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Article

# Schur-Convexity of the Mean of Convex Functions for Two Variables

by
Huan-Nan Shi
1,
Dong-Sheng Wang
2 and
Chun-Ru Fu
3,*
1
Department of Electronic Information, Teacher’s College, Beijing Union University, Beijing 100011, China
2
Basic Courses Department, Beijing Polytechnic, Beijing 100176, China
3
Applied College of Science and Technology, Beijing Union University, Beijing 102200, China
*
Author to whom correspondence should be addressed.
Axioms 2022, 11(12), 681; https://doi.org/10.3390/axioms11120681
Submission received: 17 September 2022 / Revised: 20 November 2022 / Accepted: 21 November 2022 / Published: 29 November 2022

## Abstract

:
The results of Schur convexity established by Elezovic and Pecaric for the average of convex functions are generalized relative to the case of the means for two-variable convex functions. As an application, some binary mean inequalities are given.
MSC:
26A51; 26D15; B25

## 1. Introduction

Let $R$ be a set of real numbers, g be a convex function defined on the interval $I ⊆ R → R$ and $c , d ∈ I$, $c < d$. Then
$g d + c 2 ≤ 1 d − c ∫ c d g ( t ) d t ≤ g ( d ) + g ( c ) 2 .$
This is the famous Hadamard’s inequality for convex functions.
In 2000, utilizing Hadamard’s inequality, Elezovic and Pecaric [1] researched Schur-convexity on the lower and upper limit of the integral for the mean of the convex functions and obtained the following important and profound theorem.
Theorem 1
([1]). Let I be an interval with nonempty interior on $R$ and g be a continuous function on I. Then,
$Φ ( c , d ) = 1 d − c ∫ c d g ( s ) d s , c , d ∈ I , d ≠ c g ( c ) , d = c$
is $S c h u r c o n v e x ( S c h u r c o n c a v e , r e s p . )$ on $I × I$ iff g is convex (concave, resp.) on I.
In recent years, this result attracted the attention of many scholars (see references [2,3,4,5,6,7,8,9,10,11,12] and Chapter II of the monograph [13] and its references).
In this paper, the result of theorem 1 is generalized to the case of bivariate convex functions, and some bivariate mean inequalities are established.
Theorem 2.
Let I be an interval with non-empty interior on $R$ and $g ( s , t )$ be a continuous function on $I × I$. If g is convex (or concave resp.) on $I × I$, then
$G ( u , v ) = 1 ( v − u ) 2 ∫ u v ∫ u v g ( s , t ) d s d t , ( u , v ) ∈ I × I , u ≠ v g ( u , u ) , ( u , v ) ∈ I × I , u = v$
is Schur convex (or Schur concave, resp.) on $I × I$.

## 2. Definitions and Lemmas

To prove Theorem 2, we provide the following lemmas and definitions.
Definition 1.
Let $( x 1 , x 2 )$ and $( y 1 , y 2 ) ∈ R × R$.
(1)
A set $Ω ⊂ R × R$ is said to be convex if $( x 1 , x 2 ) , ( y 1 , y 2 ) ∈ Ω$ and $0 ≤ β ≤ 1$ implies
$( β x 1 + ( 1 − β ) y 1 , β x 2 + ( 1 − β ) y 2 ) ∈ Ω .$
(2)
Let $Ω ⊂ R × R$ be convex set. A function ψ: $Ω → R$ is said to be a convex function on Ω if, for all $β ∈ [ 0 , 1 ]$ and all $( x 1 , x 2 ) , ( y 1 , y 2 ) ∈ Ω$, inequality
$ψ ( β x 1 + ( 1 − β ) y 1 , β x 2 + ( 1 − β ) y 2 ) ≤ β ψ ( x 1 , x 2 ) + ( 1 − β ) ψ ( y 1 , y 2 )$
holds. If, for all $β ∈ [ 0 , 1 ]$ and all $( x 1 , x 2 ) , ( y 1 , y 2 ) ∈ Ω$, the strict inequality in (3) holds, then ψ is said to be strictly convex. ψ is called concave (or strictly concave, resp.) iff $− ψ$ is convex (or strictly convex, resp.)
Definition 2.
([14,15]). Let $Ω ⊆ R × R , ( x 1 , x 2 )$ and $( y 1 , y 2 ) ∈ Ω$, and let $φ : Ω → R$:
(1)
$( x 1 , x 2 )$ is said to be majorized by $( y 1 , y 2 )$ (in symbols $( x 1 , x 2 ) ≺ ( y 1 , y 2 )$) if $max { x 1 , x 2 }$$≤ max { y 1 , y 2 }$ and $x 1 + x 2 = y 1 + y 2$.
(2)
ψ is said to be a Schur-convex function on Ω if $( x 1 , x 2 ) ≺ ( y 1 , y 2 )$ on $Ω$ implies $ψ ( x 1 , x 2 ) ≺ ψ ( y 1 , y 2 )$, and ψ is said to be a Schur-concave function on Ω iff $− ψ$ is a Schur-convex function.
Lemma 1
([14] (p. 5)). Let $( x 1 , x 2 ) ∈ R × R$. Then
$x 1 + x 2 2 , x 1 + x 2 2 ≺ ( x 1 , x 2 ) .$
Lemma 2
([14] (p. 5)). Let $Ω ⊆ R × R$ be symmetric set with a nonempty interior $Ω ∘$. $ψ : Ω → R$ is continuous on Ω and differentiable in $Ω ∘$. Then, function ψ is Schur convex (or Schur concave, resp.) iff ψ is symmetric on Ω and
$x 1 − x 2 ∂ ψ ∂ x 1 − ∂ ψ ∂ x 2 ≥ 0 ( o r ≤ 0 , r e s p . )$
holds for any $x 1 , x 2 ∈ Ω ∘$.
Lemma 3
([16]). Let $φ x , w$ and $∂ φ x , w ∂ w$ be continuous on
$D = ( x , w ) : a ≤ x ≤ b , c ≤ w ≤ d ; l e t$
$a ( w ) , b ( w )$ and their derivatives be continuous on $[ c , d ]$; $v ∈ [ c , d ]$ implies $a ( w ) , b ( w ) ∈ [ a , b ]$. Then,
$d d w ∫ a ( w ) b ( w ) φ ( x , w ) d x = ∫ a ( w ) b ( w ) ∂ φ ( x , w ) ∂ w d x + φ ( b ( w ) , u ) b ′ ( w ) − φ ( a ( w ) , w ) a ′ ( w ) .$
Lemma 4.
Let $g ( s , t )$ be continuous on rectangle $[ a , p ; a , q ]$, $G c , d = ∫ c d ∫ c d g ( s , t ) d s d t$. If $c = c ( b )$ and $d = d ( b )$ are differentiable with b, $a ≤ c ( b ) ≤ p$ and $a ≤ d ( b ) ≤ q$, then
$∂ G ∂ b = ∫ c d g ( s , d ) d ′ ( b ) d s − ∫ c d g ( s , c ) c ′ ( b ) d s + d ′ ( b ) ∫ c d g ( d , t ) d t − c ′ ( b ) ∫ c d g ( c , t ) d t .$
Proof.
Let $φ ( s , b ) = ∫ c d g ( s , t ) d t$. Then,
$∂ φ ( s , b ) ∂ b = g ( s , d ) d ′ ( b ) − g ( s , c ) c ′ ( b ) .$
By Lemma 3, we have
$∂ G ∂ b = d d b ∫ c d φ ( s , b ) d s = ∫ c d ∂ φ ( s , b ) ∂ b d s + φ ( d , b ) d ′ ( b ) − φ ( c , b ) c ′ ( b ) = ∫ c d g ( s , d ) d ′ ( b ) d s − ∫ c d g ( s , c ) c ′ ( b ) d s + d ′ ( b ) ∫ c d g ( d , s ) d s − c ′ ( b ) ∫ c d g ( c , s ) d s .$
Remark 1.
In passing, it is pointed out that (9) in Lemma 5 of reference [2] is incorrect and should be replaced by (4) of this paper.
Lemma 5.
Let I be an interval with nonempty interior on $R$ and $g ( s , t )$ be a continuous function on $I × I$. For $( u , v ) ∈ I × I , u ≠ v$, let $G u , v = ∫ u v ∫ u v g ( s , t ) d s d t$. Then,
$∂ G ∂ v = ∫ u v g ( s , v ) d s + ∫ u v g ( v , t ) d t ,$
$∂ G ∂ u = − ∫ u v g ( s , u ) d s + ∫ u v g ( u , t ) d t .$
Proof.
By taking $c ( b ) = a$ and $d ( b ) = b$, we have $c ′ ( b ) = 0$ and $d ′ ( b ) = 1$. By (5) in Lemma 4, we obtain (6).
Notice that $G u , v = ∫ v u ∫ v u g ( s , t ) d s d t$; from (5), we have
$∂ G ∂ u = ∫ v u g ( s , u ) d s + ∫ v u g ( u , t ) d t = − ∫ u v g ( s , u ) d s + ∫ u v g ( u , t ) d t .$
Lemma 6
([14] (p. 38, Proposition 4.3) and [15] (p. 644, B.3.d)). Let $Ω ⊂ R × R$ be an open convex set and let $ψ ( x , y ) : Ω → R$ be twice differentiable. Then, ψ is convex on Ω iff the Hessian matrix
$H ( x , y ) = ∂ 2 ψ ∂ x ∂ x ∂ 2 ψ ∂ x ∂ y ∂ 2 ψ ∂ y ∂ x ∂ 2 ψ ∂ y ∂ y$
is non-negative definite on Ω. If $H ( x )$ is positive definite on Ω, then ψ is strictly convex on Ω.

## 3. Proofs of Main Results

Proof of Theorem 2.
Let $g ( s , t )$ be convex on $I × I$. $G ( u , v )$ is evidently symmetric. By Lemma 5, we have
$∂ G ( u , v ) ∂ v = − 2 ( v − u ) 3 ∫ u v ∫ u v g ( s , t ) d s d t + 1 ( v − u ) 2 ∫ u v g ( s , v ) d s + ∫ u v g ( v , t ) d t .$
$∂ G ( u , v ) ∂ u = 2 ( v − u ) 3 ∫ u v ∫ u v g ( s , t ) d s d t − 1 ( v − u ) 2 ∫ u v g ( s , u ) d s + ∫ u v g ( u , t ) d t .$
$Δ : = ( v − u ) ∂ G ( u , v ) ∂ v − ∂ G ( u , v ) ∂ u = − 4 ( v − u ) 2 ∫ u v ∫ u v g ( s , t ) d s d t + 1 v − u ∫ u v ( g ( s , v ) + g ( s , u ) ) d s + 1 v − u ∫ u v ( g ( u , t ) + g ( v , t ) ) d t$
$2 ( v − u ) 2 ∫ u v ∫ u v g ( s , t ) d s d t = 2 v − u ∫ u v 1 v − u ∫ u v g ( s , t ) d s d t ≤ 2 v − u a ∫ u v g ( u , t ) + g ( v , t ) 2 d t = 1 v − u ∫ u v a ( g ( u , t ) + g ( v , t ) ) d t$
and
$2 ( v − u ) 2 ∫ u v ∫ u v g ( s , t ) d s d t = 2 v − u ∫ u v 1 v − u ∫ u v g ( s , t ) d t d s ≤ 2 v − u ∫ u v g ( s , u ) + g ( s , v ) 2 d s = 1 v − u ∫ u v ( g ( s , u ) + g ( s , v ) ) d s .$
Moreover, we have
$4 ( v − u ) 2 ∫ u v ∫ u v g ( s , t ) d s d t ≤ 1 v − u ∫ u v ( g ( s , v ) + g ( s , u ) ) d s + 1 v − u ∫ u v ( g ( u , t ) + g ( v , t ) ) d t .$
Therefore, $Δ ≥ 0$, so $G ( u , v )$ is Schur-convex on $I × I$.
When $g ( s , t )$ is a concave function on $I × I$, it can be proved with similar methods. □

## 4. Application on Binary Mean

Theorem 3.
Let $c > 0$ and $d > 0$. If $c ≠ d , 0 < s < 1$, then
$A ( d , c ) ≥ S s + 1 s ( d , c ) S s s − 1 ( d , c ) ≥ ( c + d ) 2 s − 1 s ( s + 1 ) ,$
where $A ( d , c ) = c + d 2$ and $S s ( d , c ) = d s − c s s ( d − c ) 1 s − 1$ are the arithmetic mean and the s-order Stolarsky mean of positive numbers c and d, respectively.
Proof.
Let $x > 0 , y > 0$ and $0 < s < 1$. From Theorem 4 in the reference [17], we know that $g ( x , y ) = x s y 1 − s$ is concave on $( 0 , + ∞ ) × ( 0 , + ∞ )$. For $c ≠ d$, by Theorem 2, from $( d + c 2 , d + c 2 ) ≺ ( c , d ) ≺ ( d + c , 0 )$, it follows that
$G ( d + c , 0 ) = 1 ( d + c − 0 ) 2 ∫ c d ∫ 0 d + c x s y 1 − s d x d y = 1 ( d + c ) 2 ∫ 0 d + c x s d x ∫ 0 d + c y 1 − s d y = 1 ( d + c ) 2 ( c + d ) s + 1 s + 1 ( c + d ) s s = ( c + d ) 2 s − 1 s ( s + 1 ) ≤ G ( c , d ) = 1 ( d − c ) 2 ∫ c d ∫ c d x s y 1 − s d x d y = 1 ( d − c ) 2 ∫ c d x s d x ∫ c d y 1 − s d y = 1 ( d − c ) 2 d s + 1 − c s + 1 s + 1 d s − c s s ≤ G d + c 2 , d + c 2 = d + c 2 ,$
That is, we obtain the following.
$( c + d ) 2 s − 1 s ( s + 1 ) ≤ S s + 1 s ( d , c ) S s s − 1 ( d , c ) = d s + 1 − c s + 1 ( s + 1 ) ( d − c ) · d s − c s s ( d − c ) ≤ d + c 2 = A ( d , c ) .$
Theorem 4.
Let $c > 0 , d > 0$. Then,
$log A ( d , c ) B ( d , c ) 2 ≥ c − d d + c 2 ,$
where $B ( d , c ) = d c$ is the geometric mean of of positive numbers c and d.
Proof.
From reference [17], we know that the function $g ( x , y ) = 1 ( x + y ) 2$ is convex on $( 0 , + ∞ ) × ( 0 , + ∞ )$. For $c > 0 , d > 0$ and $d ≠ c$, by Theorem 2, from $( d + c 2 , d + c 2 ) ≺ ( d , c )$, it follows that
$G ( c , d ) = 1 ( d − c ) 2 ∫ c d ∫ c d 1 ( x + y ) 2 d x d y = 1 ( d − c ) 2 ∫ c d 1 c + y − 1 d + y d y = 1 ( d − c ) 2 [ ( log ( d + c ) − log ( 2 c ) ) − ( log ( 2 d ) − log ( d + c ) ) ] ≥ G d + c 2 , d + c 2 = 1 ( d + c ) 2 ,$
That is, we obtain the following.
$log A ( d , c ) B ( d , c ) 2 = log ( d + c ) 2 4 d c ≥ c − d d + c 2 .$
Theorem 5.
Let $c > 0 , d > 0$. Then,
$H e ( c 2 , d 2 ) ≥ A 2 ( c , d ) ,$
where $H e ( c , d ) = c + c d + d 3$ is the Heronian mean of positive numbers c and d.
Proof.
From reference [18], we know that the function of two variables
$ψ ( x , y ) = x 2 2 r 2 + y 2 2 s 2$
is a convex function on $( 0 , + ∞ ) × ( 0 , + ∞ )$, where $s > 0$ and $r > 0$. For $d > 0 , c > 0$, and $c ≠ d$, by Theorem 2, from $( d + c 2 , d + c 2 ) ≺ ( d , c )$, it follows that
$G ( c , d ) = 1 ( d − c ) 2 ∫ c d ∫ c d x 2 2 r 2 + y 2 2 s 2 d x d y = 1 ( d − c ) 2 ∫ c d d 3 − c 3 6 r 2 + y 2 ( d − c ) 2 s 2 d y = 1 ( d − c ) 2 ( d 3 − c 3 ) ( d − c ) 6 r 2 + ( d 3 − c 3 ) ( d − c ) 6 s 2 = 1 ( d − c ) 2 · ( d 3 − c 3 ) ( d − c ) 6 1 r 2 + 1 s 2 ≥ G d + c 2 , d + c 2 = ( c + d ) 2 8 1 r 2 + 1 s 2 ,$
namely
$H e ( c 2 , d 2 ) = c 2 + c d + d 2 3 = ( d 3 − c 3 ) 3 ( d − c ) ≥ ( d + c ) 2 4 = A 2 ( d , c ) .$
Theorem 6.
Let $c > 0 , d > 0$. We have
$H e ( c 2 , d 2 ) ≥ L ( d , c ) A ( d , c ) ,$
where $L ( d , c ) = d − c log d − log c$ is the logarithmic mean of positive numbers c and d.
Proof.
Let $g ( x , y ) = y 2 x − 1 , x > 0 , y > 0$. Then,
$g x x = 2 x − 3 y 2 , g x y = − 2 x − 2 y = g y x , g y y = 2 x − 1 .$
The Hesse matrix of $g ( x , y )$ is
$H = 2 x − 3 y 2 − 2 x − 2 y − 2 x − 2 y 2 x − 1 .$
$det ( H − λ I ) = det 2 x − 3 y 2 − λ − 2 x − 2 y − 2 x − 2 y 2 x − 1 − λ = 0$
$⇒ λ ( λ − 2 x − 3 y 2 − 2 x − 1 ) = 0 ⇒ λ 1 = 0 , λ 2 = 2 x − 3 y 2 + 2 x − 1 > 0 .$
Therefore, matrix H is positive semidefinite, so it is known that $g ( x , y )$ is a convex function on $( 0 , + ∞ ) × ( 0 , + ∞ )$. For $d > 0 , c > 0$ and $d ≠ c$, by Theorem 2, from $( d + c 2 , d + c 2 ) ≺ ( d , c )$, it follows that
$G ( c , d ) = 1 ( d − c ) 2 ∫ c d ∫ c d y 2 x − 1 d x d y = log d − log c d − c · d 2 + c d + c 2 3 ≥ d + c 2 2 c + c 2 = d + c 2 ,$
which is
$H e ( c 2 , d 2 ) ≥ L ( d , c ) A ( d , c ) .$
Theorem 7.
Let $d > 0 , c > 0 , d ≠ c$. Then
$E ˜ ( d , c ) ≤ A ( d , c ) e ( d + c ) d − c e d − e c 2 ≤ A ( d , c ) ,$
where
$E ˜ ( d , c ) = c e d − d e c e d − e c + 1 , d , c ∈ I , d ≠ c c , c = d$
is exponent type mean of positive numbers c and d (see [13] (p. 134)).
Proof.
Let $g ( x , y ) = x e − ( x + y ) , y > 0 , x > 0$. From reference [19], we know that function $g ( x , y )$ is convex on $R × R$. For $d > 0 , c > 0$, and $d ≠ c$ by Theorem 2 from $( d + c 2 , d + c 2 ) ≺ ( d , c )$, it follows that
$G ( c , d ) = 1 ( c − d ) 2 ∫ c d ∫ c d x e − x − y d x d y = 1 ( c − d ) 2 ∫ c d x e − x d x ∫ c d e − y d y = 1 ( c − d ) 2 c + 1 e c − d + 1 e d · 1 e c − 1 e d = 1 ( d − c ) 2 ( c e d − d e c ) + ( e d − e c ) e ( c + d ) · e d − e c e ( c + d ) ≤ G d + c 2 , d + c 2 = c + d 2 1 e ( d + c ) ,$
which is
$c e d − d e c e d − e c + 1 ≤ d + c 2 e ( d + c ) d − c e d − e c 2 .$
For the rest, we only need to prove that
$e ( c + d ) d − c e d − e c 2 ≤ 1 .$
We write $e d = u$ and $e c = v$; then, the above inequality is equivalent to the well-known log-geometric mean inequality.
$L ( v , u ) = v − u log v − log u ≥ v u = B ( v , u ) .$

## Author Contributions

Conceptualization, H.-N.S., D.-S.W. and C.-R.F.; Methodology, H.-N.S.; Validation, C.-R.F.; Formal analysis, H.-N.S. and D.-S.W.; Investigation, D.-S.W.; Resources, C.-R.F.; Writing—original draft, D.-S.W.; Funding acquisition, C.-R.F. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

Not applicable.

Not applicable.

Not applicable.

## Acknowledgments

The authors sincerely thanks Chen Dirong and Chen Jihang for their valuable opinions and suggestions.

## Conflicts of Interest

The authors declare no conflict of interest.

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Shi, H.-N.; Wang, D.-S.; Fu, C.-R. Schur-Convexity of the Mean of Convex Functions for Two Variables. Axioms 2022, 11, 681. https://doi.org/10.3390/axioms11120681

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Shi H-N, Wang D-S, Fu C-R. Schur-Convexity of the Mean of Convex Functions for Two Variables. Axioms. 2022; 11(12):681. https://doi.org/10.3390/axioms11120681

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Shi, Huan-Nan, Dong-Sheng Wang, and Chun-Ru Fu. 2022. "Schur-Convexity of the Mean of Convex Functions for Two Variables" Axioms 11, no. 12: 681. https://doi.org/10.3390/axioms11120681

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