Condition for the Construction of a Hilbert-Type Integral Inequality Involving Upper Limit Functions

Abstract

Hilbert-type integral inequalities, which feature a symmetric structure, are a significant class of inequalities with broad applications in the study of operator theory. Hilbert-type integral inequalities involving variable upper limit integral functions are a generalized form of the classical Hilbert-type integral inequalities. In this paper, we employ the construction theorem of homogeneous kernel Hilbert-type integral inequalities and the properties of the Gamma function to discuss the conditions for constructing a Hilbert-type integral inequality involving an upper limit integral function and the best constant factor. We derive the necessary and sufficient conditions for constructing this inequality and a formula for calculating the best constant factor, thereby improving the existing results. Finally, as an application, we consider some special cases of parameters.

1. Introduction

Let $f\in L_p(0,+\infty )$ and $g\in L_q(0,+\infty )$ with $\frac{1}{p}+\frac{1}{q}=1$ ($p>1$). The well-known Hilbert integral inequality (see [1]) states:

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{1}{x+y}f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y⩽\frac{\pi }{sin(\pi /p)}{\parallel f\parallel }_p{\parallel g\parallel }_q,$$

(1)

where the constant factor is the best possible value. This is an elegant inequality with a symmetric structure. The integral operator T with the same kernel $\frac{1}{x+y}$ as (1) is given by

$$T\left(f\right)\left(y\right)={\int }_0^{+\infty }\frac{f\left(x\right)}{x+y}\mathrm{d}x,$$

then the inequality (1) is equivalent to the operator inequality

$${\parallel T\left(f\right)\parallel }_p⩽\frac{\pi }{sin(\pi /p)}{\parallel f\parallel }_p,f\in L_p(0,+\infty ),$$

which indicates that T is a bounded operator in $L_p(0,+\infty )$, and the operator norm of T is $\parallel T\parallel =\frac{\pi }{sin(\pi /p)}$. It is evident that Hilbert‘s inequality plays a significant role in the study of operators. Precisely because of the wide application of Hilbert’s inequality in operator theory and various analytical disciplines, it has attracted the attention of scholars worldwide. To further promote research, the Lebesgue space $L_p(0,+\infty )$ has been generalized to a weighted form.

Let $p>1$ and $\phi \left(x\right)>0$. The weighted Lebesgue space is defined as

$$L_p^{\phi \left(x\right)}(0,+\infty )=\left\{{f:\parallel f\parallel }_{p,\phi \left(x\right)}={\left({\int }_0^{+\infty }\phi \left(x\right){\left|f\left(x\right)\right|}^p\mathrm{d}x\right)}^{\frac{1}{p}}<+\infty \right\},$$

where $\phi \left(x\right)$ is referred to as the weight function. In the case where $\phi \left(x\right)$ is a power function, extensive research has been conducted on Hilbert-type inequalities for homogeneous kernels, quasi-homogeneous kernels, and some non-homogeneous kernels. The inequality (1) has been generalized and improved, and the results obtained have been applied to the discussion of the boundedness of operators and the estimation of operator norms. Ref. [2] obtained

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{{(x+y)}^{\lambda }}\mathrm{d}x\mathrm{d}y⩽B\left(\frac{p+\lambda -2}{p},\frac{q+\lambda -2}{q}\right){\parallel f\parallel }_{p,x^{1-\lambda }}{\parallel g\parallel }_{q,y^{1-\lambda }},$$

where $B(u,v)$ is the Beta function, and the constant factor $B\left(\frac{p+\lambda -2}{p},\frac{q+\lambda -2}{q}\right)$ is the best possible. Ref. [3] introduced two pairs of conjugate parameters $(p,q)$ and $(r,s)$, and obtained

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{x^{\lambda }+y^{\lambda }}\mathrm{d}x\mathrm{d}y\le \frac{\pi }{\lambda sin(\pi /r)}{\parallel f\parallel }_{p,\phi \left(x\right)}{\parallel g\parallel }_{q,\psi \left(y\right)},$$

where $\phi \left(x\right)=x^{p\left(1-\frac{\lambda }{r}\right)-1},\psi \left(y\right)=y^{q\left(1-\frac{\lambda }{s}\right)-1}$, and the constant factor $\frac{\pi }{\lambda sin(\pi /r)}$ is the best possible. Ref. [4] introduced another homogeneous kernel $\frac{ln(x/y)}{x^{\lambda }-y^{\lambda }}$, and obtained

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{ln(x/y)}{x^{\lambda }-y^{\lambda }}f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y\le {\left(\frac{\pi }{\lambda sin(\pi /r)}\right)}^2{\parallel f\parallel }_{p,\phi \left(x\right)}{\parallel g\parallel }_{q,\psi \left(y\right)},$$

where $\phi \left(x\right)=x^{p\left(1-\frac{\lambda }{r}\right)-1},\psi \left(y\right)=y^{q\left(1-\frac{\lambda }{s}\right)-1}$, and the constant factor ${\left(\frac{\pi }{\lambda sin(\pi /r)}\right)}^2$ is the best possible. These results are all generalizations and refinements of the Hilbert integral inequality. For such Hilbert-type inequalities involving multiple parameters, more results can be found in refs. [5,6,7]. To study the boundedness of multiple integral operators, people have also explored high-dimensional Hilbert-type inequalities. Ref. [8] introduced two sets of generalized conjugate exponents $\left(p_1,p_2,\cdots ,p_n\right)$ and $(r_1,r_2,\cdots ,r_n)$ and established a multiple Hilbert-type integral inequality:

$$\begin{array}{cc}& {\int }_0^{+\infty }\cdots {\int }_0^{+\infty }\frac{1}{{\left(x_1+x_2+\cdots +x_n\right)}^{\lambda }}\prod _{i=1}^n{f}_i\left(x_i\right)\mathrm{d}{x}_1\cdots \mathrm{d}{x}_n\hfill \\ & ⩽\frac{1}{\Gamma \left(\lambda \right)}\prod _{i=1}^n\Gamma \left(\frac{\lambda }{r_i}\right)\prod _{i=1}^n{\parallel f_i\parallel }_{p_i,x_i^{p_i\left(1-\frac{\lambda }{r_i}\right)-1}},\hfill \end{array}$$

where the constant factor $\frac{1}{\Gamma \left(\lambda \right)}{\prod }_{i=1}^n\Gamma \left(\frac{\lambda }{r_i}\right)$ is still the best possible. Subsequently, ref. [9] introduced the norm ${\parallel x\parallel }_{\alpha }={\left(x_1^{\alpha }+\cdots +x_n^{\alpha }\right)}^{1/\alpha }(\alpha >0)$ of the vector $x=\left(x_1,\cdots ,x_n\right)\in {\mathbb{R}}_{+}^n$, and obtained another form of the multiple integral Hilbert-type inequality:

$$\begin{array}{cc}& {\int }_{{\mathbb{R}}_{+}^n}{\int }_{{\mathbb{R}}_{+}^n}\frac{1}{{\left({\parallel x\parallel }_{\alpha }^{\beta }+{\parallel y\parallel }_{\alpha }^{\beta }\right)}^{\lambda }}f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y\hfill \\ & \le \frac{{\Gamma }^n(1/\alpha )}{\beta {\alpha }^{n-1}\Gamma (n/\alpha )}B\left(\frac{\lambda }{r},\frac{\lambda }{s}\right){\left({\int }_{{\mathbb{R}}_{+}^n}{\parallel x\parallel }_{\alpha }^{p\left(n-\frac{\lambda }{r}\beta \right)-n}{\left|f\left(x\right)\right|}^p\mathrm{d}x\right)}^{\frac{1}{p}}\hfill \\ & {\left({\int }_{{\mathbb{R}}_{+}^n}{\parallel y\parallel }_{\alpha }^{q\left(n-\frac{\lambda }{s}\beta \right)-n}{\left|g\left(y\right)\right|}^q\mathrm{d}y\right)}^{\frac{1}{q}},\hfill \end{array}$$

where the constant factor is the best possible. Further high-dimensional Hilbert-type inequalities can be found in [10,11,12,13]. The parameter relationship structure of the above results is relatively complex. To ensure the optimality of the inequality constant factor, which is to obtain the norm of the same kernel integral operator, it is necessary to explore the conditions satisfied by these parameters. Refs. [14,15,16,17] discussed the laws of the best matching parameters and obtained relatively ideal results. Subsequently, refs. [18,19] and others discussed the construction conditions of Hilbert-type inequalities and thus obtained the sufficient and necessary conditions for the boundedness of the corresponding integral operators and obtained the calculation formula for the operator norm. At present, the Hilbert-type inequality has formed a relatively complete theoretical system (see [20]). In recent years, another type of discrete Hilbert-type inequality involving partial sums and a Hilbert-type integral inequality involving variable upper limit integral functions have attracted attention. Refs. [21,22,23] have successively discussed Hilbert-type inequalities involving partial sums and variable upper limit integral functions for the homogeneous kernel $\frac{1}{{(m+n)}^{\lambda }}$ or $\frac{1}{{(x+y)}^{\lambda }}$, generalizing the Hilbert inequality from a new perspective, which may have a certain impact on the development of integral operator theory. In 2024, ref. [24] obtained

$$\begin{array}{cc}& {\int }_0^{+\infty }{\int }_0^{+\infty }\frac{1}{{(x+y)}^{\lambda }}f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y\hfill \\ & <\frac{\Gamma (\lambda +2)}{\Gamma \left(\lambda \right)}{B}^{\frac{1}{p}}\left({\lambda }_2+1,\lambda -{\lambda }_2+1\right)B^{\frac{1}{q}}\left({\lambda }_1+1,\lambda -{\lambda }_1+1\right){\parallel F\parallel }_{p,x^{\alpha }}{\parallel G\parallel }_{q,y^{\beta }},\hfill \end{array}$$

(2)

where $F\left(x\right)={\int }_0^{x}f\left(t\right)\mathrm{d}t,G\left(y\right)={\int }_0^{y}g\left(t\right)\mathrm{d}t,\alpha =-p{\lambda }_1-\left(\lambda -{\lambda }_1-{\lambda }_2\right)-1,\beta =-q{\lambda }_2-\left(\lambda -{\lambda }_1-{\lambda }_2\right)-1.$ It was proved that the condition for the constant factor in (2) to be optimal is

$${\lambda }_1={\lambda }_2.$$

The inequality (2) generalizes previous related results and obtains the necessary condition for the best constant, which is a very beneficial work. However, there are some shortcomings: 1. The parameter structure is quite cumbersome, which is not conducive to the application of the results; 2. The known conditions include $\underset{x\to +\infty }{lim}{e}^{-tx}F\left(x\right)=0$, $\underset{y\to +\infty }{lim}{e}^{-ty}G\left(y\right)=0$, and due to the arbitrariness of $f\left(x\right)$ and $g\left(y\right)$ in the inequality, this condition is often difficult to determine. 3. The conditions under which this inequality holds were not discussed. 4. Ref. [24] did not fully utilize the existing theoretical results of Hilbert-type inequalities, resulting in a very complex proof process. In response to these shortcomings, this paper uses the construction theorem of homogeneous kernel Hilbert-type integral inequalities from ref. [20], which not only simplifies the proof but also makes the parameter structure concise and clear and obtains the sufficient and necessary conditions for the inequality to hold and the calculation formula for the best constant factor. At the same time, it also generalizes the homogeneous kernel $\frac{1}{{(x+y)}^{\lambda }}$ in (2) to a broader quasi-homogeneous kernel $\frac{1}{{(\phi \left(x\right)+\psi \left(y\right))}^{\lambda }}$, thus generalizing and improving the results of ref. [24] from multiple perspectives.

2. Preliminary Lemmas

Lemma 1

([20]). (Construction theorem for Hilbert-type integral inequalities with homogeneous kernels) Let $\sigma ,\alpha ,\beta \in \mathbb{R}$, $\frac{1}{p}+\frac{1}{q}=1$ $(p>1)$, and $K(x,y)⩾0$ be a homogeneous function of order σ. Suppose that

$$W(\alpha ,p)={\int }_0^{+\infty }K(t,1)t^{-\frac{\alpha +1}{p}}\mathrm{d}t<+\infty .$$

$\left(i\right)$ If and only if $\frac{\alpha }{p}+\frac{\beta }{q}=1+\sigma$, there exists a constant $M>0$ such that

$${\int }_0^{+\infty }{\int }_0^{+\infty }K(x,y)f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y\le {M\parallel f\parallel }_{p,x^{\alpha }}{\parallel g\parallel }_{q,y^{\beta }},$$

(3)

where $f\in L_p^{x^{\alpha }}(0,+\infty )$ and $g\in L_q^{y^{\beta }}(0,+\infty )$.

$\left(ii\right)$ When $\frac{\alpha }{p}+\frac{\beta }{q}=1+\sigma$, i.e., when (3) holds, the best constant factor is given by

$$M_0=inf\left\{M\right\}=W(\alpha ,p)={\int }_0^{+\infty }K(t,1)t^{-\frac{\alpha +1}{p}}\mathrm{d}t.$$

Lemma 2. 

Assume that $\frac{1}{p}+\frac{1}{q}=1$ $(p>1)$, $\alpha <p-1$, $\beta <q-1$, and $\lambda >\frac{1}{q}-\frac{\alpha }{p}$.

$\left(i\right)$ If and only if $\frac{\alpha }{p}+\frac{\beta }{q}=1-\lambda$, there exists a constant $M>0$ such that

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{{(x+y)}^{\lambda }}\mathrm{d}x\mathrm{d}y⩽{M\parallel f\parallel }_{p,x^{\alpha }}{\parallel g\parallel }_{q,y^{\beta }},$$

(4)

where $f\in L_p^{x^{\alpha }}(0,+\infty )$ and $g\in L_q^{y^{\beta }}(0,+\infty )$.

$\left(ii\right)$ If $\frac{\alpha }{p}+\frac{\beta }{q}=1-\lambda$, then the best possible constant factor of (4) is  $B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right)$.

Proof. 

Denote

$$K(x,y)=\frac{1}{{(x+y)}^{\lambda }},x>0,y>0.$$

Then $K(x,y)>0$ and is a homogeneous function of order $-\lambda$, thus the condition $\frac{\alpha }{p}+\frac{\beta }{q}=1+\sigma$ can be rewritten as $\frac{\alpha }{p}+\frac{\beta }{q}=1-\lambda$.

Since $\alpha <p-1$, $\beta <q-1$ and $\lambda >\frac{1}{q}-\frac{\alpha }{p}$, it follows that $\frac{1}{q}-\frac{\alpha }{p}>0$, $\frac{1}{p}-\frac{\beta }{q}>0$ and $\lambda -(\frac{1}{q}-\frac{\alpha }{p})>0,$ then

$$\begin{array}{cc}\hfill W(\alpha ,p)& ={\int }_0^{+\infty }K(1,t)t^{-\frac{\alpha +1}{p}}\mathrm{d}t\hfill \\ & ={\int }_0^{+\infty }\frac{1}{{(1+t)}^{\lambda }}{t}^{\left(\frac{1}{q}-\frac{\alpha }{p}\right)-1}\mathrm{d}t\hfill \\ & =B\left(\frac{1}{q}-\frac{\alpha }{p},\lambda -\left(\frac{1}{q}-\frac{\alpha }{p}\right)\right)<+\infty .\hfill \end{array}$$

And when $\frac{\alpha }{p}+\frac{\beta }{q}=1-\lambda$, we have

$$\lambda -\left(\frac{1}{q}-\frac{\alpha }{p}\right)=\frac{1}{p}-\frac{\beta }{q},$$

thus,

$$W(\alpha ,p)=B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right).$$

In summary, and based on Lemma 1, Lemma 2 holds. □

Lemma 3. 

Let $\frac{1}{p}+\frac{1}{q}=1$ $(p>1)$, $\alpha <p-1$, $f\in L_p^{x^{\alpha }}(0,+\infty )$, $F\left(x\right)={\int }_0^{x}f\left(u\right)\mathrm{d}u$, and

$$\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}{x}^{\frac{1}{q}-\frac{\alpha }{p}}=0$$

for $t>0$. Then

$$\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}F\left(x\right)=0.$$

Proof. 

It follows from $f\in L_p^{x^{\alpha }}(0,+\infty )$ and Hölder’s integral inequality that

$$\begin{array}{cc}& \underset{x\to +\infty }{lim}\left|e^{-\phi \left(x\right)t}F\left(x\right)\right|\le \underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}{\int }_0^x\left|f\left(u\right)\right|\mathrm{d}u\hfill \\ & =\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}{\int }_0^x{u}^{-\frac{\alpha }{p}}\left(u^{\frac{\alpha }{p}}\left|f\left(u\right)\right|\mathrm{d}u\right)\hfill \\ & =\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}{\left({\int }_0^x{u}^{-\frac{q}{p}\alpha }\mathrm{d}u\right)}^{\frac{1}{q}}{\left({\int }_0^x{u}^{\alpha }{\left|f\left(u\right)\right|}^p\mathrm{d}u\right)}^{\frac{1}{p}}\hfill \\ & \le {\parallel f\parallel }_{p,x^{\alpha }}\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}{\left({\int }_0^x{u}^{(1-q)\alpha }\mathrm{d}u\right)}^{\frac{1}{q}}.\hfill \end{array}$$

Since $\alpha <p-1$, it implies $(1-q)\alpha +1>0$. Thus,

$${\left({\int }_0^x{u}^{(1-q)\alpha }\mathrm{d}u\right)}^{\frac{1}{p}}={\left(\frac{1}{(1-q)\alpha +1}\right)}^{\frac{1}{q}}{x}^{\frac{1}{q}-\frac{\alpha }{p}},$$

and therefore,

$$\underset{x\to +\infty }{lim}|e^{-\phi \left(x\right)t}F\left(x\right)|\le {\left(\frac{1}{(1-q)\alpha +1}\right)}^{\frac{1}{q}}{\parallel f\parallel }_{p,x^{\alpha }}\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}{x}^{\frac{1}{q}-\frac{\alpha }{p}}=0,$$

which leads to

$$\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}F\left(x\right)=0.$$

□

3. Hilbert-Type Inequalities Involving Upper Limit Functions

Let f and g be measurable functions on $(0,+\infty )$. Denote

$$F\left(x\right)={\int }_0^{x}f\left(t\right)\mathrm{d}t,G\left(y\right)={\int }_0^{y}g\left(t\right)\mathrm{d}t.$$

Then, the upper limit functions $F\left(x\right)$ and $G\left(y\right)$ are also measurable functions. We have the following construction theorem related to the Hilbert-type inequalities involving $F\left(x\right)$ and $G\left(y\right)$.

Theorem 1. 

Let $\frac{1}{p}+\frac{1}{q}=1$ $(p>1)$, $\lambda ,\alpha ,\beta \in \mathbb{R}$, $\alpha <p-1$, $\beta <q-1$, and $\lambda >\frac{1}{q}-\frac{\alpha }{p}$. Let $\phi \left(x\right)$ and $\psi \left(y\right)$ be strictly increasing differentiable functions on $(0,+\infty )$ with the same range $(0,+\infty )$. Assume that

$$\underset{x\to +\infty }{lim}{x}^{\frac{1}{q}-\frac{\alpha }{p}}{e}^{-\phi \left(x\right)t}=0,\underset{y\to +\infty }{lim}{y}^{\frac{1}{p}-\frac{\beta }{q}}{e}^{-\psi \left(y\right)t}=0(t>0).$$

Then

$\left(i\right)$ If and only if $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda$, there exists a constant $M>0$ such that the following Hilbert-type integral inequality holds:

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{{(\phi \left(x\right)+\psi \left(y\right))}^{\lambda }}\mathrm{d}x\mathrm{d}y\le {M\parallel F\parallel }_{p,{\theta }_1\left(x\right)}{\parallel G\parallel }_{q,{\theta }_2\left(y\right)},$$

(5)

where ${\theta }_1\left(x\right)={\phi }^{\alpha }\left(x\right){\phi }^{\prime }\left(x\right)$, ${\theta }_2\left(y\right)={\psi }^{\beta }\left(y\right){\psi }^{\prime }\left(y\right)$, $f\in L_p^{x^{\alpha }}(0,+\infty )$ and $g\in L_q^{y^{\beta }}(0,+\infty )$.

$\left(ii\right)$ When $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda$, i.e., when (5) holds, the optimal constant factor is given by

$$M_0=inf\left\{M\right\}=\lambda (\lambda +1)B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right).$$

Proof. 

(i) From the definition of Gamma function, we have

$$\frac{1}{{(\phi \left(x\right)+\psi \left(y\right))}^{\lambda }}=\frac{1}{\Gamma \left(\lambda \right)}{\int }_0^{+\infty }{t}^{\lambda -1}{e}^{-\left(\phi \right(x)+\psi (y\left)\right)t}\mathrm{d}t.$$

Thus,

$$\begin{array}{cc}\hfill I& :={\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{{(\phi \left(x\right)+\psi \left(y\right))}^{\lambda }}\mathrm{d}x\mathrm{d}y\hfill \\ & =\frac{1}{\Gamma \left(\lambda \right)}{\int }_0^{+\infty }{\int }_0^{+\infty }f\left(x\right)g\left(y\right)\left({\int }_0^{+\infty }{t}^{\lambda -1}{e}^{-\left(\phi \right(x)+\psi (y\left)\right)t}\mathrm{d}t\right)\mathrm{d}x\mathrm{d}y\hfill \\ & =\frac{1}{\Gamma \left(\lambda \right)}{\int }_0^{+\infty }{t}^{\lambda -1}\left({\int }_0^{+\infty }{e}^{-\phi \left(x\right)t}f\left(x\right)\mathrm{d}x\right)\left({\int }_0^{+\infty }{e}^{-\psi \left(y\right)t}g\left(y\right)\mathrm{d}y\right)\mathrm{d}t.\hfill \end{array}$$

Since $\underset{x\to +\infty }{lim}{x}^{\frac{1}{q}-\frac{\alpha }{p}}{e}^{-\phi \left(x\right)t}=0$, $\alpha <p-1$ and $f\in L_p^{x^{\alpha }}(0,+\infty )$, by Lemma 3, it follows that

$$\begin{array}{cc}\hfill {\int }_0^{+\infty }{e}^{-\phi \left(x\right)t}f\left(x\right)\mathrm{d}x& ={\int }_0^{+\infty }{e}^{-\phi \left(x\right)t}\mathrm{d}F\left(x\right)\hfill \\ & =\underset{x\to +\infty }{lim}{e}^{-\phi \left(x\right)t}F\left(x\right)+t{\int }_0^{+\infty }{\phi }^{\prime }\left(x\right)e^{-\phi \left(x\right)t}F\left(x\right)\mathrm{d}x\hfill \\ & =t{\int }_0^{+\infty }{e}^{-\phi \left(x\right)t}F\left(x\right)\mathrm{d}\phi \left(x\right)\hfill \\ & =t{\int }_0^{+\infty }{e}^{-ut}F\left({\phi }^{-1}\left(u\right)\right)\mathrm{d}u.\hfill \end{array}$$

Similarly, since $\underset{y\to +\infty }{lim}{y}^{\frac{1}{p}-\frac{\beta }{q}}{e}^{-\psi \left(y\right)t}=0$, $\beta <q-1$ and $g\in L_q^{y^{\beta }}(0,+\infty )$, it holds that

$${\int }_0^{+\infty }{e}^{-\psi \left(y\right)t}g\left(y\right)\mathrm{d}y=t{\int }_0^{+\infty }{e}^{-vt}G\left({\psi }^{-1}\left(v\right)\right)\mathrm{d}v.$$

Therefore,

$$\begin{array}{cc}\hfill I& =\frac{1}{\Gamma \left(\lambda \right)}{\int }_0^{+\infty }{t}^{\lambda -1}\left(t{\int }_0^{+\infty }{e}^{-ut}F\left({\phi }^{-1}\left(u\right)\right)\mathrm{d}u\right)\left(t{\int }_0^{+\infty }{e}^{-vt}G\left({\psi }^{-1}\left(v\right)\right)\mathrm{d}vs.\right)\mathrm{d}t\hfill \\ & =\frac{1}{\Gamma \left(\lambda \right)}{\int }_0^{+\infty }{\int }_0^{+\infty }F\left({\phi }^{-1}\left(u\right)\right)G\left({\psi }^{-1}\left(v\right)\right)\left({\int }_0^{+\infty }{t}^{\lambda +1}{e}^{-(u+v)t}\mathrm{d}t\right)\mathrm{d}u\mathrm{d}vs.\hfill \\ & =\frac{\Gamma (\lambda +2)}{\Gamma \left(\lambda \right)}{\int }_0^{+\infty }{\int }_0^{+\infty }\frac{F\left({\phi }^{-1}\left(u\right)\right)G\left({\psi }^{-1}\left(v\right)\right)}{{(u+v)}^{\lambda +2}}\mathrm{d}u\mathrm{d}v.\hfill \end{array}$$

By applying Lemma 2, if and only if $\frac{\alpha }{p}+\frac{\beta }{q}=1-(\lambda +2)=-1-\lambda$, there exists a constant $\overline{M}>0$ such that

$$\begin{array}{cc}\hfill & {\int }_0^{+\infty }{\int }_0^{+\infty }\frac{F\left({\phi }^{-1}\left(u\right)\right)G\left({\psi }^{-1}\left(v\right)\right)}{{(u+v)}^{\lambda +2}}\mathrm{d}u\mathrm{d}vs.\hfill \\ \hfill & \le \overline{M}{\left({\int }_0^{+\infty }{u}^{\alpha }{\left|F\left({\phi }^{-1}\left(u\right)\right)\right|}^p\mathrm{d}u\right)}^{\frac{1}{p}}{\left({\int }_0^{+\infty }{v}^{\beta }{\left|G\left({\psi }^{-1}\left(v\right)\right)\right|}^q\mathrm{d}vs.\right)}^{\frac{1}{q}}\hfill \\ \hfill & =\overline{M}{\left({\int }_0^{+\infty }{\phi }^{\alpha }\left(x\right){\phi }^{\prime }\left(x\right){\left|F\left(x\right)\right|}^p\mathrm{d}x\right)}^{\frac{1}{p}}{\left({\int }_0^{+\infty }{\psi }^{\beta }\left(y\right){\psi }^{\prime }\left(y\right){\left|G\left(y\right)\right|}^q\mathrm{d}y\right)}^{\frac{1}{q}}\hfill \\ \hfill & =\overline{M}{\parallel F\parallel }_{p,{\theta }_1\left(x\right)}{\parallel G\parallel }_{q,{\theta }_2\left(y\right)}.\hfill \end{array}$$

(6)

Thus, if and only if $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda$, we obtain

$$I\le \overline{M}\frac{\Gamma (\lambda +2)}{\Gamma \left(\lambda \right)}{\parallel F\parallel }_{p,{\theta }_1\left(x\right)}{\parallel G\parallel }_{q,{\theta }_2\left(y\right)}=\overline{M}{\lambda (\lambda +1)\parallel F\parallel }_{p,{\theta }_1\left(x\right)}{\parallel G\parallel }_{q,{\theta }_2\left(y\right)}.$$

Letting $M=\lambda (\lambda +1)\overline{M}$, then the desired inequality (5) is obtained.

(ii) When $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda =1-(\lambda +2)$, according to Lemma 2, the optimal constant factor of (6) is $B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right)$. Thus, the optimal constant factor of (5) becomes

$$M_0=\lambda (\lambda +1)B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right).$$

□

Taking $\phi \left(x\right)=ax,\psi \left(y\right)=by$ in Theorem 1, where $a,b>0$, then ${\theta }_1\left(x\right)={\phi }^{\alpha }\left(x\right){\phi }^{\prime }\left(x\right)=a^{\alpha +1}{x}^{\alpha },{\theta }_2\left(y\right)={\psi }^{\beta }\left(y\right){\psi }^{\prime }\left(y\right)=b^{\beta +1}{y}^{\beta }$. The following result can be deduced.

Corollary 1. 

Let $\frac{1}{p}+\frac{1}{q}=1(p>1)$, $a>0$, $b>0$, $\lambda ,\alpha ,\beta \in \mathbb{R}$, $\alpha <p-1$, $\beta <q-1$, and $\lambda >\frac{1}{q}-\frac{\alpha }{p}$. Then

$\left(i\right)$ If and only if $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda$, there exists a constant $M>0$ such that

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{{(ax+by)}^{\lambda }}\mathrm{d}x\mathrm{d}y\le {M\parallel F\parallel }_{p,x^{\alpha }}{\parallel G\parallel }_{q,y^{\beta }}$$

(7)

where $F\left(x\right)={\int }_0^{x}f\left(t\right)\mathrm{d}t$ and $G\left(y\right)={\int }_0^{y}g\left(t\right)\mathrm{d}t$, with $f\in L_p^{x^{\alpha }}(0,+\infty )$ and $g\in L_q^{y^{\beta }}(0,+\infty )$.

$\left(ii\right)$ When $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda$, the best constant factor of (7) is given by

$$M_0=\lambda (\lambda +1)a^{\frac{\alpha +1}{p}}{b}^{\frac{\beta +1}{q}}B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right).$$

Remark 1. 

In Corollary 1, by setting $a=b=1$, $\alpha =-p{\lambda }_1-(\lambda -{\lambda }_1-{\lambda }_2)-1$ and $\beta =-q{\lambda }_2-(\lambda -{\lambda }_1-{\lambda }_2)-1$, where ${\lambda }_1>0$ and ${\lambda }_2>0$, we have

$$\frac{\alpha }{p}+\frac{\beta }{q}=-{\lambda }_1-\frac{\lambda -{\lambda }_1-{\lambda }_2}{p}-\frac{1}{p}-{\lambda }_2-\frac{\lambda -{\lambda }_1-{\lambda }_2}{q}-\frac{1}{q}=-1-\lambda .$$

When ${\lambda }_1+{\lambda }_2=\lambda$, it follows that $\alpha =-p{\lambda }_1-1$, $\beta =-q{\lambda }_2-1$, $\frac{1}{q}-\frac{\alpha }{p}=1+{\lambda }_1$, $\frac{1}{p}-\frac{\beta }{q}=1+{\lambda }_2$, and

$$B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right)=B(1+{\lambda }_1,1+{\lambda }_2)=\frac{{\lambda }_1{\lambda }_2}{\lambda (\lambda +1)}B({\lambda }_1,{\lambda }_2).$$

Thus, we can derive the following result

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{{(x+y)}^{\lambda }}\mathrm{d}x\mathrm{d}y⩽{\lambda }_1{\lambda }_{2}B\left({\lambda }_1,{\lambda }_2\right){\parallel F\parallel }_{p,x^{-p{\lambda }_1-1}}{\parallel G\parallel }_{q,y^{-q{\lambda }_2-1}},$$

which corresponds to the main result in [24].

Choosing $\phi \left(x\right)=e^x-1$ and $\psi \left(y\right)=e^y-1$ in Theorem 1, then ${\theta }_1\left(x\right)={\phi }^{\alpha }\left(x\right){\phi }^{\prime }\left(x\right)=e^x{(e^x-1)}^{\alpha },{\theta }_2\left(y\right)={\psi }^{\beta }\left(y\right){\psi }^{\prime }\left(y\right)=e^y{(e^y-1)}^{\beta }.$ This leads to the following corollary.

Corollary 2. 

Let $\frac{1}{p}+\frac{1}{q}=1$ $(p>1)$, $\lambda ,\alpha ,\beta \in \mathbb{R}$, $\alpha <p-1$, $\beta <q-1$, and $\lambda >\frac{1}{q}-\frac{\alpha }{p}$. Then

$\left(i\right)$ If and only if $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda$, there exists a constant $M>0$ such that

$${\int }_0^{+\infty }{\int }_0^{+\infty }\frac{f\left(x\right)g\left(y\right)}{{(e^x+e^y-2)}^{\lambda }}\mathrm{d}x\mathrm{d}y\le {M\parallel F\parallel }_{p,{\theta }_1\left(x\right)}{\parallel G\parallel }_{q,{\theta }_2\left(y\right)},$$

(8)

where ${\theta }_1\left(x\right)=e^x{(e^x-1)}^{\alpha }$, ${\theta }_2\left(y\right)=e^y{(e^y-1)}^{\beta }$, $F\left(x\right)={\int }_0^{x}f\left(t\right)\mathrm{d}t$, $G\left(y\right)={\int }_0^{y}g\left(t\right)\mathrm{d}t$, $f\in L_p^{x^{\alpha }}(0,+\infty )$ and $g\in L_q^{y^{\beta }}(0,+\infty )$.

$\left(ii\right)$ When $\frac{\alpha }{p}+\frac{\beta }{q}=-1-\lambda$, the optimal constant factor M for inequality (8) is given by

$$M_0=\lambda (\lambda +1)B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right).$$

Corollary 3. 

Let

$$\frac{1}{p}+\frac{1}{q}=1(p>1),\lambda >0,{\lambda }_1+{\lambda }_2=0,{\lambda }_1<1+\frac{\lambda }{p},{\lambda }_2<1+\frac{\lambda }{q},$$

then

$$\begin{array}{cc}& {\int }_0^{+\infty }{\int }_0^{+\infty }\frac{1}{{(x+y)}^{\lambda }}f\left(x\right)g\left(y\right)\mathrm{d}x\mathrm{d}y\hfill \\ & ⩽\frac{1}{\Gamma \left(\lambda \right)}\Gamma \left(1+\frac{\lambda }{p}-{\lambda }_1\right)\Gamma \left(1+\frac{\lambda }{q}-{\lambda }_2\right){\parallel F\parallel }_{p,x^{{\lambda }_{1}p-\lambda -1}}{\parallel G\parallel }_{q,y^{{\lambda }_{2}q-\lambda -1}},\hfill \end{array}$$

where $f\in L_p^{x^{{\lambda }_{1}p-\lambda -1}}(0,+\infty )$, $g\in L_q^{y^{{\lambda }_{2}q-\lambda -1}}(0,+\infty )$, $F\left(x\right)={\int }_0^{x}f\left(t\right)\mathrm{d}t$, $G\left(y\right)={\int }_0^{y}g\left(t\right)\mathrm{d}t$, and the constant factor $\frac{1}{\Gamma \left(\lambda \right)}\Gamma \left(1+\frac{\lambda }{p}-{\lambda }_1\right)\Gamma \left(1+\frac{\lambda }{q}-{\lambda }_2\right)$ is the best value.

Proof. 

Let $\alpha ={\lambda }_{1}p-\lambda -1$, $\beta ={\lambda }_{2}q-\lambda -1$. Since ${\lambda }_1+{\lambda }_2=0$, we have

$$\frac{\alpha }{p}+\frac{\beta }{q}=\frac{{\lambda }_{1}p-\lambda -1}{p}+\frac{{\lambda }_{2}q-\lambda -1}{q}={\lambda }_1+{\lambda }_2-\frac{\lambda +1}{p}-\frac{\lambda +1}{q}=-1-\lambda .$$

Also, because ${\lambda }_1<1+\frac{\lambda }{p}$, ${\lambda }_2<1+\frac{\lambda }{q}$, it follows that $1+\frac{\lambda }{p}-{\lambda }_1>0$, $1+\frac{\lambda }{q}-{\lambda }_2>0$, and

$$\begin{array}{cc}\hfill M_0& =\lambda (\lambda +1)B\left(\frac{1}{q}-\frac{\alpha }{p},\frac{1}{p}-\frac{\beta }{q}\right)\hfill \\ & =\lambda (\lambda +1)B\left(\frac{1}{q}-\frac{{\lambda }_{1}p-\lambda -1}{p},\frac{1}{p}-\frac{{\lambda }_{2}q-\lambda -1}{q}\right)\hfill \\ & =\lambda (\lambda +1)B\left(1+\frac{\lambda }{p}-1,1+\frac{\lambda }{q}-1\right)\hfill \\ & =\lambda (\lambda +1)\frac{1}{\Gamma (\lambda +2)}\Gamma \left(1+\frac{\lambda }{p}-{\lambda }_1\right)\Gamma \left(1+\frac{\lambda }{q}-{\lambda }_2\right)\hfill \\ & =\frac{1}{\Gamma \left(\lambda \right)}\Gamma \left(1+\frac{\lambda }{p}-{\lambda }_1\right)\Gamma \left(1+\frac{\lambda }{q}-{\lambda }_2\right),\hfill \end{array}$$

where $B(x,y)$ is the Beta function. Therefore, by Corollary 1, Corollary 3 is established. □

Remark 2. 

By choosing various different parameter combinations, we can obtain more forms of Hilbert-type inequalities involving integral functions with variable upper limits.

4. Conclusions

Building upon the literature [24], this paper discusses a more general kernel function $\frac{1}{{(\phi \left(x\right)+\psi \left(y\right))}^{\lambda }}$, not only achieving more universal results but also establishing the sufficient and necessary conditions for constructing the corresponding inequalities. In fact, this provides a fundamental method for determining whether the inequality can hold, making our results more significant. By selecting different $\varphi \left(x\right)$ and $\psi \left(y\right)$, a multitude of specific inequalities can also be derived.