Abstract
A more accurate half-discrete Hilbert-type inequality in the whole plane with multi-parameters is established by the use of Hermite–Hadamard’s inequality and weight functions. Furthermore, some equivalent forms and some special types of inequalities and operator representations as well as reverses are considered.
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1 Introduction
For \(p>1,\) \(\frac{1}{p}+\frac{1}{q}=1,\) \(a_{m},b_{n}>0,\)
the following discrete Hardy-Hilbert inequality (cf. [10], Theorem 315, and [4,5,, 11, 36, 40]) holds true:
The constant factor \(\frac{\pi }{\sin (\pi /p)}\) is optimal.
Let f(x), g(y) \(\ge 0,\) such that
Then the following Hardy-Hilbert integral inequality with the same best possible constant factor \(\frac{\pi }{\sin (\pi /p)}\) (cf. [11], Theorem 316) is valid:
The following half-discrete Hardy–Hilbert inequality with the same best possible constant factor was recently formulated and proved (cf. [39]):
Several inequalities with homogenous kernels of degree 0 as well as with non-homogenous kernels have been proved in [7, 11,12,13,14,, 19, 37, 42, 45]. For a large variety of integral inequalities of Hilbert-type the interested reader is referred to [1,2,3,4,5,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,, 8, 12, 15, 16, 20, 22,23,24, 26, 41, 46, 47]. The above inequalities are constructed in the quarter plane of the first quadrant.
A Hilbert-type integral inequality in the whole plane was proved in [33] by Yang. Furthermore, a generalized form of a Hilbert-type integral inequality in the whole plane was considered in [34]:
The constant factor
is optimal. Additionally, in [9, 13, 14, 25, 27, 29, 30, 43, 44] several integral and discrete Hilbert-type inequalities in the whole plane where formulated and proved.
The goal of the present paper is to study the following half-discrete Hilbert-type inequality in the whole plane with parameters and a best possible constant factor, by applying the Hermite–Hadamard inequality and weight functions:
where \(\sigma ,\mu >0,\sigma +\mu =\lambda \le 1\). Moreover, a more accurate half-discrete Hilbert-type inequality with multiparameters is proved. Some equivalent forms, a few special types of inequalities as well as operator representations and reverses are studied.
2 Some lemmas
In what follows, we assume that \(\delta \in \{-1,1\},\) \(a,b\in (-1,1),\) \(\sigma ,\mu >0,\) \(\sigma +\mu =\lambda \le 1,\) \(\xi \in (-\infty ,\infty ),\eta \in [0,\frac{1}{2}].\) We set
wherefrom
Lemma 2.1
We define two weight functions \(\omega (\sigma ,n)\) and \( \varpi (\sigma ,x)\) as follows:
Then:
(i) we have
(ii) we also have
where
and
Proof
It holds that
Setting
in the above first (resp. second) integral, by simplifications, we deduce that
Hence, (8) follows.
We obtain
For \(0<\sigma <\lambda \le 1,\) since \((-1)^{i}\frac{d^{(i)}}{\mathrm{d}u^{(i)}}\frac{ \ln u}{u^{\lambda }-1}>0\) \((i=1,2)\) (cf. [36]), we find that both \( \frac{H(x,-y)}{(y+\eta )^{1-\sigma }}\) and \(\frac{H(x,y)}{(y-\eta )^{1-\sigma }}\) are strictly decreasing and strictly convex in \(y\in (\frac{1 }{2},\infty ),\) satisfying
By (11) and Hermite–Hadamard’s inequality (cf. [17]), in view of \(\eta \in [0,\frac{1}{2}],\) we have that
Setting
in the above first (resp. second) integral, by simplifications, we deduce that
By (11) and the decreasing property of series, we also have that
Setting
in the above first (resp. second) integral, by simplifications, we derive that
For some \(\kappa \in (0,\frac{\sigma }{\lambda }),\) we obtain that
and thus there exists a constant \(L>0,\) such that \(0<u^{\kappa }\frac{\ln u}{ u-1}\le L\) \((u\in (0,\infty ))\). Hence, we have
and therefore (9) and (10) follow.
This completes the proof of the lemma. \(\square \)
Lemma 2.2
For \(\varepsilon >0,\)
we have
Proof
It holds that
By (13) and the decreasing property of series, we derive that
Hence, we obtain (12) and the lemma is proved.
\(\square \)
Lemma 2.3
For \(\varepsilon >0,\) setting
we have
Proof
Setting
it follows that \(E_{\delta }=E_{\delta }^{+}\cup E_{\delta }^{-}.\) We have that
Setting \(u=[(x-\xi )(1+a)]^{\delta }\,\,(resp.\, u=[(\xi -x)(1-a)]^{\delta })\) in the above first (resp. second) integral, we obtain
Hence, we get (14) and thus the lemma is proved. \(\square \)
3 Main results
Theorem 3.1
Suppose that \(p>1,\frac{1}{p}+\frac{1}{q}=1,\)
If \(f(x),b_{n}\ge 0,\) satisfying
then we have the following equivalent inequalities:
In particular, for \(a=b=0\) we have the following equivalent inequalities:
Proof
By Hölder’s inequality (cf. [17]) and (6), we obtain that
Then by (8) and the Lebesgue term by term integration theorem (cf. [18]), in view of (7), we derive that
Hence, by (9), we deduce (17).
By Hölder’s inequality (cf. [17]), we have
Then by (17), we get (16). On the other hand, assuming that (16) is valid, we set
Then we obtain that
In view of (22), it follows that \(J_{1}<\infty .\) If \(J_{1}=0,\) then (17) is trivially valid; if \(J_{1}>0,\) then by (16), we have
namely (17) holds, which is equivalent to (16).
Similarly to as we obtained (22), by Hölder’s inequality, we have
By (9) and the Lebesgue term by term theorem, we have
Hence, by (8), we deduce (18).
We have proved that (16) is satisfied. Setting
it then follows that
and in view of (24), we obtain that \(J_{2}<\infty .\) If \(J_{2}=0,\) then ( 18) is trivially valid; if \(J_{2}>0,\) then by (16) we have
namely, (18) follows.
On the other hand, assuming that (18) is valid, by Hölder’s inequality (cf. [17]) and similarly to as we obtained (23), we have
Then by (18) we derive (16), which is equivalent to (18).
Therefore, inequalities (16), (17) and (18) are equivalent.
This completes the proof of the theorem. \(\square \)
Theorem 3.2
With regards to the assumptions of Theorem 1, the constant factor \(K_{a,b}(\sigma )\) in (16), (17) and (18 ) is the best possible.
Proof
For \(0<\varepsilon <q\sigma ,\) we set \({\widetilde{\sigma }} =\sigma -\frac{\varepsilon }{q}\,\,\ (\in (0,\lambda )),\)
and
Then by (12) and (14), we obtain that
By (9), we also have that
If the constant factor \(K_{a,b}(\sigma )\) in (16) is not the best possible, then there exists a positive number k, with \(K_{a,b}(\sigma )>k\) , such that (16) is valid when we replace \(K_{a,b}(\sigma )\) by k. Then in particular, we have \(\varepsilon {\tilde{I}}<\varepsilon k{\widetilde{I}} _{1},\) namely,
It follows that
that is,
This is a contradiction. Hence, the constant factor \(K_{a,b}(\sigma )\) in ( 16) is the best possible.
The constant factor \(K_{a,b}(\sigma )\) in (17) (resp. (18)) is still the best possible. Otherwise, we would reach a contradiction by (23) (resp. (25)) that the constant factor \(K_{a,b}(\sigma )\) in (16) is not the best possible.
This completes the proof of the theorem. \(\square \)
4 Operator expressions
Suppose that \(p>1,\frac{1}{p}+\frac{1}{q}=1.\) We set the following functions:
wherefrom,
Define the following real weight normed linear spaces:
(a) In view of Theorem 1, for \(f\in L_{p,\Phi }({\mathbf {R}}),\) setting
by (17), we have
namely, \(H^{(1)}\in l_{p,\Psi ^{1-p}}.\)
Definition 4.1
Define a Hilbert-type operator in the whole plane
as follows:
For any \(f\in L_{p,\Phi }({\mathbf {R}}),\) there exists a unique representation
satisfying
for any \(|n|\in \mathbf {N }.\)
In view of (26), it follows that
and then the operator \(T^{(1)}\) is bounded satisfying
Since the constant factor \(K_{a,b}(\sigma )\) in (26) is the best possible, we have
If we define the formal inner product of \(T^{(1)}f\) and \(b\,\,(\in l_{q,\Psi })\) as follows:
we can then rewrite (16) and (17) as follows:
(b) In view of Theorem 1, for \(b\in l_{q,\Psi },\) setting
then by (18) we have
namely \(H^{(2)}\in L_{q,\Psi ^{1-q}}({\mathbf {R}}).\)
Definition 4.2
Define a Hilbert-type operator in the whole plane
as follows:
For any \(b\in l_{q,\Psi },\) there exists a unique representation
satisfying
for any \(x\in {\mathbf {R}}\).
In view of (29), we have
and then the operator \(T^{(2)}\) is bounded satisfying
Since the constant factor \(K_{a,b}(\sigma )\) in (29) is the best possible, we have
If we define the formal inner product of \(T^{(2)}b\) and \(f\,\,(\in L_{p,\Phi }({\mathbf {R}}))\) as follows:
then we can rewrite (16) and (18) in the following manner:
Remark 4.3
(i) For \(\xi =\eta =0,\) \(\delta =1,\) (19) reduces to (5). If \(f(-x)=f(x)\) \((x>0),\) \(b_{-n}=b_{n}\,\,(n\in {\mathbf {N}}),\) then (5) reduces to the following half-discrete Hilbert-type inequality (cf. [40]):
(ii) For \(\delta =1,\) replacing \([|x-\xi |+a(x-\xi )]^{\lambda }f(x)\) to f(x), (16) reduces to the following particular inequality with homogeneous kernel of degree \(-\lambda \):
(iii) For \(\delta =-1,\) (16) reduces to the following particular inequality with nonhomogeneous kernel:
The constant factors in the above inequalities are the best possible.
5 Some equivalent reverses
In the sequel, for the cases when \(0<p<1\) and \(p<0,\) we still use \( ||b||_{q,\Phi }\) and \(||f||_{p,\Psi }\) as the formal symbols.
Theorem 5.1
Suppose that \(0<p<1,\frac{1}{p}+\frac{1}{q}=1\). If \( f(x),b_{n}\ge 0,\) satisfying \(0<||f||_{p,\Psi },\) \(||b||_{q,\Phi }<\infty ,\) then we have the following equivalent inequalities:
where the constant factor \(K_{a,b}(\sigma )\) is the best possible.
Proof
By the reverse Hölder inequality (cf. [17]) and (6), we obtain that
Then by (8) and the Lebesgue term by term integration theorem (cf. [18]), in view of (7), we derive that
Hence, by (9), we deduce (36).
By the reverse Hölder inequality (cf. [17]), we have
On the other hand, assuming that (35) is satisfied, we set
Then we obtain that
In view of (38), it follows that \(J_{1}>0.\) If \(J_{1}=\infty ,\) then (36) is trivially valid; if \(J_{1}<\infty ,\) then by (35), we have
namely (36) holds true, which is equivalent to (35).
Similarly to as we obtained (38), we derive that
Hence, by (8), we deduce (37). We have proved that (35) is valid. Setting
then it follows that
and in view of (40), we obtain that \({\widetilde{J}}_{2}>0.\) If \({\widetilde{J}} _{2}=\infty ,\) then (37) is trivially valid; if \({\widetilde{J}} _{2}<\infty ,\) then by (35), we have
namely, (37) follows.
On the other hand, assuming that (37) is satisfied, by the reverse Hölder inequality (cf. [17]), we obtain
Then by (37), we derive (15), which is equivalent to (37 ).
Therefore, inequalities (35), (36) and (37) are equivalent.
For \(0<\varepsilon <p(\lambda -\sigma ),\) we set \({\widetilde{\sigma }}=\sigma +\frac{\varepsilon }{p}\,\,(<\lambda ),\)
and
Then by (12) and (14), we find
By (9), we also have
If the constant factor \(K_{a,b}(\sigma )\) in (36) is not the best possible, then there exists a positive number k, with \(K_{a,b}(\sigma )<k\) , such that (36) is satisfied when we replace \(K_{a,b}(\sigma )\) by k. Then in particular, we have \(\varepsilon {\tilde{I}}>\varepsilon k{\widetilde{I}} _{1},\) namely
It follows that
namely
This is a contradiction. Hence, the constant factor \(K_{a,b}(\sigma )\) in ( 35) is the best possible.
The constant factor \(K_{a,b}(\sigma )\) in (36) ((37)) is still the best possible. Otherwise, we would reach a contradiction by (39) ((41)) that the constant factor \(K_{a,b}(\sigma )\) in (35) is not the best possible.
This completes the proof of the theorem. \(\square \)
Theorem 5.2
Suppose that \(p<0,\) \(\frac{1}{p}+\frac{1}{q}=1.\) If \( f(x),b_{n}\ge 0,\) \(0<||f||_{p,\Psi }, ||b||_{q,\Phi }<\infty .\) Then we have the following equivalent inequalities:
where the constant factor \(K_{a,b}(\sigma )\) is the best possible.
Proof
By the reverse Hölder’s inequality (cf. [17]) and (6), we obtain that
Then by (8) and the Lebesgue term by term integration theorem (cf. [18]), in view of (7), we deduce that
Hence, by (9), we derive (43).
By the reverse Hölder inequality (cf. [17]), we have
On the other hand, assuming that (42) is valid, we set
Then we obtain that
In view of (45), it follows that \(J_{1}>0.\) If \(J_{1}=\infty ,\) then (43) is trivially valid; if \(J_{1}<\infty ,\) then by (42), we have
namely, (43) holds true, which is equivalent to (42).
Similarly, we have
Hence, by (8), we deduce (37). We have proved that (42) is satisfied. Setting
it follows that
and in view of (47), we get \(J_{2}>0.\) If \(J_{2}=\infty ,\) then ( 44) is trivially valid; if \(J_{2}<\infty ,\) then by (42), we have
namely, (44) follows.
On the other hand, assuming that (44) is valid, by the reverse Hölder inequality (cf. [17]), we obtain
Then by (44), we deduce (22), which is equivalent to (44 ). Therefore, inequalities (42), (43) and (44) are equivalent.
For \(0<\varepsilon <|p|\sigma ,\) we set \({\widetilde{\sigma }}=\sigma +\frac{ \varepsilon }{p}\,\,(>0),\)
and
Then by (12) and (14), we obtain that
By (9), we also have
If the constant factor \(K_{a,b}(\sigma )\) in (42) is not the best possible, then there exists a positive number k, with \(K_{a,b}(\sigma )<k\) , such that (42) is satisfied when we replace \(K_{a,b}(\sigma )\) by k. Then in particular, we have \(\varepsilon {\tilde{I}}>\varepsilon k{\widetilde{I}} _{1},\) namely
It follows that
that is
This is a contradiction. Hence, the constant factor \(K_{a,b}(\sigma )\) in ( 42) is the best possible.
The constant factor \(K_{a,b}(\sigma )\) in (43) ((44)) is still the best possible. Otherwise, we would reach a contradiction by (46) ((48)) that the constant factor \(K_{a,b}(\sigma )\) in (42) is not the best possible.
This completes the proof of the theorem. \(\square \)
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Acknowledgements
B. C. Yang: This work is supported by the National Natural Science Foundation (No. 61772140), and the Characteristic innovation project of Guangdong Provincial Colleges and universities in 2020 (No. 2020KTSCX088). We are grateful for their support.
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Rassias, M.T., Yang, B. & Meletiou, G.C. A more accurate half-discrete Hilbert-type inequality in the whole plane and the reverses. Ann. Funct. Anal. 12, 50 (2021). https://doi.org/10.1007/s43034-021-00133-w
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DOI: https://doi.org/10.1007/s43034-021-00133-w
Keywords
- Hermite–Hadamard inequality
- Half-discrete Hilbert-type inequality
- Weight function
- Equivalent form
- Operator expression
- Reverse