $\, s \rightarrow \infty . \,$

Proof

Let’s prove point a). Since $d_{P} \ge d_{Q},$ it follows from $Q^{d_{P}} < P ^{d_{Q}},$ that $Q < P.$ Consequently, the sufficiency of estimate (3.3) for the relation $Q < P$ is obvious. We need to prove that estimate (3.3) follows from $Q < P.$

Suppose, on the contrary, $Q < P,$ but estimate (3.3) is not true, that is, there exists a sequence $\, \{\xi ^{s}\} \,$, such that $\, \xi ^{s} \rightarrow \infty \,$ as $\, s \rightarrow \infty$ and $\, |Q (\xi ^{s}) |^{d_{P}} / [1 + |P(\xi ^{s}) |^{d_{Q}}]$ $\rightarrow \infty . \,$ Hence $\, | Q (\xi ^{s}) |\rightarrow \infty \,$ and $Q < P$, it follows, that $\, t_{s}: = |P(\xi ^{s}) | \rightarrow \infty$ as $\, s \rightarrow \infty .$

Denote $\tau _{i}^{s}:= t_{s}^{-\lambda _{i} / d_{P}} \, \xi _{i}^{s} \,$ i.e. $\, \xi ^{s} =$ $t_{s}^{\lambda / d_{P}} \, \tau ^{s}. \,\,$ We have $P(\tau ^{s}) = 1$ $\,\, (i = 1,2,\ldots , n;\, s = 1,2, \ldots \cdot ). \,$ Due to the $\, \lambda -$homogeneity of the polynomials $\, Q \,$ and $\, P, \,$ we have

$$\begin{aligned} | Q(\xi ^{s}) |= & {} t_{s}^{d_{Q}/ d_{P}} \, | Q(\tau ^{s}) |,\\ | P(\xi ^{s}) |^{d_{Q}/ d_{P}}= & {} t_{s}^{d_{Q}} \, | P(\tau ^{s}) |^{d_{Q}/ d_{P}} = t_{s}^{d_{Q}}. \end{aligned}$$

Since $\, Q < P, \,$ from the representation above and from $P(\tau ^{s}) = 1 \,\,$ $( s = 1,2, \ldots ) \,$ we have

$$\begin{aligned} | Q(\xi ^{s}) |/ \left[ 1 +| P(\xi ^{s}) |^{d_{Q}/ d_{P}} \,\right]= & {} {t_{s}^{d_{Q}/ d_{P}} \, | Q(\tau ^{s}) |} / \left[ {1 + t_{s}^{d_{Q}} }\right] \\\le & {} c \, t_{s}^{d_{Q}/ d_{P}} \, \left[ 1 + | P (\tau ^{s}) | \,\right] / \left[ {1 + t_{s}^{d_{Q}} }\right] \le 2 \, c \, t_{s}^{d_{Q}/ d_{P} - d_{Q}} \\= & {} 2 \ c \, t_{s}^{d_{Q}(1/d_{P} -1)}. \end{aligned}$$

Since $\, d_{P} \ge 1 \,$ (see Remark 3.2), the expression $t_{s}^{d_{Q}(1/d_{P} -1)} \,$ is limited, which contradicts our assumption and proves point a) of the lemma.

Point b) follows directly from point a). When we divide both sides of the (already proved) inequality (3.3) by $\, | P(\xi ) |^{d_{P}} \,$ and assume that $\, | P (\xi ) | \,$ tends to infinity, then we have $[|Q(\xi ) | / | P(\xi ) |]^{d_{P}} \rightarrow 0.$

Now we prove point c). Assume the opposite, that for some point $\, \eta \in \Sigma (P), \,$ for a sequence $\{\eta ^{s}\}, \eta ^{s } \rightarrow \eta$ and for a number $\, \delta < d_{Q} / d_{P}$

$$\begin{aligned} | Q (\eta ^{s}) | / | P (\eta ^{s}) |^{\delta } \ge \varepsilon >0 \,\,\, (s = 1,2, \ldots ). \end{aligned}$$

(3.4)

We denote $\, t_{s}: = | P (\eta ^{s}) |^{- 1 / d_{P} }, \,\,$ $\xi ^{s}: = t_{s}^{\lambda } \, \eta ^{s} \,\, (s = 1,2, \ldots ).$ Then, with respect to the $\, \lambda -$homogeneity of the polynomials $\, Q \,$ and $\, P$, we have for all s = 1,2,...

$$\begin{aligned} | Q (\xi ^{s}) |= & {} t_{s}^{d_{Q}} \, | Q (\eta ^{s}) | \ge \varepsilon \,\, t_{s}^{d_{Q}} \, | P (\eta ^{s}) |^{\delta } = \varepsilon \, t_{s}^{d_{Q} - \delta \, d_{P}}, \end{aligned}$$

(3.5)

$$\begin{aligned} | P (\xi ^{s}) |= & {} t_{s}^{d_{P}} \, | P (\eta ^{s}) | \equiv 1. \end{aligned}$$

(3.6)

Since $\, t_{s} \rightarrow \infty \,$ $\, (P(\eta ^{s} ) \rightarrow 0 )$ and $\, d_{Q} - \delta \, d_{P} > 0 \,$, relations (3.5)–(3.6) contradict the conditions $Q < P \,$ and prove point c).

Lemma 3.1 is proved. $\square$

Based on this lemma and bearing in mind the fact, that polynomials $P^{d_{Q}}$ and $Q^{d_{P}}$ have the same $\, \lambda -$order, further on, when comparing the powers of two polynomials $P$ and $Q, \,$ whenever it is convenient for us, we assume that the polynomials that we compare have the same $\lambda -$order: $d_{P} = d_{Q} =: d.$

After such assumption, Theorem 3.1 can be rephrased in the following way

Theorem 3.1$'$ Let P and Q be $\lambda -$homogeneous polynomials of$\lambda -$orders $d: = d_{P} = d_{Q},$ where $\lambda _{j} > 0 \,\, (j = 1,\ldots ,n). \,$ Then for the relation $Q < P$ to hold, each of the following conditions 1) - 4) is necessary and the group of conditions 1) - 3) is sufficient:

• (1) $\, \Re (Q) \subset \Re (P), \,$

• (2) $\Sigma (Q) \supset \Sigma (P), \,$

• (3) for each point $\, \eta \in \Sigma (P) \,$ there exists a neighborhood $U (\eta )$ and a constant $c = c(\eta ) > 0$, such that $\, | Q (\xi ) | \le c \, | P (\xi ) | \forall \xi \in U (\eta ), \,$

• (4) $\, \Delta (\eta , Q) \ge \Delta (\eta , P) \,$ for each point $\, \eta \in \Sigma (P). \,$

Before starting the proof, we do the following remarks.

Remark 3.1$'$ Obviously, condition 3) of the theorem is equivalent to the following one: 3.a) there is a constant $\, c > 0 \,$ such that $\, |Q(\xi )| \le c \,\,$ for all $\, \xi \in {\mathcal {D}} (P)$ $: = \{ \eta \in {\mathbb {R}}^{n}; | P(\eta ) | \, = 1 \}.$

Proof

The proof is carried out by a method, different from that of Theorem 1 in [18].

We prove the necessity of condition 1). Assume the opposite, i.e. the relation $\, Q < P \,$ holds, but the condition $\, \Re (Q) \subset \Re (P) \,$ is violated. This leads to the existence of a non-zero vertex $\beta \in \Re ^{0} (Q) \setminus \Re (P). \,$ Let ${\tilde{\Re }} \,$ denote the Newton polyhedron of the set $\, [\Re ^{0} (Q) {\setminus } \{\beta \}]\cup \Re ^{0} (P)$ and a vector $\mu \in {\mathbb {R}}^{n} \,$ be an exterior normal (with respect to the polyhedron $\Re (Q) \,$) of this vertex, such that $(\mu , \beta ) > 0$ and

$$\begin{aligned} (\mu , \beta ) > \max \limits _{\alpha \in \Re (Q) \cup \Re (P); \alpha \ne \beta } (\mu , \alpha ). \end{aligned}$$

Then for any point $\eta \in {\mathbb {R}}^{n},$ $\eta ^{\beta } \ne 0$ and for $t \rightarrow \infty$ we obtain

$$\begin{aligned} |Q (t^{\mu } \eta )| / \left[ 1 + | P (t^{\mu } \eta ) | \right] = \left[ t^{ (\mu , \beta )} | \eta ^{\beta } | + o (t^{ (\mu , \beta )}) \right] / \left[ 1 + o (t^{ (\mu , \beta )}) \right] \rightarrow \infty , \end{aligned}$$

which contradicts the condition $P > Q$ and proves the necessity of condition 1).

The necessity of condition 2) is obvious, because otherwise, for some point $\, \eta \in \Sigma (Q) \setminus \Sigma (P) \,$ and for $\, t\rightarrow \infty \,$ we would have $\, P(t^{\lambda } \eta ) = t^{d} \, P(\eta ) = 0,\,$ while $\, Q(t^{\lambda } \eta ) = t^{d} \, Q(\eta ) \rightarrow \infty . \,$

To prove the necessity of condition 3) we assume the opposite i.e. $P > Q, \,$ but there exists a point $\eta \in \Sigma (P) \,$ and a sequence $\{\eta ^{s}\},$ $\, \eta ^{s} \rightarrow \eta \,$ as $s \rightarrow \infty$, such that

$$\begin{aligned} R (\eta ^{s}): = | Q (\eta ^{s}) | / |P (\eta ^{s}) |\rightarrow \infty \,\,\,\,\, as \,\,\,\, s \rightarrow \infty . \end{aligned}$$

(3.7)

If $\,\, P (\eta ^{s} ) = 0 \,\,$ for some $\, s$, then by the already proved condition 2) $\, Q(\eta ^{s}) = 0 \,\,$ for such $\, s$, and the relation $\, Q < P \,$ is obvious. Therefore, we can assume that $\,\, P (\eta ^{s} ) \ne 0 \,\,$ for all $\, s=1,2,\ldots \cdot$

We denote $\, t_{s}: = | P (\eta ^{s}) |^{ \, - \, 1 / d}, \,\,$ $\, \xi ^{s}: = t_{s}^{\lambda } \eta ^{s} (s = 1,2,\ldots ).$ On the sequence $\, \xi ^{s} \,$ we have $\, P(\xi ^{s}) = P( t_{s}^{\lambda } \eta ^{s}) = t_{s}^{d} \, P( \eta ^{s}) = 1 \,\,\, (s = 1,2,\ldots ) \,\,$ and $\, Q (\xi ^{s}) = t_{s}^{d} \,\, Q (\eta ^{s}). \,\,$

From the definition of the function $\, R \,$ it follows that $\,\, Q (\eta ^{s}) = [R (\eta ^{s})] \,\, | P (\eta ^{s}) |. \,\,$ Since $\, t_{s} \, | P (\eta ^{s}) |^{1 / d} = 1 \,\,\, (s = 1,2,\ldots ), \,\,$ we obtain (see (3.7))

$$\begin{aligned} \,\, | Q (\xi ^{s}) | = t_{s}^{d} \, | Q (\eta ^{s}) | = t_{s}^{d} \, \left[ R (\eta ^{s})\right] \,\, \, | P (\eta ^{s}) | = R (\eta ^{s}) \rightarrow \infty \end{aligned}$$

as $\, s \rightarrow \infty .$ This means that $\, Q \not < P \,$ and proves the necessity of condition 3) of the theorem.

Finally, let us prove the necessity of item 4).

By definition of $\, \Delta (\eta , Q)\,$ there exists $\, \tau \in {\mathbb {R}}^{n}$, such that $\, \sum _{(\lambda , \alpha ) = \Delta (\eta , Q) } \frac{Q^{ (\alpha )} (\eta ) }{\alpha !} \, \tau ^{\alpha } \ne 0. \,\,$

Let $\, t > 0 \,$ and $\, \xi (t) = \eta + t^{\lambda } \, \tau . \,$ Then by Taylor’s formula we obtain for sufficiently small $\, t \,$

$$\begin{aligned} Q (\xi (t))= & {} \sum \limits _{(\lambda , \alpha ) \ge \Delta (\eta , Q) } \frac{Q^{ (\alpha )} (\eta ) \, (t^{\lambda } \, \tau )^{\alpha }}{\alpha !}\nonumber \\= & {} t^{\Delta (\eta , Q) } \, \left[ \sum \limits _{(\lambda , \alpha ) = \Delta (\eta , Q) } \frac{Q^{ (\alpha )} (\eta ) \, \tau ^{\alpha }}{\alpha !} + o(1)\right] , \end{aligned}$$

(3.8)

$$\begin{aligned} P (\xi (t))= & {} t^{\Delta (\eta , P) } \,\, \left[ \sum \limits _{(\lambda , \alpha ) = \Delta (\eta , P) } \frac{P^{ (\alpha )} (\eta ) \, \tau ^{\alpha }}{\alpha !} + o(1) \right] . \end{aligned}$$

(3.9)

Due to the condition 3) of this theorem (see inequality (3.5)) from (3.8)-(3.9) for sufficiently small $\, t \,$ we have

$$\begin{aligned}{} & {} | \, t^{\Delta (\eta , Q) } \, | \sum \limits _{(\lambda , \alpha ) = \Delta (\eta , Q) } \frac{Q^{ (\alpha )} (\eta ) \, \tau ^{\alpha }}{\alpha !} + o(1) \, |\\{} & {} \quad \le t^{\Delta (\eta , P) } \,\, | \sum \limits _{(\lambda , \alpha ) = \Delta (\eta , P) } \frac{P^{ (\alpha )} (\eta ) \, \tau ^{\alpha }}{\alpha !} + o(1) |. \end{aligned}$$

From where it follows that $\, \Delta (\eta , Q) \ge \Delta (\eta , P).$

Sufficiency. We prove that conditions 1) - 3.a) (see Remark 3.1) imply $Q < P. \,$ We represent $\, {\mathbb {R}}^{n} \,$ as the union of the following two sets: $\, {\mathcal {A}}: = \{\xi \in {\mathbb {R}}^{n}, \, P(\xi ) = 0 \} \,$ and $\, {\mathcal {B}}: = {\mathbb {R}}^{n} {\setminus } {\mathcal {A}}. \,$ Due to the $\, \lambda -$homogeneity of the polynomials $\, P \,$ and $\, Q, \,$ condition 2) implies that $\, Q (\xi ) = 0 \,$ for all $\xi \in {\mathcal {A}}. \,$ Hence

$$\begin{aligned} | Q(\xi ) | \le c \, \left[ |P(\xi )| + 1\right] \,\,\, \forall \xi \in {\mathcal {A}}. \end{aligned}$$

(3.10)

Let us show that (3.10) is also valid for the set $\, {\mathcal {B}}. \,$ For $\,\xi \in {\mathcal {B}} \,$ we denote $\, \eta (\xi ): = | P(\xi ) |^{- \lambda / d}.$ It is obvious that $\, \eta (\xi ) \in {\mathcal {D}} (P) \,$ for any $\,\xi \in {\mathcal {B}}. \,$ Therefore, by virtue of condition 3.a) $| Q(\eta (\xi ) | \le c$ $\,\,\, \forall \xi \in {\mathcal {B}}. \,$ Hence, due to the $\, \lambda -$homogeneity of the polynomials $\, P \,$ and $\, Q \,$, having the same degree of homogeneity $\, d, \,$ we have

$$\begin{aligned} \left| \frac{ Q(\xi ) }{P(\xi ) } \right| = | Q (\xi \,\,| P(\xi ) |^{- \lambda / d} ) | = |Q(\eta (\xi ) ) | \le c \,\,\,\,\, \forall \xi \in {\mathcal {B}}. \end{aligned}$$

Hence, we obtain estimate (3.10) also for the set $\, {\mathcal {B}} \,$ and, therefore, for the entire space $\, {\mathbb {R}}^{n}.$

Theorem 3.1$'$ is proved. $\square$

3.2 3.3 Two - dimensional case

In this subsection, we compare the powers of generalized-homogeneous polynomials of two variables. We will consider two-dimensional case separate from the general ($n -$dimensional) case, since in this case the results obtained are final. This is due to the one lemma of B.Pini (see [34]) on the factorization of two-dimensional generalized-homogeneous polynomials. At the same time, we rephrase (and prove) the Pini lemma in terms suitable for us.

Two-dimensional case differs (in the positive sense) from the case when $\, n > 2$. In this case any $\, \lambda -$homogeneous polynomial $P(\xi _{1}, \xi _{2}) \,$ (for $\, \lambda _{1} > 0, \lambda _{2}$> 0 ) is represented as $\, P(\xi ) = \xi ^{\nu } \ p(\xi ), \,$ where $\, p(\xi ) \ne 0 \,$ at points $\, (0,a) \,$ and $\, (b,0), \,$ where $\, a\, b \ne 0, \,$ and $\, \nu \,$ is a multi-index.

Here we also describe those lower-order terms whose addition to the generalized-homogeneous polynomial $\,P \,$ preserves its power.

Remark 3.3

Note that a) in the two-dimensional case, any $\, \lambda -$homogeneous polynomial $\, R(\xi ) = R(\xi _{1},\xi _{2})$ can be represented as $\, R(\xi ) = \xi ^{\nu } R_{1} (\xi ), \,$ where $\, \nu = (\nu _{1}, \nu _{2})\,$ is a multi-index and $\, R_{1} (\xi ) \,$ is a $\lambda -$homogeneous polynomial such that $\, R_{1} (1,0) \cdot \, R_{1} (0,1) \ne 0, \,\,$ b) since 1) the numbers $\, \lambda _{1} \,$ and $\, \lambda _{2} \,$ can be chosen rational, 2) any $\, \lambda -$homogeneous polynomial is also $\, k \lambda -$homogeneous, then the number $\, k \,$ can be chosen in such way, that the fraction $\, \lambda _{1} \, / \, \lambda _{2} \,$ has the form $p \, / \, q, \,$ where $\, q \,$ is an odd number. Moreover, the following proposition holds.

Lemma 3.2

(see [18, 27, 34, 36]) Let $R (\xi ) = R (\xi _{1}, \xi _{2})$ be a $\lambda -$homogeneous polynomial of $\lambda -$order $d = d(R)$ $( \lambda _{1}> 0, \lambda _{2} > 0 )$ with real coefficients, such that $\, R (1,0) \cdot \, R (0,1) \ne 0$, then it can be represented in the form

$$\begin{aligned} R (\xi ) = r(\xi ) \prod \limits _{j = 1}^{N} (\xi _{1} - \kappa _{j} \xi _{2}^{\lambda _{1} / \lambda _{2}} )^{l_{j}} = r(\xi ) \prod \limits _{j = 1}^{N} r_{j} (\xi ), \end{aligned}$$

(3.11)

where $N = N (R)$ and $l_{j}$ are natural numbers, $\{ \kappa _{j} \}$ are non-zero, pairwise distinct numbers $(j = 1,\ldots , N),$ $r (\xi )$ is an infinitely differentiable function in ${\mathbb {R}}^{2,0}, \,\,$ while $r(\xi ) \ne 0$, when $0 \ne \xi \in {\mathbb {R}}^{2,0}.$

Proof

For $\xi _{2} = 0, \,$ the representation (3.11) is obvious. Let $\xi _{2} \ne 0,$ then

$$\begin{aligned} R (\xi )= & {} \sum \limits _{(\lambda , \alpha ) = d} \gamma _{\alpha } \xi _{1}^{\alpha _{1}} \xi _{2}^{\alpha _{2}} = \xi _{2}^{d / \lambda _{2}} \sum \limits _{(\lambda , \alpha ) = d} \gamma _{\alpha } \xi _{1}^{\alpha _{1}} \xi _{2}^{\alpha _{2} - d / \lambda _{2}} \\= & {} \xi _{2}^{d / \lambda _{2}} \sum \limits _{(\lambda , \alpha ) = d} \gamma _{\alpha } \xi _{1}^{\alpha _{1}} \xi _{2}^{- \frac{\lambda _{1}}{\lambda _{2}} \alpha _{1} } = \xi _{2}^{d / \lambda _{2}} \sum \limits _{(\lambda , \alpha ) = d} \gamma _{\alpha } x^{\alpha _{1}}, \end{aligned}$$

where $x = \xi _{1} / \xi _{2}^{\frac{\lambda _{1}}{\lambda _{2}}}, \,$ and $\, d/ \lambda _{2}\,$ is a natural number.

Let $\kappa _{1},\ldots , \kappa _{N} \,$ be the real roots of the polynomial (of one variable) $q(x): = \sum \limits _{(\lambda , \alpha ) = d} \gamma _{\alpha } x^{\alpha _{1}}$ with multiplicities respectively $l_{1},\ldots , l_{N}$ and $\, m_{1}: = \max \limits _{\alpha \in (R)} \alpha _1 = d/ \lambda _{1},$ then

$$\begin{aligned} q (x) = q_{0} (x) \prod \limits _{j = 1}^{N} (x - \kappa _{j})^{l_{j}}, \end{aligned}$$

where $\sum _{i = 1}^{N} l_{i} + ord( q_{0}) = m_{1}.$

Let us return to the old notation. Since $\frac{d}{\lambda _{2}} - m_{1} \frac{\lambda _{1}}{\lambda _{2}} = 0,$ from the last two relation we have

$$\begin{aligned} R (\xi )= & {} \xi _{2}^{d / \lambda _{2}} q(x) = \xi _{2}^{d / \lambda _{2}} \prod \limits _{j = 1}^{N} (\frac{\xi _{1}}{\xi _{2}^{\lambda _{1} / \lambda _{2}}} - \kappa _{j})^{l_{j}} q_{0} (\frac{\xi _{1}}{\xi _{2}^{\lambda _{1} / \lambda _{2}}} ) \\= & {} \xi _{2}^{ \frac{d}{\lambda _{2}} - m_{1} \frac{\lambda _{1}}{\lambda _{2}}} q_{1} (\xi ) \prod \limits _{j = 1}^{N} (\xi _{1} - \kappa _{j} \xi _{2}^{\lambda _{1} / \lambda _{2}} )^{l_{j}} \\= & {} q_{1} (\xi ) \prod \limits _{j = 1}^{ N} (\xi _{1} - \kappa _{j} \xi _{2}^{\lambda _{1} / \lambda _{2}} )^{l_{j}}: = q_{0} (\xi ) \prod \limits _{j = 1}^{N} [r_{j} (\xi )]^{l_{j}}, \end{aligned}$$

where $q_{1}(\xi ):= \xi _{2}^{m_{1} \, \lambda _{1} / \lambda _{2}} \, q_{0} (\xi _{1} / \xi _{2}^{\lambda _{1} / \lambda _{2} } ).$ This proves the representation (3.11). It remains to note, that for $\lambda _{1} = \lambda _{2}$ the functions $q_{1} (\xi )$ and $r_{j} (j = 1,\ldots ,N)$ are polynomials.

Lemma 3.2 is proved. $\square$

Remark 3.4

It follows from Remark 3.3 and Lemma 3.2, that in the two-dimensional case any $\, \lambda -$homogeneous polynomial $\, R \,$ with real coefficients can be represented as

$$\begin{aligned} R (\xi ) = \xi ^{\nu } \, r(\xi ) \prod \limits _{j = 1}^{N} (\xi _{1} - \kappa _{j} \xi _{2}^{\lambda _{1} / \lambda _{2}} )^{l_{j}} = \xi ^{\nu } \, r(\xi ) \prod \limits _{j = 1}^{N} [ r_{j} (\xi )]^{l_{j}}, \end{aligned}$$

where $\nu \,$ is a multi-index, $\, r(\xi ) \ne 0 \,$ for $\, \xi \ne 0$, and the remaining properties proved in Lemma 3.2 stay true.

V.N.Margaryan in [31] additionally proved the following: 1) since $\lambda = (\lambda _{1}, \lambda _{2}) -$homogeneous polynomial is also $k\, \lambda -$homogeneous, with the appropriate choice of the number $\, k, \,$ we can assume, that if $\lambda _{1} \ne \lambda _{2},$, then numbers $\lambda _{1} \,$ and $\, \lambda _{2} \,$ are co prime natural numbers, 2) any $\lambda = (\lambda _{1}, \lambda _{2}) -$homogeneous polynomial $\,R(\xi ) = R(\xi _{1}, \xi _{2})$ admits the following representation

$$\begin{aligned} R(\xi ) = \xi ^{\nu } \, r(\xi ) \, \prod \limits _{j = 1}^{N} (\xi _{1}^{\lambda _{2}} - \kappa _{j}\,\xi _{2}^{\lambda _{1}})^{l_{j}}, \end{aligned}$$

where $\, \nu \,$ is a multi-index, $r (\xi )$ is a $\lambda -$homogeneous semi-elliptic polynomial (i.e. $r (\xi ) \ne 0$ for $\xi \ne 0),$ $\,\, N = N (R)$, $\{ l_{j} \}$ are natural numbers, $\{ \kappa _{j} \}$ are non-zero, pairwise distinct numbers $(j = 1,\ldots , N).$

Now we prove some simple (numerical) inequalities, which we use below and which are also of independent interest.

Lemma 3.3

In order for the inequality

$$\begin{aligned} x^{a} y^{b} \le 1 + x^{c} y^{d} \end{aligned}$$

to hold for all $x \ge 1, y \in [0,1 ], \,$ it is necessary and sufficient that the positive numbers $\, a,b,c,d \,$ satisfy the relations 1) $a \le c$ and 2) $d/b \le c/a.$

Proof

The necessity of condition $a \le c$ is obvious, we prove the necessity of condition $d/b \le c/a.$

Let the condition 2) is violated, i.e. $d/b > c/a.$ Let us prove that in that case the required inequality cannot hold. Put $\, y = x^{- c/d }.$ Then

$$\begin{aligned} x^{a} y^{b} = x^{a - b \, (c / d)} = x^{b [(a/b) - (c/d)]}, \,\,\, x^{c} y^{d}= 1. \end{aligned}$$

Since by our assumption $(a/b) - (c/d) > 0, \,$ the relations obtained shows that for sufficiently large values of $\, x \,$ the required inequality can not hold.

Now we prove the sufficiency of conditions. If $b \ge d,$ then the required inequality is obvious. Let $b < d.$ Denoting $x^{a} =: u, y^{b} =: v,$ we arrive at the equivalent inequality

$$\begin{aligned} u v \le 1 + u ^{\frac{c}{a}} v^{\frac{d}{b}} \forall u \ge 1, v \in [0,1 ]. \end{aligned}$$

When $u v \le 1$ this inequality is obvious. If $u v > 1,$ then by the conditions of lemma and the assumption $b < d$ we have

$$\begin{aligned} u v \le ( u v)^{\frac{d}{b}} = u^{\frac{d}{b}} v^{\frac{d}{b}} \le u^{\frac{c}{a}} v^{\frac{d}{b}}, \end{aligned}$$

which proves the required inequality.

Lemma 3.3 is proved. $\square$

Lemma 3.4

In order for the inequality

$$\begin{aligned} x^{a} y^{b} \le 1 + C [ \sigma _{1} x^{c} y^{d} + \sigma _{2} \, x^{c - d}] \end{aligned}$$

to hold for all $x \ge 1, y \in [0,1 ] \,$ and for a pair of positive numbers $\,\sigma _{1}, \sigma _{2} \,$ with some constant $C = C(\sigma _{1}, \sigma _{2}) > 0, \,$ it is necessary and sufficient that the positive numbers $\, a,b,c,d \,\,\, (c\ge d) \,$ satisfy the relations 1) $a \le c$ and 2) $a - b \le c-d.$

Proof

The necessity of condition $a \le c$ is obvious. We prove the necessity of the condition 2). Let the condition 2) is violated, i.e. $a - b > c-d$, and let $y = x^{- 1}, \,$ then for $x \rightarrow \infty \,$ we have

$$\begin{aligned} x^{a} \, y^{b} / \{ 1 + C\, \left[ \sigma _{1} x^{c} \, y^{d} + \sigma _{2} \, x^{c - d}\right] \} = x^{a - b} /\left[ 1 + C \, (\sigma _{1} + \sigma _{2}) \, x^{c - d}\right] \rightarrow \infty , \, \end{aligned}$$

which proves the necessity of condition 2).

Now we prove the Sufficiency. When $b \ge d$ or $d/b \le c/a,$ then the required inequality is a consequence of the inequality from Lemma 3.3. If, however, $d/b > c/a \ge 1,$ then the substitution $y = t/x$ yields to the equivalent inequality

$$\begin{aligned} x^{a - b} t^{b} \le 1 + C\,\left[ \sigma _{1} x^{c - d} t^{d} + \sigma _{2} x^{c - d} \right] , \end{aligned}$$

which can be proved simply (with any constant $\, C \ge max \{ 1,\sigma _{1}, \sigma _{2} \, \} )$, if we consider separately the cases $t \ge 1$ and $t < 1. \,\,$

Lemma 3.4 is proved. $\square$

Theorem 3.2

(see [23]) Let $n = 2, \,$ $\, P \,$ and $\, Q \,$ are $\, \lambda -$homogeneous polynomials $\, (\lambda _{j} > 0, \,\, j = 1,2 )$ of $\, \lambda -$orders $d_{P} \,$ and $\, d_{Q} \,$ respectively $\, (d_{P} > d_{Q})$, presented in the form (see Lemma 3.2)

$$\begin{aligned} P(\xi )= & {} \xi ^{\nu } \, P_{1}(\xi ) = \xi ^{\nu } \, p(\xi ) \cdot \prod \limits _{j = 1}^{N_{P}} p_{j} (\xi ) = \xi ^{\nu } \, p(\xi ) \cdot \prod \limits _{j = 1}^{N_{P}} (\xi _{1} - \tau _{j} \, \xi _{2}^{\lambda _{1}/ \lambda _{2}} )^{m_{j}}, \end{aligned}$$

(3.12)

$$\begin{aligned} Q(\xi )= & {} \xi ^{\mu } \, Q_{1}(\xi ) = \xi ^{\mu } \, q(\xi ) \cdot \prod \limits _{j = 1}^{N_{Q}} q_{j} (\xi ) = \xi ^{\mu } \, q(\xi ) \cdot \prod \limits _{j = 1}^{N_{Q}} (\xi _{1} - \sigma _{j} \, \xi _{2}^{\lambda _{1}/ \lambda _{2}} )^{k_{j}}.\nonumber \\ \end{aligned}$$

(3.13)

Then $\, Q < P \,$, if and only if (possibly after some renumbering)

$$\begin{aligned}{} & {} 1) \,\, \tau _{j} = \sigma _{j } \,\, (j = 1,\ldots ,N_{P}),\\{} & {} 2) \,\, m_{j}/k_{j} \le d_{P}/d_{Q } \,\, (j = 1,\ldots ,N_{P})\\{} & {} 3) \,\, \nu _{i} \, (d_{Q} - \mu _{i}) \le \mu _{i} (d_{P} -\nu _{i}) \,\,\, (i = 1,2). \end{aligned}$$

Proof

The proof of this theorem in special cases can be found in [23, 34]. In the present general formulation, the proof of the theorem appears here for the first time.

Before starting the proof, note that a) condition 3) of the theorem is equivalent to the condition $\, \nu _{i} /\mu _{i} \le d_{P}/d_{Q }, \,\,$ $(i = 1,2),$ b) if $\nu _{j} =0 \,\,$ for $j =1$ or $j =2, \,$ then $\mu _{j} =0$ for the corresponding $\, j. \,$ Otherwise, obviously, the relation $\, Q < P \,$ could not take place.

The necessity of condition 1) is obvious. We will prove the necessity of condition 2).

Suppose, that under condition 1), condition 2) is violated for some $\, j_{0} \,\, ( 1 \le j_{0} \le N_{P} ), \,$ i.e. $m_{j_{0}}/k_{j_{0}} > d_{P}/d_{Q } \,$, and let the point $\eta \in {\mathbb {R}}^{2,0}$ is chosen in such a way, that $\, p_{j_{0}} (\eta ) = (\eta _{1} - \tau _{_{j_{0}}} \, \eta _{2}^{\lambda _{1}/ \lambda _{2}})^{m_{j_{0}}} = 0. \,$ Then $q_{j_{0}} (\eta )= 0$ and $\, p_{i} (\eta ) \ne 0$ for $i \ne j_{0}. \,$ Let $\, \delta> 0, \, t > 0 \,$ and $\, \xi _{1} = \xi _{1} (t) = t^{\lambda _{1}} (\eta _{1} + t^{- \delta }), \,\,$ $\xi _{2} = \xi _{2} (t) = t^{\lambda _{2}} \, \eta _{2}.$

For $\, t \rightarrow \infty \,$ we get

$$\begin{aligned} p_{i} (\eta _{1} + t^{- \delta }, \eta _{2} )= & {} p_{i} (\eta _{1}, \eta _{2} ) (1 + o(1)), \,\, i \ne j_{0}\\ p_{j_{0}} (\eta _{1} + t^{- \delta }, \eta _{2} )= & {} t^{- m_{j_{0}} \, \delta }. \end{aligned}$$

From these relation it follows that with some nonzero constants $\, A \,$ and $\, B \,$, when $\, t \rightarrow \infty \,$ we get

$$\begin{aligned} P(\xi )= & {} P(\xi (t)) = t^{d_{P}} P (\eta _{1} + t^{- \delta }, \eta _{2} ) = A \, t^{d_{P} - m_{j_{0}} \, \delta } \,\, (1 + o(1)), \end{aligned}$$

(3.14)

$$\begin{aligned} Q(\xi )= & {} Q(\xi (t)) = t^{d_{Q}} Q (\eta _{1} + t^{- \delta }, \eta _{2} ) = B \, t^{d_{Q} - k_{j_{0}} \, \delta } \,\, (1 + o(1)). \end{aligned}$$

(3.15)

Put $\delta = d_{P} / m_{j_{0}}. \,$ From representations (3.14)-(3.15) we obtain for $\, t \rightarrow \infty \,$

$$\begin{aligned} P(\xi )= & {} A \, (1 + o(1)), \end{aligned}$$

(3.16)

$$\begin{aligned} Q(\xi )= & {} B \, t^{d_{Q} - k_{j_{0}} \, (d_{P} / m_{j_{0}}) } \,\, (1 + o(1)). \end{aligned}$$

(3.17)

Since by our assumption $\, d_{Q} - k_{j_{0}} \, (d_{P} / m_{j_{0}}) > 0, \,$ hence $\, Q(\xi ) \rightarrow \infty \,$ for $\, t \rightarrow \infty . \,$ Then, from (3.16)-(3.17) it follows, that for $\, t \rightarrow \infty \,$ (hence $| \xi (t) | \rightarrow \infty$) we have $| Q(\xi ) | / [1 + P(\xi )] \rightarrow \infty .$ This proves the necessity of condition 2).

The proof of the necessity of condition 3) repeats the corresponding proof of the necessity of condition 2) of Lemma 3.3. We will not repeat this here.

Now we prove the sufficiency of conditions. Assume the contrary, that under the conditions of the theorem, there exists a sequence $\, \{\xi ^{s} \}$, $\,\, \xi ^{s} \rightarrow \infty$ as $s \rightarrow \infty \,$, such that

$$\begin{aligned} | Q(\xi ^{s}) | / [1 + P(\xi ^{s}) ] \rightarrow \infty . \end{aligned}$$

(3.18)

As above, denote $| \xi , \lambda |: = \sqrt{\xi _{1}^{2 / \lambda _{1}} + \xi _{2}^{2 / \lambda _{2}} }, \,$ $\eta ^{s}: = \xi ^{s} / |\xi ^{s}, \lambda |^{\lambda }, \,\,$ then $\, |\eta ^{s}, \lambda | = 1 \,\,\,$ $(s=1,2,\ldots ).$ Since the set $\, \{\eta ^{s}: |\eta ^{s}, \lambda | = 1, s= 1,2, \ldots \}$ is bounded, there exists a sub-sequence of the sequence $\, \{\xi ^{s} \} \,$ (which we also denote by $\, \{\xi ^{s} \} ) \,$ and a point $\, \eta : \,$ $\, |\eta , \lambda | = 1 \,$ such that $\, | \eta ^{s} - \eta , \lambda | \rightarrow 0 \,$ for $\, s \rightarrow \infty .$

We first consider the case when $\, \eta _{1}. \eta _{2} = 0 \,\,$ (obviously, the case $\, \eta = (0,0)$ is excluded). Let, for example, $\, \eta _{1} = 0.$ Represent the polynomials $\, P \,$ and $\, Q \,$ on the sequence $\{\xi ^{s}\}$ in the form

$$\begin{aligned} P(\xi ^{s})= & {} ( \xi ^{s})^{\nu } \, P_{1} (\xi ^{s}) = \rho _{s}^{d_{P} } ( \eta ^{s})^{\nu } \, \, P_{1} (\eta ^{s}) = [ \rho _{s}^{d_{P} } \, (\eta _{1}^{s} )^{\nu _{1}} \,] [(\eta _{2}^{s} )^{\nu _{2}} \,P_{1} (\eta ^{s}) ], \end{aligned}$$

(3.19)

$$\begin{aligned} Q(\xi ^{s})= & {} ( \xi ^{s})^{\mu } Q_{1} (\xi ^{s}) = \rho _{s}^{d_{Q} } (\eta ^{s})^{\mu } Q_{1} (\eta ^{s}) = [\rho _{s}^{d_{Q} } \, (\eta _{1}^{s} )^{\mu _{1}} ] [(\eta _{2}^{s} )^{\mu _{2}} Q_{1} (\eta _{s}) ]. \end{aligned}$$

(3.20)

Now let $s \rightarrow \infty , \,$ then $\rho _{s} \rightarrow \infty , \,$ $\, \eta ^{s} \rightarrow \eta , \,$ where $\eta _{1}^{s} \rightarrow 0, \,$ $\, \eta _{2}^{s} \rightarrow \eta _{2} \ne 0, \,$ $\, |(\eta _{2}^{s} )^{\nu _{2}} \,P_{1} (\eta _{s}) | \rightarrow A > 0, \,\, | (\eta _{2}^{s} )^{\mu _{2}} \, Q_{1} (\eta {s}) | \rightarrow B.$ Thus, we can assume, that for sufficiently large $\, s \,$ and for some constants $\, c_{1} >0 \,$ and $\, c_{2} \ge 0 \,$ we have

$$\begin{aligned} |(\eta _{2}^{s} )^{\nu _{2}} \,P_{1} (\eta ^{s})| \ge c_{1}, \,\,\, | (\eta _{2}^{s} )^{\mu _{2}} \, Q_{1} (\eta ^{s}) | \le c_{2}. \end{aligned}$$

(3.21)

From the obtained relations (3.19)–(3.21) it follows, that for comparing polynomials $\, P \,$ and $\, Q, \,$ it suffices to compare the expressions $\, \rho _{s}^{d_{P} } | (\eta _{1}^{s})^{\nu _{1}} |\,$ and $\, \rho _{s}^{d_{Q} } \, | (\eta _{1}^{s} ) ^{\mu _{1}} |.$ Namely, it suffices to prove, that for sufficiently large $\, s \,$ the inequality

$$\begin{aligned} \, \rho _{s}^{d_{Q} } \, |\eta _{1}^{s} |^{\mu _{1}} \le 1 + \rho _{s}^{d_{P} } \, |\eta _{1}^{s} |^{\nu _{1}} \end{aligned}$$

(3.22)

holds. To prove this, we apply Lemma 3.3. In this case, we denote $\, a:= d_{Q}, \,$ $\, b:= \mu _{1}, \,$ $\,c:= d_{P}, \,$ $\,d:= \nu _{1} \,$ $\, x_{s}:= \rho _{s}, \,$ $\,y_{s}:= | \eta _{1}^{s} | \,$ $\, (s = 1,2,\ldots ). \,$

It is obvious that $x_{s} \ge 1 \,$ $(s = 1,2,\ldots ), \,\,$ $\, y_{s} \in [0,1] \,$ for sufficiently large $\, s, \,$ condition $a \le c \,\,\,$ of the Lemma 3.3 follows from condition 1) of present theorem and condition $\,\, d/b \le c/a \,\,$ follows from condition 3) of present theorem.

Acting similarly in the case $\, \eta _{2} = 0, \,$ we obtain an inequality of the type (3.22) with the replacement of $\eta _{1} \,$ by $\, \eta _{2}. \,$ These inequalities together contradict our assumption (3.18) and complete the study of the case $\, \eta _{1}. \eta _{2} = 0. \,\,$

Consider the case $\, \eta _{1} \cdot \eta _{2} \ne 0 \,\$ and first let $\, P(\eta ) \ne 0.$ As a result of $\, \lambda \,-$homogeneity of the polynomials $\, P \,$ and $\, Q \,$ we have for $\, s \rightarrow \infty \,\,\,$ (i.e. for $\, \rho _{s} \rightarrow \infty$)

$$\begin{aligned} | P(\xi ^{s})| = \rho _{s}^{ d_{P} } | P(\eta ) | (1 +o(1)); \,\,\, | Q(\xi ^{s})| = \rho _{s}^{ d_{Q} } | Q(\eta ) | (1 +o(1)). \end{aligned}$$

Since $d_{Q} \le d_{P} \,$ and $\, P(\eta ) \ne 0, \,$ these representations together contradict (3.18).

It remains to consider the case, when $\, P(\eta ) = 0. \,$ This means that in expansion (3.18) $\, p_{j_{0}} (\eta ) = (\eta _{1}^{s} - \tau _{j_{0}} \, \eta _{2}^{\lambda _{1} / \lambda _{2}} )^{m_{j_{0}}} = 0 \,$ for some $\,j_{0}: \, 1 \le j_{0} \le N_{P}, \,$ moreover $\, p_{j} (\eta ) \ne 0 \,$ for $\, j \ne j_{0}. \,\,$ From condition 1) of the theorem it follows, that $\, \mu _{j_{0}} = \tau _{j_{0}}, \,\,$ hence $\, q_{j_{0}} (\eta ) = p_{j_{0}} (\eta ) = 0. \,$

We rewrite the representations (3.12)–(3.13) in the form

$$\begin{aligned} P(\xi _{s})= & {} \rho _{s}^{d_{P}} \left[ (\eta ^{s})^{\nu } \ p(\eta ^{s}) \, \prod \limits _{j = 1; j \ne j_{0} }^{N_{P}} \, p_{j} (\eta ^{s}) \right] (\eta _{1}^{s} - \tau _{j_{0}} \, (\eta _{2}^{s})^{\lambda _{1} / \lambda _{2}} )^{m_{j_{0}}}, \end{aligned}$$

(3.23)

$$\begin{aligned} Q(\xi _{s})= & {} \rho _{s}^{d_{Q}} \left[ (\eta ^{s})^{\mu } \ q(\eta ^{s}) \, \prod \limits _{j = 1; j \ne j_{0} }^{N_{Q}} \, q_{j} (\eta ^{s}) \right] (\eta _{1}^{s} - \tau _{j_{0}} \, (\eta _{2}^{s})^{\lambda _{1} / \lambda _{2}} )^{k_{j_{0}}}. \end{aligned}$$

(3.24)

Now note that when $s \rightarrow \infty \,$ 1) expressions $| (\eta ^{s})^{\nu } \ p(\eta ^{s}) \, \prod \limits _{j = 1; j \ne j_{0} }^{N_{P}} \, p_{j} (\eta ^{s}) |$ and $\, | (\eta ^{s})^{\mu } \ q(\eta ^{s}) \, \prod \limits _{j = 1; j \ne j_{0} }^{N_{Q}} \, q_{j} (\eta ^{s}) | \,$ have nonzero finite limits (let, for example A and B), $\,\, 2) \, \rho _{s} \rightarrow \infty , \,$ 3) $\, (\eta _{1}^{s} - \tau _{j_{0}} \, (\eta _{2}^{s})^{\lambda _{1} / \lambda _{2}} ) \rightarrow 0. \,$ Therefore, we can assume that $\, \rho _{s} \ge 1 \,$ and $| (\eta _{1}^{s} - \tau _{j_{0}} \, (\eta _{2}^{s})^{\lambda _{1} / \lambda _{2}}| \in [0,1] \,$ for sufficiently large $\, s. \,\,$

It remains to apply Lemma 3.3 with the following notations $\, x:= \rho _{s}, \,$ $\,\, y = | \eta _{1}^{s} - \tau _{j_{0}} \, (\eta _{2}^{s})^{\lambda _{1} / \lambda _{2}} |, \,$ $a:= d_{Q}, \,$ $b:= k_{j_{0}}, \$ $\, c:= d_{P}, \,$ $d:= m_{j_{0}}. \,\,$ Satisfaction of the conditions of Lemma 3.3 follows from condition 2) of the present theorem.

The application of Lemma 3.3 together with representations (3.23)–(3.24) contradicts assumption (3.18).

Theorem 3.2 is proved. $\square$

Examples. Compare the polynomial $P(\xi ) = (\xi _{1} - \xi _{2})^{4} \, (\xi _{1}^{6} + \xi _{2}^{6})$ with the following polynomials $Q_{1}(\xi ) = (\xi _{1} - \xi _{2})^{4} \, (\xi _{1}^{4} + \xi _{2}^{4}),$ $\,\, Q_{2}(\xi ) = (\xi _{1} - \xi _{2})^{5} \, (\xi _{1}^{2} + \xi _{2}^{2}),$ $Q_{3}(\xi ) = (\xi _{1} - \xi _{2})^{3} \, (\xi _{1}^{4} + \xi _{2}^{4}), \,\,$ $Q_{4}(\xi ) = (\xi _{1} - \xi _{2})^{2} \, (\xi _{1}^{6} + \xi _{2}^{6}).$

For this polynomials

$$\begin{aligned} d_{P}= & {} 10, \, \, \eta = (1/\sqrt{2}, 1/\sqrt{2}), \,\, \Delta (\eta , P ) = 4 \\ d_{Q_{1}}= & {} 8, \, \, \eta = (1/\sqrt{2}, 1/\sqrt{2}), \,\, \Delta (\eta , Q_{1} ) = 4 \\ d_{Q_{2}}= & {} 7, \, \, \eta = (1/\sqrt{2}, 1/\sqrt{2}), \,\, \Delta (\eta , Q_{2} ) = 5 \\ d_{Q_{3}}= & {} 7, \, \, \eta = (1/\sqrt{2}, 1/\sqrt{2}), \,\, \Delta (\eta , Q_{3} ) = 3 \\ d_{Q_{4}}= & {} 8, \, \, \eta = (1/\sqrt{2}, 1/\sqrt{2}), \,\, \Delta (\eta , Q_{4} ) = 2. \end{aligned}$$

A simple calculation shows, that for the pairs $(Q_{j}, P) \,\, (j = 1,2,3) \,$ conditions 1) and 2) of Theorem 3.2 are satisfied, but for pair $\, (Q_{4}, P) \,$ condition 2) is violated. Therefore $\, Q_{j} < P \,\, (j = 1,2,3),$ but $\, Q_{4} \not < P.$ At the same time we want to draw the reader’s attention to the fact, that in the above examples all possible relations between $\, \Delta (\eta , Q) \,$ and $\, \Delta (\eta , P) \,$ are considered. Exactly $\, \Delta (\eta , Q_{1}) = \Delta (\eta , P), \,$ $\, \Delta (\eta , Q_{2}) > \Delta (\eta , P), \,$ $\, \Delta (\eta , Q_{3}) < \Delta (\eta , P), \,$ $\, \Delta (\eta , Q_{4}) < \Delta (\eta , P). \,$

If in the above examples the expression $\, (\xi _{1} - \xi _{2}) \,$ is replaced by $\,\xi _{1} \,$ or $\,\xi _{2}, \,$ then we will already obtain examples for polynomials of the form $\, \xi ^{\nu } \, R(\xi ).$

4 Comparison of the powers of general polynomials

In this section we set ourselves the task of comparing the powers of two general (common) polynomials. Exactly, let $\, P \,$ be a given polynomial with the complete Newton polyhedron $\, \Re (P) \,$ and $\, Q \,$ be some polynomial. We need to find the conditions under which $\, Q < P.$ If polynomial $\, P \,$ is non-degenerate and $\, \Re (Q) \subset \Re (P),$ then by Theorem $2.1\,$ $\, Q < P.$ Therefore, we need to consider only the case, when polynomial $\, P \,$ is degenerate.

Before proceeding to the comparison of general polynomials we prove one simple proposition, which in some cases reduces the problem of comparing two general polynomials to the comparison of generalized homogeneous and general polynomials, which in our opinion is also of an independent interest.

Lemma 4.1

Let R be a $\lambda -$homogeneous polynomial of $\lambda -$order $\, d_{R}, \,$ and Q be a general polynomial represented by the vector $\lambda$ in the form (1.1) as a sum of $\lambda -$homogeneous polynomials, i.e.

$$\begin{aligned} Q (\xi ) = \sum \limits _{j = 1}^{N} Q_{j} (\xi ) = \sum \limits _{j = 1}^{N} \sum \limits _{(\lambda , \alpha ) = \delta _{j}} \gamma _{\alpha }^{Q} \xi ^{\alpha }, \,\,\, \delta _{1}> \delta _{2}> \ldots > \delta _{N} \ge 0. \end{aligned}$$

Then relation $R > Q$ holds, if and only if $Q_{j} < R \,\,\,$ $(j = 1,\ldots , N).$

Proof

The proof of sufficiency is obvious. We prove the necessity. Let $\, Q < R. \,$ We must prove, that $Q_{j} < R \,$ $(j = 1,\ldots , N). \,$

Since $\, Q < R, \,$ for any $t>0 \,\,\,$ $Q(t^{\lambda } \, \xi ) < R(t^{\lambda } \, \xi ) = t^{d_{R}} \, R (\xi ). \,\,$ Consequently there exists a positive number $\,c > 0 \,$ such that $| Q(t^{\lambda } \, \xi ) | \le c \, [ t^{d_{R}} \, | R (\xi ) | + 1 ] \,$ for every positive $\, t \,$ and for all $\, \xi \in {\mathbb {R}}^{n}.$

Choose (and fix) N positive numbers $t_{1},\ldots , t_{N} \,$, so that the matrix $(t_{j}^{\delta _{k}}) \,$ is non singular. We obtain from representation of Q that $Q(t^{\lambda }_{j} \, \xi ) = \sum _{k = 1}^{N} t_{j}^{\delta _{k} } \, Q_{k} (\xi ). \,$ Since the matrix $(t_{j}^{\delta _{k}}) \,$ is non singular, each of polynomials $\, Q_{k} (\xi ) \,\,$ $(k = 1,\ldots , N) \,$ is a linear combination of polynomials $Q(t^{\lambda }_{j} \, \xi ). \,$ This means that there exist numbers $\{a_{i}^{j} = a_{i}^{j}(t_{1},\ldots , t_{N}) \}_{i,j = 1}^{N}$ and $\{b_{i}^{j} = b_{i}^{j}(t_{1},\ldots , t_{N}) \}_{i,j = 1}^{N}$, such that for all $(j = 1,\ldots ,N)$

$$\begin{aligned} Q_{j} (\xi )= & {} a_{1}^{j} \, Q(t_{1}^{\lambda } \, \xi ) +\cdots + a_{N}^{j} \, Q(t_{N}^{\lambda } \, \xi ) \,\,\le b_{1}^{j} \, \left[ 1 + |R(t_{1}^{\lambda } \, \xi )| \, \right] \\{} & {} +\cdots + b_{N}^{j} (t) \left[ 1 + | R(t_{N}^{\lambda } \, \xi ) | \,\right] = b_{1}^{j} \, \left[ 1 +t_{1}^{d_{R}} \,| R(\xi ) | \, \right] +\cdots + b_{1}^{j} \, [ 1 +t_{N}^{d_{R}} \,| R(\xi ) |. \end{aligned}$$

Since the vector $\, t = (t_{1},\ldots , t_{N})$ is fixed, and denoting $\, B_{j}: = max \{b_{i}^{j}: i = 1,\ldots , N \} \,$ $\, j = 1,\ldots , N$ and $T: = max \{t_{i}^{d_{R}}, i = 1,\ldots , N\}, \,$ we obtain with some constant $\, c_{j} = c_{j} (B_{j}, T, R,Q) > 0 \,$

$$\begin{aligned} |Q_{j} (\xi ) | \le c_{j} [1 +|R(\xi )|] \,\,\, \forall \xi \in {\mathbb {R}}^{n}, \,\, j = 1,\ldots , N. \end{aligned}$$

Lemma 4.1 is proved. $\square$

To describe the set of polynomials, that are valued through a given non-homogeneous, degenerate polynomial $\, P \,$ as above (see Theorems 2.1 and $\, 2.1') \,$ we first consider the case when $P \in {I}_{n} \,$, and that only one principal face of the polyhedron $\, \Re (P) \,$ of the polynomial $\, P \,$ is degenerate, wherein 1) this is a $\, (n - 1) -$dimensional face with the outward normal $\, \mu \,$ with positive coordinates, 2) $\, P_{1} (\eta ) \ne 0 \,$ for all $\, \eta \in \Sigma (P_{0}).$

So, let us compare a degenerate polynomial $\, P \,$ satisfying the conditions of Theorem $\, 2.1 \,$ and represented by the vector $\, \mu \,$ in the form (below $d_{j} = d_{j} (\mu ) = d_{j} (P,\mu ) \,\, (j = 0,1,\ldots , M(\mu );$ $d_{0}> d_{1}> \ldots > d_{M}$)

$$\begin{aligned} P(\xi ) = \sum \limits _{j=0}^{M(\mu )} P_{j} (\xi ) = \sum \limits _{j=0}^{M(\mu )} P_{d_{j} (\mu )} (\xi ) = \sum \limits _{j=0}^{M(\mu )} \sum \limits _{(\mu , \alpha ) = d_{j} (\mu )} \gamma _{\alpha } \, \xi ^{\alpha }, \end{aligned}$$

(4.1)

and a polynomial $\, Q \,$ represented by the vector $\, \mu \,$ in the form (below $\delta _{j} = \delta _{j} (\mu ) = \delta _{j} (Q,\mu ) \,\, (j = 0,1,\ldots , N(\mu );$ $\delta _{0}> \delta _{1}> \ldots > \delta _{N}$)

$$\begin{aligned} Q (\xi ) = \sum \limits _{j = 0}^{N(\mu )} Q_{j} (\xi ) = \sum \limits _{j=0}^{N(\mu )} Q_{\delta _{j} (\mu )} (\xi ) = \sum \limits _{j = 0}^{N(\mu )} \sum \limits _{(\mu , \alpha ) = \delta _{j}} \gamma _{\alpha }^{Q} \xi ^{\alpha }. \end{aligned}$$

(4.2)

We are interested in the following question. Under what condition $\, Q < P ?$

First note the following: (1) if $\Re (Q) \subset \Re (P), \,$ and $\delta _{j_{0}} \le d_{1}\,\,$ for some number $\, j_{0}, 1 \le j_{0} \le N(\mu ),$ then by Theorem $\, 2.1 \,$ $\, Q_{j} < P \,$ for all $j = j_{0}, j_{0} + 1,\ldots ,N(\mu ).$ Therefore, it remains to consider the polynomials $\, Q_{j}$ for $j = 0,1,\ldots , j_{0} - 1.$ 2) If $\, Q_{j} < P_{0} \,$ for all $j = 0,1,\ldots , j_{0} - 1,$ then by Corollary 2.1 $\, Q_{j}< P_{0} < P \,$ for all $j = 0,1,\ldots , j_{0} - 1.$ As a result, we get that $\, Q < P.$

So, it is enough to consider the case when $\, Q_{j_{1}} \not < P_{0} \,$ for some number $j_{1}: 0 \le j_{1} \le j_{0} - 1, \,$ wherein $d_{1} < \delta _{j_{1}} \le d_{0}.$

The following theorem in a certain sense solves the problem in this case.

Theorem 4.1

• (I) Let the degenerate polynomial $\, P \,$ satisfies the conditions of Theorem $\, 2.1, \,$ let $\, Q \,$ be a $\mu -$homogeneous polynomial of $\, \mu -$order $\delta _{Q}, \mu _{j} > 0 \,\,$ $(j = 1,\ldots , n), d_{1}< \delta _{Q} < d_{0}$ and $\Re (Q) \subset \Re (P).$ Then $\, Q < P$, if and only if

  • (1) $\, \Sigma (P_{0}) \subset \Sigma (Q),$

  • (2) $(d_{0} -d_{1})/(\delta _{Q} - d_{1}) \ge {\Delta (\eta , P_{0})}/ {\Delta (\eta , Q),} \,\,\, \forall \eta \in \Sigma (P_{0}),$

  • (3) for every point $\eta \in \Sigma (P_{0}) \,$ there exists a constant $\, c = c (\eta ) >0$ and a neighborhood $U(\eta ) \,$ such that

    $$\begin{aligned} |Q(\xi )| \le c |P_{0}(\xi )|^{(\delta _{Q} - d_{1}) / (d_{0} - d_{1})} \,\,\, \forall \xi \in U(\eta ). \end{aligned}$$

• (II) Moreover, if for every point $\, \eta \in \Sigma (P_{0}) \,$ there exists a neighborhood $\, U(\eta ) \,$ such that $\, Q(\xi ) \ge 0 \,$ for all $\, \xi \in U(\eta ), \,$ then $\, P < P + Q.$

Proof

Necessity of condition I.1) is obvious.

Let’s prove the necessity of condition I.2. Assume the converse, that the condition $\, Q < P \,\,$ is satisfied, but there exists a point $\eta \in \Sigma (P_{0})$, such that

$$\begin{aligned} (d_{0} -d_{1})/(\delta _{Q} - d_{1}) < \Delta (\eta , P_{0})/ \Delta (\eta , Q). \end{aligned}$$

(4.3)

For $\,\, t> 0, \theta = ( \theta _{1},\ldots , \theta _{n}) \in {\mathbb {R}}^{n}, \,$ $\, \kappa > 0 \,\,$ set $\xi _{i} = \xi _{i} (t) = \xi _{i} (t,\theta ,\kappa ) = t^{\mu _{i}} (\eta _{i} + \theta _{i} \, t^{-\kappa \, \mu _{i}}), \,\,$ $\, i = 1,\ldots , n.$

Then by Taylor’s formula

$$\begin{aligned} Q(\xi (t))= & {} t^{\delta _{Q}} \, Q(\eta + \theta \, t^{-\kappa \, \mu }) = t^{\delta _{Q}} \, \sum \limits _{\alpha } t^{-\kappa \, (\mu , \alpha )} \, [ D^{\alpha } Q(\eta ) /(\alpha !) ] \, \theta ^{\alpha }\\= & {} t^{\delta _{Q} - \kappa \, \Delta (\eta , Q)} \,\, \sum \limits _{ (\mu , \alpha ) = \Delta (\eta , Q)} [ \, D^{\alpha } Q(\eta ) /(\alpha !)] \theta ^{\alpha } + o ( t^{\delta _{Q} - \kappa \, \Delta (\eta , Q)}). \end{aligned}$$

Choose $\theta \,$ in such a way that

$$\begin{aligned} c = c(\theta ): = \sum \limits _{ (\mu , \alpha ) = \Delta (\eta , Q)} \, [D^{\alpha } Q(\eta ) /(\alpha !)]\, \theta ^{\alpha } \ne 0. \end{aligned}$$

The existence of such a vector obviously follows from the definition of number $\Delta (\eta , Q).$ In fact, otherwise, it turned out that all the coefficients of the polynomial $c(\theta ) \,$ are equal to zero, which contradicts the definition of $\Delta (\eta , Q).$ Then (for a $\, \theta \,$, fixed in such a way), we have

$$\begin{aligned} |Q(\xi (t))| \ge c \, t^{\delta _{Q} - \kappa \, \Delta (\eta , Q)}. \end{aligned}$$

(4.4)

For the polynomials $P_{0}$ and $P_{1}$ we obviously have with a constant $\, c_{1} > 0 \,$ and for sufficiently large $\, t$

$$\begin{aligned} |P_{0} (\xi (t)) | \le c_{1} \, t^{d_{0} - \kappa \, \Delta (\eta , P_{0})}, \,\,\, |P_{1} (\xi (t)) | = t^{d_{1}} \, P_{1} (\eta )\, (1 + o(1)). \end{aligned}$$

(4.5)

Obvious geometric arguments (see also Lemma 2.2) show that as $t \rightarrow + \infty$

$$\begin{aligned} r(\xi (t)): = P(\xi (t)) - \left[ P_{0} (\xi (t)) + P_{1} (\xi (t)) \right] = o(t^{d_{1}}). \end{aligned}$$

(4.6)

We put $\kappa = (d_{0} - d_{1}) / \Delta (\eta , P_{0}), \,$ then $d_{0} - \kappa \, \Delta (\eta , P_{0}) = d_{1}, \,$ and from (4.5)-(4.6) we have with a constant $c_{2} > 0 \,$

$$\begin{aligned} |P (\xi (t)) | \le c_{2} t^{d_{1}}. \, \end{aligned}$$

(4.7)

It is easy to obtain, that from assumption (4.3) it follows that $\, d_{1} < \delta _{Q} - \kappa \, \Delta (\eta , Q).$ From this and from estimates (4.4), (4.7) it follows that $\, |Q(\xi (t))| / [1+ P(\xi (t))] \rightarrow \infty \,$ as $t \rightarrow \infty , \,$ which contradicts the condition $Q < P$ and proves the necessity of condition I.2).

Necessity of condition I.3). Assume that for some point $\eta \in \Sigma (P_{0}) \,$ there exists a sequence $\, \{\eta ^{s} \}$ such that $P_{0} (\eta ^{s}) \ne 0 \,\, (s = 1,2,\ldots ),$ $\,\,\eta ^{s} \rightarrow \eta$ as $s \rightarrow \infty \,$ and

$$\begin{aligned} R(\eta ^{s}): = | Q(\eta ^{s}) | / \left[ | P_{0} (\eta ^{s}) |^{ (\delta _{Q} - d_{1}) / (d_{0} -d_{1})} \right] \rightarrow \infty . \end{aligned}$$

(4.8)

Set $t_{s} = | P_{0} (\eta ^{s}) |^{ - 1 / (d_{0} -d_{1})}, \,\,\, \xi ^{s} = t_{s}^{\mu } \, \eta ^{s}, \,\, s = 1,2,\ldots \cdot$ Since $\, \eta ^{s} \rightarrow \eta \in \Sigma (P_{0}) \,$ we have $t_{s} \rightarrow \infty \,$ as $s \rightarrow \infty . \,$ Then as a consequence of the $\, \mu -$homogeneity of $P_{0} (\xi ),$ $\, P_{1} (\xi )$ and $Q (\xi )$ we have for sufficiently large $\, s$

$$\begin{aligned} | P_{1} (\xi ^{s}) |= & {} t_{s}^{d_{1}} \, | P_{1} (\eta ^{s}) | = t_{s}^{d_{1}} \, | P_{1} (\eta ) | \, (1 + o(1)), \end{aligned}$$

(4.9)

$$\begin{aligned} | P_{0} (\xi ^{s}) |= & {} t_{s}^{d_{0}} \, | P_{0} (\eta ^{s}) | = t_{s}^{d_{1}},\,\,\, r(\xi ) = o(t_{s}^{d_{1}} ). \end{aligned}$$

(4.10)

Representations (4.9), (4.10) show that there exists a constant $\, c_{3} > 0, \,$ such that for sufficiently large $\, s$

$$\begin{aligned} |P (\xi ^{s}) | + 1 \le c_{3} t^{d_{1}}. \end{aligned}$$

(4.11)

For $Q(\xi ) \,$ analogously we obtain

$$\begin{aligned} |Q(\xi ^{s})| = t_{s}^{\delta _{Q}} \, | Q(\eta ^{s}) | = t_{s}^{\delta _{Q}} \, R(\eta ^{s}) \, | P_{0} (\eta ^{s}) |^{(\delta _{Q} - d_{1}) / (d_{0} -d_{1})} = R(\eta ^{s}) \, t_{s}^{d_{1}}. \end{aligned}$$

(4.12)

Estimates (4.11) and (4.12) together with the assumption (4.8) show that as $s \rightarrow \infty \,$ we have

$$\begin{aligned} |Q(\xi ^{s})| / \left[ | P(\xi ^{s}) | + 1\right] \ge [1/c_{3}]\, R (\eta ^{s}) \rightarrow \infty . \end{aligned}$$

(4.13)

This proves the necessity of condition I.3) for $Q <P.$

Let us prove the Sufficiency. Assume that $Q \not < P$ under the hypotheses of Theorem 4.1, i.e. there exists a sequence $\{ \xi ^{s}\}$ such that $\xi ^{s} \rightarrow \infty$ as $s \rightarrow \infty$, and

$$\begin{aligned} |Q(\xi ^{s})| / \left[ | P(\xi ^{s}) | + 1\right] \rightarrow \infty . \end{aligned}$$

(4.14)

Proceeding as in the proof of Theorem 3.1$'$, we denote $\eta ^{s}: = \xi ^{s} / |\xi ^{s}, \lambda |^{ \lambda }, \,\,$ then $\, |\eta ^{s}, \lambda | = 1 \,\,\,$ $(s=1,2,\ldots ).$ Since the set $\, \{\eta ^{s}: |\eta ^{s}, \lambda | = 1, \,\, s= 1,2, \ldots \}$ is bounded (recall that $\, \lambda _{j} > 0 \,\, (j = 1,\ldots , n)$), there exists a sub-sequence of the sequence $\, \{\xi ^{s} \} \,$ (which we also denote by $\, \{\xi ^{s} \} ) \,$ and a point $\, \eta , \,$ $\, |\eta , \lambda | = 1, \,$ such that $\, | \eta ^{s} - \eta , \lambda | \rightarrow 0 \,$ for $\, s \rightarrow \infty .$

Let us show that $\, \eta \in \Sigma (P_{0}).$ Let, on the contrary, $\, P_{0}(\eta ) \ne 0.$ Then for $\, s \rightarrow \infty \,$ we have

$$\begin{aligned} |P(\xi ^{s})| = |\xi ^{s}, \lambda |^{d_{0}} \, |P_{0}(\eta )| (1 + o(1)), \end{aligned}$$

(4.15)

and there exists a number $\, c_{4} > 0, \,$ such that

$$\begin{aligned} |Q(\xi ^{s})| \le c_{4} \, |\xi ^{s}, \lambda |^{d_{1}} \,\,\, s = 1,2, \ldots \cdot \end{aligned}$$

(4.16)

The relations (4.15) and (4.16) together contradict our assumption (4.14) and prove that $\, \eta \in \Sigma (P_{0}).$

Since $\eta \in \Sigma (P_{0}) \,$ and $\, \eta ^{s} \rightarrow \eta \,$ for $\, s \rightarrow \infty , \,$ then, by condition 3) of the theorem, there exists a constant $\, c_{1} > 0 \,$, such that (without less of generality, we can assume that for all $s = 1,2,\ldots$ )

$$\begin{aligned} | Q(\eta ^{s}) | \le c_{1} \, |P_{0} (\eta ^{s}) |^{(\delta _{Q} - d_{1}) / (d_{0} - d_{1})}. \end{aligned}$$

(4.17)

By virtue of the condition $\, P \in {I}_{n} \,$, there exist positive constants $\, c_{2} \,$ and $\, c_{3}$, such that for sufficiently large $\, s \,$ (without loss of generality, we assume that for all $s = 1,2,\ldots$ ) we have

$$\begin{aligned} 1 + | P(\xi ^{s}) |= & {} 1 + \left| \sum \limits _{j = 0}^{M} P_{j}(\xi ^{s}) \right| = 1 + \left| \sum \limits _{j = 0}^{M} | \xi ^{s}, \lambda |^{d_{j}} \, P_{j}(\eta ^{s}) \right| \ge 1 + | \xi ^{s}, \lambda |^{d_{0}} \,| P_{0}(\eta ^{s})| \nonumber \\{} & {} + |\xi ^{s}, \lambda |^{d_{1}} \, | P_{1}(\eta ^{s})|- o ( |\xi ^{s}, \lambda |^{d_{1}})\ge c_{2} \, [| \xi ^{s}, \lambda |^{d_{0}} \,| P_{0}(\eta ^{s})| \nonumber \\{} & {} + |\xi ^{s}, \lambda |^{d_{1}} \, | P_{1}(\eta ^{s})|] \ge c_{3} \, \left[ | \xi ^{s}, \lambda |^{d_{0}} \,| P_{0}(\eta ^{s})| + |\xi ^{s}, \lambda |^{d_{1}} \, \right] . \end{aligned}$$

(4.18)

Using inequality (4.17) we also estimate $| Q(\xi ^{s}) |. \,$

$$\begin{aligned} | Q(\xi ^{s}) |= & {} | \xi ^{s}, \lambda |^{\delta _{Q}} \,| Q(\eta ^{s})| \le c_{1} \,| \xi ^{s}, \lambda |^{\delta _{Q}} \, | P_{0}(\eta ^{s})|^{ \frac{\delta _{Q} - d_{1}}{ d_{0} - d_{1} }} \\= & {} c_{1} \,\left[ | \xi ^{s}, \lambda |^{d_{0}} \, | P_{0}(\eta ^{s})| \right] ^{ \frac{\delta _{Q} - d_{1}}{ d_{0} - d_{1} }} ] \,| \xi ^{s}, \lambda |^{d_{1} \, \frac{ d_{0} - \delta _{Q} }{ d_{0} - d_{1} }}. \end{aligned}$$

Applying H$\ddot{o}$lder’s inequality for $p = \frac{d_{0} - d_{1}}{\delta _{Q} - d_{1}}$ and $q = \frac{p}{p - 1} =$ $\frac{d_{0} - d_{1}}{d_{0} - \delta _{Q}} \,$ we obtain with some constant $\, c_{4} > 0$

$$\begin{aligned} | Q(\xi ^{s}) | \le c_{4} \left[ | \xi ^{s}, \lambda |^{d_{0}} \,| P_{0}(\eta ^{s})| + |\xi ^{s}, \lambda |^{d_{1}} \right] , \,\,\, s = 1,2, \ldots \cdot \end{aligned}$$

(4.19)

From (4.18)–(4.19) we obtain with some constant $\, c_{5} > 0$

$$\begin{aligned} | Q(\xi ^{s}) | \le c_{5} \, \left[ | P(\xi ^{s})| + 1\right] , \,\,\, s = 1,2,\ldots , \end{aligned}$$

which contradicts our assumption (2.14) and proves the sufficiency part of the theorem.

Let us prove the second part.

Repeating the reasoning carried out in the proof of the sufficiency of the first part of the theorem, suppose that there exists a sequence $\, \{\xi ^{s} \}$, such that $\, |\xi ^{s}| \rightarrow \infty , \,$ $\, | \eta ^{s} - \eta , \lambda | \rightarrow 0 \,$ for $\, s \rightarrow \infty \,$ and

$$\begin{aligned} \, | P(\xi ^{s}) | / \left[ 1 + | P(\xi ^{s}) + Q(\xi ^{s}) |\right] \rightarrow \infty . \,\, \end{aligned}$$

(4.20)

Since $d_{0} > d_{1}$, in the case $\, \eta \notin \Sigma (P_{0})\,$ we obtain the following representations for polynomials $\, P \,$ and $\, Q\,$ when $\, s \rightarrow \infty : \,$ $| P(\xi ^{s}) | = | \xi ^{s}, \lambda |^{d_{0}} \, | \, | P_{0} (\eta ) | \, (1 + o(1)) \,$ and $\, |Q (\xi ^{s}) | = o (| \xi ^{s}, \lambda |^{d_{0}}), \,$ which contradicts (4.20).

If $\, P_{0} (\eta ) = 0, \,$ then by hypothesis of the theorem, $\, P_{1} (\eta ) \ne 0. \,$ In this case we obtain a contradiction with (4.20) due to the fact that $\, Q (\xi ) \ge 0, \,$ for $\, \xi \in U(\eta ) \,$ and condition $\, P \in {I}_{n} \,$ of the theorem implies that $\, P_{0}(\xi ) \ge 0,$ in a neighborhood of the point $\, \eta .$

Theorem 4.1 is proved. $\square$

Remark 4.1

Note that the conditions of Theorem 4.1 do not guarantee that $\, Q < P_{0}, \,$ which can be seen from the following example

Example 4.1

Let $\, n = 2, \,$ $\, P(\xi ):= P_{0} (\xi ) + P_{1} (\xi ) = \,$ $(\xi _{1} - \xi _{2} )^{8} + (\xi _{1}^{2} +\xi _{2}^{2})^{2}, \,\,$ $Q(\xi ) = (\xi _{1} - \xi _{2} )^{4} \, (\xi _{1}^{2} +\xi _{2}^{2}). \,$ Here $\, d_{0} = 8, d_{1} = 4, \,\,$ $\delta _{Q}= 6, \,$ $\eta = (1 / \sqrt{2}, 1 / \sqrt{2}), \,$ $\, \Delta (\eta , P_{0}) = 8,\,$ $\, \Delta (\eta , Q) = 4. \,$

It is easy to verify that all conditions of Theorem 4.1 are satisfied, hence $\, Q < P. \,$ Moreover, applying the arithmetic inequality $\,\, a\, b \le (1/ 2) (a^{2} + b^{2}), \,$ we obtain that $\, P < P + Q.$ However, in this case the (necessary) condition 4) of the Theorem 3.1 is violated, and therefore $Q \not < P_{0}. \,$ This can also be verified directly (without the help of Theorem 3.1) by taking, for example $\, \xi _{1}^{s} = s + 1, \,$ $\, \xi _{2}^{s} = s \,$, $\, s = 1,2,\ldots \cdot$

Example 4.2

Let us compare the polynomial $\,\, Q(\xi ) = ( \xi _{1} - \xi _{2} )^{2} (\xi _{1}^{6} + \xi _{2}^{6}) \,\,$ with the next two polynomials $P^{1}(\xi ):= P^{1}_{0} (\xi ) + P_{1}^{1} (\xi ) \,$ $= ( \xi _{1} - \xi _{2} )^{4} (\xi _{1}^{6} + \xi _{2}^{6}) \,\,$ $+ (\xi _{1}^{6} + \xi _{2}^{6})$ and $P^{2}(\xi ):= P^{2}_{0} (\xi ) + P_{2}^{1} (\xi ) \,$ $= ( \xi _{1} - \xi _{2} )^{4} (\xi _{1}^{6} + \xi _{2}^{6}) \,\,$ $+ (\xi _{1}^{4} + \xi _{2}^{4}).$

Simple calculations show that the pair $\, (P^{1},Q) \,$ satisfies all conditions of Theorem 4.1, while the pair $\, (P^{2},Q) \,$ does not satisfy condition 2) of this theorem. Therefore, $\, Q < P^{1},$ but $\, Q \not < P^{2}.$

5 Estimates of monomials through a given polynomial (general case)

Recall that in Theorems 2.1 we considered only the case when for the studied degenerate polynomial $\,P = P_{0}+ P_{1} +P_{2} + \ldots \,$ it was the first of the polynomials$\, \{ P_{j} \}_{j=1}^M \,$, that did not vanish at all points $\, \eta \in \Sigma (P_{0}):= \{\xi \in {\mathbb {R}}^{n}:$ $\ | \xi | \ne 0, \,$ $P_{0} (\xi )= 0 \}.$ Now we want to free ourselves from this restriction.

Namely, let, as in Theorem 2.1, $\, \Gamma := \Re _{i_{0}}^{n - 1} \,$ be the only degenerate principal face (with the outward normal $\mu ) \,$ of the complete Newton polyhedron $\, \Re (P) \,$ of polynomial $\, P \in {I}_{n} \,$, and let with respect to the vector $\mu \,$ polynomial $\, P \,$ is represented as the sum of $\, \mu -$homogeneous polynomials

$$\begin{aligned} P(\xi ) = \sum \limits _{j = 0}^{M} P_{j} (\xi ) = \sum \limits _{j = 0}^{M} \sum \limits _{(\mu , \alpha ) =d_{j}} \gamma _{\alpha }\, \xi ^{\alpha }, \end{aligned}$$

(5.1)

where $d_{0}> d_{1}> \ldots> d_{l}> \ldots > d_{M} \ge 0, \,$ $M \ge 3.$

Suppose that $\, P_{l} (\eta ) > 0 \,\, (1 \le l \le M)$ for all $\, \eta \in \Sigma (P_{0}) \,$, and each polynomial $\, P_{j} \$ vanishes at least at one point $\, \eta \in \Sigma (P_{0}) \,$ $j = 1,2\ldots , l - 1$, and denote $\Re ^{*}: =\{ \beta \in \Re , (\mu ,\beta ) \le d_{l}\}, \,$ ${\mathcal {P}} (\xi ): = P_{0} (\xi ) + P_{l} (\xi ) + P_{l+1} (\xi ) +\cdots + P_{M} (\xi ), \,$ $\, {\mathcal {P}}_{1} (\xi ): = P_{1} (\xi ) +\cdots + P_{l-1} (\xi ). \,\,$ If $l = 1, \,$ then ${\mathcal {P}} (\xi ) \equiv P(\xi )$, and it follows from Theorem 2.1, that under certain conditions on the polynomial $\, P \,$ the relation $\, \xi ^{\nu } < {\mathcal {P}} \,$ is valid for all $\, \nu \in \Re ^{*}. \,$

A natural question arises. Suppose that $l\ge 2, \,$ and polynomial ${\mathcal {P}} \,$ satisfies the conditions of Theorem 2.1, therefore, $\, \xi ^{\nu } < {\mathcal {P}} \,$ for all $\, \nu \in \Re ^{*} \,$. What conditions must the polynomials $\, P_{j} \,\, (j = 1,\ldots , l-1) \,$ satisfy, so that the relation $\, \xi ^{\nu } < P = {\mathcal {P}} + {\mathcal {P}}_{1} \,$ also holds for all $\, \nu \in \Re ^{*}? \,$

To answer to this question, we need to answer the following question (which besides of numerous applications in differential equations is also of an independent interest): which lower-order terms Q can be added to the polynomial $P = P_{0} +P_{1} + \ldots$ for the polynomials $\, P \,$ and $R: = P + Q \,$ to have the same power, i. e. $\, P< R < P \, ? \,\,$ In this case we will call the polynomial $\, Q \,$ the lower-order term with respect to the polynomial $\, P$.

It is clear, that in this case our question sounds like this: what should polynomials $P_{1}, P_{2},\ldots ,P_{l - 1} \,$ be like for the polynomials $\, P \,$ and $\, {\mathcal {P}} \,$ to have the same power, i.e. the relation $\,{\mathcal {P}}< P = {\mathcal {P}} + {\mathcal {P}}_{1} < {\mathcal {P}}\,$ took place?

Remark 5.1

Note that (as will be seen below in Theorem 5.1) for a pair of generalized-homogeneous polynomials $\, P \,$ and $\, Q \,$ (where $d_{P} > d_{Q}) \,$ the relations $\, Q < P \,$ and $P< P + Q < P \,$ are equivalent, however, generally speaking, this does not apply to general polynomials (for examples see below Examples 5.1 and 5.2). Moreover

• (1) from the Theorem 2.1$'$ and Lemma 2.2 it follows, that if polynomial $\, P \,$ with the complete Newton polyhedron $\, \Re (P) \,$ is non-degenerate, then $\,{\mathcal {P}}< P = {\mathcal {P}} + {\mathcal {P}}_{1} < {\mathcal {P}}$,

• (2) when $\, \Gamma := \Re _{i_{0}}^{n - 1} \,$ is (the only) degenerate principal face of the polyhedron $\, \Re (P), \,$ the necessary and sufficient conditions for the right-hand side of this estimate ($\, P = {\mathcal {P}} + {\mathcal {P}}_{1} < {\mathcal {P}}$) to take place are given by Theorem 4.1 (first part). It is a) $\, \Re ({\mathcal {P}} + {\mathcal {P}}_{1}) = \Re ({\mathcal {P}}), \,\,$ b) each pair of polynomials $\, ( P_{j}, {\mathcal {P}}) \, (j = 1,\ldots , l -1) \,$ must satisfy the conditions of Theorem 4.1,

• (3) in a particular case sufficient conditions for the validity of the relation $\,{\mathcal {P}} < P = {\mathcal {P}} + {\mathcal {P}}_{1} \,$ are given in the second part of Theorem 4.1.

However, in the general case Theorem 4.1 does not answer the question: when (under what conditions on the polynomials $\, P_{j} \, (j = 1,2,\ldots , l - 1) )\,\,$ $\, {\mathcal {P}} < P = {\mathcal {P}} + {\mathcal {P}}_{1} ?$ This will be the subject of the next subsection.

5.1 Adding lower-order terms to a polynomial, preserving the power of this polynomial

Let us first consider the relatively simple case when generalized-homogeneous polynomials are compared.

Theorem 5.1

Let $\, P \,$ and $\, Q \,$ be $\, \lambda -$homogeneous polynomials of $\, \lambda -$orders $d_{P} \,$ and $\, d_{Q} \,$ respectively, and $\,d_{Q} < d_{P}. \,$ Let the pair $\, \{P,Q\}$ satisfy the conditions 1) - 3) of the Theorem 3.1, i.e. $\, Q < P. \,$ Then 1) $\, |Q(\xi ) | / | P(\xi ) + Q(\xi ) | \rightarrow 0 \,$ for $\,| Q (\xi ) | \rightarrow \infty ,\,\,$ 2) $\, P < P + Q.$

Proof

To prove part 1) note, that due to the condition $\, Q < P, \,$ if $\, | Q (\xi ) | \rightarrow \infty$, then also also $\,| P (\xi ) | \rightarrow \infty .\,\,$ By the second part of Lemma 3.1, we have that under these conditions $\, |Q(\xi ) | / | P(\xi ) | \rightarrow 0.\,$ As a result, we get that

$$\begin{aligned} \, \frac{ |Q(\xi ) | }{ | P(\xi ) + Q(\xi ) | } = \frac{ \left[ |Q(\xi ) | / | P(\xi ) |\right] }{ \left[ 1 + Q(\xi ) / P(\xi ) \right] } \rightarrow 0. \end{aligned}$$

The second part follows directly from the first part.

Theorem 5.1 is proved. $\square$

The next proposition in a sense solves the question posed in the class of polynomials that we considered above.

Theorem 5.2

Suppose that a degenerate polynomial $\, P \,$ and a $\, \mu -$homogeneous polynomial $\, Q \,$ of $\, \mu -$order $\, \delta _{Q} \in (d_{1}, d_{o})$ satisfy the conditions of first part of Theorem 4.1 (consequently $\, Q < P \,$). Let for any $\, \eta \in \Sigma (P_{0}) \,$

$$\begin{aligned} \psi (\xi ): = | Q (\xi ) | / | P_{0} (\xi ) |^{(\delta _{Q} - d_{1}) / (d_{0} - d_{1}) } \rightarrow 0 \,\,\, as \,\, \xi \rightarrow \eta , P_{0} (\xi ) \ne 0. \end{aligned}$$

(5.2)

Then 1) $| Q(\xi )| / [| P (\xi ) | +1] \rightarrow 0 \,$ as $| \xi | \rightarrow \infty ,\,$ 2) $\, Q < P + Q, \,$ $\, P< P + Q < P.$

Proof

Let’s prove the statement 1). Suppose, on the contrary, that the conditions of the theorem are satisfied, but there exist a sequence $\, \{\xi ^{s}\} \,$ and a number $\, c_{1} > 0$, such that $\, \xi ^{s} \rightarrow \infty \,$ for $\, s \rightarrow \infty \,$ and

$$\begin{aligned} | Q(\xi ^{s})| / [|P(\xi ^{s})| + 1] \ge c_{1} \,, (s = 1,2,\ldots ). \end{aligned}$$

(5.3)

Reasoning as in the proof of Theorem 4.1 we obtain (for sufficiently large $\, s$) the following estimates for the polynomial $\, P$.

$$\begin{aligned} P(\xi ^{s}) = \rho _{s}^{\kappa _{1}^{s} \, (\mu , \alpha )} \, P_{0} (\rho _{s}^{h^{s}}) + \rho _{s}^{\kappa _{1}^{s} (\mu , \beta )} \, P_{1} (\rho _{s}^{h^{s}}) + o(\rho _{s}^{\kappa _{1}^{s} (\mu , \beta )}). \end{aligned}$$

Since $\, \kappa _{1}^{s} \rightarrow 1 \,$ as $\, s \rightarrow \infty , \,$ we have with some constant $\, c_{2} > 0$ and for sufficiently large $\, s$

$$\begin{aligned} |P(\xi ^{s})| + 1 \ge c_{2} \, [1+| \rho _{s}^{d_{0}} \, P_{0} (\rho _{s}^{h^{s}}) + \rho _{s}^{d_{1}} \, P_{1} (\rho _{s}^{h^{s}}) |]. \end{aligned}$$

(5.4)

Taking into account condition (5.2), for the polynomial $\, Q \,$ we have for the same values of $\, s \,$

$$\begin{aligned} |Q(\xi ^{s})| = \rho _{s}^{ \delta _{Q}} \, | Q (\rho _{s}^{h^{s}})| = \rho _{s}^{ \delta _{Q}} \, | P_{0} (\eta ^{s}) |^{(\delta _{Q} - d_{1}) / (d_{0} - d_{1}) } \,\,\, \psi (\rho _{s}^{h^{s}}). \end{aligned}$$

(5.5)

Then, from (5.4)–(5.5) we have with some constant $\, c_{3} > 0$

$$\begin{aligned} |Q(\xi ^{s})|/[|P(\xi ^{s})| +1 ] \le c_{3} \, M(\rho _{s}^{h^{s}}) \psi (\rho _{s}^{h^{s}}), \end{aligned}$$

(5.6)

where

$$\begin{aligned} M(\rho _{s}^{h^{s}}): = \rho _{s}^{\delta _{Q}} \, | P_{0} (\rho _{s}^{h^{s}}) |^{(\delta _{Q} - d_{1}) / (d_{0} - d_{1}) } / \left[ |\rho _{s}^{d_{0}} P_{0} (\rho _{s}^{h^{s}}) + \rho _{s}^{d_{1}} P_{1} (\eta ) \, | + 1 \right] . \end{aligned}$$

Let us prove the existence of some constant $\, c_{4} > 0$, for which the following inequality holds

$$\begin{aligned} M(\rho _{s}^{h^{s}}) \le c_{4} \,\,\, (s = 1,2,\ldots ). \end{aligned}$$

(5.7)

If we introduce the following notations: $x =x_{s}:= \rho _{s}, \,$ $\, y = y_{s}:= | P_{0} (\rho _{s}^{h^{s}}) |^{1/(d_{0} - d_{1})}, \,$ $a:= \delta _{Q}, \,$ $\, b:= \delta _{Q} - d_{1},$ $\, c:= d_{0}, \,$ $\,d:= d_{0} - d_{1}, \, \,$ inequality (5.7) can be rewritten as

$$\begin{aligned} x^{a} y^{b} \le 1 + c_{4} \left[ \, x^{c} y^{d} + P_{1} (\eta ) x^{c - d} \, \right] . \end{aligned}$$

(5.8)

To prove inequality (5.8), we apply Lemma 3.4, where $\sigma _{1}= 1, \,$ $\sigma _{2}= P_{1} (\eta ) > 0. \,$ The conditions of this lemma are satisfied, because 1) for sufficiently large $\, s \,$ $x_{s} >1, \,$ $y_{s} \in [0,1], \,$ $2)\, \delta _{Q} = a \le c = d_{0}, \,$ $\, a - b = c - d = d_{1}, \,$ $3) \, \sigma _{1} = 1, \,$ $\sigma _{2} = P_{1} (\eta ) > 0.$

Thus, inequality (5.7) is proved.

Since $\,\psi (\rho _{s}^{h^{s}}) \rightarrow 0 \,$ as $s \rightarrow \infty , \,$ inequalities (5.6), (5.7) together contradict the assumption (5.3) and prove the first part of the theorem.

The second part of the theorem is an immediate consequence of the first part. It is only necessary to pay attention to the fact that now the behavior of the polynomial $\, Q \,$ does not affect the behavior of $\, P + Q \,$ when $\,| \xi | \rightarrow \infty , \,$ (i.e. $\, P (\xi ) \rightarrow \infty ).$

Theorem 5.2 is proved. $\square$

Let us give an example of pair of polynomials $\, (P,Q)$, satisfying the conditions of Theorem 5.2.

Example 5.1

Let $\, n = 2, \,$ $\, P(\xi ) = (\xi _{1} - \xi _{2})^{8} + (\xi _{1}^{2} + \xi _{2}^{2})^{2}, \,\,\$ $Q(\xi ) = (\xi _{1} - \xi _{2})^{5} (\xi _{1} + \xi _{2}). \,$

Here $\, P_{0}(\xi ) = (\xi _{1} - \xi _{2})^{8}, \,\,$ $P_{1} (\xi ) = (\xi _{1}^{2} + \xi _{2}^{2})^{2}, \,\,$ $\, \Sigma (P_{0}) = \{\eta = (1 / \sqrt{2}, 1 / \sqrt{2} ), - \eta \}, \,$ $\, d_{0} = 8, d_{1} = 4, \,$ $\Delta (\eta , P_{0}) = 8, \,$ $\delta _{Q} = 6, \, \Delta (\eta , Q) = 5, \,\,$ $(\delta _{Q} - d_{1} ) / (d_{0} - d_{1}) = 1 / 2. \,$

Conditions (1)–(4) of Theorem 4.1 can be easily verified, and condition (5.2) of Theorem 5.2 is satisfied, since for any sequence $\, \{ \eta ^{s}\}$, $\eta ^{s} \rightarrow \eta \,\,$ we have

$$\begin{aligned} \psi (\eta ^{s}): = | Q(\eta ^{s}) | / \left[ | P_{0} (\eta ^{s}) \right] ^{1/2} = ((\eta _{1}^{s})^{2} - (\eta _{2}^{s})^{2}) \rightarrow 0. \end{aligned}$$

At the same time, it is obvious that $\, Q \not < P_{0}.$

As for the pair of polynomials from Example 4.1, then for any sequence $\, \{\eta ^{s}\}$, $\eta ^{s} \rightarrow (1 / \sqrt{2}, 1 / \sqrt{2} ) \,\,$ as $\,\, s \rightarrow \infty , \,$ $\, \psi (\eta ^{s}) =$ $| Q(\eta ^{s}) / |P_{0} (\eta ^{s}) |^{1 / 2}$ $= (\eta _{1}^{s})^{2} + (\eta _{2}^{s})^{2} \,$ $\rightarrow 1, \,\,$ i.e. condition (5.2) is violated. However, as we have seen above, $P < P + Q \,$ and $\, Q < P + Q \,$ (the pair $\, ( P, Q ) \,$ satisfies the condition of the second part of Theorem 4.1).

Example 5.2

If we replace the polynomial $\, Q \,$ by the polynomial $\, Q_{1} = - 2 (\xi _{1} - \xi _{2})^{4} \, (\xi _{1}^{2} +\xi _{2}^{2} )$, then despite the fact, that $\, Q_{1} < P \,$ (the pair $\,( Q_{1}, P ) \,$ also satisfies the conditions of the first part of Theorem 4.1), nevertheless a simple calculations show that $P \not < P + Q_{1} \,$ and $\, Q_{1} \not < P + Q_{1}.$ Indeed, for sufficiently large $\, s, \,$ on the sequence $\{\xi ^{s} = (s, s + 2^{1 / 4} \, s^{1 / 2} ) \}$ we have $\, 8 \, s^{4} \le |P(\xi ^{s}) | \le 29 \, s^{4}, \,\,$ $4 \, s^{4} \le |Q_{1} (\xi ^{s})| \le 6\,s^{4} \,\,$ and $| P(\xi ^{s}) + Q_{1} (\xi ^{s})|= [(\xi _{1}^{s} - \xi _{2}^{s} )^{4} - [(\xi _{1}^{s})^{2} +$ $(\xi _{2}^{s})^{2} ]]^{2} = [2\,s^{2} -2\,s^{2} - 2^{5 / 4} s \, \sqrt{s} + \sqrt{2} \, s ]^{2}$ $= O(s^{3 / 2}),$ hence $\, | P(\xi ^{s}) | / [1 + | P(\xi ^{s}) + Q_{1} (\xi ^{s}) |] \rightarrow \infty \,\,$ and $\, | Q_{1}(\xi ^{s}) | / [1 + | P(\xi ^{s}) + Q_{1} (\xi ^{s}) |] \rightarrow \infty .$

We would like to emphasize, that in Example 5.1 polynomial $\, Q \,$ takes both positive and negative values in the neighborhood of the point $\, \eta = (1 / \sqrt{2}, 1 / \sqrt{2} ). \,$

Now we are ready to turn to the question posed at the beginning of this section. Namely, let with respect to a vector $\, \mu \in {\mathbb {R}}^{n} \,$ a (general) polynomial $\, P \,$ be represented as a sum of $\, \mu -$homogeneous polynomials in the form (5.1). We need to describe those multi-indites $\, \nu \in {\mathbb {N}}_{0}^{n} \,$ for which $\, \xi ^{\nu } < P, \,$ i.e. there exists a constant $\, c = c (\nu , P) > 0 \,$, such that

$$\begin{aligned} | \xi ^{\nu } | \le c \, \left[ | P(\xi )| + 1\right] \,\, \forall \xi \in {\mathbb {R}}^{n}. \end{aligned}$$

(5.9)

Theorem 5.3

Let $\, \Re = \Re (P) \,$ be the complete Newton polyhedron of the polynomial $\, P \in I_{n}. \,$ Let all of the principal faces of $\, \Re \,$ with the exception of one $\, (n - 1) -$dimensional face $\, \Gamma :=\Re _{i_{0}}^{n - 1} \,\,$ (with the outward normal $\, \mu , \,$ $\mu _{j} > 0 \,\, (j = 1,\ldots , n) )$ be non-degenerate, and the face $\, \Gamma \,$ be degenerate.

With respect to the vector $\, \mu \,$ we represent the polynomial $\, P \,$ as a sum of $\, \mu -$homogeneous polynomials in the form (see (5.1)) $\, P = P_{0} + P_{1} +\cdots + P_{l} +\cdots + P_{M}, \,$ where $\, P_{0}(\xi ) =: P^{i_{0}, n - 1 \,\,} (\xi ), \,$ $\, P_{j} \,$ is a $\, \mu -$homogeneous polynomial of $\, \mu -$order $d_{j}$, $\, j = 0,1,\ldots , l,\ldots ,M, \,$ $\,\, d_{0}> d_{1}> \ldots> d_{l}> \ldots > d_{M} \ge 0.$

Suppose that $\, P_{l} (\eta ) \ne 0 \,$ for some number $\, l, 1 \le l \le M \,$ and for all $\, \eta \in \Sigma (P_{0}): = \{\xi \in {\mathbb {R}}^{n,0}: \,\, |\xi , \mu | = 1, \, P_{0} (\xi ) =0 \}, \,$ while every polynomial $\, P_{j} \in {\mathfrak {M}}: = \{P_{1}, P_{2},\ldots ,P_{l - 1}\}$ vanishes at all points $\, \eta \in \Sigma (P_{0}). \,$ We put ${\mathcal {P}} (\xi ): = P_{0} (\xi ) + P_{l} (\xi ), \,$ $\, {\mathcal {P}}_{1} (\xi ): = P_{1} (\xi ) +\cdots + P_{l-1} (\xi ), \,\,$ $p (\xi ): = P_{l+1} (\xi ) + \ldots + P_{M} (\xi ), \,$ $\Re ^{*}: =\{ \beta \in \Re : (\mu ,\beta ) \le d_{l}\}$, and assume that the polynomial $\, P - {\mathcal {P}}_{1} = {\mathcal {P}} + p \,$ satisfies the conditions of Theorem 2.1. Then

• (a) if $\, \nu \notin \Re ^{ *}, \,$ inequality (5.9) cannot hold,

• (b) inequality (5.9) holds for any multi-index $\, \nu \in \Re ^{*}$, if each of the polynomials $\, P_{j} \in {\mathfrak {M}}$ $\, (j = 1,\ldots , l - 1) \,$ satisfies one of the following conditions

  • (b.1) $\, P_{j} < P_{0}, \,$ i.e. for a pair $\, (P_{j}, P_{0}) \,$ of $\, ( \mu -$homogeneous) polynomials the conditions of Theorem 5.1 are fulfilled,

  • (b.2) $P_{j} < {\mathcal {P}}, \,$ i.e. for a pair of polynomials $\, (P_{j}, {\mathcal {P}}) \,$ the conditions (I)–(II) of Theorem 4.1 are fulfilled,

  • (b.3) $P_{j} < {\mathcal {P}} + P_{j}, \,\,$ ${\mathcal {P}}< {\mathcal {P}} + P_{j} < {\mathcal {P}}, i.e.$ for a pair of polynomials $\, (P_{j}, {\mathcal {P}}) \,$ the conditions of Theorem 5.2 are fulfilled.

Proof

Before proceeding to the proof of the theorem we note, that conditions b.2) and b.3) could be set not for a pair of polynomials $\, (P_{j}, {\mathcal {P}}) \,$ but for a pair $\, (P_{j}, {\mathcal {P}} + p ). \,$ On the one hand this simplifies writing and reasoning, on the other hand it is legitimate, since by Theorem 2.1 and Corollary 2.1 (see point 3 of Corollary 2.1) the polynomial $\, p(\xi ) \,$ does not affect the behavior of the polynomials $\, P \,$ and $\, {\mathcal {P}} \,$ at infinity.

Bearing in mind that for the polynomial $\, {\mathcal {P}} \,$ estimate (5.9) is valid for all $\, \nu \in \Re ^{*}, \,$ it is sufficient for us to prove that $\, {\mathcal {P}} < P = {\mathcal {P}} + {\mathcal {P}}_{1}.$

First we add to the polynomial $\, {\mathcal {P}} \,$ those polynomials from $\, {\mathfrak {M}} \,$ that (together with the polynomial $\, P_{0} \,$) satisfy condition b.1) of the theorem. Renumbering the polynomials $\, P_{1},\ldots , P_{l - 1}$ we can assume that these are polynomials ${\mathfrak {M}}_1 = \{P_{1},\ldots , P_{k_{1}} \}$, where $\, 1 \le k_{1} \le l - 1, \,$ i.e. $\, P_{j} < P_{0} \,$ $\, (j=1,\ldots ,k_{1}). \,$

Since $\, d_{j} < d_{0} \,\,$ $(j = 1,\ldots , k_{1}), \,\,$ by Lemma 3.1 $\,\, P_{j} (\xi ) = o(|\, P_{0}(\xi ) | ) \,$ for $\, |\, P_{0}(\xi ) | \rightarrow \infty \,\,$ i.e. $| P_{1}(\xi ) | + | P_{2} (\xi ) | + \ldots , | P_{k_{1}} (\xi ) | \,\,$ $= o ( |\, P_{0}(\xi ) | ) \,$ for $\, |\, P_{0}(\xi ) | \rightarrow \infty . \,\,$

Corollary 2.1 implies that $\, P_{0} < {\mathcal {P}}, \,$ hence $| P_{1}(\xi ) | + | P_{2} (\xi ) | +\cdots + | P_{k_{1}} (\xi ) | \,\,$ $= o ( |\, {\mathcal {P}} (\xi ) | ) \,$ for $\, |\, P_{0}(\xi ) | \rightarrow \infty . \,\,$ Thus, there exists a constant $\, c > 0$, such that for sufficiently large $\, |\, P_{0}(\xi ) |, \,$ the inequality $| {\mathcal {P}} (\xi ) | \le c \, [1 + | {\mathcal {P}}|+ | P_{1}(\xi ) | + | P_{2} (\xi ) | + \ldots , | P_{k_{1}} (\xi ) |]$ holds.

If $\, |\, P_{0}(\xi ) | \,$ is bounded for $\, | \xi | \rightarrow \infty \,$, then polynomials $\, \{P_{j}\}_{j = 1}^{k_{1}}$ are also bounded and $\, {\mathcal {P}} (\xi ) \rightarrow \infty , \,$ hence this inequality (perhaps with a different constant) is obvious. As a result, we get that $\, {\mathcal {P}}< {\mathcal {P}} + P_{1} + P_{2} +\cdots + P_{k_{1}} < {\mathcal {P}}.$ This means that when comparing the polynomials ${\mathcal {P}} \,$ and $\, P \,$ it suffices to compare the polynomials ${\mathcal {P}}^{1}: =$ ${\mathcal {P}} + P_{1} + P_{2} +\cdots + P_{k_{1}} \,$ and $\, P. \,$

If $k_{1} = l - 1,$ i.e. $\,\, {\mathcal {P}}^{1} (\xi ) = {\mathcal {P}} (\xi ) + {\mathcal {P}}_{1} (\xi ) =$ $P (\xi ) \,\,\,\, \forall \xi \in {\mathbb {R}}^{n}, \,$ then the theorem is proved.

Consider the case when $k_{1} < l - 1,$ i.e. ${\mathfrak {M}}_{1} \ne {\mathfrak {M}}.$

Let us now consider those polynomials $\, P_{j} \in {\mathfrak {M}} {\setminus } {\mathfrak {M}}_{1}$, which satisfy condition b.3). Let these be polynomials $\, {\mathfrak {M}}_{2}: =$ $\{ P_{k_{1} + 1}, P_{ k_{1} + 2},\ldots , P_{ k_{1} + k_{2} } \,\,$ $( \, k_{1} + k_{2} \le l - 1)\} \,\,$ i.e. $\,| P_{j} (\xi ) | / [|P(\xi ) | + 1] \rightarrow 0 \,\,$ as $\, | \xi | \rightarrow \infty$, and $\, P< P + P_{j} < P \,\,$ for all $j= k_{1} + 1,\ldots ,k_{1} + k_{2}. \,$

Arguing as in the previous case, we find that ${\mathcal {P}} < {\mathcal {P}}^{2}: = {\mathcal {P}}^{1} +$ $P_{k_{1} + 1} + P_{ k_{1} + 2} +\cdots + P_{k_{1} + k_{2} } \,\,$ $< {\mathcal {P}}, \,\,$ i.e. when comparing the polynomials ${\mathcal {P}} \,$ and $\, P \,$ it suffices to compare the polynomials ${\mathcal {P}}^{2} \,$ and $\, P. \,$

Finally, to the polynomial $\, {\mathcal {P}}^{2} \,$ we add the remaining polynomials from $\, {\mathfrak {M}} \,$ that satisfy condition b.2) of the theorem (i.e. conditions of Theorem 4.1). Let these be polynomials $\, {\mathfrak {M}}_{3}: =$ $\, \{ P_{k_{1} + k_{2} + 1}, P_{ k_{1} + k_{2} + 2},\ldots , P_{k_{1} + k_{2} + k_{3}} \}, \,\,$ $k_{1} + k_{2} + k_{3} = l - 1. \,$ Then $\, {\mathcal {P}}^{2} (\xi ) +$ $P_{k_{1} + k_{2} + 1} (\xi ) + P_{ k_{1} + k_{2} + 2} (\xi )+\cdots + P_{k_{1} + k_{2} + k_{3}} (\xi )$ $= P(\xi ) \,$ for all $\, \xi \in {\mathcal {R}}^{n}.$

As a result of the previous two cases we have that ${\mathcal {P}}< {\mathcal {P}}^{2} < {\mathcal {P}}. \,$ From Theorem 4.1 it follows that $\, {\mathcal {P}} < {\mathcal {P}} +$ $P_{k_{1} + k_{2} + 1} + P_{ k_{1} + k_{2} + 2} +\cdots + P_{k_{1} + k_{2} + k_{3}}.$ Hence $\,{\mathcal {P}}^{2} < {\mathcal {P}}^{2} +$ $P_{k_{1} + k_{2} + 1} + P_{ k_{1} + k_{2} + 2} +\cdots + P_{k_{1} + k_{2} + k_{3}} = P.$ So we obtain that $\, {\mathcal {P}}< P < {\mathcal {P}}.$

Theorem 5.3. is proved. $\square$