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Article

# A Hilbert-Type Integral Inequality in the Whole Plane Related to the Arc Tangent Function

by
Michael Th. Rassias
1,2,*,
Bicheng Yang
3 and
Andrei Raigorodskii
4,5,6,7
1
Institute of Mathematics, University of Zurich, CH-8057 Zurich, Switzerland
2
Program in Interdisciplinary Studies, Institute for Advanced Study, 1 Einstein Drive, Princeton, NJ 08540, USA
3
Department of Mathematics, Guangdong University of Education, Guangzhou 510303, China
4
Moscow Institute of Physics and Technology, Institutskiy per, d. 9, 141700 Dolgoprudny, Russia
5
Faculty of Bioengineering and Bioinformatics, Moscow State University, 119991 Moscow, Russia
6
Institute of Mathematics and Computer Science, Buryat State University, 670000 Ulan-Ude, Russia
7
Caucasus Mathematical Center, Adyghe State University, 352700 Maykop, Russia
*
Author to whom correspondence should be addressed.
Symmetry 2021, 13(2), 351; https://doi.org/10.3390/sym13020351
Submission received: 28 January 2021 / Revised: 9 February 2021 / Accepted: 17 February 2021 / Published: 21 February 2021
(This article belongs to the Special Issue Functional Equations and Analytic Inequalities)

## Abstract

:
In this work we establish a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the kernel of the arc tangent function. We prove that the constant factor, which is associated with the cosine function, is optimal. Some special cases as well as some operator expressions are also presented.
MSC:
26D15; 31A10

## 1. Introduction

If
$0 < ∫ 0 ∞ f 2 ( x ) d x < ∞ a n d 0 < ∫ 0 ∞ g 2 ( y ) d y < ∞ ,$
then we have the following well-known Hilbert integral inequality (see [1]):
where the constant factor $π$ is the best possible. Recently, using weight functions, some extensions of (1) were established in Yang’s two books (see [2,3]) and the papers [4,5,6,7,8,9]. Most of them are constructed in the quarter plane of the first quadrant.
In 2007, Yang [10] proved the following Hilbert-type integral inequality in the whole plane (namely $( x , y )$-plane) involving the exponential function:
with the best possible constant factor $B ( λ 2 , λ 2 )$, $λ > 0$, where by $B ( u , v )$ we denote the beta function). In the papers [11,12,13,14,15,16,17,18,19,20,21,22], the authors have presented some new Hilbert-type integral inequalities in the whole plane for which they have established optimal constant factors.
In 2017, Hong [23] proved two equivalent statements between a Hilbert-type inequality with the general homogenous kernel and a few parameters. This domain of research is very vibrant with many authors investigating other types of integral inequalities (cf. [24,25,26,27,28,29,30,31,32,33,34,35,36,37,38]).
In this paper, we follow the idea of Hong’s work in [23] and using techniques of real analysis as well as weight functions, we prove a few equivalent statements of a Hilbert-type integral inequality in the whole plane related to the kernel of the arc tangent function. The constant factor which is related to the cosine function is proved to be the best possible. Within this work, we also consider some particular cases of interest as well as operator expressions.

## 2. Some Lemmas

For $ρ > 0$, $0 < σ < γ$, setting $h ( u ) : = arctan ρ u γ$ $( u > 0 )$, we obtain
For $R : = ( − ∞ , ∞ ) , δ ∈ { − 1 , 1 } ,$ $α , β ∈ ( − 1 , 1 ) ,$ we set
$x α : = | x | + α x , y β : = | y | + β y ( x , y ∈ R ) , E δ : = { t ∈ R ; | t | δ ≥ 1 } , E − δ = { t ∈ R ; | t | δ ≤ 1 } .$
Lemma 1.
For $c > 0 ,$ $θ = α , β ,$ we have
For $c ≤ 0 ,$ it follows that
$∫ E δ t θ − c δ − 1 d t = ∫ E − δ t θ c δ − 1 d t = ∞ .$
Proof.
Setting
$E δ + : = { t ∈ R + ; t δ ≥ 1 } , E δ − : = { − t ∈ R + ; ( − t ) δ ≥ 1 } ,$
it follows that $E δ = E δ + ∪ E δ −$ and
Setting $u = t δ$ (or $t = u 1 δ ) ,$ we obtain
$∫ E δ + t − c δ − 1 d t = 1 | δ | ∫ 1 ∞ u 1 δ ( − c δ − 1 ) u 1 δ − 1 d u = ∫ 1 ∞ u − c − 1 d u .$
Hence, for $c > 0 ,$ Formula (4) follows and for $c ≤ 0 ,$ we get that
$∫ E δ t θ − c δ − 1 d t = ∞ .$
Since
for $c > 0 ,$ Equation (5) follows and for $c ≤ 0 ,$ we have
$∫ E − δ t θ c δ − 1 d t = ∞ .$
This completes the proof of the lemma. □
In what follows, we assume that $p > 1 ,$ $1 p + 1 q = 1 ,$ $δ ∈ { − 1 , 1 } ,$ $α , β ∈ ( − 1 , 1 ) ,$ $ρ > 0 ,$ $0 < σ < γ ,$ $σ 1 ∈ R ,$
$h ( u ) = arctan ρ u γ ( u > 0 ) ,$
$k ρ ( γ ) ( σ )$ is indicated by (3) and
$K α , β ( γ ) ( σ ) : = 2 k ρ ( γ ) ( σ ) ( 1 − α 2 ) 1 / q ( 1 − β 2 ) 1 / p .$
For $n ∈ N = { 1 , 2 , ⋯ } ,$ $E − 1 = [ − 1 , 1 ] ,$ $x ∈ E δ ,$ we define the following expressions:
Since $y β = | y | + β y = ( s g n ( y ) + β ) y ,$ where
$x α δ = ( 1 + α · s g n ( x ) ) δ | x | δ ≥ min δ ∈ { − 1 , 1 } ( 1 ± | α | ) δ ( x ∈ E δ ) ,$
and $1 − | α | ≤ ( 1 + | α | ) − 1 ≤$ $1 + | α | ≤ ( 1 − | α | ) − 1 ,$ we have
$( 1 ± β ) x α δ ≥ m α , β : = ( 1 − | β | ) ( 1 − | α | ) > 0 ( x ∈ E δ ) .$
For fixed $x ∈ E δ ,$ setting $u = x α δ y β ,$ we obtain
For $n ∈ N$, $x ∈ F δ ,$ we define the following expressions:
$J ( − ) ( x ) : = ∫ − ∞ − 1 y β σ + 1 q n − 1 arctan ρ ( x α δ y β ) γ d y , J ( + ) ( x ) : = ∫ 1 ∞ y β σ + 1 q n − 1 arctan ρ ( x α δ y β ) γ d y , J ( x ) : = ∫ E 1 y β σ + 1 q n − 1 arctan ρ ( x α δ y β ) γ d y = J ( − ) ( x ) + J ( + ) ( x ) .$
Since for $x ∈ E − δ ,$
$x α δ = ( 1 + α · s g n ( x ) ) δ | x | δ ≤ max δ ∈ { − 1 , 1 } { ( 1 ± | α | ) δ } = ( 1 − | α | ) − 1 ,$
we have
$M α , β : = ( 1 + | β | ) ( 1 − | α | ) − 1 ≥ ( 1 ± β ) x α δ ( x ∈ E − δ ) .$
For fixed $x ∈ E − δ ,$ setting $u = x α δ y β ,$ we obtain
In view of (8) and (10), we derive the lemma below:
Lemma 2.
We have the following inequalities:
$I 1 : = ∫ E δ I ( x ) x α δ ( σ 1 − 1 p n ) − 1 d x ≥ 2 1 − β 2 ∫ E δ x α − δ ( σ − σ 1 + 1 n ) − 1 d x ∫ 0 m α , β ( arctan ρ u γ ) u σ + 1 q n − 1 d u ,$
$J 1 : = ∫ E − δ J ( x ) x α δ ( σ 1 + 1 p n ) − 1 d x ≥ 2 1 − β 2 ∫ E − δ x α δ ( σ 1 − σ + 1 n ) − 1 d x ∫ M α , β ∞ ( arctan ρ u γ ) u σ + 1 q n − 1 d u .$
Lemma 3.
If there exists a constant M, such that for any nonnegative measurable functions $f ( x )$ and $g ( y )$ in $R ,$ the following inequality
holds true, then we have $σ 1 = σ .$
Proof.
If $σ 1 > σ ,$ then for $n ≥ 1 σ 1 − σ ( n ∈ N ) ,$ we consider the functions
and by (4) and (5), we obtain
By (11) and (13) (for $f = f n , g = g n$), we have
Since for any $n ≥ 1 σ 1 − σ ( n ∈ N ) , σ − σ 1 + 1 n ≤ 0 ,$ by Lemma 1 it follows that
$∫ E δ x α − δ ( σ − σ 1 + 1 n ) − 1 d x = ∞ .$
In view of
$∫ 0 m α , β ( arctan ρ u γ ) u σ + 1 q n − 1 d u > 0 ,$
we derive that $∞ ≤ M J ˜ 1 < ∞ ,$ which is a contradiction.
If $σ > σ 1 ,$ then for $n ≥ 1 σ − σ 1 ( n ∈ N ) ,$ we consider the functions
and by (4) and (5), we obtain
By (12) and (13) (for $f = f ˜ n , g = g ˜ n$), we have
Since for $n ≥ 1 σ − σ 1 ( n ∈ N ) ,$ $σ 1 − σ + 1 n ≤ 0 ,$ by Lemma 1 it follows that
$∫ F δ x α δ ( σ 1 − σ + 1 n ) − 1 d x = ∞ .$
In view of
$∫ M α , β ∞ ( arctan ρ u γ ) u σ − 1 q n − 1 d u > 0 ,$
we have $∞ ≤ M J ˜ 2 < ∞ ,$ which is a contradiction.
Hence, we conclude that $σ 1 = σ .$
This completes the proof of the lemma. □
For $σ 1 = σ ,$ we also get the lemma below:
Lemma 4.
If there exists a constant M, such that for any nonnegative measurable functions $f ( x )$ and $g ( y )$ in $R ,$ the following inequality
holds true, then we have $K α , β ( γ ) ( σ ) ≤ M .$
Proof.
For $σ 1 = σ ,$ by (8), we have
$I 1 = ∫ E δ I ( x ) x α δ ( σ − 1 p n ) − 1 d x = I 1 ( − ) + I 1 ( + ) , I 1 ( − ) : = ∫ E δ I ( − ) ( x ) x α δ ( σ − 1 p n ) − 1 d x , I 1 ( + ) : = ∫ E δ I ( + ) ( x ) x α δ ( σ − 1 p n ) − 1 d x .$
In view of the presented results, for $n > 1 q ( γ − σ ) ,$ we obtain
For $γ > σ + d$ $( d > 0 )$, we have that $( arctan ρ u γ ) u σ + d$ is continuous in $( 0 , ∞ ) ,$ and
$( arctan ρ u γ ) u σ + d → 0 ( u → ∞ ) .$
There exists a positive constant $M 1 ,$ such that
$( arctan ρ u γ ) u σ + d ≤ M 1 ( u ∈ [ m α , β , ∞ ) ) .$
By (4), it follows that
namely
$1 1 − β ∫ E δ x α − δ n − 1 ∫ ( 1 − β ) x α δ ∞ ( arctan ρ u γ ) u σ + 1 q n − 1 d u d x = O ( 1 ) ,$
and then by (15), it follows that
Similarly, we have
By (14) (for $f = f n , g = g n$), we have
For $n → ∞ ,$ by Fatou’s lemma (cf. [39]), (16) and (17), we obtain
namely
$K α , β ( γ ) ( σ ) = 2 k ρ ( γ ) ( σ ) ( 1 − α 2 ) 1 / q ( 1 − β 2 ) 1 / p ≤ M .$
This completes the proof of the lemma. □
Lemma 5.
We define the following weight functions:
Then we have
$1 − α 2 2 ω δ ( σ , y ) = 1 − β 2 2 ϖ δ ( σ , x ) = k ρ ( γ ) ( σ ) ( x , y ∈ R ∖ { 0 } ) .$
Proof.
For fixed $y ∈ ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ,$ setting $u = x α δ y β ,$ we obtain
for fixed $x ∈ ( − ∞ , 0 ) ∪ ( 0 , ∞ ) ,$ setting $u = x α δ y β ,$ it follows that
Hence, we derive (20).
This completes the proof of the lemma. □

## 3. Main Results and Some Particular Cases

Theorem 1.
If M is a constant, then the following statements (i), (ii) and (iii) are equivalent:
(i) For any $f ( x ) ≥ 0 ,$ we have the following inequality:
(ii) for any $f ( x ) , g ( y ) ≥ 0 ,$ we have the following inequality:
(iii) $σ 1 = σ ,$ and $K α , β ( γ ) ( σ ) ≤ M .$
Proof.
$( i ) ⇒ ( i i )$. By Hölder’s inequality (cf. [40]), we get
Then by (21), we have (22).
$( i i ) ⇒ ( i i i )$. By Lemma 1, we have $σ 1 = σ .$ Then by Lemma 2, we get $K α , β ( γ ) ( σ ) ≤ M .$
$( i i i ) ⇒ ( i )$. For $σ 1 = σ ,$ by Hölder’s inequality with weight (see [40]) and (18), we have
By Fubini’s theorem, (24) and (19), we derive that
For $K α , β ( γ ) ( σ ) ≤ M ,$ we have (21) (for $σ 1 = σ$).
Therefore, Statements (i), (ii) and (iii) are equivalent.
This completes the proof of the theorem. □
For $σ 1 = σ ,$ we deduce the theorem below:
Theorem 2.
If M is a constant, then the following statements (i), (ii) and (iii) are equivalent:
(i) For any $f ( x ) ≥ 0 ,$ satisfying
$0 < ∫ − ∞ ∞ x α p ( 1 − δ σ ) − 1 f p ( x ) d x < ∞ ,$
we have the following inequality:
(ii) for any $f ( x ) ≥ 0 ,$ satisfying
$0 < ∫ − ∞ ∞ x α p ( 1 − δ σ ) − 1 f p ( x ) d x < ∞ ,$
and $g ( y ) ≥ 0 ,$ satisfying
$0 < ∫ − ∞ ∞ y β q ( 1 − σ ) − 1 g q ( y ) d y < ∞ ,$
we have the following inequality:
(iii) $K α , β ( γ ) ( σ ) ≤ M .$
Moreover, if the statement (iii) holds true, then the constant factor $M = K α , β ( γ ) ( σ )$ in (25) and (26) is the best possible.
In particular:
(1) for $δ = 1 ,$ we have the following equivalent inequalities with the nonhomogeneous kernel:
where $K α , β ( γ ) ( σ )$ is the best possible constant factor;
(2) for $δ = − 1 ,$ we have the following equivalent inequalities with the homogeneous kernel of degree 0:
where $K α , β ( γ ) ( σ )$ is the best possible constant factor.
Proof.
For $σ 1 = σ$ and the assumption of statement (i), if (24) assumes the form of equality for some $y ∈ ( − ∞ , 0 ) ∪ ( 0 , ∞ )$, then (see [40]) there exist constants A and B, such that they are not both zero, and
$A y β σ − 1 x α ( δ σ − 1 ) p / q f p ( x ) = B x δ σ − 1 y β ( σ − 1 ) q / p a . e . i n R .$
We suppose that $A ≠ 0$ (otherwise $B = A = 0$). Then it follows that
$x α p ( 1 − δ σ ) − 1 f p ( x ) = y β q ( 1 − σ ) B A x α a . e . i n R .$
Since
$∫ − ∞ ∞ x α − 1 d x = ∞ ,$
$0 < ∫ − ∞ ∞ x α p ( 1 − δ σ ) − 1 f p ( x ) d x < ∞ .$
Hence, (24) takes the form of strict inequality, and so does (21). Hence, (25) and (26) are true.
In view of Theorem 1, we can establish the equivalency between the statements (i), (ii) and (iii) in Theorem 2.
In case the statement (iii) is valid, namely $K α , β ( γ ) ( σ ) ≤ M ,$ if there exists a constant $M ≤ K α , β ( γ ) ( σ ) ,$ such that (26) is satisfied, then we can derive that the constant factor $M = K α , β ( γ ) ( σ )$ in (26) is optimal.
The constant factor $M = K α , β ( γ ) ( σ )$ in (25) remains the best possible. Otherwise, by (23) (for $σ 1 = σ$), we would reach a contradiction that the constant factor $M = K α , β ( γ ) ( σ )$ in (26) is not optimal.
This completes the proof of the theorem. □

## 4. Operator Expressions

We set the following functions: $φ ( x ) : = x α p ( 1 − δ σ ) − 1 ( x ∈ R )$ and $ψ ( y ) : = y β q ( 1 − σ ) − 1 ,$ wherefrom $ψ 1 − p ( y ) = y β p σ − 1$ $( y ∈ R ) .$ Define the following real normed linear spaces:
In view of Theorem 2, for $f ∈ L p , φ ( R ) ,$ setting
by (25), we have
Definition 1.
Define a Hilbert-type integral operator with the nonhomogeneous kernel
$T :$$L p , φ ( R ) → L p , ψ 1 − p ( R )$ as follows: For any $f ∈ L p , φ ( R ) ,$ there exists a unique representation $T f = h 1 ∈ L p , ψ 1 − p ( R ) ,$ satisfying $T f ( y ) = h 1 ( y ) ,$ for any $y ∈ R$.
In view of (31), it follows that
$| | T f | | p , ψ 1 − p = | | h 1 | | p , ψ 1 − p < M | | f | | p , φ ,$
and thus the operator T is bounded satisfying
$| | T | | = sup f ( ≠ θ ) ∈ L p , φ ( R ) | | T f | | p , ψ 1 − p | | f | | p , φ ≤ M .$
If we define the formal inner product of $T f$ and g as follows:
then we can rewrite Theorem 2 as follows:
Theorem 3.
If M is a constant, then the following statements (i), (ii) and (iii) are equivalent:
(i) For any $f ( x ) ≥ 0 , f ∈ L p , φ ( R ) ,$ $| | f | | p , φ > 0 ,$ the following inequality holds true:
$| | T f | | p , ψ 1 − p < M | | f | | p , φ ;$
(ii) for any $f ( x ) , g ( y ) ≥ 0 , f ∈ L p , φ ( R ) , g ∈ L q , ψ ( R ) , | | f | | p , φ , | | g | | q , ψ > 0 ,$ the following inequality holds true:
$( T f , g ) < M | | f | | p , φ | | g | | q , ψ ;$
(iii)$K α , β ( γ ) ( σ ) ≤ M .$
Moreover, if the statement (iii) holds true, then the constant factor $M = K α , β ( γ ) ( σ )$ in (32) and (33) is optimal, i.e., $| | T | | = K α , β ( γ ) ( σ ) .$
Remark 1.
(1) In particular, for $α = β = 0$ in (27) and (28) we have the following equivalent inequalities:
where $ρ σ / γ π σ cos π σ 2 γ$ is the optimal constant factor. If $f ( − x ) = f ( x ) , g ( − y ) = g ( y )$ $( x , y ∈ R + ) ,$ then we have the following equivalent inequalities:
where $ρ σ / γ π 2 σ cos π σ 2 γ$ is the best possible constant factor.
(2) For $α = β = 0$ in (29) and (30) we have the following equivalent inequalities:
where $ρ σ / γ π σ cos π σ 2 γ$ is the best possible constant factor. If $f ( − x ) = f ( x ) , g ( − y ) = g ( y )$
$( x , y ∈ R + ) ,$ then we have the following equivalent inequalities:
where $ρ σ / γ π 2 σ cos π σ 2 γ$ is the best possible constant factor.

## 5. Conclusions

In this paper, making use of ideas of Hong [23], and by employing techniques of real analysis as well as weight functions, we obtain in Theorem 1 a few equivalent statements of a Hilbert-type integral inequality in the whole plane associated with the kernel of the arc tangent function. In Theorem 2, the constant factor associated with the cosine function is proved to be optimal. Furthermore, in Theorem 3 and Remark 1 we also consider some particular cases and operator expressions. The lemmas and theorems within this work provide an extensive account of this type of inequalities.

## Author Contributions

All authors contributed equally during all stages of the preparation of the present work. All authors have read and agreed to the published version of the manuscript.

## Funding

B. 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).

Not applicable.

Not applicable.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Rassias, M.T.; Yang, B.; Raigorodskii, A. A Hilbert-Type Integral Inequality in the Whole Plane Related to the Arc Tangent Function. Symmetry 2021, 13, 351. https://doi.org/10.3390/sym13020351

AMA Style

Rassias MT, Yang B, Raigorodskii A. A Hilbert-Type Integral Inequality in the Whole Plane Related to the Arc Tangent Function. Symmetry. 2021; 13(2):351. https://doi.org/10.3390/sym13020351

Chicago/Turabian Style

Rassias, Michael Th., Bicheng Yang, and Andrei Raigorodskii. 2021. "A Hilbert-Type Integral Inequality in the Whole Plane Related to the Arc Tangent Function" Symmetry 13, no. 2: 351. https://doi.org/10.3390/sym13020351

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