A NEW GENERALIZATION OF BOAS THEOREM FOR SOME LORENTZ SPACES Λq(ω)

Inequalities

Volume 12, Number 3 (2018), 619–633 doi:10.7153/jmi-2018-12-47

A NEW GENERALIZATION OF BOAS THEOREM FOR SOME LORENTZ SPACES Λq(ω)

AIGERIMKOPEZHANOVA, ERLANNURSULTANOV ANDLARS-ERIKPERSSON (Communicated by S. Li)

Abstract. LetΛq(ω),q>0 , denote the Lorentz space equipped with the (quasi) norm

f_Λq(ω):=

₁

0 (f^∗(t)ω(t))^qdt t

¹_q

for a function f on [0,1] and with ω positive and equipped with some additional growth properties. A generalization of Boas theorem in the form of a two-sided inequality is obtained in the case of both general regular systemΦ={ϕk}^∞k=1 and generalized LorentzΛq(ω)spaces.

1. Introduction

The following Hardy-Littlewood theorem is well known (see [26] and also [10], [4]):

THEOREMA. If f 0and f decreases, 1<p<∞,and anare the Fourier sine or cosine coefficients of f, then

∑

∞

n=1|an|^p<∞ if and only if

x^p−2f(x)^p∈L_p. This theorem can be extended as follows (see [4]):

THEOREMB. If f0 and f decreases,1<p<∞, −1/p<γ<1/p,then

∑

∞ n=1

n^−γp|an|^p<∞ converges if and only if

x^p−2x^pγ+p−2f(x)^p∈L_p. Here and in the sequel p=_(p−1)^p for p>1.

A characterization for the function f to belong to the Lorentz space L_pq was obtained by R. P. Boas in [4]. This result deals with trigonometric Fourier coefficients for the class of monotone functions and reads:

Mathematics subject classification(2010): 46E30, 42A16.

Keywords and phrases: Inequalities, two-sided inequalities, Fourier series, Boas theorem, generalized Lorentz spaces, regular systems, generalized monotone function.

c , Zagreb

Paper JMI-12-47 619

THEOREMC. If f0 and f decreases,1<p<∞, 1<q<∞,then f ∈Lpq if and only if{an} ∈l_pq.

Some other results which are related to the Hardy-Littlewood theorem for the class of monotone functions were obtained in [25], [3], [1], [24], [5], [15], [8], [9], [12] and [7].

Boas theorem was generalized and complemented in various ways also for more general Lorentz spaces Λq(ω) in 1974 by L.-E. Persson for the case when Φ={e^2πikx}^+∞_k=−∞ is trigonometric system (see. [20]–[23]). For example the following theorem was proved:

THEOREMD. Let p>0 andΦ={e^2πikt}^+∞_k=−∞be a trigonometrical system. Let ωbe a nonnegative function on[0,∞).If there exists a positive numberδ>0satisfying thatω(t)t^−δ is an increasing function of t andω(t)t^−1+δ is a decreasing function of t and if f is a nonnegative and a decreasing function on[0,¹₂],then

₁

0 (f^∗(t)ω(t))^pdt t

¹_p

<∞, if and only if

∞ k=1

∑

(kω

1 k

a^∗_k)^p1

n ¹_p

<∞,

where{a^∗_k}^∞_k=1 is the nonincreasing rerrangement of the sequence{an}^∞_k=1of Fourier coefficients of f with respect to the systemΦ.

The main aim of this paper is to derive the Boas theorem for the spaceΛq(ω)with respect to the regular system. Moreover, a new Boas type theorem for spaceΛq(ω)and for generalized monotone functions is proved and discussed.

The main results are formulated in Section 3. Note that the results in Theorem 1is obviously related to [11] but we have chosed to put also this result in this more general frame in English. The proofs can be found in Section 4 and in Section 2 we have presented some necessary preliminaries.

CONVENTIONS. The letterc(c1,c2,etc.)means a constant which does not depen- dent on the involved functions and it can be different in different occurences. Moreover, forC,D>0 the notationC∼D means that there exist positive constants a₁ and a₂ such thata₁DCa₂D.

2. Preliminaries

Let f be a measurable function on[0,1]and μ is Lebesgue measure. The nonin- creasing rerrangement f^∗ of a function f is defined as follows:

m(σ,f):=μ{x∈[0,1]: |f(x)|>σ}, f^∗(t):=in f{σ^: m(σ,f)t}.

Let 0<q∞andω be a nonnegative function on[0,1].The generalized Lorentz spacesΛq(ω)consists of the functions f on [0,1]such thatf_Λq(ω)<∞,where

f_Λq(ω):=

⎧⎪

⎨

⎪⎩

01(f^∗(t)ω(t))^{q dt}_t ¹_q

for 0<q<∞,

0t1sup f^∗(t)ω(t) forq=∞.

These spaces Λq(ω) coincide to the classical spaces L_pq in the caseω(t) =t^1p, 1<p<∞(see [16] and also e.g. [2]).

Let μ={μ(k)}k∈N be a sequence of positive number and the space λq(μ) con- sists of all sequencesa= {a_k}^∞_k=1 such thata_λq(μ)<∞,where

a_λq(μ):=

⎧⎪

⎨

⎪⎩ ∑^∞k=1

a^∗_kμ(k)_q1

k

¹_q

for 0<q<∞, sup

k a^∗_kμ(k) forq=∞.

Here, as usual,{a^∗_k}^∞_k=1is the nonincreasing rearrangement of the sequence{|a_k|}^∞_k=1. Let the function f be periodic with period 1 and integrable on [0,1]and letΦ= {ϕk}^∞_k=1be an orthonormal system on[0,1]. The numbers

a_k=a_k(f) = ₁

0 f(x)ϕk(x)dx, k∈N

are called the Fourier coefficients of the functions f with respect to the system Φ= {ϕk}^∞_k=1.

We say that the orthonormal system Φ={ϕk}^∞_k=1 is regular if there exists a con- stantB,such that

1) for every segmentefrom[0,1]andk∈Nit yields that _eϕk(x)dx

Bmin(|e|,1/k), 2) for every segmentwfromNandt∈(0,1] we have that

k∈w

∑

ϕk(·) _∗

(t)Bmin(|w|,1/t),

where(∑_k∈wϕk(·))^∗(t) as usual denotes the nonincreasing rerrangement of the func- tion∑k∈wϕk(x).

Examples of regular systems are all trigonometrical systems, the Walsh system and Prise’s system. In [17], [19], [18] some results were obtained with respect to the regular system using network space.

Let δ >0 be afixed parameter. Consider a nonnegative functionω(t) on[0,1].

We define the following classes:

A_δ :={ω(t): ω(t)t^−12−δis an increasing function and ω(t)t^−1+δis a decreasing function},

B_δ :={ω(t): ω(t)t^−δis an increasing function and ω(t)t^−1+δis a decreasing function}.

Then the classesAandBcan be defined as follows:

A=

δ>0

A_δ. and

B=

δ>0

B_δ.

For the proof of our main results we need the following Theorem:

THEOREME. LetΦ={ϕk}^∞_k=1 be a regular system and f^a.e.= ∑^∞

k=1a_kϕk. Let 1q∞.If ω belongs to the class B, then

₁ 0

f(t)ω(t)_qdt t

¹_q

c

∞

k=1

∑

(a^∗_kμ(k))^q1 k

¹_q , where f(t) =sup

ξt ξ1

ξ 0 f(s)ds

, μ(k) =kω¹_kand the constant c does not depend on f.

This is just a slight generalization of Theorem 2 in [14] (see also [11]). For the reader’s convenience we include a proof in Appendix 1.

We also need the following techniquel Lemma:

LEMMA1. Let 1q∞ and1h∞. If ω(t) belongs to the class B, then for any nonincreasing function f it yields that

⎛

⎝

∑

^∞

k=1

2^−k+1

2^−k (f(t)ω(t))^hdt t

^q_h⎞

⎠

1q

∼ ₁

0 (f(t)ω(t))^qdt t

¹_q

. (1)

Proof. First we prove the following equivalence:

⎛

⎝

∑

^∞

k=1

2^−k+1

2^−k (f(t)ω(t))^hdt t

^q_h⎞

⎠

1q

∼ ∞

k=1

∑

f(2^−k)ω(2^−k)_q¹_q

. (2)

Let

I_h:=

⎛

⎝

∑

^∞

k=1

2^−k+1

2^−k (f(t)ω(t))^hdt t

^q_h⎞

⎠

1q

=

⎛

⎝

∑

^∞

k=1

2^−k+1 2^−k

f(t)ω(t)t^−1+δt^1−δ_hdt t

^q_h⎞

⎠

1q

.

We use the fact thatω=ω(t) belong to the class B. This means that there exists δ, 0<δ <1,such thatω(t)t^−δ is an increasing function and ω(t)t^−1+δ is a decreasing function. Then we have:

I_h

⎛

⎝

∑

^∞

k=1

⎛

⎝f(2^−k)ω(2^−k)2^{−k(−1+δ)}

2^−k+1

2^−k t^(1−δ)hdt t

¹_h⎞

⎠

q⎞

⎠

1q

=c1

∞ k=1

∑

f(2^−k)ω(2^−k)2^k−kδ2^kδ−k_q¹_q

=c1

∞ k=1

∑

f(2^−k)ω(2^−k)_q¹_q . I_h=

⎛

⎝

∑

^∞

k=1

2^−k+1 2^−k

f(t)ω(t)t^−δt^δ_hdt t

^q_h⎞

⎠

1q

⎛

⎝

∑

^∞

k=1

⎛

⎝f(2^−k+1)ω(2^−k)2^kδ

2^−k+1 2^−k t^δh−1dt

¹_h⎞

⎠

q⎞

⎠

1q

=c₂ ∞

k=1

∑

f(2^−k+1)ω(2^−k)_q¹_q c₃

∞ k=1

∑

f(2^−k)ω(2^−k)_q¹_q .

Thus, (2) is proved, which, in particular means thatI_h1∼I_h2 for all h₁ andh₂.More- over, since f is nonincreasing andω∈B,it follows that

∞ k=1

∑

f(2^−k)ω(2^−k)q¹_q

∼ ₁

0 (f(t)ω(t))^qdt t

¹_q . In particular, (1) follows and the proof is complete.

3. Main results

The main results of this paper are the following generalizations of the Boas theo- rem:

THEOREM1. Let 1q∞and ω∈B.Let Φ={ϕk}^∞_k=1 be a regular system and let f^a.e.=∑^∞_k=1a_kϕk. If f is a nonnegative and a nonincreasing function, then

₁

0 (f(t)ω(t))^qdt t

¹_q

∼ ∞

k=1

∑

(a^∗_kμ(k))^q1 k

¹_q , whereμ(k) =kω(¹_k).

We say that a function f on [0,1] is generalized monotone if there exists some constantM>0 such that

|f(x)|M1 x

₀^xf(t)dt

, x>0.

Our next main result reads:

THEOREM2. Let 1q∞and ω∈A.Let Φ={ϕk}^∞_k=1 be a regular system and let f ^a.e.= ∑^∞k=1a_kϕk. If f is a nonnegative and a generalized monotone function, then

f_{Λq(ω,[0,1])}∼ ∞

k=1

∑

(a^∗_kμ(k))^q1 k

¹_q , whereμ(k) =kω(¹_k).

4. Proofs of the main results

Proof of Theorem1. The necessary part is similar to that in Theorem E. Indeed, since f is a nonincreasing function, then f(t) f(t), 0<t<1,so that

₁

0 (f(t)ω(t))^qdt t

¹_q

₁ 0

f(t)ω(t)_qdt t

¹_q

c

∞

k=1

∑

(a^∗_kμ(k))^q1 k

¹_q ,

where f(t) =sup

ξt ξ1

ξ

0 f(s)ds.We prove the sufficient condition. The conditionω(t)∈B implies that there existsδ>0 such that ω(t)t^−δ is an increasing andω(t)t^−1+δ is a decreasing function, i.e.μ(k)k^−δ is increasing andμ(k)k^−1+δ is decreasing. Then the following estimate holds:

1 k

∑

k n=1

μ^q(n)

n cμ^q(k)

k , k∈N.

Indeed,

1 k

∑

k n=1

μ^q(n)

n 1

kμ^q(k)k^−δ

∑

^k

n=1

1

n^1−δ ∼μ^q(k) k .

Next, we use Theorem 2.4.12 (ii) from [6] to conclude that the following equality holds:

λq(μ) =

λq(μ⁻¹k)

, for 1<q<∞, where

λq(μ^−1k)

is dual space for the space λq(μ). Hence, by appling the duality representation of the norm of a sequenceain the space λq(μ)(see [6]), we obtain that

a_λq(μ)= sup

b_λq(μ−1k)=1

∑

∞ k=1

a_kb_k.

Now we use Parseval’s formula andfind that a_λq(μ)= sup

b_λ

q(μ⁻¹k)=1

₁

0 f(t)g(t)dt

= sup

b_λq(μ−1k)=1

∑

∞ k=0

₂−k

2^−k−1f(t)g(t)dt

sup

b_λ

q(μ^−1k)=1

∑

∞ k=0

₂−k

2^−k−1f(t)g(t)dt

. (3) We apply the mean value theorem to the integral ₂²₋⁻k^k−1f(t)g(t)dt to conclude that there existsξ ^{from2−k−1}, 2^−k

such that

₂−k

2^−k−1 f(t)g(t)dt =

f(2^−k−1) ^ξ

2^−k−1g(t)dt f(2^−k−1)

_ξ

0 g(t)dt +

₂−k−1

0 g(t)dt

f(2^−k−1)

2^−k sup

s2^−k−2

1 s

₀^sg(t)dt

+2^−k sup

s2^−k−2

1 s

₀^sg(t)dt

=2·2^−k·f(2^−k−1)·g(2^−k−2), (4) whereg(2^−k−2) = sup

s2^−k−2

1s|₀^sg(t)dt|.

Thus, by inserting (4) in (3), we conclude that a_λq(μ)8 sup

b_λq(μ−1k)=1

∑

∞ k=0

2^−k−2f(2^−k−1)g(2^−k−2)

=8 sup

b_λ

q(μ^−1k)=1

∑

∞ k=0

2^−k−2

ω(2^−k−2)₋₁

g(2^−k−2)

·f(2^−k−1)ω(2^−k−2)

=8 sup

b_λ

q(μ⁻¹k)=1

∑

∞ k=2

2^−k

ω(2^−k)₋₁ g(2^−k)

·f(2^−k+1)ω(2^−k).

Next, by using H¨older’s inequality, we get that

a_λq(μ) c₁ sup

b_λ

q(μ^−1k)=1

∞ k=2

∑

2^−k

ω(2^−k)₋₁ g(2^−k)

_qq¹

× ∞

k=2

∑

f(2^−k+1)ω(2^−k)_q¹_q

c₁ sup

b_λ

q(μ^−1k)=1

∞ k=1

∑

2^−k

ω(2^−k)₋₁

g(2^−k+1) q_q¹

× ∞

k=1

∑

f(2^−k+1)ω(2^−k)_q¹_q

=c2 sup

b_λq(μ−1k)=1

⎛

⎝

∑

^∞

k=1

2^−k(1−δ)ω⁻¹(2^−k)

2^−k(1−δ) 2^−kg(2^−k+1) _q⎞

⎠

q1

× ∞

k=1

∑

2^kδω(2^−k)

2^kδ f(2^−k+1) ²

−k+1

2^−k

dt t

_q¹_q .

Since f(t)is a nonincreasing function for all 0<t<1 and ω(t)belongs to B, then there exists 0<δ<1 such thatω(t)t^−δ is an increasing andω(t)t^−1+δ is a decreasing function, we get that

a_λq(μ) c₃ sup

b_λ

q(μ^−1k)=1

⎛

⎝

∑

^∞

k=1

2^k(1−δ)

₂−k+1

2^−k t^1−δω⁻¹(t)g(t)dt _q⎞

⎠

q1

× ∞

k=1

∑

2^−kδ

₂−k+1

2^−k f(t)ω(t)t^−δdt t

_q¹_q

c₄ sup

b_λq(μ−1k)=1

⎛

⎝

∑

^∞

k=1

2^−k+1

2^−ktω⁻¹(t)g(t)dt t

q⎞

⎠

q1

∑

∞ k=1

2^−k+1

2^−k f(t)ω(t)dt t

q¹_q .

By now applying Lemma1, we obtain that a_λq(μ)c₅ sup

b_λq(μ−1k)=1

₁

0

tω⁻¹(t)g(t)qdt t

_q¹

· ₁

0 (f(t)ω(t))^qdt t

¹_q .

Furthermore, by using Theorem E, we obtain the following estimate

a_λq(μ) c₆ sup

b_λ

q(μ⁻¹k)=1

∞ k=1

∑

b^∗_kkμ⁻¹(k)q1 k

_q¹

· ₁

0 (f(t)ω(t))^qdt t

¹_q

=c₆ sup

b_λq(μ−1k)=1b_λq(μ−1k)· ₁

0 (f(t)ω(t))^qdt t

¹_q

=c_{6 1}

0 (f(t)ω(t))^qdt t

¹_q .

The proof is complete.

Proof of Theorem2. The conditionω(t)∈Aimplies that there existsδ >0 such thatω(t)t^−12−δ is an increasing function and ω(t)t^−1+δ is a decreasing function. The necessary condition follows in a similar way as in Theorem E. Indeed, letx>0 and

f^∗∗(x):=sup

|e|=x

1

|e|

e|f(t)|dt.

It is obvious that f^∗(x)f^∗∗(x).Since f is a generalized monotone function, it yields that

f^∗∗(x) = sup

|e|=x

1

|e|

e|f(t)|dtsup

|e|=x

1

|e|

ef(t)dt=1 x

_x

0 f(t)dt, where f(t) =sup

ξt _ξ

0 f(s)ds.

Thus, we obtain the following inequalities f_Λq(ω)f^∗∗_Λq(ω)M1

x _x

0 f(t)dt_Λq(ω). (5) We prove the following inequality

₁

0

ω(x)1

x _x

0 f(t)dt _qdx

x ¹_q

c

₁

0

f(t)ω(t)_qdt t

¹_q

. (6)

Chooseε ^{so that}−¹_q+1−δ<ε<−¹_q+1.We consider for anyx>0 _x

0 f(t)dt= ^x

0 f(t)t^εt^−εdt.

Next we use H¨older’s inequality and the fact thatε<−¹_q+1 tofind that _x

0 f(t)dt c_{1 x}

0

f(t)t^εqdt

¹_{q x}

0

t^−ε_q dt

_q¹

∼ _x

0

f(t)t^εqdt ¹_q

x^−ε+q1. Moreover,

I :=c_{2 1}

0

ω(x)x^−ε+q1−1 _{q x}

0

f(t)t^ε_q dt

dx x

¹_q

= c_{2 1}

0

f(t)t^ε_{q 1}

t x^−εq−1ω^q(x)dx x

dt

¹_q .

By now using the fact thatω(t)t^−12−δ is an increasing function, wefind that

Ic_{2 1}

0

f(t)t^εq

ω(t)t^−12−δ_{q 1}

1t

x^{εq+1−21q−δq}dx x

dt

¹_q . Taking into account thatε−¹_q+¹₂+δ,we obtain that

Ic_{3 1}

0

f(t)ω(t)_qdt t

¹_q .

Thus, we have proved the inequality (6). From (5) and (6) it follows that f_Λq(ω)c3f_Λq(ω).

By now applying Theorem E, we obtain that f_Λq(ω)c₄a_λq(μ).

Since each regular system is bounded orthonormal system, then the sufficient con- dition follows from Theorem 2 in [13].

The proof is complete.

5. Appendix 1

Proof of Theorem E. Assume that ω(t)belongs to the class B. This means that there exists δ >0 such that ω(t)t^−δ is an increasing function and ω(t)t^−1+δ is a decreasing function. Suppose that

∞ k=1

∑

(a^∗_kμ(k))^q1 k

¹_q

<∞

and f^a.e.=∑^∞_k=1a_kϕk. It yields that ₀^ξ f(s)ds

=

_ξ

0

∑

k∈N

a_kϕk(s)ds

∑

k∈N

|ak|

₀^ξϕk(s)ds

,for all ξ∈[0,1].

According to the regularity assumption we have that ₀^ξϕk(s)ds

Bmin

ξ,1 k

,k∈N.

Hence,

∑

∞ k=1|ak|

₀^ξϕk(s)ds

c₁

∑

^∞

k=1|ak|min

ξ,1 k

c₁

∑

^∞

k=1

a^∗_kmin

ξ,1 k

c₁

⎛

⎜⎝

1 ξ k=1

∑

a^∗_kξ+

∑

^∞

k=

1ξ a^∗_k1

k

⎞

⎟⎠.

Consequently,

₀^ξ f(s)ds c1

⎛

⎜⎝

1 ξ k=1

∑

a^∗_kξ+

∑

^∞

k=

ξ1

a^∗_k1 k

⎞

⎟⎠

and we have that ₁

0

f(t)ω(t)_qdt t

¹_q

c₁

⎛

⎜⎝ ₁

0

⎛

⎜⎝ω(t)sup

ξt

1 ξ

⎛

⎜⎝

1 ξ k=1

∑

a^∗_kξ+

∑

^∞

k=

ξ1

a^∗_k1 k

⎞

⎟⎠

⎞

⎟⎠

q

dt t

⎞

⎟⎠

1q

c₁

⎛

⎜⎝ ₁

0

⎛

⎜⎝ω(t)sup

ξt

1 ξ

⎛

⎜⎝

1 ξ k=1

∑

a^∗_kξ+ [¹t]

∑

k=

ξ1

a^∗_k·1 k+

∑

^∞

k=[¹t] a^∗_k1

k

⎞

⎟⎠

⎞

⎟⎠

q

dt t

⎞

⎟⎠

1q

c₁

⎛

⎜⎝ ₁

0

⎛

⎜⎝ω(t)sup

ξt

1 ξ

⎛

⎜⎝

1 ξ k=1

∑

a^∗_kξ+ [¹t]

∑

k=

ξ1

a^∗_k·ξ+

∑

^∞

k=[¹t] a^∗_k1

k

⎞

⎟⎠

⎞

⎟⎠

q

dt t

⎞

⎟⎠

1q

= c₁

⎛

⎝ ¹

0

⎛

⎝ω(t)sup

ξt

1 ξ

⎛

⎝[¹t]

k=1

∑

a^∗_kξ+

∑

^∞

k=[¹t] a^∗_k1

k

⎞

⎠

⎞

⎠

q

dt t

⎞

⎠

1q

= c₁

⎛

⎝ ¹

0

⎛

⎝ω(t)

⎛

⎝[¹t]

k=1

∑

a^∗_k+sup

ξt

1 ξ ^·

∑

∞ k=[¹t]

a^∗_k1 k

⎞

⎠

⎞

⎠

q

dt t

⎞

⎠

1q

c₁

⎛

⎝ ¹

0

⎛

⎝ω(t)

⎛

⎝[¹t]

k=1

∑

a^∗_k+1 t ·

∑

^∞

k=[¹t] a^∗_k1

k

⎞

⎠

⎞

⎠

q

dt t

⎞

⎠

1q

c₁

⎛

⎝ ¹

0

⎛

⎝ω(t)[¹t]

k=1

∑

a^∗_k

⎞

⎠

q

dt t

⎞

⎠

1q

+c₁

⎛

⎝ ¹

0

⎛

⎝ω(t)1 t

∑

∞ k=[¹t]

a^∗_k1 k

⎞

⎠

q

dt t

⎞

⎠

1q

:=c₁(I₁+I₂).

We considerfirstI₁.Choose a small numberε ^{such that 1}_q−1−δ<ε<¹_q−1.Since ω(t)t^−δ is an increasing function oft,it yields that

I₁=

⎛

⎝ ¹

0

⎛

⎝ω(t)[¹t]

k=1

∑

a^∗_k

⎞

⎠

q

dt t

⎞

⎠

1q

=

⎛

⎝ ¹

0

⎛

⎝ω(t)t^−δ t^−δ

[¹t]

k=1

∑

a^∗_k

⎞

⎠

q

dt t

⎞

⎠

1q

⎛

⎝ ¹

0

⎛

⎝t^δ[¹t]

k=1

∑

ω 1

k 1

k _−δ

a^∗_k

⎞

⎠

q

dt t

⎞

⎠

1q

=

∞ 1

t^−δ

∑

^t

k=1ω 1

k 1

k _−δ

a^∗_{k q}

dt t

¹_q

∼ ∞

n=1

∑

n^−δ

∑

ⁿ

k=1ω 1

k 1

k _−δ

a^∗_k q

1 n

¹_q . Next we use H¨older’s inequality and the fact thatε>¹_q−1−δ ^{tofind that}

I₁c₂

⎛

⎝

∑

^∞

n=1

⎛

⎝n^−δ n

k=1

∑

ω

1 k

k^−εa^∗_k

_q¹_q

∑

n k=1

k^(δ+ε)q _q¹⎞

⎠

q

1 n

⎞

⎠

1q

∼ ∞

n=1

∑

n^(−δ)qn^(δ+ε)q+qq 1 n

∑

n k=1

ω

1 k

k^−εa^∗_k

_q¹_q . Here we interchange the order of summation andfind that

I₁c₂ ∞

k=1

∑

ω

1 k

k^−εa^∗_k

_{q ∞}

n=k

∑

n^{εq+q−2 1}_q

. Furthemore, by also using thatε<¹_q−1,we have that

I1c3

∞ k=1

∑

ω

1 k

ka^∗_k

_q1 k

¹_q

=c3

∞

k=1

∑

(μ(k)a^∗_k)^q1 k

¹_q

. (7)

Next, we estimate I₂ in a similar way. Choose ε ^{such that} −1+¹_q<ε<−1+

1q+δ.By now using the growth properties ofω(t)wefind that

I₂=

⎛

⎝ ¹

0

⎛

⎝ω(t)1 t

∑

∞ k=[¹t]

a^∗_k1 k

⎞

⎠

qdt t

⎞

⎠

1q

=

⎛

⎝ ¹

0

⎛

⎝ω(t)t^−1+δ t^−1+δ

1 t

∑

∞ k=[¹t]

a^∗_k k

⎞

⎠

q

dt t

⎞

⎠

1q

⎛

⎝ ¹

0

⎛

⎝t^−δ

∑

^∞

k=[¹t]ω 1

k

k^1−δa^∗_k k

⎞

⎠

q

dt t

⎞

⎠

1q

=

∞ 1

t^δ

∑

^∞

k=tω 1

k

k^1−δa^∗_k k

_q dt

t ¹_q

∼ ∞

n=1

∑

n^δ

∑

^∞

k=nω 1

k

k^1−δa^∗_k k

q

1 n

¹_q .

Next we use H¨older’s inequality and the fact thatε<−1+¹_q+δ ^{tofind that}

I₂c₄

⎛

⎝

∑

^∞

n=1

⎛

⎝n^δ ∞

k=n

∑

a^∗_kω

1 k

k^−ε

q¹_q

∑

∞ k=n

k^(−δ+ε)q _q¹⎞

⎠

q

1 n

⎞

⎠

1q

∼ ∞

n=1

∑

n^εq+q−2

∑

^∞

k=n

a^∗_kω

1 k

k^−ε

_q¹_q

= ∞

k=1

∑

a^∗_kω

1 k

k^−ε

_{q k}

n=1

∑

n^{εq+q−2 1}_q

.

By interchanging the order of summation and using the fact that ε>−1+¹_q, we obtain that

I₂c₅ ∞

k=1

∑

(a^∗_kμ(k))^q1 k

¹_q

. (8)

To complete the proof we just combine (7) with (8).

Acknowledgement. This publication was supported by the Ministry of Education and Science of the Russian Federation (the Agreement number 02.a03.21.0008), by the Ministry of Education and Science of the Republic of Kazakhstan (the Agreement numbers AP051 32071 and AP051 32590).