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volume 1, issue 1, article 3, 2000.

Received 26 October, 1999;

accepted 7 December, 1999.

Communicated by:D.B. Hinton

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Journal of Inequalities in Pure and Applied Mathematics

A STEFFENSEN TYPE INEQUALITY

HILLEL GAUCHMAN

Department of Mathematics Eastern Illinois University Charleston, IL 61920, USA EMail:[email protected]

2000c Victoria University ISSN (electronic): 1443-5756 009-99

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A Steffensen Type Inequality Hillel Gauchman

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J. Ineq. Pure and Appl. Math.1(1) Art. 3, 2000

http://jipam.vu.edu.au

Abstract

Steffensen’s inequality deals with the comparison between integrals over a whole interval[a, b]and integrals over a subset of[a, b]. In this paper we prove an inequality which is similar to Steffensen’s inequality. The most general form of this inequality deals with integrals over a measure space. We also consider the discrete case.

2000 Mathematics Subject Classification:26A15

Key words: Steffensen inequality, upper-separating subsets

1. Introduction

The most basic inequality which deals with the comparison between integrals over a whole interval[a, b]and integrals over a subset of[a, b]is the following inequality, which was established by J.F. Steffensen in 1919, [3].

Theorem 1.1. (STEFFENSEN’S INEQUALITY) Letaandbbe real numbers such that a < b, f and g be integrable functions from [a, b] into R such that f is nonincreasing and for everyx∈[a, b],0≤g(x)≤1. Then

b

Z

b−λ

f(x)dx≤

b

Z

a

f(x)g(x)dx≤

a+λ

Z

a

f(x)dx,

whereλ=

b

R

a

g(x)dx.

The following is a discrete analogue of Steffensen’s inequality, [1]:

Theorem 1.2. (DISCRETE STEFFENSEN’S INEQUALITY). Let (x_i)ⁿ_i=1 be a nonincreasing finite sequence of nonnegative real numbers, and let (y_i)ⁿ_i=1 be a finite sequence of real numbers such that for every i, 0 ≤ y_i ≤ 1. Let k₁ k₂ ∈ {1, . . . , n}be such thatk₂ ≤

n

P

i=1

y_i ≤k₁. Then

n

X

i=n−k2+1

x_i ≤

n

X

i=1

x_iy_i ≤

k1

X

i=1

x_i.

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In section2we consider the discrete case. Our first result is the following.

Theorem 1.3. Let ` ≥ 0 be a real number, (xi)ⁿ_i=1 be a nonincreasing finite sequence of real numbers in[`,∞), and(y_i)ⁿ_i=1be a finite sequence of nonneg- ative real numbers. LetΦ : [`,∞)→[0,∞)be strictly increasing, convex, and such thatΦ(xy) ≥ Φ(x)Φ(y)for allx, y, xy ≥ `. Letk ∈ {1, . . . , n}be such thatk ≥`andΦ(k)≥Pn

i=1y_i. Then either

n

X

i=1

Φ(x_i)y_i ≤Φ

k

X

i=1

x_i

! or

k

X

i=1

y_i ≥1.

Theorem1.3takes an especially simple form ifΦ(x) = x^α, whereα≥1.

Theorem 1.4. Let (x_i)ⁿ_i=1 be a nonincreasing finite sequence of nonnegative real numbers, and let(yi)ⁿ_i=1 be a finite sequence of nonnegative real numbers.

Assume thatα≥1. Letk ∈ {1, . . . , n}be such that

k ≥

n

X

i=1

y_i

!_α¹ .

Then either

n

X

i=1

x^α_iy_i ≤

k

X

i=1

x_i

!α

or

k

X

i=1

y_i ≥1.

As an example of an application of Theorem 1.4 we obtain the following result:

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Theorem 1.5. Letα andβ be real numbers such thatα ≥ 1 +β, 0≤ β ≤ 1.

Let(xi)ⁿ_i=1 be a nonincreasing sequence of nonnegative real numbers. Assume

that n

X

i=1

x_i ≤A,

n

X

i=1

x^α_i ≥B^α,

whereAandB are positive real numbers. Letk∈ {1,2. . . , n}be such that

k ≥ A

B _α−1^β

.

Then

k

X

i=1

x^β_i ≥B^β.

Forβ = 1this is a result from [1].

The main result of section 3 is Theorem 3.2. This theorem is similar to Theorem 1.3, but it involves integrals over a measure space instead of finite sums. The key tool that we use to state and to prove Theorem3.2is the concept of separating subsets introduced and studied in [1]. If we take a measure space to be just a closed interval of the real lineR, we obtain the following simplest case of Theorem3.2:

Theorem 1.6. Let` ≥0be a real number,aandbbe real numbers such thata <

b,f andgbe integrable functions from[a, b]into[`,∞)and[0,∞)respectively, such that f is nonincreasing. Let Φ : [`,∞) → [0,∞)be strictly increasing,

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convex, and such thatΦ(xy) ≥ Φ(x)Φ(y)for allx, y, xy ≥ `. Letλbe a real number such thatΦ(λ) =Rb

ag(x)dx. Assume thatλ≤b−aand

f(a)−f(a−λ)≤

a+λ

Z

a

[f(x)−f(a+λ)]dx.

Then either

b

Z

a

(Φ◦f)g dx≤Φ





a+λ

Z

a

f dx



 or

a+λ

Z

a

g dx≥1.

Remark 1.1. In Theorems 1.3, 1.4, 1.6 and3.2 the assumption that Φ is convex can be weakened: it is enough to assume that Φis Wright-convex, where Wright-convexity means [4] thatΦ(t₂)−Φ(t₁) ≤ Φ(t₂ +δ)−Φ(t₁ +δ)for allt1, t2, δ ∈[0,∞)such thatt1 ≤t2. It is known that each convex function is Wright-convex, but the converse is not true.

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2. The Discrete Case

Proof. of Theorem1.3

n

X

i=1

Φ(x_i)y_i =

k

X

i=1

Φ(x_i)y_i+

n

X

i=k+1

Φ(x_i)y_i

≤

k

X

i=1

Φ(x_i)y_i+ Φ(x_k)

n

X

i=k+1

y_i

=

k

X

i=1

Φ(x_i)y_i+ Φ(x_k)

n

X

i=1

y_i−

k

X

i=1

y_i

!

=

k

X

i=1

y_i[Φ(x_i)−Φ(x_k)] + Φ(x_k)

n

X

i=1

y_i.

SinceΦ(k)≥ Pⁿ

i=1

yiandΦ(kxk)≥Φ(k)Φ(xk), we obtain

n

X

i=1

Φ(x_i)y_i ≤

k

X

i=1

y_i[Φ(x_i)−Φ(x_k)] + Φ(kx_k).

SinceΦis Wright-convex,

Φ(x_i)−Φ(x_k)≤Φ(x_i+ (k−1)x_k)−Φ(x_k+ (k−1)x_k)

= Φ(x_i+ (k−1)x_k)−Φ(kx_k)

≤Φ

k

X

i=1

x_i

!

−Φ(kx_k).

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Therefore

n

X

i=1

Φ(x_i)y_i ≤

"

Φ

k

X

i=1

x_i

!

−Φ(kx_k)

# _k X

i=1

y_i+ Φ(kx_k).

It follows that (2.1)

n

X

i=1

Φ(x_i)y_i−Φ

k

X

i=1

x_i

!

≤

"

Φ

k

X

i=1

x_i

!

−Φ(kx_k)

# _k X

i=1

y_i−1

! ,

since

k

X

i=1

xi ≥kxk, Φ

k

X

i=1

xi

!

−Φ(kxk)≥0.

Assume first that

Φ

k

X

i=1

x_i

!

−Φ(kx_k) = 0.

SinceΦis strictly increasing we obtain that

k

X

i=1

xi =kxk and therefore x1 =· · ·=xk.

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Then Φ

k

X

i=1

x_i

!

−

n

X

i=1

Φ(x_i)y_i ≥Φ(kx_k)−Φ(x_k)

n

X

i=1

y_i

≥Φ(k)Φ(x_k)−Φ(x_k)

n

X

i=1

y_i

= Φ(x_k) Φ(k)−

n

X

i=1

y_i

!

≥0.

Thus, in the caseΦ _k

P

i=1

x_i

−Φ(kx_k) = 0we obtain that

n

X

i=1

Φ(x_i)y_i ≤Φ

k

X

i=1

x_i

! ,

and we are done.

Assume now thatΦ _k

P

i=1

x_i

−Φ(kx_k) > 0. Then equation (2.1) implies that either

n

X

i=1

Φ(x_i)y_i ≤Φ

k

X

i=1

x_i

! or

k

X

i=1

y_i ≥1.

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Proof. of Theorem 1.5 Take x^β_i instead of x_i and ^α−1_β instead of α in Theo- rem1.4. Then we get that

k ≥

n

X

i=1

y_i

!_α−1^β

implies that either

n

X

i=1

x^α−1_i y_i ≤

k

X

k=1

x^β_i

!^α−1_β or

k

X

i=1

y_i ≥1.

Takey_i = ^x_Bⁱ fori= 1, . . . , n, then

n

X

i=1

y_i = 1 B

n

X

i=1

x_i ≤ A B.

Sincek≥ _B^A_α−1^β

, we obtain that

k ≥

n

X

i=1

y_i

!_α−1^β .

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This implies that either

k

X

i=1

x^β_i ≥

n

X

i=1

x^α−1_i y_i

!_α−1^β

= 1

B

n

X

i=1

x^α_i

!_α−1^β

≥ B^α

B _α−1^β

=B^β, or

k

X

i=1

x_i =B

k

X

i=1

y_i ≥B.

However, if

k

X

i=1

x_i ≥B,

then, since0≤β ≤1,

k

X

i=1

x^β_i ≥

k

X

i=1

x_i

!^β

≥B^β. Therefore in both cases we have that

k

X

i=1

x^β_i ≥B^β.

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Example 2.1. Let(x_i)ⁿ_i=1be a nonincreasing sequence in[0,∞)such that

k

P

i=1

x_i ≤ 400 and

k

P

i=1

x²_i ≥ 10,000. Then √

x₁ +√

x₂ ≥ 10. For a proof takeα = 2, β = ¹₂, A = 400, andB = 100in Theorem1.5. The result is the best possible since if n ≥ 16andx₁ = · · · =x₁₆ = 25,x₁₇ = · · · = x_n = 0, we have that

n

P

i=1

xi = 400,

n

P

i=1

x²_i = 10,000, and√

x1+√

x2 = 10.

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3. The Case of Integrals over a Measure Space.

Let X = (X,A, µ) be a measure space. From now on we will assume that 0< µ(X)<∞.

Definition 3.1. [1]. Let f ∈ L^◦(X), where L^◦(X)means the set of all mea- surable functions on X. Let (U, c) ∈ A ×R. We say that the pair (U, c) is upper-separating forf iff

{x∈X :f(x)> c}⊆^a U ⊆ {x^a ∈X :f(x)≥c}

whereA ⊆^a B means thatA is almost contained inB, i.e. µ(A\B) = 0. We say that a subset U of X is upper-separating for f if there existsc ∈ R such that(U, c)is an upper-separating pair forf.

It is possible to prove, [1], that ifµis continuous (for a definition of a continuous measure see, for example, [2]), then, given f ∈ L^◦(X), for any real numberλ such that0 ≤ λ ≤ µ(X), there exists an upper-separating subset U forf such thatµ(U) =λ.

Lemma 3.1. [1]. LetΦ : [0,∞)→Rbe convex and increasing. Letc∈[0,∞) and letf ∈L¹(X)have nonnegative values and satisfy the condition

(3.1) 0≤f −c≤

Z

X

(f−c)dµ a.e.

Then

Φ◦f −Φ(c)≤Φ Z

X

f dµ

−Φ (cµ(X)) a.e.

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Proof. The conclusion is trivial iff =c a.e. Suppose thatµ({x∈X :f(x)> c})>

0. Then the left inequality (3.1) implies that Z

X

(f −c)dµ > 0.

On the other hand, by integrating the right inequality (3.1), we obtain Z

X

(f−c)dµ≤



 Z

X

(f−c)dµ



µ(X),

which impliesµ(X)≥1. SinceΦis Wright-convex, we obtain that Φ◦f −Φ(c)≤Φ (f +c(µ(X)−1))−Φ (c+c(µ(X)−1))

= Φ (f −c+cµ(X))−Φ (cµ(X)) a.e.

BecauseΦis increasing it follows by (3.1) that

Φ◦f −Φ(c)≤Φ



 Z

X

(f−c)dµ+ Z

X

c dµ



−Φ (cµ(X))

= Φ(

Z

X

f dµ)−Φ (cµ(X)).

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Theorem 3.2. Let ` ≥ 0 be a real number. Let Φ : [`,∞) → R be convex strictly increasing, and such that Φ(xy) ≥ Φ(x)Φ(y)for allx, y, xy ≥ `. Let f, g ∈L⁰(X)be such thatf ≥`andg ≥0a.e.. Letλbe a real number and such thatΦ(λ) = R

Xg dµ. Assume that 0≤ λ ≤ µ(X), and let(U, c)be an upper- separating pair forf such thatµ(U) = λ. Assume thatf −c ≤ R

U

(f −c)dµ a.e. onU. Then either

Z

X

(Φ◦f)g dµ≤Φ



 Z

U

f dµ



 or Z

U

g dµ≥1.

Proof.

Z

X

(Φ◦f)g dµ= Z

U

(Φ◦f)g dµ+ Z

X\U

(Φ◦f)g dµ

≤ Z

U

(Φ◦f)g dµ+ Φ(c) Z

X\U

g dµ

= Z

U

(Φ◦f)g dµ+ Φ(c)



 Z

X

g dµ− Z

U

g dµ





= Z

U

g(Φ◦f−Φ(c))dµ+ Φ(c)Φ(λ).

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By Lemma3.1 Z

X

(Φ◦f)g dµ≤



Φ



 Z

U

f dµ



−Φ(cλ)



 Z

U

g dµ+ Φ(c)Φ(λ)

≤



Φ



 Z

U

f dµ



−Φ(cλ)



 Z

U

g dµ+ Φ(cλ).

It follows that (3.2)

Z

X

(Φ◦f)g dµ−Φ



 Z

U

f dµ



≤



Φ



 Z

U

f dµ



−Φ(cλ)







 Z

U

gdµ−1



.

Since(U, c)is upper-separating forf,f ≥conU. Hence Z

U

f dµ≥cλ and therefore Φ



 Z

U

f dµ



−Φ(cλ)≥0.

Assume first that Φ



 Z

U

f dµ



−Φ(cλ) = 0, then Φ



 Z

U

f dµ



= Φ



 Z

U

c dµ



.

SinceΦis strictly increasing, Z

U

f dµ= Z

U

c dµ, hence Z

U

(f −c)dµ= 0.

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Sincef ≥conU, we obtain thatf =ca.e. onU. Then

Φ



 Z

U

f dµ



− Z

X

(Φ◦f)g dµ= Φ



 Z

U

c dµ



− Z

X

(Φ◦f)g dµ

= Φ(cλ)− Z

X

(Φ◦f)g dµ

≥Φ(c)Φ(λ)− Z

X

(Φ◦f)g dµ.

Since(U, c) is upper-separating forf, we obtain thatf = ca.e. on U and f ≤ca.e. onX\U. Hencef ≤ca.e. onX. It follows that

Φ



 Z

U

f dµ



− Z

X

(Φ◦f)g dµ≥Φ(c)Φ(λ)− Z

X

Φ(c)g dµ

= Φ(c)



Φ(λ)− Z

X

g dµ



= 0.

This proves Theorem1.6in the caseΦ

R

U

f dµ

−Φ(cλ) = 0.

Assume now thatΦ

R

U

f dµ

−Φ(cλ)> 0, then equation3.2 implies that

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either

Z

X

(Φ◦f)g dµ−Φ



 Z

U

f dµ



≤0 or Z

U

g dµ≥1.

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References

[1] J.-C. EVARDANDH. GAUCHMAN, Steffensen type inequalities over gen- eral measure spaces, Analysis, 17 (1997), 301–322.

[2] P. HALMOS, Measure Theory, Springer-Verlag, New York, 1974.

[3] J.F. STEFFENSEN, On certain inequalities and methods of approximation, J. Inst. Actuaries, 51 (1919), 274–297.

[4] E.M. WRIGHT, An inequality for convex functions, Amer. Math. Monthly, 61 (1954), 620–622.