# A Family of Mappings Associated with Hadamard's Inequality on a Hypercube

Authors: FuLi Wang
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

This paper provides a method to generalize and refine the Hadamard's inequality on a hypercube inn-dimensional Euclidean space by establishing a family of mappings.


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## Body

## 1. Introduction

For a convex functionf defined on an open interval I of real numbers and a,b∈I with a<b, then the following inequality:
(1.1)f(a+b2)≤1b-a∫abf(x)dx≤f(a)+f(b)2
is known as the classical Hadamard’s inequality for the convex function.In [1], Dragomir established the following mapping:(1.2)H:[0,1]⟶R,H(t)=1b-a∫abf(tx+(1-t)a+b2)dx,
then the mapping H is convex, increasing on [0,1], and for all t∈[0,1], we have
(1.3)f(a+b2)=H(0)≤H(t)≤H(1)=1b-a∫abf(x)dx,
which is a refinement to the left side of (1.1).In [2], Yang and Hong established the following mapping:(1.4)G:[0,1]⟶R,G(t)=12(b-a)∫ab[f(1+t2a+1-t2x)+f(1+t2b+1-t2x)]dx,
then the mapping G is convex, increasing on [0,1], and for all t∈[0,1], we have (1.5)1b-a∫abf(x)dx=G(0)≤G(t)≤G(1)=f(a)+f(b)2,
which is a refinement to the right side of (1.1).In this paper, we first establish a mapping in Section2, defined by
(1.6)A1:[0,1]⟶R,A1(t)=12[f(1+t2a+1-t2b)+f(1-t2a+1+t2b)],
then the mapping A1 is convex, increasing on [0,1], and for all t∈[0,1] we have
(1.7)f(a+b2)=A1(0)≤A1(t)≤A1(1)=f(a)+f(b)2.
Furthermore, there is a point ξ∈(0,1) so that A1(ξ)=(1/(b-a))∫abf(x). Thus, (1.1) has been proved based on the properties of A1, and the inequality (1.7) is a refinement to the both sides of (1.1).In Section3, we also use an analogue of (1.6) to obtain the following Hadamard’s type inequality on a square:
(1.8)f(a,b)≤1h2∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy≤14h∮|x-a|=h/2|y-b|=h/2⁡f(x,y)ds,
which is a generalization of (1.1) to two-dimensional Euclidean space.In Section4, by establishing another analogue of (1.6) on a hypercube, the following Hadamard’s inequality is also pointed out:
(1.9)f(c)≤1hn∫‖x-c‖∞≤h/2f(x)dx≤12nhn-1∮{x:|xi-ci|=h/2}f(x)ds,
which is a generalization of (1.1) to n-dimensional Euclidean space.All the above Hadamard’s inequalities (1.1), (1.8), and (1.9) have a uniform format
(1.10)f(c)≤1Vol(Ω)∫Ωf(x)dx≤1Area(Ω)∮∂Ωf(x)ds.Thus, by establishing a family of the mappings, we have provided a method to get Hadamard’s inequality of the convex function on a hypercube inn-dimensional Euclidean space. The motivation for the present work is from many generalizations and refinements of the classical inequality (1.1) in [1–8].


## 2. Properties of the Mapping

The main properties of the mappingA1 defined by (1.6) are embodied in the following theorem.Theorem 2.1.
Iff is a convex function defined on an open interval I of real numbers, a,b∈I with a<b, and the mapping A1 is defined by (1.6), then, one has(i)
the mappingA1 is convex on [0,1],(ii)
one has the bounds:(2.1)inf⁡t∈[0,1]A1(t)=A1(0)=f(a+b2),sup⁡t∈[0,1]A1(t)=A1(1)=f(a)+f(b)2,(iii)
the mappingA1 is increasing on [0,1],(iv)
there is a pointξ∈(0,1) so that
(2.2)A1(ξ)=1b-a∫abf(x)dx,(v)
one obtains, the classical inequality (1.1) and has a refinement of (1.1):
(2.3)A1(0)≤A1(η)≤A1(ξ)≤A1(μ)≤A1(1),∀η∈(0,ξ),μ∈(ξ,1).Proof.
(i) Lett1,t2∈[0,1] and α,β≥0 with α+β=1, then we have
(2.4)A1(αt1+βt2)=12[f(1+αt1+βt22a+1-αt1-βt22b)+f(1-αt1-βt22a+1+αt1+βt22b)]=12{f[α(1+t12a+1-t12b)+β(1+t22a+1-t22b)]+f[α(1-t12a+1+t12b)+β(1-t22a+1+t22b)]}≤12[αf(1+t12a+1-t12b)+βf(1+t22a+1-t22b)+αf(1-t12a+1+t12b)+βf(1-t22a+1+t22b)]=12α[f(1+t12a+1-t12b)+f(1-t12a+1+t12b)]+12β[f(1+t22a+1-t22b)+f(1-t22a+1+t22b)]=αA1(t1)+βA1(t2),
which proves the convexity of A1 on [0,1].
(ii) By the convexity off on the interval [a,b], we have(2.5)A1(t)≤12[1+t2f(a)+1-t2f(b)+1-t2f(a)+1+t2f(b)]=f(a)+f(b)2=A1(1),A1(t)≥f[12(1+t2a+1-t2b+1-t2a+1+t2b)]=f(a+b2)=A1(0),
the bounds (2.1) hold.
(iii) Let0<t1<t2≤1. By the convexity of the mapping A1, we have(2.6)A1(t2)-A1(t1)t2-t1≥A1(t1)-A1(0)t1≥0.
Since we have proved that A1(t1)≥A1(0) for all t1∈(0,1] in (ii), the monotonicity of A1 is established.
(iv) We define the functionϕ on [0,1] by(2.7)ϕ(t)=∫((1+t)/2)a+((1-t)/2)b((1-t)/2)a+((1+t)/2)bf(x)dx-t∫abf(x)dx,
then
(2.8)ϕ(0)=ϕ(1)=0,ϕ′(t)=(b-a)A1(t)-∫abf(x)dx.
According to the Rolle’s mean-value theorem, there is a point ξ∈(0,1) so that
(2.9)0=ϕ′(ξ)=(b-a)A1(ξ)-∫abf(x)dx.
Thus, (2.2) holds.
(v) Considering the properties (ii), (iii), and (iv) above, we have proved the Hadamard’s inequality (1.1) and the inequality (2.3).


## 3. The Hadamard’s Inequality on a Square

We consider a convex function with two variables defined on a square in two-dimensional Euclidean space.Theorem 3.1.
Iff is a convex function defined on an open region D⊆R2, the square
(3.1){(x,y):|x-a|≤h2,|y-b|≤h2}⊆D,
and the mapping A2 is defined by
(3.2)A2:[0,1]⟶R,A2(t)=14h∮|u|=h/2|v|=h/2⁡f(a+tu,b+tv)ds,
which is a parameter curvilinear integral with respect to arc length, then one has(i)
the mappingA2 is convex on [0,1],(ii)
one has the bounds(3.3)inf⁡t∈[0,1]A2(t)=A2(0)=f(a,b),sup⁡t∈[0,1]A2(t)=A2(1)=14h∮|x-a|=h/2|y-b|=h/2⁡f(x,y)ds,(iii)
the mappingA2 is increasing on [0,1],(iv)
there is a pointξ∈(0,1) so that
(3.4)A2(ξ)=1h2∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy,(v)
one has the Hadamard’s type inequality (1.8) and a refinement
(3.5)A2(0)≤A2(η)≤A2(ξ)≤A2(μ)≤A2(1),∀η∈(0,ξ),μ∈(ξ,1).Proof.
For a fixedt∈[0,1], let u=z if |v|=h/2, and v=z if |u|=h/2, we have
(3.6)A2(t)=14h∮|u|=h/2|v|=h/2⁡f(a+tu,b+tv)ds=14h∫-h/2h/2[f(a+tz,b-th2)+f(a+tz,b+th2)+f(a-th2,b+tz)+f(a+th2,b+tz)]dz,
which indicates the mapping A2(t) is continuous on [0,1].
(i) Lett1,t2∈[0,1] and α,β≥0 with α+β=1. By the convexity of f, we have(3.7)f[a+(αt1+βt2)z,b-(αt1+βt2)h2]=f[α(a+t1z)+β(a+t2z),α(b-t1h2)+β(b-t2h2)]≤αf(a+t1z,b-t1h2)+βf(a+t2z,b-t2h2).
Similarly, we have
(3.8)f[a+(αt1+βt2)z,b+(αt1+βt2)h2]≤αf(a+t1z,b+t1h2)+βf(a+t2z,b+t2h2),f[a-(αt1+βt2)h2,b+(αt1+βt2)z]≤αf(a-t1h2,b+t1z)+βf(a-t2h2,b+t2z),f[a+(αt1+βt2)h2,b+(αt1+βt2)z]≤αf(a+t1h2,b+t1z)+βf(a+t2h2,b+t2z).
Thus,
(3.9)A2(αt1+βt2)≤αA2(t1)+βA2(t2),
which proves the convexity of A2 on [0,1].
(ii) By the convexity off and (3.6), we have(3.10)A2(t)≥12th(∫a-th/2a+th/2f(x,b)dx+∫b-th/2b+th/2f(a,y)dy)≥f(a,b)=A2(0),
the last inequality holds owing to (1.1); as the convexity of A2 we have
(3.11)A2(t)≤tA2(1)+(1-t)A2(0)≤A2(1),
and note that
(3.12)A2(1)=14h∮|u|=h/2|v|=h/2⁡f(a+u,b+v)ds=14h∮|x-a|=h/2|y-b|=h/2⁡f(x,y)ds,
the bounds (3.3) hold.
(iii) The monotonicity ofA2 follows as in the proof of Theorem 2.1, and we omit the detail.
(iv) We define the functionϕ on [0,1] by(3.13)ϕ(t)=∬|x-a|≤th/2|y-b|≤th/2⁡f(x,y)dxdy-t2∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy,
then
(3.14)ϕ(0)=ϕ(1)=0,ϕ′(t)=h2(∫a-th/2a+th/2[f(x,b+th2)+f(x,b-th2)]dx+∫b-th/2b+th/2[f(a+th2,y)+f(a-th2,y)]dy)-2t∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy,=h2∮|x-a|=th|y-b|=th⁡f(x,y)ds-2t∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy.
Let x=a+tu and y=b+tv, we have
(3.15)ϕ′(t)=th2∮|u|=h/2|v|=h/2⁡f(a+tu,b+tv)ds-2t∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy=2th2A2(t)-2t∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy.
According to the Rolle’s mean-value theorem, there is a point ξ∈(0,1) so that
(3.16)0=ϕ′(ξ)=2h2ξA2(ξ)-2ξ∬|x-a|≤h/2|y-b|≤h/2⁡f(x,y)dxdy,
thus, (3.4) holds.
(v) Considering the properties (ii), (iii), and (iv) above, we have proved the Hadamard’s type inequality (1.8) and the inequality (3.5).


## 4. Remark

In recent years, several authors have discussed the problem of extending the classical inequality (1.1) to a convex function f on a general convex body Ω of n-dimensional space [6–9]. For the left side of (1.1) (i.e., Jensen’s inequality), the following inequality was showed in [6, 9] (also see [7]):
(4.1)f(c)≤1μ(Ω)∫Ωf(x)dx,
where c is the barycenter of Ω, and μ(Ω) denotes n-dimensional Lebesgue’s measure of Ω.It is a much more delicate problem to extend the right side of (1.1) on a general convex body of n-dimensional space [7]. On a closed hyperball of the n-dimensional space, de la Cal and Cárcamo [7] obtained the inequality
(4.2)1Vol(K)∫Kf(x)dx≤1Area(K*)∮K*f(x)ds,
where Vol(K) denotes the volumes of K, and Area(K*) denotes the areas of the hypersphere K*, for n=2,3 which was early obtained by Dragomir [4, 5] based on Calculus.When the convex functionf was supposed to be differentiable on a closed convex body Ω, the following inequality was proved in [9] based on Stokes’ Formula
(4.3)1Vol(Ω)∫Ωf(x)dx≤1n+11Area(∂Ω)∮∂Ω(1+Area(∂Ω)Vol(Ω)(x-θ)π(x))f(x)ds,
where ∂Ω denotes the piecewisely smooth boundary of Ω, θ denotes the barycenter of ∂Ω, and π(x) denotes the exterior unit normal vector in the point x.For a closed hyperballΩ, the inequality (4.2) is able to be deduced from (4.3); and for a hypercube Ω, the inequality (1.9) is also able to be deduced from (4.3); however, these are on the condition that f is differentiable.Now, letf be a convex function defined on an open region D⊆Rn, the hypercube {x:∥x-c∥∞≤h/2}⊆D, the mapping An is defined by
(4.4)An:[0,1]⟶R,An(t)=12nhn-1∮‖u‖∞=h/2f(c+tu)ds,
which is a parameter surface integral with respect to area of n-1 dimension, and let(4.5)ϕ(t)=∫‖x-c‖∞≤th/2f(x)dx-tn∫‖x-c‖∞≤h/2f(x)dx,
in the same way as the proof of Theorem 3.1, by the induction we will be able to obtain the inequality (1.9).Some mappings associated to Hadamard’s inequality have been established in [1–5], which refine the classical Hadamard’s inequality. These mappings are based on Hadamard’s inequality. However, our method in this paper to establish mappings is different from theirs: we first proved (1.1) by the properties of A1 and subsequently established A2 and proved (1.8) based on (1.1). Thus, using the induction, we will be able to establish the mapping An and prove Hadamard’s inequality on a hypercube.



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*Source: 594758-2011-02-16.xml*