a method to refine the discrete jensen’s inequality for convex and mid-convex functions

Mathematical and Computer Modelling 54 (2011) 2451–2459

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Mathematical and Computer Modelling

journal homepage: www.elsevier.com/locate/mcm

A method to refine the discrete Jensen’s inequality for convex andmid-convex functionsLászló Horváth ∗

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

a r t i c l e i n f o

Article history:Received 19 May 2010Received in revised form 30 May 2011Accepted 31 May 2011

Keywords:ConvexMid-convexDiscrete Jensen’s inequalities

a b s t r a c t

A general method is developed to refine the discrete Jensen’s inequality in the convexand mid-convex cases. A number of refinements of the discrete Jensen’s inequality can beobtained by using themethod. The results generalize well known inequalities and give alsoa new treatment of them. The results are applied to some special situations.

© 2011 Elsevier Ltd. All rights reserved.

1. Introduction and the main result

The fundamental discrete Jensen’s inequalities for convex and mid-convex functions say that

Theorem A (See [1]). Let C be a convex subset of a real vector space X, let xi ∈ C, and let pi ≥ 0 (i = 1, . . . , n)with∑n

i=1 pi = 1.(a) If f : C → R is a convex function, then

f

n−

i=1

pixi

≤

n−i=1

pif (xi) . (1)

(b) If f : C → R is a mid-convex function, and pi is rational (i = 1, . . . , n), then (1) also holds.

Here the function f : C → R is called convex if

f (βx + (1 − β) y) ≤ βf (x) + (1 − β) f (y), x, y ∈ C, 0 ≤ β ≤ 1,

and mid-convex if

fx + y2

≤

12f (x) +

12f (y), x, y ∈ C .

In the paper [2] we give a refinement of Theorem A, which generalizes and unifies some previous results (see [3,4]). Thefollowing conditions are used in [2] and they will be essential in the sequel too.

(H1) Let V be a real vector space, let C be a convex subset of V , and let x1, . . . , xn ∈ C , where n ≥ 1 is a fixed integer.(H2) Let p1, . . . , pn > 0 such that

∑nj=1 pj = 1.

(H3) Let the function f : C → R be convex.(H4) Let the function f : C → R be mid-convex, and let p1, . . . , pn be rational.Now we recall the central result of [2].

∗ Tel.: +36 88 624227.E-mail address: [email protected].

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2452 L. Horváth / Mathematical and Computer Modelling 54 (2011) 2451–2459

Theorem B. Let n ≥ 1 and k ≥ 1 be fixed integers, and let Ik ⊂ {1, . . . , n}k such that

αIk,i ≥ 1, 1 ≤ i ≤ n,

where αIk,i denotes the number of occurrences of i in the sequences (i1, . . . , ik) ∈ Ik. Assume that (H1)–(H2) and either(H3) or (H4) are satisfied. Then

f

n−

r=1

prxr

≤ Ak,k ≤

n−r=1

pr f (xr), (2)

where

Ak,k = Ak,k (Ik, x1, . . . , xn, p1, . . . , pn)

:=

−(i1,...,ik)∈Ik

k−

s=1

pisαIk,is

f

k∑

s=1

pisαIk,is

xis

k∑s=1

pisαIk,is

.

Indeed, with somewhat more preliminaries it can be shown that (2) has the form

f

n−

r=1

prxr

≤ Ak,k ≤ Ak,k−1 ≤ · · · ≤ Ak,2 ≤ Ak,1 =

n−r=1

pr f (xr), (3)

but we do not define the numbers Ak,l (l = 1, . . . , k − 1) here (in this connection see [2]).The aim of this paper is to give such a generalization of Theorem B which shows the essence of the methods employed

in a lot of known results and unifies them.Let X be a set. The power set of X is denoted by P(X). |X | means the number of elements in X . For every nonnegative

integerm, let

Pm(X) := {Y ⊂ X | |Y | = m} .

In preparation for the main result we first introduce two further hypotheses:(H5) Let S1, . . . , Sn be finite, pairwise disjoint and nonempty sets, let

S :=

nj=1

Sj,

and let c be a function from S into R such that

c(s) > 0, s ∈ S, and−s∈Sj

c(s) = 1, j = 1, . . . , n. (4)

Let the function τ : S → {1, . . . , n} be defined by

τ(s) := j, if s ∈ Sj.

(H6) Suppose A ⊂ P(S) is a partition of S into pairwise disjoint and nonempty sets. Let

k := max {|A| | A ∈ A} ,

and let

Al := {A ∈ A | |A| = l} , l = 1, . . . , k.

(We note that Al (l = 1, . . . , k − 1) may be the empty set, and of course, |S| =∑k

l=1 l |Al|.)Now we are in a position to formulate the main result. The empty sum of numbers or vectors is taken to be zero.

Theorem 1. (a) If (H1)–(H3) and (H5)–(H6) are satisfied, then

f

n−

j=1

pjxj

≤ Nk ≤ Nk−1 ≤ · · · ≤ N2 ≤ N1 =

n−j=1

pjf (xj), (5)

where

Nk :=

k−l=1

−A∈Al

−s∈A

c(s)pτ(s)

f

∑s∈A

c(s)pτ(s)xτ(s)∑s∈A

c(s)pτ(s)

, (6)

L. Horváth / Mathematical and Computer Modelling 54 (2011) 2451–2459 2453

and for every 1 ≤ m ≤ k − 1 the number Nk−m is given by

Nk−m :=

m−l=1

−A∈Al

−s∈A

c(s)pτ(s)f (xτ(s))

+

k−l=m+1

m!

(l − 1) · · · (l − m)

×

−A∈Al

−B∈Pl−m(A)

−s∈B

c(s)pτ(s)

f

∑s∈B

c(s)pτ(s)xτ(s)∑s∈B

c(s)pτ(s)

. (7)

(b) Suppose (H1)–(H2) and (H4)–(H6). If the numbers c(s) (s ∈ S) are rational, then the inequality (5) remains true.

2. Discussion and examples

The first application of Theorem 1 leads to a generalization of Theorem B.

Theorem 2. Let n ≥ 1 and k ≥ 1 be fixed integers, and let Ik ⊂ {1, . . . , n}k such that

αIk,i ≥ 1, 1 ≤ i ≤ n,

where αIk,i means the number of occurrences of i in the sequences (i1, . . . , ik) ∈ Ik. For j = 1, . . . , n we introduce the sets

Sj := {((i1, . . . , ik) , l) | (i1, . . . , ik) ∈ Ik, 1 ≤ l ≤ k, il = j} . (8)

Let c be a positive function on S :=n

j=1 Sj such that−((i1,...,ik),l)∈Sj

c ((i1, . . . , ik) , l) = 1, j = 1, . . . , n. (9)

(a) Assume that (H1)–(H3) are satisfied. Then

f

n−

j=1

pjxj

≤ Nk ≤ Nk−1 ≤ · · · ≤ N2 ≤ N1 =

n−j=1

pjf (xj), (10)

where the numbers Nk−m (0 ≤ m ≤ k − 1) can be written in the following forms:

Nk :=

−(i1,...,ik)∈Ik

k−l=1

c ((i1, . . . , ik) , l) pil

f

k∑

l=1c ((i1, . . . , ik) , l) pilxil

k∑l=1

c ((i1, . . . , ik) , l) pil

,

and for every 1 ≤ m ≤ k − 1

Nk−m :=m!

(k − 1) · · · (k − m)

−(i1,...,ik)∈Ik

−1≤l1<···<lk−m≤k

k−m−j=1

c(i1, . . . , ik) , lj

pilj

× f

k−m∑l=1

c(i1, . . . , ik) , lj

pilj xilj

k−m∑l=1

c(i1, . . . , ik) , lj

pilj

.

(b) If (H1)–(H2) and (H4) are satisfied and the numbers c ((i1, . . . , ik) , l) (((i1, . . . , ik) , l) ∈ S) are rational, then the inequality(10) remains true.

An immediate consequence of the previous result is Theorem B: Choosing

c ((i1, . . . , ik) , l) =1Sj =

1αIk,j

if ((i1, . . . , ik) , l) ∈ Sj,

we can check easily that the inequality (10) corresponds to the inequality (3).


Using Theorem B, some extensions of the main results in [3,4] were obtained by Horváth and Pečarić (see Examples 2and 3 in [2]). Theorem 2 generalizes all these results: Apply it to either

Ik :=(i1, . . . , ik) ∈ {1, . . . , n}k | i1 < · · · < ik

, 1 ≤ k ≤ n,

or

Ik :=(i1, . . . , ik) ∈ {1, . . . , n}k | i1 ≤ · · · ≤ ik

, 1 ≤ k.

Now we apply Theorem 1 to some special situations which correspond to some recent results.

Example 3. Let n, m, r be fixed integers, where n ≥ 3, m ≥ 2 and 1 ≤ r ≤ n − 2. In this example, for every i = 1, 2, . . . , nand for every l = 0, 1, . . . , r the integer i + l will be identified with the uniquely determined integer j from {1, . . . , n} forwhich

l + i ≡ j (mod n). (11)

Introducing the notation

D := {1, . . . , n} × {0, . . . , r} ,

let for every j ∈ {1, . . . , n}

Sj := {(i, l) ∈ D | i + l ≡ j (mod n)}

{j} ,

and let A ⊂ P(S) (S :=n

j=1 Sj) contain the following sets:

Ai := {(i, l) ∈ D | l = 0, . . . , r} , i = 1, . . . , n

and

A := {1, . . . , n} .

Let c be a positive function on S such that−(i,l)∈Sj

c (i, l) + c (j) = 1, j = 1, . . . , n.

Acareful verification shows that the sets S1, . . . , Sn, the partitionA and the function c defined above satisfy the conditions(H5) and (H6),

τ (i, l) = i + l, (i, l) ∈ D,

(by the agreement (see (11)), i + l is identified with j)

τ (j) = j, j = 1, . . . , n,Sj = r + 2, j = 1, . . . , n,

and

|Ai| = r + 1, i = 1, . . . , n, |A| = n.

Now we suppose that either (H1)–(H3) or (H1)–(H2) and (H4) are satisfied and in the latter case the numbers c (i, l)((i, l) ∈ D) and c (j) (j = 1, . . . , n) are rational. Then by Theorem 1

f

n−

j=1

pjxj

≤ Nk =

n−i=1

r−l=0

c (i, l) pi+l

f

r∑

l=0c (i, l) pi+lxi+l

r∑l=0

c (i, l) pi+l

+

n−

j=1

c(j)pj

f

n∑

j=1c(j)pjxj

n∑j=1

c(j)pj

≤

n−j=1

pjf (xj). (12)

In case

pj :=1n, j = 1, . . . , n,

c (i, l) :=1

m (r + 1), (i, l) ∈ D, c(j) :=

m − 1m

j = 1, . . . , n,


it follows from (12) that

f

1n

n−j=1

xj

≤

1mn

n−i=1

fxi + xi+1 + · · · + xi+r

r + 1

+

m − 1m

f

1n

n−j=1

xj

≤1n

n−j=1

f (xj), (13)

which is an essential part of Theorem 2.1 in [5]. In reality, in that theorem a sequence of inequalities (similar to (5)) has beenproved. On the one hand (12) generalizes (13), on the other hand the sequence of inequalities in (5) is different from thatin (13).

Example 4. Let n and k be fixed positive integers. Let

D :=(i1, . . . , in) ∈ {1, . . . , k}n | i1 + · · · + in = n + k − 1

,

and for each j = 1, . . . , n, denote Sj the set

Sj := D × {j} .

For every (i1, . . . , in) ∈ D designate by A(i1,...,in) the set

A(i1,...,in) := {((i1, . . . , in) , l) | l = 1, . . . , n} .

It is obvious that Sj (j = 1, . . . , n) and A(i1,...,in) ((i1, . . . , in) ∈ D) are decompositions of S :=n

j=1 Sj into pairwise disjointand nonempty sets, respectively. Let c be a function on S such that

c ((i1, . . . , in) , j) > 0, ((i1, . . . , in) , j) ∈ S

and −(i1,...,in)∈D

c ((i1, . . . , in) , j) = 1, j = 1, . . . , n. (14)

In summary we have that the conditions (H5) and (H6) are valid, and

τ ((i1, . . . , in) , j) = j, ((i1, . . . , in) , j) ∈ S.

Nowwe suppose that either (H1)–(H3) or (H1)–(H2) and (H4) are satisfied and in the latter case the numbers c((i1, . . . , in), j)(((i1, . . . , in), j) ∈ S) are rational. Then by Theorem 1

f

n−

j=1

pjxj

≤ Nk =

−(i1,...,in)∈D

n−l=1

c ((i1, . . . , in) , l) pl

f

n∑

l=1c ((i1, . . . , in) , l) plxl

n∑l=1

c ((i1, . . . , in) , l) pl

≤

n−j=1

pjf (xj). (15)

If we set

pj :=1n, j = 1, . . . , n,

and

c ((i1, . . . , in) , j) :=ij

n+k−1k−1

,

then (14) holds, since by some combinatorial considerations

|D| =

n + k − 2n − 1

,

and −(i1,...,in)∈D

ij =n + k − 1

n

n + k − 2n − 1

=

n + k − 1k − 1

, j = 1, . . . , n.


In this situation (15) can be written in the form

f

1n

n−j=1

xj

≤

1n+k−2k−1

−(i1,...,in)∈D

1

n + k − 1

n−l=1

ilxl

≤

1n

n−j=1

f (xj), (16)

which inequality is contained in Theorem 1 of [6]. The inequality (16) is placed in a more general framework in [6], but thetreatment of (16) is different from our approach. Theorem 1 generalizes (16) and (5) gives a new sequence of inequalitieseven in the considered special case.

Let us close this section by deriving a sharpened version of the arithmetic mean–geometric mean inequality.

Example 5. Let n ≥ 2 be a fixed positive integer, let

Sj :=(i, j) ∈ {1, . . . , n}2 | i = 1, . . . , j

, j = 1, . . . , n,

and let

Ai :=(i, j) ∈ {1, . . . , n}2 | j = i, . . . , n

, i = 1, . . . , n.

If x1, . . . , xn are positive numbers, then it follows from Theorem 1(a) that

− lnx1 + · · · + xn

n

≤

n−i=1

−

1n

n−j=i

1j

ln

n∑j=i

xjj

n∑j=i

1j

≤ −ln (x1) + · · · + ln (xn)

n,

and therefore

(x1 · · · xn)1n ≤

n∏i=1

n∑j=i

xjj

n∑j=i

1j

1n

n∑j=i

1j

≤x1 + · · · + xn

n.

3. Preliminary results and the proofs

We confine here our attention to the proof of Theorem 1(a), so we shall suppose the conditions (H1)–(H3) and(H5)–(H6). It is easy to verify that the following results and their proofs remain valid under the hypotheses of Theorem 1(b)(Theorem A(b) can be applied in place of Theorem A(a)).

Lemma 6. If (H1)–(H3) and (H5)–(H6) are satisfied, then

f

n−

j=1

pjxj

≤ Nk.

Proof. Since {S1, . . . , Sn} and A are partitions of S into pairwise disjoint and nonempty sets, it comes from the definition ofthe function τ that

n−j=1

pjxj =

n−j=1

−s∈Sj

c(s)

pjxj =

−A∈A

−s∈A

c(s)pτ(s)xτ(s)

=

k−l=1

−A∈Al

−s∈A

c(s)pτ(s)xτ(s)

. (17)

Therefore, recalling that the numbers pj (j = 1, . . . , n) and c(s) (s ∈ S) are positive, we have

n−j=1

pjxj =

k−l=1

−A∈Al

−s∈A

c(s)pτ(s)

∑s∈A


c(s)pτ(s)

. (18)


Similar reasoning as above leads to

k−l=1

−A∈Al

−s∈A

c(s)pτ(s)

=

−A∈A

−s∈A

c(s)pτ(s)

=

n−j=1

−s∈Sj

c(s)

pj.

Then, using (H2) and (4) we get

k−l=1

−A∈Al

−s∈A

c(s)pτ(s)

= 1. (19)

It now follows from (18), (19) and Theorem A(a) that

f

n−

j=1

pjxj

≤

k−l=1

−A∈Al

−s∈A

c(s)pτ(s)

f

∑s∈A


c(s)pτ(s)

,

as was being claimed. �

Lemma 7. If (H1)–(H3) and (H5)–(H6) are satisfied, and A ∈ Al, where 2 ≤ l ≤ k, then

f

∑s∈A


c(s)pτ(s)

≤1

(l − 1)∑s∈A

c(s)pτ(s)

−B∈Pl−1(A)

−s∈B

c(s)pτ(s)

f

∑s∈B


c(s)pτ(s)

.

Proof. A simple calculation confirms that∑s∈A


c(s)pτ(s)=

−B∈Pl−1(A)

∑s∈B

c(s)pτ(s)xτ(s)

(l − 1)∑s∈A

c(s)pτ(s)

=

−B∈Pl−1(A)

∑s∈B

c(s)pτ(s)

(l − 1)∑s∈A

c(s)pτ(s)·

∑s∈B


c(s)pτ(s)

. (20)

Since

−B∈Pl−1(A)

∑s∈B

c(s)pτ(s)

(l − 1)∑s∈A

c(s)pτ(s)

= 1,

the result follows from (20) and Theorem A(a).The proof is complete. �

Proof of Theorem 1(a). The definition of the number N1 shows that

N1 =

k−1−l=1

−A∈Al

−s∈A


+

(k − 1)!(k − 1) · · · 1

−A∈Ak

−B∈P1(A)

−s∈B

c(s)pτ(s)

f

∑s∈B


c(s)pτ(s)

=

k−1−l=1

−A∈Al

−s∈A


+

−A∈Ak

−s∈A


=

k−l=1

−A∈Al

−s∈A


.

Therefore

N1 =

n−j=1

pjf (xj)

follows by an argument entirely similar to that for (17).


So according to Lemma 6 only the task of confirming the inequalities

Nk ≤ Nk−1 ≤ · · · ≤ Nk−m ≤ · · · ≤ N2 ≤ N1 (21)

remains.To this end, we suppose first that Ak = A and thus A1 = · · · = Ak−1 = ∅. By Lemma 7

Nk =

−A∈Ak

−s∈A

c(s)pτ(s)

f

∑s∈A


c(s)pτ(s)

≤

−A∈Ak

−s∈A

c(s)pτ(s)

1

(k − 1)∑s∈A

c(s)pτ(s)

−B∈Pk−1(A)

−s∈B

c(s)pτ(s)

f

∑s∈B


c(s)pτ(s)

=

1k − 1

−A∈Ak

−B∈Pk−1(A)

−s∈B

c(s)pτ(s)

f

∑s∈B


c(s)pτ(s)

= Nk−1. (22)

Suppose then that 1 ≤ m ≤ k − 2. By applying Lemma 7 again, we have

Nk−m =m!

(k − 1) · · · (k − m)

−A∈Ak

−B∈Pk−m(A)

−s∈B

c(s)pτ(s)

f

∑s∈B


c(s)pτ(s)

≤

m!

(k − 1) · · · (k − m)

−A∈Ak

−B∈Pk−m(A)

−s∈B

c(s)pτ(s)

1

(k − m − 1)∑s∈B

c(s)pτ(s)

×

−C∈Pk−m−1(B)

−s∈C

c(s)pτ(s)

f

∑s∈C

c(s)pτ(s)xτ(s)∑s∈C

c(s)pτ(s)

=

m!

(k − 1) · · · (k − m) (k − m − 1)

×

−A∈Ak

−B∈Pk−m(A)

−C∈Pk−m−1(B)

−s∈C

c(s)pτ(s)

f

∑s∈C


c(s)pτ(s)

=

(m + 1)!(k − 1) · · · (k − m) (k − m − 1)

−A∈Ak

−C∈Pk−m−1(A)

−s∈C

c(s)pτ(s)

f

∑s∈C


c(s)pτ(s)

= Nk−m−1. (23)

Together with (22) this gives (21) in the considered special case.In the general case we can majorize the members

−A∈Al

−s∈A

c(s)pτ(s)

f

∑s∈A


c(s)pτ(s)

, 2 ≤ l ≤ k

in (6) exactly as Nk in (22). Similarly, the argument employed in the proof of the inequality Nk−m ≤ Nk−m−1 in (23) can beextended to estimate the members

m!

(l − 1) · · · (l − m)

−A∈Al

−B∈Pl−m(A)

−s∈B

c(s)pτ(s)

f

∑s∈B


c(s)pτ(s)

,

3 ≤ l ≤ k, 1 ≤ m ≤ l − 2

in (7). Now (21) follows from these facts.The proof is complete. �


Proof of Theorem 2. It is obvious that the sets S1, . . . , Sn and the function c defined in the theorem satisfy the condition(H5). In this case

τ ((i1, . . . , ik) , l) = il, ((i1, . . . , ik) , l) ∈ S.

The condition (H6) is also fulfilled if

A := {{((i1, . . . , ik) , l) | l = 1, . . . , k} | (i1, . . . , ik) ∈ Ik} .

Then Ak = A and Al = ∅ (l = 1, . . . , k − 1).The result can be obtained by an application of Theorem 1 in this environment. �

Acknowledgment

Supported by Hungarian National Foundations for Scientific Research Grant No. K73274.

References

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(1988) 463–467.[5] L.-C. Wang, X.-F. Ma, L.-H. Liu, A note on some new refinements of Jensen’s inequality for convex functions, J. Inequal. Pure Appl. Math. 10 (2009) p. 6.

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a method to refine the discrete jensen’s inequality for convex and mid-convex functions

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