1 Introduction

Let ℳ 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, n1 is defined formally by

R Ω s f(x)= Ω f(y) | x y | s n dy,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)ds 0 t g 2 (s)dsimplies 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)ds. The main goal of this paper is to prove continuity of the Riesz potential operator R Ω s :EG 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 :EG 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), 1q, 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 + org 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 XY 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)|dx 0 f (t) g (t)dt, 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 dy+ | y | > | x | g ( c | y | n ) | x y | s n dy,

whence

| R Ω s f ( x ) | | x | s n 0 c | x | n g(u)du+ c | x | n | Ω | = 1 u s / n 1 g(u)du 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)du, where

Q 1 g:= t 1 g(u)du/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),gM,
(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), gM. 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)du 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, hM, 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 <1s/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 <1s/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)du,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 <1s/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 dt/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 <1s/n and let

ρ E (g)=supv(t) t 1 u s / n g (u)du/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)du. 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 =1s/n. On the other hand, if 0<u<1, then

u 1 s / n ( R Ω s f ) (u) 0 u f (v)dv+ u 1 s / n u 1 v s / n 1 f (v)dv.

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)dv+ε+ t 1 s / n δ 1 v s / n 1 f (v)dv.
(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 fE.

To prove that R s :EG 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 fG, 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)dv=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 (tw)(Ω), consisting of all f Γ (tw)(Ω) such that t f (t)0 as t0, w is slowly varying. Then β E =1. If G= Γ 1 ( t 1 s / n v) Γ (tw), 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 ).