Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 1, Issue 1, Article 3, 2000
A STEFFENSEN TYPE INEQUALITY
HILLEL GAUCHMAN
D EPARTMENT OF M ATHEMATICS , E ASTERN I LLINOIS U NIVERSITY, C HARLESTON , IL 61920, USA
cfhvg@ux1.cts.eiu.edu
Received 26 October, 1999; accepted 7 December, 1999
Communicated by D.B. Hinton
A BSTRACT. 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.
Key words and phrases: Steffensen inequality, upper-separating subsets.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
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
Zb
Zb
f (x)dx ≤
b−λ
where λ =
Rb
a
a+λ
Z
f (x)g(x)dx ≤
f (x)dx,
a
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
ISSN (electronic): 1443-5756
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009-99
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G AUCHMAN
that for every i, 0 ≤ yi ≤ 1. Let k1 k2 ∈ {1, . . . , n} be such that k2 ≤
n
P
yi ≤ k1 . Then
i=1
n
X
xi ≤
n
X
xi yi ≤
i=1
i=n−k2 +1
k1
X
xi .
i=1
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
PnΦ(xy) ≥ Φ(x)Φ(y) 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
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
! α1
n
X
k≥
yi
.
i=1
Then either
n
X
xαi yi
i=1
≤
k
X
!α
xi
or
i=1
k
X
yi ≥ 1.
i=1
As an example of an application of Theorem 1.4 we obtain the following result:
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
X
xi ≤ A,
i=1
n
X
xαi ≥ B α ,
i=1
where A and B are positive real numbers. Let k ∈ {1, 2 . . . , n} be such that
β
α−1
A
k≥
.
B
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:
J. Ineq. Pure and Appl. Math., 1(1) Art. 3, 2000
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S TEFFENSEN T YPE I NEQUALITY
3
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, convex, and such that Φ(xy) ≥ Φ(x)Φ(y) for
Rb
all x, y, xy ≥ `. Let λ be a real 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+λ

Z
(Φ ◦ f )g dx ≤ Φ 
f dx
a
a+λ
Z
g dx ≥ 1.
or
a
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.
2. T HE D ISCRETE C ASE
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
Φ(xi )yi + Φ(xk )
i=1
=
yi
yi −
i=1
k
X
n
P
!
yi
i=1
yi [Φ(xi ) − Φ(xk )] + Φ(xk )
i=1
Since Φ(k) ≥
k
X
n
X
yi .
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
Therefore
n
X
"
Φ(xi )yi ≤ Φ
i=1
J. Ineq. Pure and Appl. Math., 1(1) Art. 3, 2000
k
X
i=1
!
xi
#
− Φ(kxk )
k
X
yi + Φ(kxk ).
i=1
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It follows that
!
# k
!
! "
n
k
k
X
X
X
X
xi − Φ(kxk )
yi − 1 ,
(2.1)
Φ(xi )yi − Φ
xi ≤ Φ
i=1
i=1
i=1
i=1
since
k
X
xi ≥ kxk ,
k
X
Φ
i=1
!
− Φ(kxk ) ≥ 0.
xi
i=1
Assume first that
k
X
Φ
!
xi
− Φ(kxk ) = 0.
i=1
Since Φ is strictly increasing we obtain that
k
X
xi = kxk
x1 = · · · = xk .
and therefore
i=1
Then
k
X
Φ
!
−
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.
i=1
Thus, in the case Φ
k
P
xi − Φ(kxk ) = 0 we obtain that
i=1
n
X
Φ(xi )yi ≤ Φ
i=1
and we are done.
Assume now that Φ
k
P
k
X
!
xi ,
i=1
xi − Φ(kxk ) > 0. Then equation (2.1) implies that either
i=1
n
X
Φ(xi )yi ≤ Φ
k
X
i=1
!
xi
or
i=1
k
X
yi ≥ 1.
i=1
instead of α in Theorem 1.4. Then we get
Proof. of Theorem 1.5 Take xβi instead of xi and α−1
β
that
β
! α−1
n
X
k≥
yi
i=1
implies that either
n
X
xα−1
yi
i
i=1
J. Ineq. Pure and Appl. Math., 1(1) Art. 3, 2000
≤
k
X
k=1
! α−1
β
xβi
or
k
X
yi ≥ 1.
i=1
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S TEFFENSEN T YPE I NEQUALITY
Take yi =
xi
B
for i = 1, . . . , n, then
n
X
i=1
Since k ≥
5
A
B
β
α−1
n
1 X
A
yi =
xi ≤ .
B i=1
B
, we obtain that
k≥
n
X
β
! α−1
.
yi
i=1
This implies that either
k
X
n
X
xβi ≥
i=1
β
! α−1
xα−1
yi
i
i=1
β
! α−1
n
1 X α
=
x
B i=1 i
β
α α−1
B
≥
= Bβ ,
B
or
k
X
xi = B
i=1
k
X
yi ≥ B.
i=1
However, if
k
X
xi ≥ B,
i=1
then, since 0 ≤ β ≤ 1,
k
X
xβi
≥
i=1
k
X
!β
xi
≥ Bβ .
i=1
Therefore in both cases we have that
k
X
xβi ≥ B β .
i=1
Example 2.1. Let (xi )ni=1 be a nonincreasing sequence in [0, ∞) such that
k
P
xi ≤ 400 and
i=1
k
P
i=1
x2i ≥ 10, 000. Then
√
√
x1 + x2 ≥ 10. For a proof take α = 2, 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,
n
n
P
P
√
√
x17 = · · · = xn = 0, we have that
xi = 400,
x2i = 10, 000, and x1 + x2 = 10.
i=1
J. Ineq. Pure and Appl. Math., 1(1) Art. 3, 2000
i=1
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G AUCHMAN
3. T HE C ASE
OF I NTEGRALS OVER A
M EASURE S PACE .
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
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 .
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.
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.
X
Then
Z
f dµ − Φ (cµ(X))
Φ ◦ f − Φ(c) ≤ Φ
a.e.
X
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
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)) .
X
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 ≥ `. LetRf, g ∈ L0 (X) be 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),
J. Ineq. Pure and Appl. Math., 1(1) Art. 3, 2000
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S TEFFENSEN T YPE I NEQUALITY
7
and
R let (U, c) be an upper-separating pair for f such that µ(U ) = λ. Assume that f − c ≤
(f − c) dµ a.e. on U . Then either
U


Z
Z
(Φ ◦ f )g dµ ≤ Φ 
X
Z
f dµ
g dµ ≥ 1.
or
U
U
Proof.
Z
Z
(Φ ◦ f )g dµ =
X
Z
(Φ ◦ f )g dµ +
U
(Φ ◦ f )g dµ
X\U
Z
≤
Z
(Φ ◦ f )g dµ + Φ(c)
g dµ
U
X\U
Z


Z
Z
(Φ ◦ f )g dµ + Φ(c)  g dµ − g dµ
=
X
U
Z
U
g (Φ ◦ f − Φ(c)) dµ + Φ(c)Φ(λ).
=
U
By Lemma 3.1
 


Z
Z
(Φ ◦ f )g dµ ≤ Φ 
X
Z
f dµ − Φ(cλ)
U
U
 


Z
≤ Φ 
g dµ + Φ(c)Φ(λ)
Z
f dµ − Φ(cλ)
U
g dµ + Φ(cλ).
U
It follows that

  



Z
Z
Z
Z
(3.2)
(Φ ◦ f )g dµ − Φ  f dµ ≤ Φ  f dµ − Φ(cλ)  gdµ − 1 .
X
U
U
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
U
Assume first that


Z
Φ  f dµ − Φ(cλ) = 0,



Z
then
Φ
U
f dµ = Φ 
U
Since Φ is strictly increasing,
Z
Z
f dµ = c dµ,
U
J. Ineq. Pure and Appl. Math., 1(1) Art. 3, 2000
U

Z
c dµ .
U
Z
(f − c)dµ = 0.
hence
U
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G AUCHMAN
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.
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 either
U


Z
Z
f dµ ≤ 0 or
(Φ ◦ f )g dµ − Φ 
X
Z
U
g dµ ≥ 1.
U
R EFERENCES
[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.
J. Ineq. Pure and Appl. Math., 1(1) Art. 3, 2000
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