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A new refinement of discrete Jensen’s inequality depending on parameters

Journal of Inequalities and Applications20132013:551

https://doi.org/10.1186/1029-242X-2013-551

  • Received: 10 June 2013
  • Accepted: 23 October 2013
  • Published:

Abstract

In this paper we give a new refinement of discrete Jensen’s inequality, which generalizes a former result. The introduced sequences depend on parameters. The strict monotonicity and the convergence are investigated. We also study the behavior of the sequences when the parameters vary. One of the proofs requires an interesting convergence theorem with probability theoretical background. This result is an extension of a former result, but its proof is simpler. The results are applied to define and study some new quasi-arithmetic means.

MSC:26D07, 26A51.

Keywords

  • convex function
  • mid-convex function
  • discrete Jensen’s inequality
  • quasi-arithmetic mean

1 Introduction

We begin with discrete Jensen’s inequality which is a useful tool in different parts of mathematics. The set of nonnegative integers and that of positive integers will be denoted by and N + , respectively.

Theorem A (see [1])

Let C be a convex subset of a real vector space X, and let { x 1 , , x n } be a finite subset of C, where n N + is fixed. Let p 1 , , p n be nonnegative numbers with j = 1 n p j = 1 . If f : C R is either a convex or a mid-convex function and in the latter case the numbers p j ( 1 j n ) are rational, then
f ( j = 1 n p j x j ) j = 1 n p j f ( x j ) .
(1)
The function f : C R is called convex if
f ( α x + ( 1 α ) y ) α f ( x ) + ( 1 α ) f ( y ) , x , y C , 0 α 1 ,
(2)
and mid-convex or Jensen-convex if
f ( x + y 2 ) 1 2 f ( x ) + 1 2 f ( y ) , x , y C .

f is strictly convex if strict inequality holds in (2) whenever x y and 0 < α < 1 .

In recent years, a number of papers have appeared in which the authors constructed different refinements of discrete Jensen’s inequality. See, for example, [28] and [9]. These papers also contain a lot of interesting applications. The following parameter-dependent refinement can be found in [4].

Theorem B Let C be a convex subset of a real vector space X, and let { x 1 , , x n } be a finite subset of C, where n N + is fixed. Let p 1 , , p n be nonnegative numbers with j = 1 n p j = 1 , and let λ 1 . Suppose that f : C R is either a convex or a mid-convex function and in the latter case the numbers p j ( 1 j n ) and λ are rational. Introduce the sets
S k : = { ( i 1 , , i n ) N n | j = 1 n i j = k } , k N ,
(3)
and for k N define the numbers
C k ( λ ) = C k ( x 1 , , x n ; p 1 , , p n ; λ ) : = 1 ( n + λ 1 ) k ( i 1 , , i n ) S k k ! i 1 ! i n ! ( j = 1 n λ i j p j ) f ( j = 1 n λ i j p j x j j = 1 n λ i j p j ) .
(4)
Then:
  1. (a)
    f ( j = 1 n p j x j ) = C 0 ( λ ) C 1 ( λ ) C k ( λ ) j = 1 n p j f ( x j ) , k N .
     
  2. (b)
    Suppose that X is a normed space and f is continuous. For every fixed λ > 1 ,
    lim k C k ( λ ) = j = 1 n p j f ( x j ) .
    (5)
     

Our aim is to present a generalization of the previous result.

The sequence ( C k ( λ ) ) k = 0 depends on one parameter, but we allow the new sequence ( D k ( λ ) ) k = 0 (see (7)) to depend on n parameters. We give conditions for the strict monotonicity of ( D k ( λ ) ) k = 0 in some cases.

The proof of limit assertion (5) uses a lemma (see Lemma 15 in [4]) with a rather difficult probability theoretical proof. Essentially, a fundamental theorem from statistics, which is based on the multidimensional central limit theorem, has been applied. In this work, we obtain a convergence theorem extending (5) which will be deduced from another interesting lemma analogous to Lemma 15 in [4]. The proof of this lemma is simpler than the proof of Lemma 15 in [4], although it also has probability theoretical background, namely the strong law of large numbers is used.

The behavior of the sequence ( D k ( λ ) ) k = 0 is studied when the parameters vary. Further refinements of discrete Jensen’s inequality can be derived from these results.

As an application, some new quasi-arithmetic means are constructed, and the monotonicity and convergence of these means are investigated.

2 The main results and some preliminary results

The main results are given by Theorem 1 and Theorem 5. In order to render the main results as transparently as possible, we also give some preparatory lemmas and theorems.

The first result generalizes Theorem B.

Theorem 1 Let C be a convex subset of a real vector space X, and let { x 1 , , x n } be a finite subset of C, where n N + is fixed. Let p 1 , , p n be nonnegative numbers with j = 1 n p j = 1 , and let λ i > 1 ( 1 i n ). Suppose that f : C R is either a convex or a mid-convex function, and in the latter case, the numbers p i and λ i ( 1 i n ) are rational.

For k N , introduce
d ( λ ) = d ( λ 1 , , λ n ) : = j = 1 n 1 λ j 1
(6)
and define
D k ( λ ) = D k ( x 1 , , x n ; p 1 , , p n ; λ 1 , , λ n ) : = 1 ( d ( λ ) + 1 ) k ( i 1 , , i n ) S k k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) ,
(7)

where S k means the set in (3).

Then:
  1. (a)
    f ( j = 1 n p j x j ) = D 0 ( λ ) D 1 ( λ ) D k ( λ ) j = 1 n p j f ( x j ) , k N .
    (8)
     
  2. (b)
    Assuming X is a normed space and f is continuous, we have
    lim k D k ( λ ) = j = 1 n p j f ( x j ) .
    (9)
     

The result contains Theorem B as a special case ( λ = λ 1 = = λ n ) . Since f is continuous in part (b) of the previous result, f is necessarily convex in this case.

The clue of the proof of Theorem 1(b) is the following decisive assertion, which is deduced from the strong law of large numbers. This result extends Lemma 15 in [4] with a less complicated proof.

Theorem 2 Let C be a convex subset of a real normed space X, and let { x 1 , , x n } be a finite subset of C, where n N + is fixed. Let p 1 , , p n be positive numbers with j = 1 n p j = 1 , and let λ i > 1 ( 1 i n ). Suppose f : C R is a convex and continuous function. Define, for l = 1 , , n ,
D l k : = 1 ( d ( λ ) + 1 ) k ( i 1 , , i n ) S k k ! i 1 ! i n ! λ l i l j = 1 n 1 ( λ j 1 ) i j f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) .
Then we have
lim k D l k = f ( x l ) , 1 l n .

Next, we determine some of the cases of strict inequality in (8).

Theorem 3 Let C be a convex subset of a real vector space X, and let x 1 , , x n be points of C with at least two different elements, where either n = 2 or n = 3 . Let p 1 , , p n be positive numbers with j = 1 n p j = 1 , and let λ i > 1 ( 1 i n ). If f : C R is strictly convex, then every inequality is strict in (8).

Probably, the result remains true if n > 3 , but the method applied in the proof of Theorem 3 is getting more and more chaotic. Another treatment of the problem may be successful.

In the proof of Theorem 3, we shall use the following well-known result (see [1]).

Theorem C Let C be a convex subset of a real vector space X, and let x 1 , , x n be points of C with at least two different elements, where n 2 is a fixed integer. Let p 1 , , p n be positive numbers with j = 1 n p j = 1 . If f : C R is strictly convex, then the inequality is strict in (1).

The following notations and observations may help to understand the next results, and they will be used in their proofs. We motivate proceeding in this direction as follows: consider Theorem 1, and let I : = { i { 1 , , n } p i 0 } . It is easy to see that only the restriction of f to the convex hull of { x i i I } has significance.

Let C be a convex subset of a real vector space X, and let { x 1 , , x n } be a finite subset of C, where n 2 is a fixed integer. Suppose that f : C R is a convex function. Choose the integers l { 1 , , n } and 1 l 1 < l 2 < < l m n , where m n 1 and l i l ( i = 1 , , m ). Define the convex set
G m : = { ( t 1 , , t m ) R m | t i 0 ( i = 1 , , m ) , i = 1 m t i 1 } ,
and the function h l 1 , , l m on G m by
h l 1 , , l m ( t 1 , , t m ) : = f ( i = 1 m t i x l i + ( 1 i = 1 m t i ) x l ) .
(10)
Then h l 1 , , l m is well defined since the set
{ i = 1 m t i x l i + ( 1 i = 1 m t i ) x l X | ( t 1 , , t m ) G m }

is the convex hull of { x l 1 , , x l m , x l } , and hence it is a subset of C. A consequence of the convexity of f is that h l 1 , , l m is also convex. Further properties of the function h l 1 , , l m are studied in the following lemma.

Lemma 4 Let C be a convex subset of a real vector space X, and let { x 1 , , x n } be a finite subset of C, where n 2 is a fixed integer. Suppose that f : C R is a convex function. Choose the integers l { 1 , , n } and 1 l 1 < l 2 < < l m n , where m n 1 and l i l ( i = 1 , , m ). Then:
  1. (a)

    h l 1 , , l m is continuous on the interior int ( G m ) of G m .

     
  2. (b)

    h l 1 , , l m is bounded.

     
  3. (c)

    The restriction of h l 1 , , l m to int ( G m ) has a unique continuous (and also convex) extension h ¯ l 1 , , l m to G m .

     
There are some trivial cases of Theorem 1: if p l = 1 for an l { 1 , , n } , then
D k ( λ ) = f ( x l ) , k N ,

and therefore 0 p j < 1 ( 1 j n ) (thus n 2 ) will be supposed in the next main result, in which the parameter λ = ( λ 1 , , λ n ) varies.

Theorem 5 Let C be a convex subset of a real vector space X, and let { x 1 , , x n } be a finite subset of C, where n 2 is a fixed integer. Let p 1 , , p n be nonnegative numbers such that p j < 1 ( 1 j n ) and j = 1 n p j = 1 , and let λ i > 1 ( 1 i n ). Suppose that f : C R is a convex function. Choose l { 1 , , n } . Then, for every fixed λ i ( 1 i n , i l ) and for each fixed k N + :

(a)
lim λ l 1 + D k ( λ ) = f ( j = 1 n p j x j ) .
(b1) In the case p l = 0 ,
lim λ l D k ( λ ) = 1 ( n j l j = 1 1 λ j 1 + 1 ) k ( i 1 , , i n ) S k i l = 0 k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) .
(11)
(b2) In the case 0 < p l (<1), there exists an integer 1 m n 1 such that m elements of the sequence p 1 , , p l 1 , p l + 1 , , p n belong to the open interval ] 0 , 1 [ , and n 1 m members are 0. We can assume that p l 1 , , p l m ] 0 , 1 [ ( 1 l 1 < l 2 < < l m n and they are different from l) and the others are 0. By using the function h ¯ l 1 , , l m (see (10) and Lemma  4(c)), we have
lim λ l D k ( λ ) = 1 ( n j l j = 1 1 λ j 1 + 1 ) k
(12)
( ( i 1 , , i n ) S k i l = 0 k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) + p l h ¯ l 1 , , l m ( 0 , , 0 ) ( ( j = 1 j l n 1 λ j 1 + 1 ) k ( j = 1 j l n 1 λ j 1 ) k ) ) .
(13)

By (8), the limits in (11) and (13) also refine discrete Jensen’s inequality. It should be mentioned that in Theorem 1 the topology on X is needed in showing (b), but Theorem 5 is true without any topology on X.

3 Applications

As an application we can now extend the quasi-arithmetic means introduced in [4] (about means, see [10]).

Definition 6 Let I R be an interval, x j I ( 1 j n ), p 1 , , p n be nonnegative numbers such that j = 1 n p j = 1 , and let λ j > 1 ( 1 j n ). Let φ , ψ : I R be continuous and strictly monotone functions. We define the quasi-arithmetic means with respect to (7) by
M ψ , φ λ ( k ) : = ψ 1 ( 1 ( d ( λ ) + 1 ) k ( i 1 , , i n ) S k k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) ( ψ φ 1 ) ( j = 1 n λ j i j p j φ ( x j ) j = 1 n λ j i j p j ) ) , k N .
(14)

We study the monotonicity and convergence of the new means. For this, the next mean is also needed.

Definition 7 Let I R be an interval, x j I ( 1 j n ), p 1 , , p n be nonnegative numbers such that j = 1 n p j = 1 . For a continuous and strictly monotone function z : I R , we introduce the following mean:
M z : = z 1 ( j = 1 n p j z ( x j ) ) .
(15)

We now prove the monotonicity of the means (14) and give limit formulas.

Proposition 8 Let I R be an interval, x j I ( 1 j n ), p 1 , , p n be nonnegative numbers such that j = 1 n p j = 1 , and let λ j > 1 ( 1 j n ). Let φ , ψ : I R be continuous and strictly monotone functions. Then:
  1. (a)
    M φ = M ψ , φ λ ( 0 ) M ψ , φ λ ( k ) M ψ , k N ,
     
if either ψ φ 1 is convex and ψ is increasing or ψ φ 1 is concave and ψ is decreasing.
  1. (b)
    M φ = M ψ , φ λ ( 0 ) M ψ , φ λ ( k ) M ψ , k N ,
     
if either ψ φ 1 is convex and ψ is decreasing or ψ φ 1 is concave and ψ is increasing.
  1. (c)
    Moreover, in both cases
    lim k M ψ , φ λ ( k ) = M ψ .
     

Proof Theorem 1(a) can be applied to the function ψ φ 1 if it is convex ( ψ φ 1 if it is concave) and to the n-tuples ( φ ( x 1 ) , , φ ( x n ) ) , then upon taking ψ 1 , we get (a) and (b). (c) Comes from Theorem 1(b). □

As an illustration, we consider the following special case.

Example 9 If I : = ] 0 , [ , ψ : = ln and φ ( x ) : = x ( x ] 0 , [ ), then by Proposition  8(b), we have the following sharpened version of the weighted arithmetic mean - geometric mean inequality: for every x j > 0 ( 1 j n ), λ j > 1 ( 1 j n ) and k N + ,
j = 1 n p j x j ( i 1 , , i n ) S k ( j = 1 n λ i j p j x j j = 1 n λ i j p j ) 1 ( d ( λ ) + 1 ) k k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) j = 1 n x j p j .

4 Proofs and some auxiliary results

Before giving the proof of the main result, we introduce some preliminary lemmas. First, a simple inequality is established.

Lemma 10 If x , y > 1 , then
ln ( y ) y 1 x ln ( x ) x 1 < 0 .
Proof We shall show that
ln ( y ) y 1 < 1 < x ln ( x ) x 1 , x , y > 1 .
The left inequality is well known, and the right inequality is equivalent to
u ( x ) = ln ( x ) 1 + 1 x > 0 , x > 1 .
Since
u ( x ) = x 1 x 2 > 0 , x > 1 ,

u is strictly increasing on [ 1 , ) , hence u ( x ) > u ( 1 ) = 0 ( x > 1 ). □

In the second lemma, the measurability of some functions is studied.

Lemma 11 Let ( Ω , A ) be a measurable space, and let X be a real normed space. denotes the σ-algebra of Borel sets in X. If A A , v : A R is measurable, and x X is fixed, then the function
ω v ( ω ) x , ω A
(16)

is A B measurable.

Proof Since X is a normed space, the mapping p : R X , p ( c ) = c x is continuous. Now the result follows from the fact that function (16) can be written as p v . □

After these preparations, we can now prove Theorem 2.

Proof of Theorem 2 We can obviously suppose that n 2 and l = 1 .

Let
q 1 : = λ 1 ( d ( λ ) + 1 ) ( λ 1 1 ) , q j : = 1 ( d ( λ ) + 1 ) ( λ j 1 ) , 2 j n .

Then q j > 0 ( 1 j n ) and j = 1 n q j = 1 , that is, we have a discrete distribution.

Let ( Ω , A , P ) be a probability space, and let
χ k : Ω R , k N +
be identically distributed and independent random variables such that
P ( χ k = j ) = q j , 1 j n , k N + .
Define the random variables
Y j k : = t = 1 k 1 { χ t = j } , 1 j n , k N + ,
where 1 A : Ω R means the characteristic function of the set A Ω . The joint distribution of the vector random variable ( Y 1 k , , Y n k ) is a multinomial distribution
P ( Y 1 k = i 1 , , Y n k = i n ) = k ! i 1 ! i n ! q 1 i 1 q n i n , ( i 1 , , i n ) S k
(17)

for every k N + .

Lemma 11 and the continuity of f imply that the function Z k : Ω R ,
Z k : = f ( j = 1 n λ j Y j k p j x j j = 1 n λ j Y j k p j ) , k N +
is measurable, and therefore it is also a random variable. By applying Lemma 4 with l i : = i ( 1 i n 1 ), f is bounded on the convex hull of the set { x 1 , , x n } . It follows that ( Z k ) k = 1 is also bounded, and thus Z k ( k N + ) is P-integrable. By (17),
E ( Z k ) = Ω Z k d P = D 1 k , k N + .
It is a consequence of Lebesgue’s convergence theorem that
lim k D 1 k = Ω lim k Z k d P .
(18)
By the strong law of large numbers,
lim k Y j k k = q j P -a.e. , 1 j n .
Therefore, by Lemma 10,
lim k ( Y j k k ln ( λ j ) Y 1 k k ln ( λ 1 ) ) = q j ln ( λ j ) q 1 ln ( λ 1 ) < 0 P -a.e. , 2 j n ,
and this leads to
lim k λ j Y j k p j λ 1 Y 1 k p 1 = lim k p j p 1 exp ( k ( Y j k k ln ( λ j ) Y 1 k k ln ( λ 1 ) ) ) = 0 P -a.e. , 2 j n .
Consequently, we get from (18) that
lim k D 1 k = f ( x 1 ) .

The proof is now complete. □

The following lemma extends Lemma 14 in [4].

Lemma 12 Let k N and ( i 1 , , i n ) S k + 1 be fixed. If we set
z ( i 1 , , i n ) : = { l { 1 , , n } i l 0 } ,
then
l z ( i 1 , , i n ) ( k ! i 1 ! i l 1 ! ( i l 1 ) ! i l + 1 ! i n ! 1 ( λ 1 1 ) i 1 1 ( λ l 1 1 ) i l 1 1 ( λ l 1 ) i l 1 1 ( λ l + 1 1 ) i l + 1 1 ( λ n 1 ) i n 1 λ l 1 ) = ( k + 1 ) ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j .

Proof The lowest common denominator is i 1 ! i n ! . Combined with j = 1 n i j = k + 1 , the result follows. □

We can now prove the first main result.

Proof of Theorem 1 (a) The proof is divided into three steps.
  1. I.
    Since S 0 = { ( 0 , , 0 ) } ,
    D 0 ( λ ) = ( j = 1 n λ j 0 p j ) f ( j = 1 n λ j 0 p j x j j = 1 n λ j 0 p j ) = f ( j = 1 n p j x j ) .
     
  2. II.

    Next, we prove that D k ( λ ) D k + 1 ( λ ) ( k N ).

     
It is not hard to see that for every ( i 1 , , i n ) S k ,
j = 1 n λ j i j p j x j j = 1 n λ j i j p j = 1 d ( λ ) + 1 l = 1 n ( 1 λ l 1 j = 1 n λ j i j p j x j + ( λ l 1 ) λ l i l p l x l j = 1 n λ j i j p j + ( λ l 1 ) λ l i l p l j = 1 n λ j i j p j + ( λ l 1 ) λ l i l p l j = 1 n λ j i j p j ) .
(19)
According to Theorem A, this yields that
f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) 1 d ( λ ) + 1 l = 1 n ( 1 λ l 1 j = 1 n λ j i j p j + ( λ l 1 ) λ l i l p l j = 1 n λ j i j p j f ( j = 1 n λ j i j p j x j + ( λ l 1 ) λ l i l p l x l j = 1 n λ j i j p j + ( λ l 1 ) λ l i l p l ) ) .
(20)
Consequently,
D k ( λ ) 1 ( d ( λ ) + 1 ) k + 1 ( i 1 , , i n ) S k ( k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j l = 1 n ( 1 λ l 1 ( j = 1 n λ j i j p j + ( λ l 1 ) λ l i l p l ) f ( j = 1 n λ j i j p j x j + ( λ l 1 ) λ l i l p l x l j = 1 n λ j i j p j + ( λ l 1 ) λ l i l p l ) ) ) .
(21)
Bringing in Lemma 12, we find that the right-hand side of (21) can be written in the form
1 ( d ( λ ) + 1 ) k + 1 ( i 1 , , i n ) S k + 1 ( ( k + 1 ) ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) ) ,
which is just D k + 1 ( λ ) .
  1. III.
    Finally, we prove that
    D k ( λ ) j = 1 n p j f ( x j ) , k N + .
    (22)
     
It follows from Theorem A that
D k ( λ ) 1 ( d ( λ ) + 1 ) k ( i 1 , , i n ) S k ( k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j j = 1 n λ j i j p j f ( x j ) )
(23)
= 1 ( d ( λ ) + 1 ) k j = 1 n ( ( i 1 , , i n ) S k k ! i 1 ! i n ! 1 ( λ 1 1 ) i 1 1 ( λ j 1 1 ) i j 1 λ j i j ( λ j 1 ) i j 1 ( λ j + 1 1 ) i j + 1 1 ( λ n 1 ) i n ) p j f ( x j ) , k N + .
(24)
The multinomial theorem shows that
( i 1 , , i n ) S k k ! i 1 ! i n ! 1 ( λ 1 1 ) i 1 1 ( λ j 1 1 ) i j 1 λ j i j ( λ j 1 ) i j 1 ( λ j + 1 1 ) i j + 1 1 ( λ n 1 ) i n = ( d ( λ ) + 1 ) k , 1 j n ,

hence (24) implies (22).

(b) It is enough to confirm that for l = 1 , , n ,
lim k p l 1 ( d ( λ ) + 1 ) k ( i 1 , , i n ) S k k ! i 1 ! i n ! λ l i l j = 1 n 1 ( λ j 1 ) i j f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) = p l f ( x l ) .

This is evident if p l = 0 . If p l > 0 , then we have it from Theorem 2.

The proof is complete. □

Proof of Theorem 3 It is enough to prove that there is strict inequality for a fixed ( i 1 , , i n ) S k ( k N ) in (20) and in (23), respectively.

The treatment of inequality (23) is pretty immediate. Under the conditions of the theorem,
f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) < 1 j = 1 n λ j i j p j j = 1 n λ j i j p j f ( x j ) , ( i 1 , , i n ) S k ( k N )

by Theorem C. It follows that the last inequality in (8) is strict (this is true for every n 2 ).

It remains to study (20). Let k N be fixed, and let
( i 1 , , i n ) = ( k , 0 , , 0 ) S k .
Under the hypotheses, the coefficients of the vectors
j = 1 n λ j i j p j x j + ( λ l 1 ) λ l i l p l x l j = 1 n λ j i j p j + ( λ l 1 ) λ l i l p l , 1 l n
(25)

are positive in (19), and therefore by Theorem C, it is enough to prove that there exist two different vectors in (25).

We begin with the case n = 2 . Then x 1 x 2 , and (25) consists of two vectors, namely
v 1 : = λ 1 k + 1 p 1 x 1 + p 2 x 2 λ 1 k + 1 p 1 + p 2 , v 2 : = λ 1 k p 1 x 1 + λ 2 p 2 x 2 λ 1 k p 1 + λ 2 p 2 .
Elementary considerations show that v 1 = v 2 implies that
λ 1 k p 1 p 2 ( 1 λ 1 λ 2 ) x 1 = λ 1 k p 1 p 2 ( 1 λ 1 λ 2 ) x 2 ,

and thus x 1 = x 2 , which is a contradiction.

Now, we continue with the case n = 3 . Then the following three vectors belong to (25):
v 1 : = λ 1 k + 1 p 1 x 1 + p 2 x 2 + p 3 x 3 λ 1 k + 1 p 1 + p 2 + p 3 , v 2 : = λ 1 k p 1 x 1 + λ 2 p 2 x 2 + p 3 x 3 λ 1 k p 1 + λ 2 p 2 + p 3 , v 3 : = λ 1 k p 1 x 1 + p 2 x 2 + λ 3 p 3 x 3 λ 1 k p 1 + λ 2 p 2 + λ 3 p 3 .
Suppose v 1 = v 2 = v 3 . A simple but troublesome calculation confirms that
λ 1 k p 1 p 2 ( λ 1 1 ) ( x 1 x 2 ) + λ 1 k p 1 p 3 ( λ 1 λ 3 1 ) ( x 1 x 3 ) + p 2 p 3 ( λ 3 1 ) ( x 2 x 3 ) = 0
and
λ 1 k p 1 p 2 ( λ 1 λ 2 1 ) ( x 1 x 2 ) + λ 1 k p 1 p 3 ( λ 1 1 ) ( x 1 x 3 ) + p 2 p 3 ( 1 λ 2 ) ( x 2 x 3 ) = 0 .
Therefore, recalling that
x 1 x 3 = ( x 1 x 2 ) + ( x 2 x 3 ) ,
we have a homogeneous system of two linear equations with solutions x 1 x 2 and x 2 x 3 . A few easy calculations yield that the determinant of the matrix of this system is
λ 1 k p 1 p 2 p 3 ( λ 1 k + 1 p 1 + p 2 + p 3 ) ( λ 1 + λ 2 + λ 3 λ 1 λ 2 λ 3 2 ) .
(26)
λ i > 1 ( i = 1 , 2 , 3 ) implies that
λ 1 + λ 2 + λ 3 λ 1 λ 2 λ 3 2 < 0 ,
and hence (26) is negative (it is only important that different from 0). From this we get
x 1 x 2 = x 2 x 3 = 0 ,

that is, x 1 = x 2 = x 3 , which is also a contradiction.

The proof is complete. □

Proof of Lemma 4 (a) It is well known (see [11]) that a convex function on an open convex set in R m is continuous.

(b) h l 1 , , l m is bounded above since discrete Jensen’s inequality shows
h l 1 , , l m ( t 1 , , t m ) = f ( i = 1 m t i x l i + ( 1 i = 1 m t i ) x l ) i = 1 m t i f ( x l i ) + ( 1 i = 1 m t i ) f ( x l ) max ( f ( x l 1 ) , , f ( x l m ) , f ( x l ) ) , ( t 1 , , t m ) G m .
Let ( s 1 , , s m ) int ( G m ) be fixed. Since h l 1 , , l m is convex, there is a linear functional L on R m (a support of f at ( s 1 , , s m ) ) such that
h l 1 , , l m ( t 1 , , t m ) h l 1 , , l m ( s 1 , , s m ) + L ( t 1 s 1 , , t m s m ) , ( t 1 , , t m ) int ( G m ) ,
(27)

and therefore the continuity of L on the compact set ( s 1 , , s m ) + G m yields that h l 1 , , l m is bounded below on int ( G m ) . It is easy to check that inequality (27) holds for every boundary point of G m , too.

(c) Because G m is a polytope, this follows from (b) (see [12]).

The proof is complete. □

Proof of Theorem 5 (a) We can obviously suppose that l = n .

An elementary calculation gives
lim λ n 1 + 1 ( d ( λ ) + 1 ) k j = 1 n 1 ( λ j 1 ) i j = lim λ 1 1 + j = 1 n 1 ( λ j + n p j p = 1 λ j 1 λ p 1 ) i j = { 1 if  i n = k , 0 if  i n { 0 , , k 1 } , ( i 1 , , i n ) S k .
(28)
If p n = 0 , then for every ( i 1 , , i n ) S k ,
lim λ n 1 + ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) = ( j = 1 n 1 λ j i j p j ) f ( j = 1 n 1 λ j i j p j x j j = 1 n 1 λ j i j p j ) ,
and therefore, by (28)
lim λ n 1 + D k ( λ ) = f ( j = 1 n 1 p j x j ) ,

which we wanted.

In the rest of the proof, 0 < p n < 1 . From this, the existence of an integer 1 m n 1 such that m elements of the sequence p 1 , , p n 1 belong to the open interval ] 0 , 1 [ and n 1 m members are 0 follows. We can assume that p 1 , , p m ] 0 , 1 [ and p m + 1 = = p n 1 = 0 .

Consider the function h 1 , , m defined in (10).

Fix ( i 1 , , i n ) S k . Then
lim λ n 1 + λ j i j p j j = 1 n λ j i j p j = lim λ n 1 + { λ j i j p j j = 1 m λ j i j p j + p n > 0 if  1 j m , p n j = 1 m λ j i j p j + p n > 0 if  j = n ,
which shows that
lim λ n 1 + ( λ 1 i 1 p 1 j = 1 n λ j i j p j , , λ m i m p m j = 1 n λ j i j p j ) = ( λ 1 i 1 p 1 j = 1 m λ j i j p j + p n , , λ m i m p m j = 1 m λ j i j p j + p n )
is an interior point of G m . In light of the continuity of h 1 , , m on int ( G m ) (see Lemma 4), this implies that
lim λ n 1 + ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) = lim λ n 1 + ( j = 1 m λ j i j p j + λ n i n p n ) h ( λ 1 i 1 p 1 j = 1 n λ j i j p j , , λ m i m p m j = 1 n λ j i j p j ) = ( j = 1 m λ j i j p j + p n ) h ( λ 1 i 1 p 1 j = 1 m λ j i j p j + p n , , λ m i m p m j = 1 m λ j i j p j + p n ) .
(29)
The limit relations (28) and (29) entail
lim λ n 1 + 1 ( d ( λ ) + 1 ) k j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) = { f ( j = 1 m p j x j + p n x n ) if  i n = k , 0 if  i n { 0 , , k 1 } , ( i 1 , , i n ) S k ,

which gives the result.

We begin the proof of the second part of the result with some limit formulas.

It is obvious that
lim λ l 1 ( d ( λ ) + 1 ) k = 1 ( n j l j = 1 1 λ j 1 + 1 ) k .
(30)
It is also easy to calculate that for each ( i 1 , , i n ) S k ,
lim λ l j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) = { j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) , i l = 0 , p l n j l j = 1 ( λ j 1 ) i j , i l 0 .
(31)
(b1) The expression
f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j )
(32)

does not depend on λ l for each ( i 1 , , i n ) S k , from which we have (11) by (30) and (31).

(b2) If i l = 0 , then (32) is independent of λ l , hence we can have the coefficient in (12) and the first member in the sum (13) from (30) and (31).

Suppose ( i 1 , , i n ) S k such that i l 0 .

By using (31), it remains to study the existence of the limit of (32) as λ l . Under the conditions on the zero elements of the sequence p 1 , , p n , we have to consider
lim λ l h l 1 , , l m ( λ l 1 i l 1 p l 1 j = 1 m λ l j i l j p l j + λ l i l p l , , λ l m i l m p l m j = 1 m λ l j i l j p l j + λ l i l p l ) .
(33)
Since
lim λ l ( λ l 1 i l 1 p l 1 j = 1 m λ l j i l j p l j + λ l i l p l , , λ l m i l m p l m j = 1 m λ l j i l j p l j + λ l i l p l ) = ( 0 , , 0 )
and
( λ l 1 i l 1 p l 1 j = 1 m λ l j i l j p l j + λ l i l p l , , λ l m i l m p l m j = 1 m λ l j i l j p l j + λ l i l p l )
is an interior point of the domain G m of h l 1 , , l m for every λ l > 1 , Lemma 4(c) implies that the limit (33) exists and it is h ¯ l 1 , , l m ( 0 , , 0 ) . According to this and (31)
lim λ l j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) = p l h ¯ l 1 , , l m ( 0 , , 0 ) n j l j = 1 ( λ j 1 ) i j ,
showing that
lim λ l ( i 1 , , i n ) S k i l 0 k ! i 1 ! i n ! j = 1 n 1 ( λ j 1 ) i j ( j = 1 n λ j i j p j ) f ( j = 1 n λ j i j p j x j j = 1 n λ j i j p j ) = p l h ¯ l 1 , , l m ( 0 , , 0 ) ( i 1 , , i n ) S k i l 0 k ! i 1 ! i n ! 1 n j l j = 1 ( λ j 1 ) i j = p l h ¯ l 1 , , l m ( 0 , , 0 ) i l = 1 k ( ( k i l + 1 ) k i l ! i 1 + + i l 1 + i l + 1 + + i n = k i l ( k i l ) ! i 1 ! i l 1 ! i l + 1 ! i n ! 1 n j l j = 1 ( λ j 1 ) i j ) = p l h ¯ l 1 , , l m ( 0 , , 0 ) i l = 1 k ( k k i l ) ( j = 1 j l 1 λ j 1 ) k i l = p l h ¯ l 1 , , l m ( 0 , , 0 ) ( ( j = 1 j l n 1 λ j 1 + 1 ) k ( j = 1 j l n 1 λ j 1 ) k ) .

We have got the second sum in (13).

The proof is complete. □

Declarations

Acknowledgements

The author thanks the anonymous referees for their careful reading and valuable comments. This work was supported by Hungarian National Foundations for Scientific Research Grant No. K101217.

Authors’ Affiliations

(1)
Department of Mathematics, University of Pannonia, Egyetem u. 10, Veszprém, 8200, Hungary

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© Horváth; licensee Springer. 2013

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