The equivalent parameter conditions for constructing multiple integral half-discrete Hilbert-type inequalities with a class of nonhomogeneous kernels and their applications

Bing He; Yong Hong; Qiang Chen

doi:10.1515/math-2021-0023

Open Access Published by De Gruyter Open Access May 25, 2021

The equivalent parameter conditions for constructing multiple integral half-discrete Hilbert-type inequalities with a class of nonhomogeneous kernels and their applications

Bing He , Yong Hong and Qiang Chen

From the journal Open Mathematics

https://doi.org/10.1515/math-2021-0023

Abstract

In this paper, we establish equivalent parameter conditions for the validity of multiple integral half-discrete Hilbert-type inequalities with the nonhomogeneous kernel G ( n λ 1 ∥ x ∥ m , ρ λ 2 ) ( λ 1 λ 2 > 0 ) and obtain best constant factors of the inequalities in specific cases. In addition, we also discuss their applications in operator theory.

Keywords: multiple integral half-discrete Hilbert-type inequality; nonhomogeneous kernel; equivalent parameter condition; best constant factor; operator norm; bounded operator

MSC 2010: 26D15; 47A07

1 Introduction and preliminary knowledge

Suppose that 1 p + 1 q = 1 ( p > 1 ) , a ˜ = { a m } ∈ l p , b ˜ = { b n } ∈ L q , the classical Hilbert series inequality was obtained in 1925 [1]:

(1) ∑ n = 1 ∞ ∑ m = 1 ∞ a m b n m + n ≤ π sin ( π / p ) ∥ a ˜ ∥ p ∥ b ˜ ∥ q ,

where the constant factor π sin ( π / p ) is the best.

If f ( x ) ∈ L p ( 0 , + ∞ ) , g ( y ) ∈ L q ( 0 , + ∞ ) , the corresponding Hilbert integral inequality was obtained in 1934 [2]:

(2) ∫ 0 + ∞ ∫ 0 + ∞ f ( x ) g ( y ) x + y d x d y ≤ π sin ( π / p ) ∥ f ∥ p ∥ g ∥ q ,

where the constant factor π sin ( π / p ) is still the best.

Since (1) is of great significance for the study of boundedness and norm of series operator in l p and (2) is of great significance for the study of boundedness and norm of integral operator in L p ( 0 , + ∞ ) , Hilbert inequality has been widely concerned. To study the operator boundedness and operator norm from sequence space l to function space L or from function space L to sequence space l , the half-discrete Hilbert inequality has been paid more attention. In 2011, the following results were obtained [3]: If a ˜ = { a n } ∈ l p , f ( x ) ∈ L q ( 0 , + ∞ ) , then

(3) ∫ 0 + ∞ ∑ n = 1 ∞ a n f ( x ) n + x d x ≤ π sin ( π / p ) ∥ a ˜ ∥ p ∥ f ∥ q ,

the constant factor is also the best. Later on the equivalent conditions for validity of multiple integral half-discrete Hilbert-type inequality with generalized homogeneous kernel were discussed [4]. In [5], the parameter conditions for the optimal constant factor of half-discrete Hilbert-type inequality with homogeneous kernel in one dimension were established. Good results were obtained.

To further discuss the multiple integral half-discrete Hilbert-type inequality, we need to introduce the following notations: Suppose that m ∈ N + , x = ( x 1 , x 2 , … , x m ) , R + m = { x = ( x 1 , x 2 , … , x m ) : x i > 0 , i = 1 , 2 , … , m } . For ρ > 0 , the norm of x is defined by

∥ x ∥ m , ρ = ( x 1 ρ + x 2 ρ + ⋯ + x m ρ ) 1 / ρ .

Spaces l and L are defined by, respectively,

l p α = a ˜ = { a n } : ∥ a ˜ ∥ p , α = ∑ n = 1 ∞ n α ∣ a n ∣ p 1 / p < + ∞ , L q β ( R + m ) = f ( x ) : ∥ f ∥ q , β = ∫ R + m ∥ x ∥ m , ρ β ∣ f ( x ) ∣ q d x 1 / q < + ∞ .

If K ( n , ∥ x ∥ m , ρ ) = G ( n λ 1 ∥ x ∥ m , ρ λ 2 ) ≥ 0 , then K ( n , ∥ x ∥ m , ρ ) is a nonhomogeneous nonnegative function. In this paper, we will discuss the equivalent parameter conditions under which the multiple integral half-discrete Hilbert-type inequality

(4) ∑ n = 1 ∞ ∫ R + m G ( n λ 1 ∥ x ∥ m , ρ λ 2 ) a n f ( x ) d x ≤ M ∥ a ˜ ∥ p , α ∥ f ∥ q , β

can be established when λ 1 λ 2 > 0 . That is, what conditions do the parameters α , β , λ 1 , λ 2 , p , q meet if there is a constant M > 0 such that (4) holds? On the contrary, if there exists a constant M > 0 such that (4) holds, then what conditions do the parameters α , β , λ 1 , λ 2 , p , q satisfy? Such problems are undoubtedly very important theoretical problems, which have not been well solved at present. At the same time, we also discuss the best constant factor of (4) and its application in operator theory. More related literature can be found in [6,7, 8,9,10, 11,12,13, 14,15,16, 17,18,19, 20,21].

2 Some lemmas

By using the Hölder’s inequality of integral and series, we can easily get the following lemma.

Lemma 2.1

Assume that 1 p + 1 q = 1 ( p > 1 ) , a n ( x ) ≥ 0 , b n ( x ) ≥ 0 , Ω is measurable. Then, the mixed Hölder’s inequality can be obtained

∫ Ω ∑ n = 1 ∞ a n ( x ) b n ( x ) d x = ∑ n = 1 ∞ ∫ Ω a n ( x ) b n ( x ) d x ≤ ∫ Ω ∑ n = 1 ∞ a n p ( x ) d x 1 / p ∫ Ω ∑ n = 1 ∞ b n q ( x ) d x 1 / q .

Lemma 2.2

[22] If m ∈ N + , ρ > 0 , r > 0 , ψ ( u ) is measurable, then

∫ ∥ x ∥ m , ρ ≤ r ψ ( ∥ x ∥ m , ρ ) d x = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 r ψ ( u ) u m − 1 d u , ∫ ∥ x ∥ m , ρ ≥ r ψ ( ∥ x ∥ m , ρ ) d x = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ r + ∞ ψ ( u ) u m − 1 d u ,

where Γ ( t ) is the Gamma function, and ∥ x ∥ m , ρ ≤ r represents the region Ω r = { x = ( x 1 , x 2 , … , x m ) : x i ≥ 0 , ∥ x ∥ m , ρ ≤ r } .

According to Lemma 2.2, one gets

∫ R + m ψ ( ∥ x ∥ m , ρ ) d x = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 + ∞ ψ ( u ) u m − 1 d u .

Lemma 2.3

Assume that m ∈ N + , ρ > 0 , 1 p + 1 q = 1 ( p > 1 ) , λ 1 λ 2 > 0 , K ( n , ∥ x ∥ m , ρ ) = G ( n λ 1 ∥ x ∥ m , ρ λ 2 ) is nonnegative measurable, 1 λ 2 m λ 1 + α λ 2 p − λ 2 + β λ 1 q = c , and K ( t , 1 ) t − α + 1 p + c is monotonically decreasing in ( 0 , + ∞ ) . Denote that

W 1 = ∫ 0 + ∞ K ( 1 , t ) t − β + m q + m − 1 d t , W 2 = ∫ 0 + ∞ K ( t , 1 ) t − α + 1 p + c d t .

Then,

λ 1 W 2 = λ 2 W 1

and

ω 1 ( n ) = ∫ R + m K ( n , ∥ x ∥ m , ρ ) ∥ x ∥ m , ρ − β + m q d x = n λ 1 λ 2 β + m q − m Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) W 1 , ω 2 ( x ) = ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) n − α + 1 p + c ≤ ∥ x ∥ m , ρ λ 2 λ 1 α + 1 p − 1 − c W 2 .

Proof

It follows from m λ 1 + α λ 2 p − λ 2 + β λ 1 q = λ 2 c that 1 λ 1 − α λ 2 p + λ 2 q + λ 2 c − 1 = − β + m q + m − 1 , then

W 2 = ∫ 0 + ∞ K 1 , t λ 1 λ 2 t − α + 1 p + c d t = λ 2 λ 1 ∫ 0 + ∞ K ( 1 , u ) u λ 2 λ 1 − α + 1 p + c + λ 2 λ 1 − 1 d u = λ 2 λ 1 ∫ 0 + ∞ K ( 1 , u ) u 1 λ 1 − α λ 2 p + λ 2 q + λ 2 c − 1 d u = λ 2 λ 1 ∫ 0 + ∞ K ( 1 , u ) u − β + m q + m − 1 d u = λ 2 λ 1 W 1 .

Hence, λ 1 W 2 = λ 2 W 1 .

By Lemma 2.2, one gets

ω 1 ( n ) = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 + ∞ K ( n , t ) t − β + m q + m − 1 d t = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 + ∞ K 1 , t ⋅ n λ 1 λ 2 t − β + m q + m − 1 d t = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) n − λ 1 λ 2 − β + m q + m − 1 − λ 1 λ 2 ∫ 0 + ∞ K ( 1 , u ) u − β + m q + m − 1 d u = n λ 1 λ 2 β + m q − m Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) W 1 .

Note that K ( t , 1 ) t − α + 1 p + c is monotonically decreasing in ( 0 , + ∞ ) , thus

ω 2 ( x ) = ∑ n = 1 ∞ K ( n ⋅ ∥ x ∥ m , ρ λ 2 / λ 1 , 1 ) n − α + 1 p + c = ∥ x ∥ m , ρ λ 2 λ 1 α + 1 p − c ∑ n = 1 ∞ K ( n ⋅ ∥ x ∥ m , ρ λ 2 / λ 1 , 1 ) ( n ⋅ ∥ x ∥ m , ρ λ 2 / λ 1 ) − α + 1 p + c ≤ ∥ x ∥ m , ρ λ 2 λ 1 α + 1 p − c ∫ 0 + ∞ K ( t ⋅ ∥ x ∥ m , ρ λ 2 / λ 1 , 1 ) ( t ⋅ ∥ x ∥ m , ρ λ 2 / λ 1 ) − α + 1 p + c d t = ∥ x ∥ m , ρ λ 2 λ 1 α + 1 p − 1 − c ∫ 0 + ∞ K ( u , 1 ) u − α + 1 p + c d u = ∥ x ∥ m , ρ λ 2 λ 1 α + 1 p − 1 − c W 2 . □

3 Main results

Theorem 3.1

Suppose that m ∈ N + , ρ > 0 , λ 1 λ 2 > 0 , 1 p + 1 q = 1 ( p > 1 ) , K ( n , ∥ x ∥ m , ρ ) = G ( n λ 1 ∥ x ∥ m , ρ λ 2 ) is nonnegative measurable, 1 λ 2 m λ 1 + α λ 2 p − λ 2 + β λ 1 q = c , K ( t , 1 ) t − α + 1 p and K ( t , 1 ) t − α + 1 p + c are monotonically decreasing in ( 0 , + ∞ ) , and

W 0 = ∣ λ 1 ∣ ∫ 0 + ∞ K ( t , 1 ) t − α + 1 p d t < + ∞ .

Then,

The necessary and sufficient conditions for the validity of inequality
(5) ∫ R + m ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) a n f ( x ) d x ≤ M ∥ a ˜ ∥ p , α ∥ f ∥ q , β
with some constant M > 0 is c ≥ 0 , where a ˜ = { a n } ∈ l p α , f ( x ) ∈ L q β ( R + m ) .
For c = 0 , the best constant factor of (5) is
inf M = W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p .

Proof

(i) Suppose that (5) holds. If c < 0 , for ε = − c 2 ∣ λ 1 ∣ > 0 , taking

a n = n ( − α − 1 − ∣ λ 1 ∣ ε ) / p , n = 1 , 2 , …

and

f ( x ) = ∥ x ∥ m , ρ ( − β − m + ∣ λ 2 ∣ ε ) / q , 0 < ∥ x ∥ m , ρ ≤ 1 , 0 , ∥ x ∥ m , ρ > 1 .

It follows from Lemma 2.2 that

M ∥ a ˜ ∥ p , α ∥ f ∥ q , β = M ∑ n = 1 ∞ n − 1 − ∣ λ 1 ∣ ε 1 / p ∫ 0 < ∥ x ∥ m , ρ ≤ 1 ∥ x ∥ m , ρ − m + ∣ λ 2 ∣ ε d x 1 / q ≤ M 1 + ∫ 1 + ∞ t − 1 − ∣ λ 1 ∣ ε d t 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 1 t − 1 + ∣ λ 2 ∣ ε d t 1 / q = M ε ∣ λ 1 ∣ 1 / p ∣ λ 1 ∣ 1 / q ( 1 + ∣ λ 1 ∣ ε ) 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p = 2 M − c λ 1 λ 2 1 / q 1 − c 2 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p < + ∞ .

Since K ( t , 1 ) t − α + 1 p is monotonically decreasing in ( 0 , + ∞ ) , then

∫ R + m ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) a n f ( x ) d x = ∫ 0 < ∥ x ∥ m , ρ ≤ 1 ∥ x ∥ m , ρ ( − β − m + ∣ λ 2 ∣ ε ) / q ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) n − ( α + 1 + ∣ λ 1 ∣ ε ) / p d x = ∫ 0 < ∥ x ∥ m , ρ ≤ 1 ∥ x ∥ m , ρ − β − m + ∣ λ 2 ∣ ε q − λ 2 λ 1 − α − 1 − ∣ λ 1 ∣ ε p ∑ n = 1 ∞ K n ⋅ ∥ x ∥ m , ρ λ 2 λ 1 , 1 n ⋅ ∥ x ∥ m , ρ λ 2 λ 1 − α + 1 + ∣ λ 1 ∣ ε p d x ≥ ∫ 0 < ∥ x ∥ m , ρ ≤ 1 ∥ x ∥ m , ρ − β + m − ∣ λ 2 ∣ ε q + λ 2 λ 1 α + 1 + ∣ λ 1 ∣ ε p ∫ 1 ∞ K u ⋅ ∥ x ∥ m , ρ λ 2 λ 1 , 1 u ⋅ ∥ x ∥ m , ρ λ 2 λ 1 − α + 1 + ∣ λ 1 ∣ ε p d u d x = ∫ 0 < ∥ x ∥ m , ρ ≤ 1 ∥ x ∥ m , ρ − β + m − ∣ λ 2 ∣ ε q + λ 2 λ 1 α + 1 + ∣ λ 1 ∣ ε p − λ 2 λ 1 ∫ ∥ x ∥ m , ρ λ 2 λ 1 ∞ K ( t , 1 ) t − α + 1 + ∣ λ 1 ∣ ε p d t d x ≥ ∫ 0 < ∥ x ∥ m , ρ ≤ 1 ∥ x ∥ m , ρ − m + λ 2 λ 1 c + ∣ λ 2 ∣ ε ∫ 1 + ∞ K ( t , 1 ) t − α + 1 + ∣ λ 1 ∣ ε p d t d x = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 1 t − 1 + λ 2 λ 1 c + ∣ λ 2 ∣ ε d t ∫ 1 + ∞ K ( t , 1 ) t − α + 1 + ∣ λ 1 ∣ ε p d t = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 1 t − 1 + λ 2 2 λ 1 c d t ∫ 1 + ∞ K ( t , 1 ) t − α + 1 + ∣ λ 1 ∣ ε p d t .

Consequently,

(6) ∫ 0 1 t − 1 + λ 2 2 λ 1 c d t ∫ 1 + ∞ K ( t , 1 ) t − α + 1 + ∣ λ 1 ∣ ε p d t ≤ 2 M − c λ 1 λ 2 1 / q 1 − c 2 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) − 1 / q < + ∞ .

Note that λ 2 2 λ 1 c < 0 , hence ∫ 0 1 t − 1 + λ 2 2 λ 1 c d t = + ∞ , which contradicts (6). Therefore, c ≥ 0 .

On the contrary, suppose that c ≥ 0 . It follows from the mixed Hölder’s inequality and Lemma 2.3 that

∫ R + m ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) a n f ( x ) d x = ∫ R + m ∑ n = 1 ∞ n ( α + 1 − p c ) / ( p q ) ∥ x ∥ m , ρ ( β + m ) / ( p q ) ⋅ a n ∥ x ∥ m , ρ ( β + m ) / ( p q ) n ( α + 1 − p c ) / ( p q ) ⋅ f ( x ) K ( n , ∥ x ∥ m , ρ ) d x ≤ ∫ R + m ∑ n = 1 ∞ n ( α + 1 − p c ) / q ∥ x ∥ m , ρ ( β + m ) / q ⋅ ∣ a n ∣ p ⋅ K ( n , ∥ x ∥ m , ρ ) d x 1 / p × ∫ R + m ∑ n = 1 ∞ ∥ x ∥ m , ρ ( β + m ) / p n ( α + 1 − p c ) / p ⋅ ∣ f ( x ) ∣ q ⋅ K ( n , ∥ x ∥ m , ρ ) d x 1 / q = ∑ n = 1 ∞ n ( α + 1 − p c ) / q ∣ a n ∣ p ω 1 ( n ) 1 / p ∫ R + m ∥ x ∥ m , ρ ( β + m ) / p ∣ f ( x ) ∣ q ω 2 ( x ) d x 1 / q ≤ W 1 1 / p W 2 1 / q Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ∑ n = 1 ∞ n α + 1 − p c q + λ 1 λ 2 β + m q − m ∣ a n ∣ p 1 / p × ∫ R + m ∥ x ∥ m , ρ β + m p + λ 2 λ 1 α + 1 p − 1 − c ∣ f ( x ) ∣ q d x 1 / q = W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ∑ n = 1 ∞ n α − p c ∣ a n ∣ p 1 / p ∫ R + m ∥ x ∥ m , ρ β ∣ f ( x ) ∣ q d x 1 / q ≤ W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ∑ n = 1 ∞ n α ∣ a n ∣ p 1 / p ∫ R + m ∥ x ∥ m , ρ β ∣ f ( x ) ∣ q d x 1 / q = W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ∥ a ˜ ∥ p , α ∥ f ∥ q , β .

Taking M ≥ W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p arbitrarily, one can get (5).

(ii) For c = 0 , assuming the best constant factor of (5) is M 0 , then we can see from the previous proof that

M 0 ≤ W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ,

∫ R + m ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) a n f ( x ) d x ≤ M 0 ∥ a ˜ ∥ p , α ∥ f ∥ q , β .

For ε > 0 , sufficiently small and N sufficiently large, taking

a n = n ( − α − 1 − ∣ λ 1 ∣ ε ) / p , n = 1 , 2 , … f ( x ) = ∥ x ∥ m , ρ ( − β − m + ∣ λ 2 ∣ ε ) / q , 0 < ∥ x ∥ m , ρ ≤ N , 0 , ∥ x ∥ m , ρ > N .

Then,

M 0 ∥ a ˜ ∥ p , α ∥ f ∥ q , β = M 0 ∑ k = 1 ∞ n − 1 − ∣ λ 1 ∣ ε 1 / p ∫ 0 < ∥ x ∥ m , ρ ≤ N ∥ x ∥ m , ρ − m + ∣ λ 2 ∣ ε d x 1 / q ≤ M 0 1 + ∫ 1 + ∞ t − 1 − ∣ λ 1 ∣ ε d t 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 0 N t − 1 + ∣ λ 2 ∣ ε d t 1 / q = M 0 ε ∣ λ 1 ∣ 1 / p ∣ λ 2 ∣ 1 / q Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / q N ∣ λ 2 ∣ ε / q ( 1 + ∣ λ 1 ∣ ε ) 1 / p ,

∫ R + m ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) a n f ( x ) d x = ∑ n = 1 ∞ n − ( α + 1 + ∣ λ 1 ∣ ε ) / p ∫ 0 < ∥ x ∥ m , ρ ≤ N K ( n , ∥ x ∥ m , ρ ) ∥ x ∥ m , ρ ( − β − m + ∣ λ 2 ∣ ε ) / q d x = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∑ n = 1 ∞ n − ( α + 1 + ∣ λ 1 ∣ ε ) / p ∫ 0 N K ( n , t ) t − β + m − ∣ λ 2 ∣ ε q + m − 1 d t = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∑ n = 1 ∞ n − ( α + 1 + ∣ λ 1 ∣ ε ) / p ∫ 0 N K ( 1 , t ⋅ n λ 1 / λ 2 ) t − β + m − ∣ λ 2 ∣ ε q + m − 1 d t = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∑ n = 1 ∞ n − α + 1 + ∣ λ 1 ∣ ε p + λ 1 λ 2 β + m − ∣ λ 2 ∣ ε q − m + 1 − λ 1 λ 2 ∫ 0 N ⋅ n λ 1 / λ 2 K ( 1 , u ) u − β + m − ∣ λ 2 ∣ ε q + m − 1 d u = Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∑ n = 1 ∞ n − 1 − ∣ λ 1 ∣ ε ∫ 0 N ⋅ n λ 1 / λ 2 K ( 1 , u ) u − β + m − ∣ λ 2 ∣ ε q + m − 1 d u ≥ Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∑ n = 1 ∞ n − 1 − ∣ λ 1 ∣ ε ∫ 0 N K ( 1 , u ) u − β + m − ∣ λ 2 ∣ ε q + m − 1 d u ≥ Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) ∫ 1 + ∞ t − 1 − ∣ λ 1 ∣ ε d t ∫ 0 N K ( 1 , u ) u − β + m − ∣ λ 2 ∣ ε q + m − 1 d u = Γ m ( 1 / ρ ) ∣ λ 1 ∣ ε ρ m − 1 Γ ( m / ρ ) ∫ N + ∞ K ( 1 , u ) u − β + m − ∣ λ 2 ∣ ε q + m − 1 d u .

Consequently,

Γ m ( 1 / ρ ) ∣ λ 1 ∣ ε ρ m − 1 Γ ( m / ρ ) ∫ 0 N K ( 1 , u ) u − β + m − ∣ λ 2 ∣ ε q + m − 1 d u ≤ M 0 ε ∣ λ 1 ∣ 1 / p ∣ λ 2 ∣ 1 / q Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / q N ∣ λ 2 ∣ ε / q ( 1 + ∣ λ 1 ∣ ε ) 1 / p , ∣ λ 1 ∣ 1 / p ∣ λ 2 ∣ 1 / q ∣ λ 1 ∣ Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ∫ 0 N K ( 1 , u ) u − β + m − ∣ λ 2 ∣ ε q + m − 1 d u ≤ M 0 ⋅ N ∣ λ 2 ∣ ε / q ( 1 + ∣ λ 1 ∣ ε ) 1 / p .

Let ε → 0 + , then

∣ λ 1 ∣ 1 / p ∣ λ 2 ∣ 1 / q ∣ λ 1 ∣ Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ∫ 0 N K ( 1 , u ) u − β + m q + m − 1 d u ≤ M 0 .

In addition, let N → + ∞ , we have

1 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ∣ λ 2 ∣ ∫ 0 + ∞ K ( 1 , u ) u − β + m q + m − 1 d u ≤ M 0 .

It follows from Lemma 2.3 that

W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p ≤ M 0 .

Hence, the best constant factor of (5) is

M 0 = W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p . □

4 Applications in operator theory

The series operator T 1 and singular integral operator T 2 are defined by, respectively,

(7) T 1 ( a ˜ ) ( x ) = ∑ n = 1 ∞ K ( n , ∥ x ∥ m , ρ ) a n , T 2 ( f ) n = ∫ R + m K ( n , ∥ x ∥ m , ρ ) f ( x ) d x .

According to the basic theory of Hilbert-type inequality, (5) can be equivalently written as the following two expressions:

∥ T 1 ( a ˜ ) ∥ p , β ( 1 − p ) ≤ M ∥ a ˜ ∥ p , α and ∥ T 2 ( f ) ∥ q , α ( 1 − q ) ≤ M ∥ f ∥ q , β .

Thus, by Theorem 3.1, one has

Theorem 4.1

Assume that m ∈ N + , ρ > 0 , λ 1 λ 2 > 0 , 1 p + 1 q = 1 ( p > 1 ) , K ( n , ∥ x ∥ m , ρ ) = G ( n λ 1 ∥ x ∥ m , ρ λ 2 ) is nonnegative measurable, 1 λ 2 m λ 1 + α λ 2 p − λ 2 + β λ 1 q = c , K ( t , 1 ) t − α + 1 p and K ( t , 1 ) t − α + 1 p + c are monotonically decreasing in ( 0 , + ∞ ) , and

W 0 = ∣ λ 1 ∣ ∫ 0 + ∞ K ( t , 1 ) t − α + 1 p d t

is convergent. Then,

T 1 : l p α → L p β ( 1 − p ) ( R + m ) and T 2 : L q β ( R + m ) → l q α ( 1 − q ) are bounded operators if and only if c ≥ 0 ;
For c = 0 , i.e., m λ 1 + α λ 2 p = λ 2 + β λ 1 q , the operator norms of T 1 and T 2 are as follows:
∥ T 1 ∥ = ∥ T 2 ∥ = W 0 ∣ λ 1 ∣ 1 / q ∣ λ 2 ∣ 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p .

Corollary 4.1

Suppose that m ∈ N + , ρ > 0 , 1 p + 1 q = 1 ( p > 1 ) , a > 0 , λ 1 > 0 , λ 2 > 0 , 1 λ 2 m λ 1 + α λ 2 p − λ 2 + β λ 1 q = l , p q + max { p λ 1 ( b − a ) , p ( λ 1 b − 1 ) , p ( λ 1 c − 1 ) , p ( λ 1 b + l − 1 ) , p ( λ 1 c + l − 1 ) } < α < p q + p λ 1 c , and

W 0 = ∫ 0 1 1 ( 1 + t ) a t c + 1 λ 1 ( 1 − α + 1 p ) − 1 + t a − b − 1 λ 1 ( 1 − α + 1 p ) − 1 d t .

The operators T 1 and T 2 are defined by, respectively,

T 1 ( a ˜ ) ( x ) = ∑ n = 1 ∞ ( max { 1 , n λ 1 ∥ x ∥ m , ρ λ 2 } ) b ( min { 1 , n λ 1 ∥ x ∥ m , ρ λ 2 } ) c ( 1 + n λ 1 ∥ x ∥ m , ρ λ 2 ) a ⋅ a n , T 2 ( f ) n = ∫ R + m ( max { 1 , n λ 1 ∥ x ∥ m , ρ λ 2 } ) b ( min { 1 , n λ 1 ∥ x ∥ m , ρ λ 2 } ) c ( 1 + n λ 1 ∥ x ∥ m , ρ λ 2 ) a ⋅ f ( x ) d x .

Then,

T 1 is a bounded operator from l p α to L p β ( 1 − p ) ( R + m ) and T 2 is a bounded operator from L q β ( R + m ) to l q α ( 1 − q ) if and only if l ≥ 0 ;
For l = 0 , i.e., m λ 1 + α λ 2 p = λ 2 + β λ 1 q , the operator norms of T 1 and T 2 are as follows:
∥ T 1 ∥ = ∥ T 2 ∥ = W 0 λ 1 1 / q λ 2 1 / p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p .

Proof

It follows from p q + p λ 1 ( b − a ) < α < p q + p λ 1 c that c + 1 λ 1 1 − α + 1 p > 0 , a − b − 1 λ 1 1 − α + 1 p > 0 , and the integral in W 0 is convergent. Denote that

K ( n , ∥ x ∥ m , ρ ) = ( max { 1 , n λ 1 ∥ x ∥ m , ρ λ 2 } ) b ( min { 1 , n λ 1 ∥ x ∥ m , ρ λ 2 } ) c ( 1 + n λ 1 ∥ x ∥ m , ρ λ 2 ) a .

Then,

W 0 = λ 1 ∫ 0 + ∞ K ( t , 1 ) t − α + 1 p d t = λ 1 ∫ 0 + ∞ ( max { 1 , t λ 1 } ) b ( min { 1 , t λ 1 } ) c ( 1 + t λ 1 ) a ⋅ t − α + 1 p d t = ∫ 0 + ∞ ( max { 1 , u } ) b ( min { 1 , u } ) c ( 1 + u ) a ⋅ u 1 λ 1 ( 1 − α + 1 p ) − 1 d u = ∫ 0 1 u c + 1 λ 1 ( 1 − α + 1 p ) − 1 ( 1 + u ) a d u + ∫ 1 + ∞ u b + 1 λ 1 ( 1 − α + 1 p ) − 1 ( 1 + u ) a d u = ∫ 0 1 1 ( 1 + t ) a t c + 1 λ 1 1 − α + 1 p − 1 + t a − b − 1 λ 1 1 − α + 1 p − 1 d t .

According to α > p q + p ( λ 1 b − 1 ) and α > p q + p ( λ 1 c − 1 ) , one has λ 1 b − α + 1 p < 0 , λ 1 c − α + 1 p < 0 , and

K ( t , 1 ) t − α + 1 p = ( max { 1 , t λ 1 } ) b ( min { 1 , t λ 1 } ) c ( 1 + t λ 1 ) a ⋅ t − α + 1 p = 1 ( 1 + t λ 1 ) a ⋅ t λ 1 c − α + 1 p , 0 < t ≤ 1 , 1 ( 1 + t λ 1 ) a ⋅ t λ 1 b − α + 1 p , t > 1 .

Thus, K ( t , 1 ) t − α + 1 p is monotonically decreasing in ( 0 , + ∞ ) .

Note that α > p q + p ( λ 1 b + l − 1 ) and α > p q + p ( λ 1 c + l − 1 ) , we get λ 1 b − α + 1 p + l < 0 , λ 1 c − α + 1 p + l < 0 , and

K ( t , 1 ) t − α + 1 p + l = 1 ( 1 + t λ 1 ) a ⋅ t λ 1 c − α + 1 p + l , 0 < t ≤ 1 , 1 ( 1 + t λ 1 ) a ⋅ t λ 1 b − α + 1 p + l , t > 1 .

Therefore, K ( t , 1 ) t − α + 1 p + l is monotonically decreasing in ( 0 , + ∞ ) .

Finally, it follows from Theorem 4.1 that Corollary 4.1 holds.□

Taking c = b in Corollary 4.1, according to the properties of Beta function, one gets

W 0 = ∫ 0 1 1 ( 1 + t ) a t b + 1 λ 1 ( 1 − α + 1 p ) − 1 + t a − b + 1 λ 1 ( 1 − α + 1 p ) − 1 d t = B b + 1 λ 1 1 − α + 1 p , a − b + 1 λ 1 ( 1 − α + 1 p ) .

Hence, we have

Corollary 4.2

Assume that m ∈ N + , ρ > 0 , 1 p + 1 q = 1 ( p > 1 ) , a > 0 , λ 1 > 0 , λ 2 > 0 , 1 λ 2 m λ 1 + α λ 2 p − λ 2 + β λ 1 q = l , p q + max { p λ 1 ( b − a ) , p ( λ 1 b − 1 ) , p ( λ 1 b + l − 1 ) } < α < p q + p λ 1 b , operators T 1 and T 2 are defined by, respectively,

T 1 ( a ˜ ) ( x ) = ∑ n = 1 ∞ ( n λ 1 ∥ x ∥ m , ρ λ 2 ) b ( 1 + n λ 1 ∥ x ∥ m , ρ λ 2 ) a ⋅ a n , T 2 ( f ) n = ∫ R + m ( n λ 1 ∥ x ∥ m , ρ λ 2 ) b ( 1 + n λ 1 ∥ x ∥ m , ρ λ 2 ) a ⋅ f ( x ) d x .

Then,

T 1 : l p α → L p β ( 1 − p ) ( R + m ) and T 2 : L q β ( R + m ) → l q α ( 1 − q ) are bounded operators if and only if l ≥ 0 .
For l = 0 , i.e., m λ 1 + α λ 2 p = λ 2 + β λ 1 q , the operator norms of T 1 and T 2 are
∥ T 1 ∥ = ∥ T 2 ∥ = 1 λ 1 1 / q λ 2 1 / p B b + 1 λ 1 1 − α + 1 p , a − b + 1 λ 1 1 − α + 1 p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p .

We can get the following results by taking b = 0 in Corollary 4.2.

Corollary 4.3

Assume that m ∈ N + , ρ > 0 , 1 p + 1 q = 1 ( p > 1 ) , a > 0 , λ 1 > 0 , λ 2 > 0 , 1 λ 2 m λ 1 + α λ 2 p − λ 2 + β λ 1 q = l , max { p ( 1 − λ 1 a ) − 1 , − 1 , p l − 1 } < α < p − 1 , operators T 1 and T 2 are defined by, respectively,

T 1 ( a ˜ ) ( x ) = ∑ n = 1 ∞ a n ( 1 + n λ 1 ∥ x ∥ m , ρ λ 2 ) a and T 2 ( f ) n = ∫ R + m f ( x ) d x ( 1 + n λ 1 ∥ x ∥ m , ρ λ 2 ) a .

Then,

T 1 : l p α → L p β ( 1 − p ) ( R + m ) and T 2 : L q β ( R + m ) → l q α ( 1 − q ) are bounded operators if and only if l ≥ 0 .
For l = 0 , i.e., m λ 1 + α λ 2 p = λ 2 + β λ 1 q , the operator norms of T 1 and T 2 are expressed as follows:
∥ T 1 ∥ = ∥ T 2 ∥ = 1 λ 1 1 / q λ 2 1 / p B 1 λ 1 1 − α + 1 p , a − 1 λ 1 1 − α + 1 p Γ m ( 1 / ρ ) ρ m − 1 Γ ( m / ρ ) 1 / p .

In Corollary 4.1, let m = 1 , α = 1 λ 2 ( p σ − λ 1 ) and β = 1 λ 1 ( q σ − λ 2 ) , then m λ 1 + α λ 2 p = σ = λ 2 + β λ 1 q . The following results can be obtained.

Corollary 4.4

Assume that 1 p + 1 q = 1 ( p > 1 ) , a > 0 , λ 1 > 0 , λ 2 > 0 , λ 1 p + λ 2 q + max { λ 1 λ 2 ( b − a ) , λ 1 λ 2 b − λ 2 , λ 1 λ 2 c − λ 2 } < σ < λ 1 p + λ 2 q + λ 1 λ 2 c , α = 1 λ 2 ( p σ − λ 1 ) , β = 1 λ 1 ( q σ − λ 2 ) . Denote that

W 0 = ∫ 0 1 1 ( 1 + t ) a t 1 λ 1 λ 2 λ 1 p + λ 2 q + λ 1 λ 2 c − σ − 1 + t a − 1 λ 1 λ 2 λ 1 p + λ 2 q + λ 1 λ 2 b − σ − 1 d t .

Operators T 1 and T 2 are defined by, respectively,

T 1 ( a ˜ ) ( x ) = ∑ n = 1 ∞ ( max { 1 , n λ 1 x λ 2 } ) b ( min { 1 , n λ 1 x λ 2 } ) c ( 1 + n λ 1 x λ 2 ) a ⋅ a n , T 2 ( f ) n = ∫ 0 + ∞ ( max { 1 , n λ 1 x λ 2 } ) b ( min { 1 , n λ 1 x λ 2 } ) c ( 1 + n λ 1 x λ 2 ) a ⋅ f ( x ) d x .

Then, T 1 : l p α → L p β ( 1 − p ) ( 0 , + ∞ ) and T 2 : L q β ( 0 , + ∞ ) → l q α ( 1 − q ) are bounded operators, and the operator norms of T 1 and T 2 are expressed as follows:

∥ T 1 ∥ = ∥ T 2 ∥ = W 0 λ 1 1 / q λ 2 1 / p .

Funding information: This work was supported by the National Natural Science Foundation of China (No. 12071491) and the Innovation Team Construction Project of Guangdong Province (No. 2018KCXTD020).
Conflict of interest: Authors state no conflict of interest.

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Received: 2020-09-02

Revised: 2021-02-02

Accepted: 2021-02-04

Published Online: 2021-05-25

This work is licensed under the Creative Commons Attribution 4.0 International License.

The equivalent parameter conditions for constructing multiple integral half-discrete Hilbert-type inequalities with a class of nonhomogeneous kernels and their applications

Abstract

1 Introduction and preliminary knowledge

2 Some lemmas

Lemma 2.1

Lemma 2.2

Lemma 2.3

Proof

3 Main results

Theorem 3.1

Proof

4 Applications in operator theory

Theorem 4.1

Corollary 4.1

Proof

Corollary 4.2

Corollary 4.3

Corollary 4.4

References

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