# On a More Accurate Half-Discrete Mulholland-Type Inequality Involving One Multiple Upper Limit Function

Authors: Xianyong Huang; Bicheng Yang
Journal: Journal of Function Spaces
 (2021)

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## Abstract

By the use of the weight functions, the symmetry property, and Hermite-Hadamard’s inequality, a more accurate half-discrete Mulholland-type inequality involving one multiple upper limit function is given. The equivalent conditions of the best possible constant factor related to multiparameters are studied. Furthermore, the equivalent forms, several inequalities for the particular parameters, and the operator expressions are provided.


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## Body

## 1. Introduction

Suppose thatp>1,1/p+1/q=1,am,bn≥0,0<∑m=1∞amp<∞,and0<∑n=1∞bnq<∞. The following well-known Hardy-Hilbert’s inequality was provided (cf. [1], Theorem 315):
(1)∑m=1∞∑n=1∞ambnm+n<πsinπ/p∑m=1∞amp1/p∑n=1∞bnq1/q,where the constant factor π/sinπ/p is the best possible. Replacing 1/m+n and 1/m+n−1 in (1), we have a more accurate form of (1) (cf. [1], Theorem 323). With regard to the similar assumptions, the following Mulholland’s inequality was obtained (cf. [1], Theorem 343):
(2)∑m=2∞∑n=2∞ambnlnmn<πsinπ/p∑m=2∞mp−1amp1/p∑n=2∞nq−1bnq1/q.In 2006, by using Euler-Maclaurin’s summation formula and the technique of real analysis, Krnić and Pečarić [2] obtained an extension of (1) as follows:
(3)∑m=1∞∑n=1∞ambnm+nλ<Bλ1,λ2∑m=1∞mp1−λ1−1amp1/p∑n=1∞nq1−λ2−1bnq1/q,where λi∈0,2i=1,2, λ1+λ2=λ∈0,4, the constant factor Bλ1,λ2 in (3) is the best possible, and
(4)Bu,v=∫0∞tu−11+tu+vdtu,v>0is the beta function. For λ=1,λ1=1/q,λ2=1/p in (3), it reduces to (1); for p=q=2 and λ1=λ2=λ/2 in (3), it reduces to Yang’s inequality in [3].Using Abel’s partial sum formula and (2), in 2019, Adiyasuren et al. [4] provided a discrete Hardy-Hilbert’s inequality with the kernel as (3) involving two partial sums. We find that inequalities (1)–(3) played an important role in analysis and its applications (cf. [5–14]).In 2016, Hong and Wen [15] gave an equivalent condition of the best possible constant factor related to a few parameters in the general form of (1). In the same way, the other similar works of (1) and the integral analogues were given in [16–19].In 1934, a half-discrete Hilbert-type inequality was published in Hardy et al. [1] (Theorem 351) as follows: if Ktt>0 is decreasing, p>1,1/p+1/q=1,0<ϕs=∫0∞Ktts−1dt<∞, and an≥0, satisfying 0<∑n=1∞anp<∞, then, we have
(5)∫0∞xp−2∑n=1∞Knxanpdx<ϕp1q∑n=1∞anp.Some new extensions of (5) with their applications were provided by [20–24].In this article, following the way of [4, 15], by the use of the weight functions, the symmetry property, and Hermite-Hadamard’s inequality, a more accurate half-discrete Mulholland-type inequality with the kernel as 1/x+lnαn−ξλinvolving one multiple upper limit function is obtained. The equivalent conditions of the best possible constant factor related to multiparameters are given. Furthermore, the equivalent forms, several new inequalities for the particular parameters, and the operator expressions are considered. The lemmas and theorems of this paper provide an extensive account of this type of inequalities.


## 2. Some Lemmas

In what follows, we suppose thatp>1, 1/p+1/q=1, N≔1,2,⋯, ξ∈0,1/2, λ>0, λ1∈0,λ, λ2∈, 0,λ+m∩0,1/α, λ^1≔λ−λ2/p+λ1/q, λ^2≔λ−λ1/q+λ2/p. We also assume that fx≔F0x is a nonnegative Lebesgue integrable function in any interval 0,bb>0. Define a multiple upper limit function as follows: Fix≔∫0xFi−1tdtx≥0, satisfying
(6)Fi0=0,Fix=oetxt>0,x⟶∞;i=1,⋯,m,and for any Fmx,an≥0, it follows that
(7)0<∫0∞x−pm−1+λ∧1−1Fmpx<∞,0<∑n=2∞lnn−ξq1−αλ∧2−1n−ξq−1anq<∞.Lemma 1.
Assuming thats∈0,∞,s2∈0,s∩0,1/α,kss2≔Bs2,s−s2, define the following weight function:
(8)ϖs2,x≔αxs−s2∑n=2∞lnn−ξαs2−1x+lnαn−ξsn−ξx∈R+=0,∞.
The following inequality is valid:(9)ϖs2,x<kss2x∈R+.Proof.
For fixedx∈R+, the function lnt−ξαs2−1/x+lnαt−ξst−ξ is strictly decreasing and strictly convex in the interval 3/2,∞. In fact, for α∈0,1,s2∈0,1/α,ξ∈0,1/2,t∈3/2,∞, it follows that
(10)∂∂tlnt−ξαs2−1x+lnαt−ξst−ξ=−1−αs2lnt−ξαs2−2x+lnαt−ξst−ξ2−sαlnt−ξαs2+α−2x+lnαt−ξs+1t−ξ2−lnt−ξαs2−1x+lnαt−ξst−ξ2<0,∂2∂t2lnt−ξαs2−1x+lnαt−ξst−ξ=1−αs22−αs2lnt−ξαs2−3x+lnαt−ξst−ξ3+sα3−2αs2−εlnt−ξαs2+α−3x+lnαt−ξs+1t−ξ3+31−αs2lnt−ξαs2−2x+lnαt−ξst−ξ3+sα2s+1lnt−ξαs2+2α−3x+lnαt−ξs+2t−ξ3+3sαlnt−ξαs2+α−2x+lnαt−ξs+1t−ξ3+2lnt−ξαs2−1x+lnαt−ξst−ξ3>0.
Applying Hermite-Hadamard’s inequality (cf. [25]), we have
(11)∑n=2∞lnn−ξαs2−1x+lnαn−ξsn−ξ<∫3/2∞lnt−ξαs2−1x+lnαt−ξst−ξdt.
Settingv=lnαt−ξ/x1/t−ξdt=1/αx1/αv1/α−1dv, it follows that
(12)∫3/2∞lnt−ξαs2−1x+lnαt−ξst−ξdt=1αxs∫lnα3/2−ξ/x∞xv1/ααs2−11+vsx1/αv1/α−1dv≤1αxs−s2∫0∞vs2−1dv1+vs=1αxs−s2kss2.
By (8) and (11), we have the following inequality:
(13)ϖs2,x<αxs−s21αxs−s2kss2=kss2,namely, (9) follows.
The lemma is prove.Lemma 2.
Fors1∈0,s and kss1=Bs1,s−s1, if
(14)0<∫0∞xp1−s−s2/p+s1/q−1fpxdx<∞,(15)0<∑n=2∞lnn−ξq1−αs2/p+s−s1/q−1n−ξq−1anq<∞,then we have the following more accurate half-discrete Mulholland-type inequality:
(16)I0≔∫0∞∑n=2∞anx+lnαn−ξsfxdx<1αkss21/pkss11/q×∫0∞xp1−s−s2/p+s1/q−1fpxdx1/p∑n=2∞lnn−ξq1−αs2/p+s−s1/q−1n−ξq−1anq1/q.Proof.
Settingv=x/lnαn−ξ, we obtain the expression of the following another weight function:
(17)ωs1,n≔lnn−ξαs−s1∫0∞xs1−1dxx+lnαn−ξs=∫0∞vs1−1dvv+1s=kss1n∈N\1.
Using Hölder’s inequality (cf. [25]), it follows that
(18)I0=∫0∞∑n=2∞1x+lnαn−ξsx1−s1/qn−ξ−1/plnn−ξ1−αs2/pfxlnn−ξ1−αs2/px1−s1/qn−ξ−1/pandx≤∫0∞∑n=2∞n−ξ−1x+lnαn−ξsx1−s1p−1lnn−ξ1−αs2fpxdx1/p×∑n=2∞∫0∞1x+lnαn−ξslnn−ξ1−αs2q−1anqx1−s1n−ξ1−qdx1/q=1α∫0∞ϖs2,xxp1−s−s2/p+s1/q−1fpxdx1/p×∑n=2∞ωs1,nlnn−ξq1−αs−s1/q+s2/p−1n−ξq−1anq1/q.
We show that the above inequality does not keep the form of equality. Otherwise (cf. [25]), there exist constants A and B, such that they are not both zero and
(19)Ax1−s1p−1n−ξ−1lnn−ξ1−αs2fpx=Blnn−ξ1−αs2q−1anqx1−s1n−ξ1−qa.e.inR+×N\1..
Suppose thatA≠0. There exists a n∈N\1, such that
(20)xp1−αs−s2/p+s1/q−1fpx=BAlnn−ξq1−αs2n−ξqanq1x1+s−s1−s2a.e.inR+,which contradicts the fact that
(21)0<∫0∞xp1−s−s2/p+s1/q−1fpxdx<∞,based on ∫0∞1/x1+s−s1−s2dx=∞. Then, by (9) and (17), we have (16).
The lemma is proven.Remark 1.
Replacingfx by Fmx and setting s=λ+m∈m,∞, s1=λ1+m∈m,s, and s2=λ2∈0,s∩0,1/α in (16), we find λ∈0,∞, λ1∈0,λ, and λ2∈0,λ+m∩0,1/α and obtain the following more accurate half-discrete Mulholland-type inequality with new parameters:
(22)∫0∞∑n=2∞anx+lnαn−ξλ+mFmxdx<1αkλ+mλ21/pkλ+mλ1+m1/q×∫0∞x−pm−1+λ∧1−1Fmpxdx1/p∑n=2∞lnn−ξq1−αλ∧2−1n−ξq−1anq1/q.Lemma 3.
Fort>0, the following expression is valid:
(23)∫0∞e−txfxdx=tm∫0∞e−txFmxdx.Proof.
Form=0 since F0x=fx, (23) is naturally valid; for m∈N, integration by parts, since Fi0=0,Fix=oetxt>0,i=1,⋯m;x⟶∞, we find
(24)∫0∞e−txFi−1xdx=∫0∞e−txdFix=e−txFix0∞−∫0∞Fixde−tx=limx⟶∞e−txFix+t∫0∞e−txFixdx=t∫0∞e−txFixdx.
By substitution ofi=1,⋯,m in the above expression, we have (23).
The lemma is proven.


## 3. Main Results

Theorem 1.
The following more accurate half-discrete Mulholland-type inequality involving one multiple upper limit function is valid:(25)I≔∫0∞∑n=2∞anx+lnαn−ξλfxdx<Γλ+mΓλ1αkλ+mλ^21/pkλ+mλ^1+m1/q×∫0∞x−pm−1+λ⟵1−1Fmpxdx1/p∑n=2∞lnn−ξq1−αλ⟵2−1n−ξq−1anq1/q.
In particular, forλ1+λ2=λ (λ1∈0,λ,λ2∈0,λ∩0,1/α), we have
(26)0<∫0∞x−pm−1+λ1−1Fmpxdx<∞,0<∑n=2∞lnn−ξq1−αλ2−1n−ξq−1anq<∞,and the following inequality:
(27)I=∫0∞∑n=2∞anx+lnαn−ξλfxdx<Γλ+mα1/pΓλBλ1+m,λ2×∫0∞x−pm−1+λ1−1Fmpxdx1/p∑n=2∞lnn−ξq1−αλ2−1n−ξq−1anq1/q.Proof.
In view of the following expression:(28)1x+lnαn−ξλ=1Γλ∫0∞tλ−1e−x+lnαn−ξtdt,by the Lebesgue term by the term integration theorem (cf. [26]) and (23), we obtain
(29)I=1Γλ∫0∞∑n=2∞anfx∫0∞tλ−1e−x+lnαn−ξtdtdx=1Γλ∫0∞tλ−1∫0∞e−xtfxdx∑n=2∞e−tlnαn−ξandt=1Γλ∫0∞tλ−1tm∫0∞e−xtFmxdx∑n=2∞e−tlnαn−ξandt=1Γλ∫0∞∑n=2∞anFmx∫0∞tλ+m−1e−x+lnαn−ξtdtdx=Γλ+mΓλ∫0∞∑n=1∞anx+lnαn−ξλ+mFmxdx.
Then, by (22), we have (25).
The theorem is proven.Theorem 2.
Ifλ1+λ2=λλ1∈0,λ,λ2∈0,λ∩0,1/α, then the constant factor
(30)Γλ+mΓλ1αkλ+mλ21/pkλ+mλ1+m1/q
is the best possible. On the other hand, if we addλ2<λ,λ−λ1≤1/α and the same constant factor in (25) is the best possible, then we have λ1+λ2=λ.Proof.
Ifλ1+λ2=λ, then (25) reduces to (27). For any 0<ε<pλ1, we set
(31)f~x≔0,0<x<1,xλ1−ε/p−1,x≥1,a~n≔lnn−ξαλ2−ε/q−1n−ξ−qn∈N\1.
We haveF~0x=f~x and
(32)F~ix≔∫0xF~i−1tdt≤0,0<x<1,1∏j=0i−1λ1+j−ε/pxλ1+i−1−ε/p,x≥1,i∈1,⋯,m,satisfying F~i0=0,F~ix=oetxt>0,x⟶∞;i=1,⋯,m.
If there exists a positive constantM≤Γλ+m/α1/pΓλkλ+mλ1+m, such that (27) is valid when we replace Γλ+m/α1/pΓλkλ+mλ1+m by M, then in particular, it follows that
(33)I~≔∫0∞∑n=2∞a~nf~xdxx+lnαn−ξλ<M∫0∞x−pm−1+λ1−1F~mpxdx1/p∑n=2∞lnn−ξq1−αλ2−1n−ξq−1a~nq1/q.
Form=0,ε≥0, we define ∏i=0m−1λ1+i−ε/p=1. By the decreasing property of the series, we find
(34)I~<M∏i=0m−1λ1+i−ε/p∫1∞x−pm−1+λ1−1xpλ1+m−1−εdx1/p×∑n=2∞lnn−ξq1−αλ2−1n−ξq−1lnn−ξqαλ2−αε−qn−ξ−q1/q=M∏i=0m−1λ1+i−ε/p∫1∞x−ε−1dx1/p×ln2−ξ−αε−12−ξ−1+∑n=3∞lnn−ξ−αε−1n−ξ−11/q≤M∏i=0m−1λ1+i−ε/p∫1∞x−ε−1dx1/p×ln2−ξ−αε−12−ξ−1+∫2∞lnt−ξ−αε−1t−ξ−1dt1/q=Mε∏i=0m−1λ1+i−ε/pεln2−ξ−αε−12−ξ−1+1αln2−ξ−αε1/q.
By (17), setting s=λ and s1=λ~1≔λ1−ε/p∈0,λ0<λ~2≔λ2+ε/p<λ, it follows that
(35)I~=∑n=2∞lnn−ξαλ2+ε/p∫1∞xλ1−ε/p−1x+lnαn−ξλdxlnn−ξ−αε−1n−ξ−1=∑n=2∞ωλ~1,nlnn−ξ−αε−1n−ξ−1−∑n=2∞lnn−ξαλ~2−ε−1n−ξ−1∫01xλ~1−1x+lnαn−ξλdx>Bλ~1,λ~2∑n=2∞lnn−ξ−αε−1n−ξ−1−∑n=2∞lnn−ξαλ~2−ε−1n−ξ−1∫01xλ~1−1dxlnn−ξαλ≥Bλ~1,λ~2∫2∞lny−ξ−αε−1y−ξ−1dy−1λ~1∑n=2∞lnn−ξ−αλ~1+ε−1n−ξ−1=1ε1αBλ1−εp,λ2+εpln2−ξ−αε−εO1.
In virtue of the above results, we have(36)1αBλ1−εp,λ2+εpln2−ξ−αε−εO1<εI~<M∏i=0m−1λ1+i−ε/pεln2−ξ−αε−12−ξ−1+1αln2−ξ−αε1/q.
Forε⟶0+, by the continuity of the beta function, we have
(37)Γλ+mα1/pΓλBλ1+m,λ2=Γλ1+mΓλ2α1/pΓλ=1α1/pBλ1,λ2∏i=0m−1λ1+i≤M.
Hence,M=Γλ+m/α1/pΓλBλ1+m,λ2 is the best possible constant factor in (27).
On the other hand, forλ^1=λ−λ2/p+λ1/q,λ^2=λ−λ1/q+λ2/p, in view of 0<λ2<λ,λ−λ1≤1/α, we find λ^1+λ^2=λ−λ2/p+λ1/q+λ−λ1/q+λ2/p=λ, 0<λ^1,λ^2<λ/p+λ/q=λ, λ^2≤1/α/q+1/α/p=1/α, and
(38)Γλ+mα1/pΓλBλ^1+m,λ^2∈R+..
By substitution ofλ^i=λii=1,2 in (27), we find
(39)I=∫0∞∑n=2∞anx+lnαn−ξλfxdx<Γλ+mα1/pΓλkλ+mλ^1+m×∫0∞x−pm−1+λ∧1−1Fmpxdx1/p∑n=2∞lnn−ξq1−αλ∧2−1n−ξ−1anq1/q.
By means of Hölder’s inequality (cf. [25]), it follows that
(40)kλ+mλ^1+m=kλ+mλ−λ2p+λ1q+m=∫0∞11+uλ+muλ−λ2+m/p+λ1+m/q−1du=∫0∞11+uλ+muλ−λ2+m−1/puλ1+m−1/qdu≤∫0∞11+uλ+muλ−λ2+m−1du1/p∫0∞11+uλ+muλ1+m−1du1/q=∫0∞11+vλ+mvλ2−1dv1/p∫0∞11+uλ+muλ1+m−1du1/q=kλ+mλ21/pkλ+mλ1+m1/q.
SinceΓλ+m/α1/pΓλkλ+mλ21/pkλ+mλ1+m1/q is the best possible constant factor in (25), by (39), we have the following inequality:
(41)Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/q≤Γλ+mα1.pΓλkλ+mλ^1+m∈R+,namely, kλ+mλ^1+m≥kλ+mλ21/pkλ+mλ1+m1/q, and then, (40) keeps the form of equality.
We observe that (40) keeps the form of equality if and only if there exist constants A and B (cf. [25]), such that they are not both zero and Auλ−λ2=Buλ1a.e. in R+. Suppose that A≠0. We find uλ−λ2−λ1=B/Aa.e. in R+, namely, λ−λ2−λ1=0. Hence, it follows that λ1+λ2=λ.
The theorem is proven.


## 4. Equivalent Forms and Some Particular Inequalities

Theorem 3.
The following half-discrete more accurate Mulholland-type inequality involving a multiple upper limit function is equivalent to (25):
(42)J≔∑n=2∞lnn−ξpαλ∧2−1n−ξ−1∫0∞fxx+lnαn−ξλdxp1/p<Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/q∫0∞x−pm−1+λ∧1−1Fmpxdx1/p.
In particular, forλ1+λ2=λλ1∈0,λ,λ2∈0,λ∩0,1/α, we have the following Mulholland-type inequality equivalent to (27):
(43)∑n=2∞lnn−ξpαλ2−1n−ξ−1∫0∞fxx+lnαn−ξλdxp1/p<Γλ+mα1/pΓλBλ1+m,λ2∫0∞x−pm−1+λ1−1Fmpxdx1/p.Proof.
By assuming that (42) is valid, by Hölder’s inequality (cf. [25]), it follows that
(44)I=∑n=2∞lnn−ξαλ∧2−1/pn−ξ−1/p∫0∞fxdxx+lnαn−ξλ×lnn−ξ−αλ∧2+1/pn−ξ1/pan≤J∑n=2∞lnn−ξq1−αλ∧2−1n−ξ−1anq1/q.
Then, by (42), we have (25).
On the other hand, suppose that (25) is valid. We set
(45)an≔lnn−ξpαλ2−1n−ξ−1∫0∞fxx+lnαn−ξλdxp−1,n∈N\1.
IfJ=0, then (42) is naturally valid; if J=∞, then it is impossible that makes (42) valid, i.e., J<∞. By assuming that 0<J<∞, by (25), it follows that
(46)0<∑n=2∞lnn−ξq1−αλ∧2−1n−ξ−1anq=Jp=I<Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/q∫0∞x−pm−1+λ∧1−1Fmpxdx1/pJp−1<∞,J=∑n=2∞lnn−ξq1−λ∧2−1n−ξ−1anq1/p<Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/q∫0∞x−pm−1+λ∧1−1Fmpxdx1/p,namely, (42) follows, which is equivalent to (25).
The theorem is proven.Theorem 4.
Ifλ1+λ2=λλ1∈0,λ,λ2∈0,λ∩0,1/α, then the constant factor
(47)Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/qin (42) is the best possible. On the other hand, if we add λ2<λ,λ−λ1≤1/α and the same constant factor in (42) is the best possible, then we have λ1+λ2=λ.Proof.
Ifλ1+λ2=λ, then by Theorem 2, the constant factor
(48)Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/qin (25) is the best possible. By (44), the constant factor in (42) is still the best possible. Otherwise, we would reach a contradiction that the constant factor in (25) is not the best possible.
On the other hand, if the constant factor in (42) is the best possible, then by the equivalency of (42) and (25), in view of Jp=I (see the proof of Theorem 3), we still can show that the constant factor in (25) is the best possible. By the assumptions and Theorem 2, it follows that λ1+λ2=λ.
The theorem is proven.Example 1.
(i)
Forα=1, (27) and (43) reduce to the following equivalent inequalities:(49)∫0∞∑n=2∞anlnλexn−ξfxdx<×∫0∞x−pm−1+λ1−1Fmpxdx1/p∑n=2∞lnn−ξq1−λ2−1n−ξq−1anq1/q,(50)∑n=2∞lnn−ξpλ2−1n−ξ−1∫0∞fxlnλexn−ξdxp1/p<Bλ1,λ2∏i=0m−1λ1+i∫0∞x−pm−1+λ1−1Fmpxdx1/p.(ii)
Forξ=0, (27) and (43) reduce to the following equivalent inequalities:(51)∫0∞∑n=2∞anx+lnαnλfxdx<1α1/pBλ1,λ2∏i=0m−1λ1+i×∫0∞x−pm−1+λ1−1Fmpxdx1/p∑n=2∞lnnq1−αλ2−1nq−1anq1/q,(52)∑n=2∞lnnpαλ2−1n∫0∞fxx+lnαnλdxp1/p<1α1/pBλ1,λ2∏i=0m−1λ1+i∫0∞x−pm−1+λ1−1Fmpxdx1/p.
Hence, (27) (resp. (43)) is a more accurate form of (51) (resp. (52)).
(iii)
Forξ=1/2, (27) and (43) reduce to the following equivalent inequalities:(53)∫0∞∑n=2∞anx+lnαn−1/2λfxdx<1α1/pBλ1,λ2∏i=0m−1λ1+i×∫0∞x−pm−1+λ1−1Fmpxdx1/p∑n=2∞lnn−12q1−αλ2−1n−12q−1anq1/q,∑n=2∞lnn−12pαλ2−1n−12−1∫0∞fxx+lnαn−1/2λdxp1/p<1α1/pBλ1,λ2∏i=0m−1λ1+i∫0∞x−pm−1+λ1−1Fmpxdx1/p.
The constant factors in the above inequalities are still the best possible.


## 5. Operator Expressions

Settingϕmx≔x−pm−1+λ∧1−1,ψn≔lnn−ξq1−αλ∧2−1n−ξq−1, from
(54)ψ1−pn=lnn−ξpαλ∧2−1n−ξ−1x∈R+,n∈N\1,we define the following normed linear spaces:
(55)Lp,ϕmR+≔f=fx;fp,ϕm≔∫0∞ϕmxfxpdx1/p<∞,lq,ψ≔a=ann=2∞;aq,ψ≔∑n=2∞ψnanq1/q<∞,lp,ψ1−p≔c=cnn=2∞;cp,ψ1−p≔∑n=2∞ψ1−pncnp1/p<∞.For anyf=F0x∈Lp,ϕ0R+, setting
(56)c=cnn=2∞:cn≔∫0∞fxx+lnαn−ξλdx,we can rewrite (42) as follows:
(57)cp,ψ1−p<Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/qFmp,ϕm<∞,namely, c∈lp,ψ1−p.Definition 1.
Define a half-discrete Mulholland-type operatorT:Lp,ϕ0R+⟶lp,ψ1−p as follows: for any f∈Lp,ϕ0R+, there exists a unique representation c=Tf∈lp,ψ1−p, satisfying for any n∈N\1Tfn=cn. Define the formal inner product of Tf and a∈lq,ψ and the norm of T as follows:
(58)Tf,a≔∑n=2∞an∫0∞fxx+lnαn−ξλdx=I,T≔supf≠0∈Lp,ϕ0R+Tfp,ψ1−pFmp,ϕm..
By Theorem3 and Theorem 4, it follows that.Theorem 5.
Iff≥0∈Lp,ϕ0R+,a≥0∈lq,ψ,fp,ϕ0>0,aq,ψ>0, then we have the following equivalent inequalities:
(59)Tf,a<Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/qFmp,ϕmaq,ψ,(60)Tfp,ψ1−p<Γλ+mα1/pΓλkλ+mλ21/pkλ+mλ1+m1/qFmp,ϕm.
w?>Moreover, ifλ1+λ2=λ, then the constant factor Γλ+m/α1/pΓλkλ+mλ21/pkλ+mλ1+m1/q in (59) and (60) is the best possible, namely,
(61)T=Γλ+mα1/pΓλBλ1+m,λ2.
On the other hand, if we addλ2<λ,λ−λ1≤1/α and the constant factor in (59) and (60) is the best possible, then we have λ1+λ2=λ.



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*Source: 6970158-2021-12-07.xml*