Multiple Hardy–Littlewood Integral Operator Norm Inequalities

Abstract

How to obtain the sharp constant of the Hardy–Littlewood inequality remains unsolved. In this paper, the new analytical technique is to convert the exact constant factor to the norm of the corresponding operator, the multiple Hardy–Littlewood integral operator norm inequalities are proved. As its generalizations, some new integral operator norm inequalities with the radial kernel on n-dimensional vector spaces are established. The discrete versions of the main results are also given.

Mathematics Subject Classification (2000) 47A30, 26D10

1 Introduction

Throughout this paper, we write

$$\displaystyle \begin{aligned}E_{n}(\alpha)=\{ x=(x_{1},x_{2},\cdots,x_{n}):x_{k}\geq0,1\leq k\leq n, \|x\|{}_{\alpha}=(\sum_{k=1}^{n}|x_{k}|{}^{\alpha})^{1/\alpha},\alpha>0\}, \end{aligned}$$

E _n(α) is an n-dimensional vector space, when 1 ≤ α < ∞, E _n(α) is a normed vector space. In particular, E _n(2) is an n-dimensional Euclidean space $\mathbb {R}_{+}^{n}$.

$$\displaystyle \begin{aligned} \|f\|{}_{p,\omega}=(\int_{E_{n}(\alpha)}|f(x)|{}^{p}\omega(x)dx)^{1/p}, \end{aligned}$$

$$\displaystyle \begin{aligned}L^{p}(\omega)=\{f:f\,\,is \,\,measurable,\,\,and \|f\|{}_{p,\omega}<\infty\}, \end{aligned}$$

where, ω is a non-negative measurable function on E _n(α). If ω(x) ≡ 1, we will denote L ^p(ω) by L ^p(E _n(α)), and ∥f∥_p,1 by ∥f∥_p. Γ(α) is the Gamma function:

$$\displaystyle \begin{aligned}\Gamma(\alpha)=\int_{0}^{\infty}x^{\alpha-1}e^{-x}dx \,\,(\alpha>0). \end{aligned}$$

B(α, β) is the Beta function:

$$\displaystyle \begin{aligned}B(\alpha,\beta)=\int_{0}^{1}x^{\alpha-1}(1-x)^{\beta-1}dx \, \,(\alpha,\beta>0). \end{aligned}$$

The celebrated Hardy–Littlewood inequality (see [1], Theorem 401 and [2,3,4]) asserts that if f and g are non-negative, and 1 < p < ∞, 1 < q < ∞, $\frac {1}{p}+\frac {1}{q}\geq 1$, $\lambda =2-\frac {1}{p}-\frac {1}{q}$, $\delta <1-\frac {1}{p},\beta <1-\frac {1}{q}$, δ + β ≥ 0, and δ + β > 0, if $\frac {1}{p}+\frac {1}{q}=1$, then

$$\displaystyle \begin{aligned} \int_{0}^{\infty}\int_{0}^{\infty}\frac{f(x)g(y)} {x^{\delta}y^{\beta}|x-y|{}^{\lambda-\delta-\beta}}dxdy \leq c(\int_{0}^{\infty}f^{p}dx)^{1/p}(\int_{0}^{\infty}g^{q}dx)^{1/q}. {} \end{aligned} $$

(1)

Here, c denotes a positive number depending only on the parameters of the theorem (here p, q, δ, β). But Hardy was unable to fix the constant c in (1). We note that (1) is equivalent to

$$\displaystyle \begin{aligned} \|T_{0}f\|{}_{p}\leq c\|f\|{}_{p}, {} \end{aligned} $$

(2)

where,

$$\displaystyle \begin{aligned} T_{0}(f,x)=\int_{0}^{\infty}\frac{1}{x^{\delta}y^{\beta} |x-y|{}^{\lambda-\delta-\beta}}f(y)dy . {} \end{aligned} $$

(3)

Hence, c = ∥T ₀∥ in (2) is the sharp constant for (1) and (2). Under the above conditions, Hardy–Littlewood [2] proved that there exists a positive constant c ₁ such that

$$\displaystyle \begin{aligned} \int_{0}^{\infty}\int_{0}^{\infty}\frac{f(x)g(y)}{x^{\delta} y^{\beta}|x-y|{}^{\lambda-\delta-\beta}} dxdy\leq c_{1}\int_{0}^{\infty}\int_{0}^{\infty}\frac{f(x)g(y)}{|x-y|{}^{\lambda}}dxdy. {} \end{aligned} $$

(4)

The following Hardy–Littlewood–P$\acute {o}$lya inequality was proved in [5] and [6]:

Theorem 1

Let f ∈ L ^p(0, ∞), g ∈ L ^q(0, ∞), $1<p,q<\infty ,\frac {1}{p}+\frac {1}{q}>1$, $0<\lambda <1,\lambda =2-\frac {1}{p}-\frac {1}{q}$, then

$$\displaystyle \begin{aligned} \int_{0}^{\infty}\int_{0}^{\infty}\frac{f(x)g(y)}{|x-y|{}^{\lambda}}dxdy\leq c_{2}\|f\|{}_{p}\|g\|{}_{q}, {} \end{aligned} $$

(5)

where,

$$\displaystyle \begin{aligned} c_{2}=c_{2}(p,q,\lambda)=\frac{1}{1-\lambda}\{(\frac{p}{p-1})^{p(1-\frac{1}{q})} +(\frac{q}{q-1})^{q(1-\frac{1}{p})} \}. {}\end{aligned} $$

(6)

Let

$$\displaystyle \begin{aligned} T_{1}(f,x)=\int_{0}^{\infty}\frac{1}{|x-y|{}^{\lambda}}f(y)dy . {} \end{aligned} $$

(7)

Then (5) is equivalent to

$$\displaystyle \begin{aligned} \|T_{1}f\|{}_{p_{1}}\leq c_{2}\|f\|{}_{p}, {}\end{aligned} $$

(8)

where, $1<p<\infty ,1-\frac {1}{p}<\lambda <1$, $\frac {1}{p_{1}}=\frac {1}{p}+\lambda -1$, c ₂ is given by (6). For a function $f\in L^{p}(\mathbb {R}^{n})$, 1 < p < ∞, define its potential of order λ as

$$\displaystyle \begin{aligned} T_{2}(f,x)=\int_{\mathbb{R}^{n}}\frac{1}{\|x-y\|{}_{2}^{\lambda}}f(y)dy, \,\, 0<\lambda<n .{}\end{aligned} $$

(9)

Theorem 2 ([6, pp. 412–413])

There exists a constant c ₃ depending only upon n, p, and λ, such that

$$\displaystyle \begin{aligned} \|T_{2}f\|{}_{p_{2}}\leq c_{3}\|f\|{}_{p}, {}\end{aligned} $$

(10)

where, $\frac {1}{p_{2}}=\frac {1}{p}+\frac {\lambda }{n}-1$.

Theorem 3 ([7,8,9,10])

Let $f\in L^{p}(\mathbb {R}^{n})$, $g\in L^{q}(\mathbb {R}^{n})$, 1 < p, q < ∞, 0 < λ < n, $\frac {1}{p}+\frac {1}{q}+\frac {\lambda }{n}=2$, then there exists a constant c ₄ = c ₄(p, λ, n) (depending only upon n, p, and λ), such that

$$\displaystyle \begin{aligned} \int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)g(y)}{\|x-y\|{}_{2}^{\lambda}}dxdy \leq c_{4}\|f\|{}_{p}\|g\|{}_{q} , {}\end{aligned} $$

(11)

where,

$$\displaystyle \begin{aligned}c_{4}\leq\frac{n}{pq(n-\lambda)}(\frac{S_{n}}{n})^{\lambda/n} \{(\frac{\lambda/n}{1-(1/p)})^{\lambda/n}+(\frac{\lambda/n}{1-(1/q)})^{\lambda/n}\},\end{aligned} $$

and S _n is the surface areas of the unit sphere in $\mathbb {R}^{n}$ . In particular, for $p=q=\frac {2n}{2n-\lambda }$ ,

$$\displaystyle \begin{aligned}c_{4}=\pi^{\lambda/2}\frac{\Gamma(\frac{n-\lambda}{2})}{\Gamma(n-\frac{\lambda}{2})} \{ \frac{\Gamma(\frac{n}{2})}{\Gamma(n)} \}^{\frac{\lambda}{n}-1}\end{aligned} $$

is the best possible constant.

But when p ≠ q, the best possible value of c ₄ is also unknown.

In 2017, the author Kuang [14] established the norm inequality of operator T ₂.

Theorem 4 ([11])

Let $f\in L^{p}(\mathbb {R}^{n})$, $g\in L^{q}(\mathbb {R}^{n})$, 1 < p, q < ∞, 0 < λ < n, δ + β ≥ 0, $1-\frac {1}{p}-\frac {\lambda }{n}<\frac {\delta }{n}<1-\frac {1}{p}$, $\frac {1}{p}+\frac {1}{q}+\frac {\lambda +\delta +\beta }{n}=2$, then there exists a constant c ₅ = c ₅(p, δ, β, λ, n), such that

$$\displaystyle \begin{aligned} \int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)g(y)} {\|x\|{}_{2}^{\delta}\|y\|{}_{2}^{\beta} \|x-y\|{}_{2}^{\lambda}}dxdy\leq c_{5}\|f\|{}_{p}\|g\|{}_{q} .{} \end{aligned} $$

(12)

Remark 1

Inequality (12) can be given an equivalent form

$$\displaystyle \begin{aligned} \int_{\mathbb{R}^{n}}\int_{\mathbb{R}^{n}}\frac{f(x)g(y)} {\|x\|{}_{2}^{\delta}\|y\|{}_{2}^{\beta} \|x-y\|{}_{2}^{\lambda-\delta-\beta}}dxdy\leq c_{5}\|f\|{}_{p}\|g\|{}_{q}, {} \end{aligned} $$

(13)

then the conditions $1-\frac {1}{p}-\frac {\lambda }{n}<\frac {\delta }{n}<1-\frac {1}{p}$, $\frac {1}{p}+\frac {1}{q}+\frac {\lambda +\delta +\beta }{n}=2$ are replaced by

$$\displaystyle \begin{aligned}\frac{\delta}{n}<1-\frac{1}{p}<\frac{\lambda}{n}-\frac{\beta}{n}, \frac{1}{p}+\frac{1}{q}+\frac{\lambda}{n}=2. \end{aligned}$$

The multiple Hardy–Littlewood integral operator T ₃ defined by

$$\displaystyle \begin{aligned} T_{3}(f,x)=\int_{\mathbb{R}^{n}}\frac{f(y)}{\|x\|{}_{2}^{\delta}\|y\|{}_{2}^{\beta} \|x-y\|{}_{2}^{\lambda-\delta-\beta}}dy. {} \end{aligned} $$

(14)

Then (13) is equivalent to

$$\displaystyle \begin{aligned} \|T_{3}f\|{}_{p}\leq c_{5}\|f\|{}_{p} . {} \end{aligned} $$

(15)

But, the problem of determining the best possible constants in (13) and (15) remains unsolved. In this paper, the new analytical technique is to convert the exact constant factor to the norm c ₅ = ∥T ₃∥ of the corresponding operator T ₃. Hence, we consider operator norm inequality (15). Without loss of generality, we may consider that the multiple Hardy–Littlewood integral operator T ₄ defined by

$$\displaystyle \begin{aligned} T_{4}(f,x)=\int_{\mathbb{R}_{+}^{n}}\frac{f(y)}{\|x\|{}_{2}^{\delta}\|y\|{}_{2}^{\beta} \|x-y\|{}_{2}^{\lambda-\delta-\beta}}dy {} \end{aligned} $$

(16)

and f be a nonnegative measurable function on $\mathbb {R}_{+}^{n}$, thus, by the triangle inequality, we have

$$\displaystyle \begin{aligned}|\|x\|{}_{2}-\|y\|{}_{2}|\leq\|x-y\|{}_{2}\leq\|x\|{}_{2}+\|y\|{}_{2}. \end{aligned}$$

Let

$$\displaystyle \begin{aligned} \begin{array}{rcl} K_{4}(x,y)&\displaystyle =&\displaystyle (\|x\|{}_{2}^{\delta}\|y\|{}_{2}^{\beta}\|x-y\|{}_{2}^{\lambda-\delta-\beta})^{-1},\\ K_{5}(x,y)&\displaystyle =&\displaystyle (\|x\|{}_{2}^{\delta}\|y\|{}_{2}^{\beta}(\|x\|{}_{2}+\|y\|{}_{2})^{\lambda-\delta-\beta})^{-1},\\ K_{6}(x,y)&\displaystyle =&\displaystyle (\|x\|{}_{2}^{\delta}\|y\|{}_{2}^{\beta}(|\|x\|{}_{2}-\|y\|{}_{2}|)^{\lambda-\delta-\beta})^{-1}, \end{array} \end{aligned} $$

$$\displaystyle \begin{aligned} T_{j}(f,x)=\int_{\mathbb{R}_{+}^{n}}K_{j}(x,y)f(y)dy, {} \end{aligned} $$

(17)

$$\displaystyle \begin{aligned}\|T_{j}\|=\sup_{f\neq0}\frac{\|T_{j}f\|{}_{p,\omega}}{\|f\|{}_{p}},\,\, j=4,5,6, \end{aligned}$$

where, ω is a nonnegative measurable weight function on $\mathbb {R}_{+}^{n}$. If δ > 0, β > 0, λ − δ − β > 0, then

$$\displaystyle \begin{aligned}T_{5}(f,x)\leq T_{4}(f,x)\leq T_{6}(f,x), \end{aligned}$$

and therefore,

$$\displaystyle \begin{aligned} \|T_{5}\|\leq \|T_{4}\|\leq\|T_{6}\|. {} \end{aligned} $$

(18)

Thus, we may use the norms ∥T ₅∥, ∥T ₆∥ of the operator T ₅, T ₆ with the radial kernels to find the norm inequality of the multiple Hardy–Littlewood integral operator T ₄. As their generalizations, we introduce the new integral operator T defined by

$$\displaystyle \begin{aligned} T(f,x)=\int_{E_{n}(\alpha)}K(\|x\|{}_{\alpha},\|y\|{}_{\alpha})f(y)dy,\,\, x\in E_{n}(\alpha) , {} \end{aligned} $$

(19)

where, the radial kernel K(∥x∥_α, ∥y∥_α) is a nonnegative measurable function defined on E _n(α) × E _n(α), which satisfies the following condition:

$$\displaystyle \begin{aligned} K(\|x\|{}_{\alpha} , \|y\|{}_{\alpha})=\|x\|{}_{\alpha}^{-\lambda} K(1,\|y\|{}_{\alpha}\|x\|{}_{\alpha}^{-1}),\,\,x,y\in E_{n}(\alpha), \lambda>0 . {} \end{aligned} $$

(20)

Equation (19) includes many famous operators as special cases. In particular, for n = 1, we have

$$\displaystyle \begin{aligned} T(f,x)=\int_{0}^{\infty}K(x,y)f(y)dy ,\, \,x>0 , {} \end{aligned} $$

(21)

and

$$\displaystyle \begin{aligned} K(x,y)=x^{-\lambda}K(1,yx^{-1}),\,\,x,y>0,\lambda>0. {} \end{aligned} $$

(22)

The kernel in (3)

$$\displaystyle \begin{aligned}K(x,y)=\frac{1}{x^{\delta}y^{\beta}|x-y|{}^{\lambda-\delta-\beta}} \end{aligned}$$

satisfies (22). In 2016, the author Kuang [12] proved that if f ∈ L ^p(ω ₀), g ∈ L ^q(ω ₀), $1<p<\infty , \frac {1}{p}+\frac {1}{q}=1$, ω ₀(x) = x ^1−λ, and

$$\displaystyle \begin{aligned}\max\{ \frac{1}{p},\delta+\beta+\frac{1}{q} \}<\lambda< 1+\delta+\beta<1+\frac{1}{p}, \end{aligned}$$

then

$$\displaystyle \begin{aligned}\int_{0}^{\infty}\int_{0}^{\infty}\frac{f(x)g(y)} {x^{\delta}y^{\beta}|x-y|{}^{\lambda-\delta-\beta}}dxdy \leq c_{0}\|f\|{}_{p,\omega_{0}}\|g\|{}_{q,\omega_{0}}, \end{aligned}$$

which is equivalent to

$$\displaystyle \begin{aligned}\| T_{0}f \|{}_{p}\leq c_{0}\|f \|{}_{p,\omega_{0}}, \end{aligned}$$

where, T ₀ is defined by (3) and

$$\displaystyle \begin{aligned} \begin{array}{rcl} c_{0}&\displaystyle =&\displaystyle B(\lambda-\frac{1}{p},1-\lambda+\delta+\beta ) +B(\frac{1}{p}-\delta-\beta , 1-\lambda+\delta+\beta)\\ &\displaystyle &\displaystyle +B(\frac{1}{q},1-\lambda+\delta+\beta) +B(\lambda-\delta-\beta-\frac{1}{q},1-\lambda+\delta+\beta). {} \end{array} \end{aligned} $$

(23)

We define ω ₁ = x ^λ−1, then the above norm inequality is also equivalent to

$$\displaystyle \begin{aligned} \|T_{0}f\|{}_{p,\omega_{1}}\leq c_{0}\|f\|{}_{p}. {} \end{aligned} $$

(24)

The celebrated Hardy–Littlewood inequality (1) and (2) are important in analysis mathematics and its applications. In this paper, we give some new improvements and extensions of (24). As some further generalizations of the above results, the norm inequalities of the multiple integral operators with the radial kernels on n-dimensional vector spaces E _n(α) are established. In particular, using new analytical techniques, we convert the exact constant factor we are looking for into the norm of the corresponding operator, under a somewhat different hypothesis, we get lower and upper bounds of the sharp constant of the multiple Hardy–Littlewood inequality. Finally, the discrete versions of the main results are also given in Sect. 6.

2 Main Results

Our main results read as follows.

Theorem 5

Let 1 < p, q < ∞, λ ≥ n > 1, $\frac {1}{p}+\frac {1}{q}+\frac {\lambda }{n}=2$, $0\leq \delta <1-\frac {1}{q},0\leq \beta <1-\frac {1}{p}$, and

$$\displaystyle \begin{aligned}\max\{\beta+1-\frac{1}{q},\delta+n(1-\frac{1}{p})\}<\lambda< \min\{\frac{\delta+\beta}{1-(1/n)},\frac{\delta+\beta}{1-\frac{1}{pn(1-(1/q))}}\}. \end{aligned}$$

If $f\in L^{p}(\mathbb {R}_{+}^{n}),f(x)\geq 0,x\in \mathbb {R}_{+}^{n}$, $\omega (x)=\|x\|{ }_{2}^{p(\lambda -n)}$, then the multiple Hardy–Littlewood integral operator T ₄ is defined by (16): $T_{4}: L^{p}(\mathbb {R}_{+}^{n})\rightarrow L^{p}(\omega )$ exists as a bounded operator and

$$\displaystyle \begin{aligned}c_{3}\leq\|T_{4}\|\leq c_{1}^{1-(1/p)}c_{2}^{1/p}, \end{aligned}$$

where,

$$\displaystyle \begin{aligned} \begin{array}{rcl} c_{1}&\displaystyle =&\displaystyle \frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)}\{ B(\frac{n}{\lambda}(\frac{1}{q}-1-\beta)+n, 1-\frac{n}{\lambda}(\lambda-\delta-\beta))\\ &\displaystyle +&\displaystyle B(\frac{n}{\lambda}(\lambda-\delta-\frac{1}{q}+1)-n, 1-\frac{n}{\lambda}(\lambda-\delta-\beta)) \}, \\ c_{2}&\displaystyle =&\displaystyle \frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)}\{B(\frac{pn}{\lambda}(1-\frac{1}{q}) (1-\beta-\frac{1}{p}),1-\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-\beta))\\ &\displaystyle +&\displaystyle B(\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-1+\frac{1}{p}), 1-\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-\beta)) \},\\ c_{3}&\displaystyle =&\displaystyle \frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)} B(n(1-\frac{1}{p})-\beta,\lambda-\delta-n(1-\frac{1}{p})). \end{array} \end{aligned} $$

For n = 1, we have

Theorem 6

Let $1<p,q<\infty ,\lambda =2-\frac {1}{p}-\frac {1}{q}$ , $0\leq \beta <1-\frac {1}{p}, 0\leq \delta <1-\frac {1}{q}$ , and

$$\displaystyle \begin{aligned}\max\{\beta+1-\frac{1}{q},\delta+1-\frac{1}{p}\} <\lambda<\frac{\delta+\beta}{1-\frac{1}{p(1-(1/q))}}. \end{aligned}$$

If f ∈ L ^p(0, ∞), f(x) ≥ 0, x ∈ (0, ∞), ω(x) = x ^p(λ−1), then the Hardy–Littlewood integral operator T ₀ is defined by (3):T ₀ : L ^p(0, ∞) → L ^p(ω) exists as a bounded operator and

$$\displaystyle \begin{aligned}c_{3}\leq\|T_{0}\|\leq c_{1}^{1-(1/p)}c_{2}^{1/p} \end{aligned}$$

where,

$$\displaystyle \begin{aligned} \begin{array}{rcl} c_{1}&\displaystyle =&\displaystyle B(\frac{1}{\lambda}(\frac{1}{q}-1-\beta)+1,\frac{\delta+\beta}{\lambda}) +B(\frac{1}{\lambda}(1-\delta-\frac{1}{q}),\frac{\delta+\beta}{\lambda})\\ c_{2}&\displaystyle =&\displaystyle B(\frac{p}{\lambda}(1-\frac{1}{q})(1-\beta-\frac{1}{p}), 1-(1-\frac{\delta+\beta}{\lambda}) p(1-\frac{1}{q}))\\ &\displaystyle +&\displaystyle B(p(1-\frac{1}{q})(1-\frac{1}{\lambda}(\delta+1-\frac{1}{p})), 1-(1-\frac{\delta+\beta}{\lambda})p(1-\frac{1}{q})), \\ c_{3}&\displaystyle =&\displaystyle B(1-\beta-\frac{1}{p},\lambda-\delta-1+\frac{1}{p}) \end{array} \end{aligned} $$

Corollary 1

Let $1<p<\infty ,\frac {1}{p}+\frac {1}{q}=1$, $0\leq \delta <\frac {1}{p},0\leq \beta <\frac {1}{q}$, δ + β > 0, λ = 1. Iff ∈ L ^p(0, ∞), f(x) ≥ 0, x ∈ (0, ∞), then the integral operator T ₀ is defined by (3): T ₀ : L ^p(0, ∞) → L ^p(0, ∞) exists as a bounded operator and

$$\displaystyle \begin{aligned}B(\frac{1}{p}-\delta,\frac{1}{q}-\beta)\leq\|T_{0}\|\leq B(\frac{1}{p}-\delta,\delta+\beta)+B(\frac{1}{q}-\beta,\delta+\beta). \end{aligned}$$

As some further generalizations of the above results, we have

Theorem 7

Let 1 < p < ∞, 1 < q < ∞, δ, β ≥ 0, λ ≥ n, $\frac {1}{p}+\frac {1}{q}+\frac {\lambda }{n}=2$, $\omega (x)=\|x\|{ }_{\alpha }^{p(\lambda -n)}$, the radial kernel K(∥x∥_α, ∥y∥_α) satisfies (20).

• (i) If

  $$\displaystyle \begin{aligned} c_{1}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}(K(1,u))^{n/\lambda}u^{\frac{n}{\lambda}(\frac{1}{q}-1)+n-1}du<\infty, {} \end{aligned} $$

  (25)
  $$\displaystyle \begin{aligned} c_{2}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}(K(1,u))^{\frac{pn}{\lambda}(1-\frac{1}{q})} u^{\frac{n(p-1)(q-1)}{\lambda q}-1}du<\infty, {} \end{aligned} $$

  (26)

  then the integral operator T is defined by (19):T : L ^p(E _n(α)) → L ^p(ω) exists as a bounded operator and

  $$\displaystyle \begin{aligned} \|Tf\|{}_{p,\omega}\leq c\|f\|{}_{p}. {} \end{aligned} $$

  (27)

  This implies that

  $$\displaystyle \begin{aligned} \|T\|=\sup_{f\neq0}\frac{\|Tf\|{}_{p,\omega}}{\|f\|{}_{p}}\leq c, {} \end{aligned} $$

  (28)

  where,

  $$\displaystyle \begin{aligned} c=c_{1}^{(1-(1/p))}c_{2}^{1/p} . {} \end{aligned} $$

  (29)

• (ii) If

  $$\displaystyle \begin{aligned} c_{3}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}K(1,u)u^{(1-\frac{1}{p})n-1}du<\infty , {} \end{aligned} $$

  (30)

  then

  $$\displaystyle \begin{aligned} \|T\|\geq c_{3}. {} \end{aligned} $$

  (31)

In particular, for n = 1, by Theorem 7, we get

Theorem 8

Let 1 < p < ∞, 1 < q < ∞, δ, β ≥ 0, $1\leq \lambda =2-\frac {1}{p}-\frac {1}{q}$, ω(x) = x ^p(λ−1), the radial kernel K(x, y) satisfies (22).

• (i) If

  $$\displaystyle \begin{aligned} c_{1}=\int_{0}^{\infty}(K(1,u))^{1/\lambda}u^{\frac{1}{\lambda}(\frac{1}{q}-1)}du<\infty, {} \end{aligned} $$

  (32)
  $$\displaystyle \begin{aligned} c_{2}=\int_{0}^{\infty}(K(1,u))^{\frac{p}{\lambda}(1-\frac{1}{q})} u^{\frac{(p-1)(q-1)}{\lambda q}-1}du<\infty, {} \end{aligned} $$

  (33)

  then the integral operator T is defined by (21):T : L ^p(0, ∞) → L ^p(ω) exists as a bounded operator and

  $$\displaystyle \begin{aligned} \|Tf\|{}_{p,\omega}\leq c \|f\|{}_{p} . {} \end{aligned} $$

  (34)

  This implies that

  $$\displaystyle \begin{aligned} \|T\|=\sup_{f\neq0}\frac{\|Tf\|{}_{p,\omega}}{\|f\|{}_{p}}\leq c , {} \end{aligned} $$

  (35)

  where,

  $$\displaystyle \begin{aligned} c=c_{1}^{(1-(1/p))}c_{2}^{1/p}. {} \end{aligned} $$

  (36)

• (ii) If

  $$\displaystyle \begin{aligned} c_{3}=\int_{0}^{\infty}K(1,u)u^{-\frac{1}{p}}du<\infty, {} \end{aligned} $$

  (37)

  then

  $$\displaystyle \begin{aligned} \|T\|\geq c_{3}. {} \end{aligned} $$

  (38)

For λ = n, we have $\frac {1}{p}+\frac {1}{q}=1,$ and by Theorem 7, we get

Theorem 9

Let 1 < p < ∞, 1 < q < ∞, $\frac {1}{p}+\frac {1}{q}=1,\delta ,\beta \geq 0$, the radial kernel K(∥x∥_α, ∥y∥_α) satisfies:

$$\displaystyle \begin{aligned}K(\|x\|{}_{\alpha},\|y\|{}_{\alpha})=\|x\|{}_{\alpha}^{-n} K(1,\|y\|{}_{\alpha}\|x\|{}_{\alpha}^{-1}),\,\, x,y\in E_{n}(\alpha). \end{aligned}$$

(i) If

$$\displaystyle \begin{aligned} c_{1}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}K(1,u)u^{-(1/p)+n-1}du<\infty , {} \end{aligned} $$

(39)

$$\displaystyle \begin{aligned} c_{2}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}K(1,u)u^{-(1/p)}du<\infty , {} \end{aligned} $$

(40)

then the integral operator T is defined by (19): T : L ^p(E _n(α)) → L ^p(E _n(α)) exists as a bounded operator and

$$\displaystyle \begin{aligned} \|Tf\|{}_{p}\leq c\|f\|{}_{p} .{} \end{aligned} $$

(41)

This implies that

$$\displaystyle \begin{aligned} \|T\|=\sup_{f\neq0}\frac{\|Tf\|{}_{p}}{\|f\|{}_{p}}\leq c , {} \end{aligned} $$

(42)

where,

$$\displaystyle \begin{aligned} c=c_{1}^{(1/q)}c_{2}^{(1/p)}. {} \end{aligned} $$

(43)

(ii) If

$$\displaystyle \begin{aligned} c_{3}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}K(1,u)u^{(n/q)-1}du<\infty , {} \end{aligned} $$

(44)

then,

$$\displaystyle \begin{aligned} \|T\|\geq c_{3}. {} \end{aligned} $$

(45)

In particular, for n = 1, by Theorem 9, we get

$$\displaystyle \begin{aligned} c=c_{1}=c_{2}=c_{3}=\int_{0}^{\infty}K(1,u)u^{-(1/p)}du , {} \end{aligned} $$

(46)

then by (42), (45), and (46), we get

$$\displaystyle \begin{aligned} \|T\|=c=\int_{0}^{\infty}K(1,u)u^{-(1/p)}du. {} \end{aligned} $$

(47)

Thus, we get the following

Corollary 2

Let $1<p<\infty , \frac {1}{p}+\frac {1}{q}=1$, the kernel K(x, y) satisfies (22). Then the integral operator T is defined by (21): T : L ^p(0, ∞) → L ^p(0, ∞) exists as a bounded operator and

$$\displaystyle \begin{aligned} \|Tf\|{}_{p}\leq c\|f\|{}_{p}, {} \end{aligned} $$

(48)

where $\|T\|=c=\int _{0}^{\infty }K(1,u)u^{-(1/p)}du$ is the sharp constant.

3 Proofs of Theorems

We require the following lemmas to prove our results:

Lemma 1 ([4, 13])

If a _k, b _k, p _k > 0, 1 ≤ k ≤ n,f is a measurable function on (0, ∞), then

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \int_{\mathbb{R}_{+}^{n}}f\big(\sum_{k=1}^{n}(\frac{x_{k}}{a_{k}})^{b_{k}} \big)x_{1}^{p_{1}-1}\cdots x_{n}^{p_{n}-1}dx_{1}\cdots dx_{n}\\ &\displaystyle =&\displaystyle \frac{\prod_{k=1}^{n}a_{k}^{p_{k}}}{\prod_{k=1}^{n}b_{k}} \times\frac{\prod_{k=1}^{n}\Gamma(\frac{p_{k}}{b_{k}})} {\Gamma(\sum_{k=1}^{n}\frac{p_{k}}{b_{k}})} \int_{0}^{\infty}f(t)t^{(\sum_{k=1}^{n}\frac{p_{k}}{b_{k}}-1)}dt . \end{array} \end{aligned} $$

We get the following Lemma 2 by taking a _k = 1, b _k = α > 0, p _k = 1, 1 ≤ k ≤ n, in Lemma 1.

Lemma 2

Let f be a measurable function on (0, ∞), then

$$\displaystyle \begin{aligned} \int_{E_{n}(\alpha)}f(\|x\|{}_{\alpha}^{\alpha})dx= \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n}\Gamma(n/\alpha)} \int_{0}^{\infty}f(t)t^{(n/\alpha)-1}dt . {} \end{aligned} $$

(49)

Proof of Theorem 7

• (i) Let

  $$\displaystyle \begin{aligned}p_{1}=\frac{p}{p-1},\,\,q_{1}=\frac{q}{q-1}, \end{aligned}$$

  thus, we have

  $$\displaystyle \begin{aligned}\frac{1}{p_{1}}+\frac{1}{q_{1}}+(1-\frac{\lambda}{n})=1,\,\, \frac{p}{q_{1}}+p(1-\frac{\lambda}{n})=1 . \end{aligned}$$

  By Hölder’s inequality, we get

  $$\displaystyle \begin{aligned} \begin{array}{rcl} T(f,x)&\displaystyle =&\displaystyle \int_{E_{n}(\alpha)}K(\|x\|{}_{\alpha},\|y\|{}_{\alpha})f(y)dy \\ &\displaystyle =&\displaystyle \int_{E_{n}(\alpha)}\{\|y\|{}_{\alpha}^{(\frac{n}{p_{1}\lambda})}K^{n/\lambda} (\|x\|{}_{\alpha},\|y\|{}_{\alpha})f^{p}(y)\}^{1/q_{1}} \\ &\displaystyle \times&\displaystyle \{K^{n/\lambda}(\|x\|{}_{\alpha}, \|y\|{}_{\alpha})\|y\|{}_{\alpha}^{-(\frac{n}{q_{1}\lambda})}\}^{1/p_{1}} \{ f(y) \}^{p(1-\frac{\lambda}{n})}dy \\ &\displaystyle \leq&\displaystyle \{\int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{\frac{n}{p_{1}\lambda}} K^{n/\lambda}(\|x\|{}_{\alpha} , \|y\|{}_{\alpha})|f(y)|{}^{p}dy \}^{1/q_{1}} \\ &\displaystyle \times&\displaystyle \{ \int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{-(\frac{n}{q_{1}\lambda})} K^{n/\lambda}(\|x\|{}_{\alpha} , \|y\|{}_{\alpha})dy \}^{1/p_{1}} \|f\|{}_{p}^{p(1-\frac{\lambda}{n})} \\ &\displaystyle =&\displaystyle I_{1}^{1/q_{1}}\times I_{2}^{1/p_{1}}\times\|f\|{}_{p}^{p(1-\frac{\lambda}{n})},{} \end{array} \end{aligned} $$

  (50)

  where,

  $$\displaystyle \begin{aligned} \begin{array}{rcl} I_{1}&\displaystyle =&\displaystyle \int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{(\frac{n}{p_{1}\lambda})} K^{n/\lambda}(\|x\|{}_{\alpha},\|y\|{}_{\alpha})|f(y)|{}^{p}dy, \\ I_{2}&\displaystyle =&\displaystyle \int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{-(\frac{n}{q_{1}\lambda})} K^{n/\lambda}(\|x\|{}_{\alpha} , \|y\|{}_{\alpha})dy. \end{array} \end{aligned} $$

  In I ₂, by using Lemma 2, and letting $u=\|x\|{ }_{\alpha }^{-1}t^{1/\alpha }$, and use (20), (49), and (25), we get

  $$\displaystyle \begin{aligned} \begin{array}{rcl} I_{2}&\displaystyle =&\displaystyle \|x\|{}_{\alpha}^{-n}\int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{-(\frac{n}{q_{1}\lambda})} K^{n/\lambda}(1,\|y\|{}_{\alpha}\cdot \|x\|{}_{\alpha}^{-1})dy \\ &\displaystyle =&\displaystyle \|x\|{}_{\alpha}^{-n}\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n}\Gamma(n/\alpha)} \int_{0}^{\infty}t^{-(\frac{n}{q_{1}\lambda\alpha})} K^{n/\lambda}(1,t^{1/\alpha}\|x\|{}_{\alpha}^{-1})\times t^{\frac{n}{\alpha}-1}dt \\ &\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \|x\|{}_{\alpha}^{-(\frac{n}{q_{1}\lambda})} \int_{0}^{\infty}K^{\frac{n}{\lambda}}(1,u)u^{-(\frac{n}{q_{1}\lambda})+n-1}du \\ &\displaystyle =&\displaystyle c_{1}\|x\|{}_{\alpha}^{-(\frac{n}{q_{1}\lambda})}. {} \end{array} \end{aligned} $$

  (51)

  Hence, by (50) and (51), we conclude that

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \|Tf\|{}_{p,\omega}&\displaystyle =&\displaystyle (\int_{E_{n}(\alpha)}|T(f,x)|{}^{p}\omega(x)dx)^{1/p} \leq(\int_{E_{n}(\alpha)}I_{1}^{\frac{p}{q_{1}}}I_{2}^{\frac{p}{p_{1}}} \|f\|{}_{p}^{p^{2}(1-\frac{\lambda}{n})}\omega(x)dx)^{1/p} \\ &\displaystyle =&\displaystyle c_{1}^{\frac{1}{p_{1}}}\|f\|{}_{p}^{p(1-\frac{\lambda}{n})} \{\int_{E_{n}(\alpha)}(\int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{\frac{n}{p_{1}\lambda}} K^{\frac{n}{\lambda}}(\|x\|{}_{\alpha}, \|y\|{}_{\alpha})|f(y)|{}^{p}dy)^{\frac{p}{q_{1}}} \\ &\displaystyle \times&\displaystyle \|x\|{}_{\alpha}^{-\frac{pn}{p_{1}q_{1}\lambda}}\omega(x)dx \}^{1/p}. {} \end{array} \end{aligned} $$

  (52)

  Using the Minkowski’s inequality for integrals (see [3]):

  $$\displaystyle \begin{aligned}\{\int_{X}(\int_{Y}|f(x,y)|dy)^{p}\omega(x)dx \}^{1/p} \leq\int_{Y}\{ \int_{X}|f(x,y)|{}^{p}\omega(x)dx \}^{1/p}dy, \,\, 1\leq p<\infty, \end{aligned}$$

  and letting v = ∥y∥_αt ^−(1∕α), we obtain

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \|Tf\|{}_{p,\omega}&\displaystyle \leq&\displaystyle c_{1}^{\frac{1}{p_{1}}}\|f\|{}_{p}^{p(1-\frac{\lambda}{n})} \{ \int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{\frac{n}{p_{1}\lambda}}|f(y)|{}^{p} \\ &\displaystyle \times&\displaystyle (\int_{E_{n}(\alpha)}K^{\frac{pn}{q_{1}\lambda}}(\|x\|{}_{\alpha} ,\|y\|{}_{\alpha})\|x\|{}_{\alpha}^{-\frac{pn}{p_{1}q_{1}\lambda}} \omega(x)dx)^{\frac{q_{1}}{p}}dy \}^{1/q_{1}} \\ &\displaystyle =&\displaystyle c_{1}^{\frac{1}{p_{1}}}\|f\|{}_{p}^{p(1-\frac{\lambda}{n})}\{\int_{E_{n}(\alpha)} \|y\|{}_{\alpha}^{\frac{n}{p_{1}\lambda}}|f(y)|{}^{p} \\ &\displaystyle \times&\displaystyle ( \int_{E_{n}(\alpha)}K^{\frac{pn}{q_{1}\lambda}}(1,\|y\|{}_{\alpha}\cdot\|x\|{}_{\alpha}^{-1}) \|x\|{}_{\alpha}^{-\frac{pn}{q_{1}}-\frac{pn}{p_{1}q_{1}\lambda}+p(\lambda-n)}dx )^{\frac{q_{1}}{p}}dy \}^{\frac{1}{q_{1}}} \\ &\displaystyle =&\displaystyle c_{1}^{\frac{1}{p_{1}}}\|f\|{}_{p}^{p(1-\frac{\lambda}{n})} \{ \int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{\frac{n}{p_{1}\lambda}} |f(y)|{}^{p} \\ &\displaystyle \times&\displaystyle \big( \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n}\Gamma(n/\alpha)} \int_{0}^{\infty}K^{\frac{pn}{q_{1}\lambda}}(1,\|y\|{}_{\alpha}\cdot t^{-\frac{1}{\alpha}}) t^{-\frac{pn}{q_{1}\alpha}-\frac{pn}{\lambda\alpha p_{1}q_{1}}+\frac{p(\lambda-n)}{\alpha}} t^{\frac{n}{\alpha}-1}dt \big)^{\frac{q_{1}}{p}}dy \}^{\frac{1}{q_{1}}} \\ &\displaystyle =&\displaystyle c_{1}^{\frac{1}{p_{1}}}\|f\|{}_{p}^{p(1-\frac{\lambda}{n})} \{ \int_{E_{n}(\alpha)}\|y\|{}_{\alpha}^{\frac{n}{p_{1}\lambda}}|f(y)|{}^{p} \\ &\displaystyle \times&\displaystyle \big( \|y\|{}_{\alpha}^{-\frac{pn}{q_{1}p_{1}\lambda}} \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}K^{\frac{pn}{q_{1}\lambda}}(1,v)v^{\frac{pn}{p_{1}q_{1}\lambda}-1}dv \big)^{\frac{q_{1}}{p}}dy \}^{\frac{1}{q_{1}}} \\ &\displaystyle =&\displaystyle c_{1}^{1/p_{1}}c_{2}^{1/p}\|f\|{}_{p}^{p(1-\frac{\lambda}{n})}\|f\|{}_{p}^{\frac{p}{q_{1}}} =c_{1}^{(1-(1/p))}c_{2}^{1/p}\|f\|{}_{p}. \end{array} \end{aligned} $$

  Thus,

  $$\displaystyle \begin{aligned} \|Tf\|{}_{p,\omega} \leq c\|f\|{}_{p}. {} \end{aligned} $$

  (53)

• (ii) For proving (31), we take

  $$\displaystyle \begin{aligned} \begin{array}{rcl} f_{\varepsilon}(x)&\displaystyle =&\displaystyle \|x\|{}_{\alpha}^{-(n/p)-\varepsilon}\varphi_{B^{c}}(x), \\ g_{\varepsilon}(x)&\displaystyle =&\displaystyle (p\varepsilon)^{1/p_{1}}\{ \frac{\alpha^{n-1}\Gamma(n/\alpha)}{(\Gamma(1/\alpha))^{n}}\}^{1/p_{1}} \|x\|{}_{\alpha}^{-\frac{n}{p_{1}}-(p-1)\varepsilon}\varphi_{B^{c}}(x), \end{array} \end{aligned} $$

  where, ε > 0, B = B(0, 1) = {x ∈ E _n(α) : ∥x∥_α < 1}, $\varphi _{B^{c}}$ is the characteristic function of the set B ^c = {x ∈ E _n(α) : ∥x∥_α ≥ 1}, that is

  $$\displaystyle \begin{aligned} \varphi_{B^{c}}(x)=\left\{\begin{array}{c} 1,x\in B^{c}\\ 0,x\in B . \end{array} \right. \end{aligned}$$

  Thus, we get

  $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \|f_{\varepsilon}\|{}_{p}=\big(\frac{(\Gamma(1/\alpha))^{n}} {p\varepsilon\alpha^{n-1}\Gamma(n/\alpha)}\big)^{1/p}, \\ &\displaystyle &\displaystyle \|g_{\varepsilon}\|{}_{p_{1}}^{p_{1}}=(p\varepsilon) (\frac{\Gamma^{n}(1/\alpha)}{\alpha^{n-1}\Gamma(n/\alpha)})^{-1} \int_{B^{c}}\|x\|{}_{\alpha}^{-n-(p-1)p_{1}\varepsilon}dx \\ &\displaystyle =&\displaystyle (p\varepsilon)\frac{1}{\alpha}\int_{1}^{\infty}t^{-\frac{p\varepsilon}{\alpha}-1}dt=1. \end{array} \end{aligned} $$

  Using the sharpness in Hölder’s inequality (see [13]):

  $$\displaystyle \begin{aligned}\|Tf\|{}_{p,\omega}=\sup\{ |\int_{E_{n}(\alpha)}T(f,x)g(x)(\omega(x))^{1/p}dx|: \|g\|{}_{p_{1}}\leq1 \}, \end{aligned}$$

  where, $1<p<\infty ,\frac {1}{p}+\frac {1}{p_{1}}=1$, thus, if $\|g\|{ }_{p_{1}}\leq 1$, then

  $$\displaystyle \begin{aligned} |\int_{E_{n}(\alpha)}T(f,x)g(x)\{\omega(x)\}^{1/p}dx| \leq\|Tf\|{}_{p,\omega} . {} \end{aligned} $$

  (54)

  By (54) and (19), we get

  $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \|Tf_{\varepsilon}\|{}_{p,\omega}\geq\int_{E_{n}(\alpha)}T(f_{\varepsilon},x) g_{\varepsilon}(x)\{\omega(x)\}^{1/p}dx \\ &\displaystyle =&\displaystyle \int_{E_{n}(\alpha)}\int_{E_{n}(\alpha)}K(\|x\|{}_{\alpha},\|y\|{}_{\alpha}) f_{\varepsilon}(y)g_{\varepsilon}(x)\|x\|{}_{\alpha}^{\lambda-n}dydx \\ &\displaystyle =&\displaystyle (p\varepsilon)^{1/p_{1}}\{\frac{\alpha^{n-1}\Gamma(n/\alpha)} {(\Gamma(1/\alpha))^{n}}\}^{1/p_{1}} \\ &\displaystyle \times&\displaystyle \int_{B^{c}}\{\int_{B^{c}}K(\|x\|{}_{\alpha},\|y\|{}_{\alpha}) \|y\|{}_{\alpha}^{-(n/p)-\varepsilon}dy \} \|x\|{}_{\alpha}^{-\frac{n}{p_{1}}-(p-1)\varepsilon+\lambda-n}dx. {} \end{array} \end{aligned} $$

  (55)

  Letting $u=t^{1/\alpha }\|x\|{ }_{\alpha }^{-1}$, and using (20), we have

  $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \int_{B^{c}}K(\|x\|{}_{\alpha},\|y\|{}_{\alpha})\|y\|{}_{\alpha}^{-(n/p)-\varepsilon}dy \\ &\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n}\Gamma(n/\alpha)}\|x\|{}_{\alpha}^{-\lambda} \int_{1}^{\infty}K(1,t^{1/\alpha}\|x\|{}_{\alpha}^{-1}) t^{-(\frac{n}{p\alpha})-\frac{\varepsilon}{\alpha}+\frac{n}{\alpha}-1}dt \\ &\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \|x\|{}_{\alpha}^{-\lambda+\frac{n}{p_{1}}-\varepsilon} \int_{\|x\|{}_{\alpha}^{-1}}^{\infty}K(1,u)u^{\frac{n}{p_{1}}-\varepsilon-1}du. {} \end{array} \end{aligned} $$

  (56)

  We insert (56) into (55) and use Fubini’s theorem to obtain

  $$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \|Tf_{\varepsilon}\|{}_{p,\omega}\geq(p\varepsilon)^{1/p_{1}} \{\frac{(\Gamma(1/\alpha))^{n}} {\alpha^{n-1}\Gamma(n/\alpha)}\}^{1/p} \\ &\displaystyle \times&\displaystyle \int_{B^{c}}\|x\|{}_{\alpha}^{-p\varepsilon-n} (\int_{\|x\|{}_{\alpha}^{-1}}^{\infty}K(1,u)u^{\frac{n}{p_{1}}-\varepsilon-1}du)dx \\ &\displaystyle =&\displaystyle (p\varepsilon)^{1/p_{1}}\{\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \}^{1/p} \\ &\displaystyle \times&\displaystyle \int_{0}^{\infty}K(1,u)u^{\frac{n}{p_{1}}-\varepsilon-1} (\int_{\beta(u)}^{\infty}\|x\|{}_{\alpha}^{-p\varepsilon-n}dx)du \\ &\displaystyle =&\displaystyle (p\varepsilon)^{1/p_{1}}\{\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \}^{(1/p)+1}\times\frac{1}{\alpha} \\ &\displaystyle \times&\displaystyle \int_{0}^{\infty}K(1,u)u^{\frac{n}{p_{1}}-\varepsilon-1} (\int_{\beta(u)}^{\infty}t^{-(p\varepsilon)/\alpha-1}dt)du \\ &\displaystyle =&\displaystyle (p\varepsilon)^{-(1/p)}\{\frac{(\Gamma(1/\alpha))^{n}} {\alpha^{n-1}\Gamma(n/\alpha)}\}^{(1/p)+1} \\ &\displaystyle \times&\displaystyle \int_{0}^{\infty}K(1,u)u^{\frac{n}{p_{1}}-\varepsilon-1} (\beta(u))^{-(p\varepsilon)/\alpha}du, \end{array} \end{aligned} $$

  where, $\beta (u)=\max \{1,u^{-1}\}$. Thus, we get

  $$\displaystyle \begin{aligned} \begin{array}{rcl} \|T\|&\displaystyle =&\displaystyle \sup_{f\neq0}\frac{\|Tf\|{}_{p,\omega}}{\|f\|{}_{p}} \geq\frac{\|Tf_{\varepsilon}\|{}_{p,\omega}}{\|f_{\varepsilon}\|{}_{p}} \\ &\displaystyle \geq&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}K(1,u)u^{\frac{n}{p_{1}}-\varepsilon-1} (\beta(u))^{-(p\varepsilon)/\alpha}du. {} \end{array} \end{aligned} $$

  (57)

  By letting ε → 0⁺ in (57) and using Fatou lemma, we get

  $$\displaystyle \begin{aligned}\|T\|\geq\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}K(1,u)u^{\frac{n}{p_{1}}-1}du=c_{3}. \end{aligned}$$

The proof is complete.

4 Some Applications

As applications, a large number of known and new results have been obtained by proper choice of kernel K. In this section we present some model applications which display the importance of our results.

Example 1

Let $h:E_{n}(\alpha )\times E_{n}(\alpha )\rightarrow \mathbb {R}_{+}$ be a measurable function. K ₇ is defined by

$$\displaystyle \begin{aligned} K_{7}(\|x\|{}_{\alpha} ,\|y\|{}_{\alpha})= \frac{h(\|y\|{}_{\alpha}\cdot \|x\|{}_{\alpha}^{-1})}{\|x\|{}_{\alpha}^{\delta} \|y\|{}_{\alpha}^{\beta} |\|x\|{}_{\alpha}-\|y\|{}_{\alpha} |{}^{\lambda-\delta-\beta}}, {}\end{aligned} $$

(58)

and let

$$\displaystyle \begin{aligned}T_{7}(f,x)=\int_{E_{n}(\alpha)}\frac{h(\|y\|{}_{\alpha}\cdot\|x\|{}_{\alpha}^{-1})} {\|x\|{}_{\alpha}^{\delta} \|y\|{}_{\alpha}^{\beta} |\|x\|{}_{\alpha}-\|y\|{}_{\alpha}|{}^{\lambda-\delta-\beta}}f(y)dy.\end{aligned} $$

If p, q, λ, and ω satisfy the conditions of Theorem 7, and

$$\displaystyle \begin{aligned} c_{1}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}\{ \frac{h(u)}{u^{\beta}|1-u|{}^{\lambda-\delta-\beta}}\}^{\frac{n}{\lambda}} u^{\frac{n}{\lambda}(\frac{1}{q}-1)+n-1}du<\infty, {} \end{aligned} $$

(59)

$$\displaystyle \begin{aligned} c_{2}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}\{ \frac{h(u)}{u^{\beta}|1-u|{}^{\lambda-\delta-\beta}} \}^{\frac{pn}{\lambda}(1-\frac{1}{q})} u^{\frac{n}{\lambda}(p-1)(1-\frac{1}{q})-1}du<\infty, {} \end{aligned} $$

(60)

$$\displaystyle \begin{aligned} c_{3}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}\frac{h(u)}{u^{\beta}|1-u|{}^{\lambda-\delta-\beta}} u^{n(1-\frac{1}{p})-1}du<\infty, {} \end{aligned} $$

(61)

then by Theorem 7, we get

$$\displaystyle \begin{aligned} c_{3}\leq\|T_{7}\|\leq c_{1}^{(1-(1/p))}c_{2}^{1/p}.{} \end{aligned} $$

(62)

Setting h(u) = 1, we distinguish four cases:

• (i) The case n > 1. Let $0\leq \delta <1-\frac {1}{q}$, $0\leq \beta <1-\frac {1}{p}$, and

  

  then by (59), (60), and (61), we get

  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{1}&\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \{ B(\frac{n}{\lambda}(\frac{1}{q}-1-\beta)+n, 1-\frac{n}{\lambda}(\lambda-\delta-\beta)) \\ &\displaystyle +&\displaystyle B(\frac{n}{\lambda}(\lambda-\delta-\frac{1}{q}+1)-n, 1-\frac{n}{\lambda}(\lambda-\delta-\beta)) \}, {} \end{array} \end{aligned} $$

  (63)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{2}&\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \{B(\frac{pn}{\lambda}(1-\frac{1}{q})(1-\beta-\frac{1}{p}), 1-\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-\beta)) \\ &\displaystyle +&\displaystyle B(\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-1+\frac{1}{p}), 1-\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-\beta))\}, {} \end{array} \end{aligned} $$

  (64)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{3}&\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \{B(n(1-\frac{1}{p})-\beta,1-\lambda+\delta+\beta ) \\ &\displaystyle +&\displaystyle B( \lambda-\delta-n(1-\frac{1}{p}),1-\lambda+\delta+\beta ) \}.{} \end{array} \end{aligned} $$

  (65)

• (ii) The case n = 1. Let $0\leq \beta <1-\frac {1}{p}$,$0\leq \delta <1-\frac {1}{q}$,δ + β > 0, and

  $$\displaystyle \begin{aligned}\max\{\delta+1-\frac{1}{p},\beta+1-\frac{1}{q}\}<\lambda< \min\{1+\delta+\beta,\frac{\delta+\beta}{1-\frac{1}{p(1-(1/q))}}\},\end{aligned} $$

  then by (59), (60), and (61), we get

  $$\displaystyle \begin{aligned} c_{1}=B(\frac{1}{\lambda}(\frac{1}{q}-1-\beta)+1,\frac{\delta+\beta}{\lambda} )+B(\frac{1}{\lambda}(1-\delta-\frac{1}{q}),\frac{\delta+\beta}{\lambda}), {} \end{aligned} $$

  (66)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{2}&\displaystyle =&\displaystyle B(\frac{p}{\lambda}(1-\frac{1}{q})(1-\beta-\frac{1}{p}), 1-(1-\frac{\delta+\beta}{\lambda})p(1-\frac{1}{q})) \\ &\displaystyle +&\displaystyle B(p(1-\frac{1}{q})(1-\frac{1}{\lambda}(\delta+1-\frac{1}{p})), 1-(1-\frac{\delta+\beta}{\lambda})p(1-\frac{1}{q})), {} \end{array} \end{aligned} $$

  (67)
  $$\displaystyle \begin{aligned} c_{3}=B(1-\beta-\frac{1}{p},1-\lambda+\delta+\beta)+ B(\lambda-\delta-1+\frac{1}{p},1-\lambda+\delta+\beta). {} \end{aligned} $$

  (68)

• (iii) The case λ = n, this implies that $\frac {1}{p}+\frac {1}{q}=1$. Let $0\leq \delta <\min \{\frac {1}{p},n-\frac {1}{q}\}$, $0\leq \beta <\min \{\frac {1}{q},n-\frac {1}{p} \}$, n − 1 < δ + β, then by (59), (60), and (61), we get

  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{1}&\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \{ B(n-\frac{1}{p}-\beta,1-n+\delta+\beta) \\ &\displaystyle +&\displaystyle B(\frac{1}{p}-\delta,1-n+\delta+\beta)\}, {} \end{array} \end{aligned} $$

  (69)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{2}&\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)}\{ B(\frac{1}{q}-\beta,1-n+\delta+\beta) \\ &\displaystyle +&\displaystyle B(n-\delta-\frac{1}{q},1-n+\delta+\beta) \}, {} \end{array} \end{aligned} $$

  (70)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{3}&\displaystyle =&\displaystyle \frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \{B(\frac{n}{q}-\beta,1-n+\delta+\beta) \\ &\displaystyle +&\displaystyle B(\frac{n}{p}-\delta,1-n+\delta+\beta) \} . {} \end{array} \end{aligned} $$

  (71)

• (iv) The case λ = n = 1. Let $0\leq \delta <\frac {1}{p}$, $0\leq \beta <\frac {1}{q}$, δ + β > 0, then by (69), (70), and (71), we get

  $$\displaystyle \begin{aligned} \|T_{7}\|=B(\frac{1}{p}-\delta,\delta+\beta)+B(\frac{1}{q}-\beta,\delta+\beta). {} \end{aligned} $$

  (72)

Example 2

Let $h:E_{n}(\alpha )\times E_{n}(\alpha )\rightarrow \mathbb {R}_{+}$ be a measurable function. K ₈ is defined by

$$\displaystyle \begin{aligned} K_{8}(\|x\|{}_{\alpha},\|y\|{}_{\alpha})= \frac{h(\|y\|{}_{\alpha}\cdot\|x\|{}_{\alpha}^{-1})} {\|x\|{}_{\alpha}^{\delta} \|y\|{}_{\alpha}^{\beta} (\|x\|{}_{\alpha} +\|y\|{}_{\alpha})^{\lambda-\delta-\beta}}, {} \end{aligned} $$

(73)

and let

$$\displaystyle \begin{aligned}T_{8}(f,x)=\int_{E_{n}(\alpha)}\frac{h(\|y\|{}_{\alpha} \cdot\|x\|{}_{\alpha}^{-1})}{\|x\|{}_{\alpha}^{\delta} \|y\|{}_{\alpha}^{\beta} (\|x\|{}_{\alpha}+ \|y\|{}_{\alpha})^{\lambda-\delta-\beta}}f(y)dy. \end{aligned}$$

If p, q, λ, and ω satisfy the conditions of Theorem 7, and

$$\displaystyle \begin{aligned} c_{1}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}\{\frac{h(u)}{u^{\beta} (1+u)^{\lambda-\delta-\beta}}\}^{\frac{n}{\lambda}} u^{\frac{n}{\lambda}(\frac{1}{q}-1)+n-1}du<\infty, {} \end{aligned} $$

(74)

$$\displaystyle \begin{aligned} c_{2}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}\{\frac{h(u)}{u^{\beta} (1+u)^{\lambda-\delta-\beta}}\}^{\frac{pn}{\lambda}(1-\frac{1}{q})} u^{\frac{n}{\lambda}(p-1)(1-\frac{1}{q})-1}du<\infty, {} \end{aligned} $$

(75)

$$\displaystyle \begin{aligned} c_{3}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} \int_{0}^{\infty}\{\frac{h(u)}{u^{\beta}(1+u)^{\lambda-\delta-\beta}}\} u^{n(1-\frac{1}{p})-1}du<\infty , {} \end{aligned} $$

(76)

then by Theorem 7, we get

$$\displaystyle \begin{aligned}c_{3}\leq\|T_{8}\|\leq c_{1}^{(1-(1/p))}c_{2}^{1/p}. \end{aligned}$$

Setting h(u) = 1, we distinguish four cases:

• (i) The case n > 1. Let $0\leq \delta <1-\frac {1}{q},0\leq \beta <1-\frac {1}{p}$, and

  $$\displaystyle \begin{aligned}\lambda>\max\{\beta+1-\frac{1}{q},\delta+n(1-\frac{1}{p})\}, \end{aligned}$$

  then by (74), (75), and (76), we get

  $$\displaystyle \begin{aligned} c_{1}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} B(\frac{n}{\lambda}(\frac{1}{q}-1-\beta)+n,\frac{n}{\lambda} (\lambda-\delta+1-\frac{1}{q})-n), {} \end{aligned} $$

  (77)
  $$\displaystyle \begin{aligned} c_{2}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} B(\frac{pn}{\lambda}(1-\frac{1}{q})(1-\frac{1}{p}-\beta), \frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-1+\frac{1}{p}) ), {} \end{aligned} $$

  (78)
  $$\displaystyle \begin{aligned} c_{3}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} B(n(1-\frac{1}{p})-\beta, \lambda-\delta-n(1-\frac{1}{p})). {} \end{aligned} $$

  (79)

• (ii) The case n = 1. Let $0\leq \beta <1-\frac {1}{p}$, $0\leq \delta <1-\frac {1}{q}$, and $\lambda >\max \{ \delta +1-\frac {1}{p},\beta +1-\frac {1}{q} \}$, then by (74), (75), and (76), we get

  $$\displaystyle \begin{aligned} c_{1}=B(\frac{1}{\lambda}(\frac{1}{q}-1-\beta)+1,\frac{1}{\lambda}(1-\delta-\frac{1}{q})), {} \end{aligned} $$

  (80)
  $$\displaystyle \begin{aligned} c_{2}=B(\frac{p}{\lambda}(1-\frac{1}{q})(1-\frac{1}{p}-\beta), \frac{p}{\lambda}(1-\frac{1}{q})(\lambda-\delta-1+\frac{1}{p})), {} \end{aligned} $$

  (81)
  $$\displaystyle \begin{aligned} c_{3}=B(1-\beta-\frac{1}{p},\lambda-\delta-1+\frac{1}{p}).{} \end{aligned} $$

  (82)

• (iii) The case λ = n, this implies that $\frac {1}{p}+\frac {1}{q}=1$. Let $0\leq \beta <\frac {1}{q}$, $0\leq \delta <\min \{ \frac {n}{p},n-\frac {1}{q} \}$, then by (74), (75), and (76), we get

  $$\displaystyle \begin{aligned} c_{1}=c_{2}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} B(\frac{1}{q}-\beta ,n-\delta- \frac{1}{q}), {} \end{aligned} $$

  (83)
  $$\displaystyle \begin{aligned} c_{3}=\frac{(\Gamma(1/\alpha))^{n}}{\alpha^{n-1}\Gamma(n/\alpha)} B(\frac{n}{q}-\beta , \frac{n}{p}-\delta), {} \end{aligned} $$

  (84)

  and

  $$\displaystyle \begin{aligned} c_{3}\leq\|T_{8}\|\leq c_{1}. {} \end{aligned} $$

  (85)

• (iv) The case λ = n = 1. Let $0\leq \delta <\frac {1}{p}$, $0\leq \beta <\frac {1}{q}$, α + β > 0, and $\max \{ \beta +\frac {1}{p}, \delta +\frac {1}{q}\}<1$, then by (83), (84), and (85), we get

  $$\displaystyle \begin{aligned} \|T_{8}\|=B(\frac{1}{p}-\delta , \frac{1}{q}-\beta). {} \end{aligned} $$

  (86)

5 Multiple Hardy–Littlewood Integral Operator Norm Inequalities

In Examples 1 and 2, setting h(u) = 1, α = 2, thus, E _n(α) reduces to $E_{n}(2)=\mathbb {R}_{+}^{n}$, T ₇, T ₈ reduces to T ₆, T ₅, respectively. Assume $f\in L^{p}(\mathbb {R}_{+}^{n})$, $f(x)\geq 0,x\in \mathbb {R}_{+}^{n}$, 1 < p, q < ∞, λ ≥ n, $\delta ,\beta \geq 0 , \frac {1}{p}+\frac {1}{q}+\frac {\lambda }{n}=2$. The multiple Hardy–Littlewood integral operator T ₄ is defined by (16):$T_{4}:L^{p}(\mathbb {R}_{+}^{n})\rightarrow L^{p}(\omega )$, where $\omega (x)=\|x\|{ }_{2}^{p(\lambda -n)}$ and

$$\displaystyle \begin{aligned}\|T_{4}\|=\sup_{f\neq0} \frac{\|T_{4}\|{}_{p,\omega}}{\|f\|{}_{p}} . \end{aligned}$$

We distinguish four cases:

• (i) The case n > 1. Let $0\leq \delta <1-\frac {1}{q}$, $0\leq \beta <1-\frac {1}{p}$, and

  $$\displaystyle \begin{aligned}\max\{ \beta+1-\frac{1}{q} ,\delta+n(1- \frac{1}{p} )\}<\lambda< \min\{ \frac{\delta+\beta}{1-(1/n)},\frac{\delta+\beta}{1-\frac{1}{pn(1-(1/q))}}\}, \end{aligned}$$

  then by (18), (63), (64) and (79), we get

  $$\displaystyle \begin{aligned} c_{3}\leq\|T_{4}\|\leq c_{1}^{1-(1/p)}c_{2}^{1/p}, {} \end{aligned} $$

  (87)

  where

  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{1}&\displaystyle =&\displaystyle \frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)} \{B(\frac{n}{\lambda}(\frac{1}{q}-1-\beta)+n, 1-\frac{n}{\lambda}(\lambda-\delta-\beta)) \\ &\displaystyle +&\displaystyle B(\frac{n}{\lambda}(\lambda-\delta-\frac{1}{q}+1)-n,1-\frac{n}{\lambda} (\lambda-\delta-\beta)) \} , {} \end{array} \end{aligned} $$

  (88)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{2}&\displaystyle =&\displaystyle \frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)} \{B(\frac{pn}{\lambda}(1-\frac{1}{q})(1-\beta-\frac{1}{p}),1-\frac{pn}{\lambda} (1-\frac{1}{q})(\lambda-\delta-\beta) ) \\ &\displaystyle +&\displaystyle B(\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-1+\frac{1}{p}), 1-\frac{pn}{\lambda}(1-\frac{1}{q})(\lambda-\delta-\beta)) \} ,{} \end{array} \end{aligned} $$

  (89)
  $$\displaystyle \begin{aligned} c_{3}=\frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)}B(n(1-\frac{1}{p})-\beta, \lambda-\delta-n(1-\frac{1}{p})) . {} \end{aligned} $$

  (90)

• (ii) The case n = 1. Let $0\leq \beta <1-\frac {1}{p},0\leq \delta <1-\frac {1}{q}$, and

  $$\displaystyle \begin{aligned}\max\{\delta+1- \frac{1}{p} ,\beta+1-\frac{1}{q}\}<\lambda< \frac{\delta+\beta}{1-\frac{1}{p(1-(1/q))}}, \end{aligned}$$

  then by (18), (66), (67) and (82), we get

  $$\displaystyle \begin{aligned}c_{3}\leq\|T_{4}\|\leq c_{1}^{1-(1/p)}c_{2}^{1/p}, \end{aligned}$$

  where,

  $$\displaystyle \begin{aligned} c_{1}=B(\frac{1}{\lambda}(\frac{1}{q}-1-\beta)+1,\frac{\delta+\beta}{\lambda}) +B(\frac{1}{\lambda}(1-\delta-\frac{1}{q}),\frac{\delta+\beta}{\lambda}), {} \end{aligned} $$

  (91)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{2}&\displaystyle =&\displaystyle B(\frac{p}{\lambda}(1-\frac{1}{q})(1-\beta-\frac{1}{p}), 1-(1-\frac{\delta+\beta}{\lambda})p(1-\frac{1}{q})) \\ &\displaystyle +&\displaystyle B(p(1-\frac{1}{q})(1-\frac{1}{\lambda}(\delta+1-\frac{1}{p})), 1-(1-\frac{\delta+\beta}{\lambda})p(1-\frac{1}{q})),{} \end{array} \end{aligned} $$

  (92)
  $$\displaystyle \begin{aligned} c_{3}=B(1-\beta-\frac{1}{p}, \lambda-\delta-1+\frac{1}{p}). {} \end{aligned} $$

  (93)

• (iii) The case λ = n, this implies that $\frac {1}{p}+\frac {1}{q}=1$. Let $0\leq \delta <\frac {1}{p}$, $0\leq \beta <\frac {1}{q}$, and

  $$\displaystyle \begin{aligned}\max\{\beta+\frac{1}{p} ,\delta+\frac{1}{q}\}<n<1+\delta+\beta, \end{aligned}$$

  then by (18), (69), (70), and (84), we get

  $$\displaystyle \begin{aligned}c_{3}\leq\|T_{4}\|\leq c_{1}^{1-(1/p)}c_{2}^{1/p}, \end{aligned}$$

  where

  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{1}&\displaystyle =&\displaystyle \frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)}\{ B(n-\frac{1}{p}-\beta, 1-n+\delta+\beta) \\ &\displaystyle +&\displaystyle B(\frac{1}{p}-\delta,1-n+\delta+\beta) \}, {} \end{array} \end{aligned} $$

  (94)
  $$\displaystyle \begin{aligned} \begin{array}{rcl} c_{2}&\displaystyle =&\displaystyle \frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)}\{B(\frac{1}{q}-\beta, 1-n+\delta+\beta) \\ &\displaystyle +&\displaystyle B(n-\delta-\frac{1}{q},1-n+\delta+\beta) \}, {} \end{array} \end{aligned} $$

  (95)
  $$\displaystyle \begin{aligned} c_{3}=\frac{\pi^{n/2}}{2^{n-1}\Gamma(n/2)} B(\frac{n}{q}-\beta , \frac{n}{p}-\delta). {} \end{aligned} $$

  (96)

• (iv) The case λ = n = 1. Let $0\leq \delta <\frac {1}{p}$, $0\leq \beta <\frac {1}{q}$,δ + β > 0, then by (18), (72), and (86), we get

  $$\displaystyle \begin{aligned} B(\frac{1}{p}-\delta ,\frac{1}{q}-\beta)\leq \|T_{4}\| \leq B(\frac{1}{p}-\delta,\delta+\beta)+B(\frac{1}{q}-\beta,\delta+\beta). {} \end{aligned} $$

  (97)

We have thus also proved that Theorems 5 and 6 are correct.

6 The Discrete Versions of the Main Results

Let a = {a _m} be a sequence of real numbers, we define

$$\displaystyle \begin{aligned}\|a\|{}_{p,\omega}=\{\sum_{m=1}^{\infty}|a_{m}|{}^{p}\omega(m)\}^{1/p},\,\, l^{p}(\omega)=\{a=\{a_{m}\}:\|a\|{}_{p,\omega}<\infty \}. \end{aligned}$$

If ω(m) ≡ 1, we will denote l ^p(ω) by l ^p, and ∥a∥_p,1 by ∥a∥_p. Defining f, K by f(x) = a _m, K(x, y) = K(m, n)(m − 1 ≤ x < m, n − 1 ≤ y < n), respectively, we obtain the corresponding series form of (21):

$$\displaystyle \begin{aligned} T(a,m)=\sum_{n=1}^{\infty}K(m,n)a_{n} . {} \end{aligned} $$

(98)

Then by Theorem 8, we get

Theorem 10

Let 1 < p < ∞, 1 < q < ∞, δ, β ≥ 0, δ + β > 0, $1\leq \lambda =2-\frac {1}{p}-\frac {1}{q}$, ω(m) = m ^p(λ−1), the kernel K(m, n) satisfies

$$\displaystyle \begin{aligned} K(m,n)=m^{-\lambda}K(1,nm^{-1}). {} \end{aligned} $$

(99)

• (i) If

  $$\displaystyle \begin{aligned} c_{1}=\int_{0}^{\infty}(K(1,u))^{\frac{1}{\lambda}}u^{\frac{1-q}{\lambda q}}du<\infty , {} \end{aligned} $$

  (100)
  $$\displaystyle \begin{aligned} c_{2}=\int_{0}^{\infty}(K(1,u))^{\frac{p}{\lambda} (1-\frac{1}{q})}u^{\frac{(p-1)(q-1)}{\lambda q}-1}du<\infty, {} \end{aligned} $$

  (101)

  then the integral operator T is defined by (98): T : l ^p → l ^p(ω) exists as a bounded operator and

  $$\displaystyle \begin{aligned} \|Ta\|{}_{p,\omega}\leq c \|a\|{}_{p} . {} \end{aligned} $$

  (102)

  This implies that

  $$\displaystyle \begin{aligned} \|T\|=\sup_{a\neq0}\frac{\|Ta\|{}_{p,\omega}}{\|a\|{}_{p}}\leq c , {} \end{aligned} $$

  (103)

  where

  $$\displaystyle \begin{aligned} c=c_{1}^{(1-(1/p))}c_{2}^{1/p} . {} \end{aligned} $$

  (104)

• (ii) If

  $$\displaystyle \begin{aligned} c_{3}=\int_{0}^{\infty}K(1,u)u^{-\frac{1}{p}}du<\infty , {} \end{aligned} $$

  (105)

  then

  $$\displaystyle \begin{aligned} \|T\|\geq c_{3}. {} \end{aligned} $$

  (106)

For λ = 1, we have $\frac {1}{p}+\frac {1}{q}=1$ and by Theorem 10, we get

$$\displaystyle \begin{aligned} \|Ta\|{}_{p}\leq c \|a\|{}_{p} , {} \end{aligned} $$

(107)

where $c=\|T\|=\int _{0}^{\infty }K(1,u)u^{-(1/p)}du$ is the sharp constant. In particular, let

$$\displaystyle \begin{aligned}K(m,n)=\frac{1}{m^{\delta}n^{\beta}|m-n|{}^{\lambda-\delta-\beta}} , \end{aligned}$$

if $0\leq \beta <1-\frac {1}{p}, 0\leq \delta <1-\frac {1}{q},\delta +\beta >0$, and

$$\displaystyle \begin{aligned}\max\{\delta+1-\frac{1}{p},\beta+1-\frac{1}{q}\}<\lambda< \min\{1+\delta+\beta , \frac{\delta+\beta}{1-\frac{1}{p(1-(1/q))}}\}, \end{aligned}$$

then by Example 1, we get

$$\displaystyle \begin{aligned} c_{3}\leq\|T\|\leq c_{1}^{1-(1/p)}c_{2}^{1/p} , {} \end{aligned} $$

(108)

where c ₁, c ₂, and c ₃ are defined by (66), (67), and (68), respectively.

If λ = 1, that is, $0\leq \delta <\frac {1}{p}, 0\leq \beta <\frac {1}{q}, \delta +\beta >0$, then by (72), we have

$$\displaystyle \begin{aligned} \|T\|=B(\frac{1}{p}-\delta,\delta+\beta) +B(\frac{1}{q}-\beta,\delta+\beta). {} \end{aligned} $$

(109)

Remark 2

In 2016, the author Kuang [12] proved that if a = {a _m}∈ l ^p(ω ₀), b = {b _n}∈ l ^q(ω ₀), $1<p<\infty ,\frac {1}{p}+\frac {1}{q}=1$, ω ₀(m) = m ^1−λ, and

$$\displaystyle \begin{aligned}\max\{\frac{1}{p},\delta+\beta+\frac{1}{q}\}<\lambda< 1+\beta+\delta<1+\frac{1}{p} \end{aligned}$$

then

$$\displaystyle \begin{aligned} \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{a_{m}b_{n}} {m^{\delta}n^{\beta}|m-n|{}^{\lambda-\delta-\beta}}\leq c_{0}\|a\|{}_{p,\omega_{0}}\|b\|{}_{q,\omega_{0}} , {} \end{aligned} $$

(110)

where c ₀ is defined by (23). Inequality (110) is equivalent to

$$\displaystyle \begin{aligned} \|T_{9}(a)\|{}_{p}\leq c_{0}\|a\|{}_{p,\omega_{0}}, {} \end{aligned} $$

(111)

where,

$$\displaystyle \begin{aligned}T_{9}(a,m)=\sum_{n=1}^{\infty}\frac{a_{n}}{m^{\delta}n^{\beta} |m-n|{}^{\lambda-\delta-\beta}} \end{aligned}$$

is the Hardy–Littlewood operator. We define ω ₁(m) = m ^λ−1, then the above norm inequality is also equivalent to

$$\displaystyle \begin{aligned} \|T_{9}(a)\|{}_{p,\omega_{1}}\leq c_{0}\|a\|{}_{p}. {} \end{aligned} $$

(112)

Hence, (108) and (109) are new improvements and extensions of (112).