Abstract
By using the way of real analysis and estimating the weight functions, we build a new Hilbert-type integral inequality in the whole plane with a non-homogeneous kernel and a few parameters. The constant factor related to the beta function is proved to be the best possible. We also consider the equivalent forms, the reverses, and some particular cases.
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1 Introduction
If \(f(x),g(y)\geq0\), satisfying \(0<\int_{0}^{\infty}f^{2}(x)\, dx<\infty\) and \(0<\int_{0}^{\infty}g^{2}(y)\, dy<\infty\), then we have (cf. [1])
where the constant factor π is the best possible. Inequality (1) is known as Hilbert’s integral inequality, which is important in analysis and its applications (cf. [1, 2]).
In recent years, by using the way of weight functions, a number of extensions of (1) were given by Yang (cf. [3]). Noticing that inequality (1) is a homogeneous kernel of degree −1, in 2009, A survey of the study of Hilbert-type inequalities with the homogeneous kernels of degree negative numbers and some parameters is given by [4]. Recently, some inequalities with the homogeneous kernels of degree 0 and non-homogeneous kernels have been studied (cf. [5–10]). All of the above integral inequalities are built in the quarter plane.
In 2007, Yang [11] first gave a Hilbert-type integral inequality in the whole plane as follows:
where the constant factor \(B(\frac{\lambda}{2},\frac{\lambda }{2})\) (\(\lambda >0\)) is the best possible, and
is the beta function (cf. [12]). He et al. [13–24] also provided some Hilbert-type integral inequalities in the whole plane.
In this paper, by using the way of real analysis and estimating the weight functions, we build a new Hilbert-type integral inequality in the whole plane with the non-homogeneous kernel and a few parameters. The constant factor related to the beta function is proved to be the best possible. We also consider the equivalent forms, the reverses, and some particular cases.
2 Some lemmas
Lemma 1
Suppose that \(0<\alpha_{1}\leq\alpha_{2}<\pi\), \(\mu ,\sigma>0\), \(\mu+\sigma=\lambda\), \(\gamma\in\{\frac{1}{2k+1},2k-1\ (k\in \mathbf{N})\}\), \(\delta\in\{-1,1\}\). We define weight functions \(\omega (\sigma,y)\) (\(y\in\mathbf{R}\)), and \(\varpi(\sigma,x)\) (\(x\in \mathbf{R}\)) as follows:
Then for \(y,x\in\mathbf{R}\backslash\{0\}\), we have
Proof
(i) For \(\delta=1\), \(y\in\mathbf{R}\backslash\{0\}\), setting \(u=xy\), we find
Setting \(t=u^{\gamma}(1+\cos\alpha_{1})\) (\(t=u^{\gamma}(1-\cos \alpha _{2})\)) in the above first (second) integral, by (3), it follows that
(ii) For \(\delta=-1\), setting \(\frac{y}{x}\), we still can obtain \(\omega (\sigma,y)=K(\sigma)\).
Setting \(u=x^{\delta}y\), we also find
Hence we have (6). □
Note
If we replace \(\min_{i\in\{1,2\}}\) by \(\max_{i\in\{ 1,2\}}\) in (4) and (5), then we may exchange \(\alpha_{1}\) and \(\alpha_{2}\) in (6).
Lemma 2
Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\alpha _{1}\leq\alpha_{2}<\pi\), \(\mu,\sigma>0\), \(\mu+\sigma=\lambda\), \(\gamma\in \{\frac{1}{2k+1},2k-1\ (k\in\mathbf{N})\}\), \(\delta\in\{-1,1\}\). If \(K(\sigma)\) is indicated by (6), \(f(x)\) is a non-negative measurable function in \((-\infty,\infty)\), then we have
Proof
We set
By Hölder’s inequality (cf. [25]), we have
Then by (6) and the Fubini theorem (cf. [26]), it follows that
3 Main results and applications
Theorem 1
Suppose that \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(0<\alpha _{1}\leq\alpha_{2}<\pi\), \(\mu,\sigma>0\), \(\mu+\sigma=\lambda\), \(\gamma\in \{\frac{1}{2k+1},2k-1\ (k\in\mathbf{N})\}\), \(\delta\in\{-1,1\}\). If \(K(\sigma)\) is indicated by (6), \(f(x),g(y)\geq0\), satisfying \(0<\int_{-\infty}^{\infty}|x|^{p(1-\delta\sigma)-1}f^{p}(x)\,dx<\infty\) and \(0<\int_{-\infty}^{\infty}|y|^{q(1-\sigma)-1}g^{q}(y)\,dy<\infty\), then we have the following equivalent inequalities:
where the constant factors \(K(\sigma)\) and \(K^{p}(\sigma)\) are the best possible.
In particular, for \(\alpha_{1}=\alpha_{2}=\alpha\in(0,\pi)\), \(\gamma=1\) in (11) and (12), we find
and the following equivalent inequalities:
Proof
If (10) takes the form of equality for \(y\in(-\infty,0)\cup(0,\infty)\), then there exist constants A and B, such that they are not all zero, and
We suppose \(A\neq0\) (otherwise \(B=A=0\)). Then it follows that
which contradicts the fact that \(0<\int_{-\infty}^{\infty }|x|^{p(1-\delta \sigma)-1}f^{p}(x)\,dx<\infty\). Hence (10) takes the form of a strict inequality. So does (9), and we have (12).
By Hölder’s inequality (cf. [25]), we find
Then by (12), we have (11). On the other hand, suppose that (11) is valid. Setting
then it follows that \(J=\int_{-\infty}^{\infty}|y|^{q(1-\sigma )-1}g^{q}(y)\,dy\). By (9), we have \(J<\infty\). If \(J=0\), then (12) is obviously of value; if \(0< J<\infty\), then by (11), we obtain
Hence we have (12), which is equivalent to (11).
We set \(E_{\delta}:=\{x\in\mathbf{R};|x|^{\delta}\geq1\}\), and \(E_{\delta}^{+}:=E_{\delta}\cap\mathbf{R}_{+}=\{x\in\mathbf{R}_{+};x^{\delta}\geq1\}\). For \(\varepsilon>0\), we define functions \(\tilde{f}(x)\), \(\tilde{g}(y)\) as follows:
Then we obtain
We find
In fact, setting \(Y=-y\), we obtain
It follows that
Setting \(v=x^{\delta}\) in the above integral, by the Fubini theorem (cf. [26]), we find
If the constant factor \(K(\sigma)\) in (11) is not the best possible, then there exists a positive number k, with \(K(\sigma)< k\), such that (11) is valid when replacing \(K(\sigma)\) by k. Then we have \(\varepsilon \tilde{I}<\varepsilon k\tilde{L}\), and
By (7) and the Levi theorem (cf. [26]), we have
which contradicts the fact that \(k< K(\sigma)\). Hence the constant factor \(K(\sigma)\) in (11) is the best possible.
If the constant factor in (12) is not the best possible, then by (16), we may get a contradiction: that the constant factor in (11) is not the best possible. □
Theorem 2
As the assumptions of Theorem 1, replacing \(p>1\) by \(0< p<1\), we have the equivalent reverses of (11) and (12) with the same best constant factors.
Proof
By the reverse Hölder’s inequality (cf. [25]), we have the reverses of (9) and (16). It is easy to obtain the reverse of (12). In view of the reverses of (12) and (16), we obtain the reverse of (11). On the other hand, suppose that the reverse of (11) is valid. Setting the same \(g(y)\) as Theorem 1, by the reverse of (9), we have \(J>0\). If \(J=\infty\), then the reverse of (12) is obviously value; if \(J<\infty\), then by the reverse of (11), we obtain the reverses of (17) and (18). Hence we have the reverse of (12), which is equivalent to the reverse of (11).
If the constant factor \(K(\sigma)\) in the reverse of (11) is not the best possible, then there exists a positive constant k, with \(k>K(\sigma)\), such that the reverse of (11) is still valid when replacing \(K(\sigma)\) by k. By the reverse of (19), we have
For \(\varepsilon\rightarrow0^{+}\), by the Levi theorem (cf. [26]), we find that
There exists a constant \(\delta_{0}>0\), such that \(\sigma-\frac{1}{2}\delta_{0}>0\), and then \(K(\sigma-\frac{\delta_{0}}{2})<\infty\). For \(0<\varepsilon<\frac{\delta_{0}|q|}{4}\) (\(q<0\)), since \(u^{\sigma+\frac{2\varepsilon}{q}-1}\leq u^{\sigma-\frac{\delta_{0}}{2}-1}\), \(u\in(0,1]\), and
then by the Lebesgue control convergence theorem (cf. [26]), for \(\varepsilon\rightarrow0^{+}\), we have
By (20), (21), and (22), for \(\varepsilon\rightarrow 0^{+}\), we find \(K(\sigma)\geq k\), which contradicts the fact that \(k>K(\sigma)\). Hence, the constant factor \(K(\sigma)\) in the reverse of (11) is the best possible.
If the constant factor in reverse of (12) is not the best possible, then by the reverse of (16), we may get a contradiction that the constant factor in the reverse of (11) is not the best possible. □
Remarks
For \(\delta=-1\) in (11) and (12), replacing \(|x|^{\lambda}f(x)\) by \(f(x)\), we obtain the following equivalent inequalities with the homogeneous kernel and the best possible constant factors:
In particular, for \(\alpha_{1}=\alpha_{2}=\alpha\in(0,\pi)\), \(\gamma=1\) in (23) and (24), we obtain the following equivalent inequalities:
where \(k(\sigma)\) is indicated by (13).
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Acknowledgements
The authors wish to express their thanks to the referees for their careful reading of the manuscript and for their valuable suggestions. This work is supported by the National Natural Science Foundation (No. 61370186), and 2013 Knowledge Construction Special Foundation Item of Guangdong Institution of Higher Learning College and University (No. 2013KJCX0140).
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BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. ZG participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Gu, Z., Yang, B. A Hilbert-type integral inequality in the whole plane with a non-homogeneous kernel and a few parameters. J Inequal Appl 2015, 314 (2015). https://doi.org/10.1186/s13660-015-0844-8
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DOI: https://doi.org/10.1186/s13660-015-0844-8