Firstly, we obtain some inequalities of Hadamard type for exponentially (θ,h−m)–convex functions via Caputo k–fractional derivatives. Secondly, using integral identity including the (n+1)–order derivative of a given function via Caputo k-fractional derivatives we prove some of its related results. Some new results are given and known results are recaptured as special cases from our results.
Citation: Imran Abbas Baloch, Thabet Abdeljawad, Sidra Bibi, Aiman Mukheimer, Ghulam Farid, Absar Ul Haq. Some new Caputo fractional derivative inequalities for exponentially (θ,h−m)–convex functions[J]. AIMS Mathematics, 2022, 7(2): 3006-3026. doi: 10.3934/math.2022166
[1] | Hengxiao Qi, Muhammad Yussouf, Sajid Mehmood, Yu-Ming Chu, Ghulam Farid . Fractional integral versions of Hermite-Hadamard type inequality for generalized exponentially convexity. AIMS Mathematics, 2020, 5(6): 6030-6042. doi: 10.3934/math.2020386 |
[2] | Yu-Pei Lv, Ghulam Farid, Hafsa Yasmeen, Waqas Nazeer, Chahn Yong Jung . Generalization of some fractional versions of Hadamard inequalities via exponentially (α,h−m)-convex functions. AIMS Mathematics, 2021, 6(8): 8978-8999. doi: 10.3934/math.2021521 |
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[6] | Maryam Saddiqa, Saleem Ullah, Ferdous M. O. Tawfiq, Jong-Suk Ro, Ghulam Farid, Saira Zainab . k-Fractional inequalities associated with a generalized convexity. AIMS Mathematics, 2023, 8(12): 28540-28557. doi: 10.3934/math.20231460 |
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[8] | Eze R. Nwaeze, Muhammad Adil Khan, Ali Ahmadian, Mohammad Nazir Ahmad, Ahmad Kamil Mahmood . Fractional inequalities of the Hermite–Hadamard type for m-polynomial convex and harmonically convex functions. AIMS Mathematics, 2021, 6(2): 1889-1904. doi: 10.3934/math.2021115 |
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[10] | Muhammad Imran Asjad, Waqas Ali Faridi, Mohammed M. Al-Shomrani, Abdullahi Yusuf . The generalization of Hermite-Hadamard type Inequality with exp-convexity involving non-singular fractional operator. AIMS Mathematics, 2022, 7(4): 7040-7055. doi: 10.3934/math.2022392 |
Firstly, we obtain some inequalities of Hadamard type for exponentially (θ,h−m)–convex functions via Caputo k–fractional derivatives. Secondly, using integral identity including the (n+1)–order derivative of a given function via Caputo k-fractional derivatives we prove some of its related results. Some new results are given and known results are recaptured as special cases from our results.
Convex functions has applications in almost all branches of mathematics for example in mathematical analysis, optimization theory and mathematical statistics etc. A convex function can be expressed and visualized in many different ways which further provide the motivation and encouragement for defining new concepts. Let us recall its definition as follows:
A function f:J⊆R→R is said to be convex, if
f(tx+(1−t)y)≤tf(x)+(1−t)f(y), | (1.1) |
holds for all x,y∈J and t∈[0,1]. Likewise f is concave if (−f) is convex.
A convex function is generalized in different forms, one of the generalization is the exponentially (θ,h−m)-convex function. Farid and Mahreen [11] introduced exponentially (θ,h−m)-convex functions as follows:
Definition 1.1. Let J⊆R be an interval containing (0,1) and let h:J→R be a non-negative function. For fix t∈(0,1), (θ,m)∈(0,1]2 and η∈R. A function f:[0,b]→R is called exponentially (θ,h−m)-convex function, if f is non-negative and for all x,y∈[0,b] one has
f(tx+m(1−t)y)≤h(tθ)f(x)eηx+mh(1−tθ)f(y)eηy. | (1.2) |
Remark 1.2. By selecting suitable function h and particular values of parameter m and η, the above definition produces the functions comprise in the following remark:
(i) By setting η=0, (θ,h−m)-convex function [12] can be obtained.
(ii) By taking η=0 and θ=1, (h−m)-convex function can be captured.
(iii) By choosing η=0 and h(tθ)=tθ, (θ,m)-convex function can be obtained.
(iv) By setting η=0, θ=1 and m=1, h-convex function [32] can be captured.
(v) By taking η=0, θ=1 and h(t)=t, m-convex function [31] can be obtained.
(vi) By choosing η=0, θ=1, m=1 and h(t)=t, convex function can be captured.
(vii) By setting η=0, m=1, θ=1 and h(t)=1, P-function [6] can be obtained.
(viii) By taking θ=1 and h(t)=ts, exponentially (s,m)-convex function [28] can be captured.
(ix) By choosing θ=1, m=1 and h(t)=ts, exponentially s-convex function [23] can be obtained.
(x) By setting θ=1, and h(t)=t, exponentially m-convex function [29] can be captured.
(xi) By taking θ=1, m=1 and h(t)=t, exponentially convex function [3] can be obtained.
(xii) By choosing η=0, θ=1 and h(t)=ts, (s,m)-convex function [2] can be captured.
(xiii) By setting θ=1, η=0, m=1 and h(t)=ts, s-convex function [23] can be obtained.
(xiv) By taking θ=1, η=0, m=1 and h(t)=1t, Godunova-Levin function [14] can be captured.
(xv) By choosing θ=1, η=0, m=1 and h(t)=1ts, s-Godunova-Levin function of second kind can be obtained.
The following inequality, named Hermite–Hadamard inequality, is one of the most famous inequalities in the literature for convex functions.
Theorem 1.3. Let f:J⊆R→R be a convex function on J and a,b∈J with a<b. Then the following double inequality holds:
f(a+b2)≤1b−a∫baf(x)dx≤f(a)+f(b)2. | (1.3) |
Various extensions of this notion have been reported in the literature in recent years, see [1,4,7,16,17,18,21,22,26,30].
The objective of this paper is to obtain inequalities of Hadamard type via Caputo k-fractional derivatives of exponentially (θ,h−m)-convex functions. Study of integration or differentiation of fractional order is known as fractional calculus. Its history is as old as the history of calculus. A lot of work has been published since the day of Leibniz (1695) and since then has occupied great number of mathematicians of their time [15,20,25,27].
Fractional integral inequalities are in the study of several researchers, see [5,9,13] and references therein. The classical Caputo fractional derivatives are defined as follows:
Definition 1.4. [20] Let α>0 and α∉{1,2,3,…}, n=[α]+1, f∈ACn[a,b] (the set of all functions f such that f(n) are absolutely continuous on [a,b]). The Caputo fractional derivatives of order α are defined by
CDαa+f(x)=1Γ(n−α)∫xaf(n)(t)(x−t)α−n+1dt,x>a, | (1.4) |
and
CDαb−f(x)=(−1)nΓ(n−α)∫bxf(n)(t)(t−x)α−n+1dt,x<b. | (1.5) |
If α=n∈{1,2,3,…} and usual derivative of order n exists, then Caputo fractional derivative (CDαa+f)(x) coincides with f(n)(x), whereas (CDαb−f)(x) coincides with f(n)(x) with exactness to a constant multiplier (−1)n. In particular, we have
(CD0a+f)(x)=(CD0b−f)(x)=f(x), | (1.6) |
where n=1 and α=0.
In [9], Farid et al. defined Caputo k−fractional derivatives as follows:
Definition 1.5. Let α>0,k≥1 and α∉{1,2,3,…}, n=[α]+1, f∈ACn[a,b]. The Caputo k-fractional derivatives of order α are given as
CDα,ka+f(x)=1kΓk(n−αk)∫xaf(n)(t)(x−t)αk−n+1dt,x>a, | (1.7) |
and
CDα,kb−f(x)=(−1)nkΓk(n−αk)∫bxf(n)(t)(t−x)αk−n+1dt,x<b, | (1.8) |
where Γk(α) is the k-gamma function defined as
Γk(α)=∫∞0tα−1e−tkkdt. |
Also
Γk(α+k)=αΓk(α). |
Motivated by above results and literatures, the paper is organized in the following manner:
In section 2, we present some inequalities of Hadamard type for exponentially (θ,h−m)−convex functions via Caputo k−fractional derivatives. In section 3, we use the integral identity including the (n+1)-order derivative of f to establish interesting Hadamard type inequalities for exponentially (θ,h−m)−convexity via Caputo k-fractional derivatives. In section 4, a briefly conclusion will be provided as well.
In this section, we give the Caputo k–fractional derivatives inequality of Hadamard type for a function whose n-th derivatives are exponentially (θ,h−m)–convex.
Theorem 2.1. Let α>0,k≥1 and α∉{1,2,3,…}, n=[α]+1 and [a,b]⊂[0,+∞), f:[0,+∞)→R be a function such that f∈ACn[a,mb], where a<mb. Also, assume that f(n) be an exponentially (θ,h−m)-convex function with (θ,m)∈(0,1]2 and η∈R. Then the following inequalities for Caputo k-fractional derivatives hold:
1g(η)f(n)(bm+a2)≤kΓk(n−αk+k)(mb−a)n−αk×(h(1−12θ)mn−αk+1(−1)n(CDα,kb−f)(am)+h(12θ)(CDα,ka+f)(mb))≤kn−αk×{(h(1−12θ)m2f(n)(am2)eηam2+h(12θ)mf(n)(b)eηb)∫10tn−αk−1h(1−tθ)dt+(h(1−12θ)mf(n)(b)eηb+h(12θ)f(n)(a)eηa)∫10tn−αk−1h(tθ)dt}, | (2.1) |
where g(η)=1eηbforη<0 and g(η)=1eηamforη≥0.
Proof. Since f(n) is an exponentially (θ,h−m)-convex function on [a,b], then
f(n)(um+v2)≤h(1−12θ)mf(n)(u)eηu+h(12θ)f(n)(v)eηv,u,v∈[a,b]. |
By setting u=(1−t)am+tb≤b and v=m(1−t)b+ta≥a in the above inequality for t∈[0,1], then by integrating with respect to t over [0,1] after multiplying with tn−αk−1, we have
f(n)(bm+a2)∫10tn−αk−1dt≤h(1−12θ)(∫10tn−αk−1mf(n)((1−t)am+tb)eη((1−t)am+tb)dt+h(12θ)∫10tn−αk−1f(n)(m(1−t)b+ta)eη(m(1−t)b+ta)dt). |
Now, if we let w=(1−t)am+tb and z=m(1−t)b+ta in right hand side of above inequality, we get
f(n)(bm+a2)1n−αk≤h(1−12θ)(∫bam(w−amb−am)n−αk−1mf(n)(w)dweηw(b−am)+h(12θ)∫mba(mb−zmb−a)n−αk−1f(n)(z)dzeηz(mb−a)). |
Hence
f(n)(bm+a2)≤{g(η)kΓk(n−αk+k)(mb−a)n−αk(h(1−12θ)mn−αk+1(−1)n(CDα,kb−f)(am)+h(12θ)(CDα,ka+f)(mb))}. | (2.2) |
On the other hand by using exponentially (θ,h−m)-convexity of f(n), we have
mf(n)((1−t)am+tb)≤m2h(1−tθ)f(n)(am2)eηam2+mh(tθ)f(n)(b)eηb. |
By multiplying both sides of above inequality with (n−αk)h(1−12θ)tn−αk−1 and integrating with respect to t over [0,1], after some calculations we get
kΓk(n−αk+k)(mb−a)n−αk(h(1−12θ)mn−αk+1(−1)n(CDα,kb−f)(am))≤h(1−12θ)(n−αk){m2f(n)(am2)eηam2∫10tn−αk−1h(1−tθ)dt+mf(n)(b)eηb∫10tn−αk−1h(tθ)dt}. | (2.3) |
Similarly,
f(n)(m(1−t)b+ta)≤mh(1−tθ)f(n)(b)eηb+h(tθ)f(n)(a)eηa. |
By multiplying both sides of above inequality with (n−αk)h(12θ)tn−αk−1 and integrating with respect to t over [0,1], after some calculations we get
kΓk(n−αk+k)(mb−a)n−αk(h(12θ)(CDα,ka+f)(mb))≤h(12θ)(n−αk){mf(n)(b)eηb∫10tn−αk−1h(1−tθ)dt+f(n)(a)eηa∫10tn−αk−1h(tθ)dt}. | (2.4) |
By adding (2.3) and (2.4), we obtain
kΓk(n−αk+k)(mb−a)n−αk(h(1−12θ)mn−αk+1(−1)n(CDα,kb−f)(am)+h(12θ)(CDα,ka+f)(mb))≤(n−αk){(h(1−12θ)m2f(n)(am2)eηam2+h(12θ)mf(n)(b)eηb)∫10tn−αk−1h(1−tθ)dt+(h(1−12θ)mf(n)(b)eηb+h(12θ)f(n)(a)eηa)∫10tn−αk−1h(tθ)dt}. |
Combining above with (2.2), we get required result.
Corollary 2.2. By setting k=1 in inequality (2.1), the following inequalities hold for exponentially (θ,h−m)-convex functions via Caputo fractional derivatives:
1g(η)f(n)(bm+a2)≤Γ(n−α+1)(mb−a)n−α(h(1−12θ)mn−α+1(−1)n(CDαb−f)(am)+h(12θ)(CDαa+f)(mb))≤(n−α){(h(1−12θ)m2f(n)(am2)eηam2+h(12θ)mf(n)(b)eηb)∫10tn−α−1h(1−tθ)dt+(h(1−12θ)mf(n)(b)eηb+h(12θ)f(n)(a)eηa)∫10tn−α−1h(tθ)dt}. |
Corollary 2.3. Taking η=0 in (2.1), the following inequalities hold for (θ,h−m)-convex functions via Caputo k-fractional derivatives:
f(n)(bm+a2)≤kΓk(n−αk+k)(mb−a)n−αk(h(1−12θ)mn−αk+1(−1)n(CDα,kb−f)(am)+h(12θ)(CDα,ka+f)(mb))≤(kn−αk){(h(1−12θ)m2f(n)(am2)+h(12θ)mf(n)(b))∫10tn−αk−1h(1−tθ)dt+(h(1−12θ)mf(n)(b)+h(12θ)f(n)(a))∫10tn−αk−1h(tθ)dt}. |
Corollary 2.4. Choosing η=0 and θ=1 in (2.1), the following inequalities hold for (h−m)-convex functions via Caputo k-fractional derivatives defined in [[24], Theorem 2.1]:
f(n)(bm+a2)≤kΓk(n−αk+k)2(mb−a)n−αk(mn−αk+1(−1)n(CDα,kb−f)(am)+(CDα,ka+f)(mb))≤kn−α2k{(m2f(n)(am2)+mf(n)(b))∫10tn−αk−1h(1−t)dt+(mf(n)(b)+f(n)(a))∫10tn−αk−1h(t)dt}. |
Corollary 2.5. By setting η=0, θ=1 and k=1 in (2.1), the following inequalities hold for (h−m)-convex functions via Caputo fractional derivatives defined in [[24], Corollary 2.2]:
f(n)(bm+a2)≤Γ(n−α+1)2(mb−a)n−α(mn−α+1(−1)n(CDαb−f)(am)+(CDαa+f)(mb))≤n−α2{(m2f(n)(am2)+mf(n)(b))∫10tn−α−1h(1−t)dt+(mf(n)(b)+f(n)(a))∫10tn−α−1h(t)dt}. |
Corollary 2.6. Taking θ=1 and h(t)=ts in (2.1), the following inequalities hold for exponentially (s,m)-convex functions via Caputo k-fractional derivatives:
1g(η)f(n)(bm+a2)≤kΓk(n−αk+k)2s(mb−a)n−αk(mn−αk+1(−1)n(CDα,kb−f)(am)+(CDα,ka+f)(mb))≤kn−αk2s{(m2f(n)(am2)eηam2+mf(n)(b)eηb)β(kn−αk,s+1)+(mf(n)(b)eηb+f(n)(a)eηa)kkn−α+ks}, |
where β(⋅,⋅) is well–known beta function.
Corollary 2.7. Choosing η=0, θ=1, m=1 and h(t)=t in (2.1), the following inequalities hold for convex functions via Caputo k-fractional derivatives defined in [[8], Theorem 2.2]:
f(n)(a+b2)≤kΓk(n−αk+k)2(b−a)n−αk((−1)n(CDα,kb−f)(a)+(CDα,ka+f)(b))≤f(n)(a)+f(n)(b)2. |
Corollary 2.8. By setting η=0, θ=1, m=1, h(t)=t and k=1 in (2.1), the following inequalities hold for convex functions via Caputo fractional derivatives:
f(n)(a+b2)≤Γ(n−α+1)2(b−a)n−α((−1)n(CDαb−f)(a)+(CDαa+f)(b))≤f(n)(a)+f(n)(b)2. |
In the following we generalize the fractional Hadamard type inequalities for exponentially (θ,h−m)−convex function via Caputo k-fractional derivatives.
Theorem 2.9. Let α>0,k≥1 and α∉{1,2,3,…}, n=[α]+1 and [a,b]⊂[0,+∞), f:[0,+∞)→R be a function such that f∈ACn[a,mb], where a<mb. Also, assume that f(n) be an exponentially (θ,h−m)-convex function with (θ,m)∈(0,1]2 and η∈R. Then the following inequalities for Caputo k-fractional derivatives hold:
1g(η)f(n)(a+bm2)≤2(n−αk)kΓk(n−αk+k)(bm−a)n−αk(h(1−12θ)mn−αk+1(−1)(n)(CDα,k(a+bm2m)−f)(am)+h(12θ)(CDα,k(a+bm2)+f)(mb))≤(kn−αk){(h(1−12θ)m2f(n)(am2)eηam2+h(12θ)mf(n)(b)eηb)∫10tn−αk−1h(1−(t2)θ)dt+(h(1−12θ)mf(n)(b)eηb+h(12θ)f(n)(a)eηa)∫10tn−αk−1h((t2)θ)dt}, | (2.5) |
where g(η)=1eηbforη<0 and g(η)=1eηamforη≥0.
Proof. From exponentially (θ,h−m)−convexity of f(n) one can have
f(n)(um+v2)≤h(1−12θ)mf(n)(u)eηu+h(12θ)f(n)(v)eηv. |
Putting u=t2b+(2−t)2am and u=t2a+m(2−t)2b in the above inequality where t∈[0,1], and multiplying with tn−αk−1, then integrating with respect to t over [0,1] one can have
f(n)(a+bm2)∫10tn−αk−1dt≤h(1−12θ)(∫10tn−αk−1mf(n)(t2b+(2−t)2am)eη(t2b+(2−t)2am)dt+h(12θ)∫10tn−αk−1f(n)(t2a+m(2−t)2b)eη(t2a+m(2−t)2b)dt). |
By change of variables, we get
1g(η)f(n)(a+bm2)≤2(n−αk)kΓk(n−αk+k)(bm−a)n−αk×(h(1−12θ)mn−αk+1(−1)(n)(CDα,k(a+bm2m)−f)(am)+h(12θ)(CDα,k(a+bm2)+f)(mb)). | (2.6) |
Now, using the exponentially (θ,h−m)−convexity of f(n), we can write
f(n)(t2a+m(2−t)2b)≤h((t2)θ)f(n)(a)eηa+mh(1−(t2)θ)f(n)(b)eηb. |
Multiplying both sides of above inequality with (n−αk)h(12θ)tn−αk−1 and integrating with respect to t over [0,1], then by change of variables, we have
h(12θ)2(n−αk)kΓk(n−αk+k)(bm−a)n−αk((CDα,k(a+bm2)+f)(mb))≤(n−αk)h(12θ){mf(n)(b)eηb∫10tn−αk−1h(1−(t2)θ)dt+f(n)(a)eηa∫10tn−αk−1h((t2)θ)dt}. | (2.7) |
Again by using the exponentially (θ,h−m)−convexity of f(n), we can write
mf(n)(t2b+(2−t)2am)≤mh(t2)θf(n)(b)eηb+m2h(1−(t2)θ)f(n)(am2)eηam2. |
Multiplying both sides of above inequality with (n−αk)h(1−12θ)tn−αk−1 and integrating with respect to t over [0,1], then by change of variables, we have
h(1−12θ)2(n−αk)kΓk(n−αk+k)(bm−a)n−αk(mn−αk+1(−1)(n)(CDα,k(a+bm2m)−f)(am))≤(n−αk)h(1−12θ){m2f(n)(am2)eηam2∫10tn−αk−1h(1−(t2)θ)dt+mf(n)(b)eηb∫10tn−αk−1h((t2)θ)dt}. | (2.8) |
Adding (2.7) and (2.8), we get
2(n−αk)kΓk(n−αk+k)(bm−a)n−αk(h(1−12θ)mn−αk+1(−1)(n)(CDα,k(a+bm2m)−f)(am)+h(12θ)(CDα,k(a+bm2)+f)(mb))≤(n−αk){(h(1−12θ)m2f(n)(am2)eηam2+h(12θ)mf(n)(b)eηb)∫10tn−αk−1h(1−(t2)θ)dt+(h(1−12θ)mf(n)(b)eηb+h(12θ)f(n)(a)eηa)∫10tn−αk−1h((t2)θ)dt}. |
By combining above with (2.6), we get required result.
Corollary 2.10. By setting k=1 in inequality (2.5), the following inequalities hold for exponentially (θ,h−m)-convex functions via Caputo fractional derivatives:
1g(η)f(n)(a+bm2)≤2(n−α)Γ(n−α+1)(bm−a)n−αk(h(1−12θ)mn−α+1(−1)(n)(CDα(a+bm2m)−f)(am)+h(12θ)(CDα(a+bm2)+f)(mb))≤(n−α){(h(1−12θ)m2f(n)(am2)eηam2+h(12θ)mf(n)(b)eηb)∫10tn−α−1h(1−(t2)θ)dt+(h(1−12θ)mf(n)(b)eηb+h(12θ)f(n)(a)eηa)∫10tn−α−1h((t2)θ)dt}. |
Corollary 2.11. Taking η=0 in (2.5), the following inequalities hold for (θ,h−m)-convex functions via Caputo k-fractional derivatives:
f(n)(a+bm2)≤2(n−αk)kΓk(n−αk+k)(bm−a)n−αk(h(1−12θ)mn−αk+1(−1)(n)(CDα,k(a+bm2m)−f)(am)+h(12θ)(CDα,k(a+bm2)+f)(mb))≤(kn−αk){(h(1−12θ)m2f(n)(am2)+h(12θ)mf(n)(b))∫10tn−αk−1h(1−(t2)θ)dt+(h(1−12θ)mf(n)(b)+h(12θ)f(n)(a))∫10tn−αk−1h((t2)θ)dt}. |
Corollary 2.12. Choosing η=0 and θ=1 in (2.5), the following inequalities hold for (h−m)-convex functions via Caputo k-fractional derivatives defined in [[24], Theorem 2.4]:
f(n)(a+bm2)≤2(n−αk)kΓk(n−αk+k)(bm−a)n−αkh(12)(mn−αk+1(−1)(n)(CDα,k(a+bm2m)−f)(am)+(CDα,k(a+bm2)+f)(mb))≤kn−αkh(12)×{(m2f(n)(am2)+mf(n)(b))∫10tn−αk−1h(2−t2)dt+(mf(n)(b)+f(n)(a))∫10tn−αk−1h(t2)dt}. |
Corollary 2.13. By setting η=0, θ=1 and k=1 in (2.5), the following inequalities hold for (h−m)-convex functions via Caputo fractional derivatives defined in [[24], Corollary 2.5]:
f(n)(a+bm2)≤2(n−α)Γ(n−α+1)(bm−a)n−αkh(12)(mn−α+1(−1)(n)(CDα(a+bm2m)−f)(am)+(CDα(a+bm2)+f)(mb))≤(n−α)h(12)×{(m2f(n)(am2)+mf(n)(b))∫10tn−α−1h(2−t2)dt+(mf(n)(b)+f(n)(a))∫10tn−α−1h(t2)dt}. |
Corollary 2.14. Taking θ=1 and h(t)=ts in (2.5), the following inequalities hold for exponentially (s,m)-convex functions via Caputo k-fractional derivatives:
1g(η)f(n)(a+bm2)≤2(n−αk−s)kΓk(n−αk+k)(bm−a)n−αk×(mn−αk+1(−1)(n)(CDα,k(a+bm2m)−f)(am)+(CDα,k(a+bm2)+f)(mb)).≤122s{(m2f(n)(am2)eηam2+mf(n)(b)eηb)kΓk(n−αk+k)+(mf(n)(b)eηb+f(n)(a)eηa)kn−αkn−α+ks}. |
Corollary 2.15. Choosing η=0, θ=1, m=1 and h(t)=t in (2.5), the following inequalities hold for convex functions via Caputo k-fractional derivatives defined in [[10], Theorem 6]:
f(n)(a+b2)≤2(n−αk)kΓk(n−αk+k)2(b−a)n−αk×((−1)(n)(CDα,k(a+b2)−f)(a)+(CDα,k(a+b2)+f)(b))≤f(n)(a)+f(n)(b)2. |
Corollary 2.16. By setting η=0, θ=1, m=1, h(t)=t and k=1 in (2.5), the following inequalities hold for convex functions via Caputo fractional derivatives defined in [[19] Theorem 2.2]:
f(n)(a+b2)≤2(n−α)Γ(n−α+1)2(b−a)n−αk×((−1)(n)(CDα(a+b2)−f)(a)+(CDα(a+b2)+f)(b))≤f(n)(a)+f(n)(b)2. |
In the next, some other inequalities of Hadamard type for exponentially (θ,h−m)−convex function via Caputo k−fractional derivatives are given.
Theorem 2.17. Let α>0,k≥1 and α∉{1,2,3,…}, n=[α]+1 and f:[0,+∞)→R be a function such that f∈ACn[a,mb], where a<mb. Also, assume that f(n) be an exponentially (θ,h−m)-convex function with (θ,m)∈(0,1]2 and η∈R. Then the following inequalities for Caputo k-fractional derivatives hold:
kΓk(n−αk)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤(f(n)(a)eηa+f(n)(b)eηb)×∫10tn−αk−1h(tθ)dt+m(f(n)(bm)eηbm+f(n)(am)eηam)∫10tn−αk−1h(1−tθ)dt≤1(np−αkp−p+1)1p(∫10(h(tθ))qdt)1q×(f(n)(a)eηa+f(n)(b)eηb+m(f(n)(bm)eηbm+f(n)(am)eηam)), | (2.9) |
where p−1+q−1=1 and p>1.
Proof. Since f(n) is exponentially (θ,h−m)−convex on [a,b], then for (θ,m)∈(0,1]2 and t∈[0,1], we have
f(n)(ta+(1−t)b)+f(n)((1−t)a+tb)≤h(tθ)(f(n)(a)eηa+f(n)(b)eηb)+mh(1−tθ)(f(n)(bm)eηbm+f(n)(am)eηam). |
Multiplying both sides of above inequality with tn−αk−1 and integrating the above inequality with respect to t on [0,1], we have
∫10tn−αk−1(f(n)(ta+(1−t)b)+f(n)((1−t)a+tb))dt≤(f(n)(a)eηa+f(n)(b)eηb)∫10tn−αk−1h(tθ)dt+m(f(n)(bm)eηbm+f(n)(am)eηam)∫10tn−αk−1h(1−tθ)dt. |
If we set x=ta+(1−t)b in the left side of above inequality, we get the following inequality
kΓk(n−αk)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤(f(n)(a)eηa+f(n)(b)eηb)∫10tn−αk−1h(tθ)dt+m(f(n)(bm)eηbm+f(n)(am)eηam)∫10tn−αk−1h(1−tθ)dt. | (2.10) |
We get the first inequality of (2.9). The second inequality of (2.9) follows by using the Hölder's inequality
∫10tn−αk−1h(tθ)dt≤1(np−αkp−p+1)1p(∫10(h(tθ))qdt)1q. |
Combining it with (2.10) we get (2.9).
Corollary 2.18. By setting k=1 in (2.9), the following inequalities hold for exponentially (θ,h−m)-convex functions via Caputo fractional derivatives:
Γ(n−α)(b−a)n−α((CDαa+f)(b)+(−1)n(CDαb−f)(a))≤(f(n)(a)eηa+f(n)(b)eηb)×∫10tn−α−1h(tθ)dt+m(f(n)(bm)eηbm+f(n)(am)eηam)∫10tn−α−1h(1−tθ)dt≤1(np−αp−p+1)1p(∫10(h(tθ))qdt)1q×(f(n)(a)eηa+f(n)(b)eηb+m(f(n)(bm)eηbm+f(n)(am)eηam)). |
Corollary 2.19. Taking η=0 in (2.9), the following inequalities hold for (θ,h−m)-convex functions via Caputo k-fractional derivatives:
kΓk(n−αk)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤(f(n)(a)+f(n)(b))×∫10tn−αk−1h(tθ)dt+m(f(n)(bm)+f(n)(am))∫10tn−αk−1h(1−tθ)dt≤1(np−αkp−p+1)1p(∫10(h(tθ))qdt)1q×(f(n)(a)+f(n)(b)+m(f(n)(bm)+f(n)(am))). |
Corollary 2.20. Choosing η=0 and θ=1 in (2.9), the following inequalities hold for (h−m)-convex functions via Caputo k-fractional derivatives defined in [[24], Theorem 2.7]:
kΓk(n−αk)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤(f(n)(a)+f(n)(b))×∫10tn−αk−1h(t)dt+m(f(n)(bm)+f(n)(am))∫10tn−αk−1h(1−t)dt≤1(np−αkp−p+1)1p(∫10(h(t))qdt)1q×(f(n)(a)+f(n)(b)+m(f(n)(bm)+f(n)(am))). |
Corollary 2.21. By setting η=0, θ=1 and k=1 in (2.9), the following inequalities hold for (h−m)-convex functions via Caputo fractional derivatives defined in [[24], Corollary 2.8]:
Γ(n−α)(b−a)n−α((CDαa+f)(b)+(−1)n(CDαb−f)(a))≤(f(n)(a)+f(n)(b))×∫10tn−α−1h(t)dt+m(f(n)(bm)+f(n)(am))∫10tn−α−1h(1−t)dt≤1(np−αp−p+1)1p(∫10(h(t))qdt)1q×(f(n)(a)+f(n)(b)+m(f(n)(bm)+f(n)(am))). |
Corollary 2.22. Taking θ=1 and h(t)=ts in (2.9), the following inequalities hold for exponentially (s,m)-convex functions via Caputo k-fractional derivatives:
kΓk(n−αk)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤k(f(n)(a)eηa+f(n)(b)eηb)kn−α+ks+m(f(n)(bm)eηbm+f(n)(am)eηam)β(n−αk,s+1)≤k(f(n)(a)eηa+f(n)(b)eηb)kn−α+ks+m(f(n)(bm)eηbm+f(n)(am)eηam)(np−αkp−p+1)1p(qs+1)1q. |
Corollary 2.23. Choosing η=0, θ=1, m=1 and h(t)=t in (2.9), the following inequalities hold for convex functions via Caputo k-fractional derivatives:
kΓk(n−αk)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤k(f(n)(a)+f(n)(b))kn−α≤2(f(n)(a)+f(n)(b))(np−αkp−p+1)1p(q+1)1q. |
Theorem 2.24. Let α>0,k≥1 and α∉{1,2,3,…}, n=[α]+1 and [a,b]⊂[0,+∞), f:[0,+∞)→R be a function such that f∈ACn[a,mb], where a<mb and h be a superadditive function. Also, assume that f(n) be an exponentially (θ,h−m)-convex function with (θ,m)∈(0,1]2 and η∈R. Then the following inequality for Caputo k−fractional derivatives holds:
kΓk(n−αk+k)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤h(1)2((f(n)(a)eηa+f(n)(b)eηb)+m(f(n)(bm)eηbm+f(n)(am)eηam)). | (2.11) |
Proof. Since f(n) is exponentially (θ,h−m)−convex on [a,b], then for t∈[0,1], we get
f(n)(ta+(1−t)b)+f(n)((1−t)a+tb)≤(h(tθ)+h(1−tθ))2((f(n)(a)eηa+f(n)(b)eηb)+m(f(n)(bm)eηbm+f(n)(am)eηam)). |
Since h is superadditive function, then
h(tθ)+h(1−tθ)≤h(1),for allθ∈(0,1]andt∈[0,1]. |
Therefore
f(n)(ta+(1−t)b)+f(n)((1−t)a+tb)≤h(1)2((f(n)(a)eηa+f(n)(b)eηb)+m(f(n)(bm)eηbm+f(n)(am)eηam)). |
Multiplying both sides of above inequality with tn−αk−1 and integrating with respect to t over [0,1], yield the following
∫10tn−αk−1(f(n)(ta+(1−t)b)+f(n)((1−t)a+tb))dt≤h(1)2((f(n)(a)eηa+f(n)(b)eηb)+m(f(n)(bm)eηbm+f(n)(am)eηam))∫10tn−αk−1dt. |
By change of variable, we get the required result.
Corollary 2.25. By setting k=1 in (2.11), the following inequality holds for exponentially (θ,h−m)-convex functions via Caputo fractional derivatives:
Γ(n−α+1)(b−a)n−α((CDαa+f)(b)+(−1)n(CDαb−f)(a))≤h(1)2((f(n)(a)eηa+f(n)(b)eηb)+m(f(n)(bm)eηbm+f(n)(am)eηam)). |
Corollary 2.26. Taking η=0 and θ=1 in (2.11), the following inequality holds for (h−m)-convex functions via Caputo k-fractional derivatives defined in [[24], Theorem 2.9]:
kΓk(n−αk+k)(b−a)n−αk((CDα,ka+f)(b)+(−1)n(CDα,kb−f)(a))≤h(1)2((f(n)(a)+f(n)(b))+m(f(n)(bm)+f(n)(am))). |
Corollary 2.27. Choosing η=0, θ=1 and k=1 in (2.11), the following inequality holds for (h−m)-convex functions via Caputo k-fractional derivatives defined in [[24], Corollary 2.10]:
Γ(n−α+1)(b−a)n−α((CDαa+f)(b)+(−1)n(CDαb−f)(a))≤h(1)2((f(n)(a)+f(n)(b))+m(f(n)(bm)+f(n)(am))). |
Corollary 2.28. By setting η=0, θ=1, m=1, h(t)=t and k=1 in (2.11), the following inequality holds for convex functions via Caputo fractional derivatives:
Γ(n−α+1)(b−a)n−α((CDαa+f)(b)+(−1)n(CDαb−f)(a))≤f(n)(a)+f(n)(b). |
We need the following known lemma to prove our next results.
Lemma 3.1. [24] Let α>0,k≥1 and α∉{1,2,3,…}, n=[α]+1 and f:[a,mb]→R, where a,b∈[0,+∞) be a differentiable mapping on interval (a,mb), with a<mb and m∈(0,1]. If f∈ACn+1[a,mb], then the following equality for Caputo k−fractional derivatives holds:
f(n)(mb)+f(n)(b)2−kΓk(n−αk+k)2(mb−a)n−αk((CDα,ka+f)(mb)+(CDα,kmb−f)(a))=mb−a2∫10((1−t)n−αk−tn−αk)f(n+1)(m(1−t)b+ta)dt. |
Caputo k−fractional derivative inequalities of Hadamard type for exponentially (θ,h−m)−convex function in terms of the (n+1)-th derivatives in absolute, is obtained in the following theorem by using above lemma.
Theorem 3.2. Let α>0,k≥1 and α∉{1,2,3,…}, n=[α]+1 and [a,b]⊂[0,+∞), f:[0,+∞)→R be a function such that f∈ACn+1[a,mb], where a<mb. If |f(n+1)| is an exponentially (θ,h−m)−convex with (θ,m)∈(0,1]2 and η∈R, then the following inequality for Caputo k−fractional derivatives holds:
|f(n)(mb)+f(n)(a)2−kΓk(n−αk+k)2(mb−a)n−αk(CDα,kb+f)(mb)+(CDα,kmb−f)(a)|≤mb−a2((2np−αkp+1−1)1p(2np−αkp+1(np−αkp+1))1p−1)×(|f(n+1)(a)|eηa((∫120(h(tθ))qdt)1q+(∫112(h(tθ))qdt)1q)+m|f(n+1)(b)|eηb((∫120(h(1−tθ))qdt)1q+(∫112(h(1−tθ))qdt)1q)), | (3.1) |
where 1p+1q=1 and p>1.
Proof. From Lemma 3.1 and by using the properties of modulus, we get
|f(n)(mb)+f(n)(b)2−kΓk(n−αk+k)2(mb−a)n−αk((CDα,ka+f)(mb)+(CDα,kmb−f)(a))|≤mb−a2∫10|(1−t)n−αk−tn−αk||f(n+1)(m(1−t)b+ta)|dt. |
By exponentially (θ,h−m)−convexity of |f(n+1)|, we have
|f(n)(mb)+f(n)(a)2−kΓk(n−αk+k)2(mb−a)n−αk(CDα,ka+f)(mb)+(CDα,kmb−f)(a)|≤mb−a2∫120((1−t)n−αk−tn−αk)(mh(1−tθ)|f(n+1)(b)eηb|+h(tθ)|f(n+1)(a)eηa|)dt+∫112(tn−αk−(1−t)n−αk)(mh(1−tθ)|f(n+1)(b)eηb|+h(tθ)|f(n+1)(a)eηa|)dt=mb−a2{|f(n+1)(a)eηa|(∫120(1−t)n−αkh(tθ)dt−∫120tn−αkh(tθ)dt)+m|f(n+1)(b)eηb|(∫120(1−t)n−αkh(1−tθ)dt−∫120tn−αkh(1−tθ)dt)+|f(n+1)(a)eηa|(∫112tn−αkh(tθ)dt−∫112(1−t)n−αkh(tθ)dt)+m|f(n+1)(b)eηb|(∫112tn−αkh(1−tθ)dt−∫112(1−t)n−αkh(1−tθ)dt)}. | (3.2) |
Now, using the Hölder's inequality in the right hand side of (3.2), we obtain
|f(n)(mb)+f(n)(a)2−kΓk(n−αk+k)2(mb−a)n−αk(CDα,kb+f)(mb)+(CDα,kmb−f)(a)|≤mb−a2{|f(n+1)(a)eηa|((2np−αkp+1−1)1p−1(2np−αkp+1(np−αkp+1))1p(∫120(h(tθ))qdt)1q+(2np−αkp+1−1)1p−1(2np−αkp+1(np−αkp+1))1p(∫112(h(tθ))qdt)1q)+m|f(n+1)(b)eηb|((2np−αkp+1−1)1p−1(2np−αkp+1(np−αkp+1))1p(∫120(h(1−tθ))qdt)1q+(2np−αkp+1−1)1p−1(2np−αkp+1(np−αkp+1))1p(∫112(h(1−tθ))qdt)1q)}. |
After a little computation one can get inequality (3.1).
Corollary 3.3. By setting k=1 in (3.1), the following inequality holds for exponentially (θ,h−m)-convex functions via Caputo fractional derivatives:
|f(n)(mb)+f(n)(a)2−Γ(n−α+1)2(mb−a)n−α(CDαb+f)(mb)+(CDαmb−f)(a)|≤mb−a2((2np−αp+1−1)1p(2np−αp+1(np−αp+1))1p−1)×(|f(n+1)(a)|eηa((∫120(h(tθ))qdt)1q+(∫112(h(tθ))qdt)1q)+m|f(n+1)(b)|eηb((∫120(h(1−tθ))qdt)1q+(∫112(h(1−tθ))qdt)1q)). |
Corollary 3.4. Taking η=0 in (3.1), the following inequality holds for (θ,h−m)-convex functions via Caputo k-fractional derivatives:
|f(n)(mb)+f(n)(a)2−kΓk(n−αk+k)2(mb−a)n−αk(CDα,kb+f)(mb)+(CDα,kmb−f)(a)|≤mb−a2((2np−αkp+1−1)1p(2np−αkp+1(np−αkp+1))1p−1)×(|f(n+1)(a)|((∫120(h(tθ))qdt)1q+(∫112(h(tθ))qdt)1q)+m|f(n+1)(b)|((∫120(h(1−tθ))qdt)1q+(∫112(h(1−tθ))qdt)1q)). |
Corollary 3.5. Choosing η=0 and θ=1 in (3.1), the following inequality holds for (h−m)-convex functions via Caputo k-fractional derivatives defined in [[24], Theorem 3.1]:
|f(n)(mb)+f(n)(a)2−kΓk(n−αk+k)2(mb−a)n−αk(CDα,kb+f)(mb)+(CDα,kmb−f)(a)|≤(mb−a)(|f(n+1)(a)|+m|f(n+1)(b)|)[(2np−αkp+1−1)1p]2[(2np−αkp+1(np−αkp+1))1p−1]×((∫120(h(t))qdt)1q+(∫112(h(t))qdt)1q). |
Corollary 3.6. By setting η=0, θ=1 and k=1 in (3.1), the following inequality holds for (h−m)-convex functions via Caputo fractional derivatives defined in [[24], Corollary 3.2]:
|f(n)(mb)+f(n)(a)2−Γ(n−α+1)2(mb−a)n−α(CDαb+f)(mb)+(CDαmb−f)(a)|≤(mb−a)(|f(n+1)(a)|+m|f(n+1)(b)|)[(2np−αp+1−1)1p]2[(2np−αp+1(np−αp+1))1p−1]×((∫120(h(t))qdt)1q+(∫112(h(t))qdt)1q). |
Corollary 3.7. Taking θ=1 and h(t)=ts in (3.1), the following inequality holds for exponentially (s,m)-convex functions:
|f(n)(mb)+f(n)(a)2−kΓk(n−αk+k)2(mb−a)n−αk(CDα,kb+f)(mb)+(−1)n(CDα,kmb−f)(a)|≤(mb−a)(|f(n+1)(a)eηa|+m|f(n+1)(b)eηb|)2×{(2n−αk+s−12n−αk+s(n−αk+s+1)+(2np−αkp+1−1)1p−(2qs+1−1)1q(2np−αkp+1(np−αkp+1))1p(2qs+1(qs+1))1q)}. |
Corollary 3.8. Choosing η=0, θ=1, m=1, h(t)=t and k=1 in (3.1), the following inequality holds for convex functions via Caputo fractional derivatives:
|f(n)(b)+f(n)(a)2−Γ(n−α+1)2(b−a)n−α(CDαb+f)(b)+(CDαb−f)(a)|≤(b−a)(|f(n+1)(a)|+|f(n+1)(b)|)[(2np−αp+1−1)1p((2q+1−1)1q+1)]2[((2np−αp+1(np−αp+1))1p−1)(2q+1(q+1))1q]. |
In this paper, some inequalities of Hadamard type for exponentially (α,h−m)–convex functions via Caputo k–fractional derivatives are obtained. By applied integral identity including the (n+1)–order derivative of a given function via Caputo k–fractional derivatives, we given some new of its related integral inequalities results. Some new results are given and know results are recaptured as special cases from our results. Since convexity and (exponentially (α,h−m)–convexity) have large applications in many mathematical areas, they can be applied to obtain several results in convex analysis, special functions, quantum mechanics, related optimization theory, mathematical inequalities and may stimulate further research in different areas of pure and applied sciences.
The authors would like to thank the anonymous referees for their valuable comments and suggestions, which led to considerable improvement of the article.
The authors A. Mukheimer and T. Abdeljawad would like to thank Prince Sultan University for paying APC and for the support through the TAS research lab. The work was also supported by the Higher Education Commission of Pakistan.
The authors declare no conflict of interest.
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