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Research article Special Issues

Tumor necrosis factor-related apoptosis-inducing ligand regulate the accumulation of extracelluar matrix in pulmonary artery by activating the phosphorylation of Smad2/3

  • Introduction Previous studies have found that tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) was involved in the progression of pulmonary hypertension (PH), and TRAIL knocking (KO) has an inhibitory effect on PH, but its mechanism is not completely clear.
    Methods The effects of TRAIL on the accumulation of extracelluar matrix (ECM), which is one of the most important processes of vascular remodeling, were observed in mice and isolated pulmonary artery smooth muscle cells (PASMCs). In vivo, mice were divided into four groups: Control group (n = 5), hypoxia-induced PH mice group (n = 8), anti-TRAIL antibody (TRAIL-Ab) treatment group (n = 8) and IgG antibody (IgG) group (n = 8). The effects of TRAIL-Ab on ECM expression in hypoxic induced PH were researched; in vivo, PASMCs were divided into three groups: Control group, hypoxia-induced group, TRAIL-Ab group. Expressions of p-Smad2/3 and p-Smad1/5/8 were compared among the three groups.
    Results Hypoxia-induced PH mice had significant increases in right ventricle systolic pressure (RVSP) (P < 0.001), right ventricular hypertrophy (RVH) (P = 0.007), vascular stenosis (P < 0.001) compared with controls. Mice with anti-TRAIL antibody had lower levels in RVSP (P < 0.001), RVH (P < 0.001), vascular stenosis (P < 0.001) than PH mice. Besides, the TRAIL-Ab significantly inhibited the phosphorylation of Smad2/3 compared with hypoxia-induced group.
    Conclusion TRAIL regulates the accumulation of ECM in pulmonary artery by activating pSmad2/3.

    Citation: Erli Yang, Xiaobei Zhang, Qiangsheng Chen, Chandong Ding. Tumor necrosis factor-related apoptosis-inducing ligand regulate the accumulation of extracelluar matrix in pulmonary artery by activating the phosphorylation of Smad2/3[J]. Mathematical Biosciences and Engineering, 2020, 17(2): 1372-1380. doi: 10.3934/mbe.2020069

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  • Introduction Previous studies have found that tumor necrosis factor-related apoptosis-inducing ligand (TRAIL) was involved in the progression of pulmonary hypertension (PH), and TRAIL knocking (KO) has an inhibitory effect on PH, but its mechanism is not completely clear.
    Methods The effects of TRAIL on the accumulation of extracelluar matrix (ECM), which is one of the most important processes of vascular remodeling, were observed in mice and isolated pulmonary artery smooth muscle cells (PASMCs). In vivo, mice were divided into four groups: Control group (n = 5), hypoxia-induced PH mice group (n = 8), anti-TRAIL antibody (TRAIL-Ab) treatment group (n = 8) and IgG antibody (IgG) group (n = 8). The effects of TRAIL-Ab on ECM expression in hypoxic induced PH were researched; in vivo, PASMCs were divided into three groups: Control group, hypoxia-induced group, TRAIL-Ab group. Expressions of p-Smad2/3 and p-Smad1/5/8 were compared among the three groups.
    Results Hypoxia-induced PH mice had significant increases in right ventricle systolic pressure (RVSP) (P < 0.001), right ventricular hypertrophy (RVH) (P = 0.007), vascular stenosis (P < 0.001) compared with controls. Mice with anti-TRAIL antibody had lower levels in RVSP (P < 0.001), RVH (P < 0.001), vascular stenosis (P < 0.001) than PH mice. Besides, the TRAIL-Ab significantly inhibited the phosphorylation of Smad2/3 compared with hypoxia-induced group.
    Conclusion TRAIL regulates the accumulation of ECM in pulmonary artery by activating pSmad2/3.


    In 1889, Hölder [1] proved that

    nk=1xkyk(nk=1xpk)1p(nk=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,ϵR, γ>1, 1/γ+1/ν=1, and λ,ωC([a,b],R). If 0<γ<1, then (1.2) is reversed.

    The subsequent inequality, widely recognized as Minkowski's inequality, asserts that, for p1, 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 p1 and

    1<mφ(ϑ)ϖ(ϑ)M, ϑ[η,ϵ],

    then

    M+1(M1)(ϵη(φ(ϑ)ϖ(ϑ))pdϑ)1p(ϵηφp(ϑ)dϑ)1p+(ϵηϖp(ϑ)dϑ)1p(m+1m1)(ϵη(φ(ϑ)ϖ(ϑ))pdϑ)1p. (1.3)

    Sroysang [3] proved that if p1 and φ,ϖ>0 with

    0<c<mφ(ϑ)ϖ(ϑ)M, ϑ[η,ϵ],

    then

    M+1(Mc)(ϵη(φ(ϑ)cϖ(ϑ))pdϑ)1p(ϵηφp(ϑ)dϑ)1p+(ϵηϖp(ϑ)dϑ)1p(m+1mc)(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,s1,

    1sp+1sq=1,

    then

    nk=1(xkyk)1s(nk=1xpk)1sp(nk=1yqk)1sq, (1.4)

    and if s1,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 mN, p>1,

    1sp+1sq=1,

    then

    (nk=1(xk+yk)p)1mp(nk=1xpk)1mp+(nk=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{sT:s>ξ}

    and

    ρ(ξ):=sup{sT:s<ξ},

    respectively. The graininess function μ for a time scale T is defined by

    μ(ξ):=σ(ξ)ξ0,

    and for any function φ: TR, 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: TR is a function and let ξT. 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=(ξδ,ξ+δ)T

    for some δ>0, such that

    |φ(σ(ξ))φ(s)φΔ(ξ)(σ(ξ)s)|ϵ|σ(ξ)s|,   sU, sσ(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: TR is said to be - differentiable at ξT, 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 NU and

    |ψ(ρ(ξ))ψ(s)ψ(ξ)[ρ(ξ)s]|ϵ|ρ(ξ)s|,

    for all sN.

    Definition 2.3. ([6]) Let T be a time scale and Ξ(ξ) be differentiable on T in the Δ and sense. For ξT, we define the diamond α derivative Ξα(ξ) by

    Ξα(ξ)=αΞΔ(ξ)+(1α)Ξ(ξ),  0α1.

    Thus, Ξ is diamond α differentiable if, and only if, Ξ is Δ and differentiable.

    For α=1, we get that

    Ξα(ξ)=αΞΔ(ξ),

    and for α=0, we have that

    Ξα(ξ)=Ξ(ξ).

    Theorem 2.1. ([6]) Let Ξ, Ω: TR be diamond-α differentiable at ξT, then

    (1) Ξ+Ω: TR is diamond-α differentiable at ξT, with

    (Ξ+Ω)α(ξ)=Ξα(ξ)+Ωα(ξ).

    (2) Ξ.Ω: TR is diamond-α differentiable at ξT, with

    (Ξ.Ω)α(ξ)=Ξα(ξ)Ω(ξ)+αΞσ(ξ)ΩΔ(ξ)+(1α)Ξρ(ξ)Ω(ξ).

    (3) For Ω(ξ)Ωσ(ξ)Ωρ(ξ)0, Ξ/Ω: TR is diamond-α differentiable at ξT, with

    (ΞΩ)α(ξ)=Ξα(ξ)Ωσ(ξ)Ωρ(ξ)αΞσ(ξ)Ωρ(ξ)ΩΔ(ξ)(1α)Ξρ(ξ)Ωσ(ξ)Ω(ξ)Ω(ξ)Ωσ(ξ)Ωρ(ξ).

    Theorem 2.2. ([6]) Let Ξ, Ω: TR be diamond-α differentiable at ξT, then the following holds

    (1) (Ξ)αΔ(ξ)=αΞΔΔ(ξ)+(1α)ΞΔ(ξ),

    (2) (Ξ)α(ξ)=αΞΔ(ξ)+(1α)Ξ(ξ),

    (3) (Ξ)Δα(ξ)=αΞΔΔ(ξ)+(1α)ΞΔ(ξ)(Ξ)αΔ(ξ),

    (4) (Ξ)α(ξ)=αΞΔ(ξ)+(1α)Ξ(ξ)(Ξ)α(ξ),

    (5) (Ξ)αα(ξ)=α2ΞΔΔ(ξ)+α(1α)[ΞΔ(ξ)+ΞΔ(ξ)].

    Theorem 2.3. ([6]) Let a,ξT and h: TR, then the diamondα integral from a to ξ of h is defined by

    tah(s)αs=αtah(s)Δs+(1α)tah(s)s, 0α1,

    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)αh(ξ), for ξT.

    Example 2.1. [30] Let T={0, 1, 2},a=0, and h(ξ)=ξ2 for ξT. This gives

    (ξah(s)αs)α|ξ=1=1+2α(1α),

    so that the equality above holds only when α=Δ or α=.

    Theorem 2.4. ([6]) Let a,b,ξT, β,γ,cR, and Ξ and Ω be continuous functions on [a,b]T, then the following properties hold.

    (1) ba[γΞ(ξ)+βΩ(ξ)]αξ=γbaΞ(ξ)αξ+βbaΩ(ξ)αξ,

    (2) bacΞ(ξ)αξ=cbaΞ(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,bT 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)αξ0;

    (2) If Ξ(ξ)Ω(ξ) for all ξ[a,b]T, then baΞ(ξ)αξbaΩ(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, bR,

    and if T=N and m<n, then we obtain

    nmΞ(ξ)αξ=n1i=m[αΞ(i)+(1α)Ξ(i+1)], for m, nN. (2.1)

    Lemma 2.1. (Hölder's inequality [6]) If η,ϵT, 0α1, and λ,ωC([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,ϵT, ϵ>η,0α1,p>1, and Ξ,ΩC([η,ϵ]T,R+), then

    (ϵη(Ξ(τ)+Ω(τ))pατ)1p(ϵηΞp(τ)ατ)1p+(ϵηΩp(τ)ατ)1p. (2.3)

    Lemma 2.3. (Generalized Young inequality [4]) If a,b0,sp,sq>1 with

    1sp+1sq=1,

    then

    aspsp+bsqsqab. (2.4)

    Lemma 2.4. (See [31]) If x,y>0 and s>1, then

    (xs+ys)1sx+y, (2.5)

    and if 0<s<1, then

    (xs+ys)1sx+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, ϵiT,ϵi>ηi,i=1,2,,N,0α1, p,s>0,ps>1 such that

    1sp+1sq=1,

    and ϖ,Θ: TNR+ are continuous functions, then

    ϵNηNϵ1η1ϖ1s(τ)Θ1s(τ)ατ(ϵNηNϵ1η1ϖp(τ)ατ)1sp(ϵNηNϵ1η1Θq(τ)ατ)1sq. (3.1)

    Proof. Applying (2.4) with

    a=ϖ1s(ξ)(ϵNηNϵ1η1ϖp(τ)ατ)1sp

    and

    b=Θ1s(ξ)(ϵNηNϵ1η1Θq(τ)ατ)1sq,

    we get

    ϖp(ξ)spϵNηNϵ1η1ϖp(τ)ατ+Θq(ξ)sqϵNηNϵ1η1Θq(τ)ατϖ1s(ξ)Θ1s(ξ)(ϵNηNϵ1η1ϖp(τ)ατ)1sp×(ϵNηNϵ1η1Θq(τ)ατ)1sq. (3.2)

    Integrating (3.2) over ξi from ηi to ϵi,i=1,2,,N, we observe that

    ϵNηNϵ1η1ϖp(ξ)αξspϵNηNϵ1η1ϖp(τ)ατ+ϵNηNϵ1η1Θq(ξ)αξsqϵNηNϵ1η1Θq(τ)ατ(ϵNηNϵ1η1ϖp(τ)ατ)1sp(ϵNηNϵ1η1Θq(τ)ατ)1sq×ϵNηNϵ1η1ϖ1s(ξ)Θ1s(ξ)αξ,

    and then (note that 1sp+1sq=1)

    ϵNηNϵ1η1ϖ1s(τ)Θ1s(τ)ατ(ϵNηNϵ1η1ϖp(τ)ατ)1sp(ϵNηNϵ1η1Θq(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, ϵiR,ϵi>ηi,i=1,2,,N,p,s>0,ps>1 such that

    1sp+1sq=1,

    and ϖ,Θ: RNR+ are continuous functions, then

    ϵNηNϵ1η1ϖ1s(τ)Θ1s(τ)dτ(ϵNηNϵ1η1ϖp(τ)dτ)1sp(ϵNηNϵ1η1Θq(τ)dτ)1sq,

    where dτ=dτ1,,dτN.

    Remark 3.3. If T=N, N=1,η, ϵN,ϵ>η,0α1, p,s>0,ps>1 such that

    1sp+1sq=1,

    and ϖ,Θ are positive sequences, then

    ϵ1τ=η[αϖ1s(τ)Θ1s(τ)+(1α)ϖ1s(τ+1)Θ1s(τ+1)](ϵ1τ=ηαϖp(τ)+(1α)ϖp(τ+1))1sp×(ϵ1τ=ηαΘq(τ)+(1α)Θq(τ+1))1sq.

    Example 3.1. If T=R,ϵR, 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, ϵiT,α=1,ϵi>ηi,i=1,2,,N,p,s>0,ps>1 such that

    1sp+1sq=1

    and ϖ,Θ: TNR+, then

    ϵNηNϵ1η1ϖ1s(τ)Θ1s(τ)Δτ1ΔτN(ϵNηNϵ1η1ϖp(τ)Δτ1ΔτN)1sp(ϵNηNϵ1η1Θq(τ)Δτ1ΔτN)1sq. (3.8)

    Corollary 3.2. If ηi, ϵiT,α=0,ϵi>ηi,i=1,2,,N,p,s>0,ps>1 such that

    1sp+1sq=1,

    and ϖ,Θ: TNR+, then

    ϵNηNϵ1η1ϖ1s(τ)Θ1s(τ)τ1τN(ϵNηNϵ1η1ϖp(τ)τ1τN)1sp(ϵNηNϵ1η1Θq(τ)τ1τN)1sq. (3.9)

    Theorem 3.2. Assume that ηi, ϵiT,ϵi>ηi,i=1,2,,N,0α1,p,s>0,ps>1 such that

    1sp+1sq=1,

    and Ξ,Ω: TNR+ are continuous functions. If 0<mΞ/ΩM<, then

    ϵNηNϵ1η1Ξ1ps(τ)Ω1qs(τ)ατM1p2s2m1q2s2ϵNηNϵ1η1Ξ1qs(τ)Ω1ps(τ)ατ. (3.10)

    Proof. Applying (3.1) with ϖ(τ)=Ξ1p(τ) and Θ(τ)=Ω1q(τ), we see that

    ϵNηNϵ1η1Ξ1ps(τ)Ω1qs(τ)ατ(ϵNηNϵ1η1Ξ(τ)ατ)1sp(ϵNηNϵ1η1Ω(τ)ατ)1sq. (3.11)

    Since

    1sp+1sq=1,

    then (3.11) becomes

    ϵNηNϵ1η1Ξ1ps(τ)Ω1qs(τ)ατ(ϵNηNϵ1η1Ξ1ps(τ)Ξ1qs(τ)ατ)1sp×(ϵNηNϵ1η1Ω1ps(τ)Ω1qs(τ)ατ)1sq. (3.12)

    Since

    Ξ1ps(τ)M1psΩ1ps(τ)

    and

    Ω1qs(τ)m1qsΞ1qs(τ),

    we have from (3.12) that

    ϵNηNϵ1η1Ξ1ps(τ)Ω1qs(τ)ατM1p2s2m1q2s2(ϵNηNϵ1η1Ξ1qs(τ)Ω1ps(τ)ατ)1sp×(ϵNηNϵ1η1Ξ1qs(τ)Ω1ps(τ)ατ)1sq=M1p2s2m1q2s2(ϵNηNϵ1η1Ξ1qs(τ)Ω1ps(τ)ατ)1sp+1sq,

    then we have for

    1sp+1sq=1,

    that

    ϵNηNϵ1η1Ξ1ps(τ)Ω1qs(τ)ατM1p2s2m1q2s2ϵNηNϵ1η1Ξ1qs(τ)Ω1ps(τ)ατ,

    which is (3.10).

    Remark 3.4. Take T=R,ηi, ϵiR,ϵi>ηi,i=1,2,,N,p,s>0,ps>1 such that

    1sp+1sq=1

    and Ξ,Ω: RNR+ are continuous functions. If 0<mΞ/ΩM<, then

    ϵNηNϵ1η1Ξ1ps(τ)Ω1qs(τ)dτM1p2s2m1q2s2ϵNηNϵ1η1Ξ1qs(τ)Ω1ps(τ)dτ.

    Remark 3.5. Take T=N,N=1,η, ϵN,ϵ>η,0α1, p,s>0,ps>1 such that

    1sp+1sq=1,

    and Ξ,Ω are positive sequences. If 0<mΞ/ΩM<, then

    ϵ1τ=η[α(Ξ1ps(τ)Ω1qs(τ))+(1α)(Ξ1ps(τ+1)Ω1qs(τ+1))]M1p2s2m1q2s2ϵ1τ=η[α(Ξ1qs(τ)Ω1ps(τ))+(1α)(Ξ1qs(τ+1)Ω1ps(τ+1))].

    Theorem 3.3. Assume that ηi, ϵiT,ϵi>ηi,i=1,2,,N,0α1, sp<0 such that

    1sp+1sq=1,

    and ϕ,λ: TNR+ are continuous functions, then

    ϵNηNϵ1η1ϕ1s(τ)λ1s(τ)ατ(ϵNηNϵ1η1ϕp(τ)ατ)1sp(ϵNηNϵ1η1λq(τ)ατ)1sq. (3.13)

    Proof. Since sp<0, by applying (3.1) with indices

    sP=spsq=1sp>1,sQ=1sq,

    (note that 1sP+1sQ=1), ϖ(τ)=ϕsq(τ) and Θ(τ)=ϕsq(τ)λsq(τ), then

    ϵNηNϵ1η1ϖ1s(τ)Θ1s(τ)ατ(ϵNηNϵ1η1ϖP(τ)ατ)1sP(ϵNηNϵ1η1ΘQ(τ)ατ)1sQ,

    and substituting with values ϖ(τ) and Θ(τ), the last inequality gives us

    ϵNηNϵ1η1ϕq(τ)[ϕq(τ)λq(τ)]ατ(ϵNηNϵ1η1[ϕsq(τ)]psqατ)sqsp(ϵNηNϵ1η1[ϕsq(τ)λsq(τ)]1s2qατ)sq.

    Thus

    ϵNηNϵ1η1λq(τ)ατ(ϵNηNϵ1η1ϕp(τ)ατ)qp(ϵNηNϵ1η1ϕ1s(τ)λ1s(τ)ατ)sq,

    then we have for sp<0 and sq=sp/(sp1)>0 that

    ϵNηNϵ1η1ϕ1s(τ)λ1s(τ)ατ(ϵNηNϵ1η1ϕp(τ)ατ)1sp(ϵNηNϵ1η1λq(τ)ατ)1sq,

    which is (3.13).

    Remark 3.6. If T=R, ηi, ϵiR,ϵi>ηi,i=1,2,,N,0α1, sp<0 such that

    1sp+1sq=1,

    and ϕ,λ: RNR+are continuous functions, then

    ϵNηNϵ1η1ϕ1s(τ)λ1s(τ)dτ(ϵNηNϵ1η1ϕp(τ)dτ)1sp(ϵNηNϵ1η1λq(τ)dτ)1sq.

    Remark 3.7. If T=N, N=1,η, ϵN, ϵ>η,0α1, sp<0, such that

    1sp+1sq=1,

    and ϕ,λ are positive sequences, then

    ϵ1τ=η[α(ϕ1s(τ)λ1s(τ))+(1α)(ϕ1s(τ+1)λ1s(τ+1))](ϵ1τ=η[αϕp(τ)+(1α)ϕp(τ+1)])1sp(ϵ1τ=η[αλq(τ)+(1α)λq(τ+1)])1sq.

    Theorem 3.4. Assume that ηi, ϵiT,ϵi>ηi,i=1,2,,N,0α1, r>u>t>0, and Ξ,Ω: TNR+are continuous functions, then

    (ϵNηNϵ1η1Ξ(τ)Ωu(τ)ατ)rt(ϵNηNϵ1η1Ξ(τ)Ωt(τ)ατ)ru(ϵNηNϵ1η1Ξ(τ)Ωr(τ)ατ)ut. (3.14)

    Proof. Applying (3.1) with

    p=rtru,q=rtut,

    (note that s=1),

    ϖp(τ)=Ξ(τ)Ωt(τ)and Θq(τ)=Ξ(τ)Ωr(τ),

    we see that

    ϵNηNϵ1η1[Ξ(τ)Ωt(τ)]rurt[Ξ(τ)Ωr(τ)]utrtατ(ϵNηNϵ1η1Ξ(τ)Ωt(τ)ατ)rurt(ϵNηNϵ1η1Ξ(τ)Ωr(τ)ατ)utrt,

    and then

    (ϵNηNϵ1η1Ξ(τ)Ωu(τ)ατ)rt(ϵNηNϵ1η1Ξ(τ)Ωt(τ)ατ)ru(ϵNηNϵ1η1Ξ(τ)Ωr(τ)ατ)ut,

    which is (3.14).

    Remark 3.8. If T=R, ηi, ϵiT, ϵi>ηi,i=1,2,,N,r>u>t>0 and Ξ,Ω: RNR+ are continuous functions, then

    (ϵNηNϵ1η1Ξ(τ)Ωu(τ)dτ)rt(ϵNηNϵ1η1Ξ(τ)Ωt(τ)dτ)ru(ϵNηNϵ1η1Ξ(τ)Ωr(τ)dτ)ut.

    Remark 3.9. If T=N, N=1,η, ϵN, ϵ>η,0α1, r>u>t>0 and Ξ,Ω are positive sequences, then

    (ϵ1τ=η[α(Ξ(τ)Ωu(τ))+(1α)(Ξ(τ+1)Ωu(τ+1))])rt(ϵ1τ=η[α(Ξ(τ)Ωt(τ))+(1α)(Ξ(τ+1)Ωt(τ+1))])ru×(ϵ1τ=η[α(Ξ(τ)Ωr(τ))+(1α)(Ξ(τ+1)Ωr(τ+1))])ut.

    Theorem 3.5. Assume that ηi, ϵiT,ϵi>ηi,i=1,2,,N,0α1, s1,sp>1 such that

    1sp+1sq=1,

    and Ξ,Ω: TNR+ are continuous functions, then

    (ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)1sp(ϵNηNϵ1η1Ξp(τ)ατ)1sp+(ϵNηNϵ1η1Ωp(τ)ατ)1sp. (3.15)

    Proof. Note that

    (Ξ(τ)+Ω(τ))p=(Ξ(τ)+Ω(τ))1s(Ξ(τ)+Ω(τ))p1s. (3.16)

    Using the inequality (2.5) with replacing x, y by Ξ1s(τ) and Ω1s(τ), respectively, we have

    (Ξ(τ)+Ω(τ))1sΞ1s(τ)+Ω1s(τ),

    therefore, (3.16) becomes

    (Ξ(τ)+Ω(τ))pΞ1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)+Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp1),

    then

    ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατϵNηNϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ+ϵNηNϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ. (3.17)

    Applying (3.1) on the two terms of the righthand side of (3.17) with indices sp>1 and (sp)=sp/(sp1) (note 1sp+1(sp)=1), we obtain

    ϵNηNϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ(ϵNηNϵ1η1Ξp(τ)ατ)1sp(ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)sp1sp (3.18)

    and

    ϵNηNϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ(ϵNηNϵ1η1Ωp(τ)ατ)1sp(ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)sp1sp. (3.19)

    Adding (3.18) and (3.19) and substituting into (3.17), we see that

    ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ[(ϵNηNϵ1η1Ξp(τ)ατ)1sp+(ϵNηNϵ1η1Ωp(τ)ατ)1sp]×(ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)sp1sp,

    thus,

    (ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)1sp(ϵNηNϵ1η1Ξp(τ)ατ)1sp+(ϵNηNϵ1η1Ωp(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, ϵiT,ϵi>ηi,i=1,2,,N,0α1, p<0,0<s<1 and Ξ,Ω: TNR+ are continuous functions, then

    (ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)1sp(ϵNηNϵ1η1Ξp(τ)ατ)1sp+(ϵNηNϵ1η1Ωp(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))1sΞ1s(τ)+Ω1s(τ). (3.21)

    Since

    (Ξ(τ)+Ω(τ))p=(Ξ(τ)+Ω(τ))1s(Ξ(τ)+Ω(τ))p1s,

    by using (3.21), we get

    (Ξ(τ)+Ω(τ))pΞ1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)+Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp1),

    then

    ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατϵNηNϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ+ϵNηNϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ. (3.22)

    Applying (3.13) on ϵNηNϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ, with

    ϕ(τ)=Ξ(τ),λ(τ)=(Ξ(τ)+Ω(τ))sp1,

    and the indices sp<0, (sp)=sp/(sp1)(note 1sp+1(sp)=1), we see that

    ϵNηNϵ1η1Ξ1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ(ϵNηNϵ1η1Ξp(τ)ατ)1sp(ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)sp1sp. (3.23)

    Again by applying (3.13) on ϵNηNϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ with

    ϕ(τ)=Ω(τ),λ(τ)=(Ξ(τ)+Ω(τ))sp1,

    and the indices sp<0, (sp)=sp/(sp1)(note 1sp+1(sp)=1), we have that

    ϵNηNϵ1η1Ω1s(τ)(Ξ(τ)+Ω(τ))1s(sp1)ατ(ϵNηNϵ1η1Ωp(τ)ατ)1sp×(ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)sp1sp. (3.24)

    Substituting (3.23) and (3.24) into (3.22), we see that

    ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ[(ϵNηNϵ1η1Ξp(τ)ατ)1sp+(ϵNηNϵ1η1Ωp(τ)ατ)1sp]×(ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)sp1sp,

    then

    (ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pατ)1sp(ϵNηNϵ1η1Ξp(τ)ατ)1sp+(ϵNηNϵ1η1Ωp(τ)ατ)1sp,

    which is (3.20).

    Remark 3.11. If T=R, ηi, ϵiT,ϵi>ηi,i=1,2,,N,p<0,0<s<1 and Ξ,Ω: RNR+ are continuous functions, then

    (ϵNηNϵ1η1(Ξ(τ)+Ω(τ))pdτ)1sp(ϵNηNϵ1η1Ξp(τ)dτ)1sp+(ϵNηNϵ1η1Ωp(τ)dτ)1sp.

    Remark 3.12. If T=N, N=1,η, ϵN, ϵ>η,0α1, p<0,0<s<1, and Ξ,Ω are positive sequences, then

    (ϵ1τ=η[α(Ξ(τ)+Ω(τ))p+(1α)(Ξ(τ+1)+Ω(τ+1))p])1sp(ϵ1τ=η[αΞp(τ)+(1α)Ξp(τ+1)])1sp+(ϵ1τ=ηαΩp(τ)+(1α)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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