Journal of Inequalities in Pure and
Applied Mathematics
DIFFERENTIAL ESTIMATE FOR n-ARY FORMS ON
CLOSED ORTHANTS
volume 5, issue 4, article 112,
2004.
VLAD TIMOFTE
École Polytechnique Fédérale de Lausanne
Département de Mathématiques
1015 Lausanne, Switzerland.
Received 09 April, 2004;
accepted 03 December, 2004.
Communicated by: A. Lupaş
EMail: vlad.timofte@epfl.ch
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
076-04
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Abstract
We prove a differential inequality for real forms of arbitrary degree, the problem
being considered on closed orthants H ⊂ Rn . A sufficient positivity criterion
is derived. Our results allow computer implementation and contain enough
information to imply the fundamental AM − GM inequality.
2000 Mathematics Subject Classification: 26D15, 26D05.
Key words: Homogeneous polynomial, Symmetric polynomial, Estimate, Positivity.
Contents
1
2
3
Introduction and Notations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
3.1
The Non-symmetric Case . . . . . . . . . . . . . . . . . . . . . . . 8
3.2
The Symmetric Case . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
References
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1.
Introduction and Notations
Positivity criteria for real n-ary d-forms (d-homogeneous polynomials f : Rn →
R) are of practical significance, but effective results exist for lower degree only.
• For quadratic forms, Sylvester’s criterion characterizes strict positivity on
Rn \ {0n }.
• For symmetric cubics (d = 3), positivity on the first orthant Rn+ is reduced
in [2] to a finite number of tests (see Theorem 1.1 below), which are the
same for all cubics.
• For symmetric quartics (d = 4) on Rn+ or Rn , and for symmetric quintics
(d = 5) on Rn+ , positivity is expressed in [11] in terms of finite test-sets
depending on the symmetric form. For quartics, explicit discriminants and
effective related algorithms (Maple worksheets) are derived in [12].
All mentioned results provide equivalent conditions and allow computer implementation. For arbitrary degree, it is of interest to find “reasonable” sufficient
conditions for positivity on closed orthants in Rn . In our entire discussion we
require that n ∈ N, n ≥ 2.
For every k ∈ {1, . . . , n}, write 0k := (0, . . . , 0) ∈ Rk , 1k := (1, . . . , 1) ∈
k
R , and set k := (1k , 0n−k ) ∈ Rn , ¯k := k −1 k . For x ∈ Rn , it is convenient
to write xk for its kth component. Therefore, we avoid denoting vectors with
symbols with lower indexes (upper indexes will be allowed) and 0k , 1k , k , ¯k
are the only exceptions to this rule.
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For every x = (x1 , . . . , xn ) ∈ Rn , set
supp(x) := {j ∈ {1, . . . , n} | xj 6= 0},
kxk :=
n
X
|xj |.
j=1
We need the following theorem, which is known in the context of even symmetric sextics (for the original statement see [2, Th. 3.7, p. 567]).
Theorem 1.1. Let g : Rn → R be a symmetric cubic. Then
g ≥ 0 on Rn+ ⇐⇒ g(k ) ≥ 0 for every k ∈ {1, . . . , n}.
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2.
Main Results
Let us consider an arbitrary d-form f : Rn → R, with d ∈ N∗ .
For ease of exposition, let us define the order relation “” on Rn by
def.
u v ⇐⇒ uj ∈ {0, vj } for every j ∈ {1, . . . , n}
⇐⇒ uj = vj for every j ∈ supp(u).
Remark 2.1. The following properties of “” are immediate.
1) 0n u for every u ∈ Rn .
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n
2) u 1n ⇐⇒ u ∈ {0, 1} .
3) For every u ∈ Rn , the set {x ∈ Rn | x u} is finite.
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4) If H is a closed orthant in Rn and if u v ∈ H, then u ∈ H.
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5) If T : Rn → Rn is a diagonal linear isomorphism or a permutation of
coordinates or a composition of finitely many such operators, then
u v ⇐⇒ T u T v.
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The following result provides a lower estimate for f in terms of its d-th
differential f (d) . As we shall see, its symmetric variant (Theorem 2.2) contains
enough information to imply the fundamental AM − GM inequality.
1
Replacing f by −f leads to the corresponding upper estimate.
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Theorem 2.1. Let H ⊂ Rn be a closed orthant. There exist u1 , . . . , ud ∈
{−1, 0, 1}n ∩ H, such that 0n 6= u1 · · · ud and
(2.1)
f (x) ≥
kxkd f (d) (0n )(u1 , . . . , ud )
·
for every x ∈ H.
d!
ku1 k · · · kud k
In particular, if f (d) satisfies the inequality
f (d) (0n )(u1 , . . . , ud ) ≥ 0
(2.2)
for all 0n 6= u1 · · · ud ∈ {−1, 0, 1}n ∩ H, then f ≥ 0 on H.
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Theorem 2.2. Assume f to be symmetric, with d ≥ 4. Then there exist in
{1, . . . , n} integers k1 ≥ · · · ≥ kd−3 ≥ k, such that
(2.3)
(2.4)
6kxkd (d−3)
f
(¯k )(¯k1 , . . . , ¯kd−3 )
d!
kxkd (d)
=
f (0n )(¯k1 , . . . , ¯kd−3 , ¯k , ¯k , ¯k ) for every x ∈ Rn+ .
d!
f (x) ≥
In particular, we have (a) ⇒ (b) ⇒ (c), where:
(a) f (d) (0n )(k1 , . . . , kd ) ≥ 0 for all k1 ≥ · · · ≥ kd ,
(b) f (d−3) (k )(k1 , . . . , kd−3 ) ≥ 0 for all k1 ≥ · · · ≥ kd−3 ≥ k,
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(c) f ≥ 0 on Rn+ .
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Computer implementation of Theorems 2.1 and 2.2 is possible. The presence
of f (d) in the above statements poses no serious computation problem, since we
have the identity
f (d) (0n )(u1 , . . . , ud )
X
(−1)card(J) f
=
J⊂{1,...,d}
!
−
X
uj
for all u1 , . . . , ud ∈ Rn .
j∈J
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3.
3.1.
Proofs
The Non-symmetric Case
We first need the following lemma:
Lemma 3.1. There exist u ∈ {0, 1}n and y ∈ Rn+ , such that kyk = 1, supp(y) =
supp(u), and
(3.1)
f (x) ≥
f 0 (y)(u)
· kxkd for every x ∈ Rn+ .
dkuk
If f is symmetric, we can find y such that y1 ≥ · · · ≥ yn . In this case, we have
u = k for some k ∈ {1, . . . , n}.
Proof. Let us first observe that inequality (3.1) is d-homogeneous in x. Consider the compact set K := {x ∈ Rn+ |kxk = 1} and choose y ∈ K, such that
f (y) = min(f |K ). Assume for simplicity that y1 ≥ · · · ≥ yn (if f is symmetric,
we can find y with this property). It follows that supp(y) = {1, . . . , k} for some
k ≤ n. We claim that
(3.2)
kdf (y) = f 0 (y)(k ).
By Euler’s theorem on homogeneous functions we get
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(3.3)
df (y) =
k
X
j=1
yj
∂f
(y) = f 0 (y)(y).
∂xj
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We need to consider two cases:
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i) If k = 1, then y = 1 and (3.3) obviously reduces to (3.2).
ii) If k ≥ 2, then y 0 := (y1 , . . . , yk ) is a global minimum for the restriction of
the map ]0, ∞[k 3 z 7→ f (z, 0n−k ) ∈ R to the subset {z ∈ ]0, ∞[k | kzk =
1}. Thus, applying the method of Lagrange multipliers shows that
∂f
∂f
∂f
(y) =
(y) = · · · =
(y) = λ
∂x1
∂x2
∂xk
(3.4)
for some λ ∈ R. Now (3.3) and (3.4) yield λ = df (y). Using this in (3.4)
leads to
k
X
∂f
0
f (y)(k ) =
(y) = kλ = kdf (y).
∂x
j
j=1
Our claim is proved. For every x ∈ Rn+ \ {0n } we have kxk−1 x ∈ K, and so
0
0
f (x)
f (y)(k )
f (y)(k )
= f (kxk−1 x) ≥ f (y) =
=
,
d
kxk
kd
dkk k
which proves (3.1) for u = k .
Proof of Theorem 2.1.
Step 1. We first consider the particular case H = Rn+ . Let us show by induction
that for every i ∈ {1, . . . , d}, there exist y i ∈ Rn+ and u1 , . . . , ui ∈ {0, 1}n , such
that ky i k = 1, supp(y i ) = supp(ui ), ui · · · u1 , and
(3.5)
f (x) ≥
(d − i)!kxkd
· f (i) (y i )(u1 , . . . , ui ) for every x ∈ Rn+ .
1
i
d! ku k · · · ku k
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For i = 1 this is clear, by Lemma 3.1. Assuming the statement to hold for some
i < d, we will prove it for i + 1. For simplicity, we shall assume that y1i ≥ · · · ≥
yni (if f is symmetric, we can find such y i at each step of our induction). Consequently, we have supp(y i ) = {1, . . . , k} and ui = k for some k ∈ {1, . . . , n}.
Let us observe that the map Rk 3 z 7→ f (i) (z, 0n−k )(u1 , . . . , ui ) ∈ R is a (d−i)form. According to Lemma 3.1, there exist y i+1 = (ζ, 0n−k ) ∈ Rk+ × Rn−k
and
+
i+1
n
i+1
i+1
i+1
u
= (v, 0n−k ) ∈ {0, 1} , such that ky k = 1, supp(y ) = supp(u ),
and
f (i+1) (y i+1 )(u1 , . . . , ui , ui+1 )
f (i) (z, 0n−k )(u1 , . . . , ui ) ≥
· kzkd−i
(d − i)kui+1 k
for every z ∈ Rk+ .
For z = (y1i , . . . , yki ) we have (z, 0n−k ) = y i , kzk = ky i k = 1, and consequently
(3.6)
f (i+1) (y i+1 )(u1 , . . . , ui+1 )
f (y )(u , . . . , u ) ≥
.
(d − i)kui+1 k
(i)
i
1
i
We also have 0n 6= ui+1 k = ui . Now combining (3.5) with (3.6) completes our induction, as well as the proof for H = Rn+ , since f (d) is constant,
f (d) (y d ) = f (d) (0n ).
Step 2. We next turn to the general case. Clearly, we can find δ1 , . . . , δn ∈
{−1, 1}, such that the linear isomorphism T : Rn → Rn , T x = (δ1 x1 , . . . , δn xn ),
maps Rn+ onto H, that is, T (Rn+ ) = H. Applying the conclusion of Step 1 to
the d-form f ◦ T yields the existence of d vectors v 1 , . . . , v d ∈ Rn+ , such that
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0n 6= v d · · · v 1 1n and
f (T x) ≥
kxkd (f ◦ T )(d) (0n )(v 1 , . . . , v d )
·
for every x ∈ Rn+ .
1
d
d!
kv k · · · kv k
Let us observe that kT xk = kxk for every x ∈ Rn , and that
(f ◦ T )(d) (0n )(v 1 , . . . , v d ) = f (d) (0n )(T v 1 , . . . , T v d ),
0n 6= T v 1 · · · T v d T 1n ∈ {−1, 1}n .
It follows that the vectors ui := T v i are all in {−1, 0, 1}n ∩ H, and that (2.1)
holds.
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3.2.
The Symmetric Case
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The following needed lemma is a slight generalization of Theorem 1.1.
Lemma 3.2. Let g : Rn → R be a symmetric polynomial with deg(g) ≤ 3.
Then for every σ > 0 we have
min{g(x) | x ∈ Rn+ , kxk = σ} = min g(σ¯k ).
1≤k≤n
Proof. Fix σ > 0 and set α := min1≤k≤n g(σ¯k ), K := {x ∈
| kxk = σ}.
Hence,
α
=
g(σ¯
)
for
some
p
∈
{1,
.
.
.
,
n}.
We
have
for
g
the
decomposition
p
P
g = 3i=0 gi , with gi symmetric i-form for every i ∈ {0, 1, 2, 3}. Now define
the symmetric cubic
h : R → R,
h=
3
X
i=0
S 3−i gi − αS 3 ,
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P
where S(x) := σ −1 nj=1 xj . Obviously, h|K = g|K − α. By Theorem 1.1, we
have h ≥ 0 on Rn+ , and so g|K ≥ α. Since σ¯p ∈ K and g(σ¯p ) = α, we get
α = minx∈K g(x).
Proof of Theorem 2.2. As in the proof of Theorem 2.1 (Step 1), by the same
induction based on Lemma 3.1, we get y ∈ Rn+ and u1 = k1 , . . . , ud−3 = kd−3 ,
such that kyk = 1, k1 ≥ · · · ≥ kd−3 , supp(y) = supp(kd−3 ) = {1, . . . , kd−3 },
and
(3.7)
f (x) ≥
6kxkd (d−3)
·f
(y)(¯k1 , . . . , ¯kd−3 ) for every x ∈ Rn+ .
d!
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Note that the above inequality corresponds to (3.5) for i = d−3. Since kyk = 1,
supp(y) = {1, . . . , kd−3 }, and the polynomial map
R
kd−3
3 z 7→ f
(d−3)
(z, 0n−kd−3 )(¯k1 , . . . , ¯kd−3 ) ∈ R
is a symmetric 3-form, applying Lemma 3.2 shows that
(3.8)
f (d−3) (y)(¯k1 , . . . , ¯kd−3 ) ≥ f (d−3) (¯k )(¯k1 , . . . , ¯kd−3 )
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for some k ≤ kd−3 . Now combining (3.7) with (3.8) yields (2.3). As the map
g : Rk → R,
g(z) = f (d−3) (z, 0n−k )(¯k1 , . . . , ¯kd−3 )
is a 3-form, we have 6g(z) = g 000 (z)(z, z, z) = g 000 (0k )(z, z, z) for every z ∈ Rk .
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This gives
f (d−3) (¯k )(¯k1 , . . . , ¯kd−3 ) = g(1k )
g 000 (0k )(1k , 1k , 1k )
=
6
f (d) (0n )(¯k1 , . . . , ¯kd−3 , ¯k , ¯k , ¯k )
=
,
6
which proves (2.4).
Example 3.1. Proof of the fundamental AM − GM inequality by verifying
condition (a) from Theorem 2.2.
P
Q
Proof. Let n ∈ N∗ and f : Rn → R, f (x) = ni=1 xni − n ni=1 xi . We shall
prove that f ≥ 0 on Rn+ . Since Theorem 1.1 shows this for n ≤ 3, assume that
n ≥ 4. For all u1 , . . . , un ∈ Rn , we have
(3.9) f
(n)
1
n
(0n )(u , . . . , u ) =
n
X
i1 ,··· ,in
∂ nf
(0n ) pi1 (u1 ) · · · pin (un ),
∂x
·
·
·
∂x
i1
in
=1
where p1 , . . . , pn : Rn → R are the standard linear projections. For ease of
exposition, let us define the map
v : Rn → {1, . . . , n},
v(x) = card({x1 , . . . , xn }),
and consider the set A := {1, . . . , n}n . For every i = (i1 , . . . , in ) ∈ A, set


 n!, v(i) = 1
∂ nf
∂ nf
−n, v(i) = n
(3.10)
:=
≡

∂xi
∂xi1 ∂xi2 · · · ∂xin

0,
1 < v(i) < n
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the last equality being easily checked. Now fix k1 ≥ · · · ≥ kn in {1, . . . , n},
and let
B := {i ∈ A | i1 ≤ k1 , i2 ≤ k2 , . . . , in ≤ kn }.
By (3.9) and (3.10) we get
(3.11) f (n) (0n )(k1 , . . . , kn ) =
X ∂ nf
i∈B
∂xi
(0n ) = n! · card(B1 ) − n · card(Bn ),
where B1 := {i ∈ B | v(i) = 1} and Bn := {i ∈ B | v(i) = n}. Since
obviously
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B1 = {n , 2n , . . . , kn n },
Bn ⊂ {i ∈ A | in ≤ kn , v(i) = n} =: E,
we have card(B1 ) = kn and card(Bn ) ≤ card(E) = kn (n − 1)!. To prove the
last equality, let us observe that every element i ∈ E can be obtained by selecting in ∈ {1, . . . , kn } (there are kn possibilities), and then choosing pairwise
distinct i1 , . . . , in−1 ∈ {1, . . . , n} \ {in } (that is, a permutation of this set). By
(3.11) we get
f (n) (0n )(k1 , . . . , kn ) ≥ 0.
The conclusion follows by Theorem 2.2.
For Pólya’s general result on strictly positive forms, we refer the reader
to [3]. Bounds for the exponent from Pólya’s theorem are given in [5, 8]. Various symmetric inequalities can be found especially in [3, 6], but also in [1,
4, 7, 9]. Some general results on symmetric inequalities can be found in [10]
and [11].
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References
[1] J.L. BRENNER, A unified treatment and extension of some means of classical analysis. I. Comparison theorems, J. Combin. Inform. System Sci., 3
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[7] R.F. MUIRHEAD, Some methods applicable to identities and inequalities of symmetric algebraic functions of n letters, Proc. Edinburgh Math.
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[8] V. POWERS AND B. REZNICK, A new bound for Pólya’s theorem with
applications to polynomials positive on polyhedra. Effective methods in
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algebraic geometry (Bath, 2000), J. Pure Appl. Algebra, 164 (2001), 221–
229.
[9] S. ROSSET, Normalized symmetric functions, Newton’s inequalities and
a new set of stronger inequalities, Amer. Math. Monthly, 96 (1989), 815–
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[10] V. TIMOFTE, On the positivity of symmetric polynomial functions. Part
I: General results, J. Math. Anal. Appl., 284 (2003), 174–190.
[11] V. TIMOFTE, On the positivity of symmetric polynomial functions. Part
II: Lattice general results and positivity criteria for degrees 4 and 5, J.
Math. Anal. Appl., (in press).
[12] V. TIMOFTE, Discriminants for positivity of real symmetric n-ary quartics (preprint).
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