# A More Accurate Half-Discrete Hilbert Inequality with a Nonhomogeneous Kernel

Authors: Qiliang Huang; Bicheng Yang
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

By using the way of weight functions and the idea of introducing parameters, a more accurate half-discrete Hilbert inequality with a nonhomogeneous kernel and a best constant factor is presented. We also consider its best extension with the parameters, the equivalent forms, and the operator expressions as well as some reverses.


---

## Body

## 1. Introduction

Ifan,bn≥0,  a={an}n=1∞∈l2,  b={bn}n=1∞∈l2,  and  ∥a∥={∑n=1∞‍an2}1/2>0  and  ∥b∥={∑n=1∞‍bn2}1/2>0,  then we have the following discrete Hilbert inequality (cf. [1]):
(1)∑n=1∞‍∑m=1∞‍ambnm+n<π∥a∥∥b∥,
where the constant factorπis the best possible. Moreover, forf, g(≥0)∈L2(R+), ∥f∥={∫0∞‍f2(x)dx}1/2>0,  and  ∥g∥>0,  we have the following Hilbert integral inequality (cf. [2]):
(2)∬0∞f(x)g(y)x+ydxdy<π∥f∥∥g∥,
with the same best constant factorπ. Inequalities (1)-(2) are important in the analysis and its applications (cf. [3]). There are lots of improvements, generalizations, and applications of inequalities (1)-(2) for more details, refer to the literatures [4–15].We find a result on the half-discrete Hilbert-type inequality with the nonhomogeneous kernel, which were published early as follows (cf. [16, Theorem 351]):
(3)∫0∞‍xq-2(∑n=1∞‍an1+xn)qdx<Bq(1p,1q)∑n=1∞‍anq,
wherep>1,  q>1,  1/p+1/q=1, and  B(u,v)=∫0∞‍(tu-1dt/(1+t)u+v)(u,v>0) is the beta function. But the constant factors B(1/p,1/q)were not considered whether the best possible. Recently, Yang gave a half-discrete Hilbert inequality with the nonhomogeneous kernel as follows (cf. [17]):
(4)∑n=1∞‍an∫0∞‍f(x)1+xndx<π(∑n=1∞‍an2∫0∞‍f2(x)dx)1/2,
where the constant factor π is the best possible. And some half-discrete Hilbert-type inequalities with the nonhomogeneous kernel were given (cf. [18]).In this paper, by using the way of weight functions and the idea of introducing parameters and by means of Hadamard’s inequality, we give the more accurate inequalities of (3) and (4) as follows:
(5)∫0∞‍xq-2(∑n=1∞‍an1+xn-x/2)qdx<Bq(1p,1q)∑n=1∞‍anq,(6)∑n=1∞‍an∫0∞‍f(x)1+xn-x/2dx<π(∑n=1∞‍an2∫0∞‍f2(x)dx)1/2,
with the best constant factors B(1/p,1/q) and π. We also consider their best extensions with the parameters, the equivalent forms, and the operator expressions as well as some reverses.


## 2. Some Lemmas

In the following lemmas, we assumed thatα, β, λ>0,  α<λβ,  α≤min{1,2-β},  γ1∈(-∞,∞), and  0≤γ2≤1/2.Lemma 1.
Define the weight functions as follows:(7)ωα(n)≔(n-γ2)α∫γ1∞‍(x-γ1)α-1[1+(x-γ1)β(n-γ2)β]λdx(n∈N),(8)ϖα(x)≔(x-γ1)α∑n=1∞‍(n-γ2)α-1[1+(x-γ1)β(n-γ2)β]λ(x∈(γ1,∞)).
Setting k∶=k(α,β,λ)=(1/β)B(α/β,λ-α/β), one has the following inequalities:
(9)0<k(1-θ(x))<ϖα(x)<ωα(n)=k,
where θ(x)∶=(1/βk)∫0(x-γ1)β(1-γ2)β‍(uα/β-1/(1+u)λ)du∈(0,1)  and  θ(x)=O((x-γ1)α)  (x∈(γ1,∞)).Proof.
Puttingu=(x-γ1)β(n-γ2)β in (7), we have
(10)ωα(n)=1β∫0∞‍uα/β-1(1+u)λdu=1βB(αβ,λ-αβ)=k.
For fixed x∈(γ1,∞), setting
(11)f(t)∶=(x-γ1)α(t-γ2)α-1[1+(x-γ1)β(t-γ2)β]λ(t∈(γ2,∞)),
in view of the conditions, we find that f′(t)=-(1-α)(x-γ1)α(t-γ2)α-2/[1+(x-γ1)β(t-γ2)β]λ-λβ(x-γ1)α+β(t-γ2)α+β-2/[1+(x-γ1)β(t-γ2)β]λ+1<0  and  f′′(t)>0.  By the following Hadamard inequality (cf. [15]):
(12)f(n)<∫n-1/2n+1/2‍f(t)dt(n∈N),
and letting  u=(x-γ1)β(t-γ2)β,  it follows that
(13)ϖα(x)=∑n=1∞‍f(n)<∑n=1∞‍∫n-1/2n+1/2‍f(t)dt=∫1/2∞‍f(t)dt≤∫γ2∞‍f(t)dt=1β∫0∞‍uα/β-1(1+u)λdu=1βB(αβ,λ-αβ)=k,ϖα(x)=∑n=1∞‍f(n)>∫1∞‍f(t)dt=∫γ2∞‍f(t)dt-∫γ21‍f(t)dt=k-1β∫0(x-γ1)β(1-γ2)β‍uα/β-1(1+u)λdu=k(1-θ(x))>0,
where
(14)0<θ(x)=1βk∫0(x-γ1)β(1-γ2)β‍uα/β-1(1+u)λdu<1βk∫0(x-γ1)β(1-γ2)β‍uα/β-1du=(1-γ2)α(x-γ1)ααk.
Hence, we prove that (9) is valid.Lemma 2.
Suppose that1/p+1/q=1  (p≠0,1), and an≥0, and f(x)≥0 is a real measurable function in (γ1,∞). Then the following are considered. (i) Forp>1, one has the following:
(15)J∶={[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍∑n=1∞‍(n-γ2)pα-1×[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]p[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍}1/p≤k1/q{∫γ1∞‍ϖα(x)(x-γ1)p(1-α)-1fp(x)dx}1/p,(16)L1∶={[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍∫γ1∞‍(x-γ1)qα-1ϖαq-1(x)×[∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]qdx[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍}1/p≤{k∑n=1∞‍(n-γ2)q(1-α)-1anq}1/q,
where  ϖα(x) and ωα(n) are indicated by (7) and (8).
(ii) Forp<1  (p≠0), one has the reverses of (15) and (16).Proof.
(i) By (7)–(9) and Hölder’s inequality (cf. [15]), we find that
(17)[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]p={[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍∫γ1∞‍1[1+(x-γ1)β(n-γ2)β]λ×[(x-γ1)(1-α)/q(n-γ2)(1-α)/pf(x)]×[(n-γ2)(1-α)/p(x-γ1)(1-α)/q]dx[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍}p≤∫γ1∞‍1[1+(x-γ1)β(n-γ2)β]λ×(x-γ1)(1-α)(p-1)(n-γ2)1-αfp(x)dx×[[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍∫γ1∞‍1[1+(x-γ1)β(n-γ2)β]λ×(n-γ2)(1-α)(q-1)(x-γ1)1-αdx[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍]p-1=∫γ1∞‍fp(x)(x-γ1)(1-α)(p-1)[1+(x-γ1)β(n-γ2)β]λdx(n-γ2)1-α×[(n-γ2)q(1-α)-1ωα(n)]p-1=(n-γ2)1-pαkp-1×∫γ1∞‍fp(x)(x-γ1)(1-α)(p-1)[1+(x-γ1)β(n-γ2)β]λ1(n-γ2)1-αdx,(18)Jp≤kp-1∑n=1∞‍∫γ1∞‍fp(x)[1+(x-γ1)β(n-γ2)β]λ×(x-γ1)(1-α)(p-1)(n-γ2)1-αdx=kp-1∫γ1∞‍∑n=1∞‍(n-γ2)α-1[1+(x-γ1)β(n-γ2)β]λ×(x-γ1)α+p(1-α)-1fp(x)dx=kp-1∫γ1∞‍ϖα(x)(x-γ1)p(1-α)-1fp(x)dx.
Hence, (15) is valid. Using Hölder’s inequality again, we have
(19)[∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]q={[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍∑n=1∞‍1[1+(x-γ1)β(n-γ2)β]λ×[(x-γ1)(1-α)/q(n-γ2)(1-α)/p][(n-γ2)(1-α)/p(x-γ1)(1-α)/qan][∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]∑n=1∞‍}q≤[ϖα(x)(x-γ1)p(1-α)-1]q-1×∑n=1∞‍anq[1+(x-γ1)β(n-γ2)β]λ(n-γ2)(1-α)q/p(x-γ1)1-α=ϖαq-1(x)(x-γ1)1-qα×∑n=1∞‍(n-γ2)(1-α)(q-1)[1+(x-γ1)β(n-γ2)β]λ1(x-γ1)1-αanq,L1q≤∫γ1∞‍∑n=1∞‍1[1+(x-γ1)β(n-γ2)β]λ×(n-γ2)(1-α)(q-1)(x-γ1)1-αanqdx=∑n=1∞‍[(n-γ2)α∫γ1∞‍(x-γ1)α-1[1+(x-γ1)β(n-γ2)β]λdx]×(n-γ2)q(1-α)-1anq=∑n=1∞‍ωα(n)(n-γ2)q(1-α)-1anq=k∑n=1∞‍(n-γ2)q(1-α)-1anq.
Hence, (16) is valid.
(ii) For0<p<1  (q<0) or p<0(0<q<1), using the reverse Hölder inequality and in the same way, we have the reverses of (15) and (16).


## 3. Main Results

Some functions and spaces are introduced as follow:(20)ϕ(x)∶=(x-γ1)p(1-α)-1,ϕ~(x)∶=(1-θ(x))ϕ(x),ψ(n)≔(n-γ2)q(1-α)-1(x∈(γ1,∞),n∈N,θ(x)isindicatedasLemma1),Lp,ϕ(γ1,∞)∶={f;∥f∥p,ϕ={∫γ1∞‍ϕ(x)|f(x)|pdx}1/p<∞},lq,ψ∶={a={an}n=1∞;∥a∥q,ψ={∑n=1∞‍ψ(n)|an|q}1/q<∞}.Note 3.
Ifp>1, then Lp,ϕ(γ1,∞) and lq,ψ are normal spaces; if 0<p<1 or p<0, then both Lp,ϕ(γ1,∞) and lq,ψ are not normal spaces, but we still use the formal symbols in the following.Theorem 3.
Suppose thatp>1,  1/p+1/q=1,  α,β,λ>0,  α<λβ,  α≤min{1,2-β},  γ1∈(-∞,∞),  0≤γ2≤1/2,  and f(x),an≥0, such thatf∈Lp,ϕ(γ1,∞),  a={an}n=1∞∈lq,ψ,  ∥f∥p,ϕ>0,  and  ∥a∥q,ψ>0, then we have the following equivalent inequalities:
(21)I∶=∑n=1∞‍an∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx=∫γ1∞‍f(x)∑n=1∞‍andx[1+(x-γ1)β(n-γ2)β]λ<k∥f∥p,ϕ∥a∥q,ψ,(22)J={∑n=1∞‍(n-γ2)pα-1[∫γ1∞‍f(x)dx[1+(x-γ1)β(n-γ2)β]λ]p}1/p<k∥f∥p,ϕ,(23)L={∫γ1∞‍(x-γ1)qα-1[∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]qdx}1/q<k∥a∥q,ψ,
where the constant factor k=(1/β)B(α/β,λ-α/β) is the best possible.Proof.
By the Lebesgue term-by-term integration theorem, we find that there are two expressions ofI in (21). In view of (15), ϖα(x)<k, and 0<∥f∥p,ϕ<∞, we have (22). By Hölder’s inequality, we find that
(24)I=∑n=1∞‍[(n-γ2)α-1/p∫γ1∞‍f(x)dx[1+(x-γ1)β(n-γ2)β]λ]∑n=1∞‍×[(n-γ2)1/p-αan]≤J{∑n=1∞‍(n-γ2)q(1-α)-1anq}1/q=J∥a∥q,ψ.
Hence, (21) is valid by (22). On the other hand, assuming that (21) is valid and setting
(25)an∶=(n-γ2)pα-1[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]p-1(n∈N),
then we have
(26)∥a∥q,ψq=∑n=1∞‍(n-γ2)q(1-α)-1anq=Jp=I.
By (15), it follows that J<∞. If J=0, then (22) is trivially valid. If J>0, then 0<∥a∥q,ψ=Jp-1<∞. By (21), we have
(27)∥a∥q,ψq=Jp=I<k∥f∥p,ϕ∥a∥q,ψ,thatis,J=∥a∥q,ψq-1<k∥f∥p,ϕ.
Hence, (22) is valid, which is equivalent to (21).
In view of (16) and ϖα(x)<k, we obtain (23). By Hölder’s inequality again, we have
(28)I=∫γ1∞‍[(x-γ1)α-1/q∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]×[(x-γ1)1/q-αf(x)]dx≤L{∫γ1∞‍(x-γ1)p(1-α)-1fp(x)dx}1/p=L∥f∥p,ϕ.
Then (21) is valid by using (23). Supposing that (21) is valid and setting
(29)f(x)∶=(x-γ1)qα-1[∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]q-1(x∈(γ1,∞)),
then it follows that
(30)∥f∥p,ϕp=∫γ1∞‍(x-γ1)p(1-α)-1fp(x)dx=Lq=I.
In view of (16) and (9), we find L<∞. If L=0, then (23) is trivially valid; if L>0, that is,0<∥f∥p,ϕ<∞, then, by (21), we have
(31)∥f∥p,ϕp=Lq=I<k∥f∥p,ϕ∥a∥q,ψ,thatis,L=∥f∥p,ϕp-1<k∥a∥q,ψ.
Hence (23) is valid, which is equivalent to (21). Then (21), (22), and (23) are equivalent. We prove that the constant factor in (21) is the best possible. For 0<ε<qα, setting a~={a~n}n=1∞ and f~(x) as follows (p,q>1):
(32)a~n=(n-γ2)α-ε/q-1,f~(x)={(x-γ1)α+ε/p-1,x∈(γ1,1+γ1),0,x∈[1+γ1,∞),
if there exists a positive number k′≤k, such that (21) is still valid as we replace k by k′, then, substitution of a~ and f~(x), we have
(33)I~∶=∑n=1∞‍a~n∫γ1∞‍f~(x)[1+(x-γ1)β(n-γ2)β]λdx<k′∥f~∥p,ϕ∥a~∥q,ψ.
Forp, q>1, letting α~=α-ε/q, we find by Lemma 1 that
(34)I~=∫γ1γ1+1‍(x-γ1)ε-1(x-γ1)α-ε/q×∑n=1∞‍(n-γ2)α-ε/q-1dx[1+(x-γ1)β(n-γ2)β]λ=∫γ1γ1+1‍(x-γ1)ε-1ϖα~(x)dx>k(α~,β,λ)∫γ1γ1+1‍(x-γ1)ε-1[1-O((x-γ1)α~)]dx=k(α-εq,β,λ)(1ε-O(1)).
By simple calculation, we have
(35)∥a~∥q,ψq=∑n=1∞‍(n-γ2)q(1-α)-1a~nq=∑n=1∞‍(n-γ2)-1-ε<(1-γ2)-1-ε+∫1∞‍(y-γ2)-1-εdy=ε+1-γ2ε(1-γ2)ε+1,(36)∥f~∥p,ϕ={∫γ1γ1+1‍(x-γ1)ε-1dx}1/p=(1ε)1/p.
In view of (33), (34), and q>1, it follows that
(37)k(α-εq,β,λ)[1-εO(1)]<εI~<k′∥f~∥p,ϕ∥a~∥q,ψ<k′[ε+1-γ2(1-γ2)ε+1]1/q.
Then we have k(α,β,λ)=k≤k′ for ε→0+. Hence, k′=k is the best value of (21). By the equivalency, the constant factor k in (22) ((23)) is the best possible. Otherwise, we would reach a contradiction by (24) ((28)) that the constant factor in (21) is not the best possible.Remark 4.
(i) Define a half-discrete Hilbert operatorT:Lp,ϕ(γ1,∞)→lp,ψ1-p as follows. For f∈Lp,ϕ(γ1,∞), we define Tf∈lp,ψ1-p, satisfying
(38)Tf(n)=∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx(n∈N).
Then by (22), it follows that ∥Tf∥p,ψ1-p≤k∥f∥p,ϕ; that is, T is the bounded operator with ∥T∥≤k. Since the constant factorkin (22) is the best possible, then we have ∥T∥=k=(1/β)B(α/β,λ-α/β).
(ii) Define a half-discrete Hilbert operatorT~:lq,ψ→Lq,ϕ1-q(γ1,∞) in the following way. For a∈lq,ψ, we define T~a∈Lq,ϕ1-q, satisfying
(39)T~a(x)=∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ(x∈(γ1,∞)).
Then by (23), it follows that ∥T~a∥q,ϕ1-q≤k∥a∥q,ψ; that is, T~ is the bounded operator with ∥T~∥≤k. Since the constant factor k in (23) is the best possible, then we have ∥T~∥=k=(1/β)B(α/β,λ-α/β).Theorem 5.
Suppose that0<p<1, 1/p+1/q=1, α,β,λ>0, α<λβ, α≤min{1,2-β}, γ1∈(-∞,∞), 0≤γ2≤1/2, and  f(x),an≥0, such that f∈Lp,ϕ~(γ1,∞), a={an}n=1∞∈lq,ψ, ∥f∥p,ϕ~>0, and ∥a∥q,ψ>0. Then one has the following equivalent inequalities:
(40)I=∑n=1∞‍an∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx=∫γ1∞‍f(x)∑n=1∞‍andx[1+(x-γ1)β(n-γ2)β]λ>k∥f∥p,ϕ~∥a∥q,ψ,(41)J={∑n=1∞‍(n-γ2)pα-1[∫γ1∞‍f(x)[1+(x-γ1)β(n-γ2)β]λdx]p}1/p>k∥f∥p,ϕ~,(42)L~∶={∫γ1∞‍(x-γ1)qα-1[1-θ(x)]q-1[∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]qdx}1/q>k∥a∥q,ψ,
where the constant factor k=(1/β)B(α/β,λ-α/β) is the best possible.Proof.
In view of (9), the reverse of (15), and 0<∥f∥p,ϕ~<∞, we have (41). Using the reverse Hölder inequality, we obtain the reverse form of (24) as follows:
(43)I≥J∥a∥q,ψ.
Then by (41), (40) is valid.
On the other hand, if (40) is valid, setting an as in (25), then (26) still holds with 0<p<1. By (40), it follows that J>0. If J=∞, then (41) is trivially valid; if J<∞, then 0<∥a∥q,ψ=Jp-1<∞, and we have
(44)∥a∥q,ψq=Jp=I>k∥f∥p,ϕ~∥a∥q,ψ,thatis,J=∥a∥q,ψq-1>k∥f∥p,ϕ~.
Then (41) is valid, which is equivalent to (40).
By the reverse of (16) and ϖ(x)>k(1-θ(x)), and by q<0, we have
(45)L~>k(q-1)/qL1≥k(q-1)/q{k∑n=1∞‍(n-γ2)q(1-α)-1anq}1/q=k∥a∥q,ψ.
Hence, (42) is valid. By the reverse Hölder inequality again, we obtain
(46)I=∫γ1∞‍[(x-γ1)α-1/q(1-θ(x))1/p∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]×[(1-θ(x))1/p(x-γ1)1/q-αf(x)]dx≥L~∥f∥p,ϕ~.
Then (40) is valid by (42). On the other hand, if (40) is valid, setting
(47)f(x)=(x-γ1)qα-1[1-θ(x)]q-1[∑n=1∞‍an[1+(x-γ1)β(n-γ2)β]λ]q-1(x∈(γ1,∞)),
then ∥f∥p,ϕ~p=∫γ1∞‍[1-θ(x)](x-γ1)p(1-α)-1fp(x)dx=L~q=I. By the reverse of (16), we have L~>0. If L~=∞, then (42) is trivially valid; if 0<L~<∞, then, by (40), we obtain
(48)∥f∥p,ϕ~p=L~q=I>k∥f∥p,ϕ~∥a∥q,ψ,thatis,L~=∥f∥p,ϕ~p-1>k∥a∥q,ψ.
Hence, (42) is valid, which is equivalent to (40). It follows that (40), (41), and (42) are equivalent.
If there exists a positive numberk′≥k, such that (40) is still valid as we replace k by k′, then, in particular, we have
(49)I~=∑n=1∞‍a~n∫γ1∞‍f~(x)[1+(x-γ1)β(n-γ2)β]λdx>k′∥f~∥p,ϕ~∥a~∥q,ψ,
where a~={a~n}n=1∞ and f~ are taken as in (32) (0<ε<p(βλ-α)). Since, by (35), we find in
(50)∥f~∥p,ϕ~={∫γ11+γ1‍[1-O((x-γ1)α)](x-γ1)ε-1dx}1/p=(1ε-O(1))1/p,I~≤∑n=1∞‍(n-γ2)α-ε/q-1∫γ1∞‍(x-γ1)α+ε/p-1[1+(x-γ1)β(n-γ2)β]λdx=1β∑n=1∞‍(n-γ2)-ε-1∫0∞‍uα/β+ε/βp-1(1+u)λdu≤1β[ε+1-γ2ε(1-γ2)ε+1]B(αβ+εβp,λ-αβ-εβp),
then, by (35), (49), and the above results, we obtain (notice thatq<0)
(51)1β[ε+1-γ2(1-γ2)ε+1]B(αβ+εβp,λ-αβ-εβp)≥εI~>εk′(1ε-O(1))1/p(ε+1-γ2ε(1-γ2)ε+1)1/q=k′(1-εO(1))1/p(ε+1-γ2(1-γ2)ε+1)1/q.
For ε→0+, we have k=(1/β)B(α/β,λ-α/β)≥k′. Hence, k=k′ is the best value of (40). By the equivalency, the constant factor k in (41) ((42)) is the best possible. Otherwise, we would reach a contradiction by (43) ((46)) that the constant factor in (40) is not the best possible.In the same way, forp<0, we also have the following result.Theorem 6.
If the assumption ofp>1 in Theorem 3 is replaced by p<0, then the reverses of (21), (22), and (23) are valid and equivalent. Moreover, the same constant factor is the best possible.Remark 7.
(i) Ifγ1=γ2=0,  λ=β=1, and  α=1/p, thenk=B(1/p,1/q) and (23) reduces to (3). In particular, for p=q=2, (21) reduces to (4).
(ii) Forγ1=0,  γ2=1/2, and  α=1/pin (21) and (23), it follows
(52)∑n=1∞‍an∫0∞‍f(x)(1+(xn-x/2)β)λdx<k{∫0∞‍xp-2fp(x)dx}1/p{∑n=1∞‍anq}1/q,(53)∫0∞‍xq-2(∑n=1∞‍an(1+(xn-x/2)β)λ)qdx<kq∑n=1∞‍anq,
where the constant factors both k=(1/β)B(1/pβ,λ-1/pβ) and kq are the best possibles. In particular, for λ=β=1, (53) comes to (5); for λ=β=1 and  p=q=2, in (52), we obtain (6). Hence, inequality (21) is the best extension of (4) and (6), and inequality (23) is the best extension of (3) and (5) with the parameters.



---

*Source: 628250-2013-06-24.xml*