Research ArticleA Generalized Hermite-Hadamard Inequality for CoordinatedConvex Function and Some Associated Mappings

Atiq Ur Rehman,1 Gulam Farid,1 and Sidra Malik2

1COMSATS Institute of Information Technology, Attock Campus, Attock, Pakistan2Government Islamia High School, Attock, Pakistan

Correspondence should be addressed to Atiq Ur Rehman; atiq@mathcity.org

Received 25 July 2016; Accepted 25 October 2016

Academic Editor: Nan-Jing Huang

Copyright © 2016 Atiq Ur Rehman et al. This is an open access article distributed under the Creative Commons AttributionLicense, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properlycited.

We have discussed the generalization of Hermite-Hadamard inequality introduced by Lupas for convex functions on coordinatesdefined in a rectangle from the plane. Also we define that mappings are related to it and their properties are discussed.

1. Introduction

Convex functions have great importance in many areas ofMathematics. A real valued function 𝑓 : 𝐼 → R, where 𝐼 isinterval inR, is said to be convex if for 𝛼, 𝛽 ≥ 0with 𝛼+𝛽 = 1

𝑓 (𝛼𝑥 + 𝛽𝑦) ≤ 𝛼𝑓 (𝑥) + 𝛽𝑓 (𝑦) . (1)

In particular, if 𝛼 = 𝛽 = 1/2 we have that𝑓(𝑥 + 𝑦2 ) ≤ 𝑓 (𝑥) + 𝑓 (𝑦)2 . (2)

One of the most important inequalities that has attractedmany mathematician in this field in the last few decades isthe famous Hermite-Hadamard inequality, which establishesa refinement of (2) and it involves the notions of convexity.

Let 𝑓 : 𝐼 → R be a convex mapping defined on theinterval 𝐼 of real numbers and 𝑎, 𝑏 ∈ 𝐼with 𝑎 < 𝑏. The doubleinequality

𝑓(𝑎 + 𝑏2 ) ≤ 1𝑏 − 𝑎 ∫

𝑏

𝑎𝑓 (𝑥) 𝑑𝑥 ≤ 𝑓 (𝑎) + 𝑓 (𝑏)2 (3)

is known as the Hermite-Hadamard inequality. This inequal-ity was published by Hermite in 1883 and was independentlyproved byHadamard in 1893 (seeMitrinovic and Lackovic [1]for the whole history). It gives us an estimation of the mean

value of a convex function 𝑓 and it is important to note that(3) provides a refinement to the Jensen inequality. Fink [2] hasworked on a best possible Hermite-Hadamard inequality andhas deduced this inequality with least restrictions. Dragomirin [3, 4] worked on Hermite-Hadamard inequality and amapping in connection to it. Dragomir and Milosevic [5]gave some refinements of Hadamard's inequalities and itsapplications (see also [6–8] for more results). Dragomir andPearce [9] wrote amonograph on selected topics onHermite-Hadamard inequalities.

The aim of this paper is to discuss an analogue of thegeneralization of Hermite-Hadamard inequality introducedby Lupas for convex functions defined in a rectangle fromthe plane. Also some related mappings are defined and theirproperties are discussed.

In [10] Dragomir has given the concept of convex func-tions on the coordinates in a rectangle from the plane andestablished the Hermite-Hadamard inequality for it.

Definition 1. Let Δ2 = [𝑎, 𝑏] × [𝑐, 𝑑] ⊂ R with 𝑎 < 𝑏 and𝑐 < 𝑑. A function 𝑓 : Δ2 → R is called convex on coordinatesif the partial mappings 𝑓𝑦 : [𝑎, 𝑏] → R, 𝑓𝑦(𝑢) fl 𝑓(𝑢, 𝑦) and𝑓𝑥 : [𝑐, 𝑑] → R, 𝑓𝑥(V) fl 𝑓(𝑥, V) are convex, defined for all𝑦 ∈ [𝑐, 𝑑] and 𝑥 ∈ [𝑎, 𝑏].

Note that every convex mapping 𝑓 : Δ2 → R is convexon the coordinates but the converse is not generally true.

Hindawi Publishing CorporationJournal of MathematicsVolume 2016, Article ID 1631269, 9 pageshttp://dx.doi.org/10.1155/2016/1631269

2 Journal of Mathematics

Theorem 2. Suppose that 𝑓 : Δ2 → R is convex on the coordi-nates on Δ2. Then we have

𝑓(𝑎 + 𝑏2 , 𝑐 + 𝑑2 ) ≤ 12 [1

𝑏 − 𝑎 ∫𝑏

𝑎𝑓(𝑥, 𝑐 + 𝑑2 )𝑑𝑥

+ 1𝑑 − 𝑐 ∫

𝑑

𝑐𝑓(𝑎 + 𝑏2 , 𝑦) 𝑑𝑦] ≤ 1

(𝑏 − 𝑎) (𝑑 − 𝑐)⋅ ∫𝑏𝑎∫𝑑𝑐𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 ≤ 14 [

1𝑏 − 𝑎 ∫

𝑏

𝑎𝑓 (𝑥, 𝑐) 𝑑𝑥

+ 1𝑏 − 𝑎 ∫

𝑏

𝑎𝑓 (𝑥, 𝑑) 𝑑𝑥 + 1

𝑑 − 𝑐 ∫𝑑

𝑐𝑓 (𝑎, 𝑦) 𝑑𝑦

+ 1𝑑 − 𝑐 ∫

𝑑

𝑐𝑓 (𝑏, 𝑦) 𝑑𝑦]

≤ 𝑓 (𝑎, 𝑐) + 𝑓 (𝑎, 𝑑) + 𝑓 (𝑏, 𝑐) + 𝑓 (𝑏, 𝑑)4 .

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In [10], Dragomir for 𝑓 : Δ2 → R, defined a mapping𝐻 : [0, 1]2 → R as

𝐻(𝑡, 𝑠) = 1(𝑏 − 𝑎) (𝑑 − 𝑐) ∫

𝑏

𝑎∫𝑑𝑐𝑓(𝑡𝑥

+ (1 − 𝑡) 𝑎 + 𝑏2 , 𝑠𝑦 + (1 − 𝑠) 𝑐 + 𝑑2 )𝑑𝑦𝑑𝑥(5)

and proved the properties of this mapping in followingtheorem.

Theorem 3. Suppose that 𝑓 : Δ2 → R is convex on thecoordinates on Δ2.

(i) Themapping𝐻 is convex on the coordinates on [0, 1]2.(ii) We have the bounds

sup(𝑠,𝑡)∈[0,1]2

𝐻(𝑠, 𝑡)

= 1(𝑏 − 𝑎) (𝑑 − 𝑐) ∫

𝑏

𝑎∫𝑑𝑐𝑓 (𝑥, 𝑦) 𝑑𝑦 𝑑𝑥 = 𝐻 (1, 1) ,

inf(𝑠,𝑡)∈[0,1]2

𝐻(𝑠, 𝑡) = 𝑓(𝑎 + 𝑏2 , 𝑐 + 𝑑2 ) = 𝐻 (0, 0) .(6)

(iii) The mapping 𝐻 is monotonic nondecreasing on thecoordinates.

Also, in [10] Dragomir gave the following mapping,which is closely connected with Hadamard's inequality, �� :[0, 1]2 → R defined as

�� (𝑡, 𝑠) fl 1(𝑏 − 𝑎)2 (𝑑 − 𝑐)2

⋅ ∫𝑏𝑎∫𝑏𝑎∫𝑑𝑐∫𝑑𝑐𝑓 (𝑡𝑥 + (1 − 𝑡) 𝑦, 𝑠𝑧 + (1 − 𝑠) 𝑢) 𝑑𝑥 𝑑𝑦 𝑑𝑧 𝑑𝑢.

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and proved the following properties of this mapping.

Theorem 4. Suppose that 𝑓 : Δ2 → R is convex on the coordi-nates on Δ2.

(i) We have the equalities

�� (𝑡 + 12 , 𝑠) = �� (12 − 𝑡, 𝑠)∀𝑡 ∈ [0, 12] , 𝑠 ∈ [0, 1] ,

�� (𝑡, 𝑠 + 12) = �� (𝑡, 12 − 𝑠)∀𝑡 ∈ [0, 1] , 𝑠 ∈ [0, 12] ,

�� (1 − 𝑡, 𝑠) = �� (𝑡, 𝑠) ,�� (𝑡, 1 − 𝑠) = �� (𝑡, 𝑠)

∀ (𝑡, 𝑠) ∈ Δ2.

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(ii) �� is convex on the coordinates.(iii) We have the bounds

inf(𝑡,𝑠)∈[0,1]2

�� (𝑡, 𝑠) = �� (12 ,12) =

1(𝑏 − 𝑎)2 (𝑑 − 𝑐)2

⋅ ∫𝑏𝑎∫𝑏𝑎∫𝑑𝑐∫𝑑𝑐𝑓(𝑥 + 𝑦2 , 𝑧 + 𝑢2 ) 𝑑𝑥 𝑑𝑦𝑑𝑧 𝑑𝑢,

sup(𝑡,𝑠)∈[0,1]2

�� (𝑡, 𝑠) = �� (0, 0) = �� (1, 1) = 1(𝑏 − 𝑎) (𝑑 − 𝑐)

⋅ ∫𝑏𝑎∫𝑑𝑐𝑓 (𝑥, 𝑧) 𝑑𝑧 𝑑𝑥.

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(iv) The mapping ��(⋅, 𝑠) is monotonic nondecreasing on[0, 1/2) and nondecreasing on [1/2, 1] for all 𝑠 ∈ [0, 1].A similar property has the mapping ��(𝑡, ⋅) for all 𝑡 ∈[0, 1].

(v) We have the inequality

�� (𝑡, 𝑠) ≥ max {𝐻 (𝑡, 𝑠) ,𝐻 (1 − 𝑡, 𝑠) ,𝐻 (𝑡, 1 − 𝑠) ,𝐻 (1 − 𝑡, 1 − 𝑠)} . (10)

A generalized form of Hermite-Hadamard inequality isgiven by Lupas in [11] (see also [12, page 143]).

Theorem5. Let p, q be given positive numbers and 𝑎 < 𝑏.Thenthe inequality,

𝑓(𝑝𝑎 + 𝑞𝑏𝑝 + 𝑞 ) ≤ 12𝑦 ∫𝐴+𝑦

𝐴−𝑦𝑓 (𝑡) 𝑑𝑡 ≤ 𝑝𝑓 (𝑎) + 𝑞𝑓 (𝑏)𝑝 + 𝑞 , (11)

holds for𝐴 = (𝑝𝑎+𝑞𝑏)/(𝑝+𝑞),𝑦 > 0 and all continuous convexfunctions 𝑓 : [𝑎, 𝑏] → R iff 𝑦 ≤ ((𝑏 − 𝑎)/(𝑝 + 𝑞))min (𝑝, 𝑞).

Journal of Mathematics 3

2. Main Results

The following results comprise generalization of Hermite-Hadamard inequality introduced by Lupas on coordinates ina rectangle from the plane.

Theorem 6. Let Δ2 = [𝑎1, 𝑏1] × [𝑎2, 𝑏2] ⊂ R2 and 𝑝1, 𝑝2, 𝑞1,and 𝑞2 be positive real numbers and

𝐴 𝑖 = 𝑝𝑖𝑎𝑖 + 𝑞𝑖𝑏𝑖𝑝𝑖 + 𝑞𝑖 ,𝑦𝑖 > 0,

where 𝑦𝑖 ≤ 𝑏𝑖 − 𝑎𝑖𝑝𝑖 + 𝑞𝑖 min (𝑝𝑖, 𝑞𝑖) for 𝑖 = 1, 2.(12)

Also let 𝑓 : Δ2 → R be convex on the coordinates on Δ2.𝑓 (𝐴1, 𝐴2) ≤ 12 [

12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝐴2) 𝑑𝑠1

+ 12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝐴1, 𝑠2) 𝑑𝑠2] ≤ 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2

≤ 1(𝑝1 + 𝑞1) [

𝑝12𝑦1 ∫𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝑎2) 𝑑𝑠1

+ 𝑞12𝑦1 ∫𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝑏2) 𝑑𝑠1]

+ 1(𝑝2 + 𝑞2) [

𝑝22𝑦2 ∫𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑎1, 𝑠2) 𝑑𝑠2

+ 𝑞22𝑦2 ∫𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑏1, 𝑠2) 𝑑𝑠2] ≤ [𝑝1𝑝2𝑓 (𝑎1, 𝑎2)+ 𝑝1𝑞2𝑓 (𝑎1, 𝑏2) + 𝑞1𝑝2𝑓 (𝑏1, 𝑎2) + 𝑞1𝑞2𝑓 (𝑏1, 𝑏2)]× [ 1

(𝑝1 + 𝑞1) +1

(𝑝2 + 𝑞2)] .

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Proof. Since 𝑓 : Δ2 → R is convex on coordinates, it followsthat the mapping 𝑓𝑠

1

: [𝑎2, 𝑏2] → R, 𝑓𝑠1

(𝑠2) fl 𝑓(𝑠1, 𝑠2), isconvex on [𝑎2, 𝑏2] for all 𝑠1 ∈ [𝑎1, 𝑏1], so by inequality (11) onehas

𝑓𝑠1

(𝐴2) ≤ 12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓𝑠1

(𝑠2) 𝑑𝑠2

≤ 𝑝2𝑓𝑠1 (𝑎2) + 𝑞2𝑓𝑠1 (𝑏2)𝑝2 + 𝑞2 ,(14)

that is

𝑓 (𝑠1, 𝐴2) ≤ 12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠2

≤ 𝑝2𝑓 (𝑠1, 𝑎2) + 𝑞2𝑓 (𝑠1, 𝑏2)𝑝2 + 𝑞2 .(15)

Observe that for 0 < 𝑦1 ≤ ((𝑏1 − 𝑎1)/(𝑝1 + 𝑞1))min (𝑝1, 𝑞1),by considering two cases (0 < 𝑝1 ≤ 𝑞1 and 0 < 𝑞1 < 𝑝1) wecan easily verify that 𝑎1 ≤ 𝐴1 −𝑦1 < 𝐴1 +𝑦1 ≤ 𝑏1, so that 𝑓 isdefined on [𝐴1−𝑦1, 𝐴1+𝑦1]. Integrating the above inequalityon [𝐴1 − 𝑦1, 𝐴1 + 𝑦1], we have

12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝐴2) 𝑑𝑠1 ≤ 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2

≤ 1𝑝2 + 𝑞2 [

𝑝22𝑦1 ∫𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝑎2) 𝑑𝑠1

+ 𝑞22𝑦1 ∫𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝑏2) 𝑑𝑠1] .

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By a similar argument applied for themapping𝑓𝑠2

: [𝑎1, 𝑏1] →R, 𝑓𝑠

2

(𝑠1) fl 𝑓(𝑠1, 𝑠2), we get12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝐴1, 𝑠2) 𝑑𝑠2 ≤ 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2

≤ 1𝑝1 + 𝑞1 [

𝑝12𝑦2 ∫𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑎1, 𝑠2) 𝑑𝑠2

+ 𝑞12𝑦2 ∫𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑏1, 𝑠2) 𝑑𝑠2] .

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Summing the inequalities (16) and (17), we get the secondand third inequality in (13). Since𝑓 is convex on coordinates,using convexity of 𝑓 on first coordinate and inequality (11),we have

𝑓 (𝐴1, 𝐴2) ≤ 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝐴2) 𝑑𝑠1 (18)

and, using convexity of 𝑓 on second coordinate, we have

𝑓 (𝐴1, 𝐴2) ≤ 12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝐴1, 𝑠2) 𝑑𝑠2. (19)

Adding the inequalities (18) and (19)we get the first inequalityin (13). Finally, by the same Hermite-Hadamard inequalityintroduced by Lupas we can also state

12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝑎2) 𝑑𝑠1

≤ 𝑝1𝑓 (𝑎1, 𝑎2) + 𝑞1𝑓 (𝑏1, 𝑎2)𝑝1 + 𝑞1 ,(20)

12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝑏2) 𝑑𝑠1

≤ 𝑝1𝑓 (𝑎1, 𝑏2) + 𝑞1𝑓 (𝑏1, 𝑏2)𝑝1 + 𝑞1 ,(21)

4 Journal of Mathematics

12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑎1, 𝑠2) 𝑑𝑠2

≤ 𝑝2𝑓 (𝑎1, 𝑎2) + 𝑞2𝑓 (𝑎1, 𝑏2)𝑝2 + 𝑞2 ,(22)

12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑏1, 𝑠2) 𝑑𝑠2

≤ 𝑝2𝑓 (𝑏1, 𝑎2) + 𝑞2𝑓 (𝑏1, 𝑏2)𝑝2 + 𝑞2 .(23)

Multiply (20) and (21) by 𝑝2 and multiply (22) and (23) by 𝑞2then on addition we get the last inequality in (13).

3. Some Associated Mappings

In this section we will discuss some mappings associatedwith generalized Hermite-Hadamard inequality introducedby Lupas for convex mappings on coordinates.

Let 𝑝1, 𝑝2, 𝑞1, and 𝑞2 be positive real numbers and 𝐴 𝑖 =(𝑝𝑖𝑎𝑖 + 𝑞𝑖𝑏𝑖)/(𝑝𝑖 + 𝑞𝑖) and 𝑦𝑖 > 0 where 𝑦𝑖 ≤ ((𝑏𝑖 − 𝑎𝑖)/(𝑝𝑖 +𝑞𝑖))min(𝑝𝑖, 𝑞𝑖) for 𝑖 = 1, 2. For mapping 𝑓 : Δ2 = [𝑎1, 𝑏1] ×[𝑎2, 𝑏2] → R, we can define a mapping𝐻 : [0, 1]2 → R,

𝐻(𝑢, V) fl 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢𝑠1+ (1 − 𝑢)𝐴1, V𝑠2 + (1 − V) 𝐴2) 𝑑𝑠1𝑑𝑠2,

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where 𝑎1 ≤ 𝐴1 − 𝑦1 < 𝐴1 + 𝑦1 ≤ 𝑏1 and 𝑎2 ≤ 𝐴2 − 𝑦2 <𝐴2+𝑦2 ≤ 𝑏2, so that𝑓 is defined on [𝐴1−𝑦1, 𝐴1+𝑦1] and [𝐴2−𝑦2, 𝐴2 + 𝑦2]. The properties of this mapping are embodied inthe following theorem.

Theorem 7. Suppose that 𝑝1, 𝑝2, 𝑞1, and 𝑞2 are positive realnumbers and

𝐴 𝑖 = 𝑝𝑖𝑎𝑖 + 𝑞𝑖𝑏𝑖𝑝𝑖 + 𝑞𝑖 ,𝑦𝑖 > 0,

where 𝑦𝑖 ≤ 𝑏𝑖 − 𝑎𝑖𝑝𝑖 + 𝑞𝑖 min (𝑝𝑖, 𝑞𝑖) for 𝑖 = 1, 2.(25)

Also let 𝑓 : Δ2 → R be convex on the coordinates on Δ2.(i) Themapping𝐻 is convex on the coordinates on [0, 1]2.(ii) We have the bounds

sup(𝑢,V)∈[0,1]2

𝐻(𝑢, V)

= 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2= 𝐻 (1, 1) ,inf(𝑢,V)∈[0,1]2

𝐻(𝑢, V) = 𝑓 (𝐴1, 𝐴2) = 𝐻 (0, 0) .

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(iii) The mapping 𝐻 is monotonic nondecreasing on thecoordinates.

Proof. (i) Fix V ∈ [0, 1], for all 𝑢1, 𝑢2 ∈ [0, 1] and 𝛼, 𝛽 ≥ 0with 𝛼 + 𝛽 = 1; we have

𝐻(𝛼𝑢1 + 𝛽𝑢2, V) = 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝛼 (𝑢1𝑠1 + (1 − 𝑢1) 𝐴1)⋅ +𝛽 (𝑢2𝑠1 + (1 − 𝑢2) 𝐴1) , V𝑠2+ (1 − V) 𝐴2) 𝑑𝑠1 𝑑𝑠2 ≤ 𝛼

4𝑦1𝑦2⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢1𝑠1 + (1 − 𝑢1) 𝐴1, V𝑠2+ (1 − V) 𝐴2) 𝑑𝑠1 𝑑𝑠2 + 𝛽

4𝑦1𝑦2⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢2𝑠1 + (1 − 𝑢2) 𝐴1, V𝑠2+ (1 − V) 𝐴2) 𝑑𝑠1 𝑑𝑠2 = 𝛼𝐻 (𝑢1, V)+ 𝛽𝐻 (𝑢2, V) .

(27)

If 𝑢 ∈ [0, 1] is fixed then, for all V1, V2 ∈ [0, 1] and 𝛼, 𝛽 ≥ 0with 𝛼 + 𝛽 = 1, we also have𝐻(𝑢, 𝛼V1 + 𝛽V2) ≤ 𝛼𝐻(𝑢, V1) +𝛽𝐻(𝑢, V2) and the statement is proved.

(ii) Since 𝑓 is convex on coordinates, we have by Jensen'sinequality for integrals that

𝐻(𝑢, V) ≥ 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓(𝑢𝑠1 + (1 − 𝑢)𝐴1, 12𝑦2⋅ ∫𝐴2+𝑦2𝐴2−𝑦2

(V𝑠2 + (1 − V) 𝐴2) 𝑑𝑠2)𝑑𝑠1 = 12𝑦1

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

𝑓 (𝑢𝑠1 + (1 − 𝑢)𝐴1, 𝐴2) 𝑑𝑠1

≥ 𝑓( 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

(𝑢𝑠 + (1 − 𝑢)𝐴1) 𝑑𝑠1, 𝐴2)= 𝑓 (𝐴1, 𝐴2) .

(28)

By the convexity of𝐻 on the coordinates, we have

𝐻(𝑢, V) ≤ 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

[V 12𝑦2

⋅ ∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢𝑠1 + (1 − 𝑢)𝐴1, 𝑠2) 𝑑𝑠2 + (1 − V) 12𝑦2

⋅ ∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢𝑠1 + (1 − 𝑢)𝐴1, 𝐴2) 𝑑𝑠2] 𝑑𝑠1

≤ V12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

[𝑢 12𝑦1

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 + (1 − 𝑢) 12𝑦1

Journal of Mathematics 5

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

𝑓 (𝐴1, 𝑠2) 𝑑𝑠1] 𝑑𝑠2 + (1 − V) 12𝑦2

⋅ ∫𝐴2+𝑦2𝐴2−𝑦2

[𝑢 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝐴2) 𝑑𝑠1 + (1 − 𝑢)

⋅ 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝐴1, 𝐴2) 𝑑𝑠1] 𝑑𝑠2 = 𝑢V 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2 + (1 − 𝑢) V

⋅ 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝐴1, 𝑠2) 𝑑𝑠1 𝑑𝑠2+ 𝑢 (1 − V) 1

4𝑦1𝑦2⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝐴2) 𝑑𝑠1 𝑑𝑠2 + (1 − 𝑢)

⋅ (1 − V) 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝐴1, 𝐴2) 𝑑𝑠1 𝑑𝑠2 = 𝑢V 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2 + (1 − 𝑢) V 12𝑦2

⋅ ∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝐴1, 𝑠2) 𝑑𝑠2 + 𝑢 (1 − V) 12𝑦1

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

𝑓 (𝑠1, 𝐴2) 𝑑𝑠1 + (1 − 𝑢) (1 − V) 𝑓 (𝐴1, 𝐴2) .(29)

By Hadamard’s inequality introduced by Lupas (11), wealso have

𝑓 (𝐴1, 𝑠2) ≤ 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1,

𝑓 (𝑠1, 𝐴2) ≤ 12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠2.(30)

Thus by integration, we get that

12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝐴1, 𝑠2) 𝑑𝑠2

≤ 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2,12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

𝑓 (𝑠1, 𝐴2) 𝑑𝑠1

≤ 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2.

(31)

Also using the result

𝑓 (𝐴1, 𝐴2) ≤ 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2 (32)

we deduce the inequality

𝐻(𝑢, V)≤ [𝑢V + (1 − 𝑢) V + 𝑢 (1 − V) + (1 − 𝑢) (1 − V)]⋅ 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2 = 14𝑦1𝑦2

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2,

(33)

for all (𝑢, V) ∈ [0, 1]2 and the second bound in (ii) is proved.(iii) Firstly, we show that𝐻(𝑢, V) ≥ 𝐻(0, V) for all [𝑢, V] ∈

[0, 1]2. By Jensen’s inequality for integrals, we have𝐻(𝑢, V) ≥ 1

2𝑦2 ∫𝐴2+𝑦2

𝐴2−𝑦2

𝑓( 12𝑦1

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

(𝑢𝑠1 + (1 − 𝑢)𝐴1) 𝑑𝑠1, V𝑠2 + (1 − V)

⋅ 𝐴2)𝑑𝑠2 = 12𝑦2 ∫

𝐴2+𝑦2

𝐴2−𝑦2

𝑓 (𝐴1, V𝑠2+ (1 − V) 𝐴2) 𝑑𝑠2 = 𝐻 (0, V)

(34)

for all [𝑢, V] ∈ [0, 1]2.Now let 0 ≤ 𝑢1 ≤ 𝑢2 ≤ 1. By the convexity of mapping

𝐻(⋅, V) for all [𝑢, V] ∈ [0, 1]2, we have𝐻(𝑢2, V) − 𝐻 (𝑢1, V)𝑢2 − 𝑢1 ≥ 𝐻 (𝑢1, V) − 𝐻 (0, V)

𝑢1 ≥ 0. (35)

The following theorem also holds.

Theorem 8. Suppose that 𝑝1, 𝑝2, 𝑞1, and 𝑞2 are positive realnumbers and

𝐴 𝑖 = 𝑝𝑖𝑎𝑖 + 𝑞𝑖𝑏𝑖𝑝𝑖 + 𝑞𝑖 ,𝑦𝑖 > 0,

where 𝑦𝑖 ≤ 𝑏𝑖 − 𝑎𝑖𝑝𝑖 + 𝑞𝑖 min (𝑝𝑖, 𝑞𝑖) for 𝑖 = 1, 2.(36)

Also let 𝑓 : Δ2 → R be convex on the coordinates on Δ2.(i) The mapping𝐻 is convex on [0, 1]2.(ii) Define the mapping ℎ : [0, 1] → R, ℎ(𝑢) fl 𝐻(𝑢, 𝑢).

Then ℎ is convex, monotonic nondecreasing on [0, 1]and one has the bounds

6 Journal of Mathematics

sup𝑢∈[0,1]

ℎ (𝑢) = ℎ (1)

= 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2,inf𝑢∈[0,1]

ℎ (𝑢) = ℎ (0) = 𝑓 (𝐴1, 𝐴2) .(37)

Proof. (i) Let (𝑢1, 𝑢2), (V1, V2) ∈ [0, 1]2 and 𝛼, 𝛽 ≥ 0 with 𝛼 +𝛽 = 1. Since 𝑓 : Δ2 → R is convex on Δ2, we have𝐻(𝛼 (𝑢1, 𝑢2) + 𝛽 (V1, V2)) = 𝐻 (𝛼𝑢1 + 𝛽V1, 𝛼𝑢2 + 𝛽V2) = 1

4𝑦1𝑦2⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝛼 (𝑢1𝑠1 + (1 − 𝑢1) 𝐴1, 𝑢2𝑠2 + (1 − 𝑢2) 𝐴2)⋅ +𝛽 (V1𝑠1 + (1 − V1) 𝐴1, V2𝑠2 + (1 − V2) 𝐴2)) 𝑑𝑠1 𝑑𝑠2≤ 𝛼 ⋅ 1

4𝑦1𝑦2 ∫𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢1𝑠1 + (1 − 𝑢1) 𝐴1, 𝑢2𝑠2+ (1 − 𝑢2) 𝐴2) 𝑑𝑠1 𝑑𝑠2 + 𝛽 ⋅ 1

4𝑦1𝑦2⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (V1𝑠1 + (1 − V1) 𝐴1, V2𝑠2 + (1 − V2)⋅ 𝐴2) 𝑑𝑠1 𝑑𝑠2 = 𝛼𝐻 (𝑢1, 𝑢2) + 𝛽𝐻 (V1, V2) ,

(38)

which shows𝐻 is convex on [0, 1]2.(ii) Let 𝑢1, 𝑢2 ∈ [0, 1] and 𝛼, 𝛽 ≥ 0 with 𝛼 + 𝛽 = 1.

ℎ (𝛼𝑢1 + 𝛽𝑢2) = 𝐻 (𝛼𝑢1 + 𝛽𝑢2, 𝛼𝑢1 + 𝛽𝑢2)= 𝐻 (𝛼 (𝑢1, 𝑢1) + 𝛽 (𝑢2, 𝑢2))

≤ 𝛼𝐻 (𝑢1, 𝑢1) + 𝛽𝐻 (𝑢2, 𝑢2)= 𝛼ℎ (𝑢1) + 𝛽ℎ (𝑢2) ,

(39)

which shows the convexity of ℎ on [0, 1].We have, by the above theorem, that

ℎ (𝑢) = 𝐻 (𝑢, 𝑢) ≥ 𝐻 (0, 0) = 𝑓 (𝐴1, 𝐴2) ,𝑢 ∈ [0, 1] ,

ℎ (𝑢) = 𝐻 (𝑢, 𝑢) ≤ 𝐻 (1, 1)= 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠1 𝑑𝑠2,𝑢 ∈ [0, 1] ,

(40)

which proves the required bounds.

Now let 0 ≤ 𝑢1 ≤ 𝑢2 ≤ 1. By the convexity of mapping ℎ,we have that

ℎ (𝑢2) − ℎ (𝑢1)𝑢2 − 𝑢1 ≥ ℎ (𝑢1) − ℎ (0)𝑢1 ≥ 0, (41)

and the theorem is proved.

Next, for positive real numbers 𝑝1, 𝑝2, 𝑞1, 𝑞2 and 𝐴 𝑖 =(𝑝𝑖𝑎𝑖 + 𝑞𝑖𝑏𝑖)/(𝑝𝑖 + 𝑞𝑖) and 𝑦𝑖 > 0 where 𝑦𝑖 ≤ ((𝑏𝑖 − 𝑎𝑖)/(𝑝𝑖 +𝑞𝑖))min(𝑝𝑖, 𝑞𝑖) for 𝑖 = 1, 2, then for mapping 𝑓 : Δ2 =[𝑎1, 𝑏1] × [𝑎2, 𝑏2] → R we shall consider the mapping �� :[0, 1]2 → [0,∞), which is given as

�� (𝑢, V) fl 116𝑦21𝑦22 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢𝑠1 + (1 − 𝑢) 𝑡1, V𝑠2 + (1 − V) 𝑡2) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1, (42)

where 𝑎1 ≤ 𝐴1 − 𝑦1 < 𝐴1 + 𝑦1 ≤ 𝑏1 and 𝑎2 ≤ 𝐴2 − 𝑦2 <𝐴2 + 𝑦2 ≤ 𝑏2, so that 𝑓 is defined on [𝐴1 − 𝑦1, 𝐴1 + 𝑦1] and[𝐴2 − 𝑦2, 𝐴2 + 𝑦2].The next theorem contains the main properties of this

mapping.

Theorem 9. Suppose that 𝑝1, 𝑝2, 𝑞1, and 𝑞2 are positive realnumbers and

𝐴 𝑖 = 𝑝𝑖𝑎𝑖 + 𝑞𝑖𝑏𝑖𝑝𝑖 + 𝑞𝑖 ,𝑦𝑖 > 0,

where 𝑦𝑖 ≤ 𝑏𝑖 − 𝑎𝑖𝑝𝑖 + 𝑞𝑖 min (𝑝𝑖, 𝑞𝑖) for 𝑖 = 1, 2.(43)

Also let 𝑓 : Δ2 → R be convex on the coordinates on Δ2.

(i) We have the equalities

�� (𝑢 + 12 , V) = �� (12 − 𝑢, V) ,∀𝑢 ∈ [0, 12] , V ∈ [0, 1] ,

�� (𝑢, V + 12) = �� (𝑢, 12 − V) ,∀𝑢 ∈ [0, 12] , V ∈ [0, 1] ,

�� (1 − 𝑢, V) = �� (𝑢, V) ,�� (𝑢, 1 − V) = �� (𝑢, V) ,

∀ (𝑢, V) ∈ Δ2.

(44)

Journal of Mathematics 7

(ii) �� is convex on the coordinates.(iii) We have the bounds

inf(𝑢,V)∈[0,1]2

�� (𝑢, V) = �� (12 ,12) =

116𝑦21𝑦22 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓(𝑠1 + 𝑡12 , 𝑠2 + 𝑡22 ) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1,

sup(𝑢,V)∈[0,1]2

�� (𝑢, V) = �� (0, 0) = �� (1, 1) = 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠2 𝑑𝑠1.(45)

(iv) The mapping ��(⋅, V) is monotonic nonincreasing on[0, 1/2) and nondecreasing on [1/2, 1] for all V ∈ [0, 1].A similar property has the mapping ��(𝑢, ⋅) for all 𝑢 ∈[0, 1].

(v) We have the inequality

�� (𝑢, V) ≥ max {𝐻 (𝑠1, 𝑠2) ,𝐻 (1 − 𝑠1, 𝑠2) ,𝐻 (𝑠1, 1 − 𝑠2) ,𝐻 (1 − 𝑠1, 1 − 𝑠2)} .

(46)

Proof. (i) and (ii) are obvious.(iii) By the convexity of 𝑓 in the first variable, we get that

12 [𝑓 (𝑢𝑠1 + (1 − 𝑢) 𝑡1, V𝑠2 + (1 − V) 𝑡2)+ 𝑓 ((1 − 𝑢) 𝑠1 + 𝑢𝑡1, V𝑠2 + (1 − V) 𝑡2)]≥ 𝑓(𝑠1 + 𝑡12 , V𝑠2 + (1 − V) 𝑡2)

(47)

for all (𝑠1, 𝑡1) ∈ [𝑎1, 𝑏1]2, (𝑠2, 𝑡2) ∈ [𝑎2, 𝑏2]2, and (𝑢, V) ∈[0, 1]2. Integrating on [𝐴1 − 𝑦1, 𝐴1 + 𝑦1]2, we get14𝑦21 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

𝑓 (𝑢𝑠1 + (1 − 𝑢) 𝑡1, V𝑠2

+ (1 − V) 𝑡2) 𝑑𝑠1 𝑑𝑡1 ≥ 14𝑦21

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

𝑓(𝑠1 + 𝑡12 , V𝑠2+ (1 − V) 𝑡2)𝑑𝑠1 𝑑𝑡1.

(48)

Similarly,

14𝑦22 ∫

𝐴2+𝑦2

𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓(𝑠1 + 𝑡12 , V𝑠2 + (1 − V)

⋅ 𝑡2)𝑑𝑠2 𝑑𝑡2 ≥ 14𝑦22

⋅ ∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓(𝑠1 + 𝑡12 , 𝑠2 + 𝑡22 ) 𝑑𝑠2 𝑑𝑡2.

(49)

Now integrating this inequality on [𝐴1 − 𝑦1, 𝐴1 + 𝑦1]2 andtaking into account the above inequality, we deduce

�� (𝑢, V) ≥ 116𝑦21𝑦22 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓(𝑠1 + 𝑡12 , 𝑠2 + 𝑡22 ) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1 (50)

for (𝑢, V) ∈ [0, 1]2. The first bound in (iii) is proved and thesecond bound goes on likewise.(iv)Themonotonicity of ��(⋅, 𝑠) follows by a similar argumentfrom (iii).

(v) By Jensen’s integral inequality, we have successively for all(𝑢, V) ∈ [0, 1]2 that

�� (𝑢, V) ≥ 18𝑦1𝑦22 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓( 12𝑦1 ∫

𝐴1+𝑦1

𝐴1−𝑦1

(𝑢𝑠1 + (1 − 𝑢) 𝑡1) 𝑑𝑡1, V𝑠2 + (1 − V) 𝑡2)𝑑𝑡2 𝑑𝑠2 𝑑𝑠1

= 18𝑦1𝑦22 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢𝑠1 + (1 − 𝑢)𝐴1, V𝑠2 + (1 − V) 𝑡2) 𝑑𝑡2 𝑑𝑠2 𝑑𝑠1

≥ 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢𝑠1 + (1 − 𝑢)𝐴1, V𝑠2 + (1 − V) 𝐴2) 𝑑𝑠2 𝑑𝑠1 = 𝐻 (𝑠1, 𝑠2) .

(51)

8 Journal of Mathematics

Similarly, we can easily show ��(1 − 𝑢, V) ≥ 𝐻(1 − 𝑠1, 𝑠2),��(𝑢, 1−V) ≥ 𝐻(𝑠1, 1−𝑠2), and ��(1−𝑢, 1−V) ≥ 𝐻(1−𝑠1, 1−𝑠2).In addition, as ��(𝑢, V) = ��(1 − 𝑢, V) = ��(𝑢, 1 − V) =��(1 − 𝑢, 1 − V) for all (𝑢, V) ∈ [0, 1]2, so we deduce ��(𝑢, V) ≥

max{𝐻(𝑠1, 𝑠2),𝐻(1 − 𝑠1, 𝑠2),𝐻(𝑠1, 1 − 𝑠2),𝐻(1 − 𝑠1, 1 − 𝑠2)}.The theorem is thus proved.

Theorem 10. Suppose that 𝑝1, 𝑝2, 𝑞1, and 𝑞2 are positive realnumbers and

𝐴 𝑖 = 𝑝𝑖𝑎𝑖 + 𝑞𝑖𝑏𝑖𝑝𝑖 + 𝑞𝑖 ,

𝑦𝑖 > 0,where 𝑦𝑖 ≤ 𝑏𝑖 − 𝑎𝑖𝑝𝑖 + 𝑞𝑖 min (𝑝𝑖, 𝑞𝑖) for 𝑖 = 1, 2.

(52)

Also let 𝑓 : Δ2 → R be convex on the coordinates on Δ2.(i) The mapping �� is convex on Δ2.(ii) Define the mapping ℎ : [0, 1] → R, ℎ(𝑢) fl ��(𝑢, 𝑢).

Then ℎ is convex, monotonic nonincreasing on [0, 1/2]and nondecreasing on [1/2, 1] and one has the bounds

inf𝑢∈[0,1]

ℎ (𝑢) = ℎ (12) =1

16𝑦21𝑦22 ∫𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓(𝑠1 + 𝑡12 , 𝑠2 + 𝑡22 ) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1,

sup𝑢∈[0,1]

ℎ (𝑢) = ℎ (1) = ℎ (0) = 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠2 𝑑𝑠1.(53)

(iii) One has the inequality

ℎ (𝑢) ≥ max {ℎ (𝑢) , ℎ (1 − 𝑢)} . (54)

Proof. (i) Let (𝑢1, 𝑢2), (V1, V2) ∈ [0, 1] and 𝛼, 𝛽 ≥ 0 with 𝛼 +𝛽 = 1. Since 𝑓 is convex on Δ2, we have

�� (𝛼 (𝑢1, 𝑢2) + 𝛽 (V1, V2)) = �� (𝛼𝑢1 + 𝛽V1, 𝛼𝑢2 + 𝛽V2) = 116𝑦21𝑦22

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝛼 (𝑢1𝑠1 + (1 − 𝑢1) 𝑡1, 𝑢2𝑠2 + (1 − 𝑢2) 𝑡2) +𝛽 (V1𝑠1 + (1 − V1) 𝑡1, V2𝑠2 + (1 − V2) 𝑡2)) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1

≤ 𝛼 ⋅ 116𝑦21𝑦22 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑢1𝑠1 + (1 − 𝑢1) 𝑡1, 𝑢2𝑠2 + (1 − 𝑢2) 𝑡2) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1 + 𝛽 ⋅ 116𝑦21𝑦22

⋅ ∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (V1𝑠1 + (1 − V1) 𝑡1, V2𝑠2 + (1 − V2) 𝑡2) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1 = 𝛼�� (𝑢1, 𝑢2) + 𝛽�� (V1, V2) ,

(55)

which shows �� is convex on Δ2.(ii) Let 𝑢1, 𝑢2, ∈ [0, 1] and 𝛼, 𝛽 ≥ 0 with 𝛼 + 𝛽 = 1.

ℎ (𝛼𝑢1 + 𝛽𝑢2) = �� (𝛼𝑢1 + 𝛽𝑢2, 𝛼𝑢1 + 𝛽𝑢2)= �� (𝛼 (𝑢1, 𝑢1) + 𝛽 (𝑢2, 𝑢2))

≤ 𝛼�� (𝑢1, 𝑢1) + 𝛽�� (𝑢2, 𝑢2)= 𝛼ℎ (𝑢1) + 𝛽ℎ (𝑢2) ,

(56)

which shows the convexity of ℎ on [0, 1].As

ℎ (𝑢) = �� (𝑢, 𝑢) ≥ �� (12 ,12) = ℎ (

12) =

116𝑦21𝑦22 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴1+𝑦1𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓(𝑠1 + 𝑡12 , 𝑠2 + 𝑡22 ) 𝑑𝑡2 𝑑𝑠2 𝑑𝑡1 𝑑𝑠1,

ℎ (𝑢) = �� (𝑢, 𝑢) ≤ �� (0, 0) = �� (0, 0) = 14𝑦1𝑦2 ∫

𝐴1+𝑦1

𝐴1−𝑦1

∫𝐴2+𝑦2𝐴2−𝑦2

𝑓 (𝑠1, 𝑠2) 𝑑𝑠2 𝑑𝑠1,(57)

which proves the required bounds.

Journal of Mathematics 9

It is obvious that ℎ ismonotonic nonincreasing on [0, 1/2]and nondecreasing on [1/2, 1].

(iii) As we know

ℎ (𝑢) = �� (𝑢, 𝑢) ≥ max {𝐻 (𝑢, 𝑢) ,𝐻 (1 − 𝑢, 1 − 𝑢)}= max {ℎ (𝑢) , ℎ (1 − 𝑢)} (58)

and the theorem is proved.

Competing Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper.

References

[1] D. S. Mitrinovic and I. B. Lackovic, “Hermite and convexity,”Aequationes Mathematicae, vol. 28, no. 3, pp. 229–232, 1985.

[2] A. M. Fink, “A best possible Hadamard inequality,”Mathemati-cal Inequalities and Applications, vol. 1, no. 2, pp. 223–230, 1998.

[3] S. S. Dragomir, “A mapping in connection to Hadamard’sinequalities,” Osterreichische Akademie der Wissenschaften.Mathematisch-Naturwissenschaftliche, vol. 128, pp. 17–20, 1991.

[4] S. S. Dragomir, “On Hadamard’s inequalities for convex func-tions,”Mathematica Balkanica. New Series, vol. 6, no. 3, pp. 215–222, 1992.

[5] S. S. Dragomir, D. S. Milosevic, and J. Sandor, “On some refine-ments of Hadamard’s inequalities and applications,” PublikacijeElektrotehnickog Fakulteta. Serija Matematika, vol. 4, pp. 3–10,1993.

[6] G. Farid, M. Marwan, and A. U. Rehman, “Fejer-hadamardinequality for convex functions on the coordinates in a rectanglefrom the plane,” International Journal of Analysis and Applica-tions, vol. 10, no. 1, pp. 40–47, 2016.

[7] G. Farid and A. Ur Rehman, “Generalization of the Fejer-Hadamard’s Inequality for convex function on coordinates,”KoreanMathematical Society. Communications, vol. 31, no. 1, pp.53–64, 2016.

[8] G. Farid, M. Marwan, and A. U. Rehman, “New mean valuetheorems and generalization of Hadamard inequality via coor-dinated m-convex functions,” Journal of Inequalities and Appli-cations, vol. 2015, article 283, pp. 1–11, 2015.

[9] S. S. Dragomir and C. E. M. Pearce, Selected Topics on Hermite-Hadamard Inequalities, (RGMIAMonographs), VictoriaUniver-sity Press, 2000.

[10] S. S. Dragomir, “On the Hadamards inequality for convexfunction on the co-ordinates in a rectangle from the plane,”Taiwanese Journal of Mathematics, vol. 5, no. 4, pp. 775–778,2001.

[11] A. Lupas, “A generalization of Hadamard inequalities forconvex functions,” Publikacije Elektrotehnickog Fakulteta. SerijaMatematika I Fizika, vol. 544–576, pp. 115–121, 1976.

[12] J. E. Pecaric, F. Proschan, and Y. Tong,Convex Functions, PartialOrderings, and Statistical Applications, vol. 187, Academic Press,New York, NY, USA, 1992.

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