J. Nonlinear Sci. Appl. 1 (2008), no. 3, 179–188
The Journal of Nonlinear Sciences and Applications
http://www.tjnsa.com
CONVERGENCE OF NEW MODIFIED TRIGONOMETRIC
SUMS IN THE METRIC SPACE L
JATINDERDEEP KAUR
1
AND S.S. BHATIA
2∗
Abstract. We introduce here new modified cosine and sine sums as
n
n
a0 X X
+
4(aj cos jx)
2
k=1 j=k
and
n X
n
X
4(aj sin jx)
k=1 j=k
and study their integrability and L1 -convergence. The L1 -convergence of cosine
and sine series have been obtained as corollary. In this paper, we have been able
to remove the necessary and sufficient condition ak log k = o(1) as k → ∞ for
the L1 -convergence of cosine and sine series.
1. Introduction
Consider cosine and sine series
∞
a0 X
+
ak cos kx
2
k=1
and
∞
X
ak sin kx
(1.1)
(1.2)
k=1
Date: Received:March 2008 ; Revised: Dec 2008
Corresponding author.
2000 Mathematics Subject Classification. Primary 42A20; Secondary 42A32.
Key words and phrases. L1 -convergence, Dirichlet kernel, Fejer kernel, monotone sequence.
∗∗
This work is the part of UGC sponsored Major Research Project entitled
”Integrability and L1 -convergence of trigonometric Series”.
∗
179
180
JATINDERDEEP KAUR
or together
∞
X
1
AND S.S. BHATIA
2∗
ak φk (x)
(1.3)
k=1
Where φk (x) is cos kx or sin kx respectively. Let the partial sum of (1.3) be
denoted by Sn (x) and t(x) = lim Sn (x). Further, let tr (x) = lim Snr (x) where
n→∞
Snr (x) represents rth derivative of Sn (x).
n→∞
Definition 1.1. A sequence {ak }is said to convex if 42 ak P
≥ 0, where 42 ak =
4(4ak ) and 4ak = ak − ak+1 , and quasi-convex sequence if (k + 1)42 ak < ∞.
The concept of quasi-convex was generalized by Sidon [4] in the following manner:
Definition 1.2. [4] A null sequence {ak } is said to belong to class S if there
exists a sequence {Ak } such that
Ak ↓ 0, k → ∞,
∞
X
Ak < ∞,
(1.4)
(1.5)
k=0
and
|∆ak | ≤ Ak , ∀ k.
(1.6)
A quasi-convex null sequence satisfies conditions of the class S because we can
choose
∞
X
An =
|42 am |.
m=n
1
Concerning L -convergence of (1.1) and (1.2), the following theorems are known:
Theorem 1.3. ([1], p. 204) If ak ↓ 0 and {ak } is convex or even quasi-convex,
then for the convergence of the series (1.1) in the metric space L1 , it is necessary
and sufficient that ak log k = o(1), k → ∞.
This theorem is due to Kolmogorov [2]. Teljakovskii [5] generalized Theorem
1.3 for the cosine series (1.1) with coefficients {ak } satisfying the conditions of
the class S in the following form:
Theorem 1.4. If the coefficient sequence {ak } of the cosine series (1.1) belongs
to the class S, then a necessary and sufficient condition for L1 -convergence of
(1.1) is ak log k = o(1), k → ∞.
∞ ³
X
ak ´
Theorem 1.5. ([1], p. 201) If ak ↓ 0 and
< ∞, then (1.3) is a Fourier
k
k=1
series.
In the present paper, we introduce new modified cosine and sine sums as
n
n
a0 X X
fn (x) =
+
4(aj cos jx)
2
k=1 j=k
CONVERGENCE OF NEW MODIFIED TRIGONOMETRIC SUMS
and
gn (x) =
n X
n
X
181
4(aj sin jx)
k=1 j=k
and study their integrability and L1 -convergence under a new class SJ of coefficient sequences defined as follows:
Definition 1.6. A null sequence {ak } of positive numbers belongs to class SJ if
there exists a sequence {Ak } such that
Ak ↓ 0, as k → ∞,
∞
X
(1.7)
Ak < ∞,
(1.8)
k=1
¯ ³ a ´¯ A
¯
k ¯
k
∀ k.
¯4
¯≤
k
k
Clearly class SJ ⊂ class S, Since
¯ ³ a ´¯ A
¯
k
k ¯
⇒ |∆ak | ≤ Ak , ∀ k.
¯≤
¯4
k
k
Following example shows that the class SJ is proper subclass of class S.
(1.9)
Example 1.7. For k = I − {0, 1, 2}, where I is set of integers, define {ak } = k13 ,
then there exists {Ak } = k12 such that {ak } satisfies all the conditions of class
S but not class SJ. However, for k = 1, 2, 3... the sequence {bk } = k13 satisfies
conditions of class SJ as well as conditions of class S. Clearly, class SJ is proper
subclass of class S.
Now, we define a new class SJr of coefficient sequences which is an extension of
class SJ.
Definition 1.8. A null sequence {ak } of positive numbers belongs to class SJr if
there exists a sequence {Ak } such that
Ak ↓ 0, as k → ∞,
∞
X
k r Ak < ∞,
(1.10)
(r = 0, 1, 2, ...)
(1.11)
k=1
¯ ³ a ´¯ A
¯
k ¯
k
∀ k.
(1.12)
¯4
¯≤
k
k
clearly, for r = 0, SJr = SJ. It is obvious that SJr+1 ⊂ SJr , but converse is not
true.
1
Example 1.9. For k = 1, 2, 3..., define bk = kr+2
,
shall show that {bk } does not belong to SJr+1 .
1
Really, bn = r+2 → 0 as n → ∞.
n
r = 0, 1, 2, 3, ... Firstly, we
182
JATINDERDEEP KAUR
Let there exists {Ak } =
∞
X
1
k=1
k
1
,
kr+2
1
AND S.S. BHATIA
r = 0, 1, 2, 3, ... s.t.
∞
X
2∗
k r+1 Ak = k r+1
k=1
1
k r+2
=
is divergent, i.e. {bk } does not belong to SJr+1 .
∞
∞
X
X
1
1
But, Ak ↓ 0, as k → ∞, and
k r Ak = k r r+2 =
< ∞,
2
k
k
k=1
k=1
¯ ¡ ¢¯
Also ¯4 bkk ¯ ≤ Akk , ∀ k.
Therefore, {bk } belongs to SJr .
In what follows, tn (x) will represents fn (x) or gn (x).
2. Lemmas
We require the following lemmas in the proof of our result.
Lemma 2.1. [3] Let n ≥ 1 and let r be a nonnegative integer, x ∈ [², π]. Then
r
|D̃nr (x)| ≤ C² nx where C² is a positive constant depending on ², 0 < ² < π and
D̃n (x) is the conjugate Dirichlet kernel.
Lemma 2.2. [5] Let {ak } be a sequence of real numbers such that |ak | ≤ 1 for
all k. Then there exists a constant M > 0 such that for any n ≥ 1
¯
Z π ¯¯X
n
¯
¯
¯
ak D̃k (x)¯ dx ≤ M (n + 1).
¯
¯
0 ¯
k=0
Moreover by Bernstein’s inequality, for r = 0, 1, 2, 3.....
¯
Z π ¯¯X
n
¯
¯
¯
r
ak D̃k (x)¯ dx ≤ M (n + 1)r+1 .
¯
¯
0 ¯
k=0
Lemma 2.3. [3] ||D̃nr (x)||L1 = O(nr log n), r = 0, 1, 2, 3...., where D̃nr (x) represents the rth derivative of conjugate Dirichlet-Kernel.
3. Main Results
In this paper we shall prove the following main results:
Theorem 3.1. Let the coefficients of the series (1.3) belongs to class SJ, then
the series (1.3) is a Fourier series.
n ³
X
ak ´
, we get
Proof. Making Use of Abel’s transformation on
k
k=1
n ³
X
ak ´
k=1
k
=
³a ´
k
− an
k4
k
k=1
n−1
X
¶
n−1 µ
X
Ak
≤
− an
k
k
k=1
CONVERGENCE OF NEW MODIFIED TRIGONOMETRIC SUMS
But (1.3) belongs to class SJ, therefore, the series
∞ ³
X
ak ´
k
k=1
Hence the conclusion of theorem follows from Theorem (1.5.
183
converges.
¤
Theorem 3.2. Let the coefficients of the series (1.3) belongs to class SJ, then
lim tn (x) = t(x) exists for x ∈ (0, π).
(3.1)
t ∈ L1 (0, π)
(3.2)
n→∞
||t(x) − Sn (x)|| = o(1), n → ∞
(3.3)
Proof. We will consider only cosine sums as the proof for the sine sums follows
the same line.
To prove (3.1), we notice that
n
n
a0 X X
tn (x) =
+
4(aj cos jx)
2
k=1 j=k
n
a0 X
+
[ak cos kx − ak+1 cos(k + 1)x + ak+1 cos(k + 1)x
tn (x) =
2
k=1
=
n
X
a0
+
2
k=1
−ak+2 cos(k + 1)x + ..... + an cos nx − an+1 cos(n + 1)x]
n
X
ak cos kx −
an+1 cos(n + 1)x
k=1
tn (x) = Sn (x) − nan+1 cos(n + 1)x
(3.4)
∞
X
Ak < ∞, therefore, by Oliver’s theorem we
Since Ak ↓ 0, as k → ∞ and
k=1
have, kAk → 0, as k → ∞ and so
µ ¶
∞
∞
³a ´ X
X
Ak
k
2
2
k
nan = n
4
≤
= o(1)
k
k
k=n
k=n
Also cos(n + 1)x is finite in (0, π). Hence
lim tn (x) = lim Sn (x) = t(x)
n→∞
Moreover,
n→∞
Ã
!
n
a0 X
t(x) = lim tn (x) = lim Sn (x) = lim
+
ak cos kx
n→∞
n→∞
n→∞
2
k=1
à n
!
X
a0
+ lim
ak cos kx
=
n→∞
2
k=1
Use of Abel’s transformation yields
à n
#
!
" n−1
X
X ³ ak ´
a
n
lim
D̃k0 (x) + D̃n0 (x)
ak cos kx = lim
4
n→∞
n→∞
k
n
k=1
k=1
(3.5)
184
JATINDERDEEP KAUR
1
AND S.S. BHATIA
2∗
where D̃n0 (x) is the derivative of conjugate Dirichlet kernel.
∞
X
=
4
³a ´
k
k
k=1
≤
¶
∞ µ
X
Ak
k=1
k
D̃k0 (x)
D̃k0 (x)
¶
∞ µ
X
Ak
D̃k0 (x) converges.
k
k=1
Therefore, the limit t(x) exists for x ∈ (0, π) and thus (3.1) follows.
For x 6= 0, it follows from (3.4) that
By the given hypothesis and lemma 2.1, the series
∞
X
t(x) − tn (x) =
ak cos kx + nan+1 cos(n + 1)x
k=n+1
"
=
lim
m→∞
#
m ³
X
ak ´
k cos kx + nan+1 cos(n + 1)x
k
k=n+1
Applying Abel’s transformation, we have
=
∞
X
4
k=n+1
∞ µ
X
³a ´
k
k
¶
D̃k0 (x) −
an+1 0
D̃ (x) + nan+1 cos(n + 1)x
n+1 n
¡ ¢
4 akk
a
¡ Ak ¢ D̃k0 (x) + n+1 D̃n0 (x) + nan+1 cos(n + 1)x
≤
n+1
k
k=n+1
³ ´
³ ´
µ ¶X
µ
¶X
∞
k 4 aj
n 4 aj
X
j
j
Ak
An+1
³ ´ D̃j0 (x) −
³ ´ D̃j0 (x)
4
≤
Aj
Aj
k j=1
n + 1 j=1
k=n+1
Ak
k
j
+
j
an+1 0
D̃ (x) + nan+1 cos(n + 1)x
n+1 n
Thus from lemma 2.2 and 2.3, we obtain
³ ´
¯
¯
¯
µ ¶ Z π ¯X
k 4 aj
∞
X
¯
¯
j
Ak
0
¯
³ ´ D̃j (x)¯¯ dx
||t(x) − tn (x)|| ≤
4
¯
Aj
k
0 ¯ j=1
¯
k=n+1
j
³
´
¯
¯
¯
¯
µ
¶ Z π ¯X
Z π¯
n 4 aj
¯ an+1 0
¯
¯
¯
j
An+1
0
¯
¯
¯
¯
´
³
+
D̃
(x)
dx
+
D̃
(x)
j
¯ n + 1 n ¯ dx
¯
¯
Aj
n+1
0
0 ¯ j=1
¯
j
Z π
+n|an+1 |
| cos(n + 1)x| dx
0
CONVERGENCE OF NEW MODIFIED TRIGONOMETRIC SUMS
Ã
= O
∞
X
µ
2
k 4
k=n+1
Ak
k
¶!
µ
µ
+O n
Z
2
An+1
n+1
185
¶¶
π
+O(an+1 log n) + n|an+1 |
| cos(n + 1)x| dx
0
But
n
X
Ak =
k=1
n−1
X
k(k + 1)
k=1
2
µ
4
Ak
k
¶
+
n(n + 1) An
2
n
since {ak } ∈SJ, we have
Ak
= (k + 1)Ak = o(1) as k → ∞.
k
µ ¶
∞
X
Ak
2
and therefore the series
k 4
, converges.
k
k=n+1
Moreover,
Z π
Z π
Z π
2
|cos(n + 1)x| dx =
cos(n + 1)x dx −
cos(n + 1)x dx ≤
k(k + 1)
0
π
2
0
2
n+1
and since an ’s are positive, we have by (3.5) that an log n ≤ nan = o(1), f or n ≥
1.
Hence, it follows that
||t(x) − tn (x)|| = o(1) as n → ∞.
(3.6)
and since tn (x) is a polynomial, therefore t(x) ∈ L1 . This proves (3.2).
We now turn to the proof of (3.3), We have
||t − Sn || = ||t − tn + tn − Sn ||
≤ ||t − tn || + ||tn − Sn ||
= ||t − tn || + ||nan+1 cos(n + 1)x||
Z π
≤ ||t − tn || + n|an+1 |
| cos(n + 1)x| dx
0
Z
π
Further, ||t(x) − tn (x)|| = o(1), n → ∞ (by (3.6)),
| cos(n + 1)x| dx ≤
0
and {ak } is a null sequence,therefore the conclusion of theorem follows.
2
n+1
¤
Theorem 3.3. Let the coefficients of the series (1.3) belongs to class SJr , then
lim trn (x) = tr (x) exists for x ∈ (0, π).
(3.7)
tr ∈ L1 (0, π),
(3.8)
n→∞
(r = 0, 1, 2, ...)
||tr (x) − Snr (x)|| = o(1), n → ∞.
(3.9)
186
1
JATINDERDEEP KAUR
AND S.S. BHATIA
2∗
Proof. We will consider only cosine sums as the proof for the sine sums follows
the same line. As in the proof of the Theorem 3.2, we have
n
n
a0 X X
tn (x) =
+
4(aj cos jx)
2
k=1 j=k
= Sn (x) − nan+1 cos(n + 1)x
we have, then
rπ ´
− n(n + 1) an+1 cos (n + 1)x +
=
2
∞
X
Since Ak ↓ 0, as k → ∞ and
k r Ak < ∞, therefore, we have, k r+1 Ak →
k=1
0, as k → ∞ and so
n
r+1
an = n
¡
Also cos (n + 1)x +
³
r
Snr (x)
trn (x)
rπ
2
¢
r+2
∞
X
4
³a ´
k=n
k
k
≤
∞
X
µ
k
r+2
k=n
Ak
k
¶
= o(1)
(3.10)
is finite in (0, π). Hence
tr (x) =
lim trn (x)
n→∞
lim Snr (x)
!
à n
´
³
X
rπ
k r ak cos kx +
= lim
n→∞
2
k=1
=
n→∞
use of Abel’s transformation yields
à n
!
³
´
X
rπ
lim
k r ak cos kx +
=
n→∞
2
k=1
lim
n→∞
" n−1
X
4
³a ´
k
k
k=1
#
a
n
D̃kr+1 (x) + D̃nr+1 (x) ,
n
where D̃nr+1 (x) represents the (r + 1)th derivative of conjugate Dirichlet kernel.
=
≤
∞
X
4
k=1
∞ µ
X
k=1
³a ´
Ak
k
k
k
¶
D̃kr+1 (x) + lim
ha
n
n→∞
D̃kr+1 (x) + lim
n→∞
n
ha
n
n
i
D̃nr+1 (x)
i
D̃nr+1 (x)
By making use of the given hypothesis, lemma 2.1 and (3.10), the series
¶
∞ µ
X
Ak
k
k=1
r
converges. Therefore, the limit t (x) exists for x ∈ (0, π) and thus (3.7) follows.
To prove (3.8), we have
r
t (x) −
trn (x)
=
D̃kr+1 (x)
³
³
rπ ´
rπ ´
+ n(n + 1)r an+1 cos (n + 1)x +
k r ak cos kx +
2
2
k=n+1
∞
X
CONVERGENCE OF NEW MODIFIED TRIGONOMETRIC SUMS
187
Making use of Abel’s transformation, we obtain
=
∞
X
k=n+1
4
³a ´
k
k
D̃kr+1 (x) −
³
an+1 r+1
rπ ´
D̃n (x) + n(n + 1)r an+1 cos (n + 1)x +
n+1
2
¶ ¡ ¢
∞ µ
³
X
Ak 4 akk
an+1 r+1
rπ ´
r+1
r
¡
¢
≤
D̃k (x) −
D̃ (x) + n(n + 1) an+1 cos (n + 1)x +
Ak
k
n+1 n
2
k
k=n+1
³ ´
³ ´
aj
µ
µ ¶X
¶ n
∞
k 4 aj
X
j
An+1 X 4 j
Ak
r+1
³ ´ D̃j (x) −
³ ´ D̃jr+1 (x)
≤
4
Aj
Aj
k
n
+
1
j=1
j=1
k=n+1
j
j
³
´
an+1 r+1
rπ
+
D̃n (x) + nan+1 cos (n + 1)x +
n+1
2
Thus from lemma 2.2 and 2.3, we obtain
³ ´
¯
¯
¯ k 4 aj
¯
X
¯
¯
j
Ak
r+1
r
r
¯
¯ dx
³
´
4
D̃
(x)
||t (x) − tn (x)|| ≤
j
¯
¯
Aj
k
0 ¯ j=1
¯
k=n+1
j
³
´
¯
¯
¯
¯
µ
¶ Z π ¯X
Z π¯
n 4 aj
¯ an+1 r+1 ¯
¯
¯
j
An+1
r+1
¯
¯
¯
³ ´ D̃j (x)¯¯ dx +
+
¯ n + 1 D̃n (x)¯ dx
¯
Aj
n+1
0
0 ¯ j=1
¯
j
Z π
³
rπ ´
+n(n + 1)r |an+1 |
| cos (n + 1)x +
| dx
2
0
à ∞
µ
µ ¶!
µ
¶¶
X
Ak
An+1
r+2
r+2
+O n
= O
k 4
+ O(nr an+1 log n)
k
n
+
1
k=n+1
Z π
³
rπ ´
r
| dx
+n(n + 1) |an+1 |
| cos (n + 1)x +
2
0
µ
∞
X
¶Z
π
Using the argument as in the proof of theorem 3.2, it is easily shown that the
µ ¶
∞
X
Ak
r+2
series
, converges. Moreover,
k 4
k
k=n+1
Z
0
π
³
rπ ´
2
| cos (n + 1)x +
| dx ≤
2
n+1
and for n ≥ 1, nr an log n ≤ nr+1 an = o(1) by (3.10). Hence it follows that
||tr (x) − trn (x)|| = o(1) as n → ∞.
and since trn (x) is a polynomial, therefore tr (x) ∈ L1 . This proves (3.8).
(3.11)
188
JATINDERDEEP KAUR
1
AND S.S. BHATIA
2∗
We now turn to the proof of (3.9). We have
||tr − Snr || = ||tr − trn + trn − Snr ||
≤ ||tr − trn || + ||trn − Snr ||
³
rπ ´
= ||tr − tn r || + ||n(n + 1)r an+1 cos (n + 1)x +
||
2
Z π
³
rπ ´
r
r
r
|dx
≤ ||t − tn || + n(n + 1) |an+1 |
| cos (n + 1)x +
2
0
Z π
³
rπ ´
r
r
Further, ||t (x)−tn (x)|| = o(1), n → ∞ (by (3.11)) ,
| cos (n + 1)x +
| dx ≤
2
0
2
and {ak } is a null sequence, the conclusion of theorem follows.
¤
n+1
Remark 3.4. The case r = 0, in Theorem 3.3 yields the Theorem 3.2.
Acknowledgements: Authors are thankful to Professor Babu Ram (Retd.)
of M. D. University, Rohtak for his valuable suggestions during the preparation
of this paper.
References
[1] N.K. Bary,A treatise on trigonometric series, Vol II, Pergamon Press, London, (1964). 1.3,
1.5
[2] A.N. Kolmogorov, Sur l’ordere de grandeur des coefficients de la series de Fourier Lebesque, Bull. Acad. Pol. Sci. Ser. Sci. Math. Astronom. Phys., (1923), 83–86. 1
[3] S. Sheng, The extension of the theorems of C̆.V. Stanojević and V.B. Stanojević, Proc.
Amer. Math. Soc., 110, (1990), 895–904. 2.1, 2.3
[4] S. Sidon, Hinreichende Bedingungen für den Fourier-Charakter einer trigonometrischen
Reihe, J. London Math Soc., 14, (1939), 158–160. 1, 1.2
[5] S.A. Teljakovskǐi, A sufficient condition of Sidon for the integrability of trigonometric series,
Mat. Zametki, 14(3), (1973), 317–328. 1, 2.2
1
School of Mathematics & Computer Applications,
Thapar University Patiala(Pb.)-147004, INDIA.
E-mail address: jatinkaur4u@yahoo.co.in
2
School of Mathematics & Computer Applications,
Thapar University Patiala(Pb.)-147004, INDIA.
E-mail address: ssbhatia63@yahoo.com