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Journal of Inequalities and Applications

Erschienen in:

Open Access 01.12.2012 | Research

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

verfasst von: Bicheng Yang, Xindong Liu

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2012

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Abstract

By means of weight functions and the improved Euler-Maclaurin summation formula, a more accurate half-discrete Hilbert-type inequality with a non-homogeneous kernel and a best constant factor is given. A best extension, some equivalent forms, the operator expressions as well as some particular cases are also considered.

MSC:26D15, 47A07.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

BY carried out the molecular genetic studies participated in the sequence alignment and drafted the manuscript. XL conceived of the study and participated in its design and coordination. All authors read and approved the final manuscript.

1 Introduction

Assuming that

f, g \in L^{2} (R_{+})

∥ f ∥ = {\int_{0}^{\infty} f^{2} (x) d x}^{\frac{1}{2}} > 0

∥ g ∥ > 0

, we have the following Hilbert’s integral inequality (cf. [1]):

\int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{x + y} d x d y < π ∥ f ∥ ∥ g ∥,

(1)

where the constant factor π is best possible. If

a = {a_{n}}_{n = 1}^{\infty}, b = {b_{n}}_{n = 1}^{\infty} \in l^{2}

∥ a ∥ = {\sum_{n = 1}^{\infty} a_{n}^{2}}^{\frac{1}{2}} > 0

∥ b ∥ > 0

, then we have the following analogous discrete Hilbert’s inequality:

\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{a_{m} b_{n}}{m + n} < π ∥ a ∥ ∥ b ∥,

(2)

with the same best constant factor π. Inequalities (1) and (2) are important in analysis and its applications (cf. [2‐4]).

In 1998, by introducing an independent parameter

λ \in (0, 1]

, Yang [5] gave an extension of (1). For generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2) as follows. If

p > 1

\frac{1}{p} + \frac{1}{q} = 1

λ_{1} + λ_{2} = λ

k_{λ} (x, y)

is a non-negative homogeneous function of degree −λ satisfying

k (λ_{1}) = \int_{0}^{\infty} k_{λ} (t, 1) t^{λ_{1} - 1} d t \in R_{+}

ϕ (x) = x^{p (1 - λ_{1}) - 1}

ψ (x) = x^{q (1 - λ_{2}) - 1}

f (\geq 0) \in L_{p, ϕ} (R_{+}) = {f | {∥ f ∥}_{p, ϕ} : = {\int_{0}^{\infty} ϕ (x) {| f (x) |}^{p} d x}^{\frac{1}{p}} < \infty}

g (\geq 0) \in L_{q, ψ} (R_{+})

, and

{∥ f ∥}_{p, ϕ}

{∥ g ∥}_{q, ψ} > 0

, then

\int_{0}^{\infty} \int_{0}^{\infty} k_{λ} (x, y) f (x) g (y) d x d y < k (λ_{1}) {∥ f ∥}_{p, ϕ} {∥ g ∥}_{q, ψ},

(3)

where the constant factor

k (λ_{1})

is best possible. Moreover, if

k_{λ} (x, y)

is finite and

k_{λ} (x, y) x^{λ_{1} - 1} (k_{λ} (x, y) y^{λ_{2} - 1})

is decreasing for

x > 0

(

y > 0

), then for

a_{m,} b_{n} \geq 0

a = {a_{m}}_{m = 1}^{\infty} \in l_{p, ϕ} = {a | {∥ a ∥}_{p, ϕ} : = {\sum_{n = 1}^{\infty} ϕ (n) {| a_{n} |}^{p}}^{\frac{1}{p}} < \infty}

, and

b = {b_{n}}_{n = 1}^{\infty} \in l_{q, ψ}

{∥ a ∥}_{p, ϕ}

{∥ b ∥}_{q, ψ} > 0

, we have

\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} k_{λ} (m, n) a_{m} b_{n} < k (λ_{1}) {∥ a ∥}_{p, ϕ} {∥ b ∥}_{q, ψ},

(4)

where the constant

k (λ_{1})

is still the best value. Clearly, for

p = q = 2

λ = 1

k_{1} (x, y) = \frac{1}{x + y}

λ_{1} = λ_{2} = \frac{1}{2}

, (3) reduces to (1), while (4) reduces to (2).

Some other results about integral and discrete Hilbert-type inequalities can be found in [7‐16]. On half-discrete Hilbert-type inequalities with the general non-homogeneous kernels, Hardy et al. provided a few results in Theorem 351 of [1]. But they did not prove that the constant factors are best possible. In 2005, Yang [17] gave a result with the kernel

\frac{1}{{(1 + n x)}^{λ}}

by introducing a variable and proved that the constant factor is best possible. Very recently, Yang [18] and [19] gave the following half-discrete Hilbert’s inequality with the best constant factor:

\int_{0}^{\infty} f (x) \sum_{n = 1}^{\infty} \frac{a_{n}}{{(x + n)}^{λ}} d x < π ∥ f ∥ ∥ a ∥;

(5)

Chen [20] and Yang [21] gave two more accurate half-discrete Mulholland’s inequalities by using Hadamard’s inequality.

In this paper, by means of weight functions and the improved Euler-Maclaurin summation formula, a more accurate half-discrete Hilbert-type inequality with a non-homogeneous kernel and a best constant factor is given as follows. For

0 < α + β \leq 2

γ \in R

η \leq 1 - \frac{α + β}{8} (1 + \sqrt{3 + \frac{4}{α + β}})

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ6_HTML.gif

(6)

Moreover, a best extension of (6), some equivalent forms, the operator expressions as well as some particular inequalities are considered.

2 Some lemmas

Lemma 1 If

n_{0} \in N

s > n_{0}

g_{1} (y)

(

y \in [n_{0}, s)

g_{2} (y)

(

y \in [s, \infty)

) are decreasing continuous functions satisfying

g_{1} (n_{0}) - g_{1} (s - 0) + g_{2} (s) > 0

g_{2} (\infty) = 0

, define a function

g (y)

as follows:

g (y) : = {\begin{matrix} g_{1} (y), & y \in [n_{0}, s), \\ g_{2} (y), & y \in [s, \infty) . \end{matrix}

Then there exists

ε \in [0, 1]

such that

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ7_HTML.gif

(7)

where

ρ (y) = y - [y] - \frac{1}{2}

is a Bernoulli function of the first order. In particular, for

g_{1} (y) = 0

y \in [n_{0}, s)

, we have

g_{2} (s) > 0

and

\frac{- 1}{8} g_{2} (s) < \int_{s}^{\infty} ρ (y) g (y) d y < \frac{1}{8} g_{2} (s);

(8)

for

g_{2} (y) = 0

y \in [s, \infty)

, if

g_{1} (s - 0) \geq 0

, then it follows

g_{1} (n_{0}) > 0

and

\frac{- 1}{8} g_{1} (n_{0}) < \int_{n_{0}}^{s} ρ (y) g_{1} (y) d y < 0 .

(9)

Proof Define a decreasing continuous function

\tilde{g} (y)

\tilde{g} (y) : = {\begin{matrix} g_{1} (y) + g_{2} (s) - g_{1} (s - 0), & y \in [n_{0}, s), \\ g_{2} (y), & y \in [s, \infty) . \end{matrix}

Then it follows

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equc_HTML.gif

Since

\tilde{g} (n_{0}) = g_{1} (n_{0}) + g_{2} (s) - g_{1} (s - 0) > 0

\tilde{g} (y)

is a non-constant decreasing continuous function with

\tilde{g} (\infty) = g_{2} (\infty) = 0

, by the improved Euler-Maclaurin summation formula (cf. [6], Theorem 2.2.2), it follows

\frac{- 1}{8} (g_{1} (n_{0}) + g_{2} (s) - g_{1} (s - 0)) = \frac{- 1}{8} \tilde{g} (n_{0}) < \int_{n_{0}}^{\infty} ρ (y) \tilde{g} (y) d y < 0,

and then in view of the above results and by simple calculation, we have (7). □

Lemma 2 If

0 < α + β \leq 2

γ \in R

η \leq 1 - \frac{α + β}{8} (1 + \sqrt{3 + \frac{4}{α + β}})

, and

ω (n)

and

ϖ (x)

are weight functions given by

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ10_HTML.gif

(10)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ11_HTML.gif

(11)

then we have

ϖ (x) < ω (n) = \frac{4}{α + β} .

(12)

Proof Substituting

t = (x - γ) (n - η)

in (10), and by simple calculation, we have

\begin{array}{rcl} ω (n) & = & \int_{0}^{\infty} \frac{{(min {1, t})}^{β}}{{(max {1, t})}^{α}} t^{\frac{α - β}{2} - 1} d t \\ = & \int_{0}^{1} t^{β + \frac{α - β}{2} - 1} d t + \int_{1}^{\infty} t^{- α + \frac{α - β}{2} - 1} d t = \frac{4}{α + β} . \end{array}

For fixed

x > γ

, we find

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equf_HTML.gif

By the Euler-Maclaurin summation formula (cf. [6]), it follows

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ13_HTML.gif

(13)

(i)

For

0 < x - γ < \frac{1}{1 - η}

, we obtain

- \frac{1}{2} h (x, 1) = - \frac{1}{2} {(x - γ)}^{\frac{α + β}{2}} {(1 - η)}^{\frac{α + β}{2} - 1}

, and

\int_{η}^{1} h (x, y) d y = {(x - γ)}^{\frac{α + β}{2}} \int_{η}^{1} {(y - η)}^{\frac{α + β}{2} - 1} d y = \frac{2 {(1 - η)}^{\frac{α + β}{2}}}{α + β} {(x - γ)}^{\frac{α + β}{2}} .

Setting

g (y) : = - h_{y}^{'} (x, y)

, wherefrom

g_{1} (y) = (1 - \frac{α + β}{2}) {(x - γ)}^{\frac{α + β}{2}} {(y - η)}^{\frac{α + β}{2} - 2}

g_{2} (y) = (\frac{α + β}{2} + 1) {(x - γ)}^{- \frac{α + β}{2}} {(y - η)}^{- \frac{α + β}{2} - 2}

and

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equh_HTML.gif

then by (7), we find

\begin{array}{rcl} - \int_{1}^{\infty} ρ (y) h_{y}^{'} (x, y) d y & = & \int_{1}^{\infty} ρ (y) g (y) d y \\ > & \frac{- 1}{8} [g_{1} (1) + g_{2} (η + \frac{1}{x - γ}) - g_{1} ((η + \frac{1}{x - γ}) - 0)] \\ = & \frac{- 1}{8} [(1 - \frac{α + β}{2}) {(x - γ)}^{\frac{α + β}{2}} {(1 - η)}^{\frac{α + β}{2} - 2} + (α + β) {(x - γ)}^{2}] \\ > & \frac{- 1}{8} [(1 - \frac{α + β}{2}) {(1 - η)}^{\frac{α + β}{2} - 2} {(x - γ)}^{\frac{α + β}{2}} \\ + (α + β) {(1 - η)}^{\frac{α + β}{2} - 2} {(x - γ)}^{\frac{α + β}{2} - 2} {(x - γ)}^{2}] \\ = & \frac{- 1}{8} [(1 + \frac{α + β}{2}) {(1 - η)}^{\frac{α + β}{2} - 2} {(x - γ)}^{\frac{α + β}{2}}] . \end{array}

In view of (11) and the above results, since for

η \leq 1 - \frac{α + β}{8} (1 + \sqrt{3 + \frac{4}{α + β}})

, namely

1 - η \geq \frac{α + β}{8} (1 + \sqrt{3 + \frac{4}{α + β}})

, it follows

\begin{array}{rcl} R (x) & > & \frac{2}{α + β} {(1 - η)}^{\frac{α + β}{2}} {(x - γ)}^{\frac{α + β}{2}} - \frac{1}{2} {(x - γ)}^{\frac{α + β}{2}} {(1 - η)}^{\frac{α + β}{2} - 1} \\ - \frac{1}{8} (1 + \frac{α + β}{2}) {(1 - η)}^{\frac{α + β}{2} - 2} {(x - γ)}^{\frac{α + β}{2}} \\ = & [\frac{2 {(1 - η)}^{2}}{α + β} - \frac{(1 - η)}{2} - \frac{2 + α + β}{16}] \frac{{(x - γ)}^{\frac{α + β}{2}}}{{(1 - η)}^{2 - \frac{α + β}{2}}} \geq 0 . \end{array}

(ii)

For

x - γ \geq \frac{1}{1 - η}

, we obtain

- \frac{1}{2} h (x, 1) = - \frac{1}{2} {(x - γ)}^{- \frac{α + β}{2}} {(1 - η)}^{- \frac{α + β}{2} - 1}

, and

\begin{array}{rcl} \int_{η}^{1} h (x, y) d y & = & \int_{η}^{η + \frac{1}{x - γ}} \frac{{(x - γ)}^{\frac{α + β}{2}}}{{(y - η)}^{1 - \frac{α + β}{2}}} d y + \int_{η + \frac{1}{x - γ}}^{1} \frac{{(x - γ)}^{- \frac{α + β}{2}}}{{(y - η)}^{\frac{α + β}{2} + 1}} d y \\ = & \frac{4}{α + β} - \frac{2}{α + β} {(1 - η)}^{- \frac{α + β}{2}} {(x - γ)}^{- \frac{α + β}{2}} \\ \geq & \frac{4 {(1 - η)}^{- \frac{α + β}{2}}}{α + β} {(x - γ)}^{- \frac{α + β}{2}} - \frac{2 {(1 - η)}^{- \frac{α + β}{2}}}{α + β} {(x - γ)}^{- \frac{α + β}{2}} \\ = & \frac{2}{α + β} {(1 - η)}^{- \frac{α + β}{2}} {(x - γ)}^{- \frac{α + β}{2}} . \end{array}

Since for

y \geq 1

y - η \geq \frac{1}{x - γ}

, by the improved Euler-Maclaurin summation formula (cf. [6]), it follows

\begin{array}{rcl} - \int_{1}^{\infty} ρ (y) h_{y}^{'} (x, y) d y & = & (\frac{α + β}{2} + 1) {(x - γ)}^{- \frac{α + β}{2}} \int_{1}^{\infty} ρ (y) {(y - η)}^{- \frac{α + β}{2} - 2} d y \\ > & - \frac{1}{12} (\frac{α + β}{2} + 1) {(x - γ)}^{- \frac{α + β}{2}} {(1 - η)}^{- \frac{α + β}{2} - 2} . \end{array}

In view of (13) and the above results, for

1 - η \geq \frac{α + β}{8} (1 + \sqrt{3 + \frac{4}{α + β}})

, we find

\begin{array}{rcl} R (x) & > & \frac{2}{α + β} {(1 - η)}^{- \frac{α + β}{2}} {(x - γ)}^{- \frac{α + β}{2}} - \frac{1}{2} {(1 - η)}^{- \frac{α + β}{2} - 1} {(x - γ)}^{- \frac{α + β}{2}} \\ - \frac{1}{12} (\frac{α + β}{2} + 1) {(1 - η)}^{- \frac{α + β}{2} - 2} {(x - γ)}^{- \frac{α + β}{2}} \\ > & [\frac{2 {(1 - η)}^{2}}{α + β} - \frac{1 - η}{2} - \frac{2 + α + β}{16}] \frac{{(x - γ)}^{- \frac{α + β}{2}}}{{(1 - η)}^{2 + \frac{α + β}{2}}} \geq 0 . \end{array}

Hence, for

x > γ

, we have

R (x) > 0

, and then (12) follows. □

Lemma 3 Let the assumptions of Lemma 2 be fulfilled and, additionally, let

p > 1

\frac{1}{p} + \frac{1}{q} = 1

a_{n} \geq 0

n \in N

f (x)

be a non-negative measurable function in

(γ, \infty)

. Then we have the following inequalities:

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ14_HTML.gif

(14)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ15_HTML.gif

(15)

Proof Setting

k (x, n) : = \frac{{(min {1, (x - γ) (n - η)})}^{β}}{{(max {1, (x - γ) (n - η)})}^{α}}

, by Hölder’s inequality (cf. [22]) and (12), it follows

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equn_HTML.gif

Then by the Lebesgue term-by-term integration theorem (cf. [23]), we have

\begin{array}{rcl} J & \leq & {(\frac{4}{α + β})}^{\frac{1}{q}} {\sum_{n = 1}^{\infty} \int_{γ}^{\infty} k (x, n) \frac{{(x - γ)}^{(1 - \frac{α - β}{2}) (p - 1)}}{{(n - η)}^{1 - \frac{α - β}{2}}} f^{p} (x) d x}^{\frac{1}{p}} \\ = & {(\frac{4}{α + β})}^{\frac{1}{q}} {\int_{γ}^{\infty} \sum_{n = 1}^{\infty} k (x, n) \frac{{(x - γ)}^{(1 - \frac{α - β}{2}) (p - 1)}}{{(n - η)}^{1 - \frac{α - β}{2}}} f^{p} (x) d x}^{\frac{1}{p}} \\ = & {(\frac{4}{α + β})}^{\frac{1}{q}} {\int_{γ}^{\infty} ϖ (x) {(x - γ)}^{p (1 - \frac{α - β}{2}) - 1} f^{p} (x) d x}^{\frac{1}{p}} . \end{array}

Hence, (14) follows. By Hölder’s inequality again, we have

\begin{array}{rcl} {[\sum_{n = 1}^{\infty} k (x, n) a_{n}]}^{q} & = & {\sum_{n = 1}^{\infty} k (x, n) [\frac{{(x - γ)}^{(1 - \frac{α - β}{2}) / q}}{{(n - η)}^{(1 - \frac{α - β}{2}) / p}}] \\ \times [\frac{{(n - η)}^{(1 - \frac{α - β}{2}) / p} a_{n}}{{(x - γ)}^{(1 - \frac{α - β}{2}) / q}}]}^{q} \leq {\sum_{n = 1}^{\infty} k (x, n) \frac{{(x - γ)}^{(1 - \frac{α - β}{2}) (p - 1)}}{{(n - η)}^{1 - \frac{α - β}{2}}}}^{q - 1} \\ \times \sum_{n = 1}^{\infty} k (x, n) \frac{{(n - η)}^{(1 - \frac{α - β}{2}) (q - 1)}}{{(x - γ)}^{1 - \frac{α - β}{2}}} a_{n}^{q} \\ = & \frac{{[ϖ (x)]}^{q - 1}}{{(x - γ)}^{\frac{q (α - β)}{2} - 1}} \sum_{n = 1}^{\infty} k (x, n) \frac{{(n - η)}^{(1 - \frac{α - β}{2}) (q - 1)}}{{(x - γ)}^{1 - \frac{α - β}{2}}} a_{n}^{q} . \end{array}

By the Lebesgue term-by-term integration theorem, we have

\begin{array}{rcl} L_{1} & \leq & {\int_{γ}^{\infty} \sum_{n = 1}^{\infty} k (x, n) \frac{{(n - η)}^{(1 - \frac{α - β}{2}) (q - 1)}}{{(x - γ)}^{1 - \frac{α - β}{2}}} a_{n}^{q} d x}^{\frac{1}{q}} \\ = & {\sum_{n = 1}^{\infty} \int_{γ}^{\infty} k (x, n) \frac{{(n - η)}^{(1 - \frac{α - β}{2}) (q - 1)}}{{(x - γ)}^{1 - \frac{α - β}{2}}} a_{n}^{q} d x}^{\frac{1}{q}} \\ = & {\sum_{n = 1}^{\infty} ω (n) {(n - η)}^{q (1 - \frac{α - β}{2}) - 1} a_{n}^{q}}^{\frac{1}{q}}, \end{array}

and in view of (12), inequality (15) follows. □

Lemma 4 Let the assumptions of Lemma 2 be fulfilled and, additionally, let

p > 1

\frac{1}{p} + \frac{1}{q} = 1

0 < ε < \frac{p}{2} (α + β)

. Setting

\tilde{f} (x) = {(x - γ)}^{\frac{α - β}{2} + \frac{ε}{p} - 1}

x \in (γ, γ + 1)

;

\tilde{f} (x) = 0

x \in [γ + 1, \infty)

, and

{\tilde{a}}_{n} = {(n - η)}^{\frac{α - β}{2} - \frac{ε}{q} - 1}

n \in N

, then we have

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ16_HTML.gif

(16)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ17_HTML.gif

(17)

Proof We find

\begin{array}{rcl} \tilde{I} & = & \sum_{n = 1}^{\infty} {(n - η)}^{\frac{α - β}{2} - \frac{ε}{q} - 1} \\ \times \int_{γ}^{γ + 1} \frac{{(min {1, (x - γ) (n - η)})}^{β}}{{(max {1, (x - γ) (n - η)})}^{α}} {(x - γ)}^{\frac{α - β}{2} + \frac{ε}{p} - 1} d x \\ = & \sum_{n = 1}^{\infty} {(n - η)}^{\frac{α - β}{2} - \frac{ε}{q} - 1} [{(n - η)}^{β} \int_{γ}^{γ + \frac{1}{n - η}} {(x - γ)}^{\frac{α + β}{2} + \frac{ε}{p} - 1} d x \\ + \frac{1}{{(n - η)}^{α}} \int_{γ + \frac{1}{n - η}}^{γ + 1} {(x - γ)}^{- \frac{α + β}{2} + \frac{ε}{p} - 1} d x] \\ = & \frac{α + β}{{(\frac{α + β}{2})}^{2} - {(\frac{ε}{p})}^{2}} \sum_{n = 1}^{\infty} {(n - η)}^{- ε - 1} - \frac{1}{\frac{α + β}{2} - \frac{ε}{p}} \sum_{n = 1}^{\infty} {(n - η)}^{- \frac{α + β}{2} - \frac{ε}{q} - 1} \\ > & \frac{α + β}{{(\frac{α + β}{2})}^{2} - {(\frac{ε}{p})}^{2}} \int_{1}^{\infty} \frac{d y}{{(y - η)}^{ε + 1}} - \frac{1}{\frac{α + β}{2} - \frac{ε}{p}} \sum_{n = 1}^{\infty} {(n - η)}^{- \frac{α + β}{2} - \frac{ε}{q} - 1} \\ = & \frac{1}{ε} [\frac{(α + β) {(1 - η)}^{- ε}}{{(\frac{α + β}{2})}^{2} - {(\frac{ε}{p})}^{2}} - \frac{ε}{\frac{α + β}{2} - \frac{ε}{p}} \sum_{n = 1}^{\infty} {(n - η)}^{- \frac{α + β}{2} - \frac{ε}{q} - 1}], \end{array}

and then (16) is valid. We obtain

\begin{array}{rcl} \tilde{H} & = & {\int_{γ}^{γ + 1} {(x - γ)}^{ε - 1} d x}^{\frac{1}{p}} {{(1 - η)}^{- ε - 1} + \sum_{n = 2}^{\infty} {(n - η)}^{- ε - 1}}^{\frac{1}{q}} \\ < & {(\frac{1}{ε})}^{\frac{1}{p}} {{(1 - η)}^{- ε - 1} + \int_{1}^{\infty} {(y - η)}^{- ε - 1} d y}^{\frac{1}{q}} \\ = & \frac{1}{ε} {ε {(1 - η)}^{- ε - 1} + {(1 - η)}^{- ε}}^{\frac{1}{q}}, \end{array}

and so (17) is valid. □

3 Main results

We introduce the functions

Φ (x) : = {(x - γ)}^{p (1 - \frac{α - β}{2}) - 1} (x \in (γ, \infty)), Ψ (n) : = {(n - η)}^{q (1 - \frac{α - β}{2}) - 1} (n \in N),

wherefrom

{[Φ (x)]}^{1 - q} = {(x - γ)}^{q \frac{α - β}{2} - 1}

and

{[Ψ (n)]}^{1 - p} = {(n - η)}^{p \frac{α - β}{2} - 1}

Theorem 5 If

0 < α + β \leq 2

γ \in R

η \leq 1 - \frac{α + β}{8} (1 + \sqrt{3 + \frac{4}{α + β}})

p > 1

\frac{1}{p} + \frac{1}{q} = 1

f (x), a_{n} \geq 0

f \in L_{p, Φ} (γ, \infty)

a = {a_{n}}_{n = 1}^{\infty} \in l_{q, Ψ}

{∥ f ∥}_{p, Φ} > 0

and

{∥ a ∥}_{q, Ψ} > 0

, then we have the following equivalent inequalities:

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ18_HTML.gif

(18)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ19_HTML.gif

(19)

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ20_HTML.gif

(20)

where the constant

\frac{4}{α + β}

is the best possible in the above inequalities.

Proof The two expressions for I in (18) follow from the Lebesgue term-by-term integration theorem. By (14) and (12), we have (19). By Hölder’s inequality, we have

\begin{array}{rcl} I & = & \sum_{n = 1}^{\infty} [Ψ^{\frac{- 1}{q}} (n) \int_{γ}^{\infty} \frac{{(min {1, (x - γ) (n - η)})}^{β} f (x)}{{(max {1, (x - γ) (n - η)})}^{α}} d x] [Ψ^{\frac{1}{q}} (n) a_{n}] \\ \leq & J {∥ a ∥}_{q, Ψ} . \end{array}

Then by (19), we have (18). On the other hand, assume that (18) is valid. Setting

a_{n} : = {[Ψ (n)]}^{1 - p} {[\int_{γ}^{\infty} \frac{{(min {1, (x - γ) (n - η)})}^{β} f (x)}{{(max {1, (x - γ) (n - η)})}^{α}} d x]}^{p - 1}, n \in N,

where

J^{p - 1} = {∥ a ∥}_{q, Ψ}

. By (14), we find

J < \infty

. If

J = 0

, then (19) is trivially valid; if

J > 0

, then by (18) we have

{∥ a ∥}_{q, Ψ}^{q} = J^{q (p - 1)} = J^{p} = I < \frac{4}{α + β} {∥ f ∥}_{p, Φ} {∥ a ∥}_{q, Ψ},

therefore

{∥ a ∥}_{q, Ψ}^{q - 1} = J < \frac{4}{α + β} {∥ f ∥}_{p, Φ}

; that is, (19) is equivalent to (18). On the other hand, by (12) we have

{[ϖ (x)]}^{1 - q} > {(\frac{4}{α + β})}^{1 - q}

. Then in view of (15), we have (20). By Hölder’s inequality, we find

\begin{array}{rcl} I & = & \int_{γ}^{\infty} [Φ^{\frac{1}{p}} (x) f (x)] [Φ^{\frac{- 1}{p}} (x) \sum_{n = 1}^{\infty} \frac{{(min {1, (x - γ) (n - η)})}^{β} a_{n}}{{(max {1, (x - γ) (n - η)})}^{α}}] d x \\ \leq & {∥ f ∥}_{p, Φ} L . \end{array}

Then by (20), we have (18). On the other hand, assume that (18) is valid. Setting

f (x) : = {[Φ (x)]}^{1 - q} {[\sum_{n = 1}^{\infty} \frac{{(min {1, (x - γ) (n - η)})}^{β} a_{n}}{{(max {1, (x - γ) (n - η)})}^{α}}]}^{q - 1}, x \in (γ, \infty),

then

L^{q - 1} = {∥ f ∥}_{p, Φ}

. By (15), we find

L < \infty

. If

L = 0

, then (20) is trivially valid; if

L > 0

, then by (18), we have

{∥ f ∥}_{p, Φ}^{p} = L^{p (q - 1)} = I < \frac{4}{α + β} {∥ f ∥}_{p, Φ} {∥ a ∥}_{q, Ψ},

therefore

{∥ f ∥}_{p, Φ}^{p - 1} = L < \frac{4}{α + β} {∥ a ∥}_{q, Ψ}

; that is, (20) is equivalent to (18). Hence, (18), (19) and (20) are equivalent.

If there exists a positive number k (

\leq \frac{4}{α + β}

) such that (18) is valid as we replace

\frac{4}{α + β}

with k, then, in particular, it follows that

\tilde{I} < k \tilde{H}

. In view of (16) and (17), we have

\frac{(α + β) {(1 - η)}^{- ε}}{{(\frac{α + β}{2})}^{2} - {(\frac{ε}{p})}^{2}} - ε O (1) < k {[ε {(1 - η)}^{- ε - 1} + {(1 - η)}^{- ε}]}^{\frac{1}{q}},

and

\frac{4}{α + β} \leq k

(

ε \to 0^{+}

). Hence,

k = \frac{4}{α + β}

is the best value of (18).

By the equivalence of the inequalities, the constant factor

\frac{4}{α + β}

in (19) and (20) is the best possible. □

Remark 1 (i) Define the first type half-discrete Hilbert-type operator

T_{1} : L_{p, Φ} (γ, \infty) \to l_{p, Ψ^{1 - p}}

as follows. For

f \in L_{p, Φ} (γ, \infty)

, we define

T_{1} f \in l_{p, Ψ^{1 - p}}

T_{1} f (n) = \int_{γ}^{\infty} \frac{{(min {1, (x - γ) (n - η)})}^{β}}{{(max {1, (x - γ) (n - η)})}^{α}} f (x) d x, n \in N .

Then by (19),

{∥ T_{1} f ∥}_{p, Ψ^{1 - p}} \leq \frac{4}{α + β} {∥ f ∥}_{p, Φ}

and so

T_{1}

is a bounded operator with

∥ T_{1} ∥ \leq \frac{4}{α + β}

. Since by Theorem 5 the constant factor in (19) is best possible, we have

∥ T_{1} ∥ = \frac{4}{α + β}

(ii)

Define the second type half-discrete Hilbert-type operator

T_{2} : l_{q, Ψ} \to L_{q, Φ^{1 - q}} (γ, \infty)

as follows. For

a \in l_{q, Ψ}

, we define

T_{2} a \in L_{q, Φ^{1 - q}} (γ, \infty)

T_{2} a (x) = \sum_{n = 1}^{\infty} \frac{{(min {1, (x - γ) (n - η)})}^{β}}{{(max {1, (x - γ) (n - η)})}^{α}} a_{n}, x \in (γ, \infty) .

Then by (20),

{∥ T_{2} a ∥}_{q, Φ^{1 - q}} \leq \frac{4}{α + β} {∥ a ∥}_{q, Ψ}

and so

T_{2}

is a bounded operator with

∥ T_{2} ∥ \leq \frac{4}{α + β}

. Since by Theorem 5 the constant factor in (20) is best possible, we have

∥ T_{2} ∥ = \frac{4}{α + β}

Remark 2 (i) For

p = q = 2

, (18) reduces to (6). Since we find

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equad_HTML.gif

then for

η = γ = 0

in (18), we have the following inequality:

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ21_HTML.gif

(21)

Hence, (18) is a more accurate inequality of (21).

(ii)

For

β = 0

in (18), we have

0 < α \leq 2

γ \in R

η \leq 1 - \frac{α}{8} (1 + \sqrt{3 + \frac{4}{α}})

, and

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ22_HTML.gif

(22)

for

α = 0

in (18), we have

0 < β \leq 2

γ \in R

η \leq 1 - \frac{β}{8} (1 + \sqrt{3 + \frac{4}{β}})

, and

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ23_HTML.gif

(23)

for

β = α = λ

in (18), we have

0 < λ \leq 1

γ \in R

η \leq 1 - \frac{λ}{4} (1 + \sqrt{3 + \frac{2}{λ}})

, and

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2012-292/MediaObjects/13660_2012_Article_426_Equ24_HTML.gif

(24)

Acknowledgements

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

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

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Titel: A more accurate half-discrete Hilbert-type inequality with a non-homogeneous kernel
verfasst von: Bicheng Yang
Xindong Liu
Publikationsdatum: 01.12.2012
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2012
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2012-292

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

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Some lemmas

3 Main results

Acknowledgements

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Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Some lemmas

3 Main results

Acknowledgements

Competing interests

Authors’ contributions

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