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Introduction Consider cosine and sine series a0 2 + X∞ k=1 akcoskx (1.1) and X∞ k=1 aksinkx (1.2) Date: Received:March 2008

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J. Nonlinear Sci. Appl. 1 (2008), no. 3, 179–188

T

he

J

ournal of

N

onlinear

S

ciences and

A

pplications http://www.tjnsa.com

CONVERGENCE OF NEW MODIFIED TRIGONOMETRIC SUMS IN THE METRIC SPACE L

JATINDERDEEP KAUR1 AND S.S. BHATIA2∗

Abstract. We introduce here new modified cosine and sine sums as a0

2 + Xn

k=1

Xn

j=k

4(ajcosjx)

and n

X

k=1

Xn

j=k

4(ajsinjx)

and study their integrability andL1-convergence. TheL1-convergence of cosine and sine series have been obtained as corollary. In this paper, we have been able to remove the necessary and sufficient conditionaklogk=o(1) as k→ ∞for theL1-convergence of cosine and sine series.

1. Introduction Consider cosine and sine series

a0 2 +

X k=1

akcoskx (1.1)

and X

k=1

aksinkx (1.2)

Date: Received:March 2008 ; Revised: Dec 2008

Corresponding author.

2000Mathematics Subject Classification. Primary 42A20; Secondary 42A32.

Key words and phrases. L1-convergence, Dirichlet kernel, Fejer kernel, monotone sequence.

∗∗This work is the part of UGC sponsored Major Research Project entitled

”Integrability andL1-convergence of trigonometric Series”.

179

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or together

X k=1

akφk(x) (1.3)

Where φk(x) is coskx or sinkx respectively. Let the partial sum of (1.3) be denoted by Sn(x) and t(x) = lim

n→∞Sn(x). Further, let tr(x) = lim

n→∞Snr(x) where Snr(x) represents rth derivative of Sn(x).

Definition 1.1. A sequence {ak}is said to convex if 42ak 0, where 42ak = 4(4ak) and 4ak =ak−ak+1,and quasi-convex sequence ifP

(k+ 1)42ak <∞.

The concept of quasi-convex was generalized by Sidon [4] in the following man- ner:

Definition 1.2. [4] A null sequence {ak} is said to belong to class S if there exists a sequence {Ak} such that

Ak 0, k→ ∞, (1.4)

X k=0

Ak <∞, (1.5)

and

|∆ak| ≤Ak, ∀k. (1.6)

A quasi-convex null sequence satisfies conditions of the class S because we can choose

An = X m=n

|42am|.

ConcerningL1-convergence of (1.1) and (1.2), the following theorems are known:

Theorem 1.3. ([1], p. 204) If ak 0 and {ak} is convex or even quasi-convex, then for the convergence of the series (1.1) in the metric space L1, it is necessary and sufficient that aklogk =o(1), k → ∞.

This theorem is due to Kolmogorov [2]. Teljakovskii [5] generalized Theorem 1.3 for the cosine series (1.1) with coefficients {ak} satisfying the conditions of the class S in the following form:

Theorem 1.4. If the coefficient sequence {ak} of the cosine series (1.1) belongs to the class S, then a necessary and sufficient condition for L1-convergence of (1.1) is aklogk =o(1), k → ∞.

Theorem 1.5. ([1], p. 201) If ak 0and X k=1

³ak

k

´

<∞,then (1.3) is a Fourier series.

In the present paper, we introduce new modified cosine and sine sums as fn(x) = a0

2 + Xn k=1

Xn j=k

4(ajcosjx)

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and

gn(x) = Xn k=1

Xn j=k

4(ajsinjx)

and study their integrability and L1-convergence under a new class SJ of coeffi- cient sequences defined as follows:

Definition 1.6. A null sequence{ak} of positive numbers belongs to class SJ if there exists a sequence {Ak} such that

Ak 0, as k→ ∞, (1.7)

X k=1

Ak <∞, (1.8)

¯¯

¯4

³ak k

´¯¯

¯ Ak

k k. (1.9)

Clearly class SJ class S, Since

¯¯

¯4

³ak k

´¯¯¯ Ak

k ⇒ |∆ak| ≤Ak, k.

Following example shows that the class SJ is proper subclass of class S.

Example 1.7. For k=I− {0,1,2}, where I is set of integers, define{ak}= k13, then there exists {Ak} = k12 such that {ak} satisfies all the conditions of class S but not class SJ. However, for k = 1,2,3... the sequence {bk} = k13 satisfies conditions of class SJ as well as conditions of class S. Clearly, class SJ is proper subclass of class S.

Now, we define a new class SJr of coefficient sequences which is an extension of class SJ.

Definition 1.8. A null sequence{ak}of positive numbers belongs to class SJr if there exists a sequence {Ak} such that

Ak 0, as k → ∞, (1.10)

X k=1

krAk<∞, (r= 0,1,2, ...) (1.11)

¯¯

¯4³ak k

´¯¯¯ Ak

k k. (1.12)

clearly, for r = 0, SJr = SJ. It is obvious that SJr+1 SJr, but converse is not true.

Example 1.9. For k = 1,2,3..., define bk = kr+21 , r = 0,1,2,3, ... Firstly, we shall show that{bk}does not belong to SJr+1.

Really, bn= 1

nr+2 0 as n→ ∞.

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Let there exists {Ak} = kr+21 , r = 0,1,2,3, ... s.t.

X k=1

kr+1Ak =kr+1 1 kr+2 = X

k=1

1

k is divergent, i.e. {bk} does not belong to SJr+1. But, Ak0, as k → ∞,and

X k=1

krAk =kr 1 kr+2 =

X k=1

1

k2 <∞, Also¯

¯4¡b

k

k

¢¯¯ Akk, k.

Therefore,{bk} belongs to SJr.

In what follows,tn(x) will representsfn(x) or gn(x).

2. Lemmas

We require the following lemmas in the proof of our result.

Lemma 2.1. [3] Let n 1 and let r be a nonnegative integer, x [², π]. Then

|D˜nr(x)| ≤ C²nxr where C² is a positive constant depending on ², 0 < ² < π and D˜n(x) is the conjugate Dirichlet kernel.

Lemma 2.2. [5] Let {ak} be a sequence of real numbers such that |ak| ≤ 1 for all k. Then there exists a constant M >0 such that for any n≥1

Z π

0

¯¯

¯¯

¯ Xn k=0

akD˜k(x)

¯¯

¯¯

¯ dx≤M(n+ 1).

Moreover by Bernstein’s inequality, for r= 0,1,2,3...

Z π

0

¯¯

¯¯

¯ Xn k=0

akD˜kr(x)

¯¯

¯¯

¯ dx≤M(n+ 1)r+1.

Lemma 2.3. [3] ||D˜nr(x)||L1 = O(nrlogn), r = 0,1,2,3...., where D˜rn(x) repre- sents the rth derivative of conjugate Dirichlet-Kernel.

3. Main Results In this paper we shall prove the following main results:

Theorem 3.1. Let the coefficients of the series (1.3) belongs to class SJ, then the series (1.3) is a Fourier series.

Proof. Making Use of Abel’s transformation on Xn

k=1

³ak k

´

, we get Xn

k=1

³ak k

´

= Xn−1

k=1

k4

³ak k

´

−an

Xn−1 k=1

k µAk

k

−an

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But (1.3) belongs to class SJ, therefore, the series X k=1

³ak

k

´

converges.

Hence the conclusion of theorem follows from Theorem (1.5. ¤ Theorem 3.2. Let the coefficients of the series (1.3) belongs to class SJ, then

n→∞lim tn(x) = t(x) exists for x∈(0, π). (3.1)

t∈L1(0, π) (3.2)

||t(x)−Sn(x)||=o(1), n→ ∞ (3.3) Proof. We will consider only cosine sums as the proof for the sine sums follows the same line.

To prove (3.1), we notice that tn(x) = a0

2 + Xn k=1

Xn j=k

4(ajcosjx)

tn(x) = a0 2 +

Xn k=1

[ak coskx−ak+1 cos(k+ 1)x+ak+1 cos(k+ 1)x

−ak+2cos(k+ 1)x+...+ancosnx−an+1cos(n+ 1)x]

= a0 2 +

Xn k=1

akcoskx− Xn k=1

an+1cos(n+ 1)x

tn(x) = Sn(x)−nan+1cos(n+ 1)x (3.4)

Since Ak 0, as k → ∞ and X k=1

Ak < ∞, therefore, by Oliver’s theorem we have,kAk 0, as k → ∞ and so

nan=n2 X k=n

4

³ak k

´

X k=n

k2 µAk

k

=o(1) (3.5)

Also cos(n+ 1)x is finite in (0, π). Hence

n→∞lim tn(x) = lim

n→∞Sn(x) =t(x) Moreover,

t(x) = lim

n→∞tn(x) = lim

n→∞Sn(x) = lim

n→∞

à a0

2 + Xn k=1

akcoskx

!

= a0

2 + lim

n→∞

à n X

k=1

akcoskx

!

Use of Abel’s transformation yields

n→∞lim ÃXn

k=1

akcoskx

!

= lim

n→∞

"

Xn−1 k=1

4

³ak k

´D˜k0(x) + an n D˜0n(x)

#

(6)

where ˜Dn0(x) is the derivative of conjugate Dirichlet kernel.

= X

k=1

4

³ak k

´D˜0k(x)

X

k=1

µAk k

D˜0k(x)

By the given hypothesis and lemma 2.1, the series X k=1

µAk k

D˜k0(x) converges.

Therefore, the limitt(x) exists for x∈(0, π) and thus (3.1) follows.

Forx6= 0, it follows from (3.4) that t(x)−tn(x) =

X k=n+1

akcoskx+nan+1cos(n+ 1)x

= lim

m→∞

" m X

k=n+1

³ak

k

´

kcoskx

#

+nan+1cos(n+ 1)x

Applying Abel’s transformation, we have

= X k=n+1

4

³ak k

´D˜k0(x) an+1

n+ 1D˜n0(x) +nan+1cos(n+ 1)x

X k=n+1

µAk k

4¡a

k

k

¢

¡A

k

k

¢ D˜k0(x) + an+1

n+ 1D˜n0(x) +nan+1cos(n+ 1)x

X k=n+1

4 µAk

k

¶Xk

j=1

4

³aj

j

´

³Aj

j

´ D˜j0(x)

µAn+1

n+ 1

¶Xn

j=1

4

³aj

j

´

³Aj

j

´ D˜0j(x) +an+1

n+ 1D˜0n(x) +nan+1cos(n+ 1)x Thus from lemma 2.2 and 2.3, we obtain

||t(x)−tn(x)|| ≤ X k=n+1

4 µAk

k

¶ Z π

0

¯¯

¯¯

¯¯ Xk

j=1

4

³aj

j

´

³Aj

j

´ D˜j0(x)

¯¯

¯¯

¯¯ dx

+

µAn+1 n+ 1

¶ Z π

0

¯¯

¯¯

¯¯ Xn

j=1

4

³aj

j

´

³Aj

j

´ D˜j0(x)

¯¯

¯¯

¯¯ dx+ Z π

0

¯¯

¯¯an+1

n+ 1D˜0n(x)

¯¯

¯¯ dx

+n|an+1| Z π

0

|cos(n+ 1)x| dx

(7)

= O Ã

X

k=n+1

k24 µAk

k

¶!

+O µ

n2

µAn+1

n+ 1

¶¶

+O(an+1logn) +n|an+1| Z π

0

|cos(n+ 1)x| dx But

Xn k=1

Ak = Xn−1

k=1

k(k+ 1)

2 4

µAk k

+n(n+ 1) 2

An n since {ak} ∈SJ, we have

k(k+ 1)Ak

k = (k+ 1)Ak=o(1) as k→ ∞.

and therefore the series X k=n+1

k24 µAk

k

, converges.

Moreover, Z π

0

|cos(n+ 1)x| dx= Z π

2

0

cos(n+ 1)x dx Z π

π 2

cos(n+ 1)x dx 2 n+ 1 and sincean’s are positive, we have by (3.5) thatanlogn ≤nan =o(1), f or n≥ 1.

Hence, it follows that

||t(x)−tn(x)||=o(1) as n→ ∞. (3.6) and since tn(x) is a polynomial, therefore t(x)∈L1. This proves (3.2).

We now turn to the proof of (3.3), We have

||t−Sn|| = ||t−tn+tn−Sn||

≤ ||t−tn||+||tn−Sn||

= ||t−tn||+||nan+1cos(n+ 1)x||

≤ ||t−tn||+n|an+1| Z π

0

|cos(n+ 1)x| dx Further,||t(x)−tn(x)||=o(1), n→ ∞ (by (3.6)),

Z π

0

|cos(n+ 1)x| dx≤ 2 n+ 1 and {ak} is a null sequence,therefore the conclusion of theorem follows. ¤ Theorem 3.3. Let the coefficients of the series (1.3) belongs to class SJr, then

n→∞lim trn(x) =tr(x) exists for x∈(0, π). (3.7) tr ∈L1(0, π), (r= 0,1,2, ...) (3.8)

||tr(x)−Snr(x)||=o(1), n→ ∞. (3.9)

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Proof. We will consider only cosine sums as the proof for the sine sums follows the same line. As in the proof of the Theorem 3.2, we have

tn(x) = a0 2 +

Xn k=1

Xn j=k

4(ajcosjx)

= Sn(x)−nan+1cos(n+ 1)x we have, then

trn(x) = Snr(x)−n(n+ 1)ran+1cos³

(n+ 1)x+ 2

´

Since Ak 0, as k → ∞ and X k=1

krAk < ∞, therefore, we have, kr+1Ak 0, as k → ∞and so

nr+1an=nr+2 X k=n

4³ak k

´

X k=n

kr+2 µAk

k

=o(1) (3.10) Also cos¡

(n+ 1)x+2 ¢

is finite in (0, π). Hence tr(x) = lim

n→∞trn(x)

= lim

n→∞Snr(x)

= lim

n→∞

à n X

k=1

krakcos³

kx+ 2

´!

use of Abel’s transformation yields

n→∞lim à n

X

k=1

krakcos

³

kx+ 2

´!

= lim

n→∞

"n−1 X

k=1

4

³ak

k

´D˜r+1k (x) + an

n D˜nr+1(x)

# ,

where ˜Dnr+1(x) represents the (r+ 1)th derivative of conjugate Dirichlet kernel.

= X

k=1

4

³ak k

´D˜r+1k (x) + lim

n→∞

han

n D˜r+1n (x) i

X

k=1

µAk k

D˜r+1k (x) + lim

n→∞

han

n D˜r+1n (x) i

By making use of the given hypothesis, lemma 2.1 and (3.10), the series X

k=1

µAk k

D˜r+1k (x) converges. Therefore, the limit tr(x) exists for x∈(0, π) and thus (3.7) follows.

To prove (3.8), we have tr(x)−trn(x) =

X k=n+1

krakcos

³

kx+ 2

´

+n(n+ 1)ran+1cos

³

(n+ 1)x+ 2

´

(9)

Making use of Abel’s transformation, we obtain

= X k=n+1

4

³ak k

´D˜kr+1(x) an+1

n+ 1D˜r+1n (x) +n(n+ 1)ran+1cos

³

(n+ 1)x+ 2

´

X k=n+1

µAk k

4¡a

k

k

¢

¡A

k

k

¢ D˜r+1k (x) an+1

n+ 1D˜nr+1(x) +n(n+ 1)ran+1cos

³

(n+ 1)x+ 2

´

X k=n+1

4 µAk

k

¶Xk

j=1

4

³aj

j

´

³Aj

j

´ D˜r+1j (x)

µAn+1 n+ 1

¶Xn

j=1

4

³aj

j

´

³Aj

j

´ D˜jr+1(x) +an+1

n+ 1D˜nr+1(x) +nan+1cos

³

(n+ 1)x+ 2

´

Thus from lemma 2.2 and 2.3, we obtain

||tr(x)−trn(x)|| ≤ X k=n+1

4 µAk

k

¶ Z π

0

¯¯

¯¯

¯¯ Xk j=1

4

³aj

j

´

³Aj

j

´ D˜jr+1(x)

¯¯

¯¯

¯¯ dx

+

µAn+1 n+ 1

¶ Z π

0

¯¯

¯¯

¯¯ Xn

j=1

4³

aj

j

´

³Aj

j

´ D˜jr+1(x)

¯¯

¯¯

¯¯ dx+ Z π

0

¯¯

¯¯ an+1

n+ 1D˜nr+1(x)

¯¯

¯¯ dx

+n(n+ 1)r|an+1| Z π

0

|cos

³

(n+ 1)x+ 2

´

| dx

= O

à X

k=n+1

kr+24 µAk

k

¶!

+O µ

nr+2

µAn+1 n+ 1

¶¶

+O(nran+1logn) +n(n+ 1)r|an+1|

Z π

0

|cos

³

(n+ 1)x+ 2

´

| dx

Using the argument as in the proof of theorem 3.2, it is easily shown that the series

X k=n+1

kr+24 µAk

k

, converges. Moreover, Z π

0

|cos

³

(n+ 1)x+ 2

´

| dx≤ 2 n+ 1

and for n≥1, nranlogn ≤nr+1an =o(1) by (3.10). Hence it follows that

||tr(x)−trn(x)||=o(1) as n→ ∞. (3.11) and since trn(x) is a polynomial, therefore tr(x)∈L1. This proves (3.8).

(10)

We now turn to the proof of (3.9). We have

||tr−Snr|| = ||tr−trn+trn−Snr||

≤ ||tr−trn||+||trn−Snr||

= ||tr−tnr||+||n(n+ 1)ran+1cos

³

(n+ 1)x+ 2

´

||

≤ ||tr−trn||+n(n+ 1)r|an+1| Z π

0

|cos

³

(n+ 1)x+ 2

´

|dx Further,||tr(x)−trn(x)||=o(1), n→ ∞(by (3.11)) ,

Z π

0

|cos

³

(n+ 1)x+ 2

´

|dx≤ 2

n+ 1 and {ak} is a null sequence, the conclusion of theorem follows. ¤ Remark 3.4. The case r = 0, in Theorem 3.3 yields the Theorem 3.2.

Acknowledgements: Authors are thankful to Professor Babu Ram (Retd.) of M. D. University, Rohtak for his valuable suggestions during the preparation of this paper.

References

[1] N.K. Bary,A treatise on trigonometric series, Vol II, Pergamon Press, London, (1964). 1.3, 1.5

[2] A.N. Kolmogorov, Sur l’ordere de grandeur des coefficients de la series de Fourier - Lebesque, Bull. Acad. Pol. Sci. Ser. Sci. Math. Astronom. Phys., (1923), 83–86. 1

[3] S. Sheng, The extension of the theorems of ˘C.V. Stanojevi´c and V.B. Stanojevi´c, Proc.

Amer. Math. Soc., 110, (1990), 895–904. 2.1, 2.3

[4] S. Sidon, Hinreichende Bedingungen f¨ur den Fourier-Charakter einer trigonometrischen Reihe, J. London Math Soc., 14, (1939), 158–160. 1, 1.2

[5] S.A. Teljakovskˇii,A sufficient condition of Sidon for the integrability of trigonometric series, Mat. Zametki, 14(3), (1973), 317–328. 1, 2.2

1 School of Mathematics & Computer Applications, Thapar University Patiala(Pb.)-147004, INDIA.

E-mail address: [email protected]

2 School of Mathematics & Computer Applications, Thapar University Patiala(Pb.)-147004, INDIA.

E-mail address: [email protected]

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