In this paper, we present a new reverse Mulholland-type inequality with multi-parameters and deal with its equivalent forms. Based on the obtained inequalities, the equivalent statements of the best possible constant factor related to several parameters are discussed. As an application, some interesting inequalities for double series are derived from the special cases of our main results.
Citation: Bicheng Yang, Shanhe Wu, Aizhen Wang. A new reverse Mulholland-type inequality with multi-parameters[J]. AIMS Mathematics, 2021, 6(9): 9939-9954. doi: 10.3934/math.2021578
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In this paper, we present a new reverse Mulholland-type inequality with multi-parameters and deal with its equivalent forms. Based on the obtained inequalities, the equivalent statements of the best possible constant factor related to several parameters are discussed. As an application, some interesting inequalities for double series are derived from the special cases of our main results.
In 1925, Hardy [1] gave a generalization of the Hilbert's inequality by introducing one pair of conjugated exponents (p,q) which satisfies 1p+1q=1 and p>1, as follows:
If am⩾0,bn⩾0, 0<∑∞m=1apm<∞ and 0<∑∞n=1bqn<∞, then
∞∑m=1∞∑n=1ambnm+n<πsin(π/p)(∞∑m=1apm)1/p (∞∑n=1bqn)1/q, | (1) |
where the constant factor πsin(π/p) is the best possible. The inequality (1) is called the Hardy-Hilbert inequality. In particular, when p=q=2, the Hardy-Hilbert inequality reduces to the Hilbert's inequality (see [2]). As is known to us, the Hardy-Hilbert inequality plays an important role in analysis number theory, real analysis and divergent series theory (see [3]).
For the continuous case, the integral version of Hardy-Hilbert inequality can be stated as follows (see [3], Theorem 316):
If f(x)⩾0,g(y)⩾0,0<∫∞0fp(x)dx<∞ and 0<∫∞0gq(y)dy<∞, then
∫∞0∫∞0f(x)g(y)x+ydxdy<πsin(π/p)(∫∞0fp(x)dx)1/p (∫∞0gq(y)dy)1/q, | (2) |
where the constant factor πsin(π/p) is the best possible.
The Hardy-Hilbert inequalities (1) and (2) has been studied extensively, numerous variants, generalizations, and extensions can be found in the literatures (see [4,5,6,7,8]).
Motivated by the Hardy-Hilbert inequality, in 1929, Mulholland [9] proposed a similar version of inequality (1), which contains the same best possible constant factor πsin(π/p) as in (1), i.e.,
∞∑m=2∞∑n=2ambnmnlnmn<πsin(π/p)(∞∑m=21mapm)1/p (∞∑n=21nbqn)1/q. | (3) |
Obviously, the Mulholland's inequality (3) can be rewritten in an equivalent form as:
∞∑m=2∞∑n=2ambnlnmn<πsin(π/p)(∞∑m=21m1−papm)1/p (∞∑n=21n1−qbqn)1/q. | (4) |
In recent years, the Mulholland inequality has been generalized by various methods of constructing parameters, see [10,11,12,13,14,15] and the references cited therein.
Let us recall some results which are connected with the current investigation. Hong and Wen [16] and Hong [17] studied the Hilbert type series inequalities and the Hilbert type integral inequalities with homogeneous kernel, respectively. They established the necessary and sufficient condition for which the inequalities hold under the best constant factor. Subsequently, by using the quasi-homogeneous kernels instead of the homogeneous kernel in the Hilbert type integral inequalities, Hong, He and Yang [18] established the necessary and sufficient condition for which the inequalities hold under the best constant factor. Recently, with the help of the Euler-Maclaurin summation formula, Yang, Wu and Liao [19] and gave the extension of Hardy-Hilbert's inequality and its equivalent forms. In [20], Yang, Wu and Chen investigated the generalization of Hardy-Littlewood-Polya's inequality and its equivalent forms.
In this paper, following the ideas of [16,17,18,19,20], we will study the Mulholland-type inequalities. The present research objects are structurally different from Hilbert type inequalities, and this will involve new techniques in dealing with the inequalities. Specifically, we will establish a reverse Mulholland-type inequality with multi-parameters. And then, we discuss the equivalent statements of the best possible constant factor associated with several parameters. Finally, we illustrate that some new inequalities of Mulholland-type can be derived from the equivalent expressions of the reverse Mulholland-type inequality.
In this section, we present some preliminary results which are essential for establishing our main results in subsequent sections. We begin with introducing the notations k(η)λ(x,y) and kλ,η(γ) with their associated formulas.
(i) In view of the following expression (see [21])
cotx=1x+∞∑k=1(1x−πk+1x+πk)(x∈(0,π)), |
for b∈(0,1), we have
Ab:=∫∞0ub−1 1−udu=∫10ub−1 1−udu+∫∞1ub−1 1−udu=∫10ub−1 1−udu−∫10v−b1−vdv=∫10ub−1 −u−b1−udu=∫10∞∑k=0(uk+b−1 −uk−b)du=∞∑k=0∫10(uk+b−1 −uk−b)du=∞∑k=0(1k+b−1k+1−b)=π[1πb+∞∑k=1(1πb−πk+1πb+πk)]=πcotπb∈R:=(−∞,+∞). |
Moreover, it is easy to observe that Ab>0 for b∈(0,12); Ab<0 for b∈(12,1); Ab=0 for b=12.
(ii) For λ,η>0, we set the homogeneous function of order −λ as follows:
k(η)λ(x,y):=xη−yηxλ+η−yλ+η(x,y>0), |
which satisfies k(η)λ(ux,uy)=u−λk(η)λ(x,y)(u,x,y>0), and k(η)λ(v,v):=η(λ+η)vλ(v>0).
It follows that k(η)λ(x,y) is a positive and continuous function with respect to x,y>0. For x≠y, we obtain
∂∂xk(η)λ(x,y)=−xη−1 (xλ+η−yλ+η)−2φ(x,y), |
where φ(x,y):=λxλ+η−(λ+η)yηxλ+ηyλ+η(x,y>0).
We find that for 0<x<y,
∂∂xφ(x,y)=λ(λ+η)xλ−1 (xη−yη)<0; |
for x>y, ∂∂xφ(x,y)>0. It follows that φ(x,y) is strictly decreasing (resp. increasing) with respect to x<y (resp. x>y). Since φ(y,y)=minx>0φ(x,y)=0(y>0), then φ(x,y)>0(x≠y), namely, ∂∂xk(η)λ(x,y)<0 (x≠y). Note that k(η)λ(x,y) is continuous at x=y, we conform that k(η)λ(x,y) is strictly decreasing with respect to x>0. In the same way, we can show that k(η)λ(x,y) is also strictly decreasing with respect to y>0.
(iii) For γ∈(0,λ), since k(η)λ(x,y)>0, by (i), we obtain (see [22])
kλ,η(γ):=∫∞0k(η)λ(u,1)uγ−1 du=∫∞0uη−1 uλ+η−1 uγ−1 duv=uλ+η=1λ+η(∫∞0vγλ+η−1 1−vdv−∫∞0vγ+ηλ+η−1 1−vdv)=πλ+η[cot(πγλ+η)−cot(π(γ+η)λ+η)]=πλ+η[cot(πγλ+η)+cot(π(λ−γ)λ+η)]∈R+:=(0,∞). |
In what follows, we suppose that
p<0(0<q<1),1p+1q=1,α,β,η>0,0<λ1,λ2<λ,λ1⩽1α, λ2⩽1β,k(η)λ(lnαm,lnβn)=lnαηm−lnβηnlnα(λ+η)m−lnβ(λ+η)n(m,n∈N∖{1}:={2,3,⋯}); |
for γ=λ1,λ−λ2(∈(0,λ)),
kλ,η(γ)=πλ+η[cot(πγλ+η)+cot(π(λ−γ)λ+η)]. | (5) |
Also, we assume am,bn⩾0(m,n∈N∖{1}) such that
0<∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm<∞ and 0<∞∑n=2lnq[1−β(λ2p+λ−λ1q)]−1 nn1−pbqn<∞. |
Definition 1. We define the following weight functions:
ωλ,η(λ2,m):=lnα(λ−λ2)m∞∑n=2k(η)λ(lnαm,lnβn)lnβλ2−1 nn(m∈N∖{1}), | (6) |
ϖλ,η(λ1,n):=lnβ(λ−λ1)n∞∑m=2k(η)λ(lnαm,lnβn)lnαλ1−1 mm(n∈N∖{1}). | (7) |
Lemma 1. For βλ2≤1, we have
ωλ,η(λ2,m)<1βkλ,η(λ−λ2)∈R+(m∈N∖{1}); | (8) |
For αλ1⩽1, we have
0<1αkλ,η(λ1)(1−θλ,η(λ1,n))<ϖλ,η(λ1,n)<1αkλ,η(λ1)(n∈N∖{1}), | (9) |
where θλ,η(λ1,n)(>0) is indicated as
θλ,η(λ1,n):=1kλ,η(λ1)∫lnα2lnβn0k(η)λ(u,1)uλ1−1 du=O(1lnβλ1n). | (10) |
Proof. For fixed m∈N∖{1},βλ2−1≤0,k(η)λ(lnαm,lnβt)lnβ2−1tt is a strictly decreasing function with respect to t>1. By the decreasing property of series, setting u=lnαmlnβt, we find that
ωλ,η(λ2,m)<lnα(λ−λ2)m∫∞1k(η)λ(lnαm,lnβt)lnβλ2−1 ttdt=1β∫∞0k(η)λ(u,1)u(λ−λ2)−1 du=1βkλ,η(λ−λ2). |
Hence, we get the inequality (8).
For αλ1−1⩽0, it is evident that k(η)λ(lnαt,lnβn)lnαλ1−1 tt is a strictly decreasing function with respect to t>1. By the decreasing property of series, setting u=lnαtlnβn, we find that
ϖλ,η(λ1,n)<lnβ(λ−λ1)n∫∞1k(η)λ(lnαt,lnβn)lnαλ1−1 ttdt=1α∫∞0k(η)λ(u,1)uλ1−1 du=1αkλ,η(λ1). |
By the decreasing property of series and (10), we obtain
ϖλ,η(λ1,n)>lnβ(λ−λ1)n∫∞2k(η)λ(lnαt,lnβn)lnαλ1−1 ttdt=1α∫∞lnα2lnβnk(η)λ(u,1)uλ1−1 du=1αkλ,η(λ1)(1−θλ,η(λ1,n))>0, |
0<θλ,η(λ1,n)⩽1kλ,η(λ1)∫lnα2lnβn0uλ1−1 du=1λ1kλ,η(λ1)(lnα2lnβn)λ1. |
Hence, we deduce the inequality (9). The Lemma 1 is proved.
Lemma 2. We have the following inequality:
I:=∞∑n=2∞∑m=2k(η)λ(lnαm,lnβn)ambn |
>1β1/p α1/qk1pλ,η(λ−λ2)k1qλ,η (λ1){∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm}1p |
×{∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn}1q . | (11) |
Proof. By the reverse Hӧlder inequality [23], we obtain
I=∞∑n=2∞∑m=2k(η)λ(lnαm,lnβn)[ln(βλ2−1)pnn1/p ln(1−αλ1)/qmm−1/qam] |
×[ln(αλ1−1)/qmm1/qln(1−βλ2)/pnn−1/pbn]⩾{∞∑m=2[lnαλ1m∞∑n=2k(η)λ(lnαm,lnβn)lnβλ2−1 nn]lnp(1−αλ1)−1 mm1−papm}1p |
×{∞∑n=2[lnβλ2n∞∑m=2k(η)λ(lnαm,lnβn)lnαλ1−1 mm]lnq(1−βλ2)−1 nn1−qbqn}1q |
={∞∑m=2ωλ,η(λ2,m)lnp[1−α(λ−λ2p+λ1q)]−1 mm1−papm}1p{∞∑n=2ϖλ,η(λ1,n)lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn}1q . |
By (8) and (9), we get the inequality (11). This completes the proof of Lemma 2.
Remark 1. By (11), for λ1+λ2=λ, we find
0<∞∑m=2lnp(1−αλ1)−1 mm1−p apm<∞,0<∞∑n=2lnq(1−βλ2)−1 nn1−pbqn<∞, |
and obtain the following inequality:
∞∑n=2∞∑m=2k(η)λ(lnαm,lnβn)ambn |
>1β1/p α1/qkλ,η(λ1)[∞∑m=2lnp(1−αλ1)−1 mm1−p apm]1p[∞∑n=2(1−θλ,η(λ1,n))lnq(1−βλ2)−1 nn1−qbqn]1q . | (12) |
In particular, for α=β=1, we have
˜θλ,η(λ1,n):=1kλ,η(λ1)∫ln2lnn0k(η)λ(u,1)uλ1−1 du=O(1lnλ1n), |
0<∞∑m=2lnp(1−λ1)−1 mm1−p apm<∞,0<∞∑n=2lnq(1−λ2)−1 nn1−pbqn<∞, |
and the following inequality:
∞∑n=2∞∑m=2lnηm−lnηnlnλ+ηm−lnλ+ηnambn |
>kλ,η(λ1)[∞∑m=2lnp(1−λ1)−1 mm1−p apm]1p[∞∑n=2(1−˜θλ,η(λ1,n))lnq(1−λ2)−1 nn1−qbqn]1q . | (13) |
Hence, inequality (12) is an extension of inequality (13).
Lemma 3. For any ε>0, we have
L:=∞∑n=2O(1nlnβ(λ1+ε)+1n)=O(1). | (14) |
Proof. There exist constants m>0,M>0 such that
0<m∞∑n=21nlnβ(λ1+ε)+1n⩽L⩽M[12lnβ(λ1+ε)+1 2+∞∑n=31nlnβ(λ1+ε)+1 n]. |
By the decreasing property of series, it follows that
0<L⩽M[12lnβ(λ1+ε)+12+∫∞21ylnβ(λ1+ε)+1ydy]=M[12lnβ(λ1+ε)+12+1β(λ1+ε)ln−β(λ1+ε)2]⩽M(12lnβλ1+12+1βλ1ln−βλ12)<∞. |
Hence, inequality (14) follows. The proof of Lemma 3 is complete.
Lemma 4. For δ0:=12mini=1,2{λi,λ−λi},0<ε<qδ0, we have
kλ,η(λ1+εq)→kλ,η(λ1)(ε→0+). | (15) |
Proof. By Levi Theorem (see [24]), it follows that
∫10k(η)λ(u,1)uλ1+εq−1 du→∫10k(η)λ(u,1)uλ1−1 du(ε→0+). |
For 0<ε<qδ0, we have 0<εq<δ0, and then λ1+δ0<λ,λ−λ1+δ0<λ,
0⩽∫∞1k(η)λ(u,1)uλ1+εq−1 du⩽∫∞1k(η)λ(u,1)uλ1+δ0−1 du⩽kλ,η(λ1+δ0)<∞. |
By using Lebesgue control convergence theorem [24], we have
∫∞1k(η)λ(u,1)uλ1+εq−1 du→∫∞1k(η)λ(u,1)uλ1−1 du(ε→0+). |
Then we find
kλ,η(λ1+εq)=∫10k(η)λ(u,1)uλ1+εq−1 du+∫∞1k(η)λ(u,1)uλ1+εq−1 du→∫10k(η)λ(u,1)uλ1−1 du+∫∞1k(η)λ(u,1)uλ1−1 du=kλ,η(λ1)(ε→0+). |
The Lemma 4 is proved.
Lemma 5. If λ1+λ2=λ, then the constant factor kλ,η(λ1) in (12) is the best possible.
Proof. For 0<ε<qδ0, we set
˜am:=lnα(λ1−εp)−1 mm,˜bn:=lnβ(λ2−εq)−1 nn(m,n∈N∖{1}). |
If there exists a constant M⩾1β1/p α1/qkλ,η(λ1) such that the inequalities (12) is valid when replacing 1β1/p α1/qkλ,η(λ1) by M, then, in particular, we have
˜I:=∞∑n=2∞∑m=2k(η)λ(lnαm,lnβn)˜am˜bn |
>M[∞∑m=2lnp(1−αλ1)−1 mm1−p˜ apm]1p[∞∑n=2(1−θλ,η(λ1,n))lnq(1−βλ2)−1 nn1−q˜bqn]1q . |
In view of (14), we obtain
˜I>M[∞∑m=2lnp(1−αλ1)−1 mm1−plnpαλ1−αε−pmmp]1p |
×[∞∑n=2(1−θλ,η(λ1,n))lnq(1−βλ2)−1 nn1−qlnqβλ2−βε−qnnq]1q =M(ln−αε−1 22+∞∑m=3ln−αε−1 mm)1p(∞∑n=2ln−βε−1 nn−∞∑n=2O(1lnβλ1n)ln−βε−1 nn)1q >M(ln−αε−1 22+∫∞2ln−αε−1 xxdx)1p(∫∞2ln−βε−1 yydy−∞∑n=2O(1nlnβ(λ1+ε)+1n))1q =Mε(εln−αε−1 22+1αln−αε2)1p(1βln−βε2−εO(1))1q . |
By (8), for ˆλ2=λ2−εq(∈(0,1β))(ˆλ1=λ1+εq<λ), we find
˜I=∞∑m=2[lnα(λ1+εq)m∞∑n=2k(η)λ(lnαm,lnβn)lnβ(λ2−εq)−1 nn]ln−αε−1 mm=∞∑m=2ωλ,η(ˆλ2,m)ln−αε−1 mm <1βkλ,η(ˆλ1)(ln−1−αε22+∞∑m=3ln−1−αεmm) |
⩽1βkλ,η(ˆλ1)(ln−1−αε22+∫∞2ln−1−αεxxdx)=1εβαkλ,η(ˆλ1)(εαln−1−αε22+ln−αε2). |
Then we have
1βαkλ,η(λ1+εq)(εαln−1−αε22+ln−αε2) |
⩾ε˜I>M(εln−αε−1 22+1αln−αε2)1p(1βln−ε2−εO(1))1q . |
For ε→0+, by (15), we get
1βαkλ,η(λ1)⩾M(1α)1p(1β)1q , |
namely, 1β1/p α1/qkλ,η(λ1)⩾M. Hence, M= 1β1/p α1/qkλ,η(λ1) is the best possible constant factor of (12). The proof of Lemma 5 is completed.
Remark 2. Setting ˜λ1:=λ−λ2p+λ1q=λ1+λ−λ1−λ2p,˜λ2:=λ−λ1q+λ2p, we find
˜λ1+˜λ2=λ−λ2p+λ1q+λ−λ1q+λ2p=λp+λq=λ. |
For |λ−λ1−λ2|<|p|δ0, it means that −δ0<λ−λ1−λ2p<δ0, 0<˜λ1<λ,0<˜λ2=λ−˜λ1<λ, and then kλ,η(˜λ1)∈R+. We reduce (11) as follows:
I>1β1/p α1/qk1pλ,η(λ−λ2)k1qλ,η (λ1)[∞∑m=2lnp(1−α˜λ1)−1 mm1−p apm]1p |
×[∞∑n=2(1−θλ,η(λ1,n))lnq(1−β˜λ2)−1 nn1−qbqn]1q . | (16) |
Lemma 6. If |λ−λ1−λ2|<|p|δ0, and the constant factor 1β1/p α1/qk1pλ,η(λ−λ2)k1qλ,η (λ1) in (11) is the best possible, then we have λ=λ1+λ2.
Proof. If the constant factor k1pλ,η(λ−λ2)k1qλ,η(λ1) in (11) (or (16)) is the best possible, then by (12) (for λi=˜λi(i=1,2)), we have
1β1/p α1/q k1pλ,η(λ−λ2)k1qλ,η (λ1)⩾ 1β1/p α1/qkλ,η(˜λ1)(∈R+), |
namely, kλ,η(˜λ1) ⩽k1pλ,η(λ−λ2)k1qλ,η (λ1).
By the reverse Hӧlder inequality, we find
kλ,η(˜λ1)=kλ,η(λ−λ2p+λ1q)=∫∞0k(η)λ(u,1)uλ−λ2p+λ1q−1 du=∫∞0k(η)λ(u,1)(uλ−λ2−1 p)(uλ1−1 q)du⩾(∫∞0k(η)λ(u,1)uλ−λ2−1 du)1p(∫∞0k(η)λ(u,1)uλ1−1 du)1q |
=k1pλ,η(λ−λ2)k1qλ,η (λ1). | (17) |
It follows that (17) keeps the form of equality.
We observe that (17) keeps the form of equality if and only if there exist constants A and B such that they are not both zero and
Auλ−λ2−1=Buλ1−1 a.e. in R+=(0,∞). |
Assuming that A≠0, it follows that uλ−λ2−λ1=ABa.e. in R+, and then λ−λ2−λ1=0. Namely, λ=λ1+λ2. This completes the proof of Lemma 6.
Theorem 1. Inequality (11) is equivalent to the following inequalities:
J:=[∞∑n=2lnpβ(λ−λ1q+λ2p)−1 n(1−θλ,η(λ1,n))p−1 n(∞∑m=2k(η)λ(lnαm,lnβn)am)p]1p |
>1β1/p α1/qk1pλ,η(λ−λ2)k1qλ,η (λ1){∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm}1p, | (18) |
J1:=[∞∑m=2lnqα(λ−λ2p+λ1q)−1 mm(∞∑n=2k(η)λ(lnαm,lnβn)bn)q]1q |
>1β1/p α1/qk1pλ,η(λ−λ2)k1qλ,η (λ1){∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn}1q . | (19) |
Proof. Suppose that (18) is valid. By the reverse Hӧlder inequality, we have
I=∞∑n=2[ln−1 p+β(λ−λ1q+λ2p)n(1−θλ,η(λ1,n))1/qn1/p ∞∑m=2k(η)λ(lnαm,lnβn)am]×[(1−θλ,η(λ1,n))1q ln1p−β(λ−λ1q+λ2p)nn−1/pbn] |
⩾J{∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn}1q . | (20) |
Then by (18), we obtain (11). On the other hand, assuming that (11) is valid, we set
bn:=lnpβ(λ−λ1q+λ2p)−1 n(1−θλ,η(λ1,n))p−1 n(∞∑m=2k(η)λ(lnαm,lnβn)am)p−1 ,n∈N∖{1}. |
It follows that
Jp=∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn=I. | (21) |
If J=∞, then (18) is naturally valid; if J=0, then it is impossible that makes (18) valid, namely, J>0. Suppose that 0<J<∞. By (11), we find
∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn=Jp=I |
>k1pλ,η(λ−λ2)k1qλ,η (λ1){∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm}1pJp−1 , |
J={∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn}1p>k1pλ,η(λ−λ2)k1qλ,η (λ1){∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm}1p, |
thus, (18) follows, which is equivalent to (11).
Suppose that (19) is valid. By the reverse Hӧlder inequality, we have
I=∞∑m=2[ln1q−α(λ−λ2p+λ1q)mm−1/qam][ln−1 q+α(λ−λ2p+λ1q)mm1/q∞∑n=2k(η)λ(lnαm,lnβn)bn] |
⩾{∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm}1pJ1. | (22) |
Then by (19), we obtain (11). On the other hand, assuming that (11) is valid, we set
am:=lnqα(λ−λ2p+λ1q)−1 mm(∞∑n=2k(η)λ(lnαm,lnβn)bn)q−1 ,m∈N∖{1}. |
It follows that
Jq1=∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm=I. | (23) |
If J1=∞, then (19) is naturally valid; if J1=0, then it is impossible that makes (19) valid, namely, J1>0. Suppose that 0<J1<∞. By (11), we have
∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm=Jq1=I |
>k1pλ,η(λ−λ2)k1qλ,η (λ1)Jq−11{∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn}1q , |
J1={∞∑m=2lnp[1−α(λ−λ2p+λ1q)]−1 mm1−p apm}1q |
>k1pλ,η(λ−λ2)k1qλ,η (λ1){∞∑n=2(1−θλ,η(λ1,n))lnq[1−β(λ−λ1q+λ2p)]−1 nn1−qbqn}1q , |
thereby, (19) follows, which is equivalent to (11). Hence, inequalities (11), (18) and (19) are equivalent.
This completes the proof of Theorem 1.
Theorem 2. Suppose that |λ−λ1−λ2|<|p|δ0. The following statements (i), (ii), (iii), (iv) and (v) are equivalent:
(i) Both k1pλ,η(λ−λ2)k1qλ,η (λ1) and kλ,η(λ−λ2p+λ1q) are independent of p,q;
(ii) k1pλ,η(λ−λ2)k1qλ,η (λ1) =kλ,η(λ−λ2p+λ1q);
(iii) λ=λ1+λ2;
(iv) 1β1/p α1/qk1pλ,η(λ−λ2)k1qλ,η (λ1) is the best possible constant factor of (11);
(v) 1β1/p α1/qk1pλ,η(λ−λ2)k1qλ,η (λ1) is the best possible constant factor of (18) (resp. (19)).
Moreover, if the statement (iii) follows, namely, λ=λ1+λ2, then we have (12) and the following equivalent inequalities with the best possible constant factor 1β1/p α1/q kλ,η(λ1) :
[∞∑n=2lnpβλ2−1 n(1−θλ,η(λ1,n))p−1 n(∞∑m=2k(η)λ(lnαm,lnβn)am)p]1p |
>1β1/p α1/qkλ,η(λ1)[∞∑m=2lnp(1−αλ1)−1 mm1−p apm]1p, | (24) |
[∞∑m=2lnqαλ1−1 mm(∞∑n=2k(η)λ(lnαm,lnβn)bn)q]1q |
>1β1/p α1/qkλ,η(λ1)[∞∑n=2(1−θλ,η(λ1,n))lnq(1−βλ2)−1 nn1−qbqn]1q . | (25) |
Proof. (i)⇒(ii). By (i), we have
k1pλ,η(λ−λ2)k1qλ,η (λ1)=limp→−∞limq→1−k1pλ,η(λ−λ2)k1qλ,η (λ1)=kλ,η(λ1). |
Then by (i), (15) and the above result, we find
kλ,η(λ−λ2p+λ1q)=limp→−∞kλ,η(λ1+λ−λ1−λ2p)=kλ,η(λ1)= k1pλ,η(λ−λ2)k1qλ,η (λ1). |
(ii) ⇒ (iii). by (ii), (17) keeps the form of equality. In view of the proof of Lemma 6, it follows that λ=λ1+λ2.
(iii) ⇒ (i). By (iii), for λ=λ1+λ2, both k1pλ,η(λ−λ2)k1qλ,η (λ1) and kλ,η(λ−λ2p+λ1q) are equal to kλ,η(λ1), which are independent of p,q. Hence, we have (i) ⇔ (ii) ⇔ (iii).
(iii) ⇔ (iv). By Lemma 5 and Lemma 6, we have the conclusions.
(iv) ⇔ (v). If the constant factor in (11) is the best possible, then so is the constant factor in (18) (resp. (19)). Otherwise, by (20) (resp. (22)), we would reach a contradiction that the constant factor in (11) is not the best possible. On the other hand, if the constant factor in (18) (resp. (19)) is the best possible, then so is the constant factor in (11). Otherwise, by (21) (resp. (23)), we would reach a contradiction that the constant factor in (18) (resp. (19)) is not the best possible.
Hence, the statements (i), (ii), (iii), (iv) and (v) are equivalent. The proof of Theorem 2 is complete.
In this section, we illustrate that some novel inequalities of Mulholland type can be derived from our main results as special cases.
Remark 3. Taking α=β=1 in (24) and (25) respectively, we have the following inequalities which is equivalent to (13) with the best possible constant factor kλ,η(λ1) :
[∞∑n=2lnpλ2−1 n(1−˜θλ,η(λ1,n))p−1 n(∞∑m=2lnηm−lnηnlnλ+ηm−lnλ+ηnam)p]1p |
>kλ,η(λ1)[∞∑m=2lnp(1−λ1)−1 mm1−p apm]1p, | (26) |
[∞∑m=2lnqλ1−1 mm(∞∑n=2lnηm−lnηnlnλ+ηm−lnλ+ηnbn)q]1q |
>kλ,η(λ1)[∞∑n=2(1−˜θλ,η(λ1,n))lnq(1−λ2)−1 nn1−qbqn]1q . | (27) |
In particular, putting λ=1,λ1=λ2=12 in (13), (26) and (27) respectively, we have the following equivalent reverse inequalities with the best possible constant factor 2π1+ηcotπ2(1+η) :
∞∑n=2∞∑m=2lnηm−lnηnln1+ηm−ln1+ηnambn |
>2π1+ηcotπ2(1+η)(∞∑m=2lnp2−1 mm1−p apm)1p[∞∑n=2(1−˜θ1,η(12,n))lnq2−1 nn1−qbqn]1q , | (28) |
[∞∑n=2lnp2−1 n(1−˜θ1,η(12,n))p−1 n(∞∑m=2lnηm−lnηnln1+ηm−ln1+ηnam)p]1p |
>2π1+ηcotπ2(1+η) |
(∞∑m=2lnp2−1 mm1−p apm)1p, | (29) |
[∞∑m=2lnq2−1 mm(∞∑n=2lnηm−lnηnln1+ηm−ln1+ηnbn)q]1q |
>2π1+ηcotπ2(1+η) |
[∞∑n=2(1−˜θ1,η(12,n))lnq2−1 nn1−qbqn]1q , | (30) |
where ˜θ1,η(12,n):=1+η2πtanπ2(1+η)∫ln2lnn0uη−1 u1+η−1 u−12du=O(1ln1/2n).
Choosing η=1 in (28), (29) and (30) respectively, we have the reverse of inequality (3) and the equivalent forms with the best possible constant factor π :
∞∑n=2∞∑m=21lnmnambn |
>π(∞∑m=2lnp2−1 mm1−p apm)1p[∞∑n=2(1−˜θ(n))lnq2−1 nn1−qbqn]1q , | (31) |
[∞∑n=2lnp2−1 n(1−˜θ(n))p−1 n(∞∑m=21lnmnam)p]1p >π (∞∑m=2lnp2−1 mm1−p apm)1p, | (32) |
[∞∑m=2lnq2−1 mm(∞∑n=21lnmnbn)q]1q >π [∞∑n=2(1−˜θ(n))lnq2−1nn1−qbqn]1q , | (33) |
where ˜θ(n):=˜θ1,1(12,n)=1π∫ln2lnn01u+1 u−12du=O(1ln1/2n)∈(0,1).
In this paper, by the use of the weight coefficients and the idea of introducing parameters, a new reverse Mulholland-type inequality with multi-parameters and the equivalent forms were given in Lemma 2 and Theorem 1. The equivalent statements of the best possible constant factor related to several parameters were obtained in Theorem 2. Some other inequalities associated with reverse Mulholland-type inequality were established in Remarks 1 and 3. The lemmas and theorems presented in this paper provided an extensive account of this type of inequalities.
This work was supported by the National Natural Science Foundation (No. 61772140), the Characteristic Innovation Project of Guangdong Provincial Colleges and Universities in 2020 (No. 2020KTSCX088), and the Natural Science Foundation of Fujian Province of China (No. 2020J01365). The authors would like to express sincere appreciation to the anonymous reviewers for their helpful comments and suggestions. All authors contributed equally to the manuscript and read and approved the final manuscript.
The authors declare that they have no competing interests.
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1. | Ling Peng, Rahela Abd Rahim, Bicheng Yang, A new reverse half-discrete Mulholland-type inequality with a nonhomogeneous kernel, 2023, 2023, 1029-242X, 10.1186/s13660-023-03025-w |