Abstract
By means of weight coefficients and techniques of real analysis, a new Hardy-Hilbert-type inequality with multiparameters and the best possible constant factor is given. The equivalent forms, the operator expression with the norm, and the reverse and some particular inequalities with the best possible constant factors are also considered.
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1 Introduction
If \(p>1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(f(x),g(y)\geq0\), \(f\in L^{p}(\mathbf{R}_{+})\), \(g\in L^{q}(\mathbf{R}_{+})\), \(\|f\|_{p}=(\int_{0}^{\infty }f^{p}(x)\,dx)^{\frac{1}{p}}>0\), \(\|g\|_{q}>0\), then we have the following Hardy-Hilbert integral inequality [1]:
where, the constant factor \(\frac{\pi}{\sin(\pi/p)}\) is the best possible. Assuming that \(a_{m},b_{n}\geq0\), \(a=\{a_{m}\}_{m=1}^{\infty }\in l^{p}\), \(b=\{b_{n}\}_{n=1}^{\infty}\in l^{q}\), \(\|a\|_{p}=(\sum_{m=1}^{\infty }a_{m}^{p})^{\frac{1}{p}}>0\), \(\|b\|_{q}>0\), we have the following Hardy-Hilbert inequality with the same best possible constant factor \(\frac{\pi}{\sin(\pi/p)}\) [1]:
Inequalities (1) and (2) are important in analysis and its applications [1–5].
If \(\mu_{i},\upsilon_{j}>0 \) (\(i,j\in\mathbf{N}=\{1,2,\ldots\}\)),
then we have the following inequality (see Theorem 321 of [1]):
Replacing \(\mu_{m}^{1/q}a_{m}\) and \(\upsilon_{n}^{1/p}b_{n}\) by \(a_{m}\) and \(b_{n}\) in (4), respectively, we have the following equivalent form of (4):
For \(\mu_{i}=\upsilon_{j}=1\) (\(i,j\in\mathbf{N}\)), both (4) and (5) reduce to (2). We call (4) and (5) Hardy-Hilbert-type inequalities.
Note
The authors of [1] did not prove that (4) is valid with the best possible constant factor.
In 1998, by introducing an independent parameter \(\lambda\in(0,1]\) Yang [6] gave an extension of (1) with the kernel \(\frac{1}{(x+y)^{\lambda}}\) for \(p=q=2\). Following the results of [6], Yang [5] gave some best extensions of (1) and (2) as follows.
If \(\lambda_{1},\lambda_{2}\in\mathbf{R}\), \(\lambda_{1}+\lambda _{2}=\lambda\), \(k_{\lambda}(x,y)\) is a nonnegative homogeneous function of degree −λ with \(k(\lambda_{1})=\int_{0}^{\infty}k_{\lambda }(t,1)t^{\lambda_{1}-1}\,dt\in\mathbf{R}_{+}\), \(\phi(x)=x^{p(1-\lambda _{1})-1}\), \(\psi(x)=x^{q(1-\lambda_{2})-1}\), \(f(x),g(y)\geq0\),
\(g\in L_{q,\psi}(\mathbf{R}_{+})\), \(\|f\|_{p,\phi},\|g\|_{q,\psi}>0\), then
where the constant factor \(k(\lambda_{1})\) is the best possible. Moreover, if \(k_{\lambda}(x,y)\) is finite and \(k_{\lambda}(x,y)x^{\lambda _{1}-1} \) (\(k_{\lambda}(x,y)y^{\lambda_{2}-1}\)) is decreasing with respect to \(x>0 \) (\(y>0\)), then for \(a_{m},b_{n}\geq0\),
\(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\psi}\), \(\|a\|_{p,\phi },\|b\|_{q,\psi }>0\), it follows that
where the constant factor \(k(\lambda_{1})\) is still the best possible.
Clearly, for \(\lambda=1\), \(k_{1}(x,y)=\frac{1}{x+y}\), \(\lambda _{1}=\frac{1}{q}\), \(\lambda_{2}=\frac{1}{p}\), inequality (6) reduces to (1), whereas (7) reduces to (2). For \(0<\lambda_{1},\lambda _{2}\leq1\), \(\lambda_{1}+\lambda_{2}=\lambda\), we set
Then by (7) it follows that
where the constant \(B(\lambda_{1},\lambda_{2})\) is the best possible. Some other results including multidimensional Hilbert-type inequalities are provided in [7–24].
In this paper, by means of weight coefficients and techniques of real analysis, a new Hardy-Hilbert-type inequality with multiparameters and the best possible constant factor is given, which is with the kernel
similar to (4). The equivalent forms, the operator expression with the norm, the reverse and some particular inequalities with the best possible constant factors are also considered.
2 An example and some lemmas
In the following, we make appointment that \(\mu_{i},\upsilon _{j}>0\) (\(i,j\in\mathbf{N}\)), \(U_{m}\) and \(V_{n}\) are defined by (3), \(p\neq0,1\), \(\frac{1}{p}+\frac{1}{q}=1\), \(a_{m},b_{n}\geq0\) (\(m,n\in \mathbf{N}\)),
where
We also set
Note
For \(0< p<1\) or \(p<0\), we still use the formal symbols \(\|a\|_{p,\Phi_{\lambda}}\), \(\|b\|_{q,\Psi_{\lambda}}\), \(\|a\|_{p,\widetilde{\Phi}_{\lambda}}\), and \(\|b\|_{q,\widetilde{\Psi }_{\lambda}}\).
Example 1
For \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha\), \(\lambda_{1}+\lambda_{2}=\lambda\), we set
We find
Since
for \(\lambda_{2}\leq1-\alpha^{{}} \) (\(\lambda_{1}>-\alpha\)), \(k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}\) is decreasing for \(y>0\) and strictly decreasing for y large enough. Since
for \(\lambda_{1}\leq1-\alpha\) (\(\lambda_{2}>-\alpha\)), \(k_{\lambda}(x,y)\frac{1}{x^{1-\lambda_{1}}}\) is decreasing for \(x>0\) and strictly decreasing for x large enough.
In other words, for \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha \), \(k_{\lambda}(x,y)\frac{1}{y^{1-\lambda_{2}}}\) (\(k_{\lambda }(x,y)\frac{1}{x^{1-\lambda_{1}}}\)) is decreasing for \(y>0\) (\(x>0\)) and strictly decreasing for \(y(x)\) large enough, satisfying \(k(\lambda_{1})\in \mathbf{R}_{+}\).
Lemma 1
If \(g(t) \) (>0) is decreasing in \(\mathbf{R}_{+}\), strictly decreasing in \([n_{0},\infty)\) (\(n_{0}\in\mathbf{N}\)), and satisfying \(\int_{0}^{\infty}g(t)\,dt\in\mathbf{R}_{+}\), then we have
Proof
Since
it follows that
In the same way, we have
Adding these two inequalities, we have (10). □
Lemma 2
Let \(-\alpha<\lambda_{1},\lambda_{2}\leq 1-\alpha\), \(\lambda_{1}+\lambda_{2}=\lambda\), and \(k(\lambda_{1})\) be as in (9). Define the following weight coefficients:
Then, we have the following inequalities:
Proof
We set \(\mu(t):=\mu_{m}\), \(t\in(m-1,m] \) (\(m\in\mathbf{N}\)); \(\upsilon(t):=\upsilon_{n}\), \(t\in(n-1,n]\) (\(n\in\mathbf{N}\)), and
Then by (3) it follows that \(U(m)=U_{m}\), \(V(n)=V_{n} \) (\(m,n\in \mathbf{N}\)). For \(x\in(m-1,m)\), \(U^{\prime}(x)=\mu(x)=\mu_{m}\) (\(m\in \mathbf{N}\)); for \(y\in(n-1,n)\), \(V^{\prime}(y)=\upsilon(y)=\upsilon _{n} \) (\(n\in\mathbf{N}\)). Since \(V(y)\) is strictly increasing in \((n-1,n]\), \(-\alpha<\lambda_{2}\leq1-\alpha\), \(\lambda_{1}>-\alpha\), in view of Example 1 and Lemma 1, we find
Setting \(t=\frac{V(y)}{U_{m}}\), we obtain \(V^{\prime}(y)\,dy=U_{m}\,dt\) and
Hence, we have (13). In the same way, we have (14). □
Lemma 3
Let \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(\lambda_{1}+\lambda_{2}=\lambda \), \(k(\lambda_{1})\) be as in (9), \(m_{0},n_{0}\in\mathbf{N}\), \(\mu _{m}\geq\mu_{m+1}\) (\(m\in\{m_{0},m_{0}+1,\ldots\}\)), \(\upsilon _{n}\geq \upsilon_{n+1}\) (\(n\in\{n_{0},n_{0}+1,\ldots\}\)), \(U(\infty )=V(\infty)=\infty\). Then
(i) for \(m,n\in\mathbf{N,}\) we have
where, \(\theta(\lambda_{2},m)=O(\frac{1}{U_{m}^{\lambda_{2}+\alpha }})\in (0,1)\), \(\vartheta(\lambda_{1},n)=O(\frac{1}{V_{n}^{\lambda_{1}+\alpha }})\in(0,1)\);
(ii) for any \(a>0\), we have
Proof
Since \(\upsilon_{n}\geq\upsilon_{n+1} \) (\(n\geq n_{0}\)), \(-\alpha<\lambda_{2}\leq1-\alpha\), \(\lambda_{1}>-\alpha\), and \(V(\infty)=\infty\), by Lemma 1 we have
where
For \(U_{m}>V(n_{0})\), we obtain
and then \(\theta(\lambda_{2},m)=O(\frac{1}{U_{m}^{\lambda_{2}+\alpha}})\). Hence, we have (17).
In the same way, since \(\mu_{m}\geq\mu_{m+1} \) (\(m\geq m_{0}\)), \(-\alpha <\lambda_{1}\leq1-\alpha\), \(\lambda_{2}>-\alpha\), and \(U(\infty )=\infty\), we have
where
For \(V_{n}>U(m_{0})\), we obtain
Hence, we have (18).
For \(a>0\), we find
3 Equivalent inequalities and operator expressions
Theorem 4
If \(-\alpha<\lambda_{1},\lambda_{2}\leq 1-\alpha\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(k(\lambda_{1})\) is as in (9), then for \(p>1\), \(0<\|a\|_{p,\Phi_{\lambda }},\|b\|_{q,\Psi_{\lambda}}<\infty\), we have the following equivalent inequalities:
Proof
By Hölder’s inequality with weight (see [25]) we have
In view of (14), we find
By Hölder’s inequality we have
On the other hand, assuming that (23) is valid, we set
Then we find \(J^{p}=\|b\|_{q,\Psi_{\lambda}}^{q}\). If \(J=0\), then (24) is trivially valid; if \(J=\infty\), then by (26) and (13) it is impossible. Suppose that \(0< J<\infty\). By (23) it follows that
and then (24) follows, which is equivalent to (23). □
Theorem 5
With the assumptions of Theorem 4, if \(m_{0},n_{0}\in \mathbf{N}\), \(\mu_{m}\geq\mu_{m+1} \) (\(m\in\{m_{0},m_{0}+1,\ldots\} \)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in\{n_{0},n_{0}+1,\ldots \}\)), \(U(\infty)=V(\infty)=\infty\), then the constant factor \(k(\lambda _{1})\) in (23) and (24) is the best possible.
Proof
For \(\varepsilon\in(0,p(\lambda_{1}+\alpha))\), we set \(\widetilde{\lambda}_{1}=\lambda_{1}-\frac{\varepsilon}{p} \) (\(\in (-\alpha ,1-\alpha)\)), \(\widetilde{\lambda}_{2}=\lambda_{2}+\frac{\varepsilon }{p}\) (\(>-\alpha\)), and \(\widetilde{a}=\{\widetilde{a}_{m}\} _{m=1}^{\infty}\), \(\widetilde{b}=\{\widetilde{b}_{n}\}_{n=1}^{\infty}\),
Then by (19), (20), and (18) we have
If there exists a positive constant \(K\leq k(\lambda_{1})\) such that (23) is valid when replacing \(k(\lambda_{1})\) to K, then, in particular, we have \(\varepsilon\widetilde{I}<\varepsilon K\|\widetilde {a}\|_{p,\Phi _{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}\), namely,
It follows that \(k(\lambda_{1})\leq K \) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k(\lambda_{1})\) is the best possible constant factor of (23).
The constant factor \(k(\lambda_{1})\) in (24) is still the best possible. Otherwise, we would reach a contradiction by (27) that the constant factor in (23) is not the best possible. □
For \(p>1\), we find \(\Psi_{\lambda}^{1-p}(n)=\frac{\upsilon_{n}}{V_{n}^{1-p\lambda_{2}}}\) and define the following normed spaces:
Assuming that \(a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}\) and setting
we can rewrite (24) as
namely, \(c\in l_{p,\Psi_{\lambda}^{1-p}}\).
Definition 1
Define a Hardy-Hilbert-type operator \(T:l_{p,\Phi _{\lambda}}\rightarrow l_{p,\Psi_{\lambda}^{1-p}}\) as follows: For any \(a=\{a_{m}\}_{m=1}^{\infty}\in l_{p,\Phi_{\lambda}}\), there exists a unique representation \(Ta=c\in l_{p,\Psi_{\lambda}^{1-p}}\). Define the formal inner product of Ta and \(b=\{b_{n}\}_{n=1}^{\infty}\in l_{q,\Psi _{\lambda}}\) as follows:
Then we can rewrite (23) and (24) as follows:
Define the norm of the operator T as follows:
Then by (31) we find \(\|T\|\leq k(\lambda_{1})\). Since by Theorem 5 the constant factor in (31) is the best possible, we have
4 Some equivalent reverse inequalities
Theorem 6
If \(-\alpha<\lambda_{1},\lambda_{2}\leq1-\alpha\), \(\lambda_{1}+\lambda_{2}=\lambda\), \(k(\lambda_{1})\) is as in (9), \(m_{0},n_{0}\in\mathbf{N}\), \(\mu_{m}\geq\mu_{m+1} \) (\(m\in \{m_{0},m_{0}+1, \ldots\}\)), \(\upsilon_{n}\geq\upsilon_{n+1}\) (\(n\in \mathbf{\{}n_{0},n_{0}+1, \ldots\}\)), \(U(\infty)=V(\infty)=\infty\), then for \(0< p<1\), \(0<\|a\|_{p,\Phi_{\lambda}},\|b\|_{q,\Psi_{\lambda }}<\infty \), we have the following equivalent inequalities with the best possible constant factor \(k(\lambda_{1})\):
Proof
By the reverse Hölder’s inequality and (14), we have the reverses of (25), (26), and (27). Then by (17) we have (35). By (35) and the reverse of (27) we have (34).
On the other hand, assuming that (34) is valid, we set \(b_{n}\) as in Theorem 4. Then we find \(J^{p}=\|b\|_{q,\Psi_{\lambda }}^{q}\). If \(J=\infty \), then (35) is trivially valid; if \(J=0\), then by reverse of (26) and (17) it is impossible. Suppose that \(0< J<\infty\). By (34) it follows that
and then (35) follows, which is equivalent to (34).
For \(\varepsilon\in(0,p(\lambda_{1}+\alpha))\), we set \(\widetilde{\lambda}_{1}\), \(\widetilde{\lambda}_{2}\), \(\widetilde{a}_{m}\), and \(\widetilde{b}_{n}\) as (30). Then by (19), (20), and (14) we find
If there exists a constant \(K\geq k(\lambda_{1})\) such that (34) is valid when replacing \(k(\lambda_{1})\) to K, then, in particular, we have \(\varepsilon\widetilde{I}>\varepsilon K\|\widetilde{a}\|_{p,\widetilde {\Phi}_{\lambda}}\|\widetilde{b}\|_{q,\Psi_{\lambda}}\), namely,
It follows that \(k(\lambda_{1})\geq K \) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k(\lambda_{1})\) is the best possible constant factor of (34).
The constant factor \(k(\lambda_{1})\) in (35) is still the best possible. Otherwise, we would reach a contradiction by the reverse of (27) that the constant factor in (34) is not the best possible. □
Theorem 7
With the assumptions of Theorem 6, if \(p<0\), then we have the following equivalent inequalities with the best possible constant factor \(k(\lambda_{1})\):
Proof
By the reverse Hölder inequality with weight, since \(p<0\), by (18) we have
By the reverse Hölder inequality we have
On the other hand, assuming that (38) is valid, we set \(b_{n}\) as follows:
Then we find \(J_{1}^{p}=\|b\|_{q,\widetilde{\Psi}_{\lambda}}^{q}\). If \(J_{1}=\infty\), then (39) is trivially valid; if \(J_{1}=0\), then by (40) and (13) it is impossible. Suppose that \(0< J_{1}<\infty\). By (38) it follows that
and then (39) follows, which is equivalent to (38).
For \(\varepsilon\in(0,q(\lambda_{2}+\alpha))\), we set \(\widetilde{\lambda}_{1}=\lambda_{1}+\frac{\varepsilon}{q} \) (\(>-\alpha\)), \(\widetilde{\lambda}_{2}=\lambda_{2}-\frac{\varepsilon}{q} \) (\(\in(-\alpha ,1-\alpha)\)), and
Then by (19), (20), and (13) we have
If there exists a constant \(K\geq k(\lambda_{1})\) such that (38) is valid when replacing \(k(\lambda_{1})\) to K, then, in particular, we have \(\varepsilon\widetilde{I}>\varepsilon K\|\widetilde{a}\|_{p,\Phi _{\lambda }}\|\widetilde{b}\|_{q,\widetilde{\Psi}_{\lambda}}\), namely,
It follows that \(k(\lambda_{1})\geq K \) (\(\varepsilon\rightarrow0^{+}\)). Hence, \(K=k(\lambda_{1})\) is the best possible constant factor of (38).
The constant factor \(k(\lambda_{1})\) in (39) is still the best possible. Otherwise, we would reach a contradiction by (41) that the constant factor in (38) is not the best possible. □
Remark 1
(i) For \(\alpha=0\) and \(0<\lambda _{1},\lambda_{2}\leq1\) in (23) and (24), we have the following equivalent inequalities:
(ii) for \(\alpha=-\lambda\) and \(-1\leq\lambda_{1},\lambda_{2}<0\) in (23) and (24), we have the following equivalent inequalities:
(iii) for \(\lambda=0\), \(|\lambda_{1}|<\alpha \) (\(0<\alpha\leq\frac {1}{2}\)); \(|\lambda_{1}|<1-\alpha\) (\(\frac{1}{2}<\alpha\leq1\)), \(\lambda _{2}=-\lambda _{1}\) in (23) and (24), we have the following equivalent inequalities:
In view of Theorem 5, the constant factors in these inequalities with the particular kernels are all the best possible.
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Acknowledgements
This work is supported by Hunan Province Natural Science Foundation (No. 2015JJ4041) and Science Research General Foundation Item of Hunan Institution of Higher Learning College and University (No. 14C0938).
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BY carried out the mathematical studies, participated in the sequence alignment, and drafted the manuscript. 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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Shi, Y., Yang, B. A new Hardy-Hilbert-type inequality with multiparameters and a best possible constant factor. J Inequal Appl 2015, 380 (2015). https://doi.org/10.1186/s13660-015-0900-4
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DOI: https://doi.org/10.1186/s13660-015-0900-4