Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 1, Article 7, 2005
ČEBYŠEV’S INEQUALITY ON TIME SCALES
CHEH-CHIH YEH, FU-HSIANG WONG, AND HORNG-JAAN LI
D EPARTMENT OF I NFORMATION M ANAGEMENT
L ONG -H UA U NIVERSITY OF S CIENCE AND T ECHNOLOGY
K UEI S HAN TAOYUAN , 33306 TAIWAN
R EPUBLIC OF C HINA .
CCYeh@mail.lhu.edu.tw
D EPARTMENT OF M ATHEMATICS
NATIONAL TAIPEI T EACHER ’ S C OLLEGE
134, H O -P ING E. R D . S EC . 2
TAIPEI 10659, TAIWAN
R EPUBLIC OF C HINA .
wong@tea.ntptc.edu.tw
G ENERAL E DUCATION C ENTER
C HIEN KUO I NSTITUTE OF T ECHNOLOGY
C HANG -H UA , 50050 TAIWAN
R EPUBLIC OF C HINA .
hjli@ckit.edu.tw
Received 23 October, 2004; accepted 21 December, 2004
Communicated by D. Hinton
A BSTRACT. In this paper we establish some Čebyšev’s inequalities on time scales under suitable
conditions.
Key words and phrases: Time scales, Čebyšev’s Inequality, Delta differentiable.
2000 Mathematics Subject Classification. Primary 26B25; Secondary 26D15.
1. I NTRODUCTION
The purpose of this paper is to establish the well-known Čebyšev’s inequality on time scales.
To do this, we simply introduce the time scales calculus as follows:
In 1988, Hilger [7] introduced the time scales theory to unify continuous and discrete analysis. A time scale T is a closed subset of the set R of the real numbers. We assume that any time
scale has the topology that it inherits from the standard topology on R. Since a time scale may
or may not be connected, we need the concept of jump operators.
ISSN (electronic): 1443-5756
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204-04
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C HEH -C HIH Y EH , F U -H SIANG W ONG ,
AND
H ORNG -JAAN L I
Definition 1.1. Let t ∈ T, where T is a time scale. Then the two mappings
σ, ρ : T → R
satisfying
σ(t) = inf{γ > t|γ ∈ T},
ρ(t) = sup{γ < t|γ ∈ T}
are called the jump operators on T.
These jump operators classify the points {t} of a time scale T as right-dense, right-scattered,
left-dense and left-scattered according to whether σ(t) = t, σ(t) > t, ρ(t) = t or ρ(t) < t,
respectively, for t ∈ T.
Let t be the maximum element of a time scale T. If t is left-scattered, then t is called a
generate point of T. Let Tk denote the set of all non-degenerate points of T. Throughout this
paper, we suppose that
(a) T is a time scale;
(b) an interval means the intersection of a real interval with the given time scale;
(c) R = (−∞, ∞).
Definition 1.2. Let T be a time scale. Then the mapping f : T → R is called rd-continuous if
(a) f is continuous at each right-dense or maximal point of T;
(b) lim− f (s) = f (t− ) exists for each left-dense point t ∈ T.
s→t
Let Crd [T, R] denote the set of all rd-continuous mappings from T to R.
Definition 1.3. Let f : T → R, t ∈ Tk . Then we say that f has the (delta) derivative f ∆ (t) ∈ R
at t if for each > 0 there exists a neighborhood U of t such that for all s ∈ U
f (σ(t)) − f (s) − f ∆ (t)[σ(t) − s] ≤ |σ(t) − s| .
In this case, we say that f is (delta) differentiable at t.
Clearly, f ∆ is the usual derivative if T = R, and is the usual forward difference operator if
T = Z (the set of all integers).
Definition 1.4. A function F : T → R is an antiderivative of f : T → R if F ∆ (t) = f (t) for
each t ∈ Tk . In this case, we define the (Cauchy) integral of f by
Z t
f (γ) ∆γ = F (t) − F (s)
s
for all s, t ∈ T.
It follows from Theorem 1.74 of Bohner and Peterson [3] that every rd-continuous function
has an antiderivative. For further results on time scales calculus, we refer to [3, 9].
The purpose of this paper is to establish the well-known Čebyšev inequality [1, 5, 6, 8, 11]
on time scales. For other related results, we refer to [4, 10, 12, 13].
2. M AIN R ESULTS
We first establish some Čebyšev inequalities which generalize some results of Audréief [1],
Beesack and Pečarić [2], Dunkel [4], Fujimara [5, 6], Isayama [8], and Winckler [13]. For other
related results, we refer to the book of Mitrinovič [10].
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
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Č EBYŠEV ’ S I NEQUALITY ON T IME S CALES
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Theorem 2.1. Suppose that p ∈ Crd ([a, b]; [0, ∞)). Let f1 , f2 , k1 , k2 ∈ Crd ([a, b]; R) satisfy the
following two conditions:
(C1 ) f2 (x)k2 (x) > 0 on [a, b];
(x)
(x)
(C2 ) ff21 (x)
and kk21 (x)
are similarly ordered (or oppositely ordered), that is, for all x, y ∈ [a, b],
f1 (x) f1 (y)
k1 (x) k1 (y)
−
−
≥ 0 (or ≤ 0),
f2 (x) f2 (y)
k2 (x) k2 (y)
then
Z Z
f1 (x) f1 (y)
k1 (x) k1 (y)
1 b b
(2.1)
p(x)p(y)
∆x∆y
2! a a
f2 (x) f2 (y)
k2 (x) k2 (y)
Rb
Rb
p(x)f
(x)k
(x)∆x
p(x)f1 (x)k2 (x)∆x
1
1
a
= Rab
≥ 0 (or ≤ 0)
Rb
p(x)f
(x)k
(x)∆x
p(x)f
(x)k
(x)∆x
2
1
2
2
a
a
Proof. Let x, y ∈ [a, b]. Then it follows from (C1 ), (C2 ) and the identity
p(x)p(y)
f1 (x) f1 (y)
k1 (x) k1 (y)
f2 (x) f2 (y)
k2 (x) k2 (y)
= p(x)p(y)f2 (x)f2 (y)k2 (x)k2 (y)
f1 (x) f1 (y)
−
f2 (x) f2 (y)
k1 (x) k1 (y)
−
k2 (x) k2 (y)
that (2.1) holds.
Remark 2.2. Suppose that p, f, g ∈ Crd ([a, b]; R) with p(x) ≥ 0 on [a, b]. Let f and g be
similarly ordered (or oppositely ordered). Taking f1 (x) = f (x), k1 (x) = g(x) and f2 (x) =
k2 (x) = 1, (2.1) is reduced to the generalized Čebyšev inequality:
Z b
Z b
Z b
Z b
(2.2)
p(x)f (x)g(x)∆x
p(x)∆x ≥ (or ≤)
p(x)f (x)∆x
p(x)g(x)∆x,
a
a
a
a
which generalizes a Winckler’s result in [13] if a = 0 and b = x. Let T = Z, if a =
(a1 , a2 , . . . , an ) and b = (b1 , b2 , . . . , bn ) are similarly ordered (or oppositely ordered), and if
p = (p1 , p2 , . . . , pn ) is a nonnegative sequence, then (2.2) is reduced to
n
n
n
n
X
X
X
X
pi
pi ai bi ≥ (or ≤)
p i ai
p i bi .
i=1
i=1
i=1
i=1
If T = R, then (2.2) is reduced to
Z b
Z b
Z b
Z b
p(x)f (x)g(x) dx
p(x) dx ≥ (or ≤)
p(x)f (x) dx
p(x)g(x) dx.
a
a
a
a
(x)
(x)
, g(x) = gg21 (x)
and p(x) = f2 (x)g2 (x), inequality (2.2) is
Remark 2.3. Taking f (x) = ff21 (x)
reduced to
Z b
Z b
Z b
Z b
(2.3)
f1 (x)g1 (x)∆x
f2 (x)g2 (x)∆x ≥ (or ≤)
f1 (x)g2 ∆x
f2 (x)g1 ∆x,
a
a
if f2 (x)g2 (x) ≥ 0 on [a, b],
f1 (x)
f2 (x)
f1 (x)
f2 (x)
a
and
g1 (x)
g2 (x)
a
are both increasing or both decreasing (or one of the
g1 (x)
g2 (x)
functions
or
is nonincreasing and the other nondecreasing). Here f1 , f2 , g1 , g2 ∈
Crd ([a, b], R) with f2 (x)g2 (x) 6= 0 on [a, b]. Conversely, if f1 (x) = f (x)f2 (x), g1 (x) =
g(x)g2 (x) and p(x) = f2 (x)g2 (x), then inequality (2.3) is reduced to inequality (2.2).
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
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C HEH -C HIH Y EH , F U -H SIANG W ONG ,
H ORNG -JAAN L I
AND
Theorem 2.4. Let f ∈ Crd ([a, b], [0, ∞)) be decreasing (or increasing) with
Rb
0 and a p(x)f (x)∆x > 0. Then
Rb
xp(x)f 2 (x)∆x
a
Rb
a
xp(x)f (x)∆x
Rb
≤ (≥) Rab
a
p(x)f 2 (x)∆x
Rb
a
xp(x)f (x)∆x >
.
p(x)f (x)∆x
Proof. Clearly, for any x, y ∈ [a, b],
Z bZ
b
f (x)f (y)p(x)p(y)(y − x)(f (x) − f (y)∆x∆y ≥ (≤)0,
a
a
which implies that the desired result holds.
Remark 2.5. Let f ∈ Crd ([a, b], (0, ∞)) and n be a positive integer. If p and g are replaced by
p
and f n respectively, then Čebyšev’s inequality (2.2) is reduced to
f
Z
b
Z
n
b
p(x)f (x)∆x
a
a
b
Z
p(x)
∆x ≥
f (x)
b
Z
p(x)[f (x)]n−1 ∆x,
p(x)∆x
a
a
which implies
b
Z
Z
n
b
p(x)f (x)∆x
a
a
p(x)
∆x
f (x)
2
b
Z
≥
b
Z
n−1
p(x)∆x
p(x)[f (x)]
a
Z
∆x
a
Z
≥
a
2 Z
b
p(x)
∆x
f (x)
b
p(x)[f (x)]n−2 ∆x.
p(x)∆x
a
b
a
provided f and f n are similarly ordered. Continuing in this way, we get
Z
b
a
p(x)
∆x
f (x)
n Z
b
p(x)[f (x)] ∆x ≥
a
n+1
b
Z
n
p(x)∆x
,
a
which extends a result in Dunkel [4].
Remark 2.6. Let ν, p ∈ Crd ([a, b], [0, ∞)). If f and g are similarly ordered (or oppositely
ordered), then it follows from Remark 2.2 that
Z b
Z b
Z b
Z b
p(t)f (ν(t))g(ν(t))∆t
p(t)∆t ≥ (or ≤)
p(t)f (ν(t))∆t
p(t)g(ν(t))∆t,
a
a
a
a
which is a generalization of a result given in Stein [12].
Remark 2.7. Let p, fi ∈ Crd ([a, b], R) for each i = 1, 2, . . . , n. Suppose that f1 , f2 , . . . , fn are
similarly ordered and p(x) ≥ 0 on [a, b], then it follows from Remark 2.2 that
Z
n−1 Z
b
b
p(x)f1 (x)f2 (x) · · · fn (x)∆x
p(x)∆x
a
a
Z
=
n−2 Z
b
p(x)∆x
a
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
b
Z
p(x)f1 (x)f2 (x) · · · fn (x)∆x
p(x)∆x
a
b
a
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Č EBYŠEV ’ S I NEQUALITY ON T IME S CALES
n−2 Z
b
Z
≥
b
Z
p(x)∆x
a
b
≥
a
b
p(x)f2 (x) · · · fn (x)∆x
p(x)f1 (x)∆x
a
Z
5
a
Z
n−3
b
p(x)f1 (x)∆x
p(x)∆x
a
Z b
Z b
×
p(x)f2 (x)∆x
p(x)f3 (x) · · · fn (x)∆x
a
a
≥ ···
Z b
Z b
Z b
≥
p(x)f1 (x)∆x
p(x)f2 (x)∆x · · ·
p(x)fn (x)∆x ,
a
a
a
which is a generalization of a result in Dunkel [4].
In particular, if f1 (x) = f2 (x) = · · · = fn (x) = f (x), then
Z b
n−1 Z b
Z b
n
n
p(x)∆x
p(x)f (x)∆x ≥
p(x)f (x)∆x .
a
a
a
Theorem 2.8. If p(x), f (x) ∈ Crd ([a, b], [0, ∞)) with f (x) > 0 on [a, b] and n is a positive
integer, then
n Z b
Z b
n
Z b
p(x)
n
∆x
p(x)f (x)∆x ≥
p(x)∆x .
a
a
a f (x)
1
Proof. It follows from f (x) > 0 on [a, b] that f n (x) and f (x)
are oppositely ordered on [a, b].
Hence by (2.2),
n
Z b
Z b
p(x)
n
∆x
p(x)f (x)∆x
a
a f (x)
Z b
n−1 Z b
Z b
p(x)
≥
p(x)∆x
∆x
p(x)f n−1 (x)∆x
f
(x)
a
a
a
Z b
2 Z b
n−2 Z b
p(x)
p(x)∆x
∆x
p(x)f n−2 (x)∆x
≥
a
a f (x)
a
≥ ···
Z b
n
≥
p(x)∆x .
a
Theorem 2.9. Let g1 , g2 , . . . , gn ∈ Crd ([a, b], <) and p, h1 , h2 , . . . , hn−1 ∈ Crd ([a, b], (0, ∞))
with gn (x) > 0 on [a, b]. If
g1 (x)g2 (x) · · · gn−1 (x)
h1 (x)h2 (x) · · · hn−1 (x)
and
hn−1 (x)
gn (x)
are similarly ordered (or oppositely ordered), then
Z b
Z b
p(x)g1 (x)g2 (x) · · · gn−1 (x)
∆x
(2.4)
p(x)gn (x)∆x
h1 (x)h2 (x) · · · hn−2 (x)
a
a
Z b
Z
≥ (or ≤)
p(x)hn−1 (x)∆x
a
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
a
b
p(x)g1 (x)g2 (x) · · · gn (x)
∆x.
h1 (x)h2 (x) · · · hn−1 (x)
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C HEH -C HIH Y EH , F U -H SIANG W ONG ,
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Proof. Taking
f1 (x) =
g1 (x)g2 (x) · · · gn−1 (x)
, k1 (x) = hn−1 (x),
h1 (x)h2 (x) · · · hn−1 (x)
f2 (x) = 1 and k2 (x) = gn (x)
in Theorem 2.1, (2.1) is reduced to our desired result (2.4).
The following theorem is a time scales version of Theorem 1 in Beesack and Pečarić [2].
Theorem 2.10. Let
f1 , f2 , . . . , fn ∈ Crd ([a, b], [0, ∞) and
g1 , g2 , . . . , gn ∈ Crd ([a, b], (0, ∞)).
fn
If the functions f1 , fg12 , . . . , gn−1
are similarly ordered and for each pair
ordered for k = 2, 3, . . . , n, then
Z
b
p(x)f1 (x)
(2.5)
a
fk
, gk−1
gk−1
is oppositely
f2 (x)f3 (x) · · · fn (x)
∆x
g1 (x)g2 (x) · · · gn−1(x)
Rb
Rb
Rb
p(x)f1 (x)∆x a p(x)f2 (x)∆x · · · a p(x)fn (x)∆x
a
≥ Rb
.
Rb
Rb
p(x)g
(x)∆x
p(x)g
(x)∆x
·
·
·
p(x)g
(x)∆x
1
2
n−1
a
a
a
fn
Proof. Let f1 , f2 , . . . , fn be replaced by f1 , fg12 , . . . , gn−1
in Remark 2.7, we obtain
n−1 Z
b
Z
p(x)∆x
(2.6)
p(x)f1 (x)
a
Also, since
b
a
fk
gk−1
Z
f2 (x)f3 (x) · · · fn (x)
∆x
g1 (x)g2 (x) · · · gn−1 (x)
Z b
n Z b
Y
fk (x)
≥
p(x)f1 (x)∆x
p(x)
∆x.
g
k−1 (x)
a
a
k=2
and gk−1 are oppositely ordered, it follows from Remark 2.2 that
b
b
Z
b
Z
p(x)fk (x)∆x ≤
p(x)∆x
a
a
Z
b
p(x)gk−1 (x)∆x
a
p(x)
a
fk (x)
∆x.
gk−1 (x)
Thus
Z
a
b
p(x)fk (x)
∆x ≥
gk−1 (x)
Rb
a
Rb
p(x)∆x a p(x)fk (x)∆x
.
Rb
p(x)gk−1 (x)∆x
a
This and (2.6) imply (2.5) holds.
3. M ORE R ESULTS
In this section, we generalize some results in Isayama [8].
Theorem 3.1. Let f1 , f2 , . . . , fn ∈ Crd ([a, b], (0, ∞)), k1 , k2 , . . . , kn−1 ∈ Crd ([a, b], R) and
p(x) ∈ Crd ([a, b], [0, ∞)). If
f1 (x)f2 (x) · · · fi−1 (x)
k1 (x)k2 (x) · · · ki−1 (x)
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
and
ki−1 (x)
fi (x)
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Č EBYŠEV ’ S I NEQUALITY ON T IME S CALES
7
are similarly ordered (or oppositely ordered) for i = 2, . . . , n, then
Z
b
Z
b
Z
b
p(x)f2 (x)∆x · · ·
p(x)fn (x)∆x
a
Z b
Z b
≥ (or ≤)
p(x)k1 (x)∆x
p(x)k2 (x)∆x · · ·
a
a
Z b
Z b
f1 (x)f2 (x) · · · fn (x)
···
p(x)kn−1 (x)∆x
p(x)
∆x.
k1 (x)k2 (x) · · · kn−1 (x)
a
a
p(x)f1 (x)∆x
(3.1)
a
a
Proof. If f1 (x), k1 (x), f2 (x) and k2 (x) are replaced by f1 (x), 1, k1 (x) and kf21(x)
in Theorem 2.1,
(x)
then we obtain
Z b
Z b
Z b
Z b
f1 (x)f2 (x)
p(x)f1 (x)∆x
p(x)f2 (x)∆x ≥ (or ≤)
p(x)k1 (x)∆x
p(x)
∆x.
k1 (x)
a
a
a
a
Thus the theorem holds for n = 2.
Suppose that the theorem holds for n − 1, that is
Z
b
Z
p(x)f1 (x)∆x
(3.2)
a
b
Z
b
p(x)f2 (x)∆x · · ·
p(x)fn−1 (x)∆x
a
a
Z b
Z b
≥ (or ≤)
p(x)k1 (x)∆x
p(x)k2 (x)∆x
a
a
Z b
Z b
f1 (x)f2 (x) · · · fn−1 (x)
···
p(x)kn−2 (x)∆x
p(x)
∆x
k1 (x)k2 (x) · · · kn−2 (x)
a
a
if
f1 (x)f2 (x) · · · fi−1 (x)
k1 (x)k2 (x) · · · ki−1 (x)
and
ki−1 (x)
fi (x)
are similarly
R bordered (or oppositely ordered) for i = 2, 3, . . . , n − 1. Multiplying the both sides
of (3.2) by a p(x)fn (x)∆x, we see that
Z
b
Z
b
Z
b
Z
b
p(x)f2 (x)∆x · · ·
p(x)fn−1 (x)∆x
p(x)fn (x)∆x
a
a
a
Z b
Z b
≥ (or ≤)
p(x)k1 (x)∆x
p(x)k2 (x)∆x
a
a
Z b
Z b
Z b
f1 (x)f2 (x) · · · fn−1 (x)
···
p(x)kn−2 (x)∆x
p(x)
∆x
p(x)fn (x)∆x.
k1 (x)k2 (x) · · · kn−2 (x)
a
a
a
p(x)f1 (x)∆x
(3.3)
a
It follows from Theorem 2.10 that
Z
a
b
Z b
f1 (x)f2 (x) · · · fn−1 (x)
p(x)
∆x
p(x)fn (x)∆x
k1 (x)k2 (x) · · · kn−2 (x)
a
Z b
Z b
f1 (x)f2 (x) · · · fn (x)
≥ (or ≤)
p(x)
∆x
p(x)kn−1 (x)∆x.
k1 (x)k2 (x) · · · kn−1 (x)
a
a
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
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C HEH -C HIH Y EH , F U -H SIANG W ONG ,
AND
H ORNG -JAAN L I
This and (3.3) imply
Z b
Z b
Z b
p(x)f1 (x)∆x
p(x)f2 (x)∆x · · ·
p(x)fn (x)∆x
a
a
a
Z b
Z b
≥ (or ≤)
p(x)k1 (x)∆x
p(x)k2 (x)∆x
a
a
Z b
Z b
f1 (x)f2 (x) · · · fn (x)
···
p(x)kn−1 (x)∆x
p(x)
∆x.
k1 (x)k2 (x) · · · kn−1 (x)
a
a
By induction, we complete the proof.
Remark 3.2. Let kn ∈ Crd ([a, b], R). If f1 (x), f2 (x), . . . , fn (x), k1 (x), k2 (x), . . . , kn−1 (x) are
replaced by
f1 (x)f2 (x) · · · fn (x), k1 (x)k2 (x) · · · kn (x), . . . , k1 (x)k2 (x) · · · kn (x),
f1 (x)k2 (x) · · · kn (x), k1 (x)f2 (x)k3 (x) · · · kn (x), . . . , k1 (x)k2 (x) · · · kn−2 (x)fn−1 (x)kn (x)
in Theorem 3.1, respectively, then
Z b
n−1
Z b
(3.4)
p(x)f1 (x)f2 (x) · · · fn (x)∆x
p(x)k1 (x)k2 (x) · · · kn (x)∆x
a
a
Z
≥
b
Z
p(x)f1 (x)k2 (x) · · · kn (x)∆x
a
a
b
p(x)k1 (x)f2 (x)k3 (x) · · · kn (x)∆x
Z b
···
p(x)k1 (x)k2 (x) · · · kn−1 (x)fn (x)∆x
a
if
fi (x)
ki (x)
> 0 for i = 1, 2, . . . , n and k1 (x)k2 (x) · · · kn (x) > 0 on [a, b].
Remark 3.3. Letting f1 (x) = f2 (x) = · · · = fn (x) = f (x) and k1 (x) = k2 (x) = · · · =
1
kn (x) = k n−1 (x) in (3.4) with k(x) > 0 on [a, b], we obtain the Hölder inequality:
Z b
n−1 Z b
n
Z b
n
n
(3.5)
p(x)f (x)∆x
p(x)k n−1 (x)∆x
≥
p(x)f (x)k(x)∆x .
a
a
a
Remark 3.4. Let p, f, g ∈ Crd ([a, b], [0, ∞)). Taking
f1 (x) = f n (x)g(x),
f2 (x) = f3 (x) = · · · = fn (x) = g(x) and
k1 (x) = k2 (x) = · · · = kn−1 (x) = f (x)g(x),
(3.1) is reduced to Jensen’s inequality:
Z b
n−1 Z b
n
Z b
n
(3.6)
p(x)f (x)g(x)∆x
p(x)g(x)∆x
≥
p(x)f (x)g(x)∆x .
a
a
a
1
Remark 3.5. Taking k1 (x) = k2 (x) = · · · = kn−1 (x) = (f1 (x)f2 (x) · · · fn (x)) n , (3.1) is
reduced to
Z b
Z b
Z b
(3.7)
p(x)f1 (x)∆x
p(x)f2 (x)∆x · · ·
p(x)fn (x)∆x
a
a
a
n
Z b
1
n
≥
p(x) (f1 (x)f2 (x) · · · fn (x)) ∆x
a
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
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Č EBYŠEV ’ S I NEQUALITY ON T IME S CALES
if fi (x) > 0 on [a, b] for each i = 1, 2, . . . , n and
are similarly ordered.
9
1
1
[f (x)f2 (x) · · · fn (x)] n
fi (x) 1
(i = 1, 2, . . . , n)
Remark 3.6 (see also Remark 2.7). Taking k1 (x) = k2 (x) = · · · = kn−1 (x) = 1, (3.1) is
reduced to Čebyšev’s inequality:
Z b
Z b
Z b
(3.8)
p(x)f1 (x)∆x
p(x)f2 (x)∆x · · ·
p(x)fn (x)∆x
a
a
a
Z
≤
n−1 Z
b
b
p(x)f1 (x)f2 (x) · · · fn (x)∆x
p(x)∆x
a
a
if fi (x) (i = 1, 2, . . . , n) are similarly ordered and fi (x) ≥ 0 (i = 1, 2, . . . , n).
Remark 3.7. Taking f1 (x) = f2 (x) = · · · = fn (x) = 1, then (3.1) is reduced to
Z b
n Z b
Z b
p(x)∆x ≤
p(x)k1 (x)∆x
p(x)k2 (x)∆x
a
a
a
Z
b
Z
b
p(x)
∆x
a
a k1 (x)k2 (x) · · · kn−1 (x)
if ki (x) > 0 are similarly ordered for i = 1, 2, . . . , n − 1. Thus, if f1 (x), . . . , fn (x) are similarly
ordered and fi (x) > 0 on [a, b] (i = 1, 2, . . . , n), then
R
n+1
b
Z b
Z b
Z b
p(x)∆x
a
(3.9)
≤
p(x)f1 (x)∆x
p(x)f2 (x)∆x · · ·
p(x)fn (x)∆x.
Rb
p(x)
∆x
a
a
a
a f1 (x)f2 (x)···fn (x)
···
p(x)kn−1 (x)∆x
It follows from (3.8) and (3.9) that
R
n+1
b
Z b
n−1 Z b
p(x)∆x
a
≤
p(x)∆x
p(x)f1 (x)f2 (x) · · · fn (x)∆x
Rb
p(x)
∆x
a
a
a f1 (x)f2 (x)···fn (x)
if f1 (x), . . . , fn (x) are similarly ordered.
Remark 3.8. Let k1 (x) = k2 (x) = · · · = kn (x) = 1. If fi (x) is replaced by
1
[f1 (x)f2 (x) · · · fn (x)] n
,
fi (x)
then (3.1) is reduced to
n Z
Y
√
n
if
i=1
f1 (x)f2 (x)···fn (x)
fi (x)
a
b
n = 1, 2, . . . , n,
p
Z b
n
n
f1 (x)f2 (x) · · · fn (x)
∆x ≤
p(x)∆x
p(x)
fi (x)
a
(i = 1, 2, . . . , n) are similarly ordered.
Remark 3.9. Let f1 , f2 , . . . , fn ; k1 , k2 , . . . , kn−1 be replaced by f1 f2 , f3 f4 , . . . , f2n−1 f2n ;
f2 f3 , f4 f5 , . . . , f2n−2 f2n−1 , respectively. Then (3.1) is reduced to
Z b
Z b
Z b
p(x)f1 (x)f2 (x)∆x
p(x)f3 (x)f4 (x)∆x · · ·
p(x)f2n−1 (x)f2n (x)∆x
a
a
a
Z b
Z b
Z b
≥
p(x)f2 (x)f3 (x)∆x
p(x)f4 (x)f5 (x)∆x · · ·
p(x)f2n−2 (x)f2n−1 (x)∆x
a
if
fi (x)
fi+1 (x)
a
a
(i = 1, 2, . . . , 2n − 1) are similarly ordered.
J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
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10
C HEH -C HIH Y EH , F U -H SIANG W ONG ,
AND
H ORNG -JAAN L I
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J. Inequal. Pure and Appl. Math., 6(1) Art. 7, 2005
http://jipam.vu.edu.au/