Citation: Alessia E. Kogoj, Ermanno Lanconelli, Enrico Priola. Harnack inequality and Liouville-type theorems for Ornstein-Uhlenbeck and Kolmogorov operators[J]. Mathematics in Engineering, 2020, 2(4): 680-697. doi: 10.3934/mine.2020031
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In recent years, convexity theory has gained special attention by many researchers because of it engrossing properties and expedient characterizations. It has many applications in fields like biology, numerical analysis and statistics (see [1,2,3,4]). Mathematical inequalities are extensively studied with all type of convex functions (see[1,3,11,13,14,16]). One of the fundamental inequality is Hermite-Hadamard inequality. It has been discussed via different types of convexities and became the center of attention for many researchers. Recently, in 2016, Khan et al. have discussed generalizations of Hermite-Hadamard type for MT-convex functions [26]. In 2017, Khan et al. studied some new inequalities of Hermite-Hadamard types [27]. In 2019, Khurshid et al. have utilized conformable fractional integrals via preinvex functions [28]. In 2020, Khan et al. have discussed Hermite-Hadamard type inequalities via quantum calculus involving green function [29], Mohammed et al. have established a new version of Hermite-Hadamard inequality for Riemann-Liouville fractional integrals [30], Han et al. used fractional integral to generalize Hermite-Hadamard inequality for convex functions [31], Zhao et al. utilized harmonically convex functions to generalized fractional integral inequalities of Hermite-Hdamrd type [32], Awan et al. presented new inequalities of Hermite-Hdamard type for n-polynomial harmonically convex functions [33]. In 2022, Khan et al. introduced some new versions of Hermite-Hadamard integral inequalities in fuzzy fractional calculus for generalized pre-invex functions via fuzzy-interval-valued settings [34]. This reflects the importance of Hermite Hadamard type inequalities among current research.
In [9], s-convex function is given as,
Definition 1.1. A real valued function χ is called s-convex function on R, if
χ(ςρ+(1−ς)γ)≤ςsχ(ρ)+(1−ς)sχ(γ), |
for each ρ,γ∈R and ς∈(0,1) where s∈(0,1].
In [10], m-convexity is discussed as,
Definition 1.2. A real valued function χ defined on [0,b] is said to be a m-convex function for m∈[0,1], if
χ(ςρ+m(1−ς)γ)≤ςχ(ρ)+m(1−ς)χ(γ), |
holds for all ρ,γ∈[0,b] and ς∈[0,1].
(s,m)-convexity in [17] is discussed as,
Definition 1.3. A function χ:[0,b]⟶R, b>0 is said to be a (s,m)-convex function in the second sense where s,m∈(0,1]2, if
χ(ςρ+m(1−ς)γ)≤ςsχ(ρ)+m(1−ς)sχ(γ), |
holds provided that all ρ,γ∈[0,b] and ς∈[0,1].
Equivalent definition for (s,m)–convex functions:
Let ρ,α,γ∈[0,b], ρ<α<γ
χ(α)≤(γ−αγ−ρ)sχ(ρ)+m(α−ργ−ρ)sχ(γ). | (1.1) |
Hölder-İşcan Inequality [5]:
Let p>1, χ and ψ be real valued functions defined on [ρ,γ] and |χ|p,|ψ|q are integrable functions on interval [ρ,γ]
∫γρ|χ(ω)ψ(ω)|dω≤1γ−ρ(∫γρ(γ−ω)|χ(ω)|pdω)1p(∫γρ(γ−ω)|ψ(ω)|qdω)1q+1γ−ρ(∫γρ(ω−ρ)|χ(ω)|pdω)1p(∫γρ(ω−ρ)|ψ(ω)|qdω)1q, | (1.2) |
where 1p+1q=1.
Following lemma is useful to obtain our main results.
Lemma 1.4. [8] For n∈N, let χ:U⊆R⟶R be n-times differentiable mapping on U∘, where ρ,γ∈U∘, ρ<γ and χn∈L[ρ,γ], we have following identity
n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−γ∫ρχ(ω)dω=(−1)n+1n!γ∫ρωnχ(n)(ω)dω, | (1.3) |
where an empty set is understood to be nil.
In this paper, Hölder-İşcan inequality is used to modify inequalities involving functions having s-convex or s-concave derivatives at certain powers. The purpose of this paper is to establish some generalized inequalities for n-times differentiable (s,m)-convex functions. Applications of these inequalities to means are also discussed. Means are defined as,
Let 0<ρ<γ,
A(ρ,γ)=ρ+γ2, |
G(ρ,γ)=√ργ, |
Lp(ρ,γ)=(γp+1−ρp+1(p+1)(γ−ρ))1p, |
where p≠0,−1 and ρ≠γ.
Theorem 2.1. For any positive integer n, let χ:U⊆(0,∞)→R be n-times differentiable mapping on U∘, where ρ,γ∈U∘ with ρ<γ. If χ(n)∈L[ρ,γ] and |χ(n)|q for q>1 is (s,m)-convex on interval [ρ,γ] then
|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−μ∫ρχ(ω)dω|≤1n!(γ−ρ)1q([γLnpnp(ρ,γ)−Lnp+1np+1(ρ,γ)]1p[|χn(γ)|q(s+2)(s+1)+m|χn(ρ)|q(s+2)]1q+[Lnp+1np+1(ρ,γ)−ρLnpnp(ρ,γ)]1p[|χn(γ)|q(s+2)+m|χn(ρ)|q(s+1)(s+2)]1q), | (2.1) |
where 1p+1q=1.
Proof. Since |χn|q is (s,m)-convex by using inequality (1.1) for ρ<ω<γ, using Lemma 1.4 and Hölder-Işcan inequality (1.2),
|χn(ω)|q≤|χn(ω−ργ−ργ+mγ−ωγ−ρρ)|q≤(ω−ργ−ρ)s|χn(γ)|q+m(γ−ωγ−ρ)s|χn(ρ)|q,|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−γ∫ρχ(ω)dω|≤1n!γ∫ρωn|χ(n)(ω)|dω,≤1n!1γ−ρ{(γ∫ρ(γ−ω)ωnpdω)1p(γ∫ρ(γ−ω)|χ(n)(ω)|qdω)1q+(γ∫ρ(ω−ρ)ωnpdω)1p(γ∫ρ(ω−ρ)|χ(n)(ω)|qdω)1q},≤1n!1γ−ρ(γ∫ρ(γ−ω)ωnpdω)1p(γ∫ρ(γ−ω)[(ω−ργ−ρ)s|χn(γ)|q+m(γ−ωγ−ρ)s|χn(ρ)|q]dω)1q+1n!1γ−ρ(γ∫ρ(ω−ρ)ωnpdω)1p(γ∫ρ(ω−ρ)[(ω−ργ−ρ)s|χn(γ)|q+m(γ−ωγ−ρ)s|χn(ρ)|q]dω)1q, | (2.2) |
Let
I1=[γ∫ρ(γ−ω)ωnpdω]1p=[γ∫ρ(γωnp−ωnp+1)dω]1p=(γ−ρ)1p[γ(γnp+1−ρnp+1(γ−ρ)(np+1))−(γnp+2−ρnp+2(γ−ρ)(np+2))]1p=(γ−ρ)1p[γLnpnp(ρ,γ)−Lnp+1np+1(ρ,γ)]1p, |
I2=[γ∫ρ(ω−ρ)ωnpdt]1p=[γ∫ρ(ωnp+1−ρωnp)dω]1p=(γ−ρ)1p[(γnp+2−ρnp+2(γ−ρ)(np+2))−ρ(γnp+1−ρnp+1(γ−ρ)(np+1))]1p=(γ−ρ)1p[Lnp+1np+1(ρ,γ)−ρLnpnp(ρ,γ)]1p, |
I3=γ∫ρ(γ−ω)(ω−ρ)sdω=(γ−ω)(ω−ρ)s+1s+1|γρ+γ∫ρ(ω−ρ)s+1s+1dω=(γ−ρ)s+2(s+1)(s+2), |
I4=γ∫ρ(γ−ω)s+1dω=(γ−ρ)s+2s+2,I5=γ∫ρ(ω−ρ)s+1dω=(γ−ρ)s+2s+2,I6=γ∫ρ(ω−ρ)(γ−ω)sdω=(ω−ρ)(γ−ω)s+1(s+1)|γρ+γ∫ρ(γ−ω)s+1(s+1)dω=(γ−ρ)s+2(s+1)(s+2). |
Substituting integrals I1,I2,I3,I4,I5,I6 in inequality (2.2) we have,
|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−γ∫ρχ(ω)dω|≤1n!(γ−ρ)((γ−ρ)1p[γLnpnp(ρ,γ)−Lnp+1np+1(ρ,γ)]1p[(γ−ρ)2(|χn(γ)|q(s+2)(s+1)+m|χn(ρ)|q(s+2))]1q+(γ−ρ)1p[Lnp+1np+1(ρ,γ)−ρLnpnp(ρ,γ)]1p[(γ−ρ)2(|χn(γ)|q(s+2)+m|χn(ρ)|q(s+1)(s+2))]1q) |
=(γ−ρ)1p−1+2qn!([γLnpnp(ρ,γ)−Lnp+1np+1(ρ,γ)]1p[|χn(γ)|q(s+2)(s+1)+m|χn(ρ)|q(s+2)]1q+[Lnp+1np+1(ρ,γ)−ρLnpnp(ρ,γ)]1p[|χn(γ)|q(s+2)+m|χn(ρ)|q(s+1)(s+2)]1q) |
=1n!(γ−ρ)1q([γLnpnp(ρ,γ)−Lnp+1np+1(ρ,γ)]1p[|χn(γ)|q(s+2)(s+1)+m|χn(ρ)|q(s+2)]1q+[Lnp+1np+1(ρ,γ)−ρLnpnp(ρ,γ)]1p[|χn(γ)|q(s+2)+m|χn(ρ)|q(s+1)(s+2)]1q). |
which is required inequality (2.1).
For n=1 inequality (2.1) becomes,
|(χ(γ)γ−χ(ρ)ργ−ρ)−1γ−ργ∫ρχ(ω)dω|≤(γ−ρ)1q−1([γLpp(ρ,γ)−Lp+1p+1(ρ,γ)]1p[|χ′(γ)|q(s+1)(s+2)+m|χ′(ρ)|(s+2)q]1q+[Lp+1p+1(ρ,γ)−ρLpp(ρ,γ)]1p[m|χ′(ρ)|q(s+1)(s+2)+|χ′(γ)|(s+2)q]1q). | (2.3) |
Remark 2.2. For s=1 and m=1 our resulting inequality (2.1) becomes the inequality (2) of [5].
Theorem 2.3. For n∈N, let χ:U⊆(0,∞)→R be n-times differentiable mapping on U∘, where, ρ,γ∈U∘, ρ<γ, χ(n)∈L[ρ,γ] and |χ(n)|q for q>1, is (s,m)-convex on interval [ρ,γ] then following inequality holds
|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−γ∫ρχ(ω)dω|≤1s1qn!(12)1p(γ−ρ)2p−1((|χ(n)(γ)|q(γ−ρ)s−1[−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ)]+m|χ(n)(ρ)|q(γ−ρ)s−1[Lnq+2nq+2(ρ,γ)−2γLnq+1nq+1(ρ,γ)+γ2Lnqnq(ρ,γ)])1q+(|χ(n)(γ)|q(γ−ρ)s−1[Lnq+2nq+2(ρ,γ)−2ρLnq+1nq+1(ρ,γ)+ρ2Lnqnq(ρ,γ)]+m|χ(n)(ρ)|q(γ−ρ)s−1[−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ)])1q). | (2.4) |
Proof. Since |χ(n)|q for q>1 is (s,m)-convex on [ρ,γ], by using Lemma 1.4 and Hölder-İşcan inequality (1.2), since s∈(0,1], this fact can be used for ω,ρ,γ∈U⊆(0,∞),
(ω−ρ)s<(ω−ρ)s,(γ−ω)s<(γ−ω)s|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−γ∫ρχ(ω)dω|≤1n!γ∫ρ1.ωn|χ(n)(ω)|dω,≤1n!1(γ−ρ)([(γ∫ρ(γ−ω)dω)1p(γ∫ρ(γ−ω)ωnq|χ(n)(ω)|qdω)1q]+[(γ∫ρ(ω−ρ)dω)1p(γ∫ρ(ω−ρ)ωnq|χ(n)(ω)|qdω)1q]),≤1n!1(γ−ρ)(γ∫ρ(γ−ω)dω)1p(γ∫ρ(γ−ω)ωnq[(ω−ργ−ρ)s|χn(γ)|q+m(γ−ωγ−ρ)s|χn(ρ)|q]dt)1q+1n!1(γ−ρ)(γ∫ρ(ω−ρ)dt)1p(γ∫ρ(ω−ρ)ωnq[(ω−ργ−ρ)s|χn(γ)|q+m(γ−ωγ−ρ)s|χn(ρ)|q]dx)1q,≤1s1qn!1(γ−ρ)(γ∫ρ(γ−ω)dω)1p(γ∫ρ(γ−ω)ωnq[(ω−ρ)(γ−ρ)s|χn(γ)|q+m(γ−ω)(γ−ρ)s|χn(ρ)|q]dω)1q+1s1qn!1(γ−ρ)(γ∫ρ(ω−ρ)dω)1p(γ∫ρ(ω−ρ)ωnq[(ω−ρ)(γ−ρ)s|χn(γ)|q+m(γ−ω)(γ−ρ)s|χn(ρ)|q]dω)1q, I1=γ∫ρ(γ−ω)dω=(γ−ρ)22 I2=γ∫ρ(γ−ω)(ω−ρ)ωnqdω=γωnq+1nq+1−ργωnq+1nq+1−ωnq+3nq+3+ρωnq+2nq+2|γρ =−(γnq+3−ρnq+3nq+3)+ρ(γnq+2−ρnq+2nq+2)+γ(γnq+2−ρnq+2nq+2)−ργ(γnq+1−ρnq+1nq+1) =(γ−ρ)[−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ)], I3=γ∫ρ(γ−ω)2ωnqdω=γ2ωnq+1nq+1+ωnq+3nq+3−2γωnq+2nq+2|γρ =(γnq+3−ρnq+3nq+3)−2γ(γnq+2−ρnq+2nq+2)+γ2(γnq+1−ρnq+1nq+1) =(γ−ρ)[Lnq+2nq+2(ρ,γ)−2γLnq+1nq+1(ρ,γ)+γ2Lnqnq(ρ,γ)], I4=γ∫ρ(ω−ρ)2ωnqdω=ωnq+3nq+3+ρ2ωnq+1nq+1−2ρωnq+2nq+2|γρ =(γnq+3−ρnq+3nq+3)+ρ2(γnq+1−ρnq+1nq+1)−2ρ(γnq+2−ρnq+2nq+2) =(γ−ρ)[Lnq+2nq+2(ρ,γ)+ρ2Lnqnq(ρ,γ)−2ρLnq+1nq+1(ρ,γ)]. | (2.5) |
Substituting integrals I1,I2,I3,I4,I5,I6 in inequality (2.5) we have,
|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−γ∫ρχ(ω)dω|≤1s1qn!(12)1p(γ−ρ)2p−1×((|χ(n)(γ)|q(γ−ρ)s[(γ−ρ)(−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ))]+m|χ(n)(ρ)|q(γ−ρ)s[(γ−ρ)(Lnq+2nq+2(ρ,γ)−2γLnq+1nq+1(ρ,γ)+γ2Lnqnq(ρ,γ))])1q+(|χ(n)(γ)|q(γ−ρ)s[(γ−ρ)(Lnq+2nq+2(ρ,γ)−2ρLnq+1nq+1(ρ,γ)+ρ2Lnqnq(ρ,γ))]+m|χ(n)(ρ)|q(γ−ρ)s[(γ−ρ)(−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ))])1q), |
=1s1qn!(12)1p(γ−ρ)2p−1×((|χ(n)(γ)|q(γ−ρ)s−1[−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ)]+m|χ(n)(ρ)|q(γ−ρ)s−1[Lnq+2nq+2(ρ,γ)−2γLnq+1nq+1(ρ,γ)+γ2Lnqnq(ρ,γ)])1q+(|χ(n)(γ)|q(γ−ρ)s−1[Lnq+2nq+2(ρ,γ)−2ρLnq+1nq+1(ρ,γ)+ρ2Lnqnq(ρ,γ)]+m|χ(n)(ρ)|q(γ−ρ)s−1[−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ)])1q). |
For n=1, Theorem2.3 reduced to the inequality
|γχ(γ)−ρχ(ρ)(γ−ρ)−1(γ−ρ)γ∫ρχ(ω)dω|≤1s1q(12)1p(γ−ρ)2p−2((|χ(1)(γ)|q(γ−ρ)s−1[−Lq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)]+m|χ(1)(ρ)|(γ−ρ)s−1q[Lq+2q+2(ρ,γ)−2γLq+1q+1(ρ,γ)+γ2Lqq(ρ,γ)])1q+(|χ(1)(γ)|(γ−ρ)s−1q[Lq+2q+2(ρ,γ)−2ρLq+1q+1(ρ,γ)+ρ2Lqq(ρ,γ)]+m|χ(1)(ρ)|(γ−ρ)s−1q[−Lq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)])1q). | (2.6) |
Remark 2.4. For s=1 and m=1 our resulting inequality (2.4) becomes the inequality (6) of [5].
Theorem 2.5. If function χ:[0,b]⟶R, b>0 is a (s, m)-convex function in the second sense where (s,m)∈(0,1]2, holds provided that all ρ,γ∈[0,b] and ς∈[0,1], then
2sχ(ρ+mγ2)≤[1mγ−ρmγ∫ρχ(ω)dω+m2mγ−ργ∫ρmχ(l)dl]≤χ(ρ)+mχ(γ)s+1+χ(γ)+mχ(ρm2)s+1. | (2.7) |
Proof. A function χ:[0,b]⟶R, b>0 is said to be a (s,m)-convex function in the second sense where s,m∈(0,1]2, if
χ(ςρ+m(1−ς)γ)≤ςsχ(ρ)+m(1−ς)sχ(γ), |
holds provided that all ρ,γ∈[0,b] and ς∈[0,1].
Integrating w.r.t ς on [0,1],
1∫0χ(ςρ+m(1−ς)γ)dς≤1∫0ςsχ(ρ)dς+1∫0m(1−ς)sχ(γ)dς,=ςs+1s+1|10χ(ρ)−mχ(γ)(1−ς)s+1s+1|10=χ(ρ)+mχ(γ)s+1. 1∫0χ(ςρ+m(1−ς)γ)dς≤χ(ρ)+mχ(γ)s+1. | (2.8) |
and
χ(ςγ+m(1−ς)ρm2)≤ςsχ(γ)+m(1−ς)sχ(ρm2),1∫0χ(ςγ+m(1−ς)ρm2)dς≤χ(γ)+mχ(ρm2)s+1. | (2.9) |
As χ is (s,m)-convex,
χ(ρ+mγ2)=χ(ςρ+(1−ς)mγ2+m.(1−ς)ρm+ςγ2)≤(12)sχ(ςρ+(1−ς)γm)+m(12)sχ(ςγ+(1−ς)ρm), |
Integrating w.r.t ς over [0,1] and by using (2.8) and (2.9) we get,
2sχ(ρ+mγ2)≤1∫0(χ(ςρ+(1−ς)γm)dς+m1∫0χ(ςγ+(1−ς)ρm)dς≤χ(ρ)+mχ(γ)s+1+χ(γ)+mχ(ρm2)s+1. | (2.10) |
Substituting in first integral,
ςρ+(1−ς)γm=ω,
1∫0χ(ςρ+(1−ς)mγ)dς=1γm−ργm∫ρχ(ω)dω. | (2.11) |
Substituting in the second integral,
ςγ+(1−ς)ρm=l,
1∫0χ(ςγ+(1−ς)ρm)dς=mγm−ργ∫ρmχ(l)dl, | (2.12) |
Using (2.11) and (2.12) in (2.10) required inequality (2.7) obtained.
Remark 2.6. For s,m=1 inequality (2.7) becomes classical Hadamard inequality for convex functions.
Theorem 2.7. For n∈N, let χ:U⊆(0,∞)→R be n-times differentiable mapping on U∘, where, ρ,γ∈U∘, ρ<γ and χ(n)∈L[ρ,γ] and |χ(n)|q for q>1 is (s, m)-concave on interval [ρ,mγ], then
|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−mγ∫ρχ(ω)dω|≤2sq(mγ−ρ)1q|χ(n)(ρ+mγ2)|n!((γLnpnp(ρ,mγ)−Lnp+1np+1(ρ,mγ))1p+(Lnp+1np+1(ρ,mγ)−ρLnpnp(ρ,mγ))1p). | (2.13) |
Proof. |χ(n)|q for q>1 is (s,m)-concave then by using Theorem 2.5 we have,
|χ(n)(ρ)|q+m|χ(n)(γ)|qs+1+|χ(n)(γ)|q+m|χ(n)(ρm2)|qs+1−m2(mγ−ρ)γ∫ρm|χ(n)(l)|qdl≤1(mγ−ρ)mγ∫ρ|χ(n)(ω)|qdω≤2s|χ(n)(ρ+mγ2)|q, |
mγ∫ρ|χ(n)(ω)|qdω≤2s(mγ−ρ)|χ(n)(ρ+mγ2)|q, |
1(mγ−ρ)γm∫ρ(γ−ω)|χ(n)(ω)|qdω≤γm∫ρ|χ(n)(ω)|qdω≤2s(mγ−ρ)|χ(n)(ρ+mγ2)|q, |
1(mγ−ρ)γm∫ρ(γ−ω)|χ(n)(ω)|qdω≤γm∫ρ|χ(n)(ω)|qdω≤2s(mγ−ρ)|χ(n)(ρ+mγ2)|q. |
Using Lemma 1.4 and Hölder-Îşcan inequality (1.2),
|n−1∑ν=0(−1)ν(χ(ν)(γ)γν+1−χ(ν)(ρ)ρν+1(ν+1)!)−γm∫ρχ(ω)dω|≤1n!γm∫ρωn|χ(n)(ω)|dω, ≤1n!1γ−ρ{(γm∫ρ(γ−ω)ωnpdω)1p(mγ∫ρ(γ−ω)|χn(ω)|qdω)1q+(γm∫ρ(ω−ρ)ωnpdω)1p(mγ∫ρ(ω−ρ)|χn(ω)|qdω)1q}, ≤1n!1γ−ρ((γm∫ρ(γ−ω)ωnpdω)1p(2s(mγ−ρ)2|χ(n)(ρ+mγ2)|q)1q+(γm∫ρ(ω−ρ)ωnpdω)1p(2s(mγ−ρ)2|χ(n)(ρ+mγ2)|q)1q), I1=(γm∫ρ(γ−ω)ωnpdω)1p=(γωnp+1np+1|γmρ−ωnp+2np+2|γmρ)1p =(mγ−ρ)1p(γLnpnp(ρ,mγ)−Lnp+1np+1(ρ,mγ))1p, I2=(γm∫ρ(ω−ρ)ωnpdω)1p=(ωnp+2np+2|γmρ−ρωnp+1np+1|γmρ)1p =(mγ−ρ)1p(Lnp+1np+1(ρ,mγ)−ρLnpnp(ρ,mγ))1p. | (2.14) |
Substituting integrals I1,I2 in inequality (2.14) required inequality (2.13) is obtained.
For n=1 inequality (2.13) becomes,
|χ(γ)γ−ρχ(ρ)(γ−ρ)−1(γ−ρ)γm∫ρχ(ω)dω|≤2sq(mγ−ρ)1q|χ(1)(ρ+γ2)|1!((γLpp(ρ,mγ)−Lp+1p+1(ρ,mγ))1p+(Lp+1p+1(ρ,mγ)−ρLpp(ρ,mγ))1p). | (2.15) |
Remark 2.8. For s=1 and m=1 our resulting inequality becomes the inequality obtained in Theorem 4 of [5].
Proposition 2.9. Let ρ,γ∈(0,∞), where ρ<γ, q>1, n,i∈N with i≥n,
|Lii(ρ,γ)[(i+1)∑n−1ν=0(−1)νP(i,ν)(ν+1)!−1]|≤1n!(γ−ρ)1q−1×([γLnpnp(ρ,γ)−Lnp+1np+1(ρ,γ)]1p(γ(i−n)q(s+1)(s+2)+mρ(i−n)q(s+2))1q+[Lnp+1np+1(ρ,γ)−ρLnpnp(ρ,γ)]1p(mρ(i−n)q(s+1)(s+2)+γ(i−n)q(s+2))1q), | (2.16) |
where
P(i,n)={i(i−1)...(i−n+1),i>nn!,i=n1,n=0}. |
Proof. Let
χ(ω)=ωi,|χ(n)(ω)|q=|P(i,n)ωi−n|q |
Let
g(ς)=|P(i,n)(ςρ+m(1−ς)γ|(i−n)q−|P(i,n)ςsρ|(i−n)q−|mP(i,n)(1−ς)sγ|(i−n)q, |
g″(ς)=P(i,n)((i−n)q)((i−n)q−1)(ςρ+m(1−ς)γ)(i−n)q−2(ρ−mγ)2−s(s−1)ςs−2P(i,n)ρ(i−n)q−ms(s−1)(1−ς)s−2P(i,n)γ(i−n)q, |
g″(ς)≥0 means g is convex and g(1)=g(0)=0, which omplies g≤0, hence
|P(i,n)(ςρ+m(1−ς)γ)|(i−n)q≤|P(i,n)ςsρ|(i−n)q+|mP(i,n)(1−ς)sγ)|(i−n)q. |
By using Theorem 2.1 for |χn(ω)|q which is (s,m)–convex for s,m∈(0,1]2 inequality (2.16) obtained.
Remark 2.10. For s,m=1 inequality (2.16) becomes inequality (3) of [5].
Example 2.11. Taking i=2, n=1, p=q=2 in Proposition 2.9, the following is valid:
2A(ρ2,γ2)+G2(ρ,γ)≤(32√6)([A(3ρ2,γ2)+G2(ρ,γ)]12(γ2(s+1)(s+2)+mρ2(s+2))12+[A(ρ2,3γ2)+G2(ρ,γ)]12(mρ2(s+1)(s+2)+γ2(s+2))12), |
where A and G are classical arithmetic and geometric means, respectively.
Proposition 2.12. Let ρ,γ∈(0,∞), with, ρ<γ, q>1 and n∈N,
1≤(γ−ρ)1q−1([γLpp(ρ,γ)−Lp+1p+1(ρ,γ)]1p[(γ−q(s+1)(s+2)+mρ−q(s+2))]1q+[Lp+1p+1(ρ,γ)−ρLpp(ρ,γ)]1p[(mρ−q(s+1)(s+2)+γ−q(s+2))]1q), | (2.17) |
where L is classical logarithmic mean.
Proof.
χ(ω)=lnω,|χ(1)(ω)|q=|ω−1|q |
Let
g(ς)=|(ςρ+m(1−ς)γ|−q−|ςsρ|−q−|m(1−ς)sγ|−q |
g″(ς)=(−q)(−q−1)(ςρ+m(1−ς)γ)−q−2(ρ−mγ)2−s(s−1)ςs−2ρ−q−ms(s−1)(1−ς)s−2γ−q, |
g″(ς)≥0 means g is convex and g(1)=g(0)=0 which implies g≤0 as
|(ςρ+m(1−ς)γ|−q≤|ςsρ|−q+|m(1−ς)sγ|−q. |
So |χ(1)(ω)|q is (s,m)-convex. Then by using inequality (2.3) required inequality (2.17) obtained.
Remark 2.13. For s,m=1 inequality (2.17) becomes (4) of [5].
Example 2.14. For n=1 and p=q=2, Proposition 2.12 gives:
1≤1√6([A(3ρ2,γ2)+G2(ρ,γ)]1p[(γ−2(s+1)(s+2)+mρ−2(s+2))]12+[A(ρ2,3γ2)+G2(ρ,γ)]1p[(mρ−2(s+1)(s+2)+γ−2(s+2))]12). |
Proposition 2.15. Let ρ,γ∈(0,∞), ρ<γ, q>1, i∈(−∞,0]∪[1,∞)∖{−2q,−q}
then
Liq+1iq+1(ρ,γ)≤(γ−ρ)1q−1([γLpp(ρ,γ)−Lp+1p+1(ρ,γ)]1p[(γi(s+1)(s+2)+mρi(s+2))]1q+[Lp+1p+1(ρ,γ)−ρLpp(ρ,γ)]1p[(mρi(s+1)(s+2)+γi(s+2))]1q). | (2.18) |
Proof.
χ(t)=qi+qωiq+1,|χ′(ω)|q=ωi |
Let
g(ς)=|(ςρ+m(1−ς)γ|i−|ςsρ|i−|m(1−ς)sγ|i, |
g″(ς)=(i)(i−1)(ςρ+m(1−ς)γ)i−2(ρ−mγ)2−s(s−1)ςs−2ρi−ms(s−1)(1−ς)s−2γi, |
g″(ς)≥0 and g(1)=g(0) so g≤0 and |χ′(ω)|q is (s,m)-convex, by using inequality (2.3) we have (2.18).
Remark 2.16. For s,m=1 inequality (2.18) becomes (5) of [5].
Example 2.17. For i=2 and p=q=2 Proposition 2.15 reduced to
2A(ρ2,γ2)+G2(ρ,γ)≤(3√6)([A(3ρ2,γ2)+G2(ρ,γ)]12[(γ2(s+1)(s+2)+mρ2(s+2))]12+[A(ρ2,3γ2)+G2(ρ,γ)]12[(mρ2(s+1)(s+2)+γ2(s+2))]12). | (2.19) |
Proposition 2.18. Let ρ,γ∈(0,∞) with ρ<γ, q>1 and n∈N then we have
×|Lii(ρ,γ)[n−1∑ν=0(−1)νP(i,ν)(ν+1)!−1]|≤P(i,n)s1qn!(12)1p(γ−ρ)2p−1(γ(i−n)q(γ−ρ)s−1[−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ)]+mρ(i−n)q(γ−ρ)s−1[Lnq+2nq+2(ρ,γ)−2γLnq+1nq+1(κ,μ)+μ2Lnqnq(ρ,γ)])1q+P(i,n)s1qn!(12)1p(γ−ρ)2p−1(γ(i−n)q(γ−ρ)s−1[Lnq+2nq+2(ρ,γ)−2ρLnq+1nq+1(ρ,γ)+ρ2Lnqnq(ρ,γ)]+mρ(i−n)q(γ−ρ)s−1[−Lnq+2nq+2(ρ,γ)+(ρ+γ)Lnq+1nq+1(ρ,γ)−ργLnqnq(ρ,γ)])1q. | (2.20) |
Proof. Let,
χ(ω)=ωi,|χ(n)(ω)|q=[P(i,n)ωi−n]q |
As |χn(ω)|q is (s,m)-convex on (0,∞), therefore by using Theorem 2.3 required inequality (2.20) is obtained.
Remark 2.19. For s,m=1 inequality (2.20) becomes inequality obtained in Proposition 4 of [5].
Proposition 2.20. Let ρ,γ∈(0,∞) with ρ<γ q>1 and n∈N then we have,
1≤(γ−ρ)2p−2s1q.21p((γ−q(γ−ρ)s−1[−Lq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)]+mρ−q(γ−ρ)s−1[Lq+2q+2(ρ,γ)−2γLq+1q+1(ρ,γ)+γ2Lqq(ρ,γ)])1q+(γ−q(γ−ρ)s−1[Lq+2q+2(ρ,γ)−2ρLq+1q+1(ρ,γ)+ρ2Lqq(ρ,γ)]+mρ−q(γ−ρ)s−1[−Lq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)])1q), | (2.21) |
Proof.
χ(ω)=lnω,|χ(1)(ω)|q=[ω−1]q |
As |χ(1)(ω)|q is (s,m)–convex, therefore by using inequality (2.6) required (2.21) obtained.
Remark 2.21. For s,m=1 inequality (2.21) becomes inequality obtained in Proposition 5 of [5].
Proposition 2.22. Let ρ,γ∈(0,∞) with ρ<γ q>1 and i∈(−∞,0]∖{−2q,q}, then
Liq+1iq+1(ρ,γ)≤(γ−ρ)2p−2s1q.21p((γi(γ−ρ)s−1[−Lnq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)]+mρi(γ−ρ)s−1[Lq+2q+2(ρ,γ)−2γLq+1q+1(ρ,γ)+γ2Lqq(ρ,γ)])1q+(γi(γ−ρ)s−1[Lq+2q+2(ρ,γ)−2ρLq+1q+1(ρ,γ)+ρ2Lqq(ρ,γ)]+mρm(γ−ρ)s−1[−Lq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)])1q). | (2.22) |
Proof.
χ(ω)=qi+qωiq+1|χ′(ω)|q=ωi |
|χ′(w)|q is (s,m)-convex by using inequality (2.6) required (2.22) obtained.
For i=1 inequality (2.22) becomes,
L1q+11q+1(ρ,γ)≤(γ−ρ)2p−2s1q.21p((γ1(γ−ρ)s−1[−Lq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)]+mρ1(γ−ρ)s−1[Lq+2q+2(ρ,γ)−2γLq+1q+1(ρ,γ)+γ2Lqq(ρ,γ)])1q+(γ1(γ−ρ)s−1[Lq+2q+2(ρ,γ)−2ρLq+1q+1(ρ,γ)+ρ2Lqq(ρ,γ)]+mρ1(γ−ρ)s−1[−Lq+2q+2(ρ,γ)+(ρ+γ)Lq+1q+1(ρ,γ)−ργLqq(ρ,γ)])1q). | (2.23) |
Remark 2.23. For s,m=1 inequality (2.22) becomes inequality obtained in Proposition 6 of [5].
Proposition 2.24. Let ρ,γ∈(0,∞) with ρ<γ, q>1 and i∈[0,1] we have,
Liq+1iq+1(ρ,γ)≤2sq(mγ−ρ)1q1!Aiq(ρ,γ)((γLpp(ρ,mγ)−Lp+1p+1(ρ,mγ))1p+(Lp+1p+1(ρ,mγ)−ρLpp(ρ,mγ))1p). | (2.24) |
Proof.
χ(ω)=qi+qωiq+1,|χ′(ω)|q=ωi. |
As |χ′(ω)|q is (s,m)-concave by using inequality (2.15) we obtain required inequality (2.24).
Remark 2.25. For s,m=1 inequality (2.24) becomes the inequality obtained in Proposition 9 of [5].
In this paper, Hölder-Isçan inequality is utilized to prove Hermite-Hadamard type inequalities for n-times differentiable (s,m)-convex functions. The method is adequate and provide many generalizations of existing results as shown in remarks. Moreover, many other inequalities can be generalized for other types of convex functions.
This research received funding support from the NSRF via the Program Management Unit for Human Resources & Institutional Development, Research and Innovation, (grant number B05F650018)
The authors declare no conflict of interest.
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