# On Approximations by Trigonometric Polynomials of Classes of Functions Defined by Moduli of Smoothness

Authors: Nimete Sh. Berisha; Faton M. Berisha; Mikhail K. Potapov; Marjan Dema
Journal: Abstract and Applied Analysis
 (2017)

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## Abstract

In this paper, we give a characterization of Nikol’skiĭ-Besov type classes of functions, given by integral representations of moduli of smoothness, in terms of series over the moduli of smoothness. Also, necessary and sufficient conditions in terms of monotone or lacunary Fourier coefficients for a function to belong to such a class are given. In order to prove our results, we make use of certain recent reverse Copson-type and Leindler-type inequalities.


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## Body

## 1. Introduction

Letf∈Lp[0,2π], 1<p<∞, be a 2π-periodic function. We say that the function f has monotone Fourier coefficients if it has a cosine Fourier series with(1)fx~∑n=0∞ancos⁡nx,an↓0.We say that the functionf has lacunary Fourier coefficients if(2)fx~∑ν=1∞λνcos⁡νx,where(3)λν=aμ≥0forν=2μ,0forν≠2μ;that is,(4)fx~∑μ=0∞aμcos⁡2μx,aμ≥0.Byωkf,tp we denote the modulus of smoothness of order k in Lp metrics of a function f∈Lp, 1<p<∞: (5)ωkf,tp=suph≤t⁡Δhkfp,where(6)Δhkfx=∑ν=0k-1k-νkνfx+νhis the kth-order shift operator.ByEnfp we denote the best approximation in Lp metrics of a function f∈Lp, 1<p<∞, by means of trigonometric polynomials whose degree is not greater than n-1; that is, (7)Enfp=infTn-1⁡f-Tn-1p,where Tn-1=∑ν=0n-1(ανcos⁡νx+βνsin⁡νx) and αν and βν are arbitrary real numbers.We say that a2π-periodic function f belongs to the Nikol’skiĭ-Besov class N(p,θ,r,λ,φ), 1<p<∞, if the following conditions are satisfied:(1)
f∈Lp[0,2π].(2)
Numbersθ, r, and λ belong to the interval (0,∞), and k is an integer satisfying k>r+λ.(3)
The following inequality holds true:(8)∫0δt-rθ-1ωkf,tpθdt+δλθ∫δ1t-r+λθ-1ωkf,tpθdt1/θ≤Cφδ,while the function φ satisfies the following conditions:(4)
φ is a nonnegative continuous function on (0,1) and φ≠0.(5)
For everyδ1 and δ2, where 0≤δ1≤δ2≤1, φ(δ1)≤C1φ(δ2) holds.(6)
For everyδ, where 0≤δ≤1/2, φ(2δ)≤C2φ(δ) holds.Constants (without mentioning it explicitly, we will consider all the constants positive) C, C1, and C2 do not depend on δ1, δ2, and δ.A more detailed approach to the classesN(p,θ,r,λ,φ) is given in [1, 2] (see also [3]). In the paper, we give a characterization of N(p,θ,r,λ,φ) classes of functions in terms of series over their moduli of smoothness. Then we give the necessary and sufficient conditions in terms of monotone or lacunary Fourier coefficients for a function f∈Lp[0,2π] to belong to a class N(p,θ,r,λ,φ). In the process of proving the results, we make use of certain recent reverse lp-type inequalities [4], closely related to Copson’s and Leindler’s inequalities.Finally, by making use of our results, we construct an example of a function having a lacunary Fourier series, which shows thatN(p,θ,r,λ,φ) classes are properly embedded between the appropriate Nikol’skiĭ classes and Besov classes.


## 2. Statement of Results

Now we formulate our results.Theorem 1.
A functionf belongs to the class N(p,θ,r,λ,φ) if and only if (here and below we assume that the parameters θ, r, λ, and k satisfy condition (2) and the function φ satisfies conditions (4)–(6) of the definition of the class N(p,θ,r,λ,φ))(9)∑ν=n+1∞ωkf,1νpθνrθ-1+n-λθ∑ν=1nωkf,1νpθνr+λθ-11/θ≤Cφ1n,where constant C does not depend on n.Theorem 2.
For a functionf∈Lp[0,2π], 1<p<∞, such that(10)fx~∑ν=1∞aνcos⁡νx,aν↓0,to belong to the class N(p,θ,r,λ,φ), it is necessary and sufficient that its Fourier coefficients satisfy the condition(11)∑ν=n+1∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνrθ+λθ+θ-θ/p-11/θ≤Cφ1n,where constant C does not depend on n.Regarding Theorem1, a very interesting open question on its analogue for functions with general monotone Fourier coefficients generalized in the sense of [5, 6] remains.Corollary 3.
Puttingφ(δ)=δα, 0<α<λ, in the definition of the class N(p,θ,r,λ,φ), we obtain [1] the Nikol’skiĭ class Hpr+α. Thus Theorems 1 and 2 give the single coefficient condition (12)aν≤Cνr+α+1-1/pfor f∈Hpr+α, given in [7], where the function f is given by (10).Corollary 4.
Ifφ(δ)≥C, then we obtain [1] the Besov class Bpθr. Thus Theorems 1 and 2 give the necessary and sufficient condition(13)∑ν=1∞aνθνrθ+θ-θ/p-1<∞for f∈Bpθr, given in [8], where the function f is given by (10).Theorem 5.
For a functionf∈Lp, 1<p<∞, such that (14)fx~∑ν=1∞λνcos⁡νx,(15)λν=aμ≥0forν=2μ,0forν≠2μ,to belong to the class N(p,θ,r,λ,φ), it is necessary and sufficient that its Fourier coefficients satisfy the condition (here and below we assume that the parameters θ, r, λ, and k satisfy condition (2) and the function φ satisfies conditions (4)–(6).)(16)∑ν=m+1∞λνθνrθ+m-λθ∑ν=1mλνθνr+λθ1/θ≤Cφ1m,where constant C does not depend on m.Corollary 6.
Puttingφ(δ)=δα, 0<α<λ, in the definition of the class N(p,θ,r,λ,φ), we obtain [1] the Nikol’skiĭ class Hpr+α. Thus Theorem 5 gives the single coefficient condition(17)aμ≤C2-μr+αfor f∈Hpr+α, where the function f is given by (14).Corollary 7.
Ifφ(δ)=C, then we obtain [1] the Besov class Bpθr. Thus Theorem 5 gives the necessary and sufficient condition (18)∑μ=1∞aμθ2μrθ<∞for f∈Bpθr, given in [8], where the function f is given by (14).Example 8.
Let(19)fx~∑μ=0∞aμcos⁡2μx,where(20)aμ=2-μrμ+1-α+1/θ,α>0.Then, we have(21)C1n-α≤∑μ=n+1∞aμθ2μrθ1/θ≤C2n-α,C3n-α+1/θ≤2-nλθ∑μ=0naμθ2μr+λθ1/θ≤C4n-α+1/θ,thus implying (see the proof of Theorem 5) that f∈N(p,θ,r,λ,φ) for φ(δ)=ln⁡1/δ-α. This means that classes N are classes of embedding between classes H and B.


## 3. Auxiliary Statements

In order to establish our results, we use the following lemmas.Lemma 9.
Let0<α<β<∞ and aν≥0. The following inequality holds true: (22)∑ν=1naνβ1/β≤∑ν=1naνα1/α.Proof of the lemma is due to Jensen [9, p. 43].Lemma 10.
Letaνν=1∞ be a sequence of nonnegative numbers, let α>0, let λ be a real number, and let m and n be positive integers such that m<n. Then(1)
for1≤p<∞, the following equalities hold: (23)∑μ=mnμα-1∑ν=μnaννλp≤C1∑μ=mnμα-1aμμλ+1p,∑μ=mnμ-α-1∑ν=mμaννλp≤C2∑μ=mnμ-α-1aμμλ+1p;(2)
for0<p≤1, the following equalities hold:(24)∑μ=mnμα-1∑ν=μnaννλp≥C3∑μ=mnμα-1aμμλ+1p,∑μ=mnμ-α-1∑ν=mμaννλp≥C4∑μ=mnμ-α-1aμμλ+1p,where constants C1, C2, C3, and C4 depend only on numbers α, λ, and p and do not depend on m and n as well as on the sequence aνν=1∞.Proof of the lemma is given in [9, p. 308].Lemmas11 and 12 state certain lp-type inequalities which are reversed to the ones given in Lemma 10 and closely related to Copson’s and Leindler’s inequalities (see, e.g., [10–13]).We writeaν↓ if aνν=1∞ is a monotone-decreasing sequence of nonnegative numbers, that is, if aν≥aν+1≥0(ν=1,2,…).Lemma 11.
Letaν↓, let α>0, let λ be a real number, and let m and n be positive integers. Then(1)
for1≤p<∞ and n≥16m, the following equalities hold:(25)∑μ=mnμα-1∑ν=μnaννλp≥C1∑μ=8mnμα-1aμμλ+1p,∑μ=mnμ-α-1∑ν=mμaννλp≥C2∑μ=4mnμ-α-1aμμλ+1p;(2)
for0<p≤1 and n≥4m, the following equalities hold:(26)∑μ=4mnμα-1∑ν=μnaννλp≤C3∑μ=mnμα-1aμμλ+1p,∑μ=4mnμ-α-1∑ν=4mμaννλp≤C4∑μ=mnμ-α-1aμμλ+1p,where constants C1, C2, C3, and C4 depend only on numbers α, λ, and p and do not depend on m and n as well as on the sequence aνν=1∞.Proof of the lemma is given in [4].Lemma 12.
Letaν↓, let α>0, let λ be a real number, and let m and n be positive integers. For 0<p<∞, the following inequalities hold:(27)C1∑μ=1nμα-1aμμλ+1p≤∑μ=1nμα-1∑ν=μnaννλp≤C2∑μ=1nμα-1aμμλ+1p,C3∑μ=1nμ-α-1aμμλ+1p≤∑μ=1nμ-α-1∑ν=1μaννλp≤C4∑μ=1nμ-α-1aμμλ+1p,where constants C1, C2, C3, and C4 depend only on numbers α, λ, and p and do not depend on m and n as well as on the sequence aνν=1∞.The lemma is also proven in [4].Lemma 13.
Letf∈Lp[0,2π] for fixed p from the interval 1<p<∞ and let (28)fx~∑ν=1∞aνcos⁡νx,aν↓0.The following inequalities hold:(29)C11nk∑ν=1naνpνk+1p-21/p+∑ν=n+1∞aνpνp-21/p≤ωkf,1np≤C21nk∑ν=1naνpνk+1p-21/p+∑ν=n+1∞aνpνp-21/p,where constants C1 and C2 do not depend on n and f.The lemma is proven in [8].Lemma 14.
A functionf belongs to the class N(p,θ,r,λ,φ) if and only if (30)∑μ=n+1∞2μrθE2μfpθ+2-nλθ∑μ=0n2μr+λθE2μfpθ1/θ≤Cφ12n,where constant C does not depend on n.Proof of the lemma is given in [1].Lemma 15.
Letf∈Lp, 1<p<∞, and (31)fx~∑μ=0∞aμcos⁡2μx,aμ≥0.The following inequalities hold: (32)C1∑μ=0∞aμ21/2≤fp≤C2∑μ=0∞aμ21/2,where constants C2 and C1 do not depend on f.Proof of the lemma is due to Zygmund [14, vol. I, p. 326].Corollary 16.
Lemma15 yields the following estimate:(33)C1∑μ=n∞aμ21/2≤E2nfp≤C2∑μ=n∞aμ21/2,where constants C2 and C1 do not depend on n and f.


## 4. Proofs

Now we prove our results.Proof of Theorem1.
Put(34)I1=∫01/n+1t-rθ-1ωkf,tpθdt,I2=∫1/n+11t-r+λθ-1ωkf,tpθdt.We have [9, p. 55](35)I1=∫01/n+1t-rθ-1ωkf,tpθdt=∑ν=n+1∞∫1/ν+11/νt-rθ-1ωkf,tpθdt≤∑ν=n+1∞ωkf,1νpθ∫1/ν+11/νt-rθ-1dt≤C1∑ν=n+1∞ωkf,1νpθνrθ-1and, taking into account properties of modulus of smoothness [15], (36)I1≥∑ν=n+1∞ωkf,1ν+1pθ∫1/ν+11/νt-rθ-1dt≥C2∑ν=n+1∞ωkf,1νpθνrθ-1.In an analogous way, we estimate (37)I2≤∑ν=1nωkf,1νpθ∫1/ν+11/νt-r+λθ-1dt≤C3∑ν=1nωkf,1νpθνr+λθ-1,I2≥∑ν=1nωkf,1ν+1pθ∫1/ν+11/νt-r+λθ-1dt≥C4∑ν=1nωkf,1νpθνr+λθ-1.Let f∈N(p,θ,r,λ,φ). For a positive integer n, we put δ=1/n+1. Then we have (38)Iθ=I1+δλθI2≥C5∑ν=n+1∞ωkf,1νpθνrθ-1+n-λθ∑ν=1nωkf,1νpθνr+λθ-1.Hence, we obtain(39)J=∑ν=n+1∞ωkf,1νpθνrθ-1+n-λθ∑ν=1nωkf,1νpθνr+λθ-11/θ≤C6I≤C7φδ=C7φ1n+1≤C8φ1n,which proves inequality (9).
Now we suppose that inequality (9) holds. For δ∈(0,1), we choose the positive integer n satisfying 1/n+1<δ≤1/n. Then, taking into consideration the estimates from above for I1 and I2, we have(40)Iθ=∫01/n+1t-rθ-1ωkf,tpθdt+∫1/n+1δt-rθ-1ωkf,tpθdt+δλθ∫δ1t-r+λθ-1ωkf,tpθdt≤I1+δλθI2≤C9∑ν=n+1∞ωkf,1νpθνrθ-1+n-λθ∑ν=1nωkf,1νpθνr+λθ-1.Hence(41)I≤C10J≤C11φ1n≤C12φ12n≤C13φδ,implying that f∈N(p,θ,r,λ,φ).
Proof of Theorem1 is completed.Proof of Theorem2.
Theorem1 implies that the condition f∈N(p,θ,r,λ,φ) is equivalent to the condition(42)∑ν=n+1∞ωkf,1νpθνrθ-1+n-λθ∑ν=1nωkf,1νpθνr+λθ-1≤C1φ1nθ,where constant C1 does not depend on n. Lemma 13 yields that the last estimate is equivalent to the estimate (43)∑ν=n+1∞νr-kθ-1∑μ=1νaμpμk+1p-2θ/p+∑ν=n+1∞νrθ-1∑μ=ν∞aμpμp-2θ/p+n-λθ∑ν=1nνr+λ-kθ-1∑μ=1νaμpμk+1p-2θ/p+n-λθ∑ν=1nνr+λθ-1∑μ=ν∞aμpμp-2θ/p≤C2φ1nθ,where constant C2 does not depend on n. Hence, if we denote the terms on the left-hand side of the inequality by J1, J2, J3, and J4, respectively, then condition f∈N(p,θ,r,λ,φ) is equivalent to the condition (44)J1+J2+J3+J4≤C2φ1nθ.
Now we estimate the termsJ1, J2, J3, and J4 from below and above by means of expression taking part in the condition of the theorem.
First we estimateJ1 and J2 from below. We have(45)J1=∑ν=n+1∞νr-kθ-1∑μ=1νaμpμk+1p-2θ/p≥∑ν=n+1∞ν-k-rθ-1∑μ=n+1νaμpμk+1p-2θ/p.For k-r>0, making use of Lemmas 10 and 11, we obtain(46)J1≥C3∑ν=4n+1∞ν-k-rθ-1aνpνk+1p-2νθ/p=C3∑ν=4n+1∞aνθνrθ+θ-θ/p-1.In an analogous way, for rθ>0, we get(47)J2=∑ν=n+1∞νrθ-1∑μ=ν∞aμpμp-2θ/p≥C4∑ν=8n+1∞aνθνrθ+θ-θ/p-1.
We estimate the termJ2 from above: (48)J2≤C5∑ν=n+1/4∞νrθ-1aνpνp-2νθ/p=C5∑ν=n+1/4∞aνθνrθ+θ-θ/p-1.
ForJ1, we have (49)J1≤C6∑ν=n+1∞ν-k-rθ-1∑μ=n+1νaμpμk+1p-2θ/p+∑ν=n+1∞ν-k-rθ-1∑μ=1naμpμk+1p-2θ/p,and applying once more Lemmas 10 and 11, we obtain(50)J1≤C7∑ν=n+1/4∞aνθνrθ+θ-θ/p-1+n-k-rθ∑μ=1naμpμk+1p-2θ/p.Put(51)I1=n-k-rθ∑μ=1naμpμk+1p-2.Then, for(52)I2=I1nk-rθ,taking into account the fact that (k+1)p-2≥0 and aν↓0, we get(53)I2=∑μ=1naμpμk+1p-2≤∑μ=1n/2aμpμk+1p-2+an/2+1p∑μ=n/2+1nμk+1p-2≤∑μ=1n/2aμpμk+1p-2+C8nk+1p-1an/2+1p≤C9∑μ=1n/2aμpμk+1p-2.Since k-r-λ>0, we have (54)I1θ/p≤C10n-k-rθ∑μ=1n/2aμpμk+1p-2θ/p≤C11n-λθ∑ν=n/2nν-k-r-λθ-1∑μ=1νaμpμk+1p-2θ/p≤C11n-λθ∑ν=1nν-k-r-λθ-1∑μ=1νaμpμk+1p-2θ/p.Applying Lemma 12, we obtain(55)I1θ/p≤C12n-λθ∑ν=1nν-k-r-λθ-1aνpνk+1p-2νθ/p=C12n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.From (50), it follows that (56)J1≤C13∑ν=n+1/4∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.
This way, inequalities (46), (47), (48), and (56) yield (57)C14∑ν=8n+1∞aνθνrθ+θ-θ/p-1≤J1+J2≤C15∑ν=n+1/4∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.
Now we estimateJ3 and J4. Put (58)A1=nλθJ3=∑ν=1nνr+λ-kθ-1∑μ=1νaμpμk+1p-2θ/p,A2=nλθJ4=∑ν=1nνr+λθ-1∑μ=ν∞aμpμp-2θ/p.Applying Lemma 12, for r+λ-k<0, we get (59)A1≤C16∑ν=1naνθνr+λθ+θ-θ/p-1.
We estimateA2 in an analogous way:(60)A2≤C17∑ν=1nνr+λθ-1∑μ=νnaμpμp-2θ/p+∑ν=1nνr+λθ-1∑μ=n+1∞aμpμp-2θ/p≤C18∑ν=1naνθνr+λθ+θ-θ/p-1+nr+λθ∑μ=n+1∞aμpμp-2θ/p.
We estimate the series(61)B=∑μ=n+1∞aμpμp-2θ/p.
First letθ/p>1. Applying Hölder’s inequality, we have (62)∑μ=n+1∞aμpμp-2≤∑μ=n+1∞aμpμp-1+rp-p/θθ/pp/θ×∑μ=n+1∞μ-rp-p/θ+1θ/θ-pθ-p/θ.Since rp-p/θ+1θ/θ-p=rpθ/θ-p+1>1, we get (63)∑μ=n+1∞aμpμp-2≤C19n-rp∑μ=n+1∞aμθμθ-θ/p+rθ-1p/θ.So, for θ/p>1, we have proven that (64)B≤C20n-rθ∑μ=n+1∞aμθμrθ+θ-θ/p-1.
Letθ/p≤1. For given n, we choose the positive integer N such that 2N≤n+1<2N+1. Then, we have (65)B≤∑μ=2N∞aμpμp-2θ/p≤∑ν=N∞a2νp∑μ=2ν2ν+1-1μp-2θ/p≤C21∑ν=N∞a2νp2νp-1θ/p.Making use of Lemma 9, we obtain (66)B≤C21∑ν=N∞a2νθ2νθ-θ/p≤C22∑ν=N∞∑μ=2ν-12ν-1aμθμθ-θ/p-1=C22∑ν=2N-1∞aνθνθ-θ/p-1≤C22∑ν=n+1/4∞aνθνθ-θ/p-1≤C22n+14-rθ∑ν=n+1/4∞aνθνrθ+θ-θ/p-1.Since, for n≥3, n+1/4≥n/12 holds, we get(67)B≤C23n-rθ∑ν=n+1/4∞aνθνrθ+θ-θ/p-1.
This way, for0<θ/p<∞, we proved that(68)B≤C24n-rθ∑ν=n+1/4∞aνθνrθ+θ-θ/p-1.Hence, (60) yields(69)A2≤C25∑ν=1naνθνr+λθ+θ-θ/p-1+nλθ∑ν=n+1/4∞aνθνrθ+θ-θ/p-1.
Now, from (59), it follows that(70)J3+J4=n-λθA1+A2≤C26n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1+∑ν=n+1/4∞aνθνrθ+θ-θ/p-1.
Further, we estimate the series(71)A3=∑ν=n+1/4∞aνθνrθ+θ-θ/p-1=A4+∑ν=n+1∞aνθνrθ+θ-θ/p-1,where(72)A4=∑ν=n+1/4naνθνrθ+θ-θ/p-1≤C27an+1/4θnrθ+θ-θ/p≤C28n-λθ∑ν=1n+1/4aνθνr+λθ+θ-θ/p-1≤C28n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.Hence(73)A3≤C29n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1+∑ν=n+1∞aνθνrθ+θ-θ/p-1.
Making use of (73) and (70), we have(74)J3+J4≤C30n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1+∑ν=n+1∞aνθνrθ+θ-θ/p-1.Hence, applying (73) in (57), we obtain(75)J1+J2+J3+J4≤C31n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1+∑ν=n+1∞aνθνrθ+θ-θ/p-1.
Now we estimateA1 and A2 from below. Making use of Lemma 12, we get (76)A1≥C32∑ν=1naνθνr+λθ+θ-θ/p-1,and, in an analogous way,(77)A2≥∑ν=1nνr+λθ-1∑μ=νnaμpμp-2θ/p≥C33∑ν=1naνθνr+λθ+θ-θ/p-1.Hence,(78)A1+A2≥C34∑ν=1naνθνr+λθ+θ-θ/p-1.This way the following inequality holds:(79)J3+J4≥C35n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.
From (57), it follows that (80)J1+J2+J3+J4≥C36∑ν=8n+1∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.Since (81)∑ν=n+1ν=8n+1-1aνθνrθ+θ-θ/p-1≤C37anθnrθ+θ-θ/p≤C38n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1holds, we have(82)∑ν=n+1∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1≤C39∑ν=8n+1∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.
Now, estimates (80) and (75) imply that (83)C40∑ν=n+1∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1≤J1+J2+J3+J4≤C41∑ν=n+1∞aνθνrθ+θ-θ/p-1+n-λθ∑ν=1naνθνr+λθ+θ-θ/p-1.
This way we proved that condition (9) is equivalent to the condition of the theorem. Since condition (9) is equivalent to the condition f∈N(p,θ,r,λ,φ), proof of Theorem 2 is completed.Proof of Theorem5.
Considering Lemma14, condition f∈N(p,θ,r,λ,φ) is equivalent to the condition (84)∑ν=n+1∞2νrθE2νfpθ+2-nλθ∑ν=0n2νr+λθE2νfpθ≤C42φ12nθ,where constant C does not depend on n. Corollary 16 yields that the last estimate is equivalent to the estimate (85)∑ν=n+1∞2νrθ∑μ=ν∞aμ2θ/2+2-nλθ∑ν=0n2νr+λθ∑μ=ν∞aμ2θ/2≤C43φ12nθ,where constant C43 does not depend on n.
Put(86)J1=∑ν=n+1∞2νrθ∑μ=ν∞aμ2θ/2,J2=2-nλθ∑ν=0n2νr+λθ∑μ=ν∞aμ2θ/2;we estimate J1 and J2 from below and above.
Let0<θ/2≤1. Using Lemma 9, changing the order of summation, we get (87)J1≤∑ν=n+1∞2νrθ∑μ=ν∞aμθ=∑μ=n+1∞aμθ∑ν=n+1μ2νrθ.Therefrom, taking into consideration the fact that rθ>0 while computing the second sum, we obtain (88)J1≤C44∑μ=n+1∞aμθ2μrθ.
Let1≤θ/2<∞ and 0<ε<r. Applying Hölder’s inequality, we have (89)A=∑μ=ν∞aμ2≤∑μ=ν∞aμθ2μεθ2/θ∑μ=ν∞2-2μεθ′1/θ′,where 2/θ+1/θ′=1. Computing the second sum, we obtain (90)A≤C4522εν∑μ=ν∞aμθ2μεθ2/θ.Now we have(91)J1≤C46∑ν=n+1∞2νr-εθ∑μ=ν∞aμθ2μεθ=C46∑μ=n+1∞aμθ2μεθ∑ν=n+1μ2νr-εθ≤C47∑μ=n+1∞aμθ2μrθ.
This way, for0<θ<∞, we have (92)J1≤C48∑μ=n+1∞aμθ2μrθ,where constant C48 does not depend on n.
Now we estimateJ1 from below.
Let1≤θ/2<∞. Making use of Lemma 9, we get (93)J1≥∑ν=n+1∞2νrθ∑μ=ν∞aμθ=∑μ=n+1∞aμθ∑ν=n+1μ2νrθ.Computing the second sum, we get (94)J1≥C49∑μ=n+1∞aμθ2μrθ.
Let0<θ/2≤1 and ε>0. Applying Hölder’s inequality, we have (95)∑μ=ν∞aμθ2-μεθ≤∑μ=ν∞aμ2θ/2∑μ=ν∞2-μεθθ′1/θ′≤C502νεθ∑μ=ν∞aμ2θ/2,where θ/2+1/θ′=1. The last estimate implies that (96)J1≥C51∑ν=n+1∞2νr+εθ∑μ=ν∞aμθ2-μεθ.Changing the order of summation and then computing the second sum, we obtain (97)J1≥C51∑μ=n+1∞aμθ2-μεθ∑ν=n+1μ2νr+εθ≥C52∑μ=n+1∞aμθ2μrθ,where constant C52 does not depend on n.
Consequently, for every0<θ<∞, the following estimate holds: (98)C53∑μ=n+1∞aμθ2μrθ≤J1≤C54∑μ=n+1∞aμθ2μrθ,where constants C53 and C54 do not depend on n.
Now we estimateJ2. Obviously, (99)J2≥2-nλθ∑ν=0n2νr+λθ∑μ=νnaμ2θ/2.
Let1≤θ/2<∞. Applying Lemma 9, changing the order of summation and then computing the second sum, we obtain (100)J2≥2-nλθ∑ν=0n2νr+λθ∑μ=νnaμθ=2-nλθ∑μ=0naμθ∑ν=0μ2νr+λθ≥C552-nλθ∑μ=0naμθ2μr+λθ.
Let0<θ/2≤1 and ε>0. Applying Hölder’s inequality, we get (101)∑μ=νnaμθ2-μεθ≤∑μ=νnaμ2θ/2∑μ=νn2-μεθθ′1/θ′≤C562νεθ∑μ=νnaμ2θ/2,where θ/2+1/θ′=1. The last estimate implies that (102)J2≥C572-nλθ∑ν=0n2νr+λ+εθ∑μ=νnaμθ2-μεθ.Changing the order of summation and computing the second sum, we have (103)J2≥C572-nλθ∑μ=0naμθ2-μεθ∑ν=0μ2νr+λ+εθ≥C582-nλθ∑μ=0naμθ2μr+λθ.
Thus, for every0<θ<∞, the following holds: (104)J2≥C592-nλθ∑μ=0naμθ2μr+λθ.
Now we estimateJ2 from above. Taking into consideration the fact that (r+λ)θ>0, we have (105)J2≤C602-nλθ∑ν=0n2νr+λθ∑μ=νnaμ2θ/2+∑μ=n+1∞aμ2θ/2≤C612-nλθ∑ν=0n2νr+λθ∑μ=νnaμ2θ/2+2nrθ∑μ=n+1∞aμ2θ/2.Since(106)2nrθ∑μ=n+1∞aμ2θ/2≤∑μ=n+1∞2νrθ∑μ=n+1∞aμ2θ/2=J1holds and an upper bound for J1 is already found, we estimate from above the expression (107)J3=∑ν=0n2νr+λθ∑μ=νnaμ2θ/2.
Let0<θ/2≤1. Applying Lemma 9, we obtain (108)J3≤∑ν=0n2νr+λθ∑μ=νnaμθ=∑μ=0naμθ∑ν=0μ2νr+λθ≤C62∑μ=0naμθ2μr+λθ.
Let1≤θ/2<∞ and 0<ε<r+λ. Then, applying Hölder’s inequality, we have (109)∑μ=νnaμ2≤∑μ=νnaμθ2μεθ2/θ∑μ=νn2-2μεθ′1/θ′,where 2/θ+1/θ′=1. Using the last estimate, we get (110)J3≤∑ν=0n2νr+λθ∑μ=ν∞2-2μεθ′θ/2θ′∑μ=νnaμθ2μεθ≤C63∑ν=0n2νr+λ-εθ∑μ=νnaμθ2μεθ.Changing the order of summation and computing the second sum, we obtain (111)J3≤C63∑μ=0naμθ2μεθ∑ν=0μ2νr+λ-εθ≤C64∑μ=0naμθ2μr+λθ.
Therefore, for every0<θ<∞, the following estimate holds: (112)J3≤C65∑μ=0naμθ2μr+λθ.
Now, making use of inequalities (105) and (98), we have (113)J2≤C662-nλθ∑μ=0naμθ2μr+λθ+∑μ=n+1∞aμθ2μrθ.
This way, inequalities (98) and (104) and the last inequality imply the estimate (114)C67∑μ=n+1∞aμθ2μrθ+2-nλθ∑μ=0naμθ2μr+λθ≤J1+J2≤C68∑μ=n+1∞aμθ2μrθ+2-nλθ∑μ=0naμθ2μr+λθ,where constants C67 and C68 do not depend on n. Hence, considering condition (85), we conclude that condition f∈N(p,θ,r,λ,φ) is equivalent to the condition (115)An=∑μ=n+1∞aμθ2μrθ+2-nλθ∑μ=0naμθ2μr+λθ≤C69φ12nθ,where constant C69 does not depend on n.
We put(116)Dm=∑ν=m+1∞λνθνrθ+m-λθ∑ν=1mλνθνr+λθ.For given m, we choose the positive integer n such that 2n≤m+1<2n+1.
First we consider the case2n<m+1<2n+1. We have (117)Dm=∑ν=2n+1∞λνθνrθ+∑ν=m+12n+1-1λνθνrθ+m-λθ∑ν=12n-1λνθνr+λθ+m-λθ∑ν=2nmλνθνr+λθ.Since λν=0 for ν≠2μ, we get (118)Dm=∑ν=2n+1∞λνθνrθ+m-λθ∑ν=12n-1λνθνr+λθ+m-λθλ2nθ2nr+λθ=∑μ=n+1∞∑ν=2μ2μ+1-1λνθνrθ+m-λθ∑μ=0n-1∑ν=2μ2μ+1-1λνθνr+λθ+m-λθλ2nθ2nr+λθ=∑μ=n+1∞λ2μθ2μrθ+m-λθ∑μ=0n-1λ2μθ2μr+λθ+m-λθλ2nθ2nr+λθ.Further, since λ2μ=aμ, we get (119)Dm=∑μ=n+1∞aμθ2μrθ+m-λθ∑μ=0naμθ2μr+λθ.Hence, for 2n<m+1<2n+1, we obtain(120)C70∑μ=n+1∞aμθ2μrθ+2-nλθ∑μ=0naμθ2μr+λθ≤Dm≤C71∑μ=n+1∞aμθ2μrθ+2-nλθ∑μ=0naμθ2μr+λθ,where constants C70 and C71 do not depend on m and n.
Let us assume now thatm+1=2n. In an analogous way, we have (121)Dm=∑ν=2n∞λνθνrθ+2-nλθ∑ν=12n-1λνθνr+λθ=∑μ=n∞aμθ2μrθ+2-nλθ∑μ=0n-1aμθ2μr+λθ=∑μ=n+1∞aμθ2μrθ+2-nλθ∑μ=0naμθ2μr+λθ=An.
Thus, for2n≤m+1<2n+1, the following estimate holds: (122)C72An≤Dm≤C73An,where constants C72 and C73 do not depend on m and n. Hence, considering condition (115), we conclude that condition f∈N(p,θ,r,λ,φ) is equivalent to the condition (123)Dm≤C74φ12nθ,where constant C74 does not depend on m and n.
Since1/2n<2/m+1<2/m, we get (124)φ12n≤C75φ2m+1≤C76φ2m,where constant C76 does not depend on m and n; and since 1/2n≥1/m+1≥1/2m, we get (125)φ12n≥C77φ12m≥C78φ1m,where constant C78 does not depend on m and n. This way, condition (123) is equivalent to the condition (126)Dm≤C79φ1mθ,where constant C79 does not depend on m.
This completes the proof of Theorem5.Remark 17.
Notice that another way of proving Theorems2 and 5 is presented in [2]. Our approach here is similar to that used in [16].



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