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: cfhvg@ux1.cts.eiu.edu
volume 1, issue 1, article 3,
2000.
Received 26 October, 1999;
accepted 7 December, 1999.
Communicated by: D.B. Hinton
Abstract
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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.
A Steffensen Type Inequality
2000 Mathematics Subject Classification: 26A15
Key words: Steffensen inequality, upper-separating subsets
Contents
1
2
3
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
The Discrete Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
The Case of Integrals over a Measure Space. . . . . . . . . . . . . . . 13
References
Hillel Gauchman
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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. (S TEFFENSEN ’ S INEQUALITY ) Let a and b be real numbers such
that a < b, f and g be integrable functions from [a, b] into R such that f is
nonincreasing and for every x ∈ [a, b], 0 ≤ g(x) ≤ 1. Then
A Steffensen Type Inequality
Zb
Zb
f (x)dx ≤
a
b−λ
where λ =
Rb
a+λ
Z
f (x)g(x)dx ≤
f (x)dx,
a
Hillel Gauchman
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g(x)dx.
a
The following is a discrete analogue of Steffensen’s inequality, [1]:
Theorem 1.2. (D ISCRETE S TEFFENSEN ’ S INEQUALITY ). Let (xi )ni=1 be a
nonincreasing finite sequence of nonnegative real numbers, and let (yi )ni=1 be
a finite sequence of real numbers such that for every i, 0 ≤ yi ≤ 1. Let k1
n
P
k2 ∈ {1, . . . , n} be such that k2 ≤
yi ≤ k1 . Then
i=1
n
X
i=n−k2 +1
xi ≤
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n
X
i=1
xi yi ≤
k1
X
i=1
xi .
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In section 2 we consider the discrete case. Our first result is the following.
Theorem 1.3. Let ` ≥ 0 be a real number, (xi )ni=1 be a nonincreasing finite
sequence of real numbers in [`, ∞), and (yi )ni=1 be a finite sequence of nonnegative real numbers. Let Φ : [`, ∞) → [0, ∞) be strictly increasing, convex, and
such that Φ(xy) ≥ Φ(x)Φ(y)
Pn for all x, y, xy ≥ `. Let k ∈ {1, . . . , n} be such
that k ≥ ` and Φ(k) ≥ i=1 yi . Then either
!
n
k
k
X
X
X
Φ(xi )yi ≤ Φ
xi
or
yi ≥ 1.
i=1
i=1
i=1
A Steffensen Type Inequality
Hillel Gauchman
Theorem 1.3 takes an especially simple form if Φ(x) = xα , where α ≥ 1.
Theorem 1.4. Let (xi )ni=1 be a nonincreasing finite sequence of nonnegative
real numbers, and let (yi )ni=1 be a finite sequence of nonnegative real numbers.
Assume that α ≥ 1. Let k ∈ {1, . . . , n} be such that
k≥
n
X
! α1
yi
.
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Then either
n
X
i=1
xαi yi
≤
k
X
i=1
!α
xi
or
k
X
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yi ≥ 1.
i=1
As an example of an application of Theorem 1.4 we obtain the following
result:
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Theorem 1.5. Let α and β be real numbers such that α ≥ 1 + β, 0 ≤ β ≤ 1.
Let (xi )ni=1 be a nonincreasing sequence of nonnegative real numbers. Assume
that
n
n
X
X
xi ≤ A,
xαi ≥ B α ,
i=1
i=1
where A and B are positive real numbers. Let k ∈ {1, 2 . . . , n} be such that
β
α−1
A
.
k≥
B
A Steffensen Type Inequality
Hillel Gauchman
Then
k
X
xβi ≥ B β .
i=1
For β = 1 this 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 Theorem 3.2 is 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 line R, we obtain the following simplest
case of Theorem 3.2:
Theorem 1.6. Let ` ≥ 0 be a real number, a and b be real numbers such that a <
b, f and g be 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 that Φ(xy) ≥ Φ(x)Φ(y) for all x, y, xy ≥ `. Let λ be a real
Rb
number such that Φ(λ) = a g(x)dx. Assume that λ ≤ b − a and
a+λ
Z
f (a) − f (a − λ) ≤
[f (x) − f (a + λ)] dx.
a
Then either
Zb
a
 a+λ

Z
(Φ ◦ f )g dx ≤ Φ 
f dx
a
or
a+λ
Z
g dx ≥ 1.
A Steffensen Type Inequality
Hillel Gauchman
a
Remark 1.1. In Theorems 1.3, 1.4, 1.6 and 3.2 the assumption that Φ is convex can be weakened: it is enough to assume that Φ is Wright-convex, where
Wright-convexity means [4] that Φ(t2 ) − Φ(t1 ) ≤ Φ(t2 + δ) − Φ(t1 + δ) for
all t1 , t2 , δ ∈ [0, ∞) such that t1 ≤ 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 Theorem 1.3
n
X
Φ(xi )yi =
i=1
k
X
Φ(xi )yi +
i=1
≤
=
=
n
X
Φ(xi )yi
i=k+1
k
X
Φ(xi )yi + Φ(xk )
n
X
i=1
i=k+1
k
X
n
X
i=1
k
X
Φ(xi )yi + Φ(xk )
yi
yi −
i=1
yi [Φ(xi ) − Φ(xk )] + Φ(xk )
i=1
Since Φ(k) ≥
n
P
k
X
i=1
n
X
!
yi
Hillel Gauchman
yi .
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i=1
yi and Φ(kxk ) ≥ Φ(k)Φ(xk ), we obtain
i=1
n
X
Φ(xi )yi ≤
i=1
k
X
yi [Φ(xi ) − Φ(xk )] + Φ(kxk ).
i=1
Since Φ is Wright-convex,
Φ(xi ) − Φ(xk ) ≤ Φ(xi + (k − 1)xk ) − Φ(xk + (k − 1)xk )
= Φ(xi + (k − 1)xk ) − Φ(kxk )
!
k
X
≤Φ
xi − Φ(kxk ).
i=1
A Steffensen Type Inequality
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Therefore
n
X
"
k
X
Φ(xi )yi ≤ Φ
i=1
!
#
− Φ(kxk )
xi
i=1
k
X
yi + Φ(kxk ).
i=1
It follows that
(2.1)
n
X
Φ(xi )yi − Φ
i=1
since
k
X
!
"
≤ Φ
xi
i=1
k
X
!
xi
#
− Φ(kxk )
i=1
k
X
i=1
!
yi − 1 ,
A Steffensen Type Inequality
Hillel Gauchman
k
X
xi ≥ kxk ,
Φ
i=1
k
X
!
xi
− Φ(kxk ) ≥ 0.
i=1
Contents
Assume first that
Φ
k
X
!
xi
− Φ(kxk ) = 0.
i=1
Since Φ is strictly increasing we obtain that
k
X
i=1
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xi = kxk
and therefore
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Then
Φ
k
X
!
−
xi
i=1
n
X
Φ(xi )yi ≥ Φ(kxk ) − Φ(xk )
i=1
n
X
yi
i=1
≥ Φ(k)Φ(xk ) − Φ(xk )
n
X
yi
i=1
= Φ(xk ) Φ(k) −
n
X
!
yi
≥ 0.
A Steffensen Type Inequality
i=1
Hillel Gauchman
Thus, in the case Φ
k
P
xi − Φ(kxk ) = 0 we obtain that
i=1
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n
X
Φ(xi )yi ≤ Φ
i=1
k
X
Contents
!
xi ,
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i=1
and we are done.
Assume now that Φ
k
P
xi
− Φ(kxk ) > 0. Then equation (2.1) implies
i=1
that either
n
X
i=1
Φ(xi )yi ≤ Φ
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k
X
i=1
!
xi
or
k
X
i=1
yi ≥ 1.
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Proof. of Theorem 1.5 Take xβi instead of xi and
rem 1.4. Then we get that
k≥
n
X
α−1
β
instead of α in Theo-
β
! α−1
yi
i=1
implies that either
n
X
xα−1
yi
i
k
X
≤
i=1
Take yi =
xi
B
xβi
or
for i = 1, . . . , n, then
i=1
A
B
β
α−1
A Steffensen Type Inequality
yi ≥ 1.
Hillel Gauchman
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n
A
1 X
xi ≤ .
yi =
B i=1
B
, we obtain that
k≥
k
X
i=1
k=1
n
X
Since k ≥
! α−1
β
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X
i=1
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! α−1
yi
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This implies that either
k
X
n
X
xβi ≥
i=1
β
! α−1
xα−1
yi
i
i=1
β
! α−1
n
X
1
=
xαi
B i=1
β
α α−1
B
≥
= Bβ ,
B
or
k
X
xi = B
i=1
k
X
yi ≥ B.
Hillel Gauchman
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However, if
k
X
A Steffensen Type Inequality
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xi ≥ B,
i=1
then, since 0 ≤ β ≤ 1,
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X
xβi
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i=1
k
X
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xi
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β
≥B .
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Therefore in both cases we have that
k
X
i=1
xβi ≥ B β .
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Example 2.1. Let (xi )ni=1 be a nonincreasing sequence in [0, ∞) such that
k
P
xi ≤
i=1
400 and
k
P
x2i ≥ 10, 000. Then
√
x1 +
√
x2 ≥ 10. For a proof take α = 2,
i=1
β = 21 , A = 400, and B = 100 in Theorem 1.5. The result is the best possible
since if n ≥ 16 and x1 = · · · = x16 = 25, x17 = · · · = xn = 0, we have that
n
n
P
P
√
√
xi = 400,
x2i = 10, 000, and x1 + x2 = 10.
i=1
i=1
A Steffensen Type Inequality
Hillel Gauchman
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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 measurable functions on X. Let (U, c) ∈ A × R. We say that the pair (U, c) is
upper-separating for f iff
a
a
{x ∈ X : f (x) > c} ⊆ U ⊆ {x ∈ X : f (x) ≥ c}
a
A Steffensen Type Inequality
where A ⊆ B means that A is almost contained in B, i.e. µ(A \ B) = 0. We
say that a subset U of X is upper-separating for f if there exists c ∈ R such
that (U, c) is an upper-separating pair for f .
Hillel Gauchman
It is possible to prove, [1], that if µ is continuous (for a definition of a continuous measure see, for example, [2]), then, given f ∈ L◦ (X), for any real
number λ such that 0 ≤ λ ≤ µ(X), there exists an upper-separating subset U
for f such that µ(U ) = λ.
Contents
Lemma 3.1. [1]. Let Φ : [0, ∞) → R be convex and increasing. Let c ∈ [0, ∞)
and let f ∈ L1 (X) have nonnegative values and satisfy the condition
Z
(3.1)
0 ≤ f − c ≤ (f − c)dµ a.e.
Z
Φ ◦ f − Φ(c) ≤ Φ
f dµ − Φ (cµ(X))
X
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X
Then
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Proof. The conclusion is trivial if f = c a.e. Suppose that µ ({x ∈ X : f (x) > c}) >
0. Then the left inequality (3.1) implies that
Z
(f − c)dµ > 0.
X
On the other hand, by integrating the right inequality (3.1), we obtain


Z
Z
(f − c)dµ ≤  (f − c)dµ µ(X),
X
A Steffensen Type Inequality
Hillel Gauchman
X
which implies µ(X) ≥ 1. Since Φ is 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.
Because Φ is increasing it follows by (3.1) that


Z
Z
Φ ◦ f − Φ(c) ≤ Φ  (f − c)dµ + c dµ − Φ (cµ(X))
X
ZX
= Φ( 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 all x, y, xy ≥ `. Let
f, g ∈ L0 (X) Rbe such that f ≥ ` and g ≥ 0 a.e.. Let λ be a real number and such
that Φ(λ) = X g dµ. Assume that 0 ≤ λ ≤ µ(X), and let (U, c) be
R an upperseparating pair for f such that µ(U ) = λ. Assume that f − c ≤ (f − c) dµ
U
a.e. on U . Then either

Z

Z
(Φ ◦ f )g dµ ≤ Φ 
X
Z
f dµ
g dµ ≥ 1.
or
U
A Steffensen Type Inequality
U
Hillel Gauchman
Proof.
Z
Z
(Φ ◦ f )g dµ =
X
(Φ ◦ f )g dµ +
U
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Z
(Φ ◦ f )g dµ
X\U
Z
≤
JJ
J
Z
(Φ ◦ f )g dµ + Φ(c)
U
Contents
g dµ
X\U

Z

Z
(Φ ◦ f )g dµ + Φ(c) 
=
U
Z
X
Z
g dµ −
g dµ
U
g (Φ ◦ f − Φ(c)) dµ + Φ(c)Φ(λ).
=
U
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By Lemma 3.1
 


Z
Z
Z
(Φ ◦ f )g dµ ≤ Φ  f dµ − Φ(cλ) g dµ + Φ(c)Φ(λ)
X

U
 U
Z
Z
≤ Φ  f dµ − Φ(cλ) g dµ + Φ(cλ).
U
U
It follows that
(3.2)

  



Z
Z
Z
Z
(Φ ◦ f )g dµ − Φ  f dµ ≤ Φ  f dµ − Φ(cλ)  gdµ − 1 .
X
U
U
Contents
JJ
J
U
Assume first that


Z
Φ  f dµ − Φ(cλ) = 0,


Z
then Φ 
Z
U
(f − c)dµ = 0.
U
c dµ .
U
Z
hence

f dµ = Φ 
U
U
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
U
Since Φ is strictly increasing,
Z
Z
f dµ = c dµ,
Hillel Gauchman
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U
Since (U, c) is upper-separating for f , f ≥ c on U . Hence


Z
Z
f dµ ≥ cλ and therefore Φ  f dµ − Φ(cλ) ≥ 0.
U
A Steffensen Type Inequality
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Since f ≥ c on U , we obtain that f = c a.e. on U . Then




Z
Z
Z
Z




Φ
f dµ − (Φ ◦ f )g dµ = Φ
c dµ − (Φ ◦ f )g dµ
U
X
U
X
Z
= Φ(cλ) −
(Φ ◦ f )g dµ
X
Z
≥ Φ(c)Φ(λ) −
(Φ ◦ f )g dµ.
Hillel Gauchman
X
Since (U, c) is upper-separating for f , we obtain that f = c a.e. on U and
f ≤ c a.e. on X\U . Hence f ≤ c a.e. on X. It follows that


Z
Z
Z


Φ
f dµ − (Φ ◦ f )g dµ ≥ Φ(c)Φ(λ) − Φ(c)g dµ
U
X
X


Z
= Φ(c) Φ(λ) −
g dµ = 0.
X
R
f dµ − Φ(cλ) = 0.
This proves Theorem 1.6 in the case Φ
U
R
Assume now that Φ
f dµ − Φ(cλ) > 0, then equation 3.2 implies that
U
A Steffensen Type Inequality
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either

Z

Z
(Φ ◦ f )g dµ − Φ 
X
Z
f dµ ≤ 0 or
U
g dµ ≥ 1.
U
A Steffensen Type Inequality
Hillel Gauchman
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References
[1] J.-C. EVARD AND H. GAUCHMAN, Steffensen type inequalities over general 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.
A Steffensen Type Inequality
Hillel Gauchman
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