Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 850215, 21 pages
doi:10.1155/2010/850215
Research Article
Jensen Type Inequalities Involving
Homogeneous Polynomials
Jia-Jin Wen1 and Zhi-Hua Zhang2
1
2
College of Mathematics and Information Science, Chengdu University, Sichuan 610106, China
Department of Mathematics, Shili Senior High School in Zixing, Chenzhou, Hunan 423400, China
Correspondence should be addressed to Jia-Jin Wen, wenjiajin623@163.com
Received 4 November 2009; Revised 25 January 2010; Accepted 8 February 2010
Academic Editor: Soo Hak Sung
Copyright q 2010 J.-J. Wen and Z.-H. Zhang. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
By means of algebraic, analytical and majorization theories, and under the proper hypotheses,
we establish several Jensen type inequalities involving γth homogeneous polynomials as follows:
m
m
m
γ
γ
1/γ m
,
k1 wk fXk /fIn ≤ f k1 wk Xk /fIn k1 wk fXk /fNn ≤ f k1 wk Xk /
1/γ
m
m
fNn , and k1 wk f∗ Xk ≤ f∗ k1 wk Xk , and display their applications.
1. Introduction
The following notation and hypotheses in 1–4 will be used throughout the paper:
x x1 , x2 , . . . , xn † ,
α α1 , α2 , . . . , αn † ,
Xk xk,1 , xk,2 , . . . , xk,n † ,
R −∞, ∞,
Rn 0, ∞n ,
w w1 , w2 , . . . , wm † ,
N {0, 1, 2, . . . , n, . . .},
Rn 0, ∞n ,
n ∈ N, n ≥ 2,
Ωn {x ∈ Rn | 0 ≤ x1 ≤ x2 ≤ · · · ≤ xn }.
Also let
⎫
⎬
αj
Pγ x λα, σ xσ j λ : Bγ × Sn → R \ {0},
⎭
⎩ α,σ∈B × S
j1
γ
n
⎫
⎧
n
⎬
⎨
αj
Pγ x λα, σ xσ j λ : Bγ × Sn −→ 0, ∞ \ {0},
⎭
⎩
j1
⎧
⎨
α,σ∈Bγ × Sn
n
1.1
2
P γ x ⎧
⎨ λα
⎩ α∈B n!
per xjαi
γ
P γ x
⎧
⎨ λα
⎩ α∈B n!
per xjαi
γ
Journal of Inequalities and Applications
⎫
⎬
λ : Bγ → R \ {0},
⎭
⎫
⎬
λ : Bγ → 0, ∞ \ {0},
⎭
1.2
where
⎡
xjαi xjαi
n×n
x1α1 x2α1 x3α1 · · · xnα1
⎤
⎥
⎢ α2 α2 α2
⎢x x x · · · xnα2 ⎥
2
3
⎥
⎢ 1
⎥ ,
⎢
⎢ ..
.
.
.
..
.. . . . .. ⎥
⎥
⎢ .
⎦
⎣
αn
αn
αn
αn
x1 x2 x3 · · · xn n×n
1.3
Bγ is a nonempty and finite subset of
α∈
n
α γ , γ ∈ 0, ∞ ,
i1 i
Rn 1.4
and the permanent of n × n matrix A ai,j n×n is given by see 2, 4
per A n
aj,σj ;
1.5
σ∈Sn j1
here, the sum extends over all elements σ of the nth symmetric group Sn .
If f ∈ Pγ x, then f is called γth homogeneous polynomial; if f ∈ P γ x, then f is
called γth homogeneous symmetric polynomial see 3.
The famous Jensen inequality can be stated as follows: if f : I → R is a convex
function, then for any x ∈ I n we have
n
1
fxk ≥ f
n k1
n
1
xk .
n k1
1.6
A large number of generalizations and applications of the inequality 1.6 had been
obtained in 1 and 5–8. An interesting generalization of 1.6 was given by Chen et al., in
8: Let Bγ ⊂ Nn and f ∈ P γ x. If Xk ∈ Rn with 1 ≤ k ≤ m and 0 ≤ X1 ≤ X2 ≤ · · · ≤ Xm , then
we have the following Jensen type inequality:
m
1
fXk ≥ f
m k1
m
1
Xk .
m k1
1.7
Journal of Inequalities and Applications
3
In this paper, by means of algebraic, analytical, and majorization theories, and under
the proper hypotheses, we will establish several Jensen type inequalities involving γth
homogeneous polynomials and display their applications.
2. Jensen Type Inequalities Involving Homogeneous Polynomials
In this section, we will use the following notation see 1, 4, 9:
Q q p
∈
N
\
q
∈
N
\
,
{0},
{0}
p
γ γ
γ †
xγ x1 , x2 , . . . , xn ,
In 1, 1, . . . , 1† ,
Nn 1, 2, . . . , n† ,
φx φx1 , φx2 , . . . , φxn † ,
Ax n
1
xi ,
n i1
2.1
Δx Δx1 , Δx2 , . . . , Δxn † x1 , x2 − x1 , x3 − x2 , . . . , xn − xn−1 † .
2.1. A Jensen Type Inequality Involving Homogeneous Polynomials
We begin a Jensen type inequality involving homogeneous polynomials as follows.
Theorem 2.1. Let f ∈ Pγ x. If Xk ∈ Rn with 1 ≤ k ≤ m and w ∈ Rm
, then
⎡ ⎤1/γ
γ
m
w
X
f
k
k1
k ⎥
⎢
k1 wk fXk ≤⎣
⎦ .
fIn fIn m
2.2
The equality holds in 2.2 if there exists t ∈ 0, ∞, such that X1 X2 · · · Xm tIn .
Lemma 2.2. (Hölder’s inequality, see [1, 10]). Let ai,k ∈ 0, ∞, qi ∈ 0, ∞ with 1 ≤ i ≤ n and
1 ≤ k ≤ m. If ni1 qi ≤ 1, then
n
n
m 1
q
ai,ki ≤
m k1 i1
i1
m
1
ai,k
m k1
qi
2.3
.
The equality in 2.3 holds if ai,1 ai,2 · · · ai,m for 1 ≤ i ≤ n.
Lemma 2.3. (Power mean inequality, see [1, 10–11]). Let x ∈ Rn , μ ∈ Rn and
γ ∈ 1, ∞, then
n
γ
μi xi ≥
i1
γ
n
μi xi .
n
i1
μi 1. If
2.4
i1
The inequality is reversed for γ ∈ 0, 1. The equality in 2.4 holds if and only if γ 1, or
x1 x2 · · · xn .
4
Journal of Inequalities and Applications
Lemma 2.4. Let gx, α n
αj
j1 xσj
g
and σ ∈ Sn .
m
γ
Xk , α
If α ∈ Bγ and Xk ∈ Rn with 1 ≤ k ≤ m, then
m
≥
gXk , α
k1
γ
.
2.5
k1
The equality in 2.5 holds if α 1, 0, . . . , 0† , or there exists t ∈ 0, ∞, such that X1 X2 · · · Xm tIn .
Proof. According to α ∈ Bγ , nj1 αj /γ 1 ≤ 1 and Lemmas 2.2-2.3, we get that
g
αj
n
m
m
1
1
γ
γ
X ,α x
m k1 k
m k1 k,σ j j1
⎡ αj /γ ⎤γ
n
m
1
γ
⎦
⎣
xk,σ j
m
j1
k1
⎤γ
n
m 1
αj
⎦
≥⎣
x
m k1 j1 k,σ j ⎡
m
1
gXk , α
m k1
From
g
2.6
γ
.
m
m
1
1
γ
γ
X ,α γ g
Xk , α ,
m k1 k
m
k1
2.7
we deduce to the inequality 2.5. Lemma 2.4 is proved.
Proof of Theorem 2.1. First of all, we assume that w Im . According to γ ∈ 1, ∞, fIn α,σ∈Bγ ×Sn λα, σ and Lemmas 2.3-2.4, we find that
f
m
k1
γ
Xk
fIn α,σ∈Bγ ×Sn
m
λα, σ
γ
g
Xk , α
fIn k1
γ
m
λα, σ gXk , α
≥
fIn k1
α,σ∈Bγ ×Sn
⎤γ
⎡
m
λα, σ ≥⎣
gXk , α⎦
fIn k1
α,σ∈Bγ ×Sn
γ
m
k1 fXk .
fIn That is, the inequality 2.2 holds.
2.8
Journal of Inequalities and Applications
5
1, we have the following cases.
Secondly, for some of wk with 1 ≤ k ≤ m satisfing wk /
1 If w ∈ Nm , then the inequality 2.2 holds from the above proof.
m
2 If w ∈ Qm
, then there exists N ∈ N \ {0} that satisfies Nw ∈ N . By the result
in 1, we obtain that
m
Nwk fXk k1
fIn ⎡ m
γ
Nwk Xk
⎢f
⎢ k1
≤⎢
⎣
fIn ⎤1/γ
⎥
⎥
⎥
⎦
⎡ m
γ
wk fXk ⎢ f
wk Xk
k1
⎢ k1
≤⎢
⇐⇒
⎣
fIn fIn m
2.9
⎤1/γ
⎥
⎥
⎥
⎦
,
which implies that inequality 2.2 is also true.
i ∞
3 If w ∈ Rm
, then there exist sequences {wk }i1 , such that
i
wk ∈ Q
1 ≤ i < ∞,
i
lim wk wk
i→∞
1 ≤ k ≤ m.
2.10
We get by the case in 2 that
m
i
k1 wk fXk fIn ⎡ ⎤1/γ
i γ
m
w
X
f
k1 k
k ⎥
⎢
≤⎣
⎦ ,
fIn 2.11
and taking i → ∞ in 2.11, we can get the inequality 2.2. The proof of Theorem 2.1 is thus
completed.
2.2. Jensen Type Inequalities Involving Difference Substitution
Exchange the ith row and jth row in nth unit matrix E, then this matrix, written Ei, j, is
called nth exchange matrix. If E1 , E2 , . . . , Ep are nth exchange matrixes, then the n × n matrix
Dn Ep Ep−1 · · · E1 E0 Δn is called nth difference matrix, and the substitution x Dn y is
difference substitution, where p ∈ N, E0 E, and
⎡
1
⎢
⎢1
⎢
Δn ⎢
⎢ ..
⎢.
⎣
0 ··· 0
⎤
⎥
1 · · · 0⎥
⎥
⎥
.. . . .. ⎥
.
.
.⎥
⎦
1 1 ··· 1
.
n×n
2.12
6
Journal of Inequalities and Applications
Let f ∈ Pγ x. If fDn y ∈ Pγ y is true for any difference matrix Dn , then
fx ≥ 0 for any x ∈ Rn see 11, and the homogeneous polynomial f is called positive
semidefinite with difference substitution.
If we let
Dn {Dn | Dn be nth difference matrix},
&
!
"
#
$ %
Pγ∗ x fx ∈ Pγ x | Bγ ⊂ Nn , f Dn y ∈ Pγ y , ∀Dn ∈ Dn ,
2.13
then Dn is a finite set and the count of elements of Dn is |Dn | n!, and γ ∈ N.
We have the following Jensen type inequality involving homogeneous polynomials
and difference substitution.
n
Theorem 2.5. Let f ∈ Pγ∗ x. If w ∈ Rm
, and Xk ∈ Ω with 1 ≤ k ≤ m, then
⎡ ⎤1/γ
γ
m
w
X
f
k
k1
w
fX
k ⎥
k
⎢
k1 k
≤⎣
⎦ .
fNn fNn m
2.14
The equality holds in 2.14 if there exists t ∈ 0, ∞, such that X1 X2 · · · Xm tIn , and
fIn 0.
Lemma 2.6. (Jensen’s inequality, see [12]). For any x ∈ Rn and γ ∈ 1, ∞, we have
n
γ
xk
≥
k1
n
γ
xk .
2.15
k1
The equality in 2.33 holds if and only if γ 1, or at least n − 1 numbers equal zero among the set
{x1 , x2 , . . . , xn }.
Lemma 2.7. If γ ∈ 1, ∞ and x ∈ Ωn , then for the difference substitution x Δn y, one has the
following double inequality:
0 ≤ yγ ≤ Δxγ .
2.16
The equality yγ Δxγ holds if and only if γ 1, or x1 x2 · · · xn−1 0, or x1 x2 · · · xn .
n
Proof. From x ∈ Ωn , it is easy to know that y Δ−1
n x Δx ∈ R . By γ ∈ 1, ∞ and Lemma 2.6,
we find that
γ
γ
γ
0 ≤ y1 x1 ≤ x1 ,
γ
γ
γ
0 ≤ y2 x2 − x1 γ ≤ x2 − x1 ,
..
.
γ
γ
γ
0 ≤ yn xn − xn−1 γ ≤ xn − xn−1 .
This shows that the double inequality 2.16 holds.
2.17
Journal of Inequalities and Applications
7
Proof of Theorem 2.5. Consider the difference substitution Xk Δn Yk . Since Xk ∈ Ωn , Yk n
∗
Δ−1
n Xk ΔXk ∈ R with 1 ≤ k ≤ m. From f ∈ Pγ x, we have that fDn y ∈ Pγ y, for all
Dn ∈ Dn . Hence,
"
#
$ %
f Δn y ∈ Pγ y .
2.18
According to Theorem 2.1, we obtain that
⎡ ⎤1/γ ⎡ ⎤1/γ
γ
γ
m
m
w
Y
w
Δ
Y
f
Δ
f
n
k
k
n
k1
k1
w
fΔ
Y
k ⎥
k ⎥
n k
⎢
⎢
k1 k
≤⎣
⎦ ⎣
⎦ .
fΔn In fΔn In fNn m
2.19
In view of Yk ∈ Rn and with Lemma 2.7, we have
γ
γ
0 ≤ Yk ≤ ΔXk ,
2.20
k 1, 2, . . . , m.
By noting that fΔn y ∈ Pγ y, it implies that f
γ
respect to Yk . Thus,
m
γ
k1
wk Δn Yk is increasing with
m
m
m
γ
γ
γ
f
wk Δn Yk ≤ f
wk Δn ΔXk f
wk Xk .
k1
k1
2.21
k1
Therefore,
m
wk fXk fNn k1
m
wk fΔn Yk fΔn In k1
⎡ ⎤1/γ
γ
m
w
Δ
Y
f
k1 k n k ⎥
⎢
≤⎣
⎦
fNn 2.22
⎡ ⎤1/γ
γ
m
w
X
f
k
k1
k ⎥
⎢
≤⎣
⎦ .
fNn This evidently completes the proof of Theorem 2.5.
As an application of Theorem 2.5, we have the following.
n
Theorem 2.8. Let fx Axγ − Aγ x, γ ∈ N and γ ≥ 2. If w ∈ Rm
, Xk ∈ Ω with 1 ≤ k ≤ m,
then the inequality 2.14 holds. The equality holds in 2.14 if there exists t ∈ 0, ∞, such that
X1 X2 · · · Xm tIn .
8
Journal of Inequalities and Applications
Proof. First of all, we prove that f ∈ Pγ∗ x. If the function φ : I → R satisfies the condition
that φ : I → R is continuous, then we have the following identity:
' '
%
$
1 φ t1 xi t2 xj 1 − t1 − t2 Ax dt1 dt2 xi − xj 2 ,
A φx − φAx 2
n 1≤i<j≤n
∇
"
#
2.23
where
φ t x ∈ I n,
d2 φ
,
dt2
!
&
∇ t1 , t2 † ∈ R2 | t1 t2 ≤ 1 .
2.24
In fact,
''
∇
%
$
φ t1 xi t2 xj 1 − t1 − t2 Ax dt1 dt2
' 1−t1
'1
dt1
0
0
1
xj − Ax
1
xj − Ax
1
xj − Ax
%
$
φ t1 xi t2 xj 1 − t1 − t2 Ax dt2
'1
' 1−t1
dt1
0
'1
0
'1
% $
%
$
φ t1 xi t2 xj 1 − t1 − t2 Ax d t1 xi t2 xj 1 − t1 − t2 Ax
0
% 1−t
$
dt1 φ t1 xi t2 xj 1 − t1 − t2 Ax 0 1
)
( $
%
φ t1 xi 1 − t1 xj − φ t1 xi 1 − t1 Ax dt1
0
φt1 xi 1 − t1 xj φt1 xi 1 − t1 Ax
−
xi − xj
xi − Ax
" #
φxi − φ xj
φxi − φAx
1
−
xj − Ax
xi − xj
xi − Ax
1
xj − Ax
1
0
φAx Ax 1
1
xi 1,
"
#
#"
φxi xi − xj xj − Ax xi − Ax " #
1
φ x
x
j
j
2.25
and
' '
∇
1≤i<j≤n
%
$
φ t1 xi t2 xj 1 − t1 − t2 Ax dt1 dt2 xi − xj 2
φAx Ax 1
xi − xj
xi 1
#
"
φxi "
#
−
Ax
−
Ax
x
x
j
i
1≤i<j≤n
φ xj
xj 1
Journal of Inequalities and Applications
1 2 1≤i,j≤n
9
φAx Ax 1
1
1
φxi −
xi 1
xj − Ax xi − Ax " #
φ xj
xj 1
⎞
φAx Ax 1
φAx Ax 1
⎟
1⎜
1
1
⎜
φxi ⎝
xi 1 −
xi 1⎟
φxi ⎠
2 1≤i,j≤n xj − Ax " #
x
−
Ax
1≤i,j≤n i
" #
φ xj
φ xj
xj 1
xj 1
⎛
⎞
φAx Ax 1
φAx Ax 1
n
n
⎟
1
1
1
φxi ⎟
φx
−
⎜
1
1
x
x
i
i
i
⎠
⎝
2 j1 xj − Ax i1 " #
i1 xi − Ax j1 " #
φ xj
φ xj
xj 1
xj 1
⎛
n
⎜
n ⎞
φAx Ax 1
φAx Ax 1
⎜
⎟
n
n
n
⎜
xi 1⎟
1
1 ⎜
n
n
⎟
φxi ⎜
n
⎟
n φxi Ax 1 −
⎟
2⎜
x
−
Ax
x
−
Ax
i1
i1 i
1 φ"x # Ax 1⎠
⎝ j1 j
#
"
j
φ x
n j1
xj 1
j
⎛
n
φAx − 1 φxi 0 0
⎜
n
⎜
i1
n
n
1⎜
n
1
⎜
⎜
φxi Ax 1
2 ⎜ j1 xj − Ax n
⎝
i1
" #
φ xj
xj 1
⎛
⎞
φAx
Ax 1
⎟
n
φxi xi 1⎟
n
⎟
−
n
⎟
⎟
x
−
Ax
"
#
i
1
i1
φ xj − φAx 0 0⎠
n j1
⎫
⎧
$ "
#
%$
%
$ "
#
%
n − A φx − φAx Ax − x
n A φx − φAx
Ax − xi ⎬
n ⎨
j
−
⎭
2 ⎩ j1
xj − Ax
xi − Ax
i1
⎫
⎧
n
n
⎬
⎨
$
"
#
%
$
"
#
%
n
A φx − φAx A φx − φAx
⎭
2 ⎩ j1
i1
$ "
#
%
n2 A φx − φAx .
2.26
That is, the identity 2.23 holds.
10
Journal of Inequalities and Applications
Setting
φ : 0, ∞ −→ R,
φt tγ
2.27
in 2.23, we have that
' '
"
#$
%γ−2
1 fx 2
γ γ − 1 t1 xi t2 xj 1 − t1 − t2 Ax
dt1 dt2 xi − xj 2 .
n 1≤i<j≤n
∇
2.28
Since f ∈ P γ x, f ∈ Pγ∗ x if and only if fΔn y ∈ Pγ y. Consider the difference
substitution x Δn y. From
"
xi − xj
#2
j
2
yk
$ %
"
#2
∈ P2 y ⇒ xi − xj ∈ P2∗ x
2.29
ki1
for arbitrary i, j : 1 ≤ i < j ≤ n, it is easy to see that f ∈ P2∗ x if γ 2. If γ ≥ 3, then
"
xi − xj
$
#2
∈ P2∗ x,
%γ−2
∈ Pγ−2
t1 xi t2 xj 1 − t1 − t2 Ax
x
''
"
#$
%γ−2
γ γ − 1 t1 xi t2 xj 1 − t1 − t2 Ax
dt1 dt2 ∈ Pγ−2
⇒
x
''
⇒
2.30
∇
∇
"
#$
%γ−2
∗
γ γ − 1 t1 xi t2 xj 1 − t1 − t2 Ax
dt1 dt2 ∈ Pγ−2
x
for arbitrary i, j : 1 ≤ i < j ≤ n. Therefore, we get that f ∈ Pγ∗ x. It follows that the inequality
2.14 holds by using Theorem 2.5. Since fIn 0, the equality holds in 2.14 if there exists
t ∈ 0, ∞, such that X1 X2 · · · Xm tIn .
The proof of Theorem 2.8 is thus completed.
Remark 2.9. Theorem 2.8 has significance in the theory of matrices. Let A ai,j n×n be an
n × n positive definite Hermitian matrix and λ1 , . . . , λn its eigenvalues, let diagx be the
diagonal matrix with the components of x x1 , x2 , . . . , xn † as its diagonal elements, and
also let λ λ1 , λ2 , . . . , λn † . Then A U diagλU∗ for some unitary matrix U where U∗ is
the conjugate transpose of U and U∗ U E, see 9, 13. If γ ∈ R, then
Aγ U diag λγ U∗ ,
tr A n
i1
ai,i n
i1
λi ,
tr Aγ n
i1
γ
λi .
2.31
Journal of Inequalities and Applications
11
Write
Dγ A 1
tr Aγ −
n
1
tr A
n
γ
Aλγ − Aγ λ fλ,
2.32
then Theorem 2.8 can be rewritten as follows, let w ∈ Rm
, γ ∈ N and γ ≥ 2. If Ak are n × n
positive definite Hermitian matrix, λAk ∈ Ωn , Ai Aj Aj Ai with 1 ≤ i, j, k ≤ n, then
⎤1/γ
γ
w
A
k
k ⎥
⎢
k1 wk Dγ Ak # ⎦ .
# ≤⎣
"
"
Dγ diagNn Dγ diagNn m
⎡
Dγ
m
k1
2.33
In fact, if A, B are n × n positive definite Hermitian matrix and AB BA, there exists a
unitary matrix U such that see 13
A U diagλA U∗ ,
B U diagλB U∗ .
2.34
Thus,
#
"
#
"
Dγ A B Dγ U diagλA λB U∗ Dγ diagλA λB fλA λB .
2.35
From Ai Aj Aj Ai with 1 ≤ i, j ≤ n, we get that
Dγ Ak fλAk ,
m
m
γ
γ
wk Ak f
wk λAk ,
Dγ
k1
#
"
Dγ diagNn fNn .
2.36
k1
According to Theorem 2.8, the inequality 2.33 holds.
Remark 2.10. Theorem 2.8 has also significance in statistics. By using the same proving
method of Theorems 2.1–2.8, we can prove the following: under the hypotheses of the
Theorem 2.8, if fx Axγ , p − Aγ x, p, then f ∈ Pγ∗ x and the inequality 2.14 also
holds, where
p ∈ Rn ,
n
"
# A x, p pi xi ,
i1
n
pi 1.
2.37
i1
Let ξ be a random variable, x ∈ Ωn , let P ξ xi pi be the probability of random events
ξ xi with 1 ≤ i ≤ n. If γ 2, then
$ "
#
"
#%
"
#
2
2
Dγ ξ "
# {Eξγ − Eγ ξ} "
# A xγ , p − Aγ x, p Dγ x, p
γ γ −1
γ γ −1
2.38
12
Journal of Inequalities and Applications
is the variance of random variable ξ. The Dγ ξ is called γth variance of random variable ξ
and Dγ ξ ≥ 0 for arbitrary γ ∈ R, where
$
"
#
"
#%
D0 ξ lim Dγ ξ 2 log A x, p − A log x, p ,
γ →0
$ "
#
"
#
"
#%
D1 ξ lim Dγ ξ 2 A x log x, p − A x, p log A x, p .
2.39
γ →1
Let ξ0 be also a random variable, P ξ0 i pi with 1 ≤ i ≤ n, and let the function fk :
0, ∞ → 0, ∞ be increasing with 1 ≤ k ≤ m. Then the inequality 2.14 can be rewritten as
follows:
⎫1/γ
m
γ
% ⎧
$
⎨ Dγ
wk fk ξ ⎬
w
D
f
ξ
k1
k
γ
k
k1
≤
,
⎭
⎩
Dγ ξ0 Dγ ξ0 m
2.40
where w ∈ Rm
, γ ∈ N, γ ≥ 2.
2.3. Applications of Jensen Type Inequalities
By 1.7 and the same proving method of Theorem 2.1, we can obtain the following result.
Corollary 2.11. Let Bγ ⊂ Nn , f ∈ P γ x. If w ∈ Rm
,
0 ≤ X1 ≤ X2 ≤ · · · ≤ Xm , then
m
wk fXk ≥ f
k1
m
k1
wk 1, Xk ∈ Rn with 1 ≤ k ≤ m and
m
wk Xk .
2.41
k1
One gives several integral analogues of 2.2 and 2.41 as follows.
s
Corollary 2.12. Let E be bounded
, and let the functions w : E → 0, ∞ and
0 closed region in R
n
g : E → R be continuous, and E w dt 1. If f ∈ Pγ x and fIn 1, then
'
1 '
21/γ
" #
wf g dt ≤ f
wg γ dt
.
E
2.42
E
If Bγ ⊂ Nn , f ∈ P γ x, and gE is an ordered set, that is,
gt1 ≤ gt2 or
gt2 ≤ gt1 2.43
Journal of Inequalities and Applications
13
for arbitrary t1 , t2 : t1 ∈ E and t2 ∈ E, then
'
" #
wf g dt ≥ f
'
E
wg dt .
2.44
E
As an application of the proof of Theorem 2.8, one has the following.
m
n
n
n
Corollary 2.13. Let w ∈ Rm
and
k1 wk 1, p ∈ R and
k1 pk 1, γ ∈ 1, 2. If Xk ∈ R ,
with 1 ≤ k ≤ m, then one has the following Jensen type inequality:
m
m
wk Dγ Xk , p ≥ Dγ
wk Xk , p ,
#
"
k1
2.45
k1
where Dγ x, p 2/γγ − 1Axγ , p − Aγ x, p.
Proof. We can suppose that p 1/nIn , γ ∈ 1, 2, and
ωi,j x, t1 , t2 t1 xi t2 xj 1 − t1 − t2 Ax.
2.46
Since 0 < 2 − γ < 1, from Lemma 2.3, we get that
m
wk ωi,j Xk , t1 , t2 2
γ−2 m
wk |xk,i − xk,j |
k1
k1
m
2−γ
≤
wk ωi,j Xk , t1 , t2 −1 m
k1
2
wk |xk,i − xk,j |
k1
⎞2
⎛ m
2−γ
γ−2
m
k1 wk ωi,j Xk , t1 , t2 ωi,j Xk , t1 , t2 |xk,i − xk,j |
2−γ
⎠
wk ωi,j Xk , t1 , t2 ⎝
m
2−γ
w
ω
,
t
,
t
X
k1
k 1 2
k1 k i,j
m
2−γ
≤
wk ωi,j Xk , t1 , t2 k1
m
m
k1
2−γ
2γ−4
wk ωi,j Xk , t1 , t2 ωi,j Xk , t1 , t2 |xk,i − xk,j |2
m
2−γ
k1 wk ωi,j Xk , t1 , t2 γ−2
wk ωi,j Xk , t1 , t2 |xk,i − xk,j |2 .
k1
By using 2.28, we find that
2.47
14
Journal of Inequalities and Applications
Dγ
m
wk Xk , p
k1
⎧
⎫
γ−2
2
' ' m
m
⎬ "
#
2 ⎨
ωi,j
wk Xk , t1 , t2
dt1 dt2
w x − xk,j
2
⎭ k1 k k,i
n 1≤i<j≤n⎩
∇
k1
⎧
⎫
γ−2
2
' ' m
m
⎬ 2 ⎨
ωi,j
≤ 2
wk Xk , t1 , t2
dt1 dt2
w |x − xk,j |
⎭ k1 k k,i
n 1≤i<j≤n⎩
∇
k1
⎧
⎫
γ−2 2
' ' m
m
⎬
2 ⎨
ωi,j
2
wk Xk , t1 , t2
wk |xk,i − xk,j | dt1 dt2
⎭
n 1≤i<j≤n⎩
∇
k1
k
⎧
⎫
2
γ−2 ' ' m
m
⎨
⎬
2
2
wk ωi,j Xk , t1 , t2 wk |xk,i − xk,j | dt1 dt2
⎭
n 1≤i<j≤n⎩
∇ k1
k1
2.48
' '
m
2 γ−2
2
≤ 2
wk ωi,j Xk , t1 , t2 |xk,i − xk,j | dt1 dt2
n 1≤i<j≤n
∇ k1
⎧
⎫
m
⎨ 2 ''
⎬
"
#
γ−2
2
wk
ω
,
t
,
t
dt
−
x
x
X
dt
k
1
2
1
2
k,i
k,j
i,j
⎩ n2 1≤i<j≤n
⎭
∇
k1
m
#
"
wk Dγ Xk , p .
k1
The proof of Corollary 2.13 is thus completed.
Corollary 2.14. If Xk ∈ Ωn with 1 ≤ k ≤ m, then
m
3
nn − 1/2
det Xk i−1
j
k1
n×n
4
⎡
5
i−1 ⎤
5
m
5
⎦
6det ⎣
≤ nn−1/2
Xk
k1
j
.
2.49
n×n
Proof. The nth Vandermonde determinant is wellknown see 14:
det xji−1
n×n
"
#
xj − xi .
1≤i<j≤n
2.50
By Theorem 2.1, for arbitrary xk,i,j ∈ 0, ∞ with 1 ≤ i < j ≤ n, 1 ≤ k ≤ m, we get that
m xk,i,j
k1 1≤i<j≤n
4
5 m
5
nn−1/2
6
≤ nn−1/2
xk,i,j
.
1≤i<j≤n k1
2.51
Journal of Inequalities and Applications
15
Letting
w Im ,
xk,i,j 7
nn−1/2
xk,j − xk,i ,
1 ≤ i < j ≤ n, 1 ≤ k ≤ m
2.52
in inequality 2.51, it implies that the inequality 2.49 holds. The proof is completed.
Example 2.15. Given N-inscribed-polygon Γk Γk Ak,1 , Ak,2 , . . . , Ak,N with 1 ≤ k ≤ m.
Defining the summation of them is an N-inscribed-polygon Γ m
k1 Γk ΓA1 , A2 , . . . , AN ,
m
and its sides lengths are given by |Ai Ai1 | k1 |Ak,i Ak,i1 | with 1 ≤ i ≤ N. Also defining
Ai Aj ⇔ i ≡ jmodN, and Ak,i Ak,j ⇔ i ≡ jmodN with 1 ≤ k ≤ m.
Wen and Zhang in 15 raised a conjecture: prove that
4
5 m m 8
5 6 Γ ≥
|Γk |,
k1 k k1
2.53
where |Γ| Area Γ is the area of the N-inscribed-polygon Γ.
Now, we prove that the inequality 2.53 holds for N 3, 4 by using Theorem 2.1.
Denote
ak,i |Ak,i Ak,i1 |,
N
1
ak,i ,
2 i1
pk N
m
1
p
ai pk ,
2 i1
k1
ai |Ai Ai1 | m
ak,i ,
k1
2.54
1 ≤ i ≤ N, 1 ≤ k ≤ m.
If N 3, we have that
8
4
5
3
5
"
#
4
pk
pk − ak,i ,
|Γk | 6
i1
4
4
5 m 5 m 3 m
5 5
"
#
4
6 Γ 6
pk
pk − ak,i .
k1 k i1 k1
k1
2.55
Setting
f ∈ P4 x,
fx 4
xi ,
n 4,
i1
w Im ,
xk,1 7
4
pk ,
xk,i
7
4 pk − ak,i ,
in Theorem 2.1, then inequality 2.2 is just 2.53.
2 ≤ i ≤ 4, 1 ≤ k ≤ m
2.56
16
Journal of Inequalities and Applications
For N 4, we get that
8
4
5 4
5
"
#
4
pk − ak,i ,
|Γk | 6
i1
2.57
4
4
5 m 5 4 m
5 5
"
#
4
6 Γ 6
pk − ak,i .
k1 k i1 k1
Taking
f ∈ P4 x,
w Im ,
xk,i fx 7
4
4
n 4,
xi ,
2.58
i1
pk − ak,i ,
1 ≤ i ≤ 4, 1 ≤ k ≤ m
in Theorem 2.1, it is clear to see that inequality 2.2 deduces to 2.53.
Remark 2.16. The following result was obtained in 15. Let Γk with 1 ≤ k ≤ m and m
k1 Γk
all be N-inscribed-polygons. If p1 − a1,i ≤ p2 − a2,i ≤ · · · ≤ pm − am,i , i 1, 2, . . . , N, then for
N 3, 4, we have
2
m
m
Γk ≤ m3 |Γk |2 .
k1 k1
2.59
This inequality can also be deduced from inequality 1.7.
3. Jensen Type Inequalities Involving Homogeneous
Symmetric Polynomials
In this section, we will also use the following notation see 4, 16:
ex ex1 , ex2 , . . . , exn † ,
Ωn∗ {x ∈ Rn | x1 ≤ x2 ≤ · · · ≤ xn },
l l
l †
∈ Bγ ,
αl α1 , α2 , . . . , αn
m
k1
Xk m
k1
X1
X2
xk,1 ,
m
k1
!
&
l l
l
p max α1 , α2 , . . . , αn ,
1≤l≤N
xk,2 , . . . ,
m
†
xk,n
k1
x1,1 x1,2
x1,n
,
,...,
x2,1 x2,2
x2,n
†
.
,
3.1
Journal of Inequalities and Applications
17
Definition 3.1. see 17, 18. Bγ is called the control ordered set if
α≺β
or β ≺ α
3.2
for arbitrary α, β : α ∈ Bγ and β ∈ Bγ .
m
The well-known Chebyshev inequality states: let a, b ∈ Rm , w ∈ Rm
, and
k1 wk 1.
If a1 ≤ a2 ≤ · · · ≤ am and b1 ≤ b2 ≤ · · · ≤ bm , then
m
wk ak bk ≥
m
k1
×
wk ak
k1
m
3.3
wk bk .
k1
The inequality is reversed for b1 ≥ b2 ≥ · · · ≥ bm .
We remark here that Wen and Wang generalized the inequality 3.3 in 4: if X1 , X2 ∈
n
Ω , and α ∈ Rn , then we have the following Chebyshev type inequality:
per X1 X2 αj i
n!
≥
per X1 αj i
×
n!
per X2 αj i
n!
3.4
.
3.1. Jensen Type Inequalities Involving Homogeneous Symmetric Polynomials
In this subsection, we first present a Jensen type inequality involving homogeneous symmetric polynomials as follows.
Theorem 3.2. Let f ∈ P γ x, fIn 1, w ∈ Nm , let Bγ be a control ordered set. If Xk ∈ Ωn∗ with
1 ≤ k ≤ m, then
m
m
wk f∗ Xk ≤ f∗
wk Xk ,
k1
3.5
k1
where f∗ : Rn → R, f∗ x log fex .
Proof. By using the same proving method of Theorem 2.1, we can suppose that w Im . If
m 1, then inequality 3.5 holds. So we just need to prove the following.
Let f ∈ P γ x, fIn 1 be Bγ is a control ordered set. If Xk ∈ Ωn with 1 ≤ k ≤ m and
m ≥ 2, then
f
m
Xk
≥
k1
m
fXk .
3.6
k1
We will verify inequality 3.6 by induction.
For m 2, we find from the inequality 3.4 that
fX1 X2 λα
α∈Bγ
n!
per X1 X2 αj i
n×n
≥
α∈Bγ
λα
per X1 αj i
n!
×
per X2 αj i
n!
.
3.7
18
Journal of Inequalities and Applications
Since the control ordered set Bγ is nonempty and finite set by using Definition 3.1, we
can suppose that
&
!
Bγ α1 , α2 , . . . , αN ,
α1 ≺ α2 ≺ · · · ≺ αN .
3.8
From Hardy’s inequality see 17, page 74, we have that
1
2
1
α
per X1 j i
n!
1
per
1
α
X2 j i
1
per
≤
N
N λ αl
n!
2
n!
By means of fIn 3.3, it is easy to obtain that
≤
1
2
2
α
per X1 j i
l1
2
α
X2 j i
n!
≤ ··· ≤
2
≤ ··· ≤
1
2
N
α
per X1 j i
,
n!
3.9
1
2
N
α
per X2 j i
.
n!
λαl 1, inequality 3.7, and Chebyshev’s inequality
1
2
l
α
per X2 j i
1
2
l
αi
per
X
1 j
×
n!
n!
1
1
2
2
⎛
⎛
l ⎞
l ⎞
αi
αi
per
per
X
X
1 j
2 j
N N ⎜
⎟ ⎜
⎟
l
⎟ × ⎜ λ αl
⎟
≥⎜
⎝ λ α
⎠
⎝
⎠
n!
n!
l1
l1
fX1 X2 ≥
l1
3.10
fX1 fX2 ,
which implies that the inequality 3.6 holds for m 2.
Assume that the inequality 3.6 is true for m q ≥ 2, that is,
f
q
≥
Xk
k1
q
fXk .
3.11
k1
For m q 1, from Xq1 ∈ Ωn and X1 , X2 , . . . , Xq ∈ Ωn , we have
q
k1
Xk ∈ Ωn . Thus,
q1 q
q
"
#
Xk f Xq1
Xk ≥ f Xq1 f
Xk
f
k1
k1
k1
q
# ≥ f Xq1
fXk "
k1
q1
3.12
fXk .
k1
The inequality 3.6 is proved by induction. The proof of Theorem 3.2 is hence completed.
As an application of the inequality 3.6, we have the following.
Journal of Inequalities and Applications
19
Theorem 3.3. Let f ∈ P γ x, let Bγ be a control ordered set, that is,
&
!
Bγ α1 , α2 , . . . , αN ,
α1 ≺ α2 ≺ · · · ≺ αN .
3.13
If X1 /X2 ∈ Ωn , X 2 ∈ Ωn , and X2 ∈ Rn , then
⎡
1
X1
per⎣
n!
X2
1
αi
j
⎤
n
fX1 ⎦≤
≤
fX2 i1
n
i1
p
x1,i
γ/p
.
p
x2,i
3.14
Proof. The right-hand inequality of 3.14 is proved in 4. Now, we will give the
demonstration of the left-hand inequality in 3.14.
We can suppose that fIn 1. By means of X1 /X2 ∈ Ωn , X2 ∈ Ωn , and X2 ∈ Rn , we
find from the inequality 3.6 that
X1
f
X2
X2
fX1 X1
X1
≥f
≥f
fX2 ⇐⇒
.
X2
fX2 X2
3.15
From
&
!
Bγ α1 , α2 , . . . , αN ,
α1 ≺ α2 ≺ · · · ≺ αN
3.16
and Hardy’s inequality see 17, page 74, we obtain that
per
X1
X2
αi
j
⎡
X1
≥ per⎣
X2
1
αi
⎤
⎦,
∀α ∈ Bγ .
3.17
j
Therefore, we deduce that
fX1 X1
≥f
fX2 X2
λα
n!
α∈Bγ
αi
j
γ
⎡
X1
1
per⎣
n!
X2
The proof of Theorem 3.3 is thus completed.
1
αi
j
1
αi
j
per
X1
X2
αi
j
n×n
⎤
1
⎡
λα
X1
per⎣
≥
n!
X2
α∈B
⎡
X1
1
per⎣
n!
X2
n×n
⎤
⎦
fIn n×n
⎤
⎦
n×n
⎦
.
3.18
20
Journal of Inequalities and Applications
3.2. Remarks
Remark 3.4. If γ ∈0, ∞, then Theorems 3.2 and 3.3 are also true.
Remark 3.5. If Bγ ⊂ Nn and 1 ≤ γ ≤ 5, then Bγ is a control ordered set.
In fact,
1, 0, . . . , 0† ≺ 1, 0, . . . , 0† ;
1, 1, 0, . . . , 0† ≺ 2, 0, . . . , 0† ;
1, 1, 1, 0, . . . , 0† ≺ 2, 1, 0, . . . , 0† ≺ 3, 0, . . . , 0† ;
1, 1, 1, 1, 0, . . . , 0† ≺ 2, 1, 1, 0, . . . , 0† ≺ 2, 2, 0, . . . , 0† ≺ 3, 1, 0, . . . , 0† ≺ 4, 0, . . . , 0† ;
1, 1, 1, 1, 1, 0, . . . , 0† ≺ 2, 1, 1, 1, 0, . . . , 0† ≺ 2, 2, 1, 0, . . . , 0† ≺ 3, 1, 1, 0, . . . , 0†
≺ 3, 2, 0, . . . , 0† ≺ 4, 1, 0, . . . , 0† ≺ 5, 0, . . . , 0† .
3.19
Remark 3.6. By using the proof of Theorem 3.2 and Remark 3.5, we know the following: if
Xk ∈ Ωn with 1 ≤ k ≤ m, m ≥ 2 and
1
fx γ
n −n
n
xk
k1
γ
−
n
γ
xk
,
γ ∈ N, 1 < γ ≤ 5,
3.20
k1
then the inequality 3.6 holds.
Remark 3.7. The inequality 3.6 is also a Chebyshev type inequality involving homogeneous
symmetric polynomials.
3.3. An Open Problem
According to Theorem 3.3, we pose the following open problem.
Conjecture 3.8. Under the hypotheses of Theorem 3.3, one has
⎡
X1
1
per⎣
n!
X2
1
αi
j
1
2
1
α
n p γ/p
per X1 j i
fX1 i1 x1,i
⎦≤
≤ n p
.
1
2≤
1
fX2 αi
x
i1
2,i
per X2 j
⎤
3.21
Journal of Inequalities and Applications
21
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