Open Access

On spectral properties of the modified convolution operator

  • Ainur Jumabayeva1, 2Email author,
  • Esmukhanbet Smailov3 and
  • Nazerke Tleukhanova2
Journal of Inequalities and Applications20122012:146

https://doi.org/10.1186/1029-242X-2012-146

Received: 16 December 2011

Accepted: 6 June 2012

Published: 25 June 2012

Abstract

We investigated the s-number of the modified convolution operator and obtained the following results

c 1 sup Q ? G 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | = ? f ? M p q = c 3 sup Q ? F 1 | Q | 1 p ' + 1 q | ? Q f ( s ) d s | ,

where 1 < p < 2 < q < 8 , p ' = p p - 1 , G is a set of all segments Q from [ 0 , 1 ] , F is a set of all compacts from [ 0 , 1 ] , | Q | is the measure of a set Q.

MSC:42A45, 44A35.

Keywords

Fourier series multipliersconvolution operators-numberLorentz spaceBesov space

1 Introduction

Let 1 = p < 8 , 0 < q = 8 . We denote by S p , q the space of all compact operators A, acting in the space L 2 [ 0 , 1 ] of all 1-periodic functions square integrable on [ 0 , 1 ] for s-numbers such that the following quasinorm is finite
? A ? S p , q = ( ? m = 1 8 s m q ( A ) m q / p - 1 ) 1 / q ,
if q < 8 , and
? A ? S p , 8 = sup m m 1 p s m if q = 8 .

Recall that the sequence s m ( A ) (s-numbers of operator A) are numerated eigenvalues of the operator A * A .

We consider the convolution operator
( A f ) ( y ) = ? 0 1 K ( x - y ) f ( x ) d x
acting in L 2 [ 0 , 1 ] . Given a function f ? L 1 [ 0 , 1 ] , we consider also the modified convolution operator
( A f f ) ( y ) = ? 0 1 ( K f ) ( x - y ) f ( x ) d x .
We say that f belongs to the space M p 0 , q 0 p 1 , q 1 , if for A ? S p 0 , q 0 , A f ? S p 1 , q 1 and
? A f ? S p 1 , q 1 = c ? A ? S p 0 , q 0 ,

where c > 0 depends only on p 0 , q 0 , p 1 , q 1 .

This means that the linear operator R f defined by the equality R f ( A ) = A f is bounded from S p 0 , q 0 to S p 1 , q 1 . Let
? f ? M p 0 , q 0 p 1 , q 1 = ? R f ? S p 0 , q 0 ? S p 1 , q 1 .

Given that the eigenvalues of the operator K * f coincide with the Fourier coefficients of the kernel K with respect to the trigonometric system, in the case p 0 = p 1 = q 0 = q 1 = p this problem reduces to the well-known problem of Fourier series multipliers. Let K ? L 1 ( [ 0 , 1 ] ) and { a m ( K ) } m ? Z be the sequence of its Fourier coefficients with respect to the trigonometric system { e 2 p i k x } k ? Z . It is assumed that K is such that { a m ( K ) } m ? Z ? l p , 1 = p = 8 . Let T f = { a m ( K f ) } m ? Z ? l p . The problem is to determine conditions on the function f ensuring the boundedness of the operator T f : l p ? l p .

This problem was considered in the works of Stechkin [1], Hirschman [2], Edelstein [3], Birman and Solomyak [4], Karadzhov [5], and others.

We obtain sufficient conditions on a multiplier f ensuring that it belongs to the space M p 0 , q 0 p 1 , q 1 . These conditions are expressed in terms of Lorentz and Besov spaces. We also construct examples showing the sharpness of the obtained constants for corresponding embedding theorems.

2 Main results

Let f be a µ measurable function which takes finite values almost everywhere and let
m ( s , f ) = µ ( { x : x ? [ 0 , 1 ] , | f | > s } )
be its distribution function. The function
f * ( t ) = inf { s : m ( s , f ) = t }

is a nonincreasing rearrangement of f.

We say that a function f belongs to the Lorentz space L p , q if f is measurable and
? f ? L p , q = ( ? 0 8 ( t 1 p f * ( t ) ) q d t t ) 1 q < 8 ,
for 1 = q < 8 and
? f ? L p , 8 = sup t > 0 t 1 p f * ( t ) < 8 ,

for q = 8 .

Theorem 1 Let 1 < p 0 < 2 = p 1 , 1 = q 1 = q 0 = 8 , 1 r = 1 p 0 - 1 p 1 , 1 s = 1 q 1 - 1 q 0 and f ? L r , s [ 0 , 1 ] . If A ? S p 0 , q 0 , then A f ? S p 1 , q 1 and
? A f ? S p 1 , q 1 = c ? f ? L r , s ? A ? S p 0 , q 0 ,

i.e. L r , s [ 0 , 1 ] ? M p 0 , q 0 p 1 , q 1 .

In the following theorem the cases p = p o = q 0 , q = p 1 = q 1 are considered. The upper and the lower estimates of the norm ? f ? M p q ( M p q : = M p , p q , q ) are obtained.

Theorem 2 Let 1 < p < 2 < q < 8 , p ' = p p - 1 . Let G be a set of all segments Q from [ 0 , 1 ] , F be a set of all compacts from [ 0 , 1 ] , then
c 1 sup Q ? G 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | = ? f ? M p q = c 3 sup Q ? F 1 | Q | 1 p ' + 1 q | ? Q f ( s ) d s | ,

where | Q | is the measure of a set Q.

We shall define the class of generalized monotone functions for which the upper and the lower estimates coincide.

We say that function f is a generalized monotone function, if there exists a constant c > 0 such that for every x ? ( 0 , 1 ] the inequality
| f ( x ) | = c x | ? 0 x f ( y ) d y |

holds. The class of such functions is denoted by N .

Corollary 1 Let 1 < p < 2 < q < 8 . If f ? N , then f ? M p q if and only if
sup t > 0 t 1 p - 1 q f * ( t ) < 8 .

Moreover, ? f ? M p q ~ sup t > 0 t 1 p - 1 q f * ( t ) .

In case parameters p 0 , p 1 are both either less or greater than 2, we use the space of smooth functions.

Let 1 = p < 8 , a > 0 . We denote by B p , q a [ 0 , 1 ] the space of all measurable functions f on [ 0 , 1 ] such that
? f ? B p , q a = ( ? k = 0 8 ( 2 a k ? ? k f ? p ) q ) 1 q < 8
for 1 = q < 8 , and
? f ? B p , 8 a = sup k 2 a k ? ? k f ? p < 8

for q = 8 . Here { a m ( f ) } m ? N are the Fourier coefficients of the function f by trigonometric system { e 2 p i k x } k ? Z , ? k f = ? k f ( x ) = ? [ 2 k - 1 ] = | m | < 2 k a m ( f ) e 2 p i m x , and [ 2 k - 1 ] is the integer part of 2 k - 1 .

This class is called the Nikol’skii-Besov space.

Theorem 3 Let 1 < p 0 = p 1 < 8 , 2 ? [ p 0 , p 1 ] , 1 < q 0 = q 1 = 8 ,
a = min x ? [ 1 p 1 , 1 p 0 ] | 1 2 - x | , 1 r = max x ? [ 1 p 1 , 1 p 0 ] | 1 2 - x | , 1 - 1 s = 1 q 0 - 1 q 1

and f ? B r , s a [ 0 , 1 ] .

If A ? S p 0 , q 0 , then A f ? S p 1 , q 1 and
? A f ? S p 1 , q 1 = c ? f ? B r , s a ? A ? S p 0 , q 0 ,

i.e., B r , s a ? M p 0 , q 0 p 1 , q 1 .

In the case p 0 = p 1 = q 0 = q 1 , Karadzhov’s result (see [5]) follows from Theorem 3:
B r , 1 1 r ? M p = M p , p p , p , 1 r = | 1 p - 1 2 | .

Now consider the case 1 = q 1 < q 0 = 8 .

Theorem 4 Let 1 < p 0 < p 1 < 8 , 1 = q 1 < q 0 = 8 , 2 ? ( p 0 , p 1 ) , 1 r - a = 1 p 0 - 1 p 1 , 1 s = 1 q 1 - 1 q 0 , a > min x ? [ 1 p 1 , 1 p 0 ] | 1 2 - x | .

Then B r , s a [ 0 , 1 ] ? M p 0 , q 0 p 1 , q 1 .

3 Properties of M p 0 , q 0 p 1 , q 1 class

To prove the properties of M p 0 , q 0 p 1 , q 1 class we need the following lemma. We first define a discrete Lorentz space. l p q is called a discrete Lorentz space whose elements are sequences of numbers ? = { ? k } k = 8 8 with the only limit point 0 such that
? ? ? l p q = ( ? m = 1 8 | ? m * | q m q p - 1 ) 1 q , 1 = q < 8

where { ? m * } m = 1 8 nonincreasing rearrangement of the sequence { | ? k | } k = 8 8 .

For q = 8 ,
? ? ? l p 8 = sup m m 1 p ? m * .

Lemma 1 (See [6])

Let 1 < r , p 0 , p 1 < 8 , 1 = q 0 , q 1 , s = 8 . Then
? a * b ? l p 1 , q 1 = c ? b ? l r , s ? a ? l p 0 , q 0 ,

where 1 p 1 + 1 = 1 r + 1 p 0 , 1 q 1 = 1 s + 1 q 0 .

Let X ¯ = ( X 0 , X 1 ) , where X 0 , X 1 are Banach spaces, be a compatible pair. We define the functional K ( t , a ) for t > 0 and a ? X 0 + X 1 by the following formula:
K ( t , a ) = inf a = a 0 + a 1 ( ? a 0 ? X 0 + t ? a 1 ? X 1 ) .
We denote by X ¯ ? , q , k the space { a ? X 0 + X 1 : ? a ? ? , q , k = F ? , q ( K ( t , a ) ) } , where F ? , q is a functional defined on nonnegative functions f by formula
F ? , q ( f ( t ) ) = ( ? 0 8 ( t - ? f ( t ) ) q d t t ) 1 q , 1 = q < 8
and
F ? , 8 ( f ( t ) ) = sup t > 0 t - ? f ( t ) , q = 8 .

Let X a 1 0 , p 1 0 and X a 2 1 , p 2 1 be the spaces obtained by the method of real interpolation of Banach pairs of spaces ( X 0 1 , X 1 1 ) , ( X 0 2 , X 1 2 ) respectively.

Lemma 2 (See [7])

Let 0 < a i , ß i < 1 , 1 = p i , q i = 8 , i = 0 , 1 , a 0 ? a 1 , ß 0 ? ß 1 . If T is a bilinear operator:
T : X a 0 , p 0 × Y 0 ? Z ß 0 , q 0
and
T : X a 1 , p 1 × Y 1 ? Z ß 1 , q 1
then
T : X a , p × Y ? , r ? Z ß , q .

Here a = ( 1 - ? ) a 0 + ? a 1 , ß = ( 1 - ? ) ß 0 + ? ß 1 , 1 p + 1 r > 1 , 1 + 1 q = 1 p + 1 r + ( 1 - ? ) ( 1 q 0 - 1 p 0 ) + ? ( 1 q 1 - 1 p 1 ) + , x + = max ( x , 0 ) .

Remark Since the s-numbers of convolution operator A coincide with the modules of the Fourier coefficients of the kernel K, the problem of estimating the s-numbers of “transformed” operator A f can be reduced to the study of the following inequality
? a * b ? l p 1 , q 1 = c ? a ? l p 0 , q 0 ,
(1)

and we have to describe the class of those functions f with Fourier coefficients b = { b m } m ? Z , for which Inequality (1) holds.

Theorem 5
  1. (1)
    Let 1 = p 0 , p 1 < 8 , 1 = q 0 , q 1 = 8 , 1 p i + 1 p i ' = 1 q i + 1 q i ' = 1 , i = 0 , 1 . Then
    M p 0 , q 0 p 1 , q 1 = M p 1 ' , q 1 ' p 0 ' , q 0 ' .
     
  2. (2)
    Let 1 < p 0 < r 0 < p 1 ' < 8 , 1 p 1 + 1 p 1 ' = 1 , 1 p 1 - 1 p 0 = 1 r 1 - 1 r 0 , then
    M p 0 , q 0 p 1 , q 1 ? M r 0 , s r 1 , t ,
     

where 1 t - 1 s = ( 1 q 1 - 1 q 0 ) + .

Proof The proof of the first statement follows from Remark and from the fact that ? ( T f ) * ? = ? T f ¯ ? , where f ¯ is a complex conjugate of the function f. Now we prove (2).

Let f ? M p 0 , q 0 p 1 , q 1 , then by (1) it follows that f ? M p 1 ' , q 1 ' p 0 ' , q 0 ' , and
where 1 p i + 1 p i ' = 1 q i + 1 q i ' = 1 . According to Lemma 1, the operator T ( a , f ) = a * b
T : l p 0 , q 0 × M p 0 , q 0 p 1 , q 1 ? l p 1 , q 1
is bounded. Using (1) we have
T : l p 1 ' , q 1 ' × M p 0 , q 0 p 1 , q 1 ? l p 0 ' , q 0 ' .
Further, applying the theorem on bilinear interpolation (Lemma 2) we find that the operator
T : l r 0 , s × M p 0 , q 0 p 1 , q 1 ? l r 1 , t
is also bounded, i.e., M p 0 , q 0 p 1 , q 1 ? M r 0 , s r 1 , t , where
1 r 1 = 1 - ? p 1 + ? p 0 ' , 1 r 0 = 1 - ? p 0 + ? p 1 ' , 1 t - 1 s = ( 1 q 1 - 1 q 0 ) +
for every 0 < ? < 1 . Eliminating ? from this equation, we obtain that
1 p 1 - 1 p 0 = 1 r 1 - 1 r 0 ,

and the condition 0 < ? < 1 implies the condition 1 < p 0 < r 0 < p 1 ' < 8 , where 1 p 1 + 1 p 1 ' = 1 .

The proof is complete. ?

By (2), in particular, the following proposition follows.

Let 1 < p < r < p ' < 8 , 1 p + 1 p ' = 1 , then
M p , q ? M r , t ,

where M p , q = M p , q p , q and q , t ? [ 1 , 8 [ are any.

4 Proof of main results

For a given pair X ¯ = ( X 0 , X 1 ) we consider the space G ( X ¯ ) consisting of all functions f bounded and continuous in the strip
S = { z : 0 = Re z = 1 }
with values in X 0 + X 1 . Moreover, f are analytic in the open strip
S 0 = { z : 0 < Re z < 1 }
and such that the mapping t ? f ( j + i t ) ( j = 0 , 1 ) is a continuous function on the real axis with values in X j ( j = 0 , 1 ) which tends to 0 for | t | ? 8 . It is clear that G ( X ¯ ) is a vector space. We endow G with the norm
? f ? G = max ( sup t ? f ( i t ) ? X 0 , sup t ? f ( 1 + i t ) ? X 1 ) .
The space X ¯ [ ? ] 0 = ? = 1 consists of all elements a ? X 0 + X 1 such that a = f ( ? ) for some function f ? G ( X ¯ ) . The norm on X ¯ [ ? ] is equal to
? a ? [ ? ] = inf { ? f ? G : f ( ? ) = a , f ? G } .

In order to prove our main result, we need two lemmas in [8].

Lemma 3 (Bilinear interpolation, the complex method, see [8])

Let T be a bilinear operator such that
T : X 0 × Y 0 ? Z 0
and
T : X 1 × Y 1 ? Z 1 .
Then
T : X [ ? ] × Y [ ? ] ? Z [ ? ] ,

where X [ ? ] , Y [ ? ] , Z [ ? ] are the spaces obtained by the method of complex interpolation of Banach pairs of spaces ( X 0 , X 1 ) , ( Y 0 , Y 1 ) , ( Z 0 , Z 1 ) respectively.

Lemma 4 (Bilinear interpolation, the real method, see [8])

Let T be a bilinear operator such that
T : X 0 × Y 0 ? Z 0
and
T : X 1 × Y 1 ? Z 1
with the norms B 0 , B 1 respectively. Then
T : X ? , t 1 × Y ? , t 2 ? Z ? , s ,
where 1 s + 1 = 1 t 1 + 1 t 2 . Moreover,
? T ? = c B 0 1 - ? B 1 ? .
Proof of Theorem 1 First we prove the inequality:
? a * b ? l p 1 , q 1 = c ? f ? L r ? a ? l p 0 , q 0 ,
(2)

where b = { b k } k ? Z are Fourier coefficients of the function f.

If r = 2 , Inequality (2) follows by Lemma 1 and Hardy-Littlewood-Paley inequality [9]. Indeed, since f ? L r s , by the Hardy-Littlewood-Paley theorem, we have b ? l r ' s and the following inequality holds
? b ? l r ' s = c ? f ? L r s .
Taking s = r , we get
? b ? l r ' , r = c ? f ? L r .
Now let 2 < r < 8 . Let a ? l 2 f ~ ? k ? Z a k e 2 p i k x , then by Parseval’s equality we get
? a * b ? l 2 = ? f f ? L 2 = ? f ? L 2 ? f ? L 8 = ? f ? L 8 ? a ? l 2 ,
i.e., M 2 = L 8 . From Lemma 1, using Parseval’s equality we have
? a * b ? l p 1 , q 1 = c ? f ? L 2 ? a ? l p 0 , q 0 ,

where 1 p 1 + 1 = 1 2 + 1 p 0 1 p 1 + 1 2 = 1 p 0 1 q 1 = 1 q 0 + 1 2 .

Thus, for the bilinear operator T ( a , f ) = a * b we obtain
Applying the method of complex interpolation (Lemma 3), we obtain Inequality (2). Now we shall prove the inequality
? a * b ? l p 1 , q 1 = c ? f ? L r , s ? a ? l p 0 , q 0 ,
(3)

where 1 s = 1 q 1 - 1 q 0 .

Let q 0 = 8 and p 0 be fixed in Inequality (2). Taking 1 q 1 i = 1 r i i = 0 , 1 , choose parameters r 0 r 1 p 1 0 p 1 1 such that
1 p 0 = 1 p 1 i + 1 r i , i = 0 , 1 .
(4)
Then from Inequality (2) we have
Using Marcinkiewicz-Calderón interpolation theorem (see [8]), we get
? a * b ? l p 1 , s = ( c 1 ? a ? l p 0 , 8 ) ? ( c 2 ? a ? l p 0 , 8 ) 1 - ? ? f ? L r , s = c ? a ? l p 0 , 8 ? f ? L r , s ,
(5)

where 1 p 1 = 1 - ? p 1 0 + ? p 1 1 1 r = 1 - ? r 0 + ? r 1 , i.e., 1 p 0 - 1 p 1 = 1 r .

Now we apply Lemma 2 with fixed parameters r, s and parameters p 1 i , p 0 i , i = 0 , 1 satisfying (4) and the inequality of type (5). We have:
( L r , s , L r , s ) ? , 1 × ( l p 0 0 , 8 , l p 0 1 , 8 ) ? , q 0 ? ( l p 1 0 , s , l p 1 1 , s ) ? , q 1
or
T : L r , s × l p 0 , q 0 ? l p 1 , q 1 ,

where 1 q 1 - 1 q 0 = 1 s - 1 8 , 1 p 1 = 1 - ? p 1 0 + ? p 1 1 , 1 p 0 = 1 - ? p 0 0 + ? p 0 1 , i.e., 1 q 1 = 1 s + 1 q 0 , 1 p 0 - 1 p 1 = 1 r .

Since the parameters p 1 i , p 0 i , i = 0 , 1 are arbitrary in Inequality (5), it guarantees the arbitrary of the corresponding parameters in Inequality (4).

Thus, the following inequality holds:
? a * b ? l p 1 , q 1 = c ? a ? l p 0 , q 0 ? f ? L r , s ,

where b = { b m } m ? Z are Fourier coefficients of the function f and 1 r = 1 p 0 - 1 p 1 , 1 s = 1 q 1 - 1 q 0 . According to Remark, this inequality is equivalent to the statement of Theorem 1. ?

Proof of Theorem 2 Let f ? M p q and Q be an arbitrary segment in [ 0 , 1 ] ,
f 0 ( x ) = { 1 , x ? Q , 0 , x ? Q .
Note that by Boas theorem [10] (see also [11]) we get
? f 0 ˆ ? l p ~ ? f 0 ? L p ' , p = ( ? 0 1 ( t 1 p ' f 0 * ( t ) ) p d t t ) 1 p = | Q | 1 p ' .
(6)
Applying Theorem 5 from [12] and using (6), we obtain:
? f ? M p q = sup f ? 0 ? f f ˆ ? l q ? f ˆ ? l p = ? f 0 f ˆ ? l q ? f 0 ˆ ? l p = c | Q | 1 p ' ? 0 1 ( t 1 q ' ( sup | W | = t , W ? G 1 | W | | ? W f 0 ( x ) f ( x ) d x | ) q d t t ) 1 q = c | Q | 1 p ' sup t > 0 t 1 q ' ( sup | W | = t , W ? G 1 | W | | ? W n Q f ( x ) d x | ) = c 1 | Q | 1 p ' | Q | 1 q ' - 1 | ? Q f ( x ) d x | = c 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | .
Since the interval Q is arbitrary, we get
? f ? M p q = c 1 sup Q ? G 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | ,

where constants c and c 1 depend only on parameters p and q.

The proof of obtaining an upper estimate follows from Theorem 1 and the embedding l p , p ? l p , q , for p < q .

Indeed, from Theorem 1 it follows
L r , 8 ? M p q ,
i.e.,
? f ? M p q = c 2 sup t > 0 t 1 r f * ( t ) = c 3 sup t > 0 1 t 1 p ' + 1 q ? 0 t f * ( s ) d s = c 3 sup Q ? F 1 | Q | 1 p ' + 1 q ? Q | f ( x ) | d x ~ sup Q ? F 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | .

?

Proof of Corollary 1 Let Q be an arbitrary compact from F.

From the condition of generalized monotonicity of the function f we have
1 | Q | 1 p ' + 1 q | ? Q f ( y ) d y | = 1 | Q | 1 p ' + 1 q ? Q c y | ? 0 y f ( x ) d x | d y = c | Q | 1 p ' + 1 q sup A ? G 1 | A | 1 p ' + 1 q | ? A f ( x ) d x | ? Q d y y 1 - 1 q - 1 p ' = c 1 sup A ? G 1 | A | 1 p ' + 1 q | ? A f ( x ) d x | .
Taking into account that Q ? F is arbitrary, we have
sup Q ? F 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | = c sup Q ? G 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | .
Thus, from Theorem 2 we get
? f ? M p q ~ sup Q ? F 1 | Q | 1 p ' + 1 q | ? Q f ( x ) d x | ~ sup t > 0 t 1 p - 1 q f * ( t ) .

?

Proof of Theorem 3 Let 2 < p 0 = p 1 < 8 . For a sequence of numbers a = { a m } m ? Z and a function f ? L 1 [ 0 , 1 ] we consider the mapping T of the form T ( a , f ) = a * b , where b = { b m } m ? Z is the sequence of Fourier coefficients on the trigonometric system of functions f. This map is bilinear and from Karadzhov’s theorem [5] and Remark it follows that it is bounded from l 1 × B 2 , 1 1 2 to l 1 .

Since M 2 = L 8 , the mapping
T f : l 2 × L 8 ? l 2
is also bounded. Thus, for the operator T, the following is true
Then, by Lemma 4 on bilinear interpolation, we get
( l 1 , l 2 ) ? , q × ( B 2 , 1 1 2 , L 8 ) ? , 1 ? ( l 1 , l 2 ) ? , q ,
i.e., the operator T is bounded from l p , q × ( B 2 , 1 1 2 , L 8 ) ? , 1 to l p , q . In the paper [5] it is shown that B r , 1 1 r ? ( B 2 , 1 1 2 , L 8 ) ? , 1 , where 1 r = 1 - ? 2 . Thus, taking into account Theorem 5, we will get
T : l p , q × B r , 1 1 r ? l p , q ,
(7)

for 2 < p < 8 1 = q = 8 1 r = 1 2 - 1 p .

From Minkowski’s inequality and Parseval’s equality we get
T : l 1 × L 2 ? l 2 .
Thus, for the operator T, the following is true
Then, by Lemma 3 we get
( l p , q , l 1 ) [ ? ] × ( B r , 1 1 r , L 2 ) [ ? ] ? ( l p , q , l 2 ) [ ? ]

i.e., T is a bounded mapping from l p 0 , q 0 × B r , s a to l p 1 , q 1 , where 2 < p 0 = p 1 < 8 , 1 r = 1 2 - 1 p 1 , a = 1 2 - 1 p 0 . The arbitrary choice of parameters guarantees the arbitrary of the parameters available in the theorem. ?

The case 1 < p 0 < p 1 < 2 follows from the statements proved above and the property M p 0 , q 0 p 1 , q 1 = M p 1 ' , q 1 ' p 0 ' , q 0 ' .

Proof of Theorem 4 Let 1 < p 0 < p 1 = 2 . Let us consider the bilinear mapping T ( a , f ) = a * b , where b = { b m } m ? Z is the sequence of Fourier coefficients of the function f. The mapping
T : l p 0 , q 0 × L 2 ? l p 1 , q 1
(8)
is bounded according to Theorem 1. Here 1 p 0 - 1 p 1 = 1 2 , 1 q 1 - 1 q 0 = 1 2 , 1 < q 1 < 2 < q 0 , 1 < p 0 < 2 < p 1 . The result of Theorem 3, in the case q 0 = q 1 = 1 , p 0 = p 1 = p can be written as
T : l p , 1 × B t , 1 1 / t ? l p , 1 , 1 t = 1 p - 1 2 .
(9)
Applying Lemma 3 on the bilinear interpolation to (8) and (9), and taking into account the properties of the embedding of the spaces l p , q and B p , q a , we have:
T : l p 0 , 1 × B r , 1 a ? l p 1 , 8 ,
(10)
where parameters r, a, p 0 , p 1 satisfy the following conditions:
1 < p 0 < p 1 = 2 , 1 r - a = 1 p 0 - 1 p 1 , a > 1 p 1 - 1 2 .
(11)
Let in (11) parameter r be fixed. Using Lemma 4 on bilinear interpolation and taking into account that
( B r , 1 a 0 , B r , 1 a 1 ) ? , h = B r , h a , with a = ( 1 - ? ) a 0 + a 1 ,
we get
T : l p 0 , h 1 × B r , h 2 a ? l p , h 3 ,

where 1 h 1 + 1 = 1 h 2 + 1 h 3 , a > 1 p 1 - 1 2 = min x ? [ 1 p 1 , 1 p 0 ] | 1 2 - x | , 1 r - a = 1 p 0 - 1 p 1 .

Therefore, with fixed a ? l p 0 , 8 and r we obtain that
P a : B r , 1 a i ? l p 1 i , 8 ,
and
? P a ? B r , 1 a i ? l p 1 i , 8 = c i ? a ? l p 0 , 8 ,

where 1 r - a i = 1 p 0 - 1 p 1 i , a i > 1 p 1 i - 1 2 , i = 0 , 1 .

Using Marcinkiewicz-Calderón interpolation theorem we have
P a : B r , s a ? l p 1 , s ,
and
? P a ? B r , s a ? l p 1 , s = c ? a ? l p 0 , 8 .
Thus
T : l p 0 , 8 × B r , s a ? l p 1 , s .

To complete the proof we fix the function f and the parameters r, s, a and we choose the parameters p 0 i , p 1 i , i = 0 , 1 satisfying (11). We use Lemma 2 to get B r , s a [ 0 , 1 ] ? M p 0 , q 0 p 1 , q 1 . ?

The case 2 = p 0 < p 1 < 8 , as in the proof of Theorem 3, will follow from M p 0 , q 0 p 1 , q 1 = M p 1 ' , q 1 ' p 0 ' , q 0 ' .

5 Examples demonstrating the sharpness of the results

Proposition 1 Let 1 < p 0 < 2 = p 1 , 1 r = 1 p 0 - 1 p 1 , 1 s = ( 1 q 1 - 1 q 0 ) + . If q 1 < q 0 , then for any e > 0 there exists f 1 ? L r , s + e such that f 1 ? M p 0 , q 0 p 1 , q 1 , if q 1 = q 0 there exists f 2 ? L r - e , 8 such that f 2 ? M p 0 , q 0 p 1 , q 1 .

Proof Let e be an arbitrary positive number, and numbers ß 1 , ß 2 be such that
ß 1 > 1 s + e , ß 2 > 1 q 0 , ß 1 + ß 2 < 1 s + 1 q 0 = 1 q 1 .
Let
and
f 1 ~ ? k = - 8 + 8 b k e 2 p i k x .
Then for m ? 0
Thus, ( a * b ) m = c ( | m | + 1 ) - ( 1 r ' + 1 p 0 ) + 1 | ln ( | m | + 2 ) | - ß 1 - ß 2 . Since
? m = 0 + 8 ( ( ( m + 1 ) - ( 1 r ' + 1 p 0 ) + 1 | ln ( | m | + 2 ) | - ß 1 - ß 2 ) q 1 ( | m | + 1 ) ( q 1 p 1 - 1 ) ) = 8 ,

a * b ? l p 1 , q 1 , and therefore f 1 ? M p 0 , q 0 p 1 , q 1 . Since Fourier coefficients of f 1 are the sequence { b k } k ? Z it follows that f 1 ? L r , s .

To prove the second part of the proposition, we take s = 8 . Let numbers a 1 and a 2 be such that
a 1 > 1 ( r - e ) ' = 1 - 1 r - e , a 2 > 1 p 0 , a 1 + a 2 < 1 - 1 r + 1 p 0
(note that the last inequality does not contradict the previous two). Choosing
b k = 1 ( | k | + 1 ) a 1 , a k = 1 ( | k | + 1 ) a 2 , f 2 ~ ? k = - 8 + 8 b k e 2 p i k x ,
we can show that
a * b ~ { ( | k | + 1 ) - a 1 - a 2 + 1 } k ? Z .

Hence a * b ? l p 1 , q 1 , and therefore f 2 ? M p 0 q 0 p 1 q 1 . At the same time taking into account the monotonicity of the sequence { b k } k ? Z and Hardy-Littlewood theorem, we have that f 2 ? L r - e , 8 . The statement is proved. ?

Theorem 6 Let 1 < p 0 < p 1 < 2 , 1 < q 1 = q 0 , 1 r - a = 1 p 0 - 1 p 1 , 1 s = 1 q 1 - 1 q 0 . Then for any e > 0 there exist f 1 ? B r , 8 a - e n B r - e , 8 a and f 2 ? B r , s + e a such that f 1 ? M p 0 , q 0 p 1 , q 1 , f 2 ? M p 0 , q 0 p 1 , q 1 .

Proof Let s < 8 and numbers ß 1 , ß 2 be such that
ß 1 > 1 s + e , ß 2 > 1 q 0 , ß 1 + ß 2 < 1 s + 1 q 0 = 1 q 1 .
Let b = { b k } k ? Z and a = { a k } k ? Z , where

It is obvious that a ? l p 0 , q 0 , and f 2 ~ ? k = - 8 + 8 b k e 2 p i k x belongs to B r , s + e a .

It is easy to show that
( a * b ) m = c ( | m | + 1 ) a - 1 r + 1 p 0 ( ln ( | m | + 2 ) ) ß 1 + ß 2

and consequently, a * b ? l p 1 , q 1 . Therefore f 2 ? M p 0 , q 0 p 1 , q 1 .

To construct the function f 1 , it is sufficient to consider the sequences
b = { 1 ( | m | + 1 ) ? 1 } m ? Z , a = { 1 ( | m | + 1 ) ? 2 } m ? Z ,
where
? 1 > max ( a - e - 1 r + 1 , a - 1 r + e + 1 ) , ? 2 > 1 p 0
and
? 1 + ? 2 < a - 1 r + 1 p 0 .

f 1 ~ ? k = - 8 + 8 b k e 2 p i k x . The proof that f 1 ? B r , 8 a - e n B r - e , 8 a , f 1 ? M p 0 , q 0 p 1 , q 1 is similar to the proof of the first part. ?

Declarations

Acknowledgements

The authors thank the referees for their careful reading and valuable comments on how to improve the article. This paper was partly supported by the grant MTM 2011-27637.

Authors’ Affiliations

(1)
Centre de Recerca Matemàtica, Bellaterra, Spain
(2)
Faculty of Mechanics and Mathematics, L.N. Gumilyov Eurasian National University, Astana, Kazakhstan
(3)
Institute of Applied Mathematics, Karaganda, Kazakhstan

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