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Schur-convexity of dual form of some symmetric functions

Journal of Inequalities and Applications20132013:295

https://doi.org/10.1186/1029-242X-2013-295

Received: 16 April 2013

Accepted: 8 June 2013

Published: 17 June 2013

Abstract

By the properties of a Schur-convex function, Schur-convexity of the dual form of some symmetric functions is simply proved.

MSC:26D15, 05E05, 26B25.

Keywords

majorizationSchur-convexityinequalitysymmetric functionsdual formconvex function

1 Introduction

Throughout the article, denotes the set of real numbers, x = ( x 1 , x 2 , , x n ) denotes n-tuple (n-dimensional real vectors), the set of vectors can be written as
R n = { x = ( x 1 , , x n ) : x i R , i = 1 , , n } , R + n = { x = ( x 1 , , x n ) : x i > 0 , i = 1 , , n } .

In particular, the notations and R + denote R 1 and R + 1 , respectively.

For convenience, we introduce some definitions as follows.

Definition 1 [1, 2]

Let x = ( x 1 , , x n ) and y = ( y 1 , , y n ) R n .
  1. (i)

    x y means x i y i for all i = 1 , 2 , , n .

     
  2. (ii)

    Let Ω R n , φ : Ω R is said to be increasing if x y implies φ ( x ) φ ( y ) . φ is said to be decreasing if and only if −φ is increasing.

     

Definition 2 [1, 2]

Let x = ( x 1 , , x n ) and y = ( y 1 , , y n ) R n .
  1. (i)

    x is said to be majorized by y (in symbols x y ) if i = 1 k x [ i ] i = 1 k y [ i ] for k = 1 , 2 , , n 1 and i = 1 n x i = i = 1 n y i , where x [ 1 ] x [ n ] and y [ 1 ] y [ n ] are rearrangements of x and y in a descending order.

     
  2. (ii)

    Let Ω R n , φ : Ω R is said to be a Schur-convex function on Ω if x y on Ω implies φ ( x ) φ ( y ) . φ is said to be a Schur-concave function on Ω if and only if −φ is Schur-convex function on Ω.

     

Definition 3 [1, 2]

Let x = ( x 1 , , x n ) and y = ( y 1 , , y n ) R n .
  1. (i)

    Ω R n is said to be a convex set if x , y Ω , 0 α 1 implies α x + ( 1 α ) y = ( α x 1 + ( 1 α ) y 1 , , α x n + ( 1 α ) y n ) Ω .

     
  2. (ii)
    Let Ω R n be a convex set. A function φ : Ω R is said to be a convex function on Ω if
    φ ( α x + ( 1 α ) y ) α φ ( x ) + ( 1 α ) φ ( y )
     
for all x , y Ω and all α [ 0 , 1 ] . φ is said to be a concave function on Ω if and only if −φ is a convex function on Ω.
  1. (iii)

    Let Ω R n . A function φ : Ω R is said to be a log-convex function on Ω if the function lnφ is convex.

     

Definition 4 [1]

  1. (i)

    Ω R n is called a symmetric set, if x Ω implies P x Ω for every n × n permutation matrix P.

     
  2. (ii)

    The function φ : Ω R is called symmetric if for every permutation matrix P, φ ( P x ) = φ ( x ) for all x Ω .

     

Theorem A (Schur-convex function decision theorem [[1], p.84])

Let Ω R n be symmetric and have a nonempty interior convex set. Ω 0 is the interior of Ω. φ : Ω R is continuous on Ω and differentiable in Ω 0 . Then φ is the Schur-convex (Schur-concave) function if and only if φ is symmetric on Ω and
( x 1 x 2 ) ( φ x 1 φ x 2 ) 0 ( 0 )
(1)

holds for any x Ω 0 .

The Schur-convex functions were introduced by Schur in 1923 and have important applications in analytic inequalities, elementary quantum mechanics and quantum information theory. See [1].

In recent years, many scholars use the Schur-convex function decision theorem to determine the Schur-convexity of many symmetric functions.

Xia et al. [3] proved that the symmetric function
E k ( x 1 + x ) = 1 i 1 < < i k n j = 1 k x i j 1 + x i j , k = 1 , , n ,
(2)

is Schur-convex on R + n .

Chu et al. [4] proved that the symmetric function
E k ( x 1 x ) = 1 i 1 < < i k n j = 1 k x i j 1 x i j , k = 1 , , n ,
(3)

is Schur-convex on [ k 1 2 ( n 1 ) , 1 ) n and Schur-concave on [ 0 , k 1 2 ( n 1 ) ] n .

Xia and Chu [5] proved that the symmetric function
E k ( 1 x x ) = 1 i 1 < < i k n j = 1 k 1 x i j x i j , k = 1 , , n ,
(4)

is Schur-convex on ( 0 , 2 n k 1 2 ( n 1 ) ] n and Schur-concave on [ 2 n k 1 2 ( n 1 ) , 1 ] n .

Xia and Chu [6] also proved that the symmetric function
E k ( 1 + x 1 x ) = 1 i 1 < < i k n j = 1 k 1 + x i j 1 x i j , k = 1 , , n ,
(5)

is Schur-convex on ( 0 , 1 ) n .

Mei et al. [7] proved that the symmetric function
E k ( 1 x x ) = 1 i 1 < < i k n j = 1 k ( 1 x i j x i j ) , k = 1 , , n ,
(6)

is Schur-convex on ( 0 , 1 ) n . More results for Schur convexity of the symmetric functions, we refer the reader to [8].

In this paper, by the properties of a Schur-convex function, we study Schur-convexity of the dual form of the above symmetric functions, and we obtained the following results.

Theorem 1 The symmetric function
E k ( x 1 + x ) = 1 i 1 < < i k n j = 1 k x i j 1 + x i j , k = 1 , , n ,
(7)

is a Schur-concave function on R + n .

Theorem 2 The symmetric function
E k ( x 1 x ) = 1 i 1 < < i k n j = 1 k x i j 1 x i j , k = 1 , , n ,
(8)

is a Schur-convex function on [ 1 2 , 1 ) n .

Theorem 3 The symmetric function
E k ( 1 x x ) = 1 i 1 < < i k n j = 1 k 1 x i j x i j , k = 1 , , n ,
(9)

is a Schur-convex function on ( 0 , 1 2 ] n .

Theorem 4 The symmetric function
E k ( 1 + x 1 x ) = 1 i 1 < < i k n j = 1 k 1 + x i j 1 x i j , k = 1 , , n ,
(10)

is a Schur-convex function on ( 0 , 1 ) n .

Theorem 5 The symmetric function
E k ( 1 x x ) = 1 i 1 < < i k n j = 1 k ( 1 x i j x i j ) , k = 1 , , n ,
(11)

is a Schur-convex function on ( 0 , 5 2 ) n .

2 Lemmas

To prove the above three theorems, we need the following lemmas.

Lemma 1 ([[1], p.97], [2])

If φ is symmetric and convex (concave) on a symmetric convex set Ω, then φ is Schur-convex (Schur-concave) on Ω.

Lemma 2 [[2], p.64]

Let Ω R n , φ : Ω R + . Then logφ is Schur-convex (Schur-concave) if and only if φ is Schur-convex (Schur-concave).

Lemma 3 ([[1], p.642], [2])

Let Ω R n be an open convex set, φ : Ω R . For x , y Ω , define one variable function g ( t ) = φ ( t x + ( 1 t ) y ) on the interval ( 0 , 1 ) . Then φ is convex (concave) on Ω if and only if g is convex (concave) on [ 0 , 1 ] for all x , y Ω .

Lemma 4 Let x = ( x 1 , , x m ) and y = ( y 1 , , y m ) R + m . Then the function p ( t ) = log g ( t ) is concave on [ 0 , 1 ] , where
g ( t ) = j = 1 m t x j + ( 1 t ) y j 1 + t x j + ( 1 t ) y j .
Proof
p ( t ) = g ( t ) g ( t ) ,
where
g ( t ) = j = 1 m x j y j ( 1 + t x j + ( 1 t ) y j ) 2 , p ( t ) = g ( t ) g ( t ) ( g ( t ) ) 2 g 2 ( t ) ,
where
g ( t ) = j = 1 m 2 ( x j y j ) 2 ( 1 + t x j + ( 1 t ) y j ) 3 .
Thus,
g ( t ) g ( t ) ( g ( t ) ) 2 = ( j = 1 m 2 ( x j y j ) 2 ( 1 + t x j + ( 1 t ) y j ) 3 ) ( j = 1 m t x j + ( 1 t ) y j 1 + t x j + ( 1 t ) y j ) ( j = 1 m x j y j ( 1 + t x j + ( 1 t ) y j ) 2 ) 2 0 ,

and then p ( t ) 0 , that is, p ( t ) is concave on [ 0 , 1 ] .

The proof of Lemma 4 is completed. □

Lemma 5 Let x = ( x 1 , , x m ) and y = ( y 1 , , y m ) [ 1 2 , 1 ) m . Then the function q ( t ) = log ψ ( t ) is convex on [ 0 , 1 ] , where
ψ ( t ) = j = 1 m t x j + ( 1 t ) y j 1 t x j ( 1 t ) y j .
Proof
q ( t ) = ψ ( t ) ψ ( t ) ,
where
ψ ( t ) = j = 1 m x j y j ( 1 t x j ( 1 t ) y j ) 2 , q ( t ) = ψ ( t ) ψ ( t ) ( ψ ( t ) ) 2 ψ 2 ( t ) ,
where
ψ ( t ) = j = 1 m 2 ( x j y j ) 2 ( 1 t x j ( 1 t ) y j ) 3 .
By the Cauchy inequality, we have
ψ ( t ) ψ ( t ) ( ψ ( t ) ) 2 = ( j = 1 m 2 ( x j y j ) 2 ( 1 t x j ( 1 t ) y j ) 3 ) ( j = 1 m t x j + ( 1 t ) y j 1 t x j ( 1 t ) y j ) ( j = 1 m x j y j ( 1 t x j ( 1 t ) y j ) 2 ) 2 ( j = 1 m 2 | x j y j | ( 1 t x j ( 1 t ) y j ) 3 2 t x j + ( 1 t ) y j 1 t x j ( 1 t ) y j ) 2 ( j = 1 m x j y j ( 1 t x j ( 1 t ) y j ) 2 ) 2 = ( j = 1 m 2 | x j y j | t x j + ( 1 t ) y j ( 1 t x j ( 1 t ) y j ) 2 ) 2 ( j = 1 m x j y j ( 1 t x j ( 1 t ) y j ) 2 ) 2 .

From x j , y j [ 1 2 , 1 ) it follows that 2 t x j + ( 1 t ) y j 1 , hence ψ ( t ) ψ ( t ) ( ψ ( t ) ) 2 0 , and then q ( t ) 0 , that is, q ( t ) is convex on [ 0 , 1 ] .

The proof of Lemma 5 is completed. □

Lemma 6 Let x = ( x 1 , , x m ) and y = ( y 1 , , y m ) ( 0 , 1 2 ] m . Then the function r ( t ) = log φ ( t ) is convex on [ 0 , 1 ] , where
φ ( t ) = j = 1 m 1 t x j ( 1 t ) y j t x j + ( 1 t ) y j .
Proof
r ( t ) = φ ( t ) φ ( t ) ,
where
φ ( t ) = j = 1 m x j y j ( t x j + ( 1 t ) y j ) 2 , r ( t ) = φ ( t ) φ ( t ) ( φ ( t ) ) 2 φ 2 ( t ) ,
where
φ ( t ) = j = 1 m 2 ( x j y j ) 2 ( t x j + ( 1 t ) y j ) 3 .
By the Cauchy inequality, we have
φ ( t ) φ ( t ) ( φ ( t ) ) 2 = ( j = 1 m 2 ( x j y j ) 2 ( t x j + ( 1 t ) y j ) 3 ) ( j = 1 m 1 t x j ( 1 t ) y j t x j + ( 1 t ) y j ) ( j = 1 m x j y j ( t x j + ( 1 t ) y j ) 2 ) 2 ( j = 1 m 2 | x j y j | ( t x j + ( 1 t ) y j ) 3 2 1 t x j ( 1 t ) y j t x j + ( 1 t ) y j ) 2 ( j = 1 m x j y j ( t x j + ( 1 t ) y j ) 2 ) 2 = ( j = 1 m 2 | x j y j | 1 t x j ( 1 t ) y j ( t x j + ( 1 t ) y j ) 2 ) 2 ( j = 1 m x j y j ( t x j + ( 1 t ) y j ) 2 ) 2 .

From x j , y j ( 0 , 1 2 ] it follows that 2 1 t x j ( 1 t ) y j 1 , hence φ ( t ) φ ( t ) ( φ ( t ) ) 2 0 , and then r ( t ) 0 , that is, r ( t ) is convex on [ 0 , 1 ] .

The proof of Lemma 6 is completed. □

Lemma 7 Let x = ( x 1 , , x m ) and y = ( y 1 , , y m ) ( 0 , 1 ) m . Then the function h ( t ) = log f ( t ) is convex on [ 0 , 1 ] , where
f ( t ) = j = 1 m 1 + t x j + ( 1 t ) y j 1 t x j ( 1 t ) y j .
Proof
h ( t ) = f ( t ) f ( t ) ,
where
f ( t ) = j = 1 m 2 ( x j y j ) ( 1 t x j ( 1 t ) y j ) 2 , h ( t ) = f ( t ) f ( t ) ( f ( t ) ) 2 f 2 ( t ) ,
where
f ( t ) = j = 1 m 4 ( x j y j ) 2 ( 1 t x j ( 1 t ) y j ) 3 .
By the Cauchy inequality, we have
f ( t ) f ( t ) ( f ( t ) ) 2 = ( j = 1 m 4 ( x j y j ) 2 ( 1 t x j ( 1 t ) y j ) 3 ) ( j = 1 m 1 + t x j + ( 1 t ) y j 1 t x j ( 1 t ) y j ) ( j = 1 m 2 ( x j y j ) ( 1 t x j ( 1 t ) y j ) 2 ) 2 ( j = 1 m 2 | x j y j | ( 1 t x j ( 1 t ) y j ) 3 2 1 + t x j + ( 1 t ) y j 1 t x j ( 1 t ) y j ) 2 ( j = 1 m 2 ( x j y j ) ( 1 t x j ( 1 t ) y j ) 2 ) 2 = ( j = 1 m 2 | x j y j | 1 + t x j + ( 1 t ) y j ( 1 t x j ( 1 t ) y j ) 2 ) 2 ( j = 1 m 2 ( x j y j ) ( 1 t x j ( 1 t ) y j ) 2 ) 2 .

From x j , y j ( 0 , 1 ) it follows that 2 1 + t x j + ( 1 t ) y j 1 , hence f ( t ) f ( t ) ( f ( t ) ) 2 0 , and then h ( t ) 0 , that is, h ( t ) is convex on [ 0 , 1 ] .

The proof of Lemma 7 is completed. □

Lemma 8 Let x = ( x 1 , , x m ) and y = ( y 1 , , y m ) ( 0 , 5 2 ) m . Then the function s ( t ) = log w ( t ) is convex on [ 0 , 1 ] , where
w ( t ) = j = 1 m ( 1 t x j + ( 1 t ) y j ( t x j + ( 1 t ) y j ) ) .
Proof
s ( t ) = w ( t ) w ( t ) ,
where
w ( t ) = j = 1 m ( x j y j ) ( 1 ( t x j + ( 1 t ) y j ) 2 + 1 ) , s ( t ) = w ( t ) w ( t ) ( w ( t ) ) 2 w 2 ( t ) ,
where
w ( t ) = j = 1 m 2 ( x j y j ) 2 ( t x j + ( 1 t ) y j ) 3 .
By the Cauchy inequality, we have
w ( t ) w ( t ) ( w ( t ) ) 2 = ( j = 1 m 2 ( x j y j ) 2 ( t x j + ( 1 t ) y j ) 3 ) ( j = 1 m ( 1 t x j + ( 1 t ) y j ( t x j + ( 1 t ) y j ) ) ) ( j = 1 m ( x j y j ) ( 1 ( t x j + ( 1 t ) y j ) 2 + 1 ) ) 2 ( j = 1 m 2 | x j y j | ( t x j + ( 1 t ) y j ) 3 2 1 t x j + ( 1 t ) y j ( t x j + ( 1 t ) y j ) ) 2 ( j = 1 m ( x j y j ) ( 1 ( t x j + ( 1 t ) y j ) 2 + 1 ) ) 2 = ( j = 1 m 2 | x j y j | 1 ( t x j + ( 1 t ) y j ) 2 ( t x j + ( 1 t ) y j ) 2 ) 2 ( j = 1 m ( x j y j ) 1 + ( t x j + ( 1 t ) y j ) 2 ( t x j + ( 1 t ) y j ) 2 ) 2 .
Let u j : = t x j + ( 1 t ) y j . From x j , y j ( 0 , 5 2 ) it follows that u j 2 5 2 . Since
u j 2 5 2 ( u j 2 + 2 ) 2 5 u j 4 + 4 u j 2 1 0 2 ( 1 u j 2 ) ( 1 + u j 2 ) 2 2 1 u j 2 1 + u j 2 ,

so w ( t ) w ( t ) ( w ( t ) ) 2 0 , and then s ( t ) 0 , that is, s ( t ) is convex on [ 0 , 1 ] .

The proof of Lemma 8 is completed. □

3 Proof of main results

Proof of Theorem 4 For any 1 i 1 < < i k n , by Lemma 3 and Lemma 7, it follows that log j = 1 k 1 + x i j 1 x i j is convex on ( 0 , 1 ) k . Obviously, log j = 1 k 1 + x i j 1 x i j is also convex on ( 0 , 1 ) n , and then log E k ( 1 + x 1 x ) = 1 i 1 < < i k n log j = 1 k 1 + x i j 1 x i j is convex on ( 0 , 1 ) n . Furthermore, it is clear that log E k ( 1 + x 1 x ) is symmetric on ( 0 , 1 ) n . By Lemma 1, it follows that log E k ( 1 + x 1 x ) is Schur-convex on ( 0 , 1 ) n , and then from Lemma 2 we conclude that E k ( 1 + x 1 x ) is also Schur-convex on ( 0 , 1 ) n .

The proof of Theorem 4 is completed. □

Similar to the proof of Theorem 4, we can use Lemma 4, Lemma 5, Lemma 6 and Lemma 8 respectively to prove Theorem 1, Theorem 2, Theorem 3 and Theorem 5; therefore we omit the details of the proof.

Remark 1 Using the Schur-convex function decision theorem, Liu et al. [9] have proved Theorem 3. Xia and Chu [10] have proved that the symmetric function
E k ( 1 + x x ) = 1 i 1 < < i k n j = 1 k 1 + x i j x i j , k = 1 , , n ,
(12)

is a Schur-convex function on R + n .

The reader may wish to prove the inequality (12) by the properties of a Schur-convex function.

Declarations

Acknowledgements

The work was supported by Funding Project for Academic Human Resources Development in Institutions of Higher Learning under the Jurisdiction of Beijing Municipality (PHR (IHLB)) (PHR201108407). Thanks for the help.

Authors’ Affiliations

(1)
Department of Electronic Information, Teacher’s College, Beijing Union University, Beijing, P.R. China
(2)
Basic Courses Department, Beijing Union University, Beijing, P.R. China

References

  1. Marshall AW, Olkin I, Arnold BC: Inequalities: Theory of Majorization and Its Application. 2nd edition. Springer, New York; 2011.View ArticleMATHGoogle Scholar
  2. Wang BY: Foundations of Majorization Inequalities. Beijing Normal University Press, Beijing; 1990. (in Chinese)Google Scholar
  3. Xia WF, Wang GD, Chu YM: Schur convexity and inequalities for a class of symmetric functions. Int. J. Pure Appl. Math. 2010, 58(4):435–452.MathSciNetMATHGoogle Scholar
  4. Chu YM, Xia WF, Zhao TH: Schur convexity for a class of symmetric functions. Sci. China Math. 2010, 53(2):465–474. 10.1007/s11425-009-0188-2MathSciNetView ArticleMATHGoogle Scholar
  5. Xia WF, Chu YM: Schur convexity and Schur multiplicative convexity for a class of symmetric functions with applications. Ukr. Math. J. 2009, 61(10):1541–1555. 10.1007/s11253-010-0296-8MathSciNetView ArticleMATHGoogle Scholar
  6. Xia W-F, Chu Y-M: On Schur-convexity of some symmetric functions. J. Inequal. Appl. 2010., 2010: Article ID 543250. doi:10.1155/2010/543250Google Scholar
  7. Mei H, Bai CL, Man H: Extension of an inequality guess. J. Inn. Mong. Univ. Natl. 2006, 21(2):127–129. (in Chinese)Google Scholar
  8. Shi H-N: Theory of Majorization and Analytic Inequalities. Harbin Institute of Technology Press, Harbin; 2012. (in Chinese)Google Scholar
  9. Liu HQ, Yu Q, Zhang Y: Some properties of a class of symmetric functions and its applications. J. Hengyang Norm. Univ. 2012, 33(6):167–171. (in Chinese)MathSciNetGoogle Scholar
  10. Xia W, Chu Y: Schur convexity with respect to a class of symmetric functions and their applications. Bull. Math. Anal. Appl. 2011, 3(3):84–96. http://www.bmathaa.orgMathSciNetMATHGoogle Scholar

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© Shi and Zhang; licensee Springer 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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