SUPERQUADRACITY OF FUNCTIONS AND REARRANGEMENTS OF SETS

(1)

Superquadracity and Rearrangements Shoshana Abramovich vol. 8, iss. 2, art. 46, 2007

Title Page

Contents

JJ II

J I

Page1of 11 Go Back Full Screen

Close

SUPERQUADRACITY OF FUNCTIONS AND REARRANGEMENTS OF SETS

SHOSHANA ABRAMOVICH

Department of Mathematics University of Haifa Haifa, 31905, Israel

EMail:abramos@math.haifa.ac.il

Received: 26 December, 2006

Accepted: 29 May, 2007

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 26D15.

Key words: Superquadratic functions, Convex functions, Jensen’s inequality.

Abstract: In this paper we establish upper bounds of

n

X

i=1

fxi+xi+1

2

+f

|xi−xi+1| 2

, xn+1=x1

when the functionfis superquadratic and the set(x) = (x1, . . . , xn)is given except its arrangement.

(2)

Title Page Contents

JJ II

J I

Close

1. Introduction

We start with the definitions and results of [1] and [5] which we use in this paper.

Definition 1.1. The sets (y⁻) = y₁⁻, . . . , y_n⁻

and (⁻y) = (⁻y₁, . . . ,⁻y_n) are symmetrically decreasing rearrangements of an ordered set (y) = (y₁, . . . , y_n)ofn real numbers, if

(1.1) y⁻₁ ≤ y⁻_n ≤y⁻₂ ≤ · · · ≤ y⁻ [ⁿ⁺²2 ]

and

(1.2) ⁻yn ≤ ⁻y1 ≤ ⁻yn−1 ≤ · · · ≤ ⁻y[ⁿ⁺¹2 ].

A circular rearrangement of an ordered set (y) = (y₁, . . . , y_n)is a cyclic rearrange- ment of (y)or a cyclic rearrangement followed by inversion.

Definition 1.2. An ordered set (y) = (y₁, . . . , y_n) of n real numbers is arranged in circular symmetric order if one of its circular rearrangements is symmetrically decreasing.

Theorem A ([1]). Let F (u, v)be a symmetric function defined for α ≤ u, v ≤ β for which ^∂²_∂u∂v^F(u,v) ≥0.

Let the set (y) = (y₁, . . . , y_n), α ≤ y_i ≤ β, i = 1, . . . , n be given except its arrangement. Then

n

X

i=1

F (yi, yi+1), (yn+1 =y1) is maximal if (y)is arranged in circular symmetrical order.

(4)

JJ II

J I

Close

Definition 1.3 ([5]). A function f, defined on an interval I = [0, L] or[0,∞) is superquadratic, if for eachxinI, there exists a real numberC(x)such that

f(y)−f(x)≥C(x) (y−x) +f(|y−x|) for ally∈I.

A function is subquadratic if−f is superquadratic.

Lemma A ([5]). Letf be a superquadratic function withC(x)as in Definition1.3.

(i) Thenf(0)≤0.

(ii) Iff(0) =f⁰(0) = 0,thenC(x) = f⁰(x)wheneverf is differentiable.

(iii) Iff ≥0, thenf is convex andf(0) =f⁰(0) = 0.

The following lemma presents a Jensen’s type inequality for superquadratic functions.

Lemma B ([6, Lemma 2.3]). Suppose thatf is superquadratic. Let x_r ≥ 0,1 ≤ r≤n and letx=Pn

r=1λ_rx_r,whereλ_r ≥0,andPn

r=1λ_r= 1.Then

n

X

r=1

λ_rf(x_r)≥f(x) +

n

X

r=1

λ_rf(|x_r−x|).

If f(x)is subquadratic, the reverse inequality holds.

From Lemma B we get an immediate result which we state in the following lemma.

(5)

JJ II

J I

Close

Lemma C. Letf(x)be superquadratic on [0, L]and let x, y ∈ [0, L],0≤ λ≤1, then

λf(x) + (1−λ)f(y)

≥f(λx+ (1−λ)y) +λf((1−λ)|y−x|) + (1−λ)f(λ|y−x|)

≥f(λx+ (1−λ)y) +

t−1

X

k=0

f

2λ(1−λ)|1−2λ|^k|x−y|

+λf (1−λ)|1−2λ|^t|x−y|

+ (1−λ)f λ|1−2λ|^t|x−y|

. Iff is positive superquadratic we get that:

λf(x)+(1−λ)f(y)≥f(λx+ (1−λ)y)+

t−1

X

k=0

f

2λ(1−λ)|1−2λ|^k|x−y|

More results related to superquadracity were discussed in [2] to [6].

In this paper we refine the results in [7] by showing that for positive superquadratic functions we get better bounds than in [7].

Theorem B ([7, Thm. 1.2]). If f is a convex function andx₁, x₂, . . . , x_n lie in its domain, then

n

X

i=1

f(x_i)−f

x₁+· · ·+x_n n

≥ n−1 n

f

x₁+x₂ 2

+· · ·+f

xn−1+x_n 2

+f

x_n+x₁ 2

.

(6)

Title Page Contents

JJ II

J I

Close

Theorem C ([7, Thm. 1.4]). If f is a convex function and a₁, . . . , a_n lie in its domain, then

(n−1) [f(b₁) +· · ·+f(b_n)]≤n[f(a₁) +· · ·+f(a_n)−f(a)], wherea = ^a¹^+···+a_n ⁿ andb_i = ^na−a_n−1ⁱ, i= 1, . . . , n.

(7)

Title Page Contents

JJ II

J I

Close

2. The Main Results

Theorem 2.1. Let f(x) be a superquadratic function on [0, L]. Then for x_i ∈ [0, L], i= 1, . . . , n,wherex_n+1 =x₁,

(2.1) n−1 n

n

X

i=1

f

n

X

i=1

x_i+x_i+1 2

! +f

n

X

i=1

|x_i−x_i+1| 2

!!

≤

n

X

i=1

f(xi)

!

−f

n

X

i=1

x_i n

!

− 1 n

n

X

i=1

f

xi−

n

X

j=1

x_j n

!

holds. If f⁰⁰⁰(x)≥0too, then n−1

n

X

i=1

f

n

X

i=1

xi+xi+1

2

! +f

n

X

i=1

|xi−xi+1| 2

!!

(2.2)

≤ n−1 n

n

X

i=1

f

xb_i+bx_i+1 2

+f

|xb_i−xb_i+1| 2

≤

n

X

i=1

f(x_i)

!

−f

n

X

i=1

x_i n

!

− 1 n

n

X

i=1

f

x_i−

n

X

j=1

x_j n

! ,

where(x) = (b xb₁, . . . ,bx_n)is a circular symmetrical rearrangement of (x) = (x₁, . . . , x_n). Example 2.1. The functions

f(x) =xⁿ, n≥2, x≥0, and the function

f(x) =

x²logx, x >0,

0, x= 0

(8)

Title Page Contents

JJ II

J I

Close

are superquadratic with an increasing second derivative and therefore (2.2) holds for these functions.

Proof. Letf be a superquadratic function on[0, L].Then by Lemma Bwe get for 0≤α ≤1, 1≤k ≤nandx_i ∈[0, L], x_n+1 =x₁,

n

X

i=1

f(x_i) (2.3)

= n−k n

n

X

i=1

f(x_i) + k n

n

X

i=1

f(x_i)

= n−k n

n

X

i=1

(αf(x_i) + (1−α)f(x_i+1)) + k n

n

X

i=1

f(x_i)

≥ n−k n

n

X

i=1

f(αx_i + (1−α)x_i+1)

+ n−k n

n

X

i=1

(αf((1−α)|x_i+1−x_i|) + (1−α)f(α|x_i+1−x_i|))

+k f Pn

i=1xi

n

+

n

X

i=1

1 nf

x_i− Pn

i=1xi

n

! .

Fork = 1andα= ¹₂ we get that (2.1) holds.

Iff⁰⁰⁰(x)≥0, then ^∂²_∂u∂v^F^(u,v) ≥0,where

F (u, v) = f(u+v) +f(|u−v|), u, v ∈[0, L].

(9)

JJ II

J I

Close

Therefore according to TheoremA, the sum

n

X

i=1

f

x_i+x_i+1 2

+f

|x_i+x_i+1| 2

, x_n+1 =x₁,

is maximal for(x) = (b bx1, . . . ,bxn),which is the circular symmetric rearrangement of(x).Therefore in this case (2.2) holds as well.

Remark 1. For a positive superquadratic functionf, which according to LemmaA is also a convex function, (2.1) is a refinement of TheoremB.

Iff⁰⁰⁰(x)≥0,(2.2) is a refinement of TheoremBas well.

Remark 2. TheoremBis refined by

n

X

i=1

f(x_i)−f Pn

i=1xi

n

≥ n−1 n

n

X

i=1

f

bxi+xbi+1

2

!

≥ n−1 n

n

X

i=1

f

xi+xi+1

2

,

because a convex function f satisfies the conditions of TheoremA forF (u, v) = f(u+v).

The following inequality is a refinement of TheoremCfor a positive superquadratic functionf, which is therefore also convex. The inequality results easily from Lemma Band the identity

n

X

i=1

f(a_i) =

n

X

i=1

1 n−1

n

X

j=1

f(a_j) (1−δ_ij)

!

(whereδij = 1fori=jandδij = 0fori6=j), therefore the proof is omitted.

(10)

Title Page Contents

JJ II

J I

Close

Theorem 2.2. Let f be a superquadratic function on [0, L], and let x_i ∈ [0, L], i= 1, . . . , n. Then

n n−1

_n X

i=1

f(x_i)

!

−f(x)

!

−

n

X

i=1

f(y_i)

≥ 1 n−1

n

X

i=1 n

X

j=1

f(|y_i−x_j|) (1−δ_ij)

!

+ 1

n−1

n

X

i=1

f(|x−x_i|),

wherex=Pn i=1

xi

n, yi = ^nx−x_n−1ⁱ

, i = 1, . . . , n.

(11)

Title Page Contents

JJ II

J I

Close

References

[1] S. ABRAMOVICH, The increase of sumsand products dependent on (y₁, . . . , y_n)by rearrangement of this set, Israel J. Math., 5(3) (1967).

[2] S. ABRAMOVICH, S. BANI ´C AND M. MATI ´C, Superquadratic functions in several variables, J. Math. Anal. Appl., 327 (2007), 1444–1460.

[3] S. ABRAMOVICH, S. BANI ´C AND M. KLARICI ´C BACULA, A variant of Jensen-Steffensen’s inequality for convex and superquadratic functions, J. Ineq.

Pure & Appl. Math., 7(2) (2006), Art. 70. [ONLINE: http://jipam.vu.

edu.au/article.php?sid=687].

[4] S. ABRAMOVICH, S. BANI Ć, M. MATI Ć AND J. PE ˇCARI Ć, Jensen- Steffensen’s and related inequalities for superquadratic functions, to appear in Math. Ineq. Appl.

[5] S. ABRAMOVICH, G. JAMESONANDG. SINNAMON, Refining Jensen’s in- equality, Bull. Sci. Math. Roum., 47 (2004), 3–14.

[6] S. ABRAMOVICH, G. JAMESONANDG. SINNAMON, Inequalities for aver- ages of convex and superquadratic functions, J. Ineq. Pure & Appl. Math., 5(4) (2004), Art. 91. [ONLINE:http://jipam.vu.edu.au/article.php?

sid=444].

[7] L. BOUGOFFA, New inequalities about convex functions, J. Ineq. Pure &

Appl. Math., 7(4) (2006), Art. 148. [ONLINE: http://jipam.vu.edu.

au/article.php?sid=766].