# On Integral Inequalities of Hermite-Hadamard Type fors-Geometrically Convex Functions

Authors: Tian-Yu Zhang; Ai-Ping Ji; Feng Qi
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

The authors introduce the concept of thes-geometrically convex functions. By the well-known Hölder inequality, they establish some integral inequalities of Hermite-Hadamard type related to the s-geometrically convex functions and apply these inequalities to special means.


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## Body

## 1. Introduction

We firstly list several definitions and some known results.Definition 1.1.
A functionf:I⊂ℝ=(-∞,+∞)→ℝ is said to be convex if
(1.1)f(λx+(1-λ)y)≤λf(x)+(1-λ)f(y)
for x,y∈I and λ∈[0,1].Definition 1.2 (see [1]).
A functionf:I⊂ℝ0=[0,+∞)→ℝ0 is said to be s-convex if
(1.2)f(λx+(1-λ)y)≤λsf(x)+(1-λ)sf(y)
for some s∈(0,1], where x,y∈I, and λ∈[0,1].
Ifs=1, the s-convex function becomes a convex function on ℝ0.Theorem 1.3 ([2], Theorem 2.2).
Letf:I∘⊂ℝ→ℝ be a differentiable mapping on I∘, a,b∈I∘, a<b. (i)
If|f'(x)| is a convex function on [a,b], then
(1.3)|f(a)+f(b)2-1b-a∫abf(x)dx|≤(b-a)(|f′(a)|+|f′(b)|)8.(ii)
If|f'(x)|p/(p-1) is a convex function on [a,b], for p>1, then
(1.4)|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a2(p+1)(1/p)(|f′(a)|p/(p-1)+|f′(b)|p/(p-1)2)(p-1)/p.Theorem 1.4 ([3], Theorems 2.3 and 2.4).
Letf:I⊂ℝ→ℝ be differentiable on I∘, a,b∈I, a<b. If |f'(x)|p  is convex on [a,b], for p>1, then
(1.5)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4(4p+1)1/p(|f′(a)|+|f′(b)|),|f(a+b2)-1b-a∫abf(x)dx|≤b-a16(4p+1)1/p×[(|f′(a)|p/(p-1)+3|f′(b)|p/(p-1))(p-1)/pb-a16(4p+1)1/p+(3|f′(a)|p/(p-1)+|f′(b)|p/(p-1))(p-1)/p].Theorem 1.5 ([4], Theorem 1–4).
Letf:I⊂ℝ→ℝ be differentiable on I∘, a,b∈I, a<b. If |f'(x)|qis  s-convex on [a,b], for  q>1, then
(1.6)|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a2(12)(q-1)/q(2+(1/2)s(s+1)(s+2))1/q(|f′(a)|s+|f′(b)|s)1/q.Theorem 1.6 ([5], Theorem 4).
Letf:I→ℝ0 be differentiable on I∘, a,b∈I, a<b, and f'∈L([a,b]). If |f'(x)|q is s-convex on [a,b] for some s∈(0,1], and p,q≥1, such that (1/q)+(1/p)=1 then
(1.7)|f(a+b2)-1b-a∫abf(x)dx|≤2-1/p(|f′(a)|q+(s+1)|f'((a+b)/2)|q)1/q{(s+1)(s+2)}1/q+2-1/p(|f′(b)|q+(s+1)|f'((a+b)/2)|q)1/q{(s+1)(s+2)}1/q=2-1/p[(β(s+1,2)|f′(a)|q+β(s+2,1)|f'(a+b2)|q)1/q+(β(s+1,2)|f′(b)|q+β(s+2,1)|f'(a+b2)|q)1/q].Theorem 1.7 ([6], Theorems 2.2–2.4).
Letf:I→ℝ0 be differentiable on I∘, a,b∈I, a<b, and f'∈L([a,b]). (i)
If|f'(x)|  iss-convex on [a,b] for some s∈(0,1], then
(1.8)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4(s+1)(s+2)[|f′(a)|+2(s+1)|f'(a+b2)|+|f′(b)|]≤(22-s+1)(b-a)4(s+1)(s+2)[|f′(a)|+|f′(b)|].(ii)
If|f'(x)|p/(p-1)(p>1)is a s-convex function on [a,b] for some s∈(0,1], then
(1.9)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4(1p+1)1/p(1s+1)2/q[((21-s+s+1)|f′(a)|q+21-s|f′(b)|q)1/q+(21-s|f′(a)|q|f′(a)|p/(p-1)+(21-s+s+1)|f′(b)|q)1/q],
where 1/p+1/q=1.(iii)
If|f'(x)|q(q≥1)is  s-convex on [a,b]  for some s∈(0,1], then
(1.10)|f(a+b2)-1b-a∫abf(x)dx|≤b-a8(2(s+1)(s+2))1/q[((21-s+1)|f′(a)|q+21-s|f′(b)|q)1/q+(21-s|f′(a)|q|f′(a)|q+(21-s+1)|f′(b)|q)1/q((21-s+1)|f′(a)|q+21-s|f′(b)|q)q].Now we introduce the definition of the s-geometrically convex function.Definition 1.8.
A functionf:I⊂ℝ+=(0,+∞)→ℝ+ is said to be a geometrically convex function if
(1.11)f(xλy1-λ)≤[f(x)]λ[f(y)]1-λ
for x,y∈I and λ∈[0,1].Definition 1.9.
A functionf:I⊂ℝ+→ℝ+ is said to be a s-geometrically convex function if
(1.12)f(xλy1-λ)≤[f(x)]λs[f(y)](1-λ)s
for some s∈(0,1], where x,y∈I and λ∈[0,1].
Ifs=1, the s-geometrically convex function becomes a geometrically convex function on ℝ+.
In this paper, we will establish some integral inequalities of Hermite-Hadamard type related to thes-geometrically convex functions and then apply these inequalities to special means.


## 2. A Lemma

In order to prove our results, we need the following lemma.Lemma 2.1.
Letf:I⊂ℝ→ℝ be differentiable on I∘, and a,b∈I, with a<b. If f'∈L([a,b]), then
(2.1)f(a+b2)-1b-a∫abf(x)dx=b-a4∫01[tf'((1-t)a+ta+b2)+(t-1)f'((1-t)a+b2+tb)]dt,f(a)+f(b)2-1b-a∫abf(x)dx=b-a4∫01[(t-1)f'((1-t)a+ta+b2)+tf'((1-t)a+b2+tb)]dt.Proof.
Integrating by part and changing variables of integration yields(2.2)∫01[tf′((1-t)a+ta+b2)+(t-1)f′((1-t)a+b2+tb)]dt=2b-a[tf((1-t)a+ta+b2)|01-∫01f((1-t)a+ta+b2)dt]+2b-a[(t-1)f((1-t)a+b2+tb)|01-∫01f((1-t)a+b2+tb)dt]=4b-a(f(a+b2)-1b-a∫abf(x)dx),∫01[(t-1)f′((1-t)a+ta+b2)+tf′((1-t)a+b2+tb)]dt=2b-a[(t-1)f((1-t)a+ta+b2)|01-∫01f((1-t)a+ta+b2)dt]+2b-a[tf((1-t)a+b2+tb)|01-∫01f((1-t)a+b2+tb)dt]=4b-a(f(a)+f(b)2-1b-a∫abf(x)dx).
This completes the proof of Lemma 2.1.


## 3. Main Results

Theorem 3.1.
Letf:I⊂ℝ+→ℝ+  be differentiable on I∘,  a,b∈I, with a<b, andf'∈L([a,b]). If |f'(x)|q is  s-geometrically convex and monotonically decreasing on [a,b] for q≥1 and s∈(0,1], then
(3.1)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4(12)1-1/qG1(s,q;g1(α),g2(α))(3.2)|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a4(12)1-1/qG1(s,q;g2(α),g1(α)),
where
(3.3)g1(α)={12,α=1,αlnα-α+1[lnα]2,α≠1,g2(α)={12,α=1,α-lnα-1[lnα]2α≠1,(3.4)α(u,v)=|f′(a)|-u|f′(b)|v,u,v>0,(3.5)G1(s,q;g1(α),g2(α))={|f′(a)|s[g1(α(sq2,sq2))]1/q+|f′(a)f′(b)|s/2[g2(α(sq2,sq2))]1/q,|f′(a)|≤1,|f′(a)|1/s[g1(α(q2s,sq2))]1/q+|f′(a)|1/2s|f′(b)|s/2[g2(α(q2s,sq2))]1/q,|f′(b)|≤1≤|f'(a)|,|f′(a)|1/s[g1(α(q2s,q2s))]1/q+|f′(a)f′(b)|1/2s[g2(α(q2s,q2s))]1/q,1≤|f′(b)|.Proof.
(1) Since|f'(x)|q is s-geometrically convex and monotonically decreasing on [a,b], from Lemma 2.1 and Hölder inequality, we have
(3.6)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4∫01[t|f'((1-t)a+ta+b2)|+|t-1||f'((1-t)a+b2+tb)|]dt≤b-a4{(∫01tdt)1-1/q[∫01t|f'((1-t)a+ta+b2)|qdt]1/q+(∫01(1-t)dt)1-1/q[∫01(1-t)|f'((1-t)a+b2+tb)|qdt]1/q}≤b-a4(12)1-1/q{[∫01t|f′(a(2-t)/2bt/2)|qdt]1/q+[∫01(1-t)|f′(a(1-t)/2b(1+t)/2)|qdt]1/q}≤b-a4(12)1-1/q{[∫01t|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)sdt]1/q≤b-a4(12)1-1/q+[∫01(1-t)|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)sdt]1/q}.
If 0<μ≤1≤η,0<α,s≤1, then
(3.7)μαs≤μαs,ηαs≤ηα/s.(i)
If|f'(a)|≤1, by (3.7), we obtain that
(3.8)∫01t(|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)s)dt≤∫01t(|f′(a)|(sq/2)(2-t)|f′(b)|(sqt/2))dt=|f′(a)|sqg1(α(sq2,sq2)),∫01(1-t)(|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)s)dt≤∫01(1-t)(|f′(a)|(sq/2)(1-t)|f′(b)|(sq/2)(1+t))dt=|f′(a)f′(b)|sq/2g2(α(sq2,sq2)).(ii)
If|f'(b)|≤1≤|f'(a)|, by (3.7), we obtain that
(3.9)∫01t(|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)s)dt≤∫01t(|f′(a)|(q/2s)(2-t)|f′(b)|(sq/2)t)dt=|f′(a)|q/sg1(α(q2s,sq2)),∫01(1-t)(|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)s)dt≤∫01(1-t)(|f′(a)|(q/2s)(1-t)|f′(b)|(sq/2)(1+t))dt=|f′(a)|q/2s|f′(b)|sq/2g2(α(q2s,sq2)).(iii)
If1≤|f'(b)|, by (3.7), we obtain that
(3.10)∫01t(|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)s)dt≤∫01t(|f′(a)|(q/2s)(2-t)|f′(b)|(q/2s)t)dt=|f′(a)|q/sg1(α(q2s,q2s)),∫01(1-t)(|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)s)dt≤∫01(1-t)(|f′(a)|(q/2s)(1-t)|f′(b)|(q/2s)(1+t))dt=|f′(a)f′(b)|q/2sg2(α(q2s,q2s)).From (3.6) to (3.10),  (3.1) holds.
(2) Since|f'(x)|qis s-geometrically convex and monotonically decreasing on [a,b], from Lemma 2.1 and Hölder inequality, we have
(3.11)|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a4[(∫01(1-t)dt)1-1/q(∫01(1-t)|f'((1-t)a+ta+b2)|qdt)1/q+(∫01tdt)1-1/q(∫01t|f'((1-t)a+b2+tb)|qdt)1/q]≤b-a4(12)1-1/q{[∫01(1-t)|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)sdt]1/q+[∫01t|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)sdt]1/q}.(i)
If|f'(a)|≤1, by (3.7), we have
(3.12)∫01(1-t)(|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)s)dt≤|f′(a)|sqg2(α(sq2,sq2)),∫01t(|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)s)dt≤|f′(a)f′(b)|sq/2g1(α(sq2,sq2)).(ii)
If|f'(b)|≤1≤|f'(a)|, by (3.7), we have
(3.13)∫01(1-t)(|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)s)dt≤|f′(a)|q/sg2(α(q2s,sq2)),∫01t(|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)s)dt≤|f′(a)|q/2s|f′(b)|sq/2g1(α(q2s,sq2)).(iii)
If1≤|f'(b)|, by (3.7), we have
(3.14)∫01(1-t)(|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)s)dt≤|f′(a)|q/sg2(α(q2s,q2s)),∫01t(|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)s)dt≤|f′(a)f′(b)|q/2sg1(α(q2s,q2s)).From (3.11) to (3.14), (3.2) holds. This completes the required proof.Applying Theorem3.1 to q=1,s=1, respectively, results in the following corollary.Corollary 3.2.
Letf:I⊂ℝ+→ℝ+ be differentiable on I∘, a,b∈Iwith a<b, and f'∈L([a,b]). If |f'(x)|q is  s-geometrically convex and monotonically decreasing on [a,b]  for  s∈(0,1], then(i)
whenq=1, one has
(3.15)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4G1(s,1;g1(α),g2(α)),|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a4G1(s,1;g2(α),g1(α)).(ii)
whens=1, one has
(3.16)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4(12)1-1/qG1(1,q;g1(α),g2(α)),|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a4(12)1-1/qG1(1,q;g2(α),g1(α)),where g1(α), g2(α), α(u,v), G1(s,q;g2(α),g1(α)) are same with (3.3)–(3.5).Theorem 3.3.
Letf:I⊂ℝ+→ℝ+  be differentiable on I∘,  a,b∈I,  with a<b, and f'∈L([a,b]). If  |f'(x)|q  is  s-geometrically convex and monotonically decreasing on [a,b],  for q>1 and  s∈(0,1], then
(3.17)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4(q-12q-1)1-1/qG2(s,q;g3(α)),(3.18)|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a4(q-12q-1)1-1/qG2(s,q;g3(α)),
where
(3.19)G2(s,q;g3(α))={(|f′(a)|s+|f′(a)f′(b)|s/2)[g3(α(sq2,sq2))]1/q,|f′(a)|≤1,(|f′(a)|1/s+|f′(a)|1/2s|f′(b)|s/2)[g3(α(q2s,sq2))]1/q,|f′(b)|≤1≤|f′(a)|,(|f′(a)|1/s+|f′(a)f′(b)|1/2s)[g3(α(q2s,q2s))]1/q,1≤|f′(b)|,g3(α)={1,α=1,α-1lnα,α≠1,
and α(u,v) is the same as in (3.4).Proof.
(1) Since|f'(x)|q  is  s-geometrically convex and monotonically decreasing on [a,b], from Lemma 2.1 and Hölder inequality, we have
(3.20)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4∫01[t|f'((1-t)a+ta+b2)|+|t-1||f'((1-t)a+b2+tb)|]dt≤b-a4[(∫01tq/(q-1)dt)1-1/q(∫01|f'((1-t)a+ta+b2)|qdt)1/q+(∫01(1-t)q/(q-1)dt)1-1/q(∫01|f'((1-t)a+b2+tb)|qdt)1/q]≤b-a4(q-12q-1)1-1/q[(∫01|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)sdt)1/q+(∫01|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)sdt)1/q].(i)
If|f'(a)|≤1, we have
(3.21)∫01|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)sdt≤|f′(a)|sqg3(α(sq2,sq2)),∫01|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)sdt≤|f′(a)f′(b)|sq/2g3(α(sq2,sq2)),(ii)
If|f'(b)|≤1≤|f'(a)|, we have
(3.22)∫01|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)sdt≤|f′(a)|q/sg3(α(q2s,sq2)),∫01|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)sdt≤|f′(a)|q/2s|f′(b)|sq/2g3(α(q2s,sq2)).(iii)
If1≤|f'(b)|, we have
(3.23)∫01|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)sdt≤|f′(a)|q/sg3(α(q2s,q2s)),∫01|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)sdt≤|f′(a)f′(b)|q/2sg3(α(q2s,q2s)).From (3.20) to (3.23), (3.17) holds.
(2) Since|f'(x)|q  is  s-geometrically convex and monotonically decreasing on [a,b], from Lemma 2.1 and Hölder inequality, we have
(3.24)|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a4(q-12q-1)1-1/q[(∫01|f′(a)|q((2-t)/2)s|f′(b)|q(t/2)sdt)1/q≤b-a4(q-12q-1)1-1qb-a4+(∫01|f′(a)|q((1-t)/2)s|f′(b)|q((1+t)/2)sdt)1/q].
From (3.24) and (3.21) to (3.23), (3.18) holds. This completes the proof.If takings=1 in Theorem 3.3, we can derive the following corollary.Corollary 3.4.
Letf:I⊂ℝ+→ℝ+be differentiable on I∘,   a,b∈I,  with a<b,  and f'∈L([a,b]). If |f'(x)|q is geometrically convex and monotonically decreasing on [a,b]  for q>1, then
(3.25)|f(a+b2)-1b-a∫abf(x)dx|≤b-a4(q-12q-1)1-1/qG2(1,q;g3(α)),|f(a)+f(b)2-1b-a∫abf(x)dx|≤b-a4(q-12q-1)1-1/qG2(1,q;g3(α)),
where α(u,v), G2(s,q;g3(α)), and g3(α)are the same as in Theorem 3.3.


## 4. Application to Special Means

Let(4.1)A(a,b)=a+b2,L(a,b)=b-alnb-lna(a≠b),Lp(a,b)=(bp+1-ap+1(p+1)(b-a))1/p,a≠b,p∈R,p≠0,-1
be the arithmetic, logarithmic, generalized logarithmic means for a,b>0respectively.Letf(x)=xs/s,x∈(0,1],0<s<1,q≥1, and then the function
(4.2)|f′(x)|q=x(s-1)q
is monotonically decreasing on(0,1]. For λ∈[0,1], we have
(4.3)(s-1)q(λs-λ)≤0,(s-1)q((1-λ)s-(1-λ))≤0.
Hence, |f'(x)|qis  s-geometrically convex on (0,1] for 0<s<1.Theorem 4.1.
Let0<a<b≤1, 0<s<1, and q≥1. Then
(4.4)|[A(a,b)]s-[Ls(a,b)]s|≤(b-a)1-1/qs8(4s(1-s)qL(a,b))1/q×[a(s-1)/(2s){[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)-b(s-1)q/(2s)}1/q+b(s-1)/(2s){a(s-1)q/(2s)-[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)}1/q]|A(as,bs)-[Ls(a,b)]s|≤(b-a)1-1/qs8(4s(1-s)qL(a,b))1/q×[{a(s-1)q/(2s)-[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)}1/qb(s-1)/(2s){[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)-b(s-1)q/(2s)}1/q+a(s-1)/(2s){a(s-1)q/(2s)-[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)}1/q].
In particular, if q=1, one has
(4.5)|[A(a,b)]s-[Ls(a,b)]s|≤(b-a)s4[L(s-1)/(2s)-1(a,b)](s-1)/s-2[L(a,b)]2|A(as,bs)-[Ls(a,b)]s|≤(b-a)s4L(a,b){2[L(s-1)/s-1(a,b)](s-1)/s-1-[L(s-1)/(2s)-1(a,b)](s-1)/s-2L(a,b)}.Proof.
Letf(x)=xs/s,x∈(0,1],0<s<1. Then |f'(a)|=as-1>bs-1=|f'(b)|≥1 and
(4.6)|f′(a)|1/s[g1(α(q2s,q2s))]1/q=(b-a)-1/qa(s-1)/2s(2sL(a,b)(1-s)q)1/q{[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)-b(s-1)q/(2s)}1/q,|f′(a)f′(b)|1/(2s)[g2(α(q2s,q2s))]1/q=(b-a)-1/qb(s-1)/(2s)(2sL(a,b)(1-s)q)1/q{a(s-1)q/(2s)-[L(s-1)q/((2s)-1)(a,b)](s-1)q/(2s)-1L(a,b)}1/q,|f′(a)f′(b)|1/(2s)[g1(α(q2s,q2s))]1/q=(b-a)-1/qb(s-1)/(2s)(2sL(a,b)(1-s)q)1/q{[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)-b(s-1)q/(2s)}1/q,|f′(a)|1/s[g2(α(q2s,q2s))]1/q=(b-a)-1/qa(s-1)/(2s)(2sL(a,b)(1-s)q)1/q[a(s-1)q/(2s)-[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)]1/q.
By Theorem 3.1, Theorem 4.1 is thus proved.Theorem 4.2.
Let0<a<b≤1,s∈(0,1), and q>1. Then one has
(4.7)|[A(a,b)]s-[Ls(a,b)]s|≤(b-a)s2(q-12q-1)1-1/qA(a(s-1)/(2s),b(s-1)/(2s))×[[L(s-1)q/(2s)-1(a,b)](s-1)q/2s-1L(a,b)]1/q|A(as,bs)-[Ls(a,b)]s|≤(b-a)s2(q-12q-1)1-1/qA(a(s-1)/(2s),b(s-1)/(2s))×[[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b)]1/q.Proof.
Letf(x)=xs/s,x∈(0,1],0<s<1. Then |f'(a)|=as-1>bs-1=|f'(b)|≥1 and
(4.8)g3(α(q2s,q2s))=a-(s-1)q/(2s)[L(s-1)q/(2s)-1(a,b)](s-1)q/(2s)-1L(a,b).
Using Theorem 3.3, Theorem 4.2 is thus proved.



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*Source: 560586-2012-09-23.xml*