1 2 sinx Pn k=1 Pn j=k (4aj−1− 4aj+1) sinkx

September 2009 research paper

ON L¹-CONVERGENCE OF CERTAIN GENERALIZED MODIFIED TRIGONOMETRIC SUMS

Karanvir Singh and Kulwinder Kaur

Abstract. In this paper we define new modified generalized sine sums Knr(x) = 1

2 sinx

Pn k=1

(4^rak−1− 4^rak+1) ˜S_k^r−1(x) and study their L¹-convergence under a newly defined classK^α. Our results generalize the corresponding results of Kaur, Bhatia and Ram [6] and Kaur [7].

1. Introduction Let

K(x) = P^∞

k=1

a_kcoskx (1.1)

and

Kn(x) = 1 2 sinx

Pn k=1

Pn j=k

(4aj−1− 4aj+1) sinkx. (1.2) Using modified cosine sums

gn(x) = 1 2

Pn k=0

4ak+ Pⁿ

k=1

Pn j=k

(4aj) coskx

of Garett and Stanojevi´c [4], Kaur and Bhatia [5] proved the following theorem under the class of generalized semi-convex coefficients.

Theorem 1. If {an} is a generalized semi-convex null sequence, then gn(x) converges to K(x)in theL¹-metric if and only if4anlogn=o(1), asn→ ∞.

The above mentioned result motivated the authors [6] to define a new modified sums (1.2) and to study these sums under a different class Kof coefficients in the following way.

AMS Subject Classification: 42A20, 42A32.

Keywords and phrases:L¹−convergence; conjugate Ces`aro means; generalized sine sums.

219

Theorem 2. Let k be a positive real number. If

ak=o(1), k→ ∞ (1.3)

and P∞

k=1

k|4²ak−1− 4²ak+1|<∞, (1.4) thenKn(x) converges toK(x)in the L¹-norm.

Any sequence satisfying (1.3) and (1.4) is said to belong to the classK[6].

In particular in [6] the following corollary to Theorem 2 is proved.

Theorem 3. If {an} belongs to the classK, then the necessary and sufficient condition forL¹-convergence of the cosine series (1.1) islimn→∞anlogn= 0.

Definition. Letαbe a positive real number. If (1.3) holds and P∞

k=1

k^α|4^α+1ak−1− 4^α+1ak+1|<∞, (a0= 0), then we say that{an}belongs to the classK^α.

Forα= 1, the class K^α reduces to the classK.

Applying Abel’s transformation toKn(x), we can easily see that Kn(x) = 1

2 sinx Pn k=1

(4ak−1− 4ak+1) ˜S_k⁰(x),

where ˜S_k⁰(x) = ˜Dk(x) = sinx+ sin 2x+ sin 3x+· · ·+ sinnx. So in [7] the author studied theL¹-convergence ofKn(x) = 1

2 sinx Pn k=1

(4ak−1− 4ak+1) ˜S_k⁰(x) toK(x) by proving the following result.

Theorem 4. Let the sequence{ak}belong to classK^α. ThenKn(x)converges toK(x)in theL¹-norm.

If we takeα= 1, then this theorem reduces to Theorem 2.

It is natural to seek ways to prove theL¹-convergence of the generalized sums of the form

Knr(x) = 1 2 sinx

Pn k=1

(4^rak−1− 4^rak+1) ˜S_k^r−1(x), (1.5) whereris any real number greater than or equal to 1. It is obvious that forr= 1, Knr(x) reduces toKn(x).

The purpose of this paper is to prove theL¹-convergence of Knr(x) toK(x) under the classK^α.

2. Notation and Formulae We use the following notations [10].

Given a sequence S0, S1, S2, . . ., we define the sequence S₀^α, S₁^α, S₂^α, . . . for everyα= 0,1,2, . . . by the conditionsS⁰_n=Sn,S_n^α=S^α₀+S₁^α−1+S₂^α−1+· · ·+S^α−1_n (α= 1,2, . . .; n = 0,1,2, . . .). Similarly we define the sequence of numbersA^α₀, A^α₁, A^α₂, . . . for α = 0,1,2, . . . by the conditions A⁰_n = 1, A^α_n =A^α−1₀ +A^α−1₁ + A^α−1₂ +· · ·+A^α−1_n (α = 1,2, . . .; n = 0,1,2, . . .). Let P

an be a given infinite series. The conjugate Ces`aro sums of orderα ofP

an for any real number αare defined by

S˜_n^α(ap) = ˜S_n^α= Pⁿ

p=0

A^α_n−pap= Pⁿ

p=0

A^α−1_n−pS˜p,

where ˜Sn = ˜S_n⁰ = ˜Dn, and A^α_p denotes the binomial coefficients. The conjugate Ces`aro means ˜T_n^αof orderrofP

an will be defined by T˜_n^α= S˜_n^α

A^α_n. The following formulae will also be needed:

S˜_n^α( ˜S_p^r) = ˜S_n^α+r+1, S˜_n^α+1−S˜_n−1^α+1= ˜S_n^α, Pⁿ

p=0

A^α_n−pA^β_p =A^α+β+1_n .

The differences of orderαof the sequence{an}for any positive integerαare defined by the equations 4^αan = 4(4^α−1an), n= 0,1,2, . . . 4¹an =an−an+1. Since A^−α−1_m = 0 form≥α+ 1, we have

4^αan= P^α

m=0

A^−α−1_m an+m= P^∞

m=0

A^−α−1_m an+m. (2.1) If the series (2.1) is convergent for someαwhich is not a positive integer, then we denote the differences

4^αan = P^∞

m=0

A^−α−1_m an+m n= 0,1,2,3, . . . . The broken differences4^α_na_p are defined by

4^α_nap= ^n−pP

m=0

A^−α−1_m ap+m. By repeated partial summation of orderr, we have

Pn p=0

apbp= Pⁿ

p=0

S˜_p^α−1(ap)4^α_nbp.

3. Lemmas

We need the following lemmas for the proof of our result.

Lemma 3.1[3]Ifr≥0,p≥0, (i)²n=o(n)^−p, and

(ii) P_∞

n=0A^r+p_n |4^r+1²n|<∞, then

(iii)P_∞

n=0A^λ+p_n |4^λ+1²_n|<∞, for−1≤λ≤rand

(iv) A^λ+p_n 4^λ²n is of bounded variation for 0 ≤ λ ≤ r and tends to zero as n→ ∞.

Lemma 3.2 [1] Let r be a non-negative real number. If the sequence {²n} satisfies the conditions

(i)²n=O(1), and (ii) P^∞

n=1n^r|4^r+1²n|<∞, then4^β²n= P^∞

m=0A^r−β_m 4^r+1²n+m, for β >0.

Lemma 3.3[2]If0< δ <1 and0≤n < m, then

¯¯

¯¯ Pm i=0

A^δ−1_n−iSi

¯¯

¯¯≤ max

0≤p≤m|S_p^δ|.

Lemma 3.4[10]Let S˜_n(x)andT˜_n^αbe then^thpartial sum and Ces`aro mean of order α >0, respectively, of the series sinx+ sin 2x+ sin 3x+· · ·+ sinnx+· · ·. Then

(i)R_π

0 |S˜n(x)|dx∼logn, (ii) R_π

0 |T˜_n^α|dx remains bounded for alln.

4. Main result

The main result of this paper is the following theorem.

Theorem 4.1. Let αbe a positive real number. If a sequence{ak} belongs to the class K^α, then forα≤r≤α+ 1

(i)Knr(x)converges to K(x)pointwise for0< δ≤x≤π, and (ii) Knr(x)→K(x)in theL¹-norm.

If we take α= 1 and r= 1, then Theorem 2 is obtained as a particular case and also Theorem 4 can be deduced as a special case of Theorem 4.1 if we take r= 1 in (1.5) .

Proof. We have

K(x) = 1 2 sinx

P∞ k=1

(4^rak−1− 4^rak+1) ˜S^r−1_k (x) and

Knr(x) = 1 2 sinx

Pn k=1

(4^rak−1− 4^rak+1) ˜S_k^r−1(x).

Case 1. Let r=α+ 1. Then Knr(x) = 1

2 sinx Pn k=1

(4^α+1ak−1− 4^α+1ak+1) ˜S_k^α(x).

SoKnr(x) converges toK(x) pointwise for 0< δ≤x≤π.

Z _π Now

0

|K(x)−Knr(x)|dx

= Z _π

0

¯¯

¯¯ 1 2 sinx

P∞ k=n+1

(4^α+1ak−1− 4^α+1ak+1) ˜S^α_k(x)

¯¯

¯¯dx

≤C P^∞

k=n+1

|(4^α+1a_k−1− 4^α+1a_k+1)|R_π

0 |S˜_k^α(x)|dx

=C P^∞

k=n+1

A^α_k|(4^α+1ak−1− 4^α+1ak+1)|R_π

0 |T˜_k^α(x)|dx

≤C1

P∞ k=n+1

A^α_k|(4^α+1ak−1− 4^α+1ak+1)|=o(1),by hypothesis of Theorem 4.1.

Therefore,Knr(x) converges toK(x) asn→ ∞in the L¹−norm.

Case 2. Letα < r < α+ 1. Taker=α+ 1−δand 0< δ <1. Then K_nr(x) = 1

2 sinx Pn k=1

(4^ra_k−1− 4^ra_k+1) ˜S_k^r−1(x)

= 1

2 sinx Pn k=1

(4^α+1−δak−1− 4^α+1−δak+1) ˜S_k^α−δ(x).

Applying Abel’s transformation of order−δand using Lemma 3.2, we have 1

2 sinx Pn k=1

(4^α+1ak−1− 4^α+1ak+1) ˜S_k^α(x)

= 1

2 sinx

· _n P

k=1

S˜_k^α−δ(x)^n−kP

m=1

A^δ−1_m (4^α+1am+k−1− 4^α+1am+k+1)

¸

= 1

2 sinx

· _n P

k=1

S˜_k^α−δ(x){(4^α−δ+1ak−1− 4^α−δ+1ak+1)

− P^∞

m=n−k+1

A^δ−1_m (4^δ+1am+k−1− 4^δ+1am+k+1)}

¸

= 1 2 sinx

· _n P

k=1

S˜_k^α−δ(x)(4^α−δ+1ak−1− 4^α−δ+1ak+1)−Rn(x)

¸ , where

Rn(x) = Pⁿ

k=1

S˜_k^α−δ(x){A^δ−1_n−k+1(4^δ+1an− 4^δ+1an+2) +A^δ−1_n−k+2(4^δ+1an+1− 4^δ+1an+3) +· · · }

= Pⁿ

k=1

S˜_k^r−δ(x){A^δ−1_n−k+1(4^δ+1an+2− 4^δ+1an) + Pⁿ

k=1

S˜^r−δ_k (x)A^δ−1_n−k+2(4^δ+1an+3− 4^δ+1an+1) +· · · }.

This implies that Knr(x) = 1

2 sinx

· _n P

k=1

S˜_k^α(x)(4^α+1ak−1− 4^α+1ak+1) +Rn(x)

¸ . Hence

Z _π

0

|K(x)−Knr(x)|dx

≤C Z _π

0

¯¯

¯¯ 1

2 sinx{ P^∞

k=n+1

S˜_k^α(x)(4^α+1ak−1− 4^α+1ak+1)−Rn(x)}

¯¯

¯¯dx

≤C P^∞

k=n+1

|(4^α+1ak−1− 4^α+1ak+1)|R_π

0 |S˜_k^α(x)|dx+R_π

0 |Rn(x)|dx

=C P^∞

k=n+1

A^α_k|(4^α+1ak−1− 4^α+1ak+1)|(x)R_π

0 |T˜_k^α(x)|dx+R_π

0 |Rn(x)|dx

≤ P^∞

k=n+1

A^α_k|(4^α+1ak−1− 4^α+1ak+1)|+R_π

0 |Rn(x)|dx

=o(1) + Z _π

0

|Rn(x)|dx,by hypotheses of Theorem 4.1. (4.1) Z _πNow

0

|Rn(x)|dx

= Z _π

0

¯¯

¯¯ µ _n

P

k=1

S˜_k^α−δ(x)A^δ−1_n−k+1

¶

(4^δ+1an− 4^δ+1an+2) +

µ _n P

k=1

S˜_k^α−δ(x)A^δ−1_n−k+2

¶

(4^δ+1an+1− 4^δ+1an+3) +· · ·

¯¯

¯¯dx

≤ Z _π

0

|(4^δ+1a_n− 4^δ+1a_n+2)|

¯¯

¯¯ Xn k=1

S˜_k^α−δ(x)A^δ−1_n−k+1

¯¯

¯¯dx +

Z _π

0

|(4^δ+1an+1− 4^δ+1an+3)|

¯¯

¯¯ Pn k=1

S˜_k^α−δ(x)A^δ−1_n−k+2

¯¯

¯¯dx+· · ·

≤ Z _π

0

|(4^δ+1an− 4^δ+1an+2)| max

0≤p≤n+1|S˜_p^δ(x)|dx +

Z _π

0

|(4^δ+1an+1− 4^δ+1an+3)| max

0≤p≤n+2|S˜^δ_p(x)|dx+· · · ,by Lemma 3.3

=|(4^δ+1an− 4^δ+1an+2)|A^δ_n+1 Z _π

0

0≤p≤n+1max |T˜_p^δ(x)|dx +|(4^δ+1an+1− 4^δ+1an+3)|A^δ_n+2

Z _π

0

0≤p≤n+2max |T˜_p^δ(x)|+· · ·

=CA^δ_n+1|(4^δ+1an− 4^δ+1an+2)|+CA^δ_n+2|(4^δ+1an+1− 4^δ+1an+3)|+· · ·

=o(1),by Lemmas 3.1 and 3.4.

Thus Z _π

0

|Rn(x)|dx=o(1), n→ ∞.

Hence by (4.1)

n→∞lim Z _π

0

|K(x)−Knr(x)|dx=o(1).

Case 3. Let α=r. In this case Knr(x) = 1

2 sinx Pn k=1

(4^αak−1− 4^αak+1) ˜S_k^α−1(x).

Applying Abel’s transformation, we have Knr(x) = 1

2 sinx hPn

k=1

(4^α+1ak−1− 4^α+1ak+1) ˜S^α_k(x) + (4^αan− 4^αan+2) ˜S_n^α(x)i .

Since ˜S_k^α(x) are bounded for 0< δ≤x≤π, Knr(x)→K(x) = 1

2 sinx P∞ k=1

(4^α+1ak−1− 4^α+1ak+1) ˜S_k^α(x) pointwise for 0< δ≤x≤π. So

Z _π

0

|K(x)−Knr(x)|dx

= Z _π

0

¯¯

¯¯ 1 2 sinx

P∞ k=n+1

(4^α+1ak−1− 4^α+1ak+1) ˜S^α_k(x)−(4^αan− 4^αan+2) ˜S_n^α(x)

¯¯

¯¯dx

≤C P^∞

k=n+1

Z _π

0

|(4^α+1ak−1− 4^α+1ak+1)| |S˜_k^α(x)|dx +

Z _π

0

|(4^αan− 4^αan+2)| |S˜^α_n(x)|dx

= P^∞

k=n+1

A^α_kR_π

0 |(4^α+1ak−1− 4^α+1ak+1)||T˜_k^α(x)|dx

+A^α_n|(4^αan− 4^αan+2)|

Z _π

0

|T˜_n^α(x)|dx

≤C P^∞

k=n+1

A^α_k|(4^α+1ak−1− 4^α+1ak+1)|+C1|(4^αan− 4^αan+2)|

=o(1) +o(1) =o(1),by the hypotheses of Theorem 4.1 and Lemma 3.1.

Thus the proof is complete.

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(received 16.06.2008)

Department of Applied Sciences, GZS College of Engineering and Technology, Bathinda-151001, Punjab, India

E-mail:[email protected], [email protected]