On a generalization of a Hilbert-type integral inequality in the whole plane with a hypergeometric function

Vandanjav Adiyasuren; Tserendorj Batbold

doi:10.1186/1029-242X-2013-189

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On a generalization of a Hilbert-type integral inequality in the whole plane with a hypergeometric function

Vandanjav Adiyasuren¹ and
Tserendorj Batbold²Email author

Journal of Inequalities and Applications20132013:189

https://doi.org/10.1186/1029-242X-2013-189

Received: 17 May 2012

Accepted: 16 January 2013

Published: 19 April 2013

Abstract

By introducing some parameters, we establish a generalization of the Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree $- 2 λ$ and the best constant factor which involves the hypergeometric function.

MSC:26D15.

Keywords

Hilbert’s inequality Hölder’s inequality homogeneous kernel weight function equivalent form

1 Introduction

If

p > 1

,

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

and

f (x), g (x) \geq 0

satisfy

0 < \int_{0}^{\infty} f^{p} (x) d x < \infty and 0 < \int_{0}^{\infty} g^{q} (x) d x < \infty,

then

\int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{x + y} d x d y < \frac{π}{sin (π / p)} {\int_{0}^{\infty} f^{p} (x) d x}^{\frac{1}{p}} {\int_{0}^{\infty} g^{q} (x) d x}^{\frac{1}{q}},

(1)

where the constant factor $π / (sin π / p)$ is the best possible. Inequality (1) is called Hardy-Hilbert’s inequality [1] and is important in analysis and applications [2].

In 2001, Yang gave an extension of (1) involving beta function as (see [3]):

\begin{aligned} \int_{0}^{\infty} \int_{0}^{\infty} \frac{f (x) g (y)}{{(x + y)}^{λ}} d x d y \\ < B (\frac{p + λ - 2}{p}, \frac{q + λ - 2}{q}) {\int_{0}^{\infty} x^{1 - λ} f^{p} (x) d x}^{\frac{1}{p}} {\int_{0}^{\infty} x^{1 - λ} g^{q} (x) d x}^{\frac{1}{q}}, \end{aligned}

(2)

where the constant factor $B (\frac{p + λ - 2}{p}, \frac{q + λ - 2}{q})$ ( $λ > 2 - min {p, q}$ ) is the best possible.

Recently, some new Hilbert-type inequalities in the whole plane have been obtained [4, 5]. Xin and Yang in [5] established the following:

If

p > 1

,

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

,

| β | < 1

,

0 < α_{1} < α_{2} < π

,

f, g \geq 0

, satisfy

0 < \int_{- \infty}^{\infty} {| x |}^{- p β - 1} f^{p} (x) d x < \infty and 0 < \int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y < \infty,

then we have

\begin{aligned} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{x^{2} + 2 x y cos α_{i} + y^{2}}} f (x) g (y) d x d y \\ < k (β) {(\int_{- \infty}^{\infty} {| x |}^{- p β - 1} f^{p} (x) d x)}^{\frac{1}{p}} {(\int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y)}^{\frac{1}{q}}, \end{aligned}

(3)

and

\begin{aligned} \int_{- \infty}^{\infty} {| y |}^{p (1 - β) - 1} {(\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{x^{2} + 2 x y cos α_{i} + y^{2}}} f (x) d x)}^{p} d y \\ < k^{p} (β) \int_{- \infty}^{\infty} {| x |}^{- p β - 1} f^{p} (x) d x, \end{aligned}

(4)

where the constant factors $k (β) = \frac{π}{sin β π} (\frac{sin β α_{1}}{sin α_{1}} + \frac{sin β (π - α_{2})}{sin α_{2}})$ and $k^{p} (β)$ are the best possible. Inequalities (3) and (4) are equivalent.

By introducing some parameters, we establish generalizations of inequalities (3) and (4) with the homogeneous kernel of degree $- 2 λ$ and the best constant factor which involves the hypergeometric function.

2 Preliminary lemmas

In order to prove our assertions, we need the following lemmas.

Recall that the hypergeometric function

F (α, β; γ; x)

is defined [6] by

F (α, β; γ; x) = \sum_{r = 0}^{\infty} \frac{{(α)}_{r} {(β)}_{r}}{{(γ)}_{r}} \frac{x^{r}}{r!},

(5)

where

{(α)}_{r}

is the Pochhammer symbol defined by

{(α)}_{r} = α (α + 1) \dots (α + r - 1) = \frac{Γ (α + r)}{Γ (α)} .

It is known the series (5) converges for

| x | < 1

and diverges for

| x | > 1

. The hypergeometric function satisfies the integral representation

F (α, β; γ; x) = \frac{Γ (γ)}{Γ (β) Γ (γ - β)} \int_{0}^{1} t^{β - 1} {(1 - t)}^{γ - β - 1} {(1 - x t)}^{- α} d t, if γ > β > 0 .

Lemma 2.1 (See [7])

Suppose that

a, c > 0

,

b^{2} < a c

,

0 < α < 2 λ

. Then we have

\int_{0}^{\infty} \frac{x^{α - 1}}{{(a x^{2} + 2 b x + c)}^{λ}} d x = a^{- \frac{α}{2}} c^{\frac{α}{2} - λ} B (α, 2 λ - α) F (\frac{α}{2}, λ - \frac{α}{2}; λ + \frac{1}{2}; 1 - \frac{b^{2}}{a c}) .

Lemma 2.2 Let

a, c, λ > 0

,

b \geq 0

,

1 - 2 λ < β < 1

and

0 < α_{1} < α_{2} < π

be real parameters such that

b^{2} max {{cos}^{2} α_{1}, {cos}^{2} (π - α_{2})} < a c

. Define the weight functions

ω (x)

and

ϖ (y)

(

x, y \in (- \infty, \infty)

) as follows:

\begin{matrix} ω (x) : = \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \frac{{| x |}^{β + 2 λ - 1}}{{| y |}^{β}} d y, \\ ϖ (y) : = \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \frac{{| y |}^{1 - β}}{{| x |}^{- β - 2 λ + 2}} d x . \end{matrix}

Then we have

ω (x) = ϖ (y) = C_{λ}

(

x, y \neq 0

), where

\begin{aligned} C_{λ} = & a^{\frac{1 - β}{2} - λ} c^{\frac{1 - β}{2}} B (1 - β, 2 λ + β - 1) \\ \times [F (\frac{1 - β}{2}, λ - \frac{1 - β}{2}; λ + \frac{1}{2}; 1 - \frac{b^{2} {cos}^{2} α_{1}}{a c}) \\ + F (\frac{1 - β}{2}, λ - \frac{1 - β}{2}; λ + \frac{1}{2}; 1 - \frac{b^{2} {cos}^{2} (π - α_{2})}{a c})] . \end{aligned}

Proof For

x \in (- \infty, 0)

, setting

u = y / x

,

u = - y / x

in the following two integrals, respectively, and using Lemma 2.1, we get

\begin{array}{rcl} ω (x) & = & \int_{- \infty}^{0} \frac{1}{{(a x^{2} + 2 b x y cos α_{1} + c y^{2})}^{λ}} \frac{{(- x)}^{β + 2 λ - 1}}{{(- y)}^{β}} d y \\ + \int_{0}^{\infty} \frac{1}{{(a x^{2} + 2 b x y cos α_{2} + c y^{2})}^{λ}} \frac{{(- x)}^{β + 2 λ - 1}}{y^{β}} d y \\ = & \int_{0}^{\infty} \frac{u^{- β}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u + \int_{0}^{\infty} \frac{u^{- β}}{{(c u^{2} + 2 b u cos (π - α_{2}) + a)}^{λ}} d u \\ = & C_{λ} . \end{array}

For

x \in (0, \infty)

, setting

u = - y / x

,

u = y / x

in the following two integrals, respectively, and using Lemma 2.1, we get

\begin{array}{rcl} ω (x) & = & \int_{- \infty}^{0} \frac{1}{{(a x^{2} + 2 b x y cos α_{2} + c y^{2})}^{λ}} \frac{x^{β + 2 λ - 1}}{{(- y)}^{β}} d y \\ + \int_{0}^{\infty} \frac{1}{{(a x^{2} + 2 b x y cos α_{1} + c y^{2})}^{λ}} \frac{x^{β + 2 λ - 1}}{y^{β}} d y \\ = & \int_{0}^{\infty} \frac{u^{- β}}{{(c u^{2} + 2 b u cos (π - α_{2}) + a)}^{λ}} d u + \int_{0}^{\infty} \frac{u^{- β}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u \\ = & C_{λ} . \end{array}

By the same way, we still can find that $ω (x) = ϖ (y) = C_{λ}$ ( $x, y \neq 0$ ). The lemma is proved. □

Lemma 2.3 Let p and q be conjugate parameters with

p > 1

, and let

a, c, λ > 0

,

b \geq 0

,

1 - 2 λ < β < 1

,

0 < α_{1} < α_{2} < π

and

b^{2} max {{cos}^{2} α_{1}, {cos}^{2} (π - α_{2})} < a c

, and

f (x)

be a nonnegative measurable function in

(- \infty, \infty)

, then we have

\begin{aligned} J & : = \int_{- \infty}^{\infty} {| y |}^{p (1 - β) - 1} {(\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} f (x) d x)}^{p} d y \\ \leq C_{λ}^{p} \int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x . \end{aligned}

(6)

Proof By Lemma 2.2 and Hölder’s inequality [8], we have

\begin{aligned} {(\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} f (x) d x)}^{p} \\ = [\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \\ \times {(\frac{{| x |}^{(- β - 2 λ + 2) / q}}{{| y |}^{β / p}} f (x)) (\frac{{| y |}^{β / p}}{{| x |}^{(- β - 2 λ + 2) / q}}) d x]}^{p} \\ \leq \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \frac{{| x |}^{(1 - p) (β + 2 λ - 2)}}{{| y |}^{β}} f^{p} (x) d x \\ \times {(\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \frac{{| y |}^{(q - 1) β}}{{| x |}^{(- β - 2 λ + 2)}} d x)}^{p - 1} \\ = C_{λ}^{p - 1} {| y |}^{p (β - 1) + 1} \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \frac{{| x |}^{(1 - p) (β + 2 λ - 2)}}{{| y |}^{β}} f^{p} (x) d x . \end{aligned}

(7)

Then by the Fubini theorem, it follows that

\begin{aligned} J & \leq C_{λ}^{p - 1} \int_{- \infty}^{\infty} [\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \frac{{| x |}^{(1 - p) (β + 2 λ - 2)}}{{| y |}^{β}} f^{p} (x) d x] d y \\ = C_{λ}^{p - 1} \int_{- \infty}^{\infty} ω (x) {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x \\ = C_{λ}^{p} \int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x . \end{aligned}

The lemma is proved. □

3 Main results

Theorem 3.1 Let p and q be conjugate parameters with

p > 1

, and let

a, c, λ > 0

,

b \geq 0

,

1 - 2 λ < β < 1

,

0 < α_{1} < α_{2} < π

and

b^{2} max {{cos}^{2} α_{1}, {cos}^{2} (π - α_{2})} < a c

, and

f, g \geq 0

, satisfy

0 < \int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x < \infty

and

0 < \int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y < \infty

. Then

\begin{aligned} I & : = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} f (x) g (y) d x d y \\ < C_{λ} {(\int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x)}^{1 / p} {(\int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y)}^{1 / q}, \end{aligned}

(8)

and

\begin{aligned} J & = \int_{- \infty}^{\infty} {| y |}^{p (1 - β) - 1} {(\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} f (x) d x)}^{p} d y \\ < C_{λ}^{p} \int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x, \end{aligned}

(9)

where the constant factors $C_{λ}$ and $C_{λ}^{p}$ are the best possible and $C_{λ}$ is defined in Lemma 2.2. Inequalities (8) and (9) are equivalent.

Proof If (7) takes the form of the equality for a

y \in (- \infty, 0) \cup (0, \infty)

, then there exist constants A and B such that they are not all zero, and

A \frac{{| x |}^{(1 - p) (β + 2 λ - 2)}}{{| y |}^{β}} f^{p} (x) = B \frac{{| y |}^{(q - 1) β}}{{| x |}^{(- β - 2 λ + 2)}} a.e. in (- \infty, \infty) \times (- \infty, \infty) .

Hence, there exists a constant K such that

A {| x |}^{p (β + 2 λ - 2)} f^{p} (x) = B {| y |}^{q β} = K a.e. in (- \infty, \infty) \times (- \infty, \infty) .

We suppose $A \neq 0$ (otherwise $B = A = 0$ ). Then ${| x |}^{p (β + 2 λ - 2) - 1} f^{p} (x) = K / (A | x |)$ a.e. in $(- \infty, \infty)$ , which contradicts the fact that $0 < \int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x < \infty$ . Hence, (7) takes the form of a strict inequality, so does (6), and we have (9).

By Hölder’s inequality [8], we have

\begin{aligned} I = & \int_{- \infty}^{\infty} ({| y |}^{1 / q - β} \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} f (x) d x) \\ \times ({| y |}^{β - 1 / q} g (y)) d y \\ \leq & J^{1 / p} {(\int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y)}^{1 / q} . \end{aligned}

(10)

By (9), we have (8). On the other hand, suppose that (8) is valid. Set

g (y) = {| y |}^{p (1 - β) - 1} {(\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} f (x) d x)}^{p - 1},

then it follows

J = \int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y

. By (6), we have

J < \infty

. If

J = 0

, then (9) is obviously valid. If

0 < J < \infty

, then by (8), we obtain

\begin{array}{rcl} 0 & < & \int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y = J = I \\ < & C_{λ} {(\int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x)}^{1 / p} {(\int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y)}^{1 / q}, \end{array}

and

\begin{array}{rcl} J^{1 / p} & = & {(\int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y)}^{1 / p} \\ < & C_{λ} {(\int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} f^{p} (x) d x)}^{1 / p} . \end{array}

Hence, we have (9), which is equivalent to (8).

For

ε > 0

, define functions

\tilde{f} (x)

,

\tilde{g} (y)

as follows:

\begin{array}{rcl} \tilde{f} (x) & : = & {\begin{cases} x^{(β + 2 λ - 2) - 2 ε / p}, & x \in (1, \infty), \\ 0, & x \in [- 1, 1], \\ {(- x)}^{(β + 2 λ - 2) - 2 ε / p}, & x \in (- \infty, - 1), \end{cases} \\ \tilde{g} (y) & : = & {\begin{cases} y^{- β - 2 ε / q}, & y \in (1, \infty), \\ 0, & y \in [- 1, 1], \\ {(- y)}^{- β - 2 ε / q}, & y \in (- \infty, - 1) . \end{cases} \end{array}

Then

\tilde{L} : = {(\int_{- \infty}^{\infty} {| x |}^{- p (β + 2 λ - 2) - 1} {\tilde{f}}^{p} (x) d x)}^{1 / p} {(\int_{- \infty}^{\infty} {| y |}^{q β - 1} {\tilde{g}}^{q} (y) d y)}^{1 / q} = \frac{1}{ε},

and

\begin{array}{rcl} \tilde{I} & : = & \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{{(a x^{2} + 2 b x y cos α_{i} + c y^{2})}^{λ}}} \tilde{f} (x) \tilde{g} (y) d x d y \\ = & I_{1} + I_{2} + I_{3} + I_{4}, \end{array}

where

\begin{array}{rcl} I_{1} & : = & \int_{- \infty}^{- 1} {(- x)}^{(β + 2 λ - 2) - 2 ε / p} [\int_{- \infty}^{- 1} \frac{{(- y)}^{- β - 2 ε / q}}{{(a x^{2} + 2 b x y cos α_{1} + c y^{2})}^{λ}} d y] d x, \\ I_{2} & : = & \int_{- \infty}^{- 1} {(- x)}^{(β + 2 λ - 2) - 2 ε / p} [\int_{1}^{\infty} \frac{y^{- β - 2 ε / q}}{{(a x^{2} + 2 b x y cos α_{2} + c y^{2})}^{λ}} d y] d x, \\ I_{3} & : = & \int_{1}^{\infty} x^{(β + 2 λ - 2) - 2 ε / p} [\int_{- \infty}^{- 1} \frac{{(- y)}^{- β - 2 ε / q}}{{(a x^{2} + 2 b x y cos α_{2} + c y^{2})}^{λ}} d y] d x, \end{array}

and

I_{4} : = \int_{1}^{\infty} x^{(β + 2 λ - 2) - 2 ε / p} [\int_{1}^{\infty} \frac{y^{- β - 2 ε / q}}{{(a x^{2} + 2 b x y cos α_{1} + c y^{2})}^{λ}} d y] d x .

Taking

u = y / x

, by the Fubini theorem, we obtain

\begin{aligned} I_{1} = & I_{4} = \int_{1}^{\infty} x^{- 1 - 2 ε} \int_{1 / x}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u d x \\ = & \int_{1}^{\infty} x^{- 1 - 2 ε} (\int_{1 / x}^{1} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u + \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u) d x \\ = & \int_{0}^{1} (\int_{1 / u}^{\infty} x^{- 1 - 2 ε} d x) \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u \\ + \frac{1}{2 ε} \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u \\ = & \frac{1}{2 ε} (\int_{0}^{1} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u + \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u), \end{aligned}

and

I_{2} = I_{3} = \frac{1}{2 ε} (\int_{0}^{1} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u + \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u) .

In view of the above results, if the constant factor

C_{λ}

in (8) is not the best possible, then there exists a positive number

\tilde{C}

with

\tilde{C} < C_{λ}

such that

\begin{aligned} \int_{0}^{1} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u + \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u \\ + \int_{0}^{1} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u + \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u \\ = ε \tilde{I} < ε \tilde{C} \cdot \tilde{L} = \tilde{C} . \end{aligned}

(11)

By the Fatou lemma and (11), we have

\begin{array}{rcl} C_{λ} & = & \int_{0}^{\infty} \frac{u^{- β}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u + \int_{0}^{\infty} \frac{u^{- β}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u \\ = & \int_{0}^{1} lim_{ε \to 0^{+}} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u + \int_{1}^{\infty} lim_{ε \to 0^{+}} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u \\ + \int_{0}^{1} lim_{ε \to 0^{+}} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u + \int_{1}^{\infty} lim_{ε \to 0^{+}} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u \\ \leq & lim_{ε \to 0^{+}} [\int_{0}^{1} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u + \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} + 2 b u cos α_{1} + a)}^{λ}} d u \\ + \int_{0}^{1} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u + \int_{1}^{\infty} \frac{u^{- β - 2 ε / q}}{{(c u^{2} - 2 b u cos α_{2} + a)}^{λ}} d u] \\ \leq & \tilde{C}, \end{array}

which contradicts the fact that $\tilde{C} < C_{λ}$ . Hence, the constant factor $C_{λ}$ in (8) is the best possible.

If the constant factor in (9) is not the best possible, then by (10), we may get a contradiction that the constant factor in (8) is not the best possible. Thus the theorem is proved. □

Remark 1 Setting $λ = a = b = c = 1$ in Theorem 3.1, we have (3) and (4).

Remark 2 Setting

λ = 1 / 2

in Theorem 3.1, we have the following particular results:

\begin{aligned} \int_{- \infty}^{\infty} & \int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{\sqrt{a x^{2} + 2 b x y cos α_{i} + c y^{2}}}} f (x) g (y) d x d y \\ < & C_{1 / 2} {(\int_{- \infty}^{\infty} {| x |}^{- p (β - 1) - 1} f^{p} (x) d x)}^{1 / p} {(\int_{- \infty}^{\infty} {| y |}^{q β - 1} g^{q} (y) d y)}^{1 / q}, \end{aligned}

and

\begin{aligned} \int_{- \infty}^{\infty} & {| y |}^{p (1 - β) - 1} {(\int_{- \infty}^{\infty} min_{i \in {1, 2}} {\frac{1}{\sqrt{a x^{2} + 2 b x y cos α_{i} + c y^{2}}}} f (x) d x)}^{p} d y \\ < & C_{1 / 2}^{p} \int_{- \infty}^{\infty} {| x |}^{- p (β - 1) - 1} f^{p} (x) d x . \end{aligned}

Declarations

Authors’ Affiliations

(1)

Department of Mathematical Analysis, National University of Mongolia

(2)

Institute of Mathematics, National University of Mongolia

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