1 Introduction

Throughout this paper, ℝ 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 ℝ and R + denote R 1 and R + 1 , respectively.

Let π=(π(1),,π(n)) be a permutation of (1,,n), all permutations are totally n!. The following conclusion is proved in [[1], pp.127-129].

Theorem A Let A R k be a symmetric convex set, and let φ be a Schur-convex function defined on A with the property that for each fixed x 2 ,, x k , φ(z, x 2 ,, x k ) is convex in z on {z:(z, x 2 ,, x k )A}. Then, for any n>k,

ψ( x 1 ,, x n )= π φ( x π ( 1 ) ,, x π ( k ) )
(1)

is Schur-convex on

B= { ( x 1 , , x n ) : ( x π ( 1 ) , , x π ( k ) ) A  for all permutations  π } .

Furthermore, the symmetric function

ψ ¯ (x)= 1 i 1 < < i k n φ( x i 1 ,, x i k )
(2)

is also Schur-convex on B.

Theorem A is very effective for judgement of the Schur-convexity of the symmetric functions of the form (2), see the references [1] and [2].

The Schur geometrically convex functions were proposed by Zhang [3] in 2004. Further, the Schur harmonically convex functions were proposed by Chu and Lü [4] in 2009. The theory of majorization was enriched and expanded by using these concepts [515]. Regarding Schur geometrically convex functions and Schur harmonically convex functions, the aim of this paper is to establish the following judgement theorems which are similar to Theorem A.

Theorem 1 Let A R k be a symmetric geometrically convex set, and let φ be a Schur geometrically convex (concave) function defined on A with the property that for each fixed x 2 ,, x k , φ(z, x 2 ,, x k ) is GA convex (concave) in z on {z:(z, x 2 ,, x k )A}. Then, for any n>k,

ψ( x 1 ,, x n )= π φ( x π ( 1 ) ,, x π ( k ) )

is Schur geometrically convex (concave) on

B= { ( x 1 , , x n ) : ( x π ( 1 ) , , x π ( k ) ) A  for all permutations  π } .

Furthermore, the symmetric function

ψ ¯ (x)= 1 i 1 < < i k n φ( x i 1 ,, x i k )

is also Schur geometrically convex (concave) on B.

Theorem 2 Let A R k be a symmetric harmonically convex set, and let φ be a Schur harmonically convex (concave) function defined on A with the property that for each fixed x 2 ,, x k , φ(z, x 2 ,, x k ) is HA convex (concave) in z on {z:(z, x 2 ,, x k )A}. Then, for any n>k,

ψ( x 1 ,, x n )= π φ( x π ( 1 ) ,, x π ( k ) )

is Schur harmonically convex (concave) on

B= { ( x 1 , , x n ) : ( x π ( 1 ) , , x π ( k ) ) A  for all permutations  π } .

Furthermore, the symmetric function

ψ ¯ (x)= 1 i 1 < < i k n φ( x i 1 ,, x i k )

is also Schur harmonically convex (concave) on B.

2 Definitions and lemmas

In order to prove some further results, in this section we recall useful definitions and lemmas.

Definition 1 [1, 16]

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

  1. (i)

    We say y majorizes x (x is said to be majorized by y), denoted by xy, if i = 1 k x [ i ] i = 1 k y [ i ] for k=1,2,,n1 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 , a function φ:ΩR is said to be a Schur-convex function on Ω if xy on Ω implies φ(x) φ(y). A function φ is said to be a Schur-concave function on Ω if and only if −φ is Schur-convex function on Ω.

Definition 2 [1, 16]

Let x=( x 1 ,, x n ) and y=( y 1 ,, y n ) R n , 0α1. A set Ω R n is said to be a convex set if x,yΩ implies αx+(1α)y=(α x 1 +(1α) y 1 ,,α x n +(1α) y n )Ω.

Definition 3 [1, 16]

  1. (i)

    A set Ω R n is called a symmetric set if xΩ implies xPΩ for every n×n permutation matrix P.

  2. (ii)

    A function φ:ΩR is called symmetric if for every permutation matrix P, φ(xP)=φ(x) for all xΩ.

Definition 4 Let Ω R + n , x=( x 1 ,, x n )Ω and y=( y 1 ,, y n )Ω.

  1. (i)

    [[3], p.64] A set Ω is called a geometrically convex set if ( x 1 α y 1 β ,, x n α y n β )Ω for all x,yΩ and α,β[0,1] such that α+β=1.

  2. (ii)

    [[3], p.107] A function φ:Ω R + is said to be a Schur geometrically convex function on Ω if (log x 1 ,,log x n )(log y 1 ,,log y n ) on Ω implies φ(x) φ(y). A function φ is said to be a Schur geometrically concave function on Ω if and only if −φ is a Schur geometrically convex function.

Definition 5 [17]

Let Ω R + n .

  1. (i)

    A set Ω is said to be a harmonically convex set if xy λ x + ( 1 λ ) y Ω for every x,yΩ and λ[0,1], where xy= i = 1 n x i y i and 1 x =( 1 x 1 ,, 1 x n ).

  2. (ii)

    A function φ:Ω R + is said to be a Schur harmonically convex function on Ω if 1 x 1 y implies φ(x)φ(y). A function φ is said to be a Schur harmonically concave function on Ω if and only if −φ is a Schur harmonically convex function.

Definition 6 [18]

Let I R + , φ:I R + be continuous.

  1. (i)

    A function φ is said to be a GA convex (concave) function on I if

    φ( x y )() φ ( x ) + φ ( y ) 2

    for all x,yI.

  2. (ii)

    A function φ is said to be a HA convex (concave) function on I if

    φ ( 2 x y x + y ) () φ ( x ) + φ ( y ) 2

for all x,yI.

Lemma 1 [[16], p.57]

Let Ω R n be a symmetric convex set with a nonempty interior Ω 0 . φ:ΩR is continuous on Ω and differentiable on Ω 0 . Then φ is a 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)
(3)

holds for any x=( x 1 ,, x n ) Ω 0 .

Lemma 2 [[3], p.108]

Let Ω R + n be a symmetric geometrically convex set with a nonempty interior Ω 0 . Let φ:Ω R + be continuous on Ω and differentiable on Ω 0 . Then φ is a Schur geometrically convex (Schur geometrically concave) function if and only if φ is symmetric on Ω and

( x 1 x 2 ) ( x 1 φ x 1 x 2 φ x 2 ) 0(0)
(4)

holds for any x=( x 1 ,, x n ) Ω 0 .

Lemma 3 [17, 19]

Let Ω R + n be a symmetric harmonically convex set with a nonempty interior Ω 0 . Let φ:Ω R + be continuous on Ω and differentiable on Ω 0 . Then φ is a Schur harmonically convex (Schur harmonically concave) function if and only if φ is symmetric on Ω and

( x 1 x 2 ) ( x 1 2 φ x 1 x 2 2 φ x 2 ) 0(0)
(5)

holds for any x=( x 1 ,, x n ) Ω 0 .

Lemma 4 [18]

Let I R + be an open subinterval, and let φ:I R + be differentiable.

  1. (i)

    φ is GA-convex (concave) if and only if x φ (x) is increasing (decreasing).

  2. (ii)

    φ is HA-convex (concave) if and only if x 2 φ (x) is increasing (decreasing).

3 Proofs of main results

Proof of Theorem 1 To verify condition (4) of Lemma 2, denote by π ( i , j ) the summation over all permutations π such that π(i)=1, π(j)=2. Because φ is symmetric,

ψ ( x 1 , , x n ) = i , j k i j π ( i , j ) φ ( x 1 , x 2 , x π ( 1 ) , , x π ( i 1 ) , x π ( i + 1 ) , , x π ( j 1 ) , x π ( j + 1 ) , , x π ( k ) ) + i k < j π ( i , j ) φ ( x 1 , x π ( 1 ) , , x π ( i 1 ) , x π ( i + 1 ) , , x π ( k ) ) + j k < i π ( i , j ) φ ( x 2 , x π ( 1 ) , , x π ( j 1 ) , x π ( j + 1 ) , , x π ( k ) ) + k < i , j i j π ( i , j ) φ ( x π ( 1 ) , , x π ( k ) ) .

Then

Δ 1 : = ( x 1 ψ x 1 x 2 ψ x 2 ) ( x 1 x 2 ) = i , j k i j π ( i , j ) [ x 1 φ ( 1 ) ( x 1 , x 2 , x π ( 1 ) , , x π ( i 1 ) , x π ( i + 1 ) , , x π ( j 1 ) , x π ( j + 1 ) , , x π ( k ) ) x 2 φ ( 2 ) ( x 1 , x 2 , x π ( 1 ) , , x π ( i 1 ) , x π ( i + 1 ) , , x π ( j 1 ) , x π ( j + 1 ) , , x π ( k ) ) ] ( x 1 x 2 ) + i k < j π ( i , j ) [ x 1 φ ( 1 ) ( x 1 , x π ( 1 ) , , x π ( i 1 ) , x π ( i + 1 ) , , x π ( k ) ) x 2 φ ( 1 ) ( x 2 , x π ( 1 ) , , x π ( i 1 ) , x π ( i + 1 ) , , x π ( k ) ) ] ( x 1 x 2 ) .

Here,

( x 1 φ ( 1 ) x 2 φ ( 2 ) )( x 1 x 2 )0(0)

because φ is Schur geometrically convex (concave), and

[ x 1 φ ( 1 ) ( x 1 , z ) x 2 φ ( 1 ) ( x 2 , z ) ] ( x 1 x 2 )0(0)

because φ(z, x 2 ,, x k ) is GA convex (concave) in its first argument on {z:(z, x 2 ,, x k )A}. Accordingly, Δ 1 0 (≤0). This shows that ψ is Schur geometrically convex (concave) on

B= { ( x 1 , , x n ) : ( x π ( 1 ) , , x π ( k ) ) A  for all permutations  π } .

Notice that

ψ ¯ (x)=ψ(x)/k!(nk)!.

Of course, ψ ¯ is Schur geometrically convex (concave) whenever ψ is Schur geometrically convex (concave).

The proof of Theorem 1 is completed. □

Proof of Theorem 2 We only need to verify condition (5) of Lemma 3, the proof is similar to that of Theorem 1 and is omitted. □

Remark 1 In most applications, A has the form I k for some interval IR and in this case B= I n . Notice that the convexity of φ in its first argument also implies that φ is convex in each argument, the other arguments being fixed, because φ is symmetric.

4 Applications

Let

E k ( x 1 x ) = 1 i 1 < < i k n j = 1 k x i j 1 x i j .
(6)

In 2011, Guan and Guan [20] proved the following theorem through Lemma 2.

Theorem 3 The symmetric function E k ( x 1 x ), k=1,,n, is Schur geometrically convex on ( 0 , 1 ) n .

Now, we give a new proof of Theorem 3 by using Theorem 1. Furthermore, we prove the following theorem through Theorem 2.

Theorem 4 The symmetric function E k ( x 1 x ), k=1,,n, is Schur harmonically convex on ( 0 , 1 ) n .

Proof of Theorem 3 Let φ(z)= i = 1 k [ z i /(1 z i )]. Then

logφ(z)= i = 1 k [ log z i log ( 1 z i ) ]

and

φ ( z ) z 1 = φ ( z ) ( 1 z 1 + 1 1 z 1 ) , φ ( z ) z 2 = φ ( z ) ( 1 z 2 + 1 1 z 2 ) , Δ : = ( z 1 z 2 ) ( z 1 φ ( z ) z 1 z 2 φ ( z ) z 2 ) Δ = ( z 1 z 2 ) φ ( z ) ( z 1 1 z 1 z 2 1 z 2 ) Δ = ( z 1 z 2 ) 2 φ ( z ) 1 ( 1 z 2 ) ( 1 z 1 ) .
(7)

This shows that Δ0 when 0< z i <1, i=1,,k. According to Lemma 2, φ is Schur geometrically convex on A={z:z ( 0 , 1 ) k }. Let g(t)= t 1 t , then h(t):=t g (t)= t ( 1 t ) 2 . From t(0,1), it follows that h (t)= 1 + t ( 1 t ) 3 0. According to Lemma 4(i), φ is GA convex in its single variable on (0,1). So E k ( x 1 x ) is Schur geometrically convex on ( 0 , 1 ) n from Theorem 1. The proof of Theorem 3 is completed. □

Proof of Theorem 4 Let φ(z)= i = 1 k ( z i /1 z i ), then

logφ(z)= i = 1 k [ log z i log ( 1 z i ) ] .

From (7), we get

Δ 1 : = ( z 1 z 2 ) ( z 1 2 φ ( z ) z 1 z 2 2 φ ( z ) z 2 ) = ( z 1 z 2 ) φ ( z ) ( z 1 z 2 + z 1 2 1 z 1 z 2 2 1 z 2 ) = ( z 1 z 2 ) 2 φ ( z ) [ 1 + z 1 + z 2 z 1 z 2 ( 1 z 2 ) ( 1 z 1 ) ] .

This shows that Δ 1 0 when 0< z i <1, i=1,,k. According to Lemma 3, φ is Schur harmonically convex on A={z:z ( 0 , 1 ) k }. Let g(t)= t 1 t , then p(t):= t 2 g (t)= t 2 ( 1 t ) 2 . From t(0,1), it follows that p (t)= 2 t ( 1 t ) 3 0. According to Lemma 4(ii), φ is HA convex in its single variable on (0,1). So E k ( x 1 x ) is Schur harmonically convex on ( 0 , 1 ) n from Theorem 2. The proof of Theorem 4 is completed. □

By using Theorem A, the following conclusion is proved in [[1], p.129].

The symmetric function

ψ ¯ (x)= 1 i 1 < < i k n x i 1 + + x i k x i 1 x i k
(8)

is Schur-convex on R + n .

Now we use Theorem 1 and Theorem 2, respectively, to study Schur geometric convexity and Schur harmonic convexity of ψ ¯ (x).

Theorem 5 The symmetric function ψ ¯ (x) is Schur geometrically convex and Schur harmonically concave on R + n .

Proof Let φ(y)= i = 1 k y i / i = 1 k y i , then logφ(y)=log( i = 1 k y i ) i = 1 k log y i . Thus,

φ ( y ) y 1 = φ ( y ) ( 1 i = 1 k y i 1 y 1 ) , φ ( y ) y 2 = φ ( y ) ( 1 i = 1 k y i 1 y 2 ) , Δ : = ( y 1 y 2 ) ( y 1 φ ( y ) y 1 y 2 φ ( y ) y 2 ) Δ = ( y 1 y 2 ) φ ( y ) ( y 1 y 2 i = 1 k y i ) Δ = ( y 1 y 2 ) 2 i = 1 k y i 0 .

According to Lemma 2, φ(y) is Schur geometrically convex on R + k . Let g(z)=φ(z, x 2 ,, x k )= z + a b z = 1 b + a b z , where a= i = 2 k x i , b= i = 2 k x i , then h(z):=z g (z)= a b z . From z R + , it follows that h (z)= a b z 2 0. According to Lemma 4(i), φ is GA convex in its single variable on R + . So ψ ¯ (x) is Schur geometrically convex on R + from Theorem 1.

It is easy to check that

Δ 1 : = ( y 1 y 2 ) ( y 1 2 φ ( y ) y 1 y 2 2 φ ( y ) y 2 ) = ( y 1 y 2 ) 2 ( y 1 + y 2 i = 1 k y i ) i = 1 k y i 0 .

According to Lemma 3, φ(y) is Schur harmonically concave on R + k . Let h(z):= z 2 g (z)= a b . h (z)=0 when z R + . According to Lemma 4(ii), φ is HA concave in its single variable on R + . So ψ ¯ (x) is Schur harmonically concave on R + n from Theorem 2. □

Remark 2 Let

H= n i = 1 n 1 x i ,G= ( i = 1 n x i ) 1 n ,

where x i >0, i=1,,n. Then

(logG,,logG)(log x 1 ,,log x n ),
(9)
( 1 H , , 1 H ) ( 1 x 1 , , 1 x n ) .
(10)

From Theorem 5, it follows that

k C n k H k 1 1 i 1 < < i k n x i 1 + + x i k x i 1 x i k k C n k G k 1 .
(11)

By using Theorem A, the following conclusion is proved in [[1], p.129].

The symmetric function

ψ(x)= 1 i 1 < < i k n x i 1 x i k x i 1 + + x i k

is Schur-concave on R + n .

By applying Theorem 2, we further obtain the following result.

Theorem 6 The symmetric function ψ(x) is Schur harmonically convex on R + n .

Proof Let λ(y)= i = 1 k y i / i = 1 k y i . According to the proof of Theorem 5, φ(y) is Schur harmonically concave on R + k . Let λ(y)= 1 φ ( y ) . From the definition of Schur harmonically convex, it follows that λ(y) is Schur harmonically convex on R + k . Let g(z)=λ(z, x 2 ,, x k )= b z z + a , where a= i = 2 k x i , b= i = 2 k x i . Then h(z):= z 2 g (z)= z 2 a b ( z + a ) 2 . With the fact that h (z)= 2 z a 2 b ( z + a ) 3 0 for z R + , it follows that φ is HA convex in its single variable on R + . So, from Theorem 2, ψ(x) is Schur harmonically convex on R + n . □

Remark 3 From Theorem 6 and (10), it follows that

1 i 1 < < i k n x i 1 x i k x i 1 + + x i k H k 1 C n k k ,
(12)

where x i >0, i=1,,n.

Remark 4 It needs further discussion that ψ(x) is Schur geometrically convex on R + n .