Open Access

New proofs of Schur-concavity for a class of symmetric functions

Journal of Inequalities and Applications20122012:12

https://doi.org/10.1186/1029-242X-2012-12

Received: 24 May 2011

Accepted: 17 January 2012

Published: 17 January 2012

Abstract

By properties of the Schur-convex function, Schur-concavity for a class of symmetric functions is simply proved uniform.

2000 Mathematics Subject Classification: Primary 26D15; 05E05; 26B25.

Keywords

majorization Schur-concavity inequality symmetric functions concave functions

1. Introduction

Throughout the article, denotes the set of real numbers, x= (x1, x2, ..., x n ) denotes n-tuple (n-dimensional real vectors), the set of vectors can be written as
n = { x = ( x 1 , . . . , x n ) : x i , i = 1 , . . . , n } , + n = { x = ( x 1 , . . . , x n ) : x i > 0 , i = 1 , . . . , n } .

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

For convenience, we introduce some definitions as follows.

Definition 1. [1, 2] Let x =(x1, ..., x n ) and y =(y1, ..., y n ) n .

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

(ii) Let Ω n , φ: Ω → is said to be increasing if xy implies φ(x) ≥ φ(y ). φ is said to be decreasing if and only if is increasing.

Definition 2. [1, 2] Let x =(x1, ..., x n ) and y = (y1, ..., y n ) n .

(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.

(ii) Let Ω n , φ: Ω → 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 is Schur-convex function on Ω.

Definition 3. [1, 2] Let x = (x1, ..., x n ) and y= (y1, ..., y n ) n .

(i) Ω n is said to be a convex set if x, y Ω, 0 ≤ α ≤ 1 implies α x+ (1- α)y =(αx1 + (1 - α)y1, ...,αx n + (1- α)y n ) Ω.

(ii) Let Ω n be convex set. A function φ: Ω → 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 is convex function on Ω.

Recall that the following so-called Schur's condition is very useful for determining whether or not a given function is Schur-convex or Schur-concave.

Theorem A. [[1], p. 5] Let Ω n is symmetric and has a nonempty interior convex set. Ω0is the interior of Ω. φ: Ω → is continuous on Ω and differentiable in Ω0. 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 Ω0.

In recent years, by using Theorem A, many researchers have studied the Schur-convexity of some of symmetric functions.

Chu et al. [3] defined the following symmetric functions
F n ( x , k ) = 1 i 1 < . . . < i k n j = 1 k x i j j = 1 k ( 1 + x i j ) , k = 1 , . . . , n ,
(2)

and established the following results by using Theorem A.

Theorem B. For k = 1,..., n, F n (x , k) is an Schur-concave function on + n .

Jiang [4] are discussed the following symmetric functions
H k * ( x ) = 1 i 1 < . . . < i k n j = 1 k x i j 1 / k , k = 1 , . . . , n ,
(3)

and established the following results by using Theorem A.

Theorem C. For k = 1 , . . . , n , H k * ( x ) is an Schur-concave function on + n .

Xia and Chu [5] investigated the following symmetric functions
ϕ n ( x , k ) = 1 i 1 < . . . < i k n j = 1 k x i j 1 + x i j , k = 1 , . . . , n ,
(4)

and established the following results by using Theorem A.

Theorem D. For k = 1,..., n, F n (x , k) is an Schur-concave function on + n .

In this note, by properties of the Schur-convex function, we simply prove Theorems B, C and D uniform.

2. New proofs three theorems

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

Lemma 1. [[1], p. 67], [2]If φ is symmetric and convex (concave) on symmetric convex set Ω, then φ is Schur-convex (Schur-concave) on Ω.

Lemma 2. [[1], p. 73],[2]Let Ω n , φ: Ω → +. Then lnφ is Schur-convex (Schur-concave) if and only if φ is Schur-convex (Schur-concave).

Lemma 3. [[1], p. 446], [2]Let Ω n be open convex set, φ : Ω → . For x, y Ω, defined one variable function g(t) = φ (t x + (1 - t)y ) on 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= (x1,..., x m ) and y = (y1, ..., y m ) m . Then the following functions are concave on (0,1).

(i) f ( t ) = ln j = 1 m ( t x j + ( 1 - t ) y j ) - ln j = 1 m ( 1 + t x j + ( 1 - t ) y j ) ,

(ii) g ( t ) = ln j = 1 m ( t x j + ( 1 - t ) y j ) 1 / m ,

(iii) h ( t ) = 1 m ln ψ ( t ) , where
ψ ( t ) = j = 1 m t x j + ( 1 - t ) y j 1 + t x j + ( 1 - t ) y j .
Proof. (i) Directly calculating yields
f ( t ) = j = 1 m ( x j - y j ) 1 t x j + ( 1 - t ) y j - 1 1 + t x j + ( 1 - t ) y j
and
f ( t ) = - j = 1 m ( x j - y j ) 2 1 ( t x j + ( 1 - t ) y j ) 2 - 1 ( 1 + t x j + ( 1 - t ) y j ) 2 = - j = 1 m ( x j - y j ) 2 1 + 2 t x j + 2 ( 1 - t ) y j ( t x j + ( 1 - t ) y j ) 2 ( 1 + t x j + ( 1 - t ) y j ) 2 .

Since f''(t) ≤ 0, f(t) is concave on (0,1).

(ii) Directly calculating yields
g ( t ) = 1 m j = 1 m ( x j - y j ) 1 m - 1 j = 1 m ( t x j + ( 1 - t ) y j ) 1 / m
and
g ( t ) = - 1 m j = 1 m ( x j - y j ) 1 m - 1 2 j = 1 m ( t x j + ( 1 - t ) y j ) 2 / m .

Since g''(t) ≤ 0, f(t) is concave on (0,1)

(iii) By computing,
h ( t ) = 1 m ψ ( t ) ψ ( t ) , h ( t ) = 1 m ψ ( t ) ψ ( t ) - ( ψ ( t ) ) 2 ψ 2 ( t ) ,
where
ψ ( t ) = j = 1 m x j - y j ( 1 + t x j + ( 1 - t ) y j ) 2
and
ψ ( t ) = - j = 1 m 2 x j - y j 2 ( 1 + t x j + ( 1 - t ) y j ) 3 .
Thus,
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 0 ,

and then h'' (t) ≤ 0, so f(t) is concave on (0,1).

The proof of Lemma 4 is completed.

Proof of Theorem A: For any 1 ≤ i1 < · · · < i k n, by Lemma 3 and Lemma 4(i), it follows that ln j = 1 k x i j - ln j = 1 k ( 1 + x i j ) is concave on + n , and then ln F n ( x , k ) = 1 i 1 < < i k n ln j = 1 k x i j - ln j = 1 k ( 1 + x i j ) is concave on + n . Furthermore, it is clear that ln F n (x, k) is symmetric on + n , by Lemma 1, it follows that ln F n ( x, k) is concave on + n , and then from Lemma 2 we conclude that F n (x , k) is also concave on + n .

The proof of Theorem A is completed.

Similar to the proof of Theorem A, by Lemma 4 (ii) and Lemma 4 (iii), we can prove Theorems B and C, respectively. Omitted detailed process.

Declarations

Acknowledgements

Shi was supported in part by the Scientific Research Common Program of Beijing Municipal Commission of Education (KM201111417006). This article was typeset by using A M S - L A T E X .

Authors’ Affiliations

(1)
Department of Electronic Information, Teacher's College, Beijing Union University

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Copyright

© Shi et al; licensee Springer. 2012

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