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

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.

1. Introduction

Throughout the article, ℝ 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

$$\begin{array}{c}{ℝ}^n=\left\{x=\left(x_1,...,x_n\right):x_i\in ℝ,i=1,...,n\right\},\\ {ℝ}_{+}^n=\left\{x=\left(x_1,...,x_n\right):x_i>0,i=1,...,n\right\}.\end{array}$$

In particular, the notations ℝ and ℝ₊ denote ℝ¹ and ${ℝ}_{+}^1$ respectively.

For convenience, we introduce some definitions as follows.

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

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

(ii) Let Ω ⊂ ℝⁿ, 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 =(x₁, ..., x _n ) and y = (y₁, ..., y _n ) ∈ ℝⁿ.

(i) x is said to be majorized by y(in symbols x≺ y) if ${\sum }_{i=1}^k{x}_{\left[i\right]}\le {\sum }_{i=1}^k{y}_{\left[i\right]}$ for k = 1, 2,..., n - 1 and ${\sum }_{i=1}^n{x}_i={\sum }_{i=1}^n{y}_i$, where x_[1] ≥ · · · ≥ x_[n]and y_[1] ≥ · · · ≥ y_[n]are rearrangements of x and y in a descending order.

(ii) Let Ω ⊂ ℝⁿ, 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 = (x₁, ..., x _n ) and y= (y₁, ..., y _n ) ∈ ℝⁿ.

(i) Ω ⊆ ℝⁿis 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 ⊂ ℝⁿbe convex set. A function φ: Ω → ℝ is said to be a convex function on Ω if

$$\phi \left(\alpha x+\left(1-\alpha \right)y\right)\le \alpha \phi \left(x\right)+\left(1-\alpha \right)\phi \left(y\right)$$

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 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 ⊂ ℝⁿis symmetric and has a nonempty interior convex set. Ω⁰is the interior of Ω. φ: Ω → ℝ 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

$$\left(x_1-x_2\right)\left(\frac{\partial \phi }{\partial x_1}-\frac{\partial \phi }{\partial x_2}\right)\ge 0\left(\le 0\right)$$

(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

$$F_n\left(x,k\right)=\prod _{1\le i_1<...<i_k\le n}\frac{{\sum }_{j=1}^k{x}_{i_j}}{{\sum }_{j=1}^k\left(1+x_{i_j}\right)},k=1,...,n,$$

(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${ℝ}_{+}^n$.

Jiang [4] are discussed the following symmetric functions

$$H_k^{*}\left(x\right)=\prod _{1\le i_1<...<i_k\le n}{\sum }_{j=1}^k{x}_{i_j}^{1/k},k=1,...,n,$$

(3)

and established the following results by using Theorem A.

Theorem C. For$k=1,...,n,H_k^{*}\left(x\right)$is an Schur-concave function on${ℝ}_{+}^n$.

Xia and Chu [5] investigated the following symmetric functions

$${\varphi }_n\left(x,k\right)=\prod _{1\le i_1<...<i_k\le n}{\sum }_{j=1}^k\frac{x_{i_j}}{1+x_{i_j}},k=1,...,n,$$

(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${ℝ}_{+}^n$.

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 ⊂ ℝⁿ, φ: Ω → ℝ₊. 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 ⊂ ℝⁿbe open convex set, φ : Ω → ℝ. 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 ) ∈ ℝ^m. Then the following functions are concave on (0,1).

(i) $f\left(t\right)=\text{ln}{\sum }_{j=1}^m\left(t{x}_j+\left(1-t\right)y_j\right)-\text{ln}{\sum }_{j=1}^m\left(1+t{x}_j+\left(1-t\right)y_j\right)$,

(ii) $g\left(t\right)=\text{ln}{\sum }_{j=1}^m{\left(t{x}_j+\left(1-t\right)y_j\right)}^{1/m}$,

(iii) $h\left(t\right)=\frac{1}{m}\text{ln}\psi \left(t\right)$, where

$$\psi \left(t\right)=\sum _{j=1}^m\frac{t{x}_j+\left(1-t\right)y_j}{1+t{x}_j+\left(1-t\right)y_j}.$$

Proof. (i) Directly calculating yields

$$f^{\prime }\left(t\right)=\sum _{j=1}^m\left(x_j-y_j\right)\left[\frac{1}{t{x}_j+\left(1-t\right)y_j}-\frac{1}{1+t{x}_j+\left(1-t\right)y_j}\right]$$

and

$$\begin{array}{c}{f}^{″}\left(t\right)=-\sum _{j=1}^m{\left(x_j-y_j\right)}^2\left[\frac{1}{{\left(t{x}_j+\left(1-t\right)y_j\right)}^2}-\frac{1}{{\left(1+t{x}_j+\left(1-t\right)y_j\right)}^2}\right]\\ =-\sum _{j=1}^m{\left(x_j-y_j\right)}^2\frac{1+2t{x}_j+2\left(1-t\right)y_j}{{\left(t{x}_j+\left(1-t\right)y_j\right)}^2{\left(1+t{x}_j+\left(1-t\right)y_j\right)}^2}.\end{array}$$

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

(ii) Directly calculating yields

$$g^{\prime }\left(t\right)=\frac{\frac{1}{m}{\sum }_{j=1}^m{\left(x_j-y_j\right)}^{\frac{1}{m}-1}}{{\sum }_{j=1}^m{\left(t{x}_j+\left(1-t\right)y_j\right)}^{1/m}}$$

and

$$g^{″}\left(t\right)=-\frac{{\left[\frac{1}{m}{\sum }_{j=1}^m{\left(x_j-y_j\right)}^{\frac{1}{m}-1}\right]}^2}{{\sum }_{j=1}^m{\left(t{x}_j+\left(1-t\right)y_j\right)}^{2/m}}.$$

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

(iii) By computing,

$$\begin{array}{c}{h}^{\prime }\left(t\right)=\frac{1}{m}\frac{{\psi }^{\prime }\left(t\right)}{\psi \left(t\right)},\\ h^{″}\left(t\right)=\frac{1}{m}\frac{{\psi }^{″}\left(t\right)\psi \left(t\right)-{\left({\psi }^{\prime }\left(t\right)\right)}^2}{{\psi }^2\left(t\right)},\end{array}$$

where

$${\psi }^{\prime }\left(t\right)=\sum _{j=1}^m\frac{x_j-y_j}{{\left(1+t{x}_j+\left(1-t\right)y_j\right)}^2}$$

and

$${\psi }^{″}\left(t\right)=-\sum _{j=1}^m\frac{2{\left(x_j-y_j\right)}^2}{{\left(1+t{x}_j+\left(1-t\right)y_j\right)}^3}.$$

Thus,

$$\begin{array}{ll}\hfill {\psi }^{″}\left(t\right)\psi \left(t\right)-{\left({\psi }^{\prime }\left(t\right)\right)}^2& =-\sum _{j=1}^m\frac{2{\left(x_j-y_j\right)}^2}{{\left(1+t{x}_j+\left(1-t\right)y_j\right)}^3}\sum _{j=1}^m\frac{t{x}_j+\left(1-t\right)y_j}{1+t{x}_j+\left(1-t\right)y_j}\\ -{\left[\sum _{j=1}^m\frac{x_j-y_j}{{\left(1+t{x}_j+\left(1-t\right)y_j\right)}^2}\right]}^2\le 0,\end{array}$$

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 $\text{ln}{\sum }_{j=1}^k{x}_{i_j}-\text{ln}{\sum }_{j=1}^k\left(1+x_{i_j}\right)$ is concave on ${ℝ}_{+}^n$, and then $\text{ln}{F}_n\left(x,k\right)={\prod }_{1\le i_1<\cdots <i_k\le n}\left(\text{ln}{\sum }_{j=1}^k{x}_{i_j}-\text{ln}{\sum }_{j=1}^k\left(1+x_{i_j}\right)\right)$ is concave on ${ℝ}_{+}^n$. Furthermore, it is clear that ln F _n (x, k) is symmetric on ${ℝ}_{+}^n$, by Lemma 1, it follows that ln F _n ( x, k) is concave on ${ℝ}_{+}^n$, and then from Lemma 2 we conclude that F _n (x , k) is also concave on ${ℝ}_{+}^n$.

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.