Fractional Integral Inequalities - Universidad Rafael Urdaneta

Revista Tecnocientífica URU Universidad Rafael Urdaneta Facultad de Ingeniería Nº 6 Enero - Junio 2014 Depósito legal: ppi 201402ZU4464 ISSN: 2343 - 6360

Desigualdades integrales fraccionales y sus q-análogos Sunil Dutt Purohit1, Faruk Uçar2 and R.K. Yadavc3 Dedicated to Professor S.L. Kalla 1

Department of Basic Sciences (Mathematics), College of Technology and Engineering, M.P. University of Agriculture and Technology, Udaipur-313001, Rajasthan, India. [email protected]

2

Department of Mathematics, University of Marmara, TR-34722, Kad köy, Istanbul, Turkey. [email protected] 3

Department of Mathematics and Statistics, J. N. Vyas University. Jodhpur-342005, India. [email protected] Recibido: 24-09-2013 Aceptado: 29-01-2014

Resumen El objeto de este trabajo es establecer algunas desigualdades que envuelven operadores integrales de Saigo. Se usa el calculo q-fraccional para obtener varios resultados en la teoría de las desigualdades q-integrales. Los resultados dados anteriormente por Purohit y Raina (2013) y Sulaiman (2011) son casos especiales de los obtenidos en este trabajo. Palabras clave: Desigualdades integrales, operadores integrales fraccionales, operadores q-integrales fraccionales.

On fractional integral inequalities and their q-analogues Abstract The aim of this paper is to establish some integral inequalities involving Saigo fractional integral operators. We then use fractional q-calculus for yielding various results in the theory of q-integral inequalities. The results given earlier by Purohit and Raina (2013) and Sulaiman (2011) follow as special cases of our findings. Key words: Integral inequalities, fractional integral operators, fractional q-integral operators.

53

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Desigualdades integrales fraccionales y sus q-análogos Revista Tecnocientífica URU, Nº 6 Enero - Junio 2014 (53 - 66)

Introduction Fractional integral inequalities have many applications, the most useful ones are in establishing uniqueness of solutions in fractional boundary value problems, and in fractional partial differential equations. Further, they also provide upper and lower bounds to the solutions of the above equations. For detailed applications, one may refer to the book [1], and the recent papers [2]-[5] on the subject. I n a recent paper, Purohit and Raina [6] investigated certain Chebyshev type ([7]) integral inequalities involving the Saigo fractional integral operators, and also established the q-extensions of the main results. The aim of this paper is to establish several new integral inequalities for synchronous functions that are related to the Chebyshev functional using the Saigo fractional integral. q-Extensions of the main results are also established. Some of the results due to Purohit and Raina [6] and Sulaiman [8] follows as special cases of our results. Following definitions will be needed in the sequel. Definition 1. Two functions f and g are said to be synchronous on [a, b] , if

{( f (x) – f (y))(g(x) – g(y))} ≥ 0,

(1)

for any x, y Î [a,b]. Definition 2. A real-valued function f (t) (t > 0) is said to be in the space (C μ (μ Î R), if there exists a real number p > μ such that f (t) = t p f(t); where f(t) Î C (0, ∞). Definition 3. Let α > 0, β,η Î R, then the Saigo fractional integral I lued continuous function f (t) is defined by ([9] see also [10, p. 19], [11]):

I 0a, ,t b ,η { f (t )} =

α,β,η of order α for a real-va-

t −a − b t t  (t − t )a −1 2 F1  a + b ,−η ; a ;1 ;1 −  f (t )dt , ∫ t Γ(a ) 0 

(2)

where, the function 2 F1 (−) in the right-hand side of (2) is the Gaussian hypergeometric function defined by 2 F1 (a, b; c; t ) =

(a ) n (b) n t n ∑ (c) n! , n n=0 ∞

(3)

and (t) (a) n is the Pochhammer symbol

(a)n = a(a+1)... (a+n – 1), (a)0 = 1.

The integral operator (2) includes both the Riemann-Liouville and the Erd e′ lyi-Kober fractional integral operators given by the following relationships:

1 t Ra { f (t )} = I 0a, ,t−a ,η { f (t )} = (t − t )a −1 f (t )dt ∫ 0 Γ(a )

and

I

a ,η

a ,0,η

{ f (t )} = I 0, t

t −a −η t { f (t )} = (t − t )a −1 t η f (t )dt ∫ 0 Γ(a )

(a > 00))

(a > 0, ,ηη Î∈R R).).

(4)

(5)

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Sunil Dutt Purohit et al. Revista Tecnocientífica URU, Nº 6 Enero - Junio 2014 (53 - 66)

For f (t ) = t µ in (2), we get the known result [9]:

{ }

aα,, β b ,,η η µ I 0,t t = 0, t

Γ(μ Γ( µ +1)Γ(μ + 1) Γ( µ++11–− βb ++η) η ) μµ –− bβ t ,, Γ(μ α +η η)) ( µ +1 Γ + 1 –− β)Γ(μ b )Γ( µ++11 + a

(6)

(α > 0,min(μ, μ – β + η) > – 1,t > 0) which shall be used in the sequel.

Fractional Integral Inequalities The following theorems involving Saigo integral inequalities for the synchronous functions will be established. Theorem 1. Let f and g be two synchronous functions on [0, ∞) , h > 0 , then for all

t > 0, α > max{0,–β}, β < 1, β – 1 < η < 0). Γ(1 Γ (1 −– bβ + ηη))

a a,,βbb,η ,,ηη a ,ηη a a,,βbb,η ,,ηη α, α,,,bβb,,η α, { f (t ) g (t )h(t )} ≥ I 0,a0,t { f (t )} I 0,t {g (t )h(t )} + II0,t 0,0, tt 0, tt 0,0, tt b b β Γ(1 Γ (1 −– bβ)Γ(1 )Γ(1 + aα + ηη)) tt

α, β ,η {g(t)} I α, β ,η {f (t) h (t)} –I α, β ,η {h(t)}I α, β ,η {f (t) g (t)}. (7) I 0,t 0,t 0,t 0,t

Proof: Using Definition 1 and h > 0 , for all t , ρ ≥ 0 , we have

{( f (t ) − f ( ρ ) )(g (t ) − g ( ρ ) )(h(t ) + h( ρ ) )} ≥ 0,

(8)

which implies that

f (t ) g (t )h(t ) + f ( ρ ) g ( ρ )h( ρ ) ≥ f (t ) g ( ρ )h( ρ ) + f ( ρ ) g (t )h(t ) + g (t ) f ( ρ )h( ρ ) + g ( ρ ) f (t )h(t ) − h(t ) f ( ρ ) g ( ρ ) − h( ρ ) f (t ) g (t ).

(9)

Consider

tt−−aa −−bb (tt −−tt )aa −−11 tt   α;1 F F((tt,,tt )) == F11a a ++ bb,,−−ηη;;aa ;1;1 −−  22F ΓΓ(a a) tt  

((tt ∈ ∈((0,t);t 0,0, tt);;tt > 00) )0)

a ++ bβ)(– 1 (t − t )aa −−11 ((α )( −ηη)) (t − t )aa = + + Γ(α Γ(a ) t aa ++bb Γ(a ++ 1)1) t aa ++bb ++11

η)(– η + 1)1) (t − t )a +1 ((α a + β)(α b )( a++βb++1)(– 1)(−η )( −η + . Γ (a ++ 2) 2) Γ(α ta + b + 2

(10)

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Since each term of the above series is positive in view of the conditions stated with Theorem 1, we observe that the function F (t ,t ) remains positive, for all τ Î (0,t) (t > 0). Multiplying both sides of (9) by F (t ,t ) (defined above by (10)) and integrating with respect to t from 0 to t , and using (2), we get a ,η a ,η a ,η α,, βb ,η α,, βb ,η α,, βb ,η I 0,t { f (t ) g (t )h(t )} + f ( ρ ) g ( ρ )h( ρ ) I 0,t {1} ≥ g ( ρ )h( ρ ) I 0,t { f (t )} + 0, t 0, t 0, t

aa ,ηη aα, a,,β bb,,η ,ηη aa ,ηη α,,,bβb,,η α,,,bβb,,η {gg((tt))hh((tt))} ++ ff ((ρρ))hh((ρρ)) II0,t {gg((tt))} ++ gg((ρρ)) II0,0,t { ff ((tt))hh((tt))} −− ff ((ρρ)) II00,t ,0, tt 0,0, tt 0, tt aα,, b η aα,, b η β ,,η β ,,η {h(t )} − h( ρ ) I 00,t { f (t ) g (t )}. f ( ρ ) g ( ρ ) I 00,t , t , t

(11)

Next, multiplying both sides of (11) by F (t, ρ) (ρ Î (0,t), (t > 0), where F (t , ρ ) is given by (10), and integrating with respect to ρ from 0 to t , and using formula (6), we arrive at the desired result (7). Theorem 2. Let f and g be two synchronous functions on [0, ∞) , and h > 0 , then

Γ(1 – bb β ++ηη η))) Γ Γ(1(1 −−

Γ(1 – bβ)Γ(1 α ++ηη η)))tttβbb Γ Γ(1(1 −− b))ΓΓ(1(1 ++aa

gg ,,δδ,ζ ,,ζζ γ, II0,t + 0,0, tt { ff ((tt))gg((tt))hh((tt))} +

Γ(1 Γ Γ(1(1 −−–δδδ ++ ζζζ)))

ηη α,,,bbβ,,,η { ff ((tt))gg((tt))hh((tt))} ≥≥ II0,a0,a0,t tt δ δ Γ(1 – δδδ)Γ(1 Γ Γ(1(1 −− ))ΓΓ(1(1 ++ ggγ ++ ζζζ))) tt

α, β ,η {f (t)} I γ, δ ,ζ {g (t) h (t)}+ I α, β ,η {g (t) h (t)}I γ, δ ,ζ {f (t)} + I α, β ,η{g (t)}I γ, δ ,ζ {f (t) h (t)}+ I 0,t 0,t 0,t 0,t 0,t 0,t α, β ,η {f (t) h (t)} I γ, δ ,ζ {g (t)}– I α, β ,η {h(t)} I γ, δ ,ζ {f (t) g(t)}– I α, β ,η {f (t) g(t)} I γ, δ ,ζ {h(t)}, I 0,t 0,t 0,t 0,t 0,t 0,t (12) for all t > 0, α > max{0,– β}, γ > max{0,– δ}, β, δ < 1, β – 1 < η < 0, δ – 1 < ζ < 0. Proof: To prove the above theorem, we start with the inequality (11). On multiplying both sides of (11) by

t − g − δ (t − ρ )g −1 ρ  2 F1  g + δ ,−ζ ; g ;1 −  Γ(g ) t  

(ρ Î (0, t );tt > 0), ∈ (0,t);

and taking integration with respect to ρ from 0 to t , we get

γ, δ ,ζ {1}I α, β ,η {f (t) g (t) h(t)}+ I α, β ,η{1} I γ, δ ,ζ {f (t) g (t) h (t)}≥ I 0,t 0,t 0,t 0,t

α, β ,η {f (t)} I γ, δ ,ζ {g(t) h (t)}+ I α, β ,η {g(t) h (t)}I γ, δ ,ζ {f (t)}+ I α, β ,η {g(t)} I γ, δ ,ζ {f (t) h (t)}+ I 0,t 0,t 0,t 0,t 0,t 0,t α, β ,η {f (t) h (t)} I γ, δ ,ζ {g (t)}– I α, β ,η {h(t)} I γ, δ ,ζ {f (t) g (t)}– I α, β ,η {f (t) g (t)} I γ, δ ,ζ {h(t)}, I 0,t 0,t 0,t 0,t 0,t 0,t which on using (6) readily yields the desired result (12).

57


Remark 1. It may be noted that the inequalities (7) and (12) are reversed if the functions are asynchronous on [0, ∞), i.e.

{( f ( x) − f ( y ) )(g ( x) − g ( y) )} ≤ 0, 0,

(13)

for any x, y ∈ [0, ∞) . Remark 2. For g = a , δ = b , ζ = η , Theorem 2 immediately reduces to Theorem 1. Theorem 3. Let f , g and h be three monotonic functions on [0, ∞) satisfying the inequality

{( f (t ) − f ( ρ ) )(g (t ) − g ( ρ ) )(h(t ) − h( ρ ) )} ≥ 0, 0,

(14)

then for all , t > 0, α > max{0,– β}, γ > max{0,– δ}, β, δ < 1, β – 1 < η < 0, δ – 1 < ζ < 0.

Γ(1 – β + η)

Γ(1 – β)Γ(1 + α + η) t β

γ, δ ,ζ {f (t) g (t) h (t)} I 0,t

Γ(1 – δ + ζ)

α, β ,η{f (t) g (t) h (t)}≥ I 0,t δ Γ(1 – δ)Γ(1 + γ + ζ) t

α, β ,η{g (t) h (t)}I γ, δ ,ζ{f (t)}– I α, β ,η {f (t)}I γ, δ a,ζ{g(t) h (t)}+ I α, β ,η{f (t) h (t)} I γ, δ ,ζ {g (t)}– I 0,t 0,t 0,t 0,t 0,t 0,t α, β ,η {g (t)}I γ, δ ,ζ {f (t) h (t)}– I α, β ,η {h (t)}I γ, δ ,ζ {f (t) g (t)}+ I α, β ,η {f (t) g (t)}I γ, δ ,ζ {h(t)}. I 0,t 0,t 0,t 0,t 0,t 0,t

(15)

Proof: By applying the similar procedure as of Theorem 1 and 2, one can easily establish the above theorem. Therefore, we omit the details of the proof of this theorem. Observe that, if we set b = 0 (and δ = 0 additionally for Theorem 2), and make use of the relation (5), Theorems 1 to 3 respectively yield the following integral inequalities involving the Erd e′ lyi-Kober type fractional integral operator defined by (5): Corollary 1. Let f and g be two synchronous functions on [0, ∞) , and h > 0 , then

Γ(1 + η) I α, η {f (t) g (t) h (t)}≥ I α, η {f (t)} I α, η {g (t) h (t)}+ I α, η{g (t)}I α, η {f (t) h (t)} Γ(1 + α + η)

− I a ,η {h(t )} I a ,η { f (t ) g (t )},

(16)

for all t > 0, α > 0,– 1 < η < 0. Corollary 2. Let f and g be two synchronous functions on [0, ∞) , and h > 0 , then for all

t > 0, α, γ > 0,–1 < max (η, ζ ) < 0,

Γ(1 + η) γ, ζ Γ(1 + ζ) I {f (t) g (t) h (t)}+ I α, η {f (t) g (t) h (t)}≥ Γ(1 + α + η) Γ(1+ γ + ζ) I a ,η { f (t )} I g ,ζ {g (t )h(t )}+ I a ,η {g (t )h(t )} I g ,ζ { f (t )}+ I a ,η {g (t )} I g ,ζ { f (t )h(t )}+

58


I

a ,η

{ f (t )h(t )} I g ,ζ {g (t )} − I a ,η {h(t )} I g ,ζ { f (t ) g (t )} − I a ,η { f (t ) g (t )} I g ,ζ {h(t )}.

(17)

Corollary 3. Let f , g and h be three monotonic functions on [0, ∞) satisfying the inequality (14) then for all t > 0, α, γ > 0,–1 < max (η, ζ ) < 0,

Γ(1 + η) γ, ζ Γ(1 + ζ) I α, η {f (t) g (t) h (t)}≥ I {f (t) g (t) h (t)}– Γ(1 + α + η) Γ(1+ γ + ζ) I a ,η {g (t )h(t )} I g ,ζ { f (t )}− I a ,η { f (t )} I g ,ζ {g (t )h(t )}+ I a ,η { f (t )h(t )} I g ,ζ {g (t )} − I a ,η {g (t )} I g ,ζ { f (t )h(t )} − I a ,η {h(t )} I g ,ζ { f (t ) g (t )} + I a ,η { f (t ) g (t )} I g ,ζ {h(t )}. (18) Again, if we replace b by − a and δ by − g in Theorems 2 and 3, and make use of the relation (4), we obtain known results due to Sulaiman [8, pp. 24-25, Theorems 2.1 to 2.2].

q-Extensions of Main Results In this section, we establish q-extensions of the results derived in the previous section. We begin with the mathematical preliminaries of q-series and q-calculus. For more details of q-calculus and fractional q-calculus one can refer to [12] and [13]. The q-shifted factorial is defined for a ,q∈ C as a product of n factors by

1 ;;  ((a a;;qq))nn ==  n n − − 1 1 ... –a (1 −−–a aα)(1 ))(1(1 −− aαq) qq)) (1 (1(1 −−a aqq )) ;; (1(1

nn == 00 nn ∈ N,, ∈N

(19)

and in terms of the basic analogue of the gamma function

a

(q ; q) n =

Γq (a + n)(1 − q ) n

(n > 0), 0) ,

Γq (a )

(20)

where the q-gamma function is defined by ([12, p. 16, eqn. (1.10.1)])

Γq (t ) =

(q; q ) ∞ (1 − q )1− t (q t ; q) ∞

(0(0 < q 0, 0, 0 < qq 0 and b ,η ∈ R , a basic analogue of the Saigo’s fractional integral operator ([17, p. 172, eqn. (2.1)]) is given for t/t < 1 by

I qa , b ,η { f (t )} =

t − b −1 t (qt/t ; q )a −1 Γq (a ) ∫0

60


××

((qqaa ++bb ;;qq))mm((qq−−ηη ;;qq))mm ((ηη−−bb))mm  1)/2 tt qq (((–1) −−1)1) mm qq−−mm((mm−−1)/2 −−11 ff ((tt )) ddqqtt,,   ∑  tt mm ((qqaa ;;qq))mm((qq;;qq))mm m m==00 ∞ ∞

(31)

which in view of (27), can be written as (see [17, p. 173, eqn. (2.5)]):

I qaq ,,b ,,η { f (t )} = t −−b (1 − q)a

−−η (qa ++b ; q) m +1) 1)1) m mm (η −–−bβ ++ m (q ; q) m m q ((η ∑ a (q ; q) m m m==00 m ( q; q ) m m ∞ ∞

m; q) (q a ++m kk m) . ×∑ q f ((tq tq kk ++m ( q ; q ) kk kk ==00 (32) ∞ ∞

q kk

In the sequel, we shall be using the following image formula ([17, p. 173, eqn. (2.11)]):

{ }

I qa , b ,η t µ =

Γq ((μ µ ++ 1)1)Γ Γq ((μ µ + 11−–bβ ++ηη)) Γ Γq ( µ + 1 − b )Γq ( µ + 1 + a + η )

t µ −b ,

(33)

(α > 0,0 < q < 1, min(μ, μ – β + η ) > –1, t > 0). Now, we shall establish new q-integral inequalities for the synchronous functions involving the fractional q-integral operators, which can be treated as the q-analogues of the inequalities (7), (12) and (15). Theorem 4. Let f and g be two synchronous functions, and h(t ) > 0 on T , then

ΓΓq (1 (1 −– bβ + ηη))

I qa , b ,η { f (t ) g (t )h(t )} ≥ I qa , b ,η { f (t )} I qa , b ,η {g (t )h(t )} + b β ΓΓqq (1 (1 −– bβ)Γ )Γq (1 (1 ++aα ++ηη)) tt

I qa , b ,η {g (t )} I qa , b ,η { f (t )h(t )} − I qa , b ,η {h(t )} I qa , b ,η { f (t ) g (t )},

(34)

where t > 0,0 < q < 1, α > max {0,–β}, β < 1, β – 1 < η < 0). Proof: By the hypothesis, the functions f and g are synchronous functions on T for all τ, ρ ≥ 0, and h(t ) > 0 therefore the inequality (9) is satisfied, that is

f (t ) g (t )h(t ) + f ( ρ ) g ( ρ )h( ρ ) ≥ f (t ) g ( ρ )h( ρ ) + f ( ρ ) g (t )h(t ) + g (t ) f ( ρ )h( ρ )

+ g ( ρ ) f (t )h(t ) − h(t ) f ( ρ ) g ( ρ ) − h( ρ ) f (t ) g (t ).

Since, τ Î (0,t) (t > 0), 0 < q < 1, then tq k + m Î (0, t) for k, m Î N, therefore, on replacing τ by

tqk + m in the above inequality we get

f (tq k + m)g(tq k + m)h(tq k + m) + f (ρ) g (ρ) h (ρ) ≥ f (tq k + m) g (ρ) h (ρ) +

61


f (ρ)g(tq k+m)h(tq k+m) + g(tq k+m) f (ρ)h(ρ) + g(ρ) f (tq k+m)h(tq k+m) – h(tq k+m) f (ρ)g(ρ) – h(ρ) f (tq k+m)g(tq k+m).

(35)

Consider

H(t , q) = t − b (1 − q)a

(q a + b ; q) m (q −η ; q ) m (q a + m ; q ) k (η +1) m ++kk ( k ,m m∈ ). q (η −– bβ+1)m (k, ÎN N). a ( q ; q ) ( q ; q ) m ( q; q ) m k

(36)

Evidently, under the conditions stated with Theorem 4, we observe that the function H(t , q ) is positive for all values of k , m ∈ N . Therefore, on multiplying both sides of (35) by H(t , q ) and taking summations between the limits k = 0 to ∞, we get ∞

∞

∞

∞

m k +m t , qH) ((t,q) ft ,(tq )kff+(tq )kg+(m tq)g(tq )kh+()h(tq tm q )kh+(k+m ) gf ( ρ ) gh( ρ ) h∑ t , qH) (≥t , q) ≥ (m tq k+m )kg+(m tqk+m tq ) +) +f )(fρ+(ρ)g(ρ)h(ρ) ( ρH) (∑ ∑ H(∑

k =0 k =0

k =0 k =0

∞∞

∞

∞

k =0

k =0

k++mm k f+ (m)h(tq k)+ ) )+g (fftq(ρ) )ρ+))+∑ H(t , q) ≥ t, ,qq))ff(((tq ρ))hg(tq( ρk+m h(m gg(ρ)h(ρ) ( ρ ) h( ρ ) ∑ tqtq kk+m ( ρk +) m t , kq+) gm()tq+k+m ∑HH((t(t,q) ∑)hH((tq(t,q)g(tq kk==00

∞

∞∞

∞

k =0

k =k 0= 0

k =0

k+ +mm m k +m k +m g ( ρ )h( ρ ) ∑ H(t , q ) f (tq k +gm(fρ)(ρ)h(ρ) + )h(f ρ( ρ) ∑ ) ∑HH (t(,(t,q)g(tq tq, )qf) g(tq(tqkk+m ) )+h(g(ρ) ftq( ρk )+ ∑ )H +((t,q) t , q ) gf (tq (tq k+m )h(tqk+m )h(tq ) –) +

∞

k +m

∞∞

k +k+m km +m

g ( ρ )h( ρ ) ∑ H(t , q ) f (tq g ( ρf) (ρ)g(ρ) )+h(fρ()ρ∑ )∑ H(Ht ,(q(t,q)h(tq t ,)qf)(gtq(tq k =0

obtain

k =k0= 0

.

∞ k +m

k +m k +m ))+)–hf(h(ρ) tq( ρ ) ∑)H+((t,q) t , q ) gf (tq (tq k+m )h(tqk+m )g(tq ). ) + k =0

Now, on again taking summation from m = 0 to ∞, and then making use of the definition (32), we

I qa , b ,η { f (t ) g (t )h(t )} + f ( ρ ) g ( ρ )h( ρ ) I qa , b ,η {1} ≥ g ( ρ )h( ρ ) I qa , b ,η { f (t )} + f ( ρ ) I qa , b ,η {g (t )h(t )} + f ( ρ )h( ρ ) I qa , b ,η {g (t )} + g ( ρ ) I qa , b ,η { f (t )h(t )} − f ( ρ ) g ( ρ ) I qa ,b ,η {h(t )} − h( ρ ) I qa ,b ,η { f (t ) g (t )}.

(37)

Next, in the above inequality on replacing ρ by tqk+m , multiplying both sides of by H(t , q ) , taking summations between the limits k = 0 to ∞, and m = 0 to ∞, and then making use of the definitions (32) and (33), we arrive at the desired inequality (34). Theorem 5. Let f and g be two synchronous functions on T , and h > 0 , then for all t > 0,0 < q < 1, α > max{0,–β}, γ > max{0,–δ}, β, δ < 1, β –1< η < 0, δ –1 < ξ < 0,

62


Γq (1 – β + η)

Γq (1 – β)Γq (1+ α+η) t β

I qγ, δ ,ζ {f (t) g (t) h (t)}+

Γq (1 – δ + ζ)

I qα, β ,η{f (t) g (t) h (t)}≥ δ Γq (1 – δ)Γq (1+γ+ζ) t

I qa , b ,η { f (t )} I qg ,δ ,ζ {g (t )h(t )}+ I qa , b ,η {g (t )h(t )} I qg ,δ ,ζ { f (t )}+ I qa , b ,η {g (t )} I qg ,δ ,ζ { f (t )h(t )}+ a , b ,η { f (t )h(t )} I qg ,δ ,ζ {g (t )} − I qa , b ,η {h(t )} I qg ,δ ,ζ { f (t ) g (t )} − I qa , b ,η { f (t ) g (t )} I qg ,δ ,ζ {h(t )}. Iq

(38)

Proof: To prove the above theorem, we start with the inequality (37). On replacing ρ by tqk+m and multiplying both sides by a positive function F(t , q ) , given by

F(t , q ) = t

−δ

(1 − q )

g (q

g +δ

; q ) m (q −ζ ; q ) m (q g + m ; q ) k (ζ +1) m + + kk (k, (k ,m m∈ ), q (ζ –− δδ+1)m ÎN N). g (q; q ) k (q ; q ) m (q; q ) m

(39)

taking summations between the limits k = 0 to ∞ and m = 0 to ∞, and then making use of the definition (32), then the inequality (37) leads to

I qg ,δ ,ζ {1}I qa , b ,η { f (t ) g (t )h(t )} + I qa , b ,η {1}I qg ,δ ,ζ { f (t ) g (t )h(t )} ≥ I qa , b ,η { f (t )} I qg ,δ ,ζ {g (t )h(t )} + I qa , b ,η {g (t )h(t )} I qg ,δ ,ζ { f (t )} + I qa , b ,η {g (t )} I qg ,δ ,ζ { f (t )h(t )} +

I qa , b ,η { f (t )h(t )} I qg ,δ ,ζ {g (t )} − I qa , b ,η {h(t )} I qg ,δ ,ζ { f (t ) g (t )} − I qa , b ,η { f (t ) g (t )} I qg ,δ ,ζ {h(t )}, (40)

which yields the desired result by taking (33) into account. Remark 3. The inequalities (34) and (38) are reversed if the functions are asynchronous on T. Remark 4. Again, when γ = α, δ = β, ζ = η, then Theorem 5 leads to Theorem 4. Theorem 6. Let f , g and h be three monotonic functions on [0, ∞) satisfying the inequality (14), then for all t > 0,0 < q < 1, a > max{0,–b}, γ > max{0,–δ}, β,δ < 1, β –1 < η < 0,δ –1 0 , then

1 ++ηη)) Γq ((1 (1 ++aα + ηη)) ΓΓq (1

I qa ,η { f (t ) g (t )h(t )} ≥ I qa ,η { f (t )} I qa ,η {g (t )h(t )} + I qa ,η {g (t )} I qa ,η { f (t )h(t )} − I qa ,η {h(t )} I qa ,η { f (t ) g (t )},

(43)

for all t > 0,0 < q < 1, α > 0 and –1 < η < 0). Corollary 5. Let f and g be two synchronous functions on T , and h > 0 , then for t > 0, 0 < q < 1, α, γ > 0, such that –1 < max(η, ζ ) < 0),

) Γq (1(1++ηη) Γq (1(1 ++aα ++ηη))

I qg ,ζ { f (t ) g (t )h(t )} +

Γ q (1(1++ζζ)) ΓΓqq (1 (1 ++gγ ++ ζζ ))

I qa ,η { f (t ) g (t )h(t )} ≥

I qa ,η { f (t )} I qg ,ζ {g (t )h(t )} + I qa ,η {g (t )h(t )} I qg ,ζ { f (t )} + I qa ,η {g (t )} I qg ,ζ { f (t )h(t )} + I qa ,η { f (t )h(t )} I qg ,ζ {g (t )} − I qa ,η {h(t )} I qg ,ζ { f (t ) g (t )} − I qa ,η { f (t ) g (t )} I qg ,ζ {h(t )}.

(44)

Corollary 6. Let f , g and h be three monotonic functions on [0, ∞) satisfying the inequality (14), then for all t > 0, α, γ, > 0,–1 < max(η, ζ ) < 0,0 < q < 1

) Γq (1(1++ηη) Γq (1(1 ++aα ++ηη))

I qg ,ζ { f (t ) g (t )h(t )} +

Γq (1(1++ζζ)) ΓΓqq (1 (1 ++gγ ++ ζζ ))

I qa ,η { f (t ) g (t )h(t )} ≥

I qa ,η {g (t )h(t )} I qg ,ζ { f (t )} − I qa ,η { f (t )} I qg ,ζ {g (t )h(t )} + I qa ,η { f (t )h(t )} I qg ,ζ {g (t )} − I qa ,η {g (t )} I qg ,ζ { f (t )h(t )} − I qa ,η {h(t )} I qg ,ζ { f (t ) g (t )} + I qa ,η { f (t ) g (t )} I qg ,ζ {h(t )}. (45)

64


Further, we observe that, if we replace b by − a and δ by − g , and make use of the relation [17, p.173, eqn. (2.7)], namely

I qa ,−a ,η { f (t )} = I qa { f (t )}

(46)

I qg ,−g ,ζ { f (t )} = I qg { f (t )},

(47)

and

then, Theorems 4 to 6 reduce to the following q-integral inequalities involving the Riemann-Liouville type of fractional q-integral operators. Corollary 7. Let f and g be two synchronous functions on T , and h > 0 , then α ta I qa { f (t ) g (t )h(t )} ≥ I qa { f (t )} I qa {g (t )h(t )} + I qa {g (t )} I qa { f (t )h(t )} 1 +a Γq ((1+ α))

− I qa {h(t )} I qa { f (t ) g (t )},

(48)

for all t > 0,0 < q < 1 and a > 0 . Corollary 8. Let f

t > 0, 0 < q < 1, α, γ > 0,

and g be two synchronous functions on T , and h > 0 , then for

t aα t gγ g I q { f (t ) g (t )h(t )} + I qa { f (t ) g (t )h(t )} ≥ I qa { f (t )} I qg {g (t )h(t )} + α)) Γq (1(1+ +a Γ q (1(1+ + gγ))

I qa {g (t )h(t )} I qg { f (t )} + I qa {g (t )} I qg { f (t )h(t )} + I qa { f (t )h(t )} I qg {g (t )} −

I qa {h(t )} I qg { f (t ) g (t )} − I qa { f (t ) g (t )} I qg {h(t )}.

(49)

Corollary 9. Let f , g and h be three monotonic functions on [0, ∞) satisfying the inequality (14), then for all t > 0, α, γ > 0,0 < q < 1,

ttggγ ttαaa IIqgqg { ff ((tt))gg((tt))hh((tt))} −− IIqaqa { ff ((tt))gg((tt))hh((tt))} ≥≥ IIqaqa {gg((tt))hh((tt))} IIqgqg { ff ((tt))} −− Γ Γ ( 1 ( 1 + + ) ) Γ Γ ( 1 ( 1 + + ) ) a a g g Γqq(1+ α) qq (1+ γ)

I qa { f (t )} I qg {g (t )h(t )} + I qa { f (t )h(t )} I qg {g (t )} − I qa {g (t )} I qg { f (t )h(t )} −

I qa {h(t )} I qg { f (t ) g (t )} + I qa { f (t ) g (t )} I qg {h(t )}.

(50)

Special Cases We now, briefly consider some of the consequences of the results derived in the previous sections. If we let q → 1− , and use the limit formulas:

65


(qa ; q) n

= (a ) n

(51)

Lim Γq (a ) = Γ(a ) ,

(52)

Lim

q →1− (1 − q )

n

and

q →1−

the results of Section 3 correspond to the results obtained in Section 2. Again, in view of the above limiting cases, Corollaries 8 and 9 provide, respectively, the q-extensions of the inequalities due to Sulaiman [8, pp. 24-25, Theorems 2.1 to 2.2]. Finally, if we consider the function h as constant > 0 , the Theorems 1, 2, 4 and 5, and Corollaries 7 and 8 provide, respectively, the known results due to Purohit and Raina [6], and Öğünmez and Özkan [14].

Acknowledgements The authors are thankful to the referee for a very careful reading and valuable suggestions leading to the present improved form of the paper.

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