Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 534083, 5 pages
http://dx.doi.org/10.1155/2013/534083
Research Article
An Opial-Type Inequality on Time Scales
Qiao-Luan Li1 and Wing-Sum Cheung2
1
2
College of Mathematics and Information Science, Hebei Normal University, Shijiazhuang 050024, China
Department of Mathematics, The University of Hong Kong, Hong Kong
Correspondence should be addressed to Qiao-Luan Li; qll71125@163.com
Received 4 January 2013; Accepted 16 January 2013
Academic Editor: Allan Peterson
Copyright © 2013 Q.-L. Li and W.-S. Cheung. This is an open access article distributed under the Creative Commons Attribution
License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly
cited.
We establish some new Opial-type inequalities involving higher order delta derivatives on time scales. These extend some known
results in the continuous case in the literature and provide new estimates in the setting of time scales.
1. Introduction
Opial’s inequality appeared for the first time in 1960 in
[1] and has been receiving continual attention throughout
the years (cf., e.g., [2–7]). The inequality together with its
numerous generalizations, extensions, and discretizations
has been playing a fundamental role in the study of the
existence and uniqueness properties of solutions of initial and
boundary value problems for differential equations as well as
difference equations [8, 9]. Two excellent surveys on these
inequalities can be found in [10, 11].
In 1960, Opial established the following integral inequality.
Theorem A (see [1]). If 𝑓 ∈ 𝐶1 [0, ℎ] satisfies 𝑓(0) = 𝑓(ℎ) = 0
and 𝑓(𝑥) > 0 for all 𝑥 ∈ (0, ℎ), then
ℎ
ℎ ℎ 󵄨󵄨 󸀠 󵄨󵄨2
󵄨
󵄨
∫ 󵄨󵄨󵄨󵄨𝑓 (𝑥)𝑓󸀠 (𝑥)󵄨󵄨󵄨󵄨 𝑑𝑥 ≤
∫ 󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑑𝑥.
4 0 󵄨
0
(1)
Shortly after the publication of Opial’s paper, Olech
provided a modified version of Theorem A. His result is stated
in the following.
Theorem B (see [12]). If 𝑓 is absolutely continuous on [0, ℎ]
with 𝑓(0) = 0, then
ℎ
ℎ ℎ 󸀠 2
󵄨
󵄨
∫ 󵄨󵄨󵄨󵄨𝑓 (𝑥)𝑓󸀠 (𝑥)󵄨󵄨󵄨󵄨 𝑑𝑥 ≤
∫ 𝑓 (𝑥) 𝑑𝑥.
2 0
0
(2)
The equality in (2) holds if and only if 𝑓(𝑥) = 𝑐𝑥, where 𝑐
is a constant.
The first natural extension of Opial’s inequality (1) involving higher order derivatives 𝑥(𝑛) (𝑠) (𝑛 ≥ 1) is embodied in
the following.
Theorem C (see [10]). Let 𝑥(𝑡) ∈ 𝐶(𝑛) [0, 𝑎] be such that
𝑥(𝑖) (0) = 0, 0 ≤ 𝑖 ≤ 𝑛 − 1 (𝑛 ≥ 1). Then the following
inequality holds:
𝑎
𝑎
1
󵄨
󵄨2
󵄨
󵄨
∫ 󵄨󵄨󵄨󵄨𝑥 (𝑡) 𝑥(𝑛) (𝑡)󵄨󵄨󵄨󵄨 𝑑𝑡 ≤ 𝑎𝑛 ∫ 󵄨󵄨󵄨󵄨𝑥(𝑛) (𝑡)󵄨󵄨󵄨󵄨 𝑑𝑡.
2
0
0
(3)
In 1997, Alzer [13] considered Opial-type inequalities
which involve higher-order derivatives of two functions.
These generalize earlier results of Agarwal and Pang [14].
In this paper, we consider the Opial-type inequality which
involves higher-order delta derivatives of two functions on
time scales. Our results in special cases yield some of the
recent results on Opial’s inequality and provide some new
estimates on such types of inequalities in this general setting.
2. Main Results
Let T be a time scale; that is, T is an arbitrary nonempty
closed subset of real numbers. Let 𝑎, 𝑏 ∈ T. We suppose that
the reader is familiar with the basic features of calculus on
time scales for dynamic equations. Otherwise one can consult
2
Abstract and Applied Analysis
Bohner and Peterson’s book [15] for most of the materials
needed.
We first quote the following elementary lemma and the
delta time scales Taylor formula.
and hence
𝑝 − 1 1/𝑝
1 (1−𝑝)/𝑝
𝑎+
𝑘
𝑘
𝑝
𝑝
(4)
𝑥
󵄨
󵄨 𝑚 󵄨
󵄨󵄨 Δ𝑘
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 ≤ ∫ ℎ𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨󵄨
𝑎
𝑥
󵄨󵄨 𝑚 󵄨󵄨
= ∫ ℎ𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) 𝑝(𝜏)−1/𝑡 𝑝(𝜏)1/𝑡 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏
󵄨
󵄨
𝑎
𝑚
(T) = the set of functions
Lemma 2 (see [17]). Let 𝑓 ∈ 𝐶𝑟𝑑
that are 𝑚 times differentiable with rd-continuous derivatives
on T, 𝑚 ∈ N. Then for any 𝑎, 𝑏 ∈ T and 𝑡 ∈ [𝑎, 𝑏] ∩ T,
𝑎
1/𝑡
𝑥
󵄨󵄨 𝑚 󵄨󵄨𝑡
⋅ (∫ 𝑝 (𝜏) 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏)
󵄨
󵄨
𝑎
𝑓(𝑡) = ∑ ℎ𝑘 (𝑡, 𝑎) 𝑓Δ (𝑎)
(5)
𝑡
1−1/𝑡
𝑥
𝑡/(𝑡−1)
≤ [∫ ℎ𝑚−𝑘−1
(𝑥, 𝜎 (𝜏)) 𝑝(𝜏)1/(1−𝑡) Δ𝜏]
𝑘
𝑘=0
= 𝑃(𝑥)1−1/𝑡 𝐹(𝑥)1/𝑡 ,
𝑚
+ ∫ ℎ𝑚−1 (𝑡, 𝜎 (𝜏)) 𝑓Δ (𝜏) Δ𝜏,
𝑎
(10)
𝑡
where ℎ0 (𝑡, 𝑠) := 1, ℎ𝑛+1 (𝑡, 𝑠) := ∫𝑠 ℎ𝑛 (𝜏, 𝑠)Δ𝜏, 𝑛 ∈ N.
𝑥
𝑡/(𝑡−1)
where 𝑃(𝑥) := ∫𝑎 ℎ𝑚−𝑘−1
(𝑥, 𝜎(𝜏))𝑝(𝜏)1/(1−𝑡) Δ𝜏, 𝐹(𝑥) :=
𝑥
Our main results are given in the following theorems.
Theorem 3. Let 0 ≤ 𝑟 ≤ 𝑠 < 𝑡, 𝑠 > 0, 𝑡 > 1 be real numbers,
and let 𝑚, 𝑘 be integers with 0 ≤ 𝑘 ≤ 𝑚−1. Let 𝑝 > 0 and 𝑞 ≥ 0
be measurable functions on Υ := [𝑎, 𝑏] ∩ T. Further, let 𝑓, 𝑔 ∈
𝑖
𝑖
𝑚−1
(Υ) with 𝑓Δ (𝑎) = 𝑔Δ (𝑎) = 0, 𝑖 = 0, 1, . . . , 𝑚 − 1,
𝐶𝑟𝑑
𝑚−1
𝑚−1
and let𝑓Δ , 𝑔Δ be absolutely continuous on Υ such that the
𝑏
𝑏
𝑚
𝑚
integrals ∫𝑎 𝑝(𝑥)|𝑓Δ (𝑥)|𝑡 Δ𝑥 and ∫𝑎 𝑝(𝑥)|𝑔Δ (𝑥)|𝑡 Δ𝑥 exist.
Then one has
𝑏
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠
󵄨󵄨 𝑘
∫ 𝑞 (𝑥) [󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨
󵄨
󵄨
󵄨
󵄨
𝑎
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠
󵄨󵄨 𝑘
+󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ] Δ𝑥
󵄨󵄨
󵄨
󵄨
(6)
1−𝛼
𝑏
≤ 21−𝛼 (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
𝑎
⋅ [𝛽𝑘
𝛽−1
𝑚
∫𝑎 𝑝(𝜏)|𝑓Δ (𝜏)|𝑡 Δ𝜏.
Let
𝑥
󵄨󵄨 𝑚 󵄨󵄨𝑡
𝐺 (𝑥) := ∫ 𝑝 (𝜏) 󵄨󵄨󵄨𝑔Δ (𝜏)󵄨󵄨󵄨 Δ𝜏.
󵄨
󵄨
𝑎
𝐹 (𝑏) 𝐺 (𝑏) + (1 − 𝛽) 𝑘 (𝐹 (𝑏) + 𝐺 (𝑏))] ,
where 𝛼 := 𝑠/𝑡, 𝛽 := 𝑟/𝑠,
Then we have
󵄨𝑠
󵄨󵄨 Δ𝑚
󵄨󵄨𝑔 (𝑥)󵄨󵄨󵄨 = 𝐺Δ (𝑥)𝑠/𝑡 𝑝(𝑥)−𝑠/𝑡 .
󵄨󵄨
󵄨󵄨
So (10) together with (12) implies
󵄨󵄨 𝑘
󵄨󵄨𝑠
󵄨󵄨𝑟 󵄨󵄨 𝑚
𝑞 (𝑥) 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 ≤ ℎ (𝑥) 𝐹(𝑥)𝑟/𝑡 𝐺Δ (𝑥)𝑠/𝑡 ,
󵄨
󵄨󵄨
󵄨
(13)
where ℎ(𝑥) := 𝑞(𝑥)𝑝(𝑥)−𝑠/𝑡 𝑃(𝑥)𝑟(𝑡−1)/𝑡 . Integrating both sides
of (13) over Υ and making use of Hölder’s inequality, we
obtain
𝑏
𝑎
1/(1−𝑡)
Δ𝜏,
𝑃 (𝑥) := ∫ ℎ𝑡/(𝑡−1)
𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) 𝑝(𝜏)
𝑎
−𝛼
𝑟(𝑡−1)/𝑡
ℎ (𝑥) := 𝑞 (𝑥) 𝑝(𝑥) 𝑃(𝑥)
𝑏
≤ (∫ ℎ(𝑥)
,
󵄨󵄨 𝑚 󵄨󵄨𝑡
𝐹 (𝑥) := ∫ 𝑝 (𝜏) 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏,
󵄨
󵄨
𝑎
𝑥
(7)
𝑖
Proof. Since 𝑓Δ (𝑎) = 0, 𝑖 = 0, 1, . . . , 𝑚 − 1, we obtain from
Taylor’s theorem that for all 𝑥 ∈ Υ,
𝑚
𝑓(𝑥) = ∫ ℎ𝑚−1 (𝑥, 𝜎 (𝜏)) 𝑓Δ (𝜏) Δ𝜏,
𝑡/(𝑡−𝑠)
𝑎
1−𝑠/𝑡
Δ𝑥)
𝑏
𝑠/𝑡
𝑟/𝑠
Δ
(∫ 𝐹(𝑥) 𝐺 (𝑥) Δ𝑥) .
𝑎
(14)
Similarly, we get
󵄨󵄨 𝑚 󵄨󵄨𝑡
𝐺 (𝑥) := ∫ 𝑝 (𝜏) 󵄨󵄨󵄨𝑔Δ (𝜏)󵄨󵄨󵄨 Δ𝜏.
󵄨
󵄨
𝑎
𝑥
𝑎
(12)
≤ ∫ ℎ (𝑥) 𝐹(𝑥)𝑟/𝑡 𝐺Δ (𝑥)𝑠/𝑡 Δ𝑥
𝑥
𝑥
(11)
𝑏
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨 𝑘
󵄨󵄨𝑠
∫ 𝑞 (𝑥) 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 Δ𝑥
󵄨󵄨
󵄨
󵄨
𝑎
𝛼
𝛽
(9)
From (9) and Hölder’s inequality we get
for any 𝑘 > 0.
𝑚−1
𝑚
𝑎
Lemma 1 (see [16]). Let 𝑎 ≥ 0, 𝑝 ≥ 1 be real constants. Then
𝑎1/𝑝 ≤
𝑥
𝑘
𝑓Δ (𝑥) = ∫ ℎ𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) 𝑓Δ (𝜏) Δ𝜏.
(8)
𝑏
󵄨󵄨 𝑘
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠
∫ 𝑞 (𝑥) 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 Δ𝑥
󵄨
󵄨󵄨
󵄨
𝑎
𝑏
1−𝑠/𝑡
≤ (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
𝑎
𝑏
𝑠/𝑡
⋅ (∫ 𝐺(𝑥)𝑟/𝑠 𝐹Δ (𝑥) Δ𝑥) .
𝑎
(15)
Abstract and Applied Analysis
3
Recall the elementary inequalities
From (18) and (19), we conclude
𝑐𝛼 (𝐴 + 𝐵)𝛼 ≤ 𝐴𝛼 + 𝐵𝛼 ≤ 𝑑𝛼 (𝐴 + 𝐵)𝛼 ,
(𝐴, 𝐵 ≥ 0) ,
(16)
where
𝑏
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠
󵄨󵄨 𝑘
󵄨󵄨𝑠 󵄨󵄨 𝑘
∫ 𝑞 (𝑥) [󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ] Δ𝑥
󵄨󵄨
󵄨 󵄨
󵄨󵄨
󵄨
󵄨
𝑎
1−𝛼
𝑏
𝑐𝛼 := {
𝑑𝛼 := {
1,
21−𝛼 ,
≤ 21−𝛼 (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
0 ≤ 𝛼 ≤ 1,
𝛼 ≥ 1,
1−𝛼
2 ,
1,
𝑎
(17)
0 ≤ 𝛼 ≤ 1,
𝛼 ≥ 1.
Let 𝛽 = 𝑟/𝑠. Since 𝛼 = 𝑠/𝑡 ∈ (0, 1) and 𝐹 is nondecreasing,
from (14)–(16), we have
𝑏
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠 󵄨󵄨 𝑘
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠
󵄨󵄨 𝑘
∫ 𝑞 (𝑥) [󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ] Δ𝑥
󵄨󵄨
󵄨 󵄨
󵄨󵄨
󵄨
󵄨
𝑎
1−𝛼
𝑏
≤ (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
Δ
𝛼
𝛽
The proof is complete.
Theorem 4. Let 𝑟 ≥ 0, 𝑠 > 0, 𝑠 < 𝑡, 𝑡 > 1 be real numbers, and
let 𝑚, 𝑘 be integers with 0 ≤ 𝑘 ≤ 𝑚 − 1. Let 𝑝 > 0, and 𝑞 ≥ 0
be measurable functions on Υ := [𝑎, 𝑏] ∩ T. Further, let 𝑓, 𝑔 ∈
𝑖
𝑖
𝑚−1
(Υ) with let 𝑓Δ (𝑎) = 𝑔Δ (𝑎) = 0, 𝑖 = 0, 1, . . . , 𝑚 − 1,
𝐶𝑟𝑑
𝑚−1
𝑚−1
and 𝑓Δ , 𝑔Δ be absolutely continuous on Υ such that the
𝑏
𝑎
𝑏
𝛼
⋅ [𝛽𝑘𝛽−1 𝐹 (𝑏) 𝐺 (𝑏) + (1 − 𝛽) 𝑘𝛽 (𝐹 (𝑏) + 𝐺 (𝑏))] .
(20)
𝑏
Δ
𝛼
𝛽
𝑏
𝑚
𝑚
integrals ∫𝑎 𝑝(𝑥)|𝑓Δ (𝑥)|𝑡 Δ𝑥 and ∫𝑎 𝑝(𝑥)|𝑔Δ (𝑥)|𝑡 Δ𝑥 exist.
Then one has
⋅ [(∫ 𝐹 (𝑥) 𝐺 (𝑥) Δ𝑥) + (∫ 𝐺 (𝑥) 𝐹 (𝑥) Δ𝑥) ]
𝑎
𝑎
𝑏
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠
󵄨󵄨 𝑘
󵄨󵄨𝑠 󵄨󵄨 𝑘
∫ 𝑞 (𝑥) [󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ] Δ𝑥
󵄨󵄨
󵄨 󵄨
󵄨󵄨
󵄨
󵄨
𝑎
1−𝛼
𝑏
≤ (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
𝑎
1−𝛼
𝑏
≤ 21−𝛼 (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
𝛼
𝑏
⋅21−𝛼 (∫ [𝐹Δ (𝑥) 𝐺𝛽 (𝑥) + 𝐺Δ (𝑥) 𝐹𝛽 (𝑥)] Δ𝑥)
𝑎
𝑎
𝛼
⋅ [𝑑𝛽 Γ (𝐺 + 𝐹) − Γ (𝐺) − Γ (𝐹)] ,
1−𝛼
𝑏
≤ (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
(21)
𝑎
𝛼
𝑏
⋅ 21−𝛼 (∫ [𝐹Δ (𝑥) 𝐺𝛽 (𝑥) + 𝐺Δ (𝑥) 𝐹𝛽 (𝜎 (𝑥))] Δ𝑥) .
𝑎
(18)
𝑏
where Γ(𝐻) := ∫𝑎 𝐻𝛽 Δ𝐻, 𝛽 := 𝑟/𝑠, 𝛼 := 𝑠/𝑡, ℎ(𝑥) := 𝑞(𝑥)𝑝(𝑥)−𝛼
𝑥
𝑡/(𝑡−1)
𝑃(𝑥)𝑟(𝑡−1)/𝑡 , 𝑃(𝑥) := ∫𝑎 ℎ𝑚−𝑘−1
(𝑥, 𝜎(𝜏))𝑝(𝜏)1/(1−𝑡) Δ𝜏, and
By Lemma 1, we get
21−𝛽 ,
𝑑𝛽 := {
1,
𝑏
∫ [𝐹Δ (𝑥) 𝐺𝛽 (𝑥) + 𝐺Δ (𝑥) 𝐹𝛽 (𝜎 (𝑥))] Δ𝑥
0 ≤ 𝛽 ≤ 1,
𝛽 ≥ 1.
(22)
𝑎
𝑏
≤ ∫ [𝛽𝑘𝛽−1 𝐺 (𝑥) 𝐹Δ (𝑥) + (1 − 𝛽) 𝑘𝛽 𝐹Δ (𝑥)
Proof. Following the proof of Theorem 3, we obtain
𝑎
+ 𝛽𝑘𝛽−1 𝐹 (𝜎 (𝑥)) 𝐺Δ (𝑥) + (1 − 𝛽) 𝑘𝛽 𝐺Δ (𝑥)] Δ𝑥
𝑏
= 𝛽𝑘𝛽−1 ∫ [𝐺 (𝑥) 𝐹Δ (𝑥) + 𝐹 (𝜎 (𝑥)) 𝐺Δ (𝑥)] Δ𝑥
𝑎
𝑏
𝑏
𝑎
𝑎
= 𝛽𝑘
1−𝛼
𝑏
≤ (∫ ℎ(𝑥)𝑡/(𝑡−𝑠) Δ𝑥)
𝑎
+ (1 − 𝛽) 𝑘𝛽 [∫ 𝐹Δ (𝑥) Δ𝑥 + ∫ 𝐺Δ (𝑥) Δ𝑥]
𝛽−1
𝑏
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑟 󵄨󵄨 𝑚
󵄨󵄨𝑠
󵄨󵄨 𝑘
󵄨󵄨𝑠 󵄨󵄨 𝑘
∫ 𝑞 (𝑥) [󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑔Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ] Δ𝑥
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
󵄨
𝑎
𝑏
𝛼
⋅ 21−𝛼 (∫ [𝐹Δ (𝑥) 𝐺𝛽 𝑥 + 𝐺Δ (𝑥) 𝐹𝛽 (𝑥)] Δ𝑥) .
𝛽
𝐹 (𝑏) 𝐺 (𝑏) + (1 − 𝛽) 𝑘 [𝐹 (𝑏) + 𝐺 (𝑏)] .
𝑎
(19)
(23)
4
Abstract and Applied Analysis
Integrating both sides over 𝑥 from 𝑎 to 𝑏 and using
Cauchy-Schwarz inequality, we observe
Using (16),
𝑏
𝑏󵄨
𝑚
󵄨󵄨
󵄨 𝑘
∫ 󵄨󵄨󵄨𝑓Δ (𝑥) 𝑓Δ (𝑥)󵄨󵄨󵄨 Δ𝑥
󵄨
󵄨
𝑎
∫ [𝐹Δ (𝑥) 𝐺𝛽 (𝑥) + 𝐺Δ (𝑥) 𝐹𝛽 (𝑥)] Δ𝑥
𝑎
𝑏
= ∫ (𝐺𝛽 (𝑥) + 𝐹𝛽 (𝑥)) (𝐺Δ (𝑥) + 𝐹Δ (𝑥)) Δ𝑥
𝑎
𝑏
𝛽
Δ
𝛽
Δ
− ∫ (𝐺 (𝑥) 𝐺 (𝑥) + 𝐹 (𝑥) 𝐹 (𝑥)) Δ𝑥
𝑎
≤ 𝑑𝛽 ∫ (𝐺 (𝑥) + 𝐹 (𝑥))𝛽 Δ (𝐺 (𝑥) + 𝐹 (𝑥))
𝑎
𝑎
𝑎
𝑎
𝑏
𝑎
𝑏
𝑥
1/2
1/2
𝑏󵄨
󵄨󵄨2 𝑥 󵄨󵄨 𝑚 󵄨󵄨2
󵄨 𝑚
⋅ [∫ 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ∫ 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏Δ𝑥]
󵄨 𝑎 󵄨
󵄨
𝑎 󵄨
(24)
𝑏
𝑏
𝑏
2
≤ [∫ ∫ ℎ𝑚−𝑘−1
(𝑥, 𝜎 (𝜏)) Δ𝜏Δ𝑥]
(28)
1/2
𝑥
2
= [∫ ∫ ℎ𝑚−𝑘−1
(𝑥, 𝜎 (𝜏)) Δ𝜏Δ𝑥]
𝑎 𝑎
− ∫ 𝐺𝛽 (𝑥) Δ𝐺 (𝑥) − ∫ 𝐹𝛽 (𝑥) Δ𝐹 (𝑥)
1/2
𝑏󵄨
󵄨 𝑚 󵄨󵄨2
⋅ [∫ 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏] .
󵄨
𝑎 󵄨
= 𝑑𝛽 Γ (𝐺 + 𝐹) − Γ (𝐺) − Γ (𝐹) .
The proof is complete.
The proof is complete.
Remark 5. In the special case where T = R, Theorem 4
reduces to Theorem 1 of [13].
𝑚−1
(Υ), Υ := [𝑎, 𝑏] ∩ T be such that
Theorem 6. Let 𝑓 ∈ 𝐶𝑟𝑑
𝑚−1
Δ𝑖
𝑓 (𝑎) = 0, 0 ≤ 𝑘 ≤ 𝑚−1, let 𝑓Δ (𝑥) be absolutely continuous
𝑏
Theorem 7. Let 𝑝(𝑥) > 0, 𝑞(𝑥) be nonnegative and measur𝑚−1
able on Υ = [𝑎, 𝑏] ∩ T, and let 𝑓 ∈ 𝐶𝑟𝑑
(Υ) be such that
𝑘
𝑚−1
𝑓Δ (𝑎) = 0, 0 ≤ 𝑘 ≤ 𝑚 − 1. If 𝑓Δ (𝑥) is absolutely continuous
on Υ, then for 𝑟 > 1, 𝑟𝑘 > 0, and any 0 ≤ 𝑟𝑚 < 𝑟,
𝑏
󵄨󵄨 𝑚
󵄨󵄨𝑟𝑚 󵄨󵄨 𝑘
󵄨󵄨𝑟𝑘
∫ 𝑞 (𝑥) 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 Δ𝑥
󵄨
󵄨 󵄨
󵄨
𝑎
𝑚
on Υ, and let ∫𝑎 |𝑓Δ (𝑥)|2 Δ𝑥 < ∞. Then
𝑏
≤ [∫ 𝑞(𝑥)𝑟/(𝑟−𝑟𝑚 ) 𝑝(𝑥)−𝑟𝑚 /(𝑟−𝑟𝑚 )
𝑚
󵄨󵄨
󵄨󵄨 𝑘
∫ 󵄨󵄨󵄨𝑓Δ (𝑥) 𝑓Δ (𝑥)󵄨󵄨󵄨 Δ𝑥
󵄨
𝑎 󵄨
𝑏
𝑏
≤ (∫ ∫
𝑥
𝑎 𝑎
2
ℎ𝑚−𝑘−1
(𝑟−𝑟𝑚 )/𝑟
× 𝑄(𝑥)(𝑟−1)𝑟𝑚 /(𝑟−𝑟𝑚 ) Δ𝑥]
1/2
(𝑥, 𝜎 (𝜏)) Δ𝜏Δ𝑥)
𝑟 /𝑟
Φ(𝑦) 𝑚 ,
(25)
𝑥
𝑟/𝑟−1
where 𝑄(𝑥) := ∫𝑎 ℎ𝑚−𝑘−1
(𝑥, 𝜎(𝜏))𝑝(𝜏)−(𝑟/𝑟−1) Δ𝜏, Φ(𝑦) :=
1/2
𝑏󵄨
󵄨 𝑚 󵄨󵄨2
⋅ (∫ 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏)
󵄨
𝑎 󵄨
𝑏
𝑥
󵄨󵄨 Δ𝑘
󵄨
󵄨 𝑚 󵄨
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 ≤ ∫ ℎ𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏.
󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨󵄨
𝑎
(26)
𝑚
Multiplying (26) by |𝑓Δ (𝑥)| and using Cauchy-Schwarz
inequality, we obtain
𝑚
Proof. Following the hypotheses, it is easy to see that (26)
holds. By using Hölder’s inequality with indices 𝑟 and 𝑟/(𝑟−1),
we obtain
󵄨
󵄨󵄨 Δ𝑘
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
𝑥
󵄨󵄨 𝑚 󵄨󵄨
≤ ∫ ℎ𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) 𝑝(𝜏)−1/𝑟 𝑝(𝜏)1/𝑟 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏
󵄨
󵄨
𝑎
(𝑟−1)/𝑟
𝑥
𝑟/(𝑟−1)
≤ [∫ ℎ𝑚−𝑘−1
(𝑥, 𝜎 (𝜏)) 𝑝(𝜏)−𝑟/(𝑟−1) Δ𝜏]
(30)
𝑎
𝑚
󵄨
󵄨󵄨 Δ𝑘
󵄨󵄨𝑓 (𝑥) 𝑓Δ (𝑥)󵄨󵄨󵄨
󵄨󵄨
󵄨󵄨
󵄨󵄨 𝑥
󵄨󵄨 𝑚 󵄨󵄨
󵄨󵄨 𝑚
≤ 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ∫ ℎ𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏
󵄨
󵄨 𝑎
󵄨
󵄨
1/2
󵄨󵄨 𝑚
󵄨󵄨 𝑥
≤ 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 (∫ ℎ2𝑚−𝑘−1 (𝑥, 𝜎 (𝜏)) Δ𝜏)
󵄨
󵄨 𝑎
𝑥
∫𝑎 𝑦𝑟𝑘 /𝑟𝑚 Δ𝑦(𝑥), 𝑦(𝑥) := ∫𝑎 𝑝(𝜏)|𝑓Δ (𝜏)|𝑟 Δ𝜏.
.
Proof. From the hypotheses, we have
1/2
𝑥󵄨
󵄨 𝑚 󵄨󵄨2
⋅ (∫ 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏) .
󵄨
𝑎 󵄨
(29)
𝑎
1/𝑟
𝑥
󵄨󵄨 𝑚 󵄨󵄨𝑟
⋅ [∫ 𝑝 (𝜏) 󵄨󵄨󵄨𝑓Δ (𝜏)󵄨󵄨󵄨 Δ𝜏]
󵄨
󵄨
𝑎
(27)
= 𝑄(𝑥)(𝑟−1)/𝑟 𝑦(𝑥)1/𝑟 ,
𝑥
𝑟/(𝑟−1)
where 𝑄(𝑥) := ∫𝑎 ℎ𝑚−𝑘−1
(𝑥, 𝜎(𝜏))𝑝(𝜏)−𝑟/(𝑟−1) Δ𝜏, 𝑦(𝑥) :=
𝑥
𝑚
∫𝑎 𝑝(𝜏)|𝑓Δ (𝜏)|𝑟 Δ𝜏. So we get
󵄨󵄨 𝑚
󵄨󵄨𝑟
𝑦Δ (𝑥) = 𝑝 (𝑥) 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 ,
󵄨
󵄨
(31)
Abstract and Applied Analysis
5
and hence for any 𝑟𝑚 ,
󵄨𝑟
󵄨󵄨 Δ𝑚
󵄨󵄨𝑓 (𝑥)󵄨󵄨󵄨 𝑚 = (𝑝 (𝑥))−𝑟𝑚 /𝑟 (𝑦Δ (𝑥))𝑟𝑚 /𝑟 .
󵄨󵄨
󵄨󵄨
(32)
Thus for 𝑟𝑘 > 0,
󵄨󵄨 𝑚
󵄨󵄨𝑟𝑚 󵄨󵄨 𝑘
󵄨󵄨𝑟𝑘
𝑞 (𝑥) 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨
󵄨
󵄨 󵄨
󵄨
≤ 𝑞 (𝑥) (𝑝 (𝑥))
−𝑟𝑚 /𝑟
𝑟𝑚 /𝑟
(𝑦Δ (𝑥))
(33)
× 𝑄(𝑥)(𝑟−1)𝑟𝑘 /𝑟 𝑦(𝑥)𝑟𝑘 /𝑟 .
Integrating both sides of (33) from 𝑎 to 𝑏 and applying
Hölder’s inequality with indices 𝑟/𝑟𝑚 and 𝑟/(𝑟−𝑟𝑚 ), we obtain
𝑏
󵄨󵄨𝑟𝑚 󵄨󵄨 𝑘
󵄨󵄨𝑟𝑘
󵄨󵄨 𝑚
∫ 𝑞 (𝑥) 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 󵄨󵄨󵄨𝑓Δ (𝑥)󵄨󵄨󵄨 Δ𝑥
󵄨
󵄨
󵄨
󵄨
𝑎
𝑏
𝑟𝑚 /𝑟
≤ ∫ 𝑞 (𝑥) 𝑝(𝑥)−𝑟𝑚 /𝑟 (𝑦Δ (𝑥))
𝑄(𝑥)(𝑟−1)𝑟𝑘 /𝑟𝑦(𝑥)𝑟𝑘 /𝑟 Δ𝑥
𝑎
𝑏
≤ [∫ 𝑞(𝑥)
𝑟/(𝑟−𝑟𝑚 )
−𝑟𝑚 /(𝑟−𝑟𝑚 )
𝑝(𝑥)
𝑎
(𝑟−1)𝑟𝑘 /(𝑟−𝑟𝑚 )
𝑄(𝑥)
(𝑟−𝑟𝑚 )/𝑟
Δ𝑥]
𝑟𝑚 /𝑟
𝑏
⋅ [∫ 𝑦Δ (𝑥) 𝑦(𝑥)𝑟𝑘 /𝑟𝑚 Δ𝑥]
𝑎
𝑏
= [∫ 𝑞(𝑥)𝑟/(𝑟−𝑟𝑚 ) 𝑝(𝑥)−𝑟𝑚 /(𝑟−𝑟𝑚 )
𝑎
(𝑟−1)𝑟𝑘 /(𝑟−𝑟𝑚 )
×𝑄(𝑥)
(𝑟−𝑟𝑚 )/𝑟
Δ𝑥]
𝑟 /𝑟
Φ(𝑦) 𝑚 ,
(34)
𝑏
where Φ(𝑦) := ∫𝑎 𝑦(𝑥)𝑟𝑘 /𝑟𝑚 Δ𝑦(𝑥). The proof is complete.
Remark 8. In the special case where T = R, Theorems 6 and
7 reduce to Theorems 1 and 2 of [18].
Acknowledgments
The first author’s research was supported by NSF of China
(11071054), Natural Science Foundation of Hebei Province
(A2011205012). The second author’s research was partially
supported by an HKU URG grant.
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Hindawi Publishing Corporation
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International Journal of
Mathematics
Volume 2014
Volume 2014
Discrete Dynamics in
Nature and Society
Volume 2014
Volume 2014
Journal of
Journal of
Discrete Mathematics
Journal of
Volume 2014
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Applied Mathematics
Journal of
Function Spaces
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Volume 2014
Hindawi Publishing Corporation
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Volume 2014
Hindawi Publishing Corporation
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Volume 2014
Optimization
Hindawi Publishing Corporation
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Volume 2014
Hindawi Publishing Corporation
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Volume 2014