Parameterized discrete Hilbert type inequalities with intermediate variables

(1)

R E S E A R C H

Open Access

Parameterized discrete Hilbert-type

inequalities with intermediate variables

Ricai Luo

1*

_{and Bicheng Yang}

2

*_{Correspondence:}

[email protected]

1_{School of Mathematics and}

Statistics, Hechi University, Yizhou, P.R. China

Full list of author information is available at the end of the article

Abstract

By means of the weight coeﬃcients and the idea of introducing parameters, a discrete Hilbert-type inequality with the general homogeneous kernel and the intermediate variables is given. The equivalent form is obtained. The equivalent statements of the best possible constant factor related to some parameters, the operator expressions, and a few particular cases are considered.

MSC: 26D15

Keywords: Weight coeﬃcient; Hilbert-type inequality; Equivalent statement; Parameter; Operator expression

1 Introduction

Assuming thatp> 1, 1

p +

1

q= 1,am,bn≥0, 0 <

∞

m=1a

p

m<∞, and 0 <

∞

n=1b

q

n<∞, we have the following Hardy–Hilbert inequality with the best possible constant _sin₍π_π_/_p₎ (cf. [1], Theorem 315):

∞

m=1

∞

n=1 ambn

m+n< π

sin(π/p)

_∞

m=1 ap_m

1

p∞

n=1 bq_n

1 q

. (1)

Forp=q= 2, inequality (1) reduces to the well-known Hilbert inequality. Iff(x),g(y)≥0, 0 <₀∞fp₍_x₎_dx_<_∞_{and 0 <}∞

0 gq(y)dy<∞, then we have the following

Hardy–Hilbert integral inequality:

∞

0 ∞

0

f(x)g(y)

x+y dx dy< π

sin(π/p)

∞

0

fp(x)dx 1

p ∞

0

gq(y)dy 1 q

(2)

with the best possible constant factor π

sin(π/p)(cf. [1], Theorem 316).

In 1998, by introducing an independent parameterλ> 0, Yang [2,3] gave an extension of (2) (forp=q= 2) with the best possible constant factorB(λ

2,

λ

2) as follows:

∞

0 ∞

0

f(x)g(y)

(x+y)λ dx dy<B

λ

2,

λ

2

∞

0

x1–λf2(x)dx ∞

0

y1–λg2(y)dy 1 2

. (3)

Inequalities (1), (2), (3) and their extensions are important in analysis and its applications (cf. [4–15]).

(2)

The following half-discrete Hilbert-type inequality was provided (cf. [1], Theorem 351): IfK(x) (x> 0) is a decreasing function,p> 1, 1_p +1_q= 1, 0 <φ(s) =₀∞K(x)xs–1_dx_<_∞_,

then

∞

0 xp–2

_∞

n=1

K(nx)an

p

dx<φp

1

q

∞

n=1

ap_n. (4)

Some new extensions of (4) with their applications were provided by [16–21].

In 2016, by the use of the technique of real analysis, Hong [22] considered some equiv-alent statements of the extensions of (1) with the best possible constant factor related to a few parameters. The other similar works about the extensions of (2) and (3) were given by [23–27].

In this paper, following the way of [22], by means of the weight functions and the idea of introducing parameters, a discrete Hilbert-type inequality with the general homogeneous kernel and the intermediate variables is given, which is an extension of (1). The equivalent form is obtained. The equivalent statements of the best possible constant factor related to parameters, the operator expressions, and a few particular cases are considered.

2 Some lemmas

In what follows, we suppose thatp> 1,1_p+1_q= 1,α,β> 0,λ∈R,λ2,λ–λ1≤β1,λ1,λ–λ2≤

1

α,kλ(x,y) is a positive homogeneous function of degree –λsatisfying, for anyu,x,y> 0,

kλ(ux,uy) =u–λkλ(x,y).

Also,kλ(x,y) is strictly decreasing with respect tox,y> 0 such that, forγ =λ1,λ–λ2,

kλ(γ) :=

∞

0

kλ(u, 1)uγ–1du∈R+= (0,∞). (5)

We still assume thatam,bn≥0 (m,n∈N ={1, 2, . . .}) such that

0 <

∞

m=1 mp[1–α(

λ–λ₂

p +

λ₁

q)]–1_ap

m<∞ and 0 <

∞

n=1 nq[1–β(

λ–λ₁

q +

λ₂

p)]–1_bq

n<∞.

Definition 1 We define the following weight coefficients:

ωλ(λ2,m) :=mα(λ–λ2)

∞

n=1 kλ

mα,nβnβλ2–1 ₍_m∈_N), ₍₆₎

λ(λ1,n) :=nβ(λ–λ1)

∞

m=1 kλ

mα_,_nβ_mαλ1–1 ₍_n∈_N). ₍₇₎

Lemma 1 We have the following inequalities:

ωλ(λ2,m) <

1

βkλ(λ–λ2) (m∈N), (8)

λ(λ1,n) <

1

(3)

Proof Forβλ2– 1≤0, it is evident thatkλ(mα,yβ)yβλ2–1is strictly decreasing with respect toy> 0. By the decreasingness property, settingu=mα

yβ, we ﬁnd that

ωλ(λ2,m) <mα(λ–λ2) _∞

0 kλ

mα,yβyβλ2–1_dy

= 1

β ∞

0

kλ(u, 1)u(λ–λ2)–1du= 1

βkλ(λ–λ2).

Hence, we have (8).

Forαλ1– 1≤0, it is evident thatkλ(xα,nβ)xαλ1–1is strictly decreasing with respect to

x> 0. By the decreasingness property, settingu=xα

nβ, we ﬁnd that

λ(λ1,n) <nβ(λ–λ1) ∞

0 kλ

xα,nβxαλ1–1_dx

=1

α ∞

0

kλ(u, 1)uλ1–1du= 1

αkλ(λ1).

Hence, we have (9).

Lemma 2 We have the following inequality:

I:=

∞

n=1

∞

m=1 kλ

mα_,_nβ_a

mbn

< 1

β1/p_α1/qk

1 p

λ(λ–λ2)k

1 q

λ(λ1)

_∞

m=1

mp[1–α(λ –λ2

p +

λ1 q)]–1_ap

m

1 p

.

× _∞

n=1 nq[1–β(λ

–λ1 q +

λ2 p)]–1_bq

n

1 q

. (10)

Proof By Hölder’s inequality with weight (cf. [28]), we obtain

I=

∞

n=1

∞

m=1 kλ

mα,nβn (βλ2–1)/p

m(αλ1–1)/qam

m(αλ1–1)/q n(βλ2–1)/pbn

≤ _∞

m=1

∞

n=1 kλ

mα,nβ n

βλ2–1

m(αλ1–1)(p–1)a

p m

1

p∞

n=1

∞

m=1 kλ

mα,nβ m

αλ1–1

n(βλ2–1)(q–1)b

q n

1 q

=

_∞

m=1

ωλ(λ2,m)mp[1–α(

λ–λ2 p +

λ1 q)]–1_ap

m

1

p∞

n=1

λ(λ1,n)nq[1–β(

λ–λ1 q +

λ2 p)]–1_bq

n

1 q

.

Then, by (8) and (9), we have (10).

Remark1 (i) By (10), forλ1+λ2=λ, we ﬁnd

0 <

∞

m=1

mp(1–αλ1)–1_ap

m<∞ and 0 <

∞

n=1

nq(1–βλ2)–1_bq

(4)

and the following inequality:

∞

n=1

∞

m=1 kλ

mα_,_nβ_a

mbn

< 1

β1/p_α1/qkλ(λ1)

_∞

m=1

m

1

p∞

n=1

nq(1–βλ2)–1_bq

n

1 q

. (11)

In particular, forα=β= 1, we have

∞

n=1

∞

m=1

kλ(m,n)ambn<kλ(λ1) _∞

m=1

mp(1–λ1)–1_ap m

1

p∞

n=1

nq(1–λ2)–1_bq n

1 q

. (12)

(ii) Forλ= 1,k1(x,y) = _x1₊_y,λ1=1_q,λ2=1_p, (12) reduces to (1). Hence, (11) is an extension

of (12) and (1).

Lemma 3 The constant factor _β1/p1_α1/qkλ(λ1)in(11)is the best possible.

Proof For anyε> 0, we set

˜

am:=mα(λ1– ε

p)–1_, _b˜

n:=nβ(λ2– ε

q)–1 ₍_m_,_n∈_N).

If there exists a constant M (≤ 1

β1/p_α1/qkλ(λ1)) such that (11) is valid when replacing 1

β1/p_α1/qkλ(λ1) byM, then, in particular, we have

˜ I:=

∞

n=1

∞

m=1 kλ

mα_,_nβ_a_˜

mb˜n<M

_∞

m=1

mp(1–αλ1)–1_a˜p

m

1

p∞

n=1

nq(1–βλ2)–1_b˜q

n

1 q

.

By the decreasingness property, we obtain

˜ I<M

_∞

m=1

mp(1–αλ1)–1_mpαλ1–αε–p 1

p∞

n=1

nq(1–βλ2)–1_nqβλ2–βε–q 1

q

=M

1 +

∞

m=2 m–αε–1

1

p

1 +

∞

n=2 n–βε–1

1 q

<M

1 +

∞

1

t–αε–1_dt 1

p

1 +

∞

1

t–βε–1_dt 1 q

=M

ε

ε+1

α 1

p

ε+1

β 1 q

.

By the decreasingness property and Fubini theorem (cf. [29]), we ﬁnd

˜ I=

∞

n=1

∞

m=1 kλ

mα,nβm

αλ1–1

mεα/p ·

nβλ2–1 nεβ/q

≥ ∞

1

∞

1 kλ

xα_,_yβx

αλ1–1

xεα/p ·

yβλ2–1 yεβ/q dx

dy

u=x

α

(5)

=1

α ∞

1 y–βε–1

∞

1 yβ

kλ(u, 1)u

λ1–εp–1_du _dy

=1

α ∞

1 y–βε–1

1

1 yβ

kλ(u, 1)u

λ1–εp–1_du _dy

+ 1

α ∞

1

y–βε–1dy ∞

1

kλ(u, 1)u

λ1–pε–1_du

=1

α 1

0

∞

u–1/β

y–βε–1_dy _k

λ(u, 1)u

λ1–pε–1_du

+ 1

αβε ∞

1

kλ(u, 1)uλ1– ε

p–1_du

= 1

αβε 1

0

kλ(u, 1)uλ1+ ε

q–1_du₊ ∞

1

kλ(u, 1)uλ1– ε

p–1_du _.

Then we have

1

αβ 1

0

kλ(u, 1)u

λ1+εq–1_du₊ ∞

1

kλ(u, 1)u

λ1–εp–1_du

<M

ε+1

α 1

p

ε+ 1

β 1 q

.

Forε→0+_{, by Fatou’s lemma (cf. [}₂₉_{]), we ﬁnd}

1

αβkλ(λ1) =

1

αβ 1

0

lim

ε→0+kλ(u, 1)u

λ1+qε–1_du₊ ∞

1

lim

ε→0+kλ(u, 1)u

λ1–εp–1_du

≤ 1

αβεlim→0+ 1

0

kλ(u, 1)uλ1+ ε

q–1_du₊ ∞

1

kλ(u, 1)uλ1– ε

p–1_du

≤M lim

ε→0+

ε+1

α 1

p

ε+1

β 1 q

=M

1

α 1 p₁

β 1 q

,

namely 1

β1/p_α1/qkλ(λ1)≤M. Hence,M=_β1/p1_α1/qkλ(λ1) is the best possible constant factor

of (11).

Remark2 Settingλˆ1:=λ–_pλ2 +λ_q1,λˆ2:=λ–_qλ1 +λ_p2, we ﬁnd

ˆ λ1+λˆ2=

λ–λ2 p +

λ1 q +

λ–λ1 q +

λ2 p =

λ p+

λ q =λ,

ˆ λ1≤

1

pα+

1

qα =

1

α, λˆ2≤

1

qβ +

1

pβ =

1

β,

and by Hölder’s inequality (cf. [28]), we obtain

0 <kλ(λ–λˆ2) =kλ(λˆ1) =kλ

λ–λ2

p + λ1

q

=

_∞ 0

kλ(u, 1)u λ–λ₂

p +

λ₁

q–1_du₌ _∞

0

kλ(u, 1)

u

λ–λ_2–1

p _u

λ_1–1

(6)

≤ ∞ 0

kλ(u, 1)uλ–λ2–1du

1

p ∞

0

kλ(u, 1)uλ1–1du

1 q

=k 1 p

λ(λ–λ2)k

1 q

λ(λ1) <∞. (13)

We can reduce (10) as follows:

I< 1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1) _∞

m=1

mp(1–αλˆ1)–1_ap

m

1

p∞

n=1

nq(1–βλˆ2)–1_bq

n

1 q

. (14)

Lemma 4 If the constant factor _β1/p1_α1/qk 1 p

λ(λ–λ2)k 1 q

λ(λ1)in(10)is the best possible,then λ1+λ2=λ.

Proof If the constant factor _β1/p1_α1/qk 1 p

λ(λ–λ2)k

1 q

λ(λ1) in (10) is the best possible, then by (14) and (11) the unique best possible constant factor must be_β1/p1_α1/qkλ(λˆ1) (∈R+), namely

kλ(λˆ1) =k 1 p

λ(λ–λ2)k 1 q

λ(λ1).

We observe that (13) keeps the form of equality if and only if there exist constantsAand

Bsuch that they are not all zero and (cf. [28])

Auλ–λ2–1₌_Buλ1–1 _{a.e. in R} +.

Assuming thatA= 0 (otherwise,B=A= 0), it follows thatuλ–λ2–λ1₌B

Aa.e. in R+, and then

λ–λ2–λ1= 0, namelyλ1+λ2=λ.

3 Main results

Theorem 1 Inequality(10)is equivalent to

J:=

_∞

n=1 npβ(

λ–λ₁

q +

λ₂

p)–1 _∞

m=1 kλ

mα,nβam

p1_p

< 1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1) _∞

m=1 mp[1–α(

λ–λ₂

p +

λ₁

q)]–1_ap

m

1 p

. (15)

If the constant factor in(10)is the best possible,then so is the constant factor in(15).

Proof Suppose that (15) is valid. By Hölder’s inequality (cf. [28]), we ﬁnd

I=

∞

n=1

n–1p+β(

λ–λ₁

q +

λ₂

p)

∞

m=1 kλ

mα,nβam

n1p–β(

λ–λ₁

q +

λ₂

p)_b

n

≤J _∞

n=1 nq[1–β(

λ–λ₁

q +

λ₂

p)]–1_bq

n

1 q

. (16)

(7)

On the other hand, assuming that (10) is valid, we set

bn:=npβ( λ–λ1

q +

λ2 p)–1

_∞

m=1 kλ

mα,nβam

p–1

, n∈N.

IfJ= 0, then (15) is naturally valid; ifJ=∞, then it is impossible to make (15) valid, namely

J<∞. Suppose that 0 <J<∞. By (10), it follows that

∞

n=1 nq[1–β(

λ–λ1

q +

λ2 p)]–1_bq

n

=Jp=I< 1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1)

× _∞

m=1 mp[1–α(

λ–λ₂

p +

λ₁

q)]–1_ap

m

1

p∞

n=1 nq[1–β(

λ–λ₁

q +

λ₂

p)]–1_bq

n

1 q

,

J=

_∞

n=1 nq[1–β(

λ–λ1 q +

λ2 p)]–1_bq

n

1 p

< 1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1) _∞

m=1 mp[1–α(

λ–λ2 p +

λ1 q)]–1_ap

m

1 p

,

namely (15) follows, which is equivalent to (10).

If the constant factor in (10) is the best possible, then so is constant factor in (15). Oth-erwise, by (16), we would reach a contradiction that the constant factor in (10) is not the

best possible.

Theorem 2 The following statements(i), (ii), (iii),and(iv)are equivalent:

(i) k 1 p

λ(λ–λ2)k 1 q

λ(λ1)is independent ofp,q;

(ii) k 1 p

λ(λ–λ2)k 1 q

λ(λ1)is expressible as a single integral;

(iii) _β1/p1_α1/qk 1 p

λ(λ–λ2)k 1 q

λ(λ1)is the best possible constant factor of(10); (iv) λ1+λ2=λ.

If the statement (iv)follows,namely λ1+λ2=λ, then we have(11)and the following equivalent inequality with the best possible constant factor _β1/p1_α1/qkλ(λ1):

_∞

n=1 npβλ2–1

_∞

m=1 kλ

mα,nβam

p1_p < 1

β1/p_α1/qkλ(λ1)

_∞

m=1

m

1 p

. (17)

Proof (i)⇒(ii). Sincek 1 p

λ(λ–λ2)k 1 q

λ(λ1) is independent ofp,q, we ﬁnd

k 1 p

λ(λ–λ2)k 1 q

λ(λ1) = lim

p→∞qlim→1+k 1 p

λ(λ–λ2)k 1 q

λ(λ1) =kλ(λ1),

namelyk 1 p

λ(λ–λ2)k

1 q

λ(λ1) is expressible as a single integral

kλ(λ1) = _∞

0

(8)

(ii)⇒(iv). In (13), ifk 1 p

λ(λ–λ2)k 1 q

λ(λ1) is expressible as a single integralkλ(λ–_pλ2 +λ_q1), then (13) keeps the form of equality, from which it follows thatλ1+λ2=λ.

(iv)⇒(i). Ifλ1+λ2=λ, thenk 1 p

λ(λ–λ2)k 1 q

λ(λ1) =kλ(λ1), which is independent ofp,q.

Hence, we have (i)⇔(ii)⇔(iv).

(iii)⇒(iv). By Lemma4, we haveλ1+λ2=λ.

(iv)⇒(iii). By Lemma3, forλ1+λ2=λ,

1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1)

= 1

β1/p_α1/qkλ(λ1)

is the best possible constant factor of (10). Therefore, we ﬁnd (iii)⇔(iv).

Hence, statements (i), (ii), (iii), and (iv) are equivalent.

Remark3 (i) Forλ=α=β= 1,λ1= _q1,λ2= 1_p in (11) and (17), we have the following

equivalent inequalities with the best possible constant factork1(1_q):

∞

n=1

∞

m=1

k1(m,n)ambn<k1

1

q

∞

m=1 ap_m

1

p∞

n=1 bq_n

1 q

, (18)

_∞

n=1 _∞

m=1

k1(m,n)am

p1_p <k1

1

q

∞

m=1 ap_m

1 p

. (19)

(ii) Forλ=α=β= 1,λ1=_p1,λ2=1_q in (11) and (17), we have the following equivalent

inequalities with the best possible constant factork1(1_p):

∞

n=1

∞

m=1

k1(m,n)ambn<k1

1

p

∞

m=1 mp–2ap_m

1

p∞

n=1 nq–2bq_n

1 q

, (20)

_∞

n=1 np–2

_∞

m=1

k1(m,n)am

p1_p <k1

1

p

∞

m=1 mp–2ap_m

1 p

. (21)

(iii) Forp=q= 2, both (18) and (20) reduce to

∞

n=1

∞

m=1

k1(m,n)ambn<k1

1 2

∞

m=1 a2_m

∞

n=1 b2_n

1 2

, (22)

and both (19) and (21) reduce to the equivalent form of (22) as follows:

_∞

n=1 _∞

m=1

k1(m,n)am

21₂

<k1

1 2

∞

m=1 a2_m

1 2

. (23)

4 Operator expressions and some particular cases

We set functions

φ(m) :=mp[1–α(

λ–λ2

p +

λ1

q)]–1_, _ψ₍_n_{) :=}_nq[1–β(

λ–λ1 q +

(9)

from which

ψ1–p(n) =npβ(

λ–λ₁

q +

λ₂

p)–1 ₍_m_,_n∈_N).

Deﬁne the following real normed spaces:

lp,φ:=

a={am}∞m=1;ap,φ:=

_∞

m=1

φ(m)|am|p

1 p

<∞

,

lq,ψ:=

b={bn}∞n=1;bq,ψ:=

_∞

n=1

ψ(n)|bn|q

1 q

<∞

,

lp,ψ1–p:=

c={cn}∞n=1;cp,ψ1–p:=

_∞

n=1

ψ1–p(n)|bn|p

1 p

<∞

.

Assuming thata∈lp,φ, setting

c={cn}∞n=1, cn:=

∞

m=1 kλ

mα_,_nβ_a

m, n∈N,

we can rewrite (15) as follows:

cp,ψ1–p<

1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1)ap,φ<∞,

namelyc∈l_p_,ψ1–p.

Deﬁnition 2 Deﬁne a Hilbert-type operator T :lp,φ→lp,ψ1–p as follows: For anya∈

lp,φ,there exists a unique representationc∈lp,ψ1–p. Deﬁne the formal inner product of Taandb∈lq,ψ, and the norm ofT as follows:

(Ta,b) :=

∞

n=1 _∞

m=1 kλ

mα_,_nβ_a

m

bn,

T:= sup

a(=θ)∈lp,φ

Ta_p_,ψ1–p ap,φ

.

By Theorem1and Theorem2, we have the following.

Theorem 3 If a∈lp,φ,b∈lq,ψ,ap,φ,bq,ψ > 0,then we have the following equivalent

inequalities:

(Ta,b) < 1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1)ap,φbq,ψ, (24)

Ta_p_,ψ1–p<

1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1)ap,φ. (25)

Moreover,λ1+λ2=λif and only if the constant factor

1

β1/p_α1/qk

1 p

λ(λ–λ2)k 1 q

λ(λ1) =

1

(10)

in(24)and(25)is the best possible,namely

T= 1

β1/p_α1/qkλ(λ1). (26)

Example1 We setkλ(x,y) :=₍_cx₊1_y₎λ (c,λ> 0;x,y> 0). Then we ﬁnd

kλ

mα,nβ= 1 (cmα₊_nβ₎λ.

For 0 <λ1,λ–λ2≤ α1, 0 <λ2, λ–λ1≤

1

β,kλ(x,y) is a positive homogeneous function of degree –λsuch thatkλ(x,y) is strictly decreasing with respect tox,y> 0, and forγ =

λ1,λ–λ2,

kλ(γ) =

∞

0

uγ–1

(cu+ 1)λdu= 1

cγ

∞

0

vγ–1

(v+ 1)λdv= 1

cγB(γ,λ–γ)∈R+.

In view of Theorem3, it follows thatλ1+λ2=λif and only if

T= 1

β1/p_α1/qkλ(λ1) = 1

β1/p_α1/q · 1

cλ1B(λ1,λ2).

Example2 We setkλ(x,y) :=₍_cxln(₎cxλ_–/y_y)λ (c,λ> 0;x,y> 0). Then we ﬁnd

kλ

mα,nβ= ln(cm α_/_nβ₎

cλ_mλα_–_nλβ.

For 0 <λ1,λ–λ2≤α1, 0 <λ2,λ–λ1≤

1

β,kλ(x,y) is a positive homogeneous function of degree –λsuch thatkλ(x,y) is strictly decreasing with respect tox,y> 0 (cf. [4], Exam-ple 2.2.1), and forγ =λ1,λ–λ2,

kλ(γ) =

∞

0

uγ–1_ln₍_cu₎

(cu)λ_{– 1} du= 1

cγ_λ2 ∞

0

v(γ/λ)–1_ln_v v– 1 dv=

1

cγ

π λsin(π γ/λ)

2 ∈R+.

T= 1

β1/p_α1/qkλ(λ1) = 1

β1/p_α1/q · 1

cλ1

π λsin(π λ1/λ)

2

.

Example3 Fors∈N, we setkλ(x,y) := s 1

k=1(xλ/s+ckyλ/s) (0 <c1≤ · · · ≤cs,λ> 0;x,y> 0).

Then we ﬁnd

kλ

mα,nβ=_s 1

k=1(mαλ/s+cknβλ/s) .

For 0 <λ1,λ–λ2≤ α1, 0 <λ2, λ–λ1≤

1

β,kλ(x,y) is a positive homogeneous function of degree –λsuch thatkλ(x,y) is strictly decreasing with respect tox,y> 0, and forγ =

λ1,λ–λ2, by Example 1 of [30], we have

k_λ(s)(γ) =

∞

0

tγ–1 s

k=1(tλ/s+ck)

dt= πs

λsin(π_λsγ) s

k=1 c

sγ λ–1 k

s

j=1(j=k)

1

(11)

T= 1

β1/p_α1/qk

(s)

λ (λ1) =

1

β1/p_α1/q·

πs

λsin(πsλ1

λ )

s

k=1 c

sλ1

λ –1 k

s

j=1(j=k)

1

cj–ck .

In particular, forc1=· · ·=cs=c, we havekλ(x,y) =₍_xλ/s₊1_cyλ/s₎s and

˜

kλ(s)(λ1) := _∞

0

tλ1–1

(tλ/s₊_c₎sdt

= s

λc(1–λλ1)s

∞

0

vsλλ1–1 (v+ 1)sdv=

s

λc(1–λλ1)s

B

sλ1 λ ,

sλ2 λ .

Ifs= 1, then we havekλ(x,y) =_xλ₊1_cyλ and

T= 1

β1/p_α1/qk˜

(1)

λ (λ1) =

1

β1/p_α1/q · 1

λc1–λλ1

π

sin(π λ1

λ ) .

5 Conclusions

In this paper, by means of the weight coeﬃcients and the idea of introducing parameters, a discrete Hilbert-type inequality with the general homogeneous kernel and the interme-diate variables is obtained which is an extension of (1). The equivalent forms are given in Lemma2and Theorem1. The equivalent statements of the best possible constant fac-tor related to some parameters are considered in Theorem2. The operator expressions, some particular cases, and examples are given in Theorem3, Remark1, Remark3, and Examples1–3. The lemmas and theorems provide an extensive account of this type of inequalities.

Funding

This work is supported by the National Natural Science Foundation (No. 61772140) and Guangxi Natural Science Foundation (No. 2016GXNSFAA380125).

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. RL participated in the design of the study and performed the numerical analysis. All authors read and approved the ﬁnal manuscript.

Author details

1_{School of Mathematics and Statistics, Hechi University, Yizhou, P.R. China.}2_{Department of Mathematics, Guangdong}

University of Education, Guangzhou, China.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 17 February 2019 Accepted: 10 May 2019

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