Abstract
By using real analysis and weight functions, we obtain a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the kernel of exponent function with intermediate variables. The constant factor related to the gamma function is proved to be the best possible. We also consider some particular cases and the operator expressions.
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1 Introduction
If \(0 < \int_{0}^{\infty} f^{2}(x)\,dx < \infty \) and \(0 < \int_{0}^{\infty} g^{2}(y)\,dy < \infty \), then we have the following well-known Hilbert integral inequality (see [1]):
where the constant factor π is the best possible. In 1925, by introducing the pair of conjugate exponents \((p,q)\) (\(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1 \)), Hardy et al. gave an extension of (1) (see [1], Theorem 316). Recently, by means of weight functions, some new extensions of (1) and the Hardy’s work were given by Yang [2, 3] and in [4–9]. Most of them are built in the quarter plane of the first quadrant.
In 2007, Yang [10] provided a Hilbert-type integral inequality in the whole plane with the exponent function and intermediate variables as follows:
where the constant factor \(B(\frac{\lambda}{2},\frac{\lambda}{2}) \) is the best possible (\(\lambda > 0\), \(B(u,v) \) is the beta function). He et al. [11–19] proved some new Hilbert-type integral inequalities in the whole plane with the best possible constant factors.
In 2017, Hong [20] gave two equivalent statements between Hilbert-type inequalities with general homogenous kernel and a few parameters. A few authors continue to study this topic (see [21–25]).
In this paper, by using real analysis and weight functions we obtain a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the exponent function with intermediate variables. The constant factor related to the gamma function is proved to be the best possible. We also consider some particular cases and operator expressions.
2 Some lemmas
For \(\gamma,\rho,\sigma > 0 \), setting \(h(u): = e^{ - \rho u^{\gamma}}\) (\(u > 0 \)), we find
where \(\Gamma (s): = \int_{0}^{\infty} e^{ - v}v^{s - 1}\,dv\) (\(\operatorname{Re} s > 0 \)) is the gamma function (see [26]).
For \(\delta \in \{ - 1,1\}\), \(\alpha,\beta \in ( - 1,1) \), we set
Lemma 1
For \(c > 0\), \(\theta = \alpha, \beta \in ( - 1,1) \), we have
and for \(c \le 0 \), we have
Proof
Setting \(E_{\delta}^{ +}: = \{ t \in \mathbb{R}_{ +};t^{\delta} \ge 1\}\), \(E_{\delta}^{ -}: = \{ - t \in \mathbb{R}_{ +};( - t)^{\delta} \ge 1\} \), we find \(E_{\delta} = E_{\delta}^{ +} \cup E_{\delta}^{ -} \) and
Setting \(t = u^{\frac{1}{\delta}} \), we find
Hence, for \(c > 0 \), (4) follows, and for \(c \le 0\), \(\int_{E_{\delta}} t_{\theta}^{ - c\delta - 1} \,dt = \infty \). Since, for \(c > 0 \),
we have (5), and for \(c \le 0\), \(\int_{E_{ - \delta}} t_{\theta}^{c\delta - 1} \,dt = \infty \).
The lemma is proved. □
In the following, We further assume that \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(\delta \in \{ - 1,1\}\), \(\alpha,\beta \in ( - 1,1)\), \(\gamma,\rho,\sigma > 0\), \(\sigma_{1} \in \mathbb{R}\), \(k_{\rho}^{(\gamma )}(\sigma ) \) is given by (3), and
For \(n \in \mathbb{N} = \{ 1,2, \ldots \}\), \(E_{ - 1} = [ - 1,1]\), \(x \in E_{\delta} \), we define:
For \(y_{\beta} = (\operatorname{sgn} (y) + \beta )y \), where
and \(1 - |\alpha | \le (1 + |\alpha |)^{ - 1} \le 1 + |\alpha | \le (1 - |\alpha |)^{ - 1} \), we have
For fixed \(x \in E_{\delta} \), setting \(u = x_{\alpha}^{\delta} y_{\beta} \), we find
For \(n \in \mathbb{N} = \{ 1,2, \ldots \}\), \(x \in E_{ - \delta} \), we define:
Since, for \(x \in E_{ - \delta} \),
we have
For fixed \(x \in E_{ - \delta} \), setting \(u = x_{\alpha}^{\delta} y_{\beta} \), we find
In view of (8) and (10), we have the following:
Lemma 2
We have the following inequalities:
Lemma 3
If there exists a constant M such that, for any nonnegative measurable functions \(f(x) \) and \(g(y) \) in \(\mathbb{R} \),
then we have \(\sigma_{1} = \sigma \).
Proof
If \(\sigma_{1} > \sigma \), then for \(n \ge \frac{1}{\sigma_{1} - \sigma} \) (\(n \in \mathbb{N}\)), we define the functions:
and by (4) and (5) it follows that
By (11) and (13) (for \(f = f_{n}\), \(g = g_{n} \)) we have
Since for any \(n \ge \frac{1}{\sigma_{1} - \sigma}\), \(\sigma - \sigma_{1} + \frac{1}{n} \le 0 \), by Lemma 1 it follows that \(\int_{E_{\delta}} x_{\alpha}^{ - \delta (\sigma - \sigma_{1} + \frac{1}{n}) - 1}\,dx = \infty \). In view of \(\int_{0}^{m_{\alpha,\beta}} e^{ - \rho u^{\gamma}} u^{\sigma + \frac{1}{qn} - 1}\,du > 0 \), we find that \(\infty \le M\tilde{J}_{1} < \infty \), which is a contradiction.
If \(\sigma_{1} < \sigma \), then for \(n \ge \frac{1}{\sigma - \sigma_{1}}\) (\(n \in \mathbb{N} \)), we define the functions:
and by (4) and (5) it follows that
By (12) and (13) (for \(f = \tilde{f}_{n}\), \(g = \tilde{g}_{n} \)) we have
Since for any \(n \ge \frac{1}{\sigma - \sigma_{1}}\), \(\sigma_{1} - \sigma + \frac{1}{n} \le 0 \), by Lemma 1 it follows that \(\int_{E_{ - \delta}} x_{\alpha}^{\delta (\sigma_{1} - \sigma + \frac{1}{n}) - 1}\,dx = \infty \). In view of \(\int_{M_{\alpha,\beta}}^{\infty} e^{ - \rho u^{\gamma}} u^{\sigma - \frac{1}{qn} - 1}\,du > 0 \), we have \(\infty \le M\tilde{J}_{2} < \infty \), which is a contradiction.
Hence we conclude that \(\sigma_{1} = \sigma \).
The lemma is proved. □
Lemma 4
If there exists a constant M such that, for any nonnegative measurable functions \(f(x) \) and \(g(y) \) in \(\mathbb{R} \),
then we have \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \le M \).
Proof
For \(\sigma_{1} = \sigma \), by (8) we have
In view of the presented results, we find
Since \(e^{ - \rho u^{\gamma}} u^{2\sigma} \) is continuous in \((0,\infty ) \), and \(e^{ - \rho u^{\gamma}} u^{2\sigma} \to 0\) (\(u \to \infty \)), there exists a positive constant \(M_{1} \) such that \(e^{ - \rho u^{\gamma}} u^{2\sigma} \le M_{1}\) (\(u \in [m_{\alpha,\beta},\infty ) \)). By (4) it follows that
so that
By (15) it follows that
In the same way, we have
By (14) (for \(f = f_{n}\), \(g = g_{n} \)), we have
For \(n \to \infty \), by Fatou lemma (see [27]), (16), and (17) we find
so that \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) = \frac{2k_{\rho}^{(\gamma )}(\sigma )}{(1 - \alpha^{2})^{1/q}(1 - \beta^{2})^{1/p}} \le M \).
The lemma is proved. □
Lemma 5
We define the following weight functions:
Then we have
Proof
For fixed \(y \in \mathbb{R}\backslash \{ 0\} \), setting \(u = x_{\alpha}^{\delta} y_{\beta} \), we find
for fixed \(x \in \mathbb{R}\backslash \{ 0\} \), setting \(u = x_{\alpha}^{\delta} y_{\beta} \), it follows that
Hence we have (20).
The lemma is proved. □
3 Main results
Theorem 1
If M is a constant, then the following statements (i), (ii), and (iii) are equivalent:
-
(i)
For any \(f(x) \ge 0 \), we have:
$$ \begin{aligned}[b] J&: = \biggl\{ \int_{ - \infty}^{\infty} y_{\beta}^{p\sigma - 1} \biggl[ \int_{ - \infty}^{\infty} e^{ - \rho (x_{\alpha}^{\delta} y_{\beta} )^{\gamma}} f(x)\,dx \biggr]^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ &\le M \biggl[ \int_{ - \infty}^{\infty} x_{\alpha}^{p(1 - \delta \sigma_{1}) - 1}f^{p}(x) \,dx \biggr]^{\frac{1}{p}}. \end{aligned} $$(21) -
(ii)
For any \(f(x),g(y) \ge 0 \), we have:
$$ \begin{aligned}[b] I &= \int_{ - \infty}^{\infty} \int_{ - \infty}^{\infty} e^{ - \rho (x_{\alpha}^{\delta} y_{\beta} )^{\gamma}} f(x)g(y)\,dx \,dy \\ &\le M \biggl[ \int_{ - \infty}^{\infty} x_{\alpha}^{p(1 - \delta \sigma_{1}) - 1}f^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int_{ - \infty}^{\infty} y_{\beta}^{q(1 - \sigma ) - 1}g^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{aligned} $$(22) -
(iii)
\(\sigma_{1} = \sigma \), and \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \le M \).
Proof
(i)=>(ii). By Hölder’s inequality (see [28]) we have
(ii)=>(iii). By Lemma 1 we have \(\sigma_{1} = \sigma \). Then by Lemma 2 we have \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \le M \).
(iii)=>(i). For \(\sigma_{1} = \sigma \), by Hölder’s inequality with weight (see [28]) and (18) we have
By Fubini’s theorem (see [27]), (24), and (19) we have
For \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \le M \), we have (21) (when \(\sigma_{1} = \sigma \)).
Therefore, statements (i), (ii), and (iii) are equivalent.
The theorem is proved. □
Theorem 2
If M is a constant, then the following statements (i), (ii), and (iii) are equivalent:
-
(i)
For any \(f ( x ) \geq 0\) satisfying \(0 < \int_{ - \infty}^{\infty} x_{\alpha}^{p(1 - \delta \sigma ) - 1}f^{p}(x)\,dx < \infty \), we have:
$$ \begin{gathered}[b] \biggl\{ \int_{ - \infty}^{\infty} y_{\beta}^{p\sigma - 1} \biggl[ \int_{ - \infty}^{\infty} e^{ - \rho (x_{\alpha}^{\delta} y_{\beta} )^{\gamma}} f(x)\,dx \biggr]^{p}\,dy \biggr\} ^{\frac{1}{p}} \\ \quad < M \biggl[ \int_{ - \infty}^{\infty} x_{\alpha}^{p(1 - \delta \sigma ) - 1}f^{p}(x) \,dx \biggr]^{\frac{1}{p}}. \end{gathered} $$(25) -
(ii)
For any \(f ( x ) \geq 0\) satisfying \(0 < \int_{ - \infty}^{\infty} x_{\alpha}^{p(1 - \delta \sigma ) - 1}f^{p}(x)\,dx < \infty \), and \(g ( x ) \geq 0\) satisfying \(0 < \int_{ - \infty}^{\infty} y_{\beta}^{q(1 - \sigma ) - 1}g^{q}(y)\,dy < \infty \), we have:
$$ \begin{gathered}[b] \int_{ - \infty}^{\infty} \int_{ - \infty}^{\infty} e^{ - \rho (x_{\alpha}^{\delta} y_{\beta} )^{\gamma}} f(x)g(y)\,dx \,dy \\ \quad < M \biggl[ \int_{ - \infty}^{\infty} x_{\alpha}^{p(1 - \delta \sigma ) - 1}f^{p}(x) \,dx \biggr]^{\frac{1}{p}} \biggl[ \int_{ - \infty}^{\infty} y_{\beta}^{q(1 - \sigma ) - 1}g^{q}(y) \,dy \biggr]^{\frac{1}{q}}. \end{gathered} $$(26) -
(iii)
\(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \le M \).
Moreover, if statement (iii) holds, then the constant factor \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \) in (25) and (26) is the best possible.
In particular, (1) for \(\delta = 1\), \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \), we have the following equivalent inequalities with nonhomogeneous kernel:
where \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \) is the best possible constant factor;
(2) for \(\delta = - 1\), \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \), we have the following equivalent inequalities with homogeneous kernel of degree 0:
where \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \) is the best possible constant factor.
Proof
For \(\sigma_{1} = \sigma \), under and the assumption of statement (i), if (24) takes the form of equality for \(y \in \mathbb{R}\backslash \{ 0\} \), then there exist constants A and B such that they are not both zero and (see [28])
We suppose that \(A \ne 0 \) (otherwise, \(B = A = 0 \)). Then it follows that
Since \(\int_{ - \infty}^{\infty} x_{\alpha}^{ - 1} \,dx = \infty \), this contradicts the fact that \(0 < \int_{ - \infty}^{\infty} x_{\alpha}^{p(1 - \delta \sigma ) - 1} f^{p}(x)\,dx < \infty \). Hence (24) takes the form of strict inequality, and so does (21). Hence (25) and (26) are valid.
In view of Theorem 1, we still can conclude that statements (i), (ii), and (iii) in Theorem 2 are equivalent.
When statement (iii) holds, namely, \(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \le M \), if there exists a constant \(M( \le K_{\alpha,\beta}^{(\gamma )}(\sigma )) \) such that (26) is valid, then \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \), and we can conclude that the constant factor \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \) in (26) is the best possible.
The constant factor \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \) in (25) is still the best possible. Otherwise, by (23) (for \(\sigma_{1} = \sigma \)), we would get a contradiction that the constant factor \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \) in (26) is not the best possible.
The theorem is proved. □
4 Operator expressions
We set the following functions: \(\varphi (x): = x_{\alpha}^{p(1 - \delta \sigma ) - 1}\) (\(x \in \mathbb{R} \)), and \(\psi (y): = y_{\beta}^{q(1 - \sigma ) - 1} \), where from \(\psi^{1 - p}(y): = y_{\beta}^{p\sigma - 1}\) (\(y \in \mathbb{R} \)). Define the following real normed linear spaces:
In view of Theorem 2, for \(f \in L_{p,\varphi} (\mathbb{R}) \), setting
by (25) we have
Definition 1
Define the Hilbert-type integral operator \(T:L_{p,\varphi} (\mathbb{R}) \to L_{p,\psi^{1 - p}}(\mathbb{R}) \) as follows: For any \(f \in L_{p,\varphi} (\mathbb{R}) \), there exists a unique representation \(Tf = h_{1} \in L_{p,\psi^{1 - p}}(\mathbb{R}) \), satisfying for any \(y \in \mathbb{R}\), \(Tf(y) = h_{1}(y) \).
In view of (31), it follows that \(\Vert Tf\Vert _{p,\psi^{1 - p}} = \Vert h_{1}\Vert _{p,\psi^{1 - p}} \le M\Vert f\Vert _{p,\varphi} \), and then the operator T is bounded and satisfies
If we define the formal inner product of Tf and g as
then we can rewrite Theorem 2 as follows.
Theorem 3
If M is a constant, then the following statements (i), (ii), and (iii) are equivalent:
-
(i)
For any \(f(x) \ge 0\), \(f \in L_{p,\varphi} (\mathbb{R})\), \(\Vert f\Vert _{p,\varphi} > 0 \), we have:
$$ \Vert Tf \Vert _{p,\psi^{1 - p}} < M \Vert f \Vert _{p,\varphi}. $$(32) -
(ii)
For \(f(x),g(y) \ge 0\), \(f \in L_{p,\varphi} (\mathbb{R})\), \(g \in L_{q,\psi} (\mathbb{R}) \), \(\Vert f\Vert _{p,\varphi},\Vert g\Vert _{q,\psi} > 0 \), we have:
$$ (Tf,g) < M \Vert f \Vert _{p,\varphi} \Vert g \Vert _{q,\psi}. $$(33) -
(iii)
\(K_{\alpha,\beta}^{(\gamma )}(\sigma ) \le M \).
Moreover, if statement (iii) holds, then the constant factor \(M = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \) in (32) and (33) is the best possible, namely, \(\Vert T\Vert = K_{\alpha,\beta}^{(\gamma )}(\sigma ) \).
Remark 1
(1) In particular, for \(\alpha = \beta = 0 \) in (27) and (28), we have the following equivalent inequalities:
where \(\frac{2\Gamma (\sigma /\gamma )}{\gamma \rho^{\sigma /\gamma}} \) is the best possible constant factor.
If \(f( - x) = f(x)\), \(g( - y) = g(y)\) (\(x,y \in \mathbb{R}_{ +} \)), then we have the following equivalent inequalities:
where \(\frac{\Gamma (\sigma /\gamma )}{\gamma \rho^{\sigma /\gamma}} \) is the best possible constant factor.
(2) For \(\alpha = \beta = 0 \) in (29) and (30), we have the following equivalent inequalities:
where \(\frac{2\Gamma (\sigma /\gamma )}{\gamma \rho^{\sigma /\gamma}} \) is the best possible constant factor.
If \(f( - x) = f(x)\), \(g( - y) = g(y)\) (\(x,y \in \mathbb{R}_{ +} \)), then we have the following equivalent inequalities:
where \(\frac{\Gamma (\sigma /\gamma )}{\gamma \rho^{\sigma /\gamma}} \) is the best possible constant factor.
5 Conclusions
In this paper, by using real analysis and weight functions we obtain a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the kernel of exponent function with the intermediate variables (Theorem 1). The constant factor related to the gamma function is proved to be the best possible in Theorem 2. We also consider some particular cases and the operator expressions in Remark 1 and Theorem 3. The lemmas and theorems provide an extensive account of this type of inequalities.
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Funding
This work is supported by the National Natural Science Foundation (Nos. 61562016 and 51765012) and Science and Technology Planning Project Item of Guangzhou City (No. 201707010229). We are grateful for this help.
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BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. YZ and MH participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Zhong, Y., Huang, M. & Yang, B. A Hilbert-type integral inequality in the whole plane related to the kernel of exponent function. J Inequal Appl 2018, 234 (2018). https://doi.org/10.1186/s13660-018-1834-4
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DOI: https://doi.org/10.1186/s13660-018-1834-4