A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension

Bicheng Yang; Qiang Chen

doi:10.1186/1029-242X-2011-124

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A half-discrete Hilbert-type inequality with a homogeneous kernel and an extension

Bicheng Yang¹Email author and
Qiang Chen²

Journal of Inequalities and Applications20112011:124

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

Received: 28 August 2011

Accepted: 30 November 2011

Published: 30 November 2011

Abstract

Using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel is obtained, and a best extension with two interval variables is given. The equivalent forms, the operator expressions, the reverses and some particular cases are considered.

2000 Mathematics Subject Classification: 26D15; 47A07.

Keywords

Hilbert-type inequalityhomogeneous kernelweight functionequivalent formreverse

1 Introduction

Assuming that p > 1,

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

, f (≥ 0) ∈ L^p (R₊), g(≥ 0) ∈ L^q (R₊),

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

, || g ||_q > 0, we have the following Hardy-Hilbert's integral inequality [1]:

\int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{x + y} d x d y < \frac{π}{s i n (π / p)} | | f {| |}_{p} | | g {| |}_{q},

(1)

where the constant factor

\frac{π}{s i n (π / p)}

is the best possible. If a_m, b_n ≥ 0,

a = {a_{m}}_{m = 1}^{\infty} \in l^{p}

,

b = {b_{n}}_{n = 1}^{\infty} \in l^{q}

,

| | a | |_{p} = {\{\sum_{m = 1}^{\infty} a_{m}^{p}\}}^{\frac{1}{p}} > 0

, || b ||_q > 0, then we still have the following discrete Hardy-Hilbert's inequality with the same best constant factor

\frac{π}{s i n (π / p)}

:

\sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{a_{m} b_{n}}{m + n} < \frac{π}{s i n (π / p)} | | a {| |}_{p} | | b {| |}_{q} .

(2)

For p = q = 2, the above two inequalities reduce to the famous Hilbert's inequalities. Inequalities (1) and (2) are important in analysis and its applications [2–4].

In 1998, by introducing an independent parameter λ ∈ (0, 1], Yang [5] gave an extension of (1) for p = q = 2. Refinement and generalizing the results from [5], Yang [6] gave some best extensions of (1) and (2) as follows: If λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative homogeneous function of degree - λ satisfying for any x, y, t > 0, k_λ(tx, ty) = t^-λ k_λ (x, y),

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(≥ 0) ∈ L_q_,_ψ, || f ||_p_,_ϕ, || g ||_q_,_ψ > 0, then we have

\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(λ₁) is the 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 with respect to x > 0(y > 0), then for a_m,b_n ≥ 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}

,

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)

with the best constant factor k(λ₁). Clearly, for λ = 1, $k_{1} (x, y) = \frac{1}{x + y}$ , $λ_{1} = \frac{1}{q}$ , $λ_{2} = \frac{1}{p}$ (3) reduces to (1), and (4) reduces to (2). Some other results about Hilbert-type inequalities are provided by [7–15].

On half-discrete Hilbert-type inequalities with the 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 the best possible. And, Yang [16] gave a result by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [17] gave the following half-discrete Hilbert's inequality with the best constant factor B(λ₁, λ₂)(λ, λ₁ > 0, 0 < λ₂ ≤ 1, λ₁ + λ₂ = λ):

\int_{0}^{\infty} f (x) \sum_{n = 1}^{\infty} \frac{a_{n}}{{(x + n)}^{λ}} d x < B (λ_{1}, λ_{2}) | | f {| |}_{p, φ} | | a {| |}_{q, ψ} .

(5)

In this article, using the way of weight functions and the technique of real analysis, a half-discrete Hilbert-type inequality with a general homogeneous kernel and a best constant factor is given as follows:

\int_{0}^{\infty} f (x) \sum_{n = 1}^{\infty} k_{λ} (x, n) a_{n} d x < k (λ_{1}) | | f {| |}_{p, φ} | | a {| |}_{q, ψ},

(6)

which is a generalization of (5). A best extension of (6) with two interval variables, some equivalent forms, the operator expressions, the reverses and some particular cases are considered.

2 Some lemmas

We set 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, ∞) and strictly decreasing in an interval of (n₀ - 1, ∞).

Condition (ii) $v (y) (y \in [n_{0} - \frac{1}{2}, \infty))$ is strictly increasing with $v (n_{0} - \frac{1}{2}) \geq 0$ and for any fixed x ∈ (b, c), f(x, y) is decreasing and strictly convex for $y \in (n_{0} - \frac{1}{2}, \infty)$ .

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 fixed x ∈ (b, c), f(x, y) is piecewise smooth satisfying

R (x) : = \int_{n_{0} - β}^{n_{0}} f (x, y) d y - \frac{1}{2} f (x, n_{0}) - \int_{n_{0}}^{\infty} ρ (y) {f^{'}}_{y} (x, y) d y > 0,

where $ρ (y) (= y - [y] - \frac{1}{2})$ is Bernoulli function of the first order.

Lemma 1 If λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree - λ in

R_{+}^{2}, u (x) (x \in (b, c), - \infty \leq b < c \leq \infty)

and v(y)(y ∈ [n₀, ∞), n₀ ∈ N) are strictly increasing differential functions with u(b⁺) = 0, v(n₀) > 0, u(c^-) = v(∞) = ∞, setting K(x, y) = k_λ(u(x), v(y)), then we define weight functions ω(n) and ϖ(x) as follows:

ω (n) : = {[v (n)]}^{λ_{2}} \int_{b}^{c} K (x, n) [u (x {)]}^{λ_{1} - 1} u^{'} (x) d x, n \geq n_{0} (n \in N),

(7)

ϖ (x) : = {[u (x)]}^{λ_{1}} \sum_{n = n_{0}}^{\infty} K (x, n) [v (n {)]}^{λ_{2} - 1} v^{'} (n), x \in (b, c) .

(8)

It follows

ω (n) = k (λ_{1}) : = \int_{0}^{\infty} k_{λ} (t, 1) t^{λ_{1} - 1} d t .

(9)

Moreover, setting

f (x, y) : = {[u (x)]}^{λ_{1}} K (x, y) [v (y {)]}^{λ_{2} - 1} v^{'} (y)

, if k(λ₁) ∈ R₊ and one of the above three conditions is fulfilled, then we still have

ϖ (x) < k (λ_{1}) (x \in (b, c)) .

(10)

Proof . Setting

t = \frac{u (x)}{v (n)}

in (7), by calculation, we have (9).

(i)

If Condition (i) is fulfilled, then we have
$\begin{array}{l} ϖ (x) = \sum_{n = n_{0}}^{\infty} f (x, n) < {[u (x)]}^{λ_{1}} \int_{n_{0} - 1}^{\infty} K (x, y) [v (y {)]}^{λ_{2} - 1} v^{'} (y) d y \\ \overset{t = u (x) / v (y)}{=} \int_{0}^{\frac{u (x)}{v (n_{0} - 1)}} k_{λ} (t, 1) t^{λ_{1} - 1} d t \leq k (λ_{1}) . \end{array}$
(ii)

If Condition (ii) is fulfilled, then by Hadamard's inequality [18], we have
$\begin{gathered} ϖ (x) = \sum_{n = n_{0}}^{\infty} f (x, n) < \int_{n_{0} - \frac{1}{2}}^{\infty} f (x, y) d y \\ \overset{t = u (x) / v (y)}{=} \int_{0}^{\frac{u (x)}{v (n_{0} - \frac{1}{2})}} k_{λ} (t, 1) t^{λ_{1} - 1} d t \leq k (λ_{1}) . \end{gathered}$
(iii)

If Condition (iii) is fulfilled, then by Euler-Maclaurin summation formula [6], we have
$\begin{align} ϖ (x) & = \sum_{n = n_{0}}^{\infty} f (x, n) \\ = \int_{n_{0}}^{\infty} f (x, y) d y + \frac{1}{2} f (x, n_{0}) + \int_{n_{0}}^{\infty} ρ (y) {f^{'}}_{y} (x, y) d y \\ = \int_{n_{0} - β}^{\infty} f (x, y) d y - R (x) \\ = \int_{0}^{\frac{u (x)}{v (n_{0} - β)}} k_{λ} (t, 1) t^{λ_{1} - 1} d t - R (x) \\ \leq k (λ_{1}) - R (x) < k (λ_{1}) . \end{align}$

The lemma is proved. ■

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally, p > 0(p ≠ 1),

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

a_n ≥ 0, n ≥ n₀(n ∈ N), f (x) is a non-negative measurable function in (b, c). Then, (i) for p > 1, we have the following inequalities:

\begin{align} J_{1} : & = {\{\sum_{n = n_{0}}^{\infty} \frac{v^{'} (n)}{{[v (n)]}^{1 - p λ_{2}}} {[\int_{d}^{c} K (x, n) f (x) d x]}^{p}\}}^{\frac{1}{p}} \\ \leq {[k (λ_{1})]}^{\frac{1}{q}} {\{\int_{d}^{c} ϖ (x) \frac{{[u (x)]}^{p (1 - λ_{1}) - 1}}{{[u^{'} (x)]}^{p - 1}} f^{p} (x) d x\}}^{\frac{1}{p}}, \end{align}

(11)

\begin{align} L_{1} : & = {\{\int_{b}^{c} \frac{{[ϖ (x)]}^{1 - q} u^{'} (x)}{{[u (x)]}^{1 - q λ_{1}}} {[\sum_{n = n_{0}}^{\infty} K (x, n) a_{n}]}^{q} d x\}}^{\frac{1}{q}} \\ \leq {\{k (λ_{1}) \sum_{n = n_{0}}^{\infty} \frac{{[v (n)]}^{q (1 - λ_{2}) - 1}}{{[v^{'} (n)]}^{q - 1}} a_{n}^{q}\}}^{\frac{1}{q}}; \end{align}

(12)

(ii) for 0 < p < 1, we have the reverses of (11) and (12).

Proof(i) By Hölder's inequality with weight [18] and (9), it follows

\begin{matrix} {[\int_{b}^{c} K (x, n) f (x) d x]}^{p} = {\int_{b}^{c} K (x, n) [\frac{{[u (x)]}^{(1 - λ_{1}) / q}}{{[v (n)]}^{(1 - λ_{2}) / p}} \frac{{[v^{'} (n)]}^{1 / p}}{{[u^{'} (x)]}^{1 / q}} f (x)] \\ \times {[\frac{{[v (n)]}^{(1 - λ_{2}) / p}}{{[u (x)]}^{(1 - λ_{1}) / q}} \frac{{[u^{'} (x)]}^{1 / q}}{{[v^{'} (n)]}^{1 / p}}] d x}}^{p} \\ \leq \int_{b}^{c} K (x, n) \frac{{[u (x)]}^{(1 - λ_{1}) (p - 1)} v^{'} (n)}{{[v (n)]}^{1 - λ_{2}} {[u^{'} (x)]}^{p - 1}} f^{p} (x) d x \\ \times {\int_{b}^{c} K (x, n) \frac{{[v (n)]}^{(1 - λ_{2}) (q - 1)} u^{'} (x)}{{[u (x)]}^{1 - λ_{1}} {[v^{'} (n)]}^{q - 1}} d x}^{p - 1} \\ = {\frac{ω (n) [v (n {)]}^{q (1 - λ_{2}) - 1}}{{[v^{'} (n)]}^{q - 1}}}^{p - 1} \int_{b}^{c} K (x, n) \frac{{[u (x)]}^{(1 - λ_{1}) (p - 1)} v^{'} (n) f^{p} (x)}{{[v (n)]}^{1 - λ_{2}} {[u^{'} (x)]}^{p - 1}} d x \\ = \frac{{[k (λ_{1})]}^{p - 1}}{{[v (n)]}^{p λ_{2} - 1} v^{'} (n)} \int_{b}^{c} K (x, n) \frac{{[u (x)]}^{(1 - λ_{1}) (p - 1)} v^{'} (n)}{{[v (n)]}^{1 - λ_{2}} {[u^{'} (x)]}^{p - 1}} f^{p} (x) d x . \end{matrix}

Then, by Lebesgue term-by-term integration theorem [19], we have

\begin{align} J_{1} & \leq {[k (λ_{1})]}^{\frac{1}{q}} {\{\sum_{n = n_{0}}^{\infty} \int_{b}^{c} K (x, n) \frac{{[u (x)]}^{(1 - λ_{1}) (p - 1)} v^{'} (n) f^{p} (x)}{{[v (n)]}^{1 - λ_{2}} {[u^{'} (x)]}^{p - 1}} d x\}}^{\frac{1}{p}} \\ = {[k (λ_{1})]}^{\frac{1}{q}} {\{\int_{b}^{c} \sum_{n = n_{0}}^{\infty} K (x, n) \frac{{[u (x)]}^{(1 - λ_{1}) (p - 1)} v^{'} (n) f^{p} (x)}{{[v (n)]}^{1 - λ_{2}} {[u^{'} (x)]}^{p - 1}} d x\}}^{\frac{1}{p}} \\ = {[k (λ_{1})]}^{\frac{1}{q}} {\{\int_{b}^{c} ϖ (x) \frac{{[u (x)]}^{p (1 - λ_{1}) - 1}}{{[u^{'} (x)]}^{p - 1}} f^{p} (x) d x\}}^{\frac{1}{p}}, \end{align}

and (11) follows.

Still by Hölder's inequality, we have

\begin{align} {[\sum_{n = n_{0}}^{\infty} K (x, n) a_{n}]}^{q} & = \{\sum_{n = n_{0}}^{\infty} K (x, n) [\frac{{[u (x)]}^{(1 - λ_{1}) ∕ q}}{{[v (n)]}^{(1 - λ_{2}) ∕ p}} \frac{{[v^{'} (n)]}^{1 ∕ p}}{{[u^{'} (x)]}^{1 ∕ q}}] \\ \times {[\frac{{[v (n)]}^{(1 - λ_{2}) ∕ p}}{{[u (x)]}^{(1 - λ_{1}) ∕ q}} \frac{{[u^{'} (x)]}^{1 ∕ q}}{{[v^{'} (n)]}^{1 ∕ p}} a_{n}]\}}^{q} \\ \leq {\{\sum_{n = n_{0}}^{\infty} K (x, n) \frac{{[u (x)]}^{(1 - λ_{1}) (p - 1)}}{{[v (n)]}^{1 - λ_{2}}} \frac{v^{'} (n)}{{[u^{'} (x)]}^{p - 1}}\}}^{q - 1} \\ \times \sum_{n = n_{0}}^{\infty} K (x, n) \frac{{[v (n)]}^{(1 - λ_{2}) (q - 1)}}{{[u (x)]}^{1 - λ_{1}}} \frac{u^{'} (x)}{{[v^{'} (n)]}^{q - 1}} a_{n}^{q} \\ = \frac{{[u (x)]}^{1 - q λ_{1}}}{{[ϖ (x)]}^{1 - q} u^{'} (x)} \sum_{n = n_{0}}^{\infty} K (x, n) \frac{u^{'} (x)}{{[u (x)]}^{1 - λ_{1}}} \frac{{[v (n)]}^{(q - 1) (1 - λ_{2})}}{{[v^{'} (n)]}^{q - 1}} a_{n}^{q} . \end{align}

Then, by Lebesgue term-by-term integration theorem, we have

\begin{align} L_{1} & \leq {\{\int_{b}^{c} \sum_{n = n_{0}}^{\infty} K (x, n) \frac{u^{'} (x)}{{[u (x)]}^{1 - λ_{1}}} \frac{{[v (n)]}^{(q - 1) (1 - λ_{2})}}{{[v^{'} (n)]}^{q - 1}} a_{n}^{q} d x\}}^{\frac{1}{q}} \\ = {\{\sum_{n = n_{0}}^{\infty} [{[v (n)]}^{λ_{2}} \int_{b}^{c} K (x, n) \frac{u^{'} (x) d x}{{[u (x)]}^{1 - λ_{1}}}] \frac{{[v (n)]}^{q (1 - λ_{2}) - 1}}{{[v^{'} (n)]}^{q - 1}} a_{n}^{q}\}}^{\frac{1}{q}} \\ = {\{\sum_{n = n_{0}}^{\infty} ω (n) \frac{{[v (n)]}^{q (1 - λ_{2}) - 1}}{{[v^{'} (n)]}^{q - 1}} a_{n}^{q}\}}^{\frac{1}{q}}, \end{align}

and then in view of (9), inequality (12) follows.

(ii)

By the reverse Hölder's inequality [18] and in the same way, for q < 0, we have the reverses of (11) and (12). ■

3 Main results

We set $Φ (x) : = \frac{{[u (x)]}^{p (1 - λ_{1}) - 1}}{{[u^{'} (x)]}^{p - 1}} (x \in (b, c))$ , $Ψ (n) : = \frac{{[v (n)]}^{q (1 - λ_{2}) - 1}}{{[v^{'} (n)]}^{q - 1}} (n \geq n_{0}, n \in N)$ , wherefrom ${[Φ (x)]}^{1 - q} = \frac{u^{'} (x)}{{[u (x)]}^{1 - q λ_{1}}}$ , ${[Ψ (n)]}^{1 - p} = \frac{v^{'} (n)}{{[v (n)]}^{1 - p λ_{2}}}$ .

Theorem 1 Suppose that λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree - λ¸ in

R_{+}^{2}, u (x) (x \in (b, c), - \infty \leq b < c \leq \infty)

and v(y)(y ∈ [n₀, ∞), n₀ ∈ N are strictly increasing differential functions with u(b⁺) = 0, v(n₀) > 0, u(c^-) = v(∞) = ∞, ϖ(x) < k (λ₁) ∈ R₊(x ∈ (b, c)). If

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

, f ( x ), a_n≥ 0, f ∈ L_p_Φ ( b , c ),

a = {a_{n}}_{n = n_{0}}^{\infty} \in l_{q, Ψ},

|| f ||_p,_Φ > 0 and || a ||_q,_Ψ > 0, then we have the following equivalent inequalities:

\begin{align} I : & = \sum_{n = n_{0}}^{\infty} a_{n} \int_{b}^{c} K (x, n) f (x) d x = \int_{b}^{c} f (x) \sum_{n = n_{0}}^{\infty} K (x, n) a_{n} d x \\ < k (λ_{1}) | | f | |_{p, Φ} | | a | |_{q, Ψ}, \end{align}

(13)

J : = {\{\sum_{n = n_{0}}^{\infty} {[Ψ (n)]}^{1 - p} {[\int_{b}^{c} K (x, n) f (x) d x]}^{p}\}}^{\frac{1}{p}} < k (λ_{1}) | | f | |_{p, Φ},

(14)

L : = {\{\int_{b}^{c} {[Φ (x)]}^{1 - q} {[\sum_{n = n_{0}}^{\infty} K (x, n) a_{n}]}^{q} d x\}}^{\frac{1}{q}} < k (λ_{1}) | | a | |_{q, Ψ} .

(15)

Moreover, if $\frac{v^{'} (y)}{v (y)} (y \geq n_{0})$ is decreasing and there exist constants δ < λ₁ and M > 0, such that $k_{λ} (t, 1) \leq \frac{M}{t^{δ}} (t \in (0, \frac{1}{v (n_{0})}])$ , then the constant factor k(λ₁) in the above inequalities is the best possible.

Proof By Lebesgue term-by-term integration theorem, there are two expressions for I in (13). In view of (11), for ϖ(x) < k(λ₁) ∈ R₊, we have (14). By Hölder's inequality, we have

I = \sum_{n = n_{0}}^{\infty} [Ψ^{{\frac{- 1}{q}}^{}} (n) \int_{b}^{c} K (x, n) f (x) d x] [Ψ^{\frac{1}{q}} (n) a_{n}] \leq J | | a {| |}_{q, Ψ} .

(16)

Then, by (14), we have (13). On the other hand, assuming that (13) is valid, setting

a_{n} : = {[Ψ (n)]}^{1 - p} {[\int_{b}^{c} K (x, n) f (x) d x]}^{p - 1}, n \geq n_{0},

then J^p^-1 = || a ||_q_,Ψ . By (11), we find J < ∞. If J = 0, then (14) is naturally valid; if J > 0, then by (13), we have

| | a | |_{q, Ψ}^{q} = J^{p} = I < k (λ_{1}) | | f | |_{p, Φ} | | a | |_{q, Ψ}, | | a | |_{q, Ψ}^{q - 1} = J < k (λ_{1}) | | f | |_{p, Φ},

and we have (14), which is equivalent to (13).

In view of (12), for [ϖ(x) ]^1-q> [k(λ₁)]^1-q, we have (15). By Hölder's in equality, we find

I = \int_{b}^{c} [Φ^{\frac{1}{p}} (x) f (x)] [Φ^{\frac{- 1}{p}} (x) \sum_{n = n_{0}}^{\infty} K (x, n) a_{n}] d x \leq | | f {| |}_{p, Φ} L .

(17)

Then, by (15), we have (13). On the other hand, assuming that (13) is valid, setting

f (x) : = {[Φ (x)]}^{1 - q} {[\sum_{n = n_{0}}^{\infty} K (x, n) a_{n}]}^{q - 1}, x \in (b, c),

then L^q_-1 = || f ||_p_Φ . By (12), we find L < ∞. If L = 0, then (15) is naturally valid; if L > 0, then by (13), we have

| | f | |_{p, Φ}^{p} = L^{q} = I < k (λ_{1}) | | f | |_{p, Φ} | | a | |_{q, Ψ}, | | f | |_{p, Φ}^{p - 1} = L < k (λ_{1}) | | a | |_{q, Ψ},

and we have (15) which is equivalent to (13).

Hence, inequalities (13), (14) and (15) are equivalent.

There exists an unified constant d ∈ (b, c), satisfying u(d) = 1. For 0 < ε < p(λ₁ - δ), setting

\tilde{f} (x) = 0

, x ∈ (b, d);

\tilde{f} (x) = {[u (x)]}^{λ_{1} - \frac{ε}{p} - 1} u^{'} (x)

, x ∈ (d, c), and

ã_{n} = {[v (n)]}^{λ_{2} - \frac{ε}{q} - 1} v^{'} (n)

, n ≥ n₀, if there exists a positive number k(≤ k(λ₁)), such that (13) is valid as we replace k(λ₁) by k, then in particular, we find

\begin{align} Ĩ : & = \sum_{n = n_{0}}^{\infty} \int_{b}^{c} K (x, n) ã_{n} \tilde{f} (x) d x < k | | \tilde{f} {| |}_{p, Φ} | | ã {| |}_{q, Ψ} \\ = k {\{\int_{d}^{c} \frac{u^{'} (x)}{{[u (x)]}^{ε + 1}} d x\}}^{\frac{1}{p}} {\{\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + \sum_{n = n_{0} + 1}^{\infty} \frac{v^{'} (n)}{{[v (n)]}^{ε + 1}}\}}^{\frac{1}{q}} \\ < k {(\frac{1}{ε})}^{\frac{1}{p}} {\{\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + \int_{n_{0}}^{\infty} {[v (y)]}^{- ε - 1} v^{'} (y) d y\}}^{\frac{1}{q}} \\ = \frac{k}{ε} {\{ε \frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + {[v (n_{0})]}^{- ε}\}}^{\frac{1}{q}} \end{align}

(18)

\begin{array}{l} \tilde{I} = \sum_{n = n_{0}}^{\infty} {[v (n)]}^{λ_{2} - \frac{ε}{q} - 1} v^{'} (n) \int_{d}^{c} K (x, n) [u (x {)]}^{λ_{1} - \frac{ε}{p} - 1} u^{'} (x) d x \\ \overset{t = u (x) / v (n)}{=} \sum_{n = n_{0}}^{\infty} {[v (n)]}^{- ε - 1} v^{'} (n) \int_{1 / v (n)}^{\infty} k_{λ} (t, 1) t^{λ_{1} - \frac{ε}{p} - 1} d t \\ = k (λ_{1} - \frac{ε}{p}) \sum_{n = n_{0}}^{\infty} {[v (n)]}^{- ε - 1} v^{'} (n) - A (ε) \\ > k (λ_{1} - \frac{ε}{p}) \int_{n_{0}}^{\infty} {[v (y)]}^{- ε - 1} v^{'} (y) d y - A (ε) \\ = \frac{1}{ε} k (λ_{1} - \frac{ε}{p}) {[v (n_{0})]}^{- ε} - A (ε), \\ A (ε) : = \sum_{n = n_{0}}^{\infty} {[v (n)]}^{- ε - 1} v^{'} (n) \int_{0}^{1 / v (n)} k_{λ} (t, 1) t^{λ_{1} - \frac{ε}{p} - 1} d t . \end{array}

(19)

For

k_{λ} (t, 1) \leq M (\frac{1}{t^{δ}}) (δ < λ_{1}; t \in (0, 1 ∕ v (n_{0})])

, we find

\begin{align} 0 < A (ε) & \leq M \sum_{n = n_{0}}^{\infty} {[v (n)]}^{- ε - 1} v^{'} (n) \int_{0}^{1 ∕ v (n)} t^{λ_{1} - δ - \frac{ε}{p} - 1} d t \\ = \frac{M}{λ_{1} - δ - \frac{ε}{p}} \sum_{n = n_{0}}^{\infty} {[v (n)]}^{- λ_{1} + δ - \frac{ε}{q} - 1} v^{'} (n) \\ = \frac{M}{λ_{1} - δ - \frac{ε}{p}} [\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{λ_{1} - δ + \frac{ε}{q} + 1}} + \sum_{n = n_{0} + 1}^{\infty} \frac{v^{'} (n)}{{[v (n)]}^{λ_{1} - δ + \frac{ε}{q} + 1}}] \\ \leq \frac{M}{λ_{1} - δ - \frac{ε}{p}} [\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{λ_{1} - δ + \frac{ε}{q} + 1}} + \int_{n_{0}}^{\infty} \frac{v^{'} (y)}{{[v (y)]}^{λ_{1} - δ + \frac{ε}{q} + 1}} d y] \\ = \frac{M}{λ_{1} - δ - \frac{ε}{p}} [\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{λ_{1} - δ + \frac{ε}{q} + 1}} + \frac{{[v (n_{0})]}^{- λ_{1} + δ - \frac{ε}{q}}}{λ_{1} - δ + \frac{ε}{q}}] < \infty, \end{align}

namely A(ε) = O(1)(ε → 0⁺). Hence, by (18) and (19), it follows

k (λ_{1} - \frac{ε}{p}) {[v (n_{0})]}^{- ε} - ε O (1) < k {\{ε \frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + {[v (n_{0})]}^{- ε}\}}^{\frac{1}{q}} .

(20)

By Fatou Lemma [19], we have $k (λ_{1}) \leq {lim_{}}_{ε \to 0^{+}} k (λ_{1} - \frac{ε}{p})$ , then by (20), it follows k(λ₁) ≤ k(ε → 0⁺). Hence, k = k(λ₁) is the best value of (12).

By the equivalence, the constant factor k(λ₁) in (14) and (15) is the best possible, otherwise we can imply a contradiction by (16) and (17) that the constant factor in (13) is not the best possible. ■

Remark 1 (i) Define a half-discrete Hilbert's operator T :

L_{p, Φ} (b, c) \to l_{p, Ψ^{1 - p}}

as: for f ∈ L_p_Φ (b, c), we define

T f \in l_{p, Ψ^{1 - p}}

, satisfying

T f (n) = \int_{b}^{c} K (x, n) f (x) d x, n \geq n_{0} .

Then, by (14), it follows

| | T f | |_{p . Ψ^{1 - p}} \leq k (λ_{1}) | | f | |_{p, Φ}

and then T is a bounded operator with || T || ≤ k(λ₁). Since, by Theorem 1, the constant factor in (14) is the best possible, we have || T || = k(λ₁).

(ii)

Define a half-discrete Hilbert's operator $\tilde{T} : l_{q, Ψ} \to L_{q, Φ^{1 - q}} (b, c)$ as: for a ∈ l_q_Ψ , we define $\tilde{T} a \in L_{q, Φ^{1 - q}} (b, c)$ , satisfying
$\tilde{T} a (x) = \sum_{n = n_{0}}^{\infty} K (x, n) a_{n}, x \in (b, c) .$

Then, by (15), it follows $| | \tilde{T} a | |_{q, Φ^{1 - q}} \leq k (λ_{1}) | | a | |_{q, Ψ}$ and then $\tilde{T}$ is a bounded operator with $| | \tilde{T} | | \leq k (λ_{1})$ . Since, by Theorem 1, the constant factor in (15) is the best possible, we have $| | \tilde{T} | | = k (λ_{1})$ .

In the following theorem, for 0 < p < 1, or p < 0, we still use the formal symbols of $| | f | |_{p, \tilde{Φ}}$ and ||a||_q,_Ψ and so on. ■

Theorem 2 Suppose that λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree -λ in

R_{+}^{2}

, 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(∞) = ∞, k(λ₁) ∈ R₊, θ_λ(x) ∈ (0, 1), k(λ₁)(1 - θ_λ(x)) < ϖ(x) < k(λ₁)(x ∈ (b, c)). If 0 < p < 1,

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

, f(x), a_n ≥ 0,

\tilde{Φ} (x) : = (1 - θ_{λ} (x)) Φ (x) (x \in (b, c))

,

0 < | | f | |_{q, \tilde{Φ}} < \infty

and 0 < ||a||_q,_Ψ < ∞. Then, we have the following equivalent inequalities:

\begin{align} I : & = \sum_{n = n_{0}}^{\infty} \int_{b}^{c} K (x, n) a_{n} f (x) d x = \int_{b}^{c} \sum_{n = n_{0}}^{\infty} K (x, n) a_{n} f (x) d x \\ > k (λ_{1}) | | f | |_{p, \tilde{Φ}} | | a | |_{q, Ψ}, \end{align}

(21)

J : = {\{\sum_{n = n_{0}}^{\infty} {[Ψ (n)]}^{1 - p} {[\int_{b}^{c} K (x, n) f (x) d x]}^{p}\}}^{\frac{1}{p}} > k (λ_{1}) | | f | |_{p, \tilde{Φ}},

(22)

\tilde{L} : = {\{\int_{b}^{c} {[\tilde{Φ} (x)]}^{1 - q} {[\sum_{n = n_{0}}^{\infty} K (x, n) a_{n}]}^{q} d x\}}^{\frac{1}{q}} > k (λ_{1}) | | a | |_{q, Ψ} .

(23)

Moreover, if $\frac{v^{'} (y)}{v (y)} (y \geq n_{0})$ is decreasing and there exist constants δ, δ₀ > 0, such that $θ_{λ} (x) = O (\frac{1}{{[u (x)]}^{δ}}) (x \in (d, c))$ and k(λ₁ - δ₀) ∈ R₊, then the constant factor k(λ₁) in the above inequalities is the best possible.

Proof. In view of (9) and the reverse of (11), for ϖ(x) > k(λ₁)(1 - θ_λ(x)), we have (22). By the reverse Hölder's inequality, we have

I = \sum_{n = n_{0}}^{\infty} [Ψ^{\frac{- 1}{q}} (n) \int_{b}^{c} K (x, n) f (x) d x] [Ψ^{\frac{1}{q}} (n) a_{n}] \geq J | | a | |_{q, Ψ} .

(24)

Then, by (22), we have (21). On the other hand, assuming that (21) is valid, setting a_n as Theorem 1, then J^p^-1 = ||a||_q_Ψ. By the reverse of (11), we find J > 0. If J = ∞, then (24) is naturally valid; if J < ∞, then by (21), we have

| | a | |_{q, Ψ}^{q} = J^{p} = I > k (λ_{1}) | | f | |_{p, \tilde{Φ}} | | a | |_{q, Ψ}, | | a | |_{q, Ψ}^{q - 1} = J > k (λ_{1}) | | f | |_{p, \tilde{Φ}},

and we have (22) which is equivalent to (21).

In view of (9) and the reverse of (12), for [ϖ(x)]^1-q> [k(λ₁)(1 - θ_λ(x))]^1-q(q < 0), we have (23). By the reverse Hölder's inequality, we have

I = \int_{b}^{c} [{\tilde{Φ}}^{\frac{1}{p}} (x) f (x)] [{\tilde{Φ}}^{\frac{- 1}{p}} (x) \sum_{n = n_{0}}^{\infty} K (x, n) a_{n}] d x \geq | | f | |_{p, \tilde{Φ}} \tilde{L} .

(25)

Then, by (23), we have (21). On the other hand, assuming that (21) is valid, setting

f (x) : = {[\tilde{Φ} (x)]}^{1 - q} {[\sum_{n = n_{0}}^{\infty} K (x, n) a_{n}]}^{q - 1}, x \in (b, c),

then

{\tilde{L}}^{q - 1} = | | f | |_{p, \tilde{Φ}}

. By the reverse of (12), we find

\tilde{L} > 0

. If

\tilde{L} = \infty

, then (23) is naturally valid; if

\tilde{L} < \infty

, then by (21), we have

| | f | |_{p, \tilde{Φ}}^{p} = {\tilde{L}}^{q} = I > k (λ_{1}) | | f | |_{p, \tilde{Φ}} | | a | |_{q, Ψ}, | | f | |_{p, \tilde{Φ}}^{p - 1} = \tilde{L} > k (λ_{1}) | | a | |_{q, Ψ},

and we have (23) which is equivalent to (21). ■

Hence, inequalities (21), (22) and (23) are equivalent.

For 0 < ε < pδ₀, setting

\tilde{f} (x)

and

ã_{n}

as Theorem 1, if there exists a positive number k(≥ k(λ₁)), such that (21) is still valid as we replace k(λ₁) by k, then in particular, for q < 0, in view of (9) and the conditions, we have

\begin{align} Ĩ : & = \int_{b}^{c} \sum_{n = n_{0}}^{\infty} K (x, n) ã_{n} \tilde{f} (x) d x > k | | \tilde{f} | |_{p, \tilde{Φ}} | | ã | |_{q, Ψ} \\ = k {\{\int_{d}^{c} (1 - O (\frac{1}{{[u (x)]}^{δ}})) \frac{u^{'} (x) d x}{{[u (x)]}^{ε + 1}}\}}^{\frac{1}{p}} {\{\sum_{n = n_{0}}^{\infty} \frac{v^{'} (n)}{{[v (n)]}^{ε + 1}}\}}^{\frac{1}{q}} \\ = k {\{\frac{1}{ε} - O (1)\}}^{\frac{1}{p}} {\{\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + \sum_{n = n_{0} + 1}^{\infty} \frac{v^{'} (n)}{{[v (n)]}^{ε + 1}}\}}^{\frac{1}{q}} \\ > k {\{\frac{1}{ε} - O (1)\}}^{\frac{1}{p}} {\{\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + \int_{n_{0}}^{\infty} \frac{v^{'} (y)}{{[v (y)]}^{+ 1}} d y\}}^{\frac{1}{q}} \\ = \frac{k}{ε} {\{1 - ε O (1)\}}^{\frac{1}{p}} {\{ε \frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + {[v (n_{0})]}^{- ε}\}}^{\frac{1}{q}}, \end{align}

(26)

\begin{align} Ĩ & = \sum_{n = n_{0}}^{\infty} {[v (n)]}^{λ_{2} - \frac{ε}{q} - 1} v^{'} (n) \int_{d}^{c} K (x, n) {[u (x)]}^{λ_{1} - \frac{ε}{p} - 1} u^{'} (x) d x \\ \leq \sum_{n = n_{0}}^{\infty} {[v (n)]}^{λ_{2} - \frac{ε}{q} - 1} v^{'} (n) \int_{b}^{c} K (x, n) {[u (x)]}^{λ_{1} - \frac{ε}{p} - 1} u^{'} (x) d x \\ \overset{t = u (x) / v (n)}{=} \sum_{n = n_{0}}^{\infty} {[v (n)]}^{- ε - 1} v^{'} (n) \int_{0}^{\infty} k_{λ} (t, 1) t^{(λ_{1} - \frac{ε}{p}) - 1} d t \\ \leq k (λ_{1} - \frac{ε}{p}) [\frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + \int_{n_{0}}^{\infty} {[v (y)]}^{- ε - 1} v^{'} (y) d y] \\ = \frac{1}{ε} k (λ_{1} - \frac{ε}{p}) [ε \frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + {[v (n_{0})]}^{- ε}] . \end{align}

(27)

Since we have

k_{λ} (t, 1) t^{λ_{1} - \frac{ε}{p} - 1} \leq k_{λ} (t, 1) t^{λ_{1} - δ_{0} - 1}

, t ∈ (0, 1] and

\int_{0}^{1} k_{λ} (t, 1) t^{λ_{1} - δ_{0} - 1} d t \leq k (λ_{1} - δ_{0}) < \infty,

then by Lebesgue control convergence theorem [19], it follows

\begin{align} k (λ_{1} - \frac{ε}{p}) & \leq \int_{1}^{\infty} k_{λ} (t, 1) t^{λ_{1} - 1} d t + \int_{0}^{1} k_{λ} (t, 1) t^{λ_{1} - \frac{ε}{p} - 1} d t \\ = k (λ_{1}) + o (1) (ε \to 0^{+}) . \end{align}

By (26) and (27), we have

(k (λ_{1}) + o (1)) [ε \frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + {[v (n_{0})]}^{- ε}] > k {1 - ε O (1)}^{\frac{1}{p}} {[ε \frac{v^{'} (n_{0})}{{[v (n_{0})]}^{ε + 1}} + {[v (n_{0})]}^{- ε}]}^{\frac{1}{q}},

and then k(λ₁) ≥ k(ε → 0⁺). Hence, k = k(λ₁) is the best value of (21).

By the equivalence, the constant factor k(λ₁) in (22) and (23) is the best possible, otherwise we can imply a contradiction by (24) and (25) that the constant factor in (21) is not the best possible. ■

In the same way, for p < 0, we also have the following theorem.

Theorem 3 Suppose that λ₁, λ₂ ∈ R, λ₁ + λ₂ = λ, k_λ(x, y) is a non-negative finite homogeneous function of degree -λ¸ in

R_{+}^{2}

, 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(∞) = ∞, ϖ(x) < k(λ₁) ∈ R₊ (x ∈ (b, c)). If p < 0,

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

, f(x), a_n ≥ 0, 0 < || f ||_p_Φ < ∞ and 0 < ||a||_q,_Ψ < ∞. Then, we have the following equivalent inequalities:

\begin{gathered} I : = \sum_{n = n_{0}}^{\infty} \int_{b}^{c} K (x, n) a_{n} f (x) d x = \int_{b}^{c} \sum_{n = n_{0}}^{\infty} K (x, n) a_{n} f (x) d x \\ > k (λ_{1}) | | f | |_{p, Φ} | | a | |_{q, Ψ}, \end{gathered}

(28)

J : = {\{\sum_{n = n_{0}}^{\infty} {[Ψ (n)]}^{1 - p} {[\int_{b}^{c} K (x, n) f (x) d x]}^{p}\}}^{\frac{1}{p}} > k (λ_{1}) | | f | |_{p, Φ},

(29)

\tilde{L} : = {\{\int_{b}^{c} {[Φ (x)]}^{1 - q} {[\sum_{n = n_{0}}^{\infty} K (x, n) a_{n}]}^{q} d x\}}^{\frac{1}{q}} > k (λ_{1}) | | a | |_{q, Ψ} .

(30)

Moreover, if $\frac{v^{'} (y)}{v (y)} (y \geq n_{0})$ is decreasing and there exists constant δ₀ > 0, such that k(λ₁ + δ₀) ∈ R₊, then the constant factor k(λ₁) in the above inequalities is the best possible.

Remark 2 (i) For n₀ = 1, b = 0, c = ∞, u(x) = v(x) = x, if

ϖ (x) = x^{λ_{1}} \sum_{n = 1}^{\infty} k_{λ} (x, n) n^{λ_{2} - 1} < k (λ_{1}) \in R_{+} (x \in (0, \infty)),

then (13) reduces to (6). In particular, for

k_{λ} (x, n) = \frac{1}{{(x + n)}^{λ}} (λ = λ_{1} + λ_{2}, λ_{1} > 0, 0 < λ_{2} \leq 1)

,(6) reduces to (5).

(ii)

For n₀ = 1, b = 0, c = ∞, u(x) = v(x) = x^α(α > 0), $k_{λ} (x, y) = \frac{1}{{(m a x {x, y})}^{λ}} (λ, λ_{1} > 0, 0 < α λ_{2} \leq 1)$ , since
$f (x, y) = \frac{α x^{α λ_{1}} y^{α λ_{2} - 1}}{{(m a x {x^{α}, y^{α}})}^{λ}} = \{\begin{matrix} α x^{- α λ_{2}} y^{α λ_{2} - 1}, y \leq x \\ α x^{α λ_{1}} y^{- α λ_{1} - 1}, y > x \end{matrix}$

is decreasing for y ∈ (0, ∞) and strictly decreasing in an interval of (0, ∞), then by Condition (i), it follows

\begin{array}{l} ϖ (x) < α x^{α λ_{1}} \int_{0}^{\infty} \frac{1}{{(m a x {x^{α}, y^{α}})}^{λ}} y^{α (λ_{2} - 1)} y^{α - 1} d y \\ \overset{t = (y / x)^{α}}{=} \int_{0}^{\infty} \frac{t^{λ_{2} - 1}}{{(m a x {1, t})}^{λ}} d t = \frac{λ}{λ_{1} λ_{2}} = k (λ_{1}) . \end{array}

Since for

δ = \frac{λ_{1}}{2} < λ_{1}

,

k_{λ} (t, 1) = 1 \leq \frac{1}{t^{δ}} (t \in (0, 1])

, then by (13), we have the following inequality with the best constant factor

\frac{λ}{α λ_{1} λ_{2}}

:

\begin{gathered} \sum_{n = 1}^{\infty} a_{n} \int_{0}^{\infty} \frac{1}{{(m a x {x^{α}, n^{α}})}^{λ}} f (x) d x \\ < \frac{λ}{α λ_{1} λ_{2}} {\{\int_{0}^{\infty} x^{p (1 - α λ_{1}) - 1} f^{p} (x) d x\}}^{\frac{1}{p}} {\{\sum_{n = 1}^{\infty} n^{q (1 - α λ_{2}) - 1} a_{n}^{q}\}}^{\frac{1}{q}} . \end{gathered}

(31)

(iii)

For n₀ = 1, b = β, c = ∞, $u (x) = v (x) = x - β (0 \leq β \leq \frac{1}{2})$ , $k_{λ} (x, y) = \frac{l n (x ∕ y)}{x^{λ} - y^{λ}} (λ, λ_{1} > 0, 0 < λ_{2} \leq 1)$ , since for any fixed x ∈ (β, ∞),
$f (x, y) = (x - β)^{λ_{1}} \frac{l n [(x - β) / (y - β)]}{{(x - β)}^{λ} - {(y - β)}^{λ}} {(y - β)}^{λ_{2} - 1}$

is decreasing and strictly convex for

y \in (\frac{1}{2}, \infty)

, then by Condition (ii), it follows

\begin{array}{l} ϖ (x) < {(x - β)}^{λ_{1}} \int_{\frac{1}{2}}^{\infty} \frac{l n [(x - β) / (y - β)]}{{(x - β)}^{λ} - {(y - β)}^{λ}} {(y - β)}^{λ_{2} - 1} d y \\ \overset{t = [(y - β) / (x - β {)]}^{λ}}{=} \frac{1}{λ^{2}} \int_{{[\frac{\frac{1}{2} - β}{x - β}]}^{λ}}^{\infty} \frac{l n t}{t - 1} t^{(λ_{2} / λ) - 1} d t \\ \leq \frac{1}{λ^{2}} \int_{0}^{\infty} \frac{l n t}{t - 1} t^{(λ_{2} / λ) - 1} d t = {[\frac{π}{λ s i n (\frac{π λ_{1}}{λ})}]}^{2} . \end{array}

Since for

δ = \frac{λ_{1}}{2} < λ_{1}

,

k_{λ} (t, 1) = \frac{l n t}{t^{λ} - 1} \leq \frac{M}{t^{δ}} (t \in (0, 1])

, then by (13), we have the following inequality with the best constant factor

{[\frac{π}{λ s i n (\frac{π λ_{1}}{λ})}]}^{2}

:

\begin{gathered} \sum_{n = 1}^{\infty} a_{n} \int_{β}^{\infty} \frac{l n [(x - β) ∕ (n - β)]}{{(x - β)}^{λ} - {(n - β)}^{λ}} f (x) d x < {[\frac{π}{λ s i n (\frac{π λ_{1}}{λ})}]}^{2} \\ \times {\{\int_{β}^{\infty} {(x - β)}^{p (1 - λ_{1}) - 1} f^{p} (x) d x\}}^{\frac{1}{p}} {\{\sum_{n = 1}^{\infty} {(n - β)}^{q (1 - λ_{2}) - 1} a_{n}^{q}\}}^{\frac{1}{q}} . \end{gathered}

(32)

(iv)

For n₀ = 1, b = 1 - β = γ, c = ∞, u(x) = v(x) = (x - γ), $γ \leq 1 - \frac{1}{8} [λ + \sqrt{λ (3 λ + 4)}]$ , $k_{λ} (x, y) = \frac{1}{x^{λ} + y^{λ}} (0 < λ \leq 4)$ , $λ_{1} = λ_{2} = \frac{λ}{2}$ , we have
$\begin{align} k (\frac{λ}{2}) & = \int_{0}^{\infty} k_{λ} (t, 1) t^{\frac{λ}{2} - 1} d t = \frac{2}{λ} \int_{0}^{\infty} \frac{1}{t^{λ} + 1} d t^{\frac{λ}{2}} \\ = \frac{2}{λ} arctan t^{\frac{λ}{2}} |_{0}^{\infty} = \frac{π}{λ} \in R_{+} \end{align}$

and

f (x, y) = \frac{{(x - γ)}^{\frac{λ}{2}} {(y - γ)}^{\frac{λ}{2} - 1}}{{(x - γ)}^{λ} + {(y - γ)}^{λ}} (x, y \in (γ, \infty)) .

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

{f^{'}}_{y} (x, y) = - (1 + \frac{λ}{2}) \frac{{(x - γ)}^{\frac{λ}{2}} {(y - γ)}^{\frac{λ}{2} - 2}}{{(x - γ)}^{λ} + {(y - γ)}^{λ}} + \frac{λ {(x - γ)}^{\frac{3 λ}{2}} {(y - γ)}^{\frac{λ}{2} - 2}}{[(x - γ)^{λ} + {(y - γ^{λ}]}^{2}} .

We set

R (x) : = \int_{γ}^{1} f (x, y) d y - \frac{1}{2} f (x, 1) - \int_{1}^{\infty} ρ (y) {f^{'}}_{y} (x, y) d y .

(33)

For x ∈ (γ, ∞), 0 < λ ≤ 4, by (33) and the following improved Euler-Maclaurin summation formula [6]:

\frac{- 1}{8} g (1) < \int_{1}^{\infty} ρ (y) g (y) d y < 0 ((- {1)}^{i} g^{(i)} (y) > 0, g^{(i)} (\infty) = 0, i = 0, 1),

we have

\begin{array}{l} R (x) = \int_{γ}^{1} \frac{{(x - γ)}^{\frac{λ}{2}} {(y - γ)}^{\frac{λ}{2} - 1}}{{(x - γ)}^{λ} + {(y - γ)}^{λ}} d y - \frac{1}{2} \frac{{(x - γ)}^{\frac{λ}{2}} {(1 - γ)}^{\frac{λ}{2} - 1}}{{(x - γ)}^{λ} + {(1 - γ)}^{λ}} \\ + (1 + \frac{λ}{2}) \int_{1}^{\infty} ρ (y) \frac{{(x - γ)}^{\frac{λ}{2}} {(y - γ)}^{\frac{λ}{2} - 2}}{{(x - γ)}^{λ} + {(y - γ)}^{λ}} d y \\ - \int_{1}^{\infty} ρ (y) \frac{λ {(x - γ)}^{\frac{3 λ}{2}} {(y - γ)}^{\frac{λ}{2} - 2}}{[(x - γ)^{λ} + {(y - γ^{λ}]}^{2}} d y \\ > \frac{2}{λ} \arctan {(\frac{1 - γ}{x - γ})}^{\frac{λ}{2}} - \frac{{(1 - γ)}^{\frac{λ}{2} - 1} {(x - γ)}^{\frac{λ}{2}}}{2 [(1 - γ)^{λ} + (x - γ^{λ}]} \\ - \frac{1}{8} (1 + \frac{λ}{2}) \frac{{(1 - γ)}^{\frac{λ}{2} - 2} {(x - γ)}^{\frac{λ}{2}}}{{(1 - γ)}^{λ} + {(x - γ)}^{λ}} + 0 \\ = h (x) : = \frac{2}{λ} \arctan {(\frac{1 - γ}{x - γ})}^{\frac{λ}{2}} \\ - [\frac{1 - γ}{2} + \frac{1}{8} (1 + \frac{λ}{2})] \frac{{(1 - γ)}^{\frac{λ}{2} - 2} {(x - γ)}^{\frac{λ}{2}}}{{(1 - γ)}^{λ} + {(x - γ)}^{λ}} . \end{array}

Since for

γ \leq 1 - \frac{1}{8} [λ + \sqrt{λ (3 λ + 4)}]

, i.e.

1 - γ \geq \frac{1}{8} [λ + \sqrt{λ (3 λ + 4)}] (0 < λ \leq 4)

,

\begin{array}{l} h^{'} (x) = \frac{{(1 - γ)}^{\frac{λ}{2}} {(x - γ)}^{- \frac{λ}{2} - 1}}{(1} \end{array}