# Hermite-Hadamard and Simpson-Like Type Inequalities for Differentiable (α,m)-Convex Mappings

Authors: Jaekeun Park
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

The author establish several Hermite-Hadamard and Simpson-like
type inequalities for mappings whose first derivative in absolute value aroused
to theqth (q≥1) power are (α,m)-convex. Some applications to special means
of positive real numbers are also given.


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## Body

## 1. Introduction

Recall that, for some fixedα∈(0,1] and m∈[0,1], a mapping f:𝕀⊆[0,∞)→ℝ is said to be (α,m)-convex on an interval 𝕀 if the inequality (1.1)f(tx+m(1-t)y)≤tαf(x)+m(1-tα)f(y)
holds for x,y∈𝕀, and t∈[0,1].Denote byKmα(𝕀) the set of all (α,m)-convex mappings on 𝕀. For recent results and generalizations concerning m-convex and (α,m)-convex mappings, see [1–4].For the simplicities of notations, forf∈Kmα(𝕀), let us denote (1.2)(a-mb)Sab(f)(α,m,r)=1r{f(a)+(r-2)f(a+mb2)+f(mb)}-1mb-a∫ambf(x)dx.In [1, 3], Klaričić Bakula and Özdemir et al., proved the following Hadamard’s inequalities for mappings whose second derivative in absolute value aroused to the q-th (q≥1) power are (α,m)-convex.Theorem 1.1.
Letf:𝕀⊆[0,b*]→ℝ be a twice differentiable mapping on the interior 𝕀0 of an interval 𝕀 such that f′′∈L1([a,b]), where a,b∈𝕀 with a<b and b*>0. If |f′′|q is (α,m)-convex on [a,b] for (α,m)∈[0,1]2 and q≥1 with 1/p+1/q=1, then the following inequality holds:
(1.3)(a)|Sab(f)(α,m,2)|≤mb-a2(16)1/p{μ|f′′(a)|q+mν|f′′(b)|q}1/q,
where
(1.4)μ=1(α+2)(α+3),ν=α2+5α6(α+2)(α+3),(1.5)(b)|Sab(f)(α,m,6)|≤mb-a2(12)1/p{μ0|f′′(a)|q+mν0|f′′(b)|q}1/q,
where
(1.6)μ0=(qq+α+2)Γ(α+2)Γ(q)Γ(q+α+1),ν0=1(q+1)(q+2)-(qq+α+2)Γ(α+2)Γ(q)Γ(q+α+1),
where
(1.7)Γ(x)=∫0∞e-ttx-1dt,x>0.Theorem 1.2.
Under the same notations in Theorem2.2, if |f′′|q is (α,m)-convex on [a,b] for (α,m)∈[0,1]2 and q>1 with 1/p+1/q=1, then the following inequality holds:
(1.8)|Sab(f)(α,m,2)|≤mb-a8{Γ(p+1)Γ(p+3/2)}1/p{|f′′(a)|q1α+1+m|f′′(b)|q(αα+1)}1/q.Note that for(α,m)∈{(0,0),(α,0),(1,0),(1,m),(1,1),(α,1)} one obtains the following classes of functions: increasing, α-starshaped, starshaped, m-convex, convex, and α-convex. For the definitions and elementary properties of these classes, see [4–8].For recent years, many authors present some new results about Simpson’s inequality for(α,m)-convex mappings and have established error estimations for the Simpson’s inequality: for refinements, counterparts, generalizations, and new Simpson’s type inequalities, see [1–3, 6].In [9], Dragomir et al. proved the following theorem.Theorem 1.3.
Letf:𝕀⊂[0,∞)→ℝ be an absolutely continuous mapping on [a,b] such that f′∈Lp([a,b]), where a,b∈𝕀 with a<b. Then the following inequality holds:
(1.9)|Sab(f)(1,1,6)|≤(b-a)-1/p6{2q+1+13(q+1)}1/q‖f′‖p.The readers can estimate theerror(f) in the generalized Simpson’s formula without going through its higher derivatives which may not exist, not be bounded, or may be hard to find.In this paper, the author establishes some generalizations of Hermite-Hadamard and Simpson-like type inequalities based on differentiable(α,m)-convex mappings by using the following new identity in Lemma 2.1 and by using these results, obtain some applications to special means of positive real numbers.


## 2. Generalizations of Simpson-Like Type Inequalities onKmα(𝕀)

In order to generalize the classical Simpson-like type inequalities and prove them, we need the following lemma [6].Lemma 2.1.
Letf:𝕀⊆[0,b*]→ℝ be a differentiable mapping on the interior 𝕀0 of an interval 𝕀, where a,b∈𝕀 with 0≤a<b and b*>0. If f′∈L1([a,b]), then, for r≥2 and h∈(0,1) with 1/r≤h≤(r-1)/r, the following equality holds:
(2.1)Sab(f)(α,m,r)=∫01p(r,t)f′(ta+m(1-t)b)dt
for f∈Kmα([a,b]) and each t∈[0,1], where
(2.2)p(r,t)={t-1rt∈[0,12],t-r-1rt∈(12,1].By the similar way as Theorems1.1–1.3, we obtain the following theorems.Theorem 2.2.
Letf:𝕀⊂[0,b*]→ℝ be a differentiable mapping on 𝕀0 such that f′∈L([a,b]), where a,b∈𝕀 with 0≤a<b<∞ and b*>0. If |f′|∈Kmα([a,b]), for some (α,m)∈(0,1]2 and mb>a, then, for any r≥2, the following inequality holds:
(2.3)|Sab(f)(α,m,r)|≤{μ11+μ12+μ13+μ14}|f′(a)|+{ν11+ν12+ν13+ν14}m|f′(b)|,
where
(2.4)μ11=1(α+1)(α+2)rα+2,μ12=1(α+1)(α+2)rα+2+(α+1)r-2(α+2)2α+2(α+1)(α+2)r,μ13=(r-1)α+2(α+1)(α+2)rα+2+2(α+2)-3r2α+2(α+1)(α+2)r,μ14=(r-1)α+2(α+1)(α+2)rα+2+(α+2)-r(α+1)(α+2)r,ν11=12r2-μ11,ν12=(r-2)28r2-μ12,ν13=(r-2)28r2-μ13,ν14=12r2-μ14.Proof.
From Lemma2.1 and using the properties of the modulus, we have the following:
(2.5)|Sab(f)(α,m,r)|≤∫01/r(1r-t)|f′(ta+m(1-t)b)|dt+∫1/r1/2(t-1r)|f′(ta+m(1-t)b)|dt+∫1/2(r-1)/r(r-1r-t)|f′(ta+m(1-t)b)|dt+∫(r-1)/r1(t-r-1r)|f′(ta+m(1-t)b)|dt.
Since|f′| is (α,m)-convex on [a,b], we know that for any t∈[0,1](2.6)|f′(ta+m(1-t)b)|≤tα|f′(a)|+m(1-tα)|f′(b)|.
By (2.5) and (2.6), we get the following:
(2.7)|Sab(f)(α,m,r)|≤∫01/r(1r-t){tα|f′(a)|+m(1-tα)|f′(b)|}dt+∫1/r1/2(t-1r){tα|f′(a)|+m(1-tα)|f′(b)|}dt+∫1/2(r-1)/r(r-1r-t){tα|f′(a)|+m(1-tα)|f′(b)|}dt+∫(r-1)/r1(t-r-1r){tα|f′(a)|+m(1-tα)|f′(b)|}dt≤{∫01/2|1r-t|tαdt+∫1/21|r-1r-t|tαdt}|f′(a)|+{∫01/2|1r-t|(1-tα)dt+∫1/21|r-1r-t|(1-tα)dt}m|f′(b)|={μ11+μ12+μ13+μ14}|f′(a)|+{ν11+ν12+ν13+ν14}m|f′(b)|,
which completes the proof.Corollary 2.3.
In Theorem2.2, (i) if we choose α=1 and r=2, then we have the following:
(2.8)|(mb-a)Sab(f)(1,m,2)|=|f(a)+f(mb)2-1mb-a∫ambf(x)dx|≤mb-a8{|f′(a)|+m|f′(b)|},
and (ii) if we choose α=1 and r=6, then we have the following
(2.9)|(mb-a)Sab(f)(1,m,6)|=|16{f(a)+4f(a+mb2)+f(mb)}-1mb-a∫ambf(x)dx|≤572(mb-a){|f′(a)|+m|f′(b)|}.Theorem 2.4.
Under the same notations in Theorem2.2, if |f′|q∈Kmα([a,b]), for some (α,m)∈(0,1]2, mb>a and q>1 with 1/p+1/q=1, then, for any r≥2, the following inequality holds:
(2.10)|Sab(f)(α,m,r)|≤{12r2}1/p{(μ11|f′(a)|q+ν11m|f′(b)|q)1/q+(μ14|f′(a)|q+ν14m|f′(b)|q)1/q}+{18(r-2r)2}1/p{(μ12|f′(a)|q+ν12m|f′(b)|q)1/q+(μ13|f′(a)|q+ν13m|f′(b)|q)1/q}.Proof.
From Lemma2.1 and using the properties of modulus, we have the following:
(2.11)|Sab(f)(α,m,r)|≤∫01/r(1r-t)|f′(ta+m(1-t)b)|dt+∫1/r1/2(t-1r)|f′(ta+m(1-t)b)|dt+∫1/2(r-1)/r(r-1r-t)|f′(ta+m(1-t)b)|dt+∫(r-1)/r1(t-r-1r)|f′(ta+m(1-t)b)|dt.
Using the power-mean integral inequality and(α,m)-convexity of |f′|q for any t∈[0,1], we have the following(a)
(2.12)∫01/r(1r-t)|f′(ta+m(1-t)b)|qdt≤μ11|f′(a)|q+ν11m|f′(b)|q,(b)
(2.13)∫1/r1/2(t-1r)|f(ta+m(1-t)b)|qdt≤μ12|f′(a)|q+ν12m|f′(b)|q,(c)
(2.14)∫1/2(r-1)/r(r-1r-t)|f(ta+m(1-t)b)|qdt≤μ13|f′(a)|q+ν13m|f′(b)|q,(d)
(2.15)∫(r-1)/r1(t-r-1r)|f(ta+m(1-t)b)|qdt≤μ14|f′(a)|q+ν14m|f′(b)|q.
By the similar way as the above inequalities (a)–(d), we have the following:(a′)
(2.16)∫01/r(1r-t)|f′(ta+m(1-t)b)|dt≤{12r2}1/p{μ11|f′(a)|q+ν11m|f′(b)|q}1/q,(b′)
(2.17)∫1/r1/2(t-1r)|f′(ta+m(1-t)b)|dt≤{18(r-2r)2}1/p{μ12|f′(a)|q+ν12m|f′(b)|}1/q,(c′)
(2.18)∫1/2(r-1)/r(r-1r-t)|f′(ta+m(1-t)b)|dt≤{18(r-2r)2}1/p{μ13|f′(a)|q+ν13m|f′(b)|q}1/q,(d′)
(2.19)∫(r-1)/r1(t-r-1r)|f′(ta+m(1-t)b)|dt≤{12r2}1/p{μ14|f′(a)|q+ν14m|f′(b)|q}1/q.
By (2.11) and (2.16)–(2.19) the assertion (2.10) holds.Corollary 2.5.
In Theorem2.4, (i) if we choose α=1 and r=2, then we have that
(2.20)|f(a)+f(mb)2-1mb-a∫ambf(x)dx|≤6-1/q8(mb-a)[{|f′(a)|q+5m|f′(b)|q}1/q+{5|f′(a)|q+m|f′(b)|q}1/q],
and (ii) if we choose α=1 and r=6, then we have that
(2.21)|16{f(a)+4f(a+mb2)+f(mb)}-1mb-a∫ambf(x)dx|≤(mb-a)[(172)(118)1/q{(|f′(a)|q+17m|f′(b)|q)1/q+(17|f′(a)|q+m|f′(b)|q)1/q}+(25288)(25)2/q×{(7|f′(a)|q+11m|f′(b)|q)1/q+(11|f′(a)|q+7m|f′(b)|q)1/q}].Theorem 2.6.
Under the same notations in Theorem2.2, if |f′|q∈Kmα([a,b]), for some (α,m)∈(0,1]2, mb>a and q>1 with 1/p+1/q=1, then, for any r≥2, the following inequality holds:
(2.22)|Sab(f)(α,m,r)|≤{1rp+1}1/p{(μ21|f′(a)|q+ν21m|f′(b)|q)1/q+(μ24|f′(a)|q+ν24m|f′(b)|q)1/q}+{(r-2)p+12p+1rp+1}1/p{(μ22|f′(a)|q+ν22m|f′(b)|q)1/q+(μ23|f′(a)|q+ν23m|f′(b)|q)1/q},
where
(2.23)μ21=1rα+1(α+1),μ22=rα+1-2α+12α+1rα+1(α+1),μ23=2α+1(r-1)α+1-rα+12α+1rα+1(α+1),μ24=rα+1-(r-1)α+1rα+1(α+1),ν21=1r-μ21,ν22=r-22r+μ22,ν23=r-22r+μ23,ν24=1r-μ24.Proof.
Suppose thatq>1. From Lemma 2.1, using the Hölder integral inequality, we get the following:
(2.24)|Sab(f)(α,m,r)|≤{∫01/r(1r-t)pdt}1/p{∫01/r|f′(tb+m(1-t)a)|qdt}1/q+{∫1/r1/2(t-1r)pdt}1/p{∫1/r1/2|f′(tb+m(1-t)a)|qdt}1/q+{∫1/2(r-1)/r(r-1r-t)pdt}1/p{∫1/2(r-1)/r|f′(tb+m(1-t)a)|qdt}1/q+{∫(r-1)/r1(t-r-1r)pdt}1/p{∫(r-1)/r1|f′(tb+m(1-t)a)|qdt}1/q≤{1rp+1}1/p{∫01/r|f′(tb+m(1-t)a)|qdt}1/q+{12p+1(r-2r)p+1}1/p{∫1/r1/2|f′(tb+m(1-t)a)|qdt}1/q+{12p+1(r-2r)p+1}1/p{∫1/2(r-1)/r|f′(tb+m(1-t)a)|qdt}1/q+{1rp+1}1/p{∫(r-1)/r1|f′(tb+m(1-t)a)|qdt}1/q,
where we have used the fact that 1/2<(1/(p+1))1/p<1.
Since|f′|q∈Kαm([a,b]) for some fixed α∈(0,1] and m∈[0,1], we have the followings:
(2.25)∫01/r|f′(tb+(1-t)a)|qdt≤μ21|f′(a)|q+ν21m|f′(b)|q,∫1/r1/2|f(tb+(1-t)a)|qdt≤μ22|f′(a)|q+ν22m|f′(b)|q,∫1/2(r-1)/r|f′(tb+(1-t)a)|qdt≤μ23|f′(a)|q+ν23m|f′(b)|q,∫(r-1)/r1|f(tb+(1-t)a)|qdt≤μ24|f′(a)|q+ν24m|f′(b)|q.
Hence, if we combine the inequalities in (2.24)-(2.25), we get the desired result.Corollary 2.7.
In Theorem2.6, (i) if we choose α=1 and r=2, then we have that
(2.26)|f(a)+f(mb)2-1mb-a∫ambf(x)dx|≤(14)1+1/q(mb-a){(|f′(a)|q+3m|f′(b)|q)1/q+(3|f′(a)|q+m|f′(b)|q)1/q},
and (ii) if we choose α=1 and r=6, then we have
(2.27)|16{f(a)+4f(a+mb2)+f(mb)}-1mb-a∫ambf(x)dx|≤(mb-a)[(16)1+2/q{(|f′(a)|q+11m|f′(b)|q)1/q+(11|f′(a)|q+m|f′(b)|q)1/q}+(13)2+1/q{(|f′(a)|q+2m|f′(b)|q)1/q+(2|f′(a)|q+m|f′(b)|q)1/q}],
where we have used the fact that (1/2)1/q<1.


## 3. Applications to Special Means

Now using the results of Section2, we give some applications to the following special means of positive real numbers a,b∈ℝ+ with b≥a.(1)
The arithmetic mean:A(a,b)=(a+b)/2.(2)
The geometric mean:G(a,b)=ab.(3)
The logarithmic mean:L(a,b)=(b-a)/(lnb-lna) for a≠b.(4)
The harmonic mean:H(a,b)=2ab/(a+b).(5)
The power mean:Mr(a,b)=((ar+br)/2)1/r, r≥1,a,b∈ℝ.(6)
The generalized logarithmic mean:(3.1)Ln(a,b)=[bn+1-an+1(b-a)(n+1)]1/n,a≠b.(7)
The identric mean:(3.2)I(a,b)={aa=b1e(bbaa)1/(b-a)a≠b.Proposition 3.1.
Forn∈(-∞,0)∪[1,∞)∖{-1} and [a,b]∈[0,b*] with b*>0, we have the following inequalities:
(3.3)(a)|A(an,mnbn)-Lnn(a,mb)|≤mb-a8|n|Mn-1n-1(a,m1/(n-1)b).(b)|13A(an,mnbn)+23An(a,mb)-Lnn(a,mb)|≤(572)(mb-a)|n|Mn-1n-1(a,m1/(n-1)b).Proof.
The assertions follow from Corollary2.3 for f(x)=xn.Proposition 3.2.
For[a,b]∈[0,b*], we have the following inequalities:
(3.4)(a)|H-1(a,mb)-L-1(a,mb)|≤(mb-a8a2b2)M22(m1/2a,b),(b)|13H-1(a,mb)+23A-1(a,mb)-L-1(a,mb)|≤(5(mb-a)72a2b2)M22(m1/2a,b).Proof.
The assertions follow from Corollary2.3 for f(x)=1/x.Proposition 3.3.
Forn∈(-∞,0)∪[1,∞)∖{-1} and [a,b]∈[0,b*], we have the following inequalities:
(3.5)(a)|A(an,mnbn)-Lnn(a,mb)|≤(3-1/q8)(mb-a)|n|×{A1/q(a(n-1)q,5mb(n-1)q)1/q+A1/q(5a(n-1)q,mb(n-1)q)1/q},(b)|13A(an,mnbn)+23An(a,mb)-Lnn(a,mb)|≤(172)(19)1/q(mb-a)|n|×{A1/q(a(n-1)q,17mb(n-1)q)+A1/q(17a(n-1)q,mb(n-1)q)}+(25288)(225)1/q(mb-a)|n|×{A1/q(7a(n-1)q,11mb(n-1)q)+A1/q(11a(n-1)q,7mb(n-1)q)}.Proof.
The assertions follow from Corollary2.5 for f(x)=xn.Proposition 3.4.
Forn∈(-∞,0)∪[1,∞)∖{-1} and [a,b]∈[0,b*], we have the following inequalities:
(3.6)(a)|A(an,mnbn)-Lnn(a,b)|≤(122+1/q)|n|(mb-a)×{A1/q(a(n-1)q,3mb(n-1)q)+A1/q(3a(n-1)q,mb(n-1)q)},(b)|13A(an,mnbn)+23An(a,mb)-Lnn(a,mb)|≤(16)1+1/q(13)1/q|n|(mb-a)×{A1/q(a(n-1)q,11mb(n-1)q)+A1/q(11a(n-1)q,mb(n-1)q)}+(13)2+1/q(21/q)|n|(mb-a)×{A1/q(a(n-1)q,2mb(n-1)q)+A1/q(2a(n-1)q,mb(n-1)q)}.Proof.
The assertions follow from Corollary2.7 for f(x)=xn.Proposition 3.5.
For[a,b]∈[0,b*], we have the following inequalities:
(3.7)(a)|ln⁡I(a,b)G(a,b)|≤b-aab(141+1/q){(aq+3bq)1/q+(3aq+bq)1/q},(b)|ln⁡I(a,b)G1/3(a,b)A2/3(a,b)|≤b-aab[(16)1+2/q{(aq+11bq)1/q+(11aq+bq)1/q}+(13)2+1/q{(aq+2bq)1/q+(2aq+bq)1/q}].Proof.
The assertions follow from Corollary2.7 for f(x)=-lnx and m=1.



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