Parameterized Hilbert-Type Integral Inequalities in the Whole Plane

Abstract

By the use of the way of real analysis, we estimate the weight functions and give some new Hilbert-type integral inequalities in the whole plane with nonhomogeneous kernels and multiparameters. The constant factors related to the hypergeometric function and the beta function are proved to be the best possible. We also consider the equivalent forms, the reverses, and some particular cases in the homogeneous kernels.

1. Introduction

If f(x), g(y) ≥ 0,  0 < ∫₀ ^∞ f ²(x)dx < ∞, and 0 < ∫₀ ^∞ g ²(y)dy < ∞, then we have (cf. [1])

$$\begin{array}{c}\underset{\mathrm{0}}{\overset{\infty }{{\displaystyle \iint }}}\frac{f\left(x\right)g\left(y\right)}{x+y}dx\hspace{0.17em}dy\\ \hspace{1em}\hspace{1em}<\pi {\left(\underset{\mathrm{0}}{\overset{\infty }{{\displaystyle \int }}}{f}^{\mathrm{2}}\left(x\right)dx\underset{\mathrm{0}}{\overset{\infty }{{\displaystyle \int }}}{g}^{\mathrm{2}}\left(y\right)dy\right)}^{\mathrm{1}/\mathrm{2}},\end{array}$$ (1)

where the constant factor π is the best possible. Inequality (1) is well known as Hilbert's integral inequality, which is important in analysis and its applications (cf. [1, 2]). In recent years, by using the way of weight functions, a number of extensions of (1) were given by Yang (cf. [3]). Noticing that inequality (1) is with a homogenous kernel of degree −1, a survey of the study of Hilbert-type inequalities with the homogeneous kernels of degree negative numbers and some parameters was given by [4] in 2009. Recently, some inequalities with the homogenous kernels and nonhomogenous kernels have been studied (cf. [5–12]). All of the above integral inequalities are built in the quarter plane of the first quadrant.

In 2007, Yang [13] first gave a Hilbert-type integral inequality with the nonhomogeneous kernel in the whole plane as follows:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{{\displaystyle \iint }}}\frac{f\left(x\right)g\left(y\right)}{{\left(\mathrm{1}+e^{x+y}\right)}^{\lambda }}dx\hspace{0.17em}dy\\ \hspace{1em}<B\left(\frac{\lambda }{\mathrm{2}},\frac{\lambda }{\mathrm{2}}\right){\left(\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{e}^{-\lambda x}{f}^{\mathrm{2}}\left(x\right)dx\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{e}^{-\lambda y}{g}^{\mathrm{2}}\left(y\right)dy\right)}^{\mathrm{1}/\mathrm{2}},\end{array}$$ (2)

where the constant factor B(λ/2, λ/2)  (λ > 0) is the best possible. If 0 < λ < 1, p > 1, and (1/p) + (1/q) = 1, Yang [14] gave another new Hilbert-type integral inequality in the whole plane in 2008 as follows:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{{\displaystyle \iint }}}\frac{\mathrm{1}}{{\left|\mathrm{1}+xy\right|}^{\lambda }}f\left(x\right)g\left(y\right)dx\hspace{0.17em}dy\\ \hspace{1em}\hspace{1em}\hspace{0.17em}<k_{\lambda }{\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\left(\lambda /\mathrm{2}\right)\right)-\mathrm{1}}{f}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\hspace{1em}\hspace{1em}\times {\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\left(\lambda /\mathrm{2}\right)\right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/q},\end{array}$$ (3)

where the constant factor

$$\begin{array}{c}{k}_{\lambda }:=B\left(\frac{\lambda }{\mathrm{2}},\frac{\lambda }{\mathrm{2}}\right)+\mathrm{2}B\left(\mathrm{1}-\lambda ,\frac{\lambda }{\mathrm{2}}\right)\end{array}$$ (4)

is the best possible. He et al. [15–20] also provided some Hilbert-type integral inequalities in the whole plane by using some new methods and techniques.

In this paper, by the use of the way of real analysis, we estimate the weight functions and give some new Hilbert-type integral inequalities in the whole plane with nonhomogeneous kernels and multiparameters, which are extensions of (3). The constant factors related to the hypergeometric function and the beta function are proved to be the best possible. We also consider the equivalent forms, the reverses, and some particular inequalities with the homogeneous kernels.

2. Some Lemmas

Assuming that α > 0, we have Γ(α) = ∫₀ ^∞ x ^α−1 e ^−x dx, where Γ(α) is the Γ function (cf. [21]). For β > −1, η > 0, setting v = −η ln⁡⁡t, we find the following expression:

$$\begin{array}{c}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}{\left(-\ln t\right)}^{\beta }{t}^{\eta -\mathrm{1}}dt=\frac{\mathrm{1}}{{\eta }^{\beta +\mathrm{1}}}=\underset{\mathrm{0}}{\overset{\infty }{{\displaystyle \int }}}{v}^{\left(\beta +\mathrm{1}\right)-\mathrm{1}}{e}^{-v}dv=\frac{\mathrm{\Gamma }\left(\beta +\mathrm{1}\right)}{{\eta }^{\beta +\mathrm{1}}}.\end{array}$$ (5)

Lemma 1 . —

If β > −1, min⁡⁡{μ, σ} > −α, μ + σ = λ < 1 + β, and δ ∈ {−1,1}, define the weight functions ω _δ(σ, y) and ϖ _δ(σ, x) as follows:

$$\begin{array}{c}{\omega }_{\delta }\left(\sigma ,y\right):=\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|1+x^{\delta }y\right|}^{\lambda }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times {\left(\frac{\min \left\{1,\left|x^{\delta }y\right|\right\}}{\max \left\{1,\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\frac{{\left|y\right|}^{\sigma }}{{\left|x\right|}^{1-\delta \sigma }}dx,\\ {\varpi }_{\delta }\left(\sigma ,x\right)≔\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|1+x^{\delta }y\right|}^{\lambda }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times {\left(\frac{\min \left\{1,\left|x^{\delta }y\right|\right\}}{\max \left\{1,\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\frac{{\left|x\right|}^{\delta \sigma }}{{\left|y\right|}^{1-\sigma }}dy.\end{array}$$ (6)

Then, for y, x ∈ (−∞, 0)∪(0, +∞), we have

$$\begin{array}{c}{\omega }_{\delta }\left(\sigma ,y\right)={\varpi }_{\delta }\left(\sigma ,x\right)=K_{\beta }\left(\sigma \right)\\ ≔\mathrm{2}\mathrm{\Gamma }\left(\beta +\mathrm{1}\right){\displaystyle \sum }_{k=\mathrm{0}}^{\infty }\left(\begin{array}{c}\lambda +\mathrm{2}k-\mathrm{1}\\ \mathrm{2}k\end{array}\right)\\ \hspace{1em}\times [\frac{\mathrm{1}}{{\left(\mathrm{2}k+\sigma +\alpha \right)}^{\beta +\mathrm{1}}}\\ \hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em}+\frac{\mathrm{1}}{{\left(\mathrm{2}k+\mu +\alpha \right)}^{\beta +\mathrm{1}}}]\in {\mathbf{R}}_{+}.\end{array}$$ (7)

Proof —

(i) For δ = 1, setting u = xy, we find, for y ∈ (−∞, 0)∪(0, +∞),

$$\begin{array}{c}{\omega }_{\mathrm{1}}\left(\sigma ,y\right)=\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|u\right|\right|}^{\beta }}{{\left|\mathrm{1}+u\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|u\right|\right\}}{\max \left\{\mathrm{1},\left|u\right|\right\}}\right)}^{\alpha }{\left|u\right|}^{\sigma -\mathrm{1}}du\\ \hspace{1em}\hspace{1em}=\underset{\mathrm{0}}{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln u\right|}^{\beta }{\left(\min \left\{\mathrm{1},u\right\}\right)}^{\alpha }{u}^{\sigma -\mathrm{1}}}{{\left(\mathrm{1}+u\right)}^{\lambda }{\left(\max \left\{\mathrm{1},u\right\}\right)}^{\alpha }}du\\ \hspace{1em}\hspace{1em}\hspace{1em}+\underset{-\infty }{\overset{\mathrm{0}}{{\displaystyle \int }}}\frac{{\left|\ln \left(-u\right)\right|}^{\beta }{\left(\min \left\{\mathrm{1},-u\right\}\right)}^{\alpha }}{{\left|\mathrm{1}+u\right|}^{\lambda }{\left(\max \left\{\mathrm{1},-u\right\}\right)}^{\alpha }}{\left(-u\right)}^{\sigma -\mathrm{1}}du\\ \hspace{1em}\hspace{1em}=\underset{\mathrm{0}}{\overset{\infty }{{\displaystyle \int }}}[\frac{{\left|\ln u\right|}^{\beta }{\left(\min \left\{\mathrm{1},u\right\}\right)}^{\alpha }{u}^{\sigma -\mathrm{1}}}{{\left(\mathrm{1}+u\right)}^{\lambda }{\left(\max \left\{\mathrm{1},u\right\}\right)}^{\alpha }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+\frac{{\left|\ln u\right|}^{\beta }{\left(\min \left\{\mathrm{1},u\right\}\right)}^{\alpha }{u}^{\sigma -\mathrm{1}}}{{\left|\mathrm{1}-u\right|}^{\lambda }{\left(\max \left\{\mathrm{1},u\right\}\right)}^{\alpha }}]du\\ \hspace{1em}\hspace{1em}=\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }{u}^{\sigma +\alpha -\mathrm{1}}}{{\left(\mathrm{1}+u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }{u}^{\sigma +\alpha -\mathrm{1}}}{{\left(\mathrm{1}-u\right)}^{\lambda }}\right]du\\ \hspace{1em}\hspace{1em}\hspace{1em}+\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }{u}^{\sigma -\alpha -\mathrm{1}}}{{\left(\mathrm{1}+u\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }{u}^{\sigma -\alpha -\mathrm{1}}}{{\left(u-\mathrm{1}\right)}^{\lambda }}\right]du.\end{array}$$ (8)

By Lebesgue term-by-term integration theorem (cf. [22]), in view of (8) and (5), we find

$$\begin{array}{c}{\omega }_{\mathrm{1}}\left(\sigma ,y\right)=\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{\mathrm{1}}{{\left(\mathrm{1}+u\right)}^{\lambda }}+\frac{\mathrm{1}}{{\left(\mathrm{1}-u\right)}^{\lambda }}\right]\\ \hspace{1em}\hspace{1em}\hspace{1em}\times {\left(-\ln u\right)}^{\beta }\left(u^{\sigma +\alpha -\mathrm{1}}+u^{\mu +\alpha -\mathrm{1}}\right)du\\ =\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}{\displaystyle \sum }_{k=\mathrm{0}}^{\infty }\left(\begin{array}{c}-\lambda \\ k\end{array}\right)\left[u^k+{\left(-u\right)}^k\right]\\ \hspace{1em}\hspace{1em}\hspace{1em}\times {\left(-\ln u\right)}^{\beta }\left(u^{\sigma +\alpha -\mathrm{1}}+u^{\mu +\alpha -\mathrm{1}}\right)du\\ =\mathrm{2}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}{\displaystyle \sum }_{k=\mathrm{0}}^{\infty }\left(\begin{array}{c}\lambda +\mathrm{2}k-\mathrm{1}\\ \mathrm{2}k\end{array}\right)\\ \hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\times {\left(-\ln u\right)}^{\beta }\left(u^{\mathrm{2}k+\sigma +\alpha -\mathrm{1}}+u^{\mathrm{2}k+\mu +\alpha -\mathrm{1}}\right)du\\ =\mathrm{2}{\displaystyle \sum }_{k=\mathrm{0}}^{\infty }\left(\begin{array}{c}\lambda +\mathrm{2}k-\mathrm{1}\\ \mathrm{2}k\end{array}\right)\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times \underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}{\left(-\ln u\right)}^{\beta }\left(u^{\mathrm{2}k+\sigma +\alpha -\mathrm{1}}+u^{\mathrm{2}k+\mu +\alpha -\mathrm{1}}\right)du\\ =\mathrm{2}\mathrm{\Gamma }\left(\beta +\mathrm{1}\right){\displaystyle \sum }_{k=\mathrm{0}}^{\infty }\left(\begin{array}{c}\lambda +\mathrm{2}k-\mathrm{1}\\ \mathrm{2}k\end{array}\right)\\ \hspace{1em}\times \left[\frac{\mathrm{1}}{{\left(\mathrm{2}k+\sigma +\alpha \right)}^{\beta +\mathrm{1}}}+\frac{\mathrm{1}}{{\left(\mathrm{2}k+\mu +\alpha \right)}^{\beta +\mathrm{1}}}\right].\end{array}$$ (9)

(ii) For δ = −1, setting y/x, we still can obtain ω ₋₁(σ, y) = K _β(σ). Setting u = x ^δ y, we find

$$\begin{array}{c}{\varpi }_{\delta }\left(\sigma ,x\right)=\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|u\right|\right|}^{\beta }}{{\left|\mathrm{1}+u\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|u\right|\right\}}{\max \left\{\mathrm{1},\left|u\right|\right\}}\right)}^{\alpha }{\left|u\right|}^{\sigma -\mathrm{1}}du=K_{\beta }\left(\sigma \right).\end{array}$$ (10)

Since, for β > −1,  0 < θ ₀ < min⁡{μ + α, σ + α},

$$\begin{array}{c}{\left(\frac{-\ln u}{\mathrm{1}-u}\right)}^{\beta }{u}^{{\theta }_{\mathrm{0}}}⟶\mathrm{0}\hspace{1em}\left(u⟶{\mathrm{0}}^{+}\right),\\ {\left(\frac{-\ln u}{\mathrm{1}-u}\right)}^{\beta }{u}^{{\theta }_{\mathrm{0}}}⟶\mathrm{1}\hspace{1em}\left(u⟶{\mathrm{1}}^{-}\right),\end{array}$$ (11)

there exists a positive number L, such that ((−ln⁡u)/(1 − u))^β u ^θ₀ ≤ L(u ∈ (0,1]);   then, by (9), it follows that

$$\begin{array}{c}\mathrm{0}<K_{\beta }\left(\sigma \right)\le \mathrm{2}L\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\frac{u^{\sigma +\alpha -{\theta }_{\mathrm{0}}-\mathrm{1}}+u^{\mu +\alpha -{\theta }_{\mathrm{0}}-\mathrm{1}}}{{\left(\mathrm{1}-u\right)}^{\lambda -\beta }}du\\ =\mathrm{2}L[B\left(\mathrm{1}-\lambda +\beta ,\sigma +\alpha -{\theta }_{\mathrm{0}}\right)\\ \hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}+B\left(\mathrm{1}-\lambda +\beta ,\mu +\alpha -{\theta }_{\mathrm{0}}\right)]<\infty ,\end{array}$$ (12)

and then K _β(σ) ∈ R ₊. Hence we have (7).

Remark 2 . —

We have the following formula of the hypergeometric function F (cf. [21]). If Re(γ) > Re(θ) > 0, |arg(1 − z)|<π, then

$$\begin{array}{c}F\left(\alpha ,\theta ,\gamma ,z\right):=\frac{\mathrm{\Gamma }\left(\gamma \right)}{\mathrm{\Gamma }\left(\theta \right)\mathrm{\Gamma }\left(\gamma -\theta \right)}\\ \hspace{1em}\times \underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}{t}^{\theta -\mathrm{1}}{\left(\mathrm{1}-t\right)}^{\gamma -\theta -\mathrm{1}}{\left(\mathrm{1}-zt\right)}^{-\alpha }dt.\end{array}$$ (13)

In particular, for z = −1, γ = θ + 1  (θ > 0), it follows that

$$\begin{array}{c}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}{t}^{\theta -\mathrm{1}}{\left(\mathrm{1}+t\right)}^{-\alpha }dt=\frac{\mathrm{1}}{\theta }F\left(\alpha ,\theta ,\mathrm{1}+\theta ,-\mathrm{1}\right)\in {\mathbf{R}}_{+}.\end{array}$$ (14)

In (9), for β = 0  (λ < 1), in view of (14), we have

$$\begin{array}{c}{K}_{\mathrm{0}}\left(\sigma \right)=\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{\mathrm{1}}{{\left(\mathrm{1}+u\right)}^{\lambda }}+\frac{\mathrm{1}}{{\left(\mathrm{1}-u\right)}^{\lambda }}\right]\\ \hspace{1em}\hspace{1em}\hspace{1em}\times \left(u^{\sigma +\alpha -\mathrm{1}}+u^{\mu +\alpha -\mathrm{1}}\right)du\\ =\frac{\mathrm{1}}{\sigma +\alpha }F\left(\lambda ,\sigma +\alpha ,\mathrm{1}+\sigma +\alpha ,-\mathrm{1}\right)\\ \hspace{1em}+\frac{\mathrm{1}}{\mu +\alpha }F\left(\lambda ,\mu +\alpha ,\mathrm{1}+\mu +\alpha ,-\mathrm{1}\right)\\ \hspace{1em}+B\left(\mathrm{1}-\lambda ,\sigma +\alpha \right)+B\left(\mathrm{1}-\lambda ,\mu +\alpha \right)\in {\mathbf{R}}_{+}.\end{array}$$ (15)

Lemma 3 . —

If p > 1,  (1/p)+(1/q) = 1, β > −1,  min⁡{μ, σ} > −α, μ + σ = λ < 1 + β, δ ∈ {−1,1}, K _β(σ) is indicated by (7), and f(x) is a nonnegative measurable function in (−∞, ∞), then one has

$$\begin{array}{c}J≔\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{p\sigma -\mathrm{1}}[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\times {{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }f\left(x\right)dx]}^{p}dy\\ \le K_{\beta }^p\left(\sigma \right)\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx.\end{array}$$ (16)

Proof —

By Hölder's inequality (cf. [23]), we have

$$\begin{array}{c}{\left[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }f\left(x\right)\right]}^p\\ \hspace{1em}=[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ {\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\hspace{1em}\hspace{1em}\times \left(\frac{{\left|x\right|}^{\left(1-\delta \sigma \right)/q}}{{\left|y\right|}^{\left(1-\sigma \right)/q}}f\left(x\right)\right)\left(\frac{{\left|y\right|}^{\left(1-\sigma \right)/q}}{{\left|x\right|}^{\left(1-\delta \sigma \right)/q}}\right)dx]}^p\\ \hspace{1em}\le \underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\times \frac{{\left|x\right|}^{\left(1-\delta \sigma \right)\left(p-1\right)}}{{\left|y\right|}^{1-\sigma }}{f}^p\left(x\right)dx\\ \hspace{1em}\hspace{1em}\times (\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em}\times {\frac{{\left|y\right|}^{\left(\mathrm{1}-\sigma \right)\left(q-\mathrm{1}\right)}}{{\left|x\right|}^{\mathrm{1}-\delta \sigma }}dx)}^{p-\mathrm{1}}\\ \hspace{1em}=\frac{{\left({\omega }_{\delta }\left(\sigma ,y\right)\right)}^{p-\mathrm{1}}}{{\left|y\right|}^{p\sigma -\mathrm{1}}}\\ \hspace{1em}\hspace{1em}\times \underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\times \begin{array}{c}\frac{{\left|x\right|}^{\left(\mathrm{1}-\delta \sigma \right)\left(p-\mathrm{1}\right)}}{{\left|y\right|}^{\mathrm{1}-\sigma }}{f}^p\left(x\right)dx.\end{array}\end{array}$$ (17)

Then, by (7) and Fubini theorem (cf. [22]), it follows that

$$\begin{array}{c}J\le K_{\beta }^{p-\mathrm{1}}\left(\sigma \right)\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times \frac{{\left|x\right|}^{\left(\mathrm{1}-\delta \sigma \right)\left(p-\mathrm{1}\right)}}{{\left|y\right|}^{\mathrm{1}-\sigma }}{f}^p\left(x\right)dx]dy\\ =K_{\beta }^{p-\mathrm{1}}\left(\sigma \right)\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\varpi }_{\delta }\left(\sigma ,x\right){\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx.\end{array}$$ (18)

Hence, in view of (7), inequality (16) follows.

3. Main Results and Applications

Theorem 4 . —

If p > 1,  (1/p)+(1/q) = 1,  β > −1,  min⁡{μ, σ}>−α,  μ + σ = λ < 1 + β, and δ ∈ {−1,1},  f(x),  g(y) ≥ 0, satisfying 0 < ∫_−∞ ^∞|x|^{p(1−δσ)−1} f ^p(x)dx < ∞ and 0 < ∫_−∞ ^∞ | y|^{q(1−σ)−1} g ^q(y)dy < ∞, then one has

$$\begin{array}{c}I≔\underset{-\infty }{\overset{\infty }{{\displaystyle \iint }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\times f\left(x\right)g\left(y\right)dx\hspace{0.17em}dy\\ <K_{\beta }\left(\sigma \right){\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\times {\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/q},\end{array}$$ (19)

$$\begin{array}{c}J≔\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{p\sigma -\mathrm{1}}[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\hspace{1em}\hspace{1em}\times {{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }f\left(x\right)dx]}^{p}dy\\ <K_{\beta }^p\left(\sigma \right)\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx,\end{array}$$ (20)

where the constant factors K _β(σ) and K _β ^p(σ) are the best possible and K _β(σ) is defined by (7). Inequalities (19) and (20) are equivalent.

In particular, for δ = 1, we have the following equivalent inequalities:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{{\displaystyle \iint }}}\frac{{\left|\ln \left|xy\right|\right|}^{\beta }}{{\left|\mathrm{1}+xy\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|xy\right|\right\}}{\max \left\{\mathrm{1},\left|xy\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}\times f\left(x\right)g\left(y\right)dx\hspace{0.17em}dy\\ \hspace{1em}<K_{\beta }\left(\sigma \right){\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\hspace{1em}\times {\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/q},\\ \underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{p\sigma -\mathrm{1}}[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|xy\right|\right|}^{\beta }}{{\left|\mathrm{1}+xy\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|xy\right|\right\}}{\max \left\{\mathrm{1},\left|xy\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{0.17em}{\times f\left(x\right)dx]}^{p}dy\\ \hspace{1em}<K_{\beta }^p\left(\sigma \right)\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx.\end{array}$$ (21)

Proof —

If (17) takes the form of equality for a y ∈ (−∞, 0)∪(0, ∞), then there exist constants A and B, such that they are not all zero, and

$$\begin{array}{c}A\frac{{\left|x\right|}^{\left(\mathrm{1}-\delta \sigma \right)\left(p-\mathrm{1}\right)}}{{\left|y\right|}^{\mathrm{1}-\sigma }}{f}^p\left(x\right)\\ \hspace{1em}=B\frac{{\left|y\right|}^{\left(\mathrm{1}-\sigma \right)\left(q-\mathrm{1}\right)}}{{\left|x\right|}^{\mathrm{1}-\delta \sigma }}\hspace{0.17em}\hspace{0.17em}\text{a.e.}\hspace{0.17em}\hspace{0.17em}\text{in}\hspace{0.17em}\hspace{0.17em}\left(-\infty ,\infty \right).\end{array}$$ (22)

We suppose that A ≠ 0 (otherwise B = A = 0). Then it follows that

$$\begin{array}{c}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)={\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)}\frac{B}{A\left|x\right|}\hspace{0.17em}\hspace{0.17em}\text{a.e.}\hspace{0.17em}\hspace{0.17em}\text{in}\hspace{0.17em}\hspace{0.17em}\left(-\infty ,\infty \right),\end{array}$$ (23)

which contradicts the fact that 0 < ∫_−∞ ^∞|x|^{p(1−δσ)−1} f ^p(x)dx < ∞. Hence (17) takes the form of strict inequality and so does (16). Then we have (20). By Hölder's inequality (cf. [23]), we find

$$\begin{array}{c}I=\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}[{\left|y\right|}^{\sigma -\left(\mathrm{1}/p\right)}\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.17em}\hspace{1em}\hspace{1em}\hspace{1em}\times {\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }f\left(x\right)dx]\\ \hspace{1em}\times \left({\left|y\right|}^{\left(\mathrm{1}/p\right)-\sigma }g\left(y\right)dy\right)\\ \le J^{\mathrm{1}/p}{\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/q}.\end{array}$$ (24)

By (20), we have (19). On the other hand, suppose that (19) is valid. We set

$$\begin{array}{c}g\left(y\right)≔{\left|y\right|}^{p\sigma -\mathrm{1}}[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\hspace{1em}\times {{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }f\left(x\right)dx]}^{p-\mathrm{1}},\end{array}$$ (25)

and find J = ∫_−∞ ^∞ | y|^{q(1−σ)−1} g ^q(y)dy. By (16), we have J < ∞. If J = 0, then (20) is obviously value; if 0 < J < ∞, then, by (19), we obtain

$$\begin{array}{c}\mathrm{0}<\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy=J=I\\ <K_{\beta }\left(\sigma \right){\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\times {\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/q}<\infty ,\\ J^{\mathrm{1}/p}={\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/p}\\ <K_{\beta }\left(\sigma \right){\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}.\end{array}$$ (26)

Hence we have (20), which is equivalent to (19). We indicate two sets E _δ≔{x ∈ R; |x|^δ ≥ 1} and E _δ ⁺≔E _δ∩R ₊ = {x ∈ R ₊; x ^δ ≥ 1}. For ɛ > 0, we define two functions $\tilde{f}(x)$, $\tilde{g}(y)$ as follows:

$$\begin{array}{c}\tilde{f}\left(x\right)≔\{\begin{array}{cc}{\left|x\right|}^{\delta \left(\sigma -\left(\mathrm{2}\varepsilon /p\right)\right)-\mathrm{1}},\hfill & x\in E_{\delta },\hfill \\ \mathrm{0},\hfill & x\in \mathbf{R}\setminus E_{\delta },\hfill \end{array}\\ \tilde{g}\left(y\right)≔\{\begin{array}{cc}\mathrm{0},\hfill & y\in \left(-\infty ,-\mathrm{1}\right)\cup \left(\mathrm{1},\infty \right),\hfill \\ {\left|y\right|}^{\sigma +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}},\hfill & y\in \left[-\mathrm{1,1}\right].\hfill \end{array}\end{array}$$ (27)

Then we obtain

$$\begin{array}{c}\tilde{L}≔{\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{\tilde{f}}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\times {\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{\tilde{g}}^q\left(y\right)dy\right\}}^{\mathrm{1}/q}\\ =\mathrm{2}{\left\{{{\displaystyle \int }}_{E_{\delta }^{+}}{x}^{-\mathrm{2}\delta \varepsilon -\mathrm{1}}dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\times {\left\{\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}{y}^{\mathrm{2}\varepsilon -\mathrm{1}}dy\right\}}^{\mathrm{1}/q}=\frac{\mathrm{1}}{\varepsilon }.\end{array}$$ (28)

Since, for Y = −y, we find

$$\begin{array}{c}h\left(x\right)≔\underset{-\mathrm{1}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times {\left|y\right|}^{\sigma +\left(\mathrm{2}\epsilon /q\right)-\mathrm{1}}dy\\ \hspace{1em}\hspace{1em}=\underset{-\mathrm{1}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\frac{{\left|\ln \left|{\left(-x\right)}^{\delta }Y\right|\right|}^{\beta }}{{\left|\mathrm{1}+{\left(-x\right)}^{\delta }Y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|{\left(-x\right)}^{\delta }Y\right|\right\}}{\max \left\{\mathrm{1},\left|{\left(-x\right)}^{\delta }Y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times {\left|Y\right|}^{\sigma +\left(\mathrm{2}\epsilon /q\right)-\mathrm{1}}dY=h\left(-x\right),\end{array}$$ (29)

and h(x) is an even function, then it follows that

$$\begin{array}{c}\tilde{I}≔\underset{-\infty }{\overset{\infty }{{\displaystyle \iint }}}\frac{{\left|\ln \left|x^{\delta }y\right|\right|}^{\beta }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }}{\left(\frac{\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}{\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}}\right)}^{\alpha }\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\times \tilde{f}\left(x\right)\tilde{g}\left(y\right)dx\hspace{0.17em}dy\\ ={{\displaystyle \int }}_{E_{\delta }}{\left|x\right|}^{\delta \left(\sigma -\left(\mathrm{2}\epsilon /p\right)\right)-\mathrm{1}}h\left(x\right)dx\\ =\mathrm{2}{{\displaystyle \int }}_{E_{\delta }^{+}}{x}^{\delta \left(\sigma -\left(\mathrm{2}\epsilon /p\right)\right)-\mathrm{1}}h\left(x\right)dx\\ \stackrel{u=x^{\delta }y}{=}\mathrm{2}{{\displaystyle \int }}_{E_{\delta }^{+}}{x}^{-\mathrm{2}\delta \epsilon -\mathrm{1}}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\hspace{1em}\times \left[\underset{-x^{\delta }}{\overset{x^{\delta }}{{\displaystyle \int }}}\frac{{\left|\ln \left|u\right|\right|}^{\beta }{\left(\min \left\{\mathrm{1},\left|u\right|\right\}\right)}^{\alpha }{\left|u\right|}^{\sigma +\left(\mathrm{2}\epsilon /q\right)-\mathrm{1}}}{{\left|\mathrm{1}+u\right|}^{\lambda }{\left(\max \left\{\mathrm{1},\left|u\right|\right\}\right)}^{\alpha }}du\right]dx.\end{array}$$ (30)

Setting v = x ^δ in the above integral, by Fubini theorem (cf. [22]), we find

$$\begin{array}{c}\tilde{I}\stackrel{v=x^{\delta }}{=}\mathrm{2}\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}{v}^{-\mathrm{2}\varepsilon -\mathrm{1}}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times \left[\underset{-v}{\overset{v}{{\displaystyle \int }}}\frac{{\left|\ln \left|u\right|\right|}^{\beta }{\left(\min \left\{\mathrm{1},\left|u\right|\right\}\right)}^{\alpha }{\left|u\right|}^{\sigma +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}}{{\left|\mathrm{1}+u\right|}^{\lambda }{\left(\max \left\{\mathrm{1},\left|u\right|\right\}\right)}^{\alpha }}du\right]dv\\ =\mathrm{2}\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}{v}^{-\mathrm{2}\varepsilon -\mathrm{1}}\{\underset{\mathrm{0}}{\overset{v}{{\displaystyle \int }}}[\frac{{\left|\ln u\right|}^{\beta }{\left(\min \left\{\mathrm{1},u\right\}\right)}^{\alpha }}{{\left|\mathrm{1}-u\right|}^{\lambda }{\left(\max \left\{\mathrm{1},u\right\}\right)}^{\alpha }}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\hspace{1em}+\frac{{\left|\ln u\right|}^{\beta }{\left(\min \left\{\mathrm{1},u\right\}\right)}^{\alpha }}{{\left(\mathrm{1}+u\right)}^{\lambda }{\left(\max \left\{\mathrm{1},u\right\}\right)}^{\alpha }}]\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times u^{\sigma +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\}dv\\ =\mathrm{2}\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}{v}^{-\mathrm{2}\varepsilon -\mathrm{1}}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times \left\{\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\right\}dv\\ \hspace{1em}+\mathrm{2}\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}{v}^{-\mathrm{2}\varepsilon -\mathrm{1}}\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times \left\{\underset{\mathrm{1}}{\overset{v}{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\right\}dv\\ =\frac{\mathrm{1}}{\varepsilon }\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ \hspace{1em}+\mathrm{2}\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left(\underset{u}{\overset{\infty }{{\displaystyle \int }}}{v}^{-\mathrm{2}\varepsilon -\mathrm{1}}dv\right)\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\times \left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ =\frac{\mathrm{1}}{\varepsilon }\{\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}+\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha -\left(\mathrm{2}\varepsilon /p\right)-\mathrm{1}}du\}.\end{array}$$ (31)

If the constant factor K _β(σ) in (19) is not the best possible, then there exists a positive number k with K _β(σ) < k, such that (19) is valid when replacing K _β(σ) by k. Then we have $\tilde{I}<k\tilde{L}$, and

$$\begin{array}{c}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ \hspace{1em}\hspace{1em}+\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha -\left(\mathrm{2}\varepsilon /p\right)-\mathrm{1}}du\\ \hspace{1em}=\varepsilon \tilde{I}<\varepsilon k\tilde{L}=k.\end{array}$$ (32)

By (8) and Fatou lemma (cf. [22]), we have

$$\begin{array}{c}{K}_{\beta }\left(\sigma \right)=\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha -\mathrm{1}}du\\ \hspace{1em}+\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha -\mathrm{1}}du\\ =\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\underset{\varepsilon \to {\mathrm{0}}^{+}}{\lim }\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ \hspace{1em}+\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\underset{\varepsilon \to {\mathrm{0}}^{+}}{\lim }\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha -\left(\mathrm{2}\varepsilon /p\right)-\mathrm{1}}du\\ \le \underset{\varepsilon \to {\mathrm{0}}^{+}}{\underset{\_}{\lim }}\{\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}\hspace{0.5em}+\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times u^{\sigma -\alpha -\left(\mathrm{2}\varepsilon /p\right)-\mathrm{1}}du\}\le k,\end{array}$$ (33)

which contradicts the fact that k < K _β(σ). Hence the constant factor K _β(σ) in (19) is the best possible. If the constant factor in (20) is not the best possible, then, by (24), we may get a contradiction that the constant factor in (19) is not the best possible.

Theorem 5 . —

As the assumptions of Theorem 4, replacing p > 1 by 0 < p < 1, one has the equivalent reverses of (19) and (20) with the same best constant factors.

Proof —

By the reverse Hölder's inequality (cf. [23]), we have the reverses of (16) and (24). It is easy to obtain the reverse of (20). In view of the reverses of (20) and (24), we obtain the reverse of (19). On the other hand, suppose that the reverse of (19) is valid. Setting the same g(y) as (25) in Theorem 4, by the reverse of (16), we have J > 0. If J = ∞, then the reverse of (20) is obviously value; if J < ∞, then, by the reverse of (19), we obtain the reverses of (26). Hence we have the reverse of (20), which is equivalent to the reverse of (19). If the constant factor K _β(σ) in the reverse of (19) is not the best possible, then there exists a positive constant k, with k > K _β(σ), such that the reverse of (19) is still valid when replacing K _β(σ) by k. By the reverse of (32), we have

$$\begin{array}{c}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ \hspace{1em}+\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha -\left(\mathrm{2}\varepsilon /p\right)-\mathrm{1}}du>k.\end{array}$$ (34)

For ɛ → 0⁺, by Levi theorem (cf. [22]), we find

$$\begin{array}{c}\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha -\left(\mathrm{2}\varepsilon /p\right)-\mathrm{1}}du\\ \hspace{1em}⟶\underset{\mathrm{1}}{\overset{\infty }{{\displaystyle \int }}}\left[\frac{{\left(\ln u\right)}^{\beta }}{{\left(u-\mathrm{1}\right)}^{\lambda }}+\frac{{\left(\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma -\alpha -\mathrm{1}}du.\end{array}$$ (35)

There exists a constant δ ₀ > 0, such that σ − (1/2)δ ₀ > −α, and then 0 < K _β(σ − (δ ₀/2)) < ∞. For 0 < ɛ < δ ₀|q|/4  (q < 0), since u ^{σ+α+(2ɛ/q)−1} ≤ u ^{σ+α−(δ₀/2)−1}, u ∈ (0,1], and

$$\begin{array}{c}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha -\left({\delta }_{\mathrm{0}}/\mathrm{2}\right)-\mathrm{1}}du\le K_{\beta }\left(\sigma -\frac{{\delta }_{\mathrm{0}}}{\mathrm{2}}\right),\end{array}$$ (36)

then, by Lebesgue control convergence theorem (cf. [22]), for ɛ → 0⁺, we have

$$\begin{array}{c}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha +\left(\mathrm{2}\varepsilon /q\right)-\mathrm{1}}du\\ \hspace{1em}⟶\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\left[\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}-u\right)}^{\lambda }}+\frac{{\left(-\ln u\right)}^{\beta }}{{\left(\mathrm{1}+u\right)}^{\lambda }}\right]u^{\sigma +\alpha -\mathrm{1}}du.\end{array}$$ (37)

By (34), (35), and (37), for ɛ → 0⁺, we have K _β(σ) ≥ k, which contradicts the fact that k > K _β(σ). Hence, the constant factor K _β(σ) in the reverse of (19) is the best possible. If the constant factor in the reverse of (20) is not the best possible, then, by the reverse of (24), we may get a contradiction that the constant factor in the reverse of (19) is not the best possible.

Remark 6 . —

(i) For δ = −1 in (19) and (20), replacing |x|^λ f(x) by f(x), we obtain the following equivalent inequalities with a homogeneous kernel and the best possible constant factors:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{{\displaystyle \iint }}}\frac{{\left|\ln \left|y/x\right|\right|}^{\beta }}{{\left|x+y\right|}^{\lambda }}{\left(\frac{\min \left\{\left|x\right|,\left|y\right|\right\}}{\max \left\{\left|x\right|,\left|y\right|\right\}}\right)}^{\alpha }f\left(x\right)g\left(y\right)dx\hspace{0.17em}dy\\ \hspace{1em}<K_{\beta }\left(\sigma \right){\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\mu \right)-\mathrm{1}}{f}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\hspace{1em}\times {\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/q},\end{array}$$ (38)

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{p\sigma -\mathrm{1}}\\ \hspace{1em}\hspace{1em}\times {\left[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left|\ln \left|y/x\right|\right|}^{\beta }}{{\left|x+y\right|}^{\lambda }}{\left(\frac{\min \left\{\left|x\right|,\left|y\right|\right\}}{\max \left\{\left|x\right|,\left|y\right|\right\}}\right)}^{\alpha }f\left(x\right)dx\right]}^{p}dy\\ \hspace{1em}<K_{\beta }^p\left(\sigma \right)\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\mu \right)-\mathrm{1}}{f}^p\left(x\right)dx.\end{array}$$ (39)

(ii) For β = 0  (λ < 1) in (19) and (20), we obtain the following equivalent inequalities:

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{{\displaystyle \iint }}}\frac{{\left(\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}\right)}^{\alpha }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }{\left(\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}\right)}^{\alpha }}f\left(x\right)g\left(y\right)dx\hspace{0.17em}dy\\ \hspace{1em}<K_{\mathrm{0}}\left(\sigma \right){\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx\right\}}^{\mathrm{1}/p}\\ \hspace{1em}\hspace{1em}\times {\left\{\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{q\left(\mathrm{1}-\sigma \right)-\mathrm{1}}{g}^q\left(y\right)dy\right\}}^{\mathrm{1}/q},\end{array}$$ (40)

$$\begin{array}{c}\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|y\right|}^{p\sigma -\mathrm{1}}\\ \hspace{1em}{\hspace{1em}\hspace{1em}\times \left[\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{{\left(\min \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}\right)}^{\alpha }}{{\left|\mathrm{1}+x^{\delta }y\right|}^{\lambda }{\left(\max \left\{\mathrm{1},\left|x^{\delta }y\right|\right\}\right)}^{\alpha }}f\left(x\right)dx\right]}^{p}dy\\ \hspace{1em}<K_{\mathrm{0}}^p\left(\sigma \right)\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}{\left|x\right|}^{p\left(\mathrm{1}-\delta \sigma \right)-\mathrm{1}}{f}^p\left(x\right)dx,\end{array}$$ (41)

where K ₀(σ) is indicated by (15).

(iii) For α = 0,  σ = μ = λ/2  (0 < λ < 1) in (40), we find

$$\begin{array}{c}{K}_{\mathrm{0}}\left(\frac{\lambda }{\mathrm{2}}\right)=\underset{-\infty }{\overset{\infty }{{\displaystyle \int }}}\frac{\mathrm{1}}{{\left|\mathrm{1}+u\right|}^{\lambda }}{\left|u\right|}^{\left(\lambda /\mathrm{2}\right)-\mathrm{1}}du\\ =\underset{\mathrm{0}}{\overset{\infty }{{\displaystyle \int }}}\frac{u^{\left(\lambda /\mathrm{2}\right)-\mathrm{1}}du}{{\left(\mathrm{1}+u\right)}^{\lambda }}+\underset{-\infty }{\overset{-\mathrm{1}}{{\displaystyle \int }}}\frac{{\left(-u\right)}^{\left(\lambda /\mathrm{2}\right)-\mathrm{1}}du}{{\left(-\mathrm{1}-u\right)}^{\lambda }}\\ \hspace{1em}+\underset{-\mathrm{1}}{\overset{\mathrm{0}}{{\displaystyle \int }}}\frac{{\left(-u\right)}^{\left(\lambda /\mathrm{2}\right)-\mathrm{1}}du}{{\left(\mathrm{1}+u\right)}^{\lambda }}\\ =B\left(\frac{\lambda }{\mathrm{2}},\frac{\lambda }{\mathrm{2}}\right)+\mathrm{2}\underset{\mathrm{0}}{\overset{\mathrm{1}}{{\displaystyle \int }}}\frac{u^{\left(\lambda /\mathrm{2}\right)-\mathrm{1}}}{{\left(\mathrm{1}-u\right)}^{\lambda }}du=k_{\lambda },\end{array}$$ (42)

and then (3) follows. Hence, (40) and (19) are extensions of (3).

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.