On an extended Hardy-Littlewood-Polya’s inequality

1.   Introduction

Assuming that $p > 1, \tfrac{1}{p} + \tfrac{1}{q} = 1, {a_m}, {b_n} \geqslant 0, \; 0 < \sum\nolimits_{m = 1}^\infty {a_m^p} < {\infty ^{}}~{and~^{}}0 < \sum\nolimits_{n = 1}^\infty {b_n^q} < \infty,$ then we have the following Hardy-Hilbert's inequality with the best possible constant factor $\tfrac{\pi }{{\sin (\pi /p)}}$ :

and have the following Hardy-Littlewood-Polya's inequality:

where the constant factor pq is the best possible (cf ^[1], Theorem 315 and Theorem 341).

In 2006, by introducing parameters ${\lambda _i} \in (0, 2]\; (i = 1, 2), {\lambda _1} + {\lambda _2} = \lambda \in (0, 4],$ an extension of (1) was provided by Krnić and Pečarić ^[2], as follows:

where the constant factor $B({\lambda _1}, {\lambda _2})$ is the best possible ($B(u, v) = \int_0^\infty {\tfrac{{{t^{u - 1}}}}{{{{(1 + t)}^{u + v}}}}} d{t^{}}(u, v > 0)$ is the beta function). For $\lambda = 1, {\lambda _1} = \tfrac{1}{q}, {\lambda _2} = \tfrac{1}{p},$ inequality (3) reduces to inequality (1); for $p = q = 2, {\lambda _1} = {\lambda _2} = \tfrac{\lambda }{2},$ inequality (3) reduces to Yang's work in ^[3]. Recently, by applying inequality (2), a new inequality with the kernel $\tfrac{1}{{{{(m + n)}^\lambda }}}$ involving partial sums was given in ^[4].

If $f(x), g(y) \geqslant 0, \; 0 < \int_0^\infty {{f^p}(x)dx < \infty }$ and $0 < \int_0^\infty {{g^q}(y)dy < \infty }$, then we have the following Hardy -Hilbert's integral inequality (cf. ^[1], Theorem 316):

where the constant factor $\pi /\sin (\tfrac{\pi }{p})$ is the best possible. Inequalities (1), (2) and (3) with their extensions and reverses are important in mathematical analysis and its applications (cf. ^{[5,6,7,8,9,10,11,12,13,14,15]}).

In 1934, a half-discrete Hilbert-type inequality was given as follows (cf. ^[1], Theorem 351):

If $K{(t)^{}}(t > 0)$ is decreasing, $p > 1, \tfrac{1}{p} + \tfrac{1}{q} = 1, 0 < \phi (s) = \int_0^\infty {K(t){t^{s - 1}}} dt < \infty$, then we have

Some new extensions of inequality (5) and their reverses were provided in ^{[16,17,18,19,20]}.

In 2016, by means of the technique of real analysis, Hong and Wen ^[21] considered some equivalent statements of the extensions of (1) with the best possible constant factor related to several parameters. The other similar works concerned with inequalities (2), (4) and (5) were investigated in ^{[22,23,24,25,26,27]}.

In this paper, following the way of ^[2,21], by making use of the weight coefficients, the idea of introducing parameters and Euler-Maclaurin summation formula, an extension of inequality (2) with parameters as well as the equivalent form are provided in Lemma 2 and Theorem 1. The equivalent statements of the best possible constant factor related to several parameters and some particular cases are discussed in Theorem 2 and Remark 2. The operator expressions are considered in Theorem 3.

2.   Some lemmas

In what follows, we assume that $p > {1^{}}(q > 1), \tfrac{1}{p} + \tfrac{1}{q} = 1,$ $\lambda \in (0, 3],$ ${\lambda _i} \in (0, \tfrac{{11}}{8}] \cap (0, \lambda)$ $(i = 1, 2)$. We also assume that ${a_m}, {b_n} \geqslant 0$ $(m, n \in {\rm{N}} = {\rm{\{ 1, 2, }} \cdots {\rm{\})}}$ such that

Lemma 1. Define the weight coefficient:

Then, we have the following inequality:

Proof. For fixed $m \in {\rm{N}}$, we set function ${g_m}(t): = \tfrac{{{t^{{\lambda _2} - 1}}}}{{{{(\max \{ m, t\})}^\lambda }}}(t > 0).$ Thereby we get

(ⅰ) For ${\lambda _2} \in (0, 1],$ by the property of monotone decreasing, we obtain

Thus, in this case the inequality (7) is proved.

(ⅱ) For ${\lambda _2} \in (1, \tfrac{{11}}{8}]$, by using Euler-Maclaurin summation formula (cf. ^[20]), for $\rho (t): = t - [t] - \tfrac{1}{2},$ we have

and then it follows that

in which, for $h({\lambda _2}): = 12 - 10{\lambda _2} + \lambda _2^2,$

Since $h'({\lambda _2}): = - 10 + 2\lambda _2^{} < {0^{}}({\lambda _2} \in (1, \tfrac{{11}}{8}])$, we have

We obtain

On the other hand, we have

and then for $\tfrac{1}{{2{m^\lambda }}} - \tfrac{{{\lambda _2} - 1}}{{12{m^\lambda }}} > \tfrac{1}{{2{m^\lambda }}} - \tfrac{1}{{12{m^\lambda }}} > {0^{}}({\lambda _2} < 2)$, we obtain

Hence, in view of (ⅰ), we still have the inequality (7). This completes the proof of Lemma 1.

Lemma 2. The following extended Hardy-Littlewood-Polya's inequality holds true:

Proof. In the same way as the proof of inequality (7), under the assumption conditions $\lambda \in (0, 3],$ ${\lambda _1} \in (0, \tfrac{{11}}{8}] \cap (0, \lambda)$, we can deduce the following inequality for the weight coefficient:

By Hӧlder's inequality (cf. ^[28]), we obtain

Then by (7) and (9), we obtain inequality (8). The proof of Lemma 2 is complete.

Remark 1. By inequality (8), for, we have

and the following inequality:

Lemma 3. For ${\lambda _1} + {\lambda _2} = \lambda \in (0, \tfrac{{11}}{4}]$, the constant factor $\tfrac{\lambda }{{{\lambda _1}{\lambda _2}}}$ in (10) is the best possible.

Proof. For any $0 < \varepsilon < p{\lambda _1}$, we set

If there exists a constant $M \leqslant$ $\tfrac{\lambda }{{{\lambda _1}{\lambda _2}}}$ such that (10) is valid when replacing $\tfrac{\lambda }{{{\lambda _1}{\lambda _2}}}$ 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 decreasingness property, we obtain

By (9), setting ${\hat \lambda _1} = {\lambda _1} - \tfrac{\varepsilon }{p} \in (0, \tfrac{{11}}{8}) \cap {(0, \lambda)^{}}(0 < {\hat \lambda _2} = {\lambda _2} + \tfrac{\varepsilon }{p} = \lambda - {\hat \lambda _1} < \lambda),$ we get

Then we have

For $\varepsilon \to {0^ + }$, we find $\tfrac{\lambda }{{{\lambda _1}{\lambda _2}}} \leqslant M$. Hence, $M = \tfrac{\lambda }{{{\lambda _1}, {\lambda _2}}}$ is the best possible constant factor of (10). This completes the proof of Lemma 3.

Setting ${\tilde \lambda _1}: = \tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q}, {\tilde \lambda _2}: = \tfrac{{\lambda - {\lambda _1}}}{q} + \tfrac{{{\lambda _2}}}{p}$, we obtain

Thus we can rewrite (8) as follows:

Lemma 4. If inequality (12) is valid and the constant factor in (12) is the best possible, then we have

Proof. Note that

Hence, we have ${k_\lambda }({\tilde \lambda _1}) = \tfrac{\lambda }{{{{\tilde \lambda }_1}(\lambda - {{\tilde \lambda }_1})}} = \tfrac{\lambda }{{{{\tilde \lambda }_1}{{\tilde \lambda }_2}}} \in {{\rm{R}}_ + } = (0, \infty)$.

If the constant factor $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$ in (12) is the best possible, then in view of (10), the unique best possible constant factor must be $\tfrac{\lambda }{{{{\tilde \lambda }_1}{{\tilde \lambda }_2}}} = {k_\lambda }({\tilde \lambda _1})(\in {{\rm{R}}_ + })$, namely, ${k_\lambda }({\tilde \lambda _1})$ $= k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$.

By Hӧlder's inequality, we obtain

We observe that (13) keeps the form of equality if and only if there exist constants $A$ and $B$ (not all zero) such that (cf. ^[28])

Assuming that $A \ne 0$, we have ${u^{\lambda - {\lambda _2} - {\lambda _1}}} = {\tfrac{B}{A}^{}}a.e.{}^{}$ in ${{\rm{R}}_ + }$, and then $\lambda - {\lambda _2} - {\lambda _1} = 0$, namely, $\lambda = {\lambda _1} + {\lambda _2}$. The Lemma 4 is proved.

3.   Main results and some particular cases

Theorem 1. Inequality (8) is equivalent to the following inequality:

If the constant factor in (8) is the best possible, then so is the constant factor in (14).

Proof. Suppose that (14) is valid. By Hӧlder's inequality (cf. ^[28]), we have

Hence by (14), we obtain inequality (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, inequality (14) follows. Hence, inequality (8) is equivalent to (14).

If the constant factor in (8) is the best possible, then so is the constant factor in (14). Otherwise, by (15), we would reach a contradiction that the constant factor in (8) is not the best possible. The proof of Theorem 1 is complete.

Theorem 2. The following statements (ⅰ), (ⅱ), (ⅲ) and (ⅳ) are equivalent:

(ⅰ) $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$ is independent of $p, q$;

(ⅱ) $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$ is expressible as a single integral;

(ⅲ) $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$ in (8) is the best possible constant factor;

(ⅳ) $\lambda = {\lambda _1} + {\lambda _2}.$

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 $\tfrac{\lambda }{{{\lambda _1}{\lambda _2}}}$ :

Proof. (ⅰ) $\Rightarrow$ (ⅱ). By (ⅰ), we have

namely, $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$ is expressible as a single integral

(ⅱ) $\Rightarrow$ (ⅳ). If $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$ is expressible as a convergent single integral ${k_\lambda }(\tfrac{{\lambda - {\lambda _2}}}{p} + \tfrac{{{\lambda _1}}}{q})$, then (13) keeps the form of equality. In view of the proof of Lemma 4, it follows that $\lambda = {\lambda _1} + {\lambda _2}$.

(ⅳ)$\Rightarrow$ (ⅰ). If $\lambda = {\lambda _1} + {\lambda _2}$, then $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1}) = {k_\lambda }({\lambda _1})$, which is independent of $p, q$. Hence, it follows that (ⅰ) $\Leftrightarrow$ (ⅱ) $\Leftrightarrow$ (ⅳ).

(ⅲ) $\Rightarrow$ (ⅳ). By Lemma 4, we have $\lambda = {\lambda _1} + {\lambda _2}$.

(ⅳ)$\Rightarrow$ (ⅲ). By Lemma 3, for $\lambda = {\lambda _1} + {\lambda _2}$, $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})( = \tfrac{\lambda }{{{\lambda _1}{\lambda _2}}})$ is the best possible constant factor of (8). Therefore, we have (ⅲ) $\Leftrightarrow$ (ⅳ).

Hence, the statements (ⅰ), (ⅱ), (ⅲ) and (ⅳ) are equivalent. This completes the proof of Theorem 2.

Remark 2. (ⅰ) For $\lambda = 1, {\lambda _1} = \tfrac{1}{q}, {\lambda _2} = \tfrac{1}{p}$ in (11) and (17), we obtain the inequality (2) and the following equivalent inequality with the best possible constant factor $pq$ :

It follows that (8) and (11) are extensions of (2).

(ⅱ) For $\lambda = 1, {\lambda _1} = \tfrac{1}{p}, {\lambda _2} = \tfrac{1}{q}$ in (11) and (17), we have the following equivalent inequalities with the best possible constant factor $pq$ :

(ⅲ) For $p = q = 2,$ Both inequality (2) and inequality (18) reduce to

moreover, both inequality (17) and inequality (19) reduce to the equivalent form of (20), as follows:

4.   Operator expressions

We set functions

wherefrom, one has

Define the following real normed spaces:

Assuming that $a \in {l_{p, \phi }}$, setting

we can rewrite (14) as follows:

namely, $c \in {l_{p, {\psi ^{1 - p}}}}.$

Definition 2. Define a Hardy-Littlewood-Polya's operator $T:{l_{p, \phi }} \to {l_{p, {\psi ^{1 - p}}}}$ as follows:

For any $a \in {l_{p, \phi }},$ there exists a unique representation $c \in {l_{p, {\psi ^{1 - p}}}}$. Define the formal inner product of $Ta$ and $b \in {l_{q, \psi }}$, and the norm of T as follows:

By Theorem 1 and Theorem 2, we have

Theorem 3. If $a \in {l_{p, \phi }}, b \in {l_{q, \psi }}, ||a|{|_{p, \phi }}, ||b|{|_{q, \psi }} > 0,$ then we have the following equivalent inequalities:

Furthermore, ${\lambda _1} + {\lambda _2} = \lambda$ if and only if the constant factor $k_\lambda ^{\tfrac{1}{p}}({\lambda _2})k_\lambda ^{\tfrac{1}{q}}({\lambda _1})$ in (22) and (23) is the best possible, namely,

5.   Conclusion

Let us give a brief summary of this paper, by applying the weight coefficients, the idea of introducing parameters and Euler-Maclaurin summation formula, an extended Hardy-Littlewood-Polya's inequality and the equivalent form are given in Lemma 2 and Theorem 1. The equivalent statements of the best possible constant factor related to several parameters, and some particular cases are considered in Theorem 2 and Remark 2. The operator expressions are given in Theorem 3. The lemmas and theorems depict some essential characters of this type of inequalities.

Acknowledgments

This work is supported by the National Natural Science Foundation of China (No. 61772140), and the Science and Technology Planning Project Item of Guangzhou City (No. 201707010229). All authors contributed equally and significantly in this paper. The authors are grateful to the reviewers for their valuable comments and suggestions to improve the quality of this paper.

Conflict of interest

The authors declare that they have no competing interests.