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).

ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF 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. Introduction The following result was proved in [12]: Theorem A. Let 1 ≤ p ≤ ∞,f ∈ Lp (Rn ) and suppfˆ is bounded, where fˆ is the Fourier transform of f . Then Dα f p 1/|α| = 1. lim |α|→∞ sup |ξ α | ξ∈suppfˆ The novelty and importance of Theorem A is that it is almost impossible to calculate directly the sequence { Dα f p 1/|α| } but we can approximately calculate it by calculating the sequence {supξ∈suppfˆ |ξ α |1/|α| } and this is possible if we know all “far from the origin” points of “suppfˆ”. This result was proved for n = 1 in [11] and was studied and developed by many authors (see [1]-[16], [20], [22]-[25],[34][36]). It is natural to ask what will happen when we replace derivatives by integrals. 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 for the n-dimensional case and Orlicz spaces. 2. Definitions Let Φ : [0, +∞) → [0, +∞] be an arbitrary Young function, i.e., Φ(0) = 0, Φ(t) ≥ 0, Φ(t) ≡ 0 and Φ is convex. Denote by: ¯ = sup{ts − Φ(s)} Φ(t) s≥0 the Young function conjugate to Φ and LΦ (Rn )-the space of measurable function u such that u(x)v(x)dx < ∞ u, v = Rn Key words and phrases. Orlicz spaces, generalized functions. 2010 AMS Subject Classification. 26D10, 46E30. 1 2 HA HUY BANG & VU NHAT HUY ¯ < ∞, where for all v with ρ(v, Φ) ¯ = ρ(v, Φ) ¯ Φ(|v(x)|)dx. Rn Then LΦ (Rn ) is a Banach space with respect to the Orlicz norm u Φ = u LΦ (Rn ) u(x)v(x)dx , = sup ¯ ρ(v,Φ)≤1 Rn which is equivalent to the Luxemburg norm f (Φ) Φ(|f (x)|/λ)dx ≤ 1} < ∞. = inf{λ > 0 : Rn Note that Lebesgue spaces and their extension - Orlicz spaces play an important role in Analysis and have many applications (see [21],[27]-[33]). Recall that . (Φ) = . p where Φ(t) = tp with 1 ≤ p < ∞, and . (Φ) = . ∞ where Φ(t) = 0 for 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 |u(x)v(x)|dx ≤ u Φ v ¯. Φ Rn Lemma 2. Let u ∈ LΦ (Rn ) and v ∈ L1 (Rn ). Then u∗v Φ ≤ v Φ v 1. 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 S (Rn ) the set of all continuous linear functionals on S(Rn ). Any element h ∈ S (Rn ) is called a tempered generalized function and we write h(ϕ) = h, ϕ for all ϕ ∈ S(Rn ). Let f ∈ S (Rn ) and ej = (0, . . . , 0, 1, 0, . . . , 0) ∈ Zn+ be an unit vector such that its j th coordinate equals 1, j = 1, 2, . . . , n. The tempered generalized function I ej f is ej ej termed the eth j primitive of f if D (I f ) = f , that is, I ej f, Dej ϕ = − f, ϕ ∀ϕ ∈ S(Rn ), n where Dej f is the eth j derivative of f . Now we show that for each f ∈ S (R ) the set P ej (f ) = {h ∈ S (Rn ) : Dej h = f } is not empty. Indeed, for ϕ ∈ S(Rn ) we define the following functions +∞ ψ(x) = ϕ(x) − θ(xj ) ϕ(x1 , . . . , xj−1 , t, xj+1 , . . . , xn )dt, −∞ xj Ψ(x) = ψ(x1 , . . . , xj−1 , t, xj+1 , . . . , xn )dt, −∞ ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 3 where θ ∈ C0∞ (R), suppθ ⊂ [−1, 1], θ(t)dt = 1. R Then, it is easy to see that ψ, Ψ ∈ S(Rn ). So, for each f ∈ S (Rn ), we can define the tempered generalized function I ej f via the formula I ej f, ϕ = − f, Ψ . Then I ej f ∈ S (Rn ) and xj I ej f, Dej ϕ = − f, Dej ϕ(x1 , . . . , xj−1 , t, xj+1 , . . . , xn )dt = − f, ϕ . −∞ ej eth j Therefore, I f is the primitive of f , we have primitive of f . Furthermore, if I ej f ∈ S (Rn ) is the eth j I ej f, ϕ = I ej f (x), Dej Ψ(x) + θ(xj ) ϕ(x1 , . . . , xj−1 , t, xj+1 , . . . , xn )dt R = I ej f (x), Dej Ψ(x) + I ej f (x), θ(xj ) ϕ(x1 , . . . , xj−1 , t, xj+1 , . . . , xn ) dt R = − f, Ψ + I ej f (x), θ(xj ) ϕ(x1 , . . . , xj−1 , t, xj+1 , . . . , xn ) dt . R Hence, I ej f, ϕ = − f, Ψ + gj , ϕ , (1) where gj ∈ S (Rn ) is defined by gj , ϕ = fj (x1 , . . . , xj−1 , xj+1 , . . . , xn ), ϕ(x1 , . . . , xj−1 , t, xj+1 , . . . , xn )dt R with fj ∈ S (Rn−1 ) is defined by fj , h = I ej f (x), θ(xj )h(x1 , . . . , xj−1 , xj+1 , . . . , xn ) for each h ∈ S(Rn−1 ). Then gj is called the j th constant tempered generalized function. Note that gj becomes a constant if n = 1. Conversely, given an arbitrary the j th constant tempered generalized function gj , the functional I ej f 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 generalized ej function f ∈ S (Rn ) has in S (Rn ) the eth j primitive, which is denoted by I f , and every the eth j primitive of f is expressed by formula (1), where gj is an arbitrary the th j constant tempered generalized function. Note that the notation of primitive of a generalized function in D (a, b), a, b ∈ R can be found in [38], here we define it for tempered generalized functions in S (Rn ). We denote I 0 f = f . In the sequel, for j = 1, 2, . . . , n then I ej f denotes the eth j primitive of f ∈ S (Rn ), i.e., I ej f ∈ P ej (f ), and for any α ∈ Zn+ , |α| ≥ 1 we define the primitive of higher order by the following way: I α f = I ejα (I α−ejα f ), where jα := max{j : αj ≥ 1}, i.e., I α f ∈ P ejα (I α−ejα f ). So, Dα (I α f ) = f for all α ∈ Zn+ . Let 1 ≤ p ≤ ∞ and f ∈ LΦ (Rn ). If for any α ∈ Zn+ there exists the 4 HA HUY BANG & VU NHAT HUY ejα primitive of I α−ejα f , which belongs to LΦ (Rn ) and is denoted by I α f , we write (I α f )α∈Zn+ ⊂ LΦ (Rn ). Before giving our main results, we introduce a notation of the extended convolution of functions with different numbers of variables: Definition 3. Let n ≥ k, f be a locally integrable function in Rn , g be a locally integrable functions in Rk . If the integral f (x1 − y1 , . . . , xk − yk , xk+1 , . . . , xn ).g(y1 , . . . , yk )dy1 . . . dyk Rk exists for almost all x = (x1 , . . . , xn ), then it is called the extended convolution of f and g, and is symbolized as (f g)(x). The above definition becomes the known one if n = k. For the extended convolution we have the following result: Proposition 4. Let n, k ∈ N, n ≥ k, f ∈ LΦ (Rn ) and g ∈ L1 (Rk ). Then f LΦ (Rn ) and (2) f g LΦ (Rn ) ≤ f LΦ (Rn ) g g∈ L1 (Rk ) . Proof. Put v = (xk+1 , . . . , xn ) and denote f (x1 , . . . , xk , v) := f (x1 , . . . , xn ). Then by the definition of Orlicz norm we obtain f g LΦ (Rn ) = sup{ f g, ϕ : ϕ ¯ Φ ≤ 1}. Note that f f (x1 − y1 , . . . , xk − yk , v)g(y1 , . . . , yk )dy1 . . . dyk ϕ(x1 , . . . , xn )dx g, ϕ = Rn Rk f (x1 − y1 , . . . , xk − yk , v)ϕ(x1 , . . . , xn )dx g(y1 , . . . , yk )dy1 . . . dyk = Rk Rn F (y1 , . . . , yk )g(y1 , . . . , yk )dy1 . . . dyk , = Rk where f (x1 − y1 , . . . , xk − yk , v)ϕ(x1 , . . . , xn )dx. F (y1 , . . . , yk ) = Rn Then f(y1 ,...,yk ) (x1 , . . . , xn )ϕ(x1 , . . . , xn )dx F (y1 , . . . , yk ) = Rn with f(y1 ,...,yk ) (x1 , . . . , xn ) = f (x1 − y1 , . . . , xk − yk , v), and then it follows from Lemma 1 that |F (y1 , . . . , yk )| ≤ f(y1 ,...,yk ) ϕ LΦ (Rn ) ¯ Φ = f Therefore, f g, ϕ ≤ f f g Φ g L1 (Rk ) ϕ ¯. Φ So, LΦ (Rn ) ≤ f Φ g L1 (Rk ) . LΦ (Rn ) ϕ ¯. Φ ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 5 The proof is complete. Let f ∈ L1 (Rn ) and fˆ = F f or Fn f be its Fourier transform fˆ(ξ) = (2π)−n/2 e−ixξ f (x)dx. Rn The Fourier transform of a tempered generalized function f is defined via the formula F f, ϕ = ϕ ∈ S(Rn ). f, F ϕ , Let K be an arbitrary compact set in Rn and > 0. Denote by B(x, ) := {ξ ∈ Rn : n x−ξ = (xk − ξk )2 1/2 < } - the ball of radius > 0 with center at the point k=1 x, B[x, ] := {ξ ∈ Rn : x − ξ ≤ }, K := {ξ ∈ Rn : ∃x ∈ K : x − ξ ≤ n }, |x| = j=1 |xj |. We put (Rn , ) := {ξ ∈ Rn : min{|ξ1 |, |ξ2 |, . . . , |ξn |} > }. The extended convolution has the following property which is essential to prove our results: Proposition 5. Let n, k ∈ N, n ≥ k, f ∈ S(Rn ) and g ∈ L1 (Rk ). Then (3) (Fn (f (ξ1 , . . . , ξn )g(ξ1 , . . . , ξk )))(x1 , . . . , xn ) = (2π)−k/2 (Fn f Fk g)(x1 , . . . , xn ), (Fn−1 (f (ξ1 , . . . , ξn )g(ξ1 , . . . , ξk )))(x1 , . . . , xn ) = (2π)−k/2 (Fn−1 f Fk−1 g)(x1 , . . . , xn ). Proof. Put u = (x1 , x2 , . . . , xk ), v = (xk+1 , . . . , xn ), s = (ξ1 , ξ2 , . . . , ξk ), t = (ξk+1 , . . . , ξn ). Then x = (u, v), ξ = (s, t), dξ = dsdt, xξ = us + vt. Therefore, (Fn (f (ξ1 , . . . , ξn )g(ξ1 , . . . , ξk )))(x) = (2π)−n/2 e−i(us+vt) f (s, t)g(s)dsdt Rn = (2π)−n/2 e−ius g(s) Rk e−ivt f (s, t)dt ds Rn−k and then e−ius g(s)Gv (s)ds, (Fn (f (ξ1 , . . . , ξn )g(ξ1 , . . . , ξk )))(x) = (2π)−k/2 Rk where Gv (s) = (2π)−(n−k)/2 e−ivt f (s, t)dt. Rn−k Hence, (4) (Fn (f (ξ1 , . . . , ξn )g(ξ1 , . . . , ξk )))(x) = (2π)−k/2 (Fk Gv ) ∗ (Fk (g))(u). 6 HA HUY BANG & VU NHAT HUY We see that (Fk Gv )(u) = (2π)−k/2 e−ius Gv (s)ds Rk = (2π)−n/2 Rn Rn−k Rk e−ixξ f (ξ)dξ. e−ivt f (s, t)dt ds = (2π)−n/2 e−ius That means (5) (Fk Gv )(u) = (Fn f )(x). Combining (4)-(5), we arrive at (3). Similarly, we also have the result for the inverse Fourier transform. The proof is complete. 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 f g, h = f, h g(−x) Let f ∈ S (Rn ). Then suppfˆ is called the spectrum of f . We say that f has (O)-property if its spectrum is contained in (Rn , ∆) for some ∆ > 0. Using the extended 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, . . . , n ej there exists exactly one the eth j primitive of f , which is denoted by I f , such that I ej f ∈ LΦ (Rn ) and also has (O)-property. Moreover, suppI ej f = suppfˆ. Proof. It is sufficient to prove for the case j = 1. By the assumption, suppfˆ ⊂ (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 (6) φ(x) = 1 ∀x ∈ (−∞, −∆/2) ∪ (∆/2, +∞), (7) φ(x) = 0 ∀x ∈ (−∆/4, ∆/4). Then it follows from (7) that the function η is well defined and it is not difficult to see that supx∈R (1 + x2 )|ˆ η (x)| < ∞. So, ηˆ ∈ L1 (R) and then the following function Ψ is well defined via the formula 1 Ψ(x1 , . . . , xn ) = √ 2π f (x1 − ξ, x2 , . . . , xn )ˆ η (ξ)dξ. R ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 7 Then it follows from Proposition 4, f ∈ LΦ (Rn ) and ηˆ ∈ L1 (R) that Ψ ∈ LΦ (Rn ). For each ϕ ∈ S(Rn ), we put ψ = D3e1 ϕ. Then Ψ, ψ = Ψ(x1 , . . . , xn )ψ(x1 , . . . , xn )dx Rn 1 =√ 2π 1 =√ 2π 1 =√ 2π f (x1 − ξ, x2 , . . . , xn )ˆ η (ξ)dξ ψ(x1 , . . . , xn )dx Rn R f (ξ, x2 , . . . , xn )ˆ η (x1 − ξ)dξ ψ(x1 , . . . , xn )dx Rn R ψ(x1 , x2 , . . . , xn )ˆ η (x1 − ξ)dx1 f (ξ, x2 , . . . , xn )dξdx2 . . . , dxn Rn R = − f, g , where 1 g(ξ, x2 , . . . , xn ) = √ 2π ψ(x1 , x2 , . . . , xn )ˆ η (ξ − x1 )dx1 . R Hence, 1 g(x1 , . . . , xn ) = √ 2π ψ(x1 − ξ, x2 , . . . , xn )ˆ η (ξ)dξ. R So, applying Proposition 5, we get (8) (F −1 g)(x1 , . . . , xn ) = (F −1 ψ)(x1 , . . . , xn )η(x1 ) = (F −1 ϕ)(x1 , . . . , xn )φ(x1 ). From (6) and suppfˆ ⊂ (Rn , ∆), we have fˆ = fˆφ(x1 ). Therefore, since (8), we obtain f, g = fˆ, F −1 g = fˆ, (F −1 ϕ)(x1 , . . . , xn )φ(x1 ) = fˆφ(x1 ), (F −1 ϕ)(x1 , . . . , xn ) = fˆ, (F −1 ϕ)(x1 , . . . , xn ) = f, ϕ . This implies Ψ, D3e1 ϕ = − f, ϕ . So, D3e1 Ψ = f in the distribution sense. Hence, we have D3e1 Ψ ∈ LΦ (Rn ) and Ψ ∈ LΦ (Rn ). Therefore, D2e1 Ψ ∈ LΦ (Rn ), and it is symbolized as I e1 f . Next, we prove that suppI e1 f = suppfˆ. Indeed, Since De1 I e1 f = f , we have fˆ = ix1 I e1 f . Therefore, (9) suppfˆ ⊂ suppI e1 f ⊂ suppfˆ ∪ H1 , where H1 := {ξ ∈ Rn : ξ1 = 0}. So, to prove suppI α f = suppfˆ, it is enough to show H1 ∩ suppI e1 f = {∅}. Assume now the contrary that ∃ξ ∈ H1 ∩ I e1 f . 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 suppfˆ ⊂ (Rn , ∆), we have supphI e1 f ⊂ H1 . 8 HA HUY BANG & VU NHAT HUY Hence, there is a number N0 ∈ N such that N0 F −1 gjN0 (x2 , . . . , xn )(−ix1 )j . e1 h ∗ I f (x) = j=0 From Lemma 1 and I e1 f ∈ LΦ (Rn ), F −1 h ∈ LΦ¯ (Rn ), we get F −1 h ∗ I e1 f ∈ L∞ (Rn ). Therefore, F −1 h ∗ I e1 f (x) = g0N0 (x2 , . . . , xn ). By the same manner, we also have F −1 h ∗ Ψ(x) = g0N1 (x2 , . . . , xn ). Note that F −1 h ∗ I e1 f (x) = F −1 h ∗ D2e1 Ψ(x) = D2e1 (F −1 h ∗ Ψ(x)) = D2e1 (g0N1 (x2 , . . . , xn )) ≡ 0. So, hI e1 f = 0. On the other hand, since ξ ∈ suppI e1 f , there is a function ϕ ∈ C0∞ (B(ξ, a)) such that I e1 f , ϕ = 0. Then, since h(x) = 1 in (−a, a) × (−b − a, b + a)n−1 , we get 0 = I e1 f , ϕ = I e1 f , hϕ = hI e1 f , ϕ = 0. This is impossible. Hence, suppI e1 f = suppfˆ. n Finally, we prove the uniqueness of the eth 1 primitive of f which belongs to LΦ (R ) e1 e1 th and has (O)-property. Indeed, suppose that I f and J f are certain e1 primitives of f such that I e1 f, J e1 f ∈ LΦ (Rn ) and both of them have (O)-property. Since I e1 f, J e1 f ∈ LΦ (Rn ) and De1 (J e1 f − I e1 f ) = 0, we get J e1 f − I e1 f = g(x2 , . . . , xn ). Hence, F (J e1 f − I e1 f ), η(x1 )ϕ(x2 , . . . , xn ) = η(0) g(x2 , . . . , 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 (J e1 f − I e1 f ). On the other hand, it follows from suppF (J e1 f −I e1 f ) ⊂ suppF (J e1 f )∪ suppF (I e1 f ) that J e1 f − I e1 f has (O)-property. Therefore, suppFn−1 (g(x2 , . . . , xn )) = ∅, i.e., J e1 f − I e1 f = 0 a.e. The proof is complete. Remark 8. Let an Young function Φ : [0, ∞) → [0, ∞) and f ∈ LΦ (Rn ). Then for any j = 1, . . . , n there is at most one in LΦ (Rn ) the eth j primitive of f , which is ej denoted by I f , and moreover, there exists exactly one in LΦ (Rn ) the eth j primitive of f if f has (O)-property. Remark 9. 1. If f ∈ LΦ (Rn ) doesn’t have (O)-property then it is possible that there doesn’t exist in LΦ (Rn ) any the eth j primitive of f , j = 1, 2, . . . , n. It is illustrated in the following examples: +) LΦ (Rn ) = L∞ (Rn ) and f ≡ 1 (then suppfˆ = {0}); +) LΦ (Rn ) = L∞ (Rn ) and f (x) = nj=1 sin2 xj (then suppfˆ = {x ∈ Rn : xj ∈ ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 9 {0, −2, 2} , j = 1, 2, .., n}); +) Φ(t) = tp with 1 ≤ p < ∞ and f (x) = n j=1 sin2 xj /x2j (then suppfˆ = [−2, 2]n ). 2. Let LΦ (Rn ) = L∞ (Rn ) and f, I α f ∈ L∞ (Rn ). Put g(x) = f (x)+c, where 0 = c is a constant. Then g ∈ L∞ (Rn ) but there doesn’t exist the αth primitive I α g ∈ L∞ (Rn ). 3. Let LΦ (Rn ) = L∞ (Rn ) and f (x) = nj=1 cos xj . Then suppfˆ = {x ∈ Rn : xj ∈ {−1, 1} , j = 1, 2, . . . , n} and there exists in L∞ (Rn ) any I α f - the αth primitive of f. Theorem 10. Let f ∈ LΦ (Rn ). Then there is at most one sequence of primitives (I α f )α∈Zn+ ⊂ LΦ (Rn ). Proof. We have to show that if there exists another sequence of primitives of f , which is denoted by (J α f )α∈Zn+ , such that (J α f )α∈Zn+ ⊂ LΦ (Rn ), then J α f = I α f ∀α ∈ Zn+ . Indeed, it is sufficient to prove this fact for the case α = e1 . Since J e1 f, I e1 f, f ∈ LΦ (Rn ) and De1 (J e1 f − I e1 f ) = 0, we get J e1 f − I e1 f = g(x2 , . . . , xn ) and then De1 (J 2e1 f − I 2e1 f ) = g(x2 , . . . , xn ). Hence, J 2e1 f − I 2e1 f = (x1 + c)g(x2 , . . . , xn ), where c is some number in C. Therefore, taking account of J 2e1 f − I 2e1 f ∈ LΦ (Rn ), we get g(x2 , . . . , xn ) = 0 a.e. So, J e1 f = I e1 f . Using Theorems 7 and 10, we have the following: Theorem 11. Let f ∈ LΦ (Rn ) and f has (O)-property. Then there exists exactly one sequence of primitives (I α f )α∈Zn+ ⊂ LΦ (Rn ). Moreover, suppI α f = suppfˆ ∀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 ≡ 0, f has (O)-property and (I α f )α∈Zn+ ⊂ LΦ (Rn ). Then (10) lim |α|→∞ ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| = 1. Φ To prove Theorem 12, we need the following result: Proposition 13. Let h ∈ C ∞ (Rn ) satisfy supph ⊂ (Rn , ∆) for some ∆ > 0 and max{ Dβ h ∞ : β ∈ Zn+ , β ≤ (3, 3, . . . , 3)} ≤ C < ∞. We put A = {α ∈ Zn+ : αj ≥ 3, j = 1, . . . , n}. Then for any α ∈ A the function Hα (x) = F h(ξ)/ξ α is well defined and (11) (A) lim |α|→∞ ( inf ξ∈supp h |ξ α |) Hα 1/|α| 1 ≤ 1, 10 HA HUY BANG & VU NHAT HUY 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 ξ 2 = nj=1 ξj2 and ξ − 2 = (ξ1 − 2, . . . , ξn − 2). Since supph ⊂ (Rn , ∆) for some ∆ > 0 and α ∈ A, it follows that h(ξ)/ξ α ∈ L1 (Rn ) and then Hα is well defined. So, e−ixξ Hα (x) = (2π)−n/2 h(ξ) dξ = (2π)−n/2 ξα Rn e−ixξ h(ξ) dξ. ξα ξ∈supph Hence, for β ∈ Zn+ , β ≤ (2, 2, . . . , 2) we have the following estimate e−ixξ Dβ sup |xβ Hα (x)| ≤ (2π)−n/2 sup x∈Rn x∈Rn h(ξ) dξ ξα Rn e−ixξ Dβ = (2π)−n/2 sup x∈Rn h(ξ) dξ ξα ξ∈supph ≤ (2π)−n/2 h(ξ) dξ. ξα Dβ ξ∈supph Then it follows from the Leibniz rule that (12) sup |xβ Hα (x)| ≤ (2π)−n/2 x∈Rn γ≤β ξ∈supph ≤ (2π)−n/2 γ≤β 1 β! Dγ h(ξ)Dβ−γ α dξ γ!(β − γ)! ξ 1 β! sup |x2 Dβ−γ α | γ!(β − γ)! x∈supph x | ξ∈supph β! 1 ≤ (2π)−n/2 max sup |x2 Dθ α | θ≤(2,2,...,2) x∈supph x γ!(β − γ)! γ≤β ≤ (2π)−n/2 max 1 γ D h(ξ)|dξ ξ2 sup |x2 Dθ θ≤(2,2,...,2) x∈supph C| 1 |dξ ξ2 ξ∈(Rn ,∆) 1 C23n | . xα ∆n Since supph ⊂ (Rn , ∆), there exists a constant C1 not depending on α such that (13) sup |x2 Dθ x∈supph 1 1 | ≤ C1 |α|2n sup , α α−2 x x∈supph x ∀θ ∈ Zn+ , θ ≤ (2, 2, . . . , 2). From (12)-(13), we have (14) β −n/2 sup |x Hα (x)| ≤ (2π) x∈Rn 2n C1 |α| sup x∈supph 1 xα−2 C23n . ∆n Let 0 ≤ k ≤ n and (i1 , i2 , . . . , in ) be a permutation of (1, 2, . . . , n). We define A(i1 , . . . , in , k) = {x ∈ Rn : |xi1 | ≥ 1, . . . , |xik | ≥ 1, |xik+1 | ≤ 1, . . . , |xin | ≤ 1}. Then ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 11 it follows from (14) and Hα (x) dx ≤ Hα (x) dx (i1 ,...,in ,k)A(i ,...,i ,k) n 1 Rn sup x2i1 . . . x2ik Hα (x) ≤ (i1 ,...,in ,k) x∈Rn A(i1 ,...,in ,k) ≤ 2n x2i1 1 dx . . . x2ik sup x2i1 . . . x2ik Hα (x) (i1 ,...,in ,k) x∈Rn ≤ (2(n + 1))n sup sup x2i1 . . . x2ik Hα (x) (i1 ,...,in ,k) x∈Rn that (A) lim (15) |α|→∞ ( inf |ξ α−2 |) Hα ξ∈supph 1/|α| 1 ≤ 1. Now we prove (16) lim inf |ξ r | = inf |ξ σ |, r→σ ξ∈supph r, σ ≥ (0, 0, . . . , 0). ξ∈supph Given λ > 1. Then there is an > 0 such that λr ≥ σ for all |r − σ| < . Therefore, |ξ r | = |ξ r−(σ/λ) |(|ξ σ |)1/λ ≥ ∆|r−(σ/λ)| (|ξ σ |)1/λ ∀ξ ∈ supph. Thus we have obtained inf |ξ r | ≥ ∆|r−(σ/λ)| (|ξ σ |)1/λ . ξ∈supph Letting r → σ, we get lim inf |ξ r | ≥ ∆|σ(1−(1/λ))| (|ξ σ |)1/λ r→σ ξ∈supph Letting λ → 1+ , we have lim inf |ξ r | ≥ (17) inf |ξ σ |. ξ∈supph r→σ ξ∈supph On the other hand, for all > 0 there exists ξ ∈ supph such that inf |ξ σ | ≤ |ξ σ | ≤ ξ∈supph inf |ξ σ | + . ξ∈supph Taking account of inf |ξ r | ≤ |ξ r | ξ∈supph and letting r → σ, we have lim inf |ξ r | ≤ |ξ σ | ≤ r→σ ξ∈supph inf |ξ σ | + ξ∈supph and then (18) lim inf |ξ r | ≤ r→σ ξ∈supph by letting → 0. Combining (17) and (18), we get (16). inf |ξ σ | ξ∈supph 12 HA HUY BANG & VU NHAT HUY Now, assume the contrary that (11) is false. Then there exist an unbounded subset I ⊂ A , a number λ > 1 and a vector β ≥ 0, |β| = 1 such that α = β, (I) lim |α|→∞ |α| (I) lim (( inf |ξ α |) Hα 1 )1/|α| ≥ λ. |α|→∞ α |α|→∞ |α| Since (I) lim ξ∈supph = β and supph ⊂ (Rn , ∆), it follows from (16) that (19) inf |ξ α | (I) lim |α|→∞ α−2 |α|→∞ |α| |α|→∞ inf |ξ β |. ξ∈supph = β, supph ⊂ (Rn , ∆) and (16), we get inf |ξ α−2 | (I) lim = ξ∈supph Taking account of (I) lim (20) 1/|α| 1/|α| = ξ∈supph inf |ξ β |. ξ∈supph Combining (19)-(20), we have (I) lim |α|→∞ inf |ξ α | ξ∈supph 1/|α| = (I) lim |α|→∞ inf |ξ α−2 | 1/|α| ξ∈supph . Therefore, since (15), we obtain (I) lim |α|→∞ ( inf |ξ α |) Hα ξ∈supph 1/|α| ≤ 1. 1 This is a contradiction. So, (11) has been proved. The proof is complete. Further, we need the following result [15, 16]: Theorem B. Let f ∈ LΦ (Rn ), f ≡ 0 and suppfˆ be compact. Then lim |α|→∞ Dα f Φ sup |ξ α | 1/|α| = 1. ξ∈suppfˆ Proof of Theorem 12. Put K := suppfˆ. From Theorem 11 we have suppI α f = K (21) ∀α ∈ Zn+ . Now we show (22) Dβ (I α+β f ) = I α f ∀α, β ∈ Zn+ . Indeed, we choose a function g ∈ C ∞ (Rn ) satisfying g(x) = 1 for all x ∈ K∆/4 and g(x) = 0 for all x ∈ / K∆/2 . For any ϕ ∈ C0∞ (Rn ) it follows from (21) and suppg ⊂ (Rn , ∆/2) that (23) fˆ, (F −1 ϕ)(x)g(x)/(ix)α = (ix)α I α f , (F −1 ϕ)(x)g(x)/(ix)α = I α f , (F −1 ϕ)(x)g(x) = g(x)I α f , (F −1 ϕ)(x) = I α f, ϕ . Similarly, fˆ, (F −1 (Dβ ϕ))(x)g(x)/(ix)α+β = I α+β f, Dβ ϕ ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 13 and then I α+β f, Dβ ϕ = (−1)|β| fˆ, (F −1 ϕ)(x)g(x)/(ix)α . (24) Using (23)-(24), we have I α+β f, Dβ ϕ = (−1)|β| I α f, ϕ and then (22) have been proved. Next we prove that lim (25) ( inf |α|→∞ ξ∈supp fˆ |ξ α |). I α f 1/|α| ≤ 1. Φ To obtain (25), we divide our proof into three steps. Step 1. We show (A0 ) lim ( inf |α|→∞ ξ∈supp fˆ |ξ α |). I α f 1/|α| ≤ 1, Φ where A0 := {α ∈ Zn+ : αj ≥ 3 for all j = 1, . . . , n}. Indeed, given an arbitrary number ∈ (0, ∆/2). We choose a functions ρ(x) ∈ C0∞ ((−1, 1)n ) such that ρ(x)dx = 1 and put Rn u(x) = 1K3 /4 (x), h(x) = (u ∗ ρ /4 )(x), where ρ /4 (x) = ( 4 )n ρ( 4 x). Then the function h ∈ C ∞ (Rn ) satisfies the following conditions (26) h(x1 , . . . , xn ) = 1 ∀(x1 , . . . , xn ) ∈ K /2 , (27) h(x1 , . . . , xn ) = 0 ∀(x1 , . . . , xn ) ∈ /K, (28) max{ Dβ h ∞ β ≤ (3, 3, . . . , 3)} < C < ∞. : From (26) and the fact that fˆ = (iξ)α I α f , we get h(ξ)fˆ = (iξ)α I α f . So, fˆh(ξ)/(iξ)α = I α f and then for α ∈ A0 |I α f | = (2π)−n/2 |f ∗ F −1 (h(ξ)/ξ α )|. Therefore, it follows from Lemma 2 that for α ∈ A0 : (29) I αf Φ ≤ (2π)−n/2 f F −1 (h(ξ)/ξ α ) Φ 1 = (2π)−n/2 f Φ F (h(ξ)/ξ α ) 1 . Using (27)- (28) and Proposition 13, we have (A0 ) lim |α|→∞ 1/|α| ( inf |ξ α |) F (h(ξ)/ξ α ) ≤ 1. 1 ξ∈supph Hence, (A0 ) lim |α|→∞ ( inf |ξ α |) F (h(ξ)/ξ α ) ξ∈K 1/|α| 1 ≤ 1. Therefore, by (29), we obtain (30) (A0 ) lim |α|→∞ ( inf |ξ α |) I α f ξ∈K 1/|α| Φ ≤ 1. 14 HA HUY BANG & VU NHAT HUY Since K ⊂ (Rn , ∆), it is easy to check that 1/|α| inf |ξ α | (31) ≥ ξ∈K inf |ξ α | − . ∆ 1/|α| ∆ ξ∈K Combining (30) and (31), we get ( inf |ξ α |) I α f (A0 ) lim |α|→∞ ξ∈suppfˆ 1/|α| ≤ Φ ∆ . ∆− Letting → 0, we have ( inf |ξ α |) I α f (A0 ) lim |α|→∞ ξ∈suppfˆ 1/|α| ≤ 1. Φ Step 2. We prove that |α|→∞ where Ak := {α ∈ Zn+ : 1/|α| ( inf |ξ α |) I α f (Ak ) lim ≤ 1, Φ ξ∈suppfˆ α1 , . . . , αk ≥ 3, αk+1 = · · · = αn = 0}. Put α := (α1 , . . . , αk ), K := {ξ ∈ Rk : (ξ, u) ∈ K for some u ∈ Rn−k }, and we define a function h1 (x) ∈ C ∞ (Rk ) satisfying the following conditions (32) h1 (x1 , . . . , xk ) = 1 ∀(x1 , . . . , xk ) ∈ K /2 (33) h1 (x1 , . . . , xk ) = 0 ∀(x1 , . . . , xk ) ∈ /K (34) max{ Dβ h1 : ∞ β ∈ Zk+ , β ≤ (3, 3, . . . , 3)} < C. Using (32) and the fact that fˆ = (iξ)α I α f , we obtain h(ξ)fˆ = (iξ)α I α f . So, fˆh(ξ)/(iξ)α = I α f . Therefore, it follows from Proposition 4 and Proposition 6 we can deduce for α ∈ Ak , ϕ ∈ S(Rn ) that f Fk−1 h1 (x1 , . . . , xk )/(xα1 1 . . . , xαk k ) , ϕ = I αf , ϕ . So, |I α f | = (2π)−k/2 |f Fk−1 h1 (x1 , . . . , xk )/(xα1 1 . . . , xαk k ) |, Hence, applying Proposition 5, we get for α ∈ Ak (35) I αf Φ ≤ (2π)−k/2 f −k/2 = (2π) f Φ Fk−1 h1 (x1 , . . . , xk )/(xα1 1 . . . , xαk k ) α1 Φ Fk h1 (x1 , . . . , xk )/(x1 . . . , xαk k ) L1 (Rk ) L1 (Rk ) . Since (33) and (34), the function h1 (x1 , . . . , xk ) satisfies the conditions in the Proposition 13. So, lim ( inf |ξ α |) Fk h1 (x1 , . . . , xk )/(xα1 1 . . . , xαk k ) |α|→∞ ξ∈K 1/|α| L1 (Rk ) From this and (35) we get (Ak ) lim |α|→∞ ( inf |ξ α |) I α f ξ∈K 1/|α| Φ ≤ 1. ≤ 1. ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 15 By this and (31), we obtain (Ak ) lim ( inf |ξ α |) I α f |α|→∞ ξ∈K 1/|α| Φ ≤ ∆ . ∆− Letting → 0, we have (Ak ) lim (( inf |ξ α |) I α f |α|→∞ ξ∈suppfˆ 1/|α| Φ) ≤ 1. Step 3. We prove that (Bk,u ) lim (( inf |ξ α |) I α f |α|→∞ ξ∈suppfˆ 1/|α| Φ) ≤ 1, n where u ∈ Zn−k + , Bk,u := {α = (α1 , . . . , αk , u) ∈ Z+ : α1 ≥ 3, . . . , αk ≥ 3}. Put aα = (0, 0, . . . , 0, u) ∈ Zn+ , bα = (α1 , α2 , . . . , αk , 0, . . . , 0) ∈ Zn+ , g = I aα f . Then α = aα + bα , suppˆ g = suppfˆ, and from (22) we have I α f = I bα g. By using Step 2 we have ( inf |ξ bα |) I bα g (Ak ) lim |α|→∞ ξ∈suppˆ g 1/|bα | Φ ≤ 1. Hence, 1/|α| (Bk,u ) lim ( inf |ξ bα |) I α f Φ (Bk,u ) lim ( inf |ξ α |) I α f Φ |α|→∞ ξ∈suppfˆ ≤ 1. Therefore, |α|→∞ ξ∈suppfˆ 1/|α| ≤1 and then Step 3 has been proved. and (i1 , . . . , in ) be a permutation of (1, 2, . . . , n). We define Let 0 ≤ k ≤ n, u ∈ Zn−k + 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 α |) I α f ξ∈suppfˆ 1/|α| Φ ≤ 1. We notice that Zn+ is the finite union of Bk,u,(i1 ,...,in ) , where 0 ≤ k ≤ n, u ∈ Zn−k + ,u ≤ (3, 3, . . . , 3), (i1 , . . . , in ) is a permutation of (1, 2, . . . , n). Therefore, lim |α|→∞ ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| Φ ≤ 1. The proof of (25) is complete. Finally, we have to prove that lim (( inf |ξ α |) I α f (36) |α|→∞ ξ∈suppfˆ 1/|α| Φ) ≥ 1. Indeed, let σ ∈ suppfˆ and > 0. Then there exists a function ϕ ∈ C0∞ (B(σ, )) such that fˆ, ϕ = 0. Hence, it follows from Lemma 1 that (37) 0= f, ϕˆ = Dα (I α f ), ϕˆ = I α f, Dα ϕˆ ≤ I α f Φ Dα ϕˆ ¯. Φ 16 HA HUY BANG & VU NHAT HUY On the other hand, it follows from Theorem B that Dα ϕˆ Φ¯ sup |ξ α | lim |α|→∞ 1/|α| ≤ 1. ξ∈suppϕ Therefore, Dα ϕˆ Φ¯ sup |ξ α | lim |α|→∞ 1/|α| ≤ 1. ξ∈B(σ, ) From this and (37) we obtain 1/|α| ( sup |ξ α |) I α f lim |α|→∞ ≥ 1. Φ ξ∈B(σ, ) If ξ ∈ B(σ, ) then ∆ < |σj | and |ξj | ≤ |σj | + for j = 1, 2, . . . , n. Hence, 1/|α| ξ1 α1 ξn | . . . | |αn σ1 σn ∆ + α1 ∆ + αn 1/|α| | ...| | | ≤ |σ α | ∆ ∆ (|ξ α |)1/|α| = (|σ α |)1/|α| . | ≤ (|σ α |)1/|α| 1/|α| ∆ + . ∆ This gives sup |ξ α | ∆+ |σ α |1/|α| . ∆ 1/|α| ≤ ξ∈B(σ, ) Hence, |σ α |. I α f lim 1/|α| ≥ Φ |α|→∞ ∆ . ∆+ Letting → 0, we get |σ α |. I α f lim (38) 1/|α| ≥ 1. Φ |α|→∞ Suppose (36) is false. Then we can find an unbounded set I1 ⊂ Zn+ , λ < 1 and a vector β ≥ 0, |β| = 1 such that (I1 ) lim (39) |α|→∞ ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| ≤ λ, Φ α = β. |α|→∞ |α| (I1 ) lim We choose a number η such that 1 < η < 1 . λ α |α|→∞ |α| Because of (I1 ) lim deduce that there exists an unbounded subsequence I2 ⊂ I1 such that (40) (I2 ) inf |ξ α | ξ∈suppfˆ 1/|α| ≥ 1 inf |ξ β |. η ξ∈suppfˆ Indeed, it follows from the proof of (16) that (I1 ) lim ( inf |ξ α |)1/|α| = |α|→∞ ξ∈suppfˆ inf |ξ β |. ξ∈suppfˆ = β, we can ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 17 From this we have (40). Combining (39) - (40), we get (I2 ) lim ( inf |ξ β |)( I α f 1/|α| Φ) |α|→∞ ξ∈suppfˆ ≤ ηλ. Choose σ ∈ suppfˆ such that |σ β | ≤ η inf |ξ β |). Then we have ξ∈suppfˆ (I2 ) lim |σ β |( I α f |α|→∞ 1/|α| Φ) ≤ η 2 λ. From this and (38) we get α (I2 ) lim |σ β− |α| | ≤ η 2 λ < 1. |α|→∞ This is a contradiction. So, (36) has been proved. Combing (25), (36), we obtain lim |α|→∞ ( inf (|ξ α |) I α f ξ∈suppfˆ 1/|α| = 1. Φ The proof is complete. Using Theorems 12 and B, we have the following theorems: Theorem 14. Let f ∈ LΦ (Rn ), f ≡ 0, suppfˆ be compact, f has (O)-property and (I α f )α∈Zn+ ⊂ LΦ (Rn ). Then lim |α|→∞ Dα f Φ sup |ξ α | 1/|α| =1 ξ∈suppfˆ and lim |α|→∞ ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| = 1. Φ Theorem 15. Let (I α f )α∈Zn+ ⊂ LΦ (Rn ) and σ = (σ1 , . . . , σn ) ∈ Rn+ . Then suppfˆ ⊂ n (−∞, −σk ] ∪ [σk , +∞) if and only if k=1 lim (σ α I α f |α|→∞ 1/|α| Φ) ≤ 1. Proof. Necessary. It is clear from Theorem 12. n Sufficiency. Assume the contrary that there exists θ ∈ suppfˆ, θ ∈ (−∞, −σk ] ∪ k=1 [σk , +∞) . Then there exists j ∈ {1, 2, . . . , n} such that |θj | < σj . Therefore, 1 = lim (( inf |ξ mej |) I mej f m→∞ ξ∈suppfˆ 1/m Φ) ≤ lim (|θjm |. I mej f m→∞ 1/m Φ) |θj | |θj | lim (σjm I mej f Φ )1/m ≤ . m→∞ σj σj This is a contradiction. The proof is complete. = Remark 16. Theorem 12 is not true if f doesn’t have (O)-property. It is clear because inf |ξ α | = 0 for any α ∈ Zn+ such that αj ≥ 1, j = 1, . . . , n. ξ∈suppfˆ 18 HA HUY BANG & VU NHAT HUY 1/|α| We examine now behavior of I α f Φ 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 ), suppk fˆ := {ξ ∈ Rk : (ξ, u) ∈ suppfˆ 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 (I α f )α∈A ⊂ LΦ (Rn ). Moreover, suppI α f = suppfˆ ∀α ∈ A and (A) lim |α|→∞ ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| = 1. Φ Further, we have: α Proposition 18. Let a > 0, j ∈ {1, . . . , n} and Aj = {α ∈ Zn+ : |α|j ≥ a}. Assume that there is an element in suppfˆ, the j th coordinate of which equals 0. Then I αf (Aj ) lim |α|→∞ 1/|α| = ∞. Φ Proof. We prove for j = n. We fix an element ξ = (ξ1 , . . . , ξn−1 , 0) ∈ suppfˆ and ∈ (0, 1) then there exists a function η ∈ C0∞ (B(ξ, ))) such that fˆ, η = 0. Hence, 0= fˆ, η = Dα (I α f ), ηˆ = f, ηˆ = I α f, Dα ηˆ ≤ I α f Φ Dα ηˆ Φ¯ . Using Theorem D, we have (Aj ) lim Dα ηˆ Φ¯ n−1 αj j=1 (|ξj | + 1) αi |α|→∞ 1/|α| ≤ 1. Therefore, Dα ηˆ (Aj ) lim 1/|α| ¯ Φ n−1 j=1 (|ξj | |α|→∞ a + 1) ≤ 1. So, n−1 α (Aj ) lim ( I f 1/|α| Φ) ≥ −a |α|→∞ (|ξj | + 1)−1 . j=1 Letting → 0, we get (Aj ) lim ( I α f |α|→∞ 1/|α| Φ) = ∞. The proof is complete. Theorem 19. Assume that for any j ∈ {1, 2, . . . , n} there is an element in suppfˆ, the j th coordinate of which equals 0. Then lim |α|→∞ I αf 1/|α| Φ = ∞. Proof. For 1 ≤ j ≤ n we define Aj := {α ∈ Zn+ : αj 1 ≥ }. |α| n ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 19 Then it follows from Proposition 18 that 1/|α| I αf (Aj ) lim |α|→∞ = ∞. Φ From this and Zn+ = ∪nj=1 Aj , here we adopt the convention that I αf lim |α|→∞ 0 0 = 1, we get 1/|α| = ∞. Φ The proof is complete. We consider now the case when f doesn’t have (O)-property and not satisfies the condition in Theorem 19. The following theorem is clear from the proofs of Theorems 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 suppfˆ, the j th coordinate of which equals 0, and {ξ ∈ Rk : there exists u ∈ Rn−k such that (ξ, u) ∈ suppfˆ} ⊂ (Rk , ∆) for some ∆ > 0. Then (A) lim |α|→∞ where A = {α ∈ Zn+ : ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| = 1, Φ αk+1 = · · · = αn = 0} and (B) lim ( I α f 1/|α| Φ) |α|→∞ = ∞, here B = {α ∈ Zn+ : α1 = · · · = αk = 0}. Moreover, there doesn’t exist the limit lim ( I α f 1/|α| . Φ) |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 |ξ1α1 . . . ξkαk |) I α f ξ∈suppfˆ 1/|α| Φ ≤ 1. Therefore, I αf (A) lim |α|→∞ 1/|α| Φ ≤ ∆−1 < ∞, while (B) lim ( I α f |α|→∞ 1/|α| Φ) = ∞. The proof is complete. Let Φ be an arbitrary Young function. We write Φ ∈ ∆2 if there exists C > 0 such 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 ) ∈ n Rn , f ∈ LΦ (Rn ) and suppfˆ ⊂ (−∞, −σk ] ∪ [σk , +∞) . Then + k=1 (41) lim σ α I α f |α|→∞ Φ = 0. 20 HA HUY BANG & VU NHAT HUY To obtain Theorem 21 we need the following result [17, 26]: Bohr-Favard inequality for Orlicz spaces: Let σ > 0, f ∈ C m (R) , Dm f ∈ LΦ (R) and suppfˆ ⊂ (−∞, −σ] ∪ [σ, +∞). Then f ∈ LΦ (R) and f Φ ≤ σ −m Km Dm f Φ, where the Favard constants Km are best possible and have the following properties π 4 1 = K0 ≤ K2 < · · · < < · · · < K 3 ≤ 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 f (x) − f (λx) Φ ≤ . Put h(x) = f (λx). Then f − h Φ ≤ and n ˆ = λ suppfˆ ⊂ supph (42) ((−∞, −λσk ] ∪ [λσk , +∞)). k=1 Applying Theorem 11, we obtain suppI α h = supph and then n suppI α h ⊂ ((−∞, −λσk ] ∪ [λσk , +∞)). k=1 α α Taking account of D (I h) = h and consecutively applying the Bohr-Favard inequality for Orlicz space to each variable, we get I αh where Kα = n j=1 Φ ≤ Kα (λσ)−α h Φ, n Kαj . Since (42), suppf − h ⊂ ((−∞, −σk ] ∪ [σk , +∞)). Then k=1 it follows from the Bohr-Favard for Orlicz space inequality that I α (f − h) Φ ≤ Kα σ −α f − h Φ. Therefore, σα I αf Φ ≤ σ α I α (f − h) ≤ Kα f − h Φ ≤ Kα + Kα λ for all α ∈ Zn+ . Φ + σα I αh + Kα λ −|α| h −|α| h Φ Φ Φ n Hence, since Kα ≤ (π/2) , we have lim σ α I α f |α|→∞ Φ ≤ Kα . Letting → 0, we get (41). The proof is complete. Acknowledgement This research is funded by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.01-2011.32. A part of this work was done when the authors were working at the Vietnam Institute for Advanced Study in Mathematics (VIASM). The authors would like to thank the VIASM for providing a fruitful research environment and working condition. ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 21 References [1] L.D. Abreu, Real Paley-Wiener theorems for the Koornwinder-Swarttouw q-Hankel transform, J. Math. Anal. Appl. 334 (2007), 223-231. [2] E. Albrecht and Werner J. 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Rao and Z.D. Ren, Theory of Orlicz Spaces, Marcel Dekker, Inc., New York, 1991. [33] M.M. Rao and Z.D. Ren, Applications of Orlicz Spaces, Marcel Dekker Inc., New York, 2002. [34] V.K. Tuan, On the supports of functions, Numer. Funct. Anal. Optim. 20 (1999), 387–394. [35] V.K. Tuan, Paley-Wiener-type theorems, Fract. Calc. Appl. Anal. 2 (1999), 135–144. [36] V.K. Tuan and A. Zayed, Paley-Wiener-type theorems for a class of integral transforms, J. Math. Anal. Appl. 266 (2002), 200-226. [37] V.K. Tuan, Spectrum of signals, J. Fourier Anal. Appl. 7 (2001), 319-323 [38] V.S. Vladimirov, Methods of the theory of Generalized Functions, Taylor & Francis, London, New York, 2002. HA HUY BANG Institute of Mathematics, Vietnamese Academy of Science and Technology, 18 Hoang Quoc Viet Street, Cau Giay, Hanoi, Vietnam E-mail address: hhbang@math.ac.vn VU NHAT HUY Department of Mathematics, College of Science, Vietnam National University, 334 Nguyen Trai Street, Thanh Xuan, Hanoi, Vietnam E-mail address: nhat huy85@yahoo.com [...]... 420 - 422 [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), 163-198 [16] H.H 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 Favard for Orlicz spaces, Bull Polish Acad... | + 1)−1 j=1 Letting → 0, we get (Aj ) lim ( I α f |α|→∞ 1/|α| Φ) = ∞ The proof is complete Theorem 19 Assume that for any j ∈ {1, 2, , n} there is an element in suppfˆ, the j th coordinate of which equals 0 Then lim |α|→∞ I αf 1/|α| Φ = ∞ Proof For 1 ≤ j ≤ n we define Aj := {α ∈ Zn+ : αj 1 ≥ } |α| n ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 19 Then it follows from... 387 [18] H.H Bang and V.N Huy, Behavior of the sequence of norms of primitives of a function, J Approximation Theory 162 (2010), 1178 -1186 [19] H.H Bang and V.N Huy, Behavior of the sequences of norms of primitives of functions depending on their spectrum, Doklady Akad Nauk 440 (2011), 456 - 458 [20] J.J Betancor, J.D Betancor and J.M.R M´endez, Paley-Wiener type theorems for Ch´ebliTrim`eche transforms,... ξ∈suppfˆ Indeed, it follows from the proof of (16) that (I1 ) lim ( inf |ξ α |)1/|α| = |α|→∞ ξ∈suppfˆ inf |ξ β | ξ∈suppfˆ = β, we can ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 17 From this we have (40) Combining (39) - (40), we get (I2 ) lim ( inf |ξ β |)( I α f 1/|α| Φ) |α|→∞ ξ∈suppfˆ ≤ ηλ Choose σ ∈ suppfˆ such that |σ β | ≤ η inf |ξ β |) Then we have ξ∈suppfˆ (I2 ) lim... (Rk ) Since (33) and (34), the function h1 (x1 , , xk ) satisfies the conditions in the Proposition 13 So, lim ( inf |ξ α |) Fk h1 (x1 , , xk )/(xα1 1 , xαk k ) |α|→∞ ξ∈K 1/|α| L1 (Rk ) From this and (35) we get (Ak ) lim |α|→∞ ( inf |ξ α |) I α f ξ∈K 1/|α| Φ ≤ 1 ≤ 1 ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 15 By this and (31), we obtain (Ak ) lim ( inf |ξ... in = un−k } Then arguing similarly as in the proof of Step 3, we also have (Bk,u,(i1 , ,in ) ) lim |α|→∞ ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| Φ ≤ 1 We notice that Zn+ is the finite union of Bk,u,(i1 , ,in ) , where 0 ≤ k ≤ n, u ∈ Zn−k + ,u ≤ (3, 3, , 3), (i1 , , in ) is a permutation of (1, 2, , n) Therefore, lim |α|→∞ ( inf |ξ α |) I α f ξ∈suppfˆ 1/|α| Φ ≤ 1 The proof of (25) is complete Finally,... f, Dβ ϕ ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 13 and then I α+β f, Dβ ϕ = (−1)|β| fˆ, (F −1 ϕ)(x)g(x)/(ix)α (24) Using (23)-(24), we have I α+β f, Dβ ϕ = (−1)|β| I α f, ϕ and then (22) have been proved Next we prove that lim (25) ( inf |α|→∞ ξ∈supp fˆ |ξ α |) I α f 1/|α| ≤ 1 Φ To obtain (25), we divide our proof into three steps Step 1 We show (A0 ) lim ( inf |α|→∞... adopt the convention that I αf lim |α|→∞ 0 0 = 1, we get 1/|α| = ∞ Φ The proof is complete We consider now the case when f doesn’t have (O)-property and not satisfies the condition in Theorem 19 The following theorem is clear from the proofs of Theorems 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 suppfˆ, the. .. environment and working condition ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 21 References [1] L.D Abreu, Real Paley-Wiener theorems for the Koornwinder-Swarttouw q-Hankel transform, J Math Anal Appl 334 (2007), 223-231 [2] E Albrecht and Werner J Ricker, Functional calculi and decomposability of unbounded multiplier operators in Lp (RN ), Proc Edinburgh Math Soc 38... |ξ r | ≥ (17) inf |ξ σ | ξ∈supph r→σ ξ∈supph On the other hand, for all > 0 there exists ξ ∈ supph such that inf |ξ σ | ≤ |ξ σ | ≤ ξ∈supph inf |ξ σ | + ξ∈supph Taking account of inf |ξ r | ≤ |ξ r | ξ∈supph and letting r → σ, we have lim inf |ξ r | ≤ |ξ σ | ≤ r→σ ξ∈supph inf |ξ σ | + ξ∈supph and then (18) lim inf |ξ r | ≤ r→σ ξ∈supph by letting → 0 Combining (17) and (18), we get (16) inf |ξ σ | ξ∈supph ... Behavior of the sequence of norms of primitives of a function, J Approximation Theory 162 (2010), 1178 -1186 [19] H.H Bang and V.N Huy, Behavior of the sequences of norms of primitives of functions. .. then the following function Ψ is well defined via the formula Ψ(x1 , , xn ) = √ 2π f (x1 − ξ, x2 , , xn )ˆ η (ξ)dξ R ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES. .. which equals Then lim |α|→∞ I αf 1/|α| Φ = ∞ Proof For ≤ j ≤ n we define Aj := {α ∈ Zn+ : αj ≥ } |α| n ESTIMATE THE SEQUENCE OF NORM OF PRIMITIVES OF FUNCTIONS IN ORLICZ SPACES 19 Then it follows