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volume 4, issue 1, article 13, 2003.

Received 3 December, 2002;

accepted 19 December, 2002.

Communicated by:A.M. Rubinov

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Journal of Inequalities in Pure and Applied Mathematics

HADAMARD-TYPE INEQUALITIES FOR QUASICONVEX FUNCTIONS

N. HADJISAVVAS

Department of Product and Systems Design Engineering University of the Aegean

84100 Hermoupolis, Syros GREECE.

EMail:nhad@aegean.gr

URL:http://www.syros.aegean.gr/users/nhad/

c

2000Victoria University ISSN (electronic): 1443-5756 133-02

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Hadamard-type Inequalities for Quasiconvex Functions

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J. Ineq. Pure and Appl. Math. 4(1) Art. 13, 2003

http://jipam.vu.edu.au

Abstract

Recently Hadamard-type inequalities for nonnegative, evenly quasiconvex functions which attain their minimum have been established. We show that these inequalities remain valid for the larger class containing all nonnegative quasiconvex functions, and show equality of the corresponding Hadamard constants in case of a symmetric domain.

2000 Mathematics Subject Classification:26D15, 26B25, 39B62.

Key words: Quasiconvex function, Hadamard inequality.

1. Introduction

The well-known Hadamard inequality for convex functions has been recently generalized to include other types of functions. For instance, Pearce and Rubi- nov [2], generalized an earlier result of Dragomir and Pearce [1] by showing that for any nonnegative quasiconvex function defined on [0,1]and any u ∈ [0,1], the following inequality holds:

f(u)≤ 1

min (u,1−u) Z 1

0

f(x)dx.

In a subsequent paper, Rubinov and Dutta [3] extended the result to then- dimensional space, by imposing the restriction that the nonnegative function f attains its minimum and is not just quasiconvex, but evenly quasiconvex. The purpose of this note is to establish the inequality without these restrictions, and to obtain a simpler expression of the “Hadamard constant” which appears mul- tiplied to the integral. To be precise, given a convex subset X ofRⁿ, a Borel measure µon X and an element u ∈ X, we show that any nonnegative quasiconvex function satisfies an inequality of the form f(u) ≤ γR

Xf dµ where γ is a constant. An analogous inequalityf(u) ≤ γ∗

R

Xf dµ is obtained for all quasiconvex nonnegative functions for which f(0) = 0(under the assumption that0∈X). We obtain simple expressions for the constantsγ andγ∗ and show that they are equal, under a symmetry assumption.

In what follows,X is a convex, Borel subset ofRⁿ,µis a finite Borel measure on X, and λis the Lebesgue measure. As usual, µ λ means thatµis absolutely continuous with respect to λ. The open (closed) ball with centeru and radiusrwill be denoted byB(u, r)(B(u, r)). We denote byS the sphere

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{x∈Rⁿ :kxk= 1}and set, for eachv ∈S,u∈R, (1.1) X_v,u ={x∈X :hv, x−ui>0}.

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2. Inequality for Quasiconvex Functions

The following proposition shows that the Hadamard-type inequality for nonnegative evenly quasiconvex functions that attain their minimum, established in [3], is true for all nonnegative quasiconvex, Borel measurable functions.

Proposition 2.1. Let f :X → R∪ {+∞}be a Borel measurable, nonnegative quasiconvex function. Then for everyu∈X, the following inequality holds:

(2.1) inf

v∈Sµ(X_v,u)f(u)≤ Z

X

f dµ.

Proof. LetL={x∈X :f(x)< f(u)}. ThenLis convex andudoes not be- long to the relative interior ofL. We can thus separateuandLby a hyperplane, i.e., there exists v ∈ S such that ∀x ∈ L, hx, vi ≤ hu, vi. Hence, for every x∈X_v,u,f(x)≥f(u). Consequently,

µ(X_v,u)f(u)≤ Z

Xv,u

f dµ≤ Z

X

f dµ from which follows relation (2.1).

Note that ifµ=λ, then we do not have to assumef to be Borel measurable.

Indeed, any convex subset ofRⁿis Lebesgue measurable since it can be written as the union of its interior and a subset of its boundary; the latter is a Lebesgue null set, thus is Lebesgue measurable. Consequently, every quasiconvex function is Lebesgue measurable since by definition its level sets are convex.

It is possible thatinfv∈Sµ(X_v,u) = 0. In this case relation (2.1) does not say much. We can avoid this ifu∈intX andµdoes not vanish on sets of nonzero Lebesgue measure:

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Proposition 2.2. Assumptions as in Proposition2.1.

(i) Ifu∈intX andλµ, theninfv∈Sµ(X_v,u)>0.

(ii) Ifu /∈intX andµλ, theninfv∈Sµ(Xv,u) = 0.

Proof.

(i) Letε >0be such thatB(u, ε)⊆X. For eachv ∈Sandx∈B u+^ε₂v,^ε₂ , the triangle inequality yields

kx−uk ≤

x− u+^ε₂v +

^ε₂v < ε hencex∈X. Also,

hv, x−ui=

v, x− u+^ε₂v +

v,^ε₂v

≥ − kvk

x− u+^ε₂v

+^ε₂ >0.

Consequently,x∈X_v,u, i.e.,B u+ ^ε₂v,^ε₂

⊆X_v,u. Hence,

v∈Sinf λ(X_v,u)≥λ B u+ ^ε₂v,^ε₂

=λ B 0,^ε₂

>0.

By absolute continuity,infv∈Sµ(Xv,u)>0.

(ii) Since u /∈ intX, we can separate u and X by a hyperplane. It follows that for some v ∈ S, the set X_v,u is a subset of this hyperplane, hence λ(Xv,u) = 0which entails thatµ(Xv,u) = 0.

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Let us set

(2.2) γ = 1

inf_v∈Sµ(X_v,u),

where we make the convention ¹₀ = +∞. Then we can write (2.1) in the form

(2.3) f(u)≤γ

Z

X

f dµ.

The following Lemma will be useful for obtaining alternative expressions of

“Hadamard constants” such as γ and showing their sharpness. In particular, it shows thatX_v,u could have been defined (see relation (1.1)) by using≥instead of>. Let

(2.4) Xv,u ={x∈X :hv, x−ui ≥0}

be the closure ofX_v,u inX.

Lemma 2.3. Ifµλ, then (i) µ(X_v,u) =µ X_v,u

;

(ii) The functionv →µ(Xv,u)is continuous onS.

Proof.

(i) We know thatλ({x∈Rⁿ:hv, x−ui= 0}) = 0; consequently, λ X_v,u\X_v,u

= 0and this entails thatµ X_v,u\X_v,u

= 0.

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(ii) Suppose that(v_n)is a sequence inS, converging tov. Letε >0be given.

Chooser > 0large enough so that µ X\B(u, r)

< ε/2. Let us show that

n→∞lim λ X_v_n_,u∩B(u, r)

=λ X_v,u∩B(u, r) .

For this it is sufficient to show thatlimn→∞λ(X_n) = 0where X_n is the symmetric difference((Xv,u\Xvn,u)∪(Xvn,u\Xv,u))∩B(u, r). Ifx∈Xn

thenkx−uk ≤rand

(2.5) hv, x−ui>0≥ hvn, x−ui or

(2.6) hv_n, x−ui>0≥ hv, x−ui.

If, say, (2.6) is true, thenhv, x−ui ≤0<hvn−v, x−ui+hv, x−ui ≤ kv_n−vkr+hv, x−uithus|hv, x−ui| ≤ kv_n−vkr. The same can be deduced if (2.5) is true. Thus the projection ofX_nonv can be arbitrarily small; sinceXnis contained inB(u, r)this means thatlimn→∞λ(Xn) = 0as claimed.

By absolute continuity,limn→∞µ X_v_n_,u∩B(u, r)

=µ X_v,u∩B(u, r) . Since

|µ(X_v_n_,u)−µ(X_v,u)| ≤

µ X_v_n_,u∩B(u, r)

−µ X_v,u∩B(u, r) +

µ X_v_n_,u\B(u, r) +

µ X_v,u\B(u, r)

≤

µ Xvn,u∩B(u, r)

−µ Xv,u∩B(u, r) +ε,

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it follows that limn→∞|µ(X_v_n_,u)−µ(X_v,u)| ≤ ε. This is true for all ε >0, hencelimn→∞µ(Xvn,u) = µ(Xv,u).

We now obtain an alternative expression for the “Hadamard constant” γ, analogous to the one in [3]. Foru∈X define

A⁺_u ={(v, x₀)∈Rⁿ×X :hv, u−x₀i ≥1}.

Further, givenv ∈Rⁿandx₀ ∈Xset¹

X_v,x⁺ ₀ ={x∈X :hv, x−x₀i>1}.

Proposition 2.4. The following equality holds for everyu∈intX:

γ = 1

inf_(v,x

0)∈A⁺u µ X_v,x⁺

0

.

Proof. For every (v, x₀) ∈ A⁺_u, we set v⁰ = v/kvk. For each x ∈ X_v⁰_,u, hv, x−ui > 0 holds. Besides, (v, x₀) ∈ A⁺_u implies that hv, u−x₀i ≥ 1.

Hence, hv, x−x₀i = hv, x−ui+hv, u−x₀i > 1thusx ∈ X_v,x⁺ ₀. It follows thatX_v⁰_,u ⊆X_v,x⁺

0; consequently,

(2.7) inf

(v,x0)∈A⁺_u

µ X_v,x⁺ ₀

≥ inf

v∈Sµ(X_v,u).

1There is sometimes a change in notation with respect to [3].

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To show the reverse inequality, let v ∈ S be given. Sinceu ∈ intX, we may find x0 ∈ X such thathv, u−x0i > 0. Chooset > 0so that for v⁰ = tv one hashv⁰, u−x₀i= 1. The following equivalences hold:

x∈X_v⁺0,x0 ⇔ hv⁰, x−x₀i>1

⇔ hv⁰, x−ui+hv⁰, u−x₀i>1

⇔ hv⁰, x−ui>0

⇔ hv, x−ui>0

⇔x∈X_v,u.

Thus, for everyv ∈Sthere exists(v⁰, x₀)∈A⁺_u such thatX_v,u =X_v⁺0,x0. Hence equality holds in (2.7).

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3. Inequality for Quasiconvex Functions such that f (0) = 0.

Whenever 0 ∈ X andf(0) = 0, another Hadamard-type inequality has being obtained in [3], assuming that f is nonnegative and evenly quasiconvex. We generalize this result to nonnegative quasiconvex functions and compare with the previous findings. Let h : R⁺ → R⁺ be increasing withh(c) > 0for all c > 0andλ_h := sup_c>0 _h(c)^c <+∞(we follow the notation of [3]).

Proposition 3.1. Let f : X → R∪ {+∞}be Borel measurable, nonnegative and quasiconvex. If0∈Xandf(0) = 0, then for everyu∈X,

(3.1) inf

v∈S,hv,ui≥0µ(X_v,u)f(u)≤λ_h Z

X

h(f(x))dµ.

Proof. Iff(u) = 0we have nothing to prove. Suppose thatf(u)>0. Coming back to the proof of Proposition2.1, we know that there existsv ∈S such that

∀x ∈ X_v,u, f(x) ≥ f(u); hence h(f(x)) ≥ h(f(u)), from which it follows that

µ(Xv,u)h(f(u))≤ Z

Xv,u

h(f(x))dµ≤ Z

X

h(f(x))dµ.

Note that0∈/ X_v,u becausef(0) < f(u); thus,hv, ui ≥0. Consequently, inf

v∈S,hv,ui≥0µ(X_v,u)h(f(u))≤ Z

X

h(f(x))dµ.

Finally, note that by definition ofλ_h, f(u) ≤ λ_hh(f(u)) from which follows (3.1).

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Note that relation (3.1) is only interesting if u 6= 0 since otherwise it is trivially true. Let us defineγ∗ by

(3.2) γ_∗ =







1

infv∈S,hv,ui≥0µ(X_v,u) ifu6= 0

0 ifu= 0.

We obtain an alternative expression for γ∗, similar to that in [3]. Given u ∈ X\ {0}, setB_u = {v ∈Rⁿ:hv, ui ≥1}, and for anyv ∈ Rⁿ, setX_v⁺ = {x∈X :hv, xi>1}.

Proposition 3.2. The following equality holds for everyu∈X\ {0}:

v∈Binfu

µ X_v⁺

= inf

v∈S,hv,ui>0µ(Xv,u).

Proof. For everyv ∈B_uwe setv⁰ =v/kvkand show thatX_v⁰_,u ⊆X_v⁺. Indeed, ifx∈X_v⁰_,u, then we havehv, x−ui>0hencehv, xi=hv, ui+hv, x−ui>

1, i.e.,x∈X_v⁺. Sincehv⁰, ui>0, it follows that

(3.3) inf

v∈Bu

µ X_v⁺

≥ inf

v∈S,hv,ui>0µ(X_v,u).

To show equality, letv ∈S be such thathv, ui>0. Chooset >0such that thv, ui= 1and setv⁰ =tv. For everyx∈X_v⁺0 one hashv⁰, xi>1, hence,

hv⁰, x−ui=hv⁰, xi − hv⁰, ui>0.

It follows thathv, x−ui>0, i.e.,x∈X_v,u. Thus,X_v⁺0 ⊆X_v,uandv⁰ ∈B_u. This shows that in (3.3) equality holds.

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Proposition 3.3. Ifµλthen we also have the equalities

γ∗ = 1

infv∈S,hv,ui>0µ(X_v,u) = 1

minv∈S,hv,ui≥0µ Xv,u

(ifu6= 0)

γ = 1

min_v∈Sµ X_v,u. (3.4)

Proof. We first observe that, according to Lemma 2.3, µ X_v,u

= µ(X_v,u).

The same lemma entails that infv∈S,hv,ui>0µ(Xv,u) = infv∈S,hv,ui≥0µ(Xv,u) and that this infimum is attained, since the set{v ∈S :hv, ui ≥0}is compact.

In the same way, the infimum in (2.2) is attained.

Wheneverµ λ, the constantγ is sharp, in the sense that givenu ∈ X, there exists a nonnegative quasiconvex function f such thatf(u) = γR

Xf dµ.

Indeed, since the minimum in (3.4) is attained for somev₀ ∈ S, it is sufficient to take f to be the characteristic function of Xv0,u (see Corollary 2 of [3]).

Analogous considerations can be made forγ∗ (see Corollary 4 of [3]).

We now show the equality ofγandγ∗under a symmetry assumption:

Corollary 3.4. Suppose that X has0 as center of symmetry, u ∈ intX\ {0}

andµλ. Ifµ(A) =µ(−A)for every BorelA⊆X, thenγ =γ∗.

Proof. For every v ∈ S such that hv, ui < 0, set v⁰ = −v and Y = {x ∈ X : hv, x+ui > 0}. Since 0 is a center of symmetry, one can check that Y =−X_v⁰_,u.

If x ∈ Y then hv, x−ui = hv, x+ui −2hv, ui > 0. Thus, Y ⊆ X_v,u andµ(X_v,u)≥ µ(Y) = µ(X_v⁰_,u). It follows that the minimum in (3.4) can be restricted tov ∈Ssuch thathv, ui ≥0. Thus,γ =γ_∗.

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References

[1] S.S. DRAGOMIR AND C.E.M. PEARCE, Quasi-convex functions and Hadamard’s inequality, Bull. Austral. Math. Soc., 57 (1998), 377–385.

[2] C.E.M. PEARCE AND A.M. RUBINOV, P–functions, quasiconvex func- tions and Hadamard-type inequalities, J. Math. Anal. Appl., 240 (1999), 92–104.

[3] A.M. RUBINOV AND J. DUTTA, Hadamard type inequalities for quasi- convex functions in higher dimensions, J. Math. Anal. Appl., 270 (2002), 80–91.