1 Introduction

(1)

Volume 6, Number 1 (2015), 76 – 84

BOCHKAREV INEQUALITY FOR THE FOURIER TRANSFORM OF FUNCTIONS IN THE LORENTZ SPACES L2,r(R)

G.K. Mussabayeva, N.T. Tleukhanova Communicated by N.A. Bokayev Key words: Lorentz spaces, Fourier transform.

AMS Mathematics Subject Classification: 42B10.

Abstract. In this article we prove an analogue of Hardy -Littlewood- Stein inequality for the Fourier transform in Lorentz space L_2,r(R).

Let f(x) be a scalar-valued µ measurable function which is finite almost everywhere, we introduce the distribution function m(σ, f) defined by

m(σ, f) = µ({x:|f(x)|> σ}) and f^∗ its non-increasing rearrangement of the function f.

f^∗(t) = inf{σ :m(σ, f)6t}. (1.1) In [1], the Lorentz space L_p,q is defined that f ∈L_p,q,16p6∞,if and only if

kfk_L_p,q =



 Z

R

t¹^pf^∗(t)q dt t





1 q

<∞

when 16q <∞,

kfkLp,∞ = sup

t

t¹^pf^∗(t) when q =∞.

Well-known classical theorem Hausdorff-Young [11], whereby iff(x)is a 1- periodic function and satisfying

1

Z

0

|f(x)|^pdx <∞, 1< p 62, then following relation holds

+∞

X

n=−∞

|c_n|^p⁰

!_p¹0

6





1

Z

0

|f(x)|^pdx





1 p

, (1.2)

(2)

where

c_n=

1

Z

0

f(x)e^−2πinxdx, n ∈Z

are the Fourier coefficients of f(x) and p⁰ = _p−1^p .

An analogue of theorem Hausdorff-Young for the Fourier transform was proved by Titchmarsh [9],[10] and the inequality of Titchmarsh for Fourier integrals





∞

Z

−∞

fb(ξ)

p⁰

dξ





1 p0

6





∞

Z

−∞

|f(x)|^pdx





1 p

, (1.3)

where

fb(ξ) =

∞

Z

−∞

f(x)e^−2πixξdx

is Fourier transform of f(x).

The Hardy-Littlewood inequalities for the Lorentz spaces L_p,q(R) was proved by Stein [3]-[8]:

Let1< p <2, 16q 6∞and p⁰ = _p−1^p , then

∞

X

k=1

k^p¹⁰c^∗_kq 1 k

!¹_q

6Ckfk_L_p,q_[0,1], (1.4)

where c^∗_k is a non-increasing rearrangement of the Fourier coefficients.

kfbk_L_p0,q(R) 6Ckfk_L_p,q_(R), (1.5) Unlike inequalities (1.2) and (1.3), relations (1.4) and (1.5) don’t cover the case ofp= 2.

S.V.Bochkarev in [2] for the Fourier transform on the torus has shown that the Hardy -Littlewood- Stein inequality is not the case.

Theorem A. Let {ϕ_n}^∞_n=1 be an orthonormal system of complex-valued functions on [0,1], such that

kϕ_nk∞6M, n= 1,2, ....

let f ∈L_2,r[0,1], 2< r6∞. Then the inequality

sup

n∈N

1

|n|¹²(log(n+ 1))¹²⁻¹^r

n

X

m=1

a^∗_m 6Ckfk_L_2,r_[0,1] (1.6) holds, where an are the Fourier coefficients of the system {ϕn}^∞_n=1.

(3)

2 Main result

The aim of this paper is to obtain an analogue of Bochkarev theorem for the Fourier transform. The main result of this paper is the following.

Theorem 2.1. Let <_N = {A =

N

S

i=1

A_i, where A_i are segments in R}. Then for any function f ∈L_2,r(R) and 2< r <∞ we have the inequality

sup

N>8

sup

A⊂<N

1

|A|¹² log₂(1 +N)¹²⁻¹^r Z

A

fˆ(ξ)dξ

623kfk_L_2,r. (2.1)

To prove Theorem 2.1, we establish the following lemmas.

Lemma 2.1. Let ⁴₃ < q < 2 < r < 4 and f ∈ L_q,r(R). Then for any measurable set A of finite measure of <_N, the inequality

1

|A|¹^q Z

A

f(ξ)dξˆ

62 q

2−q

_q¹0(^r−2r )

kfk_L_q,r₍_R₎ (2.2)

holds.

Proof.

Z

A

f(ξ)dξˆ

6 Z

A

∞

Z

−∞

f(t)e^−iξtdtdξ

6|A|kfk_L₁

and from the Plancherel’s theorem, we have

Z

A

fˆ(ξ)dξ

6|A|¹²kfk_L₂.

Let τ ∈(0,+∞). Define

f₁(x) =

f(x), if |f(x)|6f^∗(τ), 0, otherwise,

f₀(x) =f(x)−f₁(x).

Then, for any0< τ <∞, we have

Z

A

f(ξ)dξˆ

6 Z

A

fˆ₀(ξ)dξ

+ Z

A

fˆ₁(ξ)dξ

6|A|

τ

Z

0

f^∗(s)ds+|A|¹²





∞

Z

τ

(f^∗(s))²ds





1 2

.

(4)

Using Holder’s inequality, we obtain

Z

A

fˆ(ξ)dξ

6





|A|





τ

Z

0

t¹^qf^∗(t) r dt

t





1

r 



τ

Z

0

t^r

0 q0dt

t





1 r0

+|A|¹²





∞

Z

τ

t¹^qf^∗(t)r dt t





1

r 



∞

Z

τ

t¹⁻²^q_r−2^r dt t





r−2 2r







6kfk_L_q,r |A|τ¹⁻¹^q

1−1 q

¹_r⁻¹

+|A|¹²τ¹²⁻¹^q

r(2−q) q(r−2)

¹_r⁻¹₂! .

Now let τ = _q(r−2)

r(2−q)

^r−2_r

q q−1

^2(1−r)_r

/|A| and using the ⁴₃ < q < 2 < r < 4, we can estimate

Z

A

f(ξ)dξˆ

62|A|¹^q q

2−q

_q¹0(^r−2r )

kfk_L_q,r.

The method of the proof in this paper is based on a study of the constants in inequalities of the form (2.2), namely their dependence on the relevant parameters.

Lemma 2.2. Let 2< p < 4, p⁰ = _p−1^p , 2< r <∞, f ∈L_p,r and A=

N

S

i=1

A_i, where A_i are segments in R. Then we have

1

|A|¹^pN^p¹⁰⁻¹^p Z

A

f(ξ)dξˆ

6Ckfk_L_p,r,

where

C = 4√ 2

r(p−2) p(r−2)

¹_p(^2−r_r )

. Proof. Let f ∈C₀^∞.

Z

A

fˆ(ξ)dξ

= Z

A

∞

Z

−∞

f(t)e^−iξtdtdξ

6

N

X

i=1

Z

Ai

∞

Z

−∞

f(t)e^−iξtdtdξ

6

N

X

i=1

∞

Z

−∞

|f(t)|

Z

Ai

e^−iξtdξ

dt=

N

X

i=1

∞

Z

−∞

|f(t)|

bi

Z

ai

e^−iξtdξ

dt=

N

X

i=1

∞

Z

−∞

|f(t)|

2 sin^(bⁱ^−a₂ ⁱ⁾t t

dt

6

N

X

i=1

∞

Z

−∞

|f(t)|min

(b_i−a_i),2 t

dt62

N

X

i=1

∞

Z

−∞

|f(t)|min

|A_i|,1 t

dt. (2.3)

(5)

From the Plancherel’s theorem we get Z

A

fˆ(ξ)dξ

6|A|¹²kfk_L₂. (2.4)

Consider the function f₁(x) =

f(x), |f(x)|6f^∗(τ), x∈R\

−^τ₂,^τ₂ 0, otherwise,

f₀(x) =f(x)−f₁(x).

Then using the (2.3) and (2.4) we get

Z

A

f(ξ)dξˆ

6 Z

A

fˆ0(ξ)dξ

+ Z

A

fˆ1(ξ)dξ

6|A|¹²kf₀k_L₂ + 2

N

X

i=1

∞

Z

−∞

|f₁(x)|min

|A_i|, 1

|x|

dx

=B(I₁+I₂) We have

I₁ =





∞

Z

−∞

|f₀(x)|²dx





1 2

=







τ 2

Z

−^τ

2

|f(x)|²dx+ Z

τ 26|x|<∞

|f₀(x)|²dx







1 2

62





τ

Z

0

|f^∗(t)|²dt





1 2

I₂ = 2

N

X

i=1

∞

Z

−∞

|f₁(x)|min

|A_i|, 1

|x|

= 2

N

X

i=1

Z

τ 26|x|<∞

|f₁(x)|min

|A_i|, 1

|x|

62N

τ

Z

0

f₁^∗(t) 1 t+τdt=

= 2N

∞

Z

0

f^∗(t+τ) dt

t+τ 62N

∞

Z

τ

f^∗(t)dt t . Therefore,

Z

A

fˆ(ξ)dξ

62|A|¹²





τ

Z

0

(f^∗(t))²dt





1 2

+ 2N

∞

Z

τ

f^∗(t)dt t .

(6)

Applying H´older’s inequality, we obtain

Z

A

f(ξ)dξˆ

62|A|¹²





τ

Z

0

(f^∗(t)t¹^p)^rdt t





1

r 



τ

Z

0

t¹⁻²^p_r−2^r dt t





r−2 2r

+2N





∞

Z

τ

f^∗(t)t^p¹ r dt

t





1

r 



∞

Z

τ

t⁻^r

0 p dt

t





1 r0

6kfkLp,r 2|A|¹²

r(p−2) p(r−2)

^2−r_2r

τ¹²⁻¹^p + 2N p^r−1^r τ⁻¹^p

! .

Now let τ = ^N²

|A|(^p−2p )^2−r^r ^p²(^1−rr ) and using that 2 < p < 4 we have the following estimate:

Z

A

f(ξ)dξˆ

64√

2|A|^1/p

p−2 p

¹_p(²r−1)

N^p¹⁰⁻¹^pkfk_L_p,r.

Lemma 2.3. Let A =

N

S

i=1

A_i, where A_i are segments. Then for any 2 < p < 4 and 2< r <∞ the inequality

1

|A|¹² Z

A

f(ξ)dξˆ

6Ckfk_L_2,r (2.5)

holds, where

C= 8 p

p−2

¹_p(¹⁻²_r) p+ 2

p ¹₄

N¹²⁻¹^p. Proof. Let τ >0,

f1(x) =

f(x), |f(x)|6f^∗(τ) 0, otherwise, f₀(x) =f(x)−f₁(x).

Then, using Lemmas 2.1, 2.2 and 2< r <∞, we obtain

Z

A

fˆ(ξ)dξ

6 Z

A

fˆ₀(ξ)dξ

+ Z

A

fˆ₁(ξ)dξ

62kf₀k_L_p0,r

p⁰ 2−p⁰

¹_p(^r−2r )

|A|^p¹⁰

+4√

2kf₁k_L_p,r

p−2 p

¹_p(^2−r_r )

|A|¹^pN^p¹⁰⁻^p¹

(7)

62





τ

Z

0

t^p¹⁰f^∗(t)r dt t





1 r

|A|^p¹⁰ p⁰

2−p⁰

¹_p(^r−2_r )

+4√ 2





∞

Z

0

t¹^pf^∗(t+τ)rdt t





1 r

|A|¹^p

p−2 p

_p¹(^2−r_r )

N^p¹⁰⁻¹^p. Further, we estimate each integral separately





τ

Z

0

t^p¹⁰f^∗(t)rdt t





1 r

=





τ

Z

0

t¹⁻¹^pf^∗(t)r dt t





1 r

6





τ

Z

0

t¹²f^∗(t)rdt t





1 r

sup

06t6τ

t¹²⁻¹^p 6τ(¹2−¹_p)kfk_L_2,r.





∞

Z

0

t^p¹f^∗(t+τ) r dt

t





1 r

6





τ

Z

0

t¹^p(t+τ)⁻¹²rdt t

sup

0<s<∞

s¹²f^∗(s) r

+

∞

Z

τ

(t¹^pf^∗(t+τ))^rdt t





1 r

6





τ

Z

0

t¹^pτ⁻¹²rdt t sup

0<s<∞

r 2

s

Z

0

ξ^r²⁻¹f^∗(ξ)^rdξ+

∞

Z

τ

t¹^pf^∗(t)r dt t





1 r

6



τ(¹_p⁻¹₂)^rp

2kfk^r_L_2,r +τ(¹_p⁻¹₂)^r

∞

Z

τ

t¹²f^∗(t)r dt t





1 r

6τ(¹_p⁻¹₂) p+ 2

2 ¹_r

kfk_L_2,r.

Therefore given the 2< r <∞, we have

Z

A

fˆ(ξ)dξ

62|A|¹⁻¹^p p

p−2

¹_p(¹⁻²r)

τ(¹₂⁻¹_p)kfk_L_2,r

+4√ 2|A|¹^p

p−2 p

¹_p(^2−rr ) p+ 2

2 ¹₂

N^p¹⁰⁻¹^pτ(¹p−¹₂)kfk_L_2,r.

Taking τ(¹2−¹_p) = N

|A|

_p¹0−¹

p p+2

2

¹₂ ¹₂

, we have

Z

A

f(ξ)dξˆ

68|A|¹²N¹²⁻¹^p p

p−2

¹_p(¹⁻²r) p+ 2

2 ¹₄

kfkL2,r.

(8)

The proof of Theorem 2.1. Let A⊂ <_N and |A|=N 68.

In the right part of the constant (2.5) we put p= 3. Then

8 p

p−2

¹_p(¹⁻²_r) p+ 2

2 ¹₄

N¹²⁻¹^p 620.

Therefore, from Lemma 2.3, we get 1

N¹² log₂(1 +N)¹²⁻¹^r Z

A

f(ξ)dξˆ

6 1 N¹²

Z

A

fˆ(ξ)dξ

620kfk_L_2,r.

Now let A⊂ <_N so that N >8. Putp= _log^{2 log}²^N

2N−2, i.e. ^p−2_2p = _log¹

2N, then p66.

Note also thatN¹²⁻¹^p =N^log2¹^N 62.

Using Lemma 2.3, we obtain 1

N¹² Z

A

fˆ(ξ)dξ

623 log₂(1 +N)

1 2− ¹

log2N

(¹⁻²r)kfk_L_2,r

623 log₂(1 +N)¹²⁻¹^rkfk_L_2,r. Taking the top exact bound over all N of N we obtain

sup

N>8

sup

A⊂<_N

1

N¹² log₂(1 +N)¹²⁻¹^r Z

A

fˆ(ξ)dξ

623kfk_L_2,r

Theorem is proved.

Acknowledgments

This work was supported by the grant of the Ministry of Education and Science of the Republic of Kazakhstan (project 3311/GF4)

(9)

References

[1] J. Bergh, J. L´ofstr´om,Interpolation Spaces, New York. 1976, 210p.

[2] S.V. Bochkarev,Hausdorff-Young-Riesz theorem in Lorentz spaces and multiplicative inequalities, Proc. Steclov Inst. Math. 219 (1997), 96-107.

[3] A.N. Kopezhanova, E.D. Nursultanov, L.-E. Persson,On inequalities for the Fourier transform of functions from Lorentz spaces, Mathematical notes. 90 (2011), no. 5, 785-788.

[4] E.D. Nursultanov,Net spaces and the Fourier transform, (Russian) Dokl. Akad. Nauk 361 (1998), no 5, 597–599.

[5] E.D Nursultanov, T.U. Aubakirov, Hardy-Littlewood theorem for Fourier series, Mathematical notes. 73 (2003), 340-347.

[6] E.D. Nursultanov,Interpolation properties of some anisotropic spaces and Hardy–Littlewood type inequalities,East J. Approx. 4 (1998), no. 2, 277-290.

[7] E.D. Nursultanov,Application of interpolational methods to the study of properties of functions of several variables, Mathematical Notes. 75 (2004), no.3, 341-351.

[8] M. Stein Elias,Interpolation of linear operators, Trans. Amer. Math. Soc. 83 (1956), 482-492.

[9] E.C. Titchmarsh,A contribution to the theory of Fourier transforms, Proc. London Math. Soc.

23 (1924), no. 2, 279-289.

[10] E.C. Titchmarsh,Introduction to the theory of Fourier integrals, Clarendon Press, Oxford 1937;

Russian transl. Gostekhizdat, Moscow-Leningrad 1948.

[11] A. Zygmund,Trigonometrical series, 2nd ed. Vols. I, II, Cambridge Univ. Press, New York 1959;

Russian transl. Mir, Moscow 1965.

Guliya Mussabayeva

L.N. Gumilyov Eurasian National University 2 Satpayev st,

010010 Astana, Kazakhstan E-mail: [email protected] Nazerke Tleukhanova

L.N. Gumilyov Eurasian National University 2 Satpayev st,

010010 Astana, Kazakhstan E-mail: [email protected]

Received: 03.02.2015