Technology (IJRASET)
Integrability of Trigonometric Series with
Coefficients
Satisfying Certain Conditions
Dr. Manisha Sharma
1Department of Applied Sciences, Krishna Engineering College, Ghaziabad, Uttar Pradesh-201007INDIA)
Abstract: Let
1
P
and
1
P
P
1
, suppose that {a
n} is a sequence of ,Numbers such thata
n
A
j orj
n
A
a
and
P P n n
P P
a
n
L
n
/ 1
1
2
)
)(
(
, then we will prove that 1/
1
f
(
x
)
L
(
P
,
)
x
L
P
And
n Pn
P P P
P P
a
n
L
n
j
P
B
x
f
x
L
1
(
)
,
,
(
)(
)
1
2
, /
1
& ii)Let {
a
n} be a sequence of numbers such thata
n
A
j ora
n
A
j. If1
P
and
1
P
P
1
, then anecessary and sufficient condition that 1/
1
f
(
x
)
L
(
P
,
)
x
L
P
is that
P
n n
P P
a
n
L
n
(
)
1
2
.
I. INTRODUCTION
A. Definitions
A function
(
x
)
is said to belong to classL
P
,
if
0
,
)
(sin
)
(
x
Px
Pdx
is a real number and P>0, it is easy to seethat
P
,
L
for
0
L
P And
,
0
L
P
for
L
PAnd
L
P
,
L
Pif
0
.
We define norm of a function
x
L
P
,
as:
P P P
P
x
x
dx
x
/ 1
0
,
sin
A positive continuous function
L
(
x
)
is said to be “slowly increasing”, in the sense of Karamata [4] if1
)
(
)
(
lim
L
x
kx
L
x
For everyk
0
.A sequence
a
n of non-negative number is said to be quasi-monotone [7, 9] if for some constant
0
n
a
a
n n
1
1 for all
n
n
0
.An equivalent definition is that
n
a
n
0
for some
0
.We shall say that the coefficients of trigonometric series,
1 0
cos
2
nn
nx
a
a
x
f
and
1
sin
)
(
n
n
nx
a
x
Technology (IJRASET)
decreases and to the class
A
j if, for somej
0
, the number n ja
n
,a
n
0
increases. The coefficients decrease monotonicallyto zero belongs to the class
A
0.B. Some known results
Theorem If
0
1
,
a
n
0
,
then 1
f
(
x
)
L
(
0
,
)
x
L
x
if and only if n
n
a
n
L
n
(
)
1 1
is convergent.
Theorem If
a
n
0
,P
1
, and
1
0
, then the necessary and sufficient condition that nPn
P P
a
n
L
n
(
)
1 1
should
converge, is that 1
1
f
(
x
)
L
(
0
,
)
x
L
x
P
P
.
Theorem Let {
a
n} is quasi-monotone if
1
and such that0
M
1
n
L
1(
n
)
a
n
M
2
with
0
, ifP
1
and
1
P
1
. Then,)
(
1
)
,
,
(
2f
x
x
L
x
P
L
f
P
is integrable in(
0
,
)
if and only if
P
n n
P
a
n
L
n
2(
)
1 2
, where
L
1 andL
2are slowlyincreasing function in the sense of Karamata.
Theorem Let
a
n be positive and tends to zero. Leta
nn
k be monotonically decreasing for some non-negative integer k.Let
1
P
and
1
P
P
1
, then a necessary and sufficient condition thatf
x
L
P
,
, where
1
cos
~
n
n
nx
a
x
f
is that
P
n n
P P
a
n
1
2
.
Theorem Let {
a
n} be a positive sequence tending to zero and {a
nn
k} be monotonically decreasing for some non-negativeinteger k. If
1
P
and
1
P
P
1
then the necessary and sufficient condition that 1/1
f
(
x
)
L
(
P
,
)
x
L
P
is
that,
P
n n
P P
a
n
L
n
(
)
1
2
, where
1
cos
~
n
n
nx
a
x
f
.C. Our theorems
We shall prove the following theorems
Theorem Let
1
P
and
1
P
P
1
, suppose that {a
n} is a sequence of numbers such thata
n
A
j ora
n
A
jand
P P n n
P P
a
n
L
n
/ 1
1
2
)
)(
(
,
Then, 1/
1
f
(
x
)
L
(
P
,
)
x
L
P
And
n Pn
P P P
P P
a
n
L
n
j
P
B
x
f
x
L
1
(
)
,
,
(
)(
)
1
2
, /
1
Technology (IJRASET)
Theorem Let {
a
n} be a sequence of numbers such thata
n
A
j ora
n
A
j. If1
P
and
1
P
P
1
, then anecessary and sufficient condition that 1/
1
f
(
x
)
L
(
P
,
)
x
L
P
is that
P n n P P
a
n
L
n
(
)
1
2
.
D. Lemmas
The following lemmas will be required for the proof of our theorems
1) Lemma: Let
f
(
x
)
0
for(
x
)
0
and f(x) be the integral of f(x). If1
P
q
and r>-1 thenq q qr
dt
t
t
f
t
L
t
/ 1 01
1
(
)
P Pdt
t
f
t
L
t
B
/ 1 0 Pr 1)
(
1
q q qrdt
t
t
f
t
L
t
/ 1 01
(
)
)
(
P Pdt
t
f
t
L
t
B
/ 1 0 Pr 1)
(
)
(
2) Lemm : Let
1
n n
a
be a series of positive terms andA
n=
n k
k
a
and suppose that),
1
,
1
(
,
)
)(
(
, 1
c
P
na
n
L
n
n p nc
then
n
L
n
A
nPB
n c
)
(
1,
)
)(
(
, 1 p n n cna
n
L
n
where B is some constant
depending upon c and P.
3) Lemma: For any non-negative
andn
, we have
j k n x j k n x n n k kA
a
if
a
j
A
a
if
a
j
kx
a
,
sin
2
,
sin
2
)
(
, 1 ) 1 ( 2 2 ), 1 ( 2 2 1 ) 1 ( 2 ) 1 ( 2 1 1
,...
2
,
1
,
0
,
2
k
k
x
Where
x
cos
x
or
x
sin
x
4) Lemma: If
a
k
A
j ora
k
A
j then
1 1 ) 1 ( 2 1 1)
,
(
)
1
(
2
n k kn
c
j
k
a
a
n
if
a
k
A
j,
1 2 ) 1 ( 2 1 0)
,
(
)
1
(
2
n k kn
c
j
k
a
a
n
if
a
k
A
j,Where,
0
2
0
2
)
,
(
1 1j
for
j
for
j
C
j
0
2
0
1
)
,
(
1 2j
for
j
for
j
C
j
Technology (IJRASET)
n
n
n
A
1
2
,
)
(
n
n k
a
n
A
2
)
(
Thenj k
n
B
j
A
n
if
a
A
na
(
)
(
)
j k
n
B
j
A
n
if
a
A
na
(
)
(
)
Where
B
(
j
)
is some positive constant depending onj
. Proof:j k j
j m j m
s k
j k
A
a
if
s
m
s
m
a
K
k
a
),
1
(
,j k j
j s j j m
s k
k
K
k
a
s
m
m
s
if
a
A
a
(
1
),
,We set
s
n
m
n
1
,
2
in (i) ands
n
,m
2
n
in (ii), we have
1
(
)
(
)
1
2
n
A
j
B
n
A
na
j jn n
n
And
na
n
B
(
j
)
A
(
n
)
.This completes the proof of the lemma.
E. Proof of Theorem
Since we are given that
P
n n
P P
a
n
L
n
(
)
1
2
, it follows by virtue of Lemma2 (putting
c
p
p
2
and ka k
k
a
)that
p
n k
k n
P P
k
a
n
L
n
(
)
1
2
, further on putting n=1, we have
1
k k
k
a
.
Put
1 ) 1 ( 2
) 1 ( 2 0 1
1
cos
cos
)
(
n
n k
k n
k k
n
x
a
kx
a
kx
R
0
), 1 ( 2 0
), 1 ( 2
2
,
,
sin
2
1
j k n
j k n
x
A
a
if
a
A
a
if
a
j
,...
2
,
1
,
0
,
2
k
k
x
by virtue of lemma3 ( choosing
x
cos
x
)Now using the particular case, when
1
of lemma 4, we obtain)
(
x
R
n2
sin
x jB
x
k
k
a
a
n k
k
n
,
2
1
1
Therefore series
1
cos
n
n
nx
Technology (IJRASET)
have
1cos
)
(
k kkx
a
x
f
=
nk k n
k
k
kx
a
kx
a
1 1cos
cos
x
B
a
j n k k
1
1 1 n k k nk
a
a
n k ka
1
x
O
a
x
O
1
n 11
n 1
k k
k
a
s
n+
x
O
a
x
O
1
n 11
n 1k k
k
a
Where
s
n=
n k k
a
1Now
2 1 2 0sin
)
(
1
sin
)
(
1
n p p n n p pdx
x
x
f
x
L
dx
x
x
f
x
L
x
dx
k
a
x
O
a
x
O
S
x
L
p p n k k n n n n n sin
1
1
1
1 1 2 1
a
x
d
x
x
L
j
p
B
dx
x
S
x
L
p
B
n p p n n p n p n n n pn
2 1 1 2 1sin
1
)
,
(
sin
1
)
(
2 1 1sin
1
)
,
(
n p p n n p n k kdx
x
k
a
x
L
j
p
B
p n n p p p n n pa
n
L
n
j
p
B
S
n
L
n
p
B
1 2 2 2 2)
(
)
,
,
(
)
(
)
,
(
+ p n k k n p pk
a
n
L
n
j
p
B
1 2 2)
(
)
,
,
(
=
j
1
j
2
j
3 (say)Now put
a
(x)
a
nfor
n
1
x
n
(
n
1
,
2
,...)
And
A
x
a
t
dt
0
)
(
)
(
then, we have,1
j
B
p
x
L
x
A
px
dx
n p n n
)
(
)
(
)
,
(
1 2 1
=dx
x
x
A
x
L
x
p
B
p p p p
(
)
(
)
)
,
(
/ 1 1 2
On applying lemma (ii) (taking
q
p
and
1
qr
p
p
2
) we get,
a
x
dx
x
L
x
p
B
Technology (IJRASET)
=
B
p
x
L
x
a
x
px
dx
n
p p n
n
)
(
)
(
)
(
)
,
(
2 2
1
p n np p
a
n
L
n
p
B
(
,
)
(
)
1 2
<
, by hypothesis of heses.By virtue of lemma 2 the theorem,
)
1
(
2
O
j
by the hypotand by the hypothesis,)
1
(
3
O
j
A similar method may be used to estimate
2
)
(sin
)
(
x
x
dx
f
P PThis finishes proof of theorem.
F. Proof of Theorem
Necessity: Suppose that 1/
1
f
(
x
)
L
(
P
,
)
x
L
P
, then we have to prove,
p
n n
p p
a
n
L
n
(
)
1 2
Let
x
du
u
f
x
f
0
1
(
)
(
)
,
x
du
u
f
x
f
0 1 2
(
)
(
)
Then
ku
du
k
a
x
f
k k
0 1
2
(
)
sin
=
x
k k
du
ku
k
a
0 1
sin
=
1
2
)
cos
1
(
k
k
kx
k
a
m
s k
k
kx
k
a
(
1
cos
)
2 (A)For any positive integers
s
and m, Case I: Whena
k
A
jNow set
1
2
n
x
andm
n
and using the inequality2
cos
1
2
nx
A
nx
for
n
1
x
4
n
4
, we have)
(
2 2x
f
Bn
A
n
, where B is some constant. By virtue of Lemma 5(i), it follows)
(
)
(
)
(
j
A
B
j
n
2f
2x
B
na
n
n
Now,
n
pn
p
na
n
L
n
(
)
1 2
Technology (IJRASET)
p n
p p
x
f
n
L
n
j
)
(
)
min(
(
))
(
21 2 2
B
x
n
n
1
4
4
dx
x
f
L
x
j
B
x pn
p p n
n
)
(
)
(
)
(sin
)
(
1 21
2 4
1 4
)
(
j
B
dx
x
x
f
L
x
p p
x p p
(sin
)
(
)
2(
)
/ 1 1 4
0
,
)
,
,
(
p
j
B
(sin
x
)
p p
(
L
(
1x))
1/pf
1(
x
)
pdx
4
0
)
,
,
(
p
j
B
x
L
pf
x
pdx
x p
)
(
)
(
)
(sin
1 1/4
0
=
B
(
,
p
,
j
)
p
p x
p
x
f
L
, 1
/ 1
)
(
This follows by lemma1 (i) (
q
p
,
p
p
1
pr
and
p
1
pr
respectively) Case II: Whena
k
A
j, we sets
n
andm
2
n
in (A) and obtain)
(
2 2x
f
Bn
A
n
For
n
1
x
8
n
8
By lemma 5(ii) we get
na
n
B
(
j
)
n
2f
2(
x
)
Now,
n
pn
p
na
n
L
n
(
)
1 2
p n
p p
x
f
n
L
n
j
)
(
)
min(
(
))
(
21 2 2
B
,
n
1
x
8
n
8
dx
x
f
L
x
j
B
x pn
p p n
n
)
(
)
(
)
(sin
)
(
1 21
2 8
1 8
)
(
j
B
dx
x
x
f
L
x
p p
x p
p
(
)
(
)
)
(sin
2/ 1 1 8
0
By some agreement as in the case I this possesses the necessity part of
theorem2.
Sufficiency: Now suppose that
p
n n
p p
na
n
L
n
(
)
1 2
. Then we have to show 1/
1
f
(
x
)
L
(
P
,
)
x
L
P
.
This follows by theorem 1 and proof of the theorem is thus completed.
II. CONCLUSION
Technology (IJRASET)
case of theorem1 while, for proof of necessity part, some miner changes are required.
For sake of convenience the theorem 1 is stated and proved otherwise theorem is essentially the same as
part of theorem 2.Our theorem 2 is not only more general then a result of Askey and Wainger [2] and theorem of Khan [5], but has a proof applicable in sine and cosine series both.
In the end I wish to express my sincere thanks to “Dr. J.P. Singh” for his kind guidance.
REFERENCES
[1] Aljancic, S R. Bojanic and M. Tomic: Sur I Integrability de certain series trigonometriques, Acad. Serbe sc. Pub1 Inst. Math, 8(1955), 67-84 [2] Askey, R and S. Wainger: Integrability theorems for Fourier series, Duke Math. Jour 33 (1966), 223-228.
[3] Igari, S: Some integrability theorems of trigonometric series and monotone decreasing functions, Tohoku math. J. (2) 12 (1960), 139-146 [4] Karamata, J: Sur Un mode de croissunce reguliers Bull. Soc. Math France 61 (1933) 55-62
[5] Khan R. S.: Intregrability of Trigonometric series Rend. Mat (3-4). (1968) 371-383
[6] Lebed, G. K: Trigonometric series with coefficients satisfying certain conditions, Math.Sbornik 74 (116)(1967) 100-118.(Russian) English translation Math SSSR Sbornik 3(1967)91-108 (1969)
[7] Shah, S. M.: A note on quasi- monotone series, The Math student 15(1947), 19-24
[8] Trigonometric series with quasi- monotone coefficients, proc. Amer. Math. Soc.13 (1962), 266-273 [9] Szasz, O: Quasi- monotone series, Amer. J. Math. 70(1948), 203-206.
