INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES
WARSZAWA 1998
GROWTH AT INFINITY OF A POLYNOMIAL WITH A COMPACT ZERO SET
J A N U S Z G W O ´ Z D Z I E W I C Z Kielce University of Technology
Al. Tysi¸ aclecia Pa´ nstwa Polskiego 7, 25-314 Kielce, Poland E-mail: matjg@tu.kielce.pl
1. Introduction. In the article we give the explicit bound for the growth at infinity of a polynomial with a compact set of zeros. Our aim is to prove the following theorem:
Theorem 1. Let F ∈ R[X
1, . . . , X
n] be a polynomial of degree d > 2 such that the set F
−1(0) is compact. Then there exist constants c, R > 0 such that
|F (x)| ≥ c|x|
d−(d−1)nfor all |x| > R.
Recall that we have a similar estimation in the complex case. Consider a polynomial map H : C
n→ C
nof degree d such that H
−1(0) is finite. Then, by Koll´ ar’s theorem,
|H(z)| ≥ const.|z|
d−dnfor |z| 1 (see [Ko]). Our theorem is a real counterpart of this inequality.
2. Two lemmas. The following lemmas will be used in the proof of the main theo- rem.
Lemma 1. Let G : R
n→ R be a polynomial of positive degree d. Then there exists a linear automorphism L : R
n→ R
nsuch that the polynomial F = G ◦ L satisfies the following conditions:
(i) All partial derivatives of F are of degree d − 1.
(ii) The sets Γ
i= {x ∈ R
n| ∂F/∂X
1(x) = . . . = ∂F/∂X
i−1(x) = ∂F/∂X
i+1(x) = . . . = ∂F/∂X
n(x) = 0, ∂F/∂X
i(x) 6= 0} (1 ≤ i ≤ n) are one-dimensional submanifolds of R
nwhenever they are not empty,
(iii) For every x ∈ Γ
i(1 ≤ i ≤ n) the differentials d
x(∂F/∂X
1), . . . , d
x(∂F/∂X
i−1), d
x(∂F/∂X
i+1), . . . , d
x(∂F/∂X
n) are linearly independent.
Research supported by KBN grant 8 P03A 07908.
1991 Mathematics Subject Classification: 14P10.
Received by the editors: December 3, 1996; in the revised form: June 2, 1998.
The paper is in final form and no version of it will be published elsewhere.
[123]
P r o o f. Let GL(n) be the set of linear automorphisms of R
n. We claim that {L ∈ GL(n) | deg ∂G ◦ L
∂X
1= . . . = deg ∂G ◦ L
∂X
n= d − 1}
is a dense subset of GL(n). Let G
d(X
1, . . . , X
n) be the leading form of the polyno- mial G that is the homogeneous polynomial of degree d for which deg(G − G
d) < d.
Consider a substitution G
d◦ L where L(X
1, . . . , X
n) = ( P
ni=1
l
i1X
i, . . . , P
ni=1
l
niX
i). We have (G
d◦ L)(X
1, . . . , X
n) = G
d( P
ni=1
l
i1X
i, . . . , P
ni=1
l
inX
i) = G
d(l
11, . . . , l
1n)X
1d+ . . . . . . + G
d(l
n1, . . . , l
nn)X
nd+ other monomials. If G
d(l
11, . . . , l
1n) 6= 0, . . . , G
d(l
n1, . . . , l
nn) 6= 0, then all partial derivatives of G ◦ L are of degree d − 1. Since the set {L ∈ GL(n) | G
d(l
11, . . . , l
1n) 6= 0, . . . , G
d(l
1n, . . . , l
nn) 6= 0} is a complement of a proper algebraic set, it is open and dense in GL(n). This proves the claim.
For any x = (x
1, . . . , x
n) from R
n\ {0} we denote by [x] the corresponding point [x
1, . . . , x
n] of the projective space RP
n−1. Consider the map
[grad G] : R
n\ (grad G)
−1(0) → RP
n−1.
From the semialgebraic version of Sard’s lemma (see [BR], page 82) it follows that the set of regular values of this map contains an open subset U ⊂ RP
n−1. The set V = {(v
1, . . . , v
n) ∈ R
n× . . . × R
n| det(v
ji) 6= 0, [v
i] ∈ U for i = 1, . . . , n} is an open subset of R
n×. . .×R
n. Each n-tuple v = (v
1, . . . , v
n) from this set yields a linear automorphism A
v: R
n→ R
n, A
v(x) = (hv
1, xi, . . . , hv
n, xi). Hence the set {A
v∈ GL(n) | v ∈ V } is open in GL(n). Since GL(n) 3 A → A
−1∈ GL(n) is an open map, {A
−1v∈ GL(n) | v ∈ V } is also an open subset of GL(n). Thus, there exists v = (v
1, . . . , v
n) ∈ V such that the automorphism L = A
−1vsatisfies (i).
Let us define the polynomial F = G ◦ L. Since G = F ◦ A
v, grad G = A
Tv◦ grad F ◦ A
v, where A
Tvis the adjoint of A
v. From this equation it follows that for any w ∈ R
n\ {0}, [w] is a regular value of [grad F ] if and only if [A
Tv(w)] is a regular value of [grad G]. Let e
1= (1, . . . , 0), . . . , e
n= (0, . . . , 1) form the standard basis of R
n. Since A
Tv(e
i) = v
ifor i = 1, . . . , n, we conclude that [e
1], . . . , [e
n] are regular values of [grad F ]. Applying the implicit function theorem to the map [grad F ] we see that each of the sets Γ
i= [grad F ]
−1([e
i]) (1 ≤ i ≤ n) is either a one-dimensional submanifold of R
nor is empty.
This proves (ii).
We prove the third part of the lemma for Γ
n. To simplify the notation we write
∂
iF for ∂F/∂X
i. Because [e
n] is a regular value of [grad F ], 0 ∈ R
n−1is a regular value of the map ψ : R
n\ (∂
nF )
−1(0) → R
n−1, ψ = (∂
1F/∂
nF, . . . , ∂
n−1F/∂
nF ) that is, the map [grad F ] written in the coordinates {[x
1, . . . , x
n] ∈ RP
n−1| x
n6= 0} 3 [x
1, . . . , x
n] → (x
1/x
n, . . . , x
n−1/x
n) ∈ R
n−1. Therefore, for every x ∈ Γ
nthe differen- tials d
x(∂
1F/∂
nF ), . . . , d
x(∂
n−1F/∂
nF ) are linearly independent. On the other hand, for x ∈ Γ
nand i = 1, . . . , n − 1 we have d
x(∂
iF/∂
nF ) = (1/∂
nF )d
x(∂
iF ), therefore the differentials d
x(∂
1F ), . . . , d
x(∂
n−1F ) are also linearly independent. The proof for Γ
i, i 6= n is similar.
Further, we denote by |x| the supremum norm |x| = max{|x
1|, . . . , |x
n|} for x = (x
1, . . . , x
n). We will also use the following convention: Using notation |x| 1 we mean that the corresponding condition is satisfied for |x| > R, where R is sufficiently large.
Lemma 2. Let F ∈ R[X
1, . . . , X
n] be a polynomial with a compact set of zeros and
let K = {x ∈ R
n| ∀y ∈ R
n|y| = |x| ⇒ |F (y)| ≥ |F (x)|}. If A ⊂ K is an unbounded
semialgebraic set , then the following conditions are equivalent :
(i) |F (x)| ≥ c|x|
αfor |x| 1, (ii) |F (x)| ≥ c|x|
αfor |x| 1, x ∈ A.
P r o o f. The implication (i) ⇒ (ii) is obvious. Assume that (ii) is true. Since {|x|
x ∈ A} is an unbounded semialgebraic subset of R
+, there exists a constant R > 0 such that (R, ∞) ⊂ {|x|
x ∈ A}. By (ii) we can choose R sufficiently large so that
|F (x)| ≥ c|x|
αfor |x| ≥ R, x ∈ A. Let y ∈ R
nbe an arbitrary point with |y| > R.
Then there exists x ∈ A such that |x| = |y|. By (ii) and the definition of K we get
|F (y)| ≥ |F (x)| ≥ c|x|
α= c|y|
αwhich ends the proof.
3. Proof of Theorem 1. The proof proceeds by induction on the number of vari- ables. For polynomials in one variable the theorem is obvious. Assume that the theorem holds for polynomials in n − 1 variables. We shall check that it is true for polynomials in n variables.
We shall perform some reductions:
If the theorem is true for a polynomial F , then it holds also for F ◦ L, where L : R
n→ R
nis a linear automorphism. Therefore, we can assume that F satisfies the conditions (i), (ii) and (iii) of Lemma 1.
The set F
−1(0) is bounded. Hence F (x) 6= 0 for all |x| > R, where R is sufficiently large. Since for n ≥ 2 the set {x ∈ R
n|x| > R} is connected, a sign of F restricted to {x ∈ R
n|x| > R} does not change. Without loss of generality we can assume that F (x) > 0 for |x| > R.
Let
K = {x ∈ R
n| ∀y ∈ R
n|y| = |x| ⇒ |F (y)| ≥ |F (x)|}.
First, we prove the theorem under the additional assumption that K ∩ (grad F )
−1(0) is unbounded. Let A be an unbounded connected component of this set. Since grad F (x) = 0 for x ∈ A, we conclude that F |
A= c with some c > 0 (see [BR], Theorem 2.5.1). By Lemma 2 we get |F (x)| ≥ c|x|
0for |x| 1 which ends the proof in this case.
Hence we may assume throughout the rest of the proof that K ∩ (grad F )
−1(0) is bounded.
Let us define
A
i= {x ∈ R
n| |x
k| < |x
i| for k ∈ {1, . . . , n} \ {i}}, B
i,j= {x ∈ R
n| x
i= x
j, |x
k| ≤ |x
i| for k = 1, . . . , n}, C
i,j= {x ∈ R
n| x
i= −x
j, |x
k| ≤ |x
i| for k = 1, . . . , n}.
Since R
n= S A
i∪ S B
i,j∪ S C
i,j, at least one of the sets K ∩ S A
i, K ∩ S B
i,j, K ∩ S C
i,jis unbounded. Let us consider three cases:
Case 1: K ∩S B
i,jis unbounded. Then at least one of the sets K ∩B
i,j(1 ≤ i < j ≤ n) is unbounded. Without loss of generality we can assume that this is the set K ∩ B
n−1,n. Consider the polynomial ˜ F (X
1, . . . , X
n−1) = F (X
1, . . . , X
n−1, X
n−1) of degree ˜ d ≤ d.
By the inductive assumption we have | ˜ F (˜ x)| ≥ c|˜ x|
d−( ˜˜ d−1)n−1for ˜ x ∈ R
n−1, |˜ x| 1.
If we take any x ∈ B
n−1,n, x = (x
1, . . . , x
n−1, x
n−1) and if we set ˜ x = (x
1, . . . , x
n−1), then |˜ x| = |x| and ˜ F (˜ x) = F (x). Hence |F (x)| ≥ c|x|
d−( ˜˜ d−1)n−1for |x| 1, x ∈ B
n−1,n. By Lemma 2 and by the inequality ˜ d−( ˜ d−1)
n−1≥ d−(d−1)
nwe get |F (x)| ≥ c|x|
d−(d−1)nfor |x| 1.
Case 2: K ∩ S C
i,jis unbounded. The proof is analogous.
Case 3: K ∩ S A
iis unbounded. Then at least one of the sets K ∩ A
i(1 ≤ i ≤ n) is unbounded. Without loss of generality we can assume that this is K ∩ A
n.
Take R > 0 large enough so that F (x) > 0 for |x| > R and let y = (y
1, . . . , y
n) be an arbitrary point in K ∩ A
nwith |y| > R. Consider a function f (x
1, . . . , x
n−1) = F (x
1, . . . , x
n−1, y
n) defined for |x
i| < |y
n| (1 ≤ i < n). Taking into account two points, y = (y
1, . . . , y
n−1, y
n) and x = (x
1, . . . , x
n−1, y
n), where |x
i| < |y
n| (1 ≤ i < n), we see that |x| = |y|, therefore F (x) ≥ F (y). Hence the point (y
1, . . . , y
n−1) is a local minimum of f . Thus ∂F/∂X
1(y) = . . . = ∂F/∂X
n−1(y) = 0.
Summarizing, we see that for all y ∈ K ∩ A
n, |y| 1 we have ∂F/∂X
1(y) = . . . =
∂F/∂X
n−1(y) = 0, ∂F/∂X
n(y) 6= 0. Moreover, from Lemma 1 it follows that K ∩ A
nis a one-dimensional semialgebraic manifold in a neighborhood of infinity. We want to find a parametrization of a branch at infinity of this set. To that end we employ complex algebraic geometry.
Define H
1= ∂F/∂X
1, . . . , H
n−1= ∂F/∂X
n−1and let C = {z ∈ C
n| H
1(z) = . . . . . . = H
n−1(z) = 0}. Decompose C to the union of irreducible algebraic components C = C
1∪ . . . ∪ C
s. Treating R
nas a subset of C
nwe see that K ∩ A
n∩ C is unbounded.
Hence there exists a component C
isuch that K ∩ A
n∩ C
iis unbounded. For simplicity put Γ = C
i.
We will check that dim
CΓ = 1. By Lemma 1 there exists x ∈ K ∩ A
n∩ Γ for which the differentials d
xH
1, . . . , d
xH
n−1are linearly independent. Therefore, dim
CΓ ≤ n−rank(Γ, x) ≤ n−rank(d
xH
1, . . . , d
xH
n−1) = 1 (see [BR], pages 122–135). Furthermore, Γ is unbounded, so dim
CΓ = 1.(
1)
Next, we will check that deg Γ ≤ (d − 1)
n−1. Let us recall an invariant δ of algebraic sets introduced in Lojasiewicz’s book ([ Lo] pages 419–420): Let W = W
1∪ . . . ∪ W
sbe a decomposition of an algebraic set W to irreducible components. Then, by definition δ(W ) = P
si=1
deg W
i. We will use the inequality δ(W ∩ V ) ≤ δ(W )δ(V ). Applying this property to the set C we see that deg Γ ≤ δ(C) = δ({H
1= 0} ∩ . . . ∩ {H
n−1= 0}) ≤ Q
n−1i=1
δ({H
i= 0}) ≤ (d − 1)
n−1.
Further, we will consider C
nas a affine part of the projective space CP
n. We will use the natural identification between (x
1, . . . , x
n) ∈ C
nand [1, x
1, . . . , x
n] ∈ CP
n. With the use of this identification we can treat K, A
nand Γ as subsets of CP
n.
Since K ∩ A
n∩ Γ is an unbounded set and CP
nis compact, there exists a point a in the hyperplane at infinity {[x
0, . . . , x
n] ∈ CP
n| x
0= 0} such that a ∈ cl(K ∩ A
n∩ Γ).
The homogeneous coordinates of a can be chosen such that a = [0, a
1, . . . , a
n−1, 1].
Indeed, for all x ∈ A
nwe have |x
i| < |x
n| for 1 ≤ i < n. Since a ∈ cl(A
n), |a
i| ≤ |a
n| for 1 ≤ i < n. Therefore, the last coordinate a
ndoes not vanish and by homogeneity we can assume that a
n= 1.
Let ¯ Γ be the projective closure of the curve Γ. Since a ∈ ¯ Γ, according to [ Lo]
(pages 173–176) there exists a finite sequence of injective holomorphic parametrizations γ
i: (D, 0) → (¯ Γ, a) (1 ≤ i ≤ l), where D = {t ∈ C | |t| < δ}, such that the curve ¯ Γ is the union γ
1(D) ∪ . . . ∪ γ
l(D) in some neighborhood of a. These parametrizations are of the form
γ
i(t) = [t
di, γ
i,1(t), . . . , γ
i,n−1(t), 1].
Furthermore, we can assume that the real branches of ¯ Γ are parametrized such that (t
di, γ
i,1(t), . . . , γ
i,n−1(t)) ∈ R
nif and only if t ∈ R. This can be done by substituting
(
1) If dim
CΓ = 0, then Γ would consist of one point.
γ
i(ξ
it), where ξ
iis an appropriate d
i-th root of unity and by shrinking δ if necessary (see [Mi] or [Du] for the details).
Let H = H(X
0, . . . , X
n) be the homogenization of the polynomial ∂F/∂X
n. Recall that it means that H is a homogeneous polynomial of degree deg H = deg ∂F/∂X
nsuch that H(1, X
1, . . . , X
n) = ∂F/∂X
n(X
1, . . . , X
n). We can calculate the intersection multiplicity of the curve ¯ Γ and the hypersurface {H = 0} at a using the formula
ι
a(¯ Γ, {H = 0}) =
l
X
i=1