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Extensions of Móricz Classes and Convergence of Trigonometric Sine Series in L1-Norm

Thapar Institute of Engineering and Technology, Patiala, Punjab 147004, India
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Author to whom correspondence should be addressed.
Mathematics 2018, 6(12), 292; https://doi.org/10.3390/math6120292
Received: 11 September 2018 / Revised: 14 November 2018 / Accepted: 21 November 2018 / Published: 29 November 2018
(This article belongs to the Special Issue Harmonic Analysis)

Abstract

In this paper, the extensions of classes S ˜ ,   C ˜ and B ˜ V are made by defining the classes S ˜ r , C ˜ r and B ˜ V r , r = 0 , 1 , 2 , It is also shown that class S ˜ r is a subclass of C ˜ r B ˜ V r . Moreover, the results on L 1 -convergence of r times differentiated trigonometric sine series have been obtained by considering the r t h ( r = 0 , 1 , 2 , ) derivative of modified sine sum under the new extended class C ˜ r B ˜ V r .
Keywords: Dirichlet kernel; L1-convergence; modified sine sum Dirichlet kernel; L1-convergence; modified sine sum

1. Introduction

Consider the trigonometric sine series
k = 1 a k sin k x
where a 0 , a 1 , a 2 , are the real coefficients. The nth partial sum, S n , of Series (1) is represented as
S n ( x ) = k = 1 n a k sin k x = k = 1 n b k ( cos k x )
where the prime denotes derivatives and b k = a k k . Also, f ( x ) = lim n S n ( x ) .
Various conditions are given in the literature (see [1,2,3,4,5,6,7,8,9]), which guarantee that Series (1) is a Fourier series.
In 1984 , Teljakovskii [9] introduced a class S ˜ , as follows:
Class S ˜ [9]. A null sequence { a k } is said to belong to class S ˜ if there exists a non-increasing sequence { B k } of numbers s.t.
| Δ b k | B k k = 1 , 2 , 3 , k = 1 k B k < .
where b k = a k k , Δ b k = b k b k + 1 and proved the following result:
Theorem 1 [9].
If { a k } S ˜ , then Series (1) is the Fourier series of some function f L 1 ( 0 , π ) .
In 1989 , Móricz [5] introduced new classes B ˜ V and C ˜ of the coefficient sequences for the sine series.
Class  B ˜ V [5].
A null sequence { a k } belongs to B ˜ V if
k = 1 k | Δ b k | <
Class  C ˜ [5].
A null sequence { a k } belongs to class C ˜ if for every ε > 0 there exists δ > 0 , independent of n , and such that for all n ,
0 δ | k = n Δ b k D k ( x ) | d x ε .
Here, D k ( x ) is the first derivative of Dirichlet kernel ( D k ( x ) = sin ( k + 1 2 ) x 2 sin x 2 ) .
Equation (4) implies that, for 1 n N ,
0 δ | k = n N Δ b k D k ( x ) | d x 2 ε .
The following result was proved by Móricz [7].
Theorem 2 [5].
If { a k } B ˜ V , then
u n f 0 ( n )   i f   a n d   o n l y   i f   { a k } C ˜ .
where u n ( x ) = S n ( x ) + b n + 1 D n ( x ) .
The classes S ˜ , B ˜ V and C ˜ seem to be more appropriate for the sine series than the classes S ([7,8]) B V [10], and C [3] in the ordinary sense. Also, Móricz [5] has proved that S ˜ B ˜ V C ˜ .
Motivated by the aforesaid authors, new extended classes S ˜ r , B ˜ V r , and C ˜ r ( r = 0 , 1 , 2 , ) are defined in this paper as follows:
Class S ˜ r .
A sequence { a k } is said to belong to class S ˜ r ( r = 0 , 1 , 2 , ) if a k 0 as k , and there exists a non-increasing sequence { B k } of numbers s.t.
| Δ b k | B k k = 1 , 2 , 3 , k = 1 k r + 1 B k < , r = 0 , 1 , 2 , 3 ,
where b k = a k k , r = 0 , 1 , 2 , 3 ,
B k 0 and k = 1 k r + 1 B k < , implies that k r + 2 B k = o ( 1 ) as k   ( r = 0 , 1 , 2 , ) .
Remark 1.
For r = 0 , S ˜ r = S ˜ .
Remark 2.
Obviously, S ˜ r + 1 S ˜ r , but the converse need not be true.
Example 1.
Consider a sequence Δ b n = 1 n r + 3 , r = 0 , 1 , 2 , and   n N .
a n = n b n = n k = n Δ b k k = n k k r + 3 = k = n 1 k r + 2 0   as   n .
Choose B n = 1 n r + 3 , r = 0 , 1 , 2 , n . Clearly, B n 0 as n and | Δ b n | B n n .
Consider the series
n = 1 n r + 1 B n = n = 1 n r + 1 1 n r + 3 n = 1 1 n 2 which   is   convergent .
This implies { a n } S ˜ r .
But the series n = 1 n r + 2 B n n = 1 1 n is divergent.
This implies that { a n } does not belong to class S ˜ r + 1 .
Class B ˜ V r .
A null sequence { a k } belongs to B ˜ V r , ( r = 0 , 1 , 2 , ) if
k = 1 k r + 1 | Δ b k | <
Remark 3.
For r = 0 , B ˜ V r = B ˜ V .
Remark 4.
Clearly, B ˜ V r + 1 B ˜ V r , ( r = 0 , 1 , 2 , ) , but the converse may not be true.
Class C ˜ r .
A null sequence { a k } belongs to class C ˜ r ( r = 0 , 1 , 2 , ), if for every ε > 0 , there exists δ > 0 , independent of n , and such that for all n ,
0 δ | k = n Δ b k D k r + 1 ( x ) | d x ε
Here, D k r + 1 ( x ) is the ( r + 1 ) t h derivative of Dirichlet kernel.
Equation (4) implies, for 1 n N ,
0 δ | k = n N Δ b k D k r + 1 ( x ) | d x 2 ε
Remark 5.
For r = 0 , C ˜ r = C ˜ .
Remark 6.
It is obvious that C ˜ r + 1 C ˜ r but the converse need not be true.
Example 2.
Define Δ b n = 1 n r + 3 , r = 0 , 1 , 2 , and n = 1 , 2 , 3 ,
a n = n b n = n k = n Δ b k k = n k k r + 3 = k = n 1 k r + 2 0   as   n .
Consider, the integral
0 π | k = n Δ b k D k r + 2 ( x ) | d x = k = n 1 n r + 3 0 π | D k r + 2 ( x ) |   d x = O ( k = n 1 n r + 3 ( n r + 2 log n ) ) = O ( k = n log n n )
which is divergent.
However,
0 π | k = n Δ b k D k r + 1 ( x ) | d x = k = n 1 n r + 3 0 π | D k r + 1 ( x ) |   d x = O ( k = n 1 n r + 3 ( n r + 1 log n ) ) = O ( k = n log n n 2 )   which   is   convergent .
Therefore { a n } C ˜ r .
Lemmas related to the main results are given in Section 2. The Section 3 comprises the main results of this paper. Firstly, in this section, we have shown that the new extended class S ˜ r is a subclass of C ˜ r B ˜ V r ( r = 0 , 1 , 2 , ) . Moreover, the theorems are presented concerning the L 1 convergence of trigonometric sine series using modified sine sum [11], defined as
β n ( x ) = k = 1 n ( a k + 1 k + 1 + j = k n Δ 2 ( a j j ) ) k   sin k x
under the extended classes of numerical sequences.

2. Lemmas

Lemma 1.
[6] Let n 1 and r be a nonnegative integer x [ ε , π ] . Then, | D n r ( x ) | C n r x , where C denotes a positive absolute constant.
Lemma 2.
[6] D n r ( x ) L 1 = O ( n r log n ) , r = 0 , 1 , 2 , where D n r ( x ) represents the r t h derivative of the Dirichlet kernel.

3. Main Results

Theorem 3.
The following relation holds S ˜ r C ˜ r B ˜ V r for each r { 0 , 1 , 2 , } .
Proof. 
It is plain that S ˜ r B ˜ V r .
In order to prove that S ˜ r C ˜ r we take a sequence { a k } in S ˜ r and consider
0 π | k = n Δ b k D k r + 1 ( x ) | d x ;   where   b k = a k k
If we apply summation by parts, we obtain
0 π | k = n Δ b k D k r + 1 ( x ) | d x           lim N [ k = n N 1 Δ B k 0 π | j = 0 k Δ b j B j D j r + 1 ( x ) | d x + B N 0 π | K = 0 N Δ b k B k D k r + 1 ( x ) | d x + B n 0 π | K = 0 n 1 Δ b k B k D k r + 1 ( x ) | d x ]
Clearly | Δ b k B k | 1 . Now, if we first apply Bernstein’s inequality [12] and then Sidon Fomin’s inequality ([1,7]), we get
0 π | k = 0 n Δ b k B k D k ( r + 1 ) ( x ) | d x M ( n + 1 ) r + 2 , r = 0 , 1 , 2 ,
0 π | k = n Δ b k D k r + 1 ( x ) | d x lim N { k = n N 1 Δ B k ( k + 1 ) r + 2 +   B N ( N + 1 ) r + 2 } +   n r + 2 B n = k = n [ ( k + 1 ) r + 2 k r + 2 ] B k + n r + 2 B n = O ( k = n k r + 1 B k ) + n r + 2 B n
So, by given hypothesis, we have
0 π | k = n Δ b k D k r + 1 ( x ) | d x ε 2   if   n   is   large   enough   say   n n 0 .
For any 1 n N , we can estimate as follows:
0 δ | k = n N Δ b k D k r + 1 ( x ) | d x 0 δ | k = n n 0 Δ b k D k r + 1 ( x ) | d x + 0 δ | k = n 0 N Δ b k D k r + 1 ( x ) | d x 1 2 δ k = 1 n 0 k ( k + 1 ) r + 1 | Δ b k | + ε 2 < ε
provided δ is small enough. This proves that { a k } C ˜ r . □
Theorem 4.
Let { a k } be a sequence of numbers belonging to the class C ˜ B ˜ V and if lim n a n log n = 0 , then
β n f = o ( 1 ) ,   n .
Proof. 
The modified trigonometric sine sum is given by
β n ( x ) = k = 1 n ( a k + 1 k + 1 + j = k n Δ 2 ( a j j ) ) k   sin k x = k = 1 n a k sin k x + ( a n + 2 n + 2 a n + 1 n + 1 ) k = 1 n k sin k x = k = 1 n b k ( cos k x ) ( b n + 2 b n + 1 ) D n ( x )
By using the summation by parts, we get
β n = k = 1 n Δ b k D k ( x ) b n D n ( x ) ( b n + 2 b n + 1 ) D n ( x )
Under the given hypothesis and Lemma 1, series k = 1 n Δ b k D k ( x ) converges absolutely and b n D n ( x ) 0 as n .
Hence lim n β n ( x ) = f ( x ) exists in ( 0 , π ) .
Next, consider
f ( x ) β n ( x ) = k = n + 1 a k sin k x ( a n + 2 n + 2 a n + 1 n + 1 ) k = 1 n k sin k x = 0 π | k = n + 1 b k ( cos k x ) ( b n + 1 b n + 2 ) D n ( x ) | d x
By using Abel’s transformation, we have
= 0 π | k = n + 1 Δ b k D k ( x ) + b n + 2 D n ( x ) | d x = 0 π | k = n + 1 Δ b k D k ( x ) | d x + n n + 2 a n + 2 log n
The second term of the above equation is of o ( 1 ) as a n log n = 0   a s   n . For the remaining part, let ε > 0 , then there exists δ > 0 , such that
0 δ | k = n + 1 Δ b k D k ( x ) | d x < ε 2   for   all   n 0 .
Then
0 π | k = n + 1 Δ b k D k ( x ) | d x = 0 δ | k = n + 1 Δ b k D k ( x ) | d x + δ π | k = n + 1 Δ b k D k ( x ) | d x ϵ 2 + k = n + 1 | Δ b k | δ π | D k ( x ) | d x ϵ 2 + C k = n + 1 k   | Δ b k | δ π d x / x 2   ϵ 2 + C δ 1 k = n + 1 k   | Δ b k | ε
This proves that f ( x ) β n ( x ) = o ( 1 )   a s   n . □
Theorem 5.
Let { a k } be a sequence of numbers belonging to the class C ˜ B ˜ V , and if lim n a n log n = 0 , then
S n f = o ( 1 ) , n .
Proof. 
S n f S n β n + β n f
| b n + 1 | 0 π | D n ( x ) | d x + | b n + 2 | 0 π | D n ( x ) | d x + o ( 1 ) | a n + 1 | log n + | a n + 2 | log n ( b y   L e m m a   2 ) = o ( 1 ) ,   n
Theorem 6.
Let { a k } be a sequence of numbers belonging to the class C ˜ r B ˜ V r and if n r a n log n = 0 ,   a s n   ,   f o r   e a c h   r = 0 , 1 , 2 , Then
β n r ( x ) f r ( x ) = o ( 1 ) , n
Here, f r ( x ) is the rth derivative of f(x), where r = 0 , 1 , 2 ,
Proof. 
Consider the modified trigonometric sine sum as
β n ( x ) = k = 1 n ( a k + 1 k + 1 + j = k n Δ 2 ( a j j ) ) k   sin k x = k = 1 n a k sin k x + ( a n + 2 n + 2 a n + 1 n + 1 ) k = 1 n k sin k x
Taking r-times differentiation of β n ( x ) , we get
β n r ( x ) = S n r ( x ) + ( a n + 2 n + 2 a n + 1 n + 1 ) k = 1 n k r + 1 sin ( k x + r π 2 ) = k = 1 n k r a k sin ( k x + r π 2 ) + ( a n + 1 n + 1 a n + 2 n + 2 ) k = 1 n k r + 1 cos ( k x + ( r + 1 ) π 2 ) = k = 1 n k r + 1 b k cos ( k x + ( r + 1 ) π 2 ) + ( b n + 1 b n + 2 ) D n r + 1 ( x )
If we apply Abel’s transformation on the first term of above equation, we get
β n r ( x ) = k = 1 n 1 Δ b k D k r + 1 ( x ) b n D n r + 1 ( x ) + ( b n + 1 b n + 2 ) D n r + 1 ( x ) = k = 1 n Δ b k D k r + 1 ( x ) b n + 2 D n r + 1 ( x )
The series k = 1 Δ b k D k r + 1 ( x ) converges absolutely and b n D n r + 1 ( x ) 0 as n using Lemma 1 and given hypothesis.
Therefore lim n β n r ( x ) = f r ( x ) exists in ( 0 , π ) .
Next, consider
f ( x ) β n ( x ) = k = n + 1 a k sin k x ( a n + 2 n + 2 a n + 1 n + 1 ) k = 1 n k sin k x f r ( x ) β n r ( x ) = k = n + 1 k r a k sin ( k x + r π 2 ) ( a n + 2 n + 2 a n + 1 n + 1 ) k = 1 n k r + 1 sin ( k x + r π 2 ) = k = n + 1 k r a k sin ( k x + r π 2 ) + ( a n + 2 n + 2 a n + 1 n + 1 ) k = 1 n k r + 1 cos ( k x + ( r + 1 ) π 2 ) = 0 π | k = n + 1 k r + 1 b k cos ( k x + ( r + 1 ) π 2 ) + ( b n + 2 b n + 1 ) D n r + 1 ( x ) | d x
If we apply Abel’s transformation, we obtain
= 0 π | k = n + 1 Δ b k D k r + 1 ( x ) + b n + 1 D n r + 1 ( x ) b n + 1 D n r + 1 ( x ) + b n + 2 D n r + 1 ( x ) | d x 0 π | k = n + 1 Δ b k D k r + 1 ( x ) | d x + | b n + 2 | 0 π | D n r + 1 ( x ) | d x 0 π | k = n + 1 Δ b k D k r + 1 ( x ) | d x + a n + 2 n + 2 n r + 1 log n
The second term of the above equation are of o(1) as n r a n log n = 0 as n . For the remaining part, let ε > 0 , then there exists δ > 0 , such that 0 δ | k = n + 1 Δ b k D k r + 1 ( x ) | d x < ε / 2 for all n 0 . Then
0 π | k = n + 1 Δ b k D k r + 1 ( x ) | d x = 0 δ | k = n + 1 Δ b k D k r + 1 ( x ) | d x + δ π | k = n + 1 Δ b k D k r + 1 ( x ) | d x       ε 2 + k = n + 1 | Δ b k | δ π | D k r + 1 ( x ) | d x       ε 2 + C k = n + 1 k r + 1 | Δ b k | δ π d x / x r + 2       ε 2 + C δ ( r + 1 ) k = n + 1 k r + 1 | Δ b k | ε      ( by   given   hypothesis )
Therefore, f r ( x ) β n r ( x ) L 1 = o ( 1 ) as n . □
Remark 7.
For r = 0 , Theorem 6 reduces to Theorem 4.
Theorem 7.
Let { a k } be a sequence of numbers belonging to the class C ˜ r B ˜ V r and if n r a n log n = o ( 1 ) as n . Then
S n r ( x ) f r ( x ) = o ( 1 ) ,   n .
where r = 0 , 1 , 2 .
Proof. 
S n r f r S n r β n r + β n r f r
| b n + 2 | 0 π | D n r + 1 ( x ) | d x + | b n + 1 | 0 π | D n r + 1 ( x ) | d x + o ( 1 ) | a n + 2 | n + 2 n r + 1 log n + | a n + 1 | n + 1 n r + 1 log n      ( b y   L e m m a   2 ) = o ( 1 )   a s   n
Remark 8.
For r = 0 , Theorem 7 reduces to Theorem 5.
Remark 9.
Combining Theorem 6 and Theorem 7 with Theorem 3, the following result holds:
Corollary 1.
If { a k } S ˜ r ( r = 0 , 1 , 2 , 3 , ) and if n r a n log n = o ( 1 ) as n . Then
(i) 
β n r ( x ) f r ( x ) = o ( 1 ) ,   n .
(ii) 
S n r ( x ) f r ( x ) = o ( 1 ) ,   n .

Author Contributions

All authors have contributed in obtaining the new results presented in this article. All authors read and approved the final manuscript. Investigation, S.K.C.; Supervision, J. K. and S.S.B.

Funding

This research received no external funding.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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