General Mathematics Vol. 13, No. 1 (2005), 41–54
Rate of approximation for Bézier variant of
Bleimann-Butzer and Hahn operators
Vijay Gupta
and
Alexandru Lupaş
Dedicated to Professor Emil C. Popa on his 60th anniversary
Abstract
In the present paper we obtain the rate of convergence for the
Bézier variant of the Bleimann-Butzer and Hahn operators Ln,α , for
functions of bounded variation.
We consider the case when
0 < α < 1.
2000 Mathematics Subject Classification: 65B99
Key words and phrases: Bleimann-Butzer and Hahn operators,
Bézier variant, rate of approximation
1
Introduction
In the year 1980, Bleimann, Butzer and Hahn [5] introduced an interesting
sequence of positive linear operators defined on the space of real functions
on the infinite interval [0, ∞) by
(1)
Ln (f, x) =
∞
X
k=0
µ
pn,k (x)f
k
n−k−1
41
¶
, x ∈ [0, ∞), n ∈ N
42
Vijay Gupta and Alexandru Lupaş
where
µ ¶
n
xk
pn,k (x) =
k (1 + x)n
The approximation properties of the Bleimann-Butzer and Hahn ope-
rators (BBH operators) were studied by many researchers see e.g. [1], [2],
[3], [5] and [6] etc. Very recently Srivastava and Gupta [8], introduced the
Bézier variant of the BBH operators, which is defined as
µ
¶
∞
X
k
∞
(2)
Ln,α (f, x) =
Qn,k (x)f
, x ∈ [0, ∞), n ∈ N
n−k+1
k=0
(α)
α
α
where Qn,k (x) = Jn,k
(x) − Jn,k+1
(x) and Jn,k (x) =
α
P
pn,j (x). Some basic
j=k
properties of Jn,k (x) can be found in [8]. It is easily verified that Ln,α (f, x)
are positive linear operators. The authors [8] have obtained the rate of
convergence for functions of bounded variation for the case whenever α ≥ 1.
As a special case α = 1, Ln,α (f, x) reduce to the operators Ln,1 (f, x) ≡
Ln (f, x) defined by (1). The other case 0 < α < 1 of the operators defined
by (2) is also equally important. In the present paper we extend the study
in this direction and obtain teh rate of convergence for functions of bounded
variation, for the operators Ln,α , 0 < α < 1.
Our main theorem is stated as:
Theorem 1. Let f be a function of bounded variation on every finite subinterval of [0, ∞). Let f (t) = O(tr ) for some r ∈ N as t → ∞. Then for
x ∈ (0, ∞), 0 < α < 1 and n sufficiently large, we have
¯
¯
µ
¶
¯
¯
¯Ln,α (f, x) − 1 f (x+) − 1 − 1 f (x−)¯ ≤
α
α
¯
¯
2
2
≤
n
7(1 + x)2 X x+x/√k
|1 − x|
|f (x+) − f (x−)|+
Vn−x/√k (fx ) + p
(n + 2)x k=1
6 2π(n + 1)x
1+x
C(α, f, x, r)
+√
εn (x)|f (x) − f (x−)| +
, m>r
nm
2enx
Rate of approximation for Bézier variant...
where
(
εn (x) =
43
1, if(n + 1)px ∈ N
0, otherwise
,



 f (t) − f (x−), 0 ≤ t < x
fx (t) =
0,
t=x


 f (t) − f (x+), x < t < ∞
and Vab (fx ) is the total variation of fx on [a, b].
We recall the Lebesque-Stieltjes integral representation
Z∞
Ln,α (f, x) =
f (t)dt (Kn,α (x, t)),
0
where


Kn,α (x, t) =

P
k≤(n−k+1)t
(α)
Qn,k (x), 0 < t < ∞
0,
.
t=0
Also we define
(
Hn,α (x, t) =
2
1 − Kn,α (x, t), 0 < t < ∞
0,
t=0
.
Auxiliary Results
In this section we give certain results, which are necessary to prove the main
results.
Lemma 1. [3] For all x ∈ [0, ∞) and n ≥ 1 we have the following inequality:
Ln ((t − x)2 , x) ≤
3x(1 + x)2
.
n+2
44
Vijay Gupta and Alexandru Lupaş
Lemma 2. For x ∈ (0, ∞), we have
¯
¯
X
¯
¯
pn,k (x) −
¯
¯k/(n−k+1)>x
¯
¯
1 ¯¯
|1 − x|
.
≤ p
¯
2 ¯ 6 2π(n + 1)x
Proof. Apart from at most one term pn,k (x) when k/(n − k + 1) = x, using
the transform
¶k µ
¶n−k
µ ¶
µ ¶µ
n
xk
n
x
1
=
=
pn,k (x) =
1+x
1+x
k (1 + x)n
k
µ ¶
n k
=
y (1 − y)n−k ≡ bn,k (y),
k
with y = x/(1 + x), we have
X
pn,k (x) =
k/(n−k+1)>x
X
bn,k (y)
k>(n+1)y
this is approximately equal to
By ((n + 1)y, n − (n + 1)y + 1)
B((n + 1)y, n − (n + 1)y + 1)
with the incomplete Beta function
Zy
ta−1 (1 − t)b−1 dt, a > 0, b > 0.
By (a, b) =
0
We have
¯
¯ ¯
¯
¯
¯ ¯
¯
X
X
¯
¯ ¯
¯
¯=
¯
¯
¯1 − 2
1
−
2
b
(y)
=
p
(x)
n,k
n,k
¯
¯ ¯
¯
¯
¯ ¯
¯
k(n+1)>y
k/(n−k+1)>x
¯
¯
¯
¯
Zy
¯
¯
1
(n+1)y−1
n−(n+1)y ¯
¯
= ¯1 − 2
t
(1 − t)
dt¯
B((n + 1)y, n − (n + 1)y + 1)
¯
¯
0
Rate of approximation for Bézier variant...
45
Now we estimate the right hand side as follows:
Let y ∈ (0, 1), we have
B((n + 1)y, (n + 1)(1 − y))
Iy (n + 1)
=1−
B((n + 1)y, (n + 1)(1 − y))
B((n + 1)y, (n + 1)(1 − y))
where
Z1
Z1
t(n+1)y−1 (1 − t)(n+1)(1−y)−1 dt =
Iy (n + 1) =
y
g(t)e(n+1)hy (t) dt
y
with
g(t) = (t(1 − t))−1
hy (t) = y log t + (1 − y) log(1 − t).
Since
h0y =
−(t + y)
< 0 (y < t < 1) and h0y (y) = 0
t(1 − t)
hy is strictly decreasing on (y, 1).
h00y (t)
Furthermore
h00y (y) = −(y(1 − y))−1 6= 0.
=
−(t − y)2 + y(1 − y)
t2 (1 − t)2
<
0
and
Thus it is well known (e.g. Iy (n) meets the assumptions of [7, Th. 1,
Kap. 3]) that there holds the complete asymptotic expansion
∞
X Γ((k + 1)/2)
1
Iy (n + 1) ∼ e(n+1)hy
2
(n + 1)(k+1)/2
k=0
√
¸
a0 π
a2
a1
−2
)
+
+
+ O(n ) ,
2(n + 1)1/2 2(n + 1) 4(n + 1)3/2
·
y
1−y
(y (1 − y)
for n → ∞, with the coefficients


k+1 


µ ¶k 


1  d
t
−
y



ak = 
g(t) q


k!
dt 


hy (y) − hy (t)
t=1/y
46
Vijay Gupta and Alexandru Lupaş
By direct calculation we obtain the explicit expressions
√
2
−2(2 − 2y)
1 − y + y2
, a1 =
a0 = p
, a2 = √
3y(1 − y)
y(1 − y)
3 2(y(1 − y))3/2
Thus we have
Iy (n + 1) = (y y (1 − y)1−y )n ·
#
√
√
π
1 − 2y
π
1 − y + y2
−
· p
+ √ ·
+ O(n−2 ) .
3/2
3ny(1
−
y)
12 2 (y(1 − y))
2nπ(1 − y)
"
Also by Stirling0 s formula, we have
"
·
Γ(n + 1)
1
1
=
= √ (y y (1 − y)1−y )−n ·
B((n + 1)y, (n + 1)(1 − y))
Γ((n + 1)x)
2π
#
1 − y + y2
(1 − y + y 2 )2
+ O(n−5/2 ) .
(n + 1)y(1 − y) − p
+
3/2
3/2
12 ny(1 − y) 299n (y(1 − y))
p
Combining the both asymptotic expansions, we have
Iy (n + 1)
1
1 − 2y
= − p
+ O(n−3/2 ).
B((n + 1)y, (n + 1)(1 − y))
2 3 2π(n + 1)y(1 − y)
Thus
¯
¯
X
¯
¯
pn,k (x) −
¯
¯k/(n−k+1)>x
¯
¯
1 ¯¯
|1 − x|
≤ p
+ O(n−3/2 ).
¯
2 ¯ 6 2π(n + 1)x
Lemma 3. For all x ∈ (0, ∞), α ≤ 1 and k ∈ N, there holds
(1 + x)
(α)
.
Qn,k (x) ≤ αpn,k (x) < √
2enx
Proof. For the Bernstein functions the optimum bound was obtained by
Zeng [8], and Bastien and Rogalski [4] in a problem posed by the author,
which is as follows:
µ ¶
n k
1
y (1 − y)n−k ≤ p
, 0 < y < 1.
k
2eny(1 − y)
x , we get the required result.
Substituting y = 1 +
x
Rate of approximation for Bézier variant...
47
Lemma 4. For 0 < α < 1 and 0 < x < t < ∞, we have
Hn,α (x, t) ≤
C
,
n (t − x)m
m
where C is a positive constant that depends on x but independent of n.
¯
¯
¯
¯
k
¯
¯
−
x
¯
¯n − k + 1
Proof. Since 0 < x < t < ∞, so
≥ 1, for k ≥ nt. Thus
|t − x|
for m ∈ N, we have
Hn,α (x, t) = 1 − Kn,α (x, t) = 1 −
X
k≤(n−k+1)t



≤

¯
¯2m/α
¯
¯
k
¯
¯
¯n − k + 1¯
X
k≥(n−k+1)t
1
≤
(t − x)2m
(t − x)2m/α
X
(α)
Qn,k (x) ≤
k≥(n−k+1)t
α


pn,k (x) ≤

Ã∞ ¯
!α
¯2m/α
X¯
¯
k
¯
¯
pn,k (x) .
¯ n − k + 1 − x¯
k=0
For all conjugate p, q ≥ 1, i.e. 1/p + 1/q = 1, we have
Ã∞ ¯
!α
¯2m/α
X¯
¯
k
¯
¯
pn,k (x)
=
¯ n − k + 1 − x¯
k=0
Ã∞ ¯
!α
¯2m/α
X¯
¯
k
1/p
1/q
¯
¯
=
pn,k (x)pn,k (x)
≤
¯ n − k + 1 − x¯
k=0
!α/p
Ã∞ ¯
¯2mp/α
X¯
¯
k
¯
¯
pn,k (x)
≤
¯ n − k + 1 − x¯
k=0
since
µ∞
P
k=0
¶α/q
= 1.
pn,k (x)
(α)
Qn,k (x) ≤
48
Vijay Gupta and Alexandru Lupaş
¤
q £
Choosing p = m m
+
1
we have that 2mp/α is an even positive
α
integer. By the well known result Bn,1 (ψx2r , x) = O(n−r ), as n → ∞
(r = 1, 2, 3, ...) we obtain
Ã∞ ¯
!α/p
¯2mp/α
X¯
¯
k
¯
¯
pn,k (x)
= (Ln,1 (ψx2mp/α , x))α/p = O(n−m )
¯ n − k + 1 − x¯
k=0
as n → ∞.
This completes the proof of Lemma 4.
3
Proof of main Theorem
Our main theorem is stated as:
Proof. We have
f (t) = 2−α f (x+)+(1−2−α )f (x−)+gx (t)+2−α (f (f +)−f (x−))sign(α) (t)+
+(f (x) − 2−α f (x+) − (1 − 2−α )f (x−))δx (t),
where
sign(α) (t − x) :=

α


 2 − 1, if t > x
0,


 −1,
if t = x
if t < x
(
and δx (t)
1, if x = t
0, if x 6= t
Therefore
(3)
¯
¯
µ
¶
¯
¯
¯Ln,α (f, x) − 1 f (x+) − 1 − 1 f (x−)¯ ≤ |Ln,α (fx , x)|+
¯
¯
2α
2α
¯
¯ f (x+) − f (x−)
Ln,α (sign(α) (t − x), x)+
+ ¯¯
2α
¯
·
µ
¶
¸
¯
1
1
+ f (x) − α f (x+) − 1 − α f (x−) Ln,α (δx , x)¯¯ .
2
2
Rate of approximation for Bézier variant...
49
We first estimate
Ln,α (sign(α) (t − x), x) = 2α
X
(α)
(α)
Qn,k (x) − 1 + en (x)Qn,k0 (x) =
k>(n−k+1)x
α

= 2α 
X
(α)
pn,k (x) − 1 + εn (x)Qn,k0 (x)
k>(n−k+1)x
and
(α)
Ln,α (δx , x) = εn (x)Qn,k0 (x).
Hence, we have
¯
¯ f (x+) − f (x−)
¯
(4)
Ln,α (signα (t − x), x)+
¯
2α
¯
µ
¶
¸
·
¯
1
1
+ f (x) − α f (x+) − 1 − α f (x−) Ln,α (δx , x)¯¯ =
2
2
¯
¯
α 
 
¯
¯
X
¯
¯ f (x+) − f (x−) α
(α)
2 
pn,k (x) − 1 + [f (x) − f (x−)]εn Qn,k (x)¯¯
= ¯¯
α
2
¯
¯
k>(n−k+1)x
By mean value theorem, we have
¯
¯
¯
α
¯
¯
¯ X
X
¯
¯
¯
1
α−1
¯
¯
pn,j (x) − α ¯¯ = α(ξn,j (x))
pn,j (x) −
¯
¯
2 ¯
¯ j>(n−j+1)x
¯j>(n−j+1)x
¯
¯
1 ¯¯
2 ¯¯
P
where ξn,j (x) lies between 12 and
pn,j (x). In view of Lemma 2, it is
j>(n−j+1)x
observed that for n sufficiently large, the intermediate point ζn,j is arbitrary
close to 1/2 i.e.
1
2+ε
with an arbitrary small |ε|. Then we have
ζn,j =
α(ξn,j (x))α−1 ≤ α(2 + ε)1−α .
50
Vijay Gupta and Alexandru Lupaş
The later expression is positive and strictly increasing for α ∈ (0, 1), since
∂
α(2 + ε)1−α = (2 + ε)1−α [1 − α log(2 + ε)] > 0,
∂α
for sufficiently small |ε|. This it takes maximum value at α = 1. This
implies
α(ξn,j (x))α−1 ≤ 1.
Hence
¯
¯
α
¯
¯
X
¯
¯
1
¯
 − α ¯ ≤ p |1 − x|
p
(x)
.
n,j
¯
2 ¯¯ 6 2π(n + 1)x
¯ j>(n−j+1)x
(5)
Also we have
(α)
α
α
α−1
Qn,k0 (x) = Jn,k
pn,k0 (x),
0 (x) − Jn,k 0 +1 (x) = α(ξn,k 0 )
where Jn,k0 +1 (x) < ξn,k0 (x) < Jn,k0 (x). Thus by Lemma 3, we have
1+x
(α)
.
Qn,k0 (x) ≤ √
2enx
(6)
Combining the estimates of (4), (5) and (6), we have
¯
¯ f (x+) − f (x−)
¯
(7)
Ln,α (sign(t − x), x)+
¯
2α
¯
·
µ
¶
¸
¯
1
1
+ f (x) − α f (x+) − 1 − α f (x−) Ln,α (δx , x)¯¯ ≤
2
2
1+x
|1 − x|
|f (x+) − f (x−)| + √
εn (x)|f (x) − f (x−)|.
≤ p
2enx
6 2π(n + 1)x
Next we estimate Ln,α (fx , x). We decompose the integral into four parts as
follows:
(8)
Z∞
Ln,α (fx , x) = fx (t)dt (Kn,α (x, t)) =
0
Rate of approximation for Bézier variant...

Z
Z
=
+
I1
Z
+
I2
Z

 fx (t)dt (Kn,α (x, t)) = E1 + E2 + E3 + E4 say,
+
I3
51
I4
√
√
√
√
where I1 = [0, x − x/ n], I2 = [x − x/ n, x + x/ n], I3 = [x + x/ n, 2x]
√
√
and I4 = [2x, ∞]. We first estimate E2 . For t ∈ [x − x/ n, x + x/ n], we
have
√
x+x/ n
|fx (t)| = |fx (t) − fx (x)| ≤ Vx−x/√n (fx )
and therefore
√
x+x/
Z n
√
x+x/ n
dt (Kn,α (x, t)).
|E2 | ≤ Vx−x/√n (fx )
√
x−x/ n
Since
Rb
dt (Kn,α (x, t)) ≤ 1 for (a, b) ⊂ [0, ∞), therefore
a
(9)
|E2 | ≤
√
x+x/ n
Vx−x/√n (fx )
n
1 X x+x/√n
√ (fx ).
≤
V
n k=1 x−x/ n
√
Next, we estimate E1 , writing y = x−x/ n and using Lebesque-Stieltjes
integration by parts, we have
Zy
E1 =
Zy
fx (t)dt (Kn,α (x, t)) = fx (y)Kn,α (x, y) −
0
Kn,α (x, t)dt (fx (t))
0
Since |fx (y)| ≤ Vyx (fx ), it follows that
Zy
|E1 | ≤
Vyx (fx )Kn,α (x, y)
Kn,α (x, t)dt (−Vtx (fx )).
+
0
Applying Lemma 1, we have
(10)
Kn,α (x, t) ≤
3x(1 + x)2
, 0 ≤ t < x.
(n + 2)
52
Vijay Gupta and Alexandru Lupaş
Using the inequality (10), we get
|E1 | ≤
Vyx (fx )
3x(1 + x)2
3x(1 + x)2
+
(n + 2)
(n + 2)(x − y)2
Zy
0
1
x
2 dt (−Vt (fx )).
(x − t)
Integrating by parts the last term, we have
Zy
0
Vyx (fx )
V0x (fx )
1
x
(f
))
=
−
d
(−V
+
+2
x
t
t
(x − t)2
(x − y)2
x2
Hence

2
|E1 | ≤
3x(1 + x)
(n + 2)
x
 V0 (fx )
2
x
Zy
+2
0
Zy
Vtx (fx )
0
dt
.
(x − t)3

Vtx (fx ) 
3 .
(x − t)
√
Now replacing the variable y in the last integral by x − x/ u, we get
n
6(1 + x)2 X x √
|E1 | ≤
V
(fx ).
(n + 2)x k=1 x−x/ k
(11)
Using the similar method to estimate E3 , we get
n
6(1 + x)2 X x+x/√k
V
|E3 | ≤
(fx ).
(n + 2)x k=1 x
(12)
Finally, by the assumption that fx (t) = 0(tr ), r ∈ N, t → ∞, we have
µ
¶r
t−x
r
|fx (t)| ≤ M t ≤ M
, for t ≥ 2x.
x
Now
¯
¯∞
¯Z
¯ Z∞
¯
¯
|E4 | = ¯¯ fx (t)Kn,α (x, t)¯¯ ≤ |fx (t)|dt Kn,α (x, t) ≤
¯
¯
2x
2x
Z∞
Z∞
≤ M x−r (t − x)r dt Kn,α (x, t) ≤ −M x−r (t − x)r dt (1 − Kn,α (x, t)) =
0
2x
Rate of approximation for Bézier variant...
= −M x
−r
53
Z∞
Z∞
r
−r
(t − x) dt (Hn,α (x, t)) ≤ −M x
(t − x)r dt (1 − Kn,α (x, t)) =
2x
0


ZR
Hn,α (x, t)dt (t − x)r  =
= M x−r lim −(t − x)r Hn,α (x, t)|R
2x +
R→∞
2x


ZR
Hn,α (x, t)dt (t − x)r−1 dt =
= M x−r lim −(t − x)r Hn,α (x, t)|R
2x +
R→∞
2x

= M x−r lim −(t − x)r
R→∞
C
rC
R
m |2x + m
n (t − x)
n

ZR
(t − x)r−m−1 dt =
m
2x
=M
C
rC
+ m
, m > r.
n x
n (m − r)xm−r
m m
By combining the estimates given by (3), (7) - (9) and (11) to (13), we
obtain the desired result. This completes the proof of Theorem.
References
[1] U. Abel, M. Ivan, Some identities for the operator Bleimann-Butzer and
Hahn involving divided differences, Calcolo 36 (3), 1999, 143 - 160.
[2] U. Abel, On the asymptotic approximation with Bivariate operators of
Bleimann-Butzer and Hahn, J. Approx. Theory 97, 1999, 181 - 198.
[3] U. Abel, M. Ivan, Best constant for a Bleimann-Butzer-Hahn moment
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School of Applied Sciences
Netaji Subhas Institute of Technology
Sector 3 Dwarka
110045 New Delhi, India
”Lucian Blaga” University of Sibiu
Department of Mathematics
Str. Dr. I. Rat.iu, No. 5-7
550012 - Sibiu, Romania
E-mail address: alexandru.lupas@ulbsibiu.ro