DSpace at VNU: Behavior of the sequence of norms of primitives of a function

Behavior of the sequence of norms of primitives of a function Ha Huy Banga,∗, Vu Nhat Huyb a Institute of Mathematics, Vietnamese Academy of Science and Technology, 18 Hoang Quoc Viet St

Behavior of the sequence of norms of primitives

of a function

Ha Huy Banga,∗, Vu Nhat Huyb

a Institute of Mathematics, Vietnamese Academy of Science and Technology, 18 Hoang Quoc Viet Street, Cau Giay,

Hanoi, Viet Nam

b Department of Mathematics, College of Science, Vietnam National University, Hanoi, Vietnam 334 Nguyen Trai Street,

Thanh Xuan, Hanoi, Viet Nam Received 28 April 2009; received in revised form 8 December 2009; accepted 30 December 2009

Available online 6 January 2010 Communicated by Francisco Marcellan

Abstract

In this paper we characterize the behavior of the sequence of the Lp-norm of primitives of a function by its spectrum (the support of its Fourier transform)

c 2010 Elsevier Inc All rights reserved

Keywords: Lp-spaces; Generalized functions

1 Introduction

following limit

lim

n→∞

kf(n)k1 /n

p

and

lim

n→∞kf(n)k1 /n

∗ Corresponding author.

E-mail addresses: hhbang@math.ac.vn (H.H Bang), nhat huy85@yahoo.com (V.N Huy).

0021-9045/$ - see front matter c 2010 Elsevier Inc All rights reserved.

doi:10.1016/j.jat.2009.12.011

where ˆf is the Fourier transform of f

Theorem Ashows that the behavior of the sequence k f(n)k1 /n

It is natural to ask what will happen when we replace derivatives by integrals? Let

x

then

x

Inf ∈ L2(R) for all n, and

lim

n→∞kInf k1/n

2 =σ−1,

vanishes a.e

if D(I f ) = f , that is,

−∞

−∞

ψ(t)dt,

hI f, ϕi = − h f, Φi

Then I f ∈ S0(R) and

, ϕ0 = −

f,Z x

−∞

ϕ0(t)dt= − hf, ϕi , i.e., I f is a primitive of f Furthermore, if I f ∈ S0(R) is a primitive of f , we have

−∞

ϕ(ξ) dξ

−∞

primitive of f So, we have proved the following result: Every tempered generalized function

where C is an arbitrary constant

n, m = 0, 1, 2,

n=0⊂Lp(R)

ˆ

2π

−∞

e−ixξf(x)dx

The Fourier transform of a tempered generalized function f is defined via the formula

We now state our main result:

lim

n→∞kInf k1/n

where

σ = inf{|ξ| : ξ ∈ supp ˆf}

Proof It is well-known that

{ ϕ∈L :k ϕk ≤1}

|hf, ϕi|,

where 1/p + 1/q = 1 So, clearly, we have

{ ϕ∈S:kϕk q ≤1}

Z

R

> k f k1−

Z

|x |≤M

> k f k1−2 and put

h =χ[−M ,M]g,

ω(x) =

(

Ce

1

x 2−1 if |x|< 1,

λω(x

lim

λ→0

Z

R

R

So,

Z

R

> k f k1−3

Next, we divide our proof into two cases

Case1.σ > 0 Let us first prove

ˆ

d

Inf

Therefore,

supp ˆf ⊂supp dInf ⊂supp ˆf ∪ {0}

So, to obtain(3), it is enough to show 0 6∈ supp dInf: We choose numbers a, b: 0 < a < b < σ

supph dInf ⊂ {0}

h dInf =

N n

X

cNn

j δ( j)

F−1h ∗ Inf(x) =

N n X

j =0

cNn

F−1h ∗ Inf(x) = cN n

Note that

F−1h ∗ Inf(x) = F−1h ∗(In+1f)0(x) = F−1h ∗ In+1f(x)
0

0

0

≡0

d

Inf, ϕ 6= 0

Second we prove

lim

n→∞kInf k1/n

p ≤ 1

nf, ϕ =DIdnf, F−1ϕE=

D d

Inf, (F−1ϕ)(ξ).h(ξ)E=

D

Inf, F (F−1ϕ)(ξ).h(ξ)
E Therefore,



ξn



nf, ϕ = |hf, ψni| = 1

√

2π

f, ϕ ∗ F hξ(ξ)n



Therefore, since(2), we have for n ≥ 3:

{ ϕ∈S:kϕk q ≤1}

1

√

2π

f, ϕ ∗ F h(ξ)

ξn



{ ϕ∈S:kϕk q ≤1}

1

√ 2πk f kp ϕ ∗ F hξ(ξ)n

q

√

ξn 1

Hence,

lim

n→∞kInf k1/n

p ≤ lim

n→∞ F h(ξ)

ξn

1 /n 1

We put for n ≥ 3:

ξn

 Then for n ≥ 3:

sup

x ∈R

|(1 + x2)gn(x)|

x ∈R

1

√

2π

 Z

R

e−ixξD2 h(ξ)

ξn



dξ

Z

R

e−ixξh(ξ)

ξn dξ



x ∈R

1

√

2π

 Z

| ξ|≥γe

−ix ξ h00(ξ)

0(ξ)



dξ

+

Z

| ξ|≥γe

−ix ξh(ξ)

ξn dξ



√

2π

4

 Therefore, it follows from

Z

R

x ∈R

|(1 + x2)gn(x)|Z

R

1

x ∈R

|(1 + x2)gn(x)|

that

lim

n→∞ F h(ξ)

ξn

1 /n 1

n→∞kgnk1/n

1 ≤ 1

lim

n→∞kInf k1/n

σ − 2

Finally, we show

lim kInf k1/n

p ≥ 1

Without loss of generality we may assume that

σ = inf{ξ : 0 < ξ ∈ supp ˆf}

0 (σ − , σ + ) such thatDfˆ, ϕE6=0 Therefore,

0 6= , ˆϕ =

n(Inf), ˆϕ =

D

Inf, ˆϕ(n)E

≤ kInf kp.k ˆϕ(n)k

q So,

lim

n→∞

kInf k1/n

lim

n→∞k ˆϕ(n)k1 /n

q

We have

sup

x ∈R

|(1 + x2) ˆϕ(n)(x)| = sup

x ∈R

(1 + x2) F ϕ(ξ)ξn

(x)

x ∈R

Z

R

e−ixξϕ(ξ)ξndξ

x ∈R

Z

R

e−ixξϕ00(ξ)ξn+2nϕ0(ξ)ξn−1+n(n − 1)ϕ(ξ)ξn−2



dξ

≤

Z

| ξ|≤σ+

|ϕ(ξ)ξn|dξ

+

Z

| ξ|≤σ+



|ϕ00(ξ)ξn| +2n|ϕ0(ξ)ξn−1| +n(n − 1)|ϕ(ξ)ξn−2|



dξ

≤max{kϕk∞, kϕ0k∞, kϕ00k∞} 4(σ + )n+1

 Therefore, it follows from k ˆϕ(n)k

q ≤π supx ∈R|(1 + x2) ˆϕ(n)(x)| that lim

n→∞k ˆϕ(n)k1 /n

lim

n→∞

kInf k1/n

and then(11)

C0∞(−, ) such thatDfˆ, ϕE6=0 Then arguing just as above we obtain

lim

n→∞

kInf k1/n

lim

n→∞k ˆϕ(n)k1/n

q

. Therefore,

lim kInf k1/n

p = ∞

So we always have

lim

n→∞kInf k1/n

p =σ−1

supp ˆf = suppcI f

Let g ∈ S0(R) and m = 1, 2, Clearly, the set

0 = n0< n1< n2< · · · < nk < · · ·

0, 1, 2, there exists an element in Pn j +1 −n j(In j f), which belongs to Lp(R) and is denoted

by Inj +1f If the function f satisfies this assumption, we write(In j f)∞

j =0⊂Lp(R)

We have the following result:

lim

j →∞

kInj f k

1

n j

p =σ−1, where

σ = inf{|ξ| : ξ ∈ supp ˆf}

all Djg, j = 1, , m − 1 belong to Lp(R) (see [10,12]) So, since Dnj +1 −n jInj +1f = Inj f

we get that all D Inj +1f, D2Inj +1 f, , Dn j +1 −n j −1Inj +1f, j = 0, 1, 2, belong to Lp(R)

Acknowledgment

This work was supported by the Vietnam National Foundation for Science and Technology Development (Project No 101.01.50.09)

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Without loss of generality we may assume that

σ = inf{ξ : < ξ ∈ supp ˆf}

0...

Therefore, since(2), we have for n ≥ 3:

{ ϕ∈S:kϕk q ≤1}

1

√...

1

x 2−1 if |x|< 1,

λω(x

lim

λ→0

Z

R

R