Superquadratic functions and refinements of some classical inequalities

J. Korean Math. Soc. 45 (2008), No. 2, pp. 513–525

SUPERQUADRATIC FUNCTIONS AND REFINEMENTS OFSOME CLASSICAL INEQUALITIES

Senka Banic, Josip Pecaric, and Sanja Varosanec

Reprinted from theJournal of the Korean Mathematical Society

Vol. 45, No. 2, March 2008

c©2008 The Korean Mathematical Society

J. Korean Math. Soc. 45 (2008), No. 2, pp. 513–525

SUPERQUADRATIC FUNCTIONS AND REFINEMENTS OFSOME CLASSICAL INEQUALITIES

Senka Banic, Josip Pecaric, and Sanja Varosanec

Abstract. Using known properties of superquadratic functions we ob-tain a sequence of inequalities for superquadratic functions such as theConverse and the Reverse Jensen type inequalities, the Giaccardi and thePetrovic type inequalities and Hermite-Hadamard’s inequalities. Espe-cially, when the superquadratic function is convex at the same time, thenwe get refinements of classical known results for convex functions. Someother properties of superquadratic functions are also given.

1. Introduction

Let I be an interval in R and ϕ : I → R a convex function on I. If (x1 , . . . , xn)is any n-tuple in In (n ≥ 2) and (p1 , . . . , pn) is any non-negative n-tuple suchthat Pn =

∑ni=1 pi > 0, then the well known Jensen’s inequality

(1.1) ϕ

(1

Pn

n∑

i=1

pixi

)≤ 1

Pn

n∑

i=1

piϕ(xi)

holds.Under the same conditions as above, for xi ∈ [m, M ] ⊆ I (i = 1, . . . , n),

where −∞ < m < M < ∞, a converse of Jensen’s inequality gives an upperbound for the right side of (1.1):

(1.2)1

Pn

n∑

i=1

piϕ(xi) ≤ M − x

M −mϕ(m) +

x−m

M −mϕ(M),

where x = 1Pn

∑ni=1 pixi. This result remains valid under a weaker condition on

ϕ, i.e., it is sufficient that either (ϕ(x)−ϕ(m))/(x−m) or (ϕ(M)−ϕ(x))/(M−x)is a non-decreasing function of x on (m,M ] and [m,M), respectively (see [5,p.105]).

An integral analogue of the above inequality in the measure space (Ω,A, µ)with 0 < µ(Ω) < ∞ states that if f : Ω → [m,M ] is a function such that

Received September 12, 2006.2000 Mathematics Subject Classification. Primary 26D15.Key words and phrases. convex functions, Giaccardi’s inequality, Hermite-Hadamard’s

inequality, Jensen’s inequality, superquadratic functions, sum of power p.

c©2008 The Korean Mathematical Society

513

514 SENKA BANIC, JOSIP PECARIC, AND SANJA VAROSANEC

f, ϕ f ∈ L1(µ), then

(1.3)1

µ(Ω)

∫

Ω

(ϕ f)dµ ≤ M − f

M −mϕ(m) +

f −m

M −mϕ(M)

holds for any convex function ϕ : [m, M ] → R, where f = 1µ(Ω)

∫Ω

fdµ.

The result of the following theorem ([5, p.152]) is a simple consequence ofinequality (1.2).

Theorem A. Let (p1 , . . . , pn) be a non-negative n-tuple and (x1 , . . . , xn) be areal n-tuple such that

∑nk=1pkxk ∈ [0, a],

n∑

k=1

pkxk 6= x0, (xi − x0)

(n∑

k=1

pkxk − xi

)≥ 0, i = 1, . . . , n,

where x0 ∈ [0, a]. If (ϕ(x) − ϕ(x0))/(x − x0) is a non-decreasing function on[0, a], then

(1.4)n∑

k=1

pkϕ(xk) ≤ Aϕ

(n∑

k=1

pkxk

)+ B

(n∑

k=1

pk − 1

)ϕ(x0)

holds, where

A =∑n

k=1pk(xk − x0)∑nk=1pkxk − x0

, B =∑n

k=1pkxk∑nk=1pkxk − x0

.

The Giaccardi type inequality (1.4) can be obtained from (1.2) by settingm = x0 and M =

∑nk=1 pkxk (if x0 <

∑nk=1 pkxk, or M = x0 and m =∑n

k=1 pkxk in the reverse case).Moreover, for x0 = 0 we have A = B = 1 in (1.4), so in that case we get

Jensen-Petrovic’s inequality:

(1.5)n∑

k=1

pkϕ(xk) ≤ ϕ

(n∑

k=1

pkxk

)+

(n∑

k=1

pk − 1

)ϕ(0).

Furthermore, under the following assumptions on a real n-tuple (p1, . . . , pn):

p1 > 0, pi ≤ 0 (i = 2, . . . , n), Pn =n∑

i=1

pi > 0,

if xi ∈ I (i = 1, . . . , n) and 1Pn

∑ni=1 pixi ∈ I, the reverse of Jensen’s inequality

(1.6) ϕ

(1

Pn

n∑

i=1

pixi

)≥ 1

Pn

n∑

i=1

piϕ(xi)

holds for any convex function ϕ : I → R.In this paper we refine all the above inequalities and we also refine the

following pair of inequalities, that is called Hermite-Hadamard’s inequalities:

(1.7) f

(a + b

2

)≤ 1

b− a

∫ b

a

f (t) dt ≤ f (a) + f (b)2

,

SUPERQUADRATIC FUNCTIONS 515

which holds for any convex function f : I → R and a, b ∈ I.Now we quote some definitions and state a list of basic properties of su-

perquadratic functions established in [2] and [3].

Definition A ([2, Definition 2.1]). A function ϕ : [0,∞) → R is superquadraticprovided that for all x ≥ 0 there exists a constant C(x) ∈ R such that

(1.8) ϕ(y)− ϕ(x)− ϕ(|y − x|) ≥ C(x)(y − x)

for all y ≥ 0. We say that ϕ is subquadratic if −ϕ is a superquadratic function.

Jensen’s inequality for a superquadratic function is given in the followingtheorem:

Theorem B ([2, Theorem 2.3]). The inequality

(1.9) ϕ

(∫

Ω

fdµ

)≤

∫

Ω

(ϕ(f(s))− ϕ(|f(s)− ∫

Ωfdµ|)) dµ(s)

holds for all probability measures µ and all non-negative, µ-integrable functionsf if and only if ϕ is superquadratic.

A discrete version of the above theorem is also used in the sequel. It canbe obtained by choosing a discrete measure µ on 1, . . . , n, defined by µ(i) =pi/Pn, pi ≥ 0, Pn =

∑ni=1 pi > 0 and the function f defined by f(i) = xi :

Lemma A ([3, Lemma 2.3]). Suppose that ϕ is superquadratic. Let xi ≥ 0,i = 1, . . . , n and let x = 1

Pn

∑ni=1 pixi where pi ≥ 0 and Pn =

∑ni=1 pi > 0.

Then

(1.10)1

Pn

n∑

i=1

piϕ(xi)− ϕ(x) ≥ 1Pn

n∑

i=1

piϕ(|xi − x|) .

Lemma B ([2, Lemma 2.2]). Let ϕ be a superquadratic function with C(x) asin Definition A. Then:

(i) ϕ(0) ≤ 0,(ii) If ϕ(0) = ϕ′(0) = 0 then C(x) = ϕ′(x) whenever ϕ is differentiable at

x > 0.(iii) If ϕ ≥ 0, then ϕ is convex and ϕ(0) = ϕ′(0) = 0.

Lemma C ([2, Lemma 3.2]). Suppose ϕ is differentiable and ϕ(0) = ϕ′(0) = 0.If ϕ is superquadratic then ϕ(x)/x2 is non-decreasing on (0,∞).

It is interesting to see some examples of superquadratic functions.

Examples. (1) [2] Function ϕ(x) = xp is superquadratic for p ≥ 2 and sub-quadratic for p ∈ (0, 2].


(2) [1] Let h : (0,∞) → R be a continuous and a non-decreasing functionwith limt→0+ th(t) = 0. Then the function f : [0,∞) → R defined by

f(t) =∫ t

0

sh(s)ds,

is superquadratic.(3) [1] Let h : (0,∞) → R be a continuously differentiable and a non-

decreasing function with

limt→0+

th(t) = 0 and limt→0+

t2h′(t) = 0

such that the function t 7→ th′(t) is non-decreasing. Then the function f :[0,∞) → R defined by

f(t) = t2h(t),

is superquadratic.

The rest of this paper is divided into four parts. The first part is devotedto the Converse of Jensen’s inequality, the Giaccardi and the Jensen-Petrovicinequalities for superquadratic function and its refinements in the case whenthe considered function is convex and superquadratic simultaneously. In thesecond part the reversals of Jensen’s and Holder’s inequality for superquadraticfunction are considered. If moreover, we know that ϕ is superquadratic andϕ′(0) = ϕ(0) = 0, then ϕ(x)/x2 is non-decreasing and such kind of functionsare considered in the third part. And finally, the last part is devoted to theHermite-Hadamard inequalities and its refinements.

2. Converse of Jensen’s and related inequalities

The first theorem that we prove here is the Converse of Jensen’s inequalityfor superquadratic functions.

Theorem 1. Let (Ω,A, µ) be a measure space with 0 < µ(Ω) < ∞ and letϕ : [0,∞) → R be a superquadratic function. If f : Ω → [m,M ] ⊆ [0,∞) issuch that f, ϕ f ∈ L1(µ), then we have

(2.1)1

µ(Ω)

∫

Ω

(ϕ f)dµ + ∆c ≤ M − f

M −mϕ(m) +

f −m

M −mϕ(M),

where f = 1µ(Ω)

∫Ω

fdµ and

∆c =1

µ(Ω)1

M −m

∫

Ω

[(M − f)ϕ(f −m) + (f −m)ϕ(M − f)] dµ.

Proof. It can be easily obtained, from Definition A, that for a superquadraticfunction ϕ the following inequality

(2.2) ϕ(x) ≤ y2 − x

y2 − y1(ϕ(y1)− ϕ(x− y1)) +

x− y1

y2 − y1(ϕ(y2)− ϕ(y2 − x))


holds for all x, y1, y2 ≥ 0, such that y1 < x < y2 (see [2]). Setting x = f(t),y1 = m, y2 = M and for all t ∈ Ω, we get

ϕ(f(t)) +M − f(t)M −m

ϕ(f(t)−m) +f(t)−m

M −mϕ(M − f(t))

≤ M − f(t)M −m

ϕ(m) +f(t)−m

M −mϕ(M).

Integrating this over Ω we obtain∫

Ω

(ϕ f)dµ +1

M −m

∫

Ω

[(M − f)ϕ(f −m) + (f −m)ϕ(M − f)]dµ

≤ µ(Ω)M − ∫Ω

fdµ

M −mϕ(m) +

∫Ω

fdµ− µ(Ω)mM −m

ϕ(M),

and after dividing with µ(Ω) we get (2.1). ¤

Remark 1. Under the assumptions of the Theorem 1, and if ϕ is a non-negativeand superquadratic function (and therefore convex) we have both of the results,(1.3) and (2.1). Since the term ∆c is non-negative in this case, then the in-equality (2.1) refines the inequality (1.3).

If µ is a discrete measure we have a discrete version of the previous resulti.e., we have the following theorem.

Theorem 2. Let ϕ : [0,∞) → R be a superquadratic function. Let (x1 , . . . , xn)be an n-tuple in [m,M ]n (0 ≤ m < M < ∞) and (p1 , . . . , pn) be a non-negativen-tuple such that Pn =

∑ni=1 pi > 0. Then

(2.3)1

Pn

n∑

i=1

piϕ(xi) + ∆dc ≤ M − x

M −mϕ(m) +

x−m

M −mϕ(M),

where x = 1Pn

∑ni=1 pixi and

∆dc =1

Pn(M −m)

n∑

i=1

pi [(M − xi)ϕ(xi −m) + (xi −m)ϕ(M − xi)] .

In particular, under the assumptions of the above Theorem, for m = x0 andM =

∑nk=1 pkxk (if x0 <

∑nk=1 pkxk, or reversely in case x0 >

∑nk=1 pkxk) we

get the following Giaccardi type inequality for superquadratic function:

Corollary 1. Let ϕ : [0,∞) → R be a superquadratic function and x0 ≥ 0. Let(p1 , . . . , pn) and (x1 , . . . , xn) be non-negative n-tuples such that

n∑

k=1

pkxk 6= x0, (xi − x0)(n∑

k=1

pkxk − xi) ≥ 0, i = 1, . . . , n.

Then

(2.4)n∑

i=1

piϕ(xi) + ∆G ≤ Aϕ(n∑

k=1

pkxk) + B(n∑

k=1

pk − 1)ϕ(x0),


where

∆G =1

n∑k=1

pkxk − x0

(n∑

i=1

pi(n∑

k=1

pkxk − xi)ϕ(|xi − x0|)

+n∑

i=1

pi(xi − x0)ϕ(|n∑

k=1

pkxk − xi|))

(2.5)

and A and B are defined in Theorem A.

In the special case of Corollary 1, when x0 = 0, and so∑n

k=1 pkxk ≥ xi ≥ 0,we get Jensen-Petrovic’s type inequality for superquadratic functions:

n∑

i=1

piϕ(xi)

+1∑n

k=1pkxk

n∑

i=1

pi[(n∑

k=1

pkxk − xi)ϕ(xi) + xiϕ(n∑

k=1

pkxk − xi)]

≤(

n∑

k=1

pk − 1

)ϕ(0) + ϕ

(n∑

k=1

pkxk

).(2.6)

Remark 2. Considering the case when ϕ is a non-negative superquadratic func-tion, by Lemma B (iii) ϕ is also convex, so (1.2) is also valid for such func-tion. Since the term ∆dc in (2.3) is non-negative in this case, we concludethat inequality (2.3) refines the inequality (1.2). Similarly, a non-negative su-perquadratic function ϕ is convex and therefore

(ϕ(x)− ϕ(x0))/(x− x0)

is an increasing function, so (1.4) is also valid for that function. Under theassumptions of Corollary 1 we have

n∑k=1

pkxk − xi

n∑k=1

pkxk − x0

≥ 0 andxi − x0

n∑k=1

pkxk − x0

≥ 0, i = 1, . . . , n,

so in this case the term ∆G in (2.4) is non-negative. Hence, in this special case,inequality (2.4) represents a refinement of inequality (1.4). Finally, in the sameway as above, we conclude that if ϕ is non-negative and superquadratic, then(2.6) refines (1.5).

3. Reversal of Jensen’s inequality and Popoviciu’s type inequality

In this section we obtain Popoviciu’s type inequality. As we know thatinequality is a reverse version of Holder’s type inequalities which are given in


articles [2] and [6]. For this purpose we need one more general result which iscalled the Reversal of Jensen’s inequality.

First, we will prove a discrete Reversal of Jensen’s inequality for super-quadratic functions.

Theorem 3. Let (p1 , . . . , pn) be a real n-tuple such that

p1 > 0, pi ≤ 0 (i = 2, . . . , n), Pn =n∑

i=1

pi > 0.

If xi ≥ 0 (i = 1, . . . , n) and x = 1Pn

∑ni=1 pixi ≥ 0, then for a superquadratic

function ϕ : [0,∞) → R the following inequality

(3.1) ϕ

(1

Pn

n∑

i=1

pixi

)≥ 1

Pn

n∑

i=1

piϕ(xi)+ϕ(|x−x1|)− 1Pn

n∑

i=2

piϕ(|xi−x1|)

holds.

Proof. After making the substitutions:

p1 → Pn, x1 → x = 1Pn

n∑i=1

pixi,

pi → −pi , xi → xi , i = 2, . . . , n

in Lemma A, we get

1p1

(Pnϕ(x)−

n∑

i=2

piϕ(xi)

)− ϕ(x1)

≥ 1p1

(Pnϕ(|x− x1|)−

n∑

i=2

piϕ(|xi − x1|))

.

Multiplying the above inequality by p1/Pn we obtain (3.1). ¤Remark 3. As in the previous cases, we are considering the case of Theorem3 when ϕ is a non-negative superquadratic function. In that case ϕ is also aconvex function, so inequality (1.6) holds too. Since ϕ ≥ 0, Pn > 0 and pi ≤ 0for i = 2, . . . , n, we have

ϕ(|x− x1|)− 1Pn

n∑

i=2

piϕ(|xi − x1|) ≥ 0.

Hence, in Theorem 3 we get the following refinement of the Reversal of Jensen’sinequality (1.6):

ϕ

(1

Pn

n∑

i=1

pixi

)≥ 1

Pn

n∑

i=1

piϕ(xi) + ϕ(|x− x1|)− 1Pn

n∑

i=2

piϕ(|xi − x1|)

≥ 1Pn

n∑

i=1

piϕ(xi).


Using the previous Theorem we are able to prove an integral version of theReversal of Jensen’s inequality.

Theorem 4. Let (Ω,A, µ) be a measure space with 0 < µ(Ω) < ∞ and let ϕ :[0,∞) → R be a superquadratic function. If p, g : Ω → [0,∞) are functions anda, u ∈ [0,∞) are real numbers such that p, pg, pϕ(g), pϕ(|

RΩ pgdµRΩ pdµ

− g|) ∈ L1(µ),

0 <∫Ω

pdµ < u and ua− ∫Ω

pgdµ ≥ 0, then

(3.2) ϕ

(ua− ∫

Ωpgdµ

u− ∫Ω

pdµ

)≥ uϕ(a)− ∫

Ωpϕ(g)dµ

u− ∫Ω

pdµ+ ∆RJ ,

where

∆RJ

=1

u− ∫Ω

pdµ

[∫

Ω

pϕ

(∣∣∣∣∫Ω

pgdµ∫Ω

pdµ− g

∣∣∣∣)

dµ +(∫

Ω

pdµ

)ϕ

(∣∣∣∣∫Ω

pgdµ∫Ω

pdµ− a

∣∣∣∣)

+(u−∫

Ω

pdµ)ϕ( ∫

Ωpdµ

u− ∫Ω

pdµ

∣∣∣∣∫Ω

pgdµ∫Ω

pdµ− a

∣∣∣∣)]

.

Proof. Putting in Theorem 3 n = 2, p1 = u, p2 = − ∫Ω

pdµ, x1 = a, x2 =RΩ pgdµRΩ pdµ

we obtain the following

ϕ

(ua− ∫

Ωpgdµ

u− ∫Ω

pdµ

)≥

uϕ(a)− (∫Ω

pdµ)ϕ(RΩ pgdµRΩ pdµ

)

u− ∫Ω

pdµ

+ ϕ

( ∫Ω

pdµ

u− ∫Ω

pdµ

∣∣∣∣∫Ω

pgdµ∫Ω

pdµ− a

∣∣∣∣)

+

∫Ω

pdµ

u− ∫Ω

pdµϕ

(∣∣∣∣∫Ω

pgdµ∫Ω

pdµ− a

∣∣∣∣)

.

Using Jensen’s inequality for superquadratic function (Theorem B) we get(∫

Ω

pdµ

)ϕ

(∫Ω

pgdµ∫Ω

pdµ

)≤

∫

Ω

pϕ(g)dµ−∫

Ω

pϕ

(|g −

∫Ω

pgdµ∫Ω

pdµ|)

dµ.

Combining these two inequalities we get inequality (3.2). ¤

Considering the superquadratic function ϕ(x) = xp, p ≥ 2, and applyingresult (3.2) we obtain the Popoviciu type inequality.

Theorem 5. Let (Ω,A, µ) be as in Theorem 4. Let p ≥ 2 and q be realnumbers such that 1

p + 1q = 1. If f, g : Ω → [0,∞) are functions and f0, g0 are

positive real numbers such that fp, gq, fg, |gq/p∫Ω

fgdµ− f∫Ω

gqdµ|p ∈ L1(µ),


gq0 −

∫Ω

gqdµ > 0, fp0 −

∫Ω

fpdµ > 0 and∫Ω

gqdµ > 0 then

f0g0 −∫

Ω

fgdµ ≥[(

fp0 −

∫

Ω

fpdµ

)(gq0 −

∫

Ω

gqdµ

)p/q

+ ∆P

]1/p

(3.3)

≥(

fp0 −

∫

Ω

fpdµ

)1/p (gq0 −

∫

Ω

gqdµ

)1/q

,

where

∆P =(gq

0 −∫Ω

gqdµ)p/q

(∫Ω

gqdµ)p

∫

Ω

|gq/p

∫

Ω

fgdµ− f

∫

Ω

gqdµ|pdµ

+∣∣∣∣f0g

−q/p0

∫

Ω

gqdµ−∫

Ω

fgdµ

∣∣∣∣p

·(

1 +(gq

0 −∫Ω

gqdµ)p/q

(∫Ω

gqdµ)p/q

).

Proof. By applying the substitutions: ϕ(x) = xp, p ≥ 2,

u → gq0, p → gq, a → g

−q/p0 f0, g → g−q/pf

in (3.2) we obtain inequality (3.3), while the second inequality follows from thepositivity of ∆P . ¤

As we can see, the result of the previous theorem is, in fact, the refinementof the classical Popoviciu’s inequality, [5, p.125].

4. The case ϕ(0) = ϕ′(0) = 0

Here we state another versions of the considered inequalities for superquad-ratic functions. First we give a result which is, in special case, a refinement ofthe Converse of Jensen’s inequality (1.2) for m = 0.

Theorem 6. Let ϕ : [0, M ] → R (M > 0) be a differentiable superquadraticfunction with ϕ(0) = ϕ′(0) = 0 and let (Ω,A, µ) be a measure space with0 < µ(Ω) < ∞. If f : Ω → (0,M ] is such that f, ϕf

f ∈ L1(µ), then we have

(4.1)∫

Ω

ϕ f

fdµ ≤ ϕ(M)

M2

∫

Ω

fdµ.

Proof. Under the given assumptions on ϕ and according to Lemma C the func-tion ϕ(x)/x2 is non-decreasing on (0,M ]. So, for any t ∈ Ω we have

ϕ(f(t))f(t)2

≤ ϕ(M)M2

.

Then we obtain ∫

Ω

ϕ f

fdµ =

∫

Ω

ϕ f

f2· fdµ

≤∫

Ω

ϕ(M)M2

fdµ =ϕ(M)M2

∫

Ω

fdµ.

¤


Remark 4 (A refinement of the Converse of Jensen’s inequality). Especially, ifthe superquadratic function ϕ is additionally non-negative, then (by LemmaB) it is also a convex function with ϕ(0) = ϕ′(0) = 0. In that case, beside (4.1),the Converse of Jensen’s inequality (1.3) for m = 0 :

(4.2)1

µ(Ω)

∫

Ω

ϕ f

Mdµ ≤ f

M2ϕ(M),

holds too.Since f(t) ≤ M for all t ∈ Ω we get

∫

Ω

ϕ f

Mdµ ≤

∫

Ω

ϕ f

fdµ ≤ ϕ(M)

M2

∫

Ω

fdµ.

So, inequality (4.1) represents a refinement of the Converse of Jensen’s inequal-ity.

Let us point out a corresponding result for a discrete measure.

Theorem 7. Let ϕ : [0, M ] → R (M > 0) be a differentiable superquadraticfunction with ϕ(0) = ϕ′(0) = 0. If pi ≥ 0, xi ∈ (0,M ] (i = 1, . . . , n), and∑n

i=1 pi > 0, thenn∑

i=1

piϕ(xi)

xi≤ ϕ(M)

M2

n∑

i=1

pixi

holds.

Remark 5 (A refinement of the discrete Converse of Jensen’s inequality). Sim-ilarly as in the previous remark if the superquadratic function ϕ is additionallynon-negative, then we have a refinement of the Converse of Jensen’s inequality(1.2)

n∑

i=1

piϕ(xi)M

≤n∑

i=1

piϕ(xi)

xi≤ ϕ(M)

M2

n∑

i=1

pixi.

Remark 6 (A refinement of Jensen-Petrovic’s inequality). In a particular case,by setting M =

∑ni=1 pixi in Theorem 7, as we already have m = x0 = 0, we get

one more version of Jensen-Petrovic’s inequality for superquadratic function:n∑

i=1

piϕ(xi)

xi≤ ϕ(

∑ni=1 pixi)∑n

i=1 pixi,

and multiplying it by∑n

i=1 pixi ≥ 0, the following equivalent inequality:

(4.3)n∑

i=1

pixi

n∑

i=1

piϕ(xi)

xi≤ ϕ(

n∑

i=1

pixi).

In case when ϕ is (beside the above assumptions) also a non-negative func-tion, as we mentioned earlier, ϕ is a convex function too, and we have ϕ(0) =


ϕ′(0) = 0. So, in that case, Jensen-Petrovic’s inequality

(4.4) ϕ(n∑

i=1

pixi) ≥n∑

i=1

piϕ(xi)

holds.Let us consider the inequalities (4.3) and (4.4). They are not comparable

for any choice of weights pi, but if pi ≥ 1 (i = 1, . . . , n), then we have

n∑

i=1

pixi

n∑

i=1

piyi =n∑

i=1

p2i xiyi +

n∑

i 6=ji,j=1

pipjxiyj

≥n∑

i=1

p2i xiyi ≥

n∑

i=1

pixiyi,

where yi = ϕ(xi)/xi

So, using (4.3) we get

(4.5) ϕ(n∑

i=1

pixi) ≥n∑

i=1

pixi

n∑

i=1

piϕ(xi)

xi≥

n∑

i=1

piϕ(xi),

which is a refinement of the Jensen-Petrovic inequality (4.4).

Corollary 2. Inequalities for sums of order p Let (x1, . . . , xn) and (p1, . . . , pn)be non-negative n-tuples with pi ≥ 1 (i = 1, . . . , n). If r ≥ 2p, then

(n∑

i=1

pixri )

1/r ≤ (n∑

i=1

pixpi

n∑

i=1

pixr−pi )1/r ≤ (

n∑

i=1

pixpi )

1/p.

Proof. It is a simple consequence of the inequality (4.5) for a superquadraticfunction ϕ(x) = xr/p. ¤

It is a refinement of the well-known inequality for sums of order p, [5, p.165],which states that

(n∑

i=1

pixri )

1/r ≤ (n∑

i=1

pixpi )

1/p

for r ≥ p.

5. Hermite-Hadamard’s inequality

The following theorem gives to us an analogue of the inequalities in (1.7) forsuperquadratic functions. As we shall see, in the special case for non-negativesuperquadratic functions, we obtain the refinements of these inequalities.


Theorem 8. Let ϕ : [0,∞) → R be an integrable superquadratic function,0 ≤ a < b. Then

ϕ

(a + b

2

)+

1b− a

∫ b

a

ϕ

(∣∣∣∣t−a + b

2

∣∣∣∣)

dt

≤ 1b− a

∫ b

a

ϕ(t)dt(5.1)

≤ ϕ(a) + ϕ(b)2

− 1(b− a)2

∫ b

a

[(b− t)ϕ(t− a) + (t− a)ϕ(b− t)] dt.(5.2)

Proof. First we prove (5.1), the left Hermite-Hadamard inequality for super-quadratic functions. Since the function ϕ is superquadratic, then for f(t) = tand measure µ on Ω = [a, b], defined by dµ = 1

b−adt, inequality (1.9) showsthat

ϕ

(1

b− a

∫ b

a

tdt

)≤ 1

b− a

∫ b

a

ϕ(t)dt− 1b− a

∫ b

a

ϕ

(∣∣∣∣∣t−1

b− a

∫ b

a

sds

∣∣∣∣∣

)dt

holds. Hence, we get (5.1).Also, as ϕ is superquadratic, by setting x = t, y1 = a and y2 = b in (2.2)

we get that

ϕ(t) ≤ b− t

b− a(ϕ(a)− ϕ(t− a)) +

t− a

b− a(ϕ(b)− ϕ(b− t))

holds for all t, a, b ≥ 0 such that a < t < b. After integrating this expressionover the segment [a, b] :

∫ b

a

ϕ(t)dt ≤ 1b− a

∫ b

a

[bϕ(a)− aϕ(b) + t(ϕ(b)− ϕ(a))] dt

− 1b− a

∫ b

a

[(b− t)ϕ(t− a) + (t− a)ϕ(b− t)] dt

and dividing with (b− a) we get (5.2). ¤

Remark 7. If ϕ is superquadratic and non-negative, then by Lemma B, ϕ isalso a convex function. In this case the second term on the left side in (5.1) andthe last term in (5.2) are non-negative, so we have the following refinements ofHermite-Hadamard’s inequality (1.7):

ϕ

(a + b

2

)≤ ϕ

(a + b

2

)+

1b− a

∫ b

a

ϕ

(∣∣∣∣t−a + b

2

∣∣∣∣)

dt ≤ 1b− a

∫ b

a

ϕ(t)dt


and1

b− a

∫ b

a

ϕ(t)dt

≤ ϕ(a) + ϕ(b)2

− 1(b− a)2

∫ b

a

((b− t)ϕ(t− a) + (t− a)ϕ(b− t))dt

≤ ϕ(a) + ϕ(b)2

.

References

[1] S. Abramovich, S. Banic, and M. Matic, Superquadratic functions in several variables, J.Math. Anal. Appl. 327 (2007), no. 2, 1444–1460.

[2] S. Abramovich, G. Jameson, and G. Sinnamon, Refining Jensen’s Inequality, Bull. Math.Soc. Sci. Math. Roumanie (N.S.) 47 (95) (2004), no. 1-2, 3–14.

[3] , Inequalities for averages of convex and superquadratic functions, JIPAM. J.Inequal. Pure Appl. Math. 5 (2004), no. 4, Article 91.

[4] D. S. Mitrinovic, J. E. Pecaric, and A. M. Fink, Classical and New Inequalities in Anal-ysis, Mathematics and its Applications (East European Series), 61. Kluwer AcademicPublishers Group, Dordrecht, 1993.

[5] J. E. Pecaric, F. Proschan, and Y. L. Tong, Convex Functions, Partial Orderings, andStatistical Applications, Mathematics in Science and Engineering, 187. Academic Press,Inc., Boston, MA, 1992.

[6] G. Sinnamon, Refining the Holder and Minkowski inequalities, J. Inequal. Appl. 6 (2001),no. 6, 633–640.

Senka BanicFaculty of civil engineering and architectureUniversity of SplitMatice hrvatske 15, 21000 SplitCroatiaE-mail address: [email protected]

Josip PecaricFaculty of Textile TechnologyUniversity of ZagrebPierottijeva 6, 10000 ZagrebCroatiaE-mail address: [email protected]

Sanja VarosanecDepartment of MathematicsUniversity of ZagrebBijenicka 30, 10000 ZagrebCroatiaE-mail address: [email protected]

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Superquadratic functions and refinements of some classical inequalities