a half-discrete hilbert-type inequality with a non-homogeneous kernel and two variables

Mediterr. J. Math. 10 (2013), 677–692DOI 10.1007/s00009-012-0213-51660-5446/13/020677-16, published online July 3, 2012© 2012 Springer Basel AG

Mediterranean Journalof Mathematics

A Half-Discrete Hilbert-Type Inequalitywith a Non-Homogeneous Kernel andTwo Variables

Bicheng Yang

Abstract. By using the way of weight functions and the technique ofreal analysis, a half-discrete Hilbert-type inequality with a general non-homogeneous kernel and two interval variables, as well as a best constantfactor is given. The equivalent forms, the operator expressions, the re-verses and some particular examples are considered.

Mathematics Subject Classification (2010). Primary 26D15; Secondary47A07.Keywords. Hilbert-type inequality, non-homogeneous kernel, weightfunction, equivalent form, reverse.

1. Introduction

Assuming that p > 1, 1p + 1

q = 1, f(≥ 0) ∈ Lp(0,∞), g(≥ 0) ∈ Lq(0,∞), ||f ||p= {∫∞

0fp(x)dx} 1

p > 0, ||g||q > 0, we have the following famous Hardy-Hilbert’s integral inequality (cf. [3]):∫ ∞

0

∫ ∞

0

f(x)g(y)x + y

dxdy <π

sin(π/p)||f ||p||g||q, (1.1)

where the constant factor πsin(π/p) is the best possible. Moreover, for am, bn ≥

0, a = {am}∞m=1 ∈ lp, b = {bn}∞n=1 ∈ lq, ||a||p = {∑∞m=1 a

pm} 1

p > 0, ||b||q > 0,we have the discrete variant of the above inequality

∞∑m=1

∞∑n=1

ambnm + n

<π

sin(π/p)||a||p||b||q, (1.2)

with the same best constant factor. For p = q = 2, the above two inequalitiesreduce respectively to Hilbert’s inequalities. Inequalities (1.1) and (1.2) are

This work is supported by Guangdong Natural Science Foundation (No. 7004344).

678 B. Yang Mediterr. J. Math.

important in analysis and its applications (cf. [12], [15], [19]) and they stillrepresent the field of interest to numerous mathematicians.

In 1998, by introducing an independent parameter λ ∈ (0, 1], Yang[13] gave an extension of (1.1) for p = q = 2. By generalizing the resultsfrom [13], Yang [16] gave the following best extensions of (1.1) and (1.2)concerning the general homogeneous kernel: Let λ, λ1, λ2 ∈ R, λ1 + λ2 = λand let kλ(x, y) is a non-negative homogeneous function of degree −λ, i.e.kλ(tx, ty) = t−λkλ(x, y), for any x, y, t > 0. If k(λ1) =

∫∞0

kλ(t, 1)tλ1−1dt ∈R+, φ(x) = xp(1−λ1)−1, ψ(x) = xq(1−λ2)−1, f(≥ 0) ∈ Lp,φ(R+) = {f |||f ||p,φ := {∫∞

0φ(x)|f(x)|pdx} 1

p < ∞}, g(≥ 0) ∈ Lq,ψ(R+), ||f ||p,φ, ||g||q,ψ >0, then ∫ ∞

0

∫ ∞

0

kλ(x, y)f(x)g(y)dxdy < k(λ1)||f ||p,φ||g||q,ψ, (1.3)

where the constant factor k(λ1) is the best possible. Moreover if kλ(x, y) isfinite and kλ(x, y)xλ1−1(kλ(x, y)yλ2−1) is decreasing for x > 0(y > 0), thenfor am, bn ≥ 0, a = {am}∞m=1 ∈ lp,φ = {a|||a||p,φ := {∑∞

m=1 φ(m)|am|p} 1p <

∞}, ||a||p,φ, ||b||q,ψ > 0, we have the discrete inequality∞∑

m=1

∞∑n=1

kλ(m,n)ambn < k(λ1)||a||p,φ||b||q,ψ, (1.4)

with the same best constant factor k(λ1). Clearly, if λ = 1, k1(x, y) = 1x+y ,

λ1 = 1q , λ2 = 1

p , inequality (1.3) reduces to (1.1), while (1.4) reduces to (1.2).Some other results about Hilbert-type inequalities are provided by [20]-[11].

Recently, Yang [17] gave the following half-discrete Hilbert’s inequalitywith the best constant factor B(λ1, λ2)(λ1 > 0, 0 < λ2 ≤ 1, λ1 + λ2 = λ):∫ ∞

0

f(x)∞∑

n=1

an(x + n)λ

dx < B(λ1, λ2)||f ||p,φ||a||q,ψ. (1.5)

Yang [18] and Yang et al. [22] gave some half-discrete Hilbert-type inequal-ity with the general homogeneous kernels. A few half-discrete Hilbert-typeinequalities with the non-homogeneous kernels were provided by Hardy etal. [3] in Theorem 351. But they did not prove that the the constant fac-tors are the best possible. However Yang [14] gave a result with the kernel

1(1+nx)λ

and an interval variable as follows: If u(t) is a differentiable strictincreasing function in (n0 − 1,∞)(n0 ∈ N), such that u((n0 − 1)+) = 0,u(∞) = ∞, λ > 0, (u(t))

λ−22 u′(t)(t ∈ (n0 − 1,∞)) is decreasing, f(x), an ≥ 0,

0 <∫∞n0−1

(u(x))1−λ

u′(x) f2(x)dx < ∞ and 0 <∑∞

n=n0

(u(n))1−λ

u′(n) a2n < ∞, then∫ ∞

n0−1

f(x)∞∑

n=n0

an(1 + u(n)u(x))λ

dx

< B(λ

2,λ

2)

{∫ ∞

n0−1

(u(x))1−λ

u′(x)f2(x)dx

∞∑n=n0

(u(n))1−λ

u′(n)a2n

} 12

, (1.6)

Vol. 10 (2013) A Half-Discrete Hilbert-Type Inequality 679

where the constant factor B(λ2 ,λ2 ) is the best possible.

In this paper, by using the way of weight functions and the techniqueof real analysis, a half-discrete Hilbert-type inequality with a general non-homogeneous kernel and two interval variables, as well as well as a bestconstant factor is given, which is a generalization of (1.6) and a refinementof Theorem 351 in [3]. Some equivalent forms, the operator expressions, thereverses and a few particular examples are considered.

2. Some Lemmas

In order to obtain our main results, we need some auxiliary lemmas.

Lemma 2.1. If α ∈ R, u(x)(x ∈ (b, c),−∞ ≤ b < c ≤ ∞) and v(y)(y ∈[n0,∞), n0 ∈ N) are strictly increasing differential functions with u(b+) =0, v(n0) > 0, u(c−) = v(∞) = ∞, h(t) is a non-negative finite measurablefunction in R+, then we define two weight functions ω(n) and �(x) as fol-lows:

ω(n) := [v(n)]α∫ c

b

h(u(x)v(n))[u(x)]α−1u′(x)dx, n ≥ n0(n ∈ N), (2.1)

�(x) := [u(x)]α∞∑

n=n0

h(u(x)v(n))[v(n)]α−1v′(n), x ∈ (b, c). (2.2)

It follows

ω(n) = k(α) :=∫ ∞

0

h(t)tα−1dt, n ≥ n0(n ∈ N). (2.3)

Moreover, let f(x, y) := [u(x)]αh(u(x)v(y))[v(y)]α−1v′(y), equipped with oneof the following conditions:

Condition (i): v(y)(y ∈ [n0−1,∞)) is strictly increasing with v(n0−1) ≥0 and for any fixed x ∈ (b, c), f(x, y) is decreasing for y ∈ [n0 − 1,∞) andstrictly decreasing on a subinterval of [n0 − 1,∞).

Condition (ii): v(y)(y ∈ [n0 − 12 ,∞)) is strictly increasing with v(n0 −

12 ) ≥ 0 and for any fixed x ∈ (b, c), f(x, y) is decreasing and strict convex for

y ∈ [n0 − 12 ,∞)

Condition (iii): There exists a constant β ≥ 0, such that v(y)(y ∈[n0 − β,∞)) is strictly increasing with v(n0 − β) ≥ 0, and for any fixedx ∈ (b, c), f(x, y) is piecewise smooth satisfying

R(x) :=∫ n0

n0−β

f(x, y)dy − 12f(x, n0) −

∫ ∞

n0

ρ(y)f ′y(x, y)dy > 0,

where ρ(y)(= y − [y] − 12 ) is Bernoulli function of first order. If k(α) ∈ R+

and one of the above conditions is fulfilled, then

�(x) < k(α)(x ∈ (b, c)). (2.4)


Proof. Applying the substitution t = u(x)v(n) to relation (2.1), we get (2.3)after an easy calculation.

Further, if Condition (i) is fulfilled, in view of k(α) ∈ R+, then we have

�(x) =∞∑

n=n0

f(x, n) < [u(x)]α∫ ∞

n0−1

h(u(x)v(y))[v(y)]α−1v′(ydy

t=u(x)v(y)=

∫ ∞

u(x)v(n0−1)

h(t)tα−1dt ≤∫ ∞

0

h(t)tα−1dt = k(α).

Moreover, if Condition (ii) is satisfied, since k(α) ∈ R+, then by Hada-mard’s inequality (cf. [9]), we obtain

�(x) =∞∑

n=n0

f(x, n) <∫ ∞

n0− 12

f(x, y)dy

t=u(x)v(y)=

∫ ∞

u(x)v(n0− 12 )

h(t)tα−1dt ≤∫ ∞

0

h(t)tα−1dt = k(α).

Finally, since k(α) ∈ R+, Condition (iii) together with the Euler-Mac-laurin summation formula (cf. [16]) yields

�(x) =∞∑

n=n0

f(x, n) =∫ ∞

n0

f(x, y)dy +12f(x, n0) +

∫ ∞

n0

ρ(y)f ′y(x, y)dy

=∫ ∞

n0−β

f(x, y)dy −R(x) =∫ ∞

u(x)v(n0−β)

h(t)tα−1dt−R(x)

≤∫ ∞

0

h(t)tα−1dt−R(x) < k(α).

The lemma is proved. �

Lemma 2.2. Let the assumptions of Lemma 2.1 be fulfilled and additionally,p > 0(p �= 1), 1

p + 1q = 1, an ≥ 0, n ≥ n0(n ∈ N), f(x) is a non-negative mea-

surable function in (b, c). (i) If p > 1, then we have the following inequalities:

J1 :=

{ ∞∑n=n0

v′(n)[v(n)]1−pα

[∫ c

b

h(u(x)v(n))f(x)dx]p} 1

p

≤ [k(α)]1q

{∫ c

b

�(x)[u(x)]p(1−α)−1

[u′(x)]p−1fp(x)dx

} 1p

, (2.5)

L1 :=

{∫ c

b

[�(x)]1−qu′(x)[u(x)]1−qα

[ ∞∑n=n0

h(u(x)v(n))an

]q

dx

} 1q

≤{k(α)

∞∑n=n0

[v(n)]q(1−α)−1

[v′(n)]q−1aqn

} 1q

; (2.6)

(ii) if 0 < p < 1, then the reverse inequalities of (2.5) and (2.6) are valid.


Proof. (i) By using Holder’s inequality with weight (cf. [9]) and relation (2.3),it follows[∫ c

b

h(u(x)v(n))f(x)dx]p

=

{∫ c

b

h(u(x)v(n))

[[u(x)]

1−αq

[v(n)]1−αp

[v′(n)]1p f(x)

[u′(x)]1q

]

×[[v(n)](1−α)/p

[u(x)](1−α)/q

[u′(x)]1/q

[v′(n)]1/p

]dx

}p

≤∫ c

b

h(u(x)v(n))[u(x)](1−α)(p−1)v′(n)[v(n)]1−α[u′(x)]p−1

fp(x)dx

×{∫ c

b

h(u(x)v(n))[v(n)](1−α)(q−1)u′(x)[u(x)]1−α[v′(n)]q−1

dx

}p−1

={ω(n)[v(n)]q(1−α)−1

[v′(n)]q−1

}p−1 ∫ c

b

h(u(x)v(n))[u(x)](1−α)(p−1)v′(n)fp(x)

[v(n)]1−α[u′(x)]p−1dx

=[k(α)]p−1

[v(n)]pα−1v′(n)

∫ c

b

h(u(x)v(n))[u(x)](1−α)(p−1)v′(n)[v(n)]1−α[u′(x)]p−1

fp(x)dx.

Then by Lebesgue term by term integration theorem (cf. [8]), we have

J1 ≤ [k(α)]1q

{ ∞∑n=n0

∫ c

b

h(u(x)v(n))[u(x)](1−α)(p−1)v′(n)fp(x)

[v(n)]1−α[u′(x)]p−1dx

} 1p

= [k(α)]1q

{∫ c

b

∞∑n=n0

h(u(x)v(n))[u(x)](1−α)(p−1)v′(n)fp(x)

[v(n)]1−α[u′(x)]p−1dx

} 1p

= [k(α)]1q

{∫ c

b

�(x)[u(x)]p(1−α)−1


} 1p

,

and (2.5) holds. Yet another use of Holder’s inequality yields inequality[ ∞∑n=n0

h(u(x)v(n))an

]q

=

{ ∞∑n=n0

h(u(x)v(n))

[[u(x)]

1−αq

[v(n)]1−αp

[v′(n)]1p

[u′(x)]1q

]

×[[v(n)](1−α)/p

[u(x)](1−α)/q

[u′(x)]1/q

[v′(n)]1/pan

]}q

≤{ ∞∑

n=n0

h(u(x)v(n))[u(x)](1−α)(p−1)

[v(n)]1−α

v′(n)[u′(x)]p−1

}q−1

×∞∑

n=n0

h(u(x)v(n))[v(n)](1−α)(q−1)

[u(x)]1−α

u′(x)[v′(n)]q−1

aqn

=[u(x)]1−qα

[�(x)]1−qu′(x)

∞∑n=n0

h(u(x)v(n))u′(x)[v(n)]α

[u(x)]1−α

[v(n)]q(1−α)−1

[v′(n)]q−1aqn,


while Lebesgue term by term integration theorem provides

L1 ≤{∫ c

b

∞∑n=n0

h(u(x)v(n))u′(x)[v(n)]α

[u(x)]1−α

[v(n)]q(1−α)−1

[v′(n)]q−1aqndx

} 1q

=

{ ∞∑n=n0

[[v(n)]α

∫ c

b

h(u(x)v(n))u′(x)dx

[u(x)]1−α

][v(n)]q(1−α)−1

[v′(n)]q−1aqn

} 1q

=

{ ∞∑n=n0

ω(n)[v(n)]q(1−α)−1

[v′(n)]q−1aqn

} 1q

,

and then in view of (2.3), inequality (2.6) follows.(ii) By the reverse Holder’s inequality (cf. [9]) and taking into account

that q < 0, we establish the reverses of (2.5) and (2.6) in the same way. �

3. Main Results

Now we are ready to establish our main results. For that sake we introducethe functions

Φ(x) :=[u(x)]p(1−α)−1

[u′(x)]p−1(x ∈ (b, c)), and Ψ(n) :=

[v(n)]q(1−α)−1

[v′(n)]q−1

(n ≥ n0, n ∈ N), wherefrom we get

[Φ(x)]1−q =u′(x)

[u(x)]1−qα, and [Ψ(n)]1−p =

v′(n)[v(n)]1−pα

.

As before, we deal with the non-negative functions and sequences, hence,such types of conditions will go without saying.

Theorem 3.1. Suppose that the assumptions of Lemma 2.1 are fulfilled, letk(α) =

∫∞0

h(t)tα−1dt ∈ R+, and let p > 1, 1p + 1

q = 1, f(x), an ≥ 0,f ∈ Lp,Φ(b, c), a = {an}∞n=n0

∈ lq,Ψ, ||f ||p,Φ, ||a||q,Ψ > 0, then the followinginequalities hold and are equivalent:

I :=∞∑

n=n0

an

∫ c

b

h(u(x)v(n))f(x)dx

=∫ c

b

f(x)∞∑

n=n0

h(u(x)v(n))andx < k(α)||f ||p,Φ||a||q,Ψ, (3.1)

J :=

{ ∞∑n=n0

[Ψ(n)]1−p

[∫ c

b


p

< k(α)||f ||p,Φ, (3.2)

L :=

{∫ c

b

[Φ(x)]1−q

[ ∞∑n=n0

h(u(x)v(n))an

]q

dx

} 1q

< k(α)||a||q,Ψ. (3.3)


Moreover, if v′(y)v(y) (y ≥ n0) is decreasing and there exist the constants δ > α

and M > 0, such that h(t) ≤ Mtδ

(t ∈ [v(n0),∞)), then the constant factork(α) is the best possible in the above inequalities.

Proof. The proof consist of two parts. In the first part we prove the aboveinequalities together with their equivalence. Note also that, by Lebesgue termby term integration theorem, there are two expressions for I in (3.1).

By Lemma 2.1, �(x) < k(α), so inequality (3.2) follows immediatelyform relation (2.5). By Holder’s inequality, we have

I =∞∑

n=n0

[Ψ−1q (n)

∫ c

b

h(u(x)v(n))f(x)dx]

[Ψ1q (n)an] ≤ J ||a||q,Ψ, (3.4)

so we get (3.1). Moreover, suppose that inequality (3.1) is valid. By consid-ering the sequence

an := [Ψ(n)]1−p

[∫ c

b

h(u(x)v(n))f(x)dx]p−1

, n ≥ n0,

we have Jp−1 = ||a||q,Ψ. Further, the inequality (2.5) implies that J < ∞. IfJ = 0, then (3.2) holds trivially; if J > 0, then by (3.1), we have

||a||qq,Ψ = Jp = I < k(α)||f ||p,Φ||a||q,Ψ, i.e. ||a||q−1q,Ψ = J < k(α)||f ||p,Φ,

which means that the inequality (3.1) and (3.2) are equivalent.The equivalence of (3.1) and (3.3) is established in the same way. More

precisely, Lemma 2.1 implies that [�(x) ]1−q > [k(α)]1−q, hence (3.3) followsfrom (2.6). Now, the Holder’s inequality implies

I =∫ c

b

[Φ1p (x)f(x)]

[Φ−1p (x)

∞∑n=n0

h(u(x)v(n))an

]dx ≤ ||f ||p,ΦL, (3.5)

that is, we have (3.1). On the contrary, assuming that (3.1) is valid anddefining

f(x) := [Φ(x)]1−q

[ ∞∑n=n0

h(u(x)v(n))an

]q−1

, x ∈ (b, c),

we have Lq−1 = ||f ||p,Φ. Clearly by (2.6), we find that L < ∞. If L = 0, then(3.3) holds trivially; if L > 0, then by (3.1), we have

||f ||pp,Φ = Lq = I < k(α)||f ||p,Φ||a||q,Ψ, i.e. ||f ||p−1p,Φ = L < k(α)||a||q,Ψ,

which yields the equivalence of (3.1) and (3.3). Hence, inequalities (3.1), (3.2)and (3.3) are equivalent.

Now, we are going to prove that the constant factor k(α) is the best pos-sible in (3.1), (3.2) and (3.3). There exists an unified constant d ∈ (b, c), sat-isfying u(d) = 1. For 0 < ε < p(δ − α), setting f(x) = [u(x)]α+

εp−1u′(x), x ∈

(b, d); f(x) = 0, x ∈ [d, c), and an = [v(n)]α−εq−1v′(n), n ≥ n0, if there exists


a positive number k(≤ k(α)), such that (3.1) is valid as we replace k(α) withk, then in particular, it follows

I :=∞∑

n=n0

∫ c

b

h(u(x)v(n))anf(x)dx < k||f ||p,Φ||a||q,Ψ

= k

{∫ d

b

u′(x)[u(x)]−ε+1

dx

} 1p{

v′(n0)[v(n0)]ε+1

+∞∑

n=n0+1

v′(n)[v(n)]ε+1

} 1q

≤ k(1ε)

1p

{v′(n0)

[v(n0)]ε+1+∫ ∞

n0

v′(y)[v(y)]ε+1

dy

} 1q

=k

ε

{ε

v′(n0)[v(n0)]ε+1

+ [v(n0)]−ε

} 1q

, (3.6)

I =∞∑

n=n0

[v(n)]α−εq−1v′(n)

∫ d

b

h(u(x)v(n))[u(x)]α+εp−1u′(x)dx

t=u(x)v(n)=

∞∑n=n0

[v(n)]−ε−1v′(n)∫ v(n)

0

h(t)tα+εp−1dt

=∞∑

n=n0

[v(n)]−ε−1v′(n)

[∫ ∞

0

h(t)tα+εp−1dt−

∫ ∞

v(n)

h(t)tα+εp−1dt

]

≥ k(α +ε

p)∫ ∞

n0

[v(y)]−ε−1v′(y)dy −A(ε)

=1εk(α +

ε

p)[v(n0)]−ε −A(ε),

A(ε) :=∞∑

n=n0

[v(n)]−ε−1v′(n)∫ ∞

v(n)

h(t)tα+εp−1dt. (3.7)

Since for h(t) ≤ M( 1tδ

)(δ > α; t ∈ [v(n0),∞)), we find

0 < A(ε) ≤ M

∞∑n=n0

[v(n)]−ε−1v′(n)∫ ∞

v(n)

tα−δ+ εp−1dt

=M

δ − α− εp

∞∑n=n0

[v(n)]α−δ− εq−1v′(n)

=M

δ − α− εp

[v′(n0)

[v(n0)]δ−α+ εq+1

+∞∑

n=n0+1

v′(n)[v(n)]δ−α+ ε

q+1

]

≤ M

δ − α− εp

[v′(n0)

[v(n0)]δ−α+ εq+1

+∫ ∞

n0

v′(y)[v(y)]δ−α+ ε

q+1dy

]

=M

δ − α− εp

[v′(n0)

[v(n0)]δ−α+ εq+1

+[v(n0)]α−δ− ε

q

δ − α + εq

]< ∞,


that is, A(ε) = O(1)(ε → 0+). Hence by (3.6) and (3.7), we have

k(α +ε

p)[v(n0)]−ε − εO(1) < k

{ε

v′(n0)[v(n0)]ε+1

+ [v(n0)]−ε

} 1q

. (3.8)

By Fatou Lemma (cf. [8]), we have k(α) ≤ limε→0+k(α + εp ), then by (3.8),

it follows k(α) ≤ k(ε → 0+), a.e. k = k(α) is the best value of (2.6).Due to the equivalence, it follows easily that k(α) is also the best possible

constant factor in (3.2) and (3.3). Namely, if we suppose that k(α) is notthe best possible constant in (3.2) and (3.3), then the relation (3.4) and(3.5) imply that k(α) is not the best possible constant factor in (3.1), whichcontradicts with the previously proved facts. �

Inequalities (3.2) and (3.3) enable us to define some interesting operatorsbetween some particular functions spaces. Due to Theorem 3.1, we shall beable to determine the norm of such operators.

Remark 3.2. (i) Define a half-discrete Hilbert’s operator T : Lp,Φ(b, c) →lp,Ψ1−p in the following way: For f ∈ Lp,Φ(b, c), we define Tf ∈ lp,Ψ1−p , as

Tf(n) =∫ c

b

h(u(x)v(n))f(x)dx, n ≥ n0.

Then by (3.2), it follows ||Tf ||p.Ψ1−p ≤ k(α)||f ||p,Φ, a.e. T is the boundedoperator with ||T || ≤ k(α). Since the constant factor in (3.2) is the bestpossible, we have ||T || = k(α).

(ii) Define a half-discrete Hilbert’s operator T : lq,Ψ → Lq,Φ1−q (b, c) asfollows: for a ∈ lq,Ψ, we define T a ∈ Lq,Φ1−q (b, c) in the following way

T a(x) =∞∑

n=n0

h(u(x)v(n))an, x ∈ (b, c).

Due to (3.3), it follows ||T a||q.Φ1−q ≤ k(α)||a||q,Ψ, that is T is the boundedoperator with ||T || ≤ k(α). Since inequality (3.3) includes the best possibleconstant factor, we have ||T || = k(α).

Remark 3.3. In particular, for n0 = 1, b = 0, c = ∞, u(x) = x, v(y) = y in(3.1), (3.2) and (3.3), we have the following equivalent inequalities with thebest constant factor k(α):

∞∑n=1

an

∫ ∞

0

h(xn)f(x)dx

< k(α){∫ ∞

0

xp(1−α)−1fp(x)dx} 1

p

{ ∞∑n=1

nq(1−α)−1aqn

} 1q

, (3.9)

{ ∞∑n=1

npα−1

[∫ ∞

0

h(xn)f(x)dx]p} 1

p

< k(α){∫ ∞

0

xp(1−α)−1fp(x)dx} 1

p

,

(3.10)


{∫ ∞

0

xqα−1

[ ∞∑n=1

h(xn)an

]q

dx

} 1q

< k(α)

{ ∞∑n=1

nq(1−α)−1aqn

} 1q

, (3.11)

which are some refinements of Theorem 351 in [3].In the following theorems, for 0 < p < 1(q < 0) or p < 0(0 < q < 1), we

still use the formal symbols of ||f ||p,˜Φ and ||a||q,Ψ et al.

Theorem 3.4. Let the assumptions of Lemma 2.1 be fulfilled and additionally,k(α) ∈ R+, θλ(x) ∈ (0, 1), k(α)(1 − θλ(x)) < �(x) < k(α)(x ∈ (b, c)). If0 < p < 1, 1

p + 1q = 1, f(x), an ≥ 0, Φ(x) := (1 − θλ(x))Φ(x)(x ∈ (b, c)),

0 < ||f ||p,˜Φ < ∞ and 0 < ||a||q,Ψ < ∞, then the following inequalities hold

and are equivalent:

I =∞∑

n=n0

∫ c

b

h(u(x)v(n))anf(x)dx

=∫ c

b

∞∑n=n0

h(u(x)v(n))anf(x)dx > k(α)||f ||p,˜Φ||a||q,Ψ, (3.12)

J =

{ ∞∑n=n0

[Ψ(n)]1−p

[∫ c

b


p

> k(α)||f ||p,˜Φ, (3.13)

L :=

{∫ c

b

[Φ(x)]1−q

[ ∞∑n=n0

h(u(x)v(n))an

]q

dx

} 1q

> k(α)||a||q,Ψ. (3.14)

Moreover, if v′(y)v(y) (y ≥ n0) is decreasing and there exist constants δ′, δ0 > 0,

satisfying θλ(x) = O([u(x)]δ′)(x ∈ (b, d)), and k(α + δ0) ∈ R+, then the

constant factor k(α) is the best possible in the above inequalities.

Proof. In view of (2.3) and the reverse of (2.5), for �(x) > k(λ1)(1− θλ(x)),we have (3.13). Moreover, the reverse Holder’s inequality yields

I =∞∑

n=n0

[Ψ−1q (n)

∫ c

b

h(u(x)v(n))f(x)dx]

[Ψ1q (n)an] ≥ J ||a||q,Ψ, (3.15)

hence (3.12) holds, since (3.13) is valid. On the other-hand, assuming that(3.12) is valid and setting an as Theorem 3.1, then Jp−1 = ||a||q,Ψ. By thereverse inequality in (2.5), we find that J > 0. If J = ∞, then (3.13) holdstrivially; if J < ∞, then by (3.9), we have

||a||qq,Ψ = Jp = I > k(α)||f ||p,˜Φ||a||q,Ψ, i.e. ||a||q−1q,Ψ = J > k(α)||f ||p,˜Φ,

that is, the relation (3.12) and (3.13) are equivalent.In view of (2.3) and the reverse of (2.6), for [�(x)]1−q > [k(α)(1 −

θλ(x))]1−q (q < 0), we have (3.14). By the reverse Holder’s inequality, wehave

I =∫ c

b

[Φ1p (x)f(x)]

[Φ−1p (x)

∞∑n=n0

h(u(x)v(n))an

]dx ≥ ||f ||p,˜ΦL. (3.16)


Then by (3.14), we have (3.12). On the other-hand, assuming that (3.12) isvalid, setting

f(x) := [Φ(x)]1−q

[ ∞∑n=n0

h(u(x)v(n))an

]q−1

, x ∈ (b, c),

then Lq−1 = ||f ||p,˜Φ. By the reverse of (2.6), we find L > 0. If L = ∞, then

(3.14) holds trivially; if L < ∞, then by (3.12), we have

||f ||pp,˜Φ

= Lq = I > k(α)||f ||p,˜Φ||a||q,Ψ, i.e. ||f ||p−1

p,˜Φ= L > k(α)||a||q,Ψ,

that is, (3.12) and (3.14) are equivalent. Hence inequalities (3.12), (3.13) and(3.14) are equivalent.

For 0 < ε < min{δ′, pδ0}, setting f(x), an as Theorem 3.1, if thereexists a positive number k(≥ k(α)), such that (3.12) is valid as we replacek(α) with k, then in particular, for q < 0, in view of (2.3) and the conditions,we have

I :=∫ c

b

∞∑n=n0

h(u(x)v(n))anf(x)dx > k||f ||p,˜Φ||a||q,Ψ

= k

{∫ d

b

(1 −O([u(x)]δ

′) u′(x)dx[u(x)]ε+1

} 1p{ ∞∑

n=n0

v′(n)[v(n)]ε+1

} 1q

= k{1ε−O(1)} 1

p

{v′(n0)

[v(n0)]ε+1+

∞∑n=n0+1

v′(n)[v(n)]ε+1

} 1q

≥ k{1ε−O(1))}

1p

{v′(n0)

[v(n0)]ε+1+∫ ∞

n0

v′(y)[v(y)]ε+1

dy

} 1q

=k

ε{1 − εO(1)} 1

p {ε v′(n0)[v(n0)]ε+1

+ [v(n0)]−ε} 1q , (3.17)

I =∞∑

n=n0

[v(n)]α−εq−1v′(n)

∫ d

b


≤∞∑

n=n0

[v(n)]λ2− εq−1v′(n)

∫ c

b


t=u(x)v(n)=

∞∑n=n0

[v(n)]−ε−1v′(n)∫ ∞

0

h(t)tα+εp−1dt

= k(α +ε

p)

[v′(n0)

[v(n0)]ε+1+

∞∑n=n0+1

[v(n)]−ε−1v′(n)

]

≤ k(α +ε

p)[

v′(n0)[v(n0)]ε+1

+∫ ∞

n0

[v(y)]−ε−1v′(y)dy]


=1εk(α +

ε

p)[ε

v′(n0)[v(n0)]ε+1

+ [v(n0)]−ε

]. (3.18)

Since we have h(t)tα+εp−1 ≤ h(t)tα+δ0−1, t ∈ [1,∞) and∫ ∞

1

h(t)tα+δ0−1dt ≤ k(α + δ0) < ∞,

then by Lebesgue control convergence theorem (cf. [8]), it follows

k(α +ε

p) ≤

∫ ∞

1

h(t)tα+εp−1dt +

∫ 1

0

h(t)tα−1dt

= k(α) + o(1)(ε → 0+).

By (3.14) and (3.15), we have

(k(α) + o(1))[ε

v′(n0)[v(n0)]ε+1

+ [v(n0)]−ε

]> k {1 − εO(1)} 1

p

[ε

v′(n0)[v(n0)]ε+1

+ [v(n0)]−ε

] 1q

,

that is, we get k(α) ≥ k(ε → 0+). Hence k = k(α) is the best possibleconstant in (3.12).

Due to the equivalence, it follows that the constant factor k(α) in (3.13)and (3.14) is the best possible, Namely, if we suppose that k(α) is not the bestconstant factor in (3.13) and (3.14), then the relations (3.15) and (3.16) implythat k(α) is not the best constant factor in (3.12), which gives a contradiction.

�

For p < 0(0 < q < 1), in the same way, we have the following theorem:

Theorem 3.5. Let the assumptions of Lemma 2.1 be fulfilled and additionally,k(α) ∈ R+, �(x) < k(α)(x ∈ (b, c)). If p < 0, 1

p + 1q = 1, f(x), an ≥ 0,

0 < ||f ||p,Φ < ∞ and 0 < ||a||q,Ψ < ∞, then the following inequalities holdand are equivalent:

I =∞∑

n=n0

∫ c

b

h(u(x)v(n))anf(x)dx

=∫ c

b

∞∑n=n0

h(u(x)v(n))anf(x)dx > k(α)||f ||p,Φ||a||q,Ψ, (3.19)

J =

{ ∞∑n=n0

[Ψ(n)]1−p

[∫ c

b


p

> k(α)||f ||p,Φ, (3.20)

L :=

{∫ c

b

[Φ(x)]1−q

[ ∞∑n=n0

h(u(x)v(n))an

]q

dx

} 1q

> k(α)||a||q,Ψ. (3.21)


Moreover, if v′(y)v(y) (y ≥ n0) is decreasing and there exist constants δ0 > 0,

satisfying k(α − δ0) ∈ R+, then the constant factor k(α) is the best possiblein the above inequalities.

4. Some Examples

We conclude this paper with few applications of our main results. More pre-cisely, we consider here some examples of non-homogeneous kernels in partic-ular settings which form the function f(x, y) which satisfy different conditionsfrom Lemma 2.1. These particular results will be presented in the form (3.1),while the equivalent forms, as well as the reverse inequalities will here beomitted. In this section, we use the notations as in Lemma 2.1.

Example 4.1. For h(t) = 1(1+t)λ

(0 < λ ≤ 2), α = λ2 , setting δ = 2λ

3 > λ2 , it

follows h(t)tδ = O(1)(t ∈ [1,∞)). If �(x) < k(λ2 ) = B(λ2 ,λ2 ), then by (3.1),

we have the following inequality with the best constant factor:

I :=∞∑

n=n0

an

∫ c

b

f(x)dx(1 + u(x)v(n))λ

< B(λ

2,λ

2)

×{∫ c

b

(u(x))p(1−λ2 )−1


} 1p{ ∞∑

n=n0

(v(n))q(1−λ2 )−1

[v′(n)]q−1aqn

} 1q

. (4.1)

In particular, for b = n0 − 1, c = ∞, u(x) = v(x), p = q = 2, (4.1) reduces to(1.6).

Example 4.2. For n0 = 1, b = 0, c = ∞, u(x) = v(x) = xβ(β > 0), h(t) =1

(max{1,t})λ (λ > 0, 0 < λβ ≤ 2), α = λ2 , by Condition (i), we have �(x) <

k(λ2 ) = 4λ . Setting δ = λ > λ

2 , it follows h(t) = 1tδ

(t ∈ [1,∞)). By (3.1), wehave the following inequality with the best constant factor 4

βλ :∞∑

n=1

an

∫ ∞

0

1(max{1, xβnβ})λ f(x)dx

<4βλ

{∫ ∞

0

xp(1−λβ2 )−1fp(x)dx

} 1p

{ ∞∑n=1

nq(1−λβ2 )−1aqn

} 1q

. (4.2)

Example 4.3. For n0 = 1, b = β, c = ∞, u(x) = v(x) = (x − β)(0 ≤ β ≤12 ), h(t) = ln t

tλ−1(0 < λ ≤ 2), α = λ

2 , by Condition (ii), we have �(x) <

k(λ2 ) = (πλ )2. Setting δ = 2λ3 > λ

2 , we have h(t)tδ = O(1)(t ∈ [1,∞)). By(3.1), we have the following inequality with the best constant factor (πλ )2:

∞∑n=1

an

∫ ∞

β

ln[(x− β)(n− β)](x− β)λ(n− β)λ − 1

f(x)dx (4.3)

< (π

λ)2

{∫ ∞

β

(x− β)p(1−λ2 )−1fp(x)dx

} 1p

{ ∞∑n=1

(n− β)q(1−λ2 )−1aqn

} 1q

.


Example 4.4. For n0 = 1, b = 1 − β = γ, c = ∞, u(x) = v(x) = (x − γ), γ ≤1 − 1

8 [λ +√λ(3λ + 4)](0 < λ ≤ 4), h(t) = 1

1+tλ, α = λ

2 , we have

k(λ

2) =

∫ ∞

0

h(t)tλ2 −1dt =

2λ

∫ ∞

0

11 + tλ

dtλ2

=2λ

arctan tλ2 |∞0 =

π

λ∈ R+,

and

f(x, y) =(x− γ)

λ2 (y − γ)

λ2 −1

1 + (x− γ)λ(y − γ)λ.

Hence v(y)(y ∈ [γ,∞)) is strictly increasing with v(1 − β) = v(γ) = 0, andfor any fixed x ∈ (γ,∞), f(x, y) is smooth with

f ′y(x, y) = −(1 +

λ

2)

(x− γ)λ2 (y − γ)

λ2 −2

1 + (x− γ)λ(y − γ)λ+

λ(x− γ)λ2 (y − γ)

λ2 −2

[1 + (x− γ)λ(y − γ)λ]2.

We set

R(x) :=∫ 1

γ

f(x, y)dy − 12f(x, 1) −

∫ ∞

1

ρ(y)f ′y(x, y)dy. (4.4)

For x ∈ (γ,∞), 0 < λ ≤ 4, by (4.4) and the following improved Euler-Maclaurin summation formula (cf. [16]),

−18

g(1) <∫ ∞

1

ρ(y)g(y)dy < 0((−1)ig(i)(y) > 0, g(i)(∞) = 0, i = 0, 1),

we have

R(x) =∫ 1

γ

(x− γ)λ2 (y − γ)

λ2 −1

1 + (x− γ)λ(y − γ)λdy − 1

2(x− γ)

λ2 (1 − γ)

λ2 −1

1 + (x− γ)λ(1 − γ)λ

+ (1 +λ

2)∫ ∞

1

ρ(y)(x− γ)

λ2 (y − γ)

λ2 −2

1 + (x− γ)λ(y − γ)λdy

−∫ ∞

1

ρ(y)λ(x− γ)

λ2 (y − γ)

λ2 −2

[1 + (x− γ)λ(y − γ)λ]2dy

>2λ

arctan(1 − γ)λ2 (x− γ)

λ2 − (1 − γ)

λ2 −1(x− γ)

λ2

2[1 + (1 − γ)λ(x− γ)λ]

− 18(1 +

λ

2)

(1 − γ)λ2 −2(x− γ)

λ2

1 + (1 − γ)λ(x− γ)λ+ 0

= h(x) :=2λ

arctan(1 − γ)λ2 (x− γ)

λ2

−[1 − γ

2+

18(1 +

λ

2)]

(1 − γ)λ2 −2(x− γ)

λ2

1 + (1 − γ)λ(x− γ)λ.


Since for γ ≤ 1− 18 [λ+

√λ(3λ + 4)], i.e. 1−γ ≥ 1

8 [λ+√

λ(3λ + 4)](0 < λ ≤ 4),

h′(x) =(1 − γ)

λ2 (x− γ)

λ2 −1

1 + (1 − γ)λ(x− γ)λ−[1 − γ

2+

18(1 +

λ

2)]

×{λ(1 − γ)

λ2 −2(x− γ)

λ2 −1

2[1 + (1 − γ)λ(x− γ)λ]− λ(1 − γ)

3λ2 −2(x− γ)

3λ2 −1

[1 + (1 − γ)λ(x− γ)λ]2

}

=[(1 − γ)2 − λ

4(1 − γ) − λ

16(1 +

λ

2)]

(1 − γ)λ2 −2(x− γ)

λ2 −1

1 + (1 − γ)λ(x− γ)λ

+[1 − γ

2+

18(1 +

λ

2)]λ(1 − γ)

3λ2 −2(x− γ)

3λ2 −1

[1 + (1 − γ)λ(x− γ)λ]2> 0,

then h(x) is strictly increasing and R(x) > h(x) > h(γ) = 0.Then by Condition (iii), it follows �(x) < k(λ2 ) = π

λ (x ∈ (γ,∞)). Forδ = λ > λ

2 = α, it follows

h(t) =1

1 + tλ≤ 1

tδ, t ∈ [1 − γ,∞),

and by (3.1), we have the following inequality with the best constant factor:∞∑

n=1

an

∫ ∞

γ

11 + (x− γ)λ(n− γ)λ

f(x)dx

<π

λ

{∫ ∞

γ

(x− γ)p(1−λ2 )−1fp(x)dx

} 1p

{ ∞∑n=1

(n− γ)q(1−λ2 )−1aqn

} 1q

(4.5)

where γ ≤ 1 − 18 [λ +

√λ(3λ + 4)](0 < λ ≤ 4).

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Bicheng YangDepartment of MathematicsGuangdong University of EducationHaizhuqu Xingang Zhonglu 351Guangzhou, Guangdong 510303P. R. Chinae-mail: [email protected]

Received: November 2, 2011.Revised: April 15, 2012.Accepted: April 26, 2012.

a half-discrete hilbert-type inequality with a non-homogeneous kernel and two variables

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