The main objective of this work is to establish several new alpha-conformable of Steffensen-type inequalities on time scales. Our results will be proved by using time scales calculus technique. We get several well-known inequalities due to Steffensen, if we take α=1. Some cases we get continuous inequalities when T=R and discrete inequalities when T=Z.
Citation: Ahmed A. El-Deeb, Osama Moaaz, Dumitru Baleanu, Sameh S. Askar. A variety of dynamic α-conformable Steffensen-type inequality on a time scale measure space[J]. AIMS Mathematics, 2022, 7(6): 11382-11398. doi: 10.3934/math.2022635
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The main objective of this work is to establish several new alpha-conformable of Steffensen-type inequalities on time scales. Our results will be proved by using time scales calculus technique. We get several well-known inequalities due to Steffensen, if we take α=1. Some cases we get continuous inequalities when T=R and discrete inequalities when T=Z.
Riemann-Liouville fractional integral given by
Iαa+ξ(℘)=1Γ(α)∫χa(χ−℘)α−1ξ(℘)dt. |
Many different concepts of fractional derivative maybe found in [9,10,11]. In [12] studied a conformable derivative:
℘αf(℘)=limϵ→0f(℘+ϵ℘1−α)−f(℘)ϵ. |
The time scale conformable derivatives was introduced by Benkhettou et al. [17].
Further, in recent years, numerous mathematicians claimed that non-integer order derivatives and integrals are well suited to describing the properties of many actual materials, such as polymers. Fractional derivatives are a wonderful tool for describing memory and learning. a variety of materials and procedures inherited properties is one of the most significant benefits of fractional ownership. For more concepts and definition on time scales see [13,14,15,16,17,18,19,33,34,35].
Continuous version of Steffensen's inequality [7] is written as: For 0≤g(℘)≤1 on ∈[a,b]. Then
∫bb−λf(℘)dt≤∫baf(℘)g(℘)dt≤∫a+λaf(℘)dt, | (1.1) |
where λ=∫bag(℘)dt.
Supposing f is nondecreasing gets the reverse of (1.1).
Also, the discrete inequality of Steffensen [6] is: For λ2≤∑nℓ=1g(ℓ)≤λ1. Then
n∑ℓ=n−λ2+1f(ℓ)≤n∑ℓ=1f(ℓ)g(ℓ)≤λ1∑ℓ=1f(ℓ). | (1.2) |
Recently, a large number of dynamic inequalities on time scales have been studied by a small number of writers who were inspired by a few applications (see [1,2,3,4,8,28,29,30,31,32,36,37,40,41,42,44,48,49,50,51,52,53]).
In [5] Jakšetić et al. proved that, if ˆμ([c,d])=∫[a,b]g(℘)dˆμ(℘), where [c,d]⊆[a,b]. Then
∫[a,b]f(℘)g(℘)dˆμ(℘)≤∫[c,d]f(℘)g(℘)dˆμ(℘)+∫[a,c](f(℘)−f(d))g(℘)dˆμ(℘), |
and
∫[c,d]f(℘)dˆμ(℘)−∫[d,b](f(c)−f(℘))g(℘)dˆμ(℘)≤∫[a,b]f(℘)g(℘)dˆμ(℘). |
Anderson, in [3], studied the inequality:
∫bb−λϕ(℘)∇℘≤∫baϕ(℘)ψ(℘)∇℘≤∫a+λaϕ(℘)∇℘, | (1.3) |
In [47] the authors have proved, for
∫m+λ1mζ(℘)d℘=∫kmζ(℘)g(℘)d℘, |
and
∫nn−λ2ζ(℘)d℘=∫nkζ(℘)g(℘)d℘. |
If there exists a constant A such that r(℘)/ζ(℘)−At is monotonic on the intervals [m,k], [k,n], and
∫nmtq(℘)g(℘)d℘=∫m+λ1mtq(℘)d℘+∫nn−λ2tq(℘)d℘, |
then
∫nmr(℘)g(℘)d℘≤∫m+λ1mr(℘)d℘+∫nn−λ2r(℘)d℘. |
In particularly, Anderson [3] proved
∫nn−λr(℘)∇℘≤∫nmr(℘)g(℘)∇℘≤∫m+λmr(℘)∇℘. |
where m,n∈Tκ with m<n, r, g:[m,n]T→R are ∇-integrable functions such that r is of one sign and nonincreasing and 0≤g(℘)≤1 on [m,n]T and λ=∫nmg(℘)∇℘, n−λ,m+λ∈T.
We prove the next two needed results:
Theorem 1.1. Assume q>0 with 0≤g(℘)≤ζ(℘) ∀℘∈[m,n]T and λ is given from ∫nmg(℘)Δα℘=∫m+λmζ(℘)Δα℘, then
∫nmr(℘)g(℘)Δα℘≤∫m+λmr(℘)ζ(℘)Δα℘. | (1.4) |
Also, provided with 0≤g(℘)≤ζ(℘) and ∫nn−λζ(℘)Δα℘=∫nmg(℘)Δα℘, we have
∫nn−λr(℘)ζ(℘)Δα℘≤∫nmr(℘)g(℘)Δα℘. | (1.5) |
We get the reverse inequalities of (1.4) and (1.5) when assuming r/ζ is nondecreasing.
Theorem 1.2. Assume ψ is integrable on time scales interval [m,n], with ζ(℘)−ψ(℘)≥g(℘)≥ψ(℘)≥0∀℘∈[m,n]T and ∫m+λmζ(℘)Δα℘=∫nmg(℘)Δα℘=∫nn−λζ(℘)Δα℘ and g, r and ζ are Δα-integrable functions, ζ(℘)≥g(℘)≥0, we have
∫nn−λr(℘)ζ(℘)Δα℘+∫nm|(r(℘)−r(n−λ))ψ(℘)|Δα℘≤∫nmr(℘)g(℘)Δα℘≤∫m+λmr(℘)ζ(℘)Δα℘−∫nm|(r(℘)−r(m+λ))ψ(℘)|Δα℘, | (1.6) |
and
∫nn−λr(℘)ζ(℘)Δα℘≤∫nn−λ[r(℘)ζ(℘)−(r(℘)−r(n−λ))][ζ(℘)−g(℘)]Δα℘≤∫nmr(℘)g(℘)Δα℘≤∫m+λm[r(℘)ζ(℘)−(r(℘)−r(m+λ))][ζ(℘)−g(℘)]Δα℘≤∫m+λmr(℘)ζ(℘)Δα℘. | (1.7) |
Proof. The proof techniques of Theorems 1.6 and 1.7 are like to that in [4] and is removed.
Several authors proved conformable Hardy's inequality [20,21], conformable Hermite-Hadamard's inequality [22,23,24], conformable inequality of Opial's [26,27] and conformable inequality of Steffensen's [25]. In [45] Anderson proved the followong results:
Theorem 1.3. [45] Suppose α∈(0,1] and r1, r2∈R such that 0≤r1≤r2. Suppose ∏:[r1,r2]→[0,∞) and Γ:[r1,r2]→[0,1] are α-fractional integrable functions on [r1,r2] with Π is decreasing, we get
∫r2r2−ℵΠ(ζ)dαζ≤∫r2r1Π(ζ)Γ(ζ)dαζ≤∫r1+ℵr1Π(ζ)dαζ, |
where ℵ=α(r2−r1)rα2−rα1∫r2r1Γ(ζ)dαζ∈[0,r2−r1].
In [46] the authors gave an extension for Theorem 1.8:
Theorem 1.4. Assume α∈(0,1] and r1, r2∈R such that 0≤r1≤r2. Suppose ∏,Γ,Σ:[r1,r2]→[0,∞) are integrable on [r1,r2] with the decreasing function Π and 0≤Γ≤Σ, we get
∫r2r2−ℵΣ(ζ)Π(ζ)dαζ≤∫r2r1Π(ζ)Γ(ζ)dαζ≤∫r1+ℵr1Σ(ζ)Π(ζ)dαζ, |
where ℵ=(r2−r1)∫r2r1Σ(ζ)dαζ∫r2r1Γ(ζ)dαζ∈[0,r2−r1].
In this paper, we prove and explore several novel speculations of the Steffensen inequality obtained in [47] through the conformable integral containing time scale concept. We furthermore recover certain known results as special cases of our results.
Lemma 2.1. Assume ζ>0 is rd-continuous function on [m,n]∩T, g, r be rd-continuous on [m,n]∩T such that r/ζ nonincreasing function and 0≤g(℘)≤1 ∀℘∈[m,n]∩T. Then
(Λ1)
∫nmr(℘)g(℘)Δα℘≤∫m+λmr(℘)Δα℘, | (2.1) |
where λ is given by
∫nmζ(℘)g(℘)Δα℘=∫m+λmζ(℘)Δα℘. |
(Λ2)
∫nn−λr(℘)Δα℘≤∫nmr(℘)g(℘)Δα℘, | (2.2) |
such that
∫nn−λζ(℘)Δα℘=∫nmζ(℘)g(℘)Δα℘. |
(2.1) and (2.2) are reversed when r/ζ is nondecreasing.
Proof. Putting g(℘)↦ζ(℘)g(℘) and r(℘)↦r(℘)/ζ(℘) in (1.4), (1.5) to get (Λ1) and (Λ2) simultaneously.
Lemma 2.2. Under the same hypotheses of Lemma 2.1. with ψ be integrable functions on [m,n]∩T and 0≤ψ(℘)≤g(℘)≤1−ψ(℘) for all ℘∈[m,n]T. Then
∫nn−λr(℘)Δα℘+∫nm|(r(℘)ζ(℘)−r(n−λ)ζ(n−λ))ζ(℘)ψ(℘)|Δα℘≤∫nmr(℘)g(℘)Δα℘≤∫m+λmr(℘)Δα℘−∫nm|(r(℘)ζ(℘)−r(m+λ)ζ(m+λ))ζ(℘)ψ(℘)|Δα℘, |
where λ is obtained from
∫m+λmh(℘)Δα℘=∫nmζ(℘)g(℘)Δα℘=∫nn−λζ(℘)Δα℘. |
Proof. Putting g(℘)↦ζ(℘)g(℘), r(℘)↦r(℘)/h(℘) and ψ(℘)↦ζ(℘)ψ(℘) in (1.6).
Lemma 2.3. Under the same conditions of Lemma 2.1. Then
∫nn−λr(℘)Δα℘≤∫nn−λ(r(℘)−[r(℘)ζ(℘)−r(n−λ)ζ(n−λ)]ζ(℘)[1−g(℘)])Δα℘≤∫nmr(℘)g(℘)Δα℘≤∫m+λm(r(℘)−[r(℘)ζ(℘)−r(a+λ)ζ(m+λ)]ζ(℘)[1−g(℘)])Δα℘≤∫m+λmr(℘)Δα℘, |
where λ is obtained from
∫m+λmζ(℘)Δα℘=∫nmg(℘)Δα℘=∫nn−λζ(℘)Δα℘. |
Proof. Taking g(℘)↦ζ(℘)g(℘) and r(℘)↦r(℘)/ζ(℘) in (1.7).
Theorem 2.1. Under the same conditions of Lemma 2.3 such that k∈(m,n) and λ1, λ2 are given from
(Λ3)
∫m+λ1mζ(℘)Δα℘=∫kmζ(℘)g(℘)Δα℘, |
∫nn−λ2ζ(℘)Δα℘=∫nkζ(℘)g(℘)Δα℘. |
If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫m+λ1mϕ(℘)ζ(℘)Δα℘+∫nn−λ2ϕ(℘)ζ(℘)Δα℘, | (2.3) |
then
∫nmrσ(℘)g(℘)Δα℘≤∫m+λ1mrσ(℘)Δα℘+∫nn−λ2rσ(℘)Δα℘. | (2.4) |
(2.4) is reversed if rσ/ζ∈AHk2[m,n] and (2.3).
(Λ4)
∫kk−λ1ζ(℘)Δα℘=∫kmζ(℘)g(℘)Δα℘, |
∫k+λ2kζ(℘)Δα℘=∫nkζ(℘)g(℘)Δα℘. |
If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫k+λ2k−λ1ϕ(℘)ζ(℘)Δα℘, | (2.5) |
then
∫nmrσ(℘)g(℘)Δα℘≥∫k+λ2k−λ1rσ(℘)Δα℘. | (2.6) |
If rσ/ζ∈AHk2[m,n] and (2.5) satisfied, then we reverse (2.6).
(Λ5) If λ1, λ2 be the same as in (Λ3) and rσ/ζ∈AHk1[m,n] so that
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫m+λ1m(ϕ(℘)ζ(℘)−[ϕ(℘)−m−λ1]ζ(℘)[1−g(℘)])Δα℘+∫nn−λ2(ϕ(℘)ζ(℘)−[ϕ(℘)−n+λ2]ζ(℘)[1−g(℘)])Δα℘, | (2.7) |
then
∫nmrσ(℘)g(℘)Δα℘≤∫m+λ1m(rσ(℘)−|rσ(℘)ζ(℘)−rσ(m+λ1)ζ(m+λ1)|ζ(℘)[1−g(℘)])Δα℘+∫nn−λ2(rσ(℘)−|rσ(℘)ζ(℘)−rσ(n−λ2)ζ(n−λ2)|ζ(℘)[1−g(℘)])Δα℘. | (2.8) |
If rσ/ζ∈AHk2[m,n] and (2.7) satisfied, the inequality in (2.8) is reversed.
(Λ6) If λ1, λ2 be defined as in (Λ4) and rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫kk−λ1(ϕ(℘)ζ(℘)−[ϕ(℘)−k+λ1]ζ(℘)[1−g(℘)])Δα℘=∫m+λ1m(ϕ(℘)ζ(℘)−[ϕ(℘)−k+λ2]ζ(℘)[1−g(℘)])Δα℘, | (2.9) |
then
∫nmrσ(℘)g(℘)Δα℘≥∫kk−λ1(rσ(℘)−[rσ(℘)ζ(℘)−rσ(k−λ1)ζ(k−λ1)]ζ(℘)[1−g(℘)])Δα℘+∫k+λ2k(rσ(℘)−[rσ(℘)ζ(℘)−rσ(k+λ2)ζ(k+λ2)]ζ(℘)[1−g(℘)])Δα℘. | (2.10) |
If rσ/ζ∈AHk2[m,n] and (2.9) satisfied, we reverse (2.10).
Proof. (Λ3) Consider rσ/ζ∈AHk1[m,n], and R1(ℓ)=rσ(ℓ)−Aϕ(ℓ)ζ(ℓ), since A is given in Definition 2.1. Since R1/ζ:[m,k]∩T→R, using Lemma 2.1(Λ1), we deduce
0≤∫m+λ1mR1(℘)Δα℘−∫kmR1(℘)g(℘)Δα℘=∫m+λ1mrσ(℘)Δα℘−∫kmrσ(℘)g(℘)Δα℘−A(∫m+λ1mϕ(℘)ζ(℘)Δα℘−∫kmϕ(℘)ζ(℘)g(℘)Δα℘). | (2.11) |
As R1/ζ:[k,n]∩T→R is nondecreasing, using Lemma 2.1(Λ2), we obtain
0≥∫nkR1(℘)g(℘)Δα℘−∫nn−λ2R1(℘)Δα℘=∫nkrσ(℘)g(℘)Δα℘−∫nn−λ2rσ(℘)Δα℘−A(∫nkϕ(℘)ζ(℘)g(℘)Δα℘−∫nn−λ2ϕ(℘)ζ(℘)Δα℘). | (2.12) |
(2.11) and (2.12) imply that
∫m+λ1mrσ(℘)Δα℘+∫nn−λ2rσ(℘)Δα℘−∫nmrσ(℘)g(℘)Δα℘≥A(∫m+λ1mϕ(℘)ζ(℘)Δα℘+∫nn−λ2ϕ(℘)ζ(℘)Δα℘−∫nmϕ(℘)ζ(℘)g(℘)Δα℘) |
Hence, if (2.3) is hold, then (2.4) holds. For rσ/ζ∈AHk2[m,n], we get the some steps.
(Λ4) Let rσ/ζ∈AHk1[m,n], also R1(x)=rσ(x)−Aϕ(x)ζ(x), where A as in Definition 2.1. R1/ζ:[m,k]∩T→R is nonincreasing, so from Lemma 2.1(Λ1) we obtain
0≤∫kmrσ(℘)g(℘)Δα℘−∫kk−λ1rσ(℘)Δα℘−A(∫kmϕ(℘)h(℘)g(℘)Δα℘−∫kc−λ1ϕ(℘)ζ(℘)Δα℘). | (2.13) |
Using Lemma 2.1(Λ1) we have
0≥∫k+λ2krσ(℘)Δα℘−∫nkrσ(℘)g(℘)Δα℘−A(∫k+λ2kϕ(℘)ζ(℘)Δα℘−∫nkϕ(℘)ζ(℘)g(℘)Δα℘). | (2.14) |
Thus, from (2.13), (2.14), we get
∫nmrσ(℘)g(℘)Δα℘−∫k+λ2k−λ1rσ(℘)Δα℘≥A(∫nmϕ(℘)ζ(℘)g(℘)Δα℘−∫k+λ2k−λ1ϕ(℘)ζ(℘)Δα℘) |
Therefore, if ∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫k+λ2k−λ1ϕ(℘)ζ(℘)Δα℘ is satisfied, then (2.8) holds. Follow the same steps for rσ/ζ∈AHk2[m,n].
Using Lemma 2.3 and repeat the steps of Theorem 2.1(Λ3) and Theorem 2.1(Λ4) in the proof of (Λ5) and (Λ6) respectively.
Corollary 2.1. The inequalities (2.4), (2.6), (2.8) and (2.10) of Theorem 2.1 letting T=R takes
(i)∫nmfσ(℘)g(℘)dα℘≤∫m+λ1mrσ(℘)dα℘+∫nn−λ2rσ(℘)dα℘. | (2.15) |
(ii)∫nmrσ(℘)g(℘)dα℘≥∫k+λ2k−λ1rσ(℘)dα℘. | (2.16) |
(iii)∫nmrσ(℘)g(℘)dα℘≤∫m+λ1m(rσ(℘)−[rσ(℘)ζ(℘)−rσ(m+λ1)ζ(m+λ1)]ζ(℘)[1−g(℘)])dα℘+∫nn−λ2(rσ(℘)−[rσ(℘)ζ(℘)−rσ(n−λ2)ζ(n−λ2)]ζ(℘)[1−g(℘)])dα℘. | (2.17) |
(iv)∫nmrσ(℘)g(℘)dα℘≥∫kk−λ1(rσ(℘)−[rσ(℘)ζ(℘)−rσ(k−λ1)ζ(k−λ1)]ζ(℘)[1−g(℘)])dα℘+∫k+λ2k(rσ(℘)−[rσ(℘)ζ(℘)−rσ(k+λ2)ζ(k+λ2)]ζ(℘)[1−g(℘)])dα℘. | (2.18) |
Corollary 2.2. We get [47,Theorems 8,10,21 and 22], if we put α=1 and ϕ(℘)=℘ in Corollary 2.1 [(i),(ii),(iii),(iv)] simultaneously.
Corollary 2.3. In Corollary 2.1 taking T=Z, the results (2.15)–(2.18) will be equivalent to
(i)n−1∑℘=mr(℘+1)g(℘)℘α−1≤m+λ1−1∑℘=mr(℘+1)+n−1∑℘=n−λ2r(℘+1)℘α−1. |
(ii)n−1∑℘=mr(℘+1)g(℘)℘α−1≥k+λ2−1∑℘=k−λ1r(℘+1)℘α−1. |
(iii)n−1∑℘=mr(℘+1)g(℘)℘α−1≤m+λ1−1∑℘=m(r(℘+1)−[r(℘+1)ζ(℘)−r(a+λ1+1)ζ(m+λ1)]ζ(℘)[1−g(℘)])℘α−1+n−1∑℘=n−λ2(r(℘+1)−[r(℘+1)ζ(℘)−r(n−λ2+1)ζ(n−λ2)]ζ(℘)[1−g(℘)])℘α−1. |
(iv)n−1∑℘=mr(℘+1)g(℘))℘α−1≥k−1∑℘=k−λ1(r(℘+1)−[r(℘+1)ζ(℘)−r(k−λ1+1)ζ(k−λ1)]ζ(℘)[1−g(℘)]))℘α−1+k+λ2−1∑℘=k(r(℘+1)−[r(℘+1)ζ(℘)−r(k+λ2+1)ζ(k+λ2)]ζ(℘)[1−g(℘)]))℘α−1. |
Theorem 2.2. Under the assumptions in Lemma 2.1 with 0≤g(℘)≤ζ(℘) and λ1, λ2 be defined as
(Λ7)
∫m+λ1mζ(℘)Δα℘=∫kmg(℘)Δα℘, |
∫nn−λ2ζ(℘)Δα℘=∫nkg(℘)Δα℘. |
If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)g(℘)Δα℘=∫m+λ1mϕ(℘)ζ(℘)Δα℘+∫nn−λ2ϕ(℘)ζ(℘)Δα℘, | (2.19) |
then
∫nmrσ(℘)g(℘)Δα℘≤∫m+λ1mrσ(℘)ζ(℘)Δα℘+∫nn−λ2rσ(℘)ζ(℘)Δα℘. | (2.20) |
(Λ8)
∫kk−λ1ζ(℘)Δα℘=∫kmg(℘)Δα℘, |
∫k+λ2kζ(℘)Δα℘=∫nkg(℘)Δα℘. |
If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)g(℘)Δα℘=∫k+λ2k−λ1ϕ(℘)ζ(℘)Δα℘, | (2.21) |
then
∫nmrσ(℘)g(℘)Δα℘≥∫k+λ2k−λ1rσ(℘)ζ(℘)Δα℘. | (2.22) |
If rσ/ζ∈AHk2[m,n] and (2.19), (2.21) satisfied, we get the reverse of (2.20) and (2.22).
Proof. By using Theorem 2.1 [(Λ3),(Λ4)] and by putting g↦g/h and f↦fh, we get the proof of (Λ7) and (Λ8).
Corollary 2.4. In Theorem 2.2 [(Λ7),(Λ8)], assuming T=R, the following results obtains:
(i)∫nmrσ(℘)g(℘)dα℘≤∫m+λ1mrσ(℘)ζ(℘)dα℘+∫nn−λ2rσ(℘)ζ(℘)dα℘. | (2.23) |
(ii)∫nmrσ(℘)g(℘)dα℘≥∫k+λ2k−λ1rσ(℘)ζ(℘)dα℘. | (2.24) |
Corollary 2.5. In Corollary 2.4 [(i),(ii)], when we put α=1 and ϕ(℘)=℘ then [47,Theorems 16 and 17] gotten.
Corollary 2.6. In (2.23) and (2.24) letting T=Z, gets
(i)n−1∑℘=mr(℘+1)g(℘)℘α−1≤m+λ1−1∑℘=mr(℘+1)h(℘)+n−1∑℘=n−λ2r(℘+1)h(℘)℘α−1. |
(ii)n−1∑℘=mr(℘+1)g(℘)℘α−1≥k+λ2−1∑℘=k−λ1r(℘+1)ζ(℘)℘α−1. |
Theorem 2.3. Using the same conditions in Lemma 2.3. Letting w:[m,n]∩T→R be integrable with 0≤g(℘)≤w(℘) ∀℘∈[m,n]∩T and
(Λ9)∫m+λ1mw(℘)ζ(℘)Δα℘=∫kmζ(℘)g(℘)Δα℘, |
∫nn−λ2w(℘)ζ(℘)Δα℘=∫nkζ(℘)g(℘)Δα℘. |
If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫m+λ1mϕ(℘)w(℘)ζ(℘)Δα℘+∫nn−λ2ϕ(℘)w(℘)ζ(℘)Δα℘, | (2.25) |
then
∫nmrσ(℘)g(℘)Δα℘≤∫m+λ1mrσ(℘)w(℘)Δα℘+∫nn−λ2rσ(℘)w(℘)Δα℘. | (2.26) |
(Λ10)∫kk−λ1w(℘)ζ(℘)Δα℘=∫kmζ(℘)g(℘)Δα℘, |
∫k+λ2kw(℘)ζ(℘)Δα℘=∫nkζ(℘)g(℘)Δα℘. |
If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫k+λ2k−λ1ϕ(℘)w(℘)ζ(℘)Δα℘, | (2.27) |
∫nmrσ(℘)g(℘)Δα℘≥∫k+λ2k−λ1rσ(℘)w(℘)Δα℘. | (2.28) |
The inequalities in (2.26) and (2.28) are reversible if rσ/ζ∈AHc2[a,b] and (2.25), (2.27) hold.
Proof. In Theorem 2.1 [(Λ3),(Λ4)], ζ changes wq, g changes g/w and r changes rw.
Corollary 2.7. In (2.26) and (2.28). Letting T=R, we have
(i)∫nmrσ(℘)g(℘)dα℘≤∫m+λ1mrσ(℘)w(℘)dα℘+∫nn−λ2rσ(℘)w(℘)dα℘. | (2.29) |
(ii)∫nmrσ(℘)g(℘)dα℘≥∫k+λ2k−λ1rσ(℘)w(℘)dα℘. | (2.30) |
Corollary 2.8. In Corollary 2.7 [(i),(ii)], letting α=1 and ϕ(℘)=℘ we get [47,Theorems 18 and 19].
Corollary 2.9. In (2.29) and (2.30), crossing T=Z, gets
(i)n−1∑℘=mr(℘+1)g(℘)℘α−1≤m+λ1−1∑℘=mr(℘+1)w(℘)+n−1∑℘=n−λ2r(℘+1)w(℘)℘α−1. |
(ii)n−1∑℘=mr(℘+1)g(℘)℘α−1≥k+λ2−1∑℘=k−λ1r(℘+1)w(℘)℘α−1. |
Theorem 2.4. Using the same conditions in Lemma 2.1, and Theorem 2.1 [(Λ3),(Λ4)] with ψ:[m,n]∩T→R be a integrable: 0≤ψ(℘)≤g(℘)≤1−ψ(℘).
(Λ11) If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫m+λ1mϕ(℘)ζ(℘)Δα℘−∫km|ϕ(℘)−m−λ1|ζ(℘)ψ(℘)Δα℘+∫nn−λ2ϕ(℘)ζ(℘)Δα℘+∫nk|ϕ(℘)−n+λ2|ζ(℘)ψ(℘)Δα℘, | (2.31) |
then
∫nmrσ(℘)g(℘)Δα℘≤∫m+λ1mrσ(℘)Δα℘−∫km|rσ(℘)ζ(℘)−rσ(m+λ1)ζ(m+λ1)|ζ(℘)ψ(℘)Δα℘+∫nn−λ2rσ(℘)Δα℘+∫nk|rσ(℘)ζ(℘)−rσ(n−λ2)ζ(n−λ2)|ζ(℘)ψ(℘)Δα℘. | (2.32) |
(Λ12) If rσ/ζ∈AHk1[m,n] and
∫nmϕ(℘)ζ(℘)g(℘)Δα℘=∫kk−λ1ϕ(℘)ζ(℘)Δα℘−∫km|ϕ(℘)−k+λ1|ζ(℘)ψ(℘)Δα℘+∫nk|ϕ(℘)−k−λ1|ζ(℘)ψ(℘)Δα℘, | (2.33) |
then
∫nmrσ(℘)g(℘)Δα℘≥∫k+λ2k−λ1rσ(℘)Δα℘+∫km|rσ(℘)ζ(℘)−rσ(k−λ1)ζ(k−λ1)|ζ(℘)ψ(℘)Δα℘−∫nk|rσ(℘)ζ(℘)−rσ(k+λ2)ζ(k+λ2)|ζ(℘)ψ(℘)Δα℘. | (2.34) |
If rσ/ζ∈AHk2[m,n] and (2.31) and (2.33) satisfied, we get the reverse of (2.32) and (2.34).
Proof. The same steps of Theorem 2.1 [(Λ3),(Λ4)] with Lemma 2.1, R1/ζ:[m,k]∩T→R nonincreasing, R1/ζ:[k,n]∩T→R nondecreasing.
Corollary 2.10. In Theorem 2.4 [(Λ11),(Λ12)], letting T=R we get:
(i)∫nmrσ(℘)g(℘)dα℘≤∫m+λ1mrσ(℘)dα℘−∫km|rσ(℘)ζ(℘)−rσ(m+λ1)ζ(m+λ1)|ζ(℘)ψ(℘)dα℘+∫nn−λ2rσ(℘)dα℘+∫nk|rσ(℘)ζ(℘)−rσ(n−λ2)ζ(n−λ2)|ζ(℘)ψ(℘)dα℘. | (2.35) |
(ii)∫nmrσ(℘)g(℘)dα℘≥∫k+λ2k−λ1rσ(℘)dα℘+∫km|rσ(℘)ζ(℘)−rσ(k−λ1)ζ(k−λ1)|ζ(℘)ψ(℘)dα℘−∫nk|rσ(℘)ζ(℘)−rσ(k+λ2)ζ(k+λ2)|ζ(℘)ψ(℘)dα℘. | (2.36) |
Corollary 2.11. In (2.35) and (2.36), we put α=1, with ϕ(℘)=℘ we get [47,Theorems 23 and 24].
Corollary 2.12. Our results (2.35) and (2.36), by using T=Z gets
(i)n−1∑℘=mr(℘+1)g(℘)℘α−1≤m+λ1−1∑℘=mr(℘+1)℘α−1−k−1∑℘=m|r(℘+1)ζ(℘)−r(m+λ1+1)ζ(m+λ1)|ζ(℘)ψ(℘)ˆ∇℘+n−1∑℘=n−λ2r(℘+1)℘α−1+n−1∑℘=k|r(℘+1)ζ(℘)−r(n−λ2+1)ζ(n−λ2)|ζ(℘)ψ(℘)℘α−1. |
(ii)n−1∑℘=mr(℘+1)g(℘)℘α−1≥k+λ2−1∑℘=k−λ1r(℘+1)℘α−1+k−1∑℘=m|r(℘+1)ζ(℘)−r(k−λ1+1)ζ(k−λ1)|ζ(℘)ψ(℘)℘α−1−n−1∑℘=k|r(℘+1)ζ(℘)−r(k+λ2+1)ζ(k+λ2)|h(℘)ψ(℘)℘α−1. |
In this work, we explore new generalizations of the integral Steffensen inequality given in [38,39,43] by the utilization of the α-conformable derivatives and integrals, A few of these results are generalised to time scales. We also obtained the discrete and continuous case of our main results, in order to gain some fresh inequalities as specific cases.
The authors extend their appreciation to the Research Supporting Project number (RSP-2022/167), King Saud University, Riyadh, Saudi Arabia.
The authors declare no conflict of interest.
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