THE MINKOWSKI'S INEQUALITIES UTILIZING NEWLY DEFINED GENERALIZED FRACTIONAL INTEGRAL OPERATORS

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C om mun. Fac. Sci. U niv. A nk. Ser. A 1 M ath. Stat. Volum e 68, N umb er 1, Pages 686–701 (2019) D O I: 10.31801/cfsuasm as.463983

ISSN 1303–5991 E-ISSN 2618-6470

http://com munications.science.ankara.edu.tr/index.php?series= A 1

THE MINKOWSKI’S INEQUALITIES UTILIZING NEWLY DEFINED GENERALIZED FRACTIONAL INTEGRAL

OPERATORS

FUAT USTA, HÜSEYIN BUDAK, FATMA ERTUGRAL, AND MEHMET ZEKI SARIKAYA

Abstract. Motivated by the recent generalized fractional integral operators proposed by Tunc et. al. [22], we establish a generalization of the reverse Minkowski’s inequalities. Within this context, we provide new upper bounds of inequalities utilizing generalized fractional integral operators and show and state other inequalities related to this fractional integral operator.

1. Introduction

Recently, a number of scientist in the …eld of mathematics have introduced di¤er-ent results about the fractional derivatives and integrals such as Riemann-Liouville fractional derivative, Riemann-Liouville fractional integral operator, Hadamard in-tegral operator, Saigo fractional inin-tegral operator and some other, and applied them to some well-know inequalities with applications [1]-[22]. In this paper the authors will provide the some reverse Minkowski’s inequalities by means of the generalized fractional integral operators.

The overall structure of the study takes the form of four sections including in-troduction. The remaining part of the paper proceeds as follows: In Section 2, we introduce generalized k-fractional integrals of a function with respect to the another function which generalizes di¤erent types of fractional integrals, including Riemann-Lioville fractional, Hadamard fractional integrals, Katugampola fractional integral, (k; s)-fractional integral operators and many others. In section 3, we provide the main results involving the reverse Minkowski’s inequality with the help of frac-tional integral operators while in section 4 discussing other inequalities using this fractional integral operators. Finally concluding remarks summarize the article.

Received by the editors: September 12, 2017; Accepted: April 06, 2018. 2010 Mathematics Subject Classi…cation. 26D15, 26A33, 26B25, 26D10.

Key words and phrases. Fractional integral operators, Hermite-Hadamard inequality, midpoint inequality, convex function.

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2. New Generalized Fractional Integral Operators

In this section we will review the concept of the generalized k-fractional integrals of a function with respect to the another function introduced by Tunc et.al.[22]. De…nition 1. In [8] Diaz and Pariguan have de…ned k -gamma function k that

is generalization of the classical gamma. k is given by formula k(x) = lim

n!1

n!kn(nk)xk 1

(x)n;k

k > 0:

It has shown that Mellin transform of the exponential function e tkk is the k-gamma

function, clearly given by

k( ) := Z ₁ 0 e tkkt 1dt: Obviously, k(x + k) = x k(x) ; (x) = lim k!1 k(x) and k(x) = k x k 1 (x k):

De…nition 2. Let de…ne the function F ;;k(x) := 1 X m=0 (m) k k( km + ) xm ( ; _{> 0; jxj < R) ;}

where the coe¢ cients _{(m) (m 2 N}0= N[ f0g) is a bounded sequence of positive

real numbers and R is the set of real numbers.

De…nition 3. _{For k > 0; let g : [a; b] ! R be an increasing and positive monotone} function having a continuous derivative g0_{(x) on (a; b) : The left and right sided}

generalized k-fractional integrals of f with respect to the function g on [a; b] are de…ned, respectively, as follows:

J ; ;a+;!;k;g f (x) = x Z a g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ] f (t)dt; x > a; (2.1) and J ; ;b ;!;k;g f (x) = b Z x g0_(t) (g(t) g(x))1 kF ;k ; [! (g(t) g(x)) ] f (t)dt; x < b; (2.2) where ; _{> 0; ! 2 R.}

Remark 1. The signi…cant special cases of the integral operators (2.1) and (2.2) are mentioned below:

1) For k = 1; operator in (2.1) leads to generalized fractional integral of f with respect to the function g on [a; b] : This relation is given by

J ; ;a+;!;g f (x) = x Z a g0_(t) (g(x) g(t))1 F ; [! (g(x) g(t)) ] f (t)dt; x > a:

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2) For g(t) = t; operator in (2.1) leads to generalized k-fractional integral of f . This relation is given by

J ; ;a+;!;k f (x) = x Z a (x t)k 1_F ;k ; [! (x t) ] f (t)dt; x > a:

3) For g(t) = ln t; operator in (2.1) leads to generalized Hadamard k-fractional integral of f . This relation is given by

H; ;a+;!;k f (x) = x Z a lnx t k 1 F ;;k h ! lnx t i f (t)dt t ; x > a: 4) For g(t) = ts+1

s+1; s 2 R f 1g operator in (2.1) leads to generalized (k;

s)-fractional integral of f . This relation is given by

s J ; ;a+;!;k f (x) = (s+1) 1 _k x Z a xs+1 ts+1 k 1_tr F;;k ! xs+1 _ts+1 s + 1 f (t)dt; x > a: Remark 2. Similarly, all above special cases can also be seen for operator (2.2).

Remark 3. For k = 1 and g(t) = t; operators in (2.1) and (2.2) reduce to the following generalized fractional integral operators de…ned by Raina [21] and Agarwal et. al [1], respectively: J ; ;a+;!f (x) = Z x a (x t) 1_F ; [! (x t) ] f (t)dt; x > a; (2.3) J ; ;b ;!f (x) = Z b x (t x) 1_F _; [! (t x) ] f (t)dt; x < b; (2.4) Remark 4. One can obtain other new generalized fractional integral operators with di¤ erent choices of g.

Remark 5. For = ; (0) = 1; w = 0 in De…nition 3, then we have the generalized fractional operators de…ned by Akkurt et al. in [3].

Remark 6. Let = ; (0) = 1; w = 0 in De…nition 3.

1) Choosing k = 1; then we have fractional integrals of a function f with respect to function g: [12].

2) Choosing g(t) = t, then we have k-fractional integrals [15].

3) Choosing k = 1 and g(t) = ln t, then we have Hadamard fractional integrals [12].

4) Choosing g(t) = t_s+1s+1_{; s 2 R} _{f 1g, then we have (k; s)-fractional integral} operators [18].

5) Choosing k = 1 and g(t) = t_s+1s+1_{; s 2 R} _{f 1g, then we have Katugampola} fractional integral operators [9].

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6) Choosing k = 1 and g(t) = t, then we have Riemann-Lioville fractional integral operators [12].

3. Reverse Minkowski Fractional Integral Inequality new generalized fractional integral operators

Theorem 1. Let u; v 2 Xp

c(a; x) two positive functions in [0; 1), such that 8x >

a; J ; ;a+;!;k;g up(x) < 1 and J ;k;g

; ;a+;!vp(x) < 1. Let g : [a; b] ! R be an increasing

and positive monotone function having a continuous derivative g0_{(x) on (a; b) : If}

0 < m u(t)_v(t) _{M and 8t 2 [0; x] ; then we have the following reverse Minkowski’s} inequality associated with the generalized k-fractional integrals with respect to the function g h J ; ;a+;!;k;g u p_(x)i 1 p +h_J _{; ;a+;!};k;g vp(x)i 1 p C1 h J ; ;a+;!;k;g (u + v) p_(x)i 1 p where C1=M (m+1)+M +1_{(M +1)(m+1)} and p 1; ; > 0; ! 2 R.

Proof. Since u(t)_v(t) _{M; t 2 [a; x] ; we deduce that} u(t) M [u(t) + v(t)] M u(t) which yields up(t) M M + 1 p [u(t) + v(t)]p: (3.1) Then multiplying by g0(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (3.1) and integrating on [a; x], we get

x Z a g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ] u p_(t)dt M M + 1 pZx a g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ] [u(t) + v(t)] p dt: As a result, we deduce that

h J ; ;a+;!;k;g u p_(x)i 1 p M M + 1 h J ; ;a+;!;k;g (u + v) p_(x)i 1 p : (3.2)

On the other hand, as m u(t)_v(t)_{; t 2 [a; x], we have}

vp(t) 1

m + 1

p

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Similarly, multiplying by g0_(t)

(g(x) g(t))1 kF

;k

; [! (g(x) g(t)) ]

both sides of (3.3) and integrating on [a; x], we get h J ; ;a+;!;k;g v p_(x)i 1 p 1 m + 1 h J ; ;a+;!;k;g (u + v) p_(x)i 1 p : (3.4)

Then adding the inequalities (3.2) and (3.4), the desired result has been obtained.

Corollary 1. We assume that the conditions of Theorem 1 hold.

1) For k = 1 in Theorem 1, we have the following reverse Minkowski’s inequality associated with the generalized fractional integrals with respect to the function g

sh J ; ;a+;!;g u p_(x)i 1 p +h_J _{; ;a+;!};g vp(x)i 1 p C1 h J ; ;a+;!;g (u + v) p_(x)i 1 p : 2) For g(t) = t in Theorem 1, we have the following reverse Minkowski’s inequality associated with the generalized k-fractional integrals

sh J ; ;a+;!;k u p_(x)i 1 p + h J ; ;a+;!;k v p_(x)i 1 p C1 h J ; ;a+;!;k (u + v) p_(x)i 1 p : 3) For g(t) = ln t in Theorem 1, we have the following reverse Minkowski’s inequal-ity associated with the generalized Hadamard k-fractional integrals

sh H ; ;a+;!;k up(x) i1 p + h H ; ;a+;!;k vp(x) i1 p C1 h H ; ;a+;!;k (u + v)p(x) i1 p : 4) For g(t) = ts+1

s+1; s 2 R f 1g in Theorem 1, we have the following reverse

Minkowski’s inequality associated with the generalized (k; s)-fractional integrals

shs J ; ;a+;!;k u p_(x)i 1 p +hs_J _{; ;a+;!};k vp(x)i 1 p C1 h s J ; ;a+;!;k (u + v) p_(x)i 1 p : Theorem 2. _{Let u; v 2 X}p

a; J ; ;a+;!;k;g up(x) < 1 and J ;k;g

0 < m u(t)_v(t) _{M and 8t 2 [0; x] ; then we have the following reverse Minkowski’s} inequality associated with the generalized k-fractional integrals with respect to the function g h J ; ;a+;!;k;g u p_(x)i 2 p +h_J _{; ;a+;!};k;g vp(x)i 2 p C2 h J ; ;a+;!;k;g u p_(x)i 1 ph J ; ;a+;!;k;g v p_(x)i 1 p where C1=(M +1)(m+1)_M 2 and p 1; ; > 0; ! 2 R.

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Proof. From the inequalities (3.2) and (3.4), we have (M + 1)(m + 1) M h J ; ;a+;!;k;g up(x) i1 ph J ; ;a+;!;k;g vp(x) i1 p h J ; ;a+;!;k;g (u + v)p(x) i2 p : (3.5) Then, thanks to the Minkowski’s inequality, we get

h J ; ;a+;!;k;g (u + v)p(x) i2 p h J ; ;a+;!;k;g up(x) i1 p + h J ; ;a+;!;k;g vp(x) i1 p 2 : (3.6) Consequently, by substituting (3.6) into (3.5), we obtain the desired result. Corollary 2. We assume that the conditions of Theorem 2 hold.

1) For k = 1 in Theorem 2, we have the following reverse Minkowski’s inequality associated with the generalized fractional integrals with respect to the function g h J ; ;a+;!;g up(x) i2 p + h J ; ;a+;!;g vp(x) i2 p C2 h J ; ;a+;!;g up(x) i1 ph J ; ;a+;!;g vp(x) i1 p : 2) For g(t) = t in Theorem 2, we have the following reverse Minkowski’s inequality associated with the generalized k-fractional integrals

h J ; ;a+;!;k up(x) i2 p + h J ; ;a+;!;k vp(x) i2 p C2 h J ; ;a+;!;k up(x) i1 ph J ; ;a+;!;k vp(x) i1 p : 3) For g(t) = ln t in Theorem 2, we have the following reverse Minkowski’s inequal-ity associated with the generalized Hadamard k-fractional integrals

h H ; ;a+;!;k up(x) i2 p + h H ; ;a+;!;k vp(x) i2 p C2 h H ; ;a+;!;k up(x) i1 ph H ; ;a+;!;k vp(x) i1 p : 4) For g(t) = t_s+1s+1_{; s 2 R} _{f 1g in Theorem 2, we have the following reverse} Minkowski’s inequality associated with the generalized (k; s)-fractional integrals h s J ; ;a+;!;k u p_(x)i 2 p + h s J ; ;a+;!;k v p_(x)i 2 p C2 h s J ; ;a+;!;k u p_(x)i 1 ph_s J ; ;a+;!;k v p_(x)i 1 p : 4. Alternative Fractional Integral Inequalities with new

generalized fractional integral operators Theorem 3. _{Let u; v 2 X}p

a; J ; ;a+;!;k;g up(x) < 1 and J ;k;g

0 < m u(t)_v(t) _{M and 8t 2 [0; x] ; then we have the following inequality associated} with the generalized k-fractional integrals with respect to the function g

h J ; ;a+;!;k;g u p_(x)i 1 ph J ; ;a+;!;k;g v p_(x)i 1 p C3 h J ; ;a+;!;k;g (u + v) p_(x)i 1 p where C3= M_m 1 pq ; 1_p+1_q = 1; p 1 and ; _{> 0; ! 2 R.}

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Proof. Since u(t)_v(t) _{M; t 2 [a; x] ; we deduce that} v1q_(t) 1 M 1 q u1q_(t) which yields u1p(t)v 1 q(t) 1 M 1 q u(t): (4.1) Multiplying by g0(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.1) and integrating on [a; x], we get 1 M 1 q J ; ;a+;!;k;g u(x) J ;k;g ; ;a+;!u 1 p_(x)v1q_(x); i.e. 1 M 1 pq h J ; ;a+;!;k;g u(x) i1 p h J ; ;a+;!;k;g u 1 p_(x)v 1 q_(x) i1 p : (4.2)

More over, as m u(t)_v(t)_{; t 2 [a; x] ; we have} mp1_v 1 p_(t) _u 1 p_(t) which gives m1p_v(t) _u1p_(t)v1q_(t): _(4.3) Then, multiplying by g0(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.3) and integrating on [a; x], we get mpq1 h J ; ;a+;!;k;g v(x) i1 p h J ; ;a+;!;k;g u 1 p_(x)v 1 q_(x) i1 p : (4.4)

Considering the inequalities (4.2) and (4.4), we obtain the required result. Corollary 3. We assume that the conditions of Theorem 3 hold.

1) For k = 1 in Theorem 3, we have the following inequality associated with the generalized fractional integrals with respect to the function g

h J ; ;a+;!;g u p_(x)i 1 ph J ; ;a+;!;g v p_(x)i 1 p C3 h J ; ;a+;!;g (u + v) p_(x)i 1 p : 2) For g(t) = t in Theorem 3, we have the following inequality associated with the generalized k-fractional integrals

h J ; ;a+;!;k u p_(x)i 1 ph J ; ;a+;!;k v p_(x)i 1 p C3 h J ; ;a+;!;k (u + v) p_(x)i 1 p :

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3) For g(t) = ln t in Theorem 3, we have the following inequality associated with the generalized Hadamard k-fractional integrals

h H ; ;a+;!;k u p_(x)i 1 ph H ; ;a+;!;k v p_(x)i 1 p C3 h H ; ;a+;!;k (u + v) p_(x)i 1 p : 4) For g(t) = t_s+1s+1_{; s 2 R} _{f 1g in Theorem 3, we have the following inequality} associated with the generalized (k; s)-fractional integrals

h s J ; ;a+;!;k up(x) i1 ph_s J ; ;a+;!;k vp(x) i1 p C3 h s J ; ;a+;!;k (u + v)p(x) i1 p : Theorem 4. _{Let u; v 2 X}p

a; J ; ;a+;!;k;g up(x) < 1 and J ;k;g

J ; ;a+;!;k;g u(x)v(x) C4J ; ;a+;!;k;g (u

p_{+ v}p_{)(x) + C} 5J ; ;a+;!;k;g (u q_{+ v}q_)(x) where C4 = 2 p 1 p M M +1 p ; C5 = 2 q 1 q 1 m+1 q ; 1 p+ 1 q = 1; p 1 and ; > 0; ! 2 R. Proof. Multiplying by g0(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (3.1) and integrating on [a; x], we get J ; ;a+;!;k;g u p_(x) M M + 1 p J ; ;a+;!;k;g (u + v) p_(x): _(4.5)

As m u(t)_v(t)_{; t 2 [a; x], we have}

vq(t) 1 m + 1 q [u(t) + v(t)]q: (4.6) Similarly, multiplying by g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.6) and integrating on [a; x], we get J ; ;a+;!;k;g vq(x)

1 m + 1

q

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Applying the Young inequality, we have u(t)v(t) u p_(t) p + vq_(t) q (4.8) and multiplying by g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.8) and integrating on [a; x], we get J ; ;a+;!;k;g (uv) (x) 1 pJ ;k;g ; ;a+;!up(x) + 1 qJ ;k;g ; ;a+;!vq(x): (4.9)

Then, by substituting the inequalities (4.5) and (4.7) into (4.9), we obtain

s J_{; ;a+;!};k;g (uv) (x) 1 p M M + 1 !p J _{; ;a+;!};k;g (u + v)p(x) +1 q 1 m + 1 !q J_{; ;a+;!};k;g (u + v)q(x): ( 4 . 1 0 )

Using the fact that (a; b)r ₂r 1_(ar_{+ b}r_{); r > 1; a; b} _{0 in the right hand side of}

the inequality (4.10), we have

s J_{; ;a+;!};k;g (uv) (x) 1 p M M + 1 !p J_{; ;a+;!};k;g (u + v)p(x) +1 q 1 m + 1 !q J _{; ;a+;!};k;g (u + v)q(x) 1 p M M + 1 !p 2p 1J _{; ;a+;!};k;g (up+ vp)(x) +1 q 1 m + 1 !q 2q 1J _{; ;a+;!};k;g (uq+ vq)(x):

Thus, the proof is completed.

J ; ;a+;!;g u(x)v(x) C4J ; ;a+;!;g (up+ vp)(x) + C5J ; ;a+;!;g (uq+ vq)(x):

2) For g(t) = t in Theorem 4, we have the following inequality associated with the generalized k-fractional integrals

J ; ;a+;!;k u(x)v(x) C4J ; ;a+;!;k (u

p_{+ v}p_{)(x) + C}

5J ; ;a+;!;k (u

q_{+ v}q_)(x):

H ; ;a+;!;k u(x)v(x) C4H ; ;a+;!;k (u

p_{+ v}p_{)(x) + C}

5H ; ;a+;!;k (u

q_{+ v}q_)(x):

4) For g(t) = t_s+1s+1_{; s 2 R} _{f 1g in Theorem 4, we have the following inequality} associated with the generalized (k; s)-fractional integrals

s J ; ;a+;!;k u(x)v(x) C s 4J ;k ; ;a+;!(u p_{+ v}p_{)(x) + C}s 5J ;k ; ;a+;!(u q_{+ v}q_)(x):

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a; J ; ;a+;!;k;g up(x) < 1 and J ;k;g

0 < n < m u(t)_v(t) _{M and 8t 2 [0; x] ; then we have the following inequalities} associated with the generalized k-fractional integrals with respect to the function g

M + 1 M n h J ; ;a+;!;k;g (u(x) nv(x)) pi 1 p h J ; ;a+;!;k;g u p_(x)i 1 p +h_J _{; ;a+;!};k;g vq(x)i 1 p m + 1 m c h J ; ;a+;!;k;g (u(x) nv(x)) pi 1 p where p 1 and ; _{> 0; ! 2 R.}

Proof. From the assumption 0 < n < m u(t)_v(t) M , we have

m n u(t) nv(t) v(t) M n which yields (u(t) nv(t))p (M n)p v p_(t) (u(t) nv(t)) p (m n)p : (4.11) Similarly, we obtain 1 M v(t) u(t) 1 m ) 1 M 1 n v(t) u(t) 1 n 1 m 1 n ) m n mn u(t) nv(t) nu(t) M n M n which yields Mp (M n)p(u(t) nv(t)) p up(t) m p (m n)p(u(t) nv(t)) p : (4.12) Multiplying by g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.11) and integrating on [a; x], we get

s 1 M n h J _{; ;a+;!};k;g (u(x) nv(x))p i 1 p h_J ;k;g ; ;a+;!vq(x) i 1_p 1 m c h J _{; ;a+;!};k;g (u(x) nv(x))p i 1 p : ( 4 . 1 3 )

Following the similar steps for (4.12), we obtain

s M M n h J_{; ;a+;!};k;g (u(x) nv(x))p i 1 p h_J ;k;g ; ;a+;!uq(x) i 1_p m m c h J _{; ;a+;!};k;g (u(x) nv(x))p i 1 p : ( 4 . 1 4 )

Considering the inequalities (4.13) and (4.14), we obtain the required result. In order to validate our result we can show that _MM +1_n _{m n}m+1. That is, from the assumption 0 < n < m u(t)_v(t) M , we have

mn+m mn+M _{M n+M ) (M+1)(m n)} _{(m+1)(M n) )} M + 1

M n

m + 1 m n:

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M + 1 M n h J ; ;a+;!;g (u(x) nv(x)) pi 1 p h J ; ;a+;!;g up(x) i1 p + h J ; ;a+;!;g vq(x) i1 p m + 1 m c h J ; ;a+;!;g (u(x) nv(x)) pi 1 p : 2) For g(t) = t in Theorem 5, we have the following inequality associated with the generalized k-fractional integrals

M + 1 M n h J ; ;a+;!;k (u(x) nv(x)) pi 1 p h J ; ;a+;!;k u p_(x)i 1 p +h_J _{; ;a+;!};k vq(x)i 1 p m + 1 m c h J ; ;a+;!;k (u(x) nv(x)) pi 1 p : 3) For g(t) = ln t in Theorem 5, we have the following inequality associated with the generalized Hadamard k-fractional integrals

M + 1 M n h H; ;a+;!;k (u(x) nv(x)) pi1p h H; ;a+;!;k up(x) i1 p + h H ; ;a+;!;k vq(x) i1 p m + 1 m c h H; ;a+;!;k (u(x) nv(x)) pi 1 p : 4) For g(t) = ts+1

s+1; s 2 R f 1g in Theorem 5, we have the following inequality

associated with the generalized (k; s)-fractional integrals M + 1 M n h s J ; ;a+;!;k (u(x) nv(x)) pi 1 p h_s J ; ;a+;!;k up(x) i1 p + h s J ; ;a+;!;k vq(x) i1 p m + 1 m c h s J ; ;a+;!;k (u(x) nv(x)) pi 1 p : Theorem 6. _{Let u; v 2 X}_cp_{(a; x) two positive functions in [0; 1), such that 8x >} a; J ; ;a+;!;k;g up(x) < 1 and J

;k;g

and positive monotone function having a continuous derivative g0(x) on (a; b) : If 0 u(t) ; 0 ' v(t) _{and 8t 2 [0; x] ; then we have the following} inequalities associated with the generalized k-fractional integrals with respect to the function g h J ; ;a+;!;k;g up(x) i1 p + h J ; ;a+;!;k;g vq(x) i1 p C6 h J ; ;a+;!;k;g (u + v) p (x) i1 p where C6= ( + )+ ('+ )_{( + )('+ )} p 1 and ; > 0; ! 2 R.

Proof. From the assumptions of 0 u(t) and 0 ' v(t) , we

deduce that 1 1 v(t) 1 ' ) u(t) v(t) '

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which yields vp(t) + p (u(t) + v(t))p (4.15) and up(t) ' + p (u(t) + v(t))p: (4.16) Multiplying by g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.15) and (4.16), then integrating on [a; x], we get h J ; ;a+;!;k;g v q_(x)i 1 p + h J ; ;a+;!;k;g (u + v) p (x)i 1 p (4.17) and h J ; ;a+;!;k;g u p_(x)i 1 p ' + h J ; ;a+;!;k;g (u + v) p (x) i1 p ; (4.18)

respectively. Adding the inequalities (4.17) and (4.18), we obtain the desired result.

h J ; ;a+;!;g u p_(x)i 1 p + h J ; ;a+;!;g v q_(x)i 1 p C6 h J ; ;a+;!;g (u + v) p (x) i1 p : 2) For g(t) = t in Theorem 6, we have the following inequality associated with the generalized k-fractional integrals

h J ; ;a+;!;k u p_(x)i 1 p + h J ; ;a+;!;k v q_(x)i 1 p C6 h J ; ;a+;!;k (u + v) p (x) i1 p : 3) For g(t) = ln t in Theorem 6, we have the following inequality associated with the generalized Hadamard k-fractional integrals

h H ; ;a+;!;k u p_(x)i 1 p +h_H _{; ;a+;!};k vq(x)i 1 p C6 h H ; ;a+;!;k (u + v) p (x)i 1 p : 4) For g(t) = t_s+1s+1_{; s 2 R} _{f 1g in Theorem 6, we have the following inequality} associated with the generalized (k; s)-fractional integrals

h s J ; ;a+;!;k up(x) i1 p + h s J ; ;a+;!;k vq(x) i1 p C6 h s J ; ;a+;!;k (u + v) p (x) i1 p : Theorem 7. _{Let u; v 2 X}p

a; J ; ;a+;!;k;g up(x) < 1 and J ;k;g

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0 < m u(t)_v(t) _{M and 8t 2 [0; x] ; then we have the following inequalities associated} with the generalized k-fractional integrals with respect to the function g

1 MJ ;k;g ; ;a+;!(uv) (x) 1 (m + 1)(M + 1)J ;k;g ; ;a+;!(u + v) 2 (x) 1 mJ ;k;g ; ;a+;!(uv) (x) where ; _{> 0; ! 2 R.}

Proof. From the assumption 0 < m u(t)_v(t) M; we get

(m + 1)v(t) u(t) + v(t) (M + 1)v(t): (4.19) Also we have, M + 1 M u(t) u(t) + v(t) m + 1 m u(t): (4.20)

From the inequalities (4.19) and (4.20), we deduce that 1 Mu(t)v(t) (u(t) + v(t))2 (m + 1)(M + 1) 1 mu(t)v(t): (4.21) Multiplying by g0(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.21), then integrating on [a; x], we obtain the required result. Corollary 7. We assume that the conditions of Theorem 7 hold.

1 MJ ;g ; ;a+;!(uv) (x) 1 (m + 1)(M + 1)J ;g ; ;a+;!(u + v) 2 (x) 1 mJ ;g ; ;a+;!(uv) (x):

1 MJ ;k ; ;a+;!(uv) (x) 1 (m + 1)(M + 1)J ;k ; ;a+;!(u + v) 2 (x) 1 mJ ;k ; ;a+;!(uv) (x):

1 MH ;k ; ;a+;!(uv) (x) 1 (m + 1)(M + 1)H ;k ; ;a+;!(u + v) 2 (x) 1 mH ;k ; ;a+;!(uv) (x):

4) For g(t) = t_s+1s+1_{; s 2 R} _{f 1g in Theorem 7, we have the following inequality} associated with the generalized (k; s)-fractional integrals

1 M s J ; ;a+;!;k (uv) (x) 1 (m + 1)(M + 1) s J ; ;a+;!;k (u + v) 2 (x) 1 m s J ; ;a+;!;k (uv) (x):

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a; J ; ;a+;!;k;g up(x) < 1 and J ;k;g

h J ; ;a+;!;k;g u p_(x)i 1 p +h_J _{; ;a+;!};k;g vp(x)i 1 p 2h_J _{; ;a+;!};k;g hp(u(x); v(x))i 1 p where h(u(x); v(x)) = max M M m + 1 u(x) M v(x) ; (m + M )v(x) u(x) m with p 1 and ; _{> 0; ! 2 R.}

Proof. From the assumption 0 < m u(t)_v(t) M , we have 0 < m M + m u(t)

v(t) (4.22)

By the inequality (4.22), we get

v(t) < (m + M )v(x) u(x)

m h(u(t); v(t)): (4.23)

On the other hand, we have 1 M 1 M + 1 m v(t) u(t) which yields u(t) M M m + 1 u(x) M v(x) h(u(t); v(t)): (4.24)

Then, using the inequalities (4.23) and (4.24), we obtain

up(t) + vp(t) 2hp(u(t); v(t)): (4.25) Multiplying by g0_(t) (g(x) g(t))1 kF ;k ; [! (g(x) g(t)) ]

both sides of (4.25), then integrating on [a; x], we obtain the desired result. Corollary 8. We assume that the conditions of Theorem 8 hold.

h J ; ;a+;!;g u p_(x)i 1 p +h_J _{; ;a+;!};g vp(x)i 1 p 2h_J _{; ;a+;!};g hp(u(x); v(x))i 1 p :

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h J ; ;a+;!;k u p_(x)i 1 p +h_J _{; ;a+;!};k vp(x)i 1 p 2h_J _{; ;a+;!};k hp(u(x); v(x))i 1 p : 3) For g(t) = ln t in Theorem 8, we have the following inequality associated with the generalized Hadamard k-fractional integrals

h H ; ;a+;!;k u p_(x)i 1 p + h H ; ;a+;!;k v p_(x)i 1 p 2 h H ; ;a+;!;k h p_{(u(x); v(x))}i 1 p : 4) For g(t) = t_s+1s+1_{; s 2 R} _{f 1g in Theorem 8, we have the following inequality} associated with the generalized (k; s)-fractional integrals

h s J ; ;a+;!;k up(x) i1 p + h s J ; ;a+;!;k vp(x) i1 p 2 h s J ; ;a+;!;k hp(u(x); v(x)) i1 p : 5. Concluding Remarks

In this research we introduced the generalization of the reverse Minkowski’s in-equalities using generalized fractional integral operator. In order to validate that their generalized behavior, we show the relation of our results with previously pub-lished ones.

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Current address : Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce-TURKEY

E-mail address : fuatusta@duzce.edu.tr

ORCID Address: http://orcid.org/0000-0002-7750-6910

E-mail address : hsyn.budak@gmail.com

ORCID Address: http://orcid.org/0000-0001-8843-955X

E-mail address : fatmaertugral14@gmail.com

ORCID Address: http://orcid.org/0000-0002-7561-8388

E-mail address : sarikayamz@gmail.com