Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 4, Article 106, 2005
RATE OF CONVERGENCE OF CHLODOWSKY TYPE DURRMEYER
OPERATORS
ERTAN IBIKLI AND HARUN KARSLI
A NKARA U NIVERSITY, FACULTY OF S CIENCES ,
D EPARTMENT OF M ATHEMATICS ,
06100 TANDOGAN - A NKARA /T URKEY
ibikli@science.ankara.edu.tr
karsli@science.ankara.edu.tr
Received 16 September, 2005; accepted 23 September, 2005
Communicated by A. Lupaş
A BSTRACT. In the present paper, we estimate the rate of pointwise convergence of the Chlodowsky
type Durrmeyer Operators Dn (f, x) for functions, defined on the interval [0, bn ], (bn → ∞), extending infinity, of bounded variation. To prove our main result, we have used some methods
and techniques of probability theory.
Key words and phrases: Approximation, Bounded variation, Chlodowsky polynomials, Durrmeyer Operators, Chanturiya’s
modulus of variation, Rate of convergence.
2000 Mathematics Subject Classification. 41A25, 41A35, 41A36.
1. I NTRODUCTION
Very recently, some authors studied some linear positive operators and obtained the rate of
convergence for functions of bounded variation. For example, Bojanic R. and Vuilleumier
M. [3] estimated the rate of convergence of Fourier Legendre series of functions of bounded
variation on the interval [0, 1], Cheng F. [4] estimated the rate of convergence of Bernstein
polynomials of functions bounded variation on the interval [0, 1], Zeng and Chen [9] estimated
the rate of convergence of Durrmeyer type operators for functions of bounded variation on the
interval [0, 1].
Durrmeyer operators Mn introduced by Durrmeyer [1]. Also let us note that these operators
were introduced by Lupaş [2]. The polynomial Mn f defined by
Z 1
n
X
Mn (f ; x) = (n + 1)
Pn,k (x)
f (t)Pn,k (t) dt,
0 ≤ x ≤ 1,
k=0
where
0
n
Pn,k (x) =
(x)k (1 − x)n−k .
k
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
276-05
2
E RTAN I BIKLI AND H ARUN K ARSLI
These operators are the integral modification of Bernstein polynomials so as to approximate
Lebesgue integrable functions on the interval [0, 1]. The operators Mn were studied by several authors. Also, Guo S. [5] investigated Durrmeyer operators Mn and estimated the rate of
convergence of operators Mn for functions of bounded variation on the interval [0, 1].
Chlodowsky polynomials are given [6] by
k n−k
n
X
k
n
x
x
Cn (f ; x) =
f
bn
1−
, 0 ≤ x ≤ bn ,
n
k
b
b
n
n
k=0
where (bn ) is a positive increasing sequence with the properties lim bn = ∞ and lim bnn = 0.
n→∞
n→∞
Works on Chlodowsky operators are fewer, since they are defined on an unbounded interval
[0, ∞).
This paper generalizes Chlodowsky polynomials by incorporating Durrmeyer operators, hence
the name Chlodowsky-Durrmeyer operators: Dn : BV [0, ∞) → P,
Z bn
n
(n + 1) X
x
t
Dn (f ; x) =
Pn,k
f (t)Pn,k
dt, 0 ≤ x ≤ bn
bn k=0
bn
bn
0
where P := {P : [0, ∞) → R}, is a polynomial functions set, (bn ) is a positive increasing
sequence with the properties,
bn
lim bn = ∞ and lim
=0
n→∞
n→∞ n
and
n
Pn,k (x) =
(x)k (1 − x)n−k
k
is the Bernstein basis.
In this paper, by means of the techniques of probability theory, we shall estimate the rate of
convergence of operators Dn , for functions of bounded variation in terms of the Chanturiya’s
modulus of variation. At the points which one sided limit exist, we shall prove that operators Dn
converge to the limit 12 [f (x+) + f (x−)] on the interval [0, bn ], (n → ∞) extending infinity,
for functions of bounded variation on the interval [0, ∞).
For the sake of brevity, let the auxiliary function gx be defined by

f (t) − f (x+), x < t ≤ bn ;



0,
t = x;
gx (t) =



f (t) − f (x+), 0 ≤ t < x.
The main theorem of this paper is as follows.
Theorem 1.1. Let f be a function of bounded variation on every finite subinterval of [0, ∞).
Then for every x ∈ (0, ∞) , and n sufficiently large, we have,
1
(1.1) Dn (f ; x) − (f (x+) + f (x−))
2


bn
√−x
x+


n
k

3An (x)b2n X _
2
q
≤ 2
|{f (x+) − f (x−)}| ,
(g
)
+
x
x
nx

x (bn − x)2 
 k=1 x− √x

(1
−
)
bn
bn
k
where An (x) =
h
2 n x(bn −x)+2b2n
n2
i
and
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
b
W
(gx ) is the total variation of gx on [a, b].
a
http://jipam.vu.edu.au/
R ATE OF C ONVERGENCE
3
2. AUXILIARY R ESULTS
In this section we give certain results, which are necessary to prove our main theorem.
Lemma 2.1. If s ∈ N and s ≤ n, then
s n!
(n + 1)!bsn X s s!
·
(xbn )r .
Dn (t ; x) =
(n + s + 1)! r=0 r r! (n − r)!
s
Proof.
n
x
bn
Thus
Z
bn
t
s
Pn,k
t dt
bn
0
#
"Z bn k n−k
n
x
n
t
t
n+1X
=
Pn,k
1−
ts dt
bn k=0
bn
k
b
b
n
n
0
Z
n
1
x
n
t
n+1X
=
Pn,k
bs+1
(u)k+s (1 − u)n−k du, set u =
n
bn k=0
bn
k
bn
0
n
n+1X
x
(k + s)!
n!
=
Pn,k
bs+1
·
.
n
bn k=0
bn
k!
(n + s + 1)!
n+1X
Dn (t ; x) =
Pn,k
bn k=0
s
n
(n + 1)!bsn X
Dn (t ; x) =
Pn,k
(n + s + 1)! k=0
s
x
bn
(k + s)!
.
k!
For s ≤ n, we have
s n k n−k
n ∂s
x
x+y
1 X n
x
y
(k + s)!
= s
s
∂x
bn
bn
bn k=0 k
bn
bn
k!
and from the Leibnitz formula
s n X
n−r
s ∂s
x
x+y
s s!
n!
x+y
1
r
=
·
(xbn )
s
∂x
bn
bn
r r! (n − r)!
bn
bsn
r=0
n−r
s 1 X s s!
n!
x+y
r
= s
·
(xbn )
bn r=0 r r! (n − r)!
bn
Let x + y = bn , we have
k n−k
n−r
n s X
n
x
y
(k + s)! X s s!
n!
x+y
r
=
·
(xbn )
k
b
b
k!
r
r!
(n
−
r)!
bn
n
n
r=0
k=0
Thus the proof is complete.
By the Lemma 2.1, we get
(2.1)
Dn (1; x) = 1
bn − 2x
n+2
[4nbn − 6(n + 1)x]
2b2n
Dn (t2 ; x) = x2 +
x+
.
(n + 2)(n + 3)
(n + 2)(n + 3)
Dn (t; x) = x +
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
http://jipam.vu.edu.au/
4
E RTAN I BIKLI AND H ARUN K ARSLI
By direct computation, we get
2(n − 3)(bn − x)x
2b2n
+
(n + 2)(n + 3)
(n + 2)(n + 3)
Dn ((t − x)2 ; x) =
and hence,
Dn ((t − x)2 ; x) ≤
(2.2)
2nx(bn − x) + 2b2n
.
n2
Lemma 2.2. For all x ∈ (0, ∞) , we have
Z t
x t
x u
λn
,
=
Kn
,
du
bn bn
bn bn
0
1
2nx(bn − x) + 2b2n
≤
(2.3)
·
,
(x − t)2
n2
where
n
x u
n+1X
x
u
Kn
,
=
Pn,k
Pn,k
.
bn bn
bn k=0
bn
bn
Proof.
λn
x t
,
bn bn
Z
t
=
Kn
0
x u
,
bn bn
du
2
x u
x−u
,
du
≤
Kn
bn bn
x−t
0
Z t
1
x u
=
Kn
,
(x − u)2 du
(x − t)2 0
bn bn
1
=
Dn ((u − x)2 ; x)
(x − t)2
Z
t
By the (2.2), we have,
x t
1
,
≤
λn
bn bn
(x − t)2
1
≤
(x − t)2
2(n − 3)(bn − x)x
2b2n
+
(n + 2)(n + 3)
(n + 2)(n + 3)
2nx(bn − x) + 2b2n
·
.
n2
·
Set
(2.4)
α
Jn,j
x
bn
=
n
X
k=j
Pn,k
x
bn
!α
,
α
Jn,n+1
x
bn
=0 ,
where α ≥ 1.
Lemma 2.3. For all x ∈ (0, 1) and j = 0, 1, 2, . . . , n, we have
α
2α
α
Jn,j (x) − Jn+1,j+1
(x) ≤ p
nx(1 − x)
and
2α
α
α
Jn,j (x) − Jn+1,j
(x) ≤ p
nx(1 − x)
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
.
http://jipam.vu.edu.au/
R ATE OF C ONVERGENCE
5
Proof. The proof of this lemma is given in [9].
For α = 1, replacing the variable x with
x
bn
in Lemma 2.3 we get the following lemma:
Lemma 2.4. For all x ∈ (0, bn ) and j = 0, 1, 2, . . . , n, we have
2
x
x Jn,j
−
J
≤r
n+1,j+1
bn
bn
n bxn 1 −
x
bn
and
(2.5)
x x
Jn,j
− Jn+1,j
≤r
bn
bn n bxn
2
1−
x
bn
.
3. P ROOF O F T HE M AIN R ESULT
Now, we can prove the Theorem 1.1.
Proof. For any f ∈ BV [0, ∞), we can decompose f into four parts on [0, bn ] for sufficiently
large n,
(3.1) f (t) =
f (x+) − f (x−)
1
(f (x+) + f (x−)) + gx (t) +
sgn(t − x)
2
2
1
+ δx (t) f (x) − (f (x+) + f (x−))
2
where
(
δx (t) =
(3.2)
1, x = t
0, x 6= t.
If we applying the operator Dn the both side of equality (3.1), we have
Dn (f ; x) =
1
(f (x+) + f (x−)) Dn (1; x) + Dn (gx ; x)
2
f (x+) − f (x−)
+
Dn (sgn(t − x); x)
2
1
+ f (x) − (f (x+) + f (x−)) Dn (δx ; x).
2
Hence, since (2.1) Dn (1; x) = 1, we get,
Dn (f ; x) − 1 (f (x+) + f (x−))
2
f (x+) − f (x−) |Dn (sgn(t − x); x)|
≤ |Dn (gx ; x)| + 2
1
+ f (x) − (f (x+) + f (x−)) |Dn (δx ; x)| .
2
For operators Dn , using (3.2) we can see that Dn (δx ; x) = 0.
Hence we have
Dn (f ; x) − 1 (f (x+) + f (x−)) ≤ |Dn (gx ; x)| + f (x+) − f (x−) |Dn (sgn(t − x); x)|
2
2
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
http://jipam.vu.edu.au/
6
E RTAN I BIKLI AND H ARUN K ARSLI
In order to prove above inequality, we need the estimates for Dn (gx ; x) and Dn (sgn(t−x); x).
We first estimate |Dn (gx ; x)| as follows:
Z bn
n
n + 1 X
x
t
|Dn (gx ; x)| = Pn,k
Pn,k
gx (t)dt bn
bn
bn
0
k=0
"
!
#
Z x− √x
Z x+ bn√−x Z bn
n
n + 1 X
n
n
x
t
+
Pn,k
=
Pn,k
+
gx (t)dt bn
bn
bn
√−x
0
x− √xn
x+ bn
k=0
n
Z x− √x
n
n + 1 X
n
x
t
≤
Pn,k
gx (t)dt
Pn,k
bn
bn
bn
0
k=0
Z x+ bn√−x
n
n + 1 X
n
t
x
+
Pn,k
Pn,k
gx (t)dt
x
bn
b
b
n
n
x− √n
k=0
Z bn
n
n + 1 X
x
t
+
Pn,k
Pn,k
gx (t)dt
b
−x
bn
bn
bn
√
x+ n
k=0
n
= |I1 (n, x)| + |I2 (n, x)| + |I3 (n, x)|
We shall evaluate I1 (n, x), I2 (n, x) and I3 (n, x). To do this we first observe that I1 (n, x),
I2 (n, x) and I3 (n, x) can be written as Lebesque-Stieltjes integral,
Z
x− √xn
x t
|I1 (n, x)| = gx (t)dt λn
,
0
bn bn
Z bn −x
x+ √n
x t
|I2 (n, x)| = gx (t)dt λn
,
x− √x
bn bn
n
Z
bn
x t
|I3 (n, x)| = gx (t)dt λn
,
,
x+ bn√−x
bn bn
n
where
λn
x t
,
bn bn
Z
=
t
Kn
0
x u
,
bn bn
du
and
n
n+1X
x
t
Kn
=
Pn,k
Pn,k
.
bn k=0
bn
bn
h
i
First we estimate I2 (n, x). For t ∈ x − √xn , x + bn√−x
, we have
n
Z bn −x
x+ √n
x
t
(gx (t) − gx (x)) dt λn
,
|I2 (n, x)| = x− √x
bn bn
n
Z x+ bn√−x
n
x
t
≤
|gx (t) − gx (x)| dt λn
,
bn bn
x− √x
x t
,
bn bn
n
√−x
x+ bn
√−x
x+ bn
n
(3.3)
≤
_
x− √xn
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
(gx ) ≤
1
n−1
n
X
_n
(gx ).
k=2 x− √xn
http://jipam.vu.edu.au/
R ATE OF C ONVERGENCE
7
Next, we estimate I1 (n, x). Using partial Lebesque-Stieltjes integration, we obtain
Z x− √x
n
x t
,
I1 (n, x) =
gx (t)dt λn
bn bn
0
x x x − √n
x
λn
,
= gx x − √
bn
bn
n
Z x− √x
n
x
x t
λn
,
dt (gx (t)) .
− gx (0)λn
,0 −
bn bn
bn
0
Since
x
_
x
x
gx x − √ = gx x − √
− gx (x) ≤
(gx ),
n
n √x
x−
n
it follows that
|I1 (n, x)| ≤
x
_
x− √xn
x − √xn Z
x
+
(gx ) λn
,
bn
bn
x− √xn
λn
0
x t
,
bn bn
dt −
x
_
!
(gx ) .
t
From (2.3), it is clear that
x x x − √n
1
2 n x(bn − x) + 2b2n
λn
,
≤ 2
.
bn
bn
n2
√x
n
It follows that
|I1 (n, x)| ≤
x
_
x− √xn
(gx ) 1
√x
n
2
2nx(bn − x) + 2b2n
n2
!
x
_
2nx(bn − x) + 2b2n
dt − (gx )
+
n2
0
t
!
Z x− √x
x
x
_
_
n
1
An (x)
=
(gx ) 2 + An (x)
d − (gx ) .
2 t
(x
−
t)
x
0
x
√
√
t
Z
x−
n
x− √xn
1
(x − t)2
n
Furthermore, since
!
Z x− √x
x
_
n
1
dt − (gx )
(x − t)2
0
t
x− √x
Z x− √x
x
x
n
_
_
n
1
2
=−
(gx )
+
(gx )dt
(x − t)2 t
(x − t)3 t
0
0
Z x− √x
x
x
x
_
_
n
1
1 _
2
=− x 2
(gx ) + 2 (gx ) +
(gx )dt.
( √n )
x 0
(x − t)3 t
0
√x
x−
n
√x
u
Putting t = x −
Z x− √x
n
0
in the last integral, we get
Z
x
x
n
x
_
2
1 n _
1 X _
(gx )dt = 2
(gx )du = 2
(gx ).
(x − t)3 t
x 1
x k=1 √x
√x
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
x−
u
x−
k
http://jipam.vu.edu.au/
8
E RTAN I BIKLI AND H ARUN K ARSLI
Consequently,
x
_
An (x)
( √xn )2
x− √xn


x
n
x
x


_
X
_
_
1
1
1
(gx )
+ An (x) − x 2
(gx ) + 2 (gx ) + 2

 ( √n )
x 0
x k=1 √x
x−
x− √xn
k


x
n
x
1 _

1 X _
= An (x)
(g
)
+
(g
)
x
x
 x2 0

x2 k=1 √x
x−
k


bn
n
x


_
X
_
An (x)
=
(gx ) +
(gx ) .

x2  0
x
√
|I1 (n, x)| ≤
(3.4)
(gx )
k=1 x−
k
Using the similar method for estimating |I3 (n, x)| ,we get


x+ bn√−x


b
n
n

X _k
An (x) _
|I3 (n, x)| ≤
(g
)
+
(g
)
x
x

(bn − x)2 

x
x
k=1


x+ bn√−x


b
n
k
n


_
X _
An (x)
≤
(gx ) +
(gx ) .

(bn − x)2 
x


√
(3.5)
0
k=1 x−
k
Hence from (3.3), (3.4) and (3.5), it follows that
|Dn (gx ; x)| ≤ |I1 (n, x)| + |I2 (n, x)| + |I3 (n, x)|


bn
n
x


_
X
_
An (x)
≤
(g
)
+
(g
)
x
x

x2  0
k=1 x− √x
k


bn
bn
√−x
√−x
x+


bn
n
n x+ n

X
_k
An (x) _
1 X _
+
(gx ) +
(gx ) +
(gx ).

(bn − x)2 
n
−
1
x
x
0

k=1 x− √
k=2 x− √
k
k
Obviously,
1
1
b2n
+
=
,
x2 (bn − x)2
x2 (bn − x)2
for
x
bn
∈ [0, 1] and
x
_
x− √x
k
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
x+ bn√−x
(gx ) ≤
_k
(gx ).
x− √x
k
http://jipam.vu.edu.au/
R ATE OF C ONVERGENCE
9
Hence,
|Dn (gx ; x)| ≤
An (x)
An (x)
+
2
x
(bn − x)2


bn
_


bn
√−x
(gx ) +
n x+_k
X
k=1 x− √x
0
k



(gx )


x+ bn√−x
k
n
1 X _
+
(gx )
n − 1 k=2
x− √x
k


bn
√−x
x+ bn√−x


b
n
n x+_k
k
n


2
_
X
_
X
1
An (x) bn
(gx ) +
= 2
(gx ) +
(gx )
2

x (bn − x) 
0
 n − 1 k=2 x− √x
k=1 x− √x
k
k


bn
−x
b
−x
n
√
√
 bn

n x+_k
n x+ k

X
An (x) b2n _
1 X _
= 2
(gx ) +
(gx ) +
(gx ).

x (bn − x)2 
0
 n − 1 k=2 x− √x
k=1 x− √x
k
k
On the other hand, note that
bn
_
x+ bn√−x
(gx ) ≤
0
n
_k
X
k=1
(gx ).
x− √x
k
By (2.3), we have

bn
√−x

n x+_k
X
2 An (x) b2n
|Dn (gx ; x)| ≤ 2
x (bn − x)2 
 k=1
Note that
1
n−1
≤
An (x) b2n
,
x2 (bn −x)2
(3.6)
x− √x
k

bn
√−x

n x+ k

1 X _
(gx ) +
(gx ).

 n − 1 k=2 x− √x
k
x
bn
∈ [0, 1]. Consequently


x+ bn√−x


n
k


3 An (x) b2n X _
|Dn (gx ; x)| ≤ 2
(gx ) .

x (bn − x)2 
 k=1 x− √x

for n > 1,
k
Now secondly, we can estimate Dn (sgn(t − x); x). If we apply operator Dn to the signum
function, we get
Z bn
Z x
n
n+1X
x
t
t
Dn (sgn(t − x); x) =
Pn,k
Pn,k
dt −
Pn,k
dt
bn k=0
bn
bn
bn
x
0
Z bn
Z x
n
n+1X
x
t
t
=
Pn,k
Pn,k
dt − 2
Pn,k
dt
bn k=0
bn
bn
bn
0
0
using (2.1), we have
n
(3.7)
n+1X
Dn (sgn(t − x); x) = 1 − 2
Pn,k
bn k=0
x
bn
Z
x
Pn,k
0
t
bn
dt.
Now, we differentiate both side of the following equality
n+1
X
x
x
Jn+1,k+1
=
Pn+1,j
.
bn
b
n
j=k+1
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
http://jipam.vu.edu.au/
10
E RTAN I BIKLI AND H ARUN K ARSLI
For k = 0, 1, 2, . . . , n we get,
n+1
x
x
d X
d
Jn+1,k+1
=
Pn+1,j
dx
bn
dx j=k+1
bn
d
x
d
x
d
x
=
Pn+1,k+1
+ Pn+1,k+2
+ · · · + Pn+1,n+1
dx
bn
dx
bn
dx
bn
x
d
Jn+1,k+1
dx
bn
x
x
x
x
(n + 1)
=
Pn,k
− Pn,k+1
+ Pn,k+1
− Pn,k+2
bn
bn
bn
bn
bn
x
x
x
+ · · · + Pn,n−1
− Pn,n
+ Pn,n
bn
bn
bn
n+1
(n + 1) X
x
x
=
Pn,j−1
− Pn,j
bn j=k+1
bn
bn
(n + 1)
x
=
Pn,k
bn
bn
and Jn+1,k+1 (0) = 0. Taking the integral from zero to x, we have
Z
(n + 1) x
t
x
Pn,k
dt = Jn+1,k+1
bn
bn
bn
0
and therefore from (2.4)
Jn+1,k+1
x
bn
n+1
X
=
Pn+1,j
j=k+1
=
n+1
X
Pn+1,j
j=0
=1−
k
X
x
bn
x
bn
−
k
X
Pn+1,j
j=0
Pn+1,j
j=0
x
bn
x
bn
.
Hence
(n + 1)
bn
Z
x
Pn,k
0
t
bn
dt = 1 −
k
X
Pn+1,j
j=0
x
bn
.
From (3.7), we get
#
x
Dn (sgn(t − x); x) = 1 − 2
Pn,k
Pn+1,j
1−
bn
j=0
k=0
X
n
n
k
X
X
x
x
x
=1−2
Pn,k
+2
Pn,k
Pn+1,j
bn
bn j=0
bn
k=0
k=0
X
n
k
X
x
x
= −1 + 2
Pn,k
Pn+1,j
.
bn j=0
bn
k=0
n
X
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
x
bn
"
k
X
http://jipam.vu.edu.au/
R ATE OF C ONVERGENCE
11
Set
(2)
Qn+1,j
x
bn
=
2
Jn+1,j
x
bn
−
2
Jn+1,j+1
x
bn
.
Also note that
n X
k
X
∗=
k=0 j=0
n+1
X
(2)
Qn+1,k
k=j
x
bn
n X
n
X
∗,
j=0 k=j
=
2
Jn+1,j
x
bn
x
bn
Jn,n+1
and
= 0,
we have
Dn (sgn(t − x); x) = −1 + 2
n
X
Pn+1,j
j=0
= −1 + 2
n
X
Pn+1,j
j=0
= −1 + 2
n+1
X
Pn+1,j
j=0
=2
n+1
X
Pn+1,j
j=0
Since
n+1
P
j=0
(2)
Qn+1,j
x
bn
x
bn
x
bn
X
n
x
bn
x
bn
Pn,k
k=j
x
bn
x
bn
Jn,j
Jn,j
Jn,j
x
bn
x
bn
− 1.
= 1, thus
Dn (sgn(t − x); x) = 2
n+1
X
Pn+1,j
j=0
x
bn
Jn,j
x
bn
−
n+1
X
(2)
Qn+1,j
j=0
x
bn
.
By the mean value theorem, we have
x
x
x
(2)
2
2
Qn+1,j
= Jn+1,j
− Jn+1,j+1
bn
bn
bn
x
x
= 2Pn+1,j
γn,j
bn
bn
where
Jn+1,j+1
x
bn
< γn,j
x
bn
< Jn+1,j
x
bn
.
Hence it follows from (2.5) that
n+1
X
x
x
x
|Dn (sgn(t − x); x)| = 2
Pn+1,j
Jn,j
− γn,j
bn
bn
bn
j=0
n+1
X
x x
x ≤2
Pn+1,j
J
−
γ
n,j
n,j
bn bn
bn j=0
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
http://jipam.vu.edu.au/
12
E RTAN I BIKLI AND H ARUN K ARSLI
x Jn,j x − Jn+1,j+1 x
bn
bn
b n x
x
x
x = Jn,j
− γn,j
+ γn,j
− Jn+1,j+1
,
bn
bn
bn
bn since γn,j bxn − Jn+1,j+1 bxn > 0,then we have
Jn,j x − γn,j x ≤ Jn,j x − Jn+1,j+1 x .
bn
bn
bn
bn Hence
n+1
X
x x
x |Dn (sgn(t − x); x)| ≤ 2
Pn+1,j
Jn,j
− Jn+1,j+1
bn
bn
bn j=0
≤2
n+1
X
Pn+1,j
j=0
(3.8)
x
bn
4
=q
n bxn (1 −
x
)
bn
2
q
n bxn (1 −
x
)
bn
.
Combining (3.6) and (3.8) we get (1.1). Thus, the proof of the theorem is completed.
R EFERENCES
[1] J.L. DURRMEYER, Une formule d’inversion de la ransformee de Laplace: Applicationsa la theorie
de moments , these de 3e cycle, Faculte des sciences de 1’universite de Paris,1971.
[2] A. LUPAŞ, Die Folge Der Betaoperatoren, Dissertation, Stuttgart Universität, 1972.
[3] R. BOJANIC AND M. VUILLEUMIER, On the rate of convergence of Fourier Legendre series of
functions of bounded variation, J. Approx. Theory, 31 (1981), 67–79.
[4] F. CHENG, On the rate of convergence of Bernstein polynomials of functions of bounded variation,
J. Approx. Theory, 39 (1983), 259–274.
[5] S. GUO, On the rate of convergence of Durrmeyer operator for functions of bounded variation, J.
Approx. Theory, 51 (1987), 183–197.
[6] I. CHLODOWSKY, Sur le dĕveloppment des fonctions dĕfines dans un interval infinien sĕries de
polynŏmes de S.N. Bernstein, Compositio Math., 4 (1937), 380–392.
[7] XIAO-MING ZENG, Bounds for Bernstein basis functions and Meyer-König-Zeller basis functions, J. Math. Anal. Appl., 219 (1998), 364–376.
[8] XIAO-MING ZENG AND A. PIRIOU, On the rate of convergence of two Bernstein-Bezier type
operators for bounded variation functions, J. Approx. Theory, 95 (1998), 369–387.
[9] XIAO-MING ZENG AND W. CHEN, On the rate of convergence of the generalized Durrmeyer
type operators for functions of bounded variation, J. Approx. Theory, 102 (2000), 1–12.
[10] A.N. SHIRYAYEV, Probability, Springer-Verlag, New York, 1984.
[11] G.G. LORENTZ, Bernstein Polynomials, Univ. of Toronto Press, Toronto, 1953.
[12] Z.A. CHANTURIYA, Modulus of variation of function and its application in the theory of Fourier
series, Dokl. Akad. Nauk. SSSR, 214 (1974), 63–66.
J. Inequal. Pure and Appl. Math., 6(4) Art. 106, 2005
http://jipam.vu.edu.au/