ISSN:2146-0957 eISSN:2146-5703 Vol.9, No.1, pp.49-59 (2019)
http://doi.org/10.11121/ijocta.01.2019.00559
RESEARCH ARTICLE
Hermite-Hadamard’s inequalities for conformable fractional
integrals
Mehmet Zeki Sarıkayaa , Abdullah Akkurtb* , H¨useyin Budaka , Merve Esra Yıldırımc , H¨useyin Yıldırımb
a
Department of Mathematics, Faculty of Science and Arts, D¨uzce University, D¨uzce, Turkey
b
Department of Mathematics, Faculty of Science and Arts, University of Kahramanmara¸s S¨ut¸c¨u ˙Imam, Kahramanmara¸s, Turkey
c
Department of Mathematics, Faculty of Science and Arts, University of Cumhuriyet, Sivas, Turkey sarikayamz@gmail.com, abdullahmat@gmail.com, hsyn.budak@gmail.com, mesra@cumhuriyet.edu.tr, hyildir@ksu.edu.tr
ARTICLE INFO ABSTRACT
Article History:
Received 06 November 2017 Accepted 02 October 2018 Available 31 January 2019
In this paper, we establish the Hermite-Hadamard type inequalities for con-formable fractional integral and we will investigate some integral inequalities connected with the left and right-hand side of the Hermite-Hadamard type inequalities for conformable fractional integral. The results presented here would provide generalizations of those given in earlier works and we show that some of our results are better than the other results with respect to midpoint inequalities.
Keywords: H¨older’s inequality Fractional derivative
Confromable fractional integrals Trapezoid inequality
Midpoint inequality AMS Classification 2010: 26D15, 26A51, 26A33, 26A42
1. Introduction
The convexity property of a given function plays an important role in obtaining integral inequali-ties. Proving inequalities for convex functions has a long and rich history in mathamatics. In [1], Beckenbach, a leading expert on the theory of convex functions, wrote that the inequality (1) was proved by Hadamard in 1893 [2]. In 1974, Mitrinoviˇc found Hermite and Hadamard’s note in Mathesis .
Let f : I ⊂ R → R be a convex function define on an interval I of real numbers, and a, b ∈ I with a < b. Then, the following inequalities hold:
f a + b 2 ≤ 1 b− a b Z a f(x)dx ≤ f(a) + f (b) 2 . (1) Inequality (1) is known in the literature as Hermite-Hadamard inequality for convex map-pings. Note that some of the classical inequalities for means can be derived from (1) for appropri-ate particular selections of the mapping f. Both inequalities hold in the reversed direction if f is concave.
Over the last decade, classical inequalities have been improved and generalized in a number of ways; there have been a large number of research papers written on this subject, [3–8]
*Corresponding Author
Definition 1. The function f : [a, b] ⊂ R → R, is said to be convex if the following inequality holds f(λx + (1 − λ) y) ≤ λf (x) + (1 − λ) f (y) (2)
for all x, y ∈ [a, b] and λ ∈ [0, 1] .
In [7], Dragomir and Agarwal proved the follow-ing results connected with the right part of (1). Lemma 1. ( [7]) Let f : I◦ ⊆ R → R be a differ-entiable mapping on I◦ , a, b ∈ I◦ with a < b. If f′
∈ L [a, b] , then the following equality holds: f(a) + f (b) 2 − 1 b− a b R a f(x) dx = b− a 2 1 R 0 (1 − 2t) f′ (ta + (1 − t) b) dt. (3) Theorem 1. ( [7]) Let f : I◦ ⊆ R → R be a dif-ferentiable mapping on I◦ , a, b∈ I◦ with a < b. If |f′
| is convex on [a, b] , then the following inequal-ity holds: f(a) + f (b) 2 − 1 (b − a) Rb af(x) dx ≤ (b − a) 4 |f′ (a)| + |f′ (b)| 2 . (4)
In [6], Kırmacı gave the following results. Lemma 2. ( [6]) Let f : I◦ ⊂ R → R be a dif-ferentiable mapping on I◦ , a, b ∈ I◦ (I◦ is the interior of I) with a < b. If f′
∈ L [a, b], then the following equality holds:
1 b− a Rb af(x) dx − f a + b 2 = (b − a)hR1/2 0 tf ′ (ta + (1 − t) b) dt +R1 1/2(t − 1) f ′ (ta + (1 − t) b) dti. (5) Theorem 2. ( [6]) Let f : I◦ ⊂ R → R be a differentiable mapping on I◦ , a, b∈ I◦ (I◦ is the interior of I) with a < b. If |f′ | is convex on [a, b], then the following inequality holds:
α b− a Rb af(x) dx − f a + b 2 ≤ b− a 8 (|f ′ (a)| + |f′ (b)|) . (6)
2. Definitions and Properties of Conformable Fractional Derivative and Integral
The following definitions and theorems with re-spect to conformable fractional derivative and in-tegral were referred in [9–14].
Definition 2. (Conformable fractional de-rivative) Given a function f : [0, ∞) → R. Then the “conformable fractional derivative” of f of or-der α is defined by
Dα(f ) (t) = lim ε→0
f t+ εt1−α − f (t)
ε (7)
for all t > 0, α ∈ (0, 1] . If f is α−differentiable in some (0, a) , α > 0, lim t→0+f (α)(t) exist, then define f(α)(0) = lim t→0+f (α)(t) . (8)
We can write f(α)(t) for Dα(f ) (t) to denote the conformable fractional derivatives of f of order α. In addition, if the conformable fractional deriva-tive of f of order α exists, then we simply say f is α−differentiable.
Theorem 3. Let α ∈ (0, 1] and f, g be α−differentiable at a point t > 0. Then
i. Dα(af + bg) = aDα(f ) + bDα(g) , for all a, b ∈ R,
ii. Dα(λ) = 0, for all constant functions f (t) = λ, iii. Dα(f g) = f Dα(g) + gDα(f ) , iv. Dα f g = Dα(f ) g − Dα(g) f g2 . If f is differentiable, then Dα(f ) (t) = t1−α df dt(t) . (9) Also: 1. Dα(1) = 0 2. Dα(eax) = ax1−αeax, a∈ R 3. Dα(sin(ax)) = ax1−αcos(ax), a ∈ R 4. Dα(cos(ax)) = −ax1−αsin(ax), a ∈ R 5. Dα α1tα = 1 6. Dα sin(t α α) = cos(t α α)
7. Dα cos(t α α) = − sin(t α α) 8. Dα e(tαα) = e(tαα).
Theorem 4 (Mean value theorem for con-formable fractional differentiable functions). Let α ∈ (0, 1] and f : [a, b] → R be a continuous on [a, b] and an α-fractional differentiable map-ping on (a, b) with 0 ≤ a < b. Then, there exists c∈ (a, b), such that
Dα(f ) (c) = f(b) − f (a) bα α −a α α .
Definition 3 (Conformable fractional integral). Let α ∈ (0, 1] and 0 ≤ a < b. A function f : [a, b] → R is α-fractional integrable on [a, b] if the integral Z b a f(x) dαx:= Z b a f(x) xα−1dx (10)
exists and is finite. All α-fractional integrable on [a, b] is indicated by L1 α([a, b]) Remark 1. Iαa(f ) (t) = I1a tα−1f = Z t a f(x) x1−αdx, where the integral is the usual Riemann improper integral, and α ∈ (0, 1].
Theorem 5. Let f : (a, b) → R be differentiable and 0 < α ≤ 1. Then, for all t > a we have
IαaDαaf(t) = f (t) − f (a) . (11)
Theorem 6. (Integration by parts) Let f, g : [a, b] → R be two functions such that fg is differ-entiable. Then Rb af(x) Daα(g) (x) dαx = f g|ba−Rb ag(x) Daα(f ) (x) dαx. (12)
Theorem 7. Assume that f : [a, ∞) → R such that f(n)(t) is continuous and α ∈ (n, n+1]. Then, for all t > a we have
Daαf(t) Iαa= f (t) .
Theorem 8. Let α ∈ (0, 1] and f : [a, b] → R be a continuous on [a, b] with 0 ≤ a < b. Then,
|Iαa(f ) (x)| ≤ Iαa|f | (x) .
For more details and properties concerning the conformable integral operators, we refer, for ex-ample, to the works [15–18].
In this paper, we establish the Hermite-Hadamard type inequalities for conformable fractional inte-gral and we will investigate some inteinte-gral inequal-ities connected with the left and right hand side of the Hermite-Hadamard type inequalities for conformable fractional integral. The results pre-sented here would provide generalizations of those given in earlier works.
3. Hermite-Hadamard’s Inequalities for Conformable Fractional Integral
We will start the following important result for α-fractional differentiable mapping;
Theorem 9. Let α ∈ (0, 1] and f : [a, b] → R be an α-fractional differentiable mapping on (a, b) with 0 ≤ a < b. Then, the following conditions are equivalent:
i) f is a convex functions on [a, b]
ii) Dαf(t) is an increasing function on [a, b] iii) for any x1, x2 ∈ [a, b]
f(x2) ≥ f (x1) + (xα
2 − xα1)
α Dα(f ) (x1) . (13) Proof. i) → ii) Let x1, x2 ∈ [a, b] with x1 < x2 and we take h > 0 which is small enough such that x1− h, x2+ h ∈ [a, b] . Since x1− h < x1 < x2 < x2+ h, then we know that
f(x1) − f (x1− h) h ≤ f(x2) − f (x1) x2− x1 ≤ f(x2+ h) − f (x2) h . (14)
Multipling the inequality (14) with x1−α1 ≤ x1−α2 , for x1 < x2, α∈ (0, 1], we get x1−α1 f(x1) − f (x1− h) h ≤ x1−α2 f(x2+ h) − f (x2) h . (15)
Let us put h = εxα−11 (and h = εxα−12 ) such that h → 0, ε → 0, then the inequality (14) can be converted to
f(x1) − f (x1− εxα−11 )
ε ≤
f(x2+ εxα−12 ) − f (x2)
ε .
Since f is α-fractional differentiable mapping on (a, b) , then let ε → 0+, we obtain
Dαf(x1) ≤ Dαf(x2) (16)
this show that Dαf is increasing in [a, b].
ii) → iii) Take x1, x2∈ [a, b] with x1 < x2. Since Dαf is increasing in [a, b], then by mean value theorem for conformable fractional differentiable we get f(x2) − f (x1) = (xα2 − xα1) α Dα(f ) (c) ≥ (x α 2 − xα1) α Dα(f ) (x1) (17)
where c ∈ (x1, x2) . It is follow that
f(x2) ≥ f (x1) +
(xα2 − xα1)
α Dα(f ) (x1) . iii) → i) For any x1, x2 ∈ [a, b], we take x3 = λx1 + (1 − λ) x2 and xα3 = λxα1 + (1 − λ) xα2 for λ ∈ (0, 1) . It is easy to show that xα1 − xα3 = (1 − λ) (xα1 − xα2) and xα2 − xα3 = −λ (xα1 − xα2). Thus, by using (13), we obtain that
f(x1) ≥ f (x3) + (xα 1 − xα3) α Dα(f ) (x3) = f (x3) + (1 − λ) (xα 1 − xα2) α Dα(f ) (x3) and f(x2) ≥ f (x3) + (xα 2 − xα3) α Dα(f ) (x3) = f (x3) − λ (xα 1 − xα2) α Dα(f ) (x3) .
Both sides of the above two expressions, multi-ply by λ and (1 − λ) , repectively, and add side to side, then we have
λf(x1) + (1 − λ) f (x2) ≥ f (x3)
= f (λx1+ (1 − λ) x2)
which is show that f is a convex function. The
proof is completed.
Theorem 10. Let α ∈ (0, 1], a ≥ 0, and f : [a, b] → R is a continuous function and ϕ : [0, ∞) → R be continuous and convex function. Then, ϕ α bα− aα Rb af(x) dαx ≤ α bα− aα Rb a ϕ(f (x)) dαx. (18)
Proof. Let ϕ : [0, ∞) → R be a convex function and x0∈ [0, ∞). From the definition of convexity, there exists m ∈ R such that,
ϕ(y) − ϕ(x0) ≥ m (y − x0) . (19)
Since f is a continuous function x0 = α bα− aα Z b a f(x) dαx (20)
is well defined. The function ϕ ◦ f is also contin-uous , thus we may apply (19) with y = f (t) and (20) to obtain
ϕ(f (t)) − ϕ(x0) ≥ m (f (t) − x0) .
Integrating above inequality from a to b, we get Z b a ϕ(f (t))dαt− ϕ(x0) Z b a dαt ≥ m Z b a f(t)dαt− x0 Z b a dαt = m Z b a f(t)dαt− xα0 Z b a dαt = 0. It is obvious that the inequality (18) holds. Hermite-Hadamard’s inequalities can be repre-sented in conformable fractional integral forms as follows:
Theorem 11. Let α ∈ (0, 1] and f : I ⊂ R+→ R be a convex function and f ∈ L1α([aα, bα]) with 0 ≤ a < b. Then, the following inequality for conformable fractional integral holds:
f a α+ bα 2 ≤ α bα− aα Rb af(xα) dαx ≤ f(a α) + f (bα) 2 . (21)
Proof. Since f is a convex function on I ⊂ R+,for xα, yα∈ [aα, bα] with λ = 12,we have
f x α+ yα 2 ≤ f(x α) + f (yα) 2 (22) i.e, with xα = tαaα + (1 − tα) bα, yα = (1 − tα) aα+ tαbα, for t ∈ [0, 1] , α ∈ (0, 1] 2f a α+ bα 2 ≤ f (tαaα+ (1 − tα) bα) +f ((1 − tα) aα+ tαbα) . (23)
By integrating the resulting inequality with re-spect to t over [0, 1] , we obtain
2R1 0 f a α+bα 2 dαt ≤R1 0 f(tαaα+ (1 − tα) bα) dαt +R1 0 f((1 − tα) aα+ tαbα) dαt = bα2α−aα Rb af(xα) dαx, (24)
and the first inequality is proved. For the proof of the second inequality in (22) we first note that if f is a convex function, then, for λ ∈ [0, 1] , it yields
f(tαaα+ (1 − tα) bα) ≤ tαf(aα) + (1 − tα) f (bα)
and
f((1 − tα) aα+ tαbα) ≤ (1 − tα) f (aα) + tαf(bα) .
By adding these inequalities we have
f(tαaα+ (1 − tα) bα) + f ((1 − tα) aα+ tαbα) ≤ f (aα) + f (bα) .
(25) Integrating inequality with respect to t over [0, 1] , we obtain R1 0 f(tαaα+ (1 − tα) bα) dαt +R1 0 f((1 − tα) aα+ tαbα) dαt ≤ [f (aα) + f (bα)]R1 0 dαt i.e. 1 bα− aα Z b a f(xα) dαx≤ f(a) + f (b) 2α .
The proof is completed.
Remark 2. If we choose α = 1 in (21), then inequality (21) become inequality (1).
Theorem 12. Let α ∈ (0, 1] and f : I ⊂ R+→ R be a convex function and f ∈ L1α([aα, bα]) with 0 ≤ a < b. Then, for t ∈ [0, 1], the following in-equality for conformable fractional integral holds:
f a α+ bα 2 ≤ h (tα) ≤ α bα− aα Rb af(xα) dαx ≤ H (tα) ≤ f(a α) + f (bα) 2 (26) where h(tα) = (1 − tα) f (1 + tα) aα+ (1 − tα) bα 2 +tαf a αtα+ (2 − tα) bα 2 and H(tα) = 12[(1 − tα) f (aα) + f (tαaα+ (1 − tα) bα) + tαf(bα)] .
Proof. Since f is a convex function on I, by applying (21) on the subinterval [aα, tαaα+ (1 − tα) bα] , with t 6= 1, we have f (1 + t α) aα+ (1 − tα) bα 2 ≤ α (1 − tα) (bα− aα) (27) × Z (tαaα+(1−tα)bα) 1 α a f(xα) dαx ≤ f(a α) + f (tαaα+ (1 − tα) bα) 2 .
Now, by applying (21) on the subinterval [tαaα+ (1 − tα) bα, bα] , with t 6= 0, we have
f a αtα+ (2 − tα) bα 2 ≤ α tα(bα− aα) Z b (tαaα+(1−tα)bα)α1 f(xα) dα(28)x ≤ f(t αaα+ (1 − tα) bα) + f (bα) 2 .
Multiplying (27) by (1 − tα) , and (27) by tα,and adding the resulting inequalities, we obtain the following inequalities h(tα) ≤ α bα− aα Z b a f(xα) dαx≤ H (tα) (29)
where h (tα) and H (tα) are defined as in Thereom 12. Using the fact that f is a convex function, we get f a α+ bα 2 = f (1 − tα)(1 + t α) aα+ (1 − tα) bα 2 +tαa αtα+ (2 − tα) bα 2 (30) ≤ (1 − tα) f a α+ [tαaα+ (1 − tα) bα] 2 +tαf [a αtα+ (1 − tα) bα] + bα 2 ≤ 1 2[(1 − t α) f (aα) + f (tαaα+ (1 − tα) bα) + tαf(bα)] ≤ f(a α) + f (bα) 2 .
Therefore, by (29) and (30) we have (26).
4. Trapezoid Type Inequalities for Conformable Fractional Integral
We need the following lemma. With the help of this, we give some integral inequalities connected with the right-side of Hermite–Hadamard-type in-equalities for conformable fractional integral. Lemma 3. Let α ∈ (0, 1] and f : I ⊂ R+ → R be an α-fractional differentiable function on (a, b) with 0 ≤ a < b. If Dα(f ) be an α-fractional
integrable function on [aα, bα] ,then the following identity for conformable fractional integral holds:
α bα− aα Rb af(xα) dαx− f(aα) + f (bα) 2 = 1 2 R1 0 (1 − 2tα) ×Dα(f ) (tαaα+ (1 − tα) bα) dαt. (31)
Proof. Integrating by parts R1 0 (1 − 2tα) Dα(f ) (tαaα+ (1 − tα) bα) dαt = (1 − 2tα) f (tαaα+ (1 − tα) bα)|10 +2αR1 0 f(tαaα+ (1 − tα) bα) dαt = − [f (aα) + f (bα)] + 2α (bα− aα) Rb af(xα) dαx.
Thus, by multiplying both sides by 1
2, we have
conclusion (31).
Remark 3. If we choose α = 1 in (31), then equality (31) become equality (3).
Theorem 13. Let α ∈ (0, 1] and f : I ⊂ R+→ R be an α-fractional differentiable function on I◦ and Dα(f ) be an α-fractional integrable function on I with 0 ≤ a < b. If |f′
| be a convex function on I,then the following inequality for conformable fractional integral holds:
f(aα) + f (bα) 2 − α bα− aα Z b a f(xα) dαx ≤ α(b α− aα) 2 23α2 +6 × 2α2− 8 3α × 23α2 (32) " aα(α−1)|Dα(f ) (aα)| + bα(α−1)|Dα(f ) (bα)| 2 # .
Proof. Using Lemma 3, it follows that f(aα) + f (bα) 2 − α bα− aα Rb af(xα) dαx ≤ 1 2 R1 0 |1 − 2tα| |Dα(f ) (tαaα+ (1 − tα) bα)| dαt. Since |f′
| is a convex function, by using the properties Dα(f ◦ g) (t) = f′(g(t)) Dαg(t) and Dα(f ) (t) = t1−αf′(t), it follows that
|Dα(f ) (tαaα+ (1 − tα) bα)| ≤ α (bα− aα)htαaα(α−1)|Dα(f ) (aα)| (33) + (1 − tα) bα(α−1)|Dα(f ) (bα)| i Using (33), we have f(aα) + f (bα) 2 − α bα− aα Rb af(xα) dαx ≤ α(b α− aα) 2 R1 0 |1 − 2tα| ×tαaα(α−1)|D α(f ) (aα)| + (1 − tα) bα(α−1)|Dα(f ) (bα)| dαt = α(b α− aα) 2 ×naα(α−1)|Dα(f ) (aα)| R1 0 |1 − 2tα| tαdαt + bα(α−1)|Dα(f ) (bα)|R01|1 − 2tα| (1 − tα) dαt o where Z 1 0 |1 − 2tα| (1 − tα) dαt = Z 1 0 |1 − 2tα| tαdαt= 23α2+6 × 2α2− 8 3α × 23α2
Thus, the proof is completed.
Remark 4. If we choose α = 1 in (32), then inequality (32) become inequality (4).
Theorem 14. Let α ∈ (0, 1] and f : I ⊂ R+→ R be an α-fractional differentiable function on I◦ and Dα(f ) be an α-fractional integrable function on I with 0 ≤ a < b. If |f′
|q, q > 1, be a con-vex function on I, then the following inequality for conformable fractional integral holds:
f(aα) + f (bα) 2 − α bα− aα Rb af(xα) dαx ≤ α(b α− aα) 2 (A(α)) 1 p aqα(α−1)|Dα(f ) (a)|q+ bqα(α−1)|Dα(f ) (b)|q 2α !1q (34) where 1p +1q = 1, A(α) is given by
A(α) = 1 2α (p + 1) ( 2 − 1 − 1 2α2−1 p+1 − 1 2α2−1 − 1 p+1) .
Proof. Using Lemma 3 and H¨older’s integral in-equality, we find f(aα) + f (bα) 2 − α bα− aα Rb af(xα) dαx ≤ 1 2 R1 0 |1 − 2tα| |Dα(f ) (tαaα+ (1 − tα) bα)| dαt ≤ 1 2 R1 0 |1 − 2tα| pd αt 1p R1 0 |Dα(f ) (tαaα+ (1 − tα) bα)| qd αt 1q . Since |f′
|q is a convex function, by using the properties Dα(f ◦ g) (t) = f′(g(t)) Dαg(t) and Dα(f ) (t) = t1−αf′(t), it follows that |Dα(f ) (tαaα+ (1 − tα) bα)|q ≤ αq(bα− aα)q (35) h tαaqα(α−1)|Dα(f ) (aα)|q + (1 − tα) bqα(α−1)|Dα(f ) (bα)|q i . By using (35), we have
f(aα) + f (bα) 2 − α bα− aα Rb af(xα) dαx ≤ α(b α− aα) 2 R1 0 |1 − 2tα| pd αt 1p h R1 0 tαaqα(α−1)|Dα(f ) (aα)| q + (1 − tα) bqα(α−1)|Dα(f ) (bα)|q dαt 1q ≤ α(b α− aα) 2 R1 0 |1 − 2tα| pd αt 1p aqα(α−1)|Dα(f ) (a)|q+ bqα(α−1)|Dα(f ) (b)|q 2α !1q . It follows that Z 1 0 |1 − 2tα|pdαt = Z 2α1 0 (1 − 2tα)pdαt+ Z 1 1 2α (2tα− 1)pdαt = 1 2α (p + 1) ( 2 − 1 − 1 2α2−1 p+1 − 1 2α2−1 − 1 p+1)
which is completed the proof.
Remark 5. If we choose α = 1 in (34), then inequality (34) become Theorem 2.3. in [7].
5. Midpoint Type Inequalities for Conformable Fractional Integral
We need the following lemma. With the help of this, we give some integral inequalities connected with the left-side of Hermite–Hadamard-type in-equalities for conformable fractional integral. Lemma 4. Let α ∈ (0, 1] and f : I ⊂ R+→ R be an α-fractional differentiable function on I◦
with 0 ≤ a < b. If Dα(f ) be an α-fractional integrable function on I, then the following identity for con-formable fractional integral holds:
f a α+ bα 2 − α bα− aα Rb af(xα) dαx =R1 0 P(t)Dα(f ) (tαaα+ (1 − tα) bα) dαt (36) where P(t) = tα, 0 ≤ t < 211/α tα− 1, 1 21/α ≤ t ≤ 1.
Proof. Integrating by parts Z 1 0 P(t)Dα(f ) (tαaα+ (1 − tα) bα) dαt = Z 1 21/α 0 tαDα(f ) (tαaα+ (1 − tα) bα) dαt + Z 1 1 21/α (tα− 1)Dα(f ) (tαaα+ (1 − tα) bα) dαt = tαf(tαaα+ (1 − tα) bα)| 1 21/α 0 −α Z 1 21/α 0 f(tαaα+ (1 − tα) bα) dαt + (tα− 1)f (tαaα+ (1 − tα) bα)|1 1 21/α −α Z 1 1 21/α f(tαaα+ (1 − tα) bα) dαt = f a α+ bα 2 − α (bα− aα) Z b a f(xα) dαx.
Thus, we have conclusion (36). Remark 6. If we choose α = 1 in (36), then equality (36) become equality (5).
Theorem 15. Let α ∈ (0, 1] and f : I ⊂ R+→ R be an α-fractional differentiable function on I◦ and Dα(f ) be an α-fractional integrable function on I. If |f′
| be a convex function on I, then the following inequality for conformable fractional in-tegrals holds: α bα− aα Z b a f(xα) dαx− f aα+ bα 2 ≤ α(b α− aα) 8 (37) aα(α−1)|Dα(f ) (aα)| + bα(α−1)|Dα(f ) (bα)| α ! .
α bα− aα Rb a f(xα) dαx− f aα+ bα 2 ≤ 1 21/α R 0 tα|Dα(f ) (tαaα+ (1 − tα) bα)| dαt + 1 R 1 21/α (1 − tα) |D α(f ) (tαaα+ (1 − tα) bα)| dαt . By using (33), we have α bα− aα Rb a f(xα) dαx− f aα+ bα 2 ≤ α (bα− aα) R 1 21/α 0 tαtαaα(α−1)|Dα(f ) (aα)| + (1 − tα) bα(α−1)|Dα(f ) (bα)| dαt +R1 1 21/α (1 − tα)tαaα(α−1)|D α(f ) (aα)| + (1 − tα) bα(α−1)|D α(f ) (bα)| dαt = α(b α− aα) 8 × a α(α−1)|D α(f ) (aα)| + bα(α−1)|Dα(f ) (bα)| α ! .
Thus, the proof is completed.
Remark 7. If we choose α = 1 in (37), then inequality (37) become the inequality (6).
Theorem 16. Let α ∈ (0, 1] and f : I ⊂ R+→ R be an α-fractional differentiable function on I◦ and Dα(f ) be an α-fractional integrable function on I. If |f′
|q, q > 1,be a convex function on I, then the following inequality for conformable frac-tional integrals holds:
α bα− aα Z b a f(xα) dαx− f aα+ bα 2 (38) ≤ α (bα− aα) 1 α(p + 1) 2p+1 1/p B(α) where 1p +1q = 1, B(α) is defined by B(α) = a qα(α−1)|D α(f ) (aα)|q 8α +3b qα(α−1)|D α(f ) (bα)| 8α !1/q + 3a qα(α−1)|D α(f ) (aα)|q 8α +b qα(α−1)|D α(f ) (bα)| 8α !1/q .
Proof. Using Lemma 3 and from H¨older’s in-equality, it follows that
α bα− aα Rb af(xα) dαx− f aα+ bα 2 ≤ 1 21/α R 0 tα|Dα(f ) (tαaα+ (1 − tα) bα)| dαt + 1 R 1 21/α (1 − tα) |Dα(f ) (tαaα+ (1 − tα) bα)| dαt ≤ 1 21/α R 0 tpαdαt 1/p × 1 21/α R 0 |Dα(f ) (tαaα+ (1 − tα) bα)|qdαt 1/q + 1 R 1 21/α (1 − tα)pdαt 1/p 1 R 1 21/α |Dα(f ) (tαaα+ (1 − tα) bα)|qdαt 1/q .
α bα− aα Rb af(xα) dαx− f aα+ bα 2 ≤ α (bα− aα) 1 α(p + 1) 2p+1 1/p × 1 21/α R 0 tαaqα(α−1)|D α(f ) (aα)|q + (1 − tα) bqα(α−1)|Dα(f ) (bα)|q dαt 1/q + 1 R 1 21/α tαaqα(α−1)|D α(f ) (aα)|q + (1 − tα) bqα(α−1)|Dα(f ) (bα)|q dαt 1/qo = α (bα− aα) 1 α(p + 1) 2p+1 1/p × ( aqα(α−1)|Dα(f ) (aα)|q 8α +3b qα(α−1)|D α(f ) (bα)| 8α !1/q + 3a qα(α−1)|D α(f ) (aα)|q 8α +b qα(α−1)|D α(f ) (bα)| 8α !1/q .
Thus, the proof of completed.
Remark 8. If we choose α = 1 in (38), then in-equality (38) become the inin-equality (2.1) in The-orem 2.3. in [6].
6. Conclusion
In this work, we have obtained some new Hermite-Hadamard type integral inequalities for con-formable integrals and we will investigate some integral inequalities connected with the left and right hand side of the Hermite-Hadamard type in-equalities for conformable fractional integral. The results presented here would provide generaliza-tions of those given in earlier works and we show that some our results are better than the other results with respect to midpoint inequalities.
Acknowledgments
M.E. Yildirim was partially supported by the Scientific and Technological Research Council of Turkey (TUBITAK Programme 2228-B).
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Mehmet Zeki Sarıkaya received his BSc (Maths), MSc (Maths) and PhD (Maths) degrees from Afyon Kocatepe University, Afyonkarahisar, Turkey in 2000, 2002 and 2007 respectively. At present, he is work-ing as a professor in the Department of Mathematics at Duzce University (Turkey) and is the head of the department. Moreover, he is the founder and Editor-in-Chief of Konuralp Journal of Mathematics (KJM). He is the author or coauthor of more than 200 papers
in the field of theory of inequalities, potential theory, integral equations and transforms, special functions, time-scales.
Abdullah Akkurt holds Bachelor of Mathematics and Master of Science degrees from the University of Kahramanmara¸s S¨ut¸c¨u ˙Imam, Turkey. He is an Re-search Assistant in the Department of Mathematics in the University of Kahramanmara¸s S¨ut¸c¨u ˙Imam. His research interests are in special functions and integral inequalities. Presently, he is undertaking his Doctor of Philosophy (Ph.D) degree programme at University of Kahramanmara¸s S¨ut¸c¨u ˙Imam.
H¨useyin Budak graduated from Kocaeli University, Kocaeli, Turkey in 2010. He received his M.Sc. from Kocaeli University in 2003. Since 2014, he is a Ph.D. student and a research assistant at Duzce University. His research interests focus on functions of bounded variation and theory of inequalities.
Merve Esra Yıldırım graduated from Ankara Uni-versity in 2012. In 2013, she received a master’s degree from Ankara University. In 2014, she started her Doc-tor of Philosophy (Ph.D) degree programme at Ankara University. Since 2015, she is a Ph.D. student Kahra-manmara¸s S¨ut¸c¨u ˙Imam University. She is an Research Assistant at Sivas Cumhuriyet University since 2015.
H¨useyin Yıldırım received his BSc (Maths) degree from Atat¨urk University, Erzurum, Turkey in 1986. He received his M.Sc. degree from Van Y¨uz¨unc¨u Yıl University in 1990. In 1995, he received a Ph.D. (Maths) degrees from Ankara University. At present, he is working as a professor in the Department of Mathematics at Kahramanmara¸s S¨ut¸c¨u ˙Imam Univer-sity (Turkey) and is the head of the department. He is the author or coauthor of more than 100 papers in the field of theory of inequalities, potential theory, inte-gral equations and transforms, special functions, time-scales.
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