The Łojasiewicz Exponent at Infinity of Non-negative and Non-degenerate Polynomials

Abstract

Let f be a real polynomial, non-negative at infinity with non-compact zero-set. Suppose that f is non-degenerate in the Kushnirenko sense at infinity. In this paper we give a formula for the Łojasiewicz exponent at infinity of f and a formula for the exponent of growth of f in terms of its Newton polyhedron.

1 Introduction

Let $f : {\mathbb {R}}^n \rightarrow {\mathbb {R}}$ be a real polynomial, $f(0)=0$ and $K \subset \mathbf {\mathbb {R}}^n$ be a compact set. The well known classical inequality (see Łojasiewicz 1959) say that there exist positive constants $C,\alpha$ such that

$$\begin{aligned} |f(x)|\ge C\cdot {\text {dist}}(x,f^{-1}(0))^\alpha , \end{aligned}$$

(1)

for all $x \in K$.

If the set K is non-compact, it may happen that such $C,\alpha$ do not exist. One may check that it is impossible for polynomials

$$\begin{aligned} f(x,y,z)=x^2 (y^2 +z^4) \quad \text {and} \quad f(x,y)=x(y-1)[y^2+(xy-1)^2]. \end{aligned}$$

(2)

For this reason, some authors modify inequality (1) or its domain.

Hörmander (1958) considered a global version of inequality (1). Precisely, he proved the following

$$\begin{aligned} |f(x)|\cdot (1+|x|)^{\beta }\ge C\cdot {\text {dist}}(x,f^{-1}(0))^\alpha , \end{aligned}$$

for all $x \in \mathbf {\mathbb {R}}^n$ and some positive constants $C,\alpha ,\beta$.

In some additional assumptions another global version of inequality (1) was given in Ɖinh (2014) i.e.

$$\begin{aligned} {|}f(x)|^\alpha +|f(x)|^\beta \ge C\cdot {\text {dist}}(x,f^{-1}(0)), x \in \mathbf {\mathbb {R}}^n, \end{aligned}$$

for some positive constants $C,\alpha ,\beta$.

In turn, Hà and Duc (2010) Ha and Nguyen modified the zero set of polynomial $f :\mathbf {\mathbb {R}}^2 \rightarrow \mathbf {\mathbb {R}}$ in inequality (1):

$$\begin{aligned} |f(x)|\ge C\cdot {\text {dist}}\left( x,{f^{-1}(0)}^{\mathbf {\mathbb {R}}}\right) ^\alpha , \end{aligned}$$

in some neighborhood at infinity, where ${f^{-1}(0)}^{\mathbf {\mathbb {R}}}$ denotes real approximation at infinity of $\{x \in \mathbf {\mathbb {C}}^2 :f(x)=0 \}$.

Another modification concerned both a zero set and a domain. Indeed, in Ɖinh (2013) Kurdyka and Le Gal established

$$\begin{aligned} |f(x)|\ge C\cdot {{\text {dist}}(x,Z})^{\alpha } , \, x\in f^{-1}(-\delta ,\delta ) \end{aligned}$$

for some positive constants $\delta ,C,\alpha$, where $Z=\{x \in \mathbf {\mathbb {R}}^n :f(x)\cdot \frac{\partial f}{\partial x_1}(x) = 0\}$, and f is a monic polynomial with respect to $x_1$. In this case constants $C,\alpha$ can be computed explicitly (see Hà et al. 2015).

If the set $f^{-1}(0)$ is compact, then

$$\begin{aligned} {\text {dist}}(x,f^{-1}(0))\approx |x|. \end{aligned}$$

In this case for real polynomial $f :\mathbf {\mathbb {R}}^n \rightarrow \mathbf {\mathbb {R}}$, Gwoździewicz (1998) proved the following

$$\begin{aligned} |f(x)|\ge C\cdot |x|^{d-(d-1)^n}, \, |x|>R, \end{aligned}$$

where $d=\deg f>2$ and $C,R>0$.

Kollár (1988) gave similar result for complex polynomial mappings $F :\mathbf {\mathbb {C}}^n \rightarrow \mathbf {\mathbb {C}}^n$, $\# F^{-1}(0)<\infty$ i.e.

$$\begin{aligned} |F(x)|\ge C\cdot |x|^{d-d^n}, \, |x|>R, \end{aligned}$$

where $d=\deg F$ and $C,R>0$.

In the paper we assume that $f^{-1}(0)$ is a non-compact set. We keep the form of inequality (1) , but we restrict the domain. Namely, we examine behavior of f:

• (i) in the neighborhood of the level set $f^{-1}(0)$ at infinity i.e. in the set

  $$\begin{aligned} \{x \in \mathbf {\mathbb {R}}^n :{\text {dist}}(x,f^{-1}(0))<\varepsilon ,\, |x|>R \}, \end{aligned}$$

• (ii) or in the set

  $$\begin{aligned} \{x \in \mathbf {\mathbb {R}}^n :{\text {dist}}(x,f^{-1}(0))>R \}. \end{aligned}$$

The lack of the distinction of these cases could lead to a situation that an exponent $\alpha$ in inequality (1) does not exist in neighborhood at infinity. See for example polynomials (2). In the case (i) and (ii) we give the following definitions.

Let $f : {\mathbb {R}}^n \rightarrow {\mathbb {R}}$ be a polynomial such that $f^{-1}(0)$ is a non-compact set. We define the Łojasiewicz exponent of f at infinity as the infimum of the exponents $l \in {\mathbb {R}}_{+}$ such that

$$\begin{aligned} |f(x)|\ge C\cdot {{\text {dist}}(x,f^{-1}(0))}^l \quad \text {for all } x \text { such that } {\text {dist}}(x,f^{-1}(0))<\varepsilon , \end{aligned}$$

(3)

in some neighborhood of infinity for some $\varepsilon >0$ and $C>0$. We denote it by ${\mathcal {L}}_{\infty }(f)$. In cases where such l does not exist, we put

$$\begin{aligned} {\mathcal {L}}_{\infty }(f)=+\infty . \end{aligned}$$

In Ɖinh (2012) authors proved that there are no sequences of the first type if and only if there exist $C,\delta ,\alpha >0$ such that

$$\begin{aligned} |f(x)|^{\alpha }\ge C\cdot {{\text {dist}}(x,f^{-1}(0))} \quad \text {for all } x \in f^{-1}([-\delta ,\delta ]). \end{aligned}$$

The sequence $(x_k)_{k=1}^{\infty }\subset \mathbf {\mathbb {R}}^n$ is of the first type if $f(x_k)\rightarrow 0$ and ${\text {dist}}(x,f^{-1}(0))\not \rightarrow 0$.

It is easy to observe that if the last inequality is true for some positive $C,\delta ,\alpha$, then there exist $C, \varepsilon , l >0$ such that inequality (3) is true. Hence if there are no sequences of the first type, then $\mathcal {L}_{\infty }(f)$ exists. However, in some cases $\mathcal {L}_{\infty }(f)$ exists but there is a sequence of the first type. For example $f(x,y)=x(y-1)[y^2+(xy-1)^2]$.

In the paper we give an effective formula for the Łojasiewicz exponent at infinity in the class of non-negative and non-degenerate polynomials in terms of the Newton polyhedron (see Sect. 2). This result is a counterpart at infinity of the local result of the paper Bùi and Pham (2014).

2 Preliminaries

We denote by ${\mathbb {R}}_{+}=\{ x \in {\mathbb {R}} : x \ge 0 \}$ and ${\mathbb {Z}}_{+}={\mathbb {Z}}\cap {\mathbb {R}}_{+}$. For $x=(x_1,\ldots ,x_n) \in {\mathbb {R}}^n$ and $\alpha =(\alpha _1,\ldots ,\alpha _n) \in {\mathbb {Z}}_{+}^n$, we denote by $x^{\alpha }$ the monomial $x_1^{\alpha _1}\ldots \ x_n^{\alpha _n}$ and put $|\alpha |=\alpha _1+\cdots +\alpha _n$, and $|x|= \max _{i=1}^n |x_i|$.

Let $f(x)= \sum \nolimits _{\alpha \in {\mathbb {Z}}_{+}^n} c_{\alpha }x^{\alpha }$. Let us define the set ${\text {supp}}(f)=\{\alpha \in {\mathbb {Z}}_{+}^n: c_{\alpha } \not = 0 \}$ and call it the support of f. Define the set $\Gamma (f)={\text {conv}}\{{\text {supp}}(f)\} \subset {\mathbb {R}}_{+}^n$ and call it the Newton polyhedron at infinity of f.

Let $q \in {\mathbb {R}}^n {\setminus } \{0\}$. Define

$$\begin{aligned} \begin{aligned} d(q,\Gamma (f))&=\min \{ \langle q,\alpha \rangle :\alpha \in \Gamma (f) \},\\ \Delta (q,\Gamma (f))&=\{ \alpha \in \Gamma (f): \langle q,\alpha \rangle = d(q,\Gamma (f))\}, \end{aligned} \end{aligned}$$

where $\langle \cdot \,,\cdot \rangle$ denotes the standard inner product in ${\mathbb {R}}^n \times {\mathbb {R}}^n$. We say that $\Delta \subset \Gamma (f)$ is a face of $\Gamma (f)$, if there exists $q \in {\mathbb {R}}^n {\setminus } \{0\}$ such that $\Delta =\Delta (q,\Gamma (f))$. By a dimension of a face $\Delta$ we mean the minimum of the dimensions of the affine subspace containing $\Delta$. By a vertice of $\Gamma (f)$ we mean the 0-dimensional faces of $\Gamma (f)$. We define Newton boundary at infinity of f as the set of faces $\Delta \subset \Gamma (f)$ such that: if q is defining a vector for $\Delta$ then $q_i<0$ for some $i \in \{1, \ldots , n\}$ and we denote it by $\Gamma _{\infty }(f)$. Denote by $\Gamma ^k _{\infty }(f)$ the set of k-dimensional faces of $\Gamma _{\infty }(f)$, $k=0, \ldots , n-1$. For $\Delta \in \Gamma _{\infty }(f)$ we define the polynomial

$$\begin{aligned} f_{\Delta }= \sum _{\alpha \in \Delta } c_{\alpha }x^{\alpha }. \end{aligned}$$

and call it the principal part of f at infinity with respect to face $\Delta$.

We say that f is Kushnirenko non-degenerate at infinity on the face $\Delta \in \Gamma _{\infty }(f)$ if the system of equations

$$\begin{aligned} \frac{\partial f_{\Delta }}{\partial x_1}(x)= \cdots =\frac{\partial f_{\Delta }}{\partial x_n}(x)=0 \end{aligned}$$

has no solution in $({\mathbb {R}} {\setminus } \{0\})^n {\setminus } K$, where $K\subset {\mathbb {R}}^n$ is a compact set. We say that f is Kushnirenko non-degenerate at infinity (shortly non-degenerate) if f is Kushnirenko non-degenerate at infinity on each face $\Delta \in \Gamma _{\infty }(f)$.

We say that f is non-negative at infinity (shortly non-negative) if there exists a compact set $K \subset {\mathbb {R}}^n$ such that $f(x)\ge 0$ for $x \in {\mathbb {R}}^n {\setminus } K$.

One of the main tool which we use in the paper is the following

Lemma 1

(Curve Selection Lemma at infinity, Ɖinh (2014), Lemma 1) Let $A\subset {\mathbb {R}}^n$ be a semi-algebraic set, and let $F:=(f_1,\ldots ,f_p):{\mathbb {R}}^n\rightarrow {\mathbb {R}}^p$ be a semi-algebraic map. Assume that there exists a sequence $x^k \in A$ such that $\lim _{k\rightarrow \infty } |x^k|=\infty$ and $\lim _{k\rightarrow \infty } F(x^k)=y \in (\overline{{\mathbb {R}}})^p$, where $\overline{{\mathbb {R}}}:={\mathbb {R}}\cup \{\pm \infty \}$. Then there exists an analytic curve $\varphi :(0,\epsilon )\rightarrow A$ of the form

$$\begin{aligned} \varphi (t)=a^0 t^q + a^1 t^{q+1} + \ldots \quad , \end{aligned}$$

such that $a^0 \in {\mathbb {R}}^n{\setminus } \{0\}$, $q<0$, $q \in {\mathbb {Z}}$, and $\lim _{t\rightarrow 0} F(\varphi (t))=y.$

Let $A \subset \mathbf {\mathbb {N}}^n$ be a finite set. Put

$$\begin{aligned} N_{A}(x)= \max _{\alpha \in A} |x^{\alpha }|. \end{aligned}$$

Let V be the set of vertices of $\Gamma (f)$. Denote

$$\begin{aligned} N_{\Gamma }= N_V. \end{aligned}$$

We recall two simple lemmas which will be used in the rest of the paper.

Lemma 2

(Ɖinh (2014), Lemma 11) There exist some subset $J_1,\ldots ,J_s$ of $\{1,\ldots ,n\}$, with $J_i \not \subseteq J_j$ for $i \not = j$, such that

$$\begin{aligned} N_{\Gamma }^{-1}(0)= \bigcup _{k=1}^s Z_k , \end{aligned}$$

where $Z_k:=\{x \in {\mathbb {R}}^n: x_j=0, j \in J_k \}$.

For a given subset $J\subset \{1,\ldots ,n\}$ we define

$$\begin{aligned} {\mathbb {R}}^J := \{\alpha =(\alpha _1,\ldots ,\alpha _n) \in {\mathbb {R}}^n : \alpha _j =0 \quad \text {for} \quad j\not \in J \}. \end{aligned}$$

Lemma 3

(Ɖinh (2014), Lemma 12) Let $J_1,\ldots ,J_s$ be as in Lemma 2. For every $(j_1,\ldots ,j_s) \in J_1\times \cdots \times J_s$, we have $V\cap {\mathbb {R}}^J \not = \emptyset$, where $J=\{j_1,\ldots ,j_s \}$.

3 The Main Theorem

Let $J_1,\ldots ,J_s$ be as in Lemma 2 and let

$$\begin{aligned} \mathcal {P}= \{I \subset \{1, \ldots ,n \} :I \not = \emptyset \wedge I\cap J_k = \emptyset \quad \text {for some} \quad k \in \{1, \ldots ,s\} \}. \end{aligned}$$

Observe that $\mathcal {P} \not =\emptyset$ i.e. $J_{k_0} \not =\{1, \ldots ,n \}$ for some $k_0 \in \{1, \ldots ,s \}$. Indeed, suppose to the contrary that $J_{k} =\{1, \ldots ,n \}$ for any $k \in \{1, \ldots ,s \}$. If $s>1$, then by Lemma 2 it is not possible. Therefore $s=1$. Hence $J_{1} =\{1, \ldots ,n \}$ and $N_\Gamma ^{-1}(0)=\{0\}$. By Lemma 10

$$\begin{aligned} f^{-1}(0)\cap (\mathbf {\mathbb {R}}^n {\setminus } K) = N_\Gamma ^{-1}(0)\cap (\mathbf {\mathbb {R}}^n {\setminus } K) = \emptyset , \end{aligned}$$

for some compact set K. This gives a contradiction to the assumption that the set $f^{-1}(0)$ is not compact.

Let us fix $I \in \mathcal {P}$. We define $\varphi ^I(x)=(\varphi _1 ^I(x), \ldots ,\varphi _n ^I(x))$, where

$$\begin{aligned} \varphi _i ^I(x)=\Big \{ \begin{array}{ccc} 1&{}\text{ for }&{} i \in I,\\ x_i&{}\text{ for }&{}i \not \in I,\\ \end{array} \end{aligned}$$

for $i=1, \ldots ,n$ and define $N_\Gamma ^I=N_\Gamma \circ {\varphi ^I }$.

Observe that

$$\begin{aligned} (N^I_{\Gamma })^{-1}(0)= \bigcup _{k=1}^{s(I)} Z_k^I=\bigcup _{k=1}^{s(I)} \{x \in {\mathbb {R}}^{I'}: x_j=0, j \in J_k^I \}, \quad J_l^I\not \subseteq J_m^I, \, l\not =m, \end{aligned}$$

where $I'=\{1, \ldots , n \}{\setminus } I$. Put

$$\begin{aligned} \alpha ^I_{\max } = \max \left\{ \alpha ^{\min }_{J}: J \in J^I_1\times \cdots \times J^I_{s(I)} \right\} , \end{aligned}$$

where

$$\begin{aligned} \alpha ^{\min }_{J}:= \min \left\{ |\alpha | : \alpha \in V^{I'} \cap {\mathbb {R}}^J \right\} , \end{aligned}$$

and $V^{I'}$ denotes the projection of the set V onto ${\mathbb {R}}^{I'}$. Observe that $N_\Gamma ^I=N_{V^{I^{'}}}$.

Now, we give the main result of the paper.

Theorem 4

Let $f : {\mathbb {R}}^n \rightarrow {\mathbb {R}}$, $f(0)=0$, be a non-negative and non-degenerate polynomial. Then

$$\begin{aligned} {\mathcal {L}}_{\infty }(f)= \max \left\{ \alpha ^I_{\max }: I \in \mathcal {P} \right\} . \end{aligned}$$

(4)

Remark 5

One can check that the assertions of the above theorems are also true if we assume Mikhailov–Gindikin non-degeneracy (see Ɖinh (2014), Section 5).

To illustrate the above theorems we give the following

Example 6

Let $f(x,y,z)=x^8(y^4+z^6).$ It is easy to see that f is non-degenerate and non-negative. We have $V=\{(8,4,0),(8,0,6)\}$ and

$$\begin{aligned} N_\Gamma (x,y,z)=\max \{x^8 y^4,x^8 z^6\}, \quad (N_\Gamma )^{-1}(0)=\{x=0\}\cup \{y=z=0\}. \end{aligned}$$

Hence $J_1=\{1\}$, $J_2=\{2,3\}$.

We calculate ${\mathcal {L}}_{\infty }(f)$. We have $\mathcal {P}=\{\{1\},\{2\},\{3\},\{2,3\}\}$. For $I=\{1\}$ we obtain $I'=\{2,3\}$ and

$$\begin{aligned} N_\Gamma ^I(y,z)=\max \{y^4,z^6\}, \quad (N_\Gamma ^I )^{-1}(0)=\{(0,0)\}, \quad J_1^I=\{2,3\}. \end{aligned}$$

Hence $J=\{2\}$ or $J=\{3\}$ and

$$\begin{aligned} \alpha ^{\min }_2=\min \{|\alpha | :\alpha \in \{4,0\}\}=4, \quad \alpha ^{\min }_3=\min \{|\alpha | :\alpha \in \{0,6\}\}=6. \end{aligned}$$

Therefore

$$\begin{aligned} \alpha ^{\{1\}}_{\max }=\max \{4,6\}=6. \end{aligned}$$

Similarly we calculate

$$\begin{aligned} \alpha ^{\{2\}}_{\max }=\alpha ^{\{3\}}_{\max }=\alpha ^{\{2,3\}}_{\max }=8. \end{aligned}$$

Finally we have

$$\begin{aligned} {\mathcal {L}}_{\infty }(f)= \max \left\{ \alpha ^I_{\max }: I \in \mathcal {P} \right\} =\max \{6,8\}=8. \end{aligned}$$

4 Auxiliary Results

The following lemmas will be used in the proof of Lemma 10. The proof of Lemma 7 is a simple transfer of its local counterpart [see Bùi and Pham (2014), Lemma 3.1]. We give it for a convenience of the reader.

Lemma 7

Suppose that f is non-negative polynomial. Then for any face $\Delta \in \Gamma _{\infty } (f)$ we have $f_{\Delta }(x) \ge 0 \,\, \text {for} \,\, x \in ({\mathbb {R}}{\setminus } \{0\})^n {\setminus } K$, where K is a compact set.

Proof

Since f is non-negative there exists a compact set K such that $f(x)\ge 0$ for $x \in {\mathbb {R}}^n {\setminus } K$. Suppose to the contrary that there exists a face $\Delta \in \Gamma _{\infty } (f)$ and there exists a point $x^0 \in ({\mathbb {R}}{\setminus } \{0\})^n {\setminus } K$ such that $f_{\Delta }(x^0)<0$. Let J be the smallest subset of $\{1, \ldots ,n \}$ such that $\Delta \subset {\mathbb {R}}^J$. Hence, there exists a non-zero vector $a \in {\mathbb {R}}^n$, with $a_j <0$ for some $j \in J$ and $a_j =0$ for $j \not \in J$ such that

$$\begin{aligned} \Delta =\{\nu \in \Gamma (f)\cap {\mathbb {R}}^J : \langle a,\nu \rangle = d(a,\Gamma (f)) \}. \end{aligned}$$

Define monomial curve $\varphi :(0,1)\rightarrow {\mathbb {R}}^n$, $t\mapsto (\varphi _1 (t), \ldots , \varphi _n (t))$, by

$$\begin{aligned} \varphi _j(t)=\Big \{ \begin{array}{ccc} x_j^0 t^{a_j}&{}\text{ for }&{} j \in J,\\ 0&{}\text{ for }&{}j \not \in J.\\ \end{array} \end{aligned}$$

Put $d:= d(a,\Gamma (f))$. Now, we may write f in the form:

$$\begin{aligned} f(\varphi (t))=f_{\Delta } (x^0) t^d \quad + \quad \text {higher order terms in}\,\, t. \end{aligned}$$

Since $f_{\Delta }(x^0)<0$, we have

$$\begin{aligned} f(\varphi (t))<0 \quad \text {for all sufficiently small} \,\, t. \end{aligned}$$

This gives a contradiction.

However counterpart of equivalence Bùi and Pham (2014, Lemma 3.2) is not true at infinity. The simple implication is the only one that holds.

Lemma 8

If f is non-negative and non-degenerate, then for any face $\Delta \in \Gamma _{\infty } (f)$ we have $f_{\Delta }>0$ on $({\mathbb {R}}{\setminus } \{0\})^n {\setminus } K$.

Proof

Using Lemma 7 we obtain $f_{\Delta }(x)\ge 0$ for all $x \in ({\mathbb {R}}{\setminus } \{0\})^n {\setminus } K$, where K is a suitably chosen compact set. Suppose to the contrary that there exists a point $x^0 \in ({\mathbb {R}}{\setminus } \{0\})^n {\setminus } K$ such that $f_{\Delta }(x^0)=0$. Therefore the function $f_{\Delta }$ attains a local minimum at the point $x^0$. Hence ${\text {grad}}f_{\Delta } (x^0)=0$. This gives a contradiction to non-degeneracy of f.

The following lemma will be also applied in the proof of the Lemma 10.

Lemma 9

Gindikin (1974, Lemma 1) Let $\nu \in \mathbf {\mathbb {R}}_+^n$, $\nu \in {\text {conv}}\{ \nu ^1, \ldots , \nu ^k \}$. Then

$$\begin{aligned} |x^\nu | \le \sum _{j=1}^k |x^{\nu _j}|. \end{aligned}$$

The next lemma plays a crucial role in the proof of the main theorem. Its proof is a substantially analogous to the proof of Lemma 3.3 of the paper Bùi and Pham (2014). However we prove the second inequality in (5) without assumption of non-degeneracy and non-negativity, using Lemma 9.

Lemma 10

If f is non-negative and non-degenerate then there exist some positive constants $C_1$ and $C_2$ such that

$$\begin{aligned} C_1 N_{\Gamma }(x)\le f(x)\le C_2 N_{\Gamma }(x), \quad \text {for all} \quad x \in {\mathbb {R}}^n {\setminus } K, \end{aligned}$$

(5)

for some compact set $K \subset {\mathbb {R}}^n$.

Proof

We will prove the first inequality. Suppose to the contrary that there exists a sequence $\{x^k\}\subset {\mathbb {R}}^n$ with $|x^k|>k$ and such that

$$\begin{aligned} f(x^k)<\frac{1}{k}N_{\Gamma }(x^k) \end{aligned}$$

for all k. By Lemma 1, there exist an analytic curves $\varphi :(0,\epsilon )\rightarrow \mathbf {\mathbb {R}}^n$, $t\mapsto (\varphi _1 (t), \ldots ,\varphi _n(t))$ and $\psi :(0,\epsilon )\rightarrow \mathbf {\mathbb {R}}_{+}$ such that

$$\begin{aligned} |\varphi (t)|\rightarrow \infty , \quad |\psi (t)|\rightarrow 0 \quad \text {as} \quad t\rightarrow 0^{+}, \end{aligned}$$

(6)

and

$$\begin{aligned} f(\varphi (t))<\psi (t) N_{\Gamma }(\varphi (t)). \end{aligned}$$

(7)

Let $J=\{j : \varphi _j \not \equiv 0 \}\subset \{1, \ldots ,n \}$. For $j \in J$ we can expand coordinate function $\varphi _j$, say

$$\begin{aligned} \varphi _j(t)= x_j^0 t^{a_j} \quad + \quad \text {higher order terms in}\,\, t, \end{aligned}$$

where $x_j^0 \not =0$ and $a_j \in \mathbb {N}$. From Condition (6), there exists $j \in J$ such that $a_j<0$. If $\Gamma (f)\cap {\mathbb {R}}^{J}=\emptyset$, then for any vertex $\alpha \in V$, there exists $j \not \in J$ such that $\alpha _j>0$ ($V \subset \Gamma (f)$) and hence $(\varphi _j (t))^{\alpha _j}\equiv 0$. Then $(\varphi (t))^{\alpha _j}\equiv 0$. Hence

$$\begin{aligned} N_{\Gamma }(\varphi (t))= \max _{\alpha \in V} |\varphi (t)^{\alpha }|\equiv 0. \end{aligned}$$

This gives a contradiction to (7).

Therefore, $\Gamma (f)\cap {\mathbb {R}}^{J} \not =\emptyset$. Put

$$\begin{aligned} d&=\min \left\{ \sum _{j \in J} a_j \alpha _j : \alpha \in \Gamma (f)\cap {\mathbb {R}}^J \right\} , \\ \Delta&=\left\{ \alpha \in \Gamma (f)\cap {\mathbb {R}}^J : \sum _{j \in J} a_j \alpha _j = d \right\} . \end{aligned}$$

We can write

$$\begin{aligned} f(\varphi (t))=f_{\Delta } (x^0) t^d \quad + \quad \text {higher order terms in}\,\, t, \end{aligned}$$

where $x^0=(x_1^0, \ldots ,x_n^0)$ and $x_j^0=1$ for $j \not \in J$. We will show that $f_{\Delta }(x^0)>0$. Indeed, since f is non-negative and non-degenerate, it follows from Lemma 8 we have that $f_{\Delta }(x)>0$ for $x \in ({\mathbb {R}}{\setminus } \{0\})^n {\setminus } K$, where K is a suitably chosen compact set. Therefore by quasi-homogeneity of $f_{\Delta }$ we have

$$\begin{aligned} f_{\Delta }(x^0)=\frac{f_{\Delta }((s^{a_j} x_{j}^0)_{j \in J})}{s^d}>0, \end{aligned}$$

where s is a positive number such that $s^{a_j} \cdot x_j^0$ is large enough for some $j \in J$. Hence

$$\begin{aligned} f(\varphi (t)) \quad \text {and} \quad t^d \end{aligned}$$

(8)

are of the same order if $t\rightarrow 0^{+}$.

On the other hand, we have

$$\begin{aligned} N_{\Gamma }(\varphi (t))=\max _{\alpha \in V} |\varphi (t)^{\alpha }|= \max _{\alpha \in \Delta } |{(x^0)}^{\alpha }|t^d \quad + \quad \text {higher order terms in}\,\, t. \end{aligned}$$

Hence and by (8) we have a contradiction to (7).

Now we prove the second inequality in (b). Let $|x|\ge R\ge 1$, where R is sufficiently large. By Lemma 9 we have

$$\begin{aligned} f(x)&=\sum _{\nu \in {\text {supp}}f} c_\nu x^\nu \le \max _{\nu \in {\text {supp}}f}|c_\nu |\cdot \sum _{\nu \in {\text {supp}}f} |x^\nu |\le \\&\le C_2\cdot \max _{\nu \in V} |x^\nu |=C_2\cdot N_\Gamma (x), \end{aligned}$$

where $C_2$ is a some positive constant.$\square$

Let $A\subset \mathbf {\mathbb {N}}^n$ be a finite set. Put

$$\begin{aligned} {\mathcal {L}} (N_A)= \inf \{ l \in {\mathbb {R}}_{+} :\exists _{C>0} |N_A(x)|\ge C\cdot {{\text {dist}}(x,N_A^{-1}(0))}^l,\, {\text {dist}}(x,N_A^{-1}(0))<1 \}. \end{aligned}$$

Now we give an effective formula to compute $\mathcal {L}(N_A)$.

Proposition 11

We have

$$\begin{aligned} {\mathcal {L}} (N_A)= \max \left\{ \alpha ^{\min }_{J}: J \in J_1\times \cdots \times J_s \right\} , \end{aligned}$$

where

$$\begin{aligned} \alpha ^{\min }_{J}:= \min \left\{ |\alpha | : \alpha \in A \cap {\mathbb {R}}^J \right\} . \end{aligned}$$

Proof

We first show that ${\mathcal {L}} (N_A)\le \max \{ \alpha ^{\min }_{J} : J \in J_1\times \cdots \times J_s \}$. Let us fix an arbitrary $x \in \mathbf {\mathbb {R}}^n$ such that

$$\begin{aligned} {\text {dist}}(x,N_A ^{-1}(0))=\delta <1. \end{aligned}$$

It is easy to check that

$$\begin{aligned} {\text {dist}}(x,N_A ^{-1}(0))=\min _{k=1}^s \max _{j \in J_k} |x_j|. \end{aligned}$$

Hence

$$\begin{aligned} \max _{j \in J_k} |x_j|\ge \delta \quad \text {for any} \quad k=1, \ldots ,s. \end{aligned}$$

This means that for each $k=1, \ldots ,s$ there exists $j_k \in J_k$ such that

$$\begin{aligned} |x_{j_k}|\ge \delta . \end{aligned}$$

Put $J=\{j_1, \ldots ,j_s\}$. By Lemma 3 we have that $A\cap {\mathbb {R}}^J \not =\emptyset$. Let us choose $\alpha = (\alpha _1, \ldots ,\alpha _s) \in A\cap {\mathbb {R}}^J$ such that

$$\begin{aligned} |\alpha |=\alpha ^{\min }_{J}. \end{aligned}$$

Hence

$$\begin{aligned} N_A (x)= & {} \max \{ |x_{j_1}^{\alpha _j} \ldots x_{j_s}^{\alpha _s}|, \ldots \}\ge \delta ^{\alpha _j} \ldots \delta ^{\alpha _s}\\= & {} \delta ^{\alpha ^{\min }_{J}} \ge \delta ^{\max \{ \alpha ^{\min }_{J} : J \in J_1\times \cdots \times J_s \}}\\= & {} {{\text {dist}}(x,N_A ^{-1}(0))}^{\max \{ \alpha ^{\min }_{J} : J \in J_1\times \cdots \times J_s \}}. \end{aligned}$$

This means that ${\mathcal {L}} (N_A)\le \max \{ \alpha ^{\min }_{J} : J \in J_1\times \cdots \times J_s \}$.

Now, we show that ${\mathcal {L}} (N_A)\ge \max \{ \alpha ^{\min }_{J} : J \in J_1\times \cdots \times J_s \}$.

Let $(j_1, \ldots ,j_s)\in J_1\times \cdots \times J_s$ be such that realized the above maximum and let $J \subset \{1, \ldots ,n\}$ be the minimal set such that $j_k \in J$, $k=1, \ldots ,s$. Put $A_J = \mathbf {\mathbb {R}}^{J}\cap A$. By Lemma 3 we have that $A_J \not =\emptyset$. Take the following parametrization $\varphi (t) = (\varphi _1(t), \ldots ,\varphi _n(t))$, $|t|<1$, where

$$\begin{aligned} \varphi _i (t)= {\left\{ \begin{array}{ll} t&{}\text {for}\quad i \in J\\ 0&{}\text {for}\quad i \not \in J, \end{array}\right. } \end{aligned}$$

(9)

for $i =1, \ldots ,n$. We have

$$\begin{aligned} N_A (\varphi (t))= \max _{\nu \in A_J} |\varphi (t)^\nu | = |t|^{\min \{ |\nu | :\nu \in A_J \}}=|t|^{ \alpha ^{\min }_{J}}={\text {dist}}(\varphi (t),N_A ^{-1}(0))^{\alpha ^{\min }_{J}}. \end{aligned}$$

Hence ${\mathcal {L}} (N_A)\ge \alpha ^{\min }_{J}.$ This ends the proof.

One can observe that the above proof in comparison with the proof of (Bùi and Pham 2014, Proposition 3.1) is more elementary.

5 Proof of the Main Theorem

Now, we are ready to give the proof of the main result.

Proof of Theorem 4

Since f is non-negative and non-degenerate polynomial, then by Lemma 10 there exist some positive constants $C_1$ and $C_2$ such that

$$\begin{aligned} C_1 N_{\Gamma }(x)\le f(x)\le C_2 N_{\Gamma }(x), \end{aligned}$$

(10)

for all $x \in {\mathbb {R}}^n {\setminus } K$ and some compact set $K \subset {\mathbb {R}}^n$. Hence

$$\begin{aligned} f^{-1}(0)\cap (\mathbf {\mathbb {R}}^n {\setminus } K)=N_{\Gamma }^{-1}(0)\cap (\mathbf {\mathbb {R}}^n {\setminus } K). \end{aligned}$$

(11)

We will show that there exist some positive constants $D_1$ and $D_2$ such that

$$\begin{aligned} D_1 {\text {dist}}(x,N_{\Gamma }^{-1}(0))\le {\text {dist}}(x,f^{-1}(0))\le D_2 {\text {dist}}(x,N_{\Gamma }^{-1}(0)), \end{aligned}$$

for all $x \in {\mathbb {R}}^n {\setminus } K_1$ and some compact set $K_1 \subset {\mathbb {R}}^n$, $K\subset K_1$. First, observe that

$$\begin{aligned} {\text {dist}}(x,K)\le |x|\le 2{\text {dist}}(x,K),\quad x \in {\mathbb {R}}^n {\setminus } K_1, \end{aligned}$$

(12)

for some compact set $K_1 \subset {\mathbb {R}}^n$, $K\subset K_1$. By (11), (12) and since $0 \in N_{\Gamma }^{-1}(0)$ we have

$$\begin{aligned} {\text {dist}}(x,f^{-1}(0))= & {} \min \{{\text {dist}}(x,f^{-1}(0){\setminus } K), {\text {dist}}(x,f^{-1}(0)\cap K)\}\\\ge & {} \min \{{\text {dist}}(x,N_{\Gamma }^{-1}(0){\setminus } K), {\text {dist}}(x,K)\}\\\ge & {} \min \{{\text {dist}}(x,N_{\Gamma }^{-1}(0)), \frac{1}{2}|x|\}\\\ge & {} \min \{{\text {dist}}(x,N_{\Gamma }^{-1}(0)), \frac{1}{2}{\text {dist}}(x,N_{\Gamma }^{-1}(0))\}\\= & {} \frac{1}{2} {\text {dist}}(x,N_{\Gamma }^{-1}(0)), \end{aligned}$$

for $x \in {\mathbb {R}}^n {\setminus } K_1$. Analogously, by (11), (12) and since $0 \in f^{-1}(0)$ we get

$$\begin{aligned} {\text {dist}}(x,N_{\Gamma }^{-1}(0))\ge \frac{1}{2} {\text {dist}}(x,f^{-1}(0)), \end{aligned}$$

for $x \in {\mathbb {R}}^n {\setminus } K_1$. Summing up we obtain

$$\begin{aligned} \frac{1}{2} {\text {dist}}(x,N_{\Gamma }^{-1}(0))\le {\text {dist}}(x,f^{-1}(0))\le 2 {\text {dist}}(x,N_{\Gamma }^{-1}(0)), \end{aligned}$$

(13)

for $x \in {\mathbb {R}}^n {\setminus } K_1$. By (10) and (13) it follows that

$$\begin{aligned} {\mathcal {L}}_{\infty }(N_\Gamma )={\mathcal {L}}_{\infty }(f). \end{aligned}$$

(14)

By (14), it is enough to prove formula (4) for $N_\Gamma$. We first show that

$$\begin{aligned} {\mathcal {L}}_{\infty }(N_\Gamma )\le \max \left\{ {\mathcal {L}}(N_{\Gamma }^I): I \in \mathcal {P} \right\} . \end{aligned}$$

(15)

Let $x \in \mathbf {\mathbb {R}}^n {\setminus } K$, where K is the same as in Lemma 10 and ${\text {dist}}(x,N_{\Gamma }^{-1}(0))<\varepsilon <1$. It can be assumed that

$$\begin{aligned} \{x \in \mathbf {\mathbb {R}}^n :|x|<1 \}\subset K. \end{aligned}$$

Let $I \not =\emptyset$ be such that

$$\begin{aligned} |x_i|\ge 1 , i \in I \quad \text {and} \quad |x_i|<1 , i \not \in I. \end{aligned}$$

It is easy to check that $I \in \mathcal {P}$. Since

$$\begin{aligned} (N_{\Gamma }^I)^{-1}(0)=N_{\Gamma }^{-1}(0)\cap \{x \in \mathbf {\mathbb {R}}^n :x_i=1 \quad \text {for} \quad i \in I \} \subset N_{\Gamma }^{-1}(0), \end{aligned}$$

we have

$$\begin{aligned} {\text {dist}}(x_I,(N_{\Gamma }^I)^{-1}(0))\ge {\text {dist}}(x_I,N_{\Gamma }^{-1}(0)). \end{aligned}$$

(16)

It is easy to check that

$$\begin{aligned} {\text {dist}}(x_I,N_{\Gamma }^{-1}(0))={\text {dist}}(x,N_{\Gamma }^{-1}(0)). \end{aligned}$$

(17)

By (16), (17) we obtain

$$\begin{aligned} |N_\Gamma (x)|\ge & {} |N_\Gamma ^I (x_I)|\ge C_I {\text {dist}}(x_I,(N_{\Gamma }^I)^{-1}(0))^{{\mathcal {L}}(N_{\Gamma }^I)}\\\ge & {} C_I {\text {dist}}(x_I,N_{\Gamma }^{-1}(0))^{{\mathcal {L}}(N_{\Gamma }^I)}= C_I {\text {dist}}(x,N_{\Gamma }^{-1}(0))^{{\mathcal {L}}(N_{\Gamma }^I)}\\\ge & {} \min \{C_I :I \in \mathcal {P} \} {\text {dist}}(x,N_{\Gamma }^{-1}(0))^{\max \{{\mathcal {L}}(N_{\Gamma }^I) :I \in \mathcal {P}\}}. \end{aligned}$$

This gives (15).

Now we show that

$$\begin{aligned} {\mathcal {L}}_{\infty }(N_\Gamma ) \ge \max \left\{ {\mathcal {L}}(N_{\Gamma }^I): I \in \mathcal {P} \right\} . \end{aligned}$$

(18)

First we choose $I \in \mathcal {P}$ such that realizes the above maximum. Take the parametrization $\varphi :\mathbf {\mathbb {R}}{\setminus } \{0\} \rightarrow \mathbf {\mathbb {R}}^{\{1, \ldots ,n \} {\setminus } I}$ defined by formula (9) such that it realizes ${\mathcal {L}}(N_\Gamma ^I)$. Let $\varepsilon >0$. Let $(\varphi _\varepsilon )_i :\mathbf {\mathbb {R}}{\setminus } \{0\} \rightarrow \mathbf {\mathbb {R}}^n$ be defined in the following way

$$\begin{aligned} (\varphi _\varepsilon (t))_i =\Big \{ \begin{array}{ccc} \varphi _i (t)&{}\text{ for }&{} i \not \in I,\\ t^{-\varepsilon }&{}\text{ for }&{}i \in I,\\ \end{array} \end{aligned}$$

Observe that

$$\begin{aligned} {\text {dist}}(\varphi _\varepsilon (t),N_{\Gamma }^{-1}(0))=t. \end{aligned}$$

Indeed, let $K=\{k \in \{1, \ldots ,s \} :J_k \cap I = \emptyset \}=\{k_1, \ldots , k_r \}$. We have

$$\begin{aligned} {\text {dist}}(\varphi _\varepsilon (t),N_{\Gamma }^{-1}(0))= & {} \min _{k=1}^s \max _{j \in J_k} |\varphi _{\varepsilon ,j} (t)|\\= & {} \min _{k=1}^s \{ \max \{ \max _{j \in J_k \cap I} |\varphi _{\varepsilon ,j} (t)|, \max _{j \in J_k {\setminus } I} |\varphi _{\varepsilon ,j} (t)| \}\}\\= & {} \min _{k=1}^s \{ \max \{|t^{-\varepsilon }|, \max _{j \in J_k {\setminus } I} |\varphi _{\varepsilon ,j} (t)| \}\} = \min _{k \in K} \{ \max _{j \in J_k} |\varphi _{j} (t)| \}. \end{aligned}$$

Now, it is enough to show that

$$\begin{aligned} \min _{k \in K} \{ \max _{j \in J_k} |\varphi _{j} (t)| \}=t. \end{aligned}$$

(19)

Observe that

$$\begin{aligned} (N_{\Gamma }^I)^{-1}(0)= & {} N_{\Gamma }^{-1}(0)\cap \{x \in \mathbf {\mathbb {R}}^n :x_i=1 , \, i \in I \}\\= & {} \bigcup _{k=1}^s Z_k \cap \{x \in \mathbf {\mathbb {R}}^n :x_i=1 ,\, i \in I \}\\= & {} \bigcup _{k=1}^s \{x \in \mathbf {\mathbb {R}}^n :x_i=0 \,\, \text {for} \,\, i \in J_k , \, x_i=1 \,\, \text {for} \,\, i \in I\}\\= & {} \bigcup _{k \in K} \{x \in \mathbf {\mathbb {R}}^n :x_i=0 \,\, \text {for} \,\, i \in J_k , \, x_i=1 \,\, \text {for} \,\, i \in I\}\\= & {} \bigcup _{l=1}^r \{x \in \mathbf {\mathbb {R}}^n :x_i=0 \,\, \text {for} \,\, i \in J_{k_l} , \, x_i=1 \,\, \text {for} \,\, i \in I\}. \end{aligned}$$

Let $(j_{k_1}, \ldots ,j_{k_r})\in J_{k_1}\times \cdots \times J_{k_r}$, $J=\{j_{k_1}, \ldots ,j_{k_r}\}$ be the same as in definition of $\varphi$. It is obvious that $J_{k_l} \cap J \not = \emptyset$, $l=1, \ldots ,r$. Therefore

$$\begin{aligned} \max _{j \in J_{k_l}} |\varphi _{j} (t)| =t, \quad l=1, \ldots , r. \end{aligned}$$

This gives (19).

Let $\nu _I$ be the system of these coordinates of $\nu$ which are in I and $\nu _{I'}$ - system of the remaining ones. We have

$$\begin{aligned} |(N_\Gamma \circ \varphi _\varepsilon )(t)|= & {} \max _{\nu \in V} |\varphi _\varepsilon (t)^\nu |=\max _{\nu \in V} \{|t^{-\varepsilon }|^{\nu _I} \cdot | \varphi _{I'} (t)^{\nu _{I'}}|\} \\\le & {} |t^{-\varepsilon }|^{\max _{\nu \in V} |\nu _I|} \cdot \max _{\nu \in V} | \varphi _{I'} (t)^{\nu _{I'}}|=|t^{-\varepsilon }|^{\max _{\nu \in V} |\nu _I|} \cdot N_\Gamma (\varphi (t))\\= & {} |t^{-\varepsilon }|^{\max _{\nu \in V} |\nu _I|} \cdot |t|^{\mathcal {L}( N_\Gamma ^I)}=|t|^{\mathcal {L}( N_\Gamma ^I)-\varepsilon \max _{\nu \in V} |\nu _I|}\\= & {} {\text {dist}}(\varphi _\varepsilon (t),N_{\Gamma }^{-1}(0))^{\mathcal {L}( N_\Gamma ^I)-\varepsilon \max _{\nu \in V} |\nu _I|}. \end{aligned}$$

It can be assumed that $\varepsilon$ is such that $\mathcal {L}( N_\Gamma ^I)-\varepsilon \max _{\nu \in V} |\nu _I| >0$. Hence

$$\begin{aligned} {\mathcal {L}}_{\infty }(N_\Gamma )\ge \mathcal {L}( N_\Gamma ^I)-\varepsilon \max _{\nu \in V} |\nu _I|. \end{aligned}$$

By arbitrary choice of $\varepsilon$ and I we obtain (18). Summing up we obtain

$$\begin{aligned} {\mathcal {L}}_{\infty }(N_\Gamma ) = \max \left\{ {\mathcal {L}}(N_{\Gamma }^I): I \in \mathcal {P} \right\} . \end{aligned}$$

By Proposition 11 we have ${\mathcal {L}}(N_{\Gamma }^I)=\alpha ^I_{\max }$ and hence we get the formula (4) for $N_\Gamma$. This ends the proof. $\square$

6 Formula of Exponent of Growth

We also define the exponent of growth of fat infinity as the supremum of the exponents $l \in {\mathbb {R}}_{+}$ such that

$$\begin{aligned} |f(x)|\ge C\cdot {{\text {dist}}(x,f^{-1}(0))}^l \quad \text {for all } x \text { such that } {\text {dist}}(x,f^{-1}(0))>R, \end{aligned}$$

in some neighborhood of infinity for some $R >0$ and $C>0$. We denote it by ${\mathcal {E}}_{\infty }(f)$. In the case that such l does not exist, we put

$$\begin{aligned} {\mathcal {E}}_{\infty }(f)=-\infty . \end{aligned}$$

The second result is a formula of exponent of growth of polynomial f at infinity.

Theorem 12

Let $f : {\mathbb {R}}^n \rightarrow {\mathbb {R}}$ be a non-negative and non-degenerate polynomial. Then

$$\begin{aligned} {\mathcal {E}}_{\infty }(f)= \min \left\{ \alpha ^{\max }_{J}: J \in J_1\times \cdots \times J_s \right\} , \end{aligned}$$

where

$$\begin{aligned} \alpha ^{\max }_{J}:= \max \left\{ |\alpha | : \alpha \in V \cap {\mathbb {R}}^J \right\} . \end{aligned}$$

Proof

By Lemma 10 we have ${\mathcal {E}}_{\infty }(f)={\mathcal {E}}_{\infty }(N_\Gamma )$. Therefore it is enough to prove this formula for $N_\Gamma$. We first show that ${\mathcal {E}}_{\infty }(N_\Gamma ) \ge \min \left\{ \alpha ^{\max }_{J}: J \in J_1\times \cdots \times J_s \right\}$. Let us fix arbitrary $x \in \mathbf {\mathbb {R}}^n {\setminus } K$ such that

$$\begin{aligned} {\text {dist}}(x,N_\Gamma ^{-1}(0))=\delta >1. \end{aligned}$$

Since

$$\begin{aligned} {\text {dist}}(x,N_\Gamma ^{-1}(0))=\min _{k=1}^s \max _{j \in J_k} |x_j|, \end{aligned}$$

hence we get

$$\begin{aligned} \max _{j \in J_k} |x_j|\ge \delta \quad \text {for any} \quad k=1, \ldots ,s. \end{aligned}$$

This means that for each $k=1, \ldots ,s$ there exists $j_k \in J_k$ such that

$$\begin{aligned} |x_{j_k}|\ge \delta . \end{aligned}$$

Put $J=\{j_1, \ldots ,j_s\}$. By Lemma 3 we have $V\cap {\mathbb {R}}^J \not =\emptyset$. Let us choose $\alpha = (\alpha _1, \ldots ,\alpha _s) \in V\cap {\mathbb {R}}^J$ such that

$$\begin{aligned} |\alpha |=\alpha ^{\max }_{J}. \end{aligned}$$

Hence

$$\begin{aligned} N_\Gamma (x)= & {} \max \{ |x_{j_1}^{\alpha _j} \ldots x_{j_s}^{\alpha _s}|, \ldots \}\ge \delta ^{\alpha _j} \ldots \delta ^{\alpha _s}\\= & {} \delta ^{\alpha ^{\max }_{J}} \ge \delta ^{\min \{ \alpha ^{\max }_{J} : J \in J_1\times \cdots \times J_s \}}\\= & {} {{\text {dist}}(x,N_\Gamma ^{-1}(0))}^{\min \{ \alpha ^{\max }_{J} : J \in J_1\times \cdots \times J_s \}}. \end{aligned}$$

This means that ${\mathcal {E}}_{\infty }(N_\Gamma ) \ge \min \left\{ \alpha ^{\max }_{J}: J \in J_1\times \cdots \times J_s \right\}$.

Now, we show that ${\mathcal {E}}_{\infty }(N_\Gamma ) \le \min \left\{ \alpha ^{\max }_{J}: J \in J_1\times \cdots \times J_s \right\}$.

Let $(j_1, \ldots ,j_s)\in J_1\times \cdots \times J_s$ and let $J \subset \{1, \ldots ,n\}$ be the minimal set such that $j_k \in J$, $k=1, \ldots ,s$. Put $V_J = \mathbf {\mathbb {R}}^{J}\cap V$. By Lemma 3 we have that $V_J \not =\emptyset$. Take the following parametrization $\varphi (t) = (\varphi _1(t), \ldots ,\varphi _n(t))$, $|t|>1$, where

$$\begin{aligned} \varphi _i (t)={\left\{ \begin{array}{ll} t&{}\text {for}\quad i \in J\\ 0&{}\text {for}\quad i \not \in J, \end{array}\right. } \end{aligned}$$

for $i =1, \ldots ,n$. We have

$$\begin{aligned} N_\Gamma (\varphi (t))= \max _{\nu \in V_J} |\varphi (t)^\nu | = |t|^{\max \{ |\nu | :\nu \in V_J \}}=|t|^{ \alpha ^{\max }_{J}}={\text {dist}}(\varphi (t),N_\Gamma ^{-1}(0))^{\alpha ^{\max }_{J}}. \end{aligned}$$

Hence ${\mathcal {E}}_{\infty }(N_\Gamma )\le \alpha ^{\max }_{J}$ and by arbitrary choice of $(j_1, \ldots ,j_s)\in J_1\times \cdots \times J_s$ we have

$$\begin{aligned} {\mathcal {E}}_{\infty }(N_\Gamma ) \le \min \left\{ \alpha ^{\max }_{J}: J \in J_1\times \cdots \times J_s \right\} . \end{aligned}$$

This ends the proof.$\square$

Example 13

Let again $f(x,y,z)=x^8(y^4+z^6)$ and $V=\{(8,4,0),(8,0,6)\}$ and $J_1=\{1\}$, $J_2=\{2,3\}$.

We calculate ${\mathcal {E}}_{\infty }(f)$. Take $J \in J_1\times J_2$. Then $J=\{1,2\}$ or $J=\{1,3\}$. We calculate

$$\begin{aligned} \alpha ^{\max }_{\{1,2\}}=\max \{|\alpha | :\alpha \in V \cap \mathbf {\mathbb {R}}^{\{1,2\}}\}=\max \{|\alpha | :\alpha \in \{(8,4)\}\}=12 \end{aligned}$$

and

$$\begin{aligned} \alpha ^{\max }_{\{1,3\}}=\max \{|\alpha | :\alpha \in V \cap \mathbf {\mathbb {R}}^{\{1,3\}}\}=\max \{|\alpha | :\alpha \in \{(8,6)\}\}=14. \end{aligned}$$

Finally

$$\begin{aligned} {\mathcal {E}}_{\infty }(f)=\min \left\{ \alpha ^{\max }_{\{1,2\}},\alpha ^{\max }_{\{1,3\}}\right\} =\min \{12,14\}=12. \end{aligned}$$