Journal of Approximation Theory 162 (2010) 1178–1186 www.elsevier.com/locate/jat Behavior of the sequence of norms of primitives of a function Ha Huy Bang a,∗ , Vu Nhat Huy b 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 December 2009; accepted 30 December 2009 Available online January 2010 Communicated by Francisco Marcellan Abstract In this paper we characterize the behavior of the sequence of the L p -norm of primitives of a function by its spectrum (the support of its Fourier transform) c 2010 Elsevier Inc All rights reserved Keywords: L p -spaces; Generalized functions Introduction The following result was proved in [5]: Theorem A Let ≤ p ≤ ∞ and f (n) ∈ L p (R), n = 0, 1, 2, Then there always exists the following limit lim f (n) 1/n p lim f (n) 1/n p n→∞ and n→∞ = σ f = sup{|ξ | : ξ ∈ supp fˆ}, ∗ 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 H.H Bang, V.N Huy / Journal of Approximation Theory 162 (2010) 1178–1186 1179 where fˆ is the Fourier transform of f Theorem A shows that the behavior of the sequence f (n) p is wholly characterized by the spectrum of f This result was studied and developed by many authors (see [1–4,6–9,11,13–17]) It is natural to ask what will happen when we replace derivatives by integrals? Let 1/n ∞ (I f )(x) := f (y)dy x be the improper indefinite Riemann integral, and I n f = I (I n−1 f ) If f (x)(1 + |x|) ∈ L (R), then ∞ (I n f )(x) = x (y − x)n−1 f (y)dy =: I+n f (x), (n − 1)! I+α where is the Weyl fractional integral operator of order α But in general, I n f as an n-fold primitive of f may exist while I+n f does not converge as an improper Riemann integral For p = 2, V.K Tuan proved the following result in [15]: Theorem B Let f ∈ L (R) and σ := inf{|ξ | : ξ ∈ supp fˆ} > Then there exists I n f , I n f ∈ L (R) for all n, and lim n→∞ In f 1/n = σ −1 , where by supp fˆ, or the spectrum of f we denote the smallest closed set outside of which fˆ vanishes a.e In this paper, we solve the problem for the general case ≤ p ≤ ∞ and σ ≥ First we need a notation of the primitive of a tempered generalized function: Denote by S(R) the Schwartz space of rapidly decreasing functions and by S (R) the set of all continuous linear functionals on S(R) Any element h ∈ S (R) is called a tempered generalized function and h(ϕ) =: h, ϕ for all ϕ ∈ S(R) Let f ∈ S (R) The tempered generalized function I f is termed the primitive of f if D(I f ) = f , that is, I f, ϕ = − f, ϕ , ∀ϕ ∈ S(R) Now we show that for each f ∈ S (R) the set P( f ) = {h ∈ S (R) : Dh = f } is not empty Indeed, ∀ϕ ∈ S(R) we put x +∞ ψ(x) = ϕ(x) − ρ(x) ϕ(t)dt, Φ(x) = −∞ where ρ ∈ C0∞ (−1, 1), ψ(t)dt, −∞ R ρ(x)dx = Then ψ(x) = ϕ(x), ∀|x| > 1, and it is easy to see that ψ(x), Φ(x) ∈ S(R) So, for each f ∈ S (R), we can put I f, ϕ = − f, Φ 1180 H.H Bang, V.N Huy / Journal of Approximation Theory 162 (2010) 1178–1186 Then I f ∈ S (R) and x I f, ϕ = − f, ϕ (t)dt = − f, ϕ , −∞ i.e., I f is a primitive of f Furthermore, if I f ∈ S (R) is a primitive of f , we have +∞ I f, ϕ = − f, Φ + C ϕ(ξ ) dξ −∞ where C = I f, ρ So, +∞ I f, ϕ = − f, Φ + C ϕ(ξ ) dξ (*) −∞ Conversely, given an arbitrary constant C, the functional I f defined on S(R) by (*) defines the primitive of f So, we have proved the following result: Every tempered generalized function f ∈ S (R) has on S (R) a primitive I f and every primitive of f is expressed by formula (*), where C is an arbitrary constant Note that the notation of primitive of a generalized function in D (a, b), a, b ∈ R can be found in [18], here we give it for tempered generalized functions in a similar manner We denote I f = f In the sequel, I f denotes a primitive of f ∈ S (R), i.e., I f ∈ P( f ), and I n+1 f = I (I n f )., i.e., I n+1 f ∈ P(I n f ), n = 0, 1, 2, So, D n I n+m f = I m f for n, m = 0, 1, 2, Let ≤ p ≤ ∞ and f ∈ L p (R) From now on we assume that for any n = 0, 1, 2, there exists a primitive of I n f , which belongs to L p (R) and is denoted by I n+1 f If the function f satisfies this assumption, we write (I n f )∞ n=0 ⊂ L p (R) Behavior of the sequence of the L p -norm of primitives Let f ∈ L (R) and fˆ = F f be its Fourier transform +∞ fˆ(ξ ) = √ 2π e−ixξ f (x)dx −∞ The Fourier transform of a tempered generalized function f is defined via the formula F f, ϕ = f, Fϕ , ϕ ∈ S(R) We now state our main result: Theorem Let ≤ p ≤ ∞ and (I n f )∞ n=0 ⊂ L p (R) Then lim n→∞ In f 1/n p = σ −1 , where σ = inf{|ξ | : ξ ∈ supp fˆ} Proof It is well-known that f p = | f, ϕ |, sup {ϕ∈L q : ϕ q ≤1} (1) 1181 H.H Bang, V.N Huy / Journal of Approximation Theory 162 (2010) 1178–1186 where 1/ p + 1/q = So, clearly, we have f p | f, ϕ | sup = {ϕ∈S: ϕ (2) q ≤1} Indeed, if ≤ q < ∞ then (2) follows from the density of S in L q Further, if q = ∞ then for any > there exists a function g ∈ L ∞ (R), g R f (x)g(x)dx > f ∞ ≤ such that − We choose a number M > such that f (x)g(x)dx > f −2 |x|≤M and put h = χ[−M,M] g, x −1 ω(x) = Ce if |x| < 1, if |x| ≥ 1, where the constant C satisfies ω ψλ = h ∗ ωλ , with ωλ (x) = lim λ→0 R x λ ω( λ ) L (R) = 1, λ > 0, Then ψλ ∈ S, ψλ f (x)ψλ (x)dx = R ∞ ≤ h ∞ ωλ ≤ and f (x)h(x)dx So, R f (x)ψλ (x)dx > f −3 for some λ > Thus we have proved (2) Next, we divide our proof into two cases Case σ > Let us first prove supp I n f = supp fˆ , n = 1, 2, , (3) where fˆ = F f is the Fourier transform of f In fact, since D n I n f = f , we have fˆ = (i x)n I n f Therefore, supp fˆ ⊂ supp I n f ⊂ supp fˆ ∪ {0} So, to obtain (3), it is enough to show ∈ supp I n f : We choose numbers a, b: < a < b < σ and a function h ∈ C0∞ (−b, b) such that h(x) = in (−a, a) Then supph I n f ⊂ {0} Suppose supph I n f = {0}, then there is a number Nn ∈ N such that Nn hIn f = j=0 c Nj n δ ( j) 1182 H.H Bang, V.N Huy / Journal of Approximation Theory 162 (2010) 1178–1186 Hence, Nn c Nj n (−i x) j F −1 h ∗ I n f (x) = j=0 Since I n f ∈ L p (R) and F −1 h ∈ L q (R), we get F −1 h ∗ I n f ∈ L ∞ (R) Therefore, F −1 h ∗ I n f (x) = c0Nn , n = 1, 2, Note that N F −1 h ∗ I n f (x) = F −1 h ∗ (I n+1 f ) (x) = F −1 h ∗ I n+1 f (x) = c0 n+1 ≡ So, h I n f = Assume now the contrary that ∈ supp I n f Then there is a function ϕ ∈ C0∞ (−a, a) such that I n f , ϕ = So, since h(x) = in (−a, a), we get = I n f , ϕ = I n f , hϕ = h I n f , ϕ = 0, which is impossible Thus we have proved (3) Since (3) we have supp I n f ⊂ R \ (−σ, σ ) (4) Second we prove lim n→∞ In f Indeed, for any h(ξ ) = h(ξ ) = σ > 0, < σ/2 we choose a function h ∈ C ∞ (R) satisfying 1/n p ≤ (5) ∀ξ ∈ −(σ − ), σ − , ∀ξ ∈ −(σ − ), σ − Let ϕ be an arbitrary element in S(R) Then it follows from (4) that I n f, ϕ = I n f , F −1 ϕ = I n f , (F −1 ϕ)(ξ ).h(ξ ) = I n f, F (F −1 ϕ)(ξ ).h(ξ ) Therefore, I n f, ϕ = I n f, ψ , where ψ = F (6) (F −1 ϕ)(ξ ).h(ξ ) ψn = F (F −1 ϕ)(ξ ) We put h(ξ ) ξn Then ψn ∈ S(R) and | f, ψn | = | D n (I n f ), ψn | = | I n f, D n (ψn ) | = | I n f, ψ | (7) Combining (6) and (7), we get for n ≥ 3: I n f, ϕ = | f, ψn | = √ 2π f, ϕ ∗ F h(ξ ) ξn (8) 1183 H.H Bang, V.N Huy / Journal of Approximation Theory 162 (2010) 1178–1186 Therefore, since (2), we have for n ≥ 3: In f p = sup √ 2π q ≤1} f, ϕ ∗ F sup √ 2π q ≤1} f {ϕ∈S: ϕ ≤ {ϕ∈S: ϕ h(ξ ) ξn ϕ∗F p h(ξ ) ξn q ≤√ 2π f p F h(ξ ) ξn Hence, 1/n p In f lim n→∞ ≤ lim n→∞ 1/n h(ξ ) ξn F (9) We put for n ≥ 3: ∞, C1 := max{ h ∞, h h γ =σ −2 , ∞ }, and gn = F h(ξ ) ξn Then for n ≥ 3: sup |(1 + x )gn (x)| x∈R ≤ sup √ x∈R 2π = sup √ x∈R 2π e−ixξ D R e−ixξ |ξ |≥γ e−ixξ + |ξ |≥γ C1 ≤√ 2π h(ξ ) dξ + ξn e−ixξ R h(ξ ) dξ ξn h(ξ ) h (ξ ) h (ξ ) − 2n n+1 + n(n + 1) n+2 dξ ξn ξ ξ h(ξ ) dξ ξn 2n 4 + n+1 + γ n−1 (n − 1) γ n γ Therefore, it follows from R |gn (x)|dx ≤ sup |(1 + x )gn (x)| x∈R dξ = π sup |(1 + x )gn (x)| + ξ R x∈R that lim n→∞ F 1/n h(ξ ) ξn = lim n→∞ gn 1/n ≤ γ (10) Combining (9) and (10), we get lim n→∞ In f 1/n p ≤ and then (5) by letting Finally, we show lim n→∞ In f 1/n p σ −2 → ≥ σ (11) 1184 H.H Bang, V.N Huy / Journal of Approximation Theory 162 (2010) 1178–1186 Without loss of generality we may assume that σ = inf{ξ : < ξ ∈ supp fˆ} Hence, there exists a function ϕ(x) ∈ C0∞ (σ − , σ + ) such that fˆ, ϕ = Therefore, I n f, ϕˆ (n) ≤ I n f f, ϕˆ = D n (I n f ), ϕˆ = 0= p ϕˆ (n) q So, In f lim 1/n p ≥ ϕˆ (n) lim n→∞ n→∞ 1/n q (12) We have sup |(1 + x )ϕˆ (n) (x)| = sup (1 + x ) F ϕ(ξ )ξ n (x) x∈R x∈R ≤ sup R x∈R e−ixξ ϕ(ξ )ξ n dξ + sup x∈R R e−ixξ ϕ (ξ )ξ n + 2nϕ (ξ )ξ n−1 + n(n − 1)ϕ(ξ )ξ n−2 dξ |ϕ(ξ )ξ n |dξ ≤ |ξ |≤σ + |ϕ (ξ )ξ n | + 2n|ϕ (ξ )ξ n−1 | + n(n − 1)|ϕ(ξ )ξ n−2 | dξ + |ξ |≤σ + ≤ max{ ϕ ∞, ϕ ∞, ϕ ∞} Therefore, it follows from ϕˆ (n) lim n→∞ ϕˆ (n) 1/n q q 4(σ + )n+1 + 4(σ + )n + 2n(σ + )n−1 n+1 ≤ π supx∈R |(1 + x )ϕˆ (n) (x)| that ≤σ+ So, by (12) we obtain lim In f 1/n p ≥ n→∞ σ+ ∀ >0 and then (11) Combining (5) and (11), we arrive at (1) Case σ = We have ∈ supp fˆ Hence, for any > there is a function ϕ(x) ∈ ∞ C0 (− , ) such that fˆ, ϕ = Then arguing just as above we obtain lim In f 1/n p n→∞ ≥ lim n→∞ Therefore, lim n→∞ In f 1/n p = ∞ 1/n ϕˆ (n) q ≥ H.H Bang, V.N Huy / Journal of Approximation Theory 162 (2010) 1178–1186 1185 So we always have lim n→∞ In f 1/n p = σ −1 The proof is complete Remark Theorem B was stated only for the case σ > and proved by a technique specific for p = 2, which cannot be applied to the general case ≤ p ≤ ∞ From the proof of Theorem 1, we have the following result: Theorem Let ≤ p ≤ ∞ and f, I f, I f ∈ L p (R) Then supp fˆ = supp I f Let g ∈ S (R) and m = 1, 2, Clearly, the set Pm (g) = {h ∈ S (R) : D m h = g} is not empty Further, let ≤ p ≤ ∞, f ∈ L p (R) and = n0 < n1 < n2 < · · · < nk < · · · be some sequence of natural numbers We denote I n f = f and assume that for any j = 0, 1, 2, there exists an element in Pn j+1 −n j (I n j f ), which belongs to L p (R) and is denoted by I n j+1 f If the function f satisfies this assumption, we write (I n j f )∞ j=0 ⊂ L p (R) We have the following result: Theorem Let ≤ p ≤ ∞ and (I n j f )∞ j=0 ⊂ L p (R) Then lim j→∞ Inj f nj p = σ −1 , where σ = inf{|ξ | : ξ ∈ supp fˆ} Proof Let m ≥ It is known that if g and its generalized derivative D m g belong to L p (R) then all D j g, j = 1, , m − belong to L p (R) (see [10,12]) So, since D n j+1 −n j I n j+1 f = I n j f we get that all D I n j+1 f, D I n j+1 f, , D n j+1 −n j −1 I n j+1 f , j = 0, 1, 2, belong to L p (R) Therefore, it remains to apply Theorem Acknowledgment This work was supported by the Vietnam National Foundation for Science and Technology Development (Project No 101.01.50.09) References 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