On a more accurate half-discrete mulholland's inequality and an extension

Qiang Chen; Bicheng Yang

doi:10.1186/1029-242X-2012-70

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On a more accurate half-discrete mulholland's inequality and an extension

Qiang Chen¹ and
Bicheng Yang²Email author

Journal of Inequalities and Applications20122012:70

https://doi.org/10.1186/1029-242X-2012-70

Received: 14 November 2011

Accepted: 26 March 2012

Published: 26 March 2012

Abstract

By using the way of weight functions and Jensen-Hadamard's inequality, a more accurate half-discrete Mulholland's inequality with a best constant factor is given. The extension with multi-parameters, the equivalent forms as well as the operator expressions are considered.

Mathematics Subject Classication 2000: 26D15; 47A07.

Keywords

Mulholland's inequality weight function equivalent form operator expression

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 the best possible. Moreover, for

a = {\{a_{m}\}}_{m = 1}^{\infty} \in l^{2}, b = {\{b_{n}\}}_{n = 1}^{\infty} \in l^{2}, ∥a∥ = {\{\sum_{m = 1}^{\infty} a_{m}^{2}\}}^{\frac{1}{2}} < 0, ∥b∥ > 0

, we still have the following 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]) and they still represent the field of interest to numerous mathematicians. Also we have the following Mulholland's inequality with the same best constant factor (cf. [1, 5]):

\sum_{m = 2}^{\infty} \sum_{n = 2}^{\infty} \frac{a_{m} b_{n}}{ln m n} < π {\{\sum_{m = 2}^{\infty} m a_{m}^{2} \sum_{n = 2}^{\infty} n b_{n}^{2}\}}^{\frac{1}{2}} .

(3)

In 1998, by introducing an independent parameter λ ∈ (0, 1], Yang [6] gave an extension of (1). By generalizing the results from [6], Yang [7] 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_{+}), {‖ 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},

(4)

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 for x > 0(y > 0), then for

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},

(5)

where, k(λ₁) is still the best value. Clearly, for $p = q = 2, λ = 1, k_{1} (x, y) = \frac{1}{x + y}, λ_{1} = λ_{2} = \frac{1}{2}$ , inequality (4) reduces to (1), while (5) reduces to (2). Some other results about Hilbert-type inequalities are provided by [8–16].

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 the constant factors in the inequalities are the best possible. However Yang [17] gave a result with the kernel

\frac{1}{{(1 + n x)}^{λ}}

by introducing an interval variable and proved that the constant factor is the best possible. Recently, Yang [18] gave the following half-discrete Hilbert's inequality with the best constant factor B(λ₁, λ₂)(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} .

(6)

In this article, by using the way of weight functions and Jensen-Hadamard's inequality, a more accurate half-discrete Mulholland's inequality with a best constant factor similar to (6) is given as follows:

\int_{\frac{3}{2}}^{\infty} f (x) \sum_{n = 2}^{\infty} \frac{a_{n}}{ln \frac{4}{9} x n} d x < π {\{\int_{\frac{3}{2}}^{\infty} x f^{2} (x) d x \sum_{n = 2}^{\infty} n a_{n}^{2}\}}^{\frac{1}{2}} .

(7)

Moreover, a best extension of (7) with multi-parameters, some equivalent forms as well as the operator expressions are also considered.

2 Some lemmas

Lemma 1 If

λ_{1} > 0, 0 < λ_{2} \leq 1, λ_{1} + λ_{2} = λ, α \geq \frac{4}{9}

, setting weight functions ω(n) and ϖ(x) as follows:

ω (n) : = {(ln \sqrt{α} n)}^{λ_{2}} \int_{\frac{1}{\sqrt{α}}}^{\infty_{}} \frac{{(ln \sqrt{α} x)}^{λ_{1} - 1}}{x {(ln α x n)}^{λ}} d x, n \in N \ \{1\},

(8)

ϖ (x) : = {(ln \sqrt{α} x)}^{λ_{1}} \sum_{n = 2}^{\infty} \frac{{(ln \sqrt{α} n)}^{λ_{2} - 1}}{n {(ln α x n)}^{λ}}, x \in (\frac{1}{\sqrt{α}}, \infty),

(9)

then we have

ϖ (x) < ω (n) = B (λ_{1}, λ_{2}) .

(10)

Proof. Applying the substitution

t = \frac{ln \sqrt{α} x}{ln \sqrt{α} n}

to (8), we obtain

ω (n) = \int_{0}^{\infty} \frac{1}{{(1 + t)}^{λ}} t^{λ_{1} - 1} d t = B (λ_{1}, λ_{2}) .

Since by the conditions and for fixed

x \geq \frac{1}{\sqrt{α}}

,

h (x, y) : = \frac{{(ln \sqrt{α} y)}^{λ_{2} - 1}}{y {(ln α x y)}^{λ}} = \frac{1}{y {(ln \sqrt{α} x + ln \sqrt{α} y)}^{λ} {(ln \sqrt{α} y)}^{1 - λ_{2}}}

is decreasing and strictly convex in

y \in (\frac{3}{2}, \infty)

, then by Jensen-Hadamard's inequality (cf. [1]), we find

\begin{gathered} ϖ (x) < {(\sqrt{α} ln x)}^{λ_{1}} \int_{\frac{3}{2}}^{\infty} \frac{1}{y {(ln α x y)}^{λ}} {(ln \sqrt{α} y)}^{λ_{2} - 1} d y \\ \underset{=}{t = (ln \sqrt{α} y)/(ln \sqrt{α} x)} \int_{\frac{ln (3 \sqrt{α} / 2)}{ln \sqrt{α} x}}^{\infty} \frac{t^{λ_{2} - 1} d t}{{(1 + t)}^{λ}} \leq B (λ_{2}, λ_{1}) B (λ_{1}, λ_{2}), \end{gathered}

namely, (10) follows. □ ▪

Lemma 2 Let the assumptions of Lemma 1 be fulfilled and additionally,

p > 1, \frac{1}{p} + \frac{1}{q} = 1, a_{n} \geq 0, \in N \ \{1\}, f (x)

is a non-negative measurable function in

(\frac{1}{\sqrt{α}}, \infty)

. Then we have the following inequalities:

\begin{gathered} J : = {\{\sum_{n = 2}^{\infty} \frac{{(ln \sqrt{α} n)}^{p λ_{2} - 1}}{n} {[\int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{f (x)}{{(ln α x n)}^{λ}} d x]}^{p}\}}^{\frac{1}{p}} \\ \leq {[B (λ_{1}, λ_{2})]}^{\frac{1}{q}} {\{\int_{\frac{1}{\sqrt{α}}}^{\infty} ϖ (x) x^{p - 1} {(ln \sqrt{α} x)}^{p (1 - λ_{1})} f^{p} (x) d x\}}^{\frac{1}{p}}, \end{gathered}

(11)

\begin{gathered} L_{1} : = {\{\int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{{(ln \sqrt{α} x)}^{q λ_{1} - 1}}{x {[ϖ (x)]}^{q - 1}} {[\sum_{n = 2}^{\infty} \frac{a_{n}}{{(ln α x n)}^{λ}}]}^{q} d x\}}^{\frac{1}{q}} \\ \leq {\{B (λ_{1}, λ_{2}) \sum_{n = 2}^{\infty} n^{q - 1} {(ln \sqrt{α} n)}^{q (1 - λ_{2}) - 1} a_{n}^{q}\}}^{\frac{1}{q}} . \end{gathered}

(12)

Proof. By Hälder's inequality cf. [1] and (10), it follows

\begin{gathered} {[\int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{f (x) d x}{{(ln α x n)}^{λ}}]}^{p} = \{\int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{1}{{(ln α x n)}^{λ}} [\frac{{(ln \sqrt{α} x)}^{(1 - λ_{1}) / q} x^{1 / q}}{{(ln \sqrt{α} n)}^{(1 - λ_{2}) / p} n^{1 / p}} f (x)] \\ {\times [\frac{(ln \sqrt{α} n^{(1 - λ_{2}) / p} n^{1 / p})}{(ln \sqrt{α} x^{(1 - λ_{1}) / q} x^{1 / q})}] d x\}}^{p} \leq \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{x^{p - 1}}{{(ln α x n)}^{λ}} \frac{{(ln \sqrt{α} x)}^{(1 - λ_{1}) (p - 1)}}{n {(ln \sqrt{α} n)}^{1 - λ_{2}}} f^{p} (x) d x \\ \times {\{\int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{1}{{(ln α x n)}^{λ}} \frac{n^{q - 1} {(ln \sqrt{α} n)}^{(1 - λ_{2}) (q - 1)}}{x {(ln \sqrt{α} x)}^{1 - λ_{1}}} d x\}}^{p - 1} \\ = {\{ω (n) \frac{{(ln \sqrt{α} n)}^{q (1 - λ_{2}) - 1}}{n^{1 - q}}\}}^{p - 1} \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{x^{p - 1} {(ln \sqrt{α} x)}^{(1 - λ_{1}) (p - 1)}}{n {(ln α x n)}^{λ} {(ln \sqrt{α} n)}^{1 - λ_{2}}} f^{p} (x) d x \\ = \frac{{[B (λ_{1}, λ_{2})]}^{p - 1} n}{{(ln \sqrt{α} n)}^{p λ_{2} - 1}} \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{x^{p - 1} {(ln \sqrt{α} x)}^{(1 - λ_{1}) (p - 1)}}{n {(ln \sqrt{α} x n)}^{λ} {(ln \sqrt{α} n)}^{1 - λ_{2}}} f^{p} (x) d x . \end{gathered}

Then by Beppo Levi's theorem (cf. [19]), we have

\begin{gathered} J \leq {[B (λ_{1}, λ_{2})]}^{\frac{1}{q}} {\{\sum_{n = 2}^{\infty} \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{x^{p - 1} {(ln \sqrt{α} x)}^{(1 - λ_{1}) (p - 1)}}{n {(ln α x n)}^{λ} {(ln \sqrt{α} n)}^{1 - λ_{2}}} f^{p} (x) d x\}}^{\frac{1}{p}} \\ = {[B (λ_{1}, λ_{2})]}^{\frac{1}{q}} {\{\int_{\frac{1}{\sqrt{α}}}^{\infty} \sum_{n = 2}^{\infty} \frac{x^{p - 1} {(ln \sqrt{α} x)}^{(1 - λ_{1}) (p - 1)}}{n {(ln α x n)}^{λ} {(ln \sqrt{α} n)}^{1 - λ_{2}}} f^{p} (x) d x\}}^{\frac{1}{p}} \\ = {[B (λ_{1}, λ_{2})]}^{\frac{1}{q}} {\{\int_{\frac{1}{\sqrt{α}}}^{\infty} ϖ (x) x^{p - 1} {(ln \sqrt{α} x)}^{p (1 - λ_{1}) - 1} f^{p} (x) d x\}}^{\frac{1}{p}}, \end{gathered}

that is, (11) follows. Still by Hölder's inequality, we have

\begin{gathered} {[\sum_{n = 2}^{\infty} \frac{a}{{(ln α x n)}^{λ}}]}^{q} = \{\sum_{n = 2}^{\infty} \frac{1}{{(ln α x n)}^{λ}} [\frac{{(ln \sqrt{α} x)}^{(1 - λ_{1}) / q} x^{1 / q}}{{(ln \sqrt{α} n)}^{(1 - λ_{2}) / p} n^{1 / p}}] \\ {\times [\frac{{(ln \sqrt{α} n)}^{(1 - λ_{2}) / p} n^{1 / p}}{{(ln \sqrt{α} x)}^{(1 - λ_{1}) / q} x^{1 / p}} a_{n}]\}}^{q} \leq {\{\sum_{n = 2}^{\infty} \frac{x^{p - 1}}{{(ln α x n)}^{λ}} \frac{{(ln \sqrt{α} x)}^{(1 - λ_{1}) (p - 1)}}{n {(ln \sqrt{α} n)}^{1 - λ_{2}}}\}}^{q - 1} \\ \times \sum_{n = 2}^{\infty} \frac{1}{{(ln α x n)}^{λ}} \frac{n^{q - 1} {(ln \sqrt{α} n)}^{(1 - λ_{2}) (q - 1)}}{x {(ln \sqrt{α} x)}^{1 - λ_{1}}} a_{n}^{q} \\ = \frac{x {[ϖ (x)]}^{q - 1}}{{(ln \sqrt{α} x)}^{q λ_{1} - 1}} \sum_{n = 2}^{\infty} \frac{{(ln \sqrt{α} x)}^{λ_{1} - 1}}{x {(ln α x n)}^{λ}} n^{q - 1} {(ln \sqrt{α} n)}^{(q - 1) (1 - λ_{2})} a_{n}^{q} . \end{gathered}

Then by Beppo Levi's theorem, we have

\begin{gathered} L_{1} \leq {\{\int_{\frac{1}{\sqrt{α}}}^{\infty} \sum_{n = 2}^{\infty} \frac{{(ln \sqrt{α} x)}^{λ_{1} - 1}}{x {(ln α x n)}^{λ}} n^{q - 1} {(ln \sqrt{α} n)}^{(q - 1) (1 - λ_{2})} a_{n}^{q} d x\}}^{\frac{1}{q}} \\ = {\{\sum_{n = 2}^{\infty} [{(ln \sqrt{α} n)}^{λ_{2}} \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{{(ln \sqrt{α} x)}^{λ_{1} - 1}}{x {(ln α x n)}^{λ}} d x] n^{q - 1} {(ln \sqrt{α} n)}^{q (1 - λ_{2}) - 1} a_{n}^{q}\}}^{\frac{1}{q}} \\ = {\{\sum_{n = 2}^{\infty} ω (n) n^{q - 1} {(ln \sqrt{α} n)}^{q (1 - λ_{2}) - 1} a_{n}^{q}\}}^{\frac{1}{q}}, \end{gathered}

and then in view of (10), inequality (12) follows. □ ▪

3 Main results

We introduce two functions

\begin{gathered} Φ (x) : = x^{p - 1} {(ln \sqrt{α} x)}^{p (1 - λ_{1}) - 1} (x \in (\frac{1}{\sqrt{α}}, \infty)), and \\ Ψ (n) : = n^{q - 1} {(ln \sqrt{α} n)}^{q (1 - λ_{2}) - 1} (n \in N \ \{1\}), \end{gathered}

wherefrom, ${[Φ (x)]}^{1 - q} = \frac{1}{x} {(ln \sqrt{α} x)}^{q λ_{1} - 1}$ , and ${[Ψ (n)]}^{1 - p} = \frac{1}{n} {(ln \sqrt{α} n)}^{p λ_{2} - 1}$ .

Theorem 3 If

p > 1, \frac{1}{p} + \frac{1}{q} = 1, λ_{1} > 0, 0 < λ_{2} \leq 1, λ_{1} + λ_{2} = λ, α \geq \frac{4}{9}, f (x), a_{n} \geq 0, f \in L_{p, Φ} (\frac{1}{\sqrt{α}}, \infty), a = {\{a_{n}\}}_{n = 2}^{\infty} \in l_{q, Ψ}, {∥f∥}_{p, Φ} > 0

, then we have the following equivalent inequalities:

I : = \sum_{n = 2}^{\infty} \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{a_{n} f (x) d x}{{(ln α x n)}^{λ}} = \int_{\frac{1}{\sqrt{α}}}^{\infty} \sum_{n = 2}^{\infty} \frac{f (x) a_{n} d x}{{(ln α x n)}^{λ}} < B (λ_{1}, λ_{2}) {∥f∥}_{p, Φ} {∥a∥}_{q, Ψ},

(13)

J = {\{\sum_{n = 2}^{\infty} {[Ψ (n)]}^{1 - p} {[\int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{f (x) d x}{{(ln α x n)}^{λ}}]}^{p}\}}^{\frac{1}{p}} < B (λ_{1}, λ_{2}) {∥f∥}_{p, Φ},

(14)

L : = {\{\int_{\frac{1}{\sqrt{α}}}^{\infty} {[Φ (x)]}^{1 - q} {[\sum_{n = 2}^{\infty} \frac{a_{n}}{{(ln α x n)}^{λ}}]}^{q} d x\}}^{\frac{1}{q}} < B (λ_{1}, λ_{2}) {∥a∥}_{q, Ψ},

(15)

where the constant B(λ₁, λ₂) is the best possible in the above inequalities.

Proof. By Beppo Levi's theorem (cf. [19]), there are two expressions for I in (13). In view of (11), for ϖ(x) < B(λ₁, λ₂), we have (14). By Hälder's inequality, we have

I = \sum_{n = 2}^{\infty} [Ψ^{\frac{- 1}{q}} (n) \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{1}{{(ln α x n)}^{λ}} 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_{\frac{1}{\sqrt{α}}}^{\infty} \frac{1}{{(ln α x n)}^{λ}} f (x) d x]}^{p - 1}, n \in N \ \{1\},

then J^p^-1 = ||a||_q, _Ψ. By (11), we find J < ∝. If J = 0, then (14) is valid trivially; if J > 0, then by (13), we have

\begin{gathered} {∥a∥}_{q, Ψ}^{q} = J^{p} = I < B (λ_{1}, λ_{2}) {∥f∥}_{p, Ψ} {∥a∥}_{q, Ψ}, i .e . \\ {∥a∥}_{q, Ψ}^{q - 1} = J < B (λ_{1}, λ_{2}) {∥f∥}_{p, Φ}, \end{gathered}

that is, (14) is equivalent to (13). By (12), since [ϖ(x)]^1-q>[B(λ₁, λ₂)]^1-q, we have (15). By Hälder's inequality, we find

I = \int_{\frac{1}{\sqrt{α}}}^{\infty} [Φ^{\frac{1}{p}} (x) f (x)] [Φ^{\frac{- 1}{p}} (x) \sum_{n = 2}^{\infty} \frac{a_{n}}{{(ln α x n)}^{λ}}] 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 = 2}^{\infty} \frac{a_{n}}{{(ln α x n)}^{λ}}]}^{q - 1}, x \in (\frac{1}{\sqrt{α}}, \infty),

then L^q^-1 = ║f║_p, _Φ.. By (12), we find L < ∝. If L = 0, then (15) is valid trivially; if L > 0, then by (13), we have

\begin{gathered} {∥f∥}_{p, Φ}^{p} = L^{q} = I < B (λ_{1}, λ_{2}) {∥f∥}_{p, Φ} {∥a∥}_{q, Ψ}, i .e . \\ {∥f∥}_{p, Φ}^{p - 1} = L < B (λ_{1}, λ_{2}) {∥a∥}_{q, Ψ}, \end{gathered}

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

For 0 < ε <pλ₁, setting

ã_{n} = \frac{1}{n} {(ln \sqrt{α} n)}^{λ_{2} - \frac{\in}{q} - 1}, n \in N \ {1}

, and

\tilde{f} (x) : = \{\begin{gathered} 0, x \in (\frac{1}{\sqrt{α}}, \frac{e}{\sqrt{α}}) \\ \frac{1}{x} {(ln \sqrt{α} x)}^{λ_{1} - \frac{\in}{p} - 1}, x \in [\frac{e}{\sqrt{α}}, \infty) \end{gathered},

if there exists a positive number k(≤ B(λ₁, λ₂)), such that (13) is valid as we replace B(λ₁, λ₂) with k, then in particular, it follows

\begin{array}{l} Ĩ : & = \sum_{n = 2}^{\infty} \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{1}{{(ln α x n)}^{λ}} ã_{n} \tilde{f} (x) d x < k {∥\tilde{f}∥}_{p, Φ} {∥ã∥}_{q, Ψ} \\ = k {\{\int_{\frac{e}{\sqrt{α}}}^{\infty} \frac{d x}{x {(ln \sqrt{α} x)}^{ε + 1}}\}}^{\frac{1}{p}} {\{\frac{1}{2 {(ln 2 \sqrt{α})}^{ε + 1}} + \sum_{n = 3}^{\infty} \frac{1}{n {(ln \sqrt{α} n)}^{ε + 1}}\}}^{\frac{1}{q}} \\ < k {(\frac{1}{ε})}^{\frac{1}{p}} {\{\frac{1}{2 {(ln 2 \sqrt{α})}^{ε + 1}} + \int_{2}^{\infty} \frac{1}{x {(ln \sqrt{α} x)}^{ε + 1}} d x\}}^{\frac{1}{q}} \\ = \frac{k}{ε} {\{\frac{ε}{2 {(ln 2 \sqrt{α})}^{ε + 1}} + \frac{1}{{(ln 2 \sqrt{α})}^{ε}}\}}^{\frac{1}{q}}, \end{array}

(18)

\begin{array}{l} Ĩ & = \sum_{n = 2}^{\infty} {(ln \sqrt{α} n)}^{λ_{2} - \frac{ε}{q} - 1} \frac{1}{n} \int_{\frac{e}{\sqrt{α}}}^{\infty} \frac{1}{x {(ln α x n)}^{λ}} {(ln \sqrt{α} x)}^{λ_{1} - \frac{ε}{p} - 1} d x \\ \underset{=}{t = (ln \sqrt{α} x) / (ln \sqrt{α} n)} \sum_{n = 2}^{\infty} \frac{1}{n {(ln \sqrt{α} n)}^{ε + 1}} \int_{1 / ln \sqrt{α} n}^{\infty} \frac{t^{λ_{1} - \frac{ε}{p} - 1}}{{(t + 1)}^{λ}} d t \\ = B (λ_{1} - \frac{ε}{p}, λ_{2} + \frac{ε}{p}) \sum_{n = 2}^{\infty} \frac{1}{n {(ln \sqrt{α} n)}^{ε + 1}} - A (ε) \\ > B (λ_{1} - \frac{ε}{p}, λ_{2} + \frac{ε}{p}) \int_{2}^{\infty} \frac{1}{y {(ln \sqrt{α} y)}^{ε + 1}} d y - A (ε) \\ = \frac{1}{ε {(ln 2 \sqrt{α})}^{ε}} B (λ_{1} - \frac{ε}{p}, λ_{2} + \frac{ε}{p}) - A (ε), \\ A (ε) & : = \sum_{n = 2}^{\infty} \frac{1}{n {(ln \sqrt{α} n)}^{ε + 1}} \int_{0}^{1 / ln \sqrt{α} n} \frac{1}{{(t + 1)}^{λ}} t^{λ_{1} - \frac{ε}{p} - 1} d t . \end{array}

(19)

W find

\begin{gathered} 0 < A (ε) \leq \sum_{n = 2}^{\infty} \frac{1}{n {(ln \sqrt{α} n)}^{ε + 1}} \int_{0}^{1 / ln \sqrt{α} n} t^{λ_{1} - \frac{ε}{p} - 1} d t \\ = \frac{1}{λ_{1} - \frac{ε}{p}} \sum_{n = 2}^{\infty} \frac{1}{n {(ln \sqrt{α} n)}^{λ_{1} + \frac{ε}{q} + 1}} < \infty, \end{gathered}

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

\frac{B (λ_{1} - \frac{ε}{p}, λ_{2} + \frac{ε}{p})}{{(ln 2 \sqrt{α})}^{ε}} - ε O (1) < k {\{\frac{ε}{2 {(ln 2 \sqrt{α})}^{ε + 1}} + \frac{1}{{(ln 2 \sqrt{α})}^{ε}}\}}^{\frac{1}{q}},

(20)

and B(λ₁, λ₂) ≤ k(ε → 0⁺). Hence, k = B(λ₁, λ₂) is the best value of (13).

Due to the equivalence, the constant factor B(λ₁, λ₂) 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 the first type half-discrete Mulholland's operator

T : L_{p, Φ} (\frac{1}{\sqrt{α}}, \infty) \to l_{p, Ψ^{1 - p}}

as follows: for

f \in L_{p, Φ} (\frac{1}{\sqrt{α}}, \infty)

, we define

T f \in l_{p, Ψ^{1 - p}}

as

T f (n) = \int_{\frac{1}{\sqrt{α}}}^{\infty} \frac{1}{{(ln α x n)}^{λ}} f (x) d x, n \in N \ {1} .

Then by (14), it follows ${∥T f∥}_{p, Ψ^{1 - p}} \leq B (λ_{1}, λ_{2}) {∥f∥}_{p, Φ}$ and then T is a bounded operator with ║T║ ≤ B(λ₁, λ₂). Since by Theorem 1, the constant factor in (14) is the best possible, we have ║T║ = B(λ₁, λ₂).

(ii) Define the second type half-discrete Mulholland's operator

\tilde{T} : l_{q, Ψ} \to L_{q, Φ^{1 - q}} (\frac{1}{\sqrt{α}}, \infty)

as follows: For a ∈ l_q,_ψ, define

\tilde{T} a \in L_{q, Φ^{1 - q}} (\frac{1}{\sqrt{α}}, \infty)

as

\tilde{T} a (x) = \sum_{n = 2}^{\infty} \frac{1}{{(ln α x n)}^{λ}} a_{n}, x \in (\frac{1}{\sqrt{α}}, \infty) .

Then by (15), it follows ${∥\tilde{T} a∥}_{q . Φ^{1 - q}} \leq B (λ_{1}, λ_{2}) {∥a∥}_{q, Ψ}$ and then $\tilde{T}$ is a bounded operator with $∥\tilde{T}∥ \leq B (λ_{1}, λ_{2})$ . Since by Theorem 1, the constant factor in (15) is the best possible, we have $∥\tilde{T}∥ = B (λ_{1}, λ_{2})$ .

Remark 2 We set

p = q = 2, λ = 1, λ_{1} = λ_{2} = \frac{1}{2}

in (13), (14) and (15). (i) if

α = \frac{4}{9}

, then we deduce (7) and the following equivalent inequalities:

\sum_{n = 2}^{\infty} \frac{1}{n} {[\int_{\frac{3}{2}}^{\infty} \frac{f (x)}{ln \frac{4}{9} x n} d x]}^{2} < π^{2} \int_{\frac{3}{2}}^{\infty} x f^{2} (x) d x,

(21)

\int_{\frac{3}{2}}^{\infty} \frac{1}{x} {[\sum_{n = 2}^{\infty} \frac{a_{n}}{ln \frac{4}{9} x n}]}^{2} d x < π^{2} \sum_{n = 2}^{\infty} n a_{n}^{2};

(22)

(ii) if α = 1, then we have the following half-discrete Mulholland's inequality and its equivalent forms:

\int_{1}^{\infty} {f (x) \sum_{n = 2}^{\infty} \frac{a_{n}}{ln x n} d x < π \{\int_{1}^{\infty} x f^{2} (x) d x \sum_{n = 2}^{\infty} n a_{n}^{2}\}}^{\frac{1}{2}},

(23)

\sum_{n = 2}^{\infty} {\frac{1}{n} [\int_{1}^{\infty} \frac{f (x)}{ln x n} d x]}^{2} < π^{2} \int_{1}^{\infty} x f^{2} (x) d x,

(24)

\int_{1}^{\infty} \frac{1}{x} {[\sum_{n = 2}^{\infty} \frac{a_{n}}{ln x n}]}^{2} d x < π^{2} \sum_{n = 2}^{\infty} n a_{n}^{2} .

(25)

Declarations

Acknowledgements

This work is supported by the Guangdong Science and Technology Plan Item (No. 2010B010600018), and the Guangdong Modern Information Service industry Develop Particularly item 2011 (No. 13090).

Authors’ Affiliations

(1)

Department of Computer Science, Guangdong University of Education

(2)

Department of Mathematics, Guangdong University of Education

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