Abstract
By the use of the weight coefficients, the idea of introducing parameters and the Euler–Maclaurin summation formula, a reverse extended Hardy–Hilbert inequality and the equivalent forms are given. The equivalent statements of the best possible constant factor related to a few parameters and some particular cases are also considered.
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1 Introduction
Assuming that \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(a_{m},b_{n} \ge 0\), \(0 < \sum_{m = 1}^{\infty } a_{m}^{p} < \infty\) and \(0 < \sum_{n = 1}^{\infty } b_{n}^{q} < \infty \), we have the following Hardy–Hilbert inequality with the best possible constant factor \(\frac{\pi }{\sin (\pi /p)}\) (cf. [1], Theorem 315):
In 2006, by introducing the parameters \(\lambda _{i} \in (0,2]\) (\(i = 1,2\)), \(\lambda _{1} + \lambda _{2} = \lambda \in (0,4]\), an extension of (1) was provided by [2] as follows:
where the constant factor \(B(\lambda _{1},\lambda _{2})\) is the best possible (\(B(u,v) = \int _{0}^{\infty } \frac{t^{u - 1}}{(1 + t)^{u + v}} \,dt\) (\(u,v > 0\)) is the beta function). For \(\lambda = 1\), \(\lambda _{1} = \frac{1}{q}\), \(\lambda _{2} = \frac{1}{p}\), inequality (2) reduces to (1); for \(p = q = 2\), \(\lambda _{1} = \lambda _{2} = \frac{\lambda }{2}\), (2) reduces to Yang’s work in [3]. Recently, applying (2), [4] gave a new inequality with the kernel \(\frac{1}{(m + n)^{\lambda }} \) involving partial sums.
If \(f(x),g(y) \ge 0\), \(0 < \int _{0}^{\infty } f^{p}(x)\,dx < \infty \), and \(0 < \int _{0}^{\infty } g^{q}(y)\,dy < \infty \), then we still have the following Hardy–Hilbert integral inequality (cf. [1], Theorem 316):
where the constant factor \(\pi /\sin (\frac{\pi }{p})\) is the best possible. Inequalities (1) and (3) with their extensions and reverses are important in analysis and its applications (cf. [5–15]).
In 1934, a half-discrete Hilbert-type inequality was given as follows (cf. [1], Theorem 351): If \(K(t)\) (\(t > 0\)) is decreasing, \(p > 1\), \(\frac{1}{p} + \frac{1}{q} = 1\), \(0 < \phi (s) = \int _{0}^{\infty } K(t)t^{s - 1} \,dt < \infty \), \(a_{n} \ge 0\), \(0 < \sum_{n = 1}^{\infty } a_{n}^{p} < \infty \), then we have
In the last ten years, some extensions of (4) with their applications and the reverses were provided by [16–20].
In 2016, by means of the techniques of real analysis, Hong et al. [21] considered some equivalent statements of the extensions of (1) with the best possible constant factor related to a few parameters. Similar work about Hilbert-type integral inequalities is in [22–26].
In this paper, following the way of [2, 21], by the use of the weight coefficients, the idea of introduced parameters and Euler–Maclaurin summation formula, a reverse extended Hardy–Hilbert inequality as well as the equivalent forms are given in Lemma 2 and Theorem 1. The equivalent statements of the best possible constant factor related to a few parameters and some particular cases are considered in Theorem 2 and Remark 1–2.
2 Some lemmas
In what follows, we assume that \(0 < p < 1\) (\(q < 0\)), \(\frac{1}{p} + \frac{1}{q} = 1\), \(\mathrm{N} = \{ 1,2, \ldots \}\), \(\lambda \in (0,6]\), \(\lambda _{i} \in (0,2] \cap (0,\lambda )\) (\(i = 1,2\)),
We also assume that \(a_{m},b_{n} \ge 0\), such that
Lemma 1
Define the following weight coefficient:
We have the following inequality:
Proof
For fixed \(m \in \mathrm{N}\), we set function \(g(m,t): = \frac{t^{\lambda _{2} - 1}}{(m + t)^{\lambda }}\) (\(t > 0\)). Using the Euler–Maclaurin summation formula (cf. [2, 3]), for \(\rho (t): = t - [t] - \frac{1}{2}\), we have
We obtain \(- \frac{1}{2}g(m,1) = \frac{ - 1}{2(m + 1)^{\lambda }} \),
We find
and for \(0 < \lambda _{2} \le 2\), \(\lambda _{2} < \lambda \le 6\), it follows that
Still by the Euler–Maclaurin summation formula (cf. [2, 3]), we obtain
and
Hence, we have \(h(m) > \frac{h_{1}}{(m + 1)^{\lambda }} + \frac{\lambda h_{2}}{(m + 1)^{\lambda + 1}} + \frac{\lambda (\lambda + 1)h_{3}}{(m + 1)^{\lambda + 2}}\), where
and \(h_{3}: = \frac{1}{\lambda _{2}(\lambda _{2} + 1)(\lambda _{2} + 2)} - \frac{\lambda + 2}{720}\).
For \(\lambda \in (0,6]\), \(\frac{\lambda }{720} < \frac{1}{24}\), \(\lambda _{2} \in (0,2]\), we find
In fact, setting \(g(\sigma ): = 24 - 20\sigma + 7\sigma ^{2} - \sigma ^{3}\) (\(\sigma \in (0,2]\)), we obtain
and then
We obtain \(h_{2} > \frac{1}{6} - \frac{1}{12} - \frac{18}{360} = \frac{1}{30} > 0\), and \(h_{3} \ge \frac{1}{24} - \frac{10}{720} = \frac{1}{36} > 0\). Hence, we have \(h(m) > 0\), and then setting \(t = mu\), it follows that
On the other hand, we also have
We have obtained \(\frac{1}{2}g(m,1) = \frac{1}{2(m + 1)^{\lambda }} \) and
For \(\lambda _{2} \in (0,2] \cap (0,\lambda )\), \(0 < \lambda \le 6\), by the Euler–Maclaurin summation formula, we obtain
Hence, we have \(H(m) > \frac{H_{1}}{(m + 1)^{\lambda }} + \frac{\lambda H_{2}(m)}{(m + 1)^{\lambda + 1}}\), where
For \(\lambda _{2} \in (0,2] \cap (0,\lambda )\), \(0 < \lambda \le 6\), we find \(H_{1} > \frac{5}{12} - \frac{42}{720} > 0\), and
It follows that \(H(m) > 0\), and then
By the integral mid-value theorem, we find
namely, (7) follows. □
Lemma 2
We have the following reverse extended Hardy–Hilbert inequality:
Proof
In the same way as obtaining (7), for \(n \in \mathbf{N}\), we obtain the following inequality of the weight coefficient:
By the reverse Hölder inequality (cf. [27]), we obtain
Then, by (7) and (9), in view of \(0 < p < 1\), \(q < 0\), we have (8). □
Remark 1
By (8), for \(\lambda _{1} + \lambda _{2} = \lambda \in (0,4]\), \(0 < \lambda _{i} \le 2\) (\(i = 1,2\)), we find
and the following reverse inequality:
Lemma 3
For any\(\varepsilon > 0\), we have
Proof
There exists a constant \(M > 0\), such that
By the decreasing property of the series, it follows that
Hence, Eq. (11) follows. □
Lemma 4
For\(\lambda _{1} + \lambda _{2} = \lambda \in (0,4]\), the constant factor\(B(\lambda _{1},\lambda _{2})\)in (10) is the best possible.
Proof
For any \(0 < \varepsilon < p\lambda _{1}\), we set
If there exists a constant \(M \ge B(\lambda _{1},\lambda _{2})\), such that (10) is valid when replacing \(B(\lambda _{1},\lambda _{2})\) by M, then in particular, substitution of \(a_{m} = \tilde{a}_{m}\) and \(b_{n} = \tilde{b}_{n}\) in (10), we have
By (11) and the decreasing property of series, we obtain
By (9), setting \(\hat{\lambda }_{1} = \lambda _{1} - \frac{\varepsilon }{p} \in (0,2) \cap (0,\lambda )\) (\(0 < \hat{\lambda }_{2} = \lambda _{2} + \frac{\varepsilon }{p} < \lambda \)), we find
Then we have
For \(\varepsilon \to 0^{ +} \), in view of the continuity of the beta function, we find \(B(\lambda _{1},\lambda _{2}) \ge M\). Hence, \(M = B(\lambda _{1},\lambda _{2})\) is the best possible constant factor of (10). □
Setting \(\tilde{\lambda }_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q}\), \(\tilde{\lambda }_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\), we find
and we can reduce (8) to the following:
Lemma 5
If\(\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))\), the constant factor\(B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\)in (12) is the best possible, then we have\(\lambda - \lambda _{1} - \lambda _{2} = 0\), namely, \(\lambda = \lambda _{1} + \lambda _{2}\).
Proof
For \(\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))\), we obtain
Hence, we have \(B(\tilde{\lambda }_{1},\tilde{\lambda }_{2}) \in \mathrm{R}_{ +} = (0,\infty )\).
If the constant factor \(B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\) in (12) is the best possible, then in view of (10), the unique best possible constant factor must be \(B(\tilde{\lambda }_{1},\tilde{\lambda }_{2})\) (\(\in \mathrm{R}_{ +} \)), namely,
By the reverse Hölder inequality, we find
We observe that (13) keeps the form of equality if and only if there exist constants A and B, such that they are not all zero and (cf. [27])
Assuming that \(A \ne 0\), it follows that
and then \(\lambda - \lambda _{2} - \lambda _{1} = 0\), namely, \(\lambda = \lambda _{1} + \lambda _{2}\). □
3 Main results and some particular cases
Theorem 1
Inequality (8) is equivalent to the following inequalities:
If the constant factor in (8) is the best possible, then so is the constant factor in (14) and (15).
Proof
Suppose that (14) is valid. By the Hölder inequality, we have
Then, by (14), we obtain (8). On the other hand, assuming that (8) is valid, we set
If \(J = \infty \), then (14) is naturally valid; if \(J = 0\), then it is impossible to make (14) valid, namely, \(J > 0\). Suppose that \(0 < J < \infty \). By (8), we have
namely, (14) follows. Hence, inequality (8) is equivalent to (14).
Suppose that (15) is valid. By the Hölder inequality, we have
Then, by (15), we obtain (8). On the other hand, assuming that (8) is valid, we set
If \(J_{1} = \infty \), then (15) is naturally valid; if \(J_{1} = 0\), then it is impossible to make (15) valid, namely, \(J_{1} > 0\). Suppose that \(0 < J_{1} < \infty \). By (8), we have
namely, (15) follows. Hence, inequality (8) is equivalent to (15) and then inequalities (8), (14) and (15) are equivalent.
If the constant factor in (8) is the best possible, then so is the constant factor in (14) and (15). Otherwise, by (16) (or (17)), we would reach a contradiction that the constant factor in (8) is not the best possible. □
Theorem 2
The following statements (i), (ii), (iii) and (iv) are equivalent:
- (i)
\(B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\)is independent ofp, q;
- (ii)
\(B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\)is expressible as a single integral;
- (iii)
\(B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\)in (8) is the best possible constant factor;
- (iv)
if\(\lambda - \lambda _{1} - \lambda _{2} \in ( - p\lambda _{1},p(\lambda - \lambda _{1}))\), then\(\lambda = \lambda _{1} + \lambda _{2}\).
If the statement (iv) follows, namely, \(\lambda = \lambda _{1} + \lambda _{2}\), then we have (10) and the following equivalent inequalities with the best possible constant factor\(B(\lambda _{1},\lambda _{2})\):
Proof
(i) ⇒ (ii). By (i), we have
namely, \(B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\) is expressible as a single integral
(ii) ⇒ (iv). If \(B^{\frac{1}{p}}(\lambda _{2},\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda - \lambda _{1})\) is expressible as a convergent single integral
then (13) keeps the form of equality. In view of the proof of Lemma 5, it follows that \(\lambda = \lambda _{1} + \lambda _{2}\).
(iv) ⇒ (i). If \(\lambda = \lambda _{1} + \lambda _{2}\), then
which is independent of p, q. Hence, it follows that (i) ⇔ (ii) ⇔ (iv).
(iii) ⇒ (iv). By Lemma 5, we have \(\lambda = \lambda _{1} + \lambda _{2}\).
(iv) ⇒ (iii). By Lemma 4, for \(\lambda = \lambda _{1} + \lambda _{2}\),
is the best possible constant factor of (8). Therefore, we have (iii) ⇔ (iv).
Hence, the statements (i), (ii), (iii) and (iv) are equivalent. □
Remark 2
For \(\lambda _{1} = \lambda _{2} = \frac{\lambda }{2} \in (0,2]\) (\(0 < \lambda \le 4\)) in (10), (18) and (19), we have the following equivalent inequalities with the best possible constant factor \(B(\frac{\lambda }{2},\frac{\lambda }{2})\):
In particular, (i) for \(\lambda = 2\), we have the following equivalent inequalities:
(ii) for \(\lambda = 4\), we have the following equivalent inequalities:
4 Conclusions
In this paper, by the use of the weight coefficients, the idea of introducing parameters and the Euler–Maclaurin summation formula, a reverse extended Hardy–Hilbert inequality as well as the equivalent forms are given in Lemma 2 and Theorem 1. The equivalent statements of the best possible constant factor related to a few parameters and some particular cases are considered in Theorem 2 and Remark 1, 2. The lemmas and theorems provide an extensive account of this type of inequalities.
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Acknowledgements
The authors thank the referee for useful proposals to reform the paper.
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This work is supported by the National Natural Science Foundation (No. 61772140), and Science and Technology Planning Project Item of Guangzhou City (No. 201707010229).
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BY carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. ZH and YS participated in the design of the study and performed the numerical analysis. All authors read and approved the final manuscript.
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Huang, Z., Shi, Y. & Yang, B. On a reverse extended Hardy–Hilbert’s inequality. J Inequal Appl 2020, 68 (2020). https://doi.org/10.1186/s13660-020-02333-9
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DOI: https://doi.org/10.1186/s13660-020-02333-9