Iran J Sci Technol Trans Sci https://doi.org/10.1007/s40995-017-0352-4

RESEARCH PAPER

Hermite–Hadamard–Feje´r Type Inequalities for p-Convex Functions via Fractional Integrals Mehmet Kunt1 • ˙Imdat ˙Is¸can2 Received: 4 October 2016 / Accepted: 23 November 2017 Ó Shiraz University 2017

Abstract In this paper, firstly, Hermite–Hadamard–Feje´r type inequalities for p-convex functions in fractional integral forms are built. Secondly, an integral identity and some Hermite–Hadamard–Feje´r type integral inequalities for p-convex functions in fractional integral forms are obtained. Finally, some Hermite–Hadamard and Hermite–Hadamard–Feje´r inequalities for convex, harmonically convex and p-convex functions are given. Many results presented here for p-convex functions provide extensions of others given in earlier works for convex, harmonically convex and p-convex functions. Keywords Hermite–Hadamard inequalities  Hermite–Hadamard–Feje´r inequalities  Fractional integrals  Convex functions  Harmonically convex functions  p-Convex functions Mathematics Subject Classification 26A51  26A33  26D10

1 Introduction Let f : I  R ! R be a convex function defined on the interval I of real numbers and a; b 2 I with a\b. The inequality   Z b aþb 1 f ðaÞ þ f ðbÞ ð1Þ f ; f ðxÞdx   2 ba a 2 is well known in the literature as Hermite–Hadamard’s inequality (Hadamard 1893; Hermite 1883). The most well-known inequalities related to the integral mean of a convex function f are the Hermite–Hadamard inequalities or its weighted versions, the so-called Hermite–Hadamard–Feje´r inequalities. Feje´r (1906) established the following Feje´r inequality which is the weighted generalization of Hermite–Hadamard inequality (1): & Mehmet Kunt [email protected] ˙Imdat ˙Is¸ can [email protected]; [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

Theorem 1 Let f : ½a; b! R be convex function. Then the inequality  Z b Z b aþb f wðxÞdx  f ðxÞwðxÞdx 2 a a ð2Þ Z f ðaÞ þ f ðbÞ b wðxÞdx;  2 a holds, where w : ½a; b! R is nonnegative, integrable and symmetric to ða þ bÞ=2: For some results which generalize, improve, and extend the inequalities (1) and (2), see Bombardelli and Varosˇanec (2009), Chen and Wu (2014), Dragomir and Agarwal (1998), Fang and Shi (2014), I˙s¸ can (2013, 2014c, d, 2016b, c), Mihai et al. (2015), Noor et al. (2016), Pearce and Pecaric (2000), Sarıkaya (2012) and Tseng et al. (2011). We will now give definitions of the right-hand side and left-hand side Riemann–Liouville fractional integrals which are used throughout this paper. Definition 1 (Kilbas et al. 2006). Let f 2 L½a; b. The right-hand side and left-hand side Riemann–Liouville a a fractional integrals Jaþ f and Jb f of order a [ 0 with b [ a  0 are defined by

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a Jaþ f ðxÞ ¼ a f ðxÞ ¼ Jb

1 CðaÞ 1 CðaÞ

Z Z

x

ð x  tÞa1 f ðtÞdt; x [ a and

a b

ðt  xÞa1 f ðtÞdt; x\b;

x

respectively, where CðaÞ is the Gamma function defined by R1 CðaÞ ¼ 0 et ta1 dt. Because of the wide application of Hermite–Hadamard type inequalities and fractional integrals, many researchers extend their studies to Hermite–Hadamard type inequalities involving fractional integrals not limited to integer integrals. Recently, more and more Hermite–Hadamard inequalities involving fractional integrals have been obtained for different classes of functions; see Dahmani (2010), ˙Is¸ can (2014a, b, 2015), ˙Is¸ can and Wu (2014), I˙s¸ can et al. (2016d), Sarıkaya et al. (2013) and Wang et al. (2012, 2013). ˙Is¸ can (2014d) gave the definition of harmonically convex function and established the following Hermite– Hadamard type inequality for harmonically convex functions as follows: Definition 2 Let I  Rnf0g be a real interval. A function f : I ! R is said to be harmonically convex if   xy f  tf ð yÞ þ ð1  tÞf ð xÞ; ð3Þ tx þ ð1  tÞy for all x; y 2 I and t 2 ½0; 1. If the inequality in (3) is reversed, then f is said to be harmonically concave. Theorem 2 (I˙s¸ can 2014d). Let f : I  Rnf0g ! R be a harmonically convex function and a; b 2 I with a\b. If f 2 L½a; b; then the following inequalities hold:   Z b 2ab ab f ð xÞ f ð aÞ þ f ð bÞ ð4Þ f : dx   aþb b  a a x2 2

Chen and Wu (2014) presented Hermite–Hadamard– Feje´r inequality for harmonically convex functions as follows: Theorem 3 Let f : I  Rnf0g ! R be a harmonically convex function and a; b 2 I with a\b. If f 2 L½a; b and w : ½a; b  Rnf0g ! R is nonnegative, integrable and 2ab harmonically symmetric with respect to aþb , then  Z b Z b 2ab w ð xÞ f ð xÞwð xÞ f dx  dx a þ b a x2 x2 a ð5Þ Z b f ð aÞ þ f ð bÞ wð xÞ dx:  2 x2 a

123

Sarıkaya et al. (2013) presented Hermite–Hadamard inequality for convex functions via fractional integrals as follows: Theorem 4 Let f : ½a; b ! R be a positive function with 0  a\b and f 2 L½a; b. If f is a convex function on ½a; b, then the following inequalities for fractional integrals hold:    aþb Cða þ 1Þ  a a  f a Jaþ f ðbÞ þ Jb f ðaÞ 2 2ðb  aÞ ð6Þ f ðaÞ þ f ðbÞ  ; 2 with a [ 0. I˙s¸ can and Wu (2014) presented Hermite–Hadamard inequality for harmonically convex functions via fractional integrals as follows: Theorem 5 Let f : I  ð0; 1Þ ! R be a function such that f 2 L½a; b, where a; b 2 I with a\b. If f is a harmonically convex function on ½a; b, then the following inequalities for fractional integrals hold:     2ab Cða þ 1Þ ab a h a f J1=a ð f gÞð1=bÞ  aþb 2 ba ð7Þ i f ðaÞ þ f ðbÞ a þ J1=bþ ; ð f gÞð1=aÞ  2 1 1 1 with a [ 0 and gð xÞ ¼ x, x 2 b ; a . ˙Is¸ can (2015) presented Hermite–Hadamard–Feje´r inequality for convex functions via fractional integrals as follows: Theorem 6 Let f : ½a; b ! R be a convex function with a\b and f 2 L½a; b. If w is nonnegative, integrable and symmetric to ða þ bÞ=2, then the following inequalities for fractional integrals hold:    aþb  a a f wðaÞ Jaþ wðbÞ þ Jb 2  a  a ð8Þ  Jaþ ð fwÞðbÞ þ Jb ð fwÞðaÞ 

 f ðaÞ þ f ðbÞ  a a Jaþ wðbÞ þ Jb w ð aÞ ; 2

with a [ 0. I˙s¸ can et al. (2016d) presented Hermite–Hadamard–Feje´r inequality for harmonically convex functions via fractional integrals as follows: Theorem 7 Let f : ½a; b! R be a harmonically convex function with a\b and f 2 L½a; b. If w : ½a; b! R is nonnegative, integrable and harmonically symmetric with respect to 2ab=a þ b, then the following inequalities for fractional integrals holds:

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 i 2ab h a a ðw gÞð1=bÞ J1=bþ ðw gÞð1=aÞ þ J1=a aþb h i a a  J1=bþ ð fw gÞð1=aÞ þ J1=a ð fw gÞð1=bÞ i f ðaÞ þ f ðbÞ h a a J1=bþ ðw gÞð1=aÞ þ J1=a  ðw gÞð1=bÞ ; 2 ð9Þ   with a [ 0 and gðxÞ ¼ 1x, x 2 1b ; 1a . f

Zhang and Wan (2007) gave the definition of p-convex function on I  R, ˙Is¸ can (2016c) gave a different definition of p-convex function on I  ð0; 1Þ as follows:

Definition 4 Let p 2 Rnf0g. A function w : ½a; b  ð0; 1Þ ! R is said to be p-symmetric with respect to ap þbp 1=p if 2   wð xÞ ¼ w ½ap þ bp  xp 1=p ; holds for all x 2 ½a; b. Lemma 1 Let p 2 Rnf0g, a [ 0 and w : ½a; b  ð0; 1Þ ! R is integrable, p-symmetric with respect to ap þbp 1=p , then 2 (i)

Jaap þ ðw gÞðbp Þ ¼ Jbap  ðw gÞðap Þ  1 ¼ Jaap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ ; 2

Definition 3 Let I  ð0; 1Þ be a real interval and p 2 Rnf0g. A function f : I ! R is said to be p-convex, if   f ½txp þ ð1  tÞyp 1=p  tf ð xÞ þ ð1  tÞf ð yÞ ð10Þ for all x; y 2 I and t 2 ½0; 1.

1

(ii)

with gð xÞ ¼ xp , x 2 ½ap ; bp . If p\0, Jbap þ ðw gÞðap Þ ¼ Jaap  ðw gÞðbp Þ  1 ¼ Jbap þ ðw gÞðap Þ þ Jaap  ðw gÞðbp Þ ; 2

It can be easily seen that for p ¼ 1 and p ¼ 1, pconvexity reduces to ordinary convexity and harmonically convexity of functions defined on I  ð0; 1Þ, respectively. In Fang and Shi (2014), Theorem 5, if we take I  ð0; 1Þ, p 2 Rnf0g and hðtÞ ¼ t, then we have the following theorem. Theorem 8 Let f : I  ð0; 1Þ ! R be a p-convex function, p 2 Rnf0g, and a; b 2 I with a\b. If f 2 L½a; b then the following inequalities hold:  p p 1=p ! Z b a þb p f ð xÞ f ðaÞþf ðbÞ f : dx   p p 1p b a a x 2 2

If p [ 0,

1

with gð xÞ ¼ xp , x 2 ½bp ; ap . Proof (i)

Let p [ 0. Since w is p-symmetric with respect to ap þbp 1=p

 , using Definition 4 we have w x1=p ¼ 2   w ½ap þ bp  x1=p for all x 2 ½ap ; bp . Hence in the following integral setting t ¼ ap þ bp  x and dt ¼ dx gives

ð11Þ

Jaap þ ðw gÞðbp Þ ¼

For some results related to p-convex functions and its generalizations, we refer the reader to see Fang and Shi (2014), I˙s¸ can (2016a, b, c), Mihai et al. (2015), Noor et al. (2016) and Zhang and Wan (2007). In this paper, we built Hermite–Hadamard–Feje´r type inequalities for p-convex functions in fractional integral forms. We obtain an integral identity and some Hermite– Hadamard–Feje´r type integral inequalities for p-convex functions in fractional integral forms. We give some Hermite–Hadamard and Hermite–Hadamard–Feje´r inequalities for convex, harmonically convex and p-convex functions.

1 CðaÞ

1 ¼ CðaÞ 1 ¼ CðaÞ

(ii)

Z

bp

Z

ap bp

Z

ap bp ap

  ðbp xÞa1 w x1=p dx   ðbp xÞa1 w ½ap þbp x1=p dx   ð xap Þa1 w x1=p dx ¼ Jbap  ðw gÞðap Þ:

This completes the proof of i. The proof is similar to i.

Theorem 9 Let f : I  ð0; 1Þ ! R be a p-convex function, p 2 Rnf0g, a [ 0 and a; b 2 I with a\b. If f 2 L½a; b and w : ½a; b ! R is nonnegative, integrable and p p p 1=p symmetric with respect to a þb , then the following 2 inequalities for fractional integrals hold:

2 Main Results Throughout this section, kwk1 ¼ supt2½a;b jwðtÞj, for the continuous function w : ½a; b! R.

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(i)

    f ½tap þ ð1  tÞbp 1=p þ f ½tbp þ ð1  tÞap 1=p

If p [ 0,

 p !  a þ bp 1=p  a Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ f 2  a   Jap þ ð fw gÞðbp Þ þ Jbap  ð fw gÞðap Þ  f ðaÞ þ f ðbÞ  a  Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ ; 2

2 f ðaÞ þ f ðbÞ :  2 ð15Þ Multiplying both sides of (15) by   2t w ½tap þ ð1  tÞbp 1=p and integrating with

ð12Þ (ii)

a1

with gðxÞ ¼ x1=p , x 2 ½ap ; bp . If p\0,

!  ap þ bp 1=p  a Jbp þ ðw gÞðap Þ þ Jaap  ðw gÞðbp Þ f 2    Jbap þ ð fw gÞðap Þ þ Jaap  ð fw gÞðbp Þ  f ðaÞ þ f ðbÞ  a Jbp þ ðw gÞðap Þ þ Jaap  ðw gÞðbp Þ ;  2

respect to t over ½0; 1, using Lemma 1-i, we get  a  Jap þ ð fw gÞðbp ÞþJbap  ð fw gÞðap Þ  f ðaÞþf ðbÞ  a Jap þ ðw gÞðbp ÞþJbap  ðw gÞðap Þ ;  2



ð13Þ

(ii)

the right hand side of (12). This completes the proof of i. The proof is similar to i. h

with gðxÞ ¼ x1=p , x 2 ½bp ; ap . Remark 1 Proof (i)

Let p [ 0. Since f : I  ð0; 1Þ ! R is a p-convex function, we have, for all x; y 2 I (with t ¼ 12 in the inequality (10))  p ! x þ yp 1=p f ð xÞ þ f ð yÞ : f  2 2 Choosing

x ¼ ½tap þ ð1  tÞbp 1=p

p

p 1=p

and

y¼

½tb þ ð1  tÞa  , we get  p p 1=p ! a þb f 2     f ½tap þ ð1tÞbp 1=p þf ½tbp þ ð1tÞap 1=p :  2 ð14Þ Multiplying both sides of (14) by   a1 2t w ½tap þ ð1  tÞbp 1=p and integrating with respect to t over ½0; 1, using Lemma 1-i, we get  p !  a þ bp 1=p  a f Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ 2  a   Jap þ ð fw gÞðbp Þ þ Jbap  ð fw gÞðap Þ ; the left hand side of (12). For the proof of the second inequality in (12), we first note that if f is a p-convex function, then, for all t 2 ½0; 1, it yields

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(1) (2) (3) (4) (5) (6) (7) (8) (9)

In Theorem 9, one can see the following.

If one takes p ¼ 1, one has (8). If one takes p ¼ 1 and wð xÞ ¼ 1, one has (6). If one takes p ¼ 1 and a ¼ 1, one has (2). If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has (1). If one takes p ¼ 1, one has (9). If one takes p ¼ 1 and wð xÞ ¼ 1, one has (7), If one takes p ¼ 1 and a ¼ 1, one has (5). If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has (4). If one takes a ¼ 1 and wð xÞ ¼ 1, one has (11).

Lemma 2 Let f : I  ð0; 1Þ ! R be a differentiable function on I and a; b 2 I with a\b, p 2 Rnf0g and a [ 0. If f 0 2 L½a; b and w : ½a; b ! R is integrable and p p p 1=p symmetric with respect to a þb , then the following 2 equalities for fractional integrals hold: (i)

If p [ 0,  f ðaÞ þ f ðbÞ  a Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp Þ þ Jbap  ð fw gÞðap Þ " # Z bp R t p a1 ðw gÞðsÞds 1 ap ðb  sÞ ð f gÞ0 ðtÞdt; ¼ R p CðaÞ ap  b ðs  ap Þa1 ðw gÞðsÞds t

ð16Þ with gðxÞ ¼ x1=p , x 2 ½ap ; bp .

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(ii)

If p\0,  f ðaÞ þ f ðbÞ  a Jbp þ ðw gÞðap Þ þ Jaap  ðw gÞðbp Þ 2    Jbap þ ð fw gÞðap Þ þ Jaap  ð fw gÞðbp Þ " # R Z ap t a1 p ðw gÞðsÞds 1 bp ða  sÞ ¼ ð f gÞ0 ðtÞdt; R p CðaÞ bp  a ðs  bp Þa1 ðw gÞðsÞds t

(ii)

Remark 2 (1)

ð17Þ (2)

with gðxÞ ¼ x1=p , x 2 ½bp ; ap .

(3) Proof (i)

(4)

Let p [ 0. It suffices to note that # Z bp " R t p a1 ðw gÞðsÞds 1 a p ðb  sÞ ð f gÞ0 ðtÞdt I¼ R p CðaÞ ap  b ðs  ap Þa1 ðw gÞðsÞds t Z bp Z t 1 ðbp  sÞa1 ðw gÞðsÞds ð f gÞ0 ðtÞdt ¼ CðaÞ ap ap Z bp Z bp 1 ðs  ap Þa1 ðw gÞðsÞds ð f gÞ0 ðtÞdt  CðaÞ ap t

(5) (6) (7) (8)

¼ I1  I2 :

ð18Þ By integration by parts and using Lemma 1-i, we have Z t  bp 1 ð f gÞ ð t Þ ðbp  sÞa1 ðw gÞðsÞds I1 ¼ C ð aÞ ap ap Z bp 1  ðbp  tÞa1 ð fw gÞðtÞdt C ð aÞ a p Z bp 1 ¼ f ð bÞ ðbp  sÞa1 ðw gÞðsÞds CðaÞ ap Z bp 1 ðbp  tÞa1 ð fw gÞðtÞdt  C ð aÞ a p  f ð bÞ  a Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ ¼ 2  Jaap þ ð fw gÞðbp Þ

A combination of (18), (19) and (20) gives (16). This completes the proof of i. The proof is similar to i. In Lemma 2, one can see the following.

If one takes p ¼ 1, one has I˙s¸ can (2015), Lemma 2.4. If one takes p ¼ 1 and wð xÞ ¼ 1, one has Sarıkaya et al. (2013), Lemma 2. If one takes p ¼ 1 and a ¼ 1, one has Sarıkaya (2012), Lemma 2.6. If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has Dragomir and Agarwal (1998), Lemma 2.1. If one takes p ¼ 1, one has ˙Is¸ can et al. (2016d), Lemma 3. If one takes p ¼ 1 and wð xÞ ¼ 1, one has ˙Is¸ can and Wu (2014), Lemma 3. If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has ˙Is¸ can (2014d), 2.5. Lemma. If one takes a ¼ 1 and wð xÞ ¼ 1, one has Noor et al. (2016), Lemma 2.4.

Theorem 10 Let f : I  ð0; 1Þ ! R be a differentiable function on I such that f 0 2 L½a; b, where a; b 2 I and a\b. If jf 0 j is p-convex function on ½a; b for p 2 Rnf0g and a [ 0, w : ½a; b ! R is continuous and p-symmetric  p p 1=p with respect to a þb , then the following inequality for 2 fractional integrals hold: (i)

If p [ 0,

f ðaÞ þ f ðbÞ  a  Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ 2    J ap ð fw gÞðbp Þ þ J ap ð fw gÞðap Þ a þ

b 

kwk1 ðbp  ap Þaþ1 ½C1 ða; pÞjf 0 ðaÞj þ C2 ða; pÞjf 0 ðbÞj;  Cða þ 1Þ

where

ð19Þ C1 ða; pÞ ¼

and similarly

Z

p½uap þ ð1  uÞbp 1ð1=pÞ

0

Z bp  bp 1 p a1 I2 ¼ ðsa Þ ðw gÞðsÞds ð f gÞðtÞ CðaÞ t ap Z bp 1 þ ðt ap Þa1 ð fw gÞðtÞdt CðaÞ ap Z bp 1 ¼ f ðaÞ ðsap Þa1 ðw gÞðsÞds CðaÞ ap Z bp 1 þ ðt ap Þa1 ð fw gÞðtÞdt CðaÞ ap  f ðaÞ  a Jap þ ðw gÞðbp ÞþJbap  ðw gÞðap Þ ¼ 2 þJbap  ð fw gÞðap Þ:

¼

(ii)

Z

jð1  uÞa ua j

1

1

0 p½uap

jð1  uÞa ua j þ ð1  uÞbp 1ð1=pÞ

udu

and

C2 ða; pÞ

ð1  uÞdu

with gðxÞ ¼ x1=p , x 2 ½ap ; bp . If p\0, f ðaÞ þ f ðbÞ  a  Jbp þ ðw gÞðap Þ þ Jaap  ðw gÞðbp Þ 2    J ap ð fw gÞðap Þ þ J ap ð fw gÞðbp Þ b þ

a 

kwk1 ðap  bp Þaþ1  ½C3 ða; pÞjf 0 ðaÞj þ C4 ða; pÞjf 0 ðbÞj; Cð a þ 1 Þ

ð20Þ

where

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C3 ða; pÞ ¼ ¼

Z

1

jð1  uÞa ua j

0 p½uap

Z

0

1

þ ð1 

uÞbp 1ð1=pÞ

jð1  uÞa ua j p½uap þ ð1  uÞbp 1ð1=pÞ

udu

and

C4 ða; pÞ

ð1  uÞdu

with gðxÞ ¼ x1=p , x 2 ½bp ; ap . Proof (i)

Let p [ 0. Using Lemma 2-i, it follows that f ðaÞ þ f ðbÞ  a  Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp Þ þ Jbap  ð fw gÞðap Þ R Z bp t p a1 1 ap ðb  sÞ ðw gÞðsÞds 0  ð f gÞ ðtÞ dt: R bp CðaÞ ap  ðs  ap Þa1 ðw gÞðsÞds t

ð21Þ  p p 1=p Since w is p-symmetric with respect to a þb , 2

1=p  using Definition 4 we have w x ¼   w ½ap þ bp  x1=p for all x 2 ½ap ; bp :

A combination of (21) and (22) gives f ðaÞþf ðbÞ  a  Jap þ ðw gÞðbp ÞþJbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp ÞþJbap  ð fw gÞðap Þ 2 R ap þbp R p p  3 a þb t 2 ðsap Þa1 ðw gÞðsÞ ds ð f gÞ0 ðtÞ dt t 1 6 ap 7   4  5 0 CðaÞ þ Rabppþbp R t p a1 ds ð f g Þ dt ð sa Þ ð w g Þ ð s Þ ð t Þ ap þbp t 2 2 R ap þbp R p p  3 a þb t 2 ðsap Þa1 ds ð f gÞ0 ðtÞ dt t kwk1 6 ap 7    4 5 CðaÞ þ Rabppþbp R t p a1 ð f gÞ0 ðtÞ dt ð sa Þ ds p p a þb t 2 2 R ap þbp R p p  1   3 a þb t 2 ðsap Þa1 ds 1ð1=pÞ f 0 t1=p dt ap t 7 kwk1 6 pt 6    1   7 5 CðaÞ 4 R bp R t a1 0 1=p p þ ap þbp ap þbp t ðsa Þ ds 1ð1=pÞ f t dt 2 pt 2 3 R ap þbp ðbp tÞa ðt ap Þa 0  1=p  2 f t dt 7 p 6 a 1 ð 1=p Þ kwk1 6 pt 7  6 7: Cðaþ1Þ 4 R bp ðt ap Þa ðbp tÞa 0  1=p  5 þ ap þbp f t dt 2 pt1ð1=pÞ

Setting t ¼ uap þ ð1  uÞbp and dt ¼ ðap  bp Þdu gives

Z t Z bp ðbp  sÞa1 ðw gÞðsÞds  ðs  ap Þa1 ðw gÞðsÞds p t a Z bp Z t ¼ ðs  ap Þa1 ðw gÞðsÞds þ ðs  ap Þa1 ðw gÞðsÞds ap þbp t bp Z t ¼ ðs  ap Þa1 ðw gÞðsÞds p p 8 a þb t  p p R ap þbp t > > ðs  ap Þa1 ðw gÞðsÞ ds; t 2 ap ; a þ b > < t 2  p :  Rt > a þ bp p > p a1 > ;b : ap þbp t ðs  a Þ ðw gÞðsÞ ds; t 2 2

ð22Þ

f ð aÞ þ f ð bÞ  a   a  p a p p a p Jap þ ðw gÞðb Þ þ Jbp  ðw gÞða Þ  Jap þ ð fw gÞðb Þ þ Jbp  ð fw gÞða Þ 2 2 R1 3   ð1  uÞa ua 0 p p 1=p 2 f ½ ua þ ð 1  u Þb  du 0 7 p½uap þ ð1  uÞbp 1ð1=pÞ kwk1 ðbp  ap Þaþ1 6 6 7  6 7 a   a 4 R1 5 Cða þ 1Þ u  ð 1  uÞ 0 p p 1=p f ½ ua þ ð 1  u Þb  þ 1 du 1 ð 1=p Þ 2 p½uap þ ð1  uÞbp  Z aþ1   1 jð1  uÞa ua j kw k1 ð bp  ap Þ 0 p p 1=p ¼ f ½ ua þ ð 1  u Þb  du: 1ð1=pÞ C ð a þ 1Þ 0 p½uap þ ð1  uÞbp 

123

ð23Þ

Iran J Sci Technol Trans Sci

Since jf 0 j is p-convex function on ½a; b, we have

(3)

If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has the following Hermite–Hadamard inequality for harmonically convex functions: Z b f ð aÞ þ f ð bÞ ab f ð xÞ  dx 2 b  a a x2   ba  ½C3 ð1; 1Þjf 0 ðaÞj þ C4 ð1; 1Þjf 0 ðbÞj: ab

(4)

If one takes p ¼ 1 and a ¼ 1, one has the following Hermite–Hadamard–Feje´r inequality for harmonically convex functions:

  0 1=p  ujf 0 ðaÞj þ ð1  uÞjf 0 ðbÞj: f ½uap þ ð1  uÞbp 

ð24Þ A combination of (23) and (24) gives

f ðaÞ þ f ðbÞ  a  Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp Þ þ Jbap  ð fw gÞðap Þ kwk1 ðbp  ap Þaþ1 Cða þ 1Þ 2 3 R1 jð1  uÞa ua j 0 udu f ð a Þ j j 0 6 7 p½uap þ ð1  uÞbp 1ð1=pÞ 6 7 6 7 a a 4 R1 5 jð1  uÞ u j ð1  uÞdujf 0 ðbÞj þ 0 1 ð 1=p Þ p p p½ua þ ð1  uÞb  

¼

(ii)

(2) (3) (4)

(2)

(5)

˙Is¸ can If one takes p ¼ 1, one has (2015), Theorem 2.6. If one takes p ¼ 1 and wð xÞ ¼ 1, one has Sarıkaya et al. (2013), Theorem 3. If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has Dragomir and Agarwal (1998), Theorem 2.2. If one takes a ¼ 1 and wð xÞ ¼ 1, one has Noor et al. (2016), Theorem 3.1. In Theorem 10, one can see the following.

If one takes p ¼ 1 and a ¼ 1, one has the following Hermite–Hadamard–Feje´r inequality for convex functions: Z Z b 1 f ð aÞ þ f ð bÞ b 1 wð xÞdx  f ð xÞwð xÞdx b  a 2 ba a a kwk1 ðb  aÞ  ½C1 ð1; 1Þjf 0 ðaÞj þ C2 ð1; 1Þjf 0 ðbÞj: 2

If one takes p ¼ 1 and wð xÞ ¼ 1, one has the following Hermite–Hadamard inequality for harmonically convex functions via fractional integrals:

Theorem 11 Let f : I  ð0; 1Þ ! R be a differentiable function on I such that f 0 2 L½a; b, where a; b 2 I and a\b. If jf 0 jq , q  1, is p-convex function on ½a; b for p 2 Rnf0g, a [ 0, w : ½a; b ! R is continuous and p p p 1=p symmetric with respect to a þb , then the following 2 inequality for fractional integrals holds: (i)

If p [ 0, f ð aÞ þ f ð bÞ  a  Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp Þ þ Jbap  ð fw gÞðap Þ kwk1 ðbp  ap Þaþ1 11q C5 ða;pÞ Cða þ 1Þ  q q 1

C1 ða;pÞjf 0 ðaÞj þC2 ða;pÞjf 0 ðbÞj q ;



If one takes p ¼ 1, one has the following Hermite– Hadamard–Feje´r inequality for harmonically convex functions via fractional integrals: i f ðaÞþf ðbÞ h a a J1=bþ ðw gÞð1=aÞþJ1=a ðw gÞð1=bÞ 2 h i a a  J1=bþ ð fw gÞð1=aÞþJ1=a ð fw gÞð1=bÞ   kwk1 abðbaÞ ba a  ½C3 ða;1Þjf 0 ðaÞjþC4 ða;1Þjf 0 ðbÞj: Cðaþ1Þ ab

kwk1 ðb  aÞ2 ½C3 ð1; 1Þjf 0 ðaÞj þ C4 ð1; 1Þjf 0 ðbÞj: 2

f ðaÞ þ f ðbÞ Cða þ 1Þ h i a a  1 1a J1=bþ ð f gÞð1=aÞ þ J1=a ð f gÞð1=bÞ 2 2 ba   ba  ½C3 ða;1Þjf 0 ðaÞj þ C4 ða;1Þjf 0 ðbÞj: ab

In Theorem 10, one can see the following.

Corollary 1 (1)



This completes the proof of i. The proof is similar to i.

Remark 3 (1)

kwk1 ðbp  ap Þaþ1 ½C1 ða; pÞjf 0 ðaÞj þ C2 ða; pÞjf 0 ðbÞj: Cða þ 1Þ

Z b Z f ðaÞ þ f ðbÞ b wðxÞ f ðxÞwðxÞ dx  dx 2 x2 x2 a a

where C1 ða; pÞ, C2 ða; pÞ are the same in Theorem 10, Z 1 jð1  uÞa ua j C5 ða; pÞ ¼ du 1ð1=pÞ 0 p½uap þ ð1  uÞbp  (ii)

with gðxÞ ¼ x1=p , x 2 ½ap ; bp . If p\0,

123

Iran J Sci Technol Trans Sci

f ð aÞ þ f ð bÞ  a  Jbp þ ðw gÞðap Þ þ Jaap  ðw gÞðbp Þ 2    Jbap þ ð fw gÞðap Þ þ Jaap  ð fw gÞðbp Þ kwk1 ðap  bp Þaþ1 11q  C6 ða; pÞ Cða þ 1Þ  q q 1 C3 ða; pÞjf 0 ðaÞj þC4 ða; pÞjf 0 ðbÞj q ;

with gðxÞ ¼ x

p

If one takes p ¼ 1 and a ¼ 1, one has the following Hermite–Hadamard–Feje´r inequality for convex functions: Z Z b 1 f ðaÞþf ðbÞ b 1 wð xÞdx f ð xÞwð xÞdx ba 2 ba a a kwk1 ðbaÞ 11q  C5 ð1;1Þ 2  q q 1 C1 ð1;1Þjf 0 ðaÞj þC2 ð1;1Þjf 0 ðbÞj q :

(3)

If one takes p ¼ 1, one has the following Hermite– Hadamard–Feje´r inequality for harmonically convex functions via fractional integrals: i f ðaÞþf ðbÞ h a a J ð w g Þ ð 1=a ÞþJ ð w g Þ ð 1=b Þ 1=bþ 1=a 2 h i a a  J1=bþ ð fw gÞð1=aÞþJ1=a ð fw gÞð1=bÞ  a kwk1 abðbaÞ ba  Cðaþ1Þ ab  11q q q 1 C6 ða;1Þ C3 ða;1Þjf 0 ðaÞj þC4 ða;1Þjf 0 ðbÞj q :

(4)

If one takes p ¼ 1 and a ¼ 1, one has the following Hermite–Hadamard–Feje´r inequality for harmonically convex functions: Z Z b f ð aÞ þ f ð bÞ b w ð x Þ f ð xÞwð xÞ dx  dx 2 x2 x2 a a

, x 2 ½b ; a .

Let p [ 0. Using (23), power mean inequality and the p-convexity of jf 0 jq ; it follows that

kwk1 ðbp ap Þaþ1  Cðaþ1Þ

Z

1

Z

jð1uÞa ua j

1 0

p½uap þ ð1uÞbp 1ð1=pÞ

!11q du

!1q   q 0 p p 1=p f ½ua þ ð1uÞb  du 1ð1=pÞ

jð1uÞa ua j

p½uap þ ð1uÞbp  !11q Z 1 kwk1 ðbp ap Þaþ1 jð1uÞa ua j  du 1ð1=pÞ Cðaþ1Þ 0 p½uap þ ð1uÞbp  ! " Z 1 jð1uÞa ua j q

udu jf 0 ðaÞj 1ð1=pÞ p 0 p½ua þ ð1uÞbp  ! #1q Z 1 jð1uÞa ua j q 0 ð 1u Þdu f ð b Þ þ j j 1ð1=pÞ 0 p½uap þ ð1uÞbp  0

¼

(ii)

This completes the proof of i. The proof is similar to i.

Remark 4 (1) (2) (3) (4)

 kwk1 ðbp ap Þaþ1 11q q q 1 C5 ða;pÞ C1 ða;pÞjf 0 ðaÞj þC2 ða;pÞjf 0 ðbÞj q : Cðaþ1Þ

In Theorem 11, one can see the following.

˙Is¸ can If one takes p ¼ 1, one has (2015), Theorem 2.8. If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has Pearce and Pecaric (2000), Theorem 1. If one takes p ¼ 1 and wð xÞ ¼ 1, one has ˙Is¸ can and Wu (2014), Theorem 5. If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has ˙Is¸ can (2014d), 2.6. Theorem.

123

If one takes p ¼ 1 and wð xÞ ¼ 1, one has the following Hermite–Hadamard inequality for convex functions via fractional integrals:

(2)

p

f ðaÞþf ðbÞ  a  Jap þ ðw gÞðbp ÞþJbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp ÞþJbap  ð fw gÞðap Þ Z kwk1 ðbp ap Þaþ1 1 jð1uÞa ua j  1ð1=pÞ p Cðaþ1Þ 0 p½ua þ ð1uÞbp    0 p p 1=p f ½ua þ ð1uÞb  du

In Theorem 11, one can see the following.

f ðaÞ þ f ðbÞ Cða þ 1Þ  a  a  J f ðbÞ  Jb f ðaÞ 2 2ðb  aÞa aþ  ðb  aÞ 11q q q 1  C5 ða; 1Þ C1 ða; 1Þjf 0 ðaÞj þC2 ða; 1Þjf 0 ðbÞj q : 2

Proof (i)

If one takes a ¼ 1 and wð xÞ ¼ 1, one has Noor et al. (2016), Theorem 3.2.

Corollary 2 (1)

where C3 ða; pÞ, C4 ða; pÞ are the same in Theorem 10, Z 1 jð1  uÞa ua j du C6 ða; pÞ ¼ 1ð1=pÞ 0 p½uap þ ð1  uÞbp  1=p

(5)



kwk1 ðb  aÞ2 11q C6 ð1; 1Þ 2  q q 1 C3 ð1; 1Þjf 0 ðaÞj þC4 ð1; 1Þjf 0 ðbÞj q :

Theorem 12 Let f : I  ð0; 1Þ ! R be a differentiable function on I such that f 0 2 L½a; b, where a; b 2 I and a\b. If jf 0 jq , q [ 1, is p-convex function on ½a; b for p 2 Rnf0g, a [ 0, 1q þ 1r ¼ 1, w : ½a; b ! R is continuous  p p 1=p , then the foland p-symmetric with respect to a þb 2 lowing inequality for fractional integrals holds:

Iran J Sci Technol Trans Sci

(i)

If p [ 0,





0

p½uap þ ð1  uÞbp 1ð1=pÞ

Z

1

C7 ða; p; r Þ ¼

Z

1 0

!r

a

jð1  uÞ ua j p½uap þ ð1  uÞbp 1ð1=pÞ

¼ du

0

q

1q

ujf ðaÞj þð1  uÞjf ðbÞj du

¼

where Z

1 0

jð1  uÞa ua j p½uap þ ð1  uÞbp 1ð1=pÞ

!r

!r

!1r du

Remark 5

Proof

(1) (2)

Let p [ 0. Using (23), Ho¨lder’s inequality and the p-convexity of jf 0 jq ; it follows that

f ðaÞþf ðbÞ  a  Jap þ ðw gÞðbp ÞþJbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp ÞþJbap  ð fw gÞðap Þ Z kwk1 ðbp ap Þaþ1 1 jð1uÞa ua j  1ð1=pÞ Cðaþ1Þ 0 p½uap þ ð1uÞbp    kwk1 ðbp ap Þaþ1

f 0 ½uap þ ð1uÞbp 1=p du Cðaþ1Þ !r !1r Z 1 a jð1uÞ ua j

du p 0 p½ua þ ð1uÞbp 1ð1=pÞ Z 1   q 1q 0 p p 1=p

du f ½ua þ ð1uÞb 

(3)

kwk1 ðbp  ap Þaþ1 1r C7 ða; p; r Þ C ða þ 1 Þ  0 1 jf ðaÞjq þjf 0 ðbÞjq q

: 2

This completes the proof of i. The proof is similar to i.

(ii)

du

with gðxÞ ¼ x1=p , x 2 ½bp ; ap .

0

kwk1 ðbp  ap Þaþ1 C ða þ 1 Þ Z 1 jð1  uÞa ua j p½uap þ ð1  uÞbp 1ð1=pÞ  0 1 jf ðaÞjq þjf 0 ðbÞjq q

2

f ðaÞ þ f ðbÞ  a  Jbp þ ðw gÞðap Þ þ Jaap  ðw gÞðbp Þ 2    Jbap þ ð fw gÞðap Þ þ Jaap  ð fw gÞðbp Þ  0 1 kwk1 ðap  bp Þaþ1 1r jf ðaÞjq þjf 0 ðbÞjq q  ; C8 ða; p; r Þ Cða þ 1Þ 2

(i)

q

du

0

with gðxÞ ¼ x1=p , x 2 ½ap ; bp . If p\0,

C8 ða; p; rÞ ¼

0

!1r

!r

0

where

(ii)

kwk1 ðbp  ap Þaþ1 Cða þ 1Þ Z 1 jð1  uÞa ua j

f ðaÞ þ f ðbÞ  a  Jap þ ðw gÞðbp Þ þ Jbap  ðw gÞðap Þ 2    Jaap þ ð fw gÞðbp Þ þ Jbap  ð fw gÞðap Þ  0 1 kwk1 ðbp  ap Þaþ1 1r jf ðaÞjq þjf 0 ðbÞjq q  C7 ða; p; r Þ ; C ða þ 1 Þ 2

In Theorem 12, one can see the following.

If one takes p ¼ 1, one has I˙s¸ can (2015), Theorem 2.9-i. If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has Dragomir and Agarwal (1998), Theorem 2.3. If one takes p ¼ 1 and a ¼ 1, one has Sarıkaya (2012), Theorem 2.8.

Corollary 3

In Theorem 12, one can see the following.

(1)

If one takes p ¼ 1 and wð xÞ ¼ 1, one has the following Hermite–Hadamard inequality for convex functions via fractional integrals: f ðaÞ þ f ðbÞ Cða þ 1Þ  a  a  J f ðbÞ  Jb f ðaÞ 2 2ðb  aÞa aþ  0 1 b  a 1r jf ðaÞjq þjf 0 ðbÞjq q  C7 ða; 1; r Þ : 2 2

(2)

If one takes p ¼ 1 and wð xÞ ¼ 1, one has the following Hermite–Hadamard inequality for harmonically convex functions via fractional integrals:

123

Iran J Sci Technol Trans Sci f ðaÞ þ f ðbÞ Cða þ 1Þ h i a a  1 1a J1=bþ ð f gÞð1=aÞ þ J1=a ð f gÞð1=bÞ 2 2 ba 1    0  b  a 1r jf ðaÞjq þjf 0 ðbÞjq q C8 ða; 1; r Þ :  ab 2

(3)

(4)

If one takes p ¼ 1, a ¼ 1 and wð xÞ ¼ 1, one has the following Hermite–Hadamard inequality for harmonically convex functions: Z b f ð aÞ þ f ð bÞ ab f ð xÞ  dx 2 b  a a x2    0 1 b  a 1r jf ðaÞjq þjf 0 ðbÞjq q  : C8 ð1; 1; r Þ ab 2 If one takes p ¼ 1, one has the following Hermite– Hadamard–Feje´r inequality for harmonically convex functions via fractional integrals: i f ðaÞ þ f ðbÞ h a a J1=bþ ðw gÞð1=aÞ þ J1=a ðw gÞð1=bÞ 2 h i a a ð fw gÞð1=aÞ þ J1=a ð fw gÞð1=bÞ j  J1=bþ  a  0 1 1 kwk1 abðb  aÞ b  a jf ða Þjq þ jf 0 ðb Þjq q C8r ða; 1; r Þ :  Cða þ 1Þ ab 2

(5)

If one takes p ¼ 1 and a ¼ 1, one has the following Hermite–Hadamard–Feje´r inequality for harmonically convex functions: Z Z b f ðaÞ þ f ðbÞ b wð xÞ f ð xÞwð xÞ dx  dx 2 x2 x2 a a  0 1 kwk1 ðb  aÞ2 1r jf ðaÞjq þjf 0 ðbÞjq q  C8 ð1; 1; r Þ : 2 2

(6)

If one takes a ¼ 1 and wð xÞ ¼ 1, one has the following Hermite–Hadamard inequality for p-convex functions:  0 1 Z b f ðaÞ þ f ðbÞ p f ð xÞ jf ðaÞjq þjf 0 ðbÞjq q  dx  2 bp  ap a x1p 2 8 p p 1 ð b  a Þ > > < C7r ð1; p; r Þ; p[0 2 :

p p 1 > ða  b Þ r > : C8 ð1; p; r Þ; p\0 2

3 Conclusion In Theorem 9, Hermite–Hadamard–Feje´r type inequalities for p-convex functions in fractional integral forms are built. In Lemma 2, an integral identity, and in Theorems 10, 11 and 12, some Hermite–Hadamard–Feje´r type integral inequalities for p-convex functions in fractional integral

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forms are obtained. In Corollaries 1, 2 and 3, some Hermite–Hadamard and Hermite–Hadamard–Feje´r inequalities for convex, harmonically convex and p-convex functions are given. Some results presented in Remarks 3, 4 and 5 provide extensions of others given in earlier works for convex, harmonically convex and p-convex functions.

Compliance with ethical standards Conflict of interest The authors declare that they have no competing interests.

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