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volume 3, issue 1, article 10, 2002.

Received 22 September, 1999;

received in final form 17 October, 2001;

accepted 18 October, 2001.

Communicated by:A.G. Babenko

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Journal of Inequalities in Pure and Applied Mathematics

SOME RESULTS ON L¹ APPROXIMATION OF THE r-TH DERIVATE OF FOURIER SERIES

Z. TOMOVSKI

Faculty of Mathematical and Natural Sciences P.O. Box 162

91000 Skopje MACEDONIA

EMail:tomovski@iunona.pmf.ukim.edu.mk

2000c Victoria University ISSN (electronic): 1443-5756 005-99

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Some Results on L¹-Approximation of ther-th

Derivate of Fourier Series Živorad Tomovski

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J. Ineq. Pure and Appl. Math. 3(1) Art. 10, 2002

http://jipam.vu.edu.au

Abstract

In this paper we obtain the conditions for L¹-convergence of ther-th deriva- tives of the cosine and sine trigonometric series. These results are extensions of corresponding Sidon’s and Telyakovskii’s theorems for trigonometric series (case:r= 0).

2000 Mathematics Subject Classification:26D15, 42A20.

Key words:L¹-approximation, Fourier series, Sidon-Telyakovskii class, Telyakovskii inequality

The author would like to thank the referee for his suggestions and encouragement.

1. Introduction

Let

f(x) = a₀ 2 +

∞

X

n=1

a_ncosnx , (1.1)

g(x) =

∞

X

n=1

a_nsinnx (1.2)

be the cosine and sine trigonometric series with real coefficients.

Let∆a_n=a_n−a_n+1,n ∈ {0,1,2,3, . . .}. The Dirichlet’s kernel, conjugate Dirichlet’s kernel and modified Dirichlet’s kernel are denoted respectively by

D_n(t) = 1 2 +

n

X

k=1

coskt= sin n+ ¹₂ t 2 sin₂^t , D˜_n(t) =

n

X

k=1

sinkt= cos₂^t−cos n+ ¹₂ t 2 sin₂^t , Dn(t) = −1

2ctgt

2+ ˜Dn(t) =−cos n+ ¹₂ t 2 sin₂^t . Let

E_n(t) = 1 2 +

n

X

k=1

e^ikt and E−n(t) = 1 2+

n

X

k=1

e^−ikt.

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Then ther-th derivativesDn^(r)(t)andD˜^(r)n (t)can be written as 2D_n^(r)(t) = E_n^(r)(t) +E_−n^(r)(t), (1.3)

2iD˜_n^(r)(t) = E_n^(r)(t)−E_−n^(r)(t). (1.4)

In [2], Sidon proved the following theorem.

Theorem 1.1. Let {α_n}^∞_n=1 and{p_n}^∞_n=1 be sequences such that|α_n| ≤ 1, for everynand letP∞

n=1|p_n|converge. If

(1.5) a_n=

∞

X

k=n

p_k k

k

X

l=n

α_l, n∈N then the cosine series (1.1) is the Fourier series of its sumf.

Several authors have studied the problem ofL¹−convergence of the series (1.1) and (1.2).

In [4] Telyakovskii defined the following class ofL¹-convergence of Fourier series. A sequence {a_k}^∞_k=0 belongs to the class S, or{a_k} ∈ S ifa_k → 0as k → ∞ and there exists a monotonically decreasing sequence {A_k}^∞_k=0 such thatP∞

k=0A_k<∞and|∆a_k| ≤A_k for allk.

The importance of Telyakovskii’s contributions are twofold. Firstly, he ex- pressed Sidon’s conditions (1.5) in a succinct equivalent form, and secondly, he showed that the class S is also a class ofL¹-convergence. Thus, the classSis usually called the Sidon–Telyakovskii class.

In the same paper, Telyakovskii proved the following two theorems.

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Theorem 1.2. [4]. Let the coefficients of the seriesf(x)belong to the classS.

Then the series is a Fourier series and the following inequality holds:

Z π 0

|f(x)|dx≤M

∞

X

n=0

A_n, whereM is a positive constant, independent onf.

Theorem 1.3. [4]. Let the coefficients of the seriesg(x)belong to the classS.

Then the following inequality holds forp= 1,2,3, . . . Z π

π/(p+1)

|g(x)|dx=

p

X

n=1

|an| n +O

∞

X

n=1

A_n

! . In particular,g(x)is a Fourier series iffP∞

n=1

|a_n| n <∞.

In [5], we extended the Sidon–Telyakovskii class S = S₀, i.e., we defined the class S_r, r = 1,2,3, . . . as follows: {a_k}^∞_k=1 ∈ S_r if a_k → 0 as k →

∞ and there exists a monotonically decreasing sequence {A_k}^∞_k=1 such that P∞

k=1k^rA_k <∞and|∆a_k| ≤A_kfor allk.

We note that byA_k ↓0andP∞

k=1k^rA_k <∞,we get

(1.6) k^r+1A_k =o(1), k → ∞.

It is trivially to see thatSr+1 ⊂Sr for allr = 1,2,3, .... Now, let{an}^∞_n=1 ∈ S₁. For arbitrary real numbera₀, we shall prove that sequence{a_n}^∞_n=0 belongs to S₀. We defineA₀ = max(|∆a₀|, A₁). Then|∆a₀| ≤ A₀, i.e. |∆a_n| ≤ A_n, for alln ∈ {0,1,2, ...}and{An}^∞_n=0 is monotonically decreasing sequence.

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On the other hand,

∞

X

n=0

A_n≤A₀+

∞

X

n=1

nA_n <∞.

Thus, {a_n}^∞_n=0 ∈ S₀, i.e. S_r+1 ⊂ S_r, for allr = 0,1,2, . . .. The next example verifies that the implication

{a_n} ∈S_r+1 ⇒ {a_n} ∈S_r, r = 0,1,2, . . . is not reversible.

Example 1.1. For n = 0,1,2,3, ...definea_n = P∞ k=n+1

1

k². Then a_n → 0as n → ∞and for n = 0,1,2,3, . . ., ∆a_n = _(n+1)¹ 2. Firstly we shall show that {a_n}^∞_n=1 ∈/ S₁.

Let{An}^∞_n=1 is an arbitrary positive sequence such thatA ↓ 0and∆an =

|∆a_n| ≤A_n. However,P∞

n=1nA_n≥P∞ n=1

n

(n+1)² is divergent, i.e. {a_n}∈/ S₁. Now, for alln = 0,1,2, . . .letAn = _(n+1)¹ 2. ThenAn↓ 0,|∆an| ≤ An and P∞

n=0A_n =P∞ n=1

1

n² <∞, i.e.{a_n}^∞_n=0 ∈S₀.

Our next example will show that there exists a sequence{a_n}^∞_n=1 such that {a_n}^∞_n=1 ∈S_r but{a_n}^∞_n=1 ∈/ S_r+1, for allr= 1,2,3, . . ..

Namely, for alln= 1,2,3, . . .leta_n =P∞ k=n

1

k^r+2. Thena_n →0asn → ∞ and for n = 1,2,3, . . . , ∆a_n = _nr+2¹ . Let {A_n}^∞_n=1 is an arbitrary positive sequence such thatAn↓0and∆an =|∆an| ≤An. However,

∞

X

n=1

n^r+1An ≥

∞

X

n=1

n^r+1 1 n^r+2 =

∞

X

n=1

1 n

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is divergent, i.e. {a_n} ∈/ S_r+1. On the other hand, for all n = 1,2, . . . let An = _nr+2¹ . ThenAn ↓ 0, |∆an| ≤ An andP∞

n=1n^rAn = P∞ n=1

1

n² < ∞, i.e.

{a_n} ∈S_r.

In the same paper [5] we proved the following theorem.

Theorem 1.4. [5]. Let the coefficients of the series (1.1) belong to the class S_r, r = 0,1,2, .... Then the r−th derivative of the series (1.1) is a Fourier series of somef^(r) ∈L¹(0, π)and the following inequality holds:

Z π 0

f^(r)(x)

dx≤M

∞

X

n=1

n^rAn, where0< M =M(r)<∞.

This is an extension of the Telyakovskii Theorem1.2.

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2. Results

In this paper, we shall prove the following main results.

Theorem 2.1. A null sequence{an}belongs to the classSr, r = 0,1,2, . . . if and only if it can be represented as

(2.1) a_n=

∞

X

k=n

p_k k

k

X

l=n

α_l, n ∈N

where{α_n}^∞_n=1 and{p_n}^∞_n=1 are sequences such that|α_n| ≤1,for allnand

∞

X

n=1

n^r|pn|<∞.

Corollary 2.2. Let{α_n}^∞_n=1 and{p_n}^∞_n=1 be sequences such that|α_n| ≤ 1,for everynand letP∞

n=1n^r|p_n|<∞,r= 0,1,2, . . .. If a_n=

∞

X

k=n

p_k k

k

X

l=n

α_l, n∈N

then ther−th derivate of the series (1.1) is a Fourier series of somef^(r) ∈L¹. Theorem 2.3. Let the coefficients of the seriesg(x)belong to the classSr, r= 0,1,2, . . . Then ther-th derivate of the series (1.2) converges to a function and form= 1,2,3, . . .the following inequality holds:

(∗)

Z π π/(m+1)

g^(r)(x)

dx≤M

m

X

n=1

|an| ·n^r−1+

∞

X

n=1

n^rAn

! ,

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where

0< M =M(r)<∞. Moreover, if P∞

n=1n^r−1|a_n|< ∞, then ther-th derivate of the series (1.2) is a Fourier series of someg^(r) ∈L¹(0, π)and

Z π 0

g^(r)(x)

dx≤M

∞

X

n=1

|a_n| ·n^r−1+

∞

X

n=1

n^rA_n

!

Corollary 2.4. Let the coefficients of the series g(x) belong to the class S_r, r ≥1. Then the following inequality holds:

Z π 0

g^(r)(x)

dx ≤M

∞

X

n=1

n^rA_n, where 0< M =M(r)<∞.

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3. Lemmas

For the proof of our new theorems we need the following lemmas.

The following lemma proved by Sheng, can be reformulated in the following way.

Lemma 3.1. [1] Letr be a nonnegative integer and x ∈ (0, π],where n ≥ 1.

Then

D_n^(r)(x) =

r

X

k=0

n+ ¹₂_k sin

n+ ¹₂

x+ ^kπ₂

sin ^x₂r+1−k ϕ_k(x),

where ϕ_r ≡ ¹₂ and ϕ_k, k = 0,1,2, . . . , r − 1 denotes various entire 4π – periodic functions ofx,independent ofn. More precisely,ϕk,k = 0,1,2, . . . , r are trigonometric polynomials of ^x₂.

Lemma 3.2. Let {αj}^k_j=0 be a sequence of real numbers. Then the following relation holds forν = 0,1,2, . . . , randr= 0,1,2, . . .

U_k = Z π

π/(k+1)

k

X

j=0

α_j j+ ¹₂ν

sin

j+ ¹₂

x+^ν+3₂ π sin ^x₂r+1−ν

dx

= O



(k+ 1)^r−ν+¹²

k

X

j=0

α²_j(j+ 1)^2ν

!^1/2

. Proof. Applying first Cauchy–Buniakowski inequality, yields

U_k ≤

"

Z π π/(k+1)

dx

sin ^x₂2(r+1−ν)

#1/2

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×





 Z π

π/(k+1)

" _k X

j=0

α_j

j +1 2

ν

sin

j+1 2

x+ (ν+ 3)π 2

#² dx







1/2

. Since

Z π π/(k+1)

dx

sin ^x₂2(r+1−ν) ≤ π^2(r+1−ν) Z π

π/(k+1)

dx x^2(r+1−ν)

≤ π(k+ 1)^{2(r+1−ν)−1} 2(r+ 1−ν)−1

≤ π(k+ 1)^{2(r+1−ν)−1}, we have

U_k ≤

π(k+ 1)^{2(r+1−ν)−1}^1/2

×





 Z π

0

" _k X

j=0

α_j

j+1 2

ν

sin

j+ 1 2

x+ ν+ 3 2 π

#² dx







1/2

≤

2π(k+ 1)^{2(r+1−ν)−1}1/2

×





 Z 2π

0

" _k X

j=0

α_j

j+ 1 2

ν

sin

(2j+ 1)t+ν+ 3 2 π

#² dt







1/2

. Then, applying Parseval’s equality, we obtain:

U_k≤h

2π(k+ 1)^{2(r+1−ν)−1}i1/2" _k X

j=0

|α_j|²(j + 1)^2ν

#1/2

.

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Finally,

U_k =O



(k+ 1)^r−ν+¹²

k

X

j=0

α²_j(j+ 1)^2ν

!^1/2

.

Lemma 3.3. Letr ∈ {0,1,2,3, . . .}and{α_k}ⁿ_k=0 be a sequence of real num- bers such that |αk| ≤ 1, for all k. Then there exists a finite constant M = M(r)>0such that for anyn ≥0

(∗∗)

Z π π/(n+1)

n

X

k=0

αkD^(r)_k (x)

dx≤M ·(n+ 1)^r+1.

Proof. Similar to Lemma3.1it is not difficult to proof the following equality D^(r)_n (x) =

r

X

k=0

n+¹₂_k sin

n+¹₂

x+^k+3₂ π

sin ^x₂^r+1−k ϕ_k(x),

whereϕ_kdenotes the same various4π-periodic functions ofx,independent of n.

Now, we have:

Z π π/(n+1)

n

X

k=0

α_kD^(r)_k (x)

dx

≤ Z π

π/(n+1)

n

X

j=0

α_j

r

X

ν=0

j+ ¹₂_ν sin

j+ ¹₂

x+^ν⁺³₂ π sin ^x₂r+1−ν ϕ_ν(x)

!

dx.

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Sinceϕ_ν are bounded, we have:

Z π π/(n+1)

n

X

j=0

α_j j+¹₂ν

sin

j+ ¹₂

x+ ^ν+3₂ π sin ^x₂r+1−ν ϕ_ν(x)

dx ≤K U_n, whereU_nis the integral as in Lemma3.2, andK =K(r)is a positive constant.

Applying Lemma3.2, to the last integral, we obtain:

Z π π/(n+1)

n

X

j=0

α_j j +¹₂ν

sin

j+¹₂

x+^ν+3₂ π sin ^x₂r+1−ν ϕ_ν(x)

dx

=O



(n+ 1)^r−ν+¹²

n

X

j=0

α_j²(j + 1)^2ν

!1/2



=O

(n+ 1)^r−ν+¹²(n+ 1)^ν+¹²

=O (n+ 1)^r+1 . Finally the inequality (∗∗) is satisfied.

Remark 3.1. Forr = 0,we obtain the Telyakovskii type inequality, proved in [4].

Lemma 3.4. Let rbe a non-negative integer. Then for all0 < |t| ≤ π and all n ≥1the following estimates hold:

(i)

E_−n^(r)(t)

≤ ⁴ⁿ_|t|^r^π, (ii)

D˜n^(r)(t)

≤ ⁴ⁿ_|t|^r^π,

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(iii)

D^(r)_n (t)

≤ ⁴ⁿ_|t|^r^π +O 1

|t|^r+1

.

Proof. (i) The caser = 0is trivial. Really,

|E_n(t)| ≤ |D_n(t)|+|D˜_n(t)| ≤ π 2|t| + π

|t| = 3π 2|t| < 4π

|t|,

|E−n(t)|=|En(−t)|< 4π

|t| . Letr ≥1. Applying the Abel’s transformation, we have:

E_n^(r)(t) =i^r

n

X

k=1

k^re^ikt=i^r

"_n−1 X

k=1

∆(k^r)

E_k(t)−1 2

+n^r

E_n(t)−1 2

#

|E_n^(r)(t)| ≤

n−1

X

k=1

[(k+ 1)^r−k^r] 1

2+|E_k(t)|

+n^r

1

2+|E_n(t)|

≤ π

2|t| + 3π 2|t|

(_n−1 X

k=1

[(k+ 1)^r−k^r] +n^r )

= 4πn^r

|t| . SinceE_−n^(r)(t) =En^(r)(−t), we obtain

E_−n^(r)(t)

≤ ⁴ⁿ_|t|^r^π. (ii) Applying the inequality(i), we obtain

D˜_n^(r)(t) =

iD˜^(r)_n (t) ≤ 1

2

E_n^(r)(t) +1

2

E_−n^(r)(t)

≤ 4n^rπ

|t| .

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(iii) We note that

ctg₂^t(r) = O

1

|t|^r+1

. Applying the inequality (ii), we obtain

|D^(r)_n (t)| ≤ |D˜^(r)_n (t)|+1 2

ctgt

2 (r)

≤ 4n^rπ

|t| +O 1

|t|^r+1

.

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4. Proofs of the Main Results

Proof of Theorem2.1. Let (2.1) hold. Then

∆a_k=α_k

∞

X

m=k

p_m m , and we denote

A_k =

∞

X

m=k

|p_m| m . Since|α_k| ≤1,we get

|∆a_k| ≤ |α_k|

∞

X

m=k

|p_m|

m ≤A_k,for allk . However,

∞

X

k=1

k^rA_k =

∞

X

k=1

k^r

∞

X

m=k

|p_m|

m =

∞

X

m=1

|p_m| m

m

X

k=1

k^r≤

∞

X

m=1

m^r|p_m|<∞, andA_k ↓0i.e.{a_k} ∈S_r.

Now, if{a_k} ∈S_r,we putα_k = ^∆a_A^k

k andp_k =k(A_k−A_k+1). Hence|αk| ≤1,and by (1.6) we get:

∞

X

k=1

k^r|p_k|=

∞

X

k=1

k^r+1(A_k−A_k+1)≤

∞

X

k=1

(r+ 1)k^rA_k<∞.

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Finally,

ak =

∞

X

i=k

∆ai =

∞

X

i=k

αiAi =

∞

X

i=k

αi

∞

X

m=i

∆Am

=

∞

X

i=k

α_i

∞

X

m=i

p_m m =

∞

X

m=k

p_m m

m

X

i=k

α_i, i.e. (2.1) holds.

Proof of Corollary2.2. The proof of this corollary follows from Theorems1.4 and2.1.

Proof of Theorem2.3. We suppose thata₀ = 0andA₀ = max (|a₁|, A₁). Applying the Abel’s transformation, we have:

(4.1) g(x) =

∞

X

k=0

∆a_kD_k(x), x∈(0, π].

Applying Lemma 3.4 (iii), we have that the series P∞

k=1∆akD^(r)_k (x) is uni- formly convergent on any compact subset of[ε, π], whereε >0.

Thus, representation (4.1) implies that

g^(r)(x) =

∞

X

k=0

∆a_kD^(r)_k (x).

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Then,

π

Z

π/(m+1)

|g^(r)(x)|dx≤

m

X

j=1

Z π/j π/(j+1)

j−1

X

k=0

∆a_kD^(r)_k (x)

dx

+O

m

X

j=1

Z π/j π/(j+1)

∞

X

k=j

∆a_kD^(r)_k (x)

dx

! . Let

I₁ =

m

X

j=1

Z π/j π/(j+1)

j−1

X

k=0

∆a_kD^(r)_k (x)

dx ,

I₂ =

m

X

j=1

Z π/j π/(j+1)

∞

X

k=j

∆a_kD^(r)_k (x)

dx . Sincectg^x₂ = _x²+P∞

n=1 4x

x²−4n²π² (see [3]) it is not difficult to proof the following estimate

ctgx

2 (r)

= 2(−1)^rr!

x^r+1 +O(1), x∈(0, π] . Thus

D^(r)_n (x) = (−1)^r+1r!

x^r+1 +O (n+ 1)^r+1

, x∈(0, π]

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Hence,

I₁ =r!

m

X

j=1

j−1

X

k=0

∆a_k

Z π/j π/(j+1)

dx x^r+1 +O

m

X

j=1

"_j−1 X

k=0

|∆a_k|(k+ 1)^r+1

#Z π/j π/(j+1)

dx

!

=O_r

m

X

j=1

|a_j|j^r−1

! +O

m

X

j=1 j−1

X

k=0

(k+ 1)^r+1|∆ak| j(j+ 1)

! , whereO_rdepends onr. However,

m

X

j=1 j−1

X

k=0

(k+ 1)^r+1|∆ak| j(j+ 1) =

m

X

j=1

1 j(j+ 1)

j−1

X

k=0

(k+ 1)^r+1|∆a_k|

≤

∞

X

k=0

(k+ 1)^r+1|∆a_k|

∞

X

j=k+1

1 j(j+ 1)

=

∞

X

k=0

(k+ 1)^r|∆ak|

= |∆a₀|+

∞

X

k=1

(k+ 1)^r|∆a_k|

≤ |a₁|+ 2^r

∞

X

k=1

k^r|∆a_k|

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≤

∞

X

k=1

|∆a_k|+ 2^r

∞

X

k=1

k^rA_k≤(1 + 2^r)

∞

X

k=1

k^rA_k. Thus,

m

X

j=1 j−1

X

k=0

|∆a_k|(k+ 1)^r+1 j(j + 1) =Or

∞

X

k=1

k^rAk

! , whereO_rdepends onr.

Therefore,

I₁ =O_r

m

X

j=1

|a_j|j^r−1

! +O_r

∞

X

k=1

k^rA_k

! . Application of Abel’s transformation, yields

∞

X

k=j

∆a_kD^(r)_k (x) =

∞

X

k=j

∆A_k

k

X

i=0

∆a_i

A_i D^(r)_i (x)−A_j

j−1

X

i=0

∆a_i

A_i D^(r)_i (x). Let us estimate the second integral:

I₂ ≤

m

X

j=1

" _∞ X

k=j

(∆A_k) Z π

π/(j+1)

k

X

i=0

∆a_i Ai

D^(r)_i (x)

+A_j Z π/j

π/(j+1)

j−1

X

i=0

∆a_i Ai

D^(r)_i (x)

dx

# .

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Applying the Lemma3.3, we have:

(4.2) J_k = Z π

π/(j+1)

k

X

i=0

∆a_i

A_i D^(r)_i (x)

dx=O_r (k+ 1)^r+1 , whereO_rdepends onr. Then, by Lemma3.4(iii),

Z π/j π/(j+1)

j−1

X

i=0

∆a_i Ai

D^(r)_i (x)

dx|

=O j^r

j−1

X

i=0

|∆a_i| A_i

Z π/j π/(j+1)

dx x

!!

+O

j−1

X

i=0

|∆a_i| A_i

Z π/j π/(j+1)

dx x^r+1

!

=O(j^r) +O_r(j^r) = O_r(j^r) (4.3)

whereO_rdepends onr. However, by (4.2), (4.3) and (1.6), we have I₂ ≤

∞

X

k=1

(∆A_k)J_k+O_r

∞

X

j=1

j^rA_j

!

= Or(1)

∞

X

k=1

(∆Ak)(k+ 1)^r+1+Or

∞

X

j=1

j^rAj

!

= O_r

∞

X

j=1

j^rA_j

! . Finally, the inequality (∗) is satisfied.

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Proof of Corollary2.4. By the inequalities

m

X

n=1

|an| ·n^r−1 ≤

∞

X

n=1

n^r−1

∞

X

k=n

|∆ak|

≤

∞

X

n=1

n^r−1

∞

X

k=n

A_k

=

∞

X

k=1

A_k

k

X

n=1

n^r−1 ≤

∞

X

k=1

k^rA_k,

and Theorem2.3, we obtain:

Z π 0

|g^(r)(x)|dx≤M

∞

X

n=1

n^rA_n

! , where0< M =M(r)<∞.

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References

[1] SHENG SHUYUN, The extensions of the theorem of C. V. Stanojevic and V.B. Stanojevic, Proc. Amer. Math. Soc., 110 (1990), 895–904.

[2] S. SIDON, Hinrehichende Bedingungen fur den Fourier-Charakter einer tri- gonometrischen Reihe, J. London Math. Soc., 72 (1939), 158–160.

[3] M.R. SPIEGEL, Theory and Problems of Complex Variables, Singapore (1988), p.175, 192.

[4] S.A. TELYAKOVSKII, On a sufficient condition of Sidon for the integra- bility of trigonometric series, Math. Notes, 14 (1973), 742–748.

[5] Ž. TOMOVSKI, An extension of the Sidon–Fomin inequality and applica- tions, Math. Ineq. & Appl., 4(2) (2001), 231–238.

[6] Ž. TOMOVSKI, Some results onL¹-approximation of the r-th derivate of Fourier series, RGMIA Research Report Collection, 2(5) (1999), 709–717.

ONLINE:http://rgmia.vu.edu.au/v2n5.html