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Some inequalities for (h,m)-convex functions

Abstract

In the paper, the authors give some inequalities of Jensen type and Popoviciu type for (h,m)-convex functions and apply these inequalities to special means.

MSC: 26A51, 26D15, 26E60.

1 Introduction

The following definition is well known in the literature.

Definition 1 A function f:IR=(,)R is said to be convex if

f ( t x + ( 1 t ) y ) tf(x)+(1t)f(y)
(1)

holds for all x,yI and t[0,1].

We cite the following inequalities for convex functions.

Theorem 1 ([[1], p.6])

If f is a convex function on I and x 1 , x 2 , x 3 I, then

f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + f ( x 1 + x 2 + x 3 3 ) 4 3 [ f ( x 1 + x 2 2 ) + f ( x 2 + x 3 2 ) + f ( x 3 + x 1 2 ) ] .
(2)

Theorem 2 ([[2], Popoviciu inequality])

If f is a convex function on I and x 1 , x 2 ,, x n I with n3, then

i = 1 n f( x i )+ n n 2 f ( 1 n i = 1 n x i ) 2 n 2 i < j f ( x i + x j 2 ) .
(3)

Theorem 3 ([[2], Generalized Popoviciu inequality])

If f is a convex function on I and a 1 , a 2 ,, a n I for n3, then

(n1) i = 1 n f( b i )n(n2)f(a)+ i = 1 n f( a i ),
(4)

where a= 1 n i = 1 n a i and b i = n a a i n 1 for i=1,2,,n.

The above inequalities were generalized as follows.

Theorem 4 ([3])

If f is a convex function on I and x 1 , x 2 ,, x n I for n3, then

i = 1 n f( x i )f ( 1 n k = 1 n x k ) n 1 n i = 1 n f ( x i + x i + 1 2 )
(5)

and

(n1) i = 1 n f( b i )n [ i = 1 n f ( a i ) f ( a ) ] ,
(6)

where x n + 1 = x 1 , a= 1 n i = 1 n a i , and b i = n a a i n 1 for i=1,2,,n.

Definition 2 ([4])

Let s(0,1]. A function f: R 0 =[0,) R 0 is said to be s-convex in the second sense if

f ( λ x + ( 1 λ ) y ) λ s f(x)+ ( 1 λ ) s f(y)
(7)

holds for all x,yI and λ[0,1].

The following inequalities for s-convex functions were established.

Theorem 5 ([[5], Theorem 4.2])

If f is nonnegative and s-convex in the second sense on I and if x 1 , x 2 ,, x n I for n3, then

i = 1 n f( x i )f ( 1 n i = 1 n x i ) 2 s 1 ( n s 1 ) n i = 1 n f ( x i + x i + 1 2 ) ,
(8)

where x 1 = x n + 1 .

Theorem 6 ([[5], Theorem 4.4])

If f is nonnegative and s-convex in the second sense on I and a 1 , a 2 ,, a n I for n3, then

( n s 1 ) i = 1 n b i n s [ i = 1 n f ( a i ) f ( a ) ] ,
(9)

where a= 1 n i = 1 n a i and b i = n a a i n 1 for i=1,2,,n.

The concept of h-convex functions below was innovated as follows.

Definition 3 ([[6], Definition 4])

Let I,JR be intervals, (0,1)J, and h:J R 0 be a nonnegative function. A function f:I R 0 is called h-convex, or as we say, f belongs to the class SX(h,I), if f is nonnegative and

f ( t x + ( 1 t ) y ) h(t)f(x)+h(1t)f(y)
(10)

for all x,yI and t[0,1].

Definition 4 ([[6], Section 3])

A function h:JR is said to be a super-multiplicative on an interval J if

h(xy)h(x)h(y)
(11)

is valid for all x,yJ. If the inequality (11) reverses, then f is said to be a sub-multiplicative function on J.

The following inequalities were established for fSX(h,I).

Theorem 7 ([[7], Theorem 6])

Let w 1 ,, w n for n2 be positive real numbers. If h is a nonnegative and super-multiplicative function and if fSX(h,I) and x 1 ,, x n I, then

f ( 1 W n i = 1 n w i x i ) i = 1 n h ( w i W n ) f( x i ),
(12)

where W n = i = 1 n w i . If h is sub-multiplicative and fSV(h,I), then the inequality (12) is reversed.

Theorem 8 ([[8], Theorem 11])

Let h be a nonnegative and super-multiplicative function. If fSX(h,I) and x 1 ,, x n I, then

i = 1 n f( x i )f ( 1 n i = 1 n x i ) 1 h ( 1 / n ) 2 h ( 1 / 2 ) i = 1 n f ( x i + x i + 1 2 ) ,
(13)

where x n + 1 = x 1 . The inequality (13) is reversed if fSV(h,I).

Theorem 9 ([[8], Theorem 12])

Let h be a nonnegative and super-multiplicative function. If fSX(h,I) and x 1 ,, x n I, then

[ 1 h ( 1 n ) ] i = 1 n f( b i )(n1)h ( 1 n 1 ) [ i = 1 n f ( a i ) f ( a ) ] ,
(14)

where a= 1 n i = 1 n a i and b i = n a a i n 1 for i=1,2,,n and n3. The inequality (14) is reversed if fSV(h,I).

Two new kinds of convex functions were introduced as follows.

Definition 5 ([9])

For f:[0,b]R and m(0,1], if

f ( t x + m ( 1 t ) y ) tf(x)+m(1t)f(y)
(15)

is valid for all x,y[0,b] and t[0,1], then we say that f(x) is an m-convex function on [0,b].

Definition 6 ([10])

Let JR be an interval, (0,1)J, h:JR be a nonnegative function. We say that f:[0,b]R is an (h,m)-convex function, or say, f belongs to the class SMX((h,m),[0,b]), if f is nonnegative and, for all x,y[0,b] and t[0,1] and for some m(0,1], we have

f ( t x + m ( 1 t ) y ) h(t)f(x)+mh(1t)f(y).
(16)

If the inequality (16) is reversed, then f is said to be (h,m)-concave and denoted by fSMV((h,m),[0,b]).

Recently the h- and (h,m)-convex functions were generalized and some properties and inequalities for them were obtained in [11, 12].

The aim of this paper is to find some inequalities of Jensen type and Popoviciu type for (h,m)-convex functions.

2 Inequalities of Jensen type and Popoviciu type

Now we are in a position to establish some inequalities of Jensen type and Popoviciu type for (h,m)-convex functions.

Theorem 10 Let h:[0,1] R 0 be a super-multiplicative function and m(0,1]. If fSMX((h,m),[0,b]), then for all x i [0,b] and w i >0 with i=1,2,,n and n2, we have

f ( 1 W n i = 1 n m i 1 w i x i ) i = 1 n m i 1 h ( w i W n ) f( x i ),
(17)

where W n = i = 1 n w i .

If h is sub-multiplicative and fSMV((h,m),[0,b]), then the inequality (17) is reversed.

Proof Assume that w i = w i W n for i=1,2,,n.

When n=2, taking t= w 1 and 1t= w 2 in Definition 6 gives the inequality (17) clearly.

Suppose that the inequality (17) holds for n=k, i.e.,

f ( i = 1 k m i 1 w i x i ) i = 1 k m i 1 h ( w i ) f( x i ).
(18)

When n=k+1, letting Δ k = i = 2 k + 1 w i and making use of (18) result in

f ( i = 1 k + 1 m i 1 w i x i ) = f ( w 1 x 1 + m Δ k i = 2 k + 1 m i 2 w i Δ k x i ) h ( w 1 ) f ( x 1 ) + m h ( Δ k ) f ( i = 2 k + 1 m i 2 w i Δ k x i ) h ( w 1 ) f ( x 1 ) + m h ( Δ k ) i = 2 k + 1 m i 2 h ( w i Δ k ) f ( x i ) .

Since h is a super-multiplicative function, it follows that

h( Δ k )h ( w i Δ k ) h ( w i )

for i=1,2,,n. Namely, when n=k+1, the inequality (17) holds. By induction, Theorem 10 is proved. □

Corollary 1 Under the conditions of Theorem  10,

  1. 1.

    if W n =1, we have

    f ( i = 1 n m i 1 w i x i ) i = 1 n m i 1 h( w i )f( x i );
    (19)
  2. 2.

    if w 1 = w 2 == w n , we have

    f ( 1 n i = 1 n m i 1 w i x i ) h ( 1 n ) i = 1 n m i 1 f( x i );
    (20)
  3. 3.

    if h is sub-multiplicative and fSMV((h,m),[0,b]), then the inequalities (19) and (20) are reversed.

Corollary 2 For m(0,1] and s(0,1], the assertion fSMX(( t s ,m),[0,b]) is valid if and only if for all x i [0,b] and w i >0 with i=1,2,,n and n2

f ( 1 W n i = 1 n m i 1 x i ) i = 1 n m i 1 ( w i W n ) s f( x i ),
(21)

where W n = i = 1 n w i .

Corollary 3 Under the conditions of Corollary  1, if h(t)= t s for s(0,1], then

f ( 1 n i = 1 n m i 1 x i ) 1 n s i = 1 n m i 1 f( x i ).
(22)

If fSMV((h,m),[0,b]), then the inequality (22) is reversed.

Theorem 11 Let h:[0,1] R 0 be a super-multiplicative function, m(0,1], and n2. If fSMX((h,m),[0, b m n 1 ]), then for all x i [0,b] and w i >0 with i=1,2,,n,

f ( 1 W n i = 1 n w i x i ) i = 1 n m i 1 h ( w i W n ) f ( x i m i 1 ) ,
(23)

where W n = i = 1 n w i .

If h is sub-multiplicative and fSMV((h,m),[0, b m n 1 ]), then the inequality (23) is reversed.

Proof Putting y i = x i m i 1 for i=1,2,,n, then from inequality (17), we have

f ( 1 W n i = 1 n w i x i ) = f ( 1 W n i = 1 n m i 1 w i y i ) i = 1 n m i 1 h ( w i W n ) f ( y i ) = i = 1 n m i 1 h ( w i W n ) f ( x i m i 1 ) .

The proof of Theorem 11 is complete. □

Corollary 4 For m(0,1], s(0,1], and n2, the assertion fSMX(( t s ,m),[0, b m n 1 ]) is valid if and only if for all x i [0,b] and w i >0 with i=1,2,,n the inequality

f ( 1 W n i = 1 n w i x i ) i = 1 n m i 1 ( w i W n ) s f ( x i m i 1 )
(24)

is valid, where W n = i = 1 n w i .

Corollary 5 Under the conditions of Theorem  11,

  1. 1.

    if W n =1, then

    f ( i = 1 n w i x i ) i = 1 n m i 1 h( w i )f ( x i m i 1 ) ;
    (25)
  2. 2.

    if w 1 = w 2 == w n , then

    f ( 1 n i = 1 n x i ) h ( 1 n ) i = 1 n m i 1 f ( x i m i 1 ) ;
    (26)
  3. 3.

    if h is sub-multiplicative and fSMV((h,m),[0, b m n 1 ]), then the inequalities (25) and (26) are reversed.

Corollary 6 Under the conditions of Corollary  5,

  1. 1.

    if h(t)= t s for s(0,1], then

    f ( 1 n i = 1 n x i ) 1 n s i = 1 n m i 1 f ( x i m i 1 ) ;
    (27)
  2. 2.

    if fSMV((h,m),[0, b m n 1 ]), then the inequality (27) is reversed.

Theorem 12 Let h:[0,1][0,1] be a super-multiplicative function and let m(0,1] and n3. If fSMX((h,m),[0,b]), then for all x i [0,b] with i=1,2,,n and 2kn, we have

i = 1 n f ( x i ) ( j = 0 n 1 m j ) 1 i = 1 n f ( 1 n j = i n + i 1 m j i x j ) 1 h ( 1 / n ) h ( 1 / k ) ( j = 0 k 1 m j ) 1 i = 1 n f ( 1 k j = i k + i 1 m j i x j ) ,
(28)

where x n + 1 = x 1 , …, x 2 n 1 = x n 1 .

If h is sub-multiplicative and fSMV((h,m),[0,b]), then the inequality (28) is reversed.

Proof By using the inequality (20), we have

i = 1 n f ( 1 k j = i k + i 1 m j i x j ) h ( 1 k ) i = 1 n j = i k + i 1 m j i f( x j )=h ( 1 k ) ( j = 0 k 1 m j ) i = 1 n f( x i )
(29)

and

i = 1 n f ( 1 n j = i n + i 1 m j i x j ) h ( 1 n ) i = 1 n j = i n + i 1 m j i f ( x j ) = h ( 1 n ) ( j = 0 n 1 m j ) i = 1 n f ( x i ) .
(30)

If h( 1 n )=1, then, from the inequality (30), the inequality (28) holds. If h( 1 n )1, it is easy to see that

i = 1 n f ( 1 k j = i k + i 1 m j i x j ) h ( 1 k ) ( j = 0 k 1 m j ) i = 1 n f ( x i ) = h ( 1 / k ) 1 h ( 1 / n ) ( j = 0 k 1 m j ) [ i = 1 n f ( x i ) h ( 1 n ) i = 1 n f ( x i ) ] h ( 1 / k ) 1 h ( 1 / n ) ( j = 0 k 1 m j ) [ i = 1 n f ( x i ) ( j = 0 n 1 m j ) 1 i = 1 n f ( 1 n j = i n + i 1 m j i x j ) ] .

The proof of Theorem 12 is complete. □

Corollary 7 Under the conditions of Theorem  12, let x ¯ n = 1 n i = 1 n x i .

  1. 1.

    When m=1, we have

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) 1 h ( 1 / n ) k h ( 1 / k ) i = 1 n f ( 1 k j = i k + i 1 x j ) .
    (31)
  2. 2.

    When m=1 and k=2, we have

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) 1 h ( 1 / n ) 2 h ( 1 / 2 ) i = 1 n f ( x i + x i + 1 2 ) .
    (32)
  3. 3.

    When m=1 and k=n1, we have

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) 1 h ( 1 / n ) ( n 1 ) h ( 1 / ( n 1 ) ) i = 1 n f ( n x ¯ n x i n 1 ) .
    (33)
  4. 4.

    If h is sub-multiplicative and fSMV((h,m),[0,b]), then the inequalities (31) to (33) are reversed.

Remark 1 The inequality (14) can be deduced from applying (33) to a i = x i for i=1,2,,n, a= 1 n i = 1 n a i , and b i = n a a i n 1 for i=1,2,,n.

Corollary 8 Under the conditions of Theorem  12,

  1. 1.

    if h(t)= t s for s(0,1], then

    i = 1 n f ( x i ) ( j = 0 n 1 m j ) 1 i = 1 n f ( 1 n j = i n + i 1 m j i x j ) k s ( n s 1 ) n s ( j = 0 k 1 m j ) 1 i = 1 n f ( 1 k j = i k + i 1 m j i x j ) ;
    (34)
  2. 2.

    if h(t)= t s for s(0,1] and m=1, then

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) k s 1 ( n s 1 ) n s i = 1 n f ( 1 k j = i k + i 1 x j ) ;
    (35)
  3. 3.

    if h(t)=t and m=1, then

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) n 1 n i = 1 n f ( 1 k j = i k + i 1 x j ) ;
    (36)
  4. 4.

    if fSMV((h,m),[0,b]), then the inequalities (34) to (36) are reversed.

Theorem 13 Let h:[0,1][0,1] be a super-multiplicative function and let m(0,1] and n3. If fSMX((h,m),[0, b m n 1 ]), then for all x i [0,b] with i=1,2,,n and 2kn and for 1 ,, k N, we have

i = 1 n f ( x i ) ( j = 0 n 1 m j ) 1 i = 1 n f ( 1 n j = i n + i 1 m j i x j ) 1 h ( 1 / n ) ( n 1 k 1 ) h ( 1 / k ) ( j = 0 k 1 m j ) 1 1 1 < < k n i = 1 k f ( 1 k j = i k + i 1 m j i x j ) ,
(37)

where k + 1 = 1 , …, 2 k 1 = k 1 .

If h is sub-multiplicative and fSMV((h,m),[0,b]), then the inequality (37) is reversed.

Proof By the inequality (20), we have

1 1 < < k n i = 1 k f ( 1 k j = i k + i 1 m j i x j ) h ( 1 k ) 1 1 < < k n i = 1 k j = i k + i 1 m j i f ( x j ) = h ( 1 k ) ( j = 0 k 1 m j ) 1 1 < < k n i = 1 k f ( x j ) = ( n 1 k 1 ) h ( 1 k ) ( j = 0 k 1 m j ) i = 1 n f ( x i ) .
(38)

If h( 1 n )=1, then, from the inequality (30), the inequality (28) holds. If h( 1 n )1, using (38) and (30), we have

1 1 < < k n i = 1 k f ( 1 k j = i k + i 1 m j i x j ) ( n 1 k 1 ) h ( 1 k ) ( j = 0 k 1 m j ) i = 1 n f ( x i ) = ( n 1 k 1 ) h ( 1 / k ) 1 h ( 1 / n ) ( j = 0 k 1 m j ) [ i = 1 n f ( x i ) h ( 1 n ) i = 1 n f ( x i ) ] ( n 1 k 1 ) h ( 1 / k ) 1 h ( 1 / n ) ( j = 0 k 1 m j ) [ i = 1 n f ( x i ) ( j = 0 n 1 m j ) 1 i = 1 n f ( 1 n j = i n + i 1 m j i x j ) ] .

The proof of Theorem 13 is complete. □

Corollary 9 Under the conditions of Theorem  13, let x ¯ n = 1 n i = 1 n x i .

  1. 1.

    When m=1, we have

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) 1 h ( 1 / n ) ( n 1 k 1 ) h ( 1 / k ) 1 1 < < k n f ( 1 k j = 1 k x j ) .
    (39)
  2. 2.

    When m=1 and k=2, we have

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) 1 h ( 1 / n ) ( n 1 ) h ( 1 / 2 ) 1 i < j n f ( x i + x j 2 ) .
    (40)
  3. 3.

    When m=1 and k=n1, we have

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) 1 h ( 1 / n ) ( n 1 ) h ( 1 / ( n 1 ) ) i = 1 n ( n x ¯ n x i n 1 ) .
    (41)
  4. 4.

    If h is sub-multiplicative and fSMV((h,m),[0,b]), then the inequalities (39) to (41) are reversed.

Corollary 10 Under the conditions of Theorem  13,

  1. 1.

    if h(t)= t s for s(0,1], then

    i = 1 n f ( x i ) ( j = 0 n 1 m j ) 1 i = 1 n f ( 1 n j = i n + i 1 m j i x j ) k s ( n s 1 ) ( n 1 k 1 ) n s ( j = 0 k 1 m j ) 1 1 1 < < k n i = 1 k f ( 1 k j = i k + i 1 m j i x j ) ;
    (42)
  2. 2.

    if m=1 and h(t)= t s for s(0,1], we have

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) k s ( n s 1 ) ( n 1 k 1 ) n s 1 1 < < k n f ( 1 k j = 1 k x j ) ;
    (43)
  3. 3.

    if m=1 and h(t)=t, then

    i = 1 n f( x i )f ( 1 n i = 1 n x i ) k ( n 1 ) ( n 1 k 1 ) n 1 1 < < k n f ( 1 k j = 1 k x j ) ;
    (44)
  4. 4.

    if fSMV((h,m),[0,b]), then the inequalities (42) to (44) are reversed.

3 Applications to means

In what follows we will apply the theorems and corollaries in the above section to establish inequalities for some special means.

For rR, r0, and m,s(0,1], let f(x)= x r for x R + and h(t)= t s for t[0,1]. Then

  1. 1.

    if r1 and 0<m1, or if r<0 and m=1, we have

    ( t x + m ( 1 t ) y ) r t x r +(1t) ( m y ) r t s x r +m ( 1 t ) s y r

for x,y R + ;

  1. 2.

    if 0<r1, 0<m1, and s=1, we have

    ( t x + m ( 1 t ) y ) r t x r +(1t) ( m y ) r t x r +m(1t) y r

for x,y R + .

Using Definition 6 yields the following:

  1. 1.

    if r1 and 0<m1, or if r<0 and m=1, the function f(x)= x r SMX(( t s ,m), R + );

  2. 2.

    if 0<r1, 0<m1, and s=1, the function f(x)= x r SMV((t,m), R + ).

By virtue of Corollary 10, we obtain the following results.

Theorem 14 Let n3 and x i R + for i=1,2,,n, let rR with r0 and m,s(0,1], and let 1 ,, k N for 2kn and k + 1 = 1 , …, 2 k 1 = k 1 .

  1. 1.

    If r1 and 0<m1, or if r<0 and m=1, then we have

    i = 1 n x i r ( j = 0 n 1 m j ) 1 i = 1 n ( 1 n j = i n + i 1 m j i x j ) r k s ( n s 1 ) ( n 1 k 1 ) n s ( j = 0 k 1 m j ) 1 1 1 < < k n i = 1 k ( 1 k j = i k + i 1 m j i x j ) r ;
    (45)
  2. 2.

    if r1 or r<0 and if m=1, we have

    i = 1 n x i r ( 1 n i = 1 n x i ) r k s ( n s 1 ) ( n 1 k 1 ) n s 1 1 < < k n ( 1 k j = 1 k x j ) r ;
    (46)
  3. 3.

    if r1 or r<0 and if m=s=1, then

    i = 1 n x i r ( 1 n i = 1 n x i ) r k ( n 1 ) ( n 1 k 1 ) n 1 1 < < k n ( 1 k j = 1 k x j ) r ;
    (47)
  4. 4.

    if 0<r1, 0<m1, and s=1, then the inequality (47) are reversed.

Corollary 11 Under the conditions of Theorem  14, when k + 1 = 1 , …, 2 k 1 = k 1 , we have the following conclusions.

  1. 1.

    If r=2, we have

    i = 1 n x i 2 ( j = 0 n 1 m j ) 1 i = 1 n ( 1 n j = i n + i 1 m j i x j ) 2 k s ( n s 1 ) ( n 1 k 1 ) n s ( j = 0 k 1 m j ) 1 1 1 < < k n i = 1 k ( 1 k j = i k + i 1 m j i x j ) 2 ;
    (48)
  2. 2.

    if r=2 and m=1, we have

    i = 1 n x i 2 ( 1 n i = 1 n x i ) 2 k s ( n s 1 ) ( n 1 k 1 ) n s 1 1 < < k n ( 1 k j = 1 k x j ) 2 ;
    (49)
  3. 3.

    if r=2 and m=s=1, then

    i = 1 n x i 2 ( 1 n i = 1 n x i ) 2 k ( n 1 ) ( n 1 k 1 ) n 1 1 < < k n ( 1 k j = 1 k x j ) 2 .
    (50)

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Acknowledgements

The authors appreciate anonymous referees for their valuable comments on and careful corrections to the original version of this paper. This work was partially supported by the NNSF under Grant No. 11361038 of China and by the Foundation of the Research Program of Science and Technology at Universities of Inner Mongolia Autonomous Region under Grant No. NJZY13159, China.

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Xi, BY., Wang, SH. & Qi, F. Some inequalities for (h,m)-convex functions. J Inequal Appl 2014, 100 (2014). https://doi.org/10.1186/1029-242X-2014-100

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  • DOI: https://doi.org/10.1186/1029-242X-2014-100

Keywords

  • convex function
  • (h,m)-convex function
  • Jensen inequality
  • Popoviciu inequality