A half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor

Xin and Yang Journal of Inequalities and Applications 2012, 2012:184http://www.journalofinequalitiesandapplications.com/content/2012/1/184

RESEARCH Open Access

A half-discrete Hilbert-type inequality withthe non-monotone kernel and the bestconstant factorDongmei Xin* and Bicheng Yang

*Correspondence:[email protected] of Mathematics,Guangdong Education University,Guangzhou, Guangdong 510303,People’s Republic of China

AbstractBy introducing two pairs of conjugate exponents and using the improvedEuler-Maclaurin summation formula, we estimate the weight functions and obtain ahalf-discrete Hilbert-type inequality with the non-monotone kernel and the bestconstant factor. We also consider its equivalent forms.MSC: 26D15

Keywords: Hilbert-type inequality; conjugate exponent; Hölder’s inequality; bestconstant factor; equivalent form

1 IntroductionIf an,bn ≥ , such that <

∑∞n= an < ∞ and <

∑∞n= bn < ∞, then we have the famous

Hilbert’s inequality as follows (cf. []):

∞∑n=

∞∑m=

ambnm + n

< π

( ∞∑n=

an∞∑n=

bn

)

, ()

where the constant factor π is the best possible.Under the same condition of (), Xin et al. [] gave the following inequality:

∞∑n=

∞∑m=

| ln(m/n)|m + n

ambn < c

( ∞∑n=

an∞∑n=

bn

)

, ()

where the constant factor c = ∑∞

n=(–)n–(n–) = .

+ is the best possible. And Yang []gave the integral analogues of ().In , Hardy et al. [] established a few results on the half-discrete Hilbert-type in-

equalities with the non-homogeneous kernel (see Theorem ). But they did not provethat the constant factors are the best possible. However, Yang [] gave a result by introduc-ing an interval variable and proved that the constant factor is the best possible. Recently,Yang et al. [–] gave some half-discrete Hilbert-type inequalities and their reverses withthe monotone kernels and best constant factors.

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Xin and Yang Journal of Inequalities and Applications 2012, 2012:184 Page 2 of 8http://www.journalofinequalitiesandapplications.com/content/2012/1/184

Recently, Yang [] gave the following half-discreteHilbert-type inequality with the non-monotone kernel and the best constant factor :

∞∑n=

∫ ∞

| ln(x/n)|anmax{x,n} f (x)dx <

( ∞∑n=

an∫ ∞

f (x)dx

)

. ()

Obviously, for a half-discrete Hilbert-type inequality with the monotone kernel, it is easyto build the relating inequality by estimating the series formand the integral formofweightfunctions. However, for a half-discrete Hilbert-type inequality with the non-monotonekernel, it is much more difficult to prove.In this paper, by using the way of weight functions, we give a new half-discrete Hilbert-

type inequality with the non-monotone kernel as follows:

∞∑n=

an∫ ∞

| ln( xn )|f (x)x + n

dx < ∞∑k=

(–)k

(k + )

( ∞∑n=

an∫ ∞

f (x)

)

, ()

where the constant factor ∑∞

k=(–)k(k+) is the best possible. The main objective of this

paper is to build the best extension of () with parameters and equivalent forms.

2 Some lemmasLemma. If x ∈ R, n ∈ Z (Z is the set of non-negative integers), [x] = n, ρ(y) = y–[y]–

(y ∈ R) is the Bernoulli function of first order [], then we have (cf. [])

∫ x

nρ(y)dy = –ε

(ε ∈ [, ]

). ()

Lemma . If r > , r +

s = , f (x, y) :=

| ln( xy )|x+y ( xy )

r (x, y ∈ (,∞)), N is the set of positive

integers, define the weight functions as follows:

ω(n) :=∫ ∞

| ln( xn )|x + n

(nx

) sdx (n ∈N), ()

� (x) :=∞∑n=

| ln( xn )|x + n

(xn

) r (

x ∈ [,∞)). ()

Then we have

ω(n) < cr :=∞∑k=

(–)k[

(k +

r )+

(k +

s )

], � (x) < cr . ()

Proof Setting u = xn in (), we have

ω(n) =∫ ∞

n

| lnu|u +

u– s du <

∫ ∞

| lnu|u +

u– s du = cr .

Setting u = yx , then it follows

∫

f (x, y)dy =

∫ x

(– lnu)u +

u– r du > s

∫ x

(– lnu)x +

du– r + =

x r

x + (s lnx + s

). ()



For ≤ y < x, f (x, y) = – ln( yx )x+y (

xy )

r , we have

f ′y (x, y) = x

r

[–

(y + x)y+

r+

ln( yx )(y + x)y

r+

ln( yx )r(y + x)y+

r

]

= –x r

[

(y + x)y+ r+

x ln( yx )(y – x)(y + x)y

r+

x ln( yx )r(y – x)(y + x)y+

r

]

+ x r

[ln( yx )

(y – x)(y + x)y r+

ln( yx )r(y – x)y+

r

].

For y ≥ x, f (x, y) = ln( yx )x+y (

xy )

r , we find

f ′y (x, y) = x

r

[

(y + x)y+ r–

ln( yx )(y + x)y

r–

ln( yx )r(y + x)y+

r

]

= x r

[

(y + x)y+ r+

x ln( yx )(y – x)(y + x)y

r+

x ln( yx )r(y – x)(y + x)y+

r

]

– x r

[ln( yx )

(y – x)(y + x)y r+

ln( yx )r(y – x)y+

r

].

Define two functions as follows:

g(y) =

⎧⎪⎨⎪⎩g(y) = x

r [ (y+x)y+

r+ x ln( yx )

(y–x)(y+x)yr+ x ln( yx )

r(y–x)(y+x)y+r], y < x;

g(y) = x r [ ln( yx )

(y–x)(y+x)yr+ ln( yx )

r(y–x)y+r], y≥ x,

h(y) =

⎧⎪⎨⎪⎩h(y) = x

r [ ln( yx )

(y–x)(y+x)yr+ ln( yx )

r(y–x)y+r], y < x;

h(y) = x r [

(y+x)y+r+ x ln( yx )

(y–x)(y+x)yr+ x ln( yx )

r(y–x)(y+x)y+r], y≥ x.

Then we have –f ′y (x, y) = g(y) – h(y). Setting a =

x , b = –x , then a – b =

x . Define twofunctions as follows:

g(y) =

⎧⎨⎩g(y) – a, y < x;g(y), y≥ x,

h(y) =

⎧⎨⎩h(y) – b, y < x;h(y), y ≥ x.

Since g(x – ) – a = g(x), h(x – ) – b = h(x), then both g(y) and h(y) (y ∈ [,∞)) aredecreasing and continuous. Besides y = x, we have (–)ig(i)(y)≥ , (–)ih(i)(y)≥ (i = , ),and g(∞) = h(∞) = . By the improved Euler-Maclaurin summation formula (cf. [], The-orem ..) and (), for ε ∈ [, ], εi ∈ (, ) (i = , ), it follows

–∫ ∞

ρ(y)f ′

y (x, y)dy

=∫ ∞

ρ(y)g(y)dy –

∫ ∞

ρ(y)h(y)dy

=∫ ∞

ρ(y)g(y)dy + a

∫ x

ρ(y)dy –

[∫ ∞

ρ(y)h(y)dy + b

∫ x

ρ(y)dy

]



=∫ ∞

ρ(y)g(y)dy –

∫ ∞

ρ(y)h(y)dy + (a – b)

[∫ [x]

ρ(y)dy +

∫ x

[x]ρ(y)dy

]

= –ε

g() + ε

h() – ε

(a – b) = –ε

(g() – a

)+

ε

(h() – b

)–

ε

(a – b).

Since g() – a ≥ g(x–) – a = g(x) > , h() – b ≥ h(x–) – b = h(x) > , then we have

–∫ ∞

ρ(y)f ′

y (x, y)dy > –(g() – a

)–(a – b)

= –x

r

(x + )–

x+ r lnx

(x + )(x – )–

x+ r lnx

r(x + )(x – )–

x

. ()

By the improved Euler-Maclaurin summation formula [], we have

� (x) =∞∑k=

f (x,k) =∫ ∞

f (x, y)dy +

f (x, ) +

∫ ∞

ρ(y)f ′

y (x, y)dy

=∫ ∞

f (x, y)dy –

(∫

f (x, y)dy –

f (x, ) –

∫ ∞

ρ(y)f ′

y (x, y)dy)= cr – θ (x),

where

θ (x) :=∫

f (x, y)dy –

f (x, ) –

∫ ∞

ρ(y)f ′

y (x, y)dy.

Since – f (x, ) = –

xr lnx

(x+) , in view of (), (), (i) for ≤ x < , – lnxx– ≥ –, we have

θ (x) > x r

x + (s lnx + s

)–

x r lnx

(x + )–

x r

(x + )

–x+

r lnx(x + )(x – )

–x+

r lnxr(x + )(x – )

–

x

≥ x r lnx

(x + )

(s –

)+

x r

x +

[s –

–

x(x + )

–xr

–x + x+

r

]

>x

r

(x + )

( –

–––

)= ;

(ii) for x≥ , – x– ≥ –

x , we have

θ (x) ≥ x r lnx

(x + )

[s –

–

(x + )

–r

]+

x r

x +

(s –

–

x + x+

r

)

>x

r lnx(x + )

(s –

––

r

)=(s + r – )x

r lnxr(x + )

> .

Hence, for x ≥ , we have θ (x) > , it follows � (x) < cr . The lemma is proved. �



Lemma . As the assumption of Lemma ., if < ε < pr , R = (

r –

εp )

–, S = ( s +εp )

–, thenwe have

I :=∞∑n=

n S –ε–

∫ ∞

| ln( xn )|x + n

x R– dx≥ cR

ε–O(). ()

Proof It is obvious that R > , R +

S = . Setting u =

xn , we have

I =∞∑n=

n–ε–∫ ∞

n

| lnu|u +

u R– du

=∞∑n=

n–ε–[∫ ∞

| lnu|u +

u R– du +

∫ n

lnuu +

u R– du

]

> cR∫ ∞

x–ε– dx + R

∞∑n=

n–ε–∫

n

lnudu

R

=cRε–

[R

∞∑n=

lnnn+ε+

R+ R

∞∑n=

n+ε+

R

]

=cRε–O()

(ε → +

).

The lemma is proved. �

Lemma . If p, r > , r +s =

p +

q = , an ≥ , f (x) is a non-negative measurable function,

then we have

J :=∞∑n=

nps –

[∫ ∞

| ln( xn )|f (x)x + n

dx]p

≤ cpr∫ ∞

x

ps –f p(x)dx, ()

J :=∫ ∞

x

qr –

[ ∞∑n=

| ln( xn )|anx + n

]q

dx ≤ cqr∞∑n=

nqr –aqn. ()

Proof By Hölder’s inequality [], in view of () and (), we have

[∫ ∞

| ln( xn )|f (x)x + n

dx]p

={∫ ∞

| ln( xn )|x + n

[x

sq

nrpf (x)

][n

rp

xsq

]dx

}p

≤∫ ∞

| ln( xn )|x + n

(xn

) rx

ps –f p(x)dx

[∫ ∞

| ln( xn )|x + n

(nx

) sn

qr – dx

]p–

=∫ ∞

| ln( xn )|x + n

(xn

) rx

ps –f p(x)dx

[n

qr –ω(n)

]p–

≤ n–ps +cp–r

∫ ∞

| ln( xn )|x + n

(xn

) rx

ps –f p(x)dx,

J ≤ cp–r

∞∑n=

∫ ∞

| ln( xn )|x + n

(xn

) rx

ps –f p(x)dx



= cp–r

∫ ∞

[ ∞∑n=

| ln( xn )|x + n

(xn

) r]x

ps –f p(x)dx

= cp–r

∫ ∞

� (x)x

ps –f p(x)dx ≤ cpr

∫ ∞

x

ps –f p(x)dx.

Hence we have (). Still by Hölder’s inequality [], () and (), we have

[ ∞∑n=

| ln( xn )|anx + n

]q

={ ∞∑

n=

| ln( xn )|x + n

[x

sq

nrp

][n

rp

xsqan

]}q

≤[� (x)

∞∑n=

xps –

]q– ∞∑n=

| ln( xn )|x + n

(xn

) rn

qr –aqn

≤ x–qr +cq–r

∞∑n=

| ln( xn )|x + n

(xn

) rn

qr –aqn,

J ≤ cq–r

∫ ∞

[ ∞∑n=

| ln( xn )|x + n

(xn

) rn

qr –aqn

]dx

= cq–r

∞∑n=

[∫ ∞

| ln( xn )|x + n

(xn

) rdx

]n

qr –aqn

= cq–r

∞∑n=

ω(n)nqr –aqn ≤ cqr

∞∑n=

nqr –aqn.

Then we have (). The lemma is proved. �

3 Main results and applicationsTheorem . If p, r > ,

p +q =

r +

s = , an, f (x)≥ such that <

∫ ∞ x

ps –f p(x)dx < ∞,∑∞

n= nqr –aqn <∞, then we have the following equivalent inequalities:

I :=∞∑n=

an∫ ∞

| ln( xn )|f (x)x + n

dx =∫ ∞

f (x)

∞∑n=

| ln( xn )|anx + n

dx

< cr{∫ ∞

x

ps –f p(x)dx

} p{ ∞∑

n=n

qr –aqn

} q

, ()

J =∞∑n=

nps –

[∫ ∞

| ln( xn )|f (x)x + n

dx]p

< cpr∫ ∞

x

ps –f p(x)dx, ()

J =∫ ∞

x

qr –

[ ∞∑n=

| ln( xn )|anx + n

]q

dx < cqr∞∑n=

nqr –aqn, ()

where the constant factors cr =∑∞

k=(–)k[ (k+

r )+

(k+ s )], cpr , cqr are the best possible.

Proof By the Lebesgue term-by-term integration theorem [], there are two kinds of rep-resentation in (). By the conditions of Theorem., () takes the formof a strict inequal-



ity, and we have (). By Hölder’s inequality [], we have

I =∞∑n=

[n

s –

p

∫ ∞

| ln( xn )|f (x)x + n

dx][n–

s +

p an

] ≤ Jp

{ ∞∑n=

nqr –aqn

} q

. ()

By (), we have (). On the other hand, suppose that () is valid. Setting an :=n

ps –[

∫ ∞

| ln( xn )|f (x)x+n dx]p–, n ∈N, then it follows J =

∑∞n= n

qr –aqn. By (), we have J <∞. If

J = , then () is obvious value; if < J <∞, then by (), we obtain

∞∑n=

nqr –aqn = J = I < cr

{∫ ∞

x

ps –f p(x)dx

} p{ ∞∑

n=n

qr –aqn

} q

,

Jp =

{ ∞∑n=

nqr –aqn

} p

< cr{∫ ∞

x

ps –f p(x)dx

} p.

()

Hence we have (), which is equivalent to ().By Hölder’s inequality [], we have

I =∫ ∞

[xr –

q f (x)

][x–

r +

q

∞∑n=

| ln( xn )|anx + n

]dx ≤

{∫ ∞

x

ps –f p(x)dx

} pJq . ()

By (), we have (). On the other hand, suppose that () is valid. Setting f (x) :=x

qr –[

∑∞n=

| ln( xn )|anx+n ]q–, x ∈ [,∞), then it follows J =

∫ ∞ x

ps –f p(x)dx. By (), we have

J <∞. If J = , then () is obvious value; if < J <∞, then by (), we obtain

∫ ∞

x

ps –f p(x)dx = J = I < cr

{∫ ∞

x

ps –f p(x)dx

} p{ ∞∑

n=n

qr –aqn

} q

,

Jq =

{∫ ∞

x

ps –f p(x)dx

} q< cr

{ ∞∑n=

nqr –aqn

} q

.

()

Hencewe have (), which is equivalent to (). Therefore (), () and () are equivalent.If the constant factor cr in () is not best possible, then there exists a positive number

K , with < K < cr , such that () is still valid if we replace cr by K . For < ε < ε, settingf (x) = n

r –

εp–, an = n

s –

εq– (n ∈N ), we have

I =∞∑n=

an∫ ∞

| ln( xn )|f (x)x + n

dx < K{∫ ∞

x

ps – f p(x)dx

} p{ ∞∑

n=n

qr –aqn

} q

= K(∫ ∞

x––ε dx

) p( +

∞∑n=

n––ε

) q

< K(ε

) p( +

∫ ∞

x––ε dx

) q=Kε(ε + )

q . ()

By () and (), we have cR – εO() < K(ε + )q and for ε → +, by Fatou lemma [], we

have cr ≤ limε→+(cR – εO()) ≤ K . This is a contradiction. Hence we can conclude that



the constant cr in () is the best possible. If the constant factors in () and () are not thebest possible, then we can imply a contradiction that the constant factor in () is not thebest possible by () and (). The theorem is proved. �

Remark For p = q = r = s = , () reduces to (). Inequality () is a new basic half-discreteHilbert-type inequality with the non-monotone kernel.

Competing interestsThe authors declare that they have no competing interests.

Authors’ contributionsDX carried out the study, and wrote the manuscript. BY participated in the design of the study, and reformed themanuscript. All authors read and approved the final manuscript.

AcknowledgementsThis work was supported by the Emphases Natural Science Foundation of Guangdong Institutions of Higher Learning,College and University (No. 05Z026) and the Natural Science Foundation of Guangdong (7004344).

Received: 19 December 2011 Accepted: 9 August 2012 Published: 30 August 2012

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doi:10.1186/1029-242X-2012-184Cite this article as: Xin and Yang: A half-discrete Hilbert-type inequality with the non-monotone kernel and the bestconstant factor. Journal of Inequalities and Applications 2012 2012:184.


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A half-discrete Hilbert-type inequality with the non-monotone kernel and the best constant factor