Embedding theorems of function classes, I
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EMBEDDING THEOREMS OF FUNCTION CLASSES, I B. SIMONOV, S. TIKHONOV Abstract. In this paper we study embedding theorems of function classes, which are subclasses of Lp, 1 ≤p≤ ∞. To define these classes, we use the notion of best trigonometric approximation as well as that of a (λ, β)-derivative, which is the generalization of a fractional derivative. Estimates of best approximations of transformed Fourier series are obtained. 1. Introduction. It is well-known that if f∈Lp, 1 ≤p≤ ∞, and ∞ P k=1 kr−1Ek(f)p<∞ for r∈N, then f(r)∈Lpand °°°f(r)°°°p≤C1(r)∞ X k=1 kr−1Ek(f)p. For p=∞this fact was proved by Bernstein in [Be1], for other pwe refer to [p.209, De-Lo] and [Ch. 5,6, Ti]. As a corollary (see [Steˇc]) we have the following inequality En(f(r))p≤C2(r)ÃnrEn(f)p+∞ X k=n+1 kr−1Ek(f)p!.(1) On the other hand, one can write the inverse inequality (see [p. 206, De- Lo]): nrEn(f)p≤C3(r)En(f(r))p. Thus, for α∈(0,1) and ε=©εn=n−(r+α)ª, δ ={δn=n−α}the following function classes coincide: Ep[ε] = nf∈Lp:En(f)p=O[εn]o,(2) Wr pE[δ] = nf∈Lp:f(r)∈Lp, En³f(r)´p=O[δn]o.(3) 2000 Mathematics Subject Classification. Primary 46E35; Secondary 41A50, 26A33. Key words and phrases. Embedding theorems, Best approximation, (λ, β)-derivative. 1
2 B. SIMONOV, S. TIKHONOV We shall obtain the necessary and sufficient conditions for embedding theorems of some function classes which are more general than (2) and (3). We shall use the concept of a (λ, β)-derivative, which allows us to consider f(r)as well as e f(r). As an r-th derivative we shall consider the fractional derivative in the sense of Weyl. We would like to mention earlier papers [Ha-Li], [Kr], [Mu], [Og] in which this concept was used to examine the question mentioned above. Also we mention papers [Be1], [Ch-Zh], [Ha-Sh], [Mo], [Steˇc] where the results were obtained in the necessity part. The paper is organized in the following way. Section 2 contains some definitions and preliminaries. In section 3, we present our main theorems. Section 4 contains lemmas. Sections 5 and 6 include the proofs of the main results for the cases 1 < p < ∞and p= 1,∞, respectively. Finally, we mention the paper by Stepanets [Step] where the analogues of inequality (1) for (λ, β)-derivatives were obtained. The advantage of our findings compared to the results of Stepanets is that our theorems are stronger for the case of 1 < p < ∞. 2. Definition and notation Let Lp=Lp[0,2π] (1 ≤p < ∞) be a space of 2π-periodic functions for which |f|pis integrable, and L∞≡C[0,2π] be the space of 2π-periodic continuous functions with kfk∞= max {|f(x)|,0≤x≤2π}. Let a function f(x)∈L1have the Fourier series f(x)∼σ(f) := a0(f) 2+∞ X ν=1 (aν(f) cos νx +bν(f) sin νx)≡∞ X ν=0 Aν(f, x). (4) By the transformed Fourier series of (4) we mean the series σ(f, λ, β) := ∞ X ν=1 λν·aνcos µνx +πβ 2¶+bνsin µνx +πβ 2¶¸, where β∈Rand λ={λn}is a given sequence of positive numbers. The sequence λ={λn}satisfies 42-condition if λ2n≤Cλnfor all n∈N.For λ={λn}n∈Nwe define 4λn:= λn−λn+1;42λn:= 4(4λn). Let Sn(f) denote the n-th partial sum of (4), Vn(f) denote the de la Vall´ee-Poussin sum and Kn(x) be the Fej´er kernel, i.e. Sn(f) = n X ν=0 Aν(x), Vn(f) = 1 n 2n−1 X ν=n Sν(f),
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 3 Kn(x) = 1 n+ 1 n X ν=0 Ã1 2+ ν X m=1 cos mx!. Let En(f)pbe the best approximation of a function fby trigonometric polynomials of order no more than n, i.e. En(f)p= inf αk,βk∈R°°°°°f(x)− n X k=0 (αkcos kx +βksin kx)°°°°°p . Let Φ be the class of all decreasing null-sequences. For β∈Rand λ= {λn>0}we define the following function class : Wλ,β p=nf∈Lp:∃g∈Lp, g(x)∼σ(f, λ, β)o.(5) We call the function g(x)∼σ(f, λ, β) the (λ, β)-derivative of the function f(x) and denote it by f(λ,β)(x). Also, we define for ε∈Φ Wλ,β pE[ε] = nf∈Wλ,β p:En³f(λ,β)´p=O[εn]o.(6) In the case λn≡1 and β= 0 the class Wλ,β pE[ε] coincides with the class Ep[ε] (see (2)). It is clear that if λn=nr, r > 0, β=r, then the class Wλ,β pE[ε] coincides with the class Wr pE[ε] (see (3) where f(r)denotes a fractional derivative in the Weyl sense) and if λn=nr, r > 0, β=r+ 1, then the class Wλ,β pE[ε] coincides with the class f Wr pE[ε] := nf∈Lp:e f(r)∈Lp, En³e f(r)´p=O[εn]o. Here and further, e fis a conjugate function to f. By C(s, t, ···) we denote the positive constants that are dependent only on s, t, ··· and may be different in different formulas. 3. Main Results Theorem 1. Let 1< p < ∞,θ= min(2, p),β∈R,and λ={λn}be a non-decreasing sequence of positive numbers satisfying 42-condition. Let
4 B. SIMONOV, S. TIKHONOV ε={εn}, ω ={ωn} ∈ Φ. Then Ep[ε]⊂Wλ,β p⇐⇒ ∞ X n=1 ¡λθ n+1 −λθ n¢εθ n<∞,(7) Ep[ε]⊂Wλ,β pE[ω]⇐⇒ ½∞ X ν=n+1 ¡λθ ν+1 −λθ ν¢εθ ν¾1 θ +λnεn=O[ωn],(8) Wλ,β p⊂Ep[ε]⇐⇒ 1 λn =O[εn],(9) Wλ,β pE[ω]⊂Ep[ε]⇐⇒ ωn λn =O[εn].(10) Theorem 2. Let p= 1,∞,β∈R,and λ={λn}be a non-decreasing sequence of positive numbers satisfying 42-condition. Let ε={εn}, ω = {ωn} ∈ Φ. A. If 4λn≤C4λ2nand 42λn≥0 (or ≤0), then Ep[ε]⊂Wλ,β p⇐⇒ |cos βπ 2|∞ X n=1 (λn+1 −λn)εn +|sin βπ 2|∞ X n=1 λn εn n<∞,(11) Ep[ε]⊂Wλ,β pE[ω]⇐⇒ |cos βπ 2|∞ X ν=n+1 (λν+1 −λν)εν+λnεn +|sin βπ 2|∞ X ν=n+1 λν εν ν=O[ωn].(12) B. If for β= 2k, k ∈Zthe condition 42(1/λn)≥0holds, and for β6= 2k, k ∈Zconditions 42(1/λn)≥0and ∞ P ν=n+1 1 νλν≤C λnare fulfilled, then Wλ,β p⊂Ep[ε]⇐⇒ 1 λn =O[εn],(13) Wλ,β pE[ω]⊂Ep[ε]⇐⇒ ωn λn =O[εn].(14) One can draw many conclusions from the inequalities which we use in proofs of Theorems 1 and 2. The simplest ones are Corollary 1. Let 1<p<∞,θ= min(2, p),and r > 0, A ≥0. If for
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 5 f∈Lpthe series ∞ X k=1 krθ−1lnAθ k Eθ k(f)p converges, then there exists f(λ,β)∈Lpwith λ={nrlnAn}and β∈R, and En(f(λ,β))p≤ ≤C(r, A, p) nrlnAn En(f)p+(∞ P k=n+1 krθ−1lnAθ k Eθ k(f)p)1 θ . Corollary 2. Let p= 1,∞, and r > 0, A ≥0. If for f∈Lpthe series ∞ X k=1 kr−1lnAk Ek(f)p converges, then there exist f(λ,β),˜ f(λ,β)∈Lpwith λ={nrlnAn}and β∈R, and En(f(λ,β))p+En(˜ f(λ,β))p≤ ≤C(r, A)ÃnrlnAn En(f)p+∞ P k=n+1 kr−1lnAk Ek(f)p!. Corollary 3. Let 1≤p≤ ∞, and r > 0, A ≥0. If for f∈Lpthere exist f(λ,β),˜ f(λ,β)∈Lpwith λ={nrlnAn}and β∈R, then nrlnAn En(f)p≤C(r, A, p)En(f(λ,β))p. nrlnAn En(f)p≤C(r, A, p)En(˜ f(λ,β))p. We note that if λ={nrlnAn}, then f(λ,β)is a fractional-logarithmic derivative of f(see, for example, [Ku]). 4. Auxiliary Results. Lemma 1. ([ V. 1, p. 215, Zy]) Let f(x)have the Fourier series ∞ P ν=1 (aνcos nνx+bνsin nνx),where nν+1/nν≥q > 1and ∞ P ν=1 ¡a2 ν+b2 ν¢<∞. Then for 1≤p < ∞ C1(p, q)(∞ X ν=1 ¡a2 ν+b2 ν¢)1 2 ≤ kfkp≤C2(p, q)(∞ X ν=1 ¡a2 ν+b2 ν¢)1 2 .
6 B. SIMONOV, S. TIKHONOV Lemma 2. ([V.2, p. 269, Ba2]) Let f(x)∈L∞have the Fourier series ∞ P ν=1 (aνcos nνx+bνsin nνx),where aν, bν≥0and nν+1/nν≥q > 1. Then C1(q)X nν>n (aν+bν)≤En(f)∞≤C2(q)X nν>n (aν+bν). Lemma 3. ([Steˇc]) Let f(x)∈Lp, p = 1,∞, and let ∞ P n=1 n−1En(f)p<∞ be true. Then ˜ f(x)∈Lpand En(˜ f)p≤CÃEn(f)p+∞ X k=n+1 k−1Ek(f)p!(k∈N). Lemma 4. ([V. 1, p. 182, Zy]) Let εn↓0. The condition νεν→0is both necessary and sufficient for ∞ P ν=1 ενsin νx to be the Fourier series of a continuous function. Lemma 5. ([Te]) Let f(x)∈L1have a Fourier series (4). Then En(f)1≥C∞ X ν=n+1 bν ν. Lemma 6. Let 1≤p≤ ∞ and Ep[ε]⊂Wλ,β pE[ω]. Then λnεn=O(ωn)n→ ∞.(15) Proof. We presume that (15) does not hold. Then there exists a sequence {mn}such that λmnεmn≥Cnωmnand Cn↑ ∞ as n→ ∞. One can also choose a subsequence {mnk}such that mnk+1 mnk≥2, εmnk≥1 2εmnk+εmnk+1 and λmnkεmnk≥Cnkωmnk. Let us consider the case 1 ≤p < ∞. We consider the series ∞ X k=0 ³ε2 mnk−ε2 mnk+1 ´1 2cos³(mnk+ 1)x−πβ 2´.(16) Since ∞ X k=0 ³ε2 mnk−ε2 mnk+1 ´=ε2 mn0,
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 7 by Lemma 1, the series (16) is the Fourier series of a function f0(x)∈Lp and Es(f0)p≤Cεs, i.e. f0∈E[ε]. Then f0∈Wλ,β pE[ω].On the other hand, Emnk(f(λ,β) 0)p≥Cλmnk+1 ³ε2 mnk−ε2 mnk+1 ´1 2= =Cλmnk+1 h³εmnk−εmnk+1 ´³εmnk+εmnk+1 ´i1 2≥Cλmnkεmnk≥ ≥Cnkωmnk. Thus, f0/∈Wλ,β pEp[ω].This contradiction implies (15). Let p=∞.Let us consider the series ∞ X k=0 ³εmnk−εmnk+1 ´cos³(mnk+ 1)x−πβ 2´.(17) By Lemma 2, there exists f1∈Lpwith Fourier series (17) and Es(f1)p≤ Cεs, i.e. f1∈Ep[ε]⊂Wλ,β pE[ω].On the other hand, Emnk(f(λ,β) 1)p≥Cλmnk+1 ³εmnk−εmnk+1 ´≥Cλmnkεmnk≥C Cnkωmnk, i.e. f1/∈Wλ,β pE[ω].¤ Lemma 7. ([Si-Ti]) Let p= 1,∞and {λn}be monotonic concave (or convex)sequence. Let Tn(x) = n X ν=0 aνcos νx +bνsin νx, Tn(λ, x) = n X ν=0 λν(aνcos νx +bνsin νx). Then for M > N ≥0one has kTM(λ, x)−TN(λ, x)kp≤µ(M, N)kTM(x)−TN(x)kp, where µ(M, N) = = 2M(λM−λM−1)+λN+1 −(N+ 1)(λN+2 −λN+1),if λn↑,42λn≥0; 2λM+ (N+ 1)(λN+2 −λN+1)−λN+1,if λn↑,42λn≤0; (N+ 1)(λN+1 −λN+2)+λN+1,if λn↓,42λn≥0. Lemma 8. Let p= 1,∞. Set T2n,2n+1 (x) = 2n+1 X ν=2n³cνcos νx +dνsin νx´.
8 B. SIMONOV, S. TIKHONOV Then C1°°°e T2n,2n+1 (·)°°°p≤°°T2n,2n+1 (·)°°p≤C2°°°e T2n,2n+1 (·)°°°p.(18) Proof. We rewrite T2n,2n+1 (x) in the following way T2n,2n+1 (x) = 2n+1 X ν=2n 1 ν³νcνcos νx +νdνsin νx´. Applying Lemma 7 and the Bernstein inequality we have °°T2n,2n+1 (·)°°p≤C1 2n°°°°°° 2n+1 X ν=2n³νcνcos νx +νdνsin νx´°°°°°°p =C1 2n°°°°°° 2n+1 X ν=2n−dνcos νx +cνsin νx 0°°°°°°p ≤C°°°e T2n,2n+1 (·)°°°p. The same reasoning for e T2n,2n+1 (x) implies the correctness of the left-hand side of (18). ¤ 5. Proof of Theorem 1. We divide the proof of Theorem 1 into two parts. 5.1. Proof of sufficiency. Step 1. Let us prove the sufficiency part in (7). First, if λn≡1, the Riesz inequality ([V. 1, p. 253, Zy]) k˜ fkp≤C(p)kfkpimplies kf(λ,β)kp≤C(p, β)kfkp(19) Let the series in the right part of (7) be convergent and f∈Ep[ε].We use the following representation λθ 2n=λθ 1+ n+1 X ν=2 ¡λθ 2ν−1−λθ 2ν−2¢. Applying the Minkowski’s inequality we get ³here and further 41:= A1(f, x),4n+2 := 2n+1 P ν=2n+1 Aν(f, x), n = 0,1,2,···, where Aν(f, x) is from
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 9 (4) ´ I1:= := 2π Z0"∞ X n=1 λ2 2n−142 n#p 2 dx 1 p ≤C(λ) 2π Z0(λ2 142 1+λ2 242 2+∞ X n=3 λ2 2n−242 n)p 2 dx 1 p ≤C(p, λ) 2π Z0 λ2 142 1+λ2 242 2+∞ X n=3 42 n"λθ 1+ n X ν=3 ¡λθ 2ν−2−λθ 2ν−3¢#2 θ p 2 dx 1 p ≤C(p, λ) λθ 1 2π Z0"∞ X n=1 42 n#p 2 dx θ p +∞ X s=3 ¡λθ 2s−2−λθ 2s−3¢ 2π Z0"∞ X n=s42 n#p 2 dx θ p 1 θ . By the Littlewood-Paley theorem ([V. II, p. 233, Zy]) and kf−Sn(f)kp≤C(p)En(f)p,(20) we get I1≤C(p, λ)(λθ 1Eθ 0(f)p+∞ X s=1 ¡λθ 2s−λθ 2s−1¢Eθ 2s(f)p)1 θ . Since f∈Ep[ε] we have I1<∞.Thus, by the Littlewood-Paley theorem, there exists a function g∈Lpwith Fourier series ∞ X n=1 λ2n−14n,(21) and kgkp≤C(p)I1. We rewrite series (21) in the form of ∞ P n=1 γnAn(f, x), where γi:= λi, i = 1,2 and γν:= λ2nfor 2n−1+ 1 ≤ν≤2n(n= 2,3,···). Further, we write the series ∞ X n=1 λnAn(f, x) = ∞ X n=1 γnΛnAn(f, x),(22) where Λ1:= Λ2:= 1, Λν:= λν/γn=λν/λ2nfor 2n−1+ 1 ≤ν≤2n(n= 2,3,···).The sequence {Λn}satisfies the conditions of the Marcinkiewicz
16 B. SIMONOV, S. TIKHONOV C2ωn≥ ≥En(f(λ,β) 5,n )p≥C(p)°°°f(λ,β) 5,n −Sk(f(λ,β) 5,n )°°°p ≥C(p) 2π Z0 *·³εp n−εp 2m+2−1´1 pcos(n+ 1)x¸(λ,β)+2 +∞ X ν=m+1 * ³εp 2ν+1−1−εp 2ν+2−1´1 p2ν(1 p−1) 2ν+1 X µ=2ν+1 cos µx (λ,β) +2 p 2 dx 1 p ≥C(p) ³εp n−εp 2m+2−1´λp n 2π Z0|cos(n+ 1)x|pdx 1 p + 2π Z0 ∞ X ν=m+1 * ³εp 2ν+1−1−εp 2ν+2−1´1 p2ν(1 p−1) 2ν+1 X µ=2ν+1 cos µx (λ,β)+2 p 2 dx 1 p ≥C(p)(³εp n−εp 2m+2−1´λp n+∞ X ν=m³εp 2ν+2−1−εp 2ν+3−1´λp 2ν)1 p ≥C(p)(³εp n−εp 2m+1−1´λp n+∞ X ν=m+1 ³εp 2ν−1−εp 2ν+1−1´λp 2ν)1 p ≥C(p)(εp nλp n+∞ X ν=n+1 ¡λp ν+1 −λp ν¢εp ν)1 p . This implies the necessity in (8) for 1 < p < 2. The proof of the necessity part in (8) is complete. Step 6. Now we shall prove that Wλ,β p⊂Ep[ε] implies 1 λn=O[εn]. First, we note that the last condition is equivalent to the following one: ∀γ={γn} ∈ Φ one has γn λn=O[εn]. We shall obtain only nontrivial part which is : γn λn=O[εn] implies 1 λn=O[εn]. Let us assume 1 λn=O[εn] does not hold. Then there exists a sequence {Ck↑ ∞} such that 1 λnkεnk≥Ck.Using 1 λn=O[εn] we have C γnk≥Ck. Choosing γnk:= 1 √Ck→0, we write C≥√Ck→ ∞.This contradiction
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 17 gives 1 λn =O[εn]⇐⇒ ∀ γ={γn} ∈ Φγn λn =O[εn].(29) Let us assume γn λn=O[εn] does not hold for all γ∈Φ, but Wλ,β p⊂Ep[ε]. Then there exist γ={γn} ∈ Φ and {Cn↑ ∞} such that γmn λmn≥Cnεmn. Further, we choose a subsequence {mnk}such that mnk+1 mnk≥2 and γmnk≤ 2−k.Consider the series ∞ X k=0 γmnk λmnk cos(mnk+ 1)x. (30) Since ∞ P k=0 γmnk λmnk≤1 λmn0 ∞ P k=0 1 2k<∞,there exists a function f6∈Lpwith Fourier series (30). Because ∞ P k=0 γmnk≤∞ P k=0 1 2k<∞,we have f(λ,β) 6∈Lp, i.e. f6∈Wλ,β p. By (20) and by Lemma 1, Emnk(f6)p≥C(p)°°°f6−Smnk(f6)°°°p≥C(p)Ã∞ X s=k γ2 mns λ2 mns!1 2 ≥C(p)γmnk λmnk ≥C(p)Cnkεmnk, i.e. f6/∈Ep[ε].This contradiction implies that the condition 1 λn=O[εn] is necessary for Wλ,β p⊂Ep[ε]. The proof of the necessity part in (9) is complete. Step 7. Let us prove that Wλ,β pE[ω]⊂Ep[ε] implies ωn λn=O[εn]. If the last condition does not hold, then there exists {Cn↑ ∞} such that ωmn λmn≥Cnεmn.We choose a subsequence {mnk}such that mnk+1 mnk≥2 and ωmnk≥1 2ωmnk+ωmnk+1 .Since ∞ P k=0 ω2 mnk−ω2 mnk+1 λ2 mnk≤ω2 mn0 λ2 mn0 ,by Lemma 1, the series ∞ X k=0 ³ω2 mnk−ω2 mnk+1 ´1 2 λmnk cos(mnk+ 1)x(31) is the Fourier series of a function f7∈Lp. We have also f(λ,β) 7∈Lp, En(f(λ,β) 7)p≤Cωn.
18 B. SIMONOV, S. TIKHONOV On the other hand, Emnk(f7)p≥C(p)°°°f7−Smnk(f7)°°°p ≥C(p)Ã∞ X s=k ω2 mns−ω2 mns+1 λ2 mns!1 2 =C(p)Ã∞ X s=k (ωmns−ωmns+1 )(ωmns+ωmns+1 ) λ2 mns!1 2 ≥C(p)ωmnk λmnk≥C(p)Cnkεmnk, i.e. f7/∈Ep[ε]. Therefore, ωn λn=O[εn]. The proof of the necessity part in (10) is complete. ¤ 6. Proof of Theorem 2. We divide the proof of Theorem 2 into two parts. 6.1. Proof of sufficiency. Step 1. Let us show that if the series in (11) converges and f∈Ep[ε], then f∈Wλ,β p. We consider the series cos πβ 2V1(λ, f)−sin πβ 2f V1(λ, f) (32) +∞ X n=1½cosπβ 2(V2n(λ, f)−V2n−1(λ, f))−sinπβ 2³g V2n(λ, f)− ^ V2n−1(λ, f)´¾, where V1(λ, f) := λ1A1(f, x), Vn(λ, f):= σ(λ, Vn(f)) = = n X m=1 λmAm(f, x) + 2n−1 X m=n+1 λmµ1−m−n n¶Am(f, x) (n≥2).
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 19 Let M > N > 0. From the inequality kf−Vn(f)kp≤CEn(f)p and Lemma 7, using the properties of {λn}, we get A:=°°°°° M X n=N·cosπβ 2(V2n+1 (λ, f)−V2n(λ, f))−sinπβ 2³^ V2n+1 (λ, f)−g V2n(λ, f) ´¸°°°°°p ≤ M X n=N·|cos πβ 2|kV2n+1 (f)−V2n(f)kp · 2n+2+3 X m=2n−1−1 |42λm+2|(m+ 1)+(2n+2 −1)|4λ2n+2−1| +|sin πβ 2|°°° ^ V2n+1 (f)−g V2n(f)°°°p · 2n+2+3 X m=2n−1−1|42λm+2|(m+ 1) + (2n+2 −1)|4λ2n+2−1| +|cos πβ 2|°°°°° M X n=N λ2n+1−1(V2n+1 −V2n) (f)°°°°°p +|sin πβ 2|°°°°° M X n=N λ2n+1−1³^ V2n+1 −g V2n´(f)°°°°°p ≤C(λ2N−1µ|cos πβ 2|E2N−1(f)p+|sin πβ 2|E2N−1(˜ f)p¶ +∞ X n=2N−1 (λn+1 −λn)µ|cos πβ 2|En(f)p+|sin πβ 2|En(˜ f)p¶). Further, we apply Lemma 3. Then the convergence of series in (11) and f∈Ep[ε] imply that there exists ϕ∈Lpsuch that the series (32) converges to ϕin Lp. Let us show that σ(ϕ) = σ(f(λ,β)).If Fnis the n-th partial sum of (32), then, say for cosine coefficients, an(ϕ) = an(ϕ−FN+n) + an(FN+n) = an(ϕ−FN+n) + an(f(λ,β)),and an(ϕ−FN+n) = 1 π π Z −π (ϕ−FN+n) (x) cos nx dx ≤C(p)kϕ−FN+nkp−→ 0 (N→ ∞).
20 B. SIMONOV, S. TIKHONOV The proof of the sufficiency part in (11) is complete. Step 2. Let us prove the sufficiency part in (12). Let sin πβ 2= 0. If n= 0,1, then the proof comes from (11). If 2m−1+1 ≤ n < 2m, m ∈N, we consider the best approximant T∗ n(x) = T∗ n(f, x), i.e. En(f)p=kf(·)−T∗ n(f, ·)kp. En(f(λ,β))p≤°°°f(λ,β)−T∗ n (λ,β)+V2m+2 (f(λ,β))−V2m+2 (f(λ,β))°°°p ≤°°°T∗ n (λ,β)−V2m+2 (f(λ,β))°°°p+°°°f(λ,β)−V2m+2 (f(λ,β))°°°p. By Lemma 7, we obtain °°°T∗ n (λ,β)−V2m+2 (f(λ,β))°°°p≤C(λ)λnkT∗ n−V2m+2 (f)kp≤C(λ)λnEn(f)p. Further, applying two times the Abel’s transformation and Lemma 7 we write for M > N A=°°°°° M X n=N³V2n+1 (f(λ,β))−V2n(f(λ,β))´°°°°°p ≤ M X n=NkV2n+1 (f)−V2n(f)kp· · 2n+2−3 X m=2n+1 |42λm+2|(m+ 1) + (2n+2 −1)|λ2n+2−1−λ2n+2 | +°°°°° M X n=N λ2n+2−1(V2n+1 −V2n) (f)°°°°°p ≤C(λ)(λ2N+1 E2N(f)p+∞ X n=N E2n(f)p(λ2n+2−1−λ2n+1−1)). Then °°°f(λ,β)−V2m+2 (f(λ,β))°°°p≤(33) ≤C(λ)(λnEn(f)p+∞ X ν=n+1 (λν+1 −λν)Eν(f)p).
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 21 Let sin πβ 26= 0. Then the convergence of series in (11) implies that there exist e f∈Lp,f(λ,0) ∈Lpand e f(λ,0) ∈Lp. Then En(f(λ,β))p≤Enµcos πβ 2f(λ,0) −sin πβ 2e f(λ,0)¶p ≤ |cos πβ 2|En³f(λ,0)´p+|sin πβ 2|En³e f(λ,0)´p. Applying (33), we get En(f(λ,0))p≤C(λ)(λnEn(f)p+∞ X ν=n+1 (λν+1 −λν)Eν(f)p). By Lemma 3, (33) implies En³e f(λ,0)´p≤C(λ)(λnEn(e f)p+∞ X ν=n+1 (λν+1 −λν)Eν(e f)p) ≤C(λ)(λnEn(f)p+∞ X ν=n+1 λν νEν(f)p). Therefore, for all β∈Rwe have En(f(λ,β))p≤C(λ, β)(λnEn(f)p+|cos πβ 2|∞ X ν=n+1 (λν+1 −λν)Eν(f)p +|sin πβ 2|∞ X ν=n+1 λν νEν(f)p). If f∈Ep[ε], from (12) we obtain En(f(λ,β))p=O(ωn),i.e. f∈Wλ,β pE[ω]. The proof of the sufficiency part in (12) is complete. Step 3. Let us show that if 1 λn=O(εn) and f∈Wλ,β p, then f∈Ep[ε]. Also we shall verify that if ω λn=O(εn) and f∈Wλ,β pE[ω], then f∈Ep[ε]. Set 1 λ:= n1 λ1,1 λ2,1 λ3,···o.
22 B. SIMONOV, S. TIKHONOV Let sin πβ 2= 0. Applying two times the Abel’s transformation and Lemma 7 we have for M > N A:= °°°°° M X n=N [V2n+1 (f)−V2n(f)]°°°°°p =°°°°° M X n=N·³V2n+1 (f(λ,β))´(1 λ,0) −³V2n(f(λ,β))´(1 λ,0)¸°°°°°p ≤ M X n=N°°°V2n+1 (f(λ,β))−V2n(f(λ,β))°°°p × 2n+2−3 X m=2n+1 |42λ−1 m+2|(m+ 1) + (2n+2 −1)|λ−1 2n+2−2−λ−1 2n+2−1| +°°°°° M X n=N λ−1 2n+2−1hV2n+1 (f(λ,β))−V2n(f(λ,β))i°°°°°p ≤C(λ) λ2N°°°f(λ,β)°°°p.(34) Then, from (34), one has En(f)p≤C(λ) λ2N°°°f(λ,β)°°°p.(35) Let T∗ n(x) be the best approximant for f(λ,β), i.e. En(f(λ,β))p= kf(λ,β)(·)−T∗ n(·)kp. Using (35) for f−T∗ n (1 λ,−β), we get En(f)p=En(f(λ,β)−T∗ n (1 λ,−β))p≤C(λ) λn°°°f(λ,β)(·)−T∗ n(·)°°°p =C(λ) λn En(f(λ,β))p.(36) Let sin πβ 26= 0.Then A=°°°°° M X n=N·cos πβ 2µ³V2n+1 (f(λ,β))´(1 λ,0) −³V2n(f(λ,β))´(1 λ,0)¶ + sin πβ 2µ³e V2n+1 (f(λ,β))´(1 λ,0) −³e V2n(f(λ,β))´(1 λ,0)¶¸°°°°p ≤°°°°°cos πβ 2 M X n=N·³V2n+1 (f(λ,β))´(1 λ,0) −³V2n(f(λ,β))´(1 λ,0)¸°°°°°p +°°°°°sin πβ 2 M X n=N·³e V2n+1 (f(λ,β))´(1 λ,0) −³e V2n(f(λ,β))´(1 λ,0)¸°°°°°p
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 23 To estimate the first item we apply (34) and to estimate the second one we use Lemmas 7, 8, as well as the condition ∞ P ν=n+1 1 νλν≤C λn. We get °°°°° M X n=N V2n+1 (f(λ,β))−V2n(f(λ,β))°°°°°p≤C(λ, β) λ2N°°°f(λ,β)°°°p. Repeating the argument we used in proving (36) we arrive at the inequality En(f)p≤C(λ, β) λn En(f(λ,β))p. From this it is clear that 1 λn=O(εn) and ωn λn=O(εn) are necessary conditions for Wλ,β p⊂Ep[ε] and Wλ,β pE(ω)⊂Ep[ε], respectively. The proves of sufficiency parts in (13) and (14) are complete. 6.2. Proof of necessity. Step 4. Let us show that if Ep[ε]⊂Wλ,β p, then the series in (11) converges. We suppose the inverse, i.e. that the series in (11) diverges. Step 4(a): p=∞. We start with the case sin πβ 2= 0.Then we define the series ∞ X ν=1 (εν−1−εν)³cos νx + sin νx´, which converges to the function f8∈Lp. It can be easily found: En(f8)p≤ εn,i.e. f8∈Ep[ε]⊂Wλ,β p.On the other hand, °°°f(λ,β) 8°°°p≥∞ X ν=1 λν(εν−1−εν) =λ1(ε0−ε1) + ∞ X ν=2 (εν−1−εν)"ν X m=2 (λm−λm−1) + λ1# =∞ X m=2 (λm−λm−1)εm−1+λ1ε0 =∞ X n=1 (λn+1 −λn)εn+λ1ε0=∞. This contradiction implies the convergence the series in (11).
24 B. SIMONOV, S. TIKHONOV Let sin πβ 26= 0.Since ∞ X n=1 (λn+1 −λn)εn≤C∞ X ν=0 λ2νε2ν ≤CÃ∞ X ν=1 ε2ν 2ν−1 X m=2ν−1 λm m+λ1ε1!≤C∞ X n=1 λn nεn, we have |cos βπ 2|∞ X n=1 (λn+1 −λn)εn+|sin βπ 2|∞ X n=1 λn εn n³∞ X n=1 λn εn n.(37) We consider the series ∞ P n=1 εn−1 nsin nx. Then, by Lemma 4, the sum of this series, say, f9(x) is from L∞.From [Ba1], En(f9)p≤Cεn, i.e. f9∈Ep[ε]⊂ Wλ,β p. We also have °°°f(λ,β) 9°°°p≥C∞ X n=1 εn−1 nλn≥C∞ X n=1 εn nλn=∞. Thus, the series in (11) converges. Step 4(b): p= 1. First let sin πβ 2= 0. We define the series ∞ X ν=1 (εν−1−εν)τν(x),(38) where τν+1(x) = ν+1 X j=1 αν jsin jx and αν j=(j ν+2 ,1≤j≤ν+2 2 1−j ν+2 ,ν+2 2≤j≤ν+ 1 . The series (38) converges to a f10 ∈Lpand En(f10)p≤Cεn(see [Ge]). Then f10 ∈Ep[ε]⊂Wλ,β p. One can rewrite (38) in the following way : ∞ X ν=1 bνsin νx, where bν= 2ν−2 X j=νµ1−ν j+ 1¶(εν−1−εν) + ∞ X j=2ν−1 ν j+ 1 (εν−1−εν). By Lemma 5, we get °°°f(λ,β) 10 °°°1≥C∞ X ν=1 λν bν ν≥CÃ∞ X ν=2 (λν+1 −λν)εν+λ3ε3!=∞. This contradicts the divergence of the series in (11). Thus, the series in (11) converges.
EMBEDDING THEOREMS OF FUNCTION CLASSES, I 25 Let sin πβ 26= 0.As we saw in (37), the divergence of the series in (11) in this case is equivalent to the divergence of ∞ P n=1 λn nεn. We consider the series ∞ X ν=1 (εν−1−εν)Kν(x).(39) This series is convergent in L1(see [Ge]) to a function f11(x), and En(f11)p= O(εn). Therefore, f10 ∈Ep[ε]⊂Wλ,β p. One can rewrite (39) in the following way : ∞ X ν=1 aνcos νx, where aν=εν−1−ν∞ X j=ν εj−1−εj j+ 1 . We note that {aν}is monotonic null sequence. Indeed, aν−aν+1 =∞ X j=ν εj−1−εj j+ 1 ≥0.(40) By Lemma 5, using monotonicity of {aν}and conditions on {λν}, we have °°°f(λ,β) 11 °°°1≥C∞ X ν=1 λν νaν≥C∞ X ν=1 λ2ν+1 a2ν =C∞ X ν=1 a2ν"ν X n=1 (λ2n+1 −λ2n) + λ2# ≥CÃλ1a1+∞ X n=1 (λ2n+1 −λ2n)a2n! ≥C λ1a1+∞ X n=2 a2n 2n+1−1 X ν=2n (λν+1 −λν) ≥CÃλ1a1+∞ X n=1 (λn+1 −λn)an!.(41)
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