ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES THROUGH THEIR SPECTRUM

Abstract. In this paper we characterize behavior of the sequence of norm of primitives of functions in Orlicz spaces by its spectrum (the support of its Fourier transform). Abstract. In this paper we characterize behavior of the sequence of norm of primitives of functions in Orlicz spaces by its spectrum (the support of its Fourier transform).

FUNCTIONS IN ORLICZ SPACES THROUGH THEIR

SPECTRUM

HA HUY BANG & VU NHAT HUY

Abstract In this paper we characterize behavior of the sequence of norm of

primitives of functions in Orlicz spaces by its spectrum (the support of its Fourier

transform).

1 IntroductionThe following result was proved in [12]:

Theorem A Let 1 ≤ p ≤ ∞,f ∈ Lp(Rn) and supp ˆf is bounded, where ˆf is theFourier transform of f Then

lim

|α|→∞

 kDαf kpsup

To this question, V.K Tuan answered in [37] for p = 2, n = 1, we answered for

1 ≤ p ≤ ∞, n = 1 in [18] and for 1 ≤ p ≤ ∞, n ≥ 1 in [19], and we answer now forthe n-dimensional case and Orlicz spaces

2 DefinitionsLet Φ : [0, +∞) → [0, +∞] be an arbitrary Young function, i.e., Φ(0) = 0, Φ(t) ≥

0, Φ(t) 6≡ 0 and Φ is convex Denote by:

¯Φ(t) = sup

Z

Rn

u(x)v(x)dx

< ∞

Key words and phrases Orlicz spaces, generalized functions.

2010 AMS Subject Classification 26D10, 46E30.

1

for all v with ρ(v, ¯Φ) < ∞, where

ρ(v, ¯Φ) =

Z

Rn

¯Φ(|v(x)|)dx

Then LΦ(Rn) is a Banach space with respect to the Orlicz norm

kukΦ = kukLΦ(Rn ) = sup

ρ(v, ¯ Φ)≤1

Z

Rn

u(x)v(x)dx

,which is equivalent to the Luxemburg norm

0 ≤ t ≤ 1 and Φ(t) = ∞ for t > 1

We have known the following results:

Lemma 1 Let u ∈ LΦ(Rn) and v ∈ LΦ¯(Rn) Then

Z

Rn

|u(x)v(x)|dx ≤ kukΦkvkΦ¯.Lemma 2 Let u ∈ LΦ(Rn) and v ∈ L1(Rn) Then

ku ∗ vkΦ ≤ kvkΦkvk1

3 mail results

We shall first give a notation of the primitive of a tempered generalized function:Denote by S(Rn) the Schwartz space of rapidly decreasing functions and S0(Rn) theset of all continuous linear functionals on S(Rn) Any element h ∈ S0(Rn) is called

a tempered generalized function and we write h(ϕ) = n).Let f ∈ S0(Rn) and ej = (0, , 0, 1, 0, , 0) ∈ Zn

+ be an unit vector such that its

jth coordinate equals 1, j = 1, 2, , n The tempered generalized function Ie jf istermed the eth

j primitive of f if De j(Ie jf ) = f , that is,

Z x j

−∞

ψ(x1, , xj−1, t, xj+1, , xn)dt,

Conversely, given an arbitrary the jth constant tempered generalized function gj,the functional Ie jf defined on S(Rn) by (1) determines the eth

j primitive of f So,

we have proved the following result: Let j ∈ {1, , n} Every tempered generalizedfunction f ∈ S0(Rn) has in S0(Rn) the ethj primitive, which is denoted by Iejf , andevery the ethj primitive of f is expressed by formula (1), where gj is an arbitrary the

jth constant tempered generalized function

Note that the notation of primitive of a generalized function in D0(a, b), a, b ∈ Rcan be found in [38], here we define it for tempered generalized functions in S0(Rn)

We denote I0f = f In the sequel, for j = 1, 2, , n then Ie jf denotes the eth

j

primitive of f ∈ S0(Rn), i.e., Ie jf ∈ Pe j(f ), and for any α ∈ Zn

+, |α| ≥ 1 wedefine the primitive of higher order by the following way: Iαf = Iejα(Iα−ejαf ),where jα := max{j : αj ≥ 1}, i.e., Iαf ∈ Pejα(Iα−ejαf ) So, Dα(Iαf ) = f forall α ∈ Zn

+ Let 1 ≤ p ≤ ∞ and f ∈ LΦ(Rn) If for any α ∈ Zn

+ there exists the

ejα primitive of Iα−ejαf , which belongs to LΦ(Rn) and is denoted by Iαf , we write(Iαf )α∈Zn

The above definition becomes the known one if n = k For the extended tion we have the following result:

convolu-Proposition 4 Let n, k ∈ N, n ≥ k, f ∈ LΦ(Rn) and g ∈ L1(Rk) Then f ~ g ∈

LΦ(Rn) and

kf ~ gkL Φ (R n ) ≤ kf kLΦ(Rn )kgkL1(Rk ).(2)

Proof Put v = (xk+1, , xn) and denote f (x1, , xk, v) := f (x1, , xn) Then

by the definition of Orlicz norm we obtain

kf ~ gkL Φ (R n ) = sup{

: kϕkΦ¯ ≤ 1}

≤ kf kΦkgkL1(Rk )kϕkΦ¯.So,

kf ~ gkL (R n )≤ kf kΦkgkL (Rk )

The proof is complete Let f ∈ L1(Rn) and ˆf = F f or Fnf be its Fourier transform

Proposition 5 Let n, k ∈ N, n ≥ k, f ∈ S(Rn) and g ∈ L1(Rk) Then

(Fn(f (ξ1, , ξn)g(ξ1, , ξk)))(x1, , xn) = (2π)−k/2(Fnf ~ Fkg)(x1, , xn),(3)

(Fn−1(f (ξ1, , ξn)g(ξ1, , ξk)))(x1, , xn) = (2π)−k/2(Fn−1f ~Fk−1g)(x1, , xn).Proof Put

u = (x1, x2, , xk), v = (xk+1, , xn), s = (ξ1, ξ2, , ξk), t = (ξk+1, , ξn).Then

Combining (4)-(5), we arrive at (3) Similarly, we also have the result for the inverse

It is easy to see the following property of the extended convolution:

Proposition 6 Let n, k ∈ N, n ≥ k, f ∈ LΦ(Rn), h ∈ S(Rn) and g ∈ L1(Rk).Then

Let f ∈ S0(Rn) Then supp ˆf is called the spectrum of f We say that f has(O)-property if its spectrum is contained in (Rn, ∆) for some ∆ > 0 Using theextended convolution we have the following result on the existence of primitives in

LΦ(Rn):

Theorem 7 Let f ∈ LΦ(Rn) and f has (O)-property Then for any j = 1, , nthere exists exactly one the eth

j primitive of f , which is denoted by Ie jf , such that

Ie jf ∈ LΦ(Rn) and also has (O)-property Moreover, supp dIe jf = supp ˆf

Proof It is sufficient to prove for the case j = 1 By the assumption, supp ˆf ⊂(Rn, ∆) for some ∆ > 0 We define a function η ∈ C∞(R) by

η(x) = φ(x)

(−ix)3 ,where the even function φ ∈ C∞(R) is given and satisfies the following conditions

φ(x) = 1 ∀x ∈ (−∞, −∆/2) ∪ (∆/2, +∞),(6)

f (x1− ξ, x2, , xn)ˆη(ξ)dξ

Then it follows from Proposition 4, f ∈ LΦ(Rn) and ˆη ∈ L1(R) that Ψ ∈ LΦ(Rn).For each ϕ ∈ S(Rn), we put ψ = D3e 1ϕ Then

R

ψ(x1, x2, , xn)ˆη(ξ − x1)dx1.Hence,

g(x1, , xn) = √1

2πZ

R

ψ(x1− ξ, x2, , xn)ˆη(ξ)dξ

So, applying Proposition 5, we get

(F−1g)(x1, , xn) = (F−1ψ)(x1, , xn)η(x1) = (F−1ϕ)(x1, , xn)φ(x1).(8)

From (6) and supp ˆf ⊂ (Rn, ∆), we have ˆf = ˆf φ(x1) Therefore, since (8), we obtain

f , F−1g f , (F−1ϕ)(x1, , xn)φ(x1)

= f φ(xˆ 1), (F−1ϕ)(x1, , xn) f , (F−1ϕ)(x1, , xn)

This implies

3e 1ϕ

So, D3e 1Ψ = f in the distribution sense Hence, we have D3e 1Ψ ∈ LΦ(Rn) and

Ψ ∈ LΦ(Rn) Therefore, D2e1Ψ ∈ LΦ(Rn), and it is symbolized as Ie1f

Next, we prove that supp dIe 1f = supp ˆf Indeed, Since De 1Ie 1f = f , we have

ˆ

f = ix1Ide 1f Therefore,

supp ˆf ⊂ supp dIe 1f ⊂ supp ˆf ∪ H1,(9)

where H1 := {ξ ∈ Rn : ξ1 = 0} So, to prove suppdIαf = supp ˆf , it is enough

to show H1 ∩ supp dIe 1f = {∅} Assume now the contrary that ∃ξ ∈ H1 ∩ dIe 1f Then b := max{|ξ1|, , |ξn|} > 0 We choose a number 0 < a < ∆ and a function

h ∈ C0∞((−∆, ∆) × (−b − a − 1, b + a + 1)n−1) such that h(x) = 1 in (−a, a) × (−b −

a, b + a)n−1 Since (9) and supp ˆf ⊂ (Rn, ∆), we have

supph dIe 1f ⊂ H1

Hence, there is a number N0 ∈ N such that

So, h dIe 1f = 0 On the other hand, since ξ ∈ supp dIe 1f , there is a function ϕ ∈

C0∞(B(ξ, a)) such that

d

Ie 1f , ϕ6= 0

Then, since h(x) = 1 in (−a, a) × (−b − a, b + a)n−1, we get

0 6= Ide 1f , ϕ Ide 1f , hϕ Ide 1f , ϕ= 0

This is impossible Hence, supp dIe 1f = supp ˆf

Finally, we prove the uniqueness of the eth1 primitive of f which belongs to LΦ(Rn)and has (O)-property Indeed, suppose that Ie 1f and Je 1f are certain eth

1 primitives

of f such that Ie 1f, Je 1f ∈ LΦ(Rn) and both of them have (O)-property Since

Ie 1f, Je 1f ∈ LΦ(Rn) and De 1(Je 1f − Ie 1f ) = 0, we get Je 1f − Ie 1f = g(x2, , xn).Hence,

e 1f − Ie1f ), η(x1)ϕ(x2, , xn) 2, , xn), Fn−1(ϕ(x2, , xn)),where ϕ ∈ S(Rn−1) and η(x) ∈ C0∞(Rn) So,

{(0, σ) ∈ Rn: σ ∈ suppFn−1(g(x2, , xn))} ⊂ suppF (Je1f − Ie1f )

On the other hand, it follows from suppF (Je 1f −Ie 1f ) ⊂ suppF (Je 1f )∪ suppF (Ie 1f )that Je 1f − Ie 1f has (O)-property Therefore, suppFn−1(g(x2, , xn)) = ∅, i.e.,

Remark 8 Let an Young function Φ : [0, ∞) → [0, ∞) and f ∈ LΦ(Rn) Thenfor any j = 1, , n there is at most one in LΦ(Rn) the eth

j primitive of f , which isdenoted by Ie jf , and moreover, there exists exactly one in LΦ(Rn) the eth

De1(J2e1f − I2e1f ) = g(x2, , xn).

Hence,

J2e1f − I2e1f = (x1+ c)g(x2, , xn),where c is some number in C Therefore, taking account of J2e 1f − I2e 1f ∈ LΦ(Rn),

we get g(x2, , xn) = 0 a.e So, Je 1f = Ie 1f Using Theorems 7 and 10, we have the following:

Theorem 11 Let f ∈ LΦ(Rn) and f has (O)-property Then there exists exactlyone sequence of primitives (Iαf )α∈Zn

+ ⊂ LΦ(Rn) Moreover, suppdIαf = supp ˆf ∀α ∈

Zn+

Now, we characterize behavior of the sequence of LΦ(Rn)−norm of primitives of

a function by its spectrum:

Theorem 12 Let f 6≡ 0, f has (O)-property and (Iαf )α∈Zn

+ ⊂ LΦ(Rn) Thenlim

|α|→∞

( inf

To prove Theorem 12, we need the following result:

Proposition 13 Let h ∈ C∞(Rn) satisfy supph ⊂ (Rn, ∆) for some ∆ > 0 andmax{kDβhk∞ : β ∈ Zn

ξ∈supp h|ξα|)kHαk11/|α|≤ 1,(11)

where (A) stands before the upper limit means that we take the limit only for α ∈ A.

Proof For ξ = (ξ1, ξ2, , ξn), we adopt the convention that ξ2 = Qn

Dβ h(ξ)

ξα

 dξ

Then it follows from the Leibniz rule that

xα−2

, ∀θ ∈ Zn

+, θ ≤ (2, 2, , 2).(13)

xα−2

C23n

∆n (14)

Let 0 ≤ k ≤ n and (i1, i2, , in) be a permutation of (1, 2, , n) We defineA(i1, , in, k) = {x ∈ Rn : |xi | ≥ 1, , |xi | ≥ 1, |xi | ≤ 1, , |xi | ≤ 1} Then

it follows from (14) and

x2i

1 x2i

kHα(x)

that

(A) lim

|α|→∞

( inf

lim

r→σ

inf

ξ∈supph|ξr| ≥ ∆|σ(1−(1/λ))|(|ξσ|)1/λLetting λ → 1+, we have

by letting  → 0

Combining (17) and (18), we get (16)

Now, assume the contrary that (11) is false Then there exist an unboundedsubset I ⊂ A , a number λ > 1 and a vector β ≥ 0, |β| = 1 such that

(I) lim

|α|→∞

α

|α| = β,(I) lim

ξ∈supph|ξα|1/|α| = (I) lim

|α|→∞

inf

ξ∈supph|ξα−2|1/|α|.Therefore, since (15), we obtain

(I) lim

|α|→∞

( inf

Now we show

Dβ(Iα+βf ) = Iαf ∀α, β ∈ Zn+.(22)

Indeed, we choose a function g ∈ C∞(Rn) satisfying g(x) = 1 for all x ∈ K∆/4and g(x) = 0 for all x /∈ K∆/2 For any ϕ ∈ C0∞(Rn) it follows from (21) andsuppg ⊂ (Rn, ∆/2) that

and then

α+βf, Dβϕ= (−1)|β| f , (Fˆ −1

ϕ)(x)g(x)/(ix)α (24)

Using (23)-(24), we have α+βf, Dβϕ= (−1)|β| αf, ϕ and then (22) have beenproved

Next we prove that

lim

|α|→∞

( inf

arbi-Rnρ(x)dx = 1 and put

u(x) = 1K3/4(x), h(x) = (u ∗ ρ/4)(x),where ρ/4(x) = (4)nρ(4x) Then the function h ∈ C∞(Rn) satisfies the followingconditions

h(x1, , xn) = 1 ∀(x1, , xn) ∈ K/2,(26)

h(x1, , xn) = 0 ∀(x1, , xn) /∈ K,(27)

|Iαf | = (2π)−n/2|f ∗ F−1(h(ξ)/ξα)|

Therefore, it follows from Lemma 2 that for α ∈ A0:

kIαf kΦ ≤ (2π)−n/2kf kΦkF−1(h(ξ)/ξα)k1 = (2π)−n/2kf kΦkF (h(ξ)/ξα)k1.(29)

Using (27)- (28) and Proposition 13, we have

(A0) lim

|α|→∞

( inf

ξ∈K 

|ξα|)kIαf kΦ1/|α| ≤ 1

(30)

Since K ⊂ (Rn, ∆), it is easy to check that

inf

ξ∈K 

|ξα|1/|α|≥inf

ξ∈K|ξα|1/|α|∆ − 

∆ .(31)

Combining (30) and (31), we get

(A0) lim

|α|→∞

( inf

(A0) lim

|α|→∞

( inf

ξ∈supp ˆ f

|ξα|)kIαf kΦ

1/|α|

≤ 1,where Ak := {α ∈ Zn

+ : α1, , αk ≥ 3, αk+1 = · · · = αn= 0} Put

α0 := (α1, , αk), K0 := {ξ ∈ Rk : (ξ, u) ∈ K for some u ∈ Rn−k},

and we define a function h1(x) ∈ C∞(Rk) satisfying the following conditions

h1(x1, , xk) = 1 ∀(x1, , xk) ∈ K/20(32)

h1(x1, , xk) = 0 ∀(x1, , xk) /∈ K0(33)

f h(ξ)/(iξ)α = dIαf Therefore, it follows from Proposition 4 and Proposition 6 we can deduce for α ∈ Ak,

|Iαf | = (2π)−k/2|f ~ F−1

k h1(x1, , xk)/(xα1

1 , xαk

k )
|,Hence, applying Proposition 5, we get for α ∈ Ak

ξ∈K |ξα|)kIαf kΦ

1/|α|

≤ 1

By this and (31), we obtain

ξ∈supp ˆ f

|ξα|)kIαf kΦ

1/|α|

≤ 1and then Step 3 has been proved

Let 0 ≤ k ≤ n, u ∈ Zn−k+ and (i1, , in) be a permutation of (1, 2, , n) We define

Bk,u,(i1, ,in) = {α = (α1, , αn) ∈ Zn

+ : αi1 ≥ 3, , αik ≥ 3, αik+1 = u1, , αin =

un−k} Then arguing similarly as in the proof of Step 3, we also have

(Bk,u,(i1, ,in)) lim

|α|→∞

( inf

ξ∈supp ˆ f

|ξα|)kIαf kΦ1/|α| ≤ 1

The proof of (25) is complete

Finally, we have to prove that

On the other hand, it follows from Theorem B that

sup

ξ∈B(σ,)

|ξα|1/|α| ≤ ∆ + 

∆ |σα|1/|α|.Hence,

(I1) lim

|α|→∞

( inf

From this we have (40).

ξ∈supp ˆ f

(|ξα|)kIαf kΦ

1/|α|

= 1

Using Theorems 12 and B, we have the following theorems:

Theorem 14 Let f ∈ LΦ(Rn), f 6≡ 0, supp ˆf be compact, f has (O)-property and(Iαf )α∈Zn

+ ⊂ LΦ(Rn) Then

lim

|α|→∞

 kDαf kΦsup

Proof Necessary It is clear from Theorem 12

Sufficiency Assume the contrary that there exists θ ∈ supp ˆf , θ 6∈

n

Q

k=1

(−∞, −σk] ∪[σk, +∞)
Then there exists j ∈ {1, 2, , n} such that |θj| < σj Therefore,

Remark 16 Theorem 12 is not true if f doesn’t have (O)-property It is clearbecause inf

ξ∈supp ˆ f

|ξα| = 0 for any α ∈ Zn

+ such that αj ≥ 1, j = 1, , n

We examine now behavior of kIαf kΦ

1/|α|

for functions f not having (O)-property.From the proof of Theorems 7, 11 and 12, we have the following result:

Theorem 17 Let f ∈ LΦ(Rn), suppkf := {ξ ∈ Rˆ k : (ξ, u) ∈ supp ˆf for some u ∈

Rn−k} has (O)-property (in Rk), and let A = {α ∈ Zn

+ : αk+1 = · · · = αn =0} Then there exists uniquely one sequence of primitives (Iαf )α∈A ⊂ LΦ(Rn).Moreover, suppdIαf = supp ˆf ∀α ∈ A and

(A) lim

|α|→∞

( inf

Proof We prove for j = n We fix an element ξ = (ξ1, , ξn−1, 0) ∈ supp ˆf and

 ∈ (0, 1) then there exists a function η ∈ C0∞(B(ξ, ))) such that f , η6= 0 Hence,ˆ

(Aj) lim

|α|→∞

 kDαηkˆ Φ¯

α iQn−1 j=1(|ξj| + 1)α j

(Aj) lim

|α|→∞(kIαf kΦ)1/|α|= ∞

Theorem 19 Assume that for any j ∈ {1, 2, , n} there is an element in supp ˆf ,the jth coordinate of which equals 0 Then

n}

Then it follows from Proposition 18 that

We consider now the case when f doesn’t have (O)-property and not satisfiesthe condition in Theorem 19 The following theorem is clear from the proofs ofTheorems 7, 11, 12 and Theorem 19:

Theorem 20 Let f ∈ LΦ(Rn) and 1 ≤ k < n Assume that for any j ∈ {k +

1, k + 2, , n} there is an element in supp ˆf , the jth coordinate of which equals 0,and {ξ ∈ Rk: there exists u ∈ Rn−k such that (ξ, u) ∈ supp ˆf } ⊂ (Rk, ∆) for some

∆ > 0 Then

(A) lim

|α|→∞

( inf

ξ∈supp ˆ f

|ξα|)kIαf kΦ

1/|α|

= 1,where A = {α ∈ Zn

+ : αk+1 = · · · = αn = 0} and

(B) lim

|α|→∞(kIαf kΦ)1/|α|= ∞,here B = {α ∈ Zn

+: α1 = · · · = αk = 0}

Moreover, there doesn’t exist the limit

lim

|α|→∞(kIαf kΦ)1/|α|.Proof We have only to show that the limit doesn’t exist From the proof of Theorem

12 we have

(A) lim

|α|→∞

( inf

(B) lim

|α|→∞(kIαf kΦ)1/|α|= ∞

Let Φ be an arbitrary Young function We write Φ ∈ ∆2 if there exists C > 0such that Φ(2t) ≤ CΦ(t) ∀t ∈ R+ In conclusion, we give the following result:Theorem 21 Let Φ be an arbitrary Young function, Φ ∈ ∆2, σ = (σ1, , σn) ∈

To obtain Theorem 21 we need the following result [17, 26]:

Bohr-Favard inequality for Orlicz spaces: Let σ > 0, f ∈ Cm(R) , Dmf ∈

LΦ(R) and supp ˆf ⊂ (−∞, −σ] ∪ [σ, +∞) Then f ∈ LΦ(R) and

kf kΦ ≤ σ−mKmkDmf kΦ,where the Favard constants Km are best possible and have the following properties

1 = K0 ≤ K2 < · · · < 4

π < · · · < K3 ≤ K1 = π

2.Proof of Theorem 21 Because Φ ∈ ∆2, for any  > 0, it is known that there exists

a number λ > 1 such that

σαkIαf kΦ ≤ σαkIα(f − h)kΦ+ σαkIαhkΦ

≤ Kαkf − hkΦ+ Kαλ−|α|khkΦ

≤ Kα + Kαλ−|α|khkΦfor all α ∈ Zn+ Hence, since Kα ≤ (π/2)n, we have

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[15] H.H Bang, The study of the properties of functions belonging to an Orlicz space depending on the geometry of their spectra , Izv Akad Nauk Ser Mat 61 (1997),... Bang, A study of the properties of functions depending on the geometry of their spectrum, Doklady Akad Nauk 355 (1997), 740 - 743.

[17] H.H Bang, An inequality of Bohr and... 1f Using Theorems and 10, we have the following:

Theorem 11 Let f ∈ LΦ(Rn) and f has (O)-property Then there exists exactlyone sequence of primitives (Iαf