Abstract
By means of the weight functions, the idea of introducing parameters and the technique of real analysis, a new Hardy–Hilbert-type integral inequality with the homogeneous kernel \(\frac{1}{(x + y)^{\lambda}}\ (\lambda > 0)\) involving two multiple upper-limit functions is obtained. The equivalent statements of the best possible constant factor related to the beta and gamma functions are considered. As applications, the equivalent forms and the case of a nonhomogeneous kernel are deduced. Some particular inequalities and the operator expressions are provided.
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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 \(0 < \sum_{n = 1}^{\infty} b_{n}^{q} < \infty \), the following Hardy–Hilbert inequality with the best possible constant factor \(\pi /\sin (\frac{\pi}{p})\) was provided (cf. [1], Theorem 315):
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 integral analog of (1) named in the Hardy–Hilbert integral inequality as follows (cf. [1], Theorem 316):
where the same constant factor \(\pi /\sin (\frac{\pi}{p})\) is still the best possible. Inequalities (1) and (2) played an important role in analysis and its applications (cf. [2–13]).
In 2006, by means of the Euler–Maclaurin summation formula, Krnic et al. [14] gave an extension of (1) with the kernel \(\frac{1}{(m + n)^{\lambda}}\ (0 < \lambda \le 4)\). Applying the result of [14], in 2019, Adiyasuren et al. [15] considered an extension of (1) involving partial sums, and then in 2020, Mo et al. [16] gave an extension of (2) involving two upper-limit functions. In 2016–2017, Hong et al. [17, 18] provided several equivalent statements of the extensions of (1) and (2) with multiparameters. Some similar results were given by [19–22].
In this paper, following [15] and [17], by means of the weight functions, the idea of introducing parameters and the techniques of real analysis, a new Hardy–Hilbert-type integral inequality with the kernel \(\frac{1}{(x + y)^{\lambda}}\ (\lambda > 0)\) involving two multiple upper-limit functions is provided. The equivalent statements of the best possible constant factor related to the beta and gamma functions are considered. As applications, the equivalent forms, the case of a nonhomogeneous kernel, a few particular inequalities and the operator expressions are deduced. The lemmas and theorems provide an extensive account of this type of inequality.
2 Some lemmas
In what follows, we assume that \(p > 1,\frac{1}{p} + \frac{1}{q} = 1,0 < \lambda _{i} < \lambda\ (i = 1,2),\hat{\lambda}_{1}: = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q},\hat{\lambda}_{2}: = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p},\ F_{0}(x)\) and \(G_{0}(y)\) are nonnegative Lebesgue integrable functions in any interval \((0,b] \subset R_{ +} (b > 0)\), such that \(F_{0}(u),G_{0}(u) = o(e^{tu})\ (t > 0;u \to \infty )\), the multiple upper-limit functions \(\{ F_{j}(x)\}_{j = 1}^{m},\{ G_{k}(y)\}_{k = 1}^{n}\) are defined inductively by \(F_{j}(x): = \int _{0}^{x} F_{j - 1}(t)\,dt (x \ge 0;j = 1,2, \ldots ,m)\), and
where \(F_{1}(x) = \int _{0}^{x} F_{0} (t_{0})\,dt{}_{0}, G_{1}(y) = \int _{0}^{y} G_{0} (t_{0})\,dt{}_{0}\),
For \(m,n \in \mathrm{N}_{0}: = \{ 0,1, \ldots \} \), we also suppose that \(F_{m}(x)\) and \(G_{n}(y)\) satisfy the following inequalities:
We indicate the gamma function as follows:
Satisfying \(\Gamma (\alpha + 1) = \alpha \Gamma (\alpha )(\alpha > 0)\), and define the following beta function (cf. [23]):
By (3), for \(\lambda ,x,y > 0\), we still have the following formula related the gamma function:
Lemma 1
For \(t > 0, m,n \in \mathrm{N}\), we have the following expressions:
Proof
For \(n \in \mathrm{N}: = \{ 1,2, \ldots \} \), since \(F_{1}(0) = 0\), using integration by parts, we find
If \(F{}_{1}(\infty ) =\) constant, then we have \(\lim_{x \to \infty} \frac{F_{1}(x)}{e^{tx}} = 0\); if \(F_{1}(\infty ) = \infty \), since \(F_{0}(x) = o(e^{tx})\ (t > 0;x \to \infty )\), then we obtain that \(\lim_{x \to \infty} \frac{F_{1}(x)}{e^{tx}} = \lim_{x \to \infty} \frac{F_{0}(x)}{te^{tx}} = 0\). Hence, we find
In the same way, we can obtain that for \(F_{i}(\infty ) = \infty\ (i = 1, \ldots ,k),\lim_{x \to \infty} \frac{F_{k}(x)}{e^{tx}} = \cdots \ =\lim_{x \to \infty} \frac{F_{0}(x)}{t^{k}e^{tx}} = 0\), and then
Hence, we obtain (6) inductively. In the same way, we have (7).
The lemma is proved. □
Lemma 2
Define the following weight functions:
We have the following expressions:
Proof
Setting \(u = \frac{t}{x}\), we find
namely, (10) follows. In the same way, we obtain (11).
The lemma is proved. □
Lemma 3
We have the following Hardy–Hilbert integral inequality:
Proof
By Hölder’s inequality (cf. [24]), we obtain
If (13) keeps the form of equality, then there exist constants A and B, such that they are not both zero, satisfying
Assuming that \(A \ne 0\), for fixed \(a.e.y \in (0,\infty )\), we have
Since for any \(a = \lambda - \lambda _{1} - \lambda _{2} \in \mathbf{R}\), \(\int _{0}^{\infty} x^{ - 1 - a}\,dx = \infty \), the above expression contradicts the fact that
Therefore, by (10), (11), and (13), we have (14).
The lemma is proved. □
3 Main results
Theorem 1
We have the following Hardy–Hilbert-type integral inequality involving two multiple upper-limit functions:
In particular, for \(\lambda _{1} + \lambda _{2} = \lambda \), we reduce (14) to the following:
where, the constant factor \(\frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B(\lambda _{1} + m,\lambda _{2} + n)\) is the best possible.
Proof
In view of (6), (7), and Fubini’s theorem (cf. [25]), we find
For \(\lambda _{1} + \lambda _{2} = \lambda \) in (14), by simplification, we have (15). By (3) and (4), we still have
where, we define \(\prod_{i = 1}^{0} (\lambda _{1} + i - 1) = \prod_{j = 1}^{0} (\lambda _{2} + j - 1): = 1\).
For any \(0 < \varepsilon < \min \{ p\lambda _{1},q\lambda _{2}\}\), we set the following functions:
We obtain that \(\tilde{F}_{0}(u) = \tilde{G}_{0}(u) = o(e^{tu})\ (t > 0;u \to \infty )\),
In general, by mathematical induction, we can show the following inequalities:
If there exists a positive constant \(M \le B(\lambda _{1},\lambda _{2})\prod_{i = 1}^{m} (\lambda _{1} + i - 1) \prod_{j = 1}^{n} (\lambda _{2} + j - 1)\), such that (15) is valid when we replace \(B(\lambda _{1},\lambda _{2})\prod_{i = 1}^{m} (\lambda _{1} + i - 1) \prod_{j = 1}^{n} (\lambda _{2} + j - 1)\), by M, then in particular, since
we have
In view of Fubini’s theorem (cf. [25]), it follows that
Hence, by (18) and the above results, it follows that
Putting \(\varepsilon \to 0^{ +} \) in the above inequality, in view of the continuity of the beta function, we find
namely, \(B(\lambda _{1},\lambda _{2})\prod_{i = 1}^{m} (\lambda _{1} + i - 1) \prod_{j = 1}^{n} (\lambda _{2} + j - 1) \le M\), and then
is the best possible constant factor in (15).
The theorem is proved. □
Remark 1
For \(\hat{\lambda}_{1} = \frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q},\hat{\lambda}_{2} = \frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p}\), it follows that \(\hat{\lambda}_{1} + \hat{\lambda}_{2} = \lambda \). We find \(0 < \hat{\lambda}_{1} < \frac{\lambda}{p} + \frac{\lambda}{q} = \lambda \), and then \(0 < \hat{\lambda}_{2} = \lambda - \hat{\lambda}_{1} < \lambda \). By Hölder’s inequality (cf. [24]), we can obtain
Theorem 2
If the constant factor
in (14) is the best possible, then we have \(\lambda _{1} + \lambda _{2} = \lambda \).
Proof
By (15) (for \(\lambda _{i} = \hat{\lambda}_{i}\ (i = 1,2)\)), since
is the best possible constant factor in (15), we have
namely,
It follows that (19) keeps the form of equality.
We observe that (19) keeps the form of equality if and only if there exist constants A and B, such that they are not both zero and
(cf. [24]). Assuming that \(A \ne 0\), it follows that \(u^{\lambda - \lambda _{2} - \lambda _{1}} = \frac{B}{A}\text{ a.e. in }R_{ +} \), namely, \(\lambda - \lambda _{1} - \lambda _{2} = 0\), and then \(\lambda _{1} + \lambda _{2} = \lambda \).
The theorem is proved. □
Theorem 3
The following statements (i), (ii), (iii), and (iv) are equivalent:
-
(i)
Both \(B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1})\) and \(B(\frac{\lambda - \lambda _{2}}{p} + \frac{\lambda _{1}}{q} + m,\frac{\lambda - \lambda _{1}}{q} + \frac{\lambda _{2}}{p} + n)\) are independent of \(p,q\);
-
(ii)
\(B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) = B(\hat{\lambda}_{1} + m,\hat{\lambda}_{2} + n)\);
-
(iii)
\(\lambda _{1} + \lambda _{2} = \lambda \);
-
(iv)
The constant factor
$$\begin{aligned} \frac{\Gamma (\lambda + m + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) \end{aligned}$$in (14) is the best possible.
Proof
(i) ⇒ (ii). In view of the continuity of the beta function. we find
Hence, we have
(ii) ⇒ (iii). Suppose that \(B^{\frac{1}{p}}(\lambda _{2} + n,\lambda + m - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1} + m,\lambda + n - \lambda _{1}) = B(\hat{\lambda}_{1} + m,\hat{\lambda}_{2} + n)\). Then, (19) keeps the form of equality. In view of the proof of Theorem 2, we have \(\lambda _{1} + \lambda _{2} = \lambda \).
(iii) ⇒ (iv). If \(\lambda _{1} + \lambda _{2} = \lambda \), then by Theorem 1, the constant factor
in (14) is the best possible.
(iv) ⇒ (i). By Theorem 2, we have \(\lambda _{1} + \lambda _{2} = \lambda \), and then
Both of them are independent of \(p,q\).
Hence, the statements (i), (ii), (iii), and (iv) are equivalent.
The theorem is proved. □
Remark 2
(i) For \(\lambda _{1} + \lambda _{2} = \lambda \) in (12), we have
We confirm that the constant factor \(B(\lambda {}_{1} + m,\lambda _{2} + n)\) in (20) is the best possible. Otherwise, we would reach a contradiction by (16) (for \(\lambda _{1} + \lambda _{2} = \lambda \)) that the constant factor in (15) is not the best possible.
(ii) For \(m = n = 0,\lambda = 1,\lambda _{1} = \frac{1}{q},\lambda _{2} = \frac{1}{p}\), both (20) and (15) reduce to (2).
4 Equivalent forms and some particular inequalities
For \(m = 0\) in (14), we have
In particular, for \(\lambda _{1} + \lambda _{2} = \lambda \), we have
where the constant factor \(\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)\) is the best possible.
Theorem 4
Inequality (21) is equivalent to the following:
In particular, for \(\lambda _{1} + \lambda _{2} = \lambda \), we reduce (23) to the equivalent form of (22) as follows:
where the constant factor \(\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)\) is the best possible.
Proof
Suppose that (19) is valid. By Hölder’s inequality, we have
On the other hand, assuming that (21) is valid, we set
If \(J = 0\), then, (23) is naturally valid; if \(J = \infty \), then it is impossible that (23) is valid, namely \(J < \infty \). Suppose that \(0 < J < \infty \). By (21), we have
namely, (23) follows, which is equivalent to (21).
The constant factor \(\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)\) in (24) is the best possible. Otherwise, by (25) (for \(\lambda _{1} + \lambda _{2} = \lambda \)), we would reach a contradiction that the constant factor in (22) is not the best possible.
The theorem is proved. □
Replacing x by \(\frac{1}{x}\), then replacing \(x^{\lambda - 2}F_{0}(\frac{1}{x})\) by \(F_{0}(x)\) in (21) and (23), by simplification, we have
Corollary 1
The following Hardy–Hilbert-type integral inequalities with a nonhomogeneous kernel are equivalent:
Moreover, \(\lambda _{1} + \lambda _{2} = \lambda \) if and only if the constant factor
in (26) and (27) is the best possible.
For \(\lambda _{1} + \lambda _{2} = \lambda \), we have the following equivalent inequalities with the best possible constant factor \(\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B(\lambda _{1},\lambda _{2} + n)\):
Remark 3
(i) For \(\lambda = 1,\lambda _{1} = \frac{1}{q},\lambda _{2} = \frac{1}{p}\) in (22), (24), (28), and (29), we have the following two couples of equivalent integral inequalities:
(ii) For \(\lambda = 1,\lambda _{1} = \frac{1}{p},\lambda _{2} = \frac{1}{q}\) in (22), (24), (28), and (29), we have the dual forms of (30)–(33) as follows:
(iii) For \(p = q = 2,(2n - 1)!!: = 1 \cdot 3 \cdot \cdots \cdot (2n - 1)\), both (30) and (34) reduce to
both (31) and (35) reduce to the following equivalent inequality of (38):
and both (33) and (37) reduce to the following equivalent inequality of (40):
The constant factor in the above inequalities (30)–(41) are all the best possible.
5 Operator expressions
We set functions
hence, \(\phi ^{1 - q}(x) = x^{q\hat{\lambda}_{1} - 1}(x,y \in \mathrm{R}_{ +} )\). Define the following real normed spaces:
Assuming that \(G_{0}(y)\) is a nonnegative Lebesgue integrable function in any interval \((0,b] \subset R_{ +} (b > 0)\),
setting
we can rewrite (23) as follows:
namely, \(h \in L_{q,\phi ^{1 - q}}(\mathrm{R}_{ +} )\).
Definition 1
Define a Hardy–Hilbert-type operator \(T:\tilde{L}(\mathrm{R}_{ +} ) \to L_{q,\phi ^{1 - q}}(\mathrm{R}_{ +} )\) as follows: For any \(G_{0} \in \tilde{L}(\ \mathrm{R}_{ +} ),\ G_{0}(y)\) is a nonnegative Lebesgue integrable function in any interval \((0,b] \subset R_{ +} (b > 0)\), there exists a unique representation \(h \in L_{q,\phi ^{1 - q}}(\mathrm{R}_{ +} )\), satisfying (42). Define the formal inner product of \(F_{0} \in L_{p,\phi} (\mathrm{R}_{ +} )\) and \(TG_{0}\), and the norm of T as follows:
By Theorem 4, we have
Theorem 5
If \(F_{0} \in L_{p,\phi} (\mathrm{R}_{ +} ),G_{0} \in \tilde{L}(\mathrm{R}_{ +} ),\Vert F_{0}\Vert _{p,\phi},\Vert G_{n}\Vert _{q,\psi} > 0\), (\(G_{0}(y)\) is a nonnegative Lebesgue integrable function in any interval \((0,b] \subset R_{ +} (b > 0)\)), then we have the following equivalent inequalities:
Moreover, \(\lambda _{1} + \lambda _{2} = \lambda \) if and only if the constant factor \(\frac{\Gamma (\lambda + n)}{\Gamma (\lambda )}B^{\frac{1}{p}}(\lambda _{2} + n,\lambda - \lambda _{2})B^{\frac{1}{q}}(\lambda _{1},\lambda + n - \lambda _{1})\) in (42) and (43) is the best possible, namely,
We also set functions \(\varphi (x): = x^{p(1 - \hat{\lambda}_{2}) - 1}\), hence,
and define the following real normed spaces:
Assuming that \(G_{0} \in \tilde{L}(\mathrm{R}_{ +} )\), setting
we can rewrite (27) as follows:
namely, \(H \in L_{q,\varphi ^{1 - q}}(\mathrm{R}_{ +} )\).
Definition 2
Define a Hardy–Hilbert-type operator \(T_{1}:\tilde{L}(\mathrm{R}_{ +} ) \to L_{q,\varphi ^{1 - q}}(\mathrm{R}_{ +} )\) as follows: For any \(G_{0} \in \tilde{L}(\ \mathrm{R}_{ +} )\), there exists a unique representation \(H \in L_{q,\varphi ^{1 - q}}(\mathrm{R}_{ +} )\). Define the formal inner product of \(F_{0} \in L_{p,\varphi} (\mathrm{R}_{ +} )\) and \(T_{1}G_{0}\), and the norm of \(T_{1}\) as follows:
By Corollary 1, we have
Corollary 2
If \(F_{0} \in L_{p,\varphi} (\mathrm{R}_{ +} ),G_{0} \in \tilde{L}(\mathrm{R}_{ +} ),\Vert F_{0}\Vert _{p,\varphi},\Vert G_{n}\Vert _{q,\psi} > 0\), then we have the following equivalent inequalities:
Moreover, \(\lambda _{1} + \lambda _{2} = \lambda \) if and only if the constant factor
in (44) and (45) is the best possible, namely,
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Acknowledgements
The authors thank the referee for his useful proposal to reform the paper.
Funding
This work is supported by the National Natural Science Foundation (Nos. 11961021,11561019), the Hechi University Research Foundation for Advanced Talents under Grant (2021GCC024), and the Characteristic Innovation Project of Guangdong Provincial Colleges and Universities (No. 2020KTSCX088). We are grateful for this help.
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B.Y. carried out the mathematical studies, participated in the sequence alignment and drafted the manuscript. R.L. and L.H. participated in the design of the study and performed the numerical analysis. All authors reviewed the manuscript.
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Luo, R., Yang, B. & He, L. A Hardy–Hilbert-type integral inequality involving two multiple upper-limit functions. J Inequal Appl 2023, 19 (2023). https://doi.org/10.1186/s13660-023-02931-3
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DOI: https://doi.org/10.1186/s13660-023-02931-3