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)