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A NOTE ON INDEX SUMMABILITY FACTORS
OF FOURIER SERIES
U.K. Misra1, Mahendra Misra2 and B.P. Padhy3
1Department of Mathematics Berhampur University
Berhampur-760007, Orissa (India)
e-mail: [email protected]
2Principal Government Science CollegeMlkangiri, Orissa
3Roland Institute of TechnologyGolanthara-760008, Orissa
(India)
e-mail: [email protected]
ABSTRACT
A theorem on index summability factors of Fourier Series has
been established
Key words:Summability factors, Fourier series.
AMS Classification No: 40D25
J. Comp. & Math. Sci. Vol. 1(1), 67-70 (2009).
1. INTRODUCTION
Let na be a given infinite serieswith the sequence of partial
sums ns . Let
n
p be a sequence of positive numbers such
that
(1.1) ,0
PnaspP in
r
vn
1,0 ip iThe sequence to sequence transformation
(1.2)
n
v
vvn
n
n spP
t0
1
defines the sequence of the npN, -mean
of the sequence ns generated by the sequence
of coefficients np .
The series na is said to be summable
1,, kpNkn
, if
(1.3)
1
1
1
n
k
nn
k
n
n ttp
P
In the case when 1np , for all kandn ,1
knpN, summability is same as 1,C summa-
bility. Forkn
pNk ,,1 summability is same
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[ 68 ]
as npN, - summability..
Let n be any sequence of positive
numbers. The series na is said to be summable
1,,, kpNknn
, if
(1.4)
11
1
nn
n
k
n tt .
Clearly,
kn
n
np
PpN ,, is same as 1,, kpN
kn
and1
1,, npN is npN, .
Let tf be a periodic function with period 2
and integrable in the sense of Lebesgue over , . Without loss of
generality, we mayassume that the constant term in the Fourier
series of )(tf is zero, so that
(1.5)
)(tf ~ )()sincos(11
tAntbntan
n
n
nn
2. Known Theorem :
Dealing with 1,, kpNkn
, summability
factors of Fourier series, Bor1 proved thefollowing theorem:
Theorem-A :If n is a non-negativeand non-increasing sequence
such that
nnp , where np is a sequenceof positive numbers such that
nasPn
and
n
v
nvv PtAP1
)(0)( . Then the factored
Fourier series nnn PtA )( is summable
1,, kpNk
n .
We prove an analogue theorem on
knnpN ,, - summability, 1k , in thefollowing form:
3. Main Theorem :
Theorem:Let np is a sequence of
positive numbers such that n ppP 21
np.... as n and n is a non-negative, non-increasing sequence
such that
nnp . If
(3.1) (i).
n
v
nvv PtAP1
)(0)(
(3.2) (ii).
1
1 1
11 ,0m
vn v
v
n
vnk
nP
p
P
p
mas and
(3.3) (iii). vvnvvn pP 01 ,
then the series nnn PtA )( is summable
1,,, kpNknn
. and n , a sequence ofpositive numbers.
4. Required Lemma :
We need the following Lemma for the
proof of our theorem.
Lemma [1]:If n is a non-negative
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[ 69 ]
and non-increasing sequence such that
nnp , where np is a sequence ofpositive numbers such that
nasP
n
then nasP nn )1(0 and nnP .
5. Proof of the Theorem :
Let )(xtn be the n-th npN, mean
of the series ,)(1
nnnn PxA then by definition
we have
n
v
v
r
rrrvn
n
n PxApP
xt0 0
)(1
)(
n
rv
vnr
n
r
rr
n
pPxAP
0
)(1
r
n
r
rnrr
n
PPxAP
0
)(1
.
Then
n
r
rrrrn
n
nn xAPP
Pxtxt
0
1
1)()(
n
r
rrrrn
n
xAPPP
1
0
1
1
)(1
)(1 1
1 xAPP
P
P
Pn
r
rrr
n
rn
n
rn
n
r
rrrnrnnvn
nn
xAPPPPPPP 1
11
1
)(1
1
1
11
1
1 n
r
rnrnnrn
nn
PPPPPP
1
)(r
v
vv xAP , using partial summation
formula with 0op .
1
1
11
1
1 n
r
rrnrnnrn
nn
PPpPp
PP
1
1
211
n
r
rrnrnnrn PpPPp , using
(3.1)
4,3,2,1, nnnn TTTT , say..
In order to complete the proof of the
theorem, using Minkowski's inequality, it is
sufficient to show that
1
,
1.4,3,2,1,
n
k
in
k
n iforT
Now, we have
1
2
1
2
1
1
1
1,
1 1m
n
m
n
kn
v
vvvnk
n
k
n
k
n
k
n Pp
PT
1
2
1
1
1 1m
n
n
v
k
v
k
vvn
n
kPp
P
11
1
1kn
v
vn
n
pP
, using Holder's
inequality.
1
2
11
1
1 11
m
n
k
vvvv
n
v
vn
n
k
n PPpP
O
1
1
1
1
1m
vn n
vnk
n
m
v
vvP
pPO
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m
v
vvpO1
)1( (using 3.2)
masO ),1( .
1
2
1
2
1
1
1
1
1
2.
1 1m
n
m
n
kn
v
vvvnk
n
k
n
k
n
k
n Pp
P
T
1
2
1
1
1
1
1 1m
n
n
v
k
v
k
vvn
n
k
n PpP
1
1
1
1
1
1k
n
v
vn
n
pP
1
1
2
1
1
1
1
1)(
1)1(
kvvm
n
vv
n
v
vn
n
k
n PPp
PO
m
v
m
vn n
vnk
vvP
pPO1
1
1 1
11)1(
,)1(0
1
m
v
vvp using (3.2)
mas,)1(0 .
Now,
1
2
1
2
1
1
1
1
3,
1 1m
n
km
n
vv
n
v
vnk
n
k
n
k
n
k
n PpP
T
1
1
1
1
1
2
1
1
1
1 11
kn
v
vn
n
m
n
k
vv
n
v
vn
n
k
n p
PPp
P
1
1
2
1
1
1
1 11
kvv
m
n
vv
n
v
vn
n
k
n PPpP
O
,1
11
2
1
1
1
1
m
n
vv
n
v
vn
n
k
n Pp
PO
by the lemma
n
vnm
vn
n
n
m
v
vvP
pPO 1
1
1
1
1
)1(
m
v
vvp1
)1(0 , using (3.2)
m,)1(0 .
Finally,
1
2
1
2 1
1
4,
1m
n
m
nk
n
k
n
k
nk
n
k
n
k
nPP
pT
2
1
1
k
vvnv
n
v
PP
km
n
vvvn
n
v
k
n
k
n Pp
P
1
2
1
1
11
1 1)1(0 ,
using (3.3)
1
2
1
1
11
1 1)1(0
m
n
vn
n
vn
k
n pP
1
1
1
1
1
1
kn
v
vn
n
k
vv pP
P
vv
n
v
vn
n
n
n
k
n Pp
P
1
1
1
1
1
2
1 1)1(0
mas),1(0 , as above
This completes the proof of the theorem.
REFERENCE
1. Bor Huseyin: On the absolute summability
factors of Fourier Series, Journal of Compu-
tational Analysis and Applications, Vol. 8,
No. 3, 223-227 (2006).
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