# New General Integral Inequalities for Lipschitzian Functions via Hadamard Fractional Integrals

Authors: İmdat İşcan
Journal: International Journal of Analysis
 (2014)

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## Abstract

The author obtains new estimates on generalization
of Hadamard, Ostrowski, and Simpson type inequalities for Lipschitzian functions via Hadamard fractional integrals. Some applications to special means of positive real numbers are also given.


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## Body

## 1. Introduction

Let real functionf be defined on some nonempty interval I of real line ℝ. The function f is said to be convex on I if inequality
(1)f(tx+(1-t)y)≤tf(x)+(1-t)f(y)
holds for all x,y∈I and t∈[0,1].Following are inequalities which are well known in the literature as Hermite-Hadamard inequality, Ostrowski inequality, and Simpson inequality, respectively.Theorem 1.
Letf:I⊆ℝ→ℝ be a convex function defined on the interval I of real numbers and a,b∈I with a<b. The following double inequality holds:
(2)f(a+b2)≤1b-a∫abf(x)dx≤f(a)+f(b)2.Theorem 2.
Letf:I⊆ℝ→ℝ be a mapping differentiable in I∘, the interior of I, and let a,b∈I∘ with a<b. If |f′(x)|≤M, x∈[a,b]; then the following inequality holds:
(3)|f(x)-1b-a∫abf(t)dt|≤Mb-a[(x-a)2+(b-x)22]
for all x∈[a,b].Theorem 3.
Letf:[a,b]→ℝ be four times continuously differentiable mapping on (a,b) and ∥f(4)∥∞=sup⁡x∈(a,b)|f(4)(x)|<∞. Then the following inequality holds:
(4)|13[f(a)+f(b)2+2f(a+b2)]-1b-a∫abf(x)dx|≤12880∥f(4)∥∞(b-a)4.In recent years, many authors have studied errors estimations for Hermite-Hadamard, Ostrowski, and Simpson inequalities; for refinements, counterparts, and generalization, see [1–9] and references therein.The following definitions are well known in the literature.Definition 4.
A functionf:I⊆ℝ→ℝ is called an M-Lipschitzian function on the interval I of real numbers with M≥0 if
(5)|f(x)-f(y)|≤M|x-y|
for all x,y∈I.For some recent results connected with Hermite-Hadamard type integral inequalities for Lipschitzian functions, see [10–13].Definition 5 (see [14, 15]).
A functionf:I⊆(0,∞)→ℝ is said to be GA-convex (geometric arithmetically convex) if
(6)f(xty1-t)≤tf(x)+(1-t)f(y)
for all x,y∈I and t∈[0,1].We will now give definitions of the right-sided and left-sided Hadamard fractional integrals which are used throughout this paper.Definition 6.
Letf∈L[a,b]. The right-sided and left-sided Hadamard fractional integrals Ja+αf and Jb-αf of oder α>0 with b>a≥0 are defined by
(7)Ja+αf(x)=1Γ(α)∫ax(ln⁡⁡xt)α-1f(t)dtt,a<x<b,Jb-αf(x)=1Γ(α)∫xb(ln⁡⁡tx)α-1f(t)dtt,a<x<b,
respectively, where Γ(α) is Gamma function defined by Γ(α)=∫0∞e-ttα-1dt (see [16]).In [17], Iscan established Hermite-Hadamard’s inequalities for GA-convex functions in Hadamard fractional integral forms as follows.Theorem 7.
Letf:I⊆(0,∞)→ℝ be a function such that f∈L[a,b], where a,b∈I with a<b. If f is a GA-convex function on [a,b], then the following inequalities for fractional integrals hold:
(8)f(ab)≤Γ(α+1)2(ln⁡⁡(b/a))α{Ja+αf(b)+Jb-αf(a)}≤f(a)+f(b)2
with α>0.In the inequality (8), if we take α=1, then we have the following inequality:
(9)f(ab)≤1ln⁡⁡b-ln⁡⁡a∫abf(t)tdt≤f(a)+f(b)2.Morever, in [17], Iscan obtained a generalization of Hadamard, Ostrowski, and Simpson type inequalities for quasi-geometrically convex functions via Hadamard fractional integrals as related to the inequality (8).In this paper, the author obtains new general inequalities for Lipschitzian functions via Hadamard fractional integrals as related to the inequality (8).


## 2. Main Results

Letf:I⊆(0,∞)→ℝ be a M-Lipschitzian function on I; throughout this section, we will take
(10)If(x,λ,α,a,b)=(1-λ)[ln⁡αxa+ln⁡αbx]f(x)+λ[f(a)ln⁡αxa+f(b)ln⁡αbx]-Γ(α+1)[Jx-αf(a)+Jx+αf(b)],Sf(x,y,λ,α,a,b)=λαf(x)+(1-λ)αf(y)-Γ(α+1)ln⁡α(b/a)[JC-αf(a)+JC+αf(b)],
where a,b∈I with a<b, x,y∈[a,b], λ∈[0,1], C=a1-λbλ, α>0 and Γ is Euler Gamma function.Theorem 8.
Letf:I⊆(0,∞)→ℝ be a M-Lipschitzian function on I and a,b∈I with a<b. Then for all x∈[a,b], λ∈[0,1], and α>0, we have the following inequality for Hadamard fractional integrals:
(11)|If(x,λ,α,a,b)|≤M{(ln⁡⁡bx)α+α(1-2λ)∫xb(ln⁡⁡bt)α-1dt}[(1-λ)x-λa](ln⁡⁡xa)α+α(2λ-1)×∫ax(ln⁡ta)α-1dt+[λb-(1-λ)x]×(ln⁡⁡bx)α+α(1-2λ)∫xb(ln⁡⁡bt)α-1dt}.Proof.
Using the hypothesis off, we have the following inequality:
(12)|If(x,λ,α,a,b)|=|-α[∫ax(ln⁡⁡ta)α-1f(t)dtt+xb(ln⁡⁡bt)α-1f(t)dtt]|(1-λ)[ln⁡αxa+ln⁡αbx]f(x)+λ[f(a)ln⁡αxa+f(b)ln⁡αbx]-α[∫ax(ln⁡⁡ta)α-1f(t)dtt+∫xb(ln⁡⁡bt)α-1f(t)dtt]|≤(1-λ)|f(x)ln⁡αxa-α∫ax(ln⁡⁡ta)α-1f(t)dtt+f(x)ln⁡αbx-α∫xb(ln⁡⁡bt)α-1f(t)dtt|+λ|f(a)ln⁡αxa-α∫ax(ln⁡⁡ta)α-1f(t)dtt+f(b)ln⁡αbx-α∫xb(ln⁡⁡bt)α-1f(t)dtt|≤α(1-λ)[+∫xb(ln⁡⁡bt)α-1|f(x)-f(t)|dtt]∫ax(ln⁡⁡ta)α-1|f(x)-f(t)|dtt+∫xb(ln⁡⁡bt)α-1|f(x)-f(t)|dtt]+αλ[∫xb(ln⁡⁡bt)α-1|f(b)-f(t)|dtt]∫ax(ln⁡⁡ta)α-1|f(a)-f(t)|dtt+∫xb(ln⁡⁡bt)α-1|f(b)-f(t)|dtt]≤α(1-λ)M[∫ax(ln⁡⁡ta)α-1(x-t)dtt+∫xb(ln⁡⁡bt)α-1(t-x)dtt]+αλM[∫xb(ln⁡⁡bt)α-1(b-t)dtt]∫ax(ln⁡⁡ta)α-1(t-a)dtt+∫xb(ln⁡⁡bt)α-1(b-t)dtt]≤M{+α(1-2λ)xb(ln⁡bt)α-1dt}[(1-λ)x-λa](ln⁡⁡xa)α+α(2λ-1)∫ax(ln⁡⁡ta)α-1dt+[λb-(1-λ)x](ln⁡⁡bx)α+α(1-2λ)∫xb(ln⁡bt)α-1dt}.Corollary 9.
In Theorem8, If we take λ=0, then we get
(13)|(ln⁡⁡(x/a))α+(ln⁡⁡(b/x))α(ln⁡⁡(b/a))αf(x)-Γ(α+1)(ln⁡⁡(b/a))α×[Jx-αf(a)+Jx+αf(b)]|(ln⁡⁡(x/a))α+(ln⁡⁡(b/x))α(ln⁡⁡(b/a))αf(x)-Γ(α+1)(ln⁡⁡(b/a))α|≤M(ln⁡⁡(b/a))α×{x[(ln⁡⁡xa)α-(ln⁡⁡bx)α]+α[∫xb(ln⁡⁡bt)α-1dt-∫ax(ln⁡⁡ta)α-1dt]}.
In this inequality,(i)
if we takeα=1, then
(14)|f(x)-1ln⁡⁡b-ln⁡⁡a∫abf(t)tdt|≤Mln⁡⁡b-ln⁡⁡a{xln⁡⁡x2ab+(a+b-2x)}.(ii)
If we takex=ab, then
(15)|f(ab)-2α-1Γ(α+1)(ln⁡⁡(b/a))α[Jab-αf(a)+Jab+αf(b)]|≤2α-1Mα(ln⁡⁡(b/a))α×{∫abb(ln⁡⁡bt)α-1dt-∫aab(ln⁡⁡ta)α-1dt}.(iii)
If we takex=ab and α=1, then
(16)|f(ab)-1ln⁡⁡b-ln⁡⁡a∫abf(t)tdt|≤Mln⁡⁡b-ln⁡⁡a(a+b-2ab).Corollary 10.
In Theorem8, if we take λ=1, then we get
(17)|[f(a)ln⁡α(x/a)+f(b)ln⁡α(b/x)(ln⁡⁡(b/a))α]-2α-1Γ(α+1)(ln⁡⁡(b/a))α[Jx-αf(a)+Jx+αf(b)]|[f(a)ln⁡α(x/a)+f(b)ln⁡α(b/x)(ln⁡⁡(b/a))α]-2α-1Γ(α+1)(ln⁡⁡(b/a))α|≤M(ln⁡⁡(b/a))α×{α[xb(ln⁡bt)α-1dt-ax(ln⁡ta)α-1dt]}bln⁡α⁡xa-aln⁡α⁡bx-α[∫xb(ln⁡⁡bt)α-1dt-∫ax(ln⁡⁡ta)α-1dt]}.
In this inequality, if we take x=ab, then
(18)|f(a)+f(b)2-2α-1Γ(α+1)(ln⁡⁡(b/a))α[Jab-αf(a)+Jab+αf(b)]|≤2α-1M(ln⁡⁡(b/a))α×{-α[∫abb(ln⁡⁡bt)α-1dt-∫aab(ln⁡⁡ta)α-1dt]}b-a2α(ln⁡⁡ba)α-α[∫abb(ln⁡⁡bt)α-1dt-∫aab(ln⁡⁡ta)α-1dt]}.
Specially if we takeα=1 in this inequality, then we have
(19)|f(a)+f(b)2-1ln⁡⁡b-ln⁡⁡a∫abf(t)tdt|≤Mln⁡⁡b-ln⁡⁡a{b-a2(ln⁡⁡ba)-(a+b-2ab)}.Corollary 11.
In Theorem8,(1)
if we takex=ab and λ=1/3, then
(20)|-2α-1Γ(α+1)(ln⁡(b/a))α[Jab-αf(a)+Jab+αf(b)]|13[f(a)+f(b)2+2f(ab)]-2α-1Γ(α+1)(ln⁡(b/a))α[Jab-αf(a)+Jab+αf(b)]|≤2α-1M3(ln⁡(b/a))α×{-α[abb(ln⁡bt)α-1dt-aab(ln⁡ta)α-1dt]}b-a2α(ln⁡ba)α-α[∫abb(ln⁡bt)α-1dt-∫aab(ln⁡ta)α-1dt]}.Specially, if we takeα=1 in this inequality, then we have
(21)|13[f(a)+f(b)2+2f(ab)]-1ln⁡⁡b-ln⁡⁡a∫abf(t)tdt|≤M3(ln⁡⁡b-ln⁡⁡a){b-a2(ln⁡ba)-(a+b-2ab)}.(2)
If we takex=ab and λ=1/2, then
(22)|12[f(a)+f(b)2+2f(ab)]-2α-1Γ(α+1)(ln⁡(b/a))α[Jab-αf(a)+Jab+αf(b)]|≤M(b-a)4.Specially, if we takeα=1 in this inequality, then we have
(23)|12[f(a)+f(b)2+2f(ab)]-1ln⁡b-ln⁡a∫abf(t)tdt|≤M(b-a)4.Corollary 12.
In Theorem8, If we take α=1, then
(24)|(1-λ)f(x)+λ[f(a)ln⁡⁡(x/a)+f(b)ln⁡⁡(b/x)ln⁡⁡b-ln⁡⁡a]-1ln⁡b-ln⁡a∫abf(t)tdt|(1-λ)f(x)+λ[f(a)ln⁡⁡(x/a)+f(b)ln⁡⁡(b/x)ln⁡⁡b-ln⁡⁡a]|≤Mln⁡⁡b-ln⁡⁡a{[(1-λ)x-λa](ln⁡⁡xa)+[λb-(1-λ)x](ln⁡⁡bx)+(1-2λ)(a+b-2x){[(1-λ)x-λa](ln⁡⁡xa)}.
Specially, if we take x=ab in this inequality, then we have
(25)|-1ln⁡b-ln⁡aabf(t)tdt|(1-λ)f(ab)+λ(f(a)+f(b)2)-1ln⁡⁡b-ln⁡⁡a∫abf(t)tdt|≤Mln⁡⁡b-ln⁡⁡a×[λ(b-a)2ln⁡⁡ba+(1-2λ)(a+b-2ab)].
We note that if we take λ=0, λ=1, λ=1/3, and λ=1/2 in inequality (25) we obtain inequalities (16), (19), (21), and (23), respectively.Letf:I⊆(0,∞)→ℝ be an M-Lipschitzian function. In the next theorem, let λ∈[0,1], C=a1-λbλ, x,y∈[a,b] and define Cα,λ, α>0 as follows.(1)
Ifa≤C≤x≤y≤b, then
(26)Cα,λ(x,y)=xαλα(ln⁡⁡ba)α+yα{(1-λ)α(ln⁡ba)α-2(ln⁡by)α}+∫yb(ln⁡bt)α-1dt-∫Cy(ln⁡bt)α-1dt-∫aC(ln⁡ta)α-1dt.(2)
Ifa≤x≤C≤y≤b, then
(27)Cα,λ(x,y)=xα{2(ln⁡⁡xa)α-λα(ln⁡⁡ba)α}+yα{(1-λ)α(ln⁡⁡ba)α-2(ln⁡⁡by)α}+∫xC(ln⁡⁡ta)α-1dt-∫ax(ln⁡⁡ta)α-1dt+∫yb(ln⁡⁡bt)α-1dt-∫Cy(ln⁡⁡bt)α-1dt.(3)
Ifa≤x≤y≤C≤b, then
(28)Cα,λ(x,y)=xα{2(ln⁡⁡xa)α-λα(ln⁡⁡ba)α}-yα(1-λ)α(ln⁡⁡ba)α+∫Cb(ln⁡⁡bt)α-1dt+∫xC(ln⁡ta)α-1dt-∫ax(ln⁡ta)α-1dt.Now, we shall give another result for Lipschitzian functions as follows.Theorem 13.
Letx,y,α,λ,C,Cα,λ and function f be defined as above. Then we have the following inequality for Hadamard fractional integrals:
(29)|Sf(x,y,λ,α,a,b)|≤αMCα,λ(x,y)(ln⁡⁡(b/a))α.Proof.
Using the hypothesis off, we have the following inequality:
(30)|Sf(x,y,λ,α,a,b)|=α(ln⁡(b/a))α|∫aC[f(x)-f(t)]t(ln⁡⁡ta)α-1dt+∫Cb[f(y)-f(t)]t(ln⁡⁡bt)α-1dt|≤α(ln⁡(b/a))α[∫aC|f(x)-f(t)|t(ln⁡⁡ta)α-1dt+∫Cbf|(y)-f(t)|t(ln⁡bt)α-1dt]≤αM(ln⁡(b/a))α[∫aC|x-t|t(ln⁡⁡ta)α-1dt+∫Cb|y-t|t(ln⁡⁡bt)α-1dt].
Now, using simple calculations, we obtain the following identities ∫aC‍(|x-t|/t)(ln⁡⁡(t/a))α-1dt and ∫Cb(|y-t|/t)(ln⁡⁡(b/t))α-1dt.(1)
Ifa≤C≤x≤y≤b, then
(31)∫aC|x-t|t(ln⁡⁡ta)α-1dt=xαλα(ln⁡⁡bα)α-∫aC(ln⁡⁡ta)α-1dt,∫Cb|y-t|t(ln⁡⁡bt)α-1dt=yα{(1-λ)α(ln⁡⁡ba)α-2(ln⁡⁡by)α}+∫yb(ln⁡⁡bt)α-1dt-∫Cy(ln⁡⁡bt)α-1dt.(2)
Ifa≤x≤C≤y≤b, then
(32)∫aC|x-t|t(ln⁡⁡ta)α-1dt=xα{2(ln⁡⁡xa)α-λα(ln⁡⁡ba)α}+∫xC(ln⁡⁡ta)α-1dt-∫ax(ln⁡⁡ta)α-1dt,∫Cb|y-t|t(ln⁡⁡bt)α-1dt=yα{(1-λ)α(ln⁡⁡ba)α-2(ln⁡⁡by)α}+∫yb(ln⁡⁡bt)α-1dt-∫Cy(ln⁡⁡bt)α-1dt.(3)
Ifa≤x≤y≤C≤b, then
(33)∫aC|x-t|t(ln⁡⁡ta)α-1dt=xα{2(ln⁡⁡xa)α-λα(ln⁡⁡ba)α}+∫xC(ln⁡⁡ta)α-1dt-∫ax(ln⁡⁡ta)α-1dt,∫Cb|y-t|t(ln⁡⁡bt)α-1dt=∫Cb(ln⁡⁡bt)α-1dt-yα(1-λ)α(ln⁡⁡ba)α.Using inequality (30) and the above identities ∫aC(|x-t|/t)(ln⁡⁡(t/a))α-1dt and ∫Cb(|y-t|/t)(ln⁡⁡(b/t))α-1dt, we derive inequality (29). This completes the proof.Under the assumptions of Theorem13, we have the following corollaries and remarks.Corollary 14.
In Theorem13, if we take α=1, then inequality (29) reduces the following inequality:
(34)|(1-λ)f(x)+λf(y)-1ln⁡⁡b-ln⁡⁡a∫abf(t)tdt|≤MC1,λ(x,y)ln⁡⁡(b/a).Corollary 15.
In Theorem13, let δ∈[1/2,1], x=aδb1-δ, and y=a1-δbδ. Then, we have the inequality
(35)|-Γ(α+1)(ln⁡(b/a))α[JC-αf(a)+JC+αf(b)]|λαf(aδb1-δ)+(1-λ)αf(a1-δbδ)-Γ(α+1)(ln⁡⁡(b/a))α[JC-αf(a)+JC+αf(b)]|≤ML(α,λ,δ),
as follows.(i)
Ifλ≤1-δ, then
(36)L(α,λ,δ)=aδb1-δλα+a1-δbδ[(1-λ)α-2(1-δ)α]+α(ln⁡⁡(b/a))α×{∫a1-δbδb(ln⁡⁡bt)α-1dt-∫Ca1-δbδ(ln⁡⁡bt)α-1dt-∫aC(ln⁡⁡ta)α-1dt}.(ii)
If1-δ≤λ≤δ, then
(37)L(α,λ,δ)=aδb1-δ[2(1-δ)α-λα]+a1-δbδ×[(1-λ)α-2(1-δ)α]+α(ln⁡⁡(b/a))α×{∫aδb1-δC(ln⁡⁡ta)α-1dt-∫aaδb1-δ(ln⁡⁡ta)α-1dt+∫a1-δbδb(ln⁡⁡bt)α-1dt-∫Ca1-δbδ(ln⁡⁡bt)α-1dt}.(iii)
Ifδ≤λ, then
(38)L(α,λ,δ)=aδb1-δ[2(1-δ)α-λα]-a1-δbδ(1-λ)α+α(ln⁡⁡(b/a))α{-∫aaδb1-δ(ln⁡⁡ta)α-1dt}∫Cb(ln⁡⁡bt)α-1dt+∫aδb1-δC(ln⁡⁡ta)α-1dt-∫aaδb1-δ(ln⁡⁡ta)α-1dt}.Corollary 16.
In Theorem13, if we take x=y=C, then we have the inequality
(39)|[λα+(1-λ)α]f(x)-Γ(α+1)(ln⁡(b/a))α[Jx-αf(a)+Jx+αf(b)]|≤M{[∫xb(ln⁡⁡bt)α-1dt-∫ax(ln⁡⁡ta)α-1dt][λα-(1-λ)α]x+α(ln⁡(b/a))α×[∫xb(ln⁡⁡bt)α-1dt-∫ax(ln⁡⁡ta)α-1dt]}.Remark 17.
In inequality (39), if we choose λ=1/2, C=ab, then we get inequality (15).Corollary 18.
In inequality (35), if we take δ=1, then we have the following weighted Hadamard-type inequalities for Lipschitzian functions via Hadamard fractional integrals:
(40)|λαf(a)+(1-λ)αf(b)-Γ(α+1)(ln⁡⁡(b/a))α[JC-αf(a)+JC+αf(b)]|≤M{[∫aC(ln⁡⁡ta)α-1dt-∫Cb(ln⁡⁡bt)α-1dt]}b(1-λ)α-aλα+α(ln⁡⁡(b/a))α×[∫aC(ln⁡⁡ta)α-1dt-∫Cb(ln⁡⁡bt)α-1dt]}.Remark 19.
In inequality (40), if we choose λ=1/2, C=ab, then we get inequality (18).


## 3. Application to Special Means

Let us recall the following special means of two positive numbersa, b with b>a.(1)
The arithmetic mean:(41)A=A(a,b):=a+b2.(2)
The geometric mean:(42)G=G(a,b):=ab.(3)
The harmonic mean:(43)H=H(a,b):=2aba+b.(4)
The logarithmic mean:(44)L=L(a,b):=b-aln⁡⁡b-ln⁡⁡a.(5)
The identric mean:(45)I=I(a,b)=1e(bbaa)1/(b-a).To prove the results of this section, we need the following lemma.Lemma 20 (see [12]).
Letf:[a,b]→ℝ be differentiable with ∥f′∥∞<∞. Then f is an M-Lipschitzian function on [a,b], where M=∥f′∥∞.Proposition 21.
Forb>a>0, λ∈[0,1] and n≥1, we have
(46)|(1-λ)Gn(a,b)+λA(an,bn)-L(an,bn)|≤nbn-1ln⁡⁡b-ln⁡⁡a×[λ(b-a)2ln⁡⁡ba+2(1-2λ)(A(a,b)-G(a,b))].Proof.
The proof follows by inequality (25) applied for the Lipschitzian function f(x)=xn on [a,b].Remark 22.
Letλ=0 and λ=1 in inequality (46). Then, using inequality (9), we have the following inequalities, respectively,
(47)0≤L(an,bn)-Gn(a,b)≤2nbn-1ln⁡⁡b-ln⁡⁡a(A(a,b)-G(a,b)),(48)0≤A(an,bn)-L(an,bn)≤nbn-1ln⁡⁡b-ln⁡⁡a[b-a2ln⁡⁡ba-2(A(a,b)-G(a,b))].Proposition 23.
Forb>a>0 and λ∈[0,1], we have
(49)|(1-λ)G(aea,beb)+λA(aea,beb)-L(ea,eb)L(a,b)|≤eb(b+1)ln⁡⁡b-ln⁡⁡a[λ(b-a)2ln⁡⁡ba+2(1-2λ)×(A(a,b)-G(a,b))[λ(b-a)2ln⁡ba+2(1-2λ)].Proof.
The proof follows by inequality (25) applied for the Lipschitzian function f(x)=xex on [a,b].Remark 24.
Letλ=0 and λ=1 in inequality (49). Then, using inequality (9), we have the following inequalities, respectively,
(50)0≤L(ea,eb)L(a,b)-G(aea,beb)≤2eb(b+1)ln⁡⁡b-ln⁡⁡a(A(a,b)-G(a,b)),0≤A(aea,beb)-L(ea,eb)L(a,b)≤eb(b+1)ln⁡⁡b-ln⁡⁡a[b-a2ln⁡⁡ba-2(A(a,b)-G(a,b))].Proposition 25.
Forb>a>0, λ∈[0,1], and n≥1, we have
(51)|(1-λ)G-1(a,b)+λH-1(a,b)-L(a,b)G-2(a,b)|≤1a2(ln⁡⁡b-ln⁡⁡a)×[λ(b-a)2ln⁡⁡ba+2(1-2λ)(A(a,b)-G(a,b))].Proof.
The proof follows by inequality (25) applied for the Lipschitzian function f(x)=1/x on [a,b].Remark 26.
Letλ=0 and λ=1 in inequality (51). Then, using inequality (9), we have the following inequalities, respectively,
(52)0≤L(a,b)-G(a,b)≤G2(a,b)a2(ln⁡⁡b-ln⁡⁡a)(A(a,b)-G(a,b)),0≤G2(a,b)-L(a,b)H(a,b)≤G2(a,b)H(a,b)a2(ln⁡⁡b-ln⁡⁡a)×[b-a2ln⁡⁡ba-2(A(a,b)-G(a,b))].Proposition 27.
Forb>a>0 and λ∈[0,1], we have
(53)ln⁡G(a,b)≤1a(ln⁡⁡b-ln⁡⁡a)×[λ(b-a)2ln⁡⁡ba+2(1-2λ)×(A(a,b)-G(a,b))[λ(b-a)2ln⁡⁡ba+2(1-2λ)].Proof.
The proof follows by inequality (25) applied for the Lipschitzian function f(x)=ln⁡⁡x on [a,b].Proposition 28.
Forb>a>e-1 and λ∈[0,1], we have
(54)|(1-λ)G(a,b)ln⁡⁡G(a,b)+λln⁡⁡G(aa,bb)-L(a,b)ln⁡⁡I(a,b)|(1-λ)G(a,b)ln⁡G(a,b)+λln⁡G(aa,bb)|≤1+ln⁡⁡bln⁡⁡b-ln⁡⁡a[λ(b-a)2ln⁡⁡ba+2(1-2λ)×(A(a,b)-G(a,b))[λ(b-a)2ln⁡ba+2(1-2λ)].Proof.
The proof follows by inequality (25) applied for Lipschitzian function f(x)=xln⁡⁡x on [a,b].



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*Source: 353924-2014-04-22.xml*