RACSAM DOI 10.1007/s13398-017-0402-y ORIGINAL PAPER

( p, q)-Hermite–Hadamard inequalities and ( p, q)-estimates for midpoint type inequalities via convex and quasi-convex functions ˙ ˙scan2 · Necmettin Alp3 · Mehmet Kunt1 · Imdat I¸ Mehmet Zeki Sarıkaya3

Received: 4 November 2016 / Accepted: 2 May 2017 © Springer-Verlag Italia 2017

Abstract In this paper, we prove the correct ( p, q)-Hermite–Hadamard inequality, some new ( p, q)-Hermite–Hadamard inequalities, and generalized ( p, q)-Hermite–Hadamard inequality. By using the left hand part of the correct ( p, q)-Hermite–Hadamard inequality, we have a new equality. Finally using the new equality, we give some ( p, q)-midpoint type integral inequalities through ( p, q)-differentiable convex and ( p, q)-differentiable quasi-convex functions. Many results given in this paper provide extensions of others given in previous works. Keywords Hermite–Hadamard inequality · Midpoint type inequality · q-integral inequalities · ( p, q)-integral inequalities · ( p, q)-derivative · ( p, q)-integration · Convexity · Quasi-convexity Mathematics Subject Classification 34A08 · 26A51 · 26D15

B

Mehmet Kunt [email protected] ˙Imdat ˙I¸scan [email protected]; [email protected] Necmettin Alp [email protected] Mehmet Zeki Sarıkaya [email protected]

1

Department of Mathematics, Faculty of Sciences, Karadeniz Technical University, 61080 Trabzon, Turkey

2

Department of Mathematics, Faculty of Sciences and Arts, Giresun University, 28200 Giresun, Turkey

3

Department of Mathematics, Faculty of Science and Arts, Düzce University, Düzce, Turkey

M. Kunt et al.

1 Introduction The study of calculus with no limits is known in the literature as quantum calculus (also q-calculus). The famous mathematician Euler initiated the study q-calculus in the eighteenth century by introducing the parameter q in Newton’s work of infinite series. In early twentieth century, Jackson [6] has started a symmetric study of q-calculus and introduced q-definite integrals. The subject of quantum calculus has numerous applications in various areas of mathematics and physics such as number theory, orthogonal polynomials, combinatorics, basic hypergeometric functions, quantum theory, mechanics and theory of relativity. Quantum calculus has received exclusive interest by many researchers and hence it is considered as an incorporative subject between mathematics and physics. Interested readers are referred to [3,4,8] for some current advances in the theory of quantum calculus and theory of inequalities in quantum calculus. In recent articles, Tariboon et al. [18,19] studied the concept of q-derivatives and qintegrals over the intervals of [a, b] ⊂ R and settled a number of quantum analogues of some well-known results like Hölder inequality, Hermite–Hadamard inequality and Ostrowski inequality, Cauchy–Bunyakovsky–Schwarz, Gruss, Gruss–Cebysev and other integral inequalities using classical convexity. Also, Noor et al. [11,12], Sudsutad et al. [14] and Zhuang et al. [10,20] have contributed to the ongoing research and have developed some integral inequalities which provide quantum estimates for the right part of the quantum analogue of Hermite–Hadamard inequality through q-differentiable convex and quasi-convex functions. Recently, Tunç and Göv [15–17] studied the concept of ( p, q)derivatives and ( p, q)-integrals over the intervals of [a, b] ⊂ R and settled a number of ( p, q) analogues of some well-known results like Hölder inequality, Minkowski inequality, Hermite–Hadamard inequality and Ostrowski inequality, Cauchy–Bunyakovsky–Schwarz, Gruss, Gruss–Cebysev and other integral inequalities using classical convexity. The most recently, Alp et al. [2], proved q-Hermite–Hadamard inequality, some new q-Hermite– Hadamard inequalities, and generalized q-Hermite–Hadamard inequality, also they studied some integral inequalities which provide quantum estimates for the left part of the quantum analogue of Hermite–Hadamard inequality through q-differentiable convex and quasi-convex functions. Latif et al. [9] studied some integral inequalities which provide ( p, q) estimates for the right part of the quantum analogue of Hermite–Hadamard inequality through ( p, q)differentiable convex and quasi-convex functions Let I ⊂ R is a nonempty interval. The function f : R → R called as convex, if f (ta + (1 − t) b) ≤ t f (a) + (1 − t) f (b) holds for every a, b ∈ I and t ∈ [0, 1]. The function f : R → R called a quasi-convex, if f (ta + (1 − t) b) ≤ sup { f (a) , f (b)} holds for every a, b ∈ I and t ∈ [0, 1]. In [7], Kırmacı obtained inequalities for differentiable convex mappings which are connected with midpoint type inequality, in [1], Alomari et al. obtained inequalities for differentiable quasi-convex mappings which are connected with midpoint type inequality. They are used the following lemma to prove their theorems. Lemma 1 [7] Let f : I ◦ ⊂ R → R be a differentiable mapping on I ◦ , a, b ∈ I ◦ with a < b. If f  ∈ L [a, b], then the following equality holds:

( p, q)-estimates for midpoint type inequalities

1 b−a



b f (t) dt − f a

⎡

⎢ =b−a⎢ ⎣

a+b 2



1

2

t f  (ta + (1 − t) b) dt +

0

1

⎤ ⎥ (t − 1) f  (ta + (1 − t) b) dt ⎥ ⎦ . (1.1)

1 2

In [2] Alp et al. gave some inequalities about q-differentiable convex and quasi-convex functions which are engaged with q-midpoint type inequality. The following lemma was used to prove their theorems. Lemma 2 [2] Let f : [a, b] → R be a ( p, q)-differentiable function on (a, b). If a D p,q f is continuous and integrable on [a, b], then the following identity holds: ⎡ 1   b 1+q ⎢ qa + b 1 ⎢ f f (x) a dq x = q(b − a) ⎣ t a Dq f (tb + (1 − t) a) 0 dq t − 1+q (b − a) a

0

1  +

t−

1 q

⎤

 a Dq

f (tb + (1 − t) a)

⎥ ⎥

0 dq t⎦.

1 1+q

(1.2) In [2], Alp et al. gave the q-Hermite–Hadamard inequality as follows: Theorem 1 [2] (q-Hermite–Hadamard inequality) Let f : [a, b] → R be a convex differentiable function on [a, b] and 0 < q < 1. Then we have  f

qa + b 1+q

 ≤

1 b−a

b f (x)

a dq x

≤

a

q f (a) + f (b) . 1+q

(1.3)

2 Preliminaries and definitions of ( p, q)-calculus Throughout this paper, let [a, b] ⊂ R is an interval, 0 < q < p ≤ 1 are constants. The following definitions and theorems for ( p, q)-derivative and ( p, q)-integral are given in [15,16]. Definition 1 [15,16] For a continuous function f : [a, b] → R then ( p, q)-derivative of f at t ∈ [a, b] is characterized by the expression a D p,q

f (t) =

f ( pt + (1 − p) a) − f (qt + (1 − q) a) , ( p − q) (t − a)

t  = a.

(2.1)

Since f : [a, b] → R is a continuous function, thus we have a D p,q f (a) = limt→a a D p,q f (t). The function f is said to be ( p, q)-differentiable on [a, b] if a D p,q f (t) exists for all t ∈ [a, b]. If a = 0 in (2.1), then 0 D p,q f (t) = D p,q f (t), where D p,q f (t) is familiar ( p, q)-derivative of f at t ∈ [a, b] defined by the expression (see [5,13]) D p,q f (t) =

f ( pt) − f (qt) , ( p − q) t

t  = 0.

(2.2)

M. Kunt et al.

Note also that if p = 1 in (2.2), then Dq f (x) is familiar q-derivative of f at x ∈ [a, b] defined by the expression (see [8]) f (t) − f (qt) , (1 − q) t

Dq f (t) =

t  = 0.

(2.3)

Definition 2 [15,16] Let f : [a, b] → R be a continuous function. The definite ( p, q) integral on [a, b] is delineated as t f (x)

a d p,q x

= ( p − q) (t − a)

   n  ∞  qn q qn a f t + 1 − p n+1 p n+1 p n+1

(2.4)

n=0

a

for t ∈ [a, b]. If c ∈ (a, t), then the ( p, q)-definite integral on [c, t] is expressed as t

t f (x)

a d p,q x

=

c

c f (x)

a d p,q x

a

−

f (x)

a d p,q x.

(2.5)

a

If p = 1 in (2.4), then one can get the classical q-definite integral on [a, b] defined by (see [14, Definition 2.2]) t f (x)

a dq x

= (1 − q) (t − a)

∞ 

qn f qnt + 1 − qn a .

(2.6)

n=0

a

If a = 0 in (2.4), then one can get the classical ( p, q)-definite integral defined by (see [13, Definition 4]) t

t 0 d p,q x =

f (x) 0

f (x) d p,q x = ( p − q) t

 n  ∞  q qn f t . p n+1 p n+1

(2.7)

n=0

0

Note also that if p = 1 in (2.7), then one can get the classical q-definite integral defined by (see [14, Definition 2.2]) t

t f (x)

0 dq x

=

0

f (x) dq x = (1 − q) t

∞ 

qn f qnt .

(2.8)

n=0

0

The ( p, q)-Hermite–Hadamard inequality is given as follows in [16]. Theorem 2 [16, Theorem 7] Let f : [a, b] → R be a convex function and 0 < q < p ≤ 1. Then we have  f

a+b 2



1 ≤ b−a

b f (x) a

a d p,q x

≤

( p + q − 1) f (a) + f (b) . p+q

(2.9)

In [9], Latif et al. give the following example to prove that the left hand side of (2.9) is not correct. Example 1 Let [a, b] = [0, 1]. Then the function f (t) = 1 − t is a convex continuous function on [0, 1]. Therefore the function f satisfies Theorem 2 assumptions. Then, from the inequality (2.9) the following inequality must be hold for all 0 < q < p ≤ 1.

( p, q)-estimates for midpoint type inequalities

 f

0+1 2



1 ≤ 1−0

1 f (t)

0 d p,q t

0

  ∞  qn 1 qn 1 − n+1 1 − ≤ ( p − q) (1 − 0) 2 p n+1 p n=0 ⎞ ⎛ 1 1 1 1 1 ⎠ − 2 ≤ ( p − q) ⎝ 2 p 1 − qp p 1 − q2 2 p   1 1 1 . ≤ ( p − q) − 2 2 p−q p − q2 Then we have

1 p+q −1 ≤ . 2 p+q

If we choose p = 1, q =

1 2

(2.10)

in (2.10) we have the following contradiction 1 1 ≤ . 2 3

It means that the left hand side of (2.9) is not correct. In the next sections we give the correct ( p, q)-Hermite–Hadamard inequality, some new ( p, q)-Hermite–Hadamard inequalities, and generalized ( p, q)-Hermite–Hadamard inequality. Also, we proved an equality for the ( p, q)-analogue of midpoint type inequality. By using this equality we have ( p, q)-midpoint type integral inequalities through ( p, q)-differentiable convex and quasi-convex functions.

3 ( p, q)-Hermite–Hadamard Inequalities In this section, we prove ( p, q)-Hermite–Hadamard inequality and varieties of ( p, q)Hermite–Hadamard inequalities. Theorem 3 (( p, q)-Hermite–Hadamard inequality) Let f : [a, b] → R be a convex differentiable function on [a, b] and 0 < q < p ≤ 1. Then we have  f

qa + pb p+q



1 ≤ p (b − a)

pb+(1−  p)a

f (x)

a d p,q x

a

≤

q f (a) + p f (b) . p+q

(3.1)

Proof Since f is differentiable function on [a, b], there is a tangent line for the function pb f at the point qa+ ∈ (a, b). This tangent line can be expressed as a function h (x) = p+q      qa+ pb qa+ pb qa+ pb  + f x − f p+q p+q p+q . Since f is a convex function on [a, b], than we have the following inequality  h (x) = f

qa + pb p+q



+ f



qa + pb p+q

 x−

qa + pb p+q

 ≤ f (x)

(3.2)

M. Kunt et al.

Fig. 1 Tangent and chord lines for a convex function

for all x ∈ [a, b] (see Fig. 1). ( p, q)-integrating the inequality (3.2) on [a, pb + (1 − p) a], we have pb+(1−  p)a

h (x)

a d p,q x

a

    qa + pb qa + pb qa + pb + f x− p+q p+q p+q a   qa + pb = p (b − a) f p+q ⎛ ⎞  pb+(1−   p)a qa + pb ⎜ qa + pb ⎟ +f x a d p,q x − p (b − a) ⎝ ⎠ p+q p+q pb+(1−  p)a

=



f

a

a d p,q x

   qa + pb qa + pb + f = p (b − a) f p+q p+q     ∞ n  q qn qn pb + − p) a) × ( p − q) p (b − a) 1 − a + ( (1 p n+1 p n+1 p n+1 

n=0

( p, q)-estimates for midpoint type inequalities

 qa + pb − p (b − a) p+q     qa + pb qa + pb = p (b − a) f + f p+q p+q     ∞ n  q q 2n qa + pb × ( p − q) p (b − a) a + 2n+1 (b − a) − p (b − a) p n+1 p p+q n=0   qa + pb = p (b − a) f p+q    qa + pb qa + pb qa + pb +f p (b − a) − p (b − a) p+q p+q p+q  = p (b − a) f

qa + pb p+q



pb+(1−  p)a

≤

f (x)

a d p,q x.

(3.3)

a

On the other hand, line connecting the points (a, f (a)) and (b, f (b)) can be expressed f (a) as a function k (x) = f (a) + f (b)− (x − a). Since f is a convex function on [a, b], then b−a we have the following inequality f (x) ≤ k (x) = f (a) +

f (b) − f (a) (x − a) b−a

(3.4)

for all x ∈ [a, b] (see Fig. 1). ( p, q)-integrating the inequality (3.4) on [a, pb + (1 − p) a], we have pb+(1−  p)a

k (x)

a d p,q x

a pb+(1−  p)a

= a

 f (b) − f (a) f (a) + (x − a) b−a

f (b) − f (a) = p (b − a) f (a) + b−a

a d p,q x

pb+(1−  p)a

(x − a) ⎛

f (b) − f (a) ⎜ = p (b − a) f (a) + ⎝ b−a

⎞

pb+(1−  p)a

x

a d p,q x

a

f (b) − f (a) b−a   ∞  qn qn × ( p − q) p (b − a) 1 − n+1 a p n+1 p n=0   qn + n+1 ( pb + (1 − p) a) − ap (b − a) p

= p (b − a) f (a) + 

= p (b − a) f (a) +

f (b) − f (a) b−a

a d p,q x

a

⎟ − ap (b − a)⎠

M. Kunt et al.



  ∞   qn q 2n × ( p − q) p (b − a) a + 2n+1 (b − a) − ap (b − a) p n+1 p n=0   qa + pb f (b) − f (a) p (b − a) − ap (b − a) = p (b − a) f (a) + b−a p+q   2 2 f (b) − f (a) p (b − a) = p (b − a) f (a) + b−a p+q   2 p (b − a) = p (b − a) f (a) + ( f (b) − f (a)) p+q q f (a) + p f (b) = p (b − a) ≥ p+q

pb+(1−  p)a

f (x)

a d p,q x.

(3.5)

a



A combination of (3.3) and (3.5) gives (3.1). Thus the proof is accomplished. Remark 1 In Theorem 3;

1. If we take p = 1, then we recapture Theorem 1, 2. If we take p = 1 and q → 1− , then we recapture the well known Hermite–Hadamard inequality for convex functions. Theorem 4 Let f : [a, b] → R be a convex differentiable function on [a, b] and 0 < q < p ≤ 1. Then we have  f

pa + qb p+q



( p − q) (b − a)  + f p+q



pa + qb p+q



1 ≤ p (b − a)

pb+(1−  p)a

f (x)

a d p,q x

a

q f (a) + p f (b) ≤ p+q

(3.6)

Proof Since f is differentiable function on [a, b], there is a tangent line for the function ∈ (a, b). This tangent line can be expressed as a function h 1 (x) = f at the point pa+qb p+q      pa+qb pa+qb pa+qb  f + f x − p+q p+q p+q . Since f is a convex function on [a, b], than we have the following inequality      pa + qb pa + qb pa + qb h 1 (x) = f + f x− ≤ f (x) (3.7) p+q p+q p+q for all x ∈ [a, b] (see Fig. 2). ( p, q)-integrating the inequality (3.7) on [a, pb + (1 − p) a], we have pb+(1−  p)a

h 1 (x)

a d p,q x

a

    pa + qb pa + qb pa + qb + f x− p+q p+q p+q a   pa + qb = p (b − a) f p+q pb+(1−  p)a

=



f

a d p,q x

( p, q)-estimates for midpoint type inequalities

+ qb Fig. 2 Tangent line for a convex function at the point pap + q

+f





⎛

pa + qb ⎜ ⎝ p+q

⎞

pb+(1−  p)a

x a

a d p,q x

− p (b − a)

pa + qb ⎟ ⎠ p+q

   pa + qb pa + qb + f p+q p+q    ∞ n  q qn qn × ( p − q) p (b − a) 1 − n+1 a + n+1 ( pb + (1 − p) a) p n+1 p p n=0  pa + qb − p (b − a) p+q     pa + qb pa + qb = p (b − a) f + f p+q p+q    ∞  n  q q 2n pa + qb a + 2n+1 (b − a) − p (b − a) × ( p − q) p (b − a) p n+1 p p+q n=0     pa + qb pa + qb = p (b − a) f + f p+q p+q    ∞  n  q q 2n pa + qb a + 2n+1 (b − a) − p (b − a) × ( p − q) p (b − a) p n+1 p p+q 

= p (b − a) f 

n=0

M. Kunt et al.

 pa + qb = p (b − a) f p+q    pa + qb qa + pb pa + qb +f p (b − a) − p (b − a) p+q p+q p+q     2 pa + qb p ( p − q) (b − a)  pa + qb = p (b − a) f + f p+q p+q p+q 

pb+(1−  p)a

≤

f (x)

a d p,q x.

(3.8)

a



A combination of (3.5) and (3.8) gives (3.6). Thus the proof is accomplished. Remark 2 In Theorem 4, if we take p = 1, then we recapture [2, Theorem 3].

Theorem 5 Let f : [a, b] → R be a convex differentiable function on [a, b] and 0 < q < p ≤ 1. Then we have 

a+b 2

f





( p − q) (b − a)  + f 2 ( p + q)

a+b 2



1 ≤ p (b − a)

pb+(1−  p)a

f (x)

a d p,q x

a

q f (a) + p f (b) (3.9) p+q Proof Since f is differentiable function on [a, b], there is a tangent line for the function ∈ (a, b). This tangent line can be expressed as a function h 2 (x) = f at the point a+b

2

 a+b f a+b + f x − a+b 2 2 2 . Since f is a convex function on [a, b], we have the following inequality      a+b a+b a+b h 2 (x) = f + f x− ≤ f (x). (3.10) 2 2 2 ≤

for all x ∈ [a, b] (see Fig. 3). ( p, q)-integrating the inequality (3.10) on [a, pb + (1 − p) a], we have pb+(1−  p)a

h 2 (x)

a d p,q x

a pb+(1−  p)a

=

 f

a+b 2



+ f



a+b 2



a

 = p (b − a) f 

a+b 2

 

+ f

 



x− ⎛

a+b ⎜ ⎝ 2 

a+b 2

 a d p,q x

⎞

pb+(1−  p)a

x

a d p,q x

− p (b − a)

a + b⎟ ⎠ 2

a

a+b 2    ∞ n  q qn qn × ( p − q) p (b − a) 1 − n+1 a + n+1 ( pb + (1 − p) a) p n+1 p p n=0  a+b − p (b − a) 2

= p (b − a) f 

a+b 2

+ f

( p, q)-estimates for midpoint type inequalities

b Fig. 3 Tangent line for a convex function at the point a + 2



a+b 2





a+b 2



qa + pb a+b − p (b − a) p+q 2     a+b p ( p − q) (b − a)2  a + b = p (b − a) f + f 2 2 ( p + q) 2

= p (b − a) f

+ f



p (b − a)

pb+(1−  p)a

≤

f (x)

a d p,q x.

(3.11)

a

A combination of (3.5) and (3.11) gives (3.9). Thus the proof is accomplished.



Remark 3 In Theorem 5, if we take p = 1, then we recapture [2, Theorem 4]. Theorem 6 (Generalized ( p, q)-Hermite–Hadamard inequality) Let f : [a, b] → R be a convex differentiable function on [a, b] and 0 < q < p ≤ 1. Then we have 1 max {I1 , I2 , I3 } ≤ p (b − a) where

 I1 = f

pb+(1−  p)a

f (x) a

 qa + pb , p+q

a d p,q x

≤

q f (a) + p f (b) . p+q

(3.12)

M. Kunt et al.

   pa + qb ( p − q) (b − a)  pa + qb + f , I2 = f p+q p+q p+q     a+b ( p − q) (b − a)  a + b I3 = f + f . 2 2 ( p + q) 2 

Proof A combination of (3.1), (3.6), and (3.9) gives (3.12). Thus the proof is accomplished.  Remark 4 In Theorem 6, if we take p = 1, then we recapture [2, Theorem 5].

4 Midpoint type inequalities via ( p, q)-calculus We will use the following Lemma to prove our main results. Lemma 3 Let f : [a, b] → R is a ( p, q)-differentiable function over (a, b). If a D p,q f is continuous and integrable over [a, b], then we get  f

qa + pb p+q



1 − p (b − a) ⎡

pb+(1−  p)a

f (x)

a d p,q x

a

⎤

p p+q



⎥ ⎢ t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎥ ⎢ 0 ⎥. ⎢ = q (b − a) ⎢  1  ⎥  1 ⎦ ⎣+ t−q a D p,q f (tb + (1 − t) a) 0 d p,q t

(4.1)

p p+q

Proof Using (2.1), we have that f (tb + (1 − t) a) f ( p [tb + (1 − t) a] + (1 − p) a) − f (q [tb + (1 − t) a] + (1 − q) a) = ( p − q) [tb + (1 − t) a − a] f ( ptb + (1 − pt) a) − f (qtb + (1 − qt) a) = . (4.2) t ( p − q) (b − a)

a D p,q

Calculating following integrals by using (2.4) and Definition 2 we have ⎤ p p+q  ⎥ ⎢ t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎥ ⎢ 0 ⎥ q (b − a) ⎢  1  ⎥ ⎢  ⎦ ⎣+ t − q1 a D p,q f (tb + (1 − t) a) 0 d p,q t ⎡

p p+q

⎡

⎤

p p+q



⎢ ⎥ t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎢ ⎥ 0 ⎢ ⎥ ⎢ ⎥   1  ⎢ ⎥ t − q1 ⎢+ a D p,q f (tb + (1 − t) a) 0 d p,q t ⎥ ⎢ ⎥ p ⎢ ⎥ = q (b − a) ⎢ p+q ⎥ p ⎢ ⎥ p+q  ⎢ ⎥ 1 ⎢ ⎥ −q a D p,q f (tb + (1 − t) a) 0 d p,q t ⎢ ⎥ 0 ⎢ ⎥ p ⎢ ⎥ p+q  ⎣ ⎦ 1 +q D f + − t) a) d t (tb (1 0 p,q a p,q 0

( p, q)-estimates for midpoint type inequalities

⎡

⎤ 1 t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎥ ⎢ ⎢ 0 ⎥ ⎢ ⎥ 1 ⎢ ⎥ 1 ⎢ − D f + − t) a) d t (tb (1 a p,q 0 p,q ⎥ = q (b − a) ⎢ q ⎥ 0 ⎢ ⎥ p ⎢ ⎥  ⎣ 1 p+q ⎦ +q a D p,q f (tb + (1 − t) a) 0 d p,q t 0

⎡

1

⎤

f ( ptb+(1− pt)a)− f (qtb+(1−qt)a) t( p−q)(b−a)

t 0 d p,q t ⎢ ⎥ ⎢ 0 ⎥ ⎢ ⎥ 1  f ( ptb+(1− pt)a)− f (qtb+(1−qt)a) ⎢ ⎥ 1 −q 0 d p,q t ⎥ = q (b − a) ⎢ t( p−q)(b−a) ⎢ ⎥ 0 ⎢ ⎥ p ⎢ ⎥  f ( ptb+(1− pt)a)− f (qtb+(1−qt)a) ⎣ 1 p+q ⎦ +q 0 d p,q t t( p−q)(b−a) 0

⎡ ⎢ ⎢ ⎢ ⎢ =⎢ ⎢ ⎢ ⎢ ⎣

q p−q

1 0

1

1 − p−q

+

⎤ d t 0 p,q ⎥ ⎥ ⎥ ⎥ ⎥ 0 d p,q t ⎥ ⎥ ⎥ ⎦ 0 d p,q t

f ( ptb + (1 − pt) a) − f (qtb + (1 − qt) a)

0

p p+q



1 p−q

0 ∞ 

f ( ptb+(1− pt)a) t

−

f ( ptb+(1− pt)a) t

−

f (qtb+(1−qt)a) t f (qtb+(1−qt)a) t

⎤        n+1  ∞ n q q n+1 qn a b + 1 − qpn a − f b + 1 − n+1 n+1 n+1 p p p ⎢ ⎥ ⎢ ⎥   n=0    n   n+1  ∞ ⎢ ⎥  n n+1 q q q q ⎥ a − a f b + 1 − f b + 1 − − =⎢ n n n+1 n+1 p p ⎢ ⎥ p p n=0 n=0 ⎢  ⎥         ∞ ∞ ⎣ ⎦   qn p qn p q n+1 p q n+1 p f pn ( p+q) b + 1 − pn ( p+q) a − f pn+1 ( p+q) b + 1 − pn+1 ( p+q) a + ⎡

q



qn pn+1 n=0 ∞ 

n=0



f

qn pn

n=0

 ∞    n  q qn 1 1  qn =q f b+ 1− n a − q p pn pn p n=0     qa + pb − f (a) − ( f (b) − f (a)) + f p+q  = f

1 f (b) − q

qa + pb p+q



 −

1 p (b − a)

pb+(1−  p)a

f (x)

a d p,q x.

a

Thus the proof is accomplished.



Remark 5 In Lemma 3; 1. If we take p = 1, then we recapture Lemma 2, 2. If we take p = 1 and q → 1− , then we recapture Lemma 1. We can now prove some quantum estimates of ( p, q)-midpoint type integral inequalities by using convexity and quasi-convexity of the absolute values of the ( p, q)-derivatives. Theorem 7 Let f : [a, b] → R is a ( p, q)-differentiable function over (a, b), a D p,q f is  continuous and integrable over [a, b] and 0 < q < p ≤ 1. If a D p,q f  is convex function over [a, b], then we have the following ( p, q)-midpoint type inequality

M. Kunt et al.

     pb+(1−   p)a   qa + pb 1   f (x) a d p,q x  − f   p+q p (b − a)   a   

   a D p,q f (b) M1 ( p, q) + a D p,q f (a) M2 ( p, q)   

 ≤ q (b − a) + a D p,q f (b) M3 ( p, q) + a D p,q f (a) M4 ( p, q)

(4.3)

where

p3

, ( p + q)3 p 2 + pq + q 2

p 2 p 2 + pq + q 2 − p 3

, M2 ( p, q) = ( p + q)3 p 2 + pq + q 2 M1 ( p, q) =

M3 ( p, q) =

2 p3

, ( p + q)3 p 2 + pq + q 2

M4 ( p, q) =

p4 + p3 q + p2 q 2 − 2 p3

. ( p + q)3 p 2 + pq + q 2

  Proof Taking absolute value on both sides of (4.1) and by using the convexity of a D p,q f , we have that      pb+(1−   p)a   qa + pb 1   f (x) a d p,q x  − f   p+q p (b − a)   a ⎡ p p+q ⎢   ≤ q (b − a) ⎢ t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎣ 0

1  + p p+q

⎤



  1 − t a D p,q f (tb + (1 − t) a) q ⎡

p p+q



⎥ ⎥

0 d p,q t ⎦

⎤

     t t a D p,q f (b) + (1 − t) a D p,q f (a)

⎢ ⎥ 0 d p,q t ⎢ ⎥ 0 ⎥ ≤ q (b − a) ⎢   1 ⎢ ⎥        1 ⎣+ ⎦     t a D p,q f (b) + (1 − t) a D p,q f (a) 0 d p,q t q −t ⎡

p p+q

p

  p+q  2 a D p,q f (b) t

⎢ ⎢ 0 ≤ q (b − a) ⎢  ⎢   1 1 ⎣ + a D p,q f (b)  q −t t p p+q

⎤

p

  p+q    t (1 − t) 0 d p,q t + a D p,q f (a) 0 d p,q t

  + a D p,q f (a)

0

1

p p+q



1 q

⎥ ⎥ ⎥. ⎥ − t (1 − t) 0 d p,q t ⎦ 0 d p,q t



(4.4)

( p, q)-estimates for midpoint type inequalities

One can easily evaluate the appearing definite ( p, q)-integrals as follows p

p+q

p3

= M1 ( p, q) , ( p + q)3 p 2 + pq + q 2

(4.5)



p 2 p 2 + pq + q 2 − p 3

= M2 ( p, q) , t (1 − t) 0 d p,q t = ( p + q)3 p 2 + pq + q 2

(4.6)

 1 −t t q

= M3 ( p, q) ,

(4.7)

p4 + p3 q + p2 q 2 − 2 p3

= M4 ( p, q) . ( p + q)3 p 2 + pq + q 2

(4.8)

t2

0 d p,q t

=

0 p p+q

 0

1  p p+q

1  p p+q

 1 − t (1 − t) q

0 d p,q t

=

0 d p,q t

=

( p + q)

3



2 p3 p 2 + pq + q 2

Making use of (4.5)–(4.8) in (4.4), gives us the desired result (4.3). Thus the proof is accomplished.  Corollary 1 In Theorem 7; 1. If we take p = 1, then we have the following q-midpoint type inequality       b   1 qa + b f  − f d x (x) a q   1+q (b − a)   a       3 −1 + 2q + 2q 2 a Dq f (a)



+ , ≤ q (b − a) a Dq f (b) (1 + q)3 1 + q + q 2 (1 + q)3 1 + q + q 2

(4.9) 2. If we take p = 1 and q → 1− , then we have the following midpoint type inequality for convex functions       b  (b − a)        1 ≤ f a+b −  f (b) +  f (a) . f d x (x)   2 8 (b − a)  

(4.10)

a

Remark 6 One can see the following: 1. In (4.9), we recapture the inequality in [2, Theorem 6, (4.3)], 2. In (4.10), we recapture the inequality [7, Theorem 2.2]. Theorem 8 Let f : [a, b] → R is a ( p, q)-differentiable function over  r (a, b), a D p,q f is continuous and integrable over [a, b] and 0 < q < p ≤ 1. If a D p,q f  is convex function over [a, b] for r ≥ 1, then we have the following ( p, q)-midpoint type inequality

M. Kunt et al.

     1 qa + pb  − f  p+q p (b − a)   ≤ q (b − a)

pb+(1−  p)a

1− r1

p2 ( p + q)3

a

⎡

    f (x) a d p,q x   

⎤   

1  a D p,q f (b)r M1 ( p, q) + a D p,q f (a)r M2 ( p, q) r ⎣   r r

1 ⎦ + a D p,q f (b) M3 ( p, q) + a D p,q f (a) M4 ( p, q) r (4.11)

where M1 ( p, q) − M4 ( p, q) are defined as in Theorem 7. Proof Taking absolute value  on both r sides of (4.1), applying the power mean inequality and by using the convexity of a D p,q f  for r ≥ 1, we have that      pb+(1−   p)a   qa + pb 1   f (x) a d p,q x  − f   p+q p (b − a)   a ⎤ ⎡ p p+q    ⎥ ⎢ t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎥ ⎢ 0 ⎥ ≤ q (b − a) ⎢   1 ⎥ ⎢    1 ⎦ ⎣+   0 d p,q t q − t a D p,q f (tb + (1 − t) a) ⎡

p p+q

⎡

⎛

⎤

⎤

⎞1− 1

p p+q

r

 ⎢ ⎥ ⎢ ⎝ ⎢ t 0 d p,q t ⎠ ⎥ ⎢ ⎢ ⎥ ⎢ 0 ⎢ ⎥ ⎢ ⎢ ⎢ 1 ⎥ ⎞ ⎛ ⎢ p ⎢ r ⎥ ⎢ ⎥ ⎢ p+q     ⎢ ⎣ ⎝  t t a D p,q f (b)r + (1 − t) a D p,q f (a)r  0 d p,q t ⎠ ⎦ ⎢ ⎢ 0 ⎢ ≤ q (b − a) ⎢ ⎡ ⎞1− 1 ⎛ ⎢ r  ⎢ ⎢ 1  1 ⎢ ⎢ ⎠ ⎝ 0 d p,q t ⎢ ⎢ q −t p ⎢ ⎢ p+q ⎢+⎢ ⎞1 ⎢ ⎢⎛ r ⎢ ⎢   r  r  ⎢ ⎣ 1  1 a D p,q f (b) + (1 − t) a D p,q f (a) ⎝ ⎠ ⎣ − t d t t 0 p,q q p p+q

 ≤ q (b − a)

p2 ( p + q)3

1− r1

⎡

⎛

p

 p+q   2 a D p,q f (b)r t

⎞ r1

⎢ ⎟ ⎜ 0 d p,q t ⎢ ⎟ ⎜ ⎢ 0 ⎟ ⎜ ⎢ p ⎟ ⎜ ⎢ p+q r  ⎠ ⎝  ⎢ ⎢ t (1 − t) 0 d p,q t + a D p,q f (a) ⎢ 0 ⎢ ⎢ ⎛ ⎞ r1  r 1  1  ⎢ a D p,q f (b) ⎢ − t t d t 0 p,q q ⎟ ⎢ ⎜ p ⎟ ⎢ ⎜ p+q ⎟ ⎢+⎜  ⎟ 1  ⎢ ⎜    1 ⎠ ⎣ ⎝ +  D f (a)r − t − t) d t (1 a p,q 0 p,q q

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤⎥ ⎥ ⎥ ⎥⎥ ⎥⎥ ⎥⎥ ⎥⎥ ⎥⎥ ⎥⎥ ⎥⎥ ⎦⎥ ⎦

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

p p+q

(4.12) Making use of (4.5)–(4.8) in (4.12), gives us the desired result (4.11). Thus the proof is accomplished. 

( p, q)-estimates for midpoint type inequalities

Corollary 2 In Theorem 8; 1. If we take p = 1, then we have the following q-midpoint type inequality       b   qa + b 1  f f d x − (x) a q   1+q (b − a)   a ⎡  ⎞1 ⎛    r 1 a D p,q f (b)r 3 ⎢ 2 (1+q) (1+q+q )  ⎠  ⎢ ⎝  r q ⎢ + a D p,q f (a) ⎢ 1 (1+q)2 (1+q+q 2 ) ⎢ ≤ q (b − a)  ⎞1 ⎛  3 ⎢   r 2 (1 + q)3− r ⎢ a D p,q f (b)r 3 ⎢ 2 (1+q) (1+q+q )  ⎠  ⎣+⎝  r −1+q+q 2 + a D p,q f (a) (1+q)3 (1+q+q 2 )

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ , (4.13) ⎥ ⎥ ⎥ ⎦

2. If we take p = 1 and q → 1− , then we have the following midpoint type inequality for convex functions       b   a + b 1 f  f d x − (x)   2 (b − a)   a ⎡ ⎤  1 r 1 1    r   1 ⎣  f  (b)r 24 + f (a) 12 ≤ (b − a)   

1 ⎦ .  3 23− r +  f  (b)r 1 +  f  (a)r 1 r 12

(4.14)

24

Remark 7 One can see the following: 1. In (4.13), we recapture the inequality [2, Theorem 7, (4.10)], 2. In (4.14), we recapture the inequality [2, Corollary 2, (4.12)]. Theorem 9 Let f : [a, b] → R is a ( p, q)-differentiable function over  r (a, b), a D p,q f is continuous and integrable over [a, b] and 0 < q < p ≤ 1. If a D p,q f  is convex function over [a, b] for r > 1, then we get:      pb+(1−   p)a   qa + pb 1   − f (x) a d p,q x  f   p+q p (b − a)   a ⎡ ⎛ ⎞1    p2  r  a D p,q f (b)r s+1   1s 3 ⎢ p p−q p+q) ( ⎝   ⎠   ⎢ 3 2 2 2 r p+q p s+1 −q s+1 ⎢ + a D p,q f (a) p +2 p q+ pq3 − p ⎢ ( p+q) ≤ q (b − a) ⎢ ⎛  ⎞1 ⎛ ⎞1    ⎢ r s a D p,q f (b)r 2 pq+q32 s ⎢ 1  1 ( p+q) ⎣+⎝ ⎝ ⎠   ⎠ − t d t   0 p,q 2 2 2 3 r q +q p + a D p,q f (a) p q+2 pq −2 pq−q p+q p+q)3

⎤ ⎥ ⎥ ⎥ ⎥ ⎥, ⎥ ⎥ ⎦

(

(4.15) where r −1 + s −1 = 1.

M. Kunt et al.

Proof Taking absolute value  r on both sides of (4.1), applying the Hölder inequality and using the convexity of a D p,q f  for r > 1, we have that      pb+(1−   p)a   qa + pb 1   f (x) a d p,q x  − f   p+q p (b − a)   a ⎤ ⎡ p p+q    ⎥ ⎢ t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎥ ⎢ 0 ⎥ ≤ q (b − a) ⎢  1  ⎥ ⎢  ⎣ +  1 − t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎦ q ⎡

p p+q

⎛

p p+q



s

p p+q

   a D p,q f (tb + (1 − t) a)r 0 d p,q t ⎠ ⎝

⎢ ⎝ ts ⎢ ⎢ 0 ⎢ ≤ q (b − a) ⎢ ⎛ ⎢ s ⎢ 1  1 ⎣+⎝ q −t ⎡

⎞1 ⎛

r

⎥ ⎥ ⎥ ⎥ ⎞1 ⎥ r ⎥ ⎥ ⎦ 0 d p,q t ⎠

0 d p,q t ⎠

0

⎞1 ⎛ s

0 d p,q t ⎠ ⎝

p p+q

 1  a D p,q f (tb + (1 − t) a)r

p p+q

⎛

⎞ r1

p

  p+q  a D p,q f (b)r t

⎛ p ⎞ ⎢ ⎜ ⎟ ⎢ 0 d p,q t p+q ⎜ ⎟  s ⎢ 0 ⎟ ⎝ t 0 d p,q t ⎠ ⎜ ⎢ p ⎜ ⎟ ⎢ p+q r  0 ⎝  ⎠ ⎢ ⎢ + a D p,q f (a) (1 − t) 0 d p,q t ⎢ 0 ≤ q (b − a) ⎢ ⎢ ⎛ ⎞ r1   1 ⎢ a D p,q f (b)r ⎢ ⎛ ⎞1 t d t 0 p,q s ⎜ ⎟ ⎢ p s ⎜ ⎟ ⎢ 1  1 p+q ⎜ ⎟ ⎢+⎝ ⎠ − t d t 0 p,q ⎜ ⎟ q 1 ⎢   p ⎝ +  D f (a)r  (1 − t) d t ⎠ ⎣ p+q a p,q 0 p,q 1 s

⎡

⎤

⎞1

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

p p+q

⎞1    p2  r a D p,q f (b)r 3 ⎢ p p−q ⎝   3 ( p+q) ⎠  ⎢ 2 2 2 s+1 s+1 r p+q p −q ⎢ + a D p,q f (a) p +2 p q+ pq3 − p ⎢ ( p+q) ≤ q (b − a) ⎢ ⎛ 1 ⎛ ⎞ ⎞1  r  2 pq+q 2  ⎢ r s     1 ⎢ D f (b) s  1 a p,q 3 ( p+q) ⎣+⎝ ⎠ ⎝   ⎠ − t d t   0 p,q 2 2 2 +q 3 r q p + a D p,q f (a) p q+2 pq −2 pq−q 3 

s+1 

 1s

⎛

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

( p+q)

p+q



Thus the proof is accomplished. Corollary 3 In Theorem 9; 1. If we take p = 1, then we have the following q-midpoint type inequality

      b   1 qa + b f  − f d x (x) a q   1+q (b − a)   a ⎡ 1

  1   r r s 2q+q 2 |a Dq f (a)| |a Dq f (b)|r 1−q 1 + ⎢ s+1 s+1 3 3 1−q (1+q) (1+q) (1+q) ⎢ ⎢ ⎞1 ≤ q (b − a) ⎢ ⎛ s  1



s r r ⎢ r 1  1 2q+q 2 |a Dq f (b)| −q+q 2 +q 3 |a Dq f (a)| ⎣+⎝ ⎠ − t d t + q 0 3 3 q 1 1+q

(1+q)

(1+q)

⎤ ⎥ ⎥ ⎥ ⎥, ⎥ ⎦

(4.16)

( p, q)-estimates for midpoint type inequalities

2. If we take p = 1 and q → 1− , then we have the following midpoint type inequality for convex functions       b   a + b 1 f − f (x) d x   − a) 2 (b   a ⎡ ⎤ r 1     r  1 r s     4 (b − a) 3 f (a) ⎣ f (b) + ≤ (4.17) r   r 1 ⎦ .  16 s+1   + 3 f (b) +  f (a) r Remark 8 One can see the following: 1. In (4.16), we recapture the inequality [2, Theorem 8, (4.13)], 2. In (4.17), we recapture the inequality [7, Theorem 2.3, (2.1)]. Theorem 10 Let f : [a, b] → R is a ( p, q)-differentiable function over  r (a, b), a D p,q f is continuous and integrable over [a, b] and 0 < q < p ≤ 1. If a D p,q f  is convex function over [a, b] for r > 1, the we have the following ( p, q)-midpoint type inequality      pb+(1−   p)a   qa + pb 1   − f (x) a d p,q x  f   p+q p (b − a)   a    

1    1s a D p,q f (b) M1 ( p, q) + a D p,q f (a) M2 ( p, q) r p2 ≤ q (b − a)   

1 ,  ( p + q)3 + a D p,q f (b) M3 ( p, q) + a D p,q f (a) M4 ( p, q) r (4.18) where M1 ( p, q) − M4 ( p, q) are defined as in Theorem 7 and r −1 + s −1 = 1. Proof Taking absolute value  r on both sides of (4.1), applying the Hölder inequality and using the convexity of a D p,q f  for r > 1, we have that      pb+(1−   p)a   qa + pb 1   − f (x) a d p,q x  f   p+q p (b − a)   a ⎡ ⎤ p p+q      ⎢ ⎥ t a D p,q f (tb + (1 − t) a) 0 d p,q t ⎢ ⎥ 0 ⎥ ≤ q (b − a) ⎢   1 ⎢ ⎥    1 ⎣+ a D p,q f (tb + (1 − t) a) 0 d p,q t ⎦ − t q ⎡

p p+q

⎡

⎤

p

p+q   1 1 ⎢ t s t r a D p,q f (tb + (1 − t) a) 0 d p,q t ⎢ 0 ≤ q (b − a) ⎢ 1  1  1  ⎢  s r 1 1 ⎣+  a D p,q f (tb + (1 − t) a) q −t q −t

⎥ ⎥ ⎥ ⎥ ⎦ 0 d p,q t

p p+q

⎛

p p+q



⎞1 ⎛ s

p p+q



 r t a D p,q f (tb + (1 − t) a)

⎞1 r

⎢ ⎝ t 0 d p,q t ⎠ ⎝ ⎢ 0 d p,q t ⎠ ⎢ 0 0 ⎢ ≤ q (b − a) ⎢ ⎛ ⎞1 ⎛ ⎞1 ⎢ s r   ⎢ r 1  1 1  1 ⎣+⎝   0 d p,q t ⎠ 0 d p,q t ⎠ ⎝ q −t q − t a D p,q f (tb + (1 − t) a) p p+q

p p+q

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

M. Kunt et al. ⎡

⎛ p ⎞1 r p+q r  r     ⎢ a D p,q f (b) + (1 − t) a D p,q f (a) ⎝ ⎠ t t d t ⎢ 0 p,q  1s ⎢  0 p2 ⎢ ≤ q (b − a) ⎢ ⎛ ⎞1 ⎢ r ( p + q)3   ⎢ r  r  1  1 ⎣+⎝     ⎠ − t D f + − t) D f d t t (b) (1 (a) a p,q a p,q 0 p,q q ⎡

p p+q

⎛ ⎞ r1 p  r p+q  2 ⎢   ⎜ ⎟ t 0 d p,q t ⎢ a D p,q f (b) ⎜ ⎟ ⎢ 0 ⎜ ⎟ ⎢ p ⎜ ⎟ ⎢ p+q   ⎝ ⎠  ⎢ r a D p,q f (a) 1 ⎢ + t − t) d t (1   0 p,q s ⎢ p2 0 ⎢ ≤ q (b − a) ⎢ ⎛ ⎞ r1  ( p + q)3  r 1  1 ⎢ a D p,q f (b) ⎢ − t t d t 0 p,q q ⎟ ⎢ ⎜ p ⎟ ⎢ ⎜ p+q ⎟ ⎢+⎜   ⎜ ⎟ 1 ⎢    r 1 ⎣ ⎝ +  D f (a) − t (1 − t) d t ⎠ a

p,q

p p+q

0 p,q

q

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(4.19)

Making use of (4.5)–(4.8) in (4.19), gives us the desired result (4.18). Thus the proof is accomplished.  Corollary 4 In Theorem 10; 1. If we take p = 1, then we have the following q-midpoint type inequality

      b   1  f qa + b −  f d x (x) a q   − a) 1 + q (b   a ⎡  1 ⎤    r q 3  1 a Dq f (a)r a Dq f (b)r + s 3 2 2 2 1 ⎢ ⎥ (1+q) (1+q+q ) (1+q) (1+q+q ) ≤ q (b − a) ⎣  1 ⎦,  r r 1+q r −1+q+q 2 2   + + a Dq f (b) D f (a) a q (1+q)3 (1+q+q 2 ) (1+q)3 (1+q+q 2 )

(4.20) 2. If we take p = 1 and q → 1− , then we have the following midpoint type inequality for convex functions       b   a + b 1 f  f d x − (x)   2 (b − a)   a ⎡ ⎤  1 r 1 1    3 r 1 s ⎣  f  (b)r 24 +  f  (a) 12 ≤ (b − a) (4.21) r r 1 ⎦ .   2 +  f  (b) 1 +  f  (a) 1 r 12

24

Remark 9 One can see the following: 1. In (4.20), we recapture the inequality [2, Theorem 9, (4.15)], 2. In (4.21), we recapture the inequality [2, Corollary 4, (4.17)]. Some results related to quasi-convexity are presented in the following theorems. Theorem 11 Let f : [a, b] → R is a ( p, q)-differentiable function b), a D p,q f is  over(a, r continuous and integrable over [a, b] and 0 < q < p ≤ 1. If a D p,q f  is quasi-convex function over [a, b] for r ≥ 1, then we have the following ( p, q)-midpoint type inequality

( p, q)-estimates for midpoint type inequalities

     qa + pb 1  − f  p+q p (b − a)  ≤ q (b − a)

2 p2 ( p + q)

3

pb+(1−  p)a

a

    f (x) a d p,q x   

  " ! sup a D p,q f (a) , a D p,q f (b) .

(4.22)

Proof Taking absolute value on both sides r of (4.1), applying the power mean inequality and using the quasi-convexity of a D p,q f  for r ≥ 1, we have that    1    p (b − a) 

pb+(1−  p)a

a

⎡

  p f (b) + q f (a)  f (x) a d p,q x −   p+q  p p+q



⎤

  t a D p,q f (tb + (1 − t) a)

⎢ ⎥ 0 d p,q t ⎢ ⎥ 0 ⎥ ≤ q (b − a) ⎢   1 ⎢ ⎥   1 ⎣+  a D p,q f (tb + (1 − t) a) 0 d p,q t ⎦ − t q ⎡

p p+q

⎛ p ⎞1− 1 ⎛ p ⎞1 r r p+q p+q r    ⎢   ⎝ ⎠ ⎠ ⎝ t d t t D f + − t) a) d t (tb (1 ⎢ 0 p,q 0 p,q a p,q ⎢ 0 0 ⎢ ≤ q (b − a) ⎢ ⎛ 1 ⎛ ⎞ ⎞1 1− r ⎢ r   ⎢ r 1  1 1  1 ⎣+⎝   ⎠ ⎠ ⎝ − t d t − t D f + − t) a) d t (1 0 p,q 0 p,q a p,q (tb q q p p+q

p p+q

⎛

⎡

⎛

⎞

⎞1− 1

p p+q

r



⎜ ⎟ ⎢ ⎝ t 0 d p,q t ⎠ ⎜ ⎟ ⎢ ⎜ ⎟ ⎢ 0 ⎜ ⎟ ⎢ ⎜ 1 ⎟ ⎛ ⎞ ⎢ p ⎜ r ⎟ ⎢ ⎜ ⎟    p+q ⎢ ⎝ × ⎝sup !a D p,q f (a)r , a D p,q f (b)r "  t 0 d p,q t ⎠ ⎠ ⎢ ⎢ 0 ⎢ ≤ q (b − a) ⎢ ⎛ ⎞1− 1 ⎢ ⎛ r ⎢  1  1 ⎢ ⎜ ⎝ ⎠ ⎢ ⎜ − t d t 0 p,q q ⎢ ⎜ p ⎢ ⎜ p+q ⎢+⎜ ⎛ ⎞1 ⎢ ⎜ r  ⎢ ⎜ 1  ⎣ ⎝ × ⎝sup ! D f (a)r ,  D f (b)r "  1 −t a p,q a p,q 0 d p,q t ⎠ ⎡   " ⎢ ! ≤ q (b − a) sup a D p,q f (a) , a D p,q f (b) ⎢ ⎣

p p+q

p

1 

p+q t

0 d p,q t

0

≤ q (b − a)

2 p2 ( p + q)3

q

+ p p+q

1 −t q

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎞⎥ ⎥ ⎥ ⎟⎥ ⎟⎥ ⎟⎥ ⎟⎥ ⎟⎥ ⎟⎥ ⎟⎥ ⎠⎦ ⎤



⎥ ⎥

0 d p,q t ⎦

  " ! sup a D p,q f (a) , a D p,q f (b) .

Hence the inequality (4.22) is established. Thus the proof is accomplished. Corollary 5 In Theorem 11; 1. If we take p = 1, then we have the following q-midpoint type inequality



M. Kunt et al.

    b  1  f qa + b − f (x)  1+q (b − a)  a

≤ (b − a)

2q (1 + q)

3

    a dq x  

  " ! sup a Dq f (a) , a Dq f (b) ,

(4.23)

2. If we take p = 1 and q → 1− , then we have the following midpoint type inequality for quasi-convex functions       b   (b − a)   " ! a + b 1 f − (4.24) f (x) d x  ≤ sup  f  (a) ,  f  (b) .  2 4 (b − a)   a

Remark 10 One can see the following: 1. In (4.23), we recapture the inequality [2, Theorem 10, (4.18)] , 2. In (4.24), we recapture the inequality [2, Corollary 5, (4.19)]. Theorem 12 Let f : [a, b] → R is a ( p, q)-differentiable function b), a D p,q f is  over(a, r continuous and integrable over [a, b] and 0 < q < p ≤ 1. If a D p,q f  is quasi-convex function over [a, b] for r > 1, then we have the following ( p, q)-midpoint type inequality      pb+(1−   p)a   qa + pb 1   − f (x) a d p,q x  f   p+q p (b − a)   a ⎡ ⎤  s+1   1s  1 r p p−q p ⎢ ⎥ p+q p+q p s+1 −q s+1 ⎥   "⎢ ! ⎢ ⎛ ⎥ 1 ⎞     ≤ q (b−a) sup a D p,q f (a) , a D p,q f (b) ⎢ s  1 ⎥ s ⎢ ⎥ 1  1 r q ⎣+ ⎝ ⎦ 0 d p,q t ⎠ q −t p+q p p+q

(4.25) where r −1 + s −1 = 1. Proof Taking absolute value on both sides of (4.1), applying the Hölder inequality and using r the quasi-convexity of a D p,q f  for r > 1, we have that    1    p (b − a) 

pb+(1−  p)a

a

⎡

  p f (b) + q f (a)  f (x) a d p,q x −   p+q  p p+q



  t a D p,q f (tb + (1 − t) a)

⎤

⎢ ⎥ 0 d p,q t ⎢ ⎥ 0 ⎥ ≤ q (b − a) ⎢  1  ⎢ ⎥   1 ⎣+ a D p,q f (tb + (1 − t) a) 0 d p,q t ⎦ − t q ⎡

p p+q

⎛

p p+q



s

p p+q

   a D p,q f (tb + (1 − t) a)r 0 d p,q t ⎠ ⎝

⎢ ⎝ ts ⎢ ⎢ 0 ⎢ ≤ q (b − a) ⎢ ⎛ ⎢ s ⎢ 1  1 ⎣+⎝ q −t p p+q

⎞1 ⎛

⎞1 r

⎞1 ⎛ s

0 d p,q t ⎠

⎝

 1  a D p,q f (tb + (1 − t) a)r

p p+q

⎥ ⎥ ⎥ ⎥ ⎞1 ⎥ r ⎥ ⎥ ⎦ 0 d p,q t ⎠

0 d p,q t ⎠

0

⎤

( p, q)-estimates for midpoint type inequalities

⎡  s+1   1s   r  r " p  r1 ! p p−q sup a D p,q f (a) , a D p,q f (b) p+q ⎢ s+1 −q s+1 p+q p ⎢ ⎢ ⎞1 ≤ q (b − a) ⎢ ⎛ s  s ⎢ r  r " q  r1 1  1 ! ⎣+⎝ ⎠ sup a D p,q f (a) , a D p,q f (b) p+q 0 d p,q t q −t p p+q

  ! ≤ q (b − a) sup a D p,q f (a) , a D p,q

⎡  s+1   1s  1  r p p−q p ⎢ s+1 s+1 p+q p+q p −q " ⎢ ⎢ ⎞1 f (b) ⎢ ⎛ s  1 s ⎢ 1  1 r q ⎣+⎝ ⎠ − t d t 0 p,q q p+q

⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ ⎥ ⎥ ⎥ ⎥. ⎥ ⎦

p p+q



The inequality (4.25) is proved. Corollary 6 In Theorem 12;

1. If we take p = 1, than we have the following q-midpoint type inequality       b   qa + b 1 f − f (x) a dq x   1+q (b − a)   a   " ! ≤ q (b − a) sup a Dq f (a) , a Dq f (b) ⎤ ⎡ ⎛ ⎞1 s 1 1  1 1         s ⎢ s r r⎥ ⎜ ⎟ 1 1 1 q 1−q ⎥ ⎢ ⎜ ⎟ + d t ×⎢ − t ⎥, q 0 ⎝ ⎠ ⎣ (1 + q)s+1 1 − q s+1 1+q q 1+q ⎦ 1 1+q

(4.26) 1− ,

2. If we take p = 1 and q → than we have the following midpoint type inequality for quasi-convex functions        1 b s   (b − a)   " ! a + b 1 1 f  − f (x) d x  ≤ sup  f  (a) ,  f  (b) .  − a) 2 2 s + 1 (b   a (4.27) Remark 11 One can see the following: 1. In (4.26), we recapture the inequality [2, Theorem 11, (4.20)] , 2. In (4.27), we recapture the inequality [2, Corollary 6, (4.21)]. Compliance with ethical standards Conflict of interest The authors declare that they have no competing interests. Author contributions All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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