A more accurate half-discrete reverse Hilbert-type inequality with a non-homogeneous kernel

Yang and Chen Journal of Inequalities and Applications 2014, 2014:96http://www.journalofinequalitiesandapplications.com/content/2014/1/96

RESEARCH Open Access

Amore accurate half-discrete reverseHilbert-type inequality with anon-homogeneous kernelBicheng Yang1* and Qiang Chen2

*Correspondence:[email protected] of Mathematics,Guangdong University ofEducation, Guangzhou, Guangdong510303, P.R. ChinaFull list of author information isavailable at the end of the article

AbstractBy means of weight functions and the improved Euler-Maclaurin summation formula,a more accurate half-discrete reverse Hilbert-type inequality with the kernel(min{1,(x–γ )(n–η)})β(max{1,(x–γ )(n–η)})α and a best constant factor is given. Some equivalent forms, the dualforms as well as some related homogeneous cases, are also considered.MSC: 26D15

Keywords: Hilbert-type inequality; weight function; equivalent form; reverse; dualform

1 IntroductionAssuming that f , g ∈ L(R+), ‖f ‖ = {∫ ∞

f (x)dx} > , ‖g‖ > , we have the following

Hilbert integral inequality (cf. []):

∫ ∞

∫ ∞

f (x)g(y)x + y

dxdy < π‖f ‖‖g‖, ()

where the constant factor π is best possible. If a = {an}∞n=, b = {bn}∞n= ∈ l, ‖a‖ ={∑∞

n= an} > , ‖b‖ > , then we have the following analogous discrete Hilbert inequality:

∞∑m=

∞∑n=

ambnm + n

< π‖a‖‖b‖, ()

with the same best constant factor π . Inequalities () and () are important in analysis andits applications (cf. [–]).In , by introducing an independent parameter λ ∈ (, ], Yang [] gave an extension

of (). For generalizing the results from [], Yang [] gave some best extensions of () and(): If p > , p +

q = , λ +λ = λ, kλ(x, y) is a non-negative homogeneous function of degree

–λ satisfying k(λ) =∫ ∞ kλ(t, )tλ– dt ∈ R+, φ(x) = xp(–λ)–, ψ(x) = xq(–λ)–, f (≥ ) ∈

Lp,φ(R+) = {f |‖f ‖p,φ := {∫ ∞ φ(x)|f (x)|p dx}

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

∫ ∞

kλ(x, y)f (x)g(y)dxdy < k(λ)‖f ‖p,φ‖g‖q,ψ , ()

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where the constant factor k(λ) is best possible.Moreover if the value of kλ(x, y) is finite andkλ(x, y)xλ– (kλ(x, y)yλ–) is decreasing for x > (y > ), then for am,bn ≥ , a = {am}∞m= ∈lp,φ = {a|‖a‖p,φ := {∑∞

n= φ(n)|an|p}p < ∞}, and b = {bn}∞n= ∈ lq,ψ , ‖a‖p,φ ,‖b‖q,ψ > , we

have

∞∑m=

∞∑n=

kλ(m,n)ambn < k(λ)‖a‖p,φ‖b‖q,ψ , ()

where the constant k(λ) is still best value. Clearly, for p = q = , λ = , k(x, y) = x+y , λ =

λ = , () reduces to (), while () reduces to (). The reverses of () and () as well as the

equivalent forms are also considered by [].Some other results about integral and discrete Hilbert-type inequalities can be found

in [–]. On half-discrete Hilbert-type inequalities with the general non-homogeneouskernels, Hardy et al. provided a few results in Theorem of []. But they did not provethat the constant factors are best possible. In , Yang [] gave a result with the kernel

(+nx)λ by introducing a variable and proved that the constant factor is best possible. Veryrecently, Yang [] and [] gave the following half-discrete reverse Hilbert inequality withbest constant factor: For < p < ,

p +q = , λ > , < λ ≤ , λ +λ = λ, θλ(x) =O(

xλ ) ∈(, ), φ(x) = ( – θλ(x))xp(–λ)–,

∫ ∞

f (x)

∞∑n=

an(x + n)λ

dx > B(λ,λ)‖f ‖p,φ‖a‖q,ψ . ()

In this paper, by means of weight functions and the improved Euler-Maclaurin summa-tion formula, a more accurate half-discrete reverse Hilbert-type inequality with the kernel(min{,(x–γ )(n–η)})β(max{,(x–γ )(n–η)})α similar to () and a best constant factor is given. Moreover, some equiva-lent forms, the dual forms as well as some relating homogeneous cases are also considered.

2 Some lemmasLemma If n ∈ N, s > n, g(y) (y ∈ [n, s)), g(y) (y ∈ [s,∞)) are continuous decreasingfunctions satisfying g(n) – g(s–)+ g(s) > , g(∞) = , define a function g(y) as follows:

g(y) :={g(y), y ∈ [n, s),g(y), y ∈ [s,∞).

()

Then there exists ε ∈ [, ], such that

–

[g(n) + ε

(g(s) – g(s – )

)]<

∫ ∞

nρ(y)g(y)dy < – ε

(g(s) – g(s – )

), ()

where ρ(y) = y–[y]– is the Bernoulli function of the first order. In particular, for g(y) = ,

y ∈ [n, s), we have g(s) > and

–g(s) <

∫ ∞

sρ(y)g(y)dy <

g(s); ()



for g(y) = , y ∈ [s,∞), if g(s – )≥ , then it follows g(n) > and

–g(n) <

∫ s

nρ(y)g(y)dy < . ()

Proof Define a continuous decreasing function g(y) as follows:

g(y) :={g(y) + g(s) – g(s – ), y ∈ [n, s),g(y), y ∈ [s,∞).

Then it follows that∫ ∞

nρ(y)g(y)dy =

∫ s

nρ(y)g(y)dy +

∫ ∞

sρ(y)g(y)dy

=∫ s

nρ(y)

(g(y) – g(s) + g(s – )

)dy +

∫ ∞

sρ(y)g(y)dy

=∫ ∞

nρ(y)g(y)dy –

(g(s) – g(s – )

)∫ s

nρ(y)dy,

∫ s

nρ(y)dy =

∫ [s]

nρ(y)dy +

∫ s

[s]ρ(y)dy =

∫ s

[s]

(y – [s] –

)dy

=

[(s – [s] –

)–

]=

ε –

(ε ∈ [, ]

).

Since g(n) = g(n) + g(s) – g(s – ) > , g(y) is a non-constant continuous decreasingfunction with g(∞) = g(∞) = , by the improved Euler-Maclaurin summation formula(cf. [], Theorem ..), it follows that

–

(g(n) + g(s) – g(s – )

)=–g(n) <

∫ ∞

nρ(y)g(y)dy < ,

and then in view of the above results and by simple calculation, we have (). �

Lemma If < α+β ≤ , γ ∈ R, η ≤ – α+β

(+√ +

α+β), and ω(n) and� (x) are weight

functions given by

ω(n) :=∫ ∞

γ

(min{, (x – γ )(n – η)})β(max{, (x – γ )(n – η)})α

(n – η)α–β

(x – γ )–α–β

dx, n ∈N,

� (x) :=∞∑n=

(min{, (x – γ )(n – η)})β(max{, (x – γ )(n – η)})α

(x – γ )α–β

(n – η)–α–β, x > γ , ()

then we have

<

α + β

( – θ (x)

)<� (x) < ω(n) =

α + β

, ()

θ (x) ={

( – η)

α+β (x – γ )

α+β , < x – γ ≤

–η,

– ( – η)–

α+β (x – γ )–

α+β , x – γ >

–η.

()



Proof Substituting t = (x – γ )(n – η) in (), and by simple calculation, we have

ω(n) =∫ ∞

(min{, t})β(max{, t})α t

α–β – dt =

∫

tβ+

α–β – dt +

∫ ∞

t–α+ α–β

– dt = α + β

.

For fixed x > γ , we find

h(x, y) := (x – γ )α–β

(min{, (x – γ )(y – η)})β(max{, (x – γ )(y – η)})α (y – η)

α–β –

={(x – γ )

α+β (y – η)

α+β –, η < y < η +

x–γ,

(x – γ )–α+β (y – η)–

α+β –, y ≥ η +

x–γ,

h′y(x, y) =

{–( – α+β

)(x – γ )α+β (y – η)

α+β –, η < y < η +

x–γ,

–( α+β

+ )(x – γ )–α+β (y – η)–

α+β –, y≥ η +

x–γ,∫ ∞

η

h(x, y)dy t=(x–γ )(y–η)=∫ ∞


α–β – dt =

α + β.

By the Euler-Maclaurin summation formula (cf. []), it follows that

� (x) =∞∑n=

h(x,n) =∫ ∞

h(x, y)dy +

h(x, ) +

∫ ∞

ρ(y)h′

y(x, y)dy

=∫ ∞

η

h(x, y)dy – R(x) = α + β

– R(x),

R(x) :=∫

η

h(x, y)dy – h(x, ) –

∫ ∞

ρ(y)h′

y(x, y)dy. ()

(i) For < x – γ < –η

, we obtain – h(x, ) = –

(x – γ )α+β ( – η)

α+β –, and

∫

η

h(x, y)dy = (x – γ )α+β

∫

η

(y – η)α+β – dy = ( – η)

α+β

α + β(x – γ )

α+β .

Setting g(y) := –h′y(x, y), wherefrom g(y) = ( – α+β

)(x – γ )α+β (y – η)

α+β –, g(y) = ( α+β

+)(x – γ )–

α+β (y – η)–

α+β – and

g(

η +

x – γ

)– g

((η +

x – γ

)–

)=

(α + β

+

)(x – γ ) –

( –

α + β

)(x – γ )

= (α + β)(x – γ ) > ,

then by (), we find

–∫ ∞

ρ(y)h′

y(x, y)dy =∫ ∞

ρ(y)g(y)dy

>–

[g() + g

(η +

x – γ

)– g

((η +

x – γ

)–

)]



=–

[( –

α + β

)(x – γ )

α+β ( – η)

α+β – + (α + β)(x – γ )

]>–

[( –

α + β

)( – η)

α+β –(x – γ )

α+β

+ (α + β)( – η)α+β –(x – γ )

α+β –(x – γ )

]=–

( +

α + β

)( – η)

α+β –(x – γ )

α+β .

In view of () and the above results, since for η ≤ – α+β

( +√ +

α+β), namely – η ≥

α+β

( +√ +

α+β), it follows that

R(x) >

α + β( – η)

α+β (x – γ )

α+β –

(x – γ )

α+β ( – η)

α+β –

–

( +

α + β

)( – η)

α+β –(x – γ )

α+β

=[( – η)

α + β–( – η)

– + α + β

](x – γ )

α+β

( – η)–α+β

≥ .

(ii) For x – γ ≥ –η

, we obtain – h(x, ) = –

(x – γ )–α+β ( – η)–

α+β –, and

∫

η

h(x, y)dy =∫ η+

x–γ

η

(x – γ )α+β

(y – η)–α+β

dy +∫

η+ x–γ

(x – γ )–α+β

(y – η)α+β +

dy

=

α + β–

α + β

( – η)–α+β (x – γ )–

α+β

≥ ( – η)–α+β

α + β(x – γ )–

α+β –

( – η)–α+β

α + β(x – γ )–

α+β

=

α + β( – η)–

α+β (x – γ )–

α+β .

Since for y ≥ , y–η ≥ x–γ

, by the improved Euler-Maclaurin summation formula (cf. []),it follows that

–∫ ∞

ρ(y)h′

y(x, y)dy =(

α + β

+

)(x – γ )–

α+β

∫ ∞

ρ(y)(y – η)–

α+β – dy

> –

(α + β

+

)(x – γ )–

α+β ( – η)–

α+β –.

In view of () and the above results, for – η ≥ α+β

( +√ +

α+β), we find

R(x) > α + β

( – η)–α+β (x – γ )–

α+β –

( – η)–

α+β –(x – γ )–

α+β

–

(α + β

+

)( – η)–

α+β –(x – γ )–

α+β

=[( – η)

α + β– – η

– + α + β

](x – γ )–

α+β

( – η)+α+β

≥ .



Hence for x > γ , we have R(x) > , and then � (x) < ω(n) = α+β

.On the other-hand, since h(x, y) is decreasing with respect to y > η, we find

� (x) >∫ ∞

(min{, (x – γ )(y – η)})β(max{, (x – γ )(y – η)})α

(x – γ )α–β

(y – η)–α–β

dy

t=(x–γ )(y–η)=∫ ∞

(–η)(x–γ )


α–β – dt =

α + β

( – θ (x)

),

where θ (x) := α+β

∫ (–η)(x–γ )

(min{,t})β(max{,t})α t

α–β – dt ∈ (, ).

(i) For < x – γ ≤ –η

, we obtain

θ (x) =α + β

∫ (–η)(x–γ )

tβ+

α–β – dt =

( – η)

α+β (x – γ )

α+β .

(ii) For x – γ > –η

, it follows that

θ (x) =α + β

[∫

tβ+

α–β – dt +

∫ (–η)(x–γ )

t–α+ α–β

– dt]

= –( – η)–

α+β (x – γ )–

α+β .

Hence we have () and (). �

Lemma Let the assumptions of Lemma be fulfilled and additionally, let < p < orp < ,

p +q = , an ≥ , n ∈N, f (x) be a non-negative measurable function in (γ ,∞). Then

we have the following inequalities:

J :={ ∞∑

n=(n – β)

p(α–β) –

[∫ ∞

γ

(min{, (x – γ )(n – η)})β(max{, (x – γ )(n – η)})α f (x)dx

]p}

p

≥(

α + β

) q{∫ ∞

γ

� (x)(x – γ )p(–α–β )–f p(x)dx

} p, ()

L :={∫ ∞

γ

(x – α)q(α–β)

–

[� (x)]q–

[ ∞∑n=

(min{, (x – γ )(n – η)})βan(max{, (x – γ )(n – η)})α

]q

dx}

q

≥{

α + β

∞∑n=

(n – η)q(–α–β )–aqn

} q

. ()

Proof For < p < , setting k(x,n) := (min{,(x–γ )(n–η)})β(max{,(x–γ )(n–η)})α , by the reverse Hölder inequality (cf.

[]) and (), it follows that

[∫ ∞

γ


]p

={∫ ∞

γ

k(x,n)[(x – γ )(–

α–β )/q

(n – η)(–α–β )/p

f (x)][

(n – γ )(–α–β )/p

(x – γ )(–α–β )/q

]dx

}p



≥∫ ∞

γ

k(x,n) (x – γ )(–α–β )(p–)

(n – η)–α–β

f p(x)dx

×{∫ ∞

γ

k(x,n) (n – η)(–α–β )(q–)

(x – γ )–α–β

dx}p–

={ω(n)(n – η)q(–

α–β )–}p– ∫ ∞

γ

k(x,n) (x – γ )(–α–β )(p–)

(n – η)–α–β

f p(x)dx

=(

α + β

)p–(n – η)–

p(α–β)

∫ ∞

γ

k(x,n) (x – γ )(–α–β )(p–)

(n – η)–α–β

f p(x)dx.

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

J ≥(

α + β

) q{ ∞∑

n=

∫ ∞

γ

k(x,n) (x – γ )(–α–β )(p–)

(n – η)–α–β

f p(x)dx}

p

=(

α + β

) q{∫ ∞

γ

∞∑n=

k(x,n) (x – γ )(–α–β )(p–)

(n – η)–α–β

f p(x)dx}

p

=(

α + β

) q{∫ ∞

γ

� (x)(x – γ )p(–α–β )–f p(x)dx

} p,

and then () follows. By the reverse Hölder inequality, for q < , we have[ ∞∑n=

k(x,n)an

]q

={ ∞∑

n=k(x,n)

[(x – γ )(–

α–β )/q

(n – η)(–α–β )/p

][(n – η)(–

α–β )/pan

(x – γ )(–α–β )/q

]}q

≤{ ∞∑

n=k(x,n)

(x – γ )(–α–β )(p–)

(n – η)–α–β

}q– ∞∑n=

k(x,n)(n – η)(–

α–β )(q–)

(x – γ )–α–β

aqn

=[� (x)]q–

(x – γ )q(α–β)

–

∞∑n=

k(x,n) (n – η)(–α–β )(q–)

(x – γ )–α–β

aqn.

By the Lebesgue term by term integration theorem, we have

L ≥{∫ ∞

γ

∞∑n=

k(x,n)(n – η)(–

α–β )(q–)

(x – γ )–α–β

aqn dx}

q

={ ∞∑

n=

∫ ∞

γ

k(x,n)(n – η)(–

α–β )(q–)

(x – γ )–α–β

aqn dx}

q

={ ∞∑

n=ω(n)(n – η)q(–

α–β )–aqn

} q

,

and in view of (), inequality () follows. For p < , by the same way we still have ()and (). �

Lemma Let the assumptions of Lemma be fulfilled and additionally, let < p < , p +

q = , < ε < p

(α + β). Setting f (x) = (x – γ )α–β + ε

p–, x ∈ (γ ,γ + ); f (x) = , x ∈ [γ + ,∞),



and an = (n – η)α–β – ε

q–, n ∈N, then we have

I :=∞∑n=

an∫ ∞

γ


<ε

(α + β)( α+β

) – ( εp )

[ε

( – η)ε++

( – η)ε

], ()

H :={∫ ∞

γ

(x – γ )p(–α–β )–f p(x)dx

} p{ ∞∑

n=(n – η)q(–

α–β )–aqn

} q

>ε

( – εO()

) p

{ε

( – η)ε++

( – η)ε

} q. ()

Proof We find

I =∞∑n=

(n – η)α–β – ε

q–∫ γ+

γ

(min{, (x – γ )(n – η)})β(max{, (x – γ )(n – η)})α (x – γ )

α–β + ε

p– dx

<∞∑n=

(n – η)α–β – ε

q–∫ ∞

γ

(min{, (x – γ )(n – η)})β(max{, (x – γ )(n – η)})α (x – γ )

α–β + ε

p– dx

=α + β

( α+β

) – ( εp )

[

( – η)ε++

∞∑n=

(n – η)ε+

]

<α + β

( α+β

) – ( εp )

[

( – η)ε++

∫ ∞

dy(y – η)ε+

]

=ε

(α + β)( α+β

) – ( εp )

[ε

( – η)ε++

( – η)ε

],

and then () is valid. We obtain

H ={∫ γ+

γ

[ –

( – η)

α+β (x – γ )

α+β

](x – γ )ε– dx

} p

×{

( – η)ε+

+∞∑n=

(n – η)ε+

} q

>(ε–O()

) p{

( – η)ε+

+∫ ∞

dy(y – η)ε+

} q

=ε

( – εO()

) p

{ε

( – η)ε++

( – η)ε

} q,

and so () is valid. �

3 Main resultsWe introduce the functions

�(x) := (x – γ )p(–α–β )–, �(x) =

( – θ (x)

)�(x)

(x ∈ (γ ,∞)

),

�(n) := (n – η)q(–α–β )– (n ∈N),



wherefrom [�(x)]–q = (x – γ )qα–β –, [�(x)]–q = ( – θ (x))–q(x – γ )q

α–β – and [�(n)]–p =

(n – η)pα–β –.

Theorem If < α+β ≤ , γ ∈ R, η ≤ – α+β

(+√ +

α+β), < p < ,

p +q = , f (x),an ≥

, f ∈ Lp,�(γ ,∞), a = {an}∞n= ∈ lq,� , ‖f ‖p,� > and ‖a‖q,� > , then we have the followingequivalent inequalities:

I :=∞∑n=

an∫ ∞

γ


=∫ ∞

γ

f (x)∞∑n=

(min{, (x – γ )(n – η)})βan(max{, (x – γ )(n – η)})α dx >

α + β

‖f ‖p,�‖a‖q,� , ()

J ={ ∞∑

n=

[�(n)

]–p[∫ ∞

γ

(min{, (x – γ )(n – η)})β f (x)(max{, (x – γ )(n – η)})α dx

]p}

p

>

α + β‖f ‖p,�, ()

L :={∫ ∞

γ

[�(x)

]–q[ ∞∑n=


]q

dx}

q

>

α + β‖a‖q,� , ()

where the constant α+β

is the best possible in the above inequalities.

Proof The two expressions for I in () follow from Lebesgue’s term by term integrationtheorem. By () and (), we have (). By the reverse Hölder inequality, we have

I =∞∑n=

[�

–q (n)

∫ ∞

γ


][�

q (n)an

] ≥ J‖a‖q,� .

Then by (), we have (). On the other-hand, assume that () is valid. Setting

an :=[�(n)

]–p[∫ ∞

γ


]p–, n ∈N,

it follows that Jp– = ‖a‖q,� . By (), we find J > . If J = ∞, then () is trivially valid; ifJ <∞, then by (), we have

‖a‖qq,� = Jq(p–) = Jp = I >

α + β‖f ‖p,�‖a‖q,� ,

therefore ‖a‖q–q,� = J > α+β

‖f ‖p,�, that is, () is equivalent to (). On the other-hand, by() we have [� (x)]–q > ( – θ (x))–q(

α+β)–q. Then in view of (), we have (). By the

Hölder inequality, we find

I =∫ ∞

γ

[�

p (x)f (x)

][�

–p (x)

∞∑n=


]dx ≥ ‖f ‖p,�L.

Then by (), we have (). On the other-hand, assume that () is valid. Setting

f (x) :=[�(x)

]–q[ ∞∑n=


]q–

, x ∈ (γ ,∞),



then Lq– = ‖f ‖p,�. By (), we find L > . If L =∞, then () is trivially valid; if L <∞, thenby (), we have

‖f ‖pp,� = Lp(q–) = I > α + β

‖f ‖p,�‖a‖q,� ,

therefore ‖f ‖p–p,� = L > α+β

‖a‖q,� , that is, () is equivalent to (). Hence, (), (), and() are equivalent.If there exists a positive number k (≥

α+β), such that () is valid as we replace

α+βwith

k, then in particular, it follows that I > kH . In view of () and (), we have

(α + β)( α+β

) – ( εp )

[ε

( – η)ε++

( – η)ε

]> k

( – εO()

) p

{ε

( – η)ε++

( – η)ε

} q,

and α+β

≥ k (ε → +). Hence k = α+β

is the best value of ().By the equivalence of the inequalities, the constant factor

α+βin () and () is the best

possible. �

For p < , we have the dual forms of (), (), and () as follows:

Theorem If < α +β ≤ , γ ∈ R, η ≤ – α+β

(+√ +

α+β), p < ,

p +q = , f (x),an ≥ ,

f ∈ Lp,�(γ ,∞), a = {an}∞n= ∈ lq,� , ‖f ‖p,� > and ‖a‖q,� > , then we have the followingequivalent inequalities:

∞∑n=

an∫ ∞

γ

(min{, (x – γ )(n – η)})β(max{, (x – γ )(n – η)})α f (x)dx >

α + β

‖f ‖p,�‖a‖q,� , ()

{ ∞∑n=

[�(n)

]–p[∫ ∞

γ


]p}

p

>

α + β‖f ‖p,�, ()

{∫ ∞

γ

[�(x)

]–q[ ∞∑n=


]q

dx}

q

>

α + β‖a‖q,� , ()

where the constant α+β

is the best possible in the above inequalities.

Proof By means of Lemma and the same way, we can prove that (), (), and () arevalid and equivalent. For < ε

|p| (α + β), setting f (x) and an as Lemma , if there exists a

positive number k (≥ α+β

), such that () is valid as we replace α+β

with k, then in partic-ular, by (), it follows that

α + β

( α+β

) – ( εp )

[ε

( – η)ε++

( – η)ε

]

> εI > εk{∫ γ+

γ

(x – γ )ε– dx}

p{ ∞∑

n=

(n – η)ε+

} q

> εk(ε

) p{∫ ∞

dy(y – η)ε+

} q= k

{

( – η)ε

} q,



and α+β

≥ k (ε → +). Hence k = α+β

is the best value of (). By the equivalence of theinequalities, the constant factor

α+βin () and () is the best possible. �

Remark (i) Since we find

min<α+β≤

{ –

α + β

( +

√ +

α + β

)}= –

√

= .+ > ,

then for η = γ = in (), we have

θ(x) ={

x

α+β , < x ≤ ,

– x

– α+β , x > ,

and the following inequality:

∞∑n=

an∫ ∞

(min{,xn})β(max{,xn})α f (x)dx

>

α + β

{∫ ∞

( – θ(x)

)xp(–

α–β )–f p(x)dx

} p{ ∞∑

n=nq(–

α–β )–aqn

} q

. ()

Hence () is a more accurate inequality of ().(ii) For β = in (), we have < α ≤ , γ ∈ R, η ≤ – α

( +√ +

α),

θ(x) ={

(x – γ ) α

, < x – γ ≤ –η

, –

(x – γ )– α , x – γ >

–η


∞∑n=

an∫ ∞

γ

f (x)dx(max{, (x – γ )(n – η)})α

>α

{∫ ∞

γ

( – θ(x)

)(x – γ )p(– α

)–f p(x)dx}

p{ ∞∑

n=(n – η)q(– α

)–aqn

} q

; ()

for α = in (), we have < β ≤ , γ ∈ R, η ≤ – β

( +√ +

β),

θ(x) ={

(x – γ )

β , < x – γ ≤

–η,

– (x – γ )–

β , x – γ >

–η


∞∑n=

an∫ ∞

γ

(min

{, (x – γ )(n – η)

})β f (x)dx

>β

{∫ ∞

γ

( – θ(x)

)(x – γ )p(+

β )–f p(x)dx

} p{ ∞∑

n=(n – η)q(+

β )–aqn

} q

; ()



for β = α = λ in (), we have < λ ≤ , γ ∈ R, η ≤ – λ ( +

√ +

λ),

θ(x) ={

(x – γ )λ, < x – γ ≤

–η,

– (x – γ )–λ, x – γ >

–η


∞∑n=

an∫ ∞

γ

[min{, (x – γ )(n – η)}max{, (x – γ )(n – η)}

]λ

f (x)dx

>λ

{∫ ∞

γ

( – θ(x)

)(x – γ )p–f p(x)dx

} p{ ∞∑

n=(n – η)q–aqn

} q

. ()

(iii) Setting y = x–γ

+ γ , g(y) = (y – γ )α–β–f ( y–γ

+ γ ), ϕ(y) = (y – γ )p(–α–β )– and ϕ(y) =

( – θ ( y–γ

+ γ ))ϕ(y) in (), by simplification, we obtain the following inequality with thehomogeneous kernel:

∞∑n=

an∫ ∞

γ

(min{y – γ ,n – η})β(max{y – γ ,n – η})α g(y)dy >

α + β

‖g‖p,ϕ‖a‖q,� . ()

It is evident that () is equivalent to (), and then the same constant factor α+β

in ()is still the best possible. In the same way, we can find the following inequalities equivalentto () with the same best possible constant factor

α+β:

{ ∞∑n=

[�(n)

]–p[∫ ∞

γ

(min{y – γ ,n – η})βg(y)(max{y – γ ,n – η})α dy

]p}

p

>

α + β‖f ‖p,ϕ , ()

{∫ ∞

γ

[ϕ(y)

]–q[ ∞∑n=

(min{y – γ ,n – η})βan(max{y – γ ,n – η})α

]q

dy}

q

>

α + β‖a‖q,� . ()

(iv) Applying the same way in Theorem , we still can obtain some particular dual formsas (i) and (ii) and some equivalent inequalities similar to (), (), and ().

Competing interestsThe authors declare that they have no competing interests.

Authors’ contributionsThe main idea of this paper was proposed by BY and QC. BY prepared the manuscript initially and performed all the stepsof the proofs in this research. All authors read and approved the final manuscript.

Author details1Department of Mathematics, Guangdong University of Education, Guangzhou, Guangdong 510303, P.R. China.2Department of Computer Science, Guangdong University of Education, Guangzhou, Guangdong 510303, P.R. China.

AcknowledgementsThis work is supported by the National Natural Science Foundation of China (No. 61370186), 2012 KnowledgeConstruction Special Foundation Item of Guangdong Institution of Higher Learning College and University(No. 2012KJCX0079), and Guangzhou Science and Technology Plan Item (No. 2013J4100009).

Received: 20 December 2013 Accepted: 11 February 2014 Published: 25 Feb 2014



References1. Hardy, GH, Littlewood, JE, Pólya, G: Inequalities. Cambridge University Press, Cambridge (1934)2. Mitrinovic, DS, Pecaric, JE, Fink, AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer

Academic, Boston (1991)3. Yang, B: Hilbert-Type Integral Inequalities. Bentham Science Publishers, Dubai (2009)4. Yang, B: Discrete Hilbert-Type Inequalities. Bentham Science Publishers, Dubai (2011)5. Yang, B: On Hilbert’s integral inequality. J. Math. Anal. Appl. 220, 778-785 (1998)6. Yang, B: The Norm of Operator and Hilbert-Type Inequalities. Science Press, Beijing (2009)7. Yang, B, Brnetic, I, Krnic, M, Pecaric, J: Generalization of Hilbert and Hardy-Hilbert integral inequalities. Math. Inequal.

Appl. 8(2), 259-272 (2005)8. Krnic, M, Pecaric, J: Hilbert’s inequalities and their reverses. Publ. Math. (Debr.) 67(3-4), 315-331 (2005)9. Jin, J, Debnath, L: On a Hilbert-type linear series operator and its applications. J. Math. Anal. Appl. 371, 691-704 (2010)10. Azar, L: On some extensions of Hardy-Hilbert’s inequality and applications. J. Inequal. Appl. 2009, Article ID 546829

(2009)11. Yang, B, Rassias, T: On the way of weight coefficient and research for Hilbert-type inequalities. Math. Inequal. Appl.

6(4), 625-658 (2003)12. Arpad, B, Choonghong, O: Best constant for certain multilinear integral operator. J. Inequal. Appl. 2006, Article ID

28582 (2006)13. Kuang, J, Debnath, L: On Hilbert’s type inequalities on the weighted Orlicz spaces. Pac. J. Appl. Math. 1(1), 95-103

(2007)14. Zhong, W: The Hilbert-type integral inequalities with a homogeneous kernel of –λ-degree. J. Inequal. Appl. 2008,

Article ID 917392 (2008)15. Li, Y, He, B: On inequalities of Hilbert’s type. Bull. Aust. Math. Soc. 76(1), 1-13 (2007)16. Yang, B: A mixed Hilbert-type inequality with a best constant factor. Int. J. Pure Appl. Math. 20(3), 319-328 (2005)17. Yang, B: A half-discrete Hilbert’s inequality. J. Guangdong Univ. Educ. 31(3), 1-7 (2011)18. Yang, B, Chen, Q: A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension. J. Inequal.

Appl. 2011, 124 (2011). doi:10.1186/1029-242X-2011-12419. Kuang, J: Applied Inequalities. Shangdong Science Technic Press, Jinan (2004)20. Kuang, J: Introduction to Real Analysis. Hunan Education Press, Changsha (1996)

10.1186/1029-242X-2014-96Cite this article as: Yang and Chen: Amore accurate half-discrete reverse Hilbert-type inequality with anon-homogeneous kernel. Journal of Inequalities and Applications 2014, 2014:96


http://dx.doi.org/10.1186/1029-242X-2011-124

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