Research

# On spectral properties of the modified convolution operator

Ainur Jumabayeva1,2*, Esmukhanbet Smailov3 and Nazerke Tleukhanova2

Author Affiliations

1 Centre de Recerca Matemàtica, Campus de Bellaterra, Edifici C, Barcelona, Bellaterra, 08193, Spain

2 Faculty of Mechanics and Mathematics, L.N. Gumilyov Eurasian National University, Munaitpasova 5, Astana, 010008, Kazakhstan

3 Institute of Applied Mathematics, Gogolya 38, Karaganda, 470055, Kazakhstan

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Journal of Inequalities and Applications 2012, 2012:146 doi:10.1186/1029-242X-2012-146

 Received: 16 December 2011 Accepted: 6 June 2012 Published: 25 June 2012

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### 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 φ ( x ) d x | φ M p q c 3 sup Q F 1 | Q | 1 p + 1 q | Q φ ( s ) d s | ,

where 1 < p < 2 < q < , 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 multipliers; convolution operator; s-number; Lorentz space; Besov space

### 1 Introduction

Let 1 p < , 0 < q . 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 s m q ( A ) m q / p 1 ) 1 / q ,

if q < , and

A S p , = sup m m 1 p s m if q = .

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 φ L 1 [ 0 , 1 ] , we consider also the modified convolution operator

( A φ f ) ( y ) = 0 1 ( K φ ) ( x y ) f ( x ) d x .

We say that φ belongs to the space M p 0 , q 0 p 1 , q 1 , if for A S p 0 , q 0 , A φ S p 1 , q 1 and

A φ 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 φ defined by the equality R φ ( A ) = A φ is bounded from S p 0 , q 0 to S p 1 , q 1 . Let

φ M p 0 , q 0 p 1 , q 1 = R φ 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 π i k x } k Z . It is assumed that K is such that { a m ( K ) } m Z l p , 1 p . Let T φ = { a m ( K φ ) } m Z l p . The problem is to determine conditions on the function φ ensuring the boundedness of the operator T φ : 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 φ 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 ( σ , f ) = μ ( { x : x [ 0 , 1 ] , | f | > σ } )

be its distribution function. The function

f ( t ) = inf { σ : m ( σ , 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 ( t 1 p f ( t ) ) q d t t ) 1 q < ,

for 1 q < and

f L p , = sup t > 0 t 1 p f ( t ) < ,

for q = .

Theorem 1Let 1 < p 0 < 2 p 1 , 1 q 1 q 0 , 1 r = 1 p 0 1 p 1 , 1 s = 1 q 1 1 q 0 and φ L r , s [ 0 , 1 ] . If A S p 0 , q 0 , then A φ S p 1 , q 1 and

A φ S p 1 , q 1 c φ 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 φ M p q ( M p q : = M p , p q , q ) are obtained.

Theorem 2Let 1 < p < 2 < q < , p = p p 1 . LetGbe a set of all segmentsQfrom [ 0 , 1 ] , Fbe a set of all compacts from [ 0 , 1 ] , then

c 1 sup Q G 1 | Q | 1 p + 1 q | Q φ ( x ) d x | φ M p q c 3 sup Q F 1 | Q | 1 p + 1 q | Q φ ( s ) d s | ,

where | Q | is the measure of a setQ.

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 1Let 1 < p < 2 < q < . If φ N , then φ M p q if and only if

sup t > 0 t 1 p 1 q φ ( t ) < .

Moreover, φ M p q sup t > 0 t 1 p 1 q φ ( 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 < , α > 0 . We denote by B p , q α [ 0 , 1 ] the space of all measurable functions f on [ 0 , 1 ] such that

f B p , q α = ( k = 0 ( 2 α k Δ k f p ) q ) 1 q <

for 1 q < , and

f B p , α = sup k 2 α k Δ k f p <

for q = . Here { a m ( f ) } m N are the Fourier coefficients of the function f by trigonometric system { e 2 π i k x } k Z , Δ k f = Δ k f ( x ) = [ 2 k 1 ] | m | < 2 k a m ( f ) e 2 π 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 3Let 1 < p 0 p 1 < , 2 [ p 0 , p 1 ] , 1 < q 0 q 1 ,

α = 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 φ B r , s α [ 0 , 1 ] .

If A S p 0 , q 0 , then A φ S p 1 , q 1 and

A φ S p 1 , q 1 c φ B r , s α A S p 0 , q 0 ,

i.e., B r , s α 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 .

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

Then B r , s α [ 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 ξ = { ξ k } k = with the only limit point 0 such that

ξ l p q = ( m = 1 | ξ m | q m q p 1 ) 1 q , 1 q <

where { ξ m } m = 1 nonincreasing rearrangement of the sequence { | ξ k | } k = .

For q = ,

ξ l p = sup m m 1 p ξ m .

Lemma 1 (See [6])

Let 1 < r , p 0 , p 1 < , 1 q 0 , q 1 , s . 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 ¯ θ , q , k the space { a X 0 + X 1 : a θ , q , k = Φ θ , q ( K ( t , a ) ) } , where Φ θ , q is a functional defined on nonnegative functions φ by formula

Φ θ , q ( φ ( t ) ) = ( 0 ( t θ φ ( t ) ) q d t t ) 1 q , 1 q <

and

Φ θ , ( φ ( t ) ) = sup t > 0 t θ φ ( t ) , q = .

Let X α 1 0 , p 1 0 and X α 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 < α i , β i < 1 , 1 p i , q i , i = 0 , 1 , α 0 α 1 , β 0 β 1 . IfTis a bilinear operator:

T : X α 0 , p 0 × Y 0 Z β 0 , q 0

and

T : X α 1 , p 1 × Y 1 Z β 1 , q 1

then

T : X α , p × Y θ , r Z β , q .

Here α = ( 1 θ ) α 0 + θ α 1 , β = ( 1 θ ) β 0 + θ β 1 , 1 p + 1 r > 1 , 1 + 1 q = 1 p + 1 r + ( 1 θ ) ( 1 q 0 1 p 0 ) + θ ( 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 φ 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 φ with Fourier coefficients b = { b m } m Z , for which Inequality (1) holds.

Theorem 5

(1) Let 1 p 0 , p 1 < , 1 q 0 , q 1 , 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) Let 1 < p 0 < r 0 < p 1 < , 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 φ ) = T φ ¯ , where φ ¯ is a complex conjugate of the function φ. Now we prove (2).

Let φ M p 0 , q 0 p 1 , q 1 , then by (1) it follows that φ 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 , φ ) = 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 θ p 1 + θ p 0 , 1 r 0 = 1 θ p 0 + θ p 1 , 1 t 1 s = ( 1 q 1 1 q 0 ) +

for every 0 < θ < 1 . Eliminating θ from this equation, we obtain that

1 p 1 1 p 0 = 1 r 1 1 r 0 ,

and the condition 0 < θ < 1 implies the condition 1 < p 0 < r 0 < p 1 < , 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 < , 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 , [ are any.

### 4 Proof of main results

For a given pair X ¯ = ( X 0 , X 1 ) we consider the space Γ ( 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 | . It is clear that Γ ( X ¯ ) is a vector space. We endow Γ with the norm

f Γ = max ( sup t f ( i t ) X 0 , sup t f ( 1 + i t ) X 1 ) .

The space X ¯ [ θ ] 0 θ 1 consists of all elements a X 0 + X 1 such that a = f ( θ ) for some function f Γ ( X ¯ ) . The norm on X ¯ [ θ ] is equal to

a [ θ ] = inf { f Γ : f ( θ ) = a , f Γ } .

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

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

LetTbe a bilinear operator such that

T : X 0 × Y 0 Z 0

and

T : X 1 × Y 1 Z 1 .

Then

T : X [ θ ] × Y [ θ ] Z [ θ ] ,

where X [ θ ] , Y [ θ ] , Z [ θ ] 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])

LetTbe 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 θ , t 1 × Y θ , t 2 Z θ , s ,

where 1 s + 1 = 1 t 1 + 1 t 2 . Moreover,

T c B 0 1 θ B 1 θ .

Proof of Theorem 1 First we prove the inequality:

a b l p 1 , q 1 c φ L r a l p 0 , q 0 , (2)

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

If r 2 , Inequality (2) follows by Lemma 1 and Hardy-Littlewood-Paley inequality [9]. Indeed, since φ 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 φ L r s .

Taking s = r , we get

b l r , r c φ L r .

Now let 2 < r < . Let a l 2 f k Z a k e 2 π i k x , then by Parseval’s equality we get

a b l 2 = f φ L 2 f L 2 φ L = φ L a l 2 ,

i.e., M 2 = L . From Lemma 1, using Parseval’s equality we have

a b l p 1 , q 1 c φ 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 , φ ) = 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 φ L r , s a l p 0 , q 0 , (3)

where 1 s = 1 q 1 1 q 0 .

Let q 0 = 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 , ) θ ( c 2 a l p 0 , ) 1 θ φ L r , s = c a l p 0 , φ L r , s , (5)

where 1 p 1 = 1 θ p 1 0 + θ p 1 1 1 r = 1 θ r 0 + θ 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 ) θ , 1 × ( l p 0 0 , , l p 0 1 , ) θ , q 0 ( l p 1 0 , s , l p 1 1 , s ) θ , 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 , 1 p 1 = 1 θ p 1 0 + θ p 1 1 , 1 p 0 = 1 θ p 0 0 + θ 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 φ L r , s ,

where b = { b m } m Z are Fourier coefficients of the function φ 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 φ M p q and Q be an arbitrary segment in [ 0 , 1 ] ,

f 0 ( x ) = { 1 , x Q , 0 , x Q .

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:

φ M p q = sup f 0 f φ ˆ l q f ˆ l p f 0 φ 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 ) φ ( 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 Q φ ( x ) d x | ) c 1 | Q | 1 p | Q | 1 q 1 | Q φ ( x ) d x | = c 1 | Q | 1 p + 1 q | Q φ ( x ) d x | .

Since the interval Q is arbitrary, we get

φ M p q c 1 sup Q G 1 | Q | 1 p + 1 q | Q φ ( 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 , M p q ,

i.e.,

φ M p q c 2 sup t > 0 t 1 r φ ( t ) c 3 sup t > 0 1 t 1 p + 1 q 0 t φ ( s ) d s = c 3 sup Q F 1 | Q | 1 p + 1 q Q | φ ( x ) | d x sup Q F 1 | Q | 1 p + 1 q | Q φ ( x ) d x | .

□

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

From the condition of generalized monotonicity of the function φ we have

1 | Q | 1 p + 1 q | Q φ ( y ) d y | 1 | Q | 1 p + 1 q Q c y | 0 y φ ( x ) d x | d y c | Q | 1 p + 1 q sup A G 1 | A | 1 p + 1 q | A φ ( x ) d x | Q d y y 1 1 q 1 p c 1 sup A G 1 | A | 1 p + 1 q | A φ ( x ) d x | .

Taking into account that Q F is arbitrary, we have

sup Q F 1 | Q | 1 p + 1 q | Q φ ( x ) d x | c sup Q G 1 | Q | 1 p + 1 q | Q φ ( x ) d x | .

Thus, from Theorem 2 we get

φ M p q sup Q F 1 | Q | 1 p + 1 q | Q φ ( x ) d x | sup t > 0 t 1 p 1 q φ ( t ) .

□

Proof of Theorem 3 Let 2 < p 0 p 1 < . For a sequence of numbers a = { a m } m Z and a function φ L 1 [ 0 , 1 ] we consider the mapping T of the form T ( a , φ ) = a b , where b = { b m } m Z is the sequence of Fourier coefficients on the trigonometric system of functions φ. 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 , the mapping

T φ : l 2 × L 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 ) θ , q × ( B 2 , 1 1 2 , L ) θ , 1 ( l 1 , l 2 ) θ , q ,

i.e., the operator T is bounded from l p , q × ( B 2 , 1 1 2 , L ) θ , 1 to l p , q . In the paper [5] it is shown that B r , 1 1 r ( B 2 , 1 1 2 , L ) θ , 1 , where 1 r = 1 θ 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 < 1 q 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 ) [ θ ] × ( B r , 1 1 r , L 2 ) [ θ ] ( l p , q , l 2 ) [ θ ]

i.e., T is a bounded mapping from l p 0 , q 0 × B r , s α to l p 1 , q 1 , where 2 < p 0 p 1 < , 1 r = 1 2 1 p 1 , α = 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 , φ ) = a b , where b = { b m } m Z is the sequence of Fourier coefficients of the function φ. 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 α , we have:

T : l p 0 , 1 × B r , 1 α l p 1 , , (10)

where parameters r, α, p 0 , p 1 satisfy the following conditions:

1 < p 0 < p 1 2 , 1 r α = 1 p 0 1 p 1 , α > 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 α 0 , B r , 1 α 1 ) θ , h = B r , h α , with α = ( 1 θ ) α 0 + α 1 ,

we get

T : l p 0 , h 1 × B r , h 2 α l p , h 3 ,

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

Therefore, with fixed a l p 0 , and r we obtain that

P a : B r , 1 α i l p 1 i , ,

and

P a B r , 1 α i l p 1 i , c i a l p 0 , ,

where 1 r α i = 1 p 0 1 p 1 i , α i > 1 p 1 i 1 2 , i = 0 , 1 .

Using Marcinkiewicz-Calderón interpolation theorem we have

P a : B r , s α l p 1 , s ,

and

P a B r , s α l p 1 , s c a l p 0 , .

Thus

T : l p 0 , × B r , s α l p 1 , s .

To complete the proof we fix the function φ and the parameters r, s, α 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 α [ 0 , 1 ] M p 0 , q 0 p 1 , q 1 . □

The case 2 p 0 < p 1 < , 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 1Let 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 ε > 0 there exists φ 1 L r , s + ε such that φ 1 M p 0 , q 0 p 1 , q 1 , if q 1 q 0 there exists φ 2 L r ε , such that φ 2 M p 0 , q 0 p 1 , q 1 .

Proof Let ε be an arbitrary positive number, and numbers β 1 , β 2 be such that

β 1 > 1 s + ε , β 2 > 1 q 0 , β 1 + β 2 < 1 s + 1 q 0 = 1 q 1 .

Let

and

φ 1 k = + b k e 2 π i k x .

Then for m 0

Thus, ( a b ) m c ( | m | + 1 ) ( 1 r + 1 p 0 ) + 1 | ln ( | m | + 2 ) | β 1 β 2 . Since

m = 0 + ( ( ( m + 1 ) ( 1 r + 1 p 0 ) + 1 | ln ( | m | + 2 ) | β 1 β 2 ) q 1 ( | m | + 1 ) ( q 1 p 1 1 ) ) = ,

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

To prove the second part of the proposition, we take s = . Let numbers α 1 and α 2 be such that

α 1 > 1 ( r ε ) = 1 1 r ε , α 2 > 1 p 0 , α 1 + α 2 < 1 1 r + 1 p 0

(note that the last inequality does not contradict the previous two). Choosing

b k = 1 ( | k | + 1 ) α 1 , a k = 1 ( | k | + 1 ) α 2 , φ 2 k = + b k e 2 π i k x ,

we can show that

a b { ( | k | + 1 ) α 1 α 2 + 1 } k Z .

Hence a b l p 1 , q 1 , and therefore φ 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 φ 2 L r ε , . The statement is proved. □

Theorem 6Let 1 < p 0 < p 1 < 2 , 1 < q 1 q 0 , 1 r α = 1 p 0 1 p 1 , 1 s = 1 q 1 1 q 0 . Then for any ε > 0 there exist φ 1 B r , α ε B r ε , α and φ 2 B r , s + ε α such that φ 1 M p 0 , q 0 p 1 , q 1 , φ 2 M p 0 , q 0 p 1 , q 1 .

Proof Let s < and numbers β 1 , β 2 be such that

β 1 > 1 s + ε , β 2 > 1 q 0 , β 1 + β 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 φ 2 k = + b k e 2 π i k x belongs to B r , s + ε α .

It is easy to show that

( a b ) m c ( | m | + 1 ) α 1 r + 1 p 0 ( ln ( | m | + 2 ) ) β 1 + β 2

and consequently, a b l p 1 , q 1 . Therefore φ 2 M p 0 , q 0 p 1 , q 1 .

To construct the function φ 1 , it is sufficient to consider the sequences

b = { 1 ( | m | + 1 ) γ 1 } m Z , a = { 1 ( | m | + 1 ) γ 2 } m Z ,

where

γ 1 > max ( α ε 1 r + 1 , α 1 r + ε + 1 ) , γ 2 > 1 p 0

and

γ 1 + γ 2 < α 1 r + 1 p 0 .

φ 1 k = + b k e 2 π i k x . The proof that φ 1 B r , α ε B r ε , α , φ 1 M p 0 , q 0 p 1 , q 1 is similar to the proof of the first part. □

### Competing interests

The authors declare that they have no competing interests.

### 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.

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