On Boas-Type Problem

ON BOAS-TYPE PROBLEM SERGEY TIKHONOV Abstract. R.P. Boas has found necessary and sufficient conditions of belonging of function to Lipschitz class. From his findings it turned out, that the conditions on sine and cosine coefficients for belonging of function to Lip α(0 <α<1) are the same, but for Lip 1 are different. Later his results were generalized by many authors in the viewpoint of generalization of condition on the majorant of modulus of continuity. The aim of this paper is to obtain Boas-type theorems for generalized Lipschitz classes. To define generalized Lipschitz classes we use the concept of modulus of smoothness of fractional order. 1. Introduction It is a well known fact that the local behavior of the sum of a Fourier series with nonnegative coefficients determines the asymptotic behavior of the coefficients, and vice versa. As a main reference, we would like to mention the remarkable book [5] by R.P.Boas Jr., where one can find many results on the subject, which we will discuss below. As it was shown by R.P.Boas [5, p. 45] (see also [4]), the conditions on sine and cosine coefficients for Lip α(0 < α < 1) from this point of view are the same: If λn≥0are the Fourier sine or cosine coefficients of ψ(x), then ψ∈Lip αif and only if ∞ X k=n λk=O¡n−α¢(1) or, equivalently, n X k=1 kλk=O¡n1−α¢. This result does not hold for α= 1. Firstly, ψ∈Zif and only if ∞ X k=n λk=O¡n−1¢, (2) 1 2 SERGEY TIKHONOV where Z is the Zygmund class, i.e., nψ∈C:kψ(x+h)−2ψ(x) + ψ(x−h)k= O(h)o. On the other hand (see [5, p. 46-47]) the conditions of belonging to Lip 1 for sine series and for cosine series are different: f∈Lip 1 if and only if (1). n X k=1 k2ak=O(n) (2). n X k=1 kaksin kx =O(1) uniformly in x. (3) g∈Lip 1 if and only if n X k=1 kbk=O(1) .(4) Here and further, we assume that the series ∞ X k=1 akcos kx, ak≥0 (5) and ∞ X k=1 bksin kx, bk≥0 (6) converge uniformly to functions f(x) and g(x), respectively and (5), (6) are Fourier series of f(x) and g(x). The aim of this paper is to present Boas-type results for generalized Lipschitz classes. The paper is organized in the following way. Section 2 contains some definitions and preliminaries. We start with the discussion of the well-known results and conclude this section describing the problem, which we will consider. In section 3, we present our main results. Section 4 contains lemmas that we will need and proofs of theorems. We finish by Section 5, where we provide a few remarks. 2. Definitions, Results, Comments 2.1. Basic Notations and Definitions. Let ψ(x) be a continuous function and kψ(·)k= max x∈[0,2π]|ψ(x)|. The key concept in the paper will be the modulus of smoothness of fractional order introduced by Butzer, Dyckhoff, G¨oerlich and Stens [6]. The modulus of smoothness of order β(β > 0) of function ψ∈Cis given by ωβ(ψ, t) = sup |h|≤t° ° °4β hψ(·)° ° °, ON BOAS-TYPE PROBLEM 3 where 4β hψ(x) = ∞ X ν=0 (−1)νµβ ν¶ψ(x+ (β−ν)h), and ¡β ν¢=β(β−1)···(β−ν+1) ν!for ν≥1, and ¡β ν¢= 1 for ν= 0. Let β > 0 and {γn}be a positive sequence. By Lipβγwe denote generalized Lipschitz classes, i.e., Lipβγ=½f∈C:ωβ³ψ, 1 n´=O(γn)¾. By this definition, Lip α≡Lip1γwith {γn=n−α}and Z ≡Lip2γwith ©γn=n−1ª. We will denote by En(ψ) the error of best trigonometric approximation of order nof a function ψ. Now we define several conditions on positive sequences and functions. A sequence γ:= {γn}of positive terms will be called almost increasing (almost decreasing), if there exists a constant K:= K(γ)≥1 such that Kγn≥γm(γn≤Kγm) holds for any n≥m. This concept was introduced by Bernstein1. We will say that a sequence γ:= {γn}of positive terms satisfies the condition (SQ), if there exists ε∈(0,1), such that {nεγn}is almost decreasing. And a sequence γ:= {γn}of positive terms satisfies the condition (SQβ), if there exists ε∈(0,1), such that ©nβ−εγnªis almost increasing. Also, we will give the following definition by Chan [9]. Let Y[a, b], a≤b be the collection of all positive functions γ(u) defined on [X, ∞), X > 0 such that u−aγ(u) is nondecreasing and u−bγ(u) is nonincreasing. Finally, let Φσ(σ∈R) be the set of nonnegative, bounded functions γ(u) on (0,∞), such that (a): γ(u)→0 as u→0, (b): γ(u) is nondecreasing, (c): γ(u)u−σis nonincreasing. 2.2. Results and Comments. After the Boas’s book [5] was published, there has been done a lot of work on generalization of his results (1)-(4) to more general classes of functions than Lip α. Different ways of generalization were considered by M. and S. Izumi ([10]), J. Nemeth ([13]), L.-Y. Chan ([9]), L. Leindler ([11]). It is natural to consider different conditions on a majorant of modulus of continuity ω1(f, 1 n) = O(γn) or modulus of smoothness ω2(f, 1 n) = O(γn). Recently, we have obtained ([15]) the interrelation between several well-known conditions (in particular, from [9]-[11] 1See reference to S.N. Bernstein in [2]. 4 SERGEY TIKHONOV and [13]-[14])) on a majorant. Also, we have proved theorem which generalizes all previous results for the case, when the conditions on Fourier sine or cosine coefficients are the same (i.e., the analogue of Boas’s results for Lip α, 0< α < 1,and Z). Theorem 2.1. ([15]) Let γ={γn}be a positive sequence and β > 0. Let λn≥0be the Fourier sine or cosine coefficients of ψ(x). (A). If γ∈SQ ∩SQβ, then for any function ψ∈Cconditions ∞ X k=n λk=O(γn),(7) n X k=1 kβλk=O¡nβγn¢,(8) ωβ³ψ, 1 n´=O(γn) (9) are equivalent. (B). Let γbe a non-increasing sequence. If for any function ψ∈Cconditions (7), (8) and (9) are equivalent, then γ∈SQ ∩SQβ. Now we recall the following equivalence result on order of decay of the modulus of smoothness. Theorem 2.2. ([16]) (A). If ψ(·)∈Cand β > 0, then there exists a function γ(·)∈Φβ, such that γ(t)≤ωβ(ψ, t)≤C(β)γ(t) (0 < t < ∞), where C(β)is a positive constant depending only on β. (B). If γ(·)∈Φβ, β > 0, then there exist a function ψ(·)∈Cand a constant t1>0, such that C1(β)ωβ(ψ, t)≤γ(t)≤C2(β)ωβ(ψ, t) (0 < t < t1), where C1(β), C2(β)are positive constants depending only on β. It is clear, that the class ©γ(·)∈Φβªis wider then ©{γn=γ¡1 n¢} ∈ SQ ∩SQβª. We give as examples two following majorants (i)γ=©γn=n−βξ(n)ª/∈SQβand (ii)γ=nγn= (ξ(n))−1o/∈SQ, where ξ(·) is positive nondecreasing slowly varying function2. 2By definition, a positive measurable function ξ(·) on [A, ∞), A > 0 is slowly varying if for all λ > 0 we have lim t→∞ ξ(λt) ξ(t)= 1. Note (see [3, p. 54]), that, by our assumptions on function ξ(·), it is sufficient to have lim t→∞ ξ(λ∗t) ξ(t)= 1 only for a positive λ∗6= 1. By Theorem 2.2, the natural restriction in the case (ii) is lim t→∞ ξ(t) = ∞. ON BOAS-TYPE PROBLEM 5 So, Theorem 2.1 does not provide Boas-type criterion of belonging of function to Lipβγfor such cases as (i) and (ii). These cases are of particular interest since the conditions for cosine and sine series are different (see, for instance, (3), (4)). In the case β= 1, the Boas-type result for a majorant as in the case (i) was obtained by Chan [9]. We recall his result. Theorem 2.3. (Cosine) Let γ(·)∈Y[c1, c2],0< c1≤c2<∞. Then ω1(f, δ) = O[γ(δ)] iff (1). n X k=1 k2ak=O·n2γµ1 n¶¸ (2). n X k=1 kaksin kx =O·nγ µ1 n¶¸uniformly in x. Theorem 2.4. (Sine) Let γ(·)∈Y[c1, c2],0< c1≤c2<∞. Then ω1(g, δ) = O[γ(δ)] if and only if n X k=1 kbk=O·nγ µ1 n¶¸. It is clear that γ(·)∈Y[c1, c2] implies ©γ¡1 n¢ª∈SQ ∩SQc2+ε⊂SQ for any ε > 0 but the converse is not true. The following result was obtained in [14]. Theorem 2.5. (Cosine) Let γ={γn}be the positive sequence and γ∈SQ1. Then ω1³f, 1 n´=O(γn)if and only if ∞ X k=n ak=O(γn). Naturally, for a majorant γas in the case (ii) we have γ∈SQ1. Our main idea in this paper is to prove Boas-type criterion for generalized Lipschitz classes Lipβγwhere a majorant γsatisfies either SQ or SQβcondition. So, if γ∈SQ, then, as an example, we can consider the case (i) γ=©γn=n−βξ(n)ªand we will have the generalization of Theorems 2.3 and 2.4, and classical Boas’s results (3),(4) for Lip 1 in particular. If γ∈SQβ, then, as an example, we can consider the case (ii) γ=nγn= (ξ(n))−1owe will have Theorem 2.5 for β= 1. Finally, if γ∈SQ ∩SQβthen the results below coincide Theorem 2.1 and we have Boas’s criterion (1) for Lip α(0 < α < 1) and (2) for Z in particular. 3. Main Results Theorem 3.1. (Cosine) Let γ={γn}be the positive sequence and β > 0. (A). If β6= 2l−1 (l∈N)and γ∈SQ, then for any function f∈Cwith 6 SERGEY TIKHONOV Fourier series (5) conditions ωβ³f, 1 n´=O(γn) (10) and n X k=1 kβak=O¡nβγn¢(11) are equivalent. (B). If β= 2l−1 (l∈N)and γ∈SQ, then for any function f∈Cwith Fourier series (5) conditions ωβ³f, 1 n´=O(γn) (12) and n X k=1 kβ+1ak=O¡nβ+1γn¢,(13) n X k=1 kβaksin kx =O¡nβγn¢uniformly in x (14) are equivalent. (C). If γ∈SQβ, then for any function f∈Cwith Fourier series (5) conditions ωβ³f, 1 n´=O(γn) (15) and ∞ X k=n ak=O(γn) (16) are equivalent. Here every equivalence result (items (A),(B),(C)) should be understood in the following sense: if for any function f∈Cwith Fourier series (5) condition ωβ¡f, 1 n¢=O(γn) holds, then suitable condition on coefficients {ak}are true. Conversely, if for any {ak≥0}∞ k=1 given condition holds, then the series (5) converges uniformly to the continuous function f(x) and (5) is Fourier series of f(x) and ωβ¡f, 1 n¢=O(γn).We will use the same assumptions in the Theorem 3.2 too. ON BOAS-TYPE PROBLEM 7 Theorem 3.2. (Sine) Let γ={γn}be the positive sequence and β > 0. (A). If β6= 2l(l∈N)and γ∈SQ, then for any function g∈Cwith Fourier series (6) conditions ωβ³g, 1 n´=O(γn) (17) and n X k=1 kβbk=O¡nβγn¢(18) are equivalent. (B). If β= 2l(l∈N)and γ∈SQ, then for any function g∈Cwith Fourier series (6) conditions ωβ³g, 1 n´=O(γn) (19) and n X k=1 kβ+1bk=O¡nβ+1γn¢,(20) n X k=1 kβbksin kx =O¡nβγn¢uniformly in x (21) are equivalent. (C). If γ∈SQβ, then for any function g∈Cwith Fourier series (6) condition ∞ X k=n bk=O(γn) (22) implies ωβ³g, 1 n´=O(γn).(23) 4. Proofs. The following lemmas3play the central role in the proofs of main theorems. 3See more about results of this type in [13], [17] and references there. 8 SERGEY TIKHONOV Lemma 4.1. (Cosine) If f(x)∈Chas a Fourier series (5), then n−β n X k=1 kβak≤C(β)ωβ³f, 1 n´,for β6= 2l−1, l = 1,2,· · · Proof of Lemma 4.1. If f(x)∼ ∞ P n=1 ancos nx, then 4β hf³x−βh 2´∼ ∞ X n=1 anµ2 sin kh 2¶β cos³kx +πβ 2´=: J. (24) Note, that for any ξ(x)∈Lsuch that kξk16= 0 we have k4β hf³· − βh 2´k ≥   2π Z 0 |ξ(x)|dx  −1¯¯¯¯¯¯ 2π Z 0 ξ(x)4β hf³x−βh 2´dx¯¯¯¯¯¯ . We take as ξ(x) the Fejer kernel ([19]), i.e., ξ(x) = K2n(x) = 1 2+ 2n X m=1 µ1−m 2n+ 1¶cos mx, n ∈N, 2π Z 0 |K2n(x)|dx =π. Then the Parseval equation implies k4β hf(·)k ≥ 1 π¯¯¯¯¯¯ 2π Z 0 K2n(x)J dx¯¯¯¯¯¯ = 2β¯¯¯¯¯ 2n X m=1 µ1−m 2n+ 1¶µsin mh 2¶β cos πβ 2am¯¯¯¯¯ . Therefore, for h∈(0,1 n] we get k4β hf(·)k ≥ C(β)hβ 2n X m=1 µ1−m 2n+ 1¶mβam ≥C(β)hβ n X m=1 mβam. The proof is complete. Lemma 4.2. (Sine) If g(x)∈Chas a Fourier series (6), then n−β n X k=1 kβbk≤C(β)ωβ³g, 1 n´,for β6= 2l, l = 1,2,· · · The proof of Lemma 4.2 is analogous to the previous lemma. Also we will need two auxiliary results for sequences. ON BOAS-TYPE PROBLEM 9 Lemma 4.3. ([12]). Let γ={γn}be a positive sequence such that the inequalities α1min(γ2k, γ2k+1 )≤γn≤α2max(γ2k, γ2k+1 ),0< α1≤α2<∞ hold for any 2k≤n≤2k+1, k = 1,2,· · · . Then the inequalities n X k=1 γ2k≤Cγ2n(n∈N, C ≥1) Ãor ∞ X k=n γ2k≤Cγ2n(n∈N, C ≥1)! hold if and only if the sequence {nεγn}is almost increasing (decreasing) with a certain negative (positive) number ε, respectively. Lemma 4.4. Let γ={γn>0},µ={µn≥0}and β > 0. (i). If γ∈SQβ, then ∞ X k=n µk=O(γn) (25) implies n X k=1 kβµk=O¡γnnβ¢.(26) (ii). If γ∈SQ, then (26) implies (25). Proof of Lemma 4.4. (i). It is clear that the condition (25) implies the condition 2n P k=n kβµk=O¡nβγn¢. We choose ssuch that 2s−1≤n < 2s. By Lemma 4.3, we have n X k=1 kβµk≤ n X k=2s−1 kβµk+ s−1 X k=0 2k+1−1 X ξ=2k ξβµξ =OÃ2β(s−1)γ2s−1+ s−1 X k=0 2βkγ2k! =O³2β(s−1)γ2s−1´. Since γ∈SQβ,{nβγn}is almost increasing and (26) is true. The proof of (ii) is analogous. Proof of Theorem 3.1. We will use several times the following trivial consequence of Theorem 2.2 ωβ(f, δ)≤ωβ(f, sδ)≤C(β)sβωβ(f, δ) (s≥1).(27)