On moduli of smoothness of fractional order

ON MODULI OF SMOOTHNESS OF FRACTIONAL ORDER S. TIKHONOV Abstract. In this paper we consider the properties of moduli of smoothness of fractional order. The main result of the paper describes the equivalence of the modulus of smoothness and a function from some class. 1. Introduction In 1977 P.L. Butzer, H. Dyckhoff, E. Goerlich, R.L. Stens (see [2]) and R.Tabersky (see [14]) introduced the modulus of smoothness of fractional order. This notion could be considered as a direct generalization of the classical modulus of smoothness, and it is more natural to use it for a number of problems of harmonic analysis (see, for example, [2], [5], [7], [10]). The important problem of approximation theory and theory of Fourier series is the problem of description of moduli of smoothness (see [1], [4], [8], [11]). One can consider this problem from the viewpoint of description of majorant of smoothness moduli. Recently, A. Medvedev (see [6]) has proved that for any modulus of continuity on [0,∞) there exists a concave majorant that is infinitely differentiable. In this paper, we obtain the description of the modulus of smoothness of fractional order from the viewpoint of the order of decreasing to zero of the modulus of smoothness. Let us introduce several definitions. If 1 ≤p < ∞, let Lpbe the space of 2π-periodic, measurable functions f(x) such that kfkp=µ2π R0 |f(x)|pdx¶ 1 p <∞. Similarly, let L∞be the space of 2π-periodic, continuous functions f(x) with kfk∞= max x∈[0,2π]|f(x)|. We will define the difference of fractional order β(β > 0) of function f(x) at the point x(x∈R) with increment h(h∈R) 0I would like to thank the Centre de Recerca Matem`atica for their support. 1 2 S. TIKHONOV by 4β hf(x) = ∞ X ν=0 (−1)νµβ ν¶f(x+ (β−ν)h), where ¡β ν¢=β(β−1)···(β−ν+1) ν!for ν > 1, ¡β ν¢=βfor ν= 1, ¡β ν¢= 1 for ν= 0. The modulus of smoothness of order β(β > 0) of function f∈Lp, 1≤p≤ ∞,is given by ωβ(f, t)p= sup |h|≤t° ° °4β hf(·)° ° °p(see [2],[14]). Let Φγ(γ∈R) be the set of nonnegative, bounded functions ϕ(δ) on (0,∞) such that a): ϕ(δ)→0 as δ→0, b): ϕ(δ) is nondecreasing, c): ϕ(δ)δ−γis nonincreasing. If for f∈Lpthere exists g∈Lpsuch that lim h→0+ ° ° °h−β4β hf(·)−g(·)° ° °p= 0 then gis called the Liouville-Grunwald-Letnikov derivative of order β > 0 of a function fin the Lp-norm, denoted by g=Dβf(see [2], [12]). Set Wβ p:= ©f∈Lp:Dβfexists as element in Lpª.The K-functional is given by K(f, t, Lp, Wβ p) := inf g∈Wβ p³kf−gkp+t° °Dβg° °p´. 2. Results Let f(x)∈Lp, p ∈[1,∞] and β > 0.It is clear that (see [12]) ¯¯¯¯µβ ν¶¯¯¯¯ =¯¯¯¯ β(β−1) · · · (β−ν+ 1) ν!¯¯¯¯ ≤C(β) νβ+1 , ν ∈N implies C∗(β) := ∞ P ν=0 ¯¯¯¡β ν¢¯¯¯<∞and the fractional difference 4β hf(x) is defined almost everywhere and belongs to Lp: k4β hf(·)kp≤C∗(β)kf(·)kp.(1) It is easy to write the following representation for C∗(β) (see [14]): C∗(β) =        2 k P ν=0 ¡β 2ν¢,if 2k < β ≤2k+ 1 (k= 0,1,2,· · · ), 2 k P ν=0 ¡β 2ν+1¢,if 2k+ 1 < β ≤2k+ 2 (k= 0,1,2,···). (2) The fractional differences and moduli of smoothness have some useful properties and we shall establish some of them in the following lemmas. Lemma 2.1. ([2], [14]) Let f∈Lp, p ∈[1,∞], α, β > 0; h∈R. Then (a): 4α h(4β hf(x)) = 4α+β hf(x)for almost every x; ON MODULI OF SMOOTHNESS OF FRACTIONAL ORDER 3 (b): k4α+β hf(·)kp≤C∗(α)k4β hf(·)kp; (c): lim h→0+ k4α hf(·)kp= 0. Lemma 2.2. Let f, f1, f2∈Lp, p ∈[1,∞], α, β > 0; x, h ∈R. Then (a): ωβ(f, δ)pis nondecreasing nonnegative function of δon (0,∞) with lim δ→0+ ωβ(f, δ)p= 0; (b): ωβ(f1+f2, δ)p≤ωβ(f1, δ)p+ωβ(f2, δ)p; (c): ωα+β(f, δ)p≤C∗(α)ωβ(f, δ)p; (d): if λ≥1, then ωβ(f, λδ)p≤C(β)λβωβ(f, δ)p; (e): if 0< t ≤δ, then ωβ(f, δ)pδ−β≤C(β)ωβ(f, t)pt−β. Indeed, we immediately have (a)−(c) from Lemma 2.1, (d) was proved in [2], and (d) implies (e): ωβ(f, δ)p=ωβµf, δ tt¶p ≤C(β)µδ t¶β ωβ(f, t)p. Lemma 2.3. Let f∈Lp, p ∈[1,∞], β > 0. (a): If β∈N, then ° °4β πf(·)° °p≤2[β+1 2]° ° °4β π 2f(·)° ° °p. (b): If β /∈N, then ° °4β πf(·)° °p≤2[β+1 2]+1 ° ° °4β π 2f(·)° ° °p. Corollary 2.4. For a function ϕ(t) = tα(0 ≤t≤π)to be a modulus of smoothness of order β(β > 0) of a function f(·)∈Lp,1≤p≤ ∞ it is necessary to have α≤hβ+1 2i+ 1. Theorem 2.5. Let p∈[1,∞], β > 0. (A): If f(·)∈Lp,then there exists a function ϕ(·)∈Φβsuch that ϕ(t)≤ωβ(f, t)p≤C(β)ϕ(t) (0 < t < ∞), where C(β)is a positive constant depending only on β. (B): If ϕ(·)∈Φβ, then there exist a function f(·)∈Lpand a constant t1>0such that C1(β)ωβ(f, t)p≤ϕ(t)≤C2(β)ωβ(f, t)p(0 < t < t1), where C1(β), C2(β)are positive constants depending only on β. Corollary 2.6. Let p∈[1,∞], β > 0. (A): If f(·)∈Lp,then there exists a function ϕ(·)∈Φβsuch that C1(β)ϕ(t)≤K(f, tβ, Lp, W β p)≤C2(β)ϕ(t).(3) is true. 4 S. TIKHONOV (B): If ϕ(·)∈Φβ, then there exists a function f(·)∈Lpsuch that (3) is true. Remark 2.7. 1). We can replace the condition f∈Lpby condition f∈L∞ in the part (B) of Theorem 2.5. 2). Note that theorem 2.5 for β∈Nwas proved in [11]. Also, for Hp-spaces the analogue of Corollary 2.6 for β∈R+and the analogue of theorem 2.5 for β∈Nwere proved in [5]. 3. Proofs Proof of Lemma 2.3. The first inequality was proved in [3]. Let β > 1, /∈ N. We shall use the following representation (see [14]) 4β 2hf(x−2βh) = ∞ X ν=0 µβ ν¶4β hf(x−βh −νh) for almost every x(4) By Lemma 2.1(a) and part (a) of this Lemma, it follows that ° °4β πf(·)° °p=° ° °³4[β] π(4β−[β] πf)´(·)° ° °p ≤2[[β]+1 2]° ° °³4[β] π 2(4β−[β] πf)´(·)° ° °p. Here we use (4) for h=π 2. We have ° °4β πf(·)° °p≤2[[β]+1 2]° ° ° ° ° 4[β] π 2(∞ X ν=0 µβ−[β] ν¶4β−[β] π 2f)(· − βπ 2−νπ 2)° ° ° ° °p = 2[[β]+1 2]° ° ° ° ° ∞ X ν=0 µβ−[β] ν¶³4[β] π 2(4β−[β] π 2f)´(·)° ° ° ° °p . Thus, by Lemma 2.1(a) and inequality (1), we get ° °4β πf(·)° °p≤C∗(β−[β]) 2[[β]+1 2]° ° °4β π 2f(·)° ° °p. If we combine this result with C∗(β−[β]) = 2 (see (2)) and 2[[β]+1 2]= 2[β+1 2], we obtain the required inequality. If 0 < β < 1, then we use (1) and (4). This completes the proof of Lemma. We will need the following lemma. Lemma 3.1. Let β > 0, n ∈N, δ > 0. (a): If f(x) = sin xand p∈[1,∞], then there exist t1>0and C1(β), C2(β)>0such that for any δ∈(0, t1)we have C1(β)δβ≤ωβ(f, δ)p≤C2(β)δβ.(5) ON MODULI OF SMOOTHNESS OF FRACTIONAL ORDER 5 (b): If f(x) = sin nx and p∈[1,∞], then for any δ∈¡0,π 2¤we have1 k4β δf(·)kp≤(2π)1 p(nδ)β. (c): If f(x) = sin nx, then k4β π/nf(·)k1= 2β+2. (d): If f(x) = sin nx, then for any δ∈¡0,π n¤we have k4β δf(·)k1≥ 4¡2 π¢β(δn)β. Proof of Lemma 3.1 Let Tn(x) = n P ν=−n cνeiνx,then 4β δTn(x−βδ 2) = n X ν=−nµ2isin νδ 2¶β cνeiνx. Thus, for f(x) = sin nx,n∈N,we get 4β δf(x−βδ 2) = µ2 sin nδ 2¶β sin µnx +βπ 2¶.(6) For n= 1 we obviously have C1(β)¡2¯¯sin δ 2¯¯¢β≤ k4β δsin(·)kp≤ C2(β)¡2¯¯sin δ 2¯¯¢β. If we combine this inequality with sin t≤t(t≥0) and sin t≥2t π(0 ≤t≤π 2), then we obtain (5). In the same way, by (6), we shall have the proofs of (b)−(d). This completes the proof of Lemma. Proof of Theorem 2.5. (A). Let us define ϕ(t) := tβinf 0<ξ≤t{ξ−βωβ(f, ξ)p}. We immediately have ϕ(t)∈Φβfrom [13, §2]. It is trivial, that ϕ(t)≤ ωβ(f, t)p. By Lemma 2.2(e), we have ωβ(f, t)p≤C(β)ϕ(t): ωβ(f, t)p=tβt−βωβ(f, t)p ≤C(β)tβinf 0<ξ≤t©ξ−βωβ(f, ξ)pª =C(β)ϕ(t). Therefore, for any t > 0 the following inequality ϕ(t)≤ωβ(f, t)p≤C(β)ϕ(t) holds and (A) follows. (B). 1 case. Let lim t→0 ϕ(t) tβ=C(0 ≤C < ∞). Then, by virtue of monotonicity of ϕ(t) tβ, we write (∗)ϕ(t)≤Ctβfor 0 < t ≤π; (∗∗ ) there exists t1>0 such that ϕ(t)≥Ctβ 2for 0 < t ≤t1. Indeed, (∗) is trivial like (∗∗) for C= 0. If C > 0 and lim t→0 ϕ(t) tβ=C, then for any ε > 0 there exists t1>0 such that C−ϕ(t) tβ≤εfor 0 < t ≤t1. Then ϕ(t) tβ≥C−ε, and choosing small εwe have (∗∗). 1Here 1 ∞= 0. 6 S. TIKHONOV Define f(x) = Csin x. By Lemma 3.1(a), we have ωβ(f, δ)p≥CC1(β)δβ≥C2(β)ϕ(δ) for 0 < δ ≤π, ωβ(f, δ)p≤CC3(β)δβ≤C4(β)ϕ(δ) for 0 < δ ≤t1, completing the proof in this case. 2 case. Let lim t→0 ϕ(t) tβ= +∞.Then lim t→0ϕ(t) = 0 and lim t→0 tβ ϕ(t)= 0.We fix a≥2. Then, following Oskolkov ([9]), we define the sequence {nν}∞ ν=1, where nν= 2mνare the numbers mνsuch that m1= 2, mν+1 = min ½m∈N: max µϕ(2−m) ϕ(2−mν),2mνβϕ(2−mν) 2mβϕ(2−m)¶≤1 a¾(ν∈N). From the definition of {nν}∞ ν=1 it follows that mν+1 > mν, nν+1 ≥2nνand for any ν∈Nwe have ϕµ1 nν+1 ¶≤1 aϕµ1 nν¶; (7) nβ νϕµ1 nν¶≤1 anβ ν+1ϕµ1 nν+1 ¶.(8) Let us fix κ= 2d(d∈N) such that κ>2π. Note that (7) implies ∞ P ν=1 ϕ³1 nν´≤ϕ³1 n1´∞ P ν=1 a1−ν<∞,and, therefore, we can define the function f(x) = ∞ P ν=1 ϕ³1 nν´sin(κnνx). First, we shall estimate ωβ(f, δ)pfrom above. By the inequality kfkp≤ (2π)1 pkfk∞≤2πkfk∞, p ∈[1,∞), it is enough to prove ωβ(f, δ)∞≤ C(β)ϕ(δ).Let δ∈(0,1 n1]. For all h∈(0,1 n1] we can find the number N∈Nsuch that 1 nN+1 < h ≤1 nN. Then ° ° °4β hf(x)° ° °∞≤° ° ° ° ° N X ν=1 ϕµ1 nν¶4β hsin(κnνx)° ° ° ° °∞ + +° ° ° ° ° ∞ X ν=N+1 ϕµ1 nν¶4β hsin(κnνx)° ° ° ° °∞ =: I1+I2. ON MODULI OF SMOOTHNESS OF FRACTIONAL ORDER 7 Combining Lemma 3.1(b), inequality (8), and condition (c) in the definition of Φβ, we get I1≤ N X ν=1 ϕµ1 nν¶° ° °4β hsin(κnνx)° ° °∞ ≤C(β) (κh)βϕµ1 nN¶nβ N N X ν=1 a−(N−ν) ≤C(β) (nNh)βϕµ1 nN¶ ≤C(β)ϕ(h). Inequalities (1) and (7) yield that I2≤ ∞ X ν=N+1 ϕµ1 nν¶° ° °4β hsin(κnνx)° ° °∞ ≤C(β) ∞ X ν=N+1 ϕµ1 nν¶ ≤C(β)ϕµ1 nN+1 ¶∞ X ν=N+1 aN+1−ν ≤C(β)ϕµ1 nN+1 ¶ ≤C(β)ϕ(h). Therefore, if h∈(1 nN+1 ,1 nN], N ∈N,then k4β hf(x)k∞≤C(β)ϕ(h),which implies ωβ(f, δ)∞≤C(β)ϕ(δ). Now we shall obtain the inequality ϕ(δ)≤C(β)ωβ(f, δ)p.From the inequality kfk1≤2πkfkp, p ∈[1,∞] it is sufficient to prove ϕ(δ)≤ C(β)ωβ(f, δ)1.Also, we note that if the last inequality holds for δ=π 2k, k = N, N + 1, N + 2,· · · ,where N∈N, then it holds for δ∈¡π 2k,π 2k+1 ¢. Indeed, from the monotonicity of t−βϕ(t), we see that the estimate ϕ(δ)≤ C(β)ϕ¡π 2k¢is true. By Lemma 2.2(a), we get ϕ(δ)≤C(β)ϕ³π 2k´ ≤C(β)ωβ³f, π 2k´1 ≤C(β)ωβ(f, δ)1. To go further, we suppose that δ=π 2k. 8 S. TIKHONOV Let Mbe the integer, M > 1, and, let h1=π κnM. We shall show that ° ° °4β h1f(x)° ° °1≥4ϕµ1 nM¶µ2β−πβ+1 a¶.(9) For this purpose, we shall use the following representation of a function f(x): f(x) = M−1 X ν=1 ϕµ1 nν¶sin(κnνx) + ϕµ1 nM¶sin(κnMx) + + ∞ X ν=M+1 ϕµ1 nν¶sin(κnνx) =: f1+f2+f3. Note, that sin(κnνx+πnν nM) = sin(κnνx) for ν > M, and f3(x) has the period T=h1=π κnM. We therefore obtain 4β h1f3(x) = f(x+βh1) ∞ X ξ=0 (−1)ξµβ ξ¶= 0. By Lemma 3.1(b) and (8), we have ° ° °4β h1f1(x)° ° °1≤ M−1 X ν=1 ϕµ1 nν¶° ° °4β h1sin(κnνx)° ° °1 ≤ M−1 X ν=1 2π(κnνh1)βϕµ1 nν¶ = 2πµπ nM¶βM−1 X ν=1 ϕµ1 nν¶nβ ν ≤2πµπ nM¶β ϕµ1 nM−1¶nβ M−1 M−1 X ν=1 a−(M−1−ν). Using M−1 P ν=1 a−(M−1−ν)≤2 and (8), we obtain ° ° °4β h1f1(x)° ° °1≤4πβ+1 aϕ³1 nM´. By Lemma 3.1(c), ° ° °4β h1f2(x)° ° °1=ϕ³1 nM´° ° °4β h1sin(κnMx)° ° °1=2β+2ϕ³1 nM´. ON MODULI OF SMOOTHNESS OF FRACTIONAL ORDER 9 Therefore, for h1=π κnM, the inequality |f| ≥ |f2| − |f1| − |f3|implies ° ° °4β h1f(x)° ° °1≥° ° °4β h1f2(x)° ° °1−° ° °4β h1f1(x)° ° °1−° ° °4β h1f3(x)° ° °1 =° ° °4β h1f2(x)° ° °1−° ° °4β h1f1(x)° ° °1 ≥4ϕµ1 nM¶µ2β−πβ+1 a¶, i.e. we obtain (9). Further, we choose the integer isuch that 1 ni+1 =1 2mi+1 < δ ≤1 2mi=1 ni . Note, that, by definition of mi, at the least one of the following inequalities is true: 2β(mi+1−1)ϕµ1 2mi+1−1¶< a 2βmiϕµ1 2mi¶,(10) ϕµ1 2mi+1−1¶>1 aϕµ1 2mi¶(11) Case 2(a). Let (10) be true. Using the monotonicity of ϕ(t) and (10), we get nβ i+1ϕµ1 ni+1 ¶≤2β2β(mi+1−1)ϕµ1 2mi+1−1¶ < a 2βnβ iϕµ1 ni¶.(12) We write f(x) = i−1 X ν=1 ϕµ1 nν¶sin(κnνx) + ϕµ1 ni¶sin(κnix) + + ∞ X ν=i+1 ϕµ1 nν¶sin(κnνx) =: f1+f2+f3. It is clear, that the function f3has a period T=2π κni+1 . Then, for κ= 2d>2πwe have δ=π 2r>1 ni+1 > T, therefore, f3has a period δand 4β δf3(x) = 0.