Citation: Scheitel M, Stanic M, Neuberger M. PM10, PM2.5, PM1, number and surface of particles at the child’s seat when smoking a cigarette in a car[J]. AIMS Environmental Science, 2016, 3(4): 582-591. doi: 10.3934/environsci.2016.4.582
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In 1889, Hölder [1] proved that
n∑k=1xkyk≤(n∑k=1xpk)1p(n∑k=1yqk)1q, | (1.1) |
where {xk}nk=1 and {yk}nk=1 are positive sequences, p and q are two positive numbers, such that
1/p+1/q=1. |
The inequality reverses if pq<0. The integral form of (1.1) is
∫ϵηλ(τ)ω(τ)dτ≤[∫ϵηλγ(τ)dτ]1γ[∫ϵηων(τ)dτ]1ν, | (1.2) |
where η,ϵ∈R, γ>1, 1/γ+1/ν=1, and λ,ω∈C([a,b],R). If 0<γ<1, then (1.2) is reversed.
The subsequent inequality, widely recognized as Minkowski's inequality, asserts that, for p≥1, if φ,ϖ are nonnegative continuous functions on [η,ϵ] such that
0<∫ϵηφp(τ)dτ<∞ and 0<∫ϵηϖp(τ)dτ<∞, |
then
(∫ϵη(φ(τ)+ϖ(τ))pdτ)1p≤(∫ϵηφp(τ)dτ)1p+(∫ϵηϖp(τ)dτ)1p. |
Sulaiman [2] introduced the following result pertaining to the reverse Minkowski's inequality: for any φ,ϖ>0, if p≥1 and
1<m≤φ(ϑ)ϖ(ϑ)≤M, ϑ∈[η,ϵ], |
then
M+1(M−1)(∫ϵη(φ(ϑ)−ϖ(ϑ))pdϑ)1p≤(∫ϵηφp(ϑ)dϑ)1p+(∫ϵηϖp(ϑ)dϑ)1p≤(m+1m−1)(∫ϵη(φ(ϑ)−ϖ(ϑ))pdϑ)1p. | (1.3) |
Sroysang [3] proved that if p≥1 and φ,ϖ>0 with
0<c<m≤φ(ϑ)ϖ(ϑ)≤M, ϑ∈[η,ϵ], |
then
M+1(M−c)(∫ϵη(φ(ϑ)−cϖ(ϑ))pdϑ)1p≤(∫ϵηφp(ϑ)dϑ)1p+(∫ϵηϖp(ϑ)dϑ)1p≤(m+1m−c)(∫ϵη(φ(ϑ)−cϖ(ϑ))pdϑ)1p. |
In 2023, Kirmaci [4] proved that for nonnegative sequences {xk}nk=1 and {yk}nk=1, if p>1,s≥1,
1sp+1sq=1, |
then
n∑k=1(xkyk)1s≤(n∑k=1xpk)1sp(n∑k=1yqk)1sq, | (1.4) |
and if s≥1,sp<0 and
1sp+1sq=1, |
then (1.4) is reversed. Also, the author [4] generalized Minkowski's inequality and showed that for nonnegative sequences {xk}nk=1 and {yk}nk=1, if m∈N, p>1,
1sp+1sq=1, |
then
(n∑k=1(xk+yk)p)1mp≤(n∑k=1xpk)1mp+(n∑k=1ypk)1mp. | (1.5) |
Hilger was a trailblazer in integrating continuous and discrete analysis, introducing calculus on time scales in his doctoral thesis, later published in his influential work [5]. The exploration of dynamic inequalities on time scales has garnered significant interest in academic literature. Agarwal et al.'s book [6] focuses on essential dynamic inequalities on time scales, including Young's inequality, Jensen's inequality, Hölder's inequality, Minkowski's inequality, and more. For further insights into dynamic inequalities on time scales, refer to the relevant research papers [7,8,9,10,11,12,13,14,15,16,17,18,19,20]. For the solutions of some differential equations, see, for instance, [21,22,23,24,25,26,27] and the references therein.
Expanding upon this trend, the present paper aims to broaden the scope of the inequalities (1.4) and (1.5) on diamond alpha time scales. We will establish the generalized form of Hölder's inequality and Minkowski's inequality for the forward jump operator and the backward jump operator in the discrete calculus. In addition, we prove these inequalities in diamond −α calculus, where for α=1, we can get the inequalities for the forward operator, and when α=0, we get the inequalities for the backward jump operator. Also, we can get the special cases of our results in the discrete and continuous calculus.
The paper is organized as follows: In Section 2, we present some definitions, theorems, and lemmas on time scales, which are needed to get our main results. In Section 3, we state and prove new dynamic inequalities and present their special cases in different (continuous, discrete) calculi.
In 2001, Bohner and Peterson [28] defined the time scale T as an arbitrary, nonempty, closed subset of the real numbers R. The forward jump operator and the backward jump operator are defined by
σ(ξ):=inf{s∈T:s>ξ} |
and
ρ(ξ):=sup{s∈T:s<ξ}, |
respectively. The graininess function μ for a time scale T is defined by
μ(ξ):=σ(ξ)−ξ≥0, |
and for any function φ: T→R, the notation φσ(ξ) and φρ(ξ) denote φ(σ(ξ)) and φ(ρ(ξ)), respectively. We define the time scale interval [η,ϵ]T by
[η,ϵ]T:=[η,ϵ]∩T. |
Definition 2.1. (The Delta derivative[28]) Assume that φ: T→R is a function and let ξ∈T. We define φΔ(ξ) to be the number, provided it exists, as follows: for any ϵ>0, there is a neighborhood U of ξ,
U=(ξ−δ,ξ+δ)∩T |
for some δ>0, such that
|φ(σ(ξ))−φ(s)−φΔ(ξ)(σ(ξ)−s)|≤ϵ|σ(ξ)−s|, ∀s∈U, s≠σ(ξ). |
In this case, we say φΔ(ξ) is the delta or Hilger derivative of φ at ξ.
Definition 2.2. (The nabla derivative [29]) A function λ: T→R is said to be ∇- differentiable at ξ∈T, if ψ is defined in a neighborhood U of ξ and there exists a unique real number ψ∇(ξ), called the nabla derivative of ψ at ξ, such that for each ϵ>0, there exists a neighborhood N of ξ with N⊆U and
|ψ(ρ(ξ))−ψ(s)−ψ∇(ξ)[ρ(ξ)−s]|≤ϵ|ρ(ξ)−s|, |
for all s∈N.
Definition 2.3. ([6]) Let T be a time scale and Ξ(ξ) be differentiable on T in the Δ and ∇ sense. For ξ∈T, we define the diamond −α derivative Ξ◊α(ξ) by
Ξ◊α(ξ)=αΞΔ(ξ)+(1−α)Ξ∇(ξ), 0≤α≤1. |
Thus, Ξ is diamond −α differentiable if, and only if, Ξ is Δ and ∇ differentiable.
For α=1, we get that
Ξ◊α(ξ)=αΞΔ(ξ), |
and for α=0, we have that
Ξ◊α(ξ)=Ξ∇(ξ). |
Theorem 2.1. ([6]) Let Ξ, Ω: T→R be diamond-α differentiable at ξ∈T, then
(1) Ξ+Ω: T→R is diamond-α differentiable at ξ∈T, with
(Ξ+Ω)◊α(ξ)=Ξ◊α(ξ)+Ω◊α(ξ). |
(2) Ξ.Ω: T→R is diamond-α differentiable at ξ∈T, with
(Ξ.Ω)◊α(ξ)=Ξ◊α(ξ)Ω(ξ)+αΞσ(ξ)ΩΔ(ξ)+(1−α)Ξρ(ξ)Ω∇(ξ). |
(3) For Ω(ξ)Ωσ(ξ)Ωρ(ξ)≠0, Ξ/Ω: T→R is diamond-α differentiable at ξ∈T, with
(ΞΩ)◊α(ξ)=Ξ◊α(ξ)Ωσ(ξ)Ωρ(ξ)−αΞσ(ξ)Ωρ(ξ)ΩΔ(ξ)−(1−α)Ξρ(ξ)Ωσ(ξ)Ω∇(ξ)Ω(ξ)Ωσ(ξ)Ωρ(ξ). |
Theorem 2.2. ([6]) Let Ξ, Ω: T→R be diamond-α differentiable at ξ∈T, then the following holds
(1) (Ξ)◊αΔ(ξ)=αΞΔΔ(ξ)+(1−α)Ξ∇Δ(ξ),
(2) (Ξ)◊α∇(ξ)=αΞΔ∇(ξ)+(1−α)Ξ∇∇(ξ),
(3) (Ξ)Δ◊α(ξ)=αΞΔΔ(ξ)+(1−α)ΞΔ∇(ξ)≠(Ξ)◊αΔ(ξ),
(4) (Ξ)∇◊α(ξ)=αΞ∇Δ(ξ)+(1−α)Ξ∇∇(ξ)≠(Ξ)◊α∇(ξ),
(5) (Ξ)◊α◊α(ξ)=α2ΞΔΔ(ξ)+α(1−α)[ΞΔ∇(ξ)+Ξ∇Δ(ξ)].
Theorem 2.3. ([6]) Let a,ξ∈T and h: T→R, then the diamond−α integral from a to ξ of h is defined by
∫tah(s)◊αs=α∫tah(s)Δs+(1−α)∫tah(s)∇s, 0≤α≤1, |
provided that there exist delta and nabla integrals of h on T.
It is known that
(∫ξah(s)Δs)Δ=h(ξ) |
and
(∫ξah(s)∇s)∇=h(ξ), |
but in general
(∫ξah(s)◊αs)◊α≠h(ξ), for ξ∈T. |
Example 2.1. [30] Let T={0, 1, 2},a=0, and h(ξ)=ξ2 for ξ∈T. This gives
(∫ξah(s)◊αs)◊α|ξ=1=1+2α(1−α), |
so that the equality above holds only when ◊α=Δ or ◊α=∇.
Theorem 2.4. ([6]) Let a,b,ξ∈T, β,γ,c∈R, and Ξ and Ω be continuous functions on [a,b]∪T, then the following properties hold.
(1) ∫ba[γΞ(ξ)+βΩ(ξ)]◊αξ=γ∫baΞ(ξ)◊αξ+β∫baΩ(ξ)◊αξ,
(2) ∫bacΞ(ξ)◊αξ=c∫baΞ(ξ)◊αξ,
(3) ∫baΞ(ξ)◊αξ=−∫abΞ(ξ)◊αξ,
(4) ∫caΞ(ξ)◊αξ=∫baΞ(ξ)◊αξ+∫cbΞ(ξ)◊αξ.
Theorem 2.5. ([6]) Let T be a time scale a,b∈T with a<b. Assume that Ξ and Ω are continuous functions on [a,b]T,
(1) If Ξ(ξ)≥0 for all ξ∈[a,b]T, then ∫baΞ(ξ)◊αξ≥0;
(2) If Ξ(ξ)≤Ω(ξ) for all ξ∈[a,b]T, then ∫baΞ(ξ)◊αξ≤∫baΩ(ξ)◊αξ;
(3) If Ξ(ξ)≥0 for all ξ∈[a,b]T, then Ξ(ξ)=0 if, and only if, ∫baΞ(ξ)◊αξ=0.
Example 2.2. ([6]) If T=R, then
∫baΞ(ξ)◊αξ=∫baΞ(ξ)dξ, for a, b∈R, |
and if T=N and m<n, then we obtain
∫nmΞ(ξ)◊αξ=n−1∑i=m[αΞ(i)+(1−α)Ξ(i+1)], for m, n∈N. | (2.1) |
Lemma 2.1. (Hölder's inequality [6]) If η,ϵ∈T, 0≤α≤1, and λ,ω∈C([η,ϵ]T,R+), then
∫ϵηλ(τ)ω(τ)◊ατ≤[∫ϵηλγ(τ)◊ατ]1γ[∫ϵηων(τ)◊ατ]1ν, | (2.2) |
where γ>1 and 1/γ+1/ν=1. The inequality (2.2) is reversed for 0<γ<1 or γ<0.
Lemma 2.2. (Minkowski's inequality[6]) Let η,ϵ∈T, ϵ>η,0≤α≤1,p>1, and Ξ,Ω∈C([η,ϵ]T,R+), then
(∫ϵη(Ξ(τ)+Ω(τ))p◊ατ)1p≤(∫ϵηΞp(τ)◊ατ)1p+(∫ϵηΩp(τ)◊ατ)1p. | (2.3) |
Lemma 2.3. (Generalized Young inequality [4]) If a,b≥0,sp,sq>1 with
1sp+1sq=1, |
then
aspsp+bsqsq≥ab. | (2.4) |
Lemma 2.4. (See [31]) If x,y>0 and s>1, then
(xs+ys)1s≤x+y, | (2.5) |
and if 0<s<1, then
(xs+ys)1s≥x+y. | (2.6) |
In the manuscript, we will operate under the assumption that the considered integrals are presumed to exist. Also, we denote ◊ατ=◊ατ1,⋯,◊ατN and ϖ(τ)=ϖ(τ1,⋯,τN).
Theorem 3.1. Assume that ηi, ϵi∈T,ϵi>ηi,i=1,2,⋯,N,0≤α≤1, p,s>0,ps>1 such that
1sp+1sq=1, |
and ϖ,Θ: TN→R+ are continuous functions, then
∫ϵNηN⋯∫ϵ1η1ϖ1s(τ)Θ1s(τ)◊ατ≤(∫ϵNηN⋯∫ϵ1η1ϖp(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1Θq(τ)◊ατ)1sq. | (3.1) |
Proof. Applying (2.4) with
a=ϖ1s(ξ)(∫ϵNηN⋯∫ϵ1η1ϖp(τ)◊ατ)−1sp |
and
b=Θ1s(ξ)(∫ϵNηN⋯∫ϵ1η1Θq(τ)◊ατ)−1sq, |
we get
ϖp(ξ)sp∫ϵNηN⋯∫ϵ1η1ϖp(τ)◊ατ+Θq(ξ)sq∫ϵNηN⋯∫ϵ1η1Θq(τ)◊ατ≥ϖ1s(ξ)Θ1s(ξ)(∫ϵNηN⋯∫ϵ1η1ϖp(τ)◊ατ)−1sp×(∫ϵNηN⋯∫ϵ1η1Θq(τ)◊ατ)−1sq. | (3.2) |
Integrating (3.2) over ξi from ηi to ϵi,i=1,2,⋯,N, we observe that
∫ϵNηN⋯∫ϵ1η1ϖp(ξ)◊αξsp∫ϵNηN⋯∫ϵ1η1ϖp(τ)◊ατ+∫ϵNηN⋯∫ϵ1η1Θq(ξ)◊αξsq∫ϵNηN⋯∫ϵ1η1Θq(τ)◊ατ≥(∫ϵNηN⋯∫ϵ1η1ϖp(τ)◊ατ)−1sp(∫ϵNηN⋯∫ϵ1η1Θq(τ)◊ατ)−1sq×∫ϵNηN⋯∫ϵ1η1ϖ1s(ξ)Θ1s(ξ)◊αξ, |
and then (note that 1sp+1sq=1)
∫ϵNηN⋯∫ϵ1η1ϖ1s(τ)Θ1s(τ)◊ατ≤(∫ϵNηN⋯∫ϵ1η1ϖp(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1Θq(τ)◊ατ)1sq, |
which is (3.1).
Remark 3.1. If s=1 and N=1, we get the dynamic diamond alpha Hölder's inequality (2.2).
Remark 3.2. If T=R, ηi, ϵi∈R,ϵi>ηi,i=1,2,⋯,N,p,s>0,ps>1 such that
1sp+1sq=1, |
and ϖ,Θ: RN→R+ are continuous functions, then
∫ϵNηN⋯∫ϵ1η1ϖ1s(τ)Θ1s(τ)dτ≤(∫ϵNηN⋯∫ϵ1η1ϖp(τ)dτ)1sp(∫ϵNηN⋯∫ϵ1η1Θq(τ)dτ)1sq, |
where dτ=dτ1,⋯,dτN.
Remark 3.3. If T=N, N=1,η, ϵ∈N,ϵ>η,0≤α≤1, p,s>0,ps>1 such that
1sp+1sq=1, |
and ϖ,Θ are positive sequences, then
ϵ−1∑τ=η[αϖ1s(τ)Θ1s(τ)+(1−α)ϖ1s(τ+1)Θ1s(τ+1)]≤(ϵ−1∑τ=ηαϖp(τ)+(1−α)ϖp(τ+1))1sp×(ϵ−1∑τ=ηαΘq(τ)+(1−α)Θq(τ+1))1sq. |
Example 3.1. If T=R,ϵ∈R, N=1, η=0,sp=3,sq=, ϖ(τ)=τs, and Θ(τ)=τs, then
∫ϵηϖ1s(τ)Θ1s(τ)dτ<(∫ϵηϖp(τ)dτ)1sp(∫ϵηΘq(τ)dτ)1sq. |
Proof. Note that
∫ϵηϖ1s(τ)Θ1s(τ)dτ=∫ϵ0τ2dτ=τ33|ϵ0=ϵ33. | (3.3) |
Using the above assumptions, we observe that
∫ϵηϖp(τ)dτ=∫ϵ0τspdτ=τsp+1sp+1|ϵ0=ϵsp+1sp+1, |
then
(∫ϵηϖp(τ)dτ)1sp=(ϵsp+1sp+1)1sp=ϵ1+1sp(sp+1)1sp. | (3.4) |
Also, we can get
∫ϵηΘq(τ)dτ=∫ϵ0τsq(τ)dτ=τsq+1sq+1|ϵ0=ϵsq+1sq+1, |
and so
(∫ϵηΘq(τ)dτ)1sq=(ϵsq+1sq+1)1sq=ϵ1+1sq(sq+1)1sq. | (3.5) |
From (3.4) and (3.5) (note 1sp+1sq=1), we have that
(∫ϵηϖp(τ)dτ)1sp(∫ϵηΘq(τ)dτ)1sq=ϵ1+1sp(sp+1)1spϵ1+1sq(sq+1)1sq=ϵ3(sp+1)1sp(sq+1)1sq. | (3.6) |
Since sp=3 and sq=3/2,
(sp+1)1sp(sq+1)1sq=413(52)23=413(254)13=(25)13=2.9240177<3, |
then
ϵ3(sp+1)1sp(sq+1)1sq>ϵ33. | (3.7) |
From (3.3), (3.6) and (3.7), we see that
∫ϵηϖ1s(τ)Θ1s(τ)dτ<(∫ϵηϖp(τ)dτ)1sp(∫ϵηΘq(τ)dτ)1sq. |
The proof is complete.
Corollary 3.1. If ηi, ϵi∈T,α=1,ϵi>ηi,i=1,2,⋯,N,p,s>0,ps>1 such that
1sp+1sq=1 |
and ϖ,Θ: TN→R+, then
∫ϵNηN⋯∫ϵ1η1ϖ1s(τ)Θ1s(τ)Δτ1⋯ΔτN≤(∫ϵNηN⋯∫ϵ1η1ϖp(τ)Δτ1⋯ΔτN)1sp(∫ϵNηN⋯∫ϵ1η1Θq(τ)Δτ1⋯ΔτN)1sq. | (3.8) |
Corollary 3.2. If ηi, ϵi∈T,α=0,ϵi>ηi,i=1,2,⋯,N,p,s>0,ps>1 such that
1sp+1sq=1, |
and ϖ,Θ: TN→R+, then
∫ϵNηN⋯∫ϵ1η1ϖ1s(τ)Θ1s(τ)∇τ1⋯∇τN≤(∫ϵNηN⋯∫ϵ1η1ϖp(τ)∇τ1⋯∇τN)1sp(∫ϵNηN⋯∫ϵ1η1Θq(τ)∇τ1⋯∇τN)1sq. | (3.9) |
Theorem 3.2. Assume that ηi, ϵi∈T,ϵi>ηi,i=1,2,⋯,N,0≤α≤1,p,s>0,ps>1 such that
1sp+1sq=1, |
and Ξ,Ω: TN→R+ are continuous functions. If 0<m≤Ξ/Ω≤M<∞, then
∫ϵNηN⋯∫ϵ1η1Ξ1ps(τ)Ω1qs(τ)◊ατ≤M1p2s2m1q2s2∫ϵNηN⋯∫ϵ1η1Ξ1qs(τ)Ω1ps(τ)◊ατ. | (3.10) |
Proof. Applying (3.1) with ϖ(τ)=Ξ1p(τ) and Θ(τ)=Ω1q(τ), we see that
∫ϵNηN⋯∫ϵ1η1Ξ1ps(τ)Ω1qs(τ)◊ατ≤(∫ϵNηN⋯∫ϵ1η1Ξ(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1Ω(τ)◊ατ)1sq. | (3.11) |
Since
1sp+1sq=1, |
then (3.11) becomes
∫ϵNηN⋯∫ϵ1η1Ξ1ps(τ)Ω1qs(τ)◊ατ≤(∫ϵNηN⋯∫ϵ1η1Ξ1ps(τ)Ξ1qs(τ)◊ατ)1sp×(∫ϵNηN⋯∫ϵ1η1Ω1ps(τ)Ω1qs(τ)◊ατ)1sq. | (3.12) |
Since
Ξ1ps(τ)≤M1psΩ1ps(τ) |
and
Ω1qs(τ)≤m−1qsΞ1qs(τ), |
we have from (3.12) that
∫ϵNηN⋯∫ϵ1η1Ξ1ps(τ)Ω1qs(τ)◊ατ≤M1p2s2m1q2s2(∫ϵNηN⋯∫ϵ1η1Ξ1qs(τ)Ω1ps(τ)◊ατ)1sp×(∫ϵNηN⋯∫ϵ1η1Ξ1qs(τ)Ω1ps(τ)◊ατ)1sq=M1p2s2m1q2s2(∫ϵNηN⋯∫ϵ1η1Ξ1qs(τ)Ω1ps(τ)◊ατ)1sp+1sq, |
then we have for
1sp+1sq=1, |
that
∫ϵNηN⋯∫ϵ1η1Ξ1ps(τ)Ω1qs(τ)◊ατ≤M1p2s2m1q2s2∫ϵNηN⋯∫ϵ1η1Ξ1qs(τ)Ω1ps(τ)◊ατ, |
which is (3.10).
Remark 3.4. Take T=R,ηi, ϵi∈R,ϵi>ηi,i=1,2,⋯,N,p,s>0,ps>1 such that
1sp+1sq=1 |
and Ξ,Ω: RN→R+ are continuous functions. If 0<m≤Ξ/Ω≤M<∞, then
∫ϵNηN⋯∫ϵ1η1Ξ1ps(τ)Ω1qs(τ)dτ≤M1p2s2m1q2s2∫ϵNηN⋯∫ϵ1η1Ξ1qs(τ)Ω1ps(τ)dτ. |
Remark 3.5. Take T=N,N=1,η, ϵ∈N,ϵ>η,0≤α≤1, p,s>0,ps>1 such that
1sp+1sq=1, |
and Ξ,Ω are positive sequences. If 0<m≤Ξ/Ω≤M<∞, then
ϵ−1∑τ=η[α(Ξ1ps(τ)Ω1qs(τ))+(1−α)(Ξ1ps(τ+1)Ω1qs(τ+1))]≤M1p2s2m1q2s2ϵ−1∑τ=η[α(Ξ1qs(τ)Ω1ps(τ))+(1−α)(Ξ1qs(τ+1)Ω1ps(τ+1))]. |
Theorem 3.3. Assume that ηi, ϵi∈T,ϵi>ηi,i=1,2,⋯,N,0≤α≤1, sp<0 such that
1sp+1sq=1, |
and ϕ,λ: TN→R+ are continuous functions, then
∫ϵNηN⋯∫ϵ1η1ϕ1s(τ)λ1s(τ)◊ατ≥(∫ϵNηN⋯∫ϵ1η1ϕp(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1λq(τ)◊ατ)1sq. | (3.13) |
Proof. Since sp<0, by applying (3.1) with indices
sP=−spsq=1−sp>1,sQ=1sq, |
(note that 1sP+1sQ=1), ϖ(τ)=ϕ−sq(τ) and Θ(τ)=ϕsq(τ)λsq(τ), then
∫ϵNηN⋯∫ϵ1η1ϖ1s(τ)Θ1s(τ)◊ατ≤(∫ϵNηN⋯∫ϵ1η1ϖP(τ)◊ατ)1sP(∫ϵNηN⋯∫ϵ1η1ΘQ(τ)◊ατ)1sQ, |
and substituting with values ϖ(τ) and Θ(τ), the last inequality gives us
∫ϵNηN⋯∫ϵ1η1ϕ−q(τ)[ϕq(τ)λq(τ)]◊ατ≤(∫ϵNηN⋯∫ϵ1η1[ϕ−sq(τ)]−psq◊ατ)−sqsp(∫ϵNηN⋯∫ϵ1η1[ϕsq(τ)λsq(τ)]1s2q◊ατ)sq. |
Thus
∫ϵNηN⋯∫ϵ1η1λq(τ)◊ατ≤(∫ϵNηN⋯∫ϵ1η1ϕp(τ)◊ατ)−qp(∫ϵNηN⋯∫ϵ1η1ϕ1s(τ)λ1s(τ)◊ατ)sq, |
then we have for sp<0 and sq=sp/(sp−1)>0 that
∫ϵNηN⋯∫ϵ1η1ϕ1s(τ)λ1s(τ)◊ατ≥(∫ϵNηN⋯∫ϵ1η1ϕp(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1λq(τ)◊ατ)1sq, |
which is (3.13).
Remark 3.6. If T=R, ηi, ϵi∈R,ϵi>ηi,i=1,2,⋯,N,0≤α≤1, sp<0 such that
1sp+1sq=1, |
and ϕ,λ: RN→R+are continuous functions, then
∫ϵNηN⋯∫ϵ1η1ϕ1s(τ)λ1s(τ)dτ≥(∫ϵNηN⋯∫ϵ1η1ϕp(τ)dτ)1sp(∫ϵNηN⋯∫ϵ1η1λq(τ)dτ)1sq. |
Remark 3.7. If T=N, N=1,η, ϵ∈N, ϵ>η,0≤α≤1, sp<0, such that
1sp+1sq=1, |
and ϕ,λ are positive sequences, then
ϵ−1∑τ=η[α(ϕ1s(τ)λ1s(τ))+(1−α)(ϕ1s(τ+1)λ1s(τ+1))]≥(ϵ−1∑τ=η[αϕp(τ)+(1−α)ϕp(τ+1)])1sp(ϵ−1∑τ=η[αλq(τ)+(1−α)λq(τ+1)])1sq. |
Theorem 3.4. Assume that ηi, ϵi∈T,ϵi>ηi,i=1,2,⋯,N,0≤α≤1, r>u>t>0, and Ξ,Ω: TN→R+are continuous functions, then
(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωu(τ)◊ατ)r−t≤(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωt(τ)◊ατ)r−u(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωr(τ)◊ατ)u−t. | (3.14) |
Proof. Applying (3.1) with
p=r−tr−u,q=r−tu−t, |
(note that s=1),
ϖp(τ)=Ξ(τ)Ωt(τ)and Θq(τ)=Ξ(τ)Ωr(τ), |
we see that
∫ϵNηN⋯∫ϵ1η1[Ξ(τ)Ωt(τ)]r−ur−t[Ξ(τ)Ωr(τ)]u−tr−t◊ατ≤(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωt(τ)◊ατ)r−ur−t(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωr(τ)◊ατ)u−tr−t, |
and then
(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωu(τ)◊ατ)r−t≤(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωt(τ)◊ατ)r−u(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωr(τ)◊ατ)u−t, |
which is (3.14).
Remark 3.8. If T=R, ηi, ϵi∈T, ϵi>ηi,i=1,2,⋯,N,r>u>t>0 and Ξ,Ω: RN→R+ are continuous functions, then
(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωu(τ)dτ)r−t≤(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωt(τ)dτ)r−u(∫ϵNηN⋯∫ϵ1η1Ξ(τ)Ωr(τ)dτ)u−t. |
Remark 3.9. If T=N, N=1,η, ϵ∈N, ϵ>η,0≤α≤1, r>u>t>0 and Ξ,Ω are positive sequences, then
(ϵ−1∑τ=η[α(Ξ(τ)Ωu(τ))+(1−α)(Ξ(τ+1)Ωu(τ+1))])r−t≤(ϵ−1∑τ=η[α(Ξ(τ)Ωt(τ))+(1−α)(Ξ(τ+1)Ωt(τ+1))])r−u×(ϵ−1∑τ=η[α(Ξ(τ)Ωr(τ))+(1−α)(Ξ(τ+1)Ωr(τ+1))])u−t. |
Theorem 3.5. Assume that ηi, ϵi∈T,ϵi>ηi,i=1,2,⋯,N,0≤α≤1, s≥1,sp>1 such that
1sp+1sq=1, |
and Ξ,Ω: TN→R+ are continuous functions, then
(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)1sp≤(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp+(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp. | (3.15) |
Proof. Note that
(Ξ(τ)+Ω(τ))p=(Ξ(τ)+Ω(τ))1s(Ξ(τ)+Ω(τ))p−1s. | (3.16) |
Using the inequality (2.5) with replacing x, y by Ξ1s(τ) and Ω1s(τ), respectively, we have
(Ξ(τ)+Ω(τ))1s≤Ξ1s(τ)+Ω1s(τ), |
therefore, (3.16) becomes
(Ξ(τ)+Ω(τ))p≤Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)+Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1), |
then
∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ≤∫ϵNηN⋯∫ϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ+∫ϵNηN⋯∫ϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ. | (3.17) |
Applying (3.1) on the two terms of the righthand side of (3.17) with indices sp>1 and (sp)∗=sp/(sp−1) (note 1sp+1(sp)∗=1), we obtain
∫ϵNηN⋯∫ϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ≤(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)sp−1sp | (3.18) |
and
∫ϵNηN⋯∫ϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ≤(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)sp−1sp. | (3.19) |
Adding (3.18) and (3.19) and substituting into (3.17), we see that
∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ≤[(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp+(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp]×(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)sp−1sp, |
thus,
(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)1sp≤(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp+(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp, |
which is (3.15).
Remark 3.10. If s=1 and N=1, then we get the dynamic Minkowski inequality (2.3).
In the following, we establish the reversed form of inequality (3.15).
Theorem 3.6. Assume that ηi, ϵi∈T,ϵi>ηi,i=1,2,⋯,N,0≤α≤1, p<0,0<s<1 and Ξ,Ω: TN→R+ are continuous functions, then
(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)1sp≥(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp+(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp. | (3.20) |
Proof. Applying (2.6) with replacing x,y by Ξ1s(τ) and Ω1s(τ), respectively, we have that
(Ξ(τ)+Ω(τ))1s≥Ξ1s(τ)+Ω1s(τ). | (3.21) |
Since
(Ξ(τ)+Ω(τ))p=(Ξ(τ)+Ω(τ))1s(Ξ(τ)+Ω(τ))p−1s, |
by using (3.21), we get
(Ξ(τ)+Ω(τ))p≥Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)+Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1), |
then
∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ≥∫ϵNηN⋯∫ϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ+∫ϵNηN⋯∫ϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ. | (3.22) |
Applying (3.13) on ∫ϵNηN⋯∫ϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ, with
ϕ(τ)=Ξ(τ),λ(τ)=(Ξ(τ)+Ω(τ))sp−1, |
and the indices sp<0, (sp)∗=sp/(sp−1)(note 1sp+1(sp)∗=1), we see that
∫ϵNηN⋯∫ϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ≥(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)sp−1sp. | (3.23) |
Again by applying (3.13) on ∫ϵNηN⋯∫ϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ with
ϕ(τ)=Ω(τ),λ(τ)=(Ξ(τ)+Ω(τ))sp−1, |
and the indices sp<0, (sp)∗=sp/(sp−1)(note 1sp+1(sp)∗=1), we have that
∫ϵNηN⋯∫ϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp−1)◊ατ≥(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp×(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)sp−1sp. | (3.24) |
Substituting (3.23) and (3.24) into (3.22), we see that
∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ≥[(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp+(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp]×(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)sp−1sp, |
then
(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))p◊ατ)1sp≥(∫ϵNηN⋯∫ϵ1η1Ξp(τ)◊ατ)1sp+(∫ϵNηN⋯∫ϵ1η1Ωp(τ)◊ατ)1sp, |
which is (3.20).
Remark 3.11. If T=R, ηi, ϵi∈T,ϵi>ηi,i=1,2,⋯,N,p<0,0<s<1 and Ξ,Ω: RN→R+ are continuous functions, then
(∫ϵNηN⋯∫ϵ1η1(Ξ(τ)+Ω(τ))pdτ)1sp≥(∫ϵNηN⋯∫ϵ1η1Ξp(τ)dτ)1sp+(∫ϵNηN⋯∫ϵ1η1Ωp(τ)dτ)1sp. |
Remark 3.12. If T=N, N=1,η, ϵ∈N, ϵ>η,0≤α≤1, p<0,0<s<1, and Ξ,Ω are positive sequences, then
(ϵ−1∑τ=η[α(Ξ(τ)+Ω(τ))p+(1−α)(Ξ(τ+1)+Ω(τ+1))p])1sp≥(ϵ−1∑τ=η[αΞp(τ)+(1−α)Ξp(τ+1)])1sp+(ϵ−1∑τ=ηαΩp(τ)+(1−α)Ωp(τ+1))1sp. |
In this paper, we present novel generalizations of Hölder's and Minkowski's dynamic inequalities on diamond alpha time scales. These inequalities give us the inequalities on delta calculus when α=1 and the inequalities on nabla calculus when α=0. Also, we introduced some of the continuous and discrete inequalities as special cases of our results. In addition, we added an example in our results to indicate the work.
In the future, we will establish some new generalizations of Hölder's and Minkowski's dynamic inequalities on conformable delta fractional time scales. Also, we will prove some new reversed versions of Hölder's and Minkowski's dynamic inequalities on time scales.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
The authors extend their appreciation to the Deputyship for Research & Innovation, Ministry of Education in Saudi Arabia for funding this research work through the project number: ISP23-86.
The authors declare that they have no conflicts of interest.
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