In this paper we obtain the conditions forL1-convergence of ther-th derivatives of the cosine and sine trigonometric series

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Volume 3, Issue 1, Article 10, 2002

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

ŽIVORAD TOMOVSKI

FACULTY OFMATHEMATICAL ANDNATURALSCIENCES

P.O. BOX162, 91000 SKOPJE, MACEDONIA.

tomovski@iunona.pmf.ukim.edu.mk

Received 22 September, 1999; accepted 18 October, 2001.

Communicated by A. Babenko

ABSTRACT. In this paper we obtain the conditions forL¹-convergence of ther-th derivatives 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).

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

2000 Mathematics Subject Classification. 26D15, 42A20.

1. INTRODUCTION

Let

f(x) = a₀ 2 +

∞

X

n=1

a_ncosnx , (1.1)

g(x) =

∞

X

n=1

ansinnx (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

Dn(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 ,

ISSN (electronic): 1443-5756

c 2002 Victoria University. All rights reserved.

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

005-99

D_n(t) =−1 2ctgt

2+ ˜D_n(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.

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

2iD˜^(r)_n (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 let P∞

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 se- quence{ak}^∞_k=0 belongs to the class S, or{ak} ∈ S if ak → 0as k → ∞ and there exists a monotonically decreasing sequence{A_k}^∞_k=0 such thatP∞

k=0A_k < ∞and|∆a_k| ≤ A_kfor all k.

The importance of Telyakovskii’s contributions are twofold. Firstly, he expressed Sidon’s conditions (1.5) in a succinct equivalent form, and secondly, he showed that the classS 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.

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 series g(x) belong to the class S. Then the following inequality holds forp= 1,2,3, . . .

Z π π/(p+1)

|g(x)|dx=

p

X

n=1

|a_n| n +O

∞

X

n=1

A_n

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

n=1

|an| 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_rifa_k→0ask→ ∞and there exists a monotonically decreasing sequence{A_k}^∞_k=1 such thatP∞

k=1k^rA_k<∞and|∆a_k| ≤A_k for 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 that S_r+1 ⊂ S_r for all r = 1,2,3, .... Now, let {a_n}^∞_n=1 ∈ S₁. For arbitrary real numbera₀, we shall prove that sequence{a_n}^∞_n=0 belongs toS₀. We defineA₀ =

max(|∆a₀|, A₁). Then|∆a₀| ≤ A₀, i.e. |∆a_n| ≤ A_n, for alln ∈ {0,1,2, ...}and{A_n}^∞_n=0 is monotonically decreasing sequence.

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. Forn = 0,1,2,3, ...definea_n = P∞ k=n+1

1

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

Let {A_n}^∞_n=1 is an arbitrary positive sequence such that A ↓ 0 and ∆a_n = |∆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 all n = 0,1,2, . . . letAn = _(n+1)¹ 2. ThenAn ↓ 0, |∆an| ≤ An andP∞

n=0An = 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_rbut {an}^∞_n=1 ∈/ Sr+1, for allr= 1,2,3, . . ..

Namely, for all n = 1,2,3, . . . let an = P∞ k=n

1

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

∞

X

n=1

n^r+1A_n≥

∞

X

n=1

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

∞

X

n=1

1 n

is divergent, i.e. {a_n} ∈/ S_r+1. On the other hand, for all n = 1,2, . . . letA_n = _nr+2¹ . Then A_n ↓0,|∆a_n| ≤A_nandP∞

n=1n^rA_n=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 classSr, r= 0,1,2, ....

Then ther−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^rA_n,

where0< M =M(r)<∞.

This is an extension of the Telyakovskii Theorem 1.2.

2. RESULTS

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

Theorem 2.1. A null sequence{a_n}belongs to the classS_r,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=1and{p_n}^∞_n=1are sequences such that|α_n| ≤1,for allnand

∞

X

n=1

n^r|p_n|<∞.

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

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 series g(x) belong to the class S_r, 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

! ,

where

0< M =M(r)<∞. Moreover, ifP∞

n=1n^r−1|an|<∞, 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 classS_r, r ≥ 1. Then the following inequality holds:

Z π 0

g^(r)(x)

dx≤M

∞

X

n=1

n^rA_n,

where0< M =M(r)<∞.

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] Letrbe a nonnegative integer andx∈(0, π],wheren ≥1.Then D^(r)_n (x) =

r

X

k=0

n+ ¹_2k sin

n+¹₂

x+ ^kπ₂

sin ^x₂r+1−k ϕ_k(x),

whereϕ_r ≡ ¹₂ andϕ_k,k = 0,1,2, . . . , r−1denotes various entire4π– periodic functions of x,independent ofn. More precisely,ϕk,k = 0,1,2, . . . , rare trigonometric polynomials of ^x₂.

Lemma 3.2. Let{α_j}^k_j=0be a sequence of real numbers. Then the following relation holds for ν = 0,1,2, . . . , r andr= 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 ^x₂^r+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

Uk ≤

"

Z π π/(k+1)

dx

sin ^x₂2(r+1−ν)

#1/2

×





 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)−11/2}

×





 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 .

Finally,

U_k=O



(k+ 1)^r−ν+12

k

X

j=0

α²_j (j + 1)^2ν

!^1/2

 .

Lemma 3.3. Let r ∈ {0,1,2,3, . . .} and {α_k}ⁿ_k=0 be a sequence of real numbers such that

|α_k| ≤1,for allk.Then there exists a finite constantM =M(r)>0such that for anyn ≥0 (∗∗)

Z π π/(n+1)

n

X

k=0

α_kD^(r)_k (x)

dx≤M ·(n+ 1)^r+1. Proof. Similar to Lemma 3.1 it is not difficult to proof the following equality

D^(r)_n (x) =

r

X

k=0

n+¹_2k sin

n+ ¹₂

x+^k+3₂ π

sin ^x₂r+1−k ϕ_k(x), whereϕ_kdenotes the same various4π-periodic functions ofx,independent ofn.

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+ ¹_2ν sin

j+ ¹₂

x+^ν+3₂ π sin ^x₂r+1−ν ϕ_ν(x)

!

dx.

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 Lemma 3.2, andK =K(r)is a positive constant.

Applying Lemma 3.2, 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

α²_j(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 3.4. Forr= 0,we obtain the Telyakovskii type inequality, proved in [4].

Lemma 3.5. Let r be a non-negative integer. Then for all 0 < |t| ≤ π and all n ≥ 1 the following estimates hold:

(i)

E_−n^(r)(t)

≤ ⁴ⁿ_|t|^rπ, (ii)

D˜^(r)n (t)

≤ ⁴ⁿ_|t|^rπ, (iii)

D^(r)_n (t)

≤ ⁴ⁿ_|t|^rπ +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)|=|E_n(−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|^rπ. (ii) Applying the inequality(i), we obtain

D˜_n^(r)(t) =

iD˜_n^(r)(t) ≤ 1

2

E_n^(r)(t) + 1

2

E_−n^(r)(t)

≤ 4n^rπ

|t| . (iii) We note that

ctg^t₂(r) =O

1

|t|^r+1

. Applying the inequality(ii), we obtain

|D^(r)_n (t)| ≤ |D˜_n^(r)(t)|+ 1 2

ctgt

2 (r)

≤ 4n^rπ

|t| +O 1

|t|^r+1

.

4. PROOFS OF THEMAIN RESULTS

Proof of Theorem 2.1. Let (2.1) hold. Then

∆a_k =α_k

∞

X

m=k

p_m m , and we denote

A_k =

∞

X

m=k

|pm| 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|<∞, andAk↓0i.e. {ak} ∈Sr.

Now, if{a_k} ∈S_r,we putα_k = ^∆a_A^k

k andp_k =k(A_k−A_k+1). Hence|α_k| ≤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 <∞.

Finally, a_k =

∞

X

i=k

∆a_i =

∞

X

i=k

α_iA_i =

∞

X

i=k

α_i

∞

X

m=i

∆A_m =

∞

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 Corollary 2.2. The proof of this corollary follows from Theorems 1.4 and 2.1.

Proof of Theorem 2.3. 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 3.5(iii), we have that the seriesP∞

k=1∆a_kD^(r)_k (x)is uniformly 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). 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

∆akD^(r)_k (x)

dx , I2 =

m

X

j=1

Z π/j π/(j+1)

∞

X

k=j

∆akD^(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]

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|∆a_k| j(j + 1)

! ,

whereO_rdepends onr. But

m

X

j=1 j−1

X

k=0

(k+ 1)^r+1|∆a_k| 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|∆a_k|

= |∆a₀|+

∞

X

k=1

(k+ 1)^r|∆a_k|

≤ |a₁|+ 2^r

∞

X

k=1

k^r|∆a_k|

≤

∞

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

|∆ak|(k+ 1)^r+1 j(j+ 1) =O_r

∞

X

k=1

k^rA_k

! ,

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

# .

Applying the Lemma 3.3, we have:

(4.2) J_k=

Z π π/(j+1)

k

X

i=0

∆a_i Ai

D^(r)_i (x)

dx=O_r (k+ 1)^r+1 ,

whereO_rdepends onr. Then, by Lemma 3.5(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

(∆Ak)Jk+Or

∞

X

j=1

j^rAj

!

= Or(1)

∞

X

k=1

(∆Ak)(k+ 1)^r+1+Or

∞

X

j=1

j^rAj

!

= Or

∞

X

j=1

j^rAj

! .

Finally, the inequality (∗) is satisfied.

Proof of Corollary 2.4. By the inequalities

m

X

n=1

|a_n| ·n^r−1 ≤

∞

X

n=1

n^r−1

∞

X

k=n

|∆a_k|

≤

∞

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 Theorem 2.3, we obtain:

Z π 0

|g^(r)(x)|dx≤M

∞

X

n=1

n^rA_n

! ,

where0< M =M(r)<∞.

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 trigonometrischen 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 integrability of trigonometric se- ries, Math. Notes, 14 (1973), 742–748.

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

[6] Ž. TOMOVSKI, Some results on L¹-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