Paper-9-Acta23.2010. 107KB Jun 04 2011 12:03:15 AM
Acta Universitatis Apulensis
ISSN: 1582-5329
No. 23/2010
pp. 99-105
ON L1 −CONVERGENCE OF THE R−TH DERIVATIVE OF COSINE
SERIES WITH SEMI-CONVEX COEFFICIENTS
Xhevat Z. Krasniqi and Naim L. Braha
Abstract. We study L1 -convergence of r − th derivative of modified sine sums
introduced by K. Kaur [2] and deduce some corollaries.
2000 Mathematics Subject Classification: 42A20, 42A32.
1. Introduction and Preliminaries
Let
∞
a0 X
+
ak cos kx
2
(1)
k=1
be cosine trigonometric series with its partial sums denoted by Sn (x) = a20 +
P
n
k=1 ak cos kx, and let f (x) = limn→∞ Sn (x).
For convenience, in the following of this paper we shall assume that a0 = 0.
A sequence (an ) is said to be semi-convex, or briefly (an ) ∈ (SC), if an → 0 as
n → ∞, and
∞
X
n|∆2 an−1 + ∆2 an | < ∞,
n=1
where ∆2 ak = ak − 2ak+1 + ak+2 .
We shall generalize the class of sequences (SC) in the following manner:
A sequence (an ) is said to be semi-convex of order r, (r = 0, 1, . . . ) or (an ) ∈
(SC)r , if an → 0 as n → ∞, and
∞
X
nr+1 |∆2 an−1 + ∆2 an | < ∞.
n=1
It is clear that (SC) ⊂ (SC)r , (r = 1, 2, . . . ) and (SC) ≡ (SC)0 .
99
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
R. Bala and B. Ram [1] have proved that for series (1) with semi-convex null
coefficients the following theorem holds true.
Theorem A. If (an ) is a semi-convex null sequence, then for the convergence of
the cosine series (1) in the metric L1 , it is necessary and sufficient that an−1 log n =
o(1), n → ∞.
Later on, K. Kaur [2] introduced new modified sine sums as
n
Kn (x) =
n
1 XX
(∆aj−1 − ∆aj+1 ) sin kx,
2 sin x
k=1 j=k
where ∆aj = aj − aj+1 , and studied the L1 -convergence of this modified sine sum
with semi-convex coefficients proving the following theorem.
Theorem B. Let (an ) be a semi-convex null sequence, then Kn (x) converges to
f (x) in L1 -norm.
The main goal of the present work is to study the L1 -convergence of r − th
derivative of these new modified sine sums with semi-convex null coefficients of
order r and to deduce the sufficient condition of Theorem A and Theorem B as
corollaries.
As usually with Dn (x) and D̃n (x) we shall denote the Dirichlet and its conjugate
kernels defined by
n
Dn (x) =
1 X
+
cos kx,
2
D̃n (x) =
k=1
n
X
sin kx.
k=1
Everywhere in this paper the constants in the O-expression denote positive constants and they may be different in different relations.
To prove the main results we need the following lemmas:
Lemma 1. ([3]) For the r-th derivatives of the Dirichlet’s kernels Dn (x) and
D̃n (x) the following estimates hold
(r)
(1) kDn (x)kL1 = π4 nr log n + O(nr ), r = 0, 1, 2, . . .
(r)
(2) kD̃n (x)kL1 = O (nr log n), r = 0, 1, 2, . . .
100
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
Lemma 2. If x ∈ [ǫ, π − ǫ], ǫ ∈ (0, π) and m ∈ N , then the following estimate
holds
!(r)
D̃m (x)
= Or,ǫ mr+1 , (r = 0, 1, 2 . . . )
2 sin x
where Or,ǫ depends only on r and ǫ.
Proof. By Leibniz formula we have
!(r)
(r−i)
r
(i)
X
D̃m (x)
r
1
=
D̃m (x)
i
2 sin x
2 sin x
i=0
(r−i) X
r
m
X
r
1
iπ
i
=
j sin jx +
2 sin x
2
i
i=0
j=1
(r−i)
r
X
r
1
r+1
.
= O(1)m
2 sin x
i
(2)
i=0
(τ )
τ (cos x)
1
= Psin
We shall prove by mathematical induction the equality 2 sin
τ +1 x ,
x
where Pτ is a cosine polynomial of degree τ .
′
cos x
(cos x)
1
= (−1/2)
Namely, we have 2 sin
= P1sin
2 x , so that for τ = 1 the above
x
sin2 x
equality is true.
(τ ) Pτ (cos x)
1
= sinτ +1 x holds. For the (τ +1)−th
Assume that the equality F (x) := 2 sin
x
1
derivative of 2 sin
we
get
x
′
F (x) :=
′
− sinτ +2 x Pτ (cos x) − (r + 1)Pτ (cos x) sinτ x cos x
=
sin2τ +2 x
′
(− sin2 x)Pτ (cos x) − (τ + 1)Pτ (cos x) cos x
=
sinτ +2 x
′
(cos2 x − 1)Pτ (cos x) − (τ + 1)Pτ (cos x) cos x
=
sinτ +2 x
Pτ +1 (cos x)
=
,
sinτ +2 x
where Pτ +1 (cos x), is a cosine polynomial of the degree τ + 1.
Therefore for x ∈ [ǫ, π − ǫ], ǫ > 0, from (2) dhe (3) we obtain
!(r)
r
X
D̃m (x)
r |Pr−i (cos x)|
r+1
= O(1)m
= Or,ǫ mr+1 .
2 sin x
r−i+1
i sin
x
i=0
101
(3)
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
2.Main Results
At the begining we prove the following result:
(r)
Theorem 1. Let (an ) be a semi-convex null sequence of order r, then Kn (x)
converges to g (r) (x) in L1 -norm.
Proof. We have
n
Kn (x) =
=
n
1 XX
(∆aj−1 − ∆aj+1 ) sin kx
2 sin x
k=1 j=k
" n
#
X
1
(ak−1 − ak+1 ) sin kx − (an − an+2 ) D̃n (x) .
2 sin x
k=1
Applying Abel’s transformation (see [4], p. 17), we get
n
Kn (x) =
1 X
(∆ak−1 − ∆ak+1 ) D̃k (x)
2 sin x
k=1
=
1
2 sin x
n
X
k=1
∆2 ak−1 + ∆2 ak D̃k (x).
Therefore,
Kn(r) (x)
=
n
X
2
2
∆ ak−1 + ∆ ak
k=1
D̃k (x)
2 sin x
!(r)
.
(4)
On the other side we have
n
n
k=1
k=1
1 X
1 X
Sn (x) =
ak cos kx sin x =
ak [sin(k + 1)x − sin(k − 1)x]
sin x
2 sin x
n
sin(n + 1)x
sin nx
1 X
+ an
(ak−1 − ak+1 ) sin kx + an+1
=
2 sin x
2 sin x
2 sin x
k=1
=
1
2 sin x
n
X
(∆ak−1 + ∆ak ) sin kx + an+1
k=1
102
sin(n + 1)x
sin nx
+ an
.
2 sin x
2 sin x
(5)
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
Applying Abel’s transformation to the equality (5) we get
n
X
Sn (x) =
∆2 ak−1 + ∆2 ak
k=1
+ (an − an+2 )
D̃k (x)
2 sin x
sin nx
sin(n + 1)x
D̃n (x)
+ an+1
+ an
.
2 sin x
2 sin x
2 sin x
Thus
!(r)
D̃
(x)
k
∆2 ak−1 + ∆2 ak
Sn(r) (x) =
+ (an − an+2 )
2 sin x
k=1
sin nx (r)
sin(n + 1)x (r)
+an+1
+ an
.
2 sin x
2 sin x
n
X
D̃n (x)
2 sin x
!(r)
(6)
By Lemma 1 and since (an ) is semi-convex null sequence of order r, we have
!(r)
D̃
(x)
n
(an − an+2 )
2
sin
x
!
∞
X
= Or,ǫ (n + 1)r+1 (an − an+2 ) = Or,ǫ (n + 1)r+1
(∆ak − ∆ak+2 )
k=n
= Or,ǫ
∞
X
r+1
(n
+
1)
(∆a
−
∆a
)
k−1
k+1
k=n+1
= Or,ǫ
∞
X
!
k r+1 |∆2 ak−1 + ∆2 ak |
k=n+1
!
= o(1), n → ∞.
(7)
Also after some elementary calculations and by virtue of Lemma 2 we obtain
an+1
= an+1
D̃n (x)
2 sin x
!(r)
−
sin(n + 1)x (r)
=
2 sin x
!(r)
!(r)
D̃
(x)
D̃n−1 (x)
n+1
+ an
−
2 sin x
2 sin x
sin nx
2 sin x
(r)
+ an
!(r)
D̃n (x)
2 sin x
= an+1 Or,ǫ nr+1 + (n − 1)r+1 + an Or,ǫ (n + 1)r+1 + nr+1
= Or,ǫ (n + 1)r+1 (an + an+1 )
103
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
= Or,ǫ (n + 1)r+1 [(an − an+2 ) + (an+1 − an+3 ) + (an+2 − an+4 ) + · · · ]
!
∞
∞
X
X
= Or,ǫ
k r+1 |∆2 ak−1 + ∆2 ak | +
k r+1 |∆2 ak−1 + ∆2 ak | + · · · = o(1), n → ∞.
k=n+1
k=n+2
(8)
Because of (7) and (8), when we pass on limit as n → ∞ to (4) and (6) we get
g (r) (x) =
lim Sn(r) (x)
n→∞
lim Kn(r) (x) =
=
n→∞
∞
X
∆2 ak−1 + ∆2 ak
k=1
D̃k (x)
2 sin x
!(r)
.
(9)
Using Lemma 2, from (4) and (9) we have
Z
π
−π
= 2
Z
0
= Or,ǫ
(r)
g (x) − Kn(r) (x) dx
π
!(r)
D̃k (x)
dx
2 sin x
k=n+1
!
∞
X
k r+1 ∆2 ak−1 + ∆2 ak = o(1), n → ∞,
∞
X
2
∆ ak−1 + ∆2 ak
k=n+1
which fully proves the Theorem 1.
Remark 1. If we replace r = 0 in Theorem 1 we get Theorem B.
Corollary 1. Let (an ) ∈ (SC)r , then the sufficient condition for L1 -convergence
of the r − th derivative of the series (1) is nr an log n = o(1), as n → ∞.
Proof. We have
(r)
g (x) − Sn(r) (x)
≤
g (r) (x) − Kn(r) (x)
+
Kn(r) (x) − Sn(r) (x)
104
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
!(r)
D̃
(x)
n
= o(1) +
(an − an+2 ) 2 sin x
sin nx (r)
sin(n + 1)x (r)
+
an+1
+ an
(by Theorem 1)
2 sin x
2 sin x
≤ o(1) + an
D̃n(r) (x)
(by (7))
= o(1) + O (nr an log n) = o(1).
(by Lemma 1)
Remark 2. If we replace r = 0 in Corollary 1 we get the sufficient condition of
Theorem A.
Acknowledgments: The authors would like to thank the anonymous referee
for his/her valuable comments and suggestions to improve this article.
References
[1] Bala R. and Ram B., Trigonometric series with semi-convex coefficients,
Tamkang J. Math. 18 (1) (1987), 75-84.
[2] Kaur K., Bhatia S. S. , and Ram B., Integrability and L1 -convergence of
modified sine sums, Georgian Math. J. 11, No. 1, (2004), 99-104.
[3] S. Sheng, The extension of the theorems of Č. V. Stanojević and V. B. Stanojević, Proc. Amer. Math. Soc. 110 (1990), 895-904.
[4] Bary K. N., Trigonometric series, Moscow, (1961)(in Russian).
Xhevat Z. Krasniqi
Faculty of Education
University of Prishtina
Agim Ramadani St., Prishtinë, 10000, Kosova
email: [email protected]
Naim L. Braha
Department of Mathematics and Computer Sciences
University of Prishtina
Avenue ”Mother Theresa ” 5, Prishtinë, 10000, Kosova
email: [email protected]
105
ISSN: 1582-5329
No. 23/2010
pp. 99-105
ON L1 −CONVERGENCE OF THE R−TH DERIVATIVE OF COSINE
SERIES WITH SEMI-CONVEX COEFFICIENTS
Xhevat Z. Krasniqi and Naim L. Braha
Abstract. We study L1 -convergence of r − th derivative of modified sine sums
introduced by K. Kaur [2] and deduce some corollaries.
2000 Mathematics Subject Classification: 42A20, 42A32.
1. Introduction and Preliminaries
Let
∞
a0 X
+
ak cos kx
2
(1)
k=1
be cosine trigonometric series with its partial sums denoted by Sn (x) = a20 +
P
n
k=1 ak cos kx, and let f (x) = limn→∞ Sn (x).
For convenience, in the following of this paper we shall assume that a0 = 0.
A sequence (an ) is said to be semi-convex, or briefly (an ) ∈ (SC), if an → 0 as
n → ∞, and
∞
X
n|∆2 an−1 + ∆2 an | < ∞,
n=1
where ∆2 ak = ak − 2ak+1 + ak+2 .
We shall generalize the class of sequences (SC) in the following manner:
A sequence (an ) is said to be semi-convex of order r, (r = 0, 1, . . . ) or (an ) ∈
(SC)r , if an → 0 as n → ∞, and
∞
X
nr+1 |∆2 an−1 + ∆2 an | < ∞.
n=1
It is clear that (SC) ⊂ (SC)r , (r = 1, 2, . . . ) and (SC) ≡ (SC)0 .
99
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
R. Bala and B. Ram [1] have proved that for series (1) with semi-convex null
coefficients the following theorem holds true.
Theorem A. If (an ) is a semi-convex null sequence, then for the convergence of
the cosine series (1) in the metric L1 , it is necessary and sufficient that an−1 log n =
o(1), n → ∞.
Later on, K. Kaur [2] introduced new modified sine sums as
n
Kn (x) =
n
1 XX
(∆aj−1 − ∆aj+1 ) sin kx,
2 sin x
k=1 j=k
where ∆aj = aj − aj+1 , and studied the L1 -convergence of this modified sine sum
with semi-convex coefficients proving the following theorem.
Theorem B. Let (an ) be a semi-convex null sequence, then Kn (x) converges to
f (x) in L1 -norm.
The main goal of the present work is to study the L1 -convergence of r − th
derivative of these new modified sine sums with semi-convex null coefficients of
order r and to deduce the sufficient condition of Theorem A and Theorem B as
corollaries.
As usually with Dn (x) and D̃n (x) we shall denote the Dirichlet and its conjugate
kernels defined by
n
Dn (x) =
1 X
+
cos kx,
2
D̃n (x) =
k=1
n
X
sin kx.
k=1
Everywhere in this paper the constants in the O-expression denote positive constants and they may be different in different relations.
To prove the main results we need the following lemmas:
Lemma 1. ([3]) For the r-th derivatives of the Dirichlet’s kernels Dn (x) and
D̃n (x) the following estimates hold
(r)
(1) kDn (x)kL1 = π4 nr log n + O(nr ), r = 0, 1, 2, . . .
(r)
(2) kD̃n (x)kL1 = O (nr log n), r = 0, 1, 2, . . .
100
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
Lemma 2. If x ∈ [ǫ, π − ǫ], ǫ ∈ (0, π) and m ∈ N , then the following estimate
holds
!(r)
D̃m (x)
= Or,ǫ mr+1 , (r = 0, 1, 2 . . . )
2 sin x
where Or,ǫ depends only on r and ǫ.
Proof. By Leibniz formula we have
!(r)
(r−i)
r
(i)
X
D̃m (x)
r
1
=
D̃m (x)
i
2 sin x
2 sin x
i=0
(r−i) X
r
m
X
r
1
iπ
i
=
j sin jx +
2 sin x
2
i
i=0
j=1
(r−i)
r
X
r
1
r+1
.
= O(1)m
2 sin x
i
(2)
i=0
(τ )
τ (cos x)
1
= Psin
We shall prove by mathematical induction the equality 2 sin
τ +1 x ,
x
where Pτ is a cosine polynomial of degree τ .
′
cos x
(cos x)
1
= (−1/2)
Namely, we have 2 sin
= P1sin
2 x , so that for τ = 1 the above
x
sin2 x
equality is true.
(τ ) Pτ (cos x)
1
= sinτ +1 x holds. For the (τ +1)−th
Assume that the equality F (x) := 2 sin
x
1
derivative of 2 sin
we
get
x
′
F (x) :=
′
− sinτ +2 x Pτ (cos x) − (r + 1)Pτ (cos x) sinτ x cos x
=
sin2τ +2 x
′
(− sin2 x)Pτ (cos x) − (τ + 1)Pτ (cos x) cos x
=
sinτ +2 x
′
(cos2 x − 1)Pτ (cos x) − (τ + 1)Pτ (cos x) cos x
=
sinτ +2 x
Pτ +1 (cos x)
=
,
sinτ +2 x
where Pτ +1 (cos x), is a cosine polynomial of the degree τ + 1.
Therefore for x ∈ [ǫ, π − ǫ], ǫ > 0, from (2) dhe (3) we obtain
!(r)
r
X
D̃m (x)
r |Pr−i (cos x)|
r+1
= O(1)m
= Or,ǫ mr+1 .
2 sin x
r−i+1
i sin
x
i=0
101
(3)
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
2.Main Results
At the begining we prove the following result:
(r)
Theorem 1. Let (an ) be a semi-convex null sequence of order r, then Kn (x)
converges to g (r) (x) in L1 -norm.
Proof. We have
n
Kn (x) =
=
n
1 XX
(∆aj−1 − ∆aj+1 ) sin kx
2 sin x
k=1 j=k
" n
#
X
1
(ak−1 − ak+1 ) sin kx − (an − an+2 ) D̃n (x) .
2 sin x
k=1
Applying Abel’s transformation (see [4], p. 17), we get
n
Kn (x) =
1 X
(∆ak−1 − ∆ak+1 ) D̃k (x)
2 sin x
k=1
=
1
2 sin x
n
X
k=1
∆2 ak−1 + ∆2 ak D̃k (x).
Therefore,
Kn(r) (x)
=
n
X
2
2
∆ ak−1 + ∆ ak
k=1
D̃k (x)
2 sin x
!(r)
.
(4)
On the other side we have
n
n
k=1
k=1
1 X
1 X
Sn (x) =
ak cos kx sin x =
ak [sin(k + 1)x − sin(k − 1)x]
sin x
2 sin x
n
sin(n + 1)x
sin nx
1 X
+ an
(ak−1 − ak+1 ) sin kx + an+1
=
2 sin x
2 sin x
2 sin x
k=1
=
1
2 sin x
n
X
(∆ak−1 + ∆ak ) sin kx + an+1
k=1
102
sin(n + 1)x
sin nx
+ an
.
2 sin x
2 sin x
(5)
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
Applying Abel’s transformation to the equality (5) we get
n
X
Sn (x) =
∆2 ak−1 + ∆2 ak
k=1
+ (an − an+2 )
D̃k (x)
2 sin x
sin nx
sin(n + 1)x
D̃n (x)
+ an+1
+ an
.
2 sin x
2 sin x
2 sin x
Thus
!(r)
D̃
(x)
k
∆2 ak−1 + ∆2 ak
Sn(r) (x) =
+ (an − an+2 )
2 sin x
k=1
sin nx (r)
sin(n + 1)x (r)
+an+1
+ an
.
2 sin x
2 sin x
n
X
D̃n (x)
2 sin x
!(r)
(6)
By Lemma 1 and since (an ) is semi-convex null sequence of order r, we have
!(r)
D̃
(x)
n
(an − an+2 )
2
sin
x
!
∞
X
= Or,ǫ (n + 1)r+1 (an − an+2 ) = Or,ǫ (n + 1)r+1
(∆ak − ∆ak+2 )
k=n
= Or,ǫ
∞
X
r+1
(n
+
1)
(∆a
−
∆a
)
k−1
k+1
k=n+1
= Or,ǫ
∞
X
!
k r+1 |∆2 ak−1 + ∆2 ak |
k=n+1
!
= o(1), n → ∞.
(7)
Also after some elementary calculations and by virtue of Lemma 2 we obtain
an+1
= an+1
D̃n (x)
2 sin x
!(r)
−
sin(n + 1)x (r)
=
2 sin x
!(r)
!(r)
D̃
(x)
D̃n−1 (x)
n+1
+ an
−
2 sin x
2 sin x
sin nx
2 sin x
(r)
+ an
!(r)
D̃n (x)
2 sin x
= an+1 Or,ǫ nr+1 + (n − 1)r+1 + an Or,ǫ (n + 1)r+1 + nr+1
= Or,ǫ (n + 1)r+1 (an + an+1 )
103
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
= Or,ǫ (n + 1)r+1 [(an − an+2 ) + (an+1 − an+3 ) + (an+2 − an+4 ) + · · · ]
!
∞
∞
X
X
= Or,ǫ
k r+1 |∆2 ak−1 + ∆2 ak | +
k r+1 |∆2 ak−1 + ∆2 ak | + · · · = o(1), n → ∞.
k=n+1
k=n+2
(8)
Because of (7) and (8), when we pass on limit as n → ∞ to (4) and (6) we get
g (r) (x) =
lim Sn(r) (x)
n→∞
lim Kn(r) (x) =
=
n→∞
∞
X
∆2 ak−1 + ∆2 ak
k=1
D̃k (x)
2 sin x
!(r)
.
(9)
Using Lemma 2, from (4) and (9) we have
Z
π
−π
= 2
Z
0
= Or,ǫ
(r)
g (x) − Kn(r) (x) dx
π
!(r)
D̃k (x)
dx
2 sin x
k=n+1
!
∞
X
k r+1 ∆2 ak−1 + ∆2 ak = o(1), n → ∞,
∞
X
2
∆ ak−1 + ∆2 ak
k=n+1
which fully proves the Theorem 1.
Remark 1. If we replace r = 0 in Theorem 1 we get Theorem B.
Corollary 1. Let (an ) ∈ (SC)r , then the sufficient condition for L1 -convergence
of the r − th derivative of the series (1) is nr an log n = o(1), as n → ∞.
Proof. We have
(r)
g (x) − Sn(r) (x)
≤
g (r) (x) − Kn(r) (x)
+
Kn(r) (x) − Sn(r) (x)
104
Xh. Z. Krasniqi, N. L. Braha - On L1 −convergence of the r−th derivative of...
!(r)
D̃
(x)
n
= o(1) +
(an − an+2 ) 2 sin x
sin nx (r)
sin(n + 1)x (r)
+
an+1
+ an
(by Theorem 1)
2 sin x
2 sin x
≤ o(1) + an
D̃n(r) (x)
(by (7))
= o(1) + O (nr an log n) = o(1).
(by Lemma 1)
Remark 2. If we replace r = 0 in Corollary 1 we get the sufficient condition of
Theorem A.
Acknowledgments: The authors would like to thank the anonymous referee
for his/her valuable comments and suggestions to improve this article.
References
[1] Bala R. and Ram B., Trigonometric series with semi-convex coefficients,
Tamkang J. Math. 18 (1) (1987), 75-84.
[2] Kaur K., Bhatia S. S. , and Ram B., Integrability and L1 -convergence of
modified sine sums, Georgian Math. J. 11, No. 1, (2004), 99-104.
[3] S. Sheng, The extension of the theorems of Č. V. Stanojević and V. B. Stanojević, Proc. Amer. Math. Soc. 110 (1990), 895-904.
[4] Bary K. N., Trigonometric series, Moscow, (1961)(in Russian).
Xhevat Z. Krasniqi
Faculty of Education
University of Prishtina
Agim Ramadani St., Prishtinë, 10000, Kosova
email: [email protected]
Naim L. Braha
Department of Mathematics and Computer Sciences
University of Prishtina
Avenue ”Mother Theresa ” 5, Prishtinë, 10000, Kosova
email: [email protected]
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