Journal of Inequalities in Pure and
Applied Mathematics
TIME SCALE INTEGRAL INEQUALITIES SIMILAR TO QI’S
INEQUALITY
volume 7, issue 4, article 128,
2006.
MEHMET ZEKI SARIKAYA, UMUT MUTLU OZKAN
AND HÜSEYIN YILDIRIM
Department of Mathematics
Faculty of Science and Arts
Kocatepe University
Afyon-TURKEY
EMail: sarikaya@aku.edu.tr
EMail: umutlu@aku.edu.tr
EMail: hyildir@aku.edu.tr
Received 15 July, 2006;
accepted 19 October, 2006.
Communicated by: D. Hinton
Abstract
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2000
Victoria University
ISSN (electronic): 1443-5756
188-06
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Abstract
In this study, some integral inequalities and Qi’s inequalities of which is proved
by the Bougoffa [5] – [7] are extended to the general time scale.
2000 Mathematics Subject Classification: 34B10 and 26D15.
Key words: Delta integral.
Time Scale Integral Inequalities
Similar to Qi’s inequality
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
5
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
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1.
Introduction
The unification and extension of continuous calculus, discret calculus, q−calculus,
and indeed arbitrary real-number calculus to time scale calculus was first accomplished by Hilger in his PhD. thesis [8]. The purpose of this work is to
extend some integral inequalities and Qi inequalities proved by Bougoffa [5] –
[7]. The following definitions will serve as a short primer on time scale calculus; they can be found in [1] – [4]. A time scale T is any nonempty closed
subset of R. Within that set, define the jump operators ρ, σ : T → T by
ρ(t) = sup{s ∈ T : s < t} and
σ(t) = inf{s ∈ T : s > t},
Time Scale Integral Inequalities
Similar to Qi’s inequality
where inf φ := sup T and sup φ := inf T. If ρ(t) = t and ρ(t) < t, then the
point t ∈ T is left-dense, left-scattered. If σ(t) = t and σ(t) > t, then the
point t ∈ T is right-dense, right-scattered. If T has a right-scattered minimum
m, define Tk := T−{m}; otherwise, set Tk = T. If T has a left-scattered
maximum M, define Tk := T−{M }; otherwise, set Tk = T. The so-called
graininess functions are µ(t) := σ(t) − t and v(t) := t − ρ(t).
For f : T → R and t ∈ Tk , the delta derivative in [3, 4] of f at t, denoted
∆
f (t), is the number (provided it exists) with the property that given any ε > 0,
there is a neighborhood U of t such that
f (σ(t)) − f (s) − f ∆ (t)[σ(t) − s] ≤ ε |σ(t) − s|
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
for all s ∈ U . For T = R, f ∆ = f 0 , the usual derivative; for T = Z the delta
derivative is the forward difference operator, f ∆ (t) = f (t + 1) − f (t); in the
case of q−difference equations with q > 1,
f (qt) − f (t)
f (s) − f (0)
f ∆ (t) =
,
f ∆ (0) = lim
.
s→0
(q − 1)t
s
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A function f : T → R is right-dense continuous or rd-continuous provided
it is continuous at right-dense points in T and its left-sided limits exist (finite)
at left-dense points in T. If T = R, then f is rd-continuous if and only if f
is continuous. It is known from Theorem 1.74 in [3] that if f is right-dense
continuous, there is a function F such that F ∆ (t) = f (t) and
Z b
f (t)∆t = F (b) − F (a).
a
Note that we have
µ(t) ≡ 0,
σ(t) = t,
Z
f ∆ = f 0,
b
f (t)∆t =
a
Time Scale Integral Inequalities
Similar to Qi’s inequality
b
Z
f (t)dt,
when T = R
a
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
while
σ(t) = t+1,
µ(t) ≡ 1,
f
∆
Z
= ∆f,
b
f (t)∆t =
a
b−1
X
f (t),
when T = Z.
Contents
t=a
Much more information concerning time scales and dynamic equations on
time scales can be found in the books [3, 4].
Theorem 1.1 (Hölder’s inequality on time scales [3]). Let a, b ∈ T. For rdcontinuous functions f, g : [a, b] → R we have
Z
b
Z
|f (x)g(x)| ∆x ≤
a
where p > 1 and q =
p
p1 Z
|f (x)| ∆x
a
p
.
p−1
b
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b
q
|g(x)| ∆x
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,
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a
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2.
Main Results
In this section, we will state our main results and give their proofs.
Lemma 2.1. Let a, b ∈ T, and p > 1 and q > 1 with p1 + 1q = 1. If two positive
p
functions f, g : [a, b] → R are rd-continuous and satisfying 0 < m ≤ fgq ≤
M < ∞ on the set [a, b], then we have the following inequality
p1 Z
b
Z
p
q
f ∆x
(2.1)
1q
b
≤
g ∆x
a
a
M
m
pq1 Z
b
f g∆x.
a
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
Inequality (2.1) is called the reverse Hölder inequality.
Proof. Since
fp
gq
p
1
≤ M, g ≥ M − q f q , therefore
p
1
Time Scale Integral Inequalities
Similar to Qi’s inequality
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1
f g ≥ M − q f 1+ q = M − q f p
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and so,
Z
b
p
1
p
≤M
f ∆x
(2.2)
1
pq
Z
f g∆x
a
On the other hand, since m ≤
b
1
p
.
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a
fp
,
gq
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f ≥ m p g p , hence
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Z
b
Z
f g∆x ≥
a
b
1
p
m g
a
1+ pq
∆x ≥ m
1
p
Z
b
g q ∆x
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J. Ineq. Pure and Appl. Math. 7(4) Art. 128, 2006
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and so,
1q
b
Z
≥m
f g∆x
1
pq
1q
b
Z
q
g ∆x
a
.
a
Combining with (2.2), we have the desired inequality
Z
b
f p ∆x
p1 Z
a
b
g q ∆x
1q
≤M
Z
1
pq
p1
b
f g∆x
a
m
1
− pq
a
=
M
m
b
Z
g q ∆x
1q
a
pq1 Z
b
Time Scale Integral Inequalities
Similar to Qi’s inequality
f g∆x.
a
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
Corollary 2.2. In Lemma 2.1, replacing f p and g q by f and g, respectively, we
obtain the reverse Hölder type inequality,
p1 Z
b
Z
b
≤
g∆x
f ∆x
(2.3)
1q
m − pq1
1
1
f p g q ∆x.
M
a
a
b
Z
is rd-continuous and 0 < m
following inequality
Z
b
1
f p ∆x
(2.4)
a
1
p
≤ f ≤ M
p
≥ (b − a)
1
p
1
p
+
1
q
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a
The proof of this corollary can be obtained from (2.1).
Theorem 2.3. Let a, b ∈ T, p > 1 and q > 1 with
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= 1. If f : [a, b] → R
< ∞ on [a, b], then we have the
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p+1
q
m p+1
pq
M
Z
b
f p ∆x
p1
.
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J. Ineq. Pure and Appl. Math. 7(4) Art. 128, 2006
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Proof. Putting g ≡ 1 in Lemma 2.1, we obtain
p1
b
Z
p
1
q
[b − a] ≤
f ∆x
b
m − pq1 Z
M
a
f ∆x.
a
Therefore, we get
p1
b
Z
p
≤
f ∆x
(2.5)
a
m − pq1
M
− 1q
b
Z
[b − a]
f ∆x.
Time Scale Integral Inequalities
Similar to Qi’s inequality
a
Again, substituting g ≡ 1 in Corollary 2.2 leads to
p1
b
Z
f ∆x
≤
a
m − pq1
M
[b − a]
− 1q
b
Z
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
1
f p ∆x,
a
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and so,
Contents
b
Z
f ∆x ≤
(2.6)
a
m − 1q
M
p
[b − a]− q
Z
b
p
1
f p ∆x
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.
a
Combining (2.5) with (2.6), we obtain
Z
b
1
p
f ∆x
a
p
≥ (b − a)
p+1
q
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Z
m p+1
pq
M
a
b
p
f ∆x
1
p
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.
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1
1
Corollary 2.4. If 0 < m p ≤ f ≤ M p < ∞ on [a, b] and
p > 1 , then
Z b
p Z b
p1
1
(2.7)
f p ∆x ≥
f p ∆x .
a
m
M
= [b − a]−p for
a
Remark 1. For T = R, (2.7) is Qi’s inequality [9].
Theorem 2.5. If f : [a, b] → R is rd-continuous and 0 < m ≤ f (x) ≤ M on
[a, b] , then we have the following inequality
Z b
p1 −1
Z b
1
f ∆x
,
(2.8)
f p ∆x ≥ B
a
1
where B = m(b − a)1+ q
a
m
M
pq1
and p > 1 , q > 1 with
1
p
+
1
q
= 1.
Proof. In Corollary 2.2, putting g ≡ 1 yields
Z b
p1
m − pq1 Z b 1
1
q
f ∆x [b − a] ≤
f p ∆x,
M
a
a
and so,
p1
Z b
p1 −1 Z b
Z b
m − pq1
1
1
f ∆x
f ∆x .
f p ∆x ≥
[b − a] q
M
a
a
a
Since 0 < m ≤ f (x), we have
Z b
p1 −1
Z b
m pq1
1
1
1+
f p ∆x ≥
m [b − a] q
f ∆x
.
M
a
a
This proves inequality (2.8).
Time Scale Integral Inequalities
Similar to Qi’s inequality
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
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1
p
Corollary 2.6. Let p > 1 and q > 1 with
m
m pq1
1
q
= 1. If
1
=
M
+
1
[b − a]1+ q
and 0 < m ≤ f (x) ≤ M on [a, b] , then
Z
b
f ∆x ≥
(2.9)
a
p1 −1
b
Z
1
p
f ∆x
.
Time Scale Integral Inequalities
Similar to Qi’s inequality
a
Remark 2. For T = R, (2.9) is Qi’s inequality [9].
Lemma 2.7. Let a, b ∈ T, and f, g : [a, b] → R be rd-continuous and nonnegative functions with 0 < m ≤ fg ≤ M < ∞ on [a, b]. Then for p > 1 and q > 1
with p1 + 1q = 1 we have the following inequality
Z
b
1
p
1
q
[f (x)] [g(x)] ∆x ≤ M
(2.10)
1
p2
m
a
−
1
q2
Z
b
1
q
b
[f (x)] [g(x)] ∆x
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1
p
1
q
[f (x)] [g(x)] ∆x
(2.11)
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1
p
and
Z
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
Close
a
≤M
1
p2
m
−
1
q2
Z
1q Z
b
f (x)∆x
a
p1
b
g(x)∆x
a
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.
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Proof. From Hölder’s inequality, we obtain
Z b
1q Z b
p1
Z b
1
1
[f (x)] p [g(x)] q ∆x ≤
f (x)∆x
g(x)∆x ,
a
a
a
that is,
Z b
1q Z b
p1
Z b
1
1
1
1
1
1
[f (x)] p [g(x)] q ∆x ≤
[f (x)] p [f (x)] q ∆x
[g(x)] p [g(x)] q ∆x .
a
a
1
p
1
p
a
1
p
1
q
Since [f (x)] ≤ M [g(x)] and [g(x)] ≤ m
equality it follows that
Z b
1
1
[f (x)] p [g(x)] q ∆x
− 1q
1
[f (x)] q , from the above in-
Time Scale Integral Inequalities
Similar to Qi’s inequality
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
a
≤M
1
p2
m
−
1
q2
Z
b
1
p
1q Z
1
q
b
[g(x)] [f (x)] ∆x
1
p
1
q
p1
[g(x)] [f (x)] ∆x
a
b
1
1
[f (x)] p [g(x)] q ∆x ≤ M
(2.12)
a
1
p2
m
− 12
q
Z
b
1
1
[f (x)] q [g(x)] p ∆x.
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a
Hence, the inequality (2.10) is proved.
The inequality (2.11) follows from substituting the following
Z b
1q Z b
p1
Z b
1
1
[f (x)] p [g(x)] q ∆x ≤
f (x)∆x
g(x)∆x
a
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a
and so,
Z
,
a
a
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into (2.12), which can be obtained by Hölder’s inequality on time scales.
J. Ineq. Pure and Appl. Math. 7(4) Art. 128, 2006
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Lemma 2.8. Let a, b ∈ T. For a given positive integer p ≥ 2, if f : [a, b] → R
is rd-continuous and 0 < m ≤ fg ≤ M < ∞ on [a, b] , then
b
Z
Z
1
p
1− p1
b
[f (x)] ∆x ≤
(2.13)
f (x)∆x
a
.
a
Proof. Putting g(x) ≡ 1 in (2.11) yields
Z
b
[f (x)] ∆x ≤ K
f (x)∆x
a
1
p2
1− p1
b
Z
1
p
Time Scale Integral Inequalities
Similar to Qi’s inequality
,
a
1
(b−a) p
where K = M (1− 1 )2 . From M ≤
p
m
inequality (2.13) is proved.
(p−1)2
m
(b−a)p
, we conclude that K ≤ 1. Thus the
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In the following we generalize to arbitrary time scales a result in [6].
Theorem 2.9. Let a, b ∈ T. If f, g : [a, b] → R is rd-continuous and satisfying
0 < m ≤ fg ≤ M < ∞ on [a, b], then we have the following inequality
Z
b
p
f (x)∆x
(2.14)
a
p1
Z
+
p1
b
p
≤c
a
where c =
b
(f (x) + g(x))p ∆x
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g (x)∆x
a
Z
Mehmet Zeki Sarikaya,
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,
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1
m pq
(M
) .
J. Ineq. Pure and Appl. Math. 7(4) Art. 128, 2006
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Proof. It follows from Lemma 2.1 that
Z b
(f (x) + g(x))p ∆x
a
Z b
Z b
p−1
=
(f (x) + g(x)) f (x)∆x +
(f (x) + g(x))p−1 g(x)∆x
a
≥
M
m
a
pq1 Z
b
f p (x)∆x
a
+
=
M
m
pq1
b
(f (x) + g(x))q(p−1) ∆x
1q
a
p1 Z
M
m
Z b
pq1 Z
b
g p (x)∆x
p1 Z
b
(f (x) + g(x))q(p−1) ∆x
a
a
p
1
q
1q
Time Scale Integral Inequalities
Similar to Qi’s inequality
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
(f (x) + g(x)) ∆x
a
" Z
×
p1
b
p
f (x)∆x
Z
b
p
p1 #
g (x)∆x
+
a
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.
Contents
a
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J
Therefore, we obtain
" Z
p1 #
p1 Z b
b
f p (x)∆x
+
g p (x)∆x
a
Go Back
a
≤
=
m pq1 Z
M
m pq1
M
b
p
1− 1q
(f (x) + g(x)) ∆x
Z b
p
p
(f (x) + g(x)) ∆x ,
a
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J. Ineq. Pure and Appl. Math. 7(4) Art. 128, 2006
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where q(p − 1) = p.
Example 2.1. Let T = Z. Let f (x) = 3x and g(x) = x2 on [3, 4] with M ≈
5.06 and m = 3. Taking p = 2, we see that the conditions of Lemma 2.1 are
fulfilled. Therefore, for
21
4
Z
2x
3 ∆x
=
3
4
Z
x4 ∆x
12
=
3
12
1 8
6
(3 − 3 )
= 33 ,
8
4−1
X
Time Scale Integral Inequalities
Similar to Qi’s inequality
! 12
x4
= 32
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
x=3
and
4
Z
x 2
3 x ∆x =
3
4−1
X
3x x2 = 35
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x=3
we get
Z
Contents
4
2x
12 Z
3 ∆x
3
4
4
x ∆x
3
12
= 243 ≤
5.06
3
14 Z
4
x 2
3 x ∆x ≈ 274.6.
3
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References
[1] R.P. AGARWAL AND M. BOHNER, Basic calculus on time scales and
some of its applications, Results Math., 35(1-2) (1999), 3–22.
[2] F.M. ATICI AND G.Sh. GUSEINOV, On Green’s functions and positive solutions for boundary value problems on time scales, J. Comput. Appl. Math.,
141 (2002) 75–99.
[3] M. BOHNER AND A. PETERSON, Dynamic Equations on Time Scales, an
Introduction with Applications, Birkhauser, Boston (2001).
Time Scale Integral Inequalities
Similar to Qi’s inequality
[4] M. BOHNER AND A. PETERSON, Advances in Dynamic Equations on
Time Scales, Birkhauser Boston, Massachusetts (2003).
Mehmet Zeki Sarikaya,
Umut Mutlu Ozkan and
Hüseyin Yildirim
[5] L. BOUGOFFA, Notes on Qi’s type integral inequalities, J. Inequal. Pure
and Appl. Math., 4(4) (2003), Art. 77. [ONLINE: http://jipam.vu.
edu.au/article.php?sid=318].
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[6] L. BOUGOFFA, An integral inequality similar to Qi’s inequality, J. Inequal.
Pure and Appl. Math., 6(1) (2005), Art 27. [ONLINE: http://jipam.
vu.edu.au/article.php?sid=496].
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[7] L. BOUGOFFA, On Minkowski and Hardy integral inequalities, J. Inequal.
Pure and Appl. Math., 7(2) (2006), Art. 60. [ONLINE: http://jipam.
vu.edu.au/article.php?sid=677].
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[8] S. HILGER, Ein Maßkettenkalkül mit Anwendung auf Zentrmsmannigfaltingkeiten, PhD thesis, Univarsi. Würzburg (1988).
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[9] F. QI, Several integral inequalities, J. Inequal. Pure and Appl. Math.,
1(2) (2000). [ONLINE: http://jipam.vu.edu.au/article.
php?sid=113].
Time Scale Integral Inequalities
Similar to Qi’s inequality
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Umut Mutlu Ozkan and
Hüseyin Yildirim
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