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Certain Inequalities Involving Generalized Erdélyi-Kober Fractional q-Integral Operators.
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PMID:  25295293     Owner:  NLM     Status:  In-Data-Review    
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In recent years, a remarkably large number of inequalities involving the fractional q-integral operators have been investigated in the literature by many authors. Here, we aim to present some new fractional integral inequalities involving generalized Erdélyi-Kober fractional q-integral operator due to Gaulué, whose special cases are shown to yield corresponding inequalities associated with Kober type fractional q-integral operators. The cases of synchronous functions as well as of functions bounded by integrable functions are considered.
Authors:
Praveen Agarwal; Soheil Salahshour; Sotiris K Ntouyas; Jessada Tariboon
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Type:  Journal Article     Date:  2014-09-11
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Title:  TheScientificWorldJournal     Volume:  2014     ISSN:  1537-744X     ISO Abbreviation:  ScientificWorldJournal     Publication Date:  2014  
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Journal ID (nlm-ta): ScientificWorldJournal
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ISSN: 1537-744X
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Copyright © 2014 Praveen Agarwal et al.
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Received Day: 21 Month: 6 Year: 2014
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Volume: 2014E-location ID: 174126
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DOI: 10.1155/2014/174126

Certain Inequalities Involving Generalized Erdélyi-Kober Fractional q-Integral Operators
Praveen Agarwal1 http://orcid.org/0000-0001-7556-8942
Soheil Salahshour2 http://orcid.org/0000-0003-1390-3551
Sotiris K. Ntouyas3
Jessada Tariboon4* http://orcid.org/0000-0001-8185-3539
1Department of Mathematics, Anand International College of Engineering, Jaipur 303012, India
2Department of Computer Engineering, Mashhad Branch, Islamic Azad University, Mashhad, Iran
3Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece
4Nonlinear Dynamic Analysis Research Center, Department of Mathematics, Faculty of Applied Science, King Mongkut's University of Technology North Bangkok, Bangkok 10800, Thailand
Correspondence: *Jessada Tariboon: jessadat@kmutnb.ac.th
[other] Academic Editor: Junesang Choi

1. Introduction

Let us start by considering the following functional (see [1]):

[Formula ID: EEq1.1]
(1) 
T(f,g,p,q) =∫abq(x)dx∫abp(x)f(x)g(x)dx  +∫abp(x)dx∫abq(x)f(x)g(x)dx  −(∫abq(x)f(x)dx)(∫abp(x)g(x)dx)  −(∫abp(x)f(x)dx)(∫abq(x)g(x)dx),
where f, g : [a, b] → R are two integrable functions on [a, b] and p(x) and q(x) are positive integrable functions on [a, b]. If f and g are synchronous on [a, b], that is,
[Formula ID: EEq1.2]
(2) 
(f(x)−f(y))(g(x)−g(y))≥0,
for any x, y ∈ [a, b], then we have (see, e.g., [2, 3])
[Formula ID: EEq1.3]
(3) 
T(f,g,p,q)≥0.
The inequality in (2) is reversed if f and g are asynchronous on [a, b]; that is,
[Formula ID: EEq1.4]
(4) 
(f(x)−f(y))(g(x)−g(y))≤0,
for any x, y ∈ [a, b]. If p(x) = q(x) for any x, y ∈ [a, b], we get the Chebyshev inequality (see [1]). Ostrowski [4] established the following generalization of the Chebyshev inequality.

If f and g are two differentiable and synchronous functions on [a, b] and p is a positive integrable function on [a, b] with |f′(x)|≥m and |g′(x)|≥r for x ∈ [a, b], then we have

[Formula ID: EEq1.5]
(5) 
T(f,g,p)=T(f,g,p,p)≥mrT(x−a,x−a,p)≥0.

If f and g are asynchronous on [a, b], then we have

[Formula ID: EEq1.6]
(6) 
T(f,g,p)≤mrT(x−a,x−a,p)≤0.
If f and g are two differentiable functions on [a, b] with |f′(x)|≤M and |g′(x)|≤R for x ∈ [a, b] and p is a positive integrable function on [a, b], then we have
[Formula ID: EEq1.7]
(7) 
|T(f,g,p)|≤MRT(x−a,x−a,p)≤0.
Here, it is worth mentioning that the functional (1) has attracted many researchers' attention mainly due to diverse applications in numerical quadrature, transform theory, probability, and statistical problems. Among those applications, the functional (1) has also been employed to yield a number of integral inequalities (see, e.g., [511]).

The study of the fractional integral and fractional q-integral inequalities has been of great importance due to the fundamental role in the theory of differential equations. In recent years, a number of researchers have done deep study, that is, the properties, applications, and different extensions of various fractional q-integral operators (see, e.g., [1216]).

The purpose of this paper is to find q-calculus analogs of some classical integral inequalities. In particular, we will find q-generalizations of the Chebyshev integral inequalities by using the generalized Erdélyi-Kober fractional q-integral operator introduced by Galué [17]. The main objective of this paper is to present some new fractional q-integral inequalities involving the generalized Erdélyi-Kober fractional q-integral operator. We consider the case of synchronous functions as well as the case of functions bounded by integrable functions. Some of the known and new results are as follows, as special cases of our main findings. We emphasize that the results derived in this paper are more generalized results rather than similar published results because we established all results by using the generalized Erdélyi-Kober fractional q-integral operator. Our results are general in character and give some contributions to the theory q-integral inequalities and fractional calculus.


2. Preliminaries

In the sequel, we required the following well-known results to establish our main results in the present paper. The q-shifted factorial (a; q)n is defined by

[Formula ID: EEq2.1]
(8) 
(a;q)n:={1,(n=0)∏k=0n−1(1−aqk),(n∈N),
where a, qC and it is assumed that aqm  (mN0).

The q-shifted factorial for negative subscript is defined by

[Formula ID: EEq2.2]
(9) 
(a;q)−n≔1(1−aq−1)(1−aq−2)⋯(1−aq−n)(n∈N0).
We also write
[Formula ID: EEq2.3]
(10) 
(a;q)∞≔∏k=0∞(1−aqk) (a,q∈C;  |q|<1).
It follows from (8), (9), and (10) that
[Formula ID: EEq2.4]
(11) 
(a;q)n=(a;q)∞(aqn;q)∞ (n∈Z),
which can be extended to n = αC as follows:
[Formula ID: EEq2.5]
(12) 
(a;q)α=(a;q)∞(aqα;q)∞ (α∈C;  |q|<1),
where the principal value of qα is taken.

We begin by noting that F. J. Jackson was the first to develop q-calculus in a systematic way. For 0 < q < 1, the q-derivative of a continuous function f on [0, b] is defined by

[Formula ID: EEq2.6]
(13) 
Dqf(t)≔dqdqtf(t)=f(t)−f(qt)(1−q)t, t∈(0,b],
and Dqf(0) = lim⁡t→0Dqf(t). It is noted that
[Formula ID: eq14]
(14) 
lim⁡q→1Dqf(t)=ddtf(t),
if f(t) is differentiable.

The function F(t) is a q-antiderivative of f(t) if DqF(t) = f(t). It is denoted by

[Formula ID: EEq2.7]
(15) 
∫f(t)dqt.

The Jackson integral of f(t) is thus defined, formally, by

[Formula ID: EEq2.8]
(16) 
∫f(t)dqt≔(1−q)t∑j=0∞qjf(qjt),
which can be easily generalized as follows:
[Formula ID: EEq2.9]
(17) 
∫f(t)dqg(t)=∑j=0∞f(qjt)(g(qjt)−g(qj+1t)).

Suppose that 0 < a < b. The definite q-integral is defined as follows:

[Formula ID: EEq2.10]
(18) 
∫0bf(t)dqt:=(1−q)b∑j=0∞qjf(qjb),
[Formula ID: EEq2.11]
(19) 
∫abf(t)dqt=∫0bf(t)dqt−∫0af(t)dqt.

A more general version of (18) is given by

[Formula ID: EEq2.12]
(20) 
∫0bf(t)dqg(t)=∑j=0∞f(qjb)(g(qjb)−g(qj+1b)).

The classical Gamma function Γ(z) (see, e.g., [18, Section  1.1]) was found by Euler while he was trying to extend the factorial n! = Γ(n + 1)(nN0) to real numbers. The q-factorial function [n]q!  (nN0) of n! defined by

[Formula ID: EEq2.13]
(21) 
[n]q!≔{1,if  n=0,[n]q[n−1]q⋯[2]q[1]q,if  n∈N,
can be rewritten as follows:
[Formula ID: EEq2.14]
(22) 
(1−q)−n∏k=0∞(1−qk+1)(1−qk+1+n)=(q;q)∞(qn+1;q)∞(1−q)−n≔Γq(n+1) (0<q<1).
Replacing n by a − 1 in (22), Jackson [19] defined the q-Gamma function Γq(a) by
[Formula ID: EEq2.15]
(23) 
Γq(a)≔(q;q)∞(qa;q)∞(1−q)1−a (0<q<1).

The q-analogue of (ta)n is defined by the polynomial

[Formula ID: EEq2.16]
(24) 
(t−a)(n)≔{1,(n=0)(t−a)(t−qa)⋯(t−qn−1a),(n∈N),=tn(at;q)n (n∈N0).
More generally, if γR, then
[Formula ID: eq25]
(25) 
(t−a)(γ)≔tγ∏k=0∞1−(a/t)qk1−(a/t)qγ+k, t≠0.

Definition 1 .

Let R(β), R(μ) > 0 and ηC. Then a generalized Erdélyi-Kober fractional integral Iqα,β,η for a real-valued continuous function f(t) is defined by (see, [17])

[Formula ID: EEq2.17]
(26) 
Iqη,μ,β{f}(t)  =βt−β(η+μ)Γq(μ)∫0t(tβ−τβq)(μ−1)τβ(η+1)−1f(τ)dqτ  =β(1−q1/β)(1−q)μ−1∑k=0∞(qμ;q)k(q;q)kqk(η+1)f(tqk/β).

Definition 2 .

A q-analogue of the Kober fractional integral operator is given by (see, [20])

[Formula ID: EEq2.18]
(27) 
Iqη,μ{f}(t):=(Iqη,μ,1{f}(t))=t−η−μΓq(μ)∫0t(t−τq)(μ−1)τηf(τ)dqτ,(μ>0;  η∈C;  0<q<1).

Remark 3 .

It is easy to see that

[Formula ID: EEq2.19]
(28) 
Γq(μ)>0;  (qμ;q)k>0,
for all μ > 0 and kN0. If f : [0, )→[0, ) is a continuous function, then we conclude that, under the given conditions in (26), each term in the series of generalized Erdélyi-Kober q-integral operator is nonnegative and thus
[Formula ID: EEq2.20]
(29) 
Iqη,μ,β{f}(t)≥0,
for all β, μ > 0 and ηC.

On the same way each term in the series of Kober q-integral operator (27) is also nonnegative and thus

[Formula ID: EEq2.21]
(30) 
Iqη,μ{f}(t)≥0,
for all μ > 0 and ηC.


3. Inequalities Involving a Generalized Erdélyi-Kober Fractional q-Integral Operator for Synchronous Functions

This section begins by presenting two inequalities involving generalized Erdélyi-Kober q-integral operator (26) stated in Lemmas 4 and 5 below.

Lemma 4 .

Let 0 < q < 1, let f and g be two continuous and synchronous functions on [0, ), and let u, v : [0, )→[0, ) be continuous functions. Then, the following inequality holds true:

[Formula ID: EEq3.1]
(31) 
Iqη,μ,β{u}(t)Iqη,μ,β{vfg}(t)  +Iqη,μ,β{v}(t)Iqη,μ,β{ufg}(t) ≥Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)  +Iqη,μ,β{vf}(t)Iqη,μ,β{ug}(t),
for all μ, β > 0 and ηC.

Proof

Let f and g be two continuous and synchronous functions on [0, ). Then, for all τ, ρ ∈ (0, t) with t > 0, we have

[Formula ID: EEq3.2]
(32) 
(f(τ)−f(ρ))(g(τ)−g(ρ))≥0,
or, equivalently,
[Formula ID: EEq3.3]
(33) 
f(τ)g(τ)+f(ρ)g(ρ)≥f(τ)g(ρ)+f(ρ)g(τ).

Now, multiplying both sides of (33) by (βtβ(η+μ)q(μ))(tβτβq)(μ−1)τβ(η+1)−1u(τ), integrating the resulting inequality with respect to τ from 0 to t, and using (26), we get

[Formula ID: EEq3.4]
(34) 
Iqη,μ,β{ufg}(t)+f(ρ)g(ρ)Iqη,μ,β{u}(t)  ≥g(ρ)Iqη,μ,β{uf}(t)+f(ρ)Iqη,μ,β{ug}(t).

Next, multiplying both sides of (34) by (βtβ(η+μ)q(μ))(tβρβq)(μ−1)ρβ(η+1)−1v(ρ), integrating the resulting inequality with respect to ρ from 0 to t, and using (26), we are led to the desired result (31).

Lemma 5 .

Let 0 < q < 1, let f and g be two continuous and synchronous functions on [0, ), and let u, v : [0, )→[0, ) be continuous functions. Then, the following inequality holds true:

[Formula ID: EEq3.5]
(35) 
Iqζ,ν,δ{v}(t)Iqη,μ,β{ufg}(t)  +Iqζ,ν,δ{vfg}(t)Iqη,μ,β{u}(t) ≥Iqζ,ν,δ{vg}(t)Iqη,μ,β{uf}(t)  +Iqζ,ν,δ{vf}(t)Iqη,μ,β{ug}(t),
for all μ, ν, β, δ > 0 and η, ζC.

Proof

Multiplying both sides of (34) by

[Formula ID: eq36]
(36) 
δt−δ(ζ+ν)Γq(ν)(tδ−ρδq)(ν−1)ρδ(ζ+1)−1v(ρ),
which remains nonnegative under the conditions in (35), integrating the resulting inequality with respect to ρ from 0 to t, and using (26), we get the desired result (35).

Theorem 6 .

Let 0 < q < 1, let f and g be two continuous and synchronous functions on [0, ), and let l, m, n : [0, )→[0, ) be continuous functions. Then, the following inequality holds true:

[Formula ID: EEq3.6]
(37) 
2Iqη,μ,β{l}(t)[Iqη,μ,β{m}(t)Iqη,μ,β{nfg}(t)+Iqη,μ,β{n}(t)Iqη,μ,β{mfg}(t)]  +2Iqη,μ,β{m}(t)Iqη,μ,β{n}(t)Iqη,μ,β{lfg}(t) ≥Iqη,μ,β{l}(t)[Iqη,μ,β{mf}(t)Iqη,μ,β{ng}(t)+Iqη,μ,β{nf}(t)Iqη,μ,β{mg}(t)]  +Iqη,μ,β{m}(t)[Iqη,μ,β{lf}(t)Iqη,μ,β{ng}(t)+Iqη,μ,β{nf}(t)Iqη,μ,β{lg}(t)]  +Iqη,μ,β{n}(t)[Iqη,μ,β{lf}(t)Iqη,μ,β{mg}(t)+Iqη,μ,β{mf}(t)Iqη,μ,β{lg}(t)],
for all μ, β > 0 and ηC.

Proof

By setting u = m and v = n in Lemma 4, we get

[Formula ID: EEq3.7]
(38) 
Iqη,μ,β{m}(t)Iqη,μ,β{nfg}(t)  +Iqη,μ,β{n}(t)Iqη,μ,β{mfg}(t) ≥Iqη,μ,β{mf}(t)Iqη,μ,β{ng}(t)  +Iqη,μ,β{nf}(t)Iqη,μ,β{mg}(t).
Since Iqη,μ,β{l}(t) ≥ 0 under the given conditions, multiplying both sides of (38) by Iqη,μ,β{l}(t), we have
[Formula ID: EEq3.8]
(39) 
Iqη,μ,β{l}(t)[Iqη,μ,β{m}(t)Iqη,μ,β{nfg}(t)+Iqη,μ,β{n}(t)Iqη,μ,β{mfg}(t)] ≥Iqη,μ,β{l}(t)[Iqη,μ,β{mf}(t)Iqη,μ,β{ng}(t)+Iqη,μ,β{nf}(t)Iqη,μ,β{mg}(t)].
Similarly replacing u, v by l, n and u, v by l, m, respectively, in (31) and then multiplying both sides of the resulting inequalities by Iqη,μ,β{m}(t) and Iqη,μ,β{n}(t) both of which are nonnegative under the given assumptions, respectively, we get the following inequalities:
[Formula ID: EEq3.9]
(40) 
Iqη,μ,β{m}(t)[Iqη,μ,β{l}(t)Iqη,μ,β{nfg}(t)+Iqη,μ,β{n}(t)Iqη,μ,β{lfg}(t)] ≥Iqη,μ,β{m}(t)[Iqη,μ,β{lf}(t)Iqη,μ,β{ng}(t)+Iqη,μ,β{nf}(t)Iqη,μ,β{lg}(t)],
[Formula ID: EEq3.10]
(41) 
Iqη,μ,β{n}(t)[Iqη,μ,β{l}(t)Iqη,μ,β{mfg}(t)+Iqη,μ,β{m}(t)Iqη,μ,β{lfg}(t)] ≥Iqη,μ,β{n}(t)[Iqη,μ,β{lf}(t)Iqη,μ,β{mg}(t)+Iqη,μ,β{mf}(t)Iqη,μ,β{lg}(t)].
Finally, by adding (39), (40), and (41), side by side, we arrive at the desired result (37).

Theorem 7 .

Let 0 < q < 1, let f and g be two continuous and synchronous functions on [0, ), and let l, m, n : [0, )→[0, ) be continuous functions. Then, the following inequality holds true:

[Formula ID: EEq3.11]
(42) 
Iqη,μ,β{l}(t)[2Iqη,μ,β{m}(t)Iqζ,ν,δ{nfg}(t)+Iqη,μ,β{n}(t)Iqζ,ν,δ{mfg}(t)+Iqζ,ν,δ{n}(t)Iqη,μ,β{mfg}(t)]  +Iqη,μ,β{lfg}(t)[Iqη,μ,β{m}(t)Iqζ,ν,δ{n}(t)+Iqη,μ,β{n}(t)Iqζ,ν,δ{m}(t)] ≥Iqη,μ,β{l}(t)[Iqη,μ,β{mf}(t)Iqζ,ν,δ{ng}(t)+Iqη,μ,β{mg}(t)Iqζ,ν,δ{nf}(t)]  +Iqη,μ,β{m}(t)[Iqη,μ,β{lf}(t)Iqζ,ν,δ{ng}(t)+Iqη,μ,β{lg}(t)Iqζ,ν,δ{nf}(t)]  +Iqη,μ,β{n}(t)[Iqη,μ,β{lf}(t)Iqζ,ν,δ{mg}(t)+Iqη,μ,β{lg}(t)Iqζ,ν,δ{mf}(t)],
for all μ, ν, β, δ > 0 and η, ζC.

Proof

Setting u = m and v = n in (35), we have

[Formula ID: EEq3.12]
(43) 
Iqζ,ν,δ{n}(t)Iqη,μ,β{mfg}(t)  +Iqζ,ν,δ{nfg}(t)Iqη,μ,β{m}(t) ≥Iqζ,ν,δ{ng}(t)Iqη,μ,β{mf}(t)  +Iqζ,ν,δ{nf}(t)Iqη,μ,β{mg}(t).
Multiplying both sides of (43) by Iqη,μ,β{l}(t), after a little simplification, we get
[Formula ID: EEq3.13]
(44) 
Iqη,μ,β{l}(t)[Iqζ,ν,δ{n}(t)Iqη,μ,β{mfg}(t)+Iqζ,ν,δ{nfg}(t)Iqη,μ,β{m}(t)] ≥Iqη,μ,β{l}(t)[Iqζ,ν,δ{ng}(t)Iqη,μ,β{mf}(t)+Iqζ,ν,δ{nf}(t)Iqη,μ,β{mg}(t)].
Now, by replacing u, v by l, n and u, v by l, m in (35), respectively, and then multiplying both sides of the resulting inequalities by Iqη,μ,β{m}(t) and Iqη,μ,β{n}(t), respectively, we get the following two inequalities:
[Formula ID: EEq3.14]
(45) 
Iqη,μ,β{m}(t)[Iqζ,ν,δ{n}(t)Iqη,μ,β{lfg}(t)+Iqζ,ν,δ{nfg}(t)Iqη,μ,β{l}(t)] ≥Iqη,μ,β{m}(t)[Iqζ,ν,δ{ng}(t)Iqη,μ,β{lf}(t)+Iqζ,ν,δ{nf}(t)Iqη,μ,β{lg}(t)],Iqη,μ,β{n}(t)[Iqζ,ν,δ{m}(t)Iqη,μ,β{lfg}(t)+Iqζ,ν,δ{mfg}(t)Iqη,μ,β{l}(t)] ≥Iqη,μ,β{n}(t)[Iqζ,ν,δ{mg}(t)Iqη,μ,β{lf}(t)+Iqζ,ν,δ{mf}(t)Iqη,μ,β{lg}(t)].
Finally, we find that the inequality (42) follows by adding the inequalities (44) and (45), side by side.

Remark 8 .

It may be noted that inequalities (37) and (42) in Theorems 6 and 7, respectively, are reversed if the functions are asynchronous on [0, ). The special case of (42) in Theorem 7 when β = δ, η = ζ, and μ = ν is easily seen to yield inequality (37) in Theorem 6.

Remark 9 .

We remark further that we can present a large number of special cases of our main inequalities in Theorems 6 and 7. Here, we give only two examples: setting β = 1 in (37) and β = δ = 1 in (42), we obtain interesting inequalities involving Erdélyi-Kober fractional integral operator.

Corollary 10 .

Let 0 < q < 1, let f and g be two continuous and synchronous functions on [0, ), and let l, m, n : [0, )→[0, ) be continuous functions. Then, the following inequality holds true:

[Formula ID: EEq3.16]
(46) 
2Iqη,μ{l}(t)[Iqη,μ{m}(t)Iqη,μ{nfg}(t)+Iqη,μ{n}(t)Iqη,μ{mfg}(t)]  +2Iqη,μ{m}(t)Iqη,μ{n}(t)Iqη,μ{lfg}(t) ≥Iqη,μ{l}(t)[Iqη,μ{mf}(t)Iqη,μ{ng}(t)+Iqη,μ{nf}(t)Iqη,μ{mg}(t)]  +Iqη,μ{m}(t)[Iqη,μ{lf}(t)Iqη,μ{ng}(t)+Iqη,μ{nf}(t)Iqη,μ{lg}(t)]  +Iqη,μ{n}(t)[Iqη,μ{lf}(t)Iqη,μ{mg}(t)+Iqη,μ{mf}(t)Iqη,μ{lg}(t)],
for all μ > 0 and ηC.

Corollary 11 .

Let 0 < q < 1, let f and g be two continuous and synchronous functions on [0, ), and let l, m, n : [0, )→[0, ) be continuous functions. Then, the following inequality holds true:

[Formula ID: EEq3.17]
(47) 
Iqη,μ{l}(t)[2Iqη,μ{m}(t)Iqζ,ν{nfg}(t)+Iqη,μ{n}(t)Iqζ,ν{mfg}(t)+Iqζ,ν{n}(t)Iqη,μ{mfg}(t)]  +Iqη,μ{lfg}(t)[Iqη,μ{m}(t)Iqζ,ν{n}(t)+Iqη,μ{n}(t)Iqζ,ν{m}(t)] ≥Iqη,μ{l}(t)[Iqη,μ{mf}(t)Iqζ,ν{ng}(t)+Iqη,μ{mg}(t)Iqζ,ν{nf}(t)]  +Iqη,μ{m}(t)[Iqη,μ{lf}(t)Iqζ,ν{ng}(t)+Iqη,μ{lg}(t)Iqζ,ν{nf}(t)]  +Iqη,μ{n}(t)[Iqη,μ{lf}(t)Iqγ,δ{mg}(t)+Iqη,μ{lg}(t)Iqζ,ν{mf}(t)],
for all μ, ν > 0 and η, ζC.

Remark 12 .

If we take η = 0 and β = 1 in Theorem 6 and η = ζ = 0 and β = δ = 1 in Theorem 7, then we obtain the known results due to Dahmani [21].


4. Inequalities Involving a Generalized Erdélyi-Kober Fractional q-Integral Operator for Bounded Functions

In this section we obtain some new inequalities involving Erdélyi-Kober fractional q-integral operator in the case where the functions are bounded by integrable functions and are not necessary increasing or decreasing as are the synchronous functions.

Theorem 13 .

Let 0 < q < 1, let f be an integrable function on [0, ), and let u, v : [0, )→[0, ) be continuous functions. Assume the following.

  •  (H1)   There exist two integrable functions φ1, φ2 on [0, ) such that
    [Formula ID: eq48]
    (48) 
    φ1(t)≤f(t)≤φ2(t), ∀t∈[0,∞).
Then, for t > 0, μ, β > 0, and ηC, we have
[Formula ID: EEq4.1]
(49) 
Iqη,μ,β{uφ2}(t)Iqη,μ,β{vf}(t)  +Iqη,μ,β{uf}(t)Iqη,μ,β{vφ1}(t) ≥Iqη,μ,β{uφ2}(t)Iqη,μ,β{vφ1}(t)  +Iqη,μ,β{uf}(t)Iqη,μ,β{vf}(t).

Proof

From (H1), for all τ ≥ 0 and ρ ≥ 0, we have

[Formula ID: eq50]
(50) 
(φ2(τ)−f(τ))(f(ρ)−φ1(ρ))≥0.
Therefore,
[Formula ID: EEq4.2]
(51) 
φ2(τ)f(ρ)+φ1(ρ)f(τ)  ≥φ1(ρ)φ2(τ)+f(τ)f(ρ).
Multiplying both sides of (51) by (βtβ(η+μ)q(μ))(tβτβq)(μ−1)τβ(η+1)−1u(τ), τ ∈ (0, t), and integrating both sides with respect to τ on (0, t), we obtain
[Formula ID: EEq4.3]
(52) 
Iqη,μ,β{uφ2}(t)f(ρ)+Iqη,μ,β{uf}(t)φ1(ρ)  ≥Iqη,μ,β{uφ2}(t)φ1(ρ)+Iqη,μ,β{uf}(t)f(ρ).
Multiplying both sides of (52) by (βtβ(η+μ)q(μ))(tβρβq)(μ−1)ρβ(η+1)−1v(ρ), ρ ∈ (0, t), and integrating both sides with respect to ρ on (0, t), we get inequality (49) as requested. This completes the proof.

As special cases of Theorems 13, we obtain the following results.

Corollary 14 .

Let 0 < q < 1, let f be an integrable function on [0, ) satisfying mf(t) ≤ M, for all t ∈ [0, ), let u, v : [0, )→[0, ) be continuous functions, and let  m, MR. Then, for t > 0, μ, β > 0, and ηC, we have

[Formula ID: eq53]
(53) 
MIqη,μ,β{u}(t)Iqη,μ,β{vf}(t)  +mIqη,μ,β{uf}(t)Iqη,μ,β{v}(t) ≥mMIqη,μ,β{u}(t)Iqη,μ,β{v}(t)  +Iqη,μ,β{uf}(t)Iqη,μ,β{vf}(t).

Corollary 15 .

Let 0 < q < 1, let f be an integrable function on [1, ), and let u, v : [0, )→[0, ) be continuous functions. Assume that there exists an integrable function φ(t) on [0, ) and a constant M > 0 such that

[Formula ID: eq54]
(54) 
φ(t)−M≤f(t)≤φ(t)+M,
for all t > 0, μ, β > 0, and ηC; we have
[Formula ID: eq55]
(55) 
Iqη,μ,β{uφ}(t)Iqη,μ,β{vf}(t)  +Iqη,μ,β{uf}(t)Iqη,μ,β{vφ}(t)  +MIqη,μ,β{u}(t)Iqη,μ,β{vf}(t)  +MIqη,μ,β{v}(t)Iqη,μ,β{uφ}(t)  +M2Iqη,μ,β{u}(t)Iqη,μ,β{v}(t) ≥Iqη,μ,β{uφ}(t)Iqη,μ,β{vφ}(t)  +Iqη,μ,β{uf}(t)Iqη,μ,β{vf}(t)  +MIqη,μ,β{u}(t)Iqη,μ,β{vφ}(t)  +MIqη,μ,β{uf}(t)Iqη,μ,β{v}(t).

Theorem 16 .

Let 0 < q < 1, let f be an integrable function on [0, ), let u, v : [0, )→[0, ) be continuous functions, and let θ1, θ2 > 0 satisfying 1/θ1 + 1/θ2 = 1. Suppose that (H1) holds. Then, for t > 0, μ, β > 0, and ηC, we have

[Formula ID: EEq4.4]
(56) 
1θ1Iqη,μ,β{v}(t)Iqη,μ,β{u(φ2−f)θ1}(t)  +1θ2Iqη,μ,β{u}(t)Iqη,μ,β{v(f−φ1)θ2}(t)  +Iqη,μ,β{uφ2}(t)Iqη,μ,β{vφ1}(t)  +Iqη,μ,β{uf}(t)Iqη,μ,β{vf}(t) ≥Iqη,μ,β{uφ2}(t)Iqη,μ,β{vf}(t)  +Iqη,μ,β{uf}(t)Iqη,μ,β{vφ1}(t).

Proof

According to the well-known Young inequality [3]

[Formula ID: eq57]
(57) 
1θ1xθ1+1θ2yθ2≥xy, ∀x,y≥0,θ1,θ2>0, 1θ1+1θ2=1,
and setting x = φ2(τ) − f(τ) and y = f(ρ) − φ1(ρ), τ, ρ ≥ 0, we have
[Formula ID: EEq4.5]
(58) 
1θ1(φ2(τ)−f(τ))θ1+1θ2(f(ρ)−φ1(ρ))θ2  ≥(φ2(τ)−f(τ))(f(ρ)−φ1(ρ)).
Multiplying both sides of (58) by
[Formula ID: eq59]
(59) 
β2t−2β(η+μ)Γq2(μ)(tβ−τβq)(μ−1)(tβ−ρβq)(μ−1)  ×(τρ)β(η+1)−1u(τ)v(ρ),
for τ, ρ ∈ (0, t), and integrating with respect to τ and ρ from 0 to t, we deduce the desired result in (56).

Corollary 17 .

Let 0 < q < 1, let f be an integrable function on [0, ) satisfying mf(t) ≤ M, for all t ∈ [0, ), let u, v : [0, )→[0, ) be continuous functions, and let m, MR. Then, for t > 0, μ, β > 0, and ηC, we have

[Formula ID: eq60]
(60) 
(m+M)2Iqη,μ,β{u}(t)Iqη,μ,β{v}(t)  +2Iqη,μ,β{uf}(t)Iqη,μ,β{vf}(t)  +Iqη,μ,β{vf2}(t)(Iqη,μ,β{u}(t)+Iqη,μ,β{v}(t)) ≥2(m+M)(Iqη,μ,β{uf}(t)Iqη,μ,β{v}(t)+ Iqη,μ,β{u}(t)Iqη,μ,β{vf}(t)).

Theorem 18 .

Let 0 < q < 1, let f be an integrable function on [0, ), let u, v : [0, )→[0, ) be continuous functions, and let θ1, θ2 > 0 satisfying θ1 + θ2 = 1. In addition, suppose that (H1) holds. Then, for t > 0, μ, β > 0, and ηC, we have

[Formula ID: EEq4.6]
(61) 
θ1Iqη,μ,β{uφ2}(t)Iqη,μ,β{v}(t)  +θ2Iqη,μ,β{u}(t)Iqη,μ,β{vf}(t) ≥θ1Iqη,μ,β{uf}(t)Iqη,μ,β{v}(t)  +θ2Iqη,μ,β{u}(t)Iqη,μ,β{vφ1}(t)  +Iqη,μ,β{u(φ2−f)θ1}(t)Iqη,μ,β{v(f−φ1)θ2}(t).

Proof

From the well-known Weighted AM-GM inequality [3]

[Formula ID: eq62]
(62) 
θ1x+θ2y≥xθ1yθ2, ∀x,y≥0,  θ1,θ2>0, θ1+θ2=1,
by setting x = φ2(τ) − f(τ) and y = f(ρ) − φ1(ρ), τ, ρ > 1, we have
[Formula ID: EEq4.7]
(63) 
θ1(φ2(τ)−f(τ))+θ2(f(ρ)−φ1(ρ))  ≥(φ2(τ)−f(τ))θ1(f(ρ)−φ1(ρ))θ2.
Multiplying both sides of (63) by
[Formula ID: eq64]
(64) 
β2t−2β(η+μ)Γq2(μ)(tβ−τβq)(μ−1)(tβ−ρβq)(μ−1)  × (τρ)β(η+1)−1u(τ)v(ρ),
for τ, ρ ∈ (0, t), and integrating with respect to τ and ρ from 0 to t, we deduce inequality (61).

Corollary 19 .

Let 0 < q < 1, let f be an integrable function on [0, ) satisfying mf(t) ≤ M, for all t ∈ [0, ), let u, v : [0, )→[0, ) be continuous functions, and let m, MR. Then, for t > 0, μ, β > 0, and ηC, we have

[Formula ID: eq65]
(65) 
(M−m)Iqη,μ,β{u}(t)Iqη,μ,β{v}(t)+Iqη,μ,β{u}(t)Iqη,μ,β{vf}(t) ≥Iqη,μ,β{uf}(t)Iqη,μ,β{v}(t)  +2Iqη,μ,β{uM−f}(t)Iqη,μ,β{vf−m}(t).

Lemma 20 (see [22]).

Assume that a ≥ 0, pq ≥ 0, and p ≠ 0. Then,

[Formula ID: eq66]
(66) 
aq/p≤(qpk(q−p)/pa+p−qpkq/p), for  any  k>0.

Theorem 21 .

Let 0 < q < 1, let f be an integrable function on [0, ), let u : [0, )→[0, ) be a continuous function, and let constants pq ≥ 0, p ≠ 0. In addition, assume that (H1) holds. Then, for any k > 0, t > 0, μ, β > 0, and ηC, the following two inequalities hold:

[Formula ID: EEq4.8]
(67) 
(i)  Iqη,μ,β{u(φ2−f)q/p}(t)+qpk(q−p)/pIqη,μ,β{uf}(t)  ≤qpk(q−p)/pIqη,μ,β{uφ2}(t)+p−qpkq/pIqη,μ,β{u}(t),
[Formula ID: EEq4.9]
(68) 
(ii)  Iqη,μ,β{u(f−φ1)q/p}(t)+qpk(q−p)/pIqη,μ,β{uφ1}(t)  ≤qpk(q−p)/pIqη,μ,β{uf}(t)+p−qpkq/pIqη,μ,β{u}(t).

Proof

By condition (H1) and Lemma 20, for pq ≥ 0, p ≠ 0, it follows that

[Formula ID: EEq4.10]
(69) 
(φ2(τ)−f(τ))q/p≤qpk(q−p)/p(φ2(τ)−f(τ))+p−qpkq/p,
for any k > 0. Multiplying both sides of (69) by (βtβ(η+μ)q(μ))(tβτβq)(μ−1)τβ(η+1)−1u(τ), τ ∈ (0, t), and integrating the resulting identity with respect to τ from 0 to t, one has inequality (i). Inequality (ii) is proved by setting a = f(τ) − φ1(τ) in Lemma 20.

Corollary 22 .

Let 0 < q < 1, let f be an integrable function on [0, ) satisfying mf(t) ≤ M, for all t ∈ [0, ), let u, v : [0, )→[0, ) be continuous functions, and let m, MR. Then, for t > 0, μ, β > 0, and ηC, we have

[Formula ID: eq70]
(70) 
(i)  2Iqη,μ,β{uM−f}(t)+Iqη,μ,β{uf}(t)  ≤(M+1)Iqη,μ,β{u}(t),(ii)  2Iqη,μ,β{uf−m}(t)+(m−1)Iqη,μ,β{u}(t)  ≤Iqη,μ,β{uf}(t).

Theorem 23 .

Let 0 < q < 1, let f and g be two integrable functions on [0, ), and let u, v : [0, )→[0, ) be continuous functions. Suppose that (H1) holds and moreover we assume the following.

  •    (H2) There exist ψ1 and ψ2 integrable functions on [0, ) such that
    [Formula ID: eq71]
    (71) 
    ψ1(t)≤g(t)≤ψ2(t) ∀t∈[0,∞).
Then, for t > 0, μ, β > 0, and ηC, the following inequalities hold:
[Formula ID: eq72]
(72) 
(i)  Iqη,μ,β{uφ2}(t)Iqη,μ,β{vg}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vψ1}(t)   ≥Iqη,μ,β{uφ2}(t)Iqη,μ,β{vψ1}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t),(ii)  Iqη,μ,β{uψ2}(t)Iqη,μ,β{vf}(t)    +Iqη,μ,β{ug}(t)Iqη,μ,β{vφ1}(t)   ≥Iqη,μ,β{uψ2}(t)Iqη,μ,β{vφ1}(t)    +Iqη,μ,β{ug}(t)Iqη,μ,β{vf}(t),(iii)  Iqη,μ,β{uφ2}(t)Iqη,μ,β{vψ2}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)   ≥Iqη,μ,β{uφ2}(t)Iqη,μ,β{vg}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vψ2}(t),(iv)  Iqη,μ,β{uφ1}(t)Iqη,μ,β{vψ1}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)   ≥Iqη,μ,β{uφ1}(t)Iqη,μ,β{vg}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vψ1}(t).

Proof

To prove (i), from (H1) and (H2), we have for t ∈ [0, ) that

[Formula ID: eq73]
(73) 
(φ2(τ)−f(τ))(g(ρ)−ψ1(ρ))≥0.
Therefore,
[Formula ID: EEq4.11]
(74) 
φ2(τ)g(ρ)+ψ1(ρ)f(τ)≥ψ1(ρ)φ2(τ) +f(τ)g(ρ).
Multiplying both sides of (74) by (βtβ(η+μ)q(μ))(tβτβq)(μ−1)τβ(η+1)−1u(τ), τ ∈ (0, t), and integrating both sides with respect to τ on (0, t), we obtain
[Formula ID: EEq4.12]
(75) 
g(ρ)Iqη,μ,β{uφ2}(t)+ψ1(ρ)Iqη,μ,β{uf}(t)  ≥ψ1(ρ)Iqη,μ,β{uφ2}(t)+g(ρ)Iqη,μ,β{uf}(t).
Multiplying both sides of (75) by (βtβ(η+μ)q(μ))(tβρβq)(μ−1)ρβ(η+1)−1v(ρ), ρ ∈ (0, t), and integrating both sides with respect to ρ on (0, t), we get the desired inequality (i).

To prove (ii)–(iv), we use the following inequalities:

[Formula ID: eq76]
(76) 
 (ii)  (ψ2(τ)−g(τ))(f(ρ)−φ1(ρ))≥0, (iii)  (φ2(τ)−f(τ))(g(ρ)−ψ2(ρ))≤0, (iv)  (φ1(τ)−f(τ))(g(ρ)−ψ1(ρ))≤0.

Theorem 24 .

Let f and g be two integrable functions on [0, ), let u, v : [0, )→[0, ) be continuous functions, and let θ1, θ2 > 0 satisfying 1/θ1 + 1/θ2 = 1. Suppose that (H1) and (H2) hold. Then, for t > 0, μ, β > 0, and ηC, the following inequalities hold:

[Formula ID: eq77]
(77) 
(i)  1θ1Iqη,μ,β{u(φ2−f)θ1}(t)Iqη,μ,β{v}(t)    +1θ2Iqη,μ,β{v(ψ2−g)θ2}(t)Iqη,μ,β{u}(t)    +Iqη,μ,β{uφ2}(t)Iqη,μ,β{vg}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vψ2}(t)   ≥Iqη,μ,β{uφ2}(t)Iqη,μ,β{vψ2}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t),(ii)  1θ1Iqη,μ,β{u(φ2−f)θ1}(t)Iqη,μ,β{v(ψ2−g)θ1}(t)    +1θ2Iqη,μ,β{u(ψ2−g)θ2}(t)Iqη,μ,β{v(φ2−f)θ2}(t)   ≥Iqη,μ,β{u(φ2−f)(ψ2−g)}(t)    ×Iqη,μ,β{v(ψ2−g)(φ2−f)}(t),(iii) 1θ1Iqη,μ,β{u(f−φ1)θ1}(t)Iqη,μ,β{v}(t)     +1θ2Iqη,μ,β{v(g−ψ1)θ2}(t)Iqη,μ,β{u}(t)    +Iqη,μ,β{uf}(t)Iqη,μ,β{vψ1}(t)    +Iqη,μ,β{uφ1}(t)Iqη,μ,β{vg}(t)   ≥Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)    +Iqη,μ,β{uφ1}(t)Iqη,μ,β{vψ1}(t),(iv)  1θ1Iqη,μ,β{u(f−φ1)θ1}(t)Iqη,μ,β{v(g−ψ1)θ1}(t)    +1θ2Iqη,μ,β{u(g−ψ1)θ2}(t)Iqη,μ,β{v(f−φ1)θ2}(t)   ≥Iqη,μ,β{u(f−φ1)(g−ψ1)}(t)    ×Iqη,μ,β{v(g−ψ1)(f−φ1)}(t).

Proof

The inequalities (i)–(iv) can be proved by choosing the parameters in the Young inequality [3]:

[Formula ID: eq78]
(78) 
(i)  x=φ2(τ)−f(τ),  y=ψ2(ρ)−g(ρ),(ii)  x=(φ2(τ)−f(τ))(ψ2(ρ)−g(ρ)),y=(ψ2(τ)−g(τ))(φ2(ρ)−f(ρ)),(iii)  x=f(τ)−φ1(τ),  y=g(ρ)−ψ1(ρ),(iv)  x=(f(τ)−φ1(τ))(g(ρ)−ψ1(ρ)),y=(g(τ)−ψ1(τ))(f(ρ)−φ1(ρ)).

Theorem 25 .

Let f and g be two integrable functions on [0, ), let u, v : [0, )→[0, ) be continuous functions, and let θ1, θ2 > 0 satisfying θ1 + θ2 = 1. Suppose that (H1) and (H2) hold. Then, for t > 0, μ, β > 0, and ηC, the following inequalities hold:

[Formula ID: eq79]
(79) 
(i)  θ1Iqη,μ,β{uφ2}(t)Iqη,μ,β{v}(t)    +θ2Iqη,μ,β{vψ2}(t)Iqη,μ,β{u}(t)   ≥θ1Iqη,μ,β{uf}(t)Iqη,μ,β{v}(t)    +θ2Iqη,μ,β{vg}(t)Iqη,μ,β{u}(t)    +Iqη,μ,β{u(φ2−f)θ1}(t)Iqη,μ,β{v(ψ2−g)θ2}(t),(ii)  θ1Iqη,μ,β{uφ2}(t)Iqη,μ,β{vψ2}(t)    +θ1Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)    +θ2Iqη,μ,β{uψ2}(t)Iqη,μ,β{vφ2}(t)    +θ2Iqη,μ,β{ug}(t)Iqη,μ,β{vf}(t)   ≥θ1Iqη,μ,β{uφ2}(t)Iqη,μ,β{vg}(t)    +θ1Iqη,μ,β{uf}(t)Iqη,μ,β{vψ2}(t)    +θ2Iqη,μ,β{uψ2}(t)Iqη,μ,β{vf}(t)    +θ2Iqη,μ,β{ug}(t)Iqη,μ,β{vφ2}(t)    +Iqη,μ,β{u(φ2−f)θ1(ψ2−g)θ2}(t)    ×Iqη,μ,β{v(ψ2−g)θ1(φ2−f)θ2}(t),(iii)  θ1Iqη,μ,β{uf}(t)Iqη,μ,β{v}(t)    +θ2Iqη,μ,β{vg}(t)Iqη,μ,β{u}(t)   ≥θ1Iqη,μ,β{uφ1}(t)Iqη,μ,β{v}(t)    +θ2Iqη,μ,β{vψ1}(t)Iqη,μ,β{u}(t)    +Iqη,μ,β{u(f−φ1)θ1}(t)Iqη,μ,β{v(g−ψ1)θ2}(t),(iv)  θ1Iqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)    +θ1Iqη,μ,β{uφ1}(t)Iqη,μ,β{vψ1}(t)    +θ2Iqη,μ,β{ug}(t)Iqη,μ,β{vf}(t)    +θ2Iqη,μ,β{uψ1}(t)Iqη,μ,β{vφ1}(t)   ≥θ1Iqη,μ,β{uf}(t)Iqη,μ,β{vψ1}(t)    +θ1Iqη,μ,β{uφ1}(t)Iqη,μ,β{vg}(t)    +θ2Iqη,μ,β{ug}(t)Iqη,μ,β{vφ1}(t)    +θ2Iqη,μ,β{uψ1}(t)Iqη,μ,β{vf}(t)    +Iqη,μ,β{u(f−φ1)θ1(g−ψ1)θ2}(t)    ×Iqη,μ,β{v(g−ψ1)θ1(f−φ1)θ2}(t).

Proof

The inequalities (i)–(iv) can be proved by choosing the parameters in the Weighted AM-GM [3]:

[Formula ID: eq80]
(80) 
(i)  x=φ2(τ)−f(τ),  y=ψ2(ρ)−g(ρ),(ii)  x=(φ2(τ)−f(τ))(ψ2(ρ)−g(ρ)),y=(ψ2(τ)−g(τ))(φ2(ρ)−f(ρ)),(iii)  x=f(τ)−φ1(τ),  y=g(ρ)−ψ1(ρ),(iv)  x=(f(τ)−φ1(τ))(g(ρ)−ψ1(ρ)),y=(g(τ)−ψ1(τ))(f(ρ)−φ1(ρ)).

Theorem 26 .

Let f and g be two integrable functions on [0, ), let u, v : [0, )→[0, ) be continuous functions, and let constants pq ≥ 0, p ≠ 0. Assume that (H1) and (H2) hold. Then, for any k > 0, t > 0, μ, β > 0, and ηC, the following inequalities hold:

[Formula ID: eq81]
(81) 
(i)  Iqη,μ,β{u(φ2−f)q/p(ψ2−g)q/p}(t)   +qpk(q−p)/pIqη,μ,β{uφ2g}(t)   +qpk(q−p)/pIqη,μ,β{ufψ2}(t)  ≤qpk(q−p)/pIqη,μ,β{uφ2ψ2}(t)   +qpk(q−p)/pIqη,μ,β{ufg}(t)   +p−qpkq/pIqη,μ,β{u}(t),(ii)  Iqη,μ,β{u(φ2−f)q/p}(t)Iqη,μ,β{v(ψ2−g)q/p}(t)   +qpk(q−p)/pIqη,μ,β{uφ2}(t)Iqη,μ,β{vg}(t)   +qpk(q−p)/pIqη,μ,β{uf}(t)Iqη,μ,β{vψ2}(t)  ≤qpk(q−p)/pIqη,μ,β{uφ2}(t)Iqη,μ,β{vψ2}(t)   +qpk(q−p)/pIqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)g(t)   +p−qpkq/pIqη,μ,β{u}(t)Iqη,μ,β{v}(t),(iii)  Iqη,μ,β{u(f−φ1)q/p(g−ψ1)q/p}(t)   +qpk(q−p)/pIqη,μ,β{uψ1f}(t)   +qpk(q−p)/pIqη,μ,β{uφ1g}(t)  ≤qpk(q−p)/pIqη,μ,β{ufg}(t)   +qpk(q−p)/pIqη,μ,β{uφ1ψ1}(t)   +p−qpkq/pIqη,μ,β{u}(t),(iv)  Iqη,μ,β{u(f−φ1)q/p}(t)Iqη,μ,β{v(g−ψ1)q/p}(t)   +qpk(q−p)/pIqη,μ,β{uf}(t)Iqη,μ,β{vψ1}(t)   +qpk(q−p)/pIqη,μ,β{uφ1}(t)Iqη,μ,β{vg}(t)  ≤qpk(q−p)/pIqη,μ,β{uf}(t)Iqη,μ,β{vg}(t)   +qpk(q−p)/pIqη,μ,β{uφ1}(t)Iqη,μ,β{vψ1}(t)   +p−qpkq/pIqη,μ,β{u}(t)Iqη,μ,β{v}(t).

Proof

The inequalities (i)–(iv) can be proved by choosing the parameters in Lemma 20:

[Formula ID: eq82]
(82) 
 (i)  a=(φ2(τ)−f(τ))(ψ2(τ)−g(τ)), (ii)  a=(φ2(τ)−f(τ))(ψ2(ρ)−g(ρ)), (iii)  a=(f(τ)−φ1(τ))(g(τ)−ψ1(τ)), (iv)  a=(f(τ)−φ1(τ))(g(ρ)−ψ1(ρ)).


5. Concluding Remark

We conclude our present investigation with the remark that the results derived in this paper are general in character and give some contributions to the theory of q-integral inequalities and fractional calculus. Moreover, they are expected to find some applications for establishing uniqueness of solutions in fractional boundary value problems and in fractional partial differential equations. In last, use of the generalized Erdélyi-Kober fractional q-integral operator due to Gaulué is the advantage of our results because after setting suitable parameter values in our main results, we get known results established by number of authors.


Acknowledgments

The research of J. Tariboon is supported by King Mongkut's University of Technology North Bangkok, Thailand. Sotiris K. Ntouyas is a Member of Nonlinear Analysis and Applied Mathematics (NAAM) Research Group at King Abdulaziz University, Jeddah, Saudi Arabia.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.


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