Open Access

Optimal couples of rearrangement invariant spaces for the Riesz potential on the bounded domain

  • Shin Min Kang1,
  • Arif Rafiq2,
  • Waqas Nazir2,
  • Irshaad Ahmad3,
  • Faisal Ali4 and
  • Young Chel Kwun5Email author
Journal of Inequalities and Applications20142014:60

https://doi.org/10.1186/1029-242X-2014-60

Received: 15 October 2013

Accepted: 20 January 2014

Published: 10 February 2014

Abstract

We prove continuity of the Riesz potential operator in optimal couples of rearrangement invariant function spaces defined in R n with the Lebesgue measure.

MSC:46E30, 46E35.

Keywords

Riesz potential operatorrearrangement invariant function spacesreal interpolation

1 Introduction

Let be the space of all locally integrable functions f on Ω R n with the Lebesgue measure, finite almost everywhere, and let M + be the space of all non-negative locally integrable functions on ( 0 , ) with respect to the Lebesgue measure, finite almost everywhere. We shall also need the following two subclasses of M + . The subclass M consists of those elements g of M + for which there exists an m > 0 such that t m g ( t ) is increasing. The subclass M 0 consists of those elements g of M + which are decreasing.

The Riesz potential operator R Ω s , 0 < s < n , n 1 is defined formally by
R Ω s f ( x ) = Ω f ( y ) | x y | s n d y , f M + ; | Ω | = 1 .
(1.1)
We shall consider rearrangement invariant quasi-Banach spaces E, continuously embedded in L 1 ( R n ) + L ( R n ) , such that the quasi-norm f E in E is generated by a quasi-norm ρ E , defined on M + with values in [ 0 , ] , in the sense that f E = ρ E ( f ) . In this way equivalent quasi-norms ρ E give the same space E. We suppose that E is nontrivial. Here f is the decreasing rearrangement of f, given by
f ( t ) = inf { λ > 0 : μ f ( λ ) t } , t > 0 ,
where μ f is the distribution function of f, defined by
μ f ( λ ) = | { x R n : | f ( x ) | > λ } | n ,

| | n denoting the Lebesgue n-measure.

Note that f ( t ) = 0 , if t > 1 .

There is an equivalent quasi-norm ρ p that satisfies the triangle inequality ρ p p ( g 1 + g 2 ) ρ p p ( g 1 ) + ρ p p ( g 2 ) for some p ( 0 , 1 ) that depends only on the space E (see [1]).

We say that the norm ρ E is K-monotone (cf. [2], p.84, and also [3], p.305) if
0 t g 1 ( s ) d s 0 t g 2 ( s ) d s implies ρ E ( g 1 ) ρ E ( g 2 ) , g 1 , g 2 M + .
(1.2)

Then ρ E is monotone, i.e., g 1 g 2 implies ρ E ( g 1 ) ρ E ( g 2 ) .

We use the notations a 1 a 2 or a 2 a 1 for non-negative functions or functionals to mean that the quotient a 1 / a 2 is bounded; also, a 1 a 2 means that a 1 a 2 and a 1 a 2 . We say that a 1 is equivalent to a 2 if a 1 a 2 .

We say that the norm ρ E satisfies the Minkovski inequality if for the equivalent quasi-norm ρ p ,
ρ p p ( g j ) ρ p p ( g j ) , g j M + .
(1.3)

For example, if E is a rearrangement invariant Banach function space as in [3], then by the Luxemburg representation theorem f E = ρ E ( f ) for some norm ρ E satisfying (1.2) and (1.3). More general example is given by the Riesz-Fischer monotone spaces as in [3], p.305.

Recall the definition of the lower and upper Boyd indices α E and β E . Let
h E ( u ) = sup { ρ E ( g u ) ρ E ( g ) : g M + } , g u ( t ) : = g ( t / u )
be the dilation function generated by ρ E . Then
α E : = sup 0 < t < 1 log h E ( t ) log t and β E : = inf 1 < t < log h E ( t ) log t .

If ρ E is monotone, then the function h E is submultiplicative, increasing, h E ( 1 ) = 1 , h E ( u ) h E ( 1 / u ) 1 , hence 0 α E β E . If ρ E is K-monotone, then by interpolation (analogously to [3], p.148), we see that h E ( s ) max ( 1 , s ) . Hence in this case we have also β E 1 .

Using the Minkovski inequality for the equivalent quasi-norm ρ p and monotonicity of f , we see that
ρ E ( f ) ρ E ( f ) if  β E < 1 ,
(1.4)
where f ( t ) = 1 t 0 t f ( s ) d s . The main goal of this paper is to prove continuity of the Riesz potential operator R Ω s : E G in optimal couples of rearrangement invariant function spaces E and G, where f G : = ρ G ( f ) . It is convenient to introduce the following classes of quasi-norms, where the optimality of R Ω s : E G is investigated. Let N d stand for all domain quasi-norms ρ E , which are monotone, rearrangement invariant, satisfying Minkowski’s inequality, ρ E ( χ ( 0 , 1 ) ) < and
E L 1 ( Ω ) .
(1.5)
Let N t consist of all target quasi-norms ρ G that are monotone, satisfy Minkowski’s inequality, ρ G ( χ ( 0 , 1 ) ) < , ρ G ( χ ( 1 , ) t s / n 1 ) < and
G Λ ( t 1 s / n ) ( R n ) ,
(1.6)

where χ ( a , b ) is the characteristic function of the interval ( a , b ) , 0 < a < b . Note that technically it is more convenient not to require that the target quasi-norm ρ G is rearrangement invariant. Of course, the target space G is rearrangement invariant, since f G = ρ G ( f ) . Finally, let N : = N d × N t .

Definition 1.1 (Admissible couple)

We say that the couple ( ρ E , ρ G ) N is admissible for the Riesz potential if the following estimate is valid:
ρ G ( ( R Ω s f ) ) ρ E ( f ) .
(1.7)

Moreover, ρ E ( E ) is called domain quasi-norm (domain space), and ρ G (G) is called a target quasi-norm (target space).

For example, by Theorem 2.2 below (the sufficient part), the couple E = Λ q ( t s / n w ) ( Ω ) , G = Λ q ( v ) , 1 q , is admissible if β E < 1 and v is related to w by the Muckenhoupt condition [4]:
( 0 t [ v ( s ) ] q d s / s ) 1 / q ( t [ w ( s ) ] r d s / s ) 1 / r 1 , 1 / q + 1 / r = 1 .
(1.8)

Definition 1.2 (Optimal target quasi-norm)

Given the domain quasi-norm ρ E N d , the optimal target quasi-norm, denoted by ρ G ( E ) , is the strongest target quasi-norm, i.e.,
ρ G ( g ) ρ G ( E ) ( g ) , g M + ,
(1.9)

for any target quasi-norm ρ G N t such that the couple ρ E , ρ G is admissible.

Definition 1.3 (Optimal domain quasi-norm)

Given the target quasi-norm ρ G N t , the optimal domain quasi-norm, denoted by ρ E ( G ) , is the weakest domain quasi-norm, i.e.,
ρ E ( G ) ( g ) ρ E ( g ) , g M + ,
(1.10)

for any domain quasi-norm ρ E N d such that the couple ρ E , ρ G is admissible.

Definition 1.4 (Optimal couple)

The admissible couple ρ E , ρ G is said to be optimal if ρ E = ρ E ( G ) and ρ G = ρ G ( E ) .

The optimal quasi-norms are uniquely determined up to equivalence, while the corresponding optimal quasi-Banach spaces are unique.

2 Admissible couples

Here we give a characterization of all admissible couples ( ρ E , ρ G ) N . It is convenient to define the case β E = 1 as limiting and the case β E < 1 as sublimiting.

Theorem 2.1 (General case β E 1 )

The couple ( ρ E , ρ G ) N is admissible if and only if
ρ G ( χ ( 0 , 1 ) S 1 g ) ρ E ( g ) , g M + or g M 0 ,
(2.1)
where
S 1 g ( t ) : = { t s / n 1 0 t g ( u ) d u + t 1 u s / n g ( u ) d u / u , 0 < t < 1 , 0 < s < n , n 1 , t s / n 1 0 1 g ( u ) d u , t > 1 , 0 < s < n , n 1 .
(2.2)
Proof First we prove
( R Ω s f ) S 1 f .
(2.3)
We are going to use real interpolation for quasi-Banach spaces. First we recall some basic definitions. Let ( A 0 , A 1 ) be a couple of two quasi-Banach spaces (see [2, 5]) and let
K ( t , f ) = K ( t , f ; A 0 , A 1 ) = inf f = f 0 + f 1 { f 0 A 0 + t f 1 A 1 } , f A 0 + A 1
be the K-functional of Peetre (see [2]). By definition, the K-interpolation space A Φ = ( A 0 , A 1 ) Φ has a quasi-norm
f A Φ = K ( t , f ) Φ ,
where Φ is a quasi-normed function space with a monotone quasi-norm on ( 0 , ) with the Lebesgue measure and such that min { 1 , t } Φ . Then (see [5])
A 0 A 1 A Φ A 0 + A 1 ,

where by X Y we mean that X is continuously embedded in Y. If g Φ = ( 0 t θ q g q ( t ) d t / t ) 1 / q , 0 < θ < 1 , 0 < q , we write ( A 0 , A 1 ) θ , q instead of ( A 0 , A 1 ) Φ (see [2]).

Using the Hardy-Littlewood inequality R n | f ( x ) g ( x ) | d x 0 f ( t ) g ( t ) d t , we get the well-known mapping property
R Ω s : Λ 1 ( t s / n ) ( Ω ) L ( R n )
and by the Minkovski inequality for the norm f we get
R Ω s : L 1 ( Ω ) Λ ( t 1 s / n ) ( R n ) .
Hence
t 1 s / n ( R Ω s f ) ( t ) K ( t 1 s / n , f ; L 1 ( Ω ) , Λ 1 ( t s / n ) ( Ω ) ) ,
therefore (see [2], Section 5.7)
t 1 s / n ( R Ω s f ) ( t ) { 0 t f ( u ) d u + t 1 s / n t 1 u s / n f ( u ) d u / u , 0 < t < 1 , 0 1 f ( u ) d u , t > 1 ,
implies
( R Ω s f ) ( t ) S 1 f ( t ) .

It is clear that (1.7) follows from (2.1) and (2.3).

Now we prove that (1.7) implies (2.1). To this end we choose the test function in the form f ( x ) = g ( c | x | n ) , g M + , so that f ( t ) = g ( t ) for some positive constant c (cf. [6]). Then
R Ω s f ( x ) = | y | < | x | g ( c | y | n ) | x y | s n d y + | y | > | x | g ( c | y | n ) | x y | s n d y ,
whence
| R Ω s f ( x ) | | x | s n 0 c | x | n g ( u ) d u + c | x | n | Ω | = 1 u s / n 1 g ( u ) d u χ ( 0 , 1 ) ( S 1 g ) ( c | x | n ) .
Note that χ ( 0 , 1 ) S 1 g χ ( 0 , 1 ) Q 1 T 1 g + χ ( 0 , 1 ) 0 1 g ( u ) d u , where
Q 1 g : = t 1 g ( u ) d u / u , t < 1 ,
and
T 1 g ( t ) : = { t s / n 1 0 t g ( u ) d u , 0 < t < 1 , 0 < s < n , n 1 , t s / n 1 0 1 g ( u ) d u , t > 1 , 0 < s < n , n 1 ,
hence χ ( 0 , 1 ) S 1 g is decreasing, therefore
| R Ω s f | ( t ) χ ( 0 , 1 ) S 1 g ( t ) .
(2.4)

Thus, if (1.7) is given, then (2.4) implies (2.1). □

In the sublimiting case β E < 1 we can simplify the condition (2.1), replacing S 1 by T 1 . Here
T 1 g ( t ) : = { t s / n 1 t 1 u s / n g ( u ) d u / u , 0 < t < 1 , 0 < s < n , n 1 , 0 , t > 1 .
(2.5)

Theorem 2.2 (Sublimiting case β E < 1 )

The couple ( ρ E , ρ G ) N is admissible if and only if
ρ G ( χ ( 0 , 1 ) T 1 g ) ρ E ( g ) , g M ,
(2.6)
where we recall that
M : = { g M + and t m g ( t ) is increasing for some m > 0 } .
Proof Let ρ E , ρ G be an admissible couple, then
ρ G ( χ ( 0 , 1 ) S 1 g ) ρ E ( g ) .
Since ρ G ( χ ( 0 , 1 ) T 1 g ) ρ G ( χ ( 0 , 1 ) S 1 g ) , it follows that ρ G ( χ ( 0 , 1 ) T 1 g ) ρ E ( g ) , g M . Now we need to prove sufficiency of (2.6). We have
χ ( 0 , 1 ) S 1 g χ ( 0 , 1 ) T 1 g + χ ( 0 , 1 ) g ( 1 ) ,
so
ρ G ( χ ( 0 , 1 ) S 1 g ) ρ G ( χ ( 0 , 1 ) T 1 g ) + ρ G ( χ ( 0 , 1 ) ) g ( 1 )
implies
ρ G ( χ ( 0 , 1 ) S 1 g ) ρ E ( g ) .

 □

In the subcritical case α E > s / n we have another simplification of (2.1).

Theorem 2.3 (Case α E > s / n )

The couple ( ρ E , ρ G ) N is admissible if and only if
ρ G ( χ ( 0 , 1 ) T 1 g ) ρ E ( g ) , g M 0 : = { g M + , g is decreasing } ,
(2.7)
where
T 1 g ( t ) : = { t s / n 1 0 t g ( u ) d u , 0 < t < 1 , 0 < s < n , n 1 , t s / n 1 0 1 g ( u ) d u , t > 1 , 0 < s < n , n 1 .
Proof Let ( ρ E , ρ G ) N be admissible, then
ρ G ( χ ( 0 , 1 ) S 1 g ) ρ E ( g ) , g M 0 .
As
ρ G ( χ ( 0 , 1 ) T 1 g ) ρ G ( χ ( 0 , 1 ) S 1 g ) ,
we have
ρ G ( χ ( 0 , 1 ) T 1 g ) ρ E ( g ) .
For the reverse, it is enough to check that (2.7) implies (2.1) for g M 0 , or
ρ G ( χ ( 0 , 1 ) T 1 g ) ρ E ( g ) , g M 0 .
As
χ ( 0 , 1 ) T 1 g χ ( 0 , 1 ) T 1 ( t s / n χ ( 0 , 1 ) T 1 g ) ,
so
ρ G ( χ ( 0 , 1 ) T 1 g ) ρ E ( t s / n χ ( 0 , 1 ) T 1 g ) ρ E ( t s / n Q 1 ( t s / n g ) ) ρ E ( g ) .
Here we use
ρ E ( Q 1 ( t s / n g ) ) ρ E ( t s / n g ) , g M 0 , α E > s / n , t < 1 .

 □

2.1 Optimal quasi-norms

Here we give a characterization of the optimal domain and optimal target quasi-norms. We can define an optimal target quasi-norm by using Theorem 2.1.

Definition 2.4 (Construction of the optimal target quasi-norm)

For a given domain quasi-norm ρ E N d we set
ρ G E ( χ ( 0 , 1 ) g ) : = inf { ρ E ( h ) : χ ( 0 , 1 ) g χ ( 0 , 1 ) S 1 h , h M + } , g M + .
(2.8)
Then
ρ G ( E ) ( g ) : = ρ G E ( χ ( 0 , 1 ) g ) + sup t > 1 t 1 s / n g .
Theorem 2.5 Let ρ E N d be a given domain quasi-norm. Then ρ G ( E ) N t , the couple ρ E , ρ G ( E ) is admissible and the target quasi-norm is optimal. By definition,
G ( E ) : = { f M : lim t f ( t ) = 0 , ρ G ( E ) ( f ) < } .
(2.9)
Proof To see that ρ G ( E ) is a quasi-norm, we first prove (1.6), for that we first prove
sup 0 < t < 1 t 1 s / n g ρ G E ( g ) , g M + .
(2.10)
Take g M + and consider an arbitrary h M + such that, for t < 1 , g S 1 h . By the Hardy inequality g S 1 ( h ) . Then,
t 1 s / n g K ( t 1 s / n , h ; L 1 ( Ω ) , Λ 1 ( t s / n ) ( Ω ) ) .
Hence
sup 0 < t < 1 t 1 s / n g K ( 1 , h ; L 1 ( Ω ) , Λ 1 ( t s / n ) ( Ω ) ) ρ E ( h ) .
Taking the infimum over all h such that g S 1 h , we get (2.10). Hence G E Λ ( t 1 s / n ) ( 0 , 1 ) , also ρ G ( χ ( 1 , ) g ) = sup t > 1 t 1 s / n g . And these two together give (1.6). ρ G ( E ) is indeed a quasi-norm on M + . Since χ ( 0 , 1 ) ( R Ω s f ) χ ( 0 , 1 ) S 1 f , which gives ρ G E ( χ ( 0 , 1 ) ( R Ω s f ) ) ρ E ( f ) . Also
sup t > 1 t 1 s / n ( R Ω s f ) sup t > 1 t 1 s / n S 1 f = 0 1 f ( u ) d u ρ E ( f ) .
Hence ρ E , ρ G ( E ) is admissible couple. Now we are going to prove that ρ G ( E ) is optimal. For this purpose, suppose that the couple ( ρ E , ρ G 1 ) N is admissible. Then by Theorem 2.1,
ρ G 1 ( χ ( 0 , 1 ) S 1 g ) ρ E ( g ) , g M + .
Therefore if χ ( 0 , 1 ) g χ ( 0 , 1 ) S 1 h , h M + , then
ρ G 1 ( χ ( 0 , 1 ) g ) ρ G 1 ( χ ( 0 , 1 ) S 1 h ) ρ E ( h ) ,
so taking the infimum on the right-hand side, we get
ρ G 1 ( χ ( 0 , 1 ) g ) ρ G E ( χ ( 0 , 1 ) g ) ,

hence ρ G 1 ( g ) ρ G ( E ) ( g ) . □

In the sublimiting case β E < 1 we can simplify the optimal target quasi-norm.

Theorem 2.6 If ρ E N d be a given domain quasi-norm. Then for g M + ,
ρ G E ( χ ( 0 , 1 ) g ) ρ ( χ ( 0 , 1 ) g ) , ρ ( χ ( 0 , 1 ) g ) : = inf { ρ E ( h ) : χ ( 0 , 1 ) g χ ( 0 , 1 ) T 1 h , h M } ,
(2.11)
i.e.,
ρ G ( E ) ( g ) ρ ( χ ( 0 , 1 ) g ) + sup t > 1 t 1 s / n g .
Proof If χ ( 0 , 1 ) g χ ( 0 , 1 ) T 1 h , h M , then χ ( 0 , 1 ) g χ ( 0 , 1 ) S 1 h , therefore
ρ G E ( χ ( 0 , 1 ) g ) ρ E ( h )
and taking the infimum, we get
ρ G E ( χ ( 0 , 1 ) g ) ρ ( χ ( 0 , 1 ) g ) .

Now for the reverse, let χ ( 0 , 1 ) g χ ( 0 , 1 ) S 1 h , h M + .

Then
χ ( 0 , 1 ) g χ ( 0 , 1 ) S 1 ( h ) χ ( 0 , 1 ) T 1 ( h ) + χ ( 0 , 1 ) f ( 1 ) ,
so
χ ( 0 , 1 ) g χ ( 0 , 1 ) f ( 1 ) χ ( 0 , 1 ) T 1 ( h ) ,
which gives, since h M ,
ρ ( χ ( 0 , 1 ) g χ ( 0 , 1 ) f ( 1 ) ) ρ E ( h ) ρ E ( h ) ρ E ( h ) ,
and this implies
ρ ( χ ( 0 , 1 ) g ) ρ E ( h ) + f ( 1 ) ,
which gives
ρ ( χ ( 0 , 1 ) g ) ρ E ( h ) .

Taking the infimum, we get ρ ( χ ( 0 , 1 ) g ) ρ G E ( χ ( 0 , 1 ) g ) , hence ρ ( χ ( 0 , 1 ) g ) ρ G E ( χ ( 0 , 1 ) g ) . □

A simplification of the optimal target quasi-norm is possible also in the subcritical case α E > s / n .

Theorem 2.7 Let ρ E N d be a given domain quasi-norm. Then for g M + ,
ρ G E ( χ ( 0 , 1 ) g ) ρ 1 ( χ ( 0 , 1 ) g ) , ρ 1 ( χ ( 0 , 1 ) g ) : = inf { ρ E ( h ) : χ ( 0 , 1 ) g T 1 h , h M 0 } ,
(2.12)
i.e.,
ρ G ( E ) ( g ) ρ 1 ( χ ( 0 , 1 ) g ) + sup t > 1 t 1 s / n g .
Proof If χ ( 0 , 1 ) g χ ( 0 , 1 ) T 1 h , h M 0 , then
χ ( 0 , 1 ) g χ ( 0 , 1 ) S 1 h .
Therefore
ρ G E ( χ ( 0 , 1 ) g ) ρ E ( h ) ,
and taking the infimum, we get
ρ G E ( χ ( 0 , 1 ) g ) ρ 1 ( χ ( 0 , 1 ) g ) .
For the reverse, let χ ( 0 , 1 ) g χ ( 0 , 1 ) S 1 h . Then χ ( 0 , 1 ) g χ ( 0 , 1 ) T 1 ( h ) + χ ( 0 , 1 ) T 1 ( h ) . As
χ ( 0 , 1 ) T 1 g χ ( 0 , 1 ) T 1 ( t s / n χ ( 0 , 1 ) T 1 g ) ,
we get
χ ( 0 , 1 ) g χ ( 0 , 1 ) T 1 ( h + t s / n χ ( 0 , 1 ) T 1 ( h ) ) ,
whence
ρ 1 ( χ ( 0 , 1 ) g ) ρ E ( t s / n χ ( 0 , 1 ) T 1 ( h ) ) + ρ E ( h ) ρ E ( t s / n Q 1 ( t s / n h ) ) + ρ E ( h ) ρ E ( h ) ,
where we use
ρ E ( Q 1 ( t s / n g ) ) ρ E ( t s / n g ) , g M 0 , α E > s / n , t < 1 .
Therefore, taking the infimum we arrive at
ρ 1 ( g ) ρ G E ( g ) .

 □

We can construct an optimal domain quasi-norm ρ E ( G ) by Theorem 2.1 as follows.

Definition 2.8 (Construction of an optimal domain quasi-norm)

For a given target quasi-norm ρ G N t , we construct an optimal domain quasi-norm ρ E ( G ) by
ρ E ( G ) ( g ) : = ρ G ( χ ( 0 , 1 ) S 1 g ) , g M + .
(2.13)

Theorem 2.9 If ρ G N t is a given target quasi-norm, then the domain quasi-norm ρ E ( G ) is optimal. Moreover, if β G < 1 s / n , then the couple ρ E ( G ) , ρ G is optimal.

Proof Since χ ( 0 , 1 ) S 1 g χ ( 0 , 1 ) T 1 g + χ ( 0 , 1 ) g ( 1 ) , so
ρ E ( G ) ( g ) ρ G ( χ ( 0 , 1 ) T 1 g + χ ( 0 , 1 ) g ( 1 ) ) ,
it follows that ρ E ( G ) is a quasi-norm. To prove the property (1.5), we notice that
ρ E ( G ) ( f ) = ρ G ( χ ( 0 , 1 ) S 1 f ) ρ G ( χ ( 0 , 1 ) ) ( S f ) ( 1 ) 0 1 f ( t ) d t f L 1 ( Ω ) .
The couple ρ E ( G ) , ρ G is admissible since ρ E ( G ) ( g ) = ρ G ( χ ( 0 , 1 ) S 1 g ) ρ G ( χ ( 0 , 1 ) S 1 g ) . Moreover, ρ E ( G ) is optimal, since for any admissible couple ( ρ E 1 , ρ G ) N we have ρ G ( χ ( 0 , 1 ) S 1 h ) ρ E 1 ( h ) , h M + . Therefore,
ρ E ( G ) ( g ) ρ E 1 ( g ) .
To check that if β G < 1 s / n , the couple ρ E ( G ) , ρ G is optimal, we need only to prove that ρ G is an optimal target quasi-norm, i.e., ρ ( g ) ρ G ( g ) , where ρ = ρ G ( E ( G ) ) is defined by (2.11), since β E ( G ) < 1 . We have χ ( 0 , 1 ) g ( t ) χ ( 0 , 1 ) g ( 1 ) = χ ( 0 , 1 ) T 1 h , where h ( t ) = t s / n [ g ( t ) g ( t ) ] M , t < 1 , therefore,
ρ G E ( G ) ( χ ( 0 , 1 ) g ( t ) χ ( 0 , 1 ) g ( 1 ) ) ρ E ( G ) ( h ) = ρ G ( χ ( 0 , 1 ) S 1 h )
implies
ρ G E ( G ) ( χ ( 0 , 1 ) g ( t ) ) ρ G ( χ ( 0 , 1 ) S 1 h ) + g ( 1 ) ,
since
χ ( 0 , 1 ) S 1 h = χ ( 0 , 1 ) t s / n h + χ ( 0 , 1 ) T 1 h χ ( 0 , 1 ) t s / n h + χ ( 0 , 1 ) T 1 h ,
so
ρ G E ( G ) ( χ ( 0 , 1 ) g ( t ) ) ρ G ( χ ( 0 , 1 ) t s / n h ) + ρ G ( χ ( 0 , 1 ) T 1 h ) + g ( 1 ) .
Now we define
P 1 g ( t ) : = 1 t 0 t g ( u ) d u , t < 1 .
For t < 1 , since h Q 1 h , we have h = P 1 h Q 1 P 1 h , therefore T 1 h T 1 Q 1 ( P 1 h ) T 1 ( P 1 h ) . Also T 1 ( P 1 h ) T 1 h + t s / n P 1 h and P 1 h h . Therefore,
ρ G E ( G ) ( χ ( 0 , 1 ) g ) ρ G ( χ ( 0 , 1 ) T 1 h ) + ρ G ( χ ( 0 , 1 ) t s / n h ) + g ( 1 ) ρ G ( χ ( 0 , 1 ) g ) + ρ G ( χ ( 0 , 1 ) t s / n h ) + g ( 1 ) .
For t < 1 , since h ( t ) t s / n g ( t ) we have h ( t ) t s / n g , therefore using β G < 1 s / n , Minkowski’s inequality, and monotonicity of ρ G , we have
ρ G ( χ ( 0 , 1 ) t s / n h ) ρ G ( χ ( 0 , 1 ) g ) .
Thus
ρ G E ( G ) ( χ ( 0 , 1 ) g ) ρ G ( χ ( 0 , 1 ) g ) ρ G ( χ ( 0 , 1 ) g ) ,

hence ρ ( g ) ρ G ( g ) . □

Example 2.10 If G = C 0 consists of all bounded continuous functions such that f ( ) = 0 and ρ G ( g ) = g ( 0 ) = g ( 0 ) , then α G = β G = 0 and ρ E ( G ) ( g ) 0 1 t s / n g d t / t , i.e., E = Γ 1 ( t s / n ) ( Ω ) and the couple E, G is optimal.

Example 2.11 Let G = Λ ( v ) with β G < 1 s / n and let
ρ E ( g ) = sup v ( t ) t 1 u s / n g ( u ) d u / u .

Then, the couple E, G is optimal and β E < 1 . In particular, this is true if v is slowly varying since then α G = β G = 0 and α E = β E = s / n < 1 .

2.2 Subcritical case

Here we suppose that s / n < α E .

Theorem 2.12 (Sublimiting case β E < 1 )

For a given domain quasi-norm ρ E N d with ρ E ( χ ( 0 , 1 ) ( t ) t s / n ) < , we have
ρ G E ( χ ( 0 , 1 ) g ) ρ E ( t s / n g ) ρ E ( t s / n g ) ,
(2.14)
i.e.,
ρ G ( E ) ( g ) ρ G E ( χ ( 0 , 1 ) g ) + sup t > 1 t 1 s / n g .

Moreover, the couple ρ E , ρ G ( E ) is optimal.

Proof If χ ( 0 , 1 ) g χ ( 0 , 1 ) T 1 h , h M 0 , then for t < 1 , t s / n g h , whence
ρ E ( t s / n g ) ρ E ( h ) ρ E ( h ) ρ E ( h ) .
Taking the infimum, we get
ρ E ( t s / n g ) ρ G E ( χ ( 0 , 1 ) g ) .

For the reverse, we notice that χ ( 0 , 1 ) T 1 ( t s / n g ) χ ( 0 , 1 ) g = g , hence ρ G E ( χ ( 0 , 1 ) g ) ρ E ( t s / n g ) .

It remains to prove that the domain quasi-norm ρ E is also optimal. Let ρ E 1 , ρ G ( E ) be an admissible couple in N . Then
ρ E 1 ( g ) ρ G ( E ) ( χ ( 0 , 1 ) S 1 g ) = ρ G E ( χ ( 0 , 1 ) S 1 g ) + sup t > 1 t 1 s / n χ ( 0 , 1 ) S 1 g ρ E ( t s / n χ ( 0 , 1 ) S 1 g ) + 0 ρ E ( t s / n χ ( 0 , 1 ) T 1 g ) ρ E ( χ ( 0 , 1 ) g ) ρ E ( χ ( 0 , 1 ) g ) ρ E ( g ) .

 □

Now we give an example.

Example 2.13 Let
E = Λ q ( t α w 1 ) ( Ω ) Λ r ( t β w 2 ) ( Ω ) , s / n < α < β < 1 , 0 < q , r ,
where w 1 and w 2 are slowly varying. Then we have α E = α , β E = β . Now by applying the previous theorem, we get
G ( E ) = Λ 0 q ( t α s / n w 1 ) Λ 0 r ( t β s / n w 2 ) ,

and the couple ( E , G ( E ) ) is optimal.

In the limiting case β E = 1 , the formula for the optimal target quasi-norm is more complicated.

Theorem 2.14 (Limiting case)

Let
ρ E ( g ) : = ρ H ( χ ( 0 , 1 ) g ) , ρ G 1 ( g ) : = ρ H ( t 1 sup 0 < u < t u 1 s / n g ( u ) ) ,
where ρ H is a monotone quasi-norm with α H = β H = 1 , ρ H ( χ ( 0 , 1 ) ) < , ρ H ( χ ( 1 , ) t 1 ) < and let
E : = { f M : t f ( t ) 0 as t 0 and ρ E ( f ) < } , G 1 : = { f M : sup 0 < u < t u 1 s / n f ( u ) 0 as t 0 and ρ G ( f ) < } .
Define
ρ G ( g ) : = ρ G 1 ( χ ( 0 , 1 ) g ) + sup t > 1 t 1 s / n g .

Then the couple ρ E , ρ G is optimal.

Proof Note that
E L 1 ( Ω ) .
Indeed, ρ E ( f ) = ρ H ( χ ( 0 , 1 ) g ) f ( 1 ) = 0 1 f ( u ) d u . Hence the above embedding follows. Consequently, ρ E N d . On the other hand,
ρ G ( f ) ρ H ( χ ( 1 , ) t 1 sup 0 < u < t u 1 s / n f ( u ) ) sup 0 < u < 1 u 1 s / n f ( u ) ρ H ( χ ( 1 , ) t 1 ) .
Hence G 1 Λ ( t 1 s / n ) ( 0 , 1 ) . This together with ρ G ( χ ( 1 , ) ) = sup t > 1 t 1 s / n g gives G Λ ( t 1 s / n ) . Then from the conditions on G 1 it follows that ρ G N t . Also, α E = β E = 1 and α G = β G = 1 s / n . On the other hand, if 0 < u < 1 , then
u 1 s / n ( R Ω s f ) ( u ) 0 u f ( v ) d v + u 1 s / n u 1 v s / n 1 f ( v ) d v .
For every ε > 0 , we can find a δ > 0 , such that v f ( v ) < ε for all 0 < v < δ . Then for 0 < t < 1 ,
sup 0 < u < t u 1 s / n ( R Ω s f ) ( u ) 0 t f ( v ) d v + ε + t 1 s / n δ 1 v s / n 1 f ( v ) d v .
(2.15)

Now it is easy to check that lim t 0 sup 0 < u < t u 1 s / n ( R Ω s f ) = 0 if f E .

To prove that R s : E G we need to check that the couple ρ E , ρ G is admissible. We write for t < 1 ,
T 1 g ( t ) = T 1 g ( t ) = t s / n g ( t ) , g M 0 .
Then
ρ G ( χ ( 0 , 1 ) T 1 g ) = ρ G 1 ( χ ( 0 , 1 ) T 1 g ) + sup t > 1 t 1 s / n χ ( 0 , 1 ) T 1 g = ρ H ( χ ( 0 , 1 ) t 1 sup 0 < u < t u 1 s / n T 1 g ( u ) ) + sup t > 1 t 1 s / n χ ( 0 , 1 ) T 1 g = ρ H ( χ ( 0 , 1 ) g ) = ρ E ( g ) .
To prove that the target space is optimal, notice first that
sup 0 < u < t u 1 s / n f ( u ) K ( t 1 s / n , f ; Λ ( t 1 s / n ) , L ) .
If f G , then by [2]
sup 0 < u < t u 1 s / n f ( u ) 0 t 1 s / n h 1 ( u ) d u ( where  h 1 , is decreasing ) 0 t h 1 ( v 1 s / n ) v s / n d v ( by a change of variables ) .
If h ( v ) = h 1 ( v 1 s / n ) v s / n then obviously h M 0 and
sup 0 < u < t u 1 s / n f ( u ) 0 t h ( v ) d v = t h ( t ) ,
whence
ρ E ( h ) ρ H ( χ ( 0 , 1 ) h ) ρ H ( χ ( 0 , 1 ) t 1 sup 0 < u < t u 1 s / n f ( u ) ) ρ G 1 ( χ ( 0 , 1 ) f ) .
On the other hand,
sup 0 < u < t u 1 s / n f ( u ) t h ( t )
implies t 1 s / n f ( t ) t h ( t ) , which gives f t s / n h , which implies χ ( 0 , 1 ) f χ ( 0 , 1 ) T 1 h , and therefore
ρ G E ( χ ( 0 , 1 ) f ) ρ E ( h ) ρ G 1 ( χ ( 0 , 1 ) f ) ,
proving optimality of G. To check optimality of E, we notice that
ρ E ( G ) ( h ) = ρ G ( χ ( 0 , 1 ) S 1 h ) ρ G ( χ ( 0 , 1 ) T 1 h ) ρ H ( t 1 sup 0 < u < t u 1 s / n χ ( 0 , 1 ) T 1 h ( u ) ) ρ H ( χ ( 0 , 1 ) h ) .
Hence
ρ E ( G ) ( h ) ρ E ( h ) .

 □

Example 2.15 Let E = Γ 0 ( t w ) ( Ω ) , consisting of all f Γ ( t w ) ( Ω ) such that t f ( t ) 0 as t 0 , w is slowly varying. Then β E = 1 . If G = Γ 1 ( t 1 s / n v ) Γ ( t w ) , where v ( t ) = sup u > t w ( u ) and
Γ 1 ( v ) : = { f Γ ( v ) : sup 0 < u < t u 1 s / n f ( u ) 0  as  t 0 } ,

then this couple is optimal. In particular, if w = 1 , then E = L 1 ( Ω ) and G = Γ 1 ( t 1 s / n ) .

Declarations

Acknowledgements

This study was supported by research funds from Dong-A University.

Authors’ Affiliations

(1)
Department of Mathematics and RINS, Gyeongsang National University
(2)
Department of Mathematics, Lahore Leads University
(3)
Department of Mathematics, Government College University
(4)
Centre for Advanced Studies in Pure and Applied Mathematic, Bahauddin Zakariya University
(5)
Department of Mathematics, Dong-A University

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© Kang et al.; licensee Springer. 2014

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