Cmjv02i05p0699

J. Comp. & Math. Sci. Vol.2 (5), 699-703 (2011)

On Index Summability Factors of Fourier Series B. P. PADHY1, BANITAMANI MALLIK2, U. K. MISRA3 and MAHENDRA MISRA4 1

Department of Mathematics, Roland Institute of Technology Berhampur,Odisha, India.

2 Department of Mathematics, JITM, Paralakhemundi Gajapati, Odisha, India.

1 P.G. Department of Mathematics, Berhampur University, Odisha, India.

1 Principal N.C. College (Autonomous) Jajpur,Odisha, India.

ABSTRACT A theorem on index summabilty factors of Fourier Series has been established Keywords: Summability factors, Fourier series.

AMS Classification No: 40D25.

1. INTRODUCTION Let

∑a

n

be a given infinite series

The sequence to sequence transformation (1.5)

t nα =

with the sequence of partial sums {s n } . Let

{p n }

be a sequence of positive numbers such that n

(1.1)

Pn = ∑ pv → ∞ , as n → ∞ (P−i = p− i = 0, i ≥ 1)

p nα = Pnα =

where (1.4)

α

An

p α ν sν ∑ ν n−

=0

(

{sn }

of the sequence

)

generated by the

sequence of coefficients {pn }.

The series

∑a

summable N , p nα : δ

k

n

is said to be

, k ≥ 1, δ ≥ 0 , if

n

Aα ν pν , ∑ ν =0

n−

and (1.3)

n

defines the sequence of the N, p nα -mean

r =0

Let us write (1.2)

1 Pnα

(1.6)

n

pνα , ∑ ν =0

n +α  α  , p −i = 0, i ≥ 1 . =   α 

 Pnα  α ∑ n =1  p n ∞

In

the

case

  

δk + k −1

as

C ,1

k

< ∞.

p n = 1 , for all

when

n and n = 1, N , p nα ; δ same

t n − t n −1

k

summability is

summability.

Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)

For

700

B. P. Padhy, et al., J. Comp. & Math. Sci. Vol.2 (5), 699-703 (2011)

k = 1, δ = 0, N , p nα ; δ

summability

k

is

same an N , p n -summability.

Theorem:

Let f (t ) be a periodic function with period 2π and integrable in the sense of Lebesgue over (− π , π ) . Without loss of generality, we may assume that the constant term in the Fourier series of f (t ) is zero, so that ∞

∞

n =1

n =1

f (t )~ ∑ (a n cos nt + b n sin nt )= ∑ An (t )

(1.4)

3. MAIN THEOREM

2. KNOWN THEOREM

{ } is a sequence of positive

Let p nα

numbers such that Pnα → ∞ as n → ∞

and {λn } is a non-negative , non-increasing sequence such that n

(3.1) (i).

v

v =1

(3.2) (ii).

Dealing with N , p n , k ≥ 1 ,

∑P

m +1  P α  ∑  nα n = v +1  pn

λn < ∞ . If

∑ pα n

Av (t ) = 0( Pn )    

δk + k −1  α   pα  p n −ν −1   ν   = 0  Pα  Pα   ν n −1  

and

k

summability factors of Fourier Series, Bor1 proved the following theorem:

(3.3) (iii).

increasing sequence such that

{p n }

∑p

n

λn < ∞ ,

is a sequence of positive

numbers such that Pn → ∞ as n → ∞ and n

∑P v =1

v

A v ( t ) = 0 ( Pn ) . Then the factored

Fourier series

∑ A (t ) P n

n

∑ A (t ) P α n

n

λ n is

We prove an analogue theorem on k

-summability, k ≥ 1, δ ≥ 0 , in

the following form:

summable N , p n ; δ , k ≥ 1 . k

4. REQUIRED LEMMA We need the following for the proof of our theorem.

Lemma1:

λ n is summable

N , p n k , k ≥1 .

N , p nα ; δ

)

α

If {λn } is a non-negative and nonwhere

(

Pnα−ν −1 ∆λv = 0 p nα−ν λv ,

then the series Theorem:

  , as m → ∞ 

If {λn } is a non-negative and nonincreasing sequence such that

{ }

∑ pα λ n

n

< ∞, ,

where p nα is a sequence of positive numbers such α

that

Pnα → ∞ as n → ∞

Pn λn = 0(1) as n → ∞ and

∑ Pα n

Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)

then

∆λn < ∞ .

701

B. P. Padhy, et al., J. Comp. & Math. Sci. Vol.2 (5), 699-703 (2011)

5. PROOF OF THE THEOREM

(N , p ) α

Let t n (x) be the n-th mean of the series

∞

∑

A n ( x ) Pn λ n ,

n

∑ pnα−ν v =0

1 Pnα 1 = α Pn

n

α

n =1

1 Pnα n

t n ( x) =

=

then

by definition we have

v

∑ A ( x) P α λ r

r =0

r

n

n

∑ Ar ( x) Prα λr

∑ pα ν

r =0

v=r

n

∑ A ( x) P α P α

n−

n− r

λr .

∑ A ( x) P α P α

λr

r

r =0

r

r

Then

t n ( x) − t n −1 ( x) =

1 Pnα

n

∑ Ar ( x) Prα Pnα−r λr

−

r =0

1 Pnα−1

n −1

r =0

r

r

n − r −1

 Pnα− r Pnα− r −1  α  = ∑  α − α  Pr λ r Ar ( x) Pn −1  r =1  Pn 1 = α α (Pnα−ν Pnα−1 − Pnα− r −1 Pnα ) Prα λ r Ar ( x) Pn Pn −1 n

 n −1 1  r α  α α α α ∆ P P − P P λ  ∑ Pν Av ( x)  with po = 0 ∑  n − r n − 1 n − r − 1 n r α α Pn Pn −1  r =1  v =1  n −1 1  = α α ∑ ( p nα− r Pnα−1 − p nα− r −1 Pnα ) λ r Prα Pn Pn −1  r =1

{(

=

) }

+

∑ (( p α n −1

n − r −1

r =1

Pnα−1 − Pnα− r − 2 p nα

))P

α

r

 ∆λ r  

= Tn ,1 + Tn, 2 + Tn,3 + Tn, 4 , say.

In order to complete the proof of the theorem, using Minkowski’s inequality, it is sufficient to show that

 Pnα  α ∑ n =1  p n ∞

  

δk + k −1

Tn ,i

k

< ∞, for i = 1,2,3,4.

Now, we have

 Pnα  α ∑ n=2  p n m +1

  

δk + k −1

Tn ,1

k

 Pα = ∑  nα n= 2  p n m +1

  

δk + k −1

 1  n −1 α  ∑ p n −ν Pνα λ v  α k ( Pn )  v =1 

k

Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)

702

B. P. Padhy, et al., J. Comp. & Math. Sci. Vol.2 (5), 699-703 (2011)

 Pnα ≤ ∑  α n=2  p n

  

m +1

δk + k −1

 Pnα = O(1) ∑  α n= 2  p n m +1

  

1  n −1 α k   1 α k p ( P ) λ    α ∑ n − v ν ν ( Pnα )  v =1   Pn δk + k −1

= O (1)

m

v =1

(

m +1

α ν

∑p

1  n −1 α  ∑ p n −ν Pνα λv Pνα λν α ( Pn )  v =1

 Pα = O (1)∑ P λv ∑  nα v =1 n = v +1  p n m

n −1

  

δk + k −1

)

k −1

α

n −ν

  

k −1

  

p nα−ν Pnα

pνα λv (using 3.2)

∑ v =1

= O(1), as m → ∞ .

Next

 Pnα  α ∑ n=2  p n m +1

  

δk + k −1

Tn , 2

k

 Pα = ∑  nα n=2  p n m +1

 Pα ≤ ∑  nα n=2  p n m +1

  

  

δk + k −1

δk + k −1

 Pα = O (1)∑  nα n=2  p n m +1

   

1  n −1 α   ∑ p n −ν −1 Pνα λv  α k ( Pn )  v =1 

1  n −1 α k   1  ∑ p n −ν −1 ( Pνα ) k λv   α α ( Pn −1 )  v =1   Pn −1

δk + k −1

= 0(1)

m +1

α ν

m

∑ r =1

(

1  n −1 α  ∑ p n −ν −1 Pνα λv Pνα λν ( Pnα−1 )  v =1

 Pα = O (1)∑ P λv ∑  nα v =1 n = v +1  p n m

k

  

δk + k −1

)

n −1

∑p v =1

k −1

α

n −ν −1

  

k −1

  

p nα−ν −1 , Pnα−1

pνα λv , using (3.2)

= 0(1) , as m → ∞ .

Now,

 Pnα  α ∑ n=2  p n m +1

  

δk + k −1

Tn ,3

k

 Pα = ∑  nα n=2  p n m +1

 Pα ≤ ∑  nα n=2  p n m +1

  

  

δk + k −1

δk + k −1

1  n −1 α   ∑ p n −ν −1 Pνα ∆λv  α k ( Pn )  v =1 

k

1  n −1 α  1  ∑ p n −ν −1 ( Pνα ) k (∆λv ) k   α α ( Pn )  v =1   Pn

n −1

∑p v =1

α

n −ν −1

Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)

  

k −1

703

B. P. Padhy, et al., J. Comp. & Math. Sci. Vol.2 (5), 699-703 (2011)

 Pnα = O (1) ∑  α n=2  p n

  

 Pnα = O(1)∑  α n=2  p n

  

m +1

m +1

δk + k −1

δk + k −1

k −1  1  n −1 α  ∑ p n −ν −1 Pνα ∆λv (Pνα ∆λν )  α ( Pn )  v =1 

1 Pnα

n −1

∑ pα ν

 Pα = O (1)∑ P ∆λv ∑  nα v =1 n = v +1  p n m

= 0(1)

m +1

α ν

m

∑ pνα v =1

n − −1

v =1

  

δk + k −1

Pv ∆λv , by the lemma p nα−ν −1 Pnα−1

∆λv , using (3.2)

= 0(1) , m → ∞ .

Finally,

 Pnα  α ∑ n=2  p n m +1

≤ 0(1)

  

δk + k −1

m +1

∑ n= 2

≤ 0(1)

m +1

∑ n=2

Tn , 4  Pnα  α p  n

   

 Pnα  α p  n

   

k

 Pα = ∑  nα n=2  p n

δk + k −1

δk + k −1

m +1

  

δk + k −1

( p nα ) k  n −1 α   ∑ Pn −ν − 2 Pνα ∆λv  α α k ( Pn Pn −1 )  v =1  k

p nα k  n −1 ( α α ) ∑ Pn −1 Pn  v =1 1 ( Pnα−1 ) k

n −1

∑ v =1

k

α

p n −ν −1

(

 P ∆λ v  , using (3.3)  α ν

k

p n −ν −1 P λv α

α ν

)

 1  α P  n −1

 p n −ν −1  ∑ v =1  n −1

k −1

α

δk + k −1

 Pnα  1 n −1 α = 0(1) ∑  α  p P α λv α ∑ n −ν −1 ν Pn −1 v =1 n=2  p n  = 0(1), as m → ∞ , as above. n +1

This completes the proof of the theorem.

REFERENCE 1. Bor Huseyin: On the absolute summability factors of Fourier Series,

Journal of Computational Analysis and Applications, Vol.8, No.3, 223-227, 2006.

Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)